text stringlengths 14 5.77M | meta dict | __index_level_0__ int64 0 9.97k ⌀ |
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import React, { FC, useEffect } from 'react'
import {
useHistory,
} from 'react-router-dom'
import { Button, Form, Input, DatePicker } from 'antd'
import { useBaseContext } from 'ROOT_SOURCE/hooks'
import request from 'ROOT_SOURCE/utils/request'
import { sleep } from 'ROOT_SOURCE/utils'
import Rules from 'ROOT_SOURCE/utils/validateRules'
const Add: FC<{}> = () => {
const [form] = Form.useForm()
const routeHistory = useHistory()
const CONTAINER_ROUTE_PREFIX = useBaseContext()
// 初始化调用
useEffect(() => {
form.setFieldsValue({
note: 'note'
})
// eslint-disable-next-line react-hooks/exhaustive-deps
}, [])
async function onFinish(values: object): Promise<void> {
try {
// 提交表单最好新一个事务,不受其他事务影响
await sleep()
// action
await request.post('/asset/addAsset', values)
// 提交后返回list页
routeHistory.push(`${CONTAINER_ROUTE_PREFIX}/list`)
} catch (error) {
console.log(error)
}
}
return (
<Form className='ui-background' form={form} onFinish={onFinish}>
<Form.Item name='assetName' label='资产方名称' rules={[{ required: true }]}>
<Input />
</Form.Item>
<Form.Item name='contract' label='签约主体' rules={[{ required: true }]}>
<Input />
</Form.Item>
<Form.Item name='contractDate' label='签约时间' rules={[{ required: true }]}>
<DatePicker showTime />
</Form.Item>
<Form.Item name='contacts' label='联系人'>
<Input />
</Form.Item>
<Form.Item name='contactsPhone' label='联系电话' rules={[{ pattern: Rules.phone, message: '无效' }]}>
<Input maxLength={11} />
</Form.Item>
<Form.Item className='ui-btn-group ui-align-center'>
<Button type='primary' htmlType='submit'>
提交
</Button>
<Button htmlType='button' onClick={routeHistory.goBack}>
取消/返回
</Button>
</Form.Item>
</Form>
)
}
export default Add
| {
"redpajama_set_name": "RedPajamaGithub"
} | 4,371 |
{"url":"http:\/\/math.stackexchange.com\/questions\/494571\/calculate-sizes-of-rectangles-of-fixed-proportions-that-would-fill-a-grids-heig","text":"Calculate sizes of rectangles of fixed proportions that would fill a grid's height.\n\nEach rectangle has a fixed proportion e.g. 1:2 Each rectangle has to be the same size Grid's height has to be filled Last row can have space if the number of elements is odd. Width might have space if proportions don't allow to fill it. What size each rectangle should be (the size will be the same among all rectangles)?\n\nEDIT: What I ended up doing is this:\n\nintroducing some size e.g. 400x500 for rectangles\n\nnow if e.g. area to fit in is 3000x4000 I can calculate it's proportions (height\/width)\n\nso area proportions = height \/ width\nadd rect to upper left corner\nwhile(there are still rectangles to position)\n{\ncalculate current occupied area of all rects\ncurrent occupied area proportions = occupied area height \/ width\nif area proportions < current occupied area proportions\nfill column with rectangles\nelse\nfill row with rectangles\n}\n\nareaHeightToOccupiedHeightProportion = Area.Height \/ currentOccupiedSpace.Height;\nareaWidthToOccupiedWidthProportion = Area.Width \/ currentOccupiedSpace.Width;\nif(areaHeightToOccupiedHeightProportion < areaWidthToOccupiedWidthProportion){\n\/\/scaling to maximum width would make the occupied area bigger than the area we want to position rectangles in\nvar scale = areaHeightToOccupiedHeightProportion;\n}\n{\nelse scale = areaWidthToOccupiedWidthProportion;\n}","date":"2014-03-11 22:00:59","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\": 0.8171882033348083, \"perplexity\": 4869.524458941828}, \"config\": {\"markdown_headings\": false, \"markdown_code\": true, \"boilerplate_config\": {\"ratio_threshold\": 0.18, \"absolute_threshold\": 10, \"end_threshold\": 15, \"enable\": true}, \"remove_buttons\": true, \"remove_image_figures\": true, \"remove_link_clusters\": true, \"table_config\": {\"min_rows\": 2, \"min_cols\": 3, \"format\": \"plain\"}, \"remove_chinese\": true, \"remove_edit_buttons\": true, \"extract_latex\": true}, \"warc_path\": \"s3:\/\/commoncrawl\/crawl-data\/CC-MAIN-2014-10\/segments\/1394011294162\/warc\/CC-MAIN-20140305092134-00066-ip-10-183-142-35.ec2.internal.warc.gz\"}"} | null | null |
Die Satanische Bibel (Originaltitel: The Satanic Bible) ist ein von Anton Szandor LaVey verfasstes Grundlagenwerk der Church of Satan. Es besteht aus den vier Büchern Satan, Luzifer, Belial und Leviathan. In dem Buch beschreibt LaVey seine Weltanschauung, mit der er zugleich den Status einer Religion beansprucht, die nach LaVeys Darstellung in der Walpurgisnacht 1966 begründet wurde.
Entstehung
Gemäß der Darstellung des Verfassers LaVey wurde das Werk von ihm in eben jener Walpurgisnacht verfasst.
Demgegenüber steht die Darstellung von LaVeys Tochter Zeena und Nikolas Schreck, wonach der Text unter vom Verlag ausgeübtem Zeitdruck entstanden sei. Zudem wird der Vorwurf erhoben, das Werk sei in großen Teilen ein Plagiat, welches aus ungekennzeichneten Zitaten kompletter Passagen diverser anderer Schriften besteht. Als Beispiele werden Ragnar Redbeards sozialdarwinistisches Might is Right und John Dees Enochian Keys, wohl zitiert aus Aleister Crowleys Equinox angeführt.
Die Satanische Bibel beruht unter anderem auf der Doktrin des radikalen Egoismus und Individualismus, der US-amerikanische Schriftstellerin und Philosophin Ayn Rand. Rand ist einer der wichtigsten Autoren, die in der Satanischen Bibel zitiert werden. LaVey erklärte, dass seine Religion "nur die Philosophie von Ayn Rand ist, der Zeremonien und Rituale hinzugefügt wurden" und dass der Satanismus weit mehr mit dem Objektivismus gemein hat als mit jeder anderen Religion oder Philosophie.
Inhalt
Am Anfang des Buches werden die neun satanischen Grundsätze genannt, die als die Gebote der Church of Satan zu betrachten sind.
Satan bedeutet Sinnesfreude anstatt Abstinenz.
Satan bedeutet Lebenskraft anstatt Hirngespinste.
Satan bedeutet unverfälschte Weisheit anstatt heuchlerischem Selbstbetrug.
Satan bedeutet Güte gegenüber denen, die sie verdienen, anstatt Liebe an Undankbare.
Satan bedeutet Rache anstatt Hinhalten der anderen Wange.
Satan bedeutet Verantwortung für die Verantwortungsbewussten anstatt Fürsorge für psychische Vampire.
Satan bedeutet, dass der Mensch lediglich ein Tier unter anderen Tieren ist, manchmal besser, häufig jedoch schlechter als die Vierbeiner, da er auf Grund seiner "göttlichen, geistigen und intellektuellen Entwicklung" zum bösartigsten aller Tiere geworden ist.
Satan bedeutet alle sogenannten Sünden, denn sie alle führen zu physischer, geistiger und emotionaler Erfüllung.
Satan ist der beste Freund, den die Kirche jemals gehabt hat, denn er hat sie die ganzen Jahre am Leben erhalten.
Die Satanische Bibel umfasst ferner vier Kapitel, die als Bücher bezeichnet werden und für die vier Grundelemente alles Seins gemäß der Vier-Elemente-Lehre stehen:
Das Buch Satan (Feuer)
Das Buch ist aus der Erzählerperspektive Satans geschrieben und in fünf Bereiche unterteilt. Unter anderem werden Religionen, insbesondere die christliche und jüdische, kritisiert.
Das Buch Luzifer (Luft)
Dieses Kapitel behandelt das Verhältnis eines Satanisten zu Gott und beschreibt satanischen Sex. Es beinhaltet gleichfalls satanische Feiertage. In diesem Kapitel distanziert sich der Autor davon, dass Satanisten Menschen und Tiere opfern, Orgien feiern und Blut trinken.
Das Buch Belial (Erde)
Hier werden vorwiegend die Theorie und die Praxis der satanischen Magie aufgestellt und Ritualarten genannt. Auch der Aufbau und eine Liste der Gegenstände, die verwendet werden, sind zu finden.
Das Buch Leviathan (Wasser)
Umfasst die Anrufung Satans, höllische Namen sowie die Anrufung zum Beschwören von Lust, Vernichtung und Mitleid. Auch John Dees 19 Henochische Schlüssel sind hier zu finden.
Ausgaben
Erstausgabe: The Satanic Bible. Avon Books, 1969, ISBN 0-380-01539-0.
Die deutsche Übersetzung von Ingrid Meyer wurde erstmals im Jahr 1999 vom Verlag Second Sight Books herausgegeben. Der gesamte Text wurde dabei in weißer Schrift auf schwarzem Hintergrund gedruckt. Später wurde die Satanische Bibel zusammen mit LaVeys Die satanischen Rituale herausgegeben.
Die Satanische Bibel. Second Sight Books, Berlin 1999, ISBN 3-0000-4343-8.
Die Satanische Bibel und ihre Rituale. Second Sight Books, Berlin 2003, ISBN 3-935684-05-3. Neuausgabe: Index-Verlag, Zeltingen-Rachtig 2013, ISBN 978-3-936878-26-4.
Einzelnachweise
Okkultismus
Satanismus
Esoterisches Werk
Philosophisches Werk | {
"redpajama_set_name": "RedPajamaWikipedia"
} | 3,914 |
{"url":"https:\/\/kerodon.net\/tag\/02U2","text":"# Kerodon\n\n$\\Newextarrow{\\xRightarrow}{5,5}{0x21D2}$ $\\newcommand\\empty{}$\n\nExample 4.5.9.15. Let $\\operatorname{\\mathcal{D}}$ be any simplicial set. Then the projection map $\\operatorname{\\mathcal{D}}\\rightarrow \\Delta ^{0}$ is exponentiable (this is a reformulation of Remark 4.5.3.7).","date":"2022-05-29 01:22:12","metadata":"{\"extraction_info\": {\"found_math\": true, \"script_math_tex\": 0, \"script_math_asciimath\": 0, \"math_annotations\": 0, \"math_alttext\": 0, \"mathml\": 0, \"mathjax_tag\": 0, \"mathjax_inline_tex\": 1, \"mathjax_display_tex\": 0, \"mathjax_asciimath\": 0, \"img_math\": 0, \"codecogs_latex\": 0, \"wp_latex\": 0, \"mimetex.cgi\": 0, \"\/images\/math\/codecogs\": 0, \"mathtex.cgi\": 0, \"katex\": 0, \"math-container\": 0, \"wp-katex-eq\": 0, \"align\": 0, \"equation\": 0, \"x-ck12\": 0, \"texerror\": 0, \"math_score\": 0.9920801520347595, \"perplexity\": 1022.6929919776778}, \"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-2022-21\/segments\/1652663035797.93\/warc\/CC-MAIN-20220529011010-20220529041010-00440.warc.gz\"}"} | null | null |
\section{Introduction}
Strong gravitational lensing is a phenomenon that originates when light rays propagating from a background source galaxy are deflected, on their way towards the observer, by the gravitational field of a foreground galaxy, creating multiple images, arcs and/or rings around the foreground galaxy.
Strong gravitational lensing is a unique probe for studying the (dark) matter distribution of galaxies and providing cosmological constraints.
E.g., gravitational lenses have been used to measure the Hubble constant through time delays of lensed quasar images (e.g., \citealt{suyu2010,Suyu2017,bonvin2016}) and to constrain the dark energy equation of state (e.g., \citealt{Biesiada+10,Collet2014de,Cao+12,Cao2015de}).
Gravitational lensing also allows measuring the fraction of dark matter in the central regions of galaxies (\citealt{gavazzi2007,jiang2007,covone2009,grillo2010,cardone2009,Cardone2010,Auger+10_SLACSX,tortora2010central,more2011,ruff2011,sonnenfeld2015sl2s})
and to constrain the slope of the inner mass density profile (e.g., \citealt{Treu2002MNRAS,Treu2002,koopmans2006,koopmans2003,Moore2008,barnabe2009,koopmans2009,cao2016limits,Mukherjee2018}).
Moreover, studying gravitational lenses can constrain the stellar initial mass function (e.g., \citealt{treu2010, ferreras2010, Spiniello:2011p8239, brewer2012swells,sonnenfeld2015sl2s,posacki2015stellar,Spiniello+15_IMF_vs_density,moller2007strong,Sonnenfeld2018imf}).
Strong lensing also works as a ``cosmic telescope", producing a magnified view of background objects otherwise not observable (e.g., \citealt{impellizzeri2008,swinbank2009,richard2011,deane2013,treu2015grism,Mason2016,Salmon2017,Kelly2017}).
Discovering new gravitational lenses allows placement of more precise constraints on the above-mentioned quantities (see e.g., \citealt{Vegetti2009,Barnab2011,Li2016}).
For a comprehensive review of the scientific applications of strong lensing see, e.g., \cite{schneider1992gravitational}, \cite{Schneider06_SAASFEE33} and \cite{Treu2010rev}.
Originally, gravitational lenses were found serendipitously in astronomical surveys, while currently they are considered as an important class of objects to systematically search in large sky surveys.
The most successful campaign aiming at building a homogeneous dataset of strong gravitational lenses was the Sloan Lens ACS Survey (SLACS; \citealt{SLACS2008}) with more than 100 observed lenses that were identified by analyzing spectra from the Sloan Digital Sky Survey (SDSS; \citealt{SDSS}) that exhibited the imprint of two galaxies at two different redshifts.
On-going optical surveys such as the Hyper Suprime-Cam survey (HSC; \citealt{HSCmiyazaki2012hyper}), the Kilo Degree Survey (KiDS; \citealt{deJong+15_KiDS_paperI}) and the Dark Energy Survey (DES; \citealt{DES}) are expected to provide in the coming years thousands of new lenses (see \citealt{collet2015} and \citealt{Petrillo2017}) and have already provided new lens candidates \citep{Petrillo2017,Diehl2017,Sonnenfeld2018}.
The future is bright
also in the sub-millimeter wavelength, where Herschel \citep{negrello2010} and the South Pole Telescope (\citealt{carlstrom2011}), coupled with the Atacama Large Millimeter/sub-millimeter Array, are providing
several hundreds of new lens candidates \citep{vieira2013dusty,negrello2017herschel}.
However, it is the next decade that holds a treasure trove of new gravitational lenses. It has been estimated that samples of $\sim 10^5$ strong lenses \citep{oguri2010gravitationally,pawase2012,collet2015,McKean2015} will be observed by Euclid \citep{Laureijs:2011wi}, the Large Synoptic Survey Telescope (LSST; \citealt{abell2009}) and the Square Kilometer Array\footnote{\href{https://www.skatelescope.org/}{\tt https://www.skatelescope.org/}}.
The huge number of possible new candidates, together with the difficulty of identifying them in the enormous volume of survey data, drives the growing effort in developing automatic lens-finders.
Most are based on the identification of arc-like features (e.g., \citealt{Lenzen2004,Horesh2005,Alard2006,Estrada2007,Seidel2007,Kubo2008,More2012, Maturi2014}).
Other approaches, such as described by \cite{Gavazzi2014} and \cite{Joseph2014}, focus on the analysis of the residuals after subtracting the candidate lens galaxies from the astronomical images. Both methods have been used to find lens candidates in the Canada-France-Hawaii Telescope
Legacy Survey (CFHTLS\footnote{\href{http://www.cfht.hawaii.edu/Science/CFHLS/}{\tt http://www.cfht.hawaii.edu/Science/CFHLS/}}) by \cite{Sonnenfeld2013,Gavazzi2014,Paraficz2016}.
Instead, the algorithm developed by \cite{Chan2015} is specialized in identifying lensed quasars and together with the algorithm \textsc{YattaLens} \citep{Sonnenfeld2018} has been applied to find lens candidates in HSC \citep{Sonnenfeld2018}.
Another approach, as in \cite{Brault2015}, is modeling the probability that the targets are actual lenses. \cite{Stapelberg2017} applied the same strategy to clusters and groups of galaxies. Gravitational lenses have been identified also with citizen-science experiment approaches with the Space Warps project \citep{Marshall2016,more2016} where
non professional volunteers can classify galaxy images with the help of a web applet\footnote{\href{https://spacewarps.org/}{\tt https://spacewarps.org/}}.
Most recently, \cite{Petrillo2017} and \cite{Jacobs2017} have used Convolutional Neural Network (ConvNets) for finding lens candidates in KiDS and CFHTLS, respectively. Finally, \cite{Hartley2017} have used a technique based on Support Vector Machines (SVMs) and applied it to KiDS. Instead \cite{Spiniello2018} focused on the search of lensed quasars in KiDS using 3 different morphology based methods.
In this paper we present and test our latest ConvNet lens-finders, improving on the work of \cite{Petrillo2017}. ConvNets (\citealt{fukushima1980neocognitron,lecun1998gradient}) are the state of the art and often the standard choice among machine learning algorithms for pattern recognition in digital images. The winners of the ImageNet Large Scale Visual Recognition Competition (ILSVRC; \citealt{ILSVRC15}; the most important image classification competition) in recent years have all been groups utilizing ConvNets. The advantage of the latter method with respect to other pattern recognition algorithms is that the features are not hand-crafted but are themselves extracted automatically during the training procedure, thus the algorithm decides which features are most representative for classifying the images.
The theoretical basis of ConvNets was developed in the 1980s and the 1990s. However only recently ConvNets have started to outperform other algorithms thanks to the advent of large labeled datasets, improved algorithms and faster training, especially on Graphics Processing Units (GPUs). The interested reader is referred to the Appendix for a brief introduction on ConvNets and to the reviews by \cite{schmidhuber2015deep}, \cite{lecun2015deep} and \cite{guo2016deep} for a more detailed introduction.
ConvNets have been used recently in many astronomical problems, e.g., galaxy morphology classification \citep{dieleman2015rotation,Huertas2015}, estimation of photometric redshifts \citep{Hoyle2016,D'Isanto2018}, spectra classification \citep{Hala2014,Tao2018}, identifying exoplanets \citep{Shallue2018}, transient detection \citep{Cabrera-Vives2017}, galaxy surface brightness estimation \citep{Tuccillo2018}, strong lensing parameters estimation \citep{Hezaveh2017} and star/galaxy separation \citep{Kim2016}.
More importantly, \cite{Metcalf2018} presented the results of a large international {\sl challenge} in which various methods of identifying simulated gravitational lenses were tested blindly. This challenge, the first of a series, sets out to prepare the community for finding lenses in the data of ESA's Euclid mission. Its large data volume requires fast and efficient algorithms to identify strong gravitational lenses. However, the methods were also tested on simulated KiDS data.
ConvNets and Support Vector Machines (SVMs) were recognized to be the most promising methods, among many different methods tested in the challenge.
The ConvNet lens-finders presented in this paper will be applied on $\sim$ 900 sq. deg of the KiDS survey in a forthcoming paper with the purpose of starting a systematic census of strong lenses named ``LinKS" (Lenses in KiDS Survey).
The paper is organized as follows.
In \mbox{Sect.~}\ref{SECtraining}, we illustrate our lens-finding ConvNet-based algorithms and how the training dataset is built.
In \mbox{Sect.~}\ref{SECanalysis}, we evaluate the performances of the ConvNets.
In \mbox{Sect.~}\ref{SECrealdata}, we apply the lens-finders to $\sim22000$ extracted from $\sim255$ square degrees of KiDS for testing the algorithms on real data.
Finally, in \mbox{Sect.~}\ref{SECdiscussion}, we provide a summary and the main conclusions of this work.
\section{training the CONVNETS to find strong lenses}\label{SECtraining}
A Convolutional Neural Network (ConvNet) can be seen as a sequence of non-linear functions, called layers, that create, starting from an input image, a series of increasingly abstract representations of the input called feature maps. The final layer of the ConvNet converts the input feature maps into a set of numbers that represent the outcome of the classification. Hence a ConvNet maps an image onto a single or few numbers.
In our case the output is a single number, denoted by $p$, which can vary between 0 and 1, and it is related to the probability that the input image is a lens (see \citealt{saerens2002} for a detailed discussion).
The parameters of the non-linear functions are obtained during the so called training phase where labeled images are fed to the ConvNet. In more detail, the parameters are derived by minimizing a loss function that expresses the difference between the label values of the images (1 for lenses, 0 for non-strong-lensing systems) and the output $p$ of the ConvNet.
Although in \cite{Petrillo2017} we have used a similar set-up, the aim of this work is to improve the performance of our previous lens-finder.
Currently we use a ConvNet with a \textit{ResNet}-type architecture that has 18 layers, exactly as described in \cite{he2015deep}. ResNet-type architectures are often the preferred choice in image classification tasks because of their faster convergence and higher classification accuracy with respect to other architectures. Moreover, ResNet architectures have already been tested successfully on identifying simulated lenses and they have proven to be one of the best architecture for this task \citep{Schaefer2017,lanusse2018,Metcalf2018}.
We train two different ConvNets with the same architecture except that one takes in input RGB-images composed with \textsc{HumVI}\footnote{{\tt https://github.com/drphilmarshall/HumVI}} \citep{Marshall2016}, while the other takes single \textit{r}-band images as input. We choose the \textit{r}-band as single-band input because the KiDS observing strategy reserves the best seeing conditions for this band (which is used for the weak lensing studies; \citealt{Kuijken2015,Hildebrandt2017}).
The technical details of the ConvNet and of the training procedure are described in Appendix A together with a brief introduction on ConvNets.
To produce the data used to train and validate the ConvNets, we adopt a hybrid approach similarly as done in \cite{Petrillo2017,Jacobs2017,Pourrahmani2018}, creating mock images of strong gravitational lenses using images of real galaxies from KiDS
and super-imposing simulated lensed images.
We adopt this approach because we do not have a sample of genuine KiDS lenses large enough to train a ConvNet (usually of the order of $10^6$).
\subsection{Data}
In this section we describe the dataset used to train the ConvNets, which is composed of real KiDS galaxies and simulated lensed sources.
\subsubsection{Luminous Red Galaxies}\label{sec:LRGs}
We use the sample of Luminous Red Galaxies (LRGs; \citealt{Eisenstein2001}) presented in \cite{Petrillo2017}. We choose to focus on massive early-type galaxies, because it has been estimated that these galaxies form $\sim80\%$ of the lens-galaxy population \citep{turner1984statistics,fukugita1992statistical,kochanek1996flat,chae2003cosmic,oguri2006image,moller2007strong}. Spiral galaxies form the other $\sim$20\% but are much harder to identify.
This training sample of LRGs is a random subset of 6554 galaxies from a parent sample of 21789 selected from 255 square degrees of KiDS DR3 \citep{deJong+17_KiDS_DR3} with the following criteria (see \citealt{Petrillo2017} for more details):
\noindent (i) The low-$z$ ($z<0.4$) LRG colour-magnitude selection of \cite{Eisenstein2001}, adapted to include more sources, both fainter and bluer:
\begin{equation}
\begin{split}
&r<20 \\
&|c_{\rm{perp}}| < 0.2 \\
&r<14+c_{\rm{par}}/0.3 \\
\text{where}\\
&c_{\rm{par}}=0.7(g-r)+1.2[(r-i)-0.18)]\\
&c_{\rm{perp}}=(r-i)-(g-r)/4.0-0.18
\end{split}
\end{equation}
\noindent (ii) A source size in the \textit{r}-band larger than the average FWHM of the PSF of the respective tiles, times an empirical factor to maximize the separation between stars and galaxies.
\subsubsection{Contaminants}\label{sec:contaminants}
Moreover, we have used a set of $\sim 6000$ KiDS sources to train the ConvNets to recognize sources that would likely be incorrectly classified as lenses otherwise, either because they can resemble lensing features or they are ``ghosts'', i.e. they are undetected, at least significantly, in the luminous red galaxies sample discussed in \mbox{Sect.~}\ref{sec:LRGs}.
\begin{itemize}
\item $\sim$2000 sources wrongly classified as lenses in previous tests with ConvNets identified by the authors. This is done to teach the ConvNets not to replicate previous mistakes;
\item $\sim$3000 randomly extracted KiDS sources with \textit{r}-band magnitude brighter than 21. To provide the network with general true negatives.
\item $\sim$1000 KiDS sources visually classified as spiral galaxies from an on-going new project of GalaxyZoo (\citealt{Willett2013}, Kelvin et al., in prep.). This is done to decrease the false positives due to spiral features. To select the galaxies we used a preliminary reduced version of the GAMA-KiDS Galaxy Zoo catalogue for the GAMA09 9h region (see \citealt{GAMAdriver2011} for further details). This catalogue contains $\sim10^4$ sources out to a redshift of $z=0.15$. We select galaxies for which a large majority of people replied to the question \textit{``Is the galaxy in the centre of the image simply smooth and rounded, or does it have features?"} with \textit{``it has features"} \footnote{The actual selection is done by selecting sources from the catalogue with a value of the attribute {\tt features\_features\_frac} larger than $0.6$.}
\end{itemize}
There is a non-zero probability that among the contaminants and the LRGs described in the previous Section there are actual gravitational lenses. We can estimate that the percentage would be of the order of $10^{-2}$ among the contaminants and $\sim1\%$ among the LRGs \citep{Petrillo2017}. Thus, even if real lenses are actually in the training sample, with such a small percentage they would not contaminate the training procedure.
\subsubsection{Mock lensed-sources}\label{SECsims}
We simulate $10^6$ lensed images of $101 \times 101$ pixels,
using the same spatial resolution of KiDS ($\sim0.2$ arcsec per pixel), corresponding to a $20 \times 20$ arcsec field of view.
To produce more realistic lensing systems, we add more complexity both in the source and in the lens plane with respect to the simulations in \cite{Petrillo2017}.
The distribution of the lens and source parameters that we choose for simulating the lensed images are chosen to create a wide range of realistic lensed images. They are not meant to statistically represent a real lens population, since the training set has to be populated sufficiently densely in the parameter space to allow the ConvNets to learn all the possible configurations and to recognize lenses that are rare in a real distribution (or currently even unknown). To ensure this, a more homogeneous distribution of the parameters is advantageous in order not to over-train on the most common lens configurations.
We proceed in the following way, we sample the parameters of the Singular Isothermal Ellipsoid (SIE; \citealt{KSB_SIE94}) and \cite{Sersic68} source models as listed in \mbox{Table~}\ref{TABLEmocksourceslens}. The values of the lens Einstein radius and the source effective radius are drawn from a logarithmic distribution, while the remaining parameters, listed in \mbox{Table~}\ref{TABLEmocksourceslens}, are drawn from a uniform distribution.
In this way our simulation sample contains a higher fraction of smaller rings and arcs compared to \cite{Petrillo2017} for making the new ConvNets more sensitive to this kind of objects with respect to the old one.
The source positions are chosen uniformly within the radial distance of the tangential caustics plus one effective radius of the source S\'ersic profile. This leads our training set to be mostly composed of high-magnification rings, arcs, quads, folds and cusps rather than doubles \citep{schneider1992gravitational}.
To add complexity in the lensed sources, besides the S\'ersic profile, we add between 1 and 5 small circular S\'ersic blobs. The centers of these blobs are drawn from a Gaussian probability distribution function (PDF) around the main S\'ersic source. The width of the standard deviation of the PDF is the same as the effective radius of the main S\'ersic profile. The sizes of the blobs are chosen uniformly within 1-10$\%$ of the effective radius of the main S\'ersic profile. The S\'ersic indices of the blobs are drawn using the same prescription as for the main central source. The amplitudes of the blobs are also chosen from a uniform distribution in such a way that the ratio of the amplitude of an individual blob to the amplitude of the main S\'ersic profile is at most 20$\%$.
Moreover, we add Gaussian Random Field (GRF) fluctuations to the lens potential, which, to a first order approximation, make the lens sample more realistic by adding small scale substructures \citep{SAIKAT}.
The GRF realizations we added in our simulations all follow a power law power-spectrum with a fixed exponent $-6$, which is to the first order a good approximation of substructures in lens plane in the $\Lambda$CDM paradigm \citep{hezaveh}. The variances of the realizations are drawn from a logarithmic distribution between $10^{-4} - 10^{-1}$ about mean zero in the units of square of the lensing potential. This yields both structured sources and lenses that are not perfect SIE.
For each source a realistic color is simulated to create
images in \textit{g}, \textit{r}, \textit{i}-bands.
In order to produce realistic 3-band images we extract magnitudes from ``COSMOS" models in \textsc{Le Phare} (\citealt{Arnouts+99}; \citealt{Ilbert+06}). This library of spectra, consists of 31 models, used for COSMOS photo-z (\citealt{Ilbert+09_COSMOS}). The basic ``COSMOS'' library is composed of 8 templates for elliptical/S0 galaxies, 11 for spiral types, and 12 for galaxies with star-burst ages ranging from
0.03 to 3 Gyr, allowing us to span a wide range of galaxy types and colours. In order to simulate the typical blue arcs observed in most of the observed lenses, we choose models bluer than S0 and calculate observer-frame magnitudes in the three KiDS wavebands \textit{g}, \textit{r} and \textit{i} for model spectra redshifted up to a redshift of $z = 3$ with a $0.1$ binning.
Moreover, to populate the magnitude-space more uniformly, we perturb the three magnitudes adding to each of them a random number uniformly extracted from the range $[-0.1,0.1]$ mag.
We also take into account dust extinction by considering a color excess $E(B-V)$, we extract it from a normal distribution with $\sigma=0.1$ and mean 0 considering only the positives values. In this way we obtain a small extinction correction in order to avoid very red lensed sources which, in the real universe, are much rarer than blue ones.
We adopt a typical extinction curve with $R_V=3.1$, using the relation $A_x = R_x \, E(B-V)$ where $x$ represents the value for the \textit{g}, \textit{r} and \textit{i} SDSS-filters that can be found in \mbox{Table~} 2 of \cite{Yuan13}.
Finally, we convolve the three images with an average KiDS--DR3 PSF for each different band: with a FWHM of $\sim0.86$ arcsec for \textit{g}, $\sim0.68$ arcsec for \textit{r} and $\sim0.81$ arcsec for \textit{i} \citep{deJong+17_KiDS_DR3}.
\begin{table}
\caption{Range of parameter values adopted for simulating the lensed sources. The parameters are drawn uniformly, except for Einstein and effective radius, as indicated. See \mbox{Sect.~}\ref{SECsims} for further details.}
\begin{center}
\begin{tabular}{l l c}
Parameter & Range & Unit \\
\hline
\multicolumn{3}{c}{Lens (SIE)}\\
\hline
Einstein radius & 1.0 - 5.0 (log)& arcsec\\
Axis ratio & 0.3 - 1.0 & -\\
Major-axis angle & 0.0 - 180 & degree\\
External shear & 0.0 - 0.05 & -\\
External-shear angle & 0.0 - 180 & degree\\
\hline
\multicolumn{3}{c}{Main source (S\'ersic)}\\
\hline
Effective radius ($R_{eff}$) & 0.2 - 0.6 (log)& arcsec\\
Axis ratio & 0.3 - 1.0 & -\\
Major-axis angle & 0.0 - 180 & degree\\
S\'ersic index & 0.5 - 5.0 & -\\
\hline
\multicolumn{3}{c}{S\'ersic blobs (1 up to 5)}\\
\hline
Effective radius & $(1\% - 10\%) R_{eff}$ & arcsec\\
Axis ratio & 1.0 & -\\
Major-axis angle & 0.0 & degree\\
S\'ersic index & 0.5 - 5.0 & -\\
\hline
\end{tabular}
\label{TABLEmocksourceslens}
\end{center}
\end{table}
\captionsetup[subfigure]{labelformat=empty}
\begin{figure*}
\centering
\hspace{\fill}
\subfloat[]{\includegraphics[width=35mm]{sim1.jpg}
\subfloat[]{\includegraphics[width=35mm]{sim2.jpg}
\subfloat[]{\includegraphics[width=35mm]{sim7.jpg}
\subfloat[]{\includegraphics[width=35mm]{sim4.jpg}
\subfloat[]{\includegraphics[width=35mm]{sim5.jpg}
\caption{Examples of RGB images of simulated strong lens galaxies used to train the ConvNets. The lens galaxies are observed KiDS galaxies, while the lensed sources are simulated, as described in \mbox{Sect.~}\ref{SECsims}.}
\label{FIGmocklenses}
\end{figure*}
\subsection{Creating the training set}\label{SECcreatingtrain}
The data presented above are used to build the training set which is composed of mock strong-lensing systems (labeled with a 1) and non-strong-lensing systems (labeled with a 0), i.e., objects without lensing features. In the following we outline the procedure used to build the two kinds of objects in the training set.
\\
\\
\noindent {\bf Mock strong-lensing systems}:
To create mock strong-lensing systems we carry out the following procedure:
\\
(i) We randomly choose a mock lensed source (Sect.~\ref{SECsims}) and a LRG (Sect.~\ref{sec:LRGs});
we rescale the brightness of the simulated source to the peak brightness of the LRG in the $r$-band multiplied by a factor $\alpha$ randomly drawn from the interval $[0.02, 0.3]$. This accounts for the typical lower brightness of the lensing features with respect to the lens galaxies;
\\
(ii) we stack the LRG and the mock source for each one of the three bands;
\\
(iii) for the single-band images, we clip the negative values of the pixels to zero and performing a square-root stretch of the image to emphasize lower luminosity features. Instead, we create 3-band images with \textsc{HumVI} that operates an arcsinh stretch of the image following the \cite{Lupton2004} composition algorithm;
\\
(iv) finally, we normalize the resulting images by the galaxy peak brightness (only for single-band images).
Some examples of mock strong-lensing systems obtained in this way are shown in \mbox{Fig.~}\ref{FIGmocklenses}.
\\
\\
\textbf{Non-strong-lensing systems}
To create the non-strong-lensing system sample we carry out the following procedure:
\\
(i) we choose a random galaxy from either the LRG sample (with a probability of 20\%) or from the contaminant sample (80\% probability);
\\
(ii) we clip the negative values of the pixels to zero and performing a square-root stretch of the images. We create 3-band images with \textsc{HumVI};
\\
(iii) we normalize the images by the galaxy peak brightness (only for single-band images).\\
\label{sec:Dataugm}
Finally, we augment the images, which is a standard procedure in machine learning (see e.g., \citealt{Simard2003}). It is used to avoid over-fitting by expanding artificially the training set through different transformations of the images.
Before feeding the images to the ConvNets, we apply the following transformations:
\noindent
(i) a random rotation between 0 and $2\pi$;
\\
(ii) a random shift in both $x$ and $y$ direction between -4 and +4 pixels;
\\
(iii) a $50\%$ probability of horizontally flipping the image;
\\
(iv) a rescaling with a scale factor sampled log-uniformly between $1$ and $1.1$.
\noindent
All transformations are applied to both the mock strong-lensing systems and the non-strong-lensing systems.
The final set of inputs of the ConvNets are postage stamps of 101 times 101 pixels which correspond to $\sim 20 \times 20$ arcsec.
The images are produced in real-time during the training phase. For more details on the training phase see Appendix A.
\section{Analysis}\label{SECanalysis}
After the training is completed, the ConvNets must be tested in order to assess whether the training was successful.
In this section we define the metric for evaluating the results and evaluate the performances of the ConvNets on a dataset composed by non-lenses and mock lenses in comparison to \cite{Petrillo2017}.
\begin{figure}
\begin{center}
{\includegraphics[width=90mm]{ROC_curves.png}}
\caption{ROC curves for the 1-band (blue) and 3-band (red) ConvNet. Each point of the curves is the true positive rate vs. false positive rate for different values of threshold for $p$ (decreasing from left to right; some values are shown on the curve for reference).}
\label{ROC curves}
\end{center}
\end{figure}
\begin{figure*}
\begin{center}
{\includegraphics[width=85mm,height=68mm]{foo2.png}}
{\includegraphics[width=85mm,height=68mm]{foo.png}}
\caption{Percentage of false negatives (i.e., the percentage of lenses that have been misclassified) for bins of $R_E$ and $\alpha$ defined as the ratio between the peak brightness of the lens galaxy and the lensed source.}
\label{FIG_fp_er_cont}
\end{center}
\end{figure*}
\begin{figure}
\begin{center}
{\includegraphics[width=90mm]{n_detect.png}}
\caption{Number of detections as a function of the threshold for $p$ for the 1-band (blue) and the 3-bands ConvNet (red) compared to the ConvNet of \citet{Petrillo2017}(grey).}
\label{n_detections}
\end{center}
\end{figure}
\subsection{Performance metric}
To evaluate the performances of the ConvNets we use:
\begin{itemize}
\item the true-positive rate (TPR), which measures the fraction of positive objects (in our case the lenses) detected by the algorithm. It is given by the ratio between the number of real positive (the number of real lenses that algorithm finds) and the sum of the latter and the number of false negatives (the lenses that the algorithm does not find):
\begin{equation}
\mathrm{TPR} = \dfrac{N_{\mathrm{TruePositives}}}{N_{\mathrm{TruePositives}}+N_{\mathrm{FalseNegatives}}} \in [0,1] ;
\end{equation}
\item the false-positive rate (FPR), which measures the fraction of negative objects (non-strong-lensing systems) misclassified as positives (lenses). It is given by the ratio between the number of false positive (the number of not lenses that algorithm misclassifies as lenses) and the sum of the latter and the number of true negatives (the non lenses that the algorithm classifies correctly)
\begin{equation}
\mathrm{FPR} = \dfrac{N_{\mathrm{FalsePositives}}}{N_{\mathrm{TrueNegatives}}+N_{\mathrm{FalsePositives}}} \in [0,1] ;
\end{equation}
\item these two quantities can be used to build Receiver Operating Characteristic (ROC) curves which allow to check at a glance the degree of completeness and contamination of a binary classifier. ROC curves are created by plotting $\mathrm{TPR}$ as a function of $\mathrm{FPR}$ varying the threshold of detection for $p$ between 0 and 1. This allows us to tune the value for the threshold for $p$ in order to get the desired amount of $\mathrm{TPR}$ and $\mathrm{FPR}$ for a given classification problem. In our case $p$ is the output of the ConvNet and we can tune the $p$-threshold depending how many lens candidates we desire and what level of contamination is deemed to be acceptable.
\end{itemize}
\subsection{Performance}
The ROC curves for a test-set composed of 5000 mock strong-lensing systems and 5000 non-strong-lensing systems created, as described in Sec. 2, are shown in Fig. \ref{ROC curves}. In general the 3-bands ConvNet has a better performance than the 1-band ConvNet, retrieving more mock strong-lensing systems than the 1-band ConvNet. On the contrary, the 1-band ConvNet is less contaminated by false positives at higher values of the threshold for $p$.
Since gravitational lenses are rare events, it is important to keep a low value of FPR. Otherwise a candidate sample selected from real data would be dominated by false positives and a large amount of time would be needed to discard them through a visual inspection.
In \mbox{Fig.~}\ref{FIG_fp_er_cont} we show, for a fiducial value for the threshold of the detection $p=0.8$, the percentage of false negatives (i.e., the percentage of lenses that have been misclassified) as a function of the Einstein radius, $R_E$, and the source over lens-galaxy brightness contrast, $\alpha$, defined in \mbox{Sect.~}\ref{SECcreatingtrain}. Lenses with small Einstein radii and low-contrast lensed images are, as expected, the ones with which the ConvNets struggle the most. This suggests that our mock lens training samples currently covers the range in which lenses are found most easily. Smaller lenses are effectively smeared to an unrecognizable configuration by the PSF and fainter lensed sources will be too noisy to detect.
In \mbox{Fig.~}\ref{FIG_fp_er_cont} we also see that the accuracy decreases for larger Einstein radii, possibly due to the fact that we covered the Einstein radius with logarithmic distribution, focusing slightly more on the small-separation systems, and secondly because their lensed images are more likely to blend in to the local environment and therefore harder to distinguish from nearby galaxies by the ConvNets.
However, because our goal is to efficiently select true strong-lenses in real astronomical observations, therefore it is necessary to assess the TPR and FPR when the ConvNets are applied to real data where the performance might be worse than on a simulated dataset.
\begin{figure*}
\begin{center}
{\includegraphics[width=85mm]{single_compar.png}}
{\includegraphics[width=85mm]{multi_compar.png}}
{\includegraphics[width=85mm]{dist-p_single2.png}}
{\includegraphics[width=85mm]{dist-p_multi2.png}}
\caption{The first row shows the distribution of the scores for the 56 candidates with $p>0.5$ selected in \citealt{Petrillo2017}. The scores of the 1-band (blue) and 3-bands ConvNet (red) are compared with those of \citet{Petrillo2017} (grey). The second row shows the same for a subsample of 22 candidates selected as described in \mbox{Sect.~}\ref{SECresults}.}
\label{FIGhigh_like}
\end{center}
\end{figure*}
\begin{figure*}
\begin{center}
{\includegraphics[width=38mm]{rgb5.png}}
{\includegraphics[width=38mm]{KSL327.png}}
{\includegraphics[width=38mm]{KSL627.png}}
{\includegraphics[width=38mm]{KSL376.png}}
{\includegraphics[width=38mm]{669.png}}
{\includegraphics[width=38mm]{620.png}}
{\includegraphics[width=38mm]{606.png}}
{\includegraphics[width=38mm]{086.png}}
{\includegraphics[width=38mm]{046.png}}
{\includegraphics[width=38mm]{94.png}}
{\includegraphics[width=38mm]{342.png}}
{\includegraphics[width=38mm]{13.png}}
{\includegraphics[width=38mm]{351.png}}
{\includegraphics[width=38mm]{686.png}}
{\includegraphics[width=38mm]{465.png}}
{\includegraphics[width=38mm]{spaziobianco.png}}\hspace{-2.5pt}
\caption{Sources classified by the new ConvNets with $p<0.5$ which were classified in \citealt{Petrillo2017} with $p>0.5$. On each image the score assigned by the single-band ConvNet (blue), the multi-band ConvNet (red) and \citet{Petrillo2017}'s ConvNet (grey) are reported.}
\label{FIGpm05single}
\end{center}
\end{figure*}
\section{Application to real data}\label{SECrealdata}
Testing the ConvNets on real data is fundamental, since the algorithms have been trained on a mixture of real and simulated data. It is not trivial how the method will perform on a completely real dataset, since the domain of application is slightly different with respect to the domain where the ConvNets have been trained on.
Ideally, the morphologies in the ensemble of simulated strong-lens systems and non-strong-lens systems would be a fair representation of all morphologies observed in their equivalents in real observations.
Hence, to properly analyze the performances of the ConvNets, we apply them to the full LRG sample composed of 21789 galaxies extracted from 255 square degrees as described in \mbox{Sect.~}\ref{sec:LRGs}.
Using the same LRG sample of \cite{Petrillo2017} allows us to assess whether there has been any improvement with respect to our previous work.
For each galaxy image we opt to obtain an average prediction given by the average of the $p$'s for the original image and the images obtained operating a rotation of 90, 180 and 270 degrees respectively. Generally this procedure allows to increase the accuracy of the classifications.
\subsection{Results on the LRG sample}\label{SECresults}
In Fig. \ref{n_detections} we show the number of lens candidates detected varying the threshold $p$. The 1-band ConvNet detects more lens candidates compared to the 3-band one for any given threshold for $p$.
For each of the three ConvNets it holds that the lower the threshold in p is set, the more candidates will have to be inspected visually.
In other words, one wants to set as the threshold to an as low as possible value that yields both a sufficiently large sample of candidates and a sufficiently high TPR for the purpose of the scientific project.
In \cite{Petrillo2017} we used visual inspection to select visually promising strong-lens candidates within the sample of systems assigned with a $p>0.5$ by the ConvNet. This sample contains 56 candidates.
Moreover, in \cite{Petrillo2017} we selected a subsample of 22 candidates based on the agreement between their expected Einstein radii, computed from the stellar mass or the velocity dispersion of the candidate lens galaxies, and the actual galaxy-image configurations. This does not guarantee that the 22 candidates are actual lenses but it allows us to exclude the cases with more implausible configurations. \mbox{Fig.~}\ref{FIGhigh_like} compares the $p$-values for these two samples assigned by the two new ConvNets to those assigned by \cite{Petrillo2017} ConvNet.
We note that the $p$-values of the new ConvNets have a noticeable peak at high values that becomes even more pronounced considering only the 22 candidates. In particular, the single-band ConvNet selects high-confidence candidates assigning high values of $p$.
This is a fair improvement of
the performance of the algorithm since there is a
larger clustering of the higher visually ranked candidates toward the high $p$-values.
Instead in \mbox{Fig.~}\ref{FIGpm05single} we show the subset of the 56 galaxies that the new ConvNets classify with $p<0.5$.
For the 1-band finder there are not clear candidates that would be lost: maybe a couple of galaxies could be considered acceptable lens candidates, while for the rest a low $p$-value it is the ideally desired output. In particular, three candidates (third, fourth and fifth galaxy in \mbox{Fig.~}\ref{FIGpm05single}, which have been selected as lenses in \cite{Petrillo2017} by visual inspection but after a more careful analysis have been revealed as false positives (likely a merger and two ring galaxies), are classified as non-lenses.
Thus, the new finder does cumulatively a better job in excluding contaminants and selecting lens candidates.
Instead the 3-bands lose some acceptable candidates, but more importantly misidentifies a known gravitational lens and a clear good candidate (first and second galaxy in \mbox{Fig.~}\ref{FIGpm05single}).
This needs further investigation, thus, in the following subsection, we analyze the behavior of the two lens-finders on a small sample composed by real lenses and striking lens candidates.
\subsection{Application to a small sample of clear lens candidates}\label{SECrepsample}
Additional insights on ConvNet performance can be obtained from inspecting the results on a set of real lenses and striking lens candidates.
We gather a set of 6 galaxies composed as follows (see Fig. \ref{FIGrepresentative}):
\begin{itemize}
\item The four confirmed lenses known in literature which are present in our LRG sample: J085446-012137 \citep{Limousin2010}, J114330-014427 and J1403+0006 \citep{SLACS2008},
J115252+004733 \citep{More2017};
\item Three extremely likely lenses found in \cite{Petrillo2017}, i.e. KSL713, KSL327 and KSL427;
\\
The $p$-values for each of these galaxies are shown in \mbox{Fig.~}\ref{FIGrepresentative} and \mbox{Table~}\ref{TABpvalues}. It is immediately noticeable that the 1-band ConvNet is the best-performing one: the 3 extremely likely lenses and 3 of the 4 known lenses are identified correctly with a very high $p$-value;
Instead, the 3-bands ConvNet gives more weight
to the source colours: it identifies easily blue-features but it struggles with other colours. This could be due to the larger number of training examples with blue colours.
Moreover, both the new ConvNets presented in this paper are able to pick up the quad lens J115252+004733 not selected in \citep{Petrillo2017}.
This piece of improvement is possibly due to the improved realism of the mock lensed sources adopted for this new ConvNets, collecting a larger variety of configurations with respect to \cite{Petrillo2017}.
To validate this hypothesis, we train another 1-band ConvNet with the same mock lensed sources of \cite{Petrillo2017}. This new ConvNet does not detect the quad (see \mbox{Table~}\ref{TABpvalues}), thus confirming the idea that the realism of the simulations plays a key-role in the performance of the lens-finder.
To further understand the role of the simulations, we also train a ConvNet with the same configuration of \cite{Petrillo2017} but with the simulated strong-lensing systems produced for this work (see \mbox{Sect.~}\ref{SECsims}). Also in this case the quad is detected (see \mbox{Table~}\ref{TABpvalues}) even if the performance is worse than the new 1-band ConvNet. This could be due to the different architecture of the two ConvNets.
To conclude, despite the limited size of the control sample presented in this section, the 1-band ConvNet is the one which has generally the best performance under all conditions and it seems the best one to systematically apply to real data, both in terms of purity and completeness. The 3-band set-up, instead, is generally biased toward bluer colours and sometimes has failed to select known or very promising strong-lensing systems.
\end{itemize}
\begin{table*}
\caption{Scores given by ConvNets on the sample described in \mbox{Sect.~}\ref{SECrepsample}. The 1-band and 3-band ConvNets presented in this work are compared with \citet{Petrillo2017} (old ConvNet). In the 5th column are shown the scores for a ConvNet with the same architecture of the 1-band ConvNet but trained with the mock lensed sources used in \citet{Petrillo2017} (1-band/old mocks). In the 6th column are reported the scores for a ConvNet with the same architecture of \citet{Petrillo2017} but trained with the same training set of this work (old ConvNet/new mocks).}
\begin{center}
\begin{tabular}{l c c c c c}
\hline
ID & 1-band & 3-bands & old ConvNet & 1-band/old mocks & old ConvNet/new mocks \\
\hline
\multicolumn{6}{c}{}\\
J085446-012137 & 1.000 & 0.998 & 0.696 & 0.148 & 0.996 \\
J114330-014427 & 0.998 & 0.403 & 0.997 & 0.947 & 0.605 \\
J1403+0006 & 0.360 & 0.456 & <0.5 & 0.038 & 0.419 \\
J115252+004733 & 0.992 & 0.887 & <0.5 & 0.321 & 0.904 \\
KSL713 & 0.999 & 0.611 & 0.942 & 0.990 & 0.942 \\
KSL427 & 0.999 & 0.999 & 0.943 & 0.982 & 0.752 \\
KSL327 & 0.933 & 0.012 & 0.522 & 0.587 & 0.479 \\
\hline
\end{tabular}
\label{TABpvalues}
\end{center}
\end{table*}
\captionsetup[subfigure]{labelformat=empty}
\begin{figure*}
\subfloat[J085446-012137]{\includegraphics[width=38mm]{rgb2.png}}
\hspace{\fill}
\subfloat[J114330-014427]{\includegraphics[width=38mm]{rgb5.png}}
\hspace{\fill}
\subfloat[J1403+0006]{\includegraphics[width=38mm]{otherknown.png}}
\hspace{\fill}
\subfloat[J115252+004733]{\includegraphics[width=38mm]{quad.png}}
\hspace{\fill}
\subfloat[KSL713]{\includegraphics[width=38mm]{KSL713.png}}
\hspace{\fill}
\subfloat[KSL427]{\includegraphics[width=38mm]{KSL427.png}}
\hspace{\fill}
\subfloat[KSL327]{\includegraphics[width=38mm]{KSL327.png}}
\hspace{\fill}
\subfloat[]{\includegraphics[width=38mm]{spaziobianco.png}}
\caption{RGB images of the previously known lenses in our test-sample (first row) and the extremely likely lenses found by \citealt{Petrillo2017} in the same test-sample (second row), this sample is discussed in \mbox{Sect.~}\ref{SECrepsample}. The scores assigned by the single-band ConvNet (blue), the multi-band ConvNet (red) and \citet{Petrillo2017}'s ConvNet (grey) are reported on each image.}
\label{FIGrepresentative}
\end{figure*}
\begin{figure*}
\begin{center}
{\includegraphics[width=88mm]{ROC_curves_real_1.png}}
{\includegraphics[width=88mm]{ROC_curves_real_2.png}}
{\includegraphics[width=88mm]{TP_FP_1.png}}
{\includegraphics[width=88mm]{TP_FP_2.png}}
\caption{In the first row we show ROC curves of the 3 ConvNets built for a real sample of Luminous Red Galaxies (LRGs) as described in \mbox{Sect.~}\ref{SECvisisp}.
On the left is shown the result for a sample including also more dubious candidates; on the right for a sample with solid candidates only.
On the bottom row we plot the absolute number of True Positive vs. False positives for the same samples.}
\label{FIGROCreal}
\end{center}
\end{figure*}
\subsection{Visual inspection}\label{SECvisisp}
To further analyze the performances of the ConvNets, we visually inspect the sources in the LRG sample (see \mbox{Sect.~}\ref{SECrealdata}) selected by the ConvNets with a threshold $p>0.8$. To compare the results, we visually inspect the sources selected with the same threshold, $p>0.8$, by the ConvNet in \cite{Petrillo2017}. This yields 721 sources to inspect for the 1-band ConvNet, 390 for the 3-band ConvNet and 292 sources for the ConvNet of \cite{Petrillo2017} ConvNet. The visual inspection is carried out by 3 observers who have three possible choices: ``Yes, it is a lens", ``Maybe it is a lens", ``No, it is not a lens". For each source the classifiers visualize the \textit{g}, \textit{r} and \textit{i} fits files and a \textit{g-r-i} composed image with the software STIFF\footnote{\href{http://www.astromatic.net/software/stiff}{\tt http://www.astromatic.net/software/stiff}} \citep{Bertin2012}.
After the visual inspection, we build ROC curves assigning a ground truth for each source from the result of the visual inspection. ROC curves are shown in \mbox{Fig.~}\ref{FIGROCreal} for two different ground truths: one where we assign the label ``lens" to the sources that received at least two times \textit{Maybe}, and another where ``lenses" are the sources rated with 3 times \textit{Maybe} or at least one \textit{Yes} and one \textit{Maybe}. In this way we have two different ground truths: one where also dubious lens candidates are labeled as ``lenses", and one more conservative where only more solid lens candidates are labeled as ``lenses".
In \mbox{Fig.~}\ref{FIGROCreal} we also show the absolute number of true positives versus false positives as a function of the threshold $p$ for the same ground truths.
The ConvNets presented in this paper have both higher purity and completeness with respect to \cite{Petrillo2017} and a larger number of lens candidates retrieved.
Moreover, the 1-band ConvNet has higher purity and completeness, especially for the more conservative sample, for sources with higher values of $p$ as also the results of \mbox{Sect.~}\ref{SECrepsample} seems to indicate.
In addition, in \mbox{Fig.~}\ref{FIGtp_fp_p} we show for the conservative classification the number of candidates detected by the ConvNets varying the threshold for $p$ and how many of these are False Positives.
The True Positives
and False Positives
do not grow at the same pace as a function of the threshold $p$ of detection.
Moreover, the percentage of false positives is similar for the two new ConvNets.
From the visual inspection analysis we conclude that the 1-band ConvNet would be the best choice if one wants to have a small sample of good lens candidates with few contaminants and thus less images to inspect visually.
\subsection{Prospects for Euclid}\label{SECprospects}
We have employed about 1 hour to visually inspect the 721 targets selected by the 1-band ConvNet, selecting 60 lens candidates in the more conservative case
(see previous subsection). For current surveys, investing such an amount of time in visual inspection is still reasonable for obtaining a more complete sample of lens candidates. However, for future surveys, the time dedicated to visually inspect the targets must be minimized considering the extremely higher number of resolved targets in the survey data.
Nevertheless, one can choose a higher threshold of detection for the ConvNet and spend considerably less time in visually inspecting the targets.
For example, if we select a threshold $p=0.9919$ for the 1-band ConvNet, which is the threshold to retrieve 3 of the 4 known lenses presented in \mbox{Sect.~}\ref{SECrepsample}, we obtain a total of 35 sources selected by the 1-band ConvNet. These 35 sources
are composed of
14 positives and 21 negatives, considering the more conservative result from the visual classification. This translates to roughly $\sim40\%$ purity, assuming that our visual inspection is accurate enough.
By using the code \textsc{LensPop} by \cite{collet2015}, we estimate that there are, in 255 square degree of KiDS, $\sim125$ lenses with a Einstein radius larger than 1 arcsecond and with our LRG colour cut selection (which corresponds to a cut in redshift of $z \mathrel{\rlap{\lower3.5pt\hbox{\hskip0.5pt$\sim$} 0.5$, as shown in Fig. 8 in \citealt{Petrillo2017}). Hence, we can very roughly say that a candidate sample selected by the 1-band ConvNet with the fiducial threshold $p=0.9919$ is $\sim40\%$ pure and $\sim11\%$ (i.e. 14 out of 125) complete by considering a population similar to that on which we trained our ConvNets.
To translate this result in a prediction for Euclid, assuming that ConvNets will perform at least as well on the same domain on Euclid data as they do on KiDS data, we can proceed in the following way.
\cite{collet2015} predicted that there will be $\sim170000$ discoverable lenses in Euclid. If we consider only the lenses with an Einstein radius larger than 1 arcsecond and with a redshift $z < 0.5$, the number reduces to $\sim20000$. Therefore, with the fiducial threshold of $p=0.9919$, it should be possible to select, from a sample of Euclid LRGs, $\sim2200$ good gravitational lens candidates, by visually inspecting $\sim 5500$ galaxies.
Considering that an observer needs $\sim5$ seconds to inspect a lens candidate, that would imply that $\sim8$ hours of visual inspection would be already enough to obtain a sample of lens candidates far exceeding in size to the total number of currently known gravitational lenses. Nevertheless, this is an extremely conservative estimate, since Euclid data will have better spatial sampling (0.1 arcseconds per pixel), PSF FWHM ($\sim 0.2$ arcseconds), and thus image quality, allowing algorithms and humans to better identify lensing features. Moreover, it will be possible to train the algorithms on a wider parameter space, and thus retrieving.a larger number of lenses.
We conclude however that current ConvNets, without much adaptation, can yield already enormous numbers of lenses from Euclid data without much human effort.
\section{Discussion and Conclusions}\label{SECdiscussion}
Automatic lens finders have become a standard method for finding gravitational lens candidates in survey data.
They are crucial to cope with the large set of an estimated $\sim10^6$ gravitational lens candidates that will be produced by upcoming surveys such as Euclid and LSST.
Therefore, it is important to build and validate such lens-finders in order to be ready for this anticipated data-avalanche.
It has been shown that Convolutional Neural Networks (ConvNets) are one of the most promising methods for finding lens candidates (see, e.g., \citealt{Metcalf2018}). ConvNets achieve outstanding results when the data used to train the algorithm are very similar to the data where the algorithm is intended to be applied. However, this is not necessarily the case when the domain of application is different from the domain where the algorithm is trained on. There is an active field of research investigating how to adapt the algorithms, trained on one domain to other domains (see, e.g., \citealt{csurka2017domain} for a review).
In all published cases, ConvNet lens-finders are trained using simulated mock lensed sources. Moreover, in many cases, the lens-finders are tested only on fully simulated datasets. This does not ensure that the lens-finders will perform in a satisfactory way on real survey data as shown in this paper.
It is important to conduct a thorough analysis of the performances on real data. Optimally, one would like to build a benchmark where all the lens-finders can be tested and compared against. Data from surveys such as KiDS and DES, with their identified lens candidates, could be used to build such a benchmark.
In this paper we have tested two lens-finders based on a Convolutional Neural Network (ConvNet) for selecting strong gravitational lens candidates in the Kilo-Degree Survey (KiDS). One finder just uses \textit{r}-band images while the other uses RGB images composed with \textit{g}, \textit{r}, \textit{i} images.
To train the algorithms, we have generated a large sample of simulated lensing features on top of real colour-magnitude selected galaxies from KiDS.
Both the lens-finders are able to identify real lenses and good lens candidates in the survey.
The performance of the two lens-finders is similar but the 3-bands finder seems to under-perform when the lensed sources do not exhibit a bluer colour. This is most likely due to the fact that the mock lensed sources in the training set have mostly blue colours. Although genuine lensed sources are usually blue, this could select against non-blue sources. One way of dealing with this issue could be populating the mock lensed sources with a wider selection of colours and, in addition, dividing the training set in different classes of colours. This would help not only the ConvNet to explicitly classify sources with different colours but could also improve the general classification given the new information provided.
In any case, the power of the single-band set-up is particularly encouraging in view of the Euclid mission data which will rely especially on the VIS band.
In addition, we have tested and compared the lens-finders with a similar one presented in \cite{Petrillo2017}. The lens-finders presented in this work have a better performance, i.e., they have both a better completeness and purity (\mbox{Sect.~}\ref{SECvisisp}) and also they tend to classify more probable lens candidates with higher output values (see \mbox{Sect.~}\ref{SECrepsample}).
This is a fair improvement of the performance which implies that selecting candidates with high output values we will have a purer sample which turns out to be convenient for visual inspection and/or spectroscopic follow-ups. Indeed, the larger the future survey will become (see e.g. Euclid and LSST) the more prohibiting the visual inspection of candidates will be, hence the purity of machine learning tools will be crucial for the automatic selection of sources to be set on the queue of spectrographs for their spectroscopic confirmations.
The differences between this work and \cite{Petrillo2017} can be summarized in three points:
\begin{itemize}
\item more complex mock lensed source simulations;
\item a modified ConvNet architecture;
\item a slightly larger training set.
\end{itemize}
These differences have contributed to the improvement of the performance, but our analysis presented in \mbox{Sections~}\ref{SECrepsample} indicates that the main reason is the improved mock lensed sources simulations. In this work the simulated sample is more populated with smaller sized lensed source galaxies and the sources exhibit more complexity in their structure, i.e., the presence of substructures in the lensed sources and a Gaussian Random Field perturbing the lens potential.
The ConvNet lens-finders can be tuned in terms of completeness and purity according to the specific needs of the science project.
If one wants to have a more complete sample of lenses, a low threshold of detection can be chosen and the lower purity can be corrected by visually inspecting the candidates, something still feasible for current surveys.
On the other hand, a purer (but less complete) sample of candidates can be obtained choosing a higher threshold. We have shown in \mbox{Sect.~}\ref{SECprospects} that by using a high threshold for detection will be already enough to retrieve in Euclid a sample of lenses exceeding in size the total number of currently known gravitational lenses.
A series of possible improvements can be applied to the lens-finders.
As we have shown, the performance strongly depends on the composition of the training set. Hence, making the lens simulations more realistic and using real gravitational lens candidates in the training set would probably improve the quality of the classification.
Also enlarging the training set with more KiDS galaxies would probably help, as well as adding more labels for helping the ConvNets to discriminate between different classes of galaxies (e.g., adding labels for ellipticals and spirals).
Moreover, particular care can be put in producing lens simulations where the S/N is always high enough for the lensing features to be recognizable by an observer.
Another possibility would be to specialize the ConvNets in recognizing different kinds of sources (e.g., large/small-separation systems or extended/compact-sources). This could be obtained by either training different ConvNets with specialized training sets or using a single ConvNet trained with a unique training set but with multiple labels rather than with a binary classification scheme.
Instead, on the algorithm side, a trivial improvement could be the so called \textit{ensemble averaging}, i.e., averaging the output of different ConvNets in order to possibly reduce the statistical noise of the classification. An approach experimented by, e.g., \cite{Schaefer2017} for identifying strong lens simulations.
Finally, in a forthcoming paper we will apply the algorithms to $\sim900$ square degrees of the KiDS survey starting a systematic census of strong gravitational lens candidates named \textit{``LinKS"} (Lenses in the Kilo Degree Survey).
\begin{figure*}
\begin{center}
{\includegraphics[width=88mm]{TP_FP.png}}
{\includegraphics[width=88mm]{TP_FP_3.png}}
\caption{Results of the more conservative visual classification of the candidates selected by the ConvNets (\mbox{Sect.~}\ref{SECvisisp}). On the left: number of False Positives and True Negatives for the 1-band and 3-bands ConvNet as a function of the threshold of detection $p$.
On the right: percentage of False Positives as a function of the threshold of detection $p$ for both the ConvNets.}
\label{FIGtp_fp_p}
\end{center}
\end{figure*}
\section*{Acknowledgements}
CEP thanks Leon Doddema, Ewout Helmich, Jelte de Jong and Kateryna Frantseva for help and support.
CEP, CT, GV, and LVEK are supported through an NWO-VICI grant (project number 639.043.308). The authors thank the anonymous referee for the insightful report.
SC has been financially supported by a grant (project number 614.001.206) from the Netherlands Organization for Scientific Research (NWO).
GVK acknowledges financial support from the Netherlands Research School for Astronomy (NOVA) and Target. Target is supported by Samenwerkingsverband Noord Nederland, European fund for regional development, Dutch Ministry of economic affairs, Pieken in de Delta, Provinces of Groningen and Drenthe. NRN acknowledges financial support from the European Union Horizon 2020 research and innovation programme under the Marie Sklodowska-Curie grant agreement N. 721463 to the SUNDIAL ITN network.
GAMA is a joint European-Australasian project based around a spectroscopic campaign using the Anglo-Australian Telescope. The GAMA input catalogue is based on data taken from the SDSS and the UKIDSS. Complementary imaging of the GAMA regions is being obtained by a number of independent survey programmes including GALEX MIS, VST KiDS, VISTA VIKING, WISE, Herschel-ATLAS, GMRT, and ASKAP, providing UV to radio coverage. GAMA is funded by the STFC (UK), the ARC (Australia), the AAO, and the participating institutions. The GAMA website is www.gama-survey.org.
Based on data products from observations made with ESO Telescopes at the La Silla Paranal Observatory under programme IDs 177.A-3016, 177.A-3017, and 177.A-3018, and on data products produced by Target/OmegaCEN, INAF-OACN, INAF-OAPD, and the KiDS production team, on behalf of the KiDS consortium. OmegaCEN and the KiDS production team acknowledge support by NOVA and NWO-M grants. Members of INAF-OAPD and INAF-OACN also acknowledge the support from the Department of Physics and Astronomy of the University of Padova, and of the Department of Physics of University of Federico II (Naples).
This publication has been made possible by the
participation in the Galaxy Zoo project of more than 20.000 volunteers from around the world, with almost 2 million classifications provided. Their contributions are individually acknowledged at
http://authors.galaxyzoo.org/.
The data are generated via the Zooniverse.org platform, development of which is funded by generous support, including a Global Impact Award from Google, and by a grant from the Alfred P. Sloan Foundation.
\bibliographystyle{mnras}
\section{Introduction}
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\subsection{Title}
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Keith T. Smith,$^{1}$
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\subsection{Sections}
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\section{Mathematics and symbols}
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\subsection{Equations}
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\begin{verbatim}
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\subsection{Special symbols}
\begin{table}
\caption{Additional commands for special symbols commonly used in astronomy. These can be used anywhere.}
\label{tab:anysymbols}
\begin{tabular}{lll}
\hline
Command & Output & Meaning\\
\hline
\verb'\sun' & \sun & Sun, solar\\[2pt]
\verb'\earth' & \earth & Earth, terrestrial\\[2pt]
\verb'\micron' & \micron & microns\\[2pt]
\verb'\degr' & \degr & degrees\\[2pt]
\verb'\arcmin' & \arcmin & arcminutes\\[2pt]
\verb'\arcsec' & \arcsec & arcseconds\\[2pt]
\verb'\fdg' & \fdg & fraction of a degree\\[2pt]
\verb'\farcm' & \farcm & fraction of an arcminute\\[2pt]
\verb'\farcs' & \farcs & fraction of an arcsecond\\[2pt]
\verb'\fd' & \fd & fraction of a day\\[2pt]
\verb'\fh' & \fh & fraction of an hour\\[2pt]
\verb'\fm' & \fm & fraction of a minute\\[2pt]
\verb'\fs' & \fs & fraction of a second\\[2pt]
\verb'\fp' & \fp & fraction of a period\\[2pt]
\verb'\diameter' & \diameter & diameter\\[2pt]
\verb'\sq' & \sq & square, Q.E.D.\\[2pt]
\hline
\end{tabular}
\end{table}
\begin{table}
\caption{Additional commands for mathematical symbols. These can only be used in maths mode.}
\label{tab:mathssymbols}
\begin{tabular}{lll}
\hline
Command & Output & Meaning\\
\hline
\verb'\upi' & $\upi$ & upright pi\\[2pt]
\verb'\umu' & $\umu$ & upright mu\\[2pt]
\verb'\upartial' & $\upartial$ & upright partial derivative\\[2pt]
\verb'\lid' & $\lid$ & less than or equal to\\[2pt]
\verb'\gid' & $\gid$ & greater than or equal to\\[2pt]
\verb'\la' & $\la$ & less than of order\\[2pt]
\verb'\ga' & $\ga$ & greater than of order\\[2pt]
\verb'\loa' & $\loa$ & less than approximately\\[2pt]
\verb'\goa' & $\goa$ & greater than approximately\\[2pt]
\verb'\cor' & $\cor$ & corresponds to\\[2pt]
\verb'\sol' & $\sol$ & similar to or less than\\[2pt]
\verb'\sog' & $\sog$ & similar to or greater than\\[2pt]
\verb'\lse' & $\lse$ & less than or homotopic to \\[2pt]
\verb'\gse' & $\gse$ & greater than or homotopic to\\[2pt]
\verb'\getsto' & $\getsto$ & from over to\\[2pt]
\verb'\grole' & $\grole$ & greater over less\\[2pt]
\verb'\leogr' & $\leogr$ & less over greater\\
\hline
\end{tabular}
\end{table}
Some additional symbols of common use in astronomy have been added in the MNRAS class. These are shown in tables~\ref{tab:anysymbols}--\ref{tab:mathssymbols}. The command names are -- as far as possible -- the same as those used in other major astronomy journals.
Many other mathematical symbols are also available, either built into \LaTeX\ or via additional packages. If you want to insert a specific symbol but don't know the \LaTeX\ command, we recommend using the Detexify website\footnote{\url{http://detexify.kirelabs.org}}.
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\subsection{Ions}
A new \verb'\ion{}{}' command has been added to the class file, for the correct typesetting of ionisation states.
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\section{Figures and tables}
\label{sec:fig_table}
Figures and tables (collectively called `floats') are mostly the same as built into \LaTeX.
\subsection{Basic examples}
\begin{figure}
\includegraphics[width=\columnwidth]{example}
\caption{An example figure.}
\label{fig:example}
\end{figure}
Figures are inserted in the usual way using a \verb'figure' environment and \verb'\includegraphics'. The example Figure~\ref{fig:example} was generated using the code:
\begin{verbatim}
\begin{figure}
\includegraphics[width=\columnwidth]{example}
\caption{An example figure.}
\label{fig:example}
\end{figure}
\end{verbatim}
\begin{table}
\caption{An example table.}
\label{tab:example}
\begin{tabular}{lcc}
\hline
Star & Mass & Luminosity\\
& $M_{\sun}$ & $L_{\sun}$\\
\hline
Sun & 1.00 & 1.00\\
$\alpha$~Cen~A & 1.10 & 1.52\\
$\epsilon$~Eri & 0.82 & 0.34\\
\hline
\end{tabular}
\end{table}
The example Table~\ref{tab:example} was generated using the code:
\begin{verbatim}
\begin{table}
\caption{An example table.}
\label{tab:example}
\begin{tabular}{lcc}
\hline
Star & Mass & Luminosity\\
& $M_{\sun}$ & $L_{\sun}$\\
\hline
Sun & 1.00 & 1.00\\
$\alpha$~Cen~A & 1.10 & 1.52\\
$\epsilon$~Eri & 0.82 & 0.34\\
\hline
\end{tabular}
\end{table}
\end{verbatim}
\subsection{Captions and placement}
Captions go \emph{above} tables but \emph{below} figures, as in the examples above.
The \LaTeX\ float placement commands \verb'[htbp]' are intentionally disabled.
Layout of figures and tables will be adjusted by the publisher during the production process, so authors should not concern themselves with placement to avoid disappointment and wasted effort.
Simply place the \LaTeX\ code close to where the figure or table is first mentioned in the text and leave exact placement to the publishers.
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If a figure or table is too long to fit on a single page it can be split it into several parts.
Create an additional figure or table which uses \verb'\contcaption{}' instead of \verb'\caption{}'.
This will automatically correct the numbering and add `\emph{continued}' at the start of the caption.
\begin{table}
\contcaption{A table continued from the previous one.}
\label{tab:continued}
\begin{tabular}{lcc}
\hline
Star & Mass & Luminosity\\
& $M_{\sun}$ & $L_{\sun}$\\
\hline
$\tau$~Cet & 0.78 & 0.52\\
$\delta$~Pav & 0.99 & 1.22\\
$\sigma$~Dra & 0.87 & 0.43\\
\hline
\end{tabular}
\end{table}
Table~\ref{tab:continued} was generated using the code:
\begin{verbatim}
\begin{table}
\contcaption{A table continued from the previous one.}
\label{tab:continued}
\begin{tabular}{lcc}
\hline
Star & Mass & Luminosity\\
& $M_{\sun}$ & $L_{\sun}$\\
\hline
$\tau$~Cet & 0.78 & 0.52\\
$\delta$~Pav & 0.99 & 1.22\\
$\sigma$~Dra & 0.87 & 0.43\\
\hline
\end{tabular}
\end{table}
\end{verbatim}
To produce a landscape figure or table, use the \verb'pdflscape' package and the \verb'landscape' environment.
The landscape Table~\ref{tab:landscape} was produced using the code:
\begin{verbatim}
\begin{landscape}
\begin{table}
\caption{An example landscape table.}
\label{tab:landscape}
\begin{tabular}{cccccccccc}
\hline
Header & Header & ...\\
Unit & Unit & ...\\
\hline
Data & Data & ...\\
Data & Data & ...\\
...\\
\hline
\end{tabular}
\end{table}
\end{landscape}
\end{verbatim}
Unfortunately this method will force a page break before the table appears.
More complicated solutions are possible, but authors shouldn't worry about this.
\begin{landscape}
\begin{table}
\caption{An example landscape table.}
\label{tab:landscape}
\begin{tabular}{cccccccccc}
\hline
Header & Header & Header & Header & Header & Header & Header & Header & Header & Header\\
Unit & Unit & Unit & Unit & Unit & Unit & Unit & Unit & Unit & Unit \\
\hline
Data & Data & Data & Data & Data & Data & Data & Data & Data & Data\\
Data & Data & Data & Data & Data & Data & Data & Data & Data & Data\\
Data & Data & Data & Data & Data & Data & Data & Data & Data & Data\\
Data & Data & Data & Data & Data & Data & Data & Data & Data & Data\\
Data & Data & Data & Data & Data & Data & Data & Data & Data & Data\\
Data & Data & Data & Data & Data & Data & Data & Data & Data & Data\\
Data & Data & Data & Data & Data & Data & Data & Data & Data & Data\\
Data & Data & Data & Data & Data & Data & Data & Data & Data & Data\\
\hline
\end{tabular}
\end{table}
\end{landscape}
\section{References and citations}
\subsection{Cross-referencing}
The usual \LaTeX\ commands \verb'\label{}' and \verb'\ref{}' can be used for cross-referencing within the same paper.
We recommend that you use these whenever relevant, rather than writing out the section or figure numbers explicitly.
This ensures that cross-references are updated whenever the numbering changes (e.g. during revision) and provides clickable links (if available in your compiler).
It is best to give each section, figure and table a logical label.
For example, Table~\ref{tab:mathssymbols} has the label \verb'tab:mathssymbols', whilst section~\ref{sec:packages} has the label \verb'sec:packages'.
Add the label \emph{after} the section or caption command, as in the examples in sections~\ref{sec:sections} and \ref{sec:fig_table}.
Enter the cross-reference with a non-breaking space between the type of object and the number, like this: \verb'see Figure~\ref{fig:example}'.
The \verb'\autoref{}' command can be used to automatically fill out the type of object, saving on typing.
It also causes the link to cover the whole phrase rather than just the number, but for that reason is only suitable for single cross-references rather than ranges.
For example, \verb'\autoref{tab:journal_abbr}' produces \autoref{tab:journal_abbr}.
\subsection{Citations}
\label{sec:cite}
MNRAS uses the Harvard -- author (year) -- citation style, e.g. \citet{author2013}.
This is implemented in \LaTeX\ via the \verb'natbib' package, which in turn is included via the \verb'usenatbib' package option (see section~\ref{sec:options}), which should be used in all papers.
Each entry in the reference list has a `key' (see section~\ref{sec:ref_list}) which is used to generate citations.
There are two basic \verb'natbib' commands:
\begin{description}
\item \verb'\citet{key}' produces an in-text citation: \citet{author2013}
\item \verb'\citep{key}' produces a bracketed (parenthetical) citation: \citep{author2013}
\end{description}
Citations will include clickable links to the relevant entry in the reference list, if supported by your \LaTeX\ compiler.
\defcitealias{smith2014}{Paper~I}
\begin{table*}
\caption{Common citation commands, provided by the \texttt{natbib} package.}
\label{tab:natbib}
\begin{tabular}{lll}
\hline
Command & Ouput & Note\\
\hline
\verb'\citet{key}' & \citet{smith2014} & \\
\verb'\citep{key}' & \citep{smith2014} & \\
\verb'\citep{key,key2}' & \citep{smith2014,jones2015} & Multiple papers\\
\verb'\citet[table 4]{key}' & \citet[table 4]{smith2014} & \\
\verb'\citep[see][figure 7]{key}' & \citep[see][figure 7]{smith2014} & \\
\verb'\citealt{key}' & \citealt{smith2014} & For use with manual brackets\\
\verb'\citeauthor{key}' & \citeauthor{smith2014} & If already cited in close proximity\\
\verb'\defcitealias{key}{Paper~I}' & & Define an alias (doesn't work in floats)\\
\verb'\citetalias{key}' & \citetalias{smith2014} & \\
\verb'\citepalias{key}' & \citepalias{smith2014} & \\
\hline
\end{tabular}
\end{table*}
There are a number of other \verb'natbib' commands which can be used for more complicated citations.
The most commonly used ones are listed in Table~\ref{tab:natbib}.
For full guidance on their use, consult the \verb'natbib' documentation\footnote{\url{http://www.ctan.org/pkg/natbib}}.
If a reference has several authors, \verb'natbib' will automatically use `et al.' if there are more than two authors. However, if a paper has exactly three authors, MNRAS style is to list all three on the first citation and use `et al.' thereafter. If you are using \bibtex\ (see section~\ref{sec:ref_list}) then this is handled automatically. If not, the \verb'\citet*{}' and \verb'\citep*{}' commands can be used at the first citation to include all of the authors.
\subsection{The list of references}
\label{sec:ref_list}
It is possible to enter references manually using the usual \LaTeX\ commands, but we strongly encourage authors to use \bibtex\ instead.
\bibtex\ ensures that the reference list is updated automatically as references are added or removed from the paper, puts them in the correct format, saves on typing, and the same reference file can be used for many different papers -- saving time hunting down reference details.
An MNRAS \bibtex\ style file, \verb'mnras.bst', is distributed as part of this package.
The rest of this section will assume you are using \bibtex.
References are entered into a separate \verb'.bib' file in standard \bibtex\ formatting.
This can be done manually, or there are several software packages which make editing the \verb'.bib' file much easier.
We particularly recommend \textsc{JabRef}\footnote{\url{http://jabref.sourceforge.net/}}, which works on all major operating systems.
\bibtex\ entries can be obtained from the NASA Astrophysics Data System\footnote{\label{foot:ads}\url{http://adsabs.harvard.edu}} (ADS) by clicking on `Bibtex entry for this abstract' on any entry.
Simply copy this into your \verb'.bib' file or into the `BibTeX source' tab in \textsc{JabRef}.
Each entry in the \verb'.bib' file must specify a unique `key' to identify the paper, the format of which is up to the author.
Simply cite it in the usual way, as described in section~\ref{sec:cite}, using the specified key.
Compile the paper as usual, but add an extra step to run the \texttt{bibtex} command.
Consult the documentation for your compiler or latex distribution.
Correct formatting of the reference list will be handled by \bibtex\ in almost all cases, provided that the correct information was entered into the \verb'.bib' file.
Note that ADS entries are not always correct, particularly for older papers and conference proceedings, so may need to be edited.
If in doubt, or if you are producing the reference list manually, see the MNRAS instructions to authors$^{\ref{foot:itas}}$ for the current guidelines on how to format the list of references.
\section{Appendices and online material}
To start an appendix, simply place the \verb' | {
"redpajama_set_name": "RedPajamaArXiv"
} | 6,213 |
require "rom/components"
module ROM
# Globally accessible public interface exposed via ROM module
#
# @api public
module Global
# Set base global registries in ROM constant
#
# @api private
def self.extended(rom)
super
rom.instance_variable_set("@adapters", {})
end
# An internal adapter identifier => adapter module map used by setup
#
# @return [Hash<Symbol=>Module>]
#
# @api private
attr_reader :adapters
# An internal component handler registry
#
# @return [Plugins]
#
# @api private
attr_reader :handlers
# @api public
def setup(*args, &block)
case args.first
when Setup
args.first
else
Setup.new(*args, &block)
end.finalize
end
# Register adapter namespace under a specified identifier
#
# @param [Symbol] identifier
# @param [Class,Module] adapter
#
# @return [self]
#
# @api private
def register_adapter(identifier, adapter)
adapters[identifier] = adapter
self
end
end
end
| {
"redpajama_set_name": "RedPajamaGithub"
} | 9,793 |
#include <elf/pb.h>
#include <elf/log.h>
#include <elf/memory.h>
#include <map>
#include <string>
using namespace google::protobuf;
namespace elf {
typedef std::map<std::string, pb_new> reg_map;
static reg_map s_regs;
void pb_regist(const std::string &name, pb_new init)
{
s_regs[name] = init;
}
pb_t *pb_create(const std::string &name)
{
reg_map::const_iterator itr = s_regs.find(name);
if (itr != s_regs.end()) {
pb_new init = itr->second;
return init();
}
return NULL;
}
void pb_destroy(pb_t *pb)
{
E_DELETE pb;
}
int pb_get_int(const pb_t &pb, int num)
{
const Descriptor *des = pb.GetDescriptor();
const Reflection *ref = pb.GetReflection();
const FieldDescriptor *fd = des->FindFieldByNumber(num);
return ref->GetInt32(pb, fd);
}
pb_t *pb_get_field(pb_t *pb, const std::string &key)
{
assert(pb);
const Descriptor *des = pb->GetDescriptor();
const Reflection *ref = pb->GetReflection();
const FieldDescriptor *fd = des->FindFieldByName(key);
return ref->MutableMessage(pb, fd);
}
void pb_set_field(pb_t *pb, const std::string &key,
const char *val)
{
assert(pb && val);
const Descriptor *des = pb->GetDescriptor();
const FieldDescriptor *fd = des->FindFieldByName(key);
pb_set_field(pb, fd, val);
}
void pb_set_field(pb_t *pb, const FieldDescriptor *fd,
const char *val)
{
assert(pb && fd && val);
const Reflection *ref = pb->GetReflection();
FieldDescriptor::CppType type = fd->cpp_type();
switch (type) {
case FieldDescriptor::CPPTYPE_INT32:
ref->SetInt32(pb, fd, atoi(val));
break;
case FieldDescriptor::CPPTYPE_INT64:
ref->SetInt64(pb, fd, atoll(val));
break;
case FieldDescriptor::CPPTYPE_UINT32:
ref->SetUInt32(pb, fd, atoi(val));
break;
case FieldDescriptor::CPPTYPE_UINT64:
ref->SetUInt64(pb, fd, atoll(val));
break;
case FieldDescriptor::CPPTYPE_FLOAT:
ref->SetFloat(pb, fd, atof(val));
break;
case FieldDescriptor::CPPTYPE_DOUBLE:
ref->SetDouble(pb, fd, atof(val));
break;
case FieldDescriptor::CPPTYPE_BOOL:
ref->SetBool(pb, fd, (atoi(val) > 0) ? true : false);
break;
case FieldDescriptor::CPPTYPE_STRING:
ref->SetString(pb, fd, val);
break;
default:
LOG_ERROR("pb",
"Invalid field type %d.",
type);
assert(0);
}
}
void pb_set_field(pb_t *pb, const FieldDescriptor *fd,
const char *val, int len)
{
assert(pb && fd && val);
const Reflection *ref = pb->GetReflection();
FieldDescriptor::CppType type = fd->cpp_type();
switch (type) {
case FieldDescriptor::CPPTYPE_INT32:
ref->SetInt32(pb, fd, atoi(val));
break;
case FieldDescriptor::CPPTYPE_INT64:
ref->SetInt64(pb, fd, atoll(val));
break;
case FieldDescriptor::CPPTYPE_UINT32:
ref->SetUInt32(pb, fd, atoi(val));
break;
case FieldDescriptor::CPPTYPE_UINT64:
ref->SetUInt64(pb, fd, atoll(val));
break;
case FieldDescriptor::CPPTYPE_FLOAT:
ref->SetFloat(pb, fd, atof(val));
break;
case FieldDescriptor::CPPTYPE_DOUBLE:
ref->SetDouble(pb, fd, atof(val));
break;
case FieldDescriptor::CPPTYPE_BOOL:
ref->SetBool(pb, fd, (atoi(val) > 0) ? true : false);
break;
case FieldDescriptor::CPPTYPE_STRING:
ref->SetString(pb, fd, std::string(val, len));
break;
default:
LOG_ERROR("pb",
"Invalid field type %d.",
type);
assert(0);
}
}
} // namespace elf
| {
"redpajama_set_name": "RedPajamaGithub"
} | 1,111 |
Chazyoceras ("Horn of the Chazyan") is a moderately large endocerid included in the Endoceratidae with a Nanno type apex and a ventral siphuncle with a holochoanitic (where "holo" is entire, and "choan" refers to its funnel-shaped opening) wall, characteristic of the family. The siphuncle swelling at the apex is subtriangular in longitudinal profile. The endocones are of medium length.
Chazyoceras was named by Rousseau Flower in 1958. The genotype is Chazyoceras valcourense which came from the Middle Chazyan (Lower Middle Ordovician) of Valcour Island on Lake Champlain. Chazyoceras resembles Perkinsoceras which Flower included in the Allotrioceratidae - a family which Flower named and included in the Endocerida based on the strange fossil, Allotrioceras
Both Chazyoceras and Perkinsoceras types are housed in the paleontology collection at the New Mexico Museum and Natural History and Science in Albuquerque, where they were transferred from Flower's collection at the New Mexico Bureau of Mines in Socorro.
References
Flower, R. H 1958. Some Chazyan and Mohawkian Endoceratida, Jour Paleon V 32, N.2, pp 433–458, May 1858.
Flower, R.H 1976 Some Whiterock and Chazy Endooceroids, Memoir 28, Part II, New Mexico Bureau of Mines and Mineral Resources, Socorro, New Mexico
Teichert, Curt 1964. Endoceratoidea; Treatise on Invertebrate Paleontology, Part K Mollusca 3; Geol Soc of America and University of Kansas Press
Prehistoric nautiloid genera
Taxa named by Rousseau H. Flower | {
"redpajama_set_name": "RedPajamaWikipedia"
} | 7,701 |
{"url":"http:\/\/techiemathteacher.com\/2014\/08\/01\/median-triangle\/","text":"# Median of a Triangle\n\nWhen we were in high school we never thought that there is actually a formula to solve for the median of a triangle. Probably because the derivation of this formula is by cosine law and geometry is being teach in 3rd year (grade 9).\n\nWell it\u2019s time to move on. I\u2019m done with my high school life. This topic may guide high schoolers whenever they need it for school or contest.\n\nMedian of a triangle is a line connecting from one vertex to the midpoint of the side opposite to it.\n\nmedian of a triangle\n\nIn the figure above, $d$ is the median of $\\triangle ABC$\n\nDerivation: Using the same figure above\n\nUsing cosine law in $\\triangle ABC$ we have the following\n\n$b^2=a^2+c^2-2ac\\cos B$\u00a0 (1)\n\n$c^2=a^2+b^2-2ab\\cos C$ (2)\n\nAddig (1) and (2) we have,\n\n$b^2+ c^2=a^2+c^2-2ac\\cos B+a^2+b^2-2ab\\cos C$\n\n$0=2a^2-2ac\\cos B-2ab\\cos C$\n\n$2a^2=2ac\\cos B+2ab\\cos C$\n\n$a=c\\cos B+b\\cos C$\u00a0 (3)\n\nUsing cosine law in $\\triangle ABD$,\n\n$d^2=(\\displaystyle\\frac{a}{2})^2+c^2-2\\displaystyle\\frac{a}{2}c\\cos B$\n\n$d^2=(\\displaystyle\\frac{a}{2})^2+c^2-ac\\cos B$\n\n$\\cos B=\\displaystyle\\frac{c^2+\\frac{a^2}{4}-d^2}{ac}$\u00a0 (4)\n\nUsing cosine law in $\\triangle ACD$,\n\n$d^2=(\\displaystyle\\frac{a}{2})^2+b^2-2\\displaystyle\\frac{a}{2}b\\cos C$\n\n$d^2=(\\displaystyle\\frac{a}{2})^2+b^2-ab\\cos C$\n\n$\\cos C=\\displaystyle\\frac{b^2+\\frac{a^2}{4}-d^2}{ab}$\u00a0 (5)\n\nSubstitute both (4) and (5) to (3)\n\n$a=c\\cos B+b\\cos C$\n\n$a=c\\displaystyle(\\displaystyle\\frac{c^2+\\frac{a^2}{4}-d^2}{ac})+b\\displaystyle(\\displaystyle\\frac{b^2+\\frac{a^2}{4}-d^2}{ab})$\n\n$a=\\displaystyle(\\displaystyle\\frac{c^2+\\frac{a^2}{4}-d^2}{a})+\\displaystyle(\\displaystyle\\frac{b^2+\\frac{a^2}{4}-d^2}{a})$\n\n$a^2=b^2+c^2+\\displaystyle\\frac{a^2}{2}-2d^2$\n\nSolving for median (d) we have\n\n$2d^2= b^2+c^2+\\displaystyle\\frac{a^2}{2}-a^2$\n\nSimplifying this further we have\n\n$d_a=\\displaystyle\\frac{\\sqrt{2b^2+2c^2-a^2}}{2}$\n\nGenerally, the length of the median from one vertex to the opposite side is half the square root of the difference of the sum of twice the sum of squares of adjacent sides and the opposite side. Yes! That\u2019s confusing. The latter formula can be interchanged in the following manner.\n\n$d_b=\\displaystyle\\frac{\\sqrt{2a^2+2c^2-b^2}}{2}$\n\n$d_c=\\displaystyle\\frac{\\sqrt{2b^2+2a^2-c^2}}{2}$\n\nProceed to next page for worked problems\n\nPages: 1 2","date":"2017-02-24 08:07:41","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\": 25, \"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\": 0.8475401401519775, \"perplexity\": 532.039568015076}, \"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-2017-09\/segments\/1487501171418.79\/warc\/CC-MAIN-20170219104611-00048-ip-10-171-10-108.ec2.internal.warc.gz\"}"} | null | null |
{"url":"https:\/\/www.kxy.ai\/reference\/latest\/utilities\/","text":"# Misc\u00b6\n\nkxy.api.core.utils.empirical_copula_uniform(x)\n\nEvaluate the empirical copula-uniform dual representation of x as rank(x)\/n.\n\nParameters\n\nx ((n, d) np.array) \u2013 n i.i.d. draws from a d-dimensional distribution.\n\nkxy.api.core.utils.get_x_data(x_c, x_d, n_outputs, categorical_encoding='two-split', non_monotonic_extension=True, space='dual')\nkxy.api.core.utils.get_y_data(y_c, y_d, categorical_encoding='two-split')\nkxy.api.core.utils.hqi(h, q)\n\nComputes $$\\bar{h}_q^{-1}(x)$$ where\n\n$\\bar{h}_q(a) = -a \\log a -(1-a) \\log \\left(\\frac{1-a}{q-1}\\right), ~~~~ a \\geq \\frac{1}{q}.$\nkxy.api.core.utils.one_hot_encoding(x)\n\nComputes the one hot encoding representation of the input array.\n\nThe representation used is the one where all distincts inputs are first converted to string, then sorted using Python\u2019s sorted method.\n\nThe $$i$$-th element in this sort has its $$i$$-th bit from the right set to 1 and all others to 0.\n\nParameters\n\nx ((n,) or (n, d) np.array) \u2013 Array of n inputs (typically strings) to encode and that take q distinct values.\n\nReturns\n\nBinary array representing the one-hot encoding representation of the inputs.\n\nReturn type\n\n(n, q) np.array\n\nkxy.api.core.utils.pearson_corr(x)\n\nCalculate the Pearson correlation matrix, ignoring nans.\n\nParameters\n\nx ((n, d) np.array) \u2013 Input data representing n i.i.d. draws from the d-dimensional random variable, whose Pearson correlation matrix this function calculates.\n\nReturns\n\ncorr \u2013 The Pearson correlation matrix.\n\nReturn type\n\nnp.array\n\nkxy.api.core.utils.prepare_data_for_mutual_info_analysis(x_c, x_d, y_c, y_d, non_monotonic_extension=True, categorical_encoding='two-split', space='dual')\nkxy.api.core.utils.prepare_test_data_for_prediction(test_x_c, test_x_d, train_x_c, train_x_d, train_y_c, train_y_d, non_monotonic_extension=True, categorical_encoding='two-split', space='dual')\nkxy.api.core.utils.robust_log_det(c)\n\nComputes the logarithm of the determinant of a positive definite matrix in a fashion that is more robust to ill-conditioning than taking the logarithm of np.linalg.det.\n\nNote\n\nSpecifically, we compute the SVD of c, and return the sum of the log of eigenvalues. np.linalg.det on the other hand computes the Cholesky decomposition of c, which is more likely to fail than its SVD, and takes the product of its diagonal elements, which could be subject to underflow error when diagonal elements are small.\n\nParameters\n\nc ((d, d) np.array) \u2013 Square input matrix for computing log-determinant.\n\nReturns\n\nd \u2013 Log-determinant of the input matrix.\n\nReturn type\n\nfloat\n\nkxy.api.core.utils.spearman_corr(x)\n\nCalculate the Spearman rank correlation matrix, ignoring nans.\n\nParameters\n\nx ((n, d) np.array) \u2013 Input data representing n i.i.d. draws from the d-dimensional random variable, whose Spearman rank correlation matrix this function calculates.\n\nReturns\n\ncorr \u2013 The Spearman rank correlation matrix.\n\nReturn type\n\nnp.array\n\nkxy.api.core.utils.two_split_encoding(x)\n\nAlso known as binary encoding, the two-split encoding method turns categorical data taking $$q$$ distinct values into the ordinal data $$1, \\dots, q$$, and then generates the binary representation of the ordinal data.\n\nThis encoding is more economical than one-hot encoding. Unlike the one-hot encoding methods which requires as many columns as the number of distinct inputs, two-split encoding only requires $$\\log_2 \\left\\lceil q \\right\\rceil$$ columns.\n\nEvery bit in the encoding splits the set of all $$q$$ possible categories in 2 subsets of equal size (when q is a power of 2), and the value of the bit determines which subset the category of interest belongs to. Each bit generates a different partitioning of the set of all distinct categories.\n\nThe ordinal value assigned to a categorical value is the order of its string representation among all $$q$$ distinct categorical values in the array.\n\nIn other words, while a bit in one-hot encoding determines whether the category is equal to a specific values, a bit in the two-split encoding determines whether the category belong in one of two subsets of equal size.\n\nParameters\n\nx ((n,) or (n, d) np.array) \u2013 Array of n inputs (typically strings) to encode and that take q distinct values.\n\nReturns\n\nBinary array representing the two-split encoding representation of the inputs.\n\nReturn type\n\n(n, q) np.array","date":"2021-01-16 17:41:54","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\": 0.8184602856636047, \"perplexity\": 2274.5945463643766}, \"config\": {\"markdown_headings\": true, \"markdown_code\": true, \"boilerplate_config\": {\"ratio_threshold\": 0.18, \"absolute_threshold\": 10, \"end_threshold\": 15, \"enable\": true}, \"remove_buttons\": true, \"remove_image_figures\": true, \"remove_link_clusters\": true, \"table_config\": {\"min_rows\": 2, \"min_cols\": 3, \"format\": \"plain\"}, \"remove_chinese\": true, \"remove_edit_buttons\": true, \"extract_latex\": true}, \"warc_path\": \"s3:\/\/commoncrawl\/crawl-data\/CC-MAIN-2021-04\/segments\/1610703506832.21\/warc\/CC-MAIN-20210116165621-20210116195621-00683.warc.gz\"}"} | null | null |
Q: The server does not accept clients after several reboots I'm writing a small server, which can parse XML files and send pieces of information to the clients (OS Windows Server 2008 R2).
SERVER
For passing io_service as a parameter to the methods, I use boost::shared_ptr:
typedef boost::shared_ptr<io_service> service_ptr;
...
_service_ptr = boost::make_shared<io_service>();
Clients accepting thread:
void accept_thread(ServerConfig const& config) {
try {
while(!_data_.IsValid()) {
std::cout << "Wait for complete data..." << std::endl;
boost::this_thread::sleep( millisec(500));
}
while (true) {
_pool.WaitNewClient(_service_ptr);
}
} catch(std::exception& e) {
LOG::Write(LOG_ERR, std::string("In Accept thread: ") + e.what());
std::cerr << "In Accept thread: " << e.what() << std::endl;
}
}
Accepting next client:
void CClientPool::WaitNewClient(service_ptr& serv_ptr) {
client_ptr new_( new talk_to_client(serv_ptr));
_acceptor.Accept(new_);
boost::lock_guard<boost::mutex> lk(*_mutex);// safe lock for clients
new_->reset_last_ping();
clients.push_back(new_);
LOG::Write(LOG_INF, "Connected new client.");
std::cout << "Connected new client." << std::endl;
}
And:
void CClientAcceptor::Init(int port, service_ptr& serv_ptr) {
acceptor_ptr_ = boost::make_shared<ip::tcp::acceptor>(*serv_ptr, ip::tcp::endpoint(ip::tcp::v4(), port) );
}
void CClientAcceptor::Accept(client_ptr& cl_ptr) {
acceptor_ptr_->accept( cl_ptr->sock());
}
,where acceptor_ptr_:
typedef boost::shared_ptr<ip::tcp::acceptor> acceptor_ptr;
...
acceptor_ptr acceptor_ptr_;
CLIENT
void connect(std::string const& host, int port, boost::shared_ptr<boost::asio::io_service>& service) {
if(connected_){
disconnect();
}
sock_.reset(new ip::tcp::socket(*service));
boost::system::error_code ec;
ep = boost::asio::ip::tcp::endpoint(ip::address::from_string(host), port);
// do while not connection
int attempt = 0;
do {
if(attempt++ > 0) {
std::cerr << "attempt reconnect #" << attempt << "..." << std::endl;
}
sock_->connect(ep, ec);
if(ec){
std::cerr << "Failed connection to server: " << ec.message() << std::endl;
boost::this_thread::sleep(boost::posix_time::millisec(1000));
}
} while(ec);
connected_ = true;
}
void disconnect() {
if(sock_) {
boost::system::error_code errorcode;
if(sock_->is_open()) {
sock_->shutdown(boost::asio::ip::tcp::socket::shutdown_both, errorcode);
if (errorcode) {
std::cerr << "socket.shutdown error: " << errorcode.message() << std::endl;
}
sock_->close(errorcode);
if (errorcode) {
std::cerr << "socket.close error: " << errorcode.message() << std::endl;
}
}
sock_.reset();
}
connected_ = false;
}
When I debug the program (starting and stopping the server/clients), there is a point when the server stops accepting clients. While connecting the client runs without errors, the server does not accept the client anymore. After changing the port the problem disappears. What could be the problem?
| {
"redpajama_set_name": "RedPajamaStackExchange"
} | 3,338 |
Q: Dynamic Programming Help: Binary Tree Cost Edge Given a binary tree with n leaves and a set of C colors. Each leaf node of the tree is given a unique color from the set C. Thus no leaf nodes have the same color. The internal nodes of the tree are uncolored. Every pair of colors in the set C has a cost associated with it. So if a tree edge connects two nodes of colors A and B, the edge cost is the cost of the pair (A, B). Our aim is to give colors to the internal nodes of the tree, minimizing the total edge cost of the tree.
I have working on this problem for hours now, and haven't really come up with a working solution. Any hints would be appreciated.
A: I am going to solve the problem with pseudocode, because I tried writing explanation and it was completely not understandable even for me. Hopefully the code will do the trick. The complexity of my solution is not very good - memory an druntime in O(C^2 * N).
I will need couple of arrays I will be using in dynamic approach to your task:
dp [N][C][C] -> dp[i][j][k] the maximum price you can get from a tree rooted at node i, if you paint it in color j and its parent is colored in color k
maxPrice[N][C] -> maxPrice[i][j] the maximum price you can get from a tree rooted in node i if its parent is colored in color j
color[leaf] -> the color of the leaf leaf
price[C][C] -> price[i][j] the price you get if you have pair of neighbouring nodes with colors i and j
chosenColor[N][C] -> chosenColor[i][j] the color one should choose for node i to obtain maxPrice[i][j]
Lets assume the nodes are ordered using topological sorting, i.e we will be processing first leaves. Topological sorting is very easy to do in a tree. Let the sorting have given a list of inner nodes inner_nodes
for leaf in leaves:
for i in 0..MAX_C, j in 0..MAX_C
dp[leaf][i][j] = (i != color[leaf]) ? 0 : price[i][j]
for i in 0..MAX_C,
maxPrice[leaf][i] = price[color[leaf]][i]
chosenColor[leaf][i] = color[leaf]
for node in inner_nodes
for i in 0..MAX_C, j in 0..MAX_C
dp[node][i][j] = (i != root) ? price[i][j] : 0
for descendant in node.descendants
dp[node][i][j] += maxPrice[descendant][i]
for i in 0...MAX_C
for j in 0...MAX_C
if maxPrice[node][i] < dp[node][j][i]
maxPrice[node][i] = dp[node][j][i]
chosenColor[node][i] = j
for node in inner_nodes (reversed)
color[node] = (node == root) ? chosenColor[node][0] : chosenColor[node][color[parent[node]]
A: As a starting point, you can use a greedy solution which gives you an upper bound on the total cost:
while the root is not colored
pick an uncolored node having colored descendants only
choose the color that minimizes the total cost to its descendants
| {
"redpajama_set_name": "RedPajamaStackExchange"
} | 252 |
'use strict';
// Server sends a large string. Client counts bytes and pauses every few
// seconds. Makes sure that pause and resume work properly.
const common = require('../common');
const assert = require('assert');
if (!common.hasCrypto) {
common.skip('missing crypto');
return;
}
const tls = require('tls');
const fs = require('fs');
process.stdout.write('build body...');
const body = 'hello world\n'.repeat(1024 * 1024);
process.stdout.write('done\n');
const options = {
key: fs.readFileSync(`${common.fixturesDir}/keys/agent2-key.pem`),
cert: fs.readFileSync(`${common.fixturesDir}/keys/agent2-cert.pem`)
};
const server = tls.Server(options, common.mustCall(function(socket) {
socket.end(body);
}));
let recvCount = 0;
server.listen(common.PORT, function() {
const client = tls.connect({
port: common.PORT,
rejectUnauthorized: false
});
client.on('data', function(d) {
process.stdout.write('.');
recvCount += d.length;
client.pause();
process.nextTick(function() {
client.resume();
});
});
client.on('close', function() {
console.error('close');
server.close();
clearTimeout(timeout);
});
});
function displayCounts() {
console.log('body.length: %d', body.length);
console.log(' recvCount: %d', recvCount);
}
const timeout = setTimeout(displayCounts, 10 * 1000);
process.on('exit', function() {
displayCounts();
assert.strictEqual(body.length, recvCount);
});
| {
"redpajama_set_name": "RedPajamaGithub"
} | 3,023 |
var fs = require('fs');
var df = require('dateformat');
var util = require('util');
var path = require('path');
var timers = require('timers');
var lineReader = require('line-reader');
var reader = require ("buffered-reader");
var config, client,video, navReader, vidReader, rawVideo, lastFrame;
var BinaryReader = reader.BinaryReader;
var NAV_INTERVAL = 1000/15; // Navdata sent 15/s in demo mode
var VIDEO_INTERVAL = 1000/30; // 30 fps
function replay(name, deps) {
config = deps.config;
client = deps.client;
video = client.getVideoStream();
// Open the navdata file for line-by-line read
var self = this;
var navPath = path.join(config.replay.path, 'navdata.txt');
lineReader.open(navPath, function(reader) {
navReader = reader;
});
// Open the video raw stream
var videoPath = path.join(config.replay.path, 'video.h264');
rawVideo = new BinaryReader(videoPath);
// Open the video headers stream
var headerPath = path.join(config.replay.path, 'paveHeaders.txt');
lineReader.open(headerPath, function(reader) {
vidReader = reader;
vidReader.nextLine(function(data) {
// Read the first line and send video immediately the first time
var frame = JSON.parse(data);
lastFrame = frame;
emitVideo();
});
});
// Schedule timer to simulate nav data emit
timers.setInterval(emitNav, NAV_INTERVAL);
}
function emitNav() {
if (navReader && navReader.hasNextLine()) {
navReader.nextLine(function(data) {
client.emit('navdata', JSON.parse(data));
});
}
}
function emitVideo() {
if (vidReader && vidReader.hasNextLine()) {
vidReader.nextLine(function(data) {
var frame = JSON.parse(data);
var next = frame.timestamp - lastFrame.timestamp;
lastFrame = frame;
// Read a block (based on the size in the lastFrame)
rawVideo.read(lastFrame.payload_size, function (error, bytes, bytesRead) {
if (error) throw error;
video.emit('data', bytes);
});
timers.setTimeout(emitVideo, next);
});
}
}
module.exports = replay;
| {
"redpajama_set_name": "RedPajamaGithub"
} | 586 |
Description: Pyramid Lake, Nevada is home to the World's largest cutthroat trout. The lake offers anglers miles upon miles of easily accessible waters to fish. Fish for the world's largest strain of cutthroat trout at Pyramid Lake, Nevada. It's located on the Pyramid Lake Paiute Tribal Reservation about 30 miles north of Reno; and 7.5 hours straight west from Salt Lake. The strain was extinct for about 70 years and successfully reintroduced in 2006, after being rediscovered in a small mountain creek. | {
"redpajama_set_name": "RedPajamaC4"
} | 4,193 |
package com.semmle.util.data;
import java.math.BigDecimal;
import java.nio.charset.Charset;
import java.security.MessageDigest;
import java.security.NoSuchAlgorithmException;
import java.text.DecimalFormat;
import java.text.DecimalFormatSymbols;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Collection;
import java.util.Iterator;
import java.util.List;
import java.util.Locale;
import java.util.Random;
import java.util.regex.Pattern;
import com.semmle.util.exception.CatastrophicError;
import com.semmle.util.exception.Exceptions;
public class StringUtil {
private static final Random RANDOM = new Random();
private static final DecimalFormat DOUBLE_FORMATTER;
static {
// Specify the root locale to ensure we use the "." decimal separator
DOUBLE_FORMATTER = new DecimalFormat(
"#.######",
new DecimalFormatSymbols(Locale.ROOT)
);
DecimalFormatSymbols decimalFormatSymbols = DOUBLE_FORMATTER.getDecimalFormatSymbols();
decimalFormatSymbols.setNaN("NaN");
decimalFormatSymbols.setInfinity("Infinity");
DOUBLE_FORMATTER.setDecimalFormatSymbols(decimalFormatSymbols);
}
public static String box(List<String> strings) {
List<String> lines = new ArrayList<>();
for (String s : strings)
for (String line : lines(s))
lines.add(line);
int length = 0;
for (String s : lines)
length = Math.max(length, s.length());
StringBuilder sb = new StringBuilder();
for (int i = 0; i < length + 6; i++)
sb.append('*');
for (String s : lines) {
sb.append("\n* ");
sb.append(s);
for (int i = 0; i < length - s.length(); i++)
sb.append(' ');
sb.append(" *");
}
sb.append('\n');
for (int i = 0; i < length + 6; i++)
sb.append('*');
return sb.toString();
}
public static String escapeStringLiteralForRegexp(String literal, String charsToPreserve) {
final String charsToEscape = "(){}[].^$+\\*?";
StringBuilder buf = new StringBuilder();
for(int i = 0; i < literal.length(); i++) {
char c = literal.charAt(i);
if(charsToEscape.indexOf(c) != -1 && charsToPreserve.indexOf(c) == -1) {
buf.append("\\").append(c);
}
else {
buf.append(c);
}
}
return buf.toString();
}
public static String pad(int minWidth, Padding pad, String s) {
int length = s.length();
int toPad = minWidth - length;
if (toPad > 0) {
int left;
int right;
switch (pad) {
case LEFT:
left = 0;
right = toPad;
break;
case RIGHT:
left = toPad;
right = 0;
break;
case CENTRE:
left = toPad / 2;
right = toPad - left;
break;
default:
throw new CatastrophicError("Unknown padding kind: " + pad);
}
StringBuilder sb = new StringBuilder();
for (int i = 0; i < left; i++)
sb.append(' ');
sb.append(s);
for (int i = 0; i < right; i++)
sb.append(' ');
return sb.toString();
} else
return s;
}
public static List<String> pad(Padding pad, Collection<String> strings) {
List<String> result = new ArrayList<>(strings.size());
int maxWidth = 0;
for (String s : strings)
maxWidth = Math.max(maxWidth, s.length());
for (String s : strings)
result.add(pad(maxWidth, pad, s));
return result;
}
public static List<String> pad(Padding pad, String... strings) {
List<String> result = new ArrayList<>(strings.length);
int maxWidth = 0;
for (String s : strings)
maxWidth = Math.max(maxWidth, s.length());
for (String s : strings)
result.add(pad(maxWidth, pad, s));
return result;
}
public static void padTable(List<String[]> rows, Padding... pad) {
int width = pad.length;
int[] maxLengths = new int[width];
for (String[] row : rows) {
if (row.length != width)
throw new CatastrophicError("padTable can only be used with a rectangular table. Expected " + width +
" columns but found row: " + Arrays.toString(row));
for (int i = 0; i < width; i++)
maxLengths[i] = Math.max(maxLengths[i], row[i].length());
}
for (String[] row : rows)
for (int i = 0; i < width; i++)
row[i] = pad(maxLengths[i], pad[i], row[i]);
}
public static String glue(String separator, Iterable<?> values) {
StringBuilder sb = new StringBuilder();
boolean first = true;
for (Object o : values) {
if (first)
first = false;
else
sb.append(separator);
sb.append(o == null ? "<null>" : o.toString());
}
return sb.toString();
}
public static String glue(String separator, Object[] values) {
return glue(separator, Arrays.asList(values));
}
public static String glue(String separator, String... values) {
return glue(separator, (Object[]) values);
}
public static enum Padding { LEFT, RIGHT, CENTRE }
/**
* Return a new String with any of the four characters !#:= replaced with a back-slash escaped
* equivalent, and any newline characters replaced by a back-slash n.
* <p>
* This allows the String to be used in a .properties file (assuming it does not contain any
* extended unicode characters, which must be converted to unicode escapes).
* </p>
* <p>
* Note that <b>it does not ensure that the String can be used as a <i>key</i> in a .properties
* file</b>, which requires additional escaping of any spaces.
* </p>
*
* @param string The String to escape; must not be null.
* @return The given {@code string} with a back-slash inserted before each instance of any of the
* four characters: #!:=
* @see #escapePropertiesValue(String)
*/
public static String escapePropertiesValue (String string)
{
return string.replace("!", "\\!")
.replace(":", "\\:")
.replace("#", "\\#")
.replace("=", "\\=")
.replace("\n", "\\n");
}
/**
* See {@link #escapePropertiesValue(String)}. This method also escapes spaces, so that the
* {@code string} can be used as a .properties key.
*
* @param string The String to escape; must not be null.
* @return The given {@code string} with a back-slash inserted before each instance of any of the
* four characters: #!:= or the space character, and newlines replaced by a backslash n.
*/
public static String escapePropertiesKey (String string)
{
return escapePropertiesValue(string).replace(" ", "\\ ");
}
/**
* Print a float in a locale-independent way suitable for reading with Double.valueOf().
*/
public static String printFloat(double value) {
if (Math.abs(value) > 999999999999999.0 && !Double.isInfinite(value)) {
// `DecimalFormat` for `double` loses precision on large numbers,
// printing the least significant digits as all 0.
return DOUBLE_FORMATTER.format(new BigDecimal(value));
} else {
return DOUBLE_FORMATTER.format(value);
}
}
public static String escapeHTML(String s) {
if (s == null) return null;
int length = s.length();
// initialize a StringBuilder with initial capacity of twice the size of the string,
// except when its size is zero, or when doubling the size causes overflow
StringBuilder sb = new StringBuilder(length * 2 > 0 ? length * 2 : length);
for (int i = 0; i < length; i++) {
char c = s.charAt(i);
switch (c) {
case '<':
sb.append("<");
break;
case '>':
sb.append(">");
break;
case '&':
sb.append("&");
break;
case '"':
sb.append(""");
break;
case '\'':
sb.append("'");
break;
// be careful with this one (non-breaking white space)
/*
case ' ':
sb.append(" ");
break;*/
default:
sb.append(c);
break;
}
}
return sb.toString();
}
/**
* Escape special characters in the given string like JSON.stringify does
* (see ECMAScript 5.1, Section 15.12.3).
*/
public static String escapeJSON(String str) {
if (str == null)
return null;
StringBuilder res = new StringBuilder();
for (int i=0, n=str.length(); i<n; ++i) {
char c = str.charAt(i);
switch (c) {
case '"': res.append("\\\""); break;
case '\\': res.append("\\\\"); break;
case '\b': res.append("\\b"); break;
case '\f': res.append("\\f"); break;
case '\n': res.append("\\n"); break;
case '\r': res.append("\\r"); break;
case '\t': res.append("\\t"); break;
default:
if (c < ' ')
res.append(String.format("\\u%04x", Integer.valueOf(c)));
else
res.append(c);
}
}
return res.toString();
}
// we don't include curly brackets although they're mentioned in the spec as escapable,
// because they don't actually do anything
private static final List<Character> specialMarkdownChars = Arrays.asList(
'\\', '`', '_', '*', '(', ')', '[', ']', '#', '+', '-', '.', '!');
/**
* Escape special markdown characters in the given string.
*/
public static String escapeMarkdown(String str) {
return escapeMarkdown(specialMarkdownChars, str);
}
/**
* Escape special markdown characters in the given string.
*/
public static String escapeMarkdown(List<Character> specialMarkdownChars, String str) {
if (str == null)
return null;
StringBuilder res = new StringBuilder();
boolean escapeOctothorp = true;
for (int i=0, n=str.length(); i<n; ++i) {
char c = str.charAt(i);
if (specialMarkdownChars.contains(c) && (c != '#' || escapeOctothorp))
res.append("\\").append(c);
else
res.append(c);
// If this character is an '&' and the next character is an '#' it will not be escaped.
// This is to avoid escaping it in HTML entities, e.g. ☃.
escapeOctothorp = (c != '&');
}
return res.toString();
}
/**
* Make a QL string literal for the given string, using escapes for non-printable characters
* where possible. The return value includes the surrounding double-quote characters.
*/
public static String makeQLStringLiteral(String value) {
StringBuilder sb = new StringBuilder();
sb.append('\"');
for (char c : value.toCharArray()) {
switch (c) {
case '\\':
sb.append("\\\\");
break;
case '\n':
sb.append("\\n");
break;
case '\"':
sb.append("\\\"");
break;
case '\r':
sb.append("\\r");
break;
case '\t':
sb.append("\\t");
break;
default:
sb.append(c);
break;
}
}
sb.append('\"');
return sb.toString();
}
/**
* Wrap a long text to a given number of columns. The wrapping is done
* naively: At least one word is on every line, and the first word to
* push the line length above the value of <code>cols</code> marks the
* start of a new line (and ends up on the new line). For this method,
* "word" means sequence of non-whitespace characters.
* @param text The text that should be wrapped.
* @param cols The number of characters to permit on each line; it is
* only exceeded if there are single words that are longer.
* @return The text with sequences of whitespace before words that would
* exceed the permitted width replaced with '\n'.
*/
public static String wrap(String text, int cols) {
if(text == null)
return null;
List<String> lines = new ArrayList<>();
int lineStart = -1; int wordStart = -1; int col = 0;
for (int cur = 0; cur < text.length(); cur++) {
if (text.charAt(cur) == '\n') {
// Forced new line.
if (lineStart < 0) {
// Empty new line.
lines.add("");
} else {
lines.add(text.substring(lineStart, cur).trim());
}
lineStart = -1;
wordStart = -1;
col = 0;
} else if (Character.isWhitespace(text.charAt(cur))) {
// Possible break.
if (col > cols) {
// Break is needed.
if (lineStart < 0) {
// Long run of whitespace.
continue;
} else if (wordStart < 0) {
// Sequence of whitespace went over after real word.
String line = text.substring(lineStart, cur).trim();
if (line.length() > 0) lines.add(line);
lineStart = -1;
} else if (wordStart > lineStart) {
// Word goes onto new line.
lines.add(text.substring(lineStart, wordStart - 1).trim());
lineStart = wordStart;
col = cur - lineStart + 1;
wordStart = -1;
} else {
// Word is a line on its own.
lines.add(text.substring(wordStart, cur).trim());
lineStart = -1;
wordStart = -1;
}
} else {
// No break, but new word
wordStart = -1;
}
} else {
if (lineStart < 0) {
lineStart = cur;
col = 0;
}
if (wordStart < 0)
wordStart = cur;
}
if (lineStart >= 0)
col++;
}
if (lineStart > -1)
lines.add(text.substring(lineStart).trim());
return glue("\n", lines);
}
/**
* Get the first word of the given string, delimited by whitespace. Leading whitespace
* is ignored.
*/
public static String firstWord(String s) {
s = s.trim();
int i = 0;
while (i < s.length() && !Character.isWhitespace(s.charAt(i)))
i++;
return s.substring(0, i);
}
/**
* Strip the first word (delimited by whitespace, leading whitespace ignored) and get the
* remainder of the string, trimmed.
*/
public static String stripFirstWord(String s) {
s = s.trim();
int i = 0;
while (i < s.length() && !Character.isWhitespace(s.charAt(i)))
i++;
return s.substring(i).trim();
}
/**
* Trim leading and trailing occurrences of a character from a string
* @param str the string to trim
* @param c the character to remove
* @return A string whose value is <code>str</code>, with any leading and trailing occurrences of <code>c</code> removed,
* or <code>str</code> if it has no leading or trailing occurrences of <code>c</code>.
*/
public static String trim(String str, char c) {
return trim(str, c, true, true);
}
public static String trim(String str, char c, boolean trimLeading, boolean trimTrailing) {
int begin = 0;
int end = str.length();
if (trimLeading) {
while ((begin < end) && (str.charAt(begin) == c)) {
begin++;
}
}
if (trimTrailing) {
while ((begin < end) && (str.charAt(end - 1) == c)) {
end--;
}
}
if ((begin > 0) || (end < str.length()))
str = str.substring(begin, end);
return str;
}
public static String trimTrailingWhitespace(String str) {
int begin = 0;
int end = str.length();
while((begin < end) && (Character.isWhitespace(str.charAt(end-1)))) {
end--;
}
if (end < str.length())
str = str.substring(begin, end);
return str;
}
private static final Charset UTF8_CHARSET = Charset.forName("UTF-8");
/**
* Invert the conversion performed by {@link #stringToBytes(String)}.
*/
public static String bytesToString(byte[] bytes)
{
return new String(bytes, 0, bytes.length, UTF8_CHARSET);
}
public static String bytesToString(byte[] bytes, int offset, int length) {
return new String(bytes, offset, length, UTF8_CHARSET);
}
/** Convert a String into a sequence of bytes (according to a UTF-8 representation of the String). */
public static byte[] stringToBytes (String str)
{
return str.getBytes(Charset.forName("UTF-8"));
}
/**
* Compute a SHA-1 sum for the given String.
* <p>
* The SHA-1 is obtained by first converting the String to bytes, which is Charset-dependent,
* though this method always uses {@link #stringToBytes(String)}.
* </p>
*
* @see #toHex(byte[])
*/
public static byte[] stringToSHA1 (String str)
{
MessageDigest messageDigest;
try {
messageDigest = MessageDigest.getInstance("SHA-1");
return messageDigest.digest(stringToBytes(str));
}
catch (NoSuchAlgorithmException e) {
throw new CatastrophicError("Failed to obtain MessageDigest for computing SHA-1", e);
}
}
/**
* Constructs a string that repeats the repeatee the specified number of times.
* For example, repeat("foo", 3) would return "foofoofoo".
*
* @param repeatee The string to be repeated.
* @param times The number of times to repeat it.
* @return The result of repeating the repeatee the specified number of times.
*/
public static String repeat(String repeatee, int times) {
if (times == 0)
return "";
return new String(new char[times]).replace("\0", repeatee);
}
/**
* Computes the lower-case version of the given string in a way that is independent
* of the system default locale.
* @param s A string value to lowercase.
* @return The value of {@code s} with all English letters converted to lower-case.
*/
public static String lc(String s) {
return s.toLowerCase(Locale.ENGLISH);
}
/**
* Computes the upper-case version of the given string in a way that is independent
* of the system default locale.
* @param s A string value to uppercase.
* @return The value of {@code s} with all English letters converted to upper-case.
*/
public static String uc(String s) {
return s.toUpperCase(Locale.ENGLISH);
}
public static String ucfirst(String s) {
if( s.isEmpty() || !Character.isLowerCase(s.charAt(0)))
return s;
else
return uc(s.substring(0,1))+s.substring(1);
}
private static final Pattern lineEnding = Pattern.compile("\r\n|\r|\n");
/**
* Regex to match line endings using look-behind,
* so that line separators can be included in the split lines.
* \r\n is matched eagerly, i.e. we only match on \r individually if it is not followed by \n.
*/
private static final Pattern lineEndingIncludingSeparators = Pattern.compile("(?<=(\r\n|\r(?!\n)|\n))");
/**
* Get the lines in a given string. All known style of line terminator (CRLF, CR, LF)
* are recognised. Trailing empty lines are not returned, and the resulting strings
* do not include the line separators.
*/
public static String[] lines(String s) {
return lines(s, false, true);
}
/**
* Get the lines in a given string, including the line separators.
* All known style of line terminator (CRLF, CR, LF) are recognised.
* Trailing empty strings are not returned (but lines containing only separators are).
*/
public static String[] linesWithSeparators(String s) {
return lines(s, true, true);
}
/**
* Get the lines in a given string. All known style of line terminator (CRLF, CR, LF)
* are recognised. The resulting strings do not include the line separators. If
* {@code squishTrailing} is <code>true</code>, trailing empty lines are not included
* in the result; otherwise, they will appear as empty strings.
*/
public static String[] lines(String s, boolean squishTrailing) {
return lines(s, false, squishTrailing);
}
/**
* Gets the lines in a given string. All known style of line terminator (CRLF, CR, LF)
* are recognised.
* If {@code includeSeparators} is <code>true</code>, then the line separators are included
* at the end of their corresponding lines; otherwise they are dropped.
* If {@code squishTrailing} is <code>true</code>, then trailing empty lines are not included
* in the result; otherwise, they will appear as empty strings.
*/
public static String[] lines(String s, boolean includeSeparators, boolean squishTrailing) {
if (s.length() == 0)
return new String[0];
Pattern pattern = includeSeparators ? lineEndingIncludingSeparators : lineEnding;
return pattern.split(s, squishTrailing ? 0 : -1);
}
/**
* Replace all line endings in the given string with \n
*/
public static String toUnixLineEndings(String s) {
return lineEnding.matcher(s).replaceAll("\n");
}
/**
* Get a boolean indicating whether the string contains line separators
* All known style of line terminator (CRLF, CR, LF) are recognised.
*/
public static boolean isMultiLine(String s) {
return lineEnding.matcher(s).find();
}
private static final Pattern whitespace = Pattern.compile("\\s+");
/**
* Get the words (i.e. whitespace-delimited chunks of non-whitespace) from the given string.
* Empty words are not included -- this means that the result is a zero-length array for
* an input string that consists entirely of whitespace.
*/
public static String[] words(String s) {
s = s.trim();
if (s.length() == 0)
return new String[0];
return whitespace.split(s);
}
/**
* Split a string into paragraphs (delimited by empty lines). Line endings are not preserved.
*/
public static String[] paragraphs(String s) {
List<String> paragraphs = new ArrayList<>();
StringBuilder paragraph = new StringBuilder();
boolean emptyParagraph = true;
for (String line : StringUtil.lines(s)) {
if (line.isEmpty()) {
// line only has line endings, i.e. is between paragraphs.
if (!emptyParagraph)
paragraphs.add(paragraph.toString());
paragraph = new StringBuilder();
emptyParagraph = true;
} else {
if(paragraph.length() != 0)
paragraph.append(' ');
paragraph.append(line);
emptyParagraph = false;
}
}
if (!emptyParagraph)
paragraphs.add(paragraph.toString());
return paragraphs.toArray(new String[0]);
}
private static final char[] HEX_CHARS = "0123456789abcdef".toCharArray();
/** Convert an array of bytes into an array of lower-case hex digits. */
public static String toHex (byte ... bytes)
{
StringBuilder strBldr = new StringBuilder(bytes.length * 2);
char[] hexchars = HEX_CHARS;
for (byte b : bytes) {
strBldr.append(hexchars[(b >>> 4) & 0xF]).append(hexchars[b & 0xF]);
}
return strBldr.toString();
}
/**
* Convert String of hexadecimal digits to an array of bytes.
* @throws NumberFormatException if string does not have an even length or
* contains invalid characters.
*/
public static byte[] fromHex(String string) {
int len = string.length();
if(len % 2 != 0)
throw new NumberFormatException("Hexadecimal string should have an even number of characters.");
byte[] data = new byte[len / 2];
int index = 0;
for (int i = 0; i < len; i += 2) {
int a = Character.digit(string.charAt(i), 16);
if(a == -1)
throw new NumberFormatException("Invalid character in hexadecimal string: " + string.charAt(i));
int b = Character.digit(string.charAt(i+1), 16);
if(b == -1)
throw new NumberFormatException("Invalid character in hexadecimal string: " + string.charAt(i+1));
data[index] = (byte) ((a << 4) | b);
index++;
}
return data;
}
/**
* Return a 8 character String describing the duration in {@code nanoSeconds}.
* <p>
* The duration will be scaled and expressed using the appropriate units: nano-, micro-, milli-,
* seconds, minutes, hours, days, or years.
* </p>
*/
public static String getDurationString (long nanoSeconds)
{
char sign = nanoSeconds < 0 ? '-' : '+';
nanoSeconds = nanoSeconds < 0 ? -nanoSeconds : nanoSeconds;
if (nanoSeconds < 1e4) {
return sign + getDurationString(nanoSeconds, 1, "[ns]");
}
else if (nanoSeconds < 1e7) {
return sign + getDurationString(nanoSeconds, 1e3, "[us]");
}
else if (nanoSeconds < 1e10) {
return sign + getDurationString(nanoSeconds, 1e6, "[ms]");
}
else if (nanoSeconds < 1e13) {
return sign + getDurationString(nanoSeconds, 1e9, "[s] ");
}
else if (nanoSeconds < 60e13) {
return sign + getDurationString(nanoSeconds, 60e9, "[m] ");
}
else if (nanoSeconds < 3600e13) {
return sign + getDurationString(nanoSeconds, 3600e9, "[h] ");
}
else if (nanoSeconds < 86400e13) {
return sign + getDurationString(nanoSeconds, 86400e9, "[d] ");
}
else {
return sign + getDurationString(nanoSeconds, 31536000e9, "[y] ");
}
}
/**
* Return a four character representation of the given duration in {@code nanoSeconds}, divided by
* the given {@code divisor} and suffixed with the given {@code unit}.
*
* @param nanoSeconds The duration to express; must be non-negative.
* @see #getDurationString(long)
*/
private static String getDurationString (long nanoSeconds, double divisor, String unit)
{
// Format as a 4 character floating point
String scaledStr = String.format("%-4f", nanoSeconds / divisor).substring(0, 4);
// Replace a trailing decimal with a space
return (scaledStr.endsWith(".") ? scaledStr.replace(".", " ") : scaledStr) + unit;
}
/**
* Parse an Integer from the given {@code string}, returning null if parsing failed for any
* reason.
*/
public static Integer parseInteger (String string)
{
// Quick break-out if string is null
if (string == null) {
return null;
}
// Attempt to parse an integer
try {
return Integer.parseInt(string);
}
catch(NumberFormatException nfe) {
Exceptions.ignore(nfe, "deliberate test");
return null;
}
}
/**
* Append to a given {@link StringBuilder} a sequence of Objects via their
* {@link Object#toString()} method, and return the StringBuilder.
*/
public static StringBuilder appendLine (StringBuilder strBldr, Object ... args)
{
for(Object arg : args) {
strBldr.append(arg);
}
return strBldr.append("\n");
}
/**
* Compose a new String with every line prepended with a given prefix.
* <p>
* The final portion of the {@code text} that is not terminated with a newline character will be
* prefixed if and only if it is non-empty.
* </p>
*
* @param prefix The string that shall be prefixed to every line in {@code text}.
* @param text The string to split into lines and prefix.
*/
public static String prefixLines (String prefix, String text)
{
return text.replaceAll("(?m)^", prefix);
}
/**
* Count the number of times a character occurs in a string.
*
* @param str The string to search in.
* @param ch The character to look for.
*/
public static int count (String str, char ch)
{
int r = 0;
for (int i = 0; i < str.length(); i++) {
if (str.charAt(i) == ch) {
r++;
}
}
return r;
}
/**
* Add line numbers to the start of each line of the given {@code plainText}.
*
* @param plainText - some plain text, with lines distinguished by one of the standard
* line-endings.
* @return the {@code plainText}, with 1-indexed line numbers inserted at the start of each
* line. The line numbers will be of a fixed width comprising the length of the largest
* line number.
*/
public static String addLineNumbers(String plainText) {
/*
* Add line numbers to the plain text code sample.
*/
String[] lines = StringUtil.lines(plainText, false);
// The maximum number of characters needed to represent the line number
int lineColumnWidth = Integer.toString(lines.length).length();
StringBuilder sampleWithLineNumbers = new StringBuilder();
for (int i = 0; i < lines.length; i++) {
boolean last = i == lines.length -1;
sampleWithLineNumbers.append(String.format("%" + lineColumnWidth + "s %s" + (last ? "" : "\n"), i + 1, lines[i]));
}
return sampleWithLineNumbers.toString();
}
// Pattern that matches a string of (at least one) decimal digits
private static final Pattern DIGITS_PATTERN = Pattern.compile("[0-9]+");
/** Return true iff the given string consists only of digits */
public static boolean isDigits(String s) {
return DIGITS_PATTERN.matcher(s).matches();
}
/**
* Determine whether a given {@code char} is an ASCII letter.
*/
public static boolean isAsciiLetter (char c)
{
return (c >= 'a' && c <= 'z') || (c >= 'A' && c <= 'Z');
}
/**
* A {@link String} comparison function to hopefully help mitigate against timing attacks.
* Aims to be constant time in the length of the first argument however due to the nature
* of Java this is very hard to achieve reliably. Callers should not rely on this method
* being perfectly constant time and other defenses should be introduced as necessary
* to prevent timing attacks based on how critical it is to avoid them.
* <p>
* Each argument may safely be null.
* <p>
* Note there is a unit tests that asserts the timing properties of this method which is
* committed but not run by default. If any changes are made to the implementation then
* {@code StringUtilTests#secureIsEqualTiming} must be run manually.
*/
public static boolean secureIsEqual(String a, String b) {
if (a == null) {
// Since we are aiming for constant time in the length of the
// first argument only, it is ok to bail out quickly if a is null.
return b == null;
}
byte[] aBytes = stringToBytes(a);
boolean definitelyDifferent = b == null || b.length() != a.length();
byte[] bBytes = stringToBytes(definitelyDifferent ? a : b);
byte[] randomBBytes = new byte[a.length()];
RANDOM.nextBytes(randomBBytes);
if (definitelyDifferent) {
bBytes = randomBBytes;
}
int result = 0;
for (int i = 0; i < aBytes.length; i++) {
result |= aBytes[i] ^ bBytes[i];
}
return result == 0 && !definitelyDifferent;
}
public static String lineSeparator(){
return System.getProperty("line.separator");
}
public static <T> String naturalGlue(String separator, String finalSeparator, Collection<T> values) {
StringBuilder stringBuilder = new StringBuilder();
Iterator<T> iterator = values.iterator();
boolean first = true;
if(iterator.hasNext()) {
boolean hasNext = true;
T current = iterator.next();
while(hasNext) {
hasNext = iterator.hasNext();
T next = iterator.hasNext() ? iterator.next() : null;
if(first) {
first = false;
}
else if(!hasNext) {
stringBuilder.append(finalSeparator);
}
else {
stringBuilder.append(separator);
}
stringBuilder.append(current != null ? current : "<null>");
current = next;
}
}
return stringBuilder.toString();
}
/**
* Convert a CamelCase (or camelCase) string to spinal-case. Adjacent sequences of upper-case
* characters are treated as a single word, so that "getXML" would be converted to "get-xml".
* Where a lower-case character follows an upper-case character <i>after</i> a sequence of at
* least on upper-case character, the last upper-case character in the sequence is assumed to
* be the first letter of a new word, rather than the last letter of an acronym. Thus,
* "getXMLFile" becomes "get-xml-file" rather than "get-xmlfile" or "get-xmlf-ile".
*
* @return The spinal-cased equivalent of {@code camelCaseStr}, or null if it is null.
*/
public static String camelToSpinal(String camelCaseStr) {
// Quick break-out if the string is null
if (camelCaseStr == null)
return null;
// Convert to spinal-case
String lcStr = camelCaseStr.toLowerCase(Locale.ENGLISH);
StringBuilder strBldr = new StringBuilder();
for(int i=0; i<camelCaseStr.length();) {
if (camelCaseStr.charAt(i) != lcStr.charAt(i)) {
if (i > 0) {
// Next character is upper-case, and not at the start of the string,
// so insert a preceding dash
strBldr.append("-");
}
// Consume (append in l.c.) all contiguously following u.c. characters, except that
// if a sequence of two or more u.c. characters occurs followed by a l.c. character
// assume that the last u.c. character is the first in a new word and insert a -
// preceding the new word.
//
// Thus getXML becomes get-xml, but getXMLFile becomes get-xml-file rather than
// get-xmlfile.
int numUc = 0;
while(i<camelCaseStr.length() && camelCaseStr.charAt(i) != lcStr.charAt(i)) {
if ( numUc > 0
&& i+1 < camelCaseStr.length()
&& camelCaseStr.charAt(i+1) == lcStr.charAt(i+1)
&& isAsciiLetter(lcStr.charAt(i+1))) {
strBldr.append("-").append(lcStr.charAt(i++));
break;
}
strBldr.append(lcStr.charAt(i++));
++numUc;
}
}
// Consume (append) all contiguously following l.c. characters
while(i<camelCaseStr.length() && camelCaseStr.charAt(i) == lcStr.charAt(i))
strBldr.append(lcStr.charAt(i++));
}
return strBldr.toString();
}
public static String lowerCaseFirstLetter(String s) {
if (s == null || s.isEmpty())
return s;
final char[] chars = s.toCharArray();
chars[0] = Character.toLowerCase(chars[0]);
return new String(chars);
}
/**
* Returns the string with double-quoted strings, single line comments and multiline comments
* removed.
* <p>
*
* @param ql The QL code string
* @param terminateStringsAtLineEnd If true, then strings are treated as ending at EOL;
* if false, unterminated strings result in an IllegalArgumentException.
*
* NB QL does not support multiline strings.
*/
public static String stripQlCommentsAndStrings(String ql, boolean terminateStringsAtLineEnd) {
StringBuilder returnBuilder = new StringBuilder();
// in a quoted string you must ignore both comment starters
// in a multi-line comment you must ignore the other two (single line comment and string) starters
// in a single line comment you can just eat up to the end of the line
boolean inString = false;
boolean inMultiLineComment = false;
boolean inSingleLineComment = false;
for (int i = 0; i < ql.length(); i++) {
// String
if (!inMultiLineComment && !inSingleLineComment && matchesAt(ql, i, "\"") && !isEscaped(ql, i)) {
inString = !inString;
continue;
} else if (matchesEolAt(ql, i)) {
if (terminateStringsAtLineEnd) {
inString = false; // force strings to end at EOL - multi-line strings are invalid
} else if (inString) {
throw new IllegalArgumentException("Unterminated string found.");
}
}
// Single-line comment
if (!inString && !inMultiLineComment && matchesAt(ql, i, "//")) {
inSingleLineComment = true;
continue;
} else if (inSingleLineComment && matchesEolAt(ql, i)) {
inSingleLineComment = false;
}
// Multi-line comment
if (!inString && !inSingleLineComment && matchesAt(ql, i, "/*")) {
inMultiLineComment = true;
} else if (inMultiLineComment && matchesAt(ql, i, "*/")) {
inMultiLineComment = false;
i++; // skip the next character (the '/') as well as this one
continue;
}
if (inString || inMultiLineComment || inSingleLineComment) {
continue;
}
returnBuilder.append(ql.charAt(i));
}
if (inString && !terminateStringsAtLineEnd) {
throw new IllegalArgumentException("Unterminated string found.");
}
return returnBuilder.toString();
}
/**
* Calls (@link #stripQlCommentsAndStrings(String, boolean),
* passing {@code false} for the {@code terminateStringsAtLineEnd} parameter.
*/
public static String stripQlCommentsAndStrings(String ql) {
return stripQlCommentsAndStrings(ql, false);
}
private static boolean matchesAt(String sourceString, int currentCharIndex, String subString) {
if (currentCharIndex + subString.length() > sourceString.length()) {
return false;
}
return sourceString.substring(currentCharIndex, currentCharIndex + subString.length()).equals(subString);
}
private static boolean matchesEolAt(String sourceString, int currentCharIndex ) {
return matchesOneOfAt(sourceString, currentCharIndex, "\n", "\r");
}
private static boolean matchesOneOfAt(String sourceString, int currentCharIndex, String... subStrings) {
for (String subString: subStrings) {
if (matchesAt(sourceString, currentCharIndex, subString)) {
return true;
}
}
return false;
}
private static boolean isEscaped(String theString, int currentCharIndex) {
if (currentCharIndex == 0) {
return false;
}
return previousCharacter(theString, currentCharIndex) == '\\' && !isEscaped(theString, currentCharIndex-1);
}
private static char previousCharacter(String theString, int currentCharIndex) {
if (currentCharIndex == 0) {
return Character.MIN_VALUE;
}
return theString.charAt(currentCharIndex - 1);
}
/**
* Compare two arrays of strings. The two arrays are considered equal if they
* are either both null or contain equal elements in the same order (ignoring
* case).
*
* @param a
* the first array
* @param a2
* the second array
* @return true iff the elements in the arrays are equal when ignoring case
*/
public static boolean arrayEqualsIgnoreCase(String[] a, String[] a2) {
if (a == null) return a2 == null;
if (a2 == null) return false;
if (a.length != a2.length) return false;
for (int i = 0; i < a.length; i++) {
if ((a[i] == null && a2[i] != null) || !a[i].equalsIgnoreCase(a2[i])) return false;
}
return true;
}
public static final Pattern NEWLINE_PATTERN = Pattern.compile("\n");
/**
* Convert a string into a doc comment with the content as the body.
*/
public static String toCommentString(String content) {
StringBuilder result = new StringBuilder();
result.append("/**\n");
String[] lines = StringUtil.lines(content);
for (String line: lines) {
result.append(" *" + line);
result.append("\n");
}
result.append(" */");
return result.toString();
}
/**
* Is {@code str} composed only of printable ASCII characters (excluding
* newline, carriage return and tab but including space)?
*/
public static boolean isPrintableAscii(String str) {
if (str == null)
return false;
for(int i=0; i<str.length(); ++i) {
if (!isPrintableAscii(str.charAt(i)))
return false;
}
return true;
}
/**
* Is {@code c} a printable ASCII character (excluding newline, carriage
* return and tab, but including space)?
*/
public static boolean isPrintableAscii(char c) {
return c >= 32 && c < 127;
}
/**
* Return true if {@code str} only contains characters which are either printable ASCII or ASCII whitespace
*/
public static boolean isAsciiText(String str) {
for(int i=0; i<str.length(); ++i) {
if (!isAsciiText(str.charAt(i)))
return false;
}
return true;
}
/**
* Return true if {@code c} is either a printable ASCII character or ASCII whitespace character
*/
public static boolean isAsciiText(char c) {
return (
isPrintableAscii(c) ||
c == '\t' ||
c == '\n' ||
c == '\r'
);
}
/**
* Returns true if {@code str} contains ASCII characters < 32 or == 127,
* other than \n, \t, \r
*/
public static boolean containsControlCharacters(String str) {
for (char c: str.toCharArray()) {
if (isControlCharacter(c)) {
return true;
}
}
return false;
}
/**
* "Control character" here means code point 127 (DEL),
* characters in C0 block [0, 31] excluding \t, \n, or \r
* or characters in C1 block [128, 159]
*/
public static boolean isControlCharacter(char c) {
// C1 control characters run from 128-159
// but Unicode above that is okay
if (c > 159) {
return false;
}
// Most of ASCII is okay
if (c >= 32 && c < 127) {
return false;
}
// Basic whitespace is okay
if (c == '\t' ||
c == '\n' ||
c == '\r') {
return false;
}
// If we've got this far, it must be a control character
return true;
}
}
| {
"redpajama_set_name": "RedPajamaGithub"
} | 932 |
\section{Introduction}
\label{sec:intro}
The stellar initial mass function (IMF), for $M\ge $ 1 M$_{\odot}$\ plays a central role in a wide
variety of astrophysics. The relative numbers of such stars is
essential to interpreting the stellar populations of star-forming galaxies across
cosmic time, testing and validating theories of star
formation, constraining chemical evolution models, the formation of compact objects and gravitational waves, and understanding the interplay between stars and gas \citep[e.g.,][]{schmidt1959, tinsley1968, searle1973, talbot1973, ostriker1974, audouze1976, scalo1986, kennicutt1998b, elmegreen1999, kroupa2001, massey2003, alves2007, fardal2007, mckee2007, wilkins2008, pflammAltenburg2009, kennicutt2012, narayanan2012, conroy2013, belczynski2014, krumholz2014, madau2014b}.
\subsection{Current State of the IMF Above $\sim$ 1 M$_{\odot}$}
\label{sec:currentimf}
Our best constraints on the high-mass IMF come from studies in the Milky Way (MW) and nearby galaxies. For the closest galaxies, it is possible to count individual stars and make a direct measurement of the high-mass IMF slope. While measuring the IMF from direct star counts is straightforward in principle, it is far more complicated in practice. Accurate measurements are challenging due to a variety of observational and physical effects including completeness, dynamical evolution, distance/extinction ambiguity, stellar multiplicity, the accuracy of stellar evolution models, degeneracies between star formation history (SFH) and the IMF slope, and a paucity of massive stars in the nearby universe, all of which complicate the translation of the present day mass function (MF) into the IMF \citep[e.g.,][]{lequeux1979, miller1979, scalo1986, mateo1993, massey1995, delafuentemarcos1997, maizapellaniz2005, elmegreen2006, maizapellaniz2008, ascenso2009, kruijssen2009, demarchi2010, portegieszwart2010, bastian2010}.
Observations of more distant, unresolved galaxies alleviate some of these challenges. For example, by studying an entire galaxy, effects of stellar dynamics (e.g., migration, mass segregation) and distance/extinction confusion are less important. Further, distant galaxies span a more diverse range of environments than what is available in the very local universe, offering better opportunities to measure the high-mass IMF as a function of environment \citep[metallicity, mass, redshift, etc.; e.g.,][]{baldry2003, fardal2007, dave2011, narayanan2013}. However, the unresolved nature of these observations severely limits the precision and accuracy of a high-mass IMF measurement. Most integrated observations have insufficient leverage to mitigate the SFH-IMF degeneracy, and are forced to make simplified assumptions about a galaxy's recent SFH (e.g., constant, single short burst) and dust (e.g., a single value for an entire galaxy) to provide \emph{any} constraint on the high-mass IMF \citep[e.g.,][]{miller1979, elmegreen2006}. Although studies of unresolved clusters may provide a new avenue for IMF studies \citep[e.g.,][]{calzetti2010, andrews2013}, high-mass IMF studies using integrated light typically provide coarse characterizations rather than precise measurements.
As a result of these challenges, our current knowledge of the high-mass IMF remains remarkably poor. The widespread adoption of a `universal' Salpeter \citep[$\Gamma=+1.35$;][]{salpeter1955} or Kroupa \citep[$\Gamma=+1.3$;][]{kroupa2001} IMF above 1 M$_{\odot}$, presumed not to vary with environment has given the impression that the high-mass IMF is a solved problem \citep[e.g.,][]{kennicutt1998b}. However, compilations of literature IMF studies from resolved star counts indicate an ``average value'' of $\Gamma=1.7$ with a scatter $\gtrsim$0.5 dex in IMF slope measurements that do not clearly support a `universal' IMF \citep[e.g.,][]{scalo1986, scalo1998, kroupa2002, scalo2005}\footnote{Although a `Kroupa IMF' has a high-mass slope of $\Gamma=1.3$, \citet{kroupa2002} argue that the slope should probably be close $\Gamma \sim 1.7$, as most historical IMF studies do not account for unresolved binary stars, which, in principle, cause measured MF slopes to be shallower than true IMF slopes.}. There also does not appear to be any unambiguous systematic dependence of IMF slope on environment and/or initial star-forming conditions \citep[e.g.,][]{bastian2010}, which has significant implications for how stars form in a wide variety of physical conditions \citep[e.g.,][]{elmegreen1999, mckee2007, hopkins2013a, krumholz2014}.
Controversy over the form and universality of the IMF have persisted since the first studies in the 1950s.
Following the seminal work of \citet{salpeter1955}, a handful of studies reported
similar IMFs in a number of MW clusters and field environments \citep{jaschek1957, sandage1957, vandenbergh1957}, prompting early speculation about the universality of the ``original mass function'' \citep[e.g.,][]{limber1960, vandenbergh1960}. However, this early agreement with Salpeter gave way to large scatter in IMF measurements beginning in the 1970s, as several studies with more sophisticated completeness corrections and updated stellar physics highlighted the shortcomings in the first generation of IMF studies \citep[e.g.,][]{taff1974,lequeux1979, tarrab1982}. Furthermore, \citet{miller1979} provided a comprehensive reassessment of the MW field star IMF, in light of the previously unaccounted for SFH-IMF degeneracy. They suggested a high-mass IMF of the with $\Gamma=1.5$ for stars $1\lesssim$M/M$_{\odot}$$\lesssim$10 and $\Gamma=2.3$ for stars M/M$_{\odot}$$\ga$10. However, \citet{kennicutt1983} showed that such a steep IMF slope above $\sim$10 M$_{\odot}$\ was incompatible with observations of $B-V$ colors and H$\alpha$\ equivalent widths in nearby disk galaxies, suggesting that extrapolating the value of $\Gamma=1.5$ from 1-100 M$_{\odot}$\ provided a better match to the data. This conclusion was further solidified in \citet{kennicutt1994}.
The number of observational high-mass IMF studies peaked between the late 1980s and early 2000s. This golden age followed the comprehensive IMF review by \citet{scalo1986}, which concluded that systematic uncertainties were the dominant source of confusion in our poor knowledge of the IMF. Shortly after, the proliferation of CCDs enabled significant progress in reducing these systematics, primarily by allowing for quantitative completeness corrections via artificial star tests \citep[e.g.,][]{stetson1987, mateo1993}. The 1990s saw a flurry of qualitatively new IMF studies including star counts in the LMC, spectroscopy of starburst galaxies, and IMF constraints from stellar abundance patterns (see \citealt{leitherer1998, massey1998, gilmore2001} for extensive discussion of the literature in these areas). The sheer number and richness of IMF studies from this era precludes a detailed history in this paper, and we instead refer readers to a number of excellent papers that chronicle the history of the IMF \citep[e.g.,][]{scalo1986, massey1998, kennicutt1998b, elmegreen1999, eisenhauer2001, gilmore2001, massey2003, scalo2005, elmegreen2006b, zinnecker2005, bastian2010}.
To briefly summarize, many studies from this era concluded that the high-mass IMF in the local universe was not drastically different from Salpeter over a wide range of environments \citep[e.g.,][]{massey1998, leitherer1998, gilmore2001, kroupa2001, massey2003}. However, there was no clear explanation for the observed scatter in the measured IMF slopes, which contradicts a truly universal high-mass IMF. Some have argued that measurement uncertainty and systematics are the dominant source of scatter \citep[e.g.,][]{elmegreen1999, kroupa2001, kroupa2002, massey2003}. However, without systematically re-visiting each study, it is challenging to dismiss the scatter as solely a measurement uncertainty. As a clear counter example, there are well-known studies of star clusters with some of the same physical properties (e.g., age, mass) that have been analyzed identically, but have very different high-mass IMF slopes \citep[e.g.,][]{phelps1993}, reinforcing that some of the scatter may be physical in nature. At the same time, on the whole, $>$75\% of resolved star IMF studies have significantly underreported uncertainties relative to the theoretically possible precision \citep{weisz2013a}, which underlines the difficulty in disentangling systematics from intrinsic IMF variation using literature IMF studies.
Since the early 2000s, studies of the high-mass IMF have diminished in number, as attention turned to the sub-solar regime \citep[e.g.,][]{chabrier2003, scalo2005, zinnecker2005}. However, focus on the high-mass IMF was recently rekindled following several reports of a systematic deficit of massive stars in lower-luminosity galaxies, based on observations of integrated broadband and/or H$\alpha$-to-UV flux ratios \citep[e.g.,][]{hoversten2008, boselli2009, lee2009, meurer2009}. Yet, whether these observations are indicative of a systematically varying IMF \citep[e.g.,][]{weidner2005, pflammAltenburg2009} remains an open question, as other mechanisms such as stochastic sampling of the cluster mass function and/or bursty SFHs can also explain these observed properties \citep[e.g.,][]{fumagalli2011, dasilva2012, weisz2012a}.
Thus, at present, the high-mass IMF is characterized by common usage of a universal Salpeter/Kroupa IMF, but also boasts a vast body of literature studies which does not unambiguously support this as the \textit{de facto} standard. After decades of study, the main limitation in our knowledge of the high-mass IMF remains our reliance on a rich, but heterogenous set of literature IMF studies, instead of a large, homogenous survey of individually resolved young, massive stars.
\subsection{The IMF $>$ 2--3 M$_{\odot}$\ in M31}
\label{sec:phat_imf}
A primary goal of the Panchromatic Hubble Andromeda Treasury program \citep[PHAT;][]{dalcanton2012} is a comprehensive exploration of the stellar IMF above $\sim$ 2-3 M$_{\odot}$. This 828-orbit \textit{HST} multi-cycle program acquired near-UV through the near-IR imaging of $\sim$120 million
resolved stars in $\sim$ 1/3 of M31's star-forming disk \citep{williams2014}, providing a dataset of unprecedented size and homogeneity for measuring the high-mass IMF.
The focus of the present IMF study is on M31's young star cluster population. Star clusters provide a more direct path to measuring the IMF compared to field populations, which suffer from IMF-SFH degeneracies for stars more luminous than the oldest main sequence turnoff \citep[e.g.,][]{miller1979, elmegreen2006}, Further, the differential extinction found in the disk of M31 \citep[e.g.,][]{draine2013, dalcanton2015}, and similarly sized star-forming galaxies, adds a significant layer of complexity to modeling the field stellar populations. In contrast, star clusters provide a relatively simple way to probe the IMF. Their co-eval nature genuinely mitigates the effects of SFH, while their vulnerability to foreground differential extinction is minimal due to their small sizes.
In this paper, we focus on determining the MF slopes for PHAT clusters that are $\la 25$ Myr old and $\ga10^3$ M$_{\odot}$\ in mass using resolved star color-magnitude diagrams (CMDs). For these clusters, the detectable portion of the main sequence is well-populated from $\sim$ 2M$_{\odot}$\ to $\ga 25$M$_{\odot}$\ and they are of order $\sim$ 0.1 relaxation times old, which mitigates the strongest effects of mass segregation, making them an ideal sample with which to study the high-mass IMF. We model the optical CMD of each cluster $k$ to constrain its power-law slope $\Gamma_k$ of the present day high-mass MF, ${\mathrm p}(\Gamma_k)$, marginalized over other parameters of that fit (e.g., cluster age, extinction, binarity).
Subsequently, we propose a simple model for the ensemble properties of the cluster MF, with a mean, $\Gamma$, and a scatter, $\sigma_\Gamma$, and fit it to the MF probability distribution functions (PDFs) of the individual clusters. We also use this model to search for possible dependencies of $\Gamma$ on cluster mass, age, and size. We verify this approach with an extensive set of artificial clusters that were inserted into PHAT images, processed, and analyzed identically to the real clusters.
This paper is organized as follows. In \S \ref{sec:data}, we define our cluster sample and briefly describe the PHAT photometry, artificial star tests, and artificial cluster sample. We outline our method of modeling the cluster CMDs in \S \ref{sec:methodology} and present the results for the ensembles of real and artificial clusters in \S \ref{sec:results}. Finally, in \S \ref{sec:discussion}, we discuss our findings in the context of current and historical views on the high-mass IMF and illustrate the implications of our results
\section{Data}
\label{sec:data}
\subsection{Cluster Sample}
\label{sec:sample}
Clusters in PHAT were identified as part of the Andromeda Project\footnote{\url{http://www.andromedaproject.org}}, a collaborative effort between PHAT and the Zooniverse\footnote{\url{https://www.zooniverse.org}} citizen science platform. As described in \citet{johnson2015}, cluster candidates were assigned probabilities of actually being a cluster based on citizen scientist ranking. These probabilities were verified using synthetic cluster results and comparison with the preliminary catalog, which was assembled by professional astronomers \citep{johnson2012}. Extensive details of the cluster identification methodology can be found in \citet{johnson2015}.
For the present study, we have limited our analysis to the 85 clusters with CMD-based ages $\lesssim 25$ Myr and masses $\gtrsim 10^3$ M$_{\odot}$\ \citep{beerman2015}, and half-light radii $\ga$ 2 pc \citep{johnson2015}. We focus on these young, more massive clusters in PHAT to avoid the very sparse sampling in the low-mass clusters; to limit the impact of stellar crowding in the densest clusters; to facilitate the extrapolation of the observable MF to an IMF; and to remain in the regime of high completeness ($>$90\% for our sample; see Figure 10 in \citealt{johnson2015}). We will explore each of these issues in detail in our comprehensive study of PHAT cluster MFs in a future paper.
\subsection{Cluster Photometry}
\label{sec:photometry}
Photometry of PHAT clusters was performed using DOLPHOT\footnote{\url{http://americano.dolphinsim.com}}, a version of \texttt{HSTPHOT} \citep{dolphin2000b} with HST specific modules, following the description in \citet{williams2014}\footnote{\url{http://archive.stsci.edu/missions/hlsp/phat/}}. Slight changes to the survey-wide photometry include reducing only the optical photometry (F475W, similar to SDSS g; and F814W, similar to I-band), which, for the purposes of this study provide the most information. The UV imaging is significantly shallower ($\sim$ 2 mag) than the optical, even along the main sequence, resulting in many fewer sources detected. The IR also yields far fewer main sequence stars as a result of significant crowding due to the lower angular resolution of the WFC3/IR camera. An image and optical CMD of a typical cluster used in this study are shown in Figures \ref{fig:image} and \ref{fig:cmd}.
In principle, the full spectral energy distribution of each star can increase the degree of precision to which we know its mass, and, in turn provide improved constraints on the IMF. However, the UV and IR photometry in PHAT clusters typically contains fewer than $\sim$ 50\% the number of stars detected in the optical, reducing their statistical utility for constraining the MF slope \citep[cf.][]{weisz2013a}. Further, the UV and IR bolometric corrections are not as certain as those in the optical, particularly for massive, metal rich stars, which can introduce systematics into our analysis. Finally, the modest gain (at best) in precision on the IMF slope of each cluster must be balanced against the computational cost of processing $>$10 million additional artificial star tests (ASTs). Thus, by only reducing and analyzing the optical data, we retain virtually all statistical leverage on measuring the MF slope, but save considerably on computational costs.
\begin{figure}[t!]
\epsscale{0.85}
\plotone{ap94_color.png}
\caption{The HST/ACS optical color image of AP~94, a typical cluster analyzed in this paper. The angular high-resolution of HST allows us to resolve the individual stars clusters at the distance of M31. For reference, the image is $\sim$50pc on each side.}
\label{fig:image}
\end{figure}
However, analysis of the full stellar SEDs are valuable for searching for the most massive stars in clusters. In addition to the slope, the maximum stellar mass plays a fundamental role in defining the high-mass IMF, but is unconstrained by the approach we use to measure the MF slope \citep[e.g.,][]{weisz2013a}. An investigation of the most massive stars in the PHAT area is the subject on an ongoing study and will be presented in a future paper.
For each cluster, we ran $\sim$50,000 ASTs by adding individual stars of known luminosities to PHAT images of each cluster. The ASTs were uniformly distributed over each cluster's CMD and spatially distributed approximately according to each cluster's light profile. Extensive testing showed that such a spatial scheme was necessary to accurately characterize the completeness and photometric uncertainty profiles of the clusters, as a simple uniform spatial scheme results in overly optimistic completeness and photometric uncertainty profiles.
\subsection{Artificial Clusters}
\label{sec:sample}
We use extensive sets of artificial clusters to verify the accuracy of each component of our MF determinations, including photometry, ASTs, and CMD modeling. The artificial clusters were designed to match the physical properties of our real cluster sample and span a range of physical parameters (age, mass, size, extinction) that encompass the real clusters. The artificial clusters were inserted into images at a range of galactocentric radii and environments to capture the full range of background densities in PHAT. Overall, the artificial cluster tests confirmed that we can accurately recover the true MF from clusters in PHAT imaging, along with all other cluster physical properties.
\begin{figure}[t!]
\epsscale{1.2}
\plotone{ap94_cmd.png}
\caption{Panel (a) -- The optical CMD of AP 94. The grey points are below the nominal completeness limit, and were not included in the CMD analysis. For reference, we indicate the main sequence turn off masses on the left hand side and over-plot the best fitting isochrone for this cluster (see \S \ref{sec:match} and Figure \ref{fig:triangle_match}). Panel (b) -- the background CMD constructed by selecting all stars in an annulus surrounding the cluster that has 10x times the cluster area. The black points in this CMD are used to statistically model background contribution to the cluster CMD.}
\label{fig:cmd}
\end{figure}
\section{Methodology}
\label{sec:methodology}
\subsection{Measuring the Stellar Mass Function of a Cluster}
\label{sec:match}
To measure the MF of a cluster we construct synthetic models of its optical CMD using \texttt{MATCH}, a CMD fitting program described in \citet{dolphin2002}. Briefly, for a given stellar evolution library, binary fraction, IMF, age, metallicity, extinction, and distance, \texttt{MATCH} constructs a simple stellar population (SSP) in the desired filter combinations. This synthetic SSP is then convolved with completeness and photometric biases and uncertainties as determined by the ASTs. The resulting synthetic CMD is then linearly combined with a suitable CMD of foreground/background populations, with the relative contribution of each controlled by a scale parameter. For a given set of these parameters, the synthetic Hess diagram, the binned number density of stars in the CMD, is compared to the observed CMD using a Poisson likelihood function, which is necessary to account for the effects of sparse sampling. The procedure is repeated over a large grid of parameters in order to map all of likelihood space. For this analysis, we adopted the Padova stellar evolution models \citep{girardi2002, marigo2008, girardi2010} and we have listed the other model parameters and their ranges in Table \ref{tab:matchparams}.
Of these parameters, we limited the metallicity to a fairly small range near Z$_{\odot}$. This was done because (a) the cluster CMDs are too sparsely populated to provide robust metallicity constraints and (b) the current metallicity of M31's disk is known to be approximately solar, with a very weak gradient based on H{\sc II} region and supergiant abundances \citep[e.g.,][]{venn2000, trundle2002, sanders2012, zurita2012}.
Similarly, we adopted a fixed binary fraction of 0.35, where the mass ratio of the stars is drawn from a uniform distribution. Historically, stellar multiplicity has been problematic for high-mass IMF determinations from luminosity functions \citep[e.g.,][]{maizapellaniz2005, maizapellaniz2008}. However, for CMD analysis, it is much less of an issue for two reasons. First, the addition of color information provides clear separation between the single and binary star sequences, minimizing confusion between the two. In contrast, when only luminosity information is available, the binary fraction and IMF slope are essentially degenerate parameters. Second, stellar multiplicity is observationally less important for high-mass stars compared to sub-solar mass stars. For stars above 1 M$_{\odot}$, the majority of companion stars are likely to be significantly less massive and less luminous than the primary \citep[cf.][]{kobulnicky2014}, resulting in minimal change to the CMDs.
\begin{deluxetable}{cccc}
\tablecaption{Parameters for Measuring the MF of an M31 Star Cluster}
\tablecolumns{4}
\tablehead{
\colhead{Parameter} &
\colhead{Range} &
\colhead{Grid Resolution} &
\colhead{Notes} \\
\colhead{(1)} &
\colhead{(2)} &
\colhead{(3)} &
\colhead{(4)} \\
}
\startdata
$\Gamma$ & (-1.2,5) & 0.2 dex & MF slope for stars $\ga$ 2 M$_{\odot}$ \\
Binary Fraction & 0.35 & & Fraction of Stars with a Companion \\
Log(Age)& (6.60, 9.0) & 0.1 dex & Cluster Age \\
A$_V$ & (0.0, 3.0) & 0.05 dex & Line of Sight Extinction \\
(m-M)$_0$ & 24.47 & & Adopted Distance Modulus to M31 \\
log(Z) & (-0.2, 0.1) & 0.1 dex & Metallicity
\label{tab:matchparams}
\tablecomments{Parameters and their ranges used for modeling M31 cluster CMDs. The adopted distance modulus is from the tip of the red giant branch measurement made by \citet{mcconnachie2005}.}
\end{deluxetable}
Within PHAT, we verified that our choice in binary fraction does not change our results. For example, in the case of AP~94 (Figures \ref{fig:cmd} and \ref{fig:triangle_match}), we modeled the CMD with binary fractions of 0.35 and 0.70 and found differences in the resulting IMF slope constraints to be $<$0.02 dex. Tests including different binary fractions on artificial clusters yielded similarly small changes.
However, we did find that extreme binary fractions of 0 or 1 resulted in significantly worse CMD fits. In these two cases, the model either entirely populated the single star sequence (binary fraction = 0) or only sparsely populated it (binary fraction = 1), which substantially reduced the resemblance of the model CMD to the observed CMD. Thus, for the present work, we adopted a fixed binary fraction of 0.35, which appears to be a reasonable value from LMC cluster studies \citep[e.g.,][]{elson1998}.
To model the background, we construct an empirical CMD from an annulus around the cluster that is ten times larger than the cluster. The large area is necessary to sample all parts of the CMD to accurately constrain the background scaling parameter.
Given a set of cluster model parameters (age, extinction, metallicity, and MF slope) the resulting model CMD, and the empirically derived CMD for fore-/background sources are compared to the observed CMD in a likelihood sense \citep{dolphin2002}. The resulting posterior probabilities, $\mathrm{p}(t,A_V,\Gamma )$ are illustrated in Figure \ref{fig:triangle_match} for a typical cluster in the sample.
The prior on metallicity tightly brackets solar metallicity and the cluster mass emerges as a normalization factor, rather than a fit parameter. Finally, the smoothness of the $\mathrm{p}(t,A_V,\Gamma )$ of each cluster allows us to interpolate in order to increase the resolution of our PDFs.
\begin{figure}[t!]
\epsscale{1.2}
\plotone{ap94_triangle.png}
\caption{The joint and marginalized distributions of age, MF slope, and extinction for a single young M31 cluster AP 94 that resulted from modeling the cluster's optical CMD, which is shown in Figure \ref{fig:cmd}. The dark points in each panel reflect the regions of high probability, and the pixel sizes reflect the grid resolution as indicated in Table \ref{tab:matchparams}. We use the marginalized distribution for $\Gamma$ (shaded histogram) in the subsequent modeling of the distribution of MF slopes (\S \ref{sec:inference}). }
\label{fig:triangle_match}
\end{figure}
\subsection{Inferring the Intrinsic Distribution of Mass Function Slopes From a Set of Clusters}
\label{sec:inference}
One challenge for past MF studies was how to combine measurements from individual clusters into a broader statement about the MF slope of `clusters in general'. Typically, the variance weighted mean is used to compute the ``average'' MF from an ensemble \citep[e.g.,][]{scalo1986}. However, this can potentially be a biased estimator (e.g., assumptions of normally distributed uncertainties), and does not necessarily use all available information (e.g., the probability that an object is actually a cluster).
Here, we adopt a framework that allows us to infer population wide characteristics
about the \textit{distribution of MF slopes}, given a set of noisy measurements. This model follows the approach and implementation laid out in \citet{hogg2010} and \citet{foremanmackey2014}. We briefly describe the model below and we refer the reader to those papers for more detail.
As a simple model for the \emph{distribution} of cluster MFs, we assume that it can be described by a normal distribution, $\mathcal{N}(\Gamma, \sigma_\Gamma)$, with a mean of $\Gamma$ and an intrinsic dispersion of $\sigma_\Gamma$. We initially assume that their parameters are independent of age, mass, and size of the cluster. If we assume the probability that each object is a cluster, $\ensuremath{\bvec{Q}}_k$, we can then write down a Gaussian mixture model likelihood function for the MF of clusters, where we simply assume that the MFs of falsely presumed clusters have a different $\mathcal{N}(\Gamma_{false}, \sigma_{\Gamma,false} )$. This mixture model mitigates the impact of erroneously identified clusters, without forcing a binary decision on
which clusters have been identified with sufficient confidence.
More explicitly, the mixture model can be written as
\begin{eqnarray}
p(\ensuremath{\bvec{\Gamma}}_k\, |\, \ensuremath{\bvec{\theta}}, \ensuremath{\bvec{Q}}_k) &=& \ensuremath{\bvec{Q}}_k \, \exp\left(-(\Gamma - \ensuremath{\bvec{\Gamma}}_k) / (2 \, \sigma_\Gamma^2)\right) \nonumber \\
&+& (1-\ensuremath{\bvec{Q}}_k) \, \exp\left(-(\Gamma_{false} - \ensuremath{\bvec{\Gamma}}_k) / (2 \, \sigma_{false}^2)\right) \nonumber \, \\
\label{eq:imfmodel}
\end{eqnarray}
\noindent where $\ensuremath{\bvec{\Gamma}}_k$ is the marginalized PDF for the MF of the k$^{th}$ object and $\ensuremath{\bvec{Q}}_k$ is the probability that the k$^{th}$ object is a cluster. $\Gamma$ and $\sigma_\Gamma^2$ are the mean and variance of the Gaussian of interest for the clusters and $\Gamma_{false}$ and $\sigma_{false}^2$ are nuisance parameters for possible contaminating sources. Simply put, with this model, the more probable an object is a cluster the more it contributes to the MF parameters of interest. In this model context, a universal Kroupa MF would be represented by $\Gamma=1.30$ and $\sigma_{\Gamma}^2\rightarrow0$, and universal Salpeter MF by $\Gamma=1.35$ and $\sigma_{\Gamma}^2\rightarrow0$.
\begin{figure}[t!]
\epsscale{1.1}
\plotone{Gamma_mass_IMFsample.png}
\caption{85 young, intermediate mass clusters from PHAT used to measure the high-mass IMF in M31. Both panels show the median of the probability distribution for for each cluster's MF slope vs. cluster mass. Points in the top panel are color-coded by their age, while those in the bottom panel by the probability of being a cluster. In all cases, point sizes are proportional to half-light radius and the uncertainties in $\Gamma$ reflect the 68\% confidence interval on the MF slope.}
\label{fig:gammapoints}
\end{figure}
This same framework also readily allows us to explore whether the mean MF slope depends on any of the other cluster properties such as age, mass, and size. Specifically, we can generalize the model to
\begin{eqnarray}
\Gamma (t_{k},M_k,r_k )= \overline{\Gamma} + a_m \, {\rm log} \frac{M_k}{M_c} \, + a_t \, {\rm log} \frac{t_{k}}{t_{c}} + a_r \, {\rm log} \frac{r_k}{r_c} \,\,
\\ \nonumber
\label{eq:hypergamma1}
\end{eqnarray}
\noindent where $\{t_k,M_k,r_k\}$ are the most likely age, mass, and size for the $k^{th}$ object and $\{t_c,M_c,r_c\}$ are the mean of our cluster sample in each case.
We assume priors that are flat in $\overline\Gamma,~\log{\sigma_\Gamma},~a_m,~a_t,~a_r$,
and sample these parameter's PDFs with the affine invariant ensemble Markov chain Monte Carlo (MCMC) program \texttt{emcee} \citep{foremanmackey2013}\footnote{\url{https://github.com/dfm/emcee}}.
\section{Results}
\label{sec:results}
The analysis laid out in Section \ref{sec:methodology} proceeds in two steps, the first leading to the MF slope measurements for individual clusters, $p(\Gamma_k)$, the second to the characterization of the intrinsic, i.e., error corrected, distribution of mass function slopes in M31 clusters.
The result of the first step, the marginalized PDF, $\mathrm{p}(\Gamma_k)$, for all 85 sample clusters is
illustrated in Figure \ref{fig:gammapoints}, showing the MF slope and its uncertainty for each cluster versus its mass. Points in the top panel are color-coded by age and those in the bottom panels by their probability of being a cluster. In both cases, the point size is proportional to the cluster's half-light radius. Clusters with larger half-light radii have lower crowding and thus more accurate mass function determination. From a physical standpoint, the size of a cluster may also reflect the environmental conditions in which it was born.
\begin{figure}[t!]
\epsscale{1.2}
\plotone{m31_imf_smalltriangle.png}
\caption{The joint and marginalized distributions for the distribution of cluster MFs described by Equations \ref{eq:imfmodel} and \ref{eq:hypergamma1}. In all panels, except for $\sigma_{\Gamma}$, the dashed blue lines represent the 16th, 50th, and 84th percentiles of each marginalized distribution. The distribution for $\sigma_{\Gamma}$ is highly non-Gaussian, making the median a poor point estimate. Instead, we use the mode as a point estimate, as it reflects the most probable portion of the PDF, and use the 68th percentile to represent an upper limit. For reference, the solid red lines indicate the value of a Kroupa IMF slope in the $\overline{Gamma}$ panels, the upper uncertainty on $\sigma_{\Gamma}$ for a universal IMF as determined by similar analysis of artificial clusters, and a value of zero in the remaining panels. This plots shows that the mean IMF slope in M31 clusters is steeper than Kroupa, that the recovered scatter is consistent with expectations of a universal IMF, with a small tail to larger values, and that there are no significant trends between the IMF slope and age, mass, or size of the clusters. }
\label{fig:triangle}
\end{figure}
While we have used a different methodology here than \citet{weisz2013a}, the error bars on the MF are similar in the two approaches, limited in both cases by the number of well-detected cluster members and by the range of stellar masses they span. For example, consider AP 94, whose CMD and derived physical property PDFs are shown in Figures \ref{fig:cmd} and \ref{fig:triangle_match}. Effectively, there are $\sim$120 stars (after background correction) between $\sim$ 3 and 11 M$_{\odot}$\ on the upper MS that are used for MF determination. Given these numbers, from Figure 7 in \citet{weisz2013a}, we expect a theoretical uncertainty of $\sim$$\pm$0.4 on its MF slope. In comparison, the 1-$\sigma$ uncertainty from the marginalized MF PDF in Figure \ref{fig:triangle_match} is $\pm$0.45, which is very close to theoretical expectation. Uncertainties in cluster membership, the exact mass range on the main sequence, etc., can explain the small difference in these two values. In general, the cluster MF slopes are generally less precise for older and lower mass clusters, and follow the expected precision relationship derived in \citet{weisz2013a}.
\begin{figure}[t!]
\epsscale{1.2}
\plotone{IMF_posterior_draws2.png}
\caption{A comparison between the expected IMF slope, $\Gamma$, from artificial clusters inserted into PHAT images (blue) and the real clusters (red), under the assumption that the underlying IMF slope is universal, i.e., $\sigma_{\Gamma} \rightarrow 0$. The solid lines indicate the most probable values, while the shaded regions are the 68\% confidence intervals. The uncertainties on the two measurements begin to overlap at a 2-$\sigma$ level.}
\label{fig:IMFmodel}
\end{figure}
Figure \ref{fig:gammapoints} also gives a qualitative impression of the ensemble properties of MF measurements. The majority of clusters have MF slopes that are near the Kroupa (1.30) or Salpeter (1.35) slopes; $\sim70\%$ of the clusters MF slopes are within $\sim$ 1-$\sigma$ of Kroupa and $\sim$92\% are within 2-$\sigma$. Remarkably, in this large sample of young clusters we find that none have extremely steep or flat MF slopes. At least for the parameter space covered by our cluster sample, extreme MF slope outliers ($>3-\sigma$) are quite rare.
The plots also show the lack of strong correlations between cluster MF slope and their physical properties. From visual inspection, it is clear that the MF slopes do not show significant trends as a function of age, mass, size, or probability of being a cluster.
These impressions from Figure \ref{fig:gammapoints} are quantified in the second step of our analysis, where we propose and constrain
a simple model for the cluster's distribution of MFs (see \S\ref{sec:inference}), and constrain its parameters through the comparison with
all 85 $\mathrm{p}(\Gamma_k)$.
Figure \ref{fig:triangle} illustrates the result for the five parameters in our model:
the mean MF slope, $\Gamma$, and its intrinsic dispersion $\sigma_\Gamma$, and the three coefficients that represent linear trends between the MF slope and cluster age, mass, and size. This plot shows that the mean MF slope ($\overline{\Gamma}=1.46^{+0.04}_{-0.07}$) is steeper than a Kroupa IMF, that the scatter ($\sigma_{\Gamma}=0.03^{+0.1}_{-0.00}$) is consistent with expectations from a universal IMF as determined by similar analysis of the artificial clusters, and that there are no significant trends between MF slope and cluster physical properties. The small degree of scatter is particularly interesting, giving the visual impression of large variation between single clusters Figure \ref{fig:gammapoints}. The typical cluster has an uncertainty on its MF slope of $\sim$ 0.5 dex, but that ensemble scatter scales roughly as $1/\sqrt(N_{\rm clusters})$, indicating that a large and homogenous sample of clusters are needed to statistically identify a universal IMF. Finally, the limit on $a_t$ implies that there is no evidence that the MF slope changes between 4 and 25~Myr. We list summary statistics for each distribution in Table \ref{tab:params} along with the same statistics for our analysis of the population of artificial clusters.
There are two reasons why the artificial clusters results are not an exact delta function. First, the finite grid resolution in p$(\Gamma_k)$ sets a floor of $\sigma\sim0.03$; second, as $\sigma_\Gamma$ is a positive definite quantity, its most likely estimate from noisy data must be significantly positive. However, this effect is quite small, as the lower limit on the scatter for the artificial clusters is consistent with the minimum possible value, given the resolution of our individual cluster PDFs.
\begin{deluxetable}{cccc}
\tablecaption{IMF Parameters for Ensemble of Clusters}
\tablecolumns{4}
\tablehead{
\colhead{Parameter} &
\colhead{Artificial} &
\colhead{Real} &
\colhead{Real} \\
\colhead{} &
\colhead{Clusters} &
\colhead{Clusters} &
\colhead{Clusters} \\
\colhead{(1)} &
\colhead{(2)} &
\colhead{(3)} &
\colhead{(4)}
}
\startdata
$\overline{\Gamma_1}$ & $1.29^{+0.05}_{-0.04}$ & $1.46^{+0.04}_{-0.07}$ & $1.45^{+0.03}_{-0.06}$ \\
$\sigma_{\Gamma_1}$ & $0.03^{+0.02}_{-0.00}$ & $0.03^{+0.1}_{-0.00}$ & $\cdots$ \\
$a_m$ & $0.01^{+0.08}_{-0.08}$ & $0.05^{+0.10}_{-0.10}$ & $\cdots$ \\
$a_t$ & $0.05^{+0.07}_{-0.08}$ & $0.06^{+0.12}_{-0.09}$ & $\cdots$ \\
$a_r$ & $-0.02^{+0.10}_{-0.12}$ & $-0.05^{+0.09}_{-0.11}$ & $\cdots$
\tablecomments{Column (2) -- Values of the five parameter model for the ensemble IMF from Equations \ref{eq:imfmodel} and \ref{eq:hypergamma1}. In the case of $\sigma_{\Gamma}$, we quote the mode of its marginalized PDF, along with the upper 68\% confidence interval, as the distribution is highly non-Gaussian. Column (3) -- Value of the IMF slope by fixing all other parameters to zero, i.e., the assumption that the IMF is `universal'. The listed values reflect the median and surrounding 68\% confidence interval.}
\label{tab:params}
\end{deluxetable}
Given that the scatter is consistent with expectations for a universal IMF, we can consider a model in which all the clusters have the same underlying MF i.e., $\sigma \rightarrow$ 0. We find an ensemble-wide value of $\Gamma=1.45^{+0.03}_{-0.05}$, which is consistent with results from the hierarchical model. For comparison, the same analysis applied to the artificial clusters yields a value of $\Gamma=1.30^{+0.04}_{-0.04}$. Uncertainties on the two mean values begin to overlap at the $\sim$ 2-$\sigma$ level. We plot the results of the `universal' MF model in Figure \ref{fig:IMFmodel}.
\section{Discussion}
\label{sec:discussion}
The quality of the PHAT data has enabled important progress on three issues: (1) quantifying the extent to which an MF measurement reflects an \emph{initial} MF (IMF) estimate; (2) determining accurately what the mean IMF slope is in typical solar-metallicity (cluster) populations of a large disk galaxy; and (3) assessing the extent to which the IMF is `universal', in the sense that it shows very little scatter in the slope, and no trends of the slope with cluster properties. The fundamental conceptual limitation of the present approach is that it only applies to clusters that remain grouped for longer than 4~Myr, i.e., they have not (yet) disrupted.
Our conclusions are based on measurements of the high-mass IMF that combine a principled and systematic constraint of each cluster's MF slope,
from CMD analysis, with a probabilistic modeling framework to analyze the distribution of MF slopes.
Overall, we find that the high-mass IMF for clusters in M31 is consistent with a universal IMF. That is, as discussed in \S \ref{sec:results} and illustrated in Figures \ref{fig:triangle} and \ref{fig:IMFmodel}, the favored high-mass IMF model has a steeper mean slope than Salpeter/Kroupa with a small scatter that is inline with expectations for a universal IMF. Importantly, we also show that the mean MF slope in clusters of different ages changes by $\Delta\Gamma\la 0.1$ per decade of age.This suggests that the MF slope is unchanged over the first $\sim 25$~Myrs, and provides direct observational evidence that the measured MF represents the IMF. In this sense, our result represents an important reference point for understanding the potential evolution of the cluster MF with cluster physical parameters (age, mass, size) and local star-forming conditions, both of which will be explored in \citet{weisz2015}.
Of these results, the most robust determination is the mean IMF slope. Even if we assume that all of the clusters come from the same underlying IMF (i.e., no scatter) the steep slope remains. The lack of correlations between mean IMF slope and cluster physical properties suggests that other potential sources of bias (e.g., mass segregation) contribute at a negligible level to our measurement of the mean.
\begin{deluxetable}{ccc}
\tablecaption{Select Literature High-Mass IMF Slopes}
\tablecolumns{3}
\tablehead{
\colhead{Cluster} &
\colhead{$\Gamma$} &
\colhead{M$_{\mathrm upper}$} \\
\colhead{} &
\colhead{} &
\colhead{(M$_{\odot}$)} \\
\colhead{(1)} &
\colhead{(2)} &
\colhead{(3)}
}
\startdata
& LMC & \\
\hline
LH 6 & 1.7$\pm$0.4 & 85 \\
LH 9 & 1.7$\pm$0.4 & 55 \\
LH 10 & 1.7$\pm$0.4 & 90\\
LH 38 & 1.7$\pm$0.4 & 85 \\
LH 47/48 & 1.3$\pm$0.2 & 50 \\
LH 58 & 1.4$\pm$0.2 & 50 \\
LH 73 & 1.3$\pm$0.4 & 65 \\
LH 83 & 1.3$\pm$0.5 & 50 \\
LH 114 & 1.0$\pm$0.1 & 50 \\
LH 117/118 & 1.6$\pm$0.2 & $>$120 \\
R136 & 1.3$\pm$0.1 & $>$120 \\
\hline
& MW & \\
\hline
NGC 6823 & 1.3$\pm$0.4 & 40 \\
NGC 6871 & 0.9$\pm$0.4 & 40 \\
NGC 6913 & 1.1$\pm$0.6 & 40 \\
Berkeley 86 & 1.7$\pm$0.4 & 40 \\
NGC 7280 & 1.7$\pm$0.3 & 65 \\
Cep OB5 & 2.1$\pm$0.6 & 30 \\
IC 1805 & 1.3$\pm$0.2 & 100 \\
NGC 1893 & 1.6$\pm$0.3 & 65 \\
NGC 2244 & 0.8$\pm$0.3 & 70 \\
NGC 6611 & 0.7$\pm$0.2 & 75 \\
Cyg OB2 & 0.9$\pm$0.2 & 110 \\
Tr 14/16 & 1.0$\pm$0.2 & $>$120 \\
h \& $\chi$ Per & 1.3$\pm$0.2 & 120
\tablecomments{High-mass IMF slopes from the massive star studies in the MW and LMC clusters as listed in
\citet{massey1998} and \citet{massey2003}. The dynamic mass range of the clusters extend from $>$1 M$_{\odot}$\ to the mass listed in column (3). }
\label{tab:massey}
\end{deluxetable}
In contrast, it is more challenging to interpret the significance of the upper bound on the scatter. The results of our inference suggest that a universal IMF is most probable. However, there is a tail to the distribution that extends to $\gtrsim$ 0.1 dex, which exceeds the tail of the same distribution for the artificial clusters. We believe that this scatter is likely due to sources of uncertainty not included in our modeling that may artificially enhance the scatter relative to expectations from the artificial clusters. For example, mismatches between the stellar models and real stars are not captured by the artificial cluster tests. However, the degree to which model-data mismatches contribute to the scatter is hard to quantify.
Differential effects may also contribute to the observed scatter. For example, significant cluster-to-cluster variations in mass segregation or binarity could contribute to the scatter in ways not captured by the artificial clusters or our model. Similarly, we impose a single binary fraction for all masses, which may not be accurate, given that the most massive stars may be more likely to have equal mass companions than intermediate mass stars \citep[e.g.,][]{sana2013, kobulnicky2014}. At the same time, given the modest size of the upper bound, we expect these effects are, at most, fairly small.
Overall, it is remarkable that the IMF slope is consistent with universal in this mass, metallicity, and cluster density regime. As summarized in \citet{kennicutt1998b}, quantifying the degree of scatter is one of the most critical aspects of understanding the IMF, and ultimately how stars form out of dense gas. The intrinsic scatter we find suggests that individual clusters should only minimally deviate from the fiducial value, and that extreme outliers (e.g., $\gtrsim 3-\sigma$) are quite rare. The identification of such objects may represent insight into a mode of star-formation that truly differs from what we observe in this regime.
\begin{figure}[tp!]
\epsscale{1.2}
\plotone{Relative_IMF.png}
\caption{The number of stars predicted by various common IMF models, relative to Kroupa (dashed line).
The vertical dotted lines indicate the mass range over which the high-mass IMF was measured in M31. Our median IMF (solid red line) is essentially identical to that of \citet{kennicutt1983} ($\Gamma_{\rm Kennicutt}=1.5$). However, we also find a small spread, which is indicated by the red band. Our IMF model indicates a modest deficit of stars $>$2 M$_{\odot}$\ relative to the canonical Kroupa IMF.}
\label{fig:relativeIMF}
\end{figure}
\begin{deluxetable*}{ccccccccccc}
\tablecaption{Number and Fraction of Massive Stars for Different IMFs}
\tablecolumns{11}
\tablehead{
\colhead{IMF} &
\colhead{$N{\star}\ge2 M_{\odot}$} &
\colhead{$N{\star}\ge8 M_{\odot}$} &
\colhead{$N{\star}\ge20 M_{\odot}$} &
\colhead{$N{\star}\ge50 M_{\odot}$} &
\colhead{$N{\star}\ge90 M_{\odot}$} &
\colhead{Frac. $\ge$2 M$_{\odot}$} &
\colhead{Frac. $\ge$8 M$_{\odot}$} &
\colhead{Frac. $\ge$20 M$_{\odot}$} &
\colhead{Frac. $\ge$50 M$_{\odot}$} &
\colhead{Frac. $\ge$90 M$_{\odot}$} \\
\colhead{} &
\colhead{$\frac{10^4 \rm stars}{10^6 \, M_{\odot}}$} &
\colhead{$\frac{10^3 \rm stars}{10^6 \, M_{\odot}}$} &
\colhead{$\frac{10^3 \rm stars}{10^6 \, M_{\odot}}$} &
\colhead{$\frac{10^2 \rm stars}{10^6 \, M_{\odot}}$} &
\colhead{$\frac{10^1 \rm stars}{10^6 \, M_{\odot}}$} &
\colhead{} &
\colhead{} &
\colhead{} &
\colhead{} &
\colhead{} \\
\colhead{(1)} &
\colhead{(2)} &
\colhead{(3)} &
\colhead{(4)} &
\colhead{(5)} &
\colhead{(6)} &
\colhead{(7)} &
\colhead{(8)} &
\colhead{(9)} &
\colhead{(10)} &
\colhead{(11)}
}
\startdata
Kroupa & 6.9 & $1.1$ & 3.0 & $6.2$ & $6.3$ & 1 & 1 & 1 & 1 & 1 \\
Salpeter & 6.8 & $1.0$ & $2.7$ & $5.4$ & $5.3$ & 0.97 & 0.90 & 0.87 & 0.84 & 0.83 \\
M31 & $6.7^{+0.1}_{-0.0}$ & $0.88^{+0.08}_{-0.04}$ & $2.2^{+0.3}_{-0.2}$ & $4.0^{+0.8}_{-0.4}$ & $3.8^{+0.9}_{-0.4}$ & $0.90^{+0.04}_{-0.01}$ & $0.74^{+0.09}_{-0.04}$ & $0.66^{+0.11}_{-0.06}$ & $0.60^{+0.13}_{-0.07}$ & $0.56^{+0.14}_{-0.06}$ \\
Kennicutt & 6.7 & $0.82$ & $1.9$ & $3.5$ & $3.2$ & 0.87 & 0.67 & 0.57 & 0.50 & 0.46 \\
Miller \& Scalo & 6.8 & $0.66$ & $0.82$ & $0.81$ & $0.56$ & 0.85 & 0.51 & 0.23 & 0.11 & 0.08 \\
Scalo & 5.6 & $0.20$ & $0.43$ & $0.72$ & $0.65$ & 0.54 & 0.14 & 0.10 & 0.08 & 0.07
\enddata
\tablecomments{The number of expected stars above the specified mass limit per $10^6$ M$_{\odot}$. For these calculations, we have adopted a \citet{kroupa2001} IMF below 2 M$_{\odot}$, the specified IMF above 2M$_{\odot}$, and assumed lower and upper mass bounds for the total IMF to be 0.08 M$_{\odot}$\ and 100 M$_{\odot}$. Columns (7) - (11) give the fraction of stars formed above the listed mass, relative to the standard Kroupa IMF.}
\label{fig:nstars}
\end{deluxetable*}
\subsection{Comparison with Other High-Mass IMF Slopes in the Local Group}
\label{sec:LG}
It is important to place our results into context of other high-mass IMF studies in the Local Group (LG). Among the strongest evidence for the universality of the high-mass IMF in the LG comes from a series of papers that combine spectroscopy and photometry of resolved, massive stars in clusters and OB associations in the LMC and MW \citep{massey1995, massey1995a, massey1998, massey2003}.
To summarize, these studies found (a) no drastic differences between the IMF slopes of very young, $<$ 5 Myr star forming regions in the MW and LMC, despite large differences in metallicity; and (b) that the resulting IMF slopes are broadly consistent with Salpeter/Kroupa. Combined, these studies represented a major milestone in our understanding of the high-mass IMF and its (in-)sensitivity to environment.
Although re-analyzing these studies is beyond the scope of this paper, we can quantify what the ensemble high-mass IMF slopes are in the LMC and MW using data from \citet{massey2003}. The purpose is to give a general sense of how different our M31 result is from other LG environments. To perform this analysis, we extracted the necessary data from \citet{massey2003} and have listed it in Table \ref{tab:massey}. We simply use the model from Equation \ref{eq:imfmodel} with $\ensuremath{\bvec{Q}}_k=1$ for each object, i.e., we know they are all \textit{bone fide} clusters, to infer the mean and variance for the MW and LMC clusters. We do not model possible age, mass, size dependancies.
From this exercise we find mean values of $\Gamma_{\mathrm MW}=1.16^{+0.12}_{-0.10}$ and $\Gamma_{\mathrm LMC}=1.29^{+0.11}_{-0.11}$. In both cases, $\sigma_{\Gamma} \sim 0.3-0.4$. The large scatter in the posterior distributions is likely a reflection of underestimated uncertainties and/or systematics, as opposed to true physical variation.
\citet{massey2003} emphasizes that the reported uncertainties are only statistical in nature. However, even these are likely under-estimated, as discussed in \citet{weisz2013a}. Unfortunately, not all of these literature studies list both the number of stars and mass ranges needed to estimate degree of under-estimation of the random uncertainties. There are also issues of systematic uncertainties in stellar models that can significantly change the masses of the most massive stars used in these studies \citep[e.g.,][]{massey2011}.
However, even from this cursory exercise, we see that the MW, LMC, and M31 do not appear to have\emph{ drastically} different IMF slopes. While this is quite remarkable given the diversity of environments, it should be tempered by the heterogenous nature of this comparison. If high-mass IMF variations are subtle, then homogenous and principled analyses are absolutely necessary to uncover them. At present, only our analysis in M31 is the result of a systematic study of the high-mass IMF over a large number of young clusters. Any comparison of the high-mass IMF slopes of LG galaxies must come from similarly homogenous data and principled analysis in order to minimize the systematics that have been persistent in IMF studies for the better part of six decades.
\subsection{Practical Usage of High-Mass IMF}
\label{sec:recommend}
Given the analysis presented in this paper and comparison with LG literature, we do not find strong evidence that a universal Salpeter/Kroupa IMF is the best representation of the high-mass IMF slope in the LG. Although we only rule it out at a $\sim 2-\sigma$ level, there appears to be little basis for this canonical value being correct in the first place\footnote{\citet{salpeter1955} has had remarkable staying power, despite the fact
that the power-law index of $\Gamma=1.35$
was derived assuming the MW was 6 Gyr old. A more realistic
13 Gyr age of the MW yields a value of $\Gamma=1.05$, which is considered an
extreme outlier by modern standards \citep[e.g.,][]{salpeter2005, zinnecker2005}.}. Instead, for practical purposes the high-mass IMF can be represented by a single-sloped power-law with $1.45^{+0.03}_{-0.06}$. Although this is only 0.1-0.15 dex steeper than Salpeter/Kroupa it does make a difference in the number of high-mass stars formed, and related quantities such as commonly used star formation rate (SFR) indicators. We discuss these implications further in \S \ref{sec:implications}.
For general usage, we recommend adoption of the following forms of the the IMF over all stellar masses. With the M31 high-mass IMF, a modified Chabrier IMF \citet{chabrier2003} has the form
\begin{eqnarray}
\xi(m)\, \Delta(m)\, &=& \frac{0.086}{m} \, exp^{\frac{-(log(m) - log(0.22))^2} {2 \times 0.57^2}} \nonumber \\
\nonumber \\
\xi(m)\, = c\, m^{- (\Gamma+1)}\, , \; \Gamma &=&1.45^{+0.03}_{-0.06} \, , \mbox{for } 1.0 < m < 100 \ \, M_{\odot}\, ,
\label{eqn:newchabrier}
\end{eqnarray}
\noindent and a modified \citet{kroupa2001} IMF has the form
\begin{equation}
\xi(m)\, = c\, m^{- ( \Gamma+1)} \begin{cases} \Gamma = 0.3 &\mbox{for } 0.08 < m < 0.5 \, M_{\odot} \\
\Gamma = 1.3 &\mbox{for } 0.5 < m < 1.0 \, M_{\odot} \\
\Gamma = 1.45^{+0.03}_{-0.06} \, , &\mbox{for } 1.0 < m < 100 \, M_{\odot} \, .\end{cases}
\label{eqn:newkroupa}
\end{equation}
Here, we have made two extrapolations to match conventional IMF definitions in the literature. First, we have extrapolated the M31 IMF down to 1 M$_{\odot}$. For most practical purposes, this extension from 2 M$_{\odot}$\ to 1 M$_{\odot}$\ affects the resulting stellar mass spectra at the level of a few percent.
Second, we have extrapolated the M31 IMF up to 100 M$_{\odot}$. While our data do not provide good constraints above $\sim$ 25 M$_{\odot}$, there is little other information to go on for mass spectrum of the most massive stars. Thus, as is commonplace in the literature, we suggest extrapolating the M31 IMF up to the highest masses, but recognize that systematic spectroscopic searches for the massive stars are absolutely critical for understanding the IMF of the most massive stars \citep[e.g.,][]{massey1995a, massey2003}.
Finally, we have assumed that the sub-solar IMF in M31 is similar to that of the Galaxy. Although not confirmed by direct star counts, \citet{conroy2012} have shown that spectral features sensitive to the low-mass IMF yield a result that is similar to the low-mass Galactic IMF.
\subsection{Broader Implications}
\label{sec:implications}
In Figure \ref{fig:relativeIMF} we illustrate the mass spectra of high-mass stars predicted by various commonly used IMF models. The plot shows the number of stars expected relative to a Kroupa IMF ($\Gamma=1.3$) as a function of mass. For fair comparison, the slopes have all been normalized at 2 M$_{\odot}$, the lower main sequence mass limit of the PHAT data, and extended to 100 M$_{\odot}$\ on the upper end. We chose this normalization, as opposed to 1 M$_{\odot}$\ to discuss out ability to distinguish different literature forms of the IMF over the dynamic mass range of the data, which is indicates by the vertical dashed lines.
The M31 high-mass IMF predicts the formation of fewer stars than Kroupa at all masses. The red solid line presents our median IMF model and shows that the fractional deficit varies from $\sim$0.9 at $\sim$ 10 M$_{\odot}$\ to $\sim0.7$ at 100 M$_{\odot}$. The red shaded region reflects the 68\% confidence interval in the number of predicted high-mass stars due to uncertainty on the mean high-mass IMF slope in M31. At most masses, the uncertainty deviates from the median by $\pm$$\sim$5\%, but increases to $\pm$15\% for stars $>$ 50 M$_{\odot}$.
The M31 high-mass IMF is similar to other high-mass IMF models in the literature. It is particularly close to the Kennicutt IMF \citep[$\Gamma=1.5$;][]{kennicutt1983}. It is also the same as the IMF of \citet{miller1979} over the 1-10 M$_{\odot}$\ range. The steepness of the \citet{miller1979} IMF above 10 M$_{\odot}$\ is known to be too extreme \citep[e.g.,][]{kennicutt1983, kennicutt1994}. Further, the high-mass IMF from \citet{scalo1986} has a similar slope ($\Gamma=1.6$) above $\sim$ 4 M$_{\odot}$, but a much steeper slope below it. Finally, the M31 high-mass IMF predicts fewer high-mass stars than a Salpter IMF for masses $>$ 2 M$_{\odot}$.
We further quantify differences in these IMF models in Table \ref{fig:nstars}. Here, we have computed the expected number of stars per 10$^6$ M$_{\odot}$\ formed at selected stellar masses assuming a \citet{kroupa2001} IMF below 2 M$_{\odot}$\ and the listed IMFs above 2 M$_{\odot}$. This table solidifies many of the above points regarding differences in commonly used high-mass IMF models. It also highlights an important implication for core collapse supernovae, whose progenitors are believed to have masses $\gtrsim$ 8M$_{\odot}$ \citep[e.g.,][]{smartt2009}. For example, the median M31 high-mass IMF model predicts $\sim$25\% fewer stars with masses $\gtrsim$ 8M$_{\odot}$\ compared to a Kroupa IMF.
\begin{deluxetable}{ccccc}
\tablecaption{Updated Broadband SFR Indicators}
\tablecolumns{5}
\tablehead{
\colhead{Band} &
\colhead{$L_{x}$ Units} &
\colhead{log($C_x$)} &
\colhead{log($C_x$)} &
\colhead{$\dot M_{\star}(M31) / \dot M_{\star}$(Kroupa)} \\
\colhead{} &
\colhead{} &
\colhead{(Kroupa)} &
\colhead{(M31)} &
\colhead{} \\
\colhead{(1)} &
\colhead{(2)} &
\colhead{(3)} &
\colhead{(4)} &
\colhead{(5)}
}
\startdata
FUV & erg s$^{-1}$ ($\nu L_{\nu}$) & 43.29 & 43.18$^{+0.05}_{-0.02}$ & 1.28$^{+0.06}_{-0.12}$ \\
NUV & erg s$^{-1}$ ($\nu L_{\nu}$) & 43.14 & 43.04$^{+0.03}_{-0.02}$ & 1.24$^{+0.06}_{-0.11}$ \\
H$\alpha$\ & erg s$^{-1}$ & 40.91 & 40.73$^{+0.04}_{-0.07}$ & 1.52$^{+0.13}_{-0.24}$
\enddata
\tablecomments{A comparison of commonly used SFR indicators for a Kroupa and M31 IMF values. The values of $L_x$, $C_x$, and $\dot M_{\star}$ follow \citet{kennicutt2012} and Equation \ref{eqn:sfr}.}
\label{tab:sfrindicators}
\end{deluxetable}
Finally, we consider the implications for commonly used SFR indicators \citep[e.g.,][]{kennicutt1998, kennicutt2012}. Using the Flexible Stellar Population Synthesis code \citep[FSPS;][]{conroy2009, conroy2010a}, we compute the luminosity to SFR conversion coefficients for GALEX far- and near-UV luminosities, along with H$\alpha$, which we list in Table \ref{tab:sfrindicators}. To do this, we have assumed a constant SFH over the last 1 Gyr, the indicated IMF, mass limits from 0.08 to 100 M$_{\odot}$, and case B recombination to convert the number of ionizing photons to H$\alpha$\ luminosity. From \citet{kennicutt2012} the canonical conversion from luminosity to SFR can be written as
\begin{equation}
{\rm log}\, \dot M_{\star}\, ({\rm M}_{\odot} \, {\rm yr}^{-1}) \,=\, {\rm log}\, L_x\, -\, \rm{log}\, C_x
\label{eqn:sfr}
\end{equation}
\noindent where $\dot M_{\star}$ is the SFR, $L_x$ is the luminosity over the bandpass $x$, and $C_x$ is the conversion constant over the same wavelength range. We have provided updated values for $C_x$ in Table \ref{tab:sfrindicators}.
From column (5) of Table \ref{tab:sfrindicators}, we see modest differences in the SFR indicators due to an M31 high-mass IMF. The variations are such that for a fixed luminosity, using the M31 high-mass IMF results in SFRs that are a factor $\sim$1.3--1.5 larger.
\acknowledgments
The authors would like to thank Nate Bastian, Charlie Conroy, Rob Kennicutt, and John Saclo for their useful comments on this paper and discussions about the IMF and its history. Support for this work was provided by NASA through grant number HST GO-12055 from the Space Telescope Science Institute, which is operated by AURA, Inc., under NASA contract NAS5-26555.
DRW is supported by NASA through Hubble Fellowship grant HST-HF-51331.01 awarded by the
Space Telescope Science Institute.
DRW also wishes to thank the MPIA and KITP for their generous hospitality during the writing of this paper.
This work used the Extreme Science and Engineering Discovery Environment (XSEDE), which is supported by National Science Foundation grant number ACI-1053575, was supported in part by the National Science Foundation under Grant No. NSF PHY11-25915, and made extensive use of NASA's
Astrophysics Data System Bibliographic Services.
In large part, analysis and plots presented in this paper utilized IPython and
packages from NumPy, SciPy, and Matplotlib \citep[][]{hunter2007,
oliphant2007, perez2007, astropy2013}.
| {
"redpajama_set_name": "RedPajamaArXiv"
} | 5,438 |
Nike #KissMyAirs
On the Run With Eugene Tong
One of the OG stylists to mix streetwear with menswear, Eugene Tong's philosophy has always been grounded in experimentation and integration of seemingly different elements. With an upbringing in Taiwan and majority of his adult life spent in New York, the complex of juxtapositions he's been exposed to serve as his inspiration — finding order in chaotic cities, retaining and adapting to cultures, mixing classics with the new. Tong channels all of this in his approach to his work. In the latest installment to the #KissMyAirs series, director Jonathon Lim follows Tong to get inside his head while he takes us for a run in the new Nike Air VaporMax.
As someone who has been constantly on the move, Eugene admits he feels like a guest just about everywhere he goes. However, there's an advantage to this — every place he sees, he sees with new eyes and a fresh perspective. He's able to interpret a culture differently as an outsider, and combine the best with the best of every place he's been. Tong tributes his iconic black and white street style to these experiences, finding it to be the perfect solution to blend his Eastern and Western influence and move between the two seamlessly. Keeping things simple and monochromatic for him is key to creating a lasting style, something he sees in the design of the VaporMax.
Like Tong, director Jonathon Lim has also had to move between cultures, with an upbringing in Australia and eventual move to Shanghai for work. He's had a hand in the streetwear scene working with the likes of Edison Chen and covering music, art and skate culture across Asia. In his opportunity to shoot with Tong between New York and Shanghai, he was able to find a bit of each culture in the other city – streets in New York that would feel like streets in Shanghai and vis versa.
In collaboration with Nike, we've covered three RevolutionAirs — Lil' Buck, Eugene Tong and soon, Lee Yuchun — who have shaken up their industries, be it fashion, music or art. Embodying the attitude of #KissMyAirs, they're coming together to celebrate the annual Air Max Day on March 26, alongside the global debut of the much-anticipated Nike Air VaporMax.
Their journeys to the forefront of their disciplines have been profiled by three of Asia's most promising young filmmakers, whose short films which will exclusively premiere here.
Watching. Nike – #KissMyAirs Eugene TongPrev. Loewe – Bags CollectionP Next. Nike – #KissMyAirs Lil BuckN | {
"redpajama_set_name": "RedPajamaCommonCrawl"
} | 9,259 |
/* mtalk */
/*
Copyright (c) 2012 by Jason Cheung <yujiecheung@gmail.com>
All rights reserved.
Redistribution and use in source and binary forms, with or without
modification, are permitted provided that the following conditions are met:
* Redistributions of source code must retain the above copyright
notice, this list of conditions and the following disclaimer.
* Redistributions in binary form must reproduce the above copyright
notice, this list of conditions and the following disclaimer in the
documentation and/or other materials provided with the distribution.
* Neither the name of Jason Cheung nor the
names of its contributors may be used to endorse or promote products
derived from this software without specific prior written permission.
THIS SOFTWARE IS PROVIDED BY THE COPYRIGHT HOLDERS AND CONTRIBUTORS "AS IS" AND
ANY EXPRESS OR IMPLIED WARRANTIES, INCLUDING, BUT NOT LIMITED TO, THE IMPLIED
WARRANTIES OF MERCHANTABILITY AND FITNESS FOR A PARTICULAR PURPOSE ARE
DISCLAIMED. IN NO EVENT SHALL THE COPYRIGHT OWNER OR CONTRIBUTORS BE LIABLE FOR ANY
DIRECT, INDIRECT, INCIDENTAL, SPECIAL, EXEMPLARY, OR CONSEQUENTIAL DAMAGES
(INCLUDING, BUT NOT LIMITED TO, PROCUREMENT OF SUBSTITUTE GOODS OR SERVICES;
LOSS OF USE, DATA, OR PROFITS; OR BUSINESS INTERRUPTION) HOWEVER CAUSED AND
ON ANY THEORY OF LIABILITY, WHETHER IN CONTRACT, STRICT LIABILITY, OR TORT
(INCLUDING NEGLIGENCE OR OTHERWISE) ARISING IN ANY WAY OUT OF THE USE OF THIS
SOFTWARE, EVEN IF ADVISED OF THE POSSIBILITY OF SUCH DAMAGE.
*/
#pragma comment(lib,"Ws2_32.lib")
#include <winsock2.h>
#include "mtalk.h"
#include "define.h"
#define MTALK_MAX_DATA_LEN 256
enum enum_mtalk_type
{
MTALK_TYPE_PIN_STATUS = 1
};
struct tag_mtalk_sock
{
SOCKET sock;
unsigned char data[MTALK_MAX_DATA_LEN];
};
struct tag_mtalk
{
struct tag_mtalk_sock server;
struct tag_mtalk_sock *client;
void (*receiver)(const struct tag_mtalk_pin_status *status);
int max_client_count;
int is_server;
int port;
HANDLE evExit;
HANDLE evDisconnect;
unsigned char send_data[MTALK_MAX_DATA_LEN];
char addr[MTALK_MAX_DATA_LEN];
};
static struct tag_mtalk g_mtalk;
/*------------------------------------------------------------------------------
Protocol
*/
/*
|--------------------------------|
| 1 byte | 1 byte | N bytes |
|--------------------------------|
| type | payload len | payload |
|--------------------------------|
*/
/*
pin status payload
|-----------------|
| 1 byte | 1 byte |
|-----------------|
| pin ID | status |
|-----------------|
*/
static void mtalk_send(unsigned char *data, int len)
{
if (NULL != g_mtalk.client) {
int i;
for (i = 0; i < g_mtalk.max_client_count; i++) {
if (0 != g_mtalk.client[i].sock) {
send(g_mtalk.client[i].sock, (const char *)data, len, 0);
}
}
} else {
if (0 != g_mtalk.server.sock) {
send(g_mtalk.server.sock, (const char *)data, len, 0);
}
}
}
static int mtalk_pack_pin_status(unsigned char *data, const struct tag_mtalk_pin_status *status)
{
int len = 0;
*data++ = status->pin_id;
*data = status->status;
len += 2;
return len;
}
static void mtalk_unpack_pin_status(unsigned char *data, int len)
{
struct tag_mtalk_pin_status status;
while (len > 0) {
status.pin_id = *data++;
status.status = *data++;
len -= 2;
g_mtalk.receiver(&status);
}
}
static void mtalk_unpack(unsigned char *data, int len)
{
if (NULL != g_mtalk.receiver) {
unsigned char type = *data++;
unsigned char payload_len = *data++;
if ((payload_len + 2) == len) {
if (MTALK_TYPE_PIN_STATUS == type) {
mtalk_unpack_pin_status(data, payload_len);
}
}
}
}
/*------------------------------------------------------------------------------
Private
*/
static struct tag_mtalk_sock *get_free_client(void)
{
struct tag_mtalk_sock *client = NULL;
int i;
for (i = 0; i < g_mtalk.max_client_count; i++) {
if (0 == g_mtalk.client[i].sock) {
client = &(g_mtalk.client[i]);
break;
}
}
return client;
}
static DWORD WINAPI receive_thread_proc(LPVOID param)
{
struct tag_mtalk_sock *sock = (struct tag_mtalk_sock *)param;
int recv_len;
while (1) {
recv_len = recv(sock->sock, sock->data, MTALK_MAX_DATA_LEN, 0);
if (recv_len > 0) {
mtalk_unpack(sock->data, recv_len);
} else {
sock->sock = 0;
break;
}
}
if (NULL != g_mtalk.evDisconnect) {
SetEvent(g_mtalk.evDisconnect);
}
return 0;
}
static DWORD WINAPI server_thread_proc(LPVOID param)
{
SOCKET client_sock;
HANDLE hthread;
struct tag_mtalk_sock *client;
while (1) {
client_sock = accept(g_mtalk.server.sock, NULL, NULL);
if (INVALID_SOCKET == client_sock) {
break;
}
client = get_free_client();
if (NULL != client) {
client->sock = client_sock;
hthread = CreateThread(NULL, 0, receive_thread_proc, (LPVOID)client, 0, NULL);
if (NULL != hthread) {
CloseHandle(hthread);
}
}
}
return 0;
}
static DWORD WINAPI client_thread_proc(LPVOID param)
{
HANDLE hthread;
HANDLE evObject[2];
int sock_result;
struct sockaddr_in service;
DWORD objects;
service.sin_family = AF_INET;
service.sin_addr.s_addr = inet_addr(g_mtalk.addr);
service.sin_port = htons(g_mtalk.port);
g_mtalk.evExit = CreateEvent(NULL, FALSE, FALSE, NULL);
g_mtalk.evDisconnect = CreateEvent(NULL, FALSE, FALSE, NULL);
evObject[0] = g_mtalk.evExit;
evObject[1] = g_mtalk.evDisconnect;
while (1) {
sock_result = connect(g_mtalk.server.sock, (SOCKADDR *)&service, sizeof(service));
if (SOCKET_ERROR != sock_result) {
hthread = CreateThread(NULL, 0, receive_thread_proc, (LPVOID)&g_mtalk.server, 0, NULL);
if (NULL != hthread) {
CloseHandle(hthread);
}
}
objects = WaitForMultipleObjects(2, evObject, FALSE, 0);
if (WAIT_OBJECT_0 == objects) {
break;
}
}
return 0;
}
/*------------------------------------------------------------------------------
Public
*/
int mtalk_init(void)
{
WSADATA wsaData;
int result;
memset(&g_mtalk, 0, sizeof(struct tag_mtalk));
result = WSAStartup(MAKEWORD(2, 2), &wsaData);
if (NO_ERROR != result) {
result = FAIL;
} else {
result = SUCCESS;
}
return result;
}
void mtalk_release(void)
{
WSACleanup();
}
void mtalk_stop(void)
{
if (NULL != g_mtalk.evExit) {
SetEvent(g_mtalk.evExit);
}
if (0 != g_mtalk.server.sock) {
shutdown(g_mtalk.server.sock, SD_BOTH);
closesocket(g_mtalk.server.sock);
g_mtalk.server.sock = 0;
}
}
int mtalk_server_run(const char *addr, int port, int max_client_number)
{
int sock_result;
int result = FAIL;
HANDLE hthread;
if (NULL != g_mtalk.client) {
free(g_mtalk.client);
}
g_mtalk.client = (struct tag_mtalk_sock *)malloc(sizeof(struct tag_mtalk_sock) *
max_client_number);
if (NULL != g_mtalk.client) {
memset(g_mtalk.client, 0, sizeof(struct tag_mtalk_sock) * max_client_number);
g_mtalk.max_client_count = max_client_number;
result = SUCCESS;
}
if (SUCCESS == result) {
g_mtalk.server.sock = socket(AF_INET, SOCK_STREAM, IPPROTO_TCP);
if (INVALID_SOCKET == g_mtalk.server.sock) {
result = FAIL;
}
}
if (SUCCESS == result) {
struct sockaddr_in service;
service.sin_family = AF_INET;
service.sin_addr.s_addr = inet_addr(addr);
service.sin_port = htons(port);
sock_result = bind(g_mtalk.server.sock, (SOCKADDR *)&service, sizeof(service));
if (SOCKET_ERROR == sock_result) {
result = FAIL;
}
}
if (SUCCESS == result) {
sock_result = listen(g_mtalk.server.sock, max_client_number);
if (SOCKET_ERROR == sock_result) {
result = FAIL;
}
}
if (SUCCESS == result) {
hthread = CreateThread(NULL, 0, server_thread_proc, NULL, 0, NULL);
if (NULL == hthread) {
result = FAIL;
} else {
CloseHandle(hthread);
}
}
if (SUCCESS != result) {
if (0 != g_mtalk.server.sock) {
shutdown(g_mtalk.server.sock, SD_BOTH);
closesocket(g_mtalk.server.sock);
g_mtalk.server.sock = 0;
}
}
return result;
}
int mtalk_client_run(const char *addr, int port)
{
int result = SUCCESS;
HANDLE hthread;
if (NULL != g_mtalk.evExit) {
CloseHandle(g_mtalk.evExit);
g_mtalk.evExit = NULL;
}
if (NULL != g_mtalk.evDisconnect) {
CloseHandle(g_mtalk.evDisconnect);
g_mtalk.evDisconnect = NULL;
}
g_mtalk.server.sock = socket(AF_INET, SOCK_STREAM, IPPROTO_TCP);
if (INVALID_SOCKET == g_mtalk.server.sock) {
result = FAIL;
}
if (SUCCESS == result) {
memcpy(g_mtalk.addr, addr, strlen(addr));
g_mtalk.port = port;
hthread = CreateThread(NULL, 0, client_thread_proc, NULL, 0, NULL);
if (NULL != hthread) {
CloseHandle(hthread);
}
}
if (SUCCESS != result) {
if (0 != g_mtalk.server.sock) {
shutdown(g_mtalk.server.sock, SD_BOTH);
closesocket(g_mtalk.server.sock);
g_mtalk.server.sock = 0;
}
}
return result;
}
int mtalk_send_pin_status(const struct tag_mtalk_pin_status *status, int number)
{
int result = FAIL;
int i;
int len = 0;
int packed_number = 0;
unsigned char *data;
while (packed_number < number) {
data = (unsigned char *)g_mtalk.send_data;
*data++ = MTALK_TYPE_PIN_STATUS;
*data++ = 0;
for (i = 0; i < number; i++) {
len += mtalk_pack_pin_status(data,
(const struct tag_mtalk_pin_status *)&status[i]);
data += len;
packed_number++;
if ((len + 2) >= MTALK_MAX_DATA_LEN) {
break;
}
}
g_mtalk.send_data[1] = len;
mtalk_send(g_mtalk.send_data, len + 2);
}
return result;
}
void mtalk_receive_pin_status(void (*receiver)(const struct tag_mtalk_pin_status *status))
{
g_mtalk.receiver = receiver;
}
| {
"redpajama_set_name": "RedPajamaGithub"
} | 8,343 |
Working the Land of East Cape
By M4 Interactive Collaborator | Feb 04,  2020
East Cape landowner, Jim White, owns and manages the 2000 hectares in partnership with his older brother and sister. "Most of our work is done on horseback as the land is pretty steep. It's summit country here so we get snow in the winter," he explains.
Predominantly a sheep and beef farm, Jim also keeps sections of his property as Mānuka bush as part of his partnership agreement with us. Jim is also pleased the honey is now bringing a new income opportunity for everyone in the region. "It's a beautiful lifestyle – there's the sea and the mountains - and we can make a living on the land," says Jim. Norm (our Apiary Manager East Coast) is the beekeeper, and a good friend of Jim, who manages the hives on Jim's farm. Jim is always happy to welcome Norm onto his property, and enjoys seeing him there.
"Norm was a beekeeper before it became popular," says Jim. "It's always interesting to see him."
Jim's two boys have also discovered a passion for beekeeping. While one lives and works on the East Cape, his other son is now beekeeping in Australia. "Everybody knows everybody here, so you have to get on. That's the lifestyle."
Three DIY Skin Restoring Manuka Honey Masks
Feb 04,  2020
News Report: New Zealand Manuka Group on b...
Treat your mum to a hearty breakfast
Nature's Sweetener for Cooking | {
"redpajama_set_name": "RedPajamaCommonCrawl"
} | 1,783 |
De Schans Turkeye (ook: Groot Turkeye) is een in 1604 aangelegde schans die behoorde tot de Passageule-Linie.Ze bevindt zich nabij de Passageule tussen Waterlandkerkje en IJzendijke.
Ze werd aangelegd door de Staatse troepen, nadat ze dit gebied hadden veroverd. De naamgeving houdt mogelijk verband met de goede relaties die de toenmalige Republiek met het Turkse Rijk onderhield (Turkije erkende de Republiek reeds in 1612), maar de precieze oorsprong van de naam is onzeker.
De nabijgelegen buurtschap Turkeye is naar deze schans, en de aldaar gelegen redoute Klein Turkeye, vernoemd.
De schans ligt aan de Passageule-Linie tussen de Schans Konstantinopel en de Jonkvrouwschans.
Vestingwerk van Sluis | {
"redpajama_set_name": "RedPajamaWikipedia"
} | 9,809 |
Readers love ANDREW GREY
Artistic Appeal
"Mr. Grey's amazing storytelling ability captured this reader's attention from the first word and kept me enthralled with the story and the characters within the book until the very last word."
—Night Owl Reviews
Artistic Pursuits
"...settle in for one entertaining ride."
—Love Romances and More
A Troubled Range
"This is one book that I won't forget and you shouldn't miss it."
—Fallen Angels Reviews
Helping of Love
"This story, like the others in the series, is a well-written, feel-good story that I will be reading again."
—Literary Nymphs Reviews
Legal Tender
"Buy the book and have an engaging, enjoyable day of reading."
—Randy's Book Bag Reviews
Legal Artistry
"...a mix of old-fashioned romance and European flavour."
—Elisa Rolle Reviews and Ramblings
A Foreign Range
"If you are looking for a quick romantic read about some emotionally guarded but hunky cowboys then saddle up and take a ride on the Range series."
—Guilty Indulgences
Novels by ANDREW GREY
Accompanied by a Waltz
Dutch Treat
The Good Fight • The Fight Within
Love Comes Silently
Three Fates (anthology) • Work Me Out (anthology)
ART SERIES
Legal Artistry • Artistic Appeal • Artistic Pursuits • Legal Tender
BOTTLED UP STORIES
Bottled Up • Uncorked • The Best Revenge • An Unexpected Vintage
CHILDREN OF BACCHUS STORIES
Children of Bacchus • Thursday's Child • Child of Joy
LOVE MEANS... SERIES
Love Means... No Shame • Love Means... Courage • Love Means... No Boundaries
Love Means... Freedom • Love Means... No Fear
Love Means... Family • Love Means... Renewal • Love Means... No Limits
SEVEN DAYS STORIES
Seven Days • Unconditional Love
STORIES FROM THE RANGE
A Shared Range • A Troubled Range • An Unsettled Range
A Foreign Range • An Isolated Range• A Volatile Range
TASTE OF LOVE STORIES
A Taste of Love • A Serving of Love • A Helping of Love • A Slice of Love
Novellas by ANDREW GREY
A Present in Swaddling Clothes
Shared Revelations
Snowbound in Nowhere
BY FIRE SERIES
Redemption by Fire • Strengthened by Fire • Burnished by Fire
CHILDREN OF BACCHUS STORIES
Spring Reassurance • Winter Love
LOVE MEANS... SERIES
Love Means... Healing • Love Means... Renewal
WORK OUT SERIES
Spot Me • Pump Me Up • Core Training • Crunch Time
Positive Resistance • Personal Training • Cardio Conditioning
Published by DREAMSPINNER PRESS
http://www.dreamspinnerpress.com
# Copyright
Published by
Dreamspinner Press
5032 Capital Circle SW
Ste 2, PMB# 279
Tallahassee, FL 32305-7886
USA
http://www.dreamspinnerpress.com/
This is a work of fiction. Names, characters, places, and incidents either are the product of the author's imagination or are used fictitiously, and any resemblance to actual persons, living or dead, business establishments, events, or locales is entirely coincidental.
The Fight Within
Copyright © 2013 by Andrew Grey
Cover Art by Anne Cain
annecain.art@gmail.com
All rights reserved. No part of this book may be reproduced or transmitted in any form or by any means, electronic or mechanical, including photocopying, recording, or by any information storage and retrieval system without the written permission of the Publisher, except where permitted by law. To request permission and all other inquiries, contact Dreamspinner Press, 5032 Capital Circle SW, Ste 2, PMB# 279, Tallahassee, FL 32305-7886, USA.
http://www.dreamspinnerpress.com/
ISBN: 978-1-62380-400-8
Digital ISBN: 978-1-62380-401-5
Printed in the United States of America
First Edition
March 2013
To all of America's native peoples
# Chapter 1
"SO, BRYCE, are you excited about this weekend?" Jerry, Bryce's boss, asked with a knowing grin that turned into a warm laugh. Jerry owned the small consulting firm where Bryce worked, and was a close friend. Bryce often thought his friends were as excited about the upcoming event as he was.
Bryce could hardly sit still. "Yes. I'm all set. The caterers are ready. The florist has been contacted four times and they finally understand what I want." Bryce looked up from his computer screen, temporarily giving up on trying to debug the program he was working on. It wasn't going to happen until he could concentrate, and he couldn't do that as long as he was thinking about Percy and his wedding. "Will you and John still help decorate the hall Saturday morning?" Bryce asked, and he saw Jerry roll his eyes.
"You've asked us eight times. Just relax. Everything will be just fine. You and Percy have done a great job organizing everything," John, Jerry's stunning Native American partner, said from the desk next to his. Thankfully they weren't on a particular deadline, or Jerry would crack the proverbial whip and get them all back to work. "Even the weather seems to be cooperating. The television said that it was supposed to be seventy-five and sunny on Saturday. The flowers are all blooming right now, so it should be a perfect day for an outdoor ceremony."
"Yeah," Bryce said with a grin. "It's going to be perfect." Everyone had told him he was crazy for planning an outdoor wedding in May. The weather was too changeable and uncertain, but Bryce had insisted, and he'd been prepared to order a tent if necessary, but the gardens they were using should be perfect. He had compromised and agreed to hold the reception inside, but it was important to him to hold the ceremony outside. His parents had held their wedding outside, as had his grandparents, so he very much wanted to carry on the tradition.
"You and Percy deserve it," Jerry said before his head disappeared behind his monitor and the clicking of the keyboard replaced the sound of his voice. John did the same, and Bryce forced his mind away from his wedding and onto the program.
"Got you, ya bugger," Bryce murmured as he spotted the error and corrected it, then ran the program to make sure it worked properly. Jerry was a stickler for details, so Bryce went through the specifications in detail, checking off each requirement before flagging the program as completed and moving on to the next one.
The work slowly consumed Bryce's thoughts and attention the way it always did. Code and numbers—they spoke to him. When Jerry had hired both John and him three years earlier, Bryce had been fresh out of college and very green. Jerry had taken them both under his wing, and now he could see the improvement and felt complete confidence in his work. Bryce worked straight on until lunch, and then they all broke and went inside.
Jerry had converted an old workshop into an office, which worked great for all of them. Jerry had been talking lately about hiring another associate, but there wasn't room, and Bryce was beginning to wonder where he was planning to seat them. At lunch, they went into Jerry's house and ate at his kitchen table. Bryce had asked him once why they didn't eat at their desks. "We all need a break and a change of scenery," Jerry had said, so this had become their routine. "I'm going to close the office tomorrow," Jerry told them once they'd sat at the table. "It'll give you an extra day for last-minute preparations, and I have a feeling that none of us will be up to working, anyway."
"You don't have to, Jerry," Bryce said.
"Think of it as an additional wedding present. The kids are home from school, too, some sort of in-service day, so they'll be able to help. Mato and Ichante are going to love helping their Uncle Bryce."
He couldn't stop smiling whenever he heard that. John and Jerry had adopted John's niece Ichante and his nephew Mato. Those two were incredibly special kids, and Bryce loved that they called both Percy and him uncle. It made them feel like part of the family. "Thanks. I was trying to figure out how I was going to get all the last-minute stuff done without working until midnight."
"Don't worry—we're all here for you," John said, and Bryce saw Jerry nod his agreement. "When does Percy get home?"
Bryce humphed softly. "I have to pick him up at six." Bryce took a bite of his sandwich and then set it back on the plate. "I can't believe he had to go to New York the week before our wedding."
"It was training for work, you know that," Jerry soothed.
"I know, and he couldn't help it, but that doesn't mean I have to like it," Bryce groused lightly before returning to his lunch, running over all the things he had yet to get done. The list seemed to grow longer every day, and Bryce had no idea how he was going to get everything done.
"Stop worrying. Everything will be great. You've got everything under control, so relax and enjoy it," John said, patting Bryce's hand lightly. "If you don't take it easy, the entire day will fly by and you won't even remember it. So what if the flowers aren't perfect and the chairs aren't in the exact position you want? You're sharing your love with Percy, and that's what's important." Bryce agreed with him, but the nerves didn't settle and that damned list kept running through his head.
They finished lunch and Bryce went back to work, the coding the one thing that seemed to soothe him. Late in the afternoon, Mato and Ichante got home from school, and they hurried into the office to say hello. They hugged Jerry and John before giving Bryce hugs too. Then they told John about their day. Or that's what Bryce thought they were doing, since they were speaking the Lakota language. John felt it important that both children have as close a link to their heritage as they could. Bryce had seen everything John and Jerry had gone through to get Mato and Ichante out of the state childcare system, and it was amazing how normal and well-adjusted the kids were now, considering how hurt and scared they'd been when John first brought them home.
"The office is going to be closed tomorrow, so would you two like to help get the guest favors ready?" Bryce asked.
"Ai," they both said happily. Bryce knew that was the Lakota word for "yes."
A short while later, the workday ended, and Bryce got ready to leave. He had errands to run, so after saying good-bye, he got in his car and drove to the tuxedo shop to pick up the clothes he and Percy would wear. Then he stopped by the printer and picked up the programs for the ceremony. After checking the time as he left the shop, Bryce placed the programs in the trunk and was about to head to the airport when his phone rang. It was Jerry.
"Can you come back to the house?"
"I need to get to the airport to pick up Percy," Bryce explained as he unlocked the car and slid into the seat.
"You'll have plenty of time to stop by here. It's right on your way," Jerry said insistently.
"Okay, I'm on my way," Bryce said and closed the phone, setting it on the seat next to him. He and Percy had made a mix CD of all their favorite songs. They were going to put a copy at each place at dinner. Bryce slid one of the CDs into the player and drove to Jerry's as "I'll Have To Say I Love You In A Song" played through the car speakers. This had been one of Percy's choices, and it always made him smile because Percy had sung that song to him when he'd told Bryce he loved him. Percy was nearly tone-deaf, but his rendition of this song was the most beautiful one Bryce had ever heard.
Thankfully, Sioux Falls was not very large, and Bryce pulled onto Jerry's street a few minutes later. He kept checking the dashboard clock and knew he could only spend a few minutes at Jerry's before he had to dash out to the airport.
Bryce parked out front and raced up the door before going right inside. In the living room, he saw John and Jerry each holding one of the kids, who had obviously been crying. "What happened? Is there something wrong with the kids?" Jerry shook his head, and Bryce saw tears running down his friend's face.
"You're Bryce Morton?"
He turned and saw the police officer for the first time. Bryce nodded, his mouth going dry, and he instantly began to shake. Mato slid off Jerry's lap, and Jerry guided Bryce toward a chair. Bryce sat and looked up at the officer who said, "I'm very sorry to have to tell you this, but Percival Howland was involved in an accident at Midway Airport."
Jerry hugged him and so did Mato. "Is he okay? Do I need to get there to be with him?"
The police officer's expression changed, becoming even graver. "I'm sorry, but Mr. Howland didn't survive."
Bryce shook his head. "It can't be. Percy... my Percy is dead?" The officer nodded once. Bryce swallowed and tried to get ahold of himself. This was surreal, and suddenly he felt as though he were floating outside his body. "How did he die?"
"The investigation is still underway, but from the information we were given, a driver lost control of a loaded luggage cart and barreled into people who were walking onto the tarmac to board the commuter flight to Sioux Falls. I don't know if it was a mechanical issue or not, but two other people were injured."
Bryce nodded and looked down at the floor. Percy was gone. Bryce's head throbbed and the room began to spin. He grabbed the arm of the chair, but that didn't do any good. The room continued moving. Bryce clamped his eyes closed, and the last thing he remembered was falling forward.
He didn't know how long he was out, but when he came to, he was on the sofa with paramedics hovering over him. At first he couldn't figure out why they were there, but then everything came back to him. Bryce opened his mouth, but nothing came. He gasped for breath and then heard a banshee from hell scream at the top of its lungs. It took him a few seconds to realize he'd made that sound.
"Please try to calm down, sir," one of the paramedics said, but Bryce couldn't. Tears welled in his eyes and he opened his mouth to breathe, only to whimper. As Ichante took his hand, the last of Bryce's control burst like Hoover Dam, and he wailed and cried. A lamentation went up from somewhere else in the room, followed by another and then another.
"What are they doing?" the paramedic asked.
"Guiding Percy's soul to the afterlife," Bryce heard Jerry say, and then he too joined in the howling.
Then the room became quiet except for Bryce's tears and those of the children. Ichante curled next to him on the sofa, hugging him, and Bryce continued crying. "Should we take him in?" one of the paramedics asked, and Bryce rolled his head on the sofa cushion. He didn't want to go anywhere, except maybe someplace where he could die like Percy. With every breath he took, his heart broke a little more. The tears continued coming, and he couldn't stop them even when he did try.
"We're here for you," John said softly, but the words barely registered for Bryce. Everything from outside his head seemed muffled and distant.
"I'm very sorry for your loss," the police officer said, and Bryce watched as he walked over to him. He pressed a card into Bryce's hand. "If there's anything we can do, please give us a call."
Bryce mumbled something, but for the life of him he didn't know what it was. Through watery eyes, he saw the officer leave as well as the paramedics. Then, slowly, he began sitting up. Mato and Ichante sat on either side of him. "What do I do?" His mind was barely functioning, and he was supposed to get married in two days. "Oh God," Bryce wailed, putting his hands over his face.
Eventually the tears stopped, but the shocked and bewildered feelings continued, like his mind refused to work at all. "It's going to be okay," Ichante said from next to him, and Bryce smiled at the twelve-year-old, trying to make her feel better.
"I have to call everyone and let them know...," Bryce said before breaking into quiet sobs. He had no idea how he was going to be able to call everyone and tell them that Percy had been killed. He couldn't. There was no way. He'd never make it through all those calls.
"Don't worry," Jerry told him. "I'll go over to your place and get your computer. John and I can make the calls. You stay where you are and don't worry about anything." Jerry seemed to spring into action, and he was soon gone. John and the kids sat with him, but no one said anything, which was fine with Bryce. He couldn't deal with talking.
"Why did the police come here?" Bryce asked quietly.
"The number they had for you was wrong and they called your work and I said you were on your way," John explained, and Bryce nodded blankly, only half hearing what was being said.
"Percy was really cool," Ichante eventually offered. "At school last year, he came into my class and told everyone what it was like to be a guy nurse."
"He got me a stefoscope so I could listen to my heart," Mato said and slipped off the sofa. Bryce heard him run upstairs and then he came back down with the stethoscope around his neck. "I wanna be a nurse like Uncle Percy when I grow up."
Bryce tugged Mato to him and began crying again as he held the boy. How could his Percy be gone? He was too kind and caring and certainly didn't deserve to die alone out on some airport tarmac. "I'm going to miss him so much," Bryce said. Percy had loved him deeply, Bryce knew that, and now it seemed part of the light had gone from his life. "How can I keep going without him?"
No one answered, not that Bryce expected anyone to. There was nothing anyone could do. His phone rang, and he fished it out of his pocket and handed it to John without looking at who was calling. He couldn't bear to start the phone chain that would let everyone know that Percy was gone. Maybe if they didn't know, then he'd be around for just a little bit longer.
"Bryce, it's your mother," John said as he gave him the phone. "I don't think she knows," he added in a whisper.
"What happened?" his mother asked.
"Percy's dead," was all he could say before breaking into sobs once again. John took the phone, and Bryce heard him talking briefly to his mother before hanging up.
"They're on their way over," John told him, and Bryce got himself under control. Jerry returned, and Bryce's parents showed up a short while later. There were more tears and hugs, and then tears again, before everyone else sat down and began making calls. He heard Jerry, John, and his mother and dad tell people that Percy was dead. They cried, and Bryce was sure he heard every platitude people said when they didn't know what to say. After a while, Bryce couldn't take it any longer and blankly left the house. He wandered up and down the sidewalk, wondering what his sweet Percy could have done to deserve this. He sat on the curb, watching as the sun set.
"Come inside," he heard his mother say from behind him, but Bryce shook his head. "There's nothing you can do out here."
"There's nothing I can do at all. We were supposed to get married in two days. Percy loved me enough to put up with all my arrangements and plans." Bryce turned as his mother sat down next to him.
"I can't tell you I know how you feel right now, because I don't and I never can. You've lost someone very dear to you, and the only way I could feel what you are right now would be if I lost your father or you." She took his hand, and Bryce squeezed lightly. "I do know that the next few days are going to be very difficult." Bryce nodded slowly. He had no doubt about that. "Your father and I want you to come home with us for a few days."
Bryce shook his head. "No, Mom, I'm going to go home." He needed to be where he could feel Percy close to him, and their small house was just the place he needed to be. They'd bought it together a year ago. "I need to." She nodded and didn't argue with him. Bryce leaned against her and closed his eyes, letting his mother comfort him. "I already miss him, Mom, and he's only been dead a few hours."
"I know, honey, I know," his mother said and just sat next to him.
As darkness began to surround them, they walked back to the house. The chill of the evening air worked through Bryce's clothes, but he barely noticed it. Inside, he quickly warmed, and John helped him into a chair. "We called everyone and got in touch with the caterer, florist, and photographer. They all send their condolences," John said.
"I called the gardens and the reception hall, so I think we have that covered," Jerry said, and Bryce nodded as blankly as he felt.
"Thank you."
"Do you want to stay here tonight?" Jerry asked.
Bryce shook his head. "I'm going home."
"At least let one of us take you," Jerry said, and Bryce agreed. He said good-bye to his mother and father, who promised they'd be over to see him in the morning. After hugs and more tears, they left and Jerry drove him across town. "Do you want me to go in with you?"
Bryce stared at the dark house and shook his head. "I'll call tomorrow," Bryce said, closing the car door. He saw Jerry nod, but the car didn't move. Bryce slowly walked to the front door and let himself inside. After closing the door behind him, Bryce turned on the light. Nothing was physically different, yet nothing felt the same. Bryce wandered from room to room, remembering. Eventually, unable to take anymore, Bryce ended up on the sofa, curled into a ball as he cried his eyes out.
Bryce lost all track of time, but eventually he got up off the sofa and climbed the stairs. In their room, he stood staring at the bed. Percy's pillow lay in its place, and Bryce lifted it to his nose, inhaling his scent. Unable to actually lie down, not in their bed, anyway, Bryce carried the pillow back down the stairs. He lay on the sofa, pulled the throw over him, and hugged Percy's pillow to his chest. Eventually he cried himself to sleep, wondering how his life was ever going to be the same.
# Chapter 2
BRYCE woke and rolled over toward the nightstand. Percy smiled back at him from a picture frame. Over the past year, Bryce had found he coped better if he saw Percy first thing in the morning. "I love you," Bryce said, the way he always did before getting out of bed. He knew it was probably a little odd to still be telling a picture of his dead lover that he loved him, and maybe it was, but it was how Bryce felt. Even after a year, he still needed to feel close to Percy. After shuffling to the bathroom, Bryce cleaned up and did his business before returning to his bedroom to get dressed.
The doorbell rang as he was pulling on his shoes. After tying them, he hurried down the stairs and opened the door. John, Jerry, and the two kids smiled at him from the front porch. "I thought you were leaving for the reservation this morning," Bryce said, motioning them inside.
"We are and we want you to come with us," John said in a calming tone.
Bryce shook his head. He didn't want to go anywhere. For the past year, he'd spent much of his time when he wasn't working alone at home. "Maybe next time."
"That's what you said last time, Uncle Bryce," Mato told him, his arms crossed over his chest. The kid was the spitting image of his uncle and even acted like him sometimes. "Grandma's going to cook, and she asked that you come for a visit."
"Please tell your grandma that I'll think about coming next time," Bryce said, knowing he probably wouldn't then, either. He wasn't ready to see a lot of people.
"Why don't you two go out back and see if Bryce has any new little fish in his pond?" Jerry told the kids, and they nodded, hurrying through the house and then out the back door. Percy had raised koi, and consequently he'd added a pond to the backyard and landscaped around it. Bryce kept the backyard just the way Percy had planted it. "Bryce, we're worried about you," Jerry said once the kids were gone. "You rarely leave the house except when you're working."
"Do you have any complaints about my work?" Bryce snapped, more harshly than he meant.
"No. This isn't about work. This is about you." Jerry sighed and looked around the room. "Your house is like a shrine to Percy. You haven't changed anything. If I go upstairs I bet I'll find his clothes still in his closet."
Bryce knew Jerry was right. He'd been able to get rid of very little. He'd known for a while that it was past time, but he couldn't bring himself to do it.
"We know you've had a rough time of it, and we care about you." Jerry moved closer, touching his shoulder lightly. "You need some time around people, and at the reservation you can help others."
"I don't know if I can," Bryce said, the grief welling up once again. That was the hard part. The grief and loss seemed to come back to him when he least expected it. There were times when he expected to feel it, like on the anniversary of when they met or on Percy's birthday. But once, he'd gone to the pet store to get some food for the fish, and as he passed the large pool where they kept the koi for sale, he'd stopped and stood there, crying. He remembered Percy standing in the same place, telling the salesperson exactly which fish he wanted and then waiting for him to catch that particular fish. Bryce had watched all this, and when they finally had the fish in the bag, Percy had given it to Bryce.
"I know it's hard, but you need to try to get on with your life," Jerry said. "We know you've been through more than anyone should have to go through, but isn't it time to move on?"
"But...."
"I'm not saying you should find someone to replace Percy, but it's time you got out of the house, and a change of scenery will do you good. We could really use your help. We're teaching a couple of computer classes this weekend, and one of John's relatives needs help with a new roof. You'll be busy."
Bryce found himself nodding slowly, unable to fight them any longer. Besides, maybe they were right and he did need to get away. It certainly couldn't hurt. "Okay," Bryce agreed. "Let me pack some things."
Both Jerry and John smiled, and Bryce turned to climb the stairs. Bryce went to his room and pulled a suitcase from under the bed. He did his best not to think about it, because he knew it was Percy's. He packed underwear, socks, shirts, jeans, and an extra pair of shoes. Before closing the suitcase, he placed the framed photograph of Percy between two pairs of jeans and then closed the suitcase. Last, he grabbed a jacket, then descended the stairs.
"I'll put this in the van," John said, taking the suitcase. "The kids have fed the fish for you."
Bryce nodded and began locking up the house. Once the house was secure, Bryce climbed in the backseat next to Mato, and they were off.
"I'm glad you're coming with us," Mato said, looking up at him from the center of the seat, an open book on his lap.
Bryce smiled at him. "What are you reading?"
"It's a bunch of Native American legends," Mato said with a grin before turning his attention back to the book.
Bryce wasn't really in the mood to talk much, so he simply stared out the window as Jerry guided the van through the city and then onto the freeway. They picked up speed and began their trek across the state.
"Have you heard anything from Percy's family lately?" John asked, and Bryce shook his head. "I'm sorry." John turned back toward the front and spoke with Jerry, but Bryce didn't listen.
"It hurts. The people who have the closest connection to Percy other than myself don't want anything to do with me."
"Why?" Mato asked, looking up from his book.
Bryce sighed. "Because people are strange when it comes to money, and Percy left his to me," he explained before turning to look out the window again. He'd made this trip many times over the past three years, just not since Percy's death. The land was beautiful—flat, but so very green, and full of spring. Bryce took a deep breath, then released it slowly, trying his best to let go of some of the hurt and pain.
"It's okay to be sad because Uncle Percy's gone," Mato told him. "And it's okay to be happy too. Uncle Percy knows we love him and he wants us to be happy." Mato looked so serious.
"Who told you that?"
"Uncle Jerry, right after Uncle Percy died. He said that it was okay to cry if I wanted to, but that it was okay to smile too, because Uncle Percy always made us laugh and he'd want us to remember him that way."
"He was right," Bryce admitted and smiled, hugging Mato lightly. He'd sort of forgotten that, and now he could see in his mind Percy playing with the kids, giving them pony rides on his back, everyone squealing and giggling with happiness. "Sometimes I wish the grief would just go away. I sort of figured it would pass by now," Bryce said.
"It takes time; everyone heals at their own pace," John explained from the front seat. "It took time for all of us to heal after their mother's death," John said, indicating the children, "and the ordeal we all went through, but we did, and we're all stronger because of it." John turned back to face forward. "Some time spent communing with nature and helping others will take your mind off your own cares."
Bryce nodded and then stiffened. "What do you mean communing with nature?"
"We're going camping," Ichante said with a grin. "The tents are in the back, and you're going to stay with Mato and me."
Bryce glared at Jerry in the rearview mirror. No one had said a word about camping. "We'll only be about five miles from my mother's," John told him. "We aren't disappearing into the wilderness."
"Aren't there wild animals?" Bryce asked.
"Nothing to worry about. Mato will protect you," John said, and Bryce felt so much better, protected by a preteen bodyguard. "Seriously, it's a place where others camp, so we won't be all alone. Trust us, you'll enjoy it."
Bryce didn't have the heart to tell them that what he'd really enjoy was a bed in a real house with running water and flush toilets. He humphed and said nothing, figuring he could always try it for one night, and after that, he'd see if anyone would be willing to take pity on him. He'd met John's large family—someone had to have a place he could sleep.
"Can we stop at the Corn Palace?" Mato asked when he saw the signs for Mitchell.
"Not this time," John said, and Mato went back to his book.
They continued driving for hours, only stopping for a picnic lunch and then continuing on. In Rapid City, they turned south and spent some time on main roads until they reached the reservation. Then the roads became narrower. Jerry, of course, knew where he was going, and Bryce watched out the window. Whenever he came here, he was always surprised by the poverty. In some places, homes looked almost cobbled together from scrounged building materials. Old travel trailers and mobile homes sometimes sat clumped together in the middle of open fields. Needless to say, the view didn't do much to improve Bryce's mood.
Finally, they pulled into the yard of John's parents' home. Bryce had never been so happy to step out of a vehicle in his life. It hadn't been a bad ride, but his legs and butt ached something fierce. While Bryce wandered around to get the blood flowing again, the kids raced inside. Soon John's mother, Kiya, came out to greet them.
"They convinced you to come," she said, pulling Bryce into a hug. "You'll feel better now that you're here." Bryce had quickly learned that no one argued with Kiya, so he nodded and let her lead him inside, where a table full of food had been laid out. Bryce wasn't hungry, but it appeared that half the reservation had been awaiting their arrival, so they sat down, and Bryce listened as the conversation became a din of overlapping excited voices. Everyone seemed to be sharing news and bringing everyone else up to date on gossip and happenings. "You aren't eating," Kiya said, and Bryce took a bite to appease her, smiling as he chewed. "You're way too thin."
"Mom, leave him alone," John told her with a hint of humor. "We got him to come this time, but he'll stay away if we nag."
Kiya sat back in her chair and laughed loud and full. "We only nag the ones we love," she retorted, and everyone else laughed as well. Bryce looked around the table, wondering what the joke was.
"Then we're the most loved people on earth," John countered, looking all around the table. The meal continued, and thankfully Bryce was able to just listen. He spoke up a few times, but mostly he sat. It was nice to be around a family again. After Percy died, some of those he'd thought of as his family had fought with him and then cut him off, and for all intents and purposes had disappeared from his life.
"How's your mother?" one of John's sisters asked. He couldn't remember her name, but she was always surrounded by a passel of children.
"Okay. The cancer is in remission, and they're continuing medication for now. Eventually they'll wean her off it and then we'll see what happens." For a while, the bad news had come in tsunamis. Thankfully it had stopped, because for a few months Bryce hadn't known where to turn.
"Dang," Jerry said from next to him, turning toward John. "I forgot to pick up hot dog buns for the campout."
"I can run to the trading post in a few minutes," John said, tapping Jerry's hand. "They'll probably have some, and if not, I brought a loaf of bread." Both Mato and Ichante groaned together. Obviously eating hot dogs on bread was not popular.
Bryce pushed back his chair. "I'll go in to pick them up if someone can give me directions," he offered. He was anxious to get away for a little while.
"I can go," John said, but Bryce was already asking for the keys, and Jerry handed them to him.
"Turn right out of the driveway and go to the end of the road. Turn left, and just before you reach the reservation center, the trading post is on the left," Jerry told him, and John pressed some bills into his hand. Bryce tried to give them back, but a stern look from his friend stopped him.
"I'll be back soon," Bryce said, shoving the bills into his pocket. He heard the kids ask to come with him, but Jerry quietly hushed them as Bryce left the house. He walked to the van and climbed into the driver's seat.
The directions Jerry had given him were easy to follow, and soon he was parked in front of what looked like a small, rustic grocery store. There was no one else in the lot, which Bryce thought a bit strange, but he slammed the van door, then walked to the front door and pulled it open.
Inside, the first thing he noticed was a complete lack of air-conditioning, though a number of ceiling fans stirred the air. Bryce looked around and realized this was much more than just a grocery store. It also appeared to act as post office, lending library, tackle and bait shop, as well as hardware store. And judging by the scent, a bakery.
The man behind the counter with long black hair and a stern expression that made him initially appear much older looked up from the magazine he was reading and met Bryce's gaze. Bryce shivered in response. Never in his life had he seen such a hard look from a complete stranger. "Afternoon," Bryce said, his mouth a little dry.
The man nodded but said nothing. However, his gaze never left Bryce. Figuring the best course of action was to get what he needed and get out, Bryce walked up and down the aisles, picking up hot dog buns, a case of the soda he liked, and some fruit snacks for the kids. He also couldn't help following his nose to the cinnamon-sugar doughnuts he'd been smelling since he entered. "What do you want?" the man said, the look in his eyes not softening one bit. "You some tourist here to see the injuns?" he asked mockingly.
"No. I'm a guest of the Black Ravens," Bryce said, and the man's look softened slightly. "I'm also helping to teach computer classes this weekend at the community center." Bryce forced a smile, because, after all, there was no need to meet rudeness with rudeness. He placed his purchases on the counter. "I'd also like a dozen of the cinnamon-sugar doughnuts."
The man got out a bag and counted out twelve doughnuts for him before ringing everything up. "That'll be $19.90," he said.
Bryce pulled out his wallet and handed the man a twenty. He didn't make any move to take it, and Bryce moved the bill closer. "Is something wrong?" Bryce asked, looking at the bill to make sure it was okay.
"We don't take those here," the man said.
"Take what?" Bryce asked in complete confusion.
"Bills with the Indian Hater on them. If you don't have something else, you can go." He actually sat down, so Bryce fished around in his wallet, but all he had were twenties. Then he remembered the money John had given him and pulled two tens from his pocket and placed them on the counter.
"Is that better?" Bryce asked, and the man took the bills and handed Bryce a dime without touching his hand. He bagged up the groceries and gave them to Bryce without a word. Then he sat back down and started reading his magazine again.
"You're welcome," Bryce said, turning away from the counter, as pissed off as he could ever remember being.
Bryce carried his purchases out of the store. As he opened the door, he turned back to the man and saw him staring at him. Bryce met his gaze as sort of a challenge, and then stepped outside. After placing the bag on the seat, Bryce retraced his route and drove back to Kiya's. Leaving the bag in the van for later, he went inside. It appeared that no one had moved in the time he'd been gone. Though more food had been devoured, the din of overlapping conversations hadn't lessened one bit.
"How did it go?" Jerry asked, and Bryce shrugged and rolled his eyes.
"You met Paytah Stillwater, didn't you?" John asked.
"If you mean the surliest, crankiest person on earth, then yeah," Bryce said sitting back down. "He made me pay with tens," Bryce said.
"That's why I gave them to you," John said as he shifted his gaze to his mother.
"There are many in the Native American communities who hate Andrew Jackson. They consider him the devil and a killer of our people. He stole land from native groups, including the ones that had helped him win his famous battles, and he was also responsible for the Trail of Tears, so many of our people avoid using twenty-dollar bills. The ATMs on the reservation generally only dispense tens."
"I didn't know. He looked at me like I'd killed his relatives," Bryce said, still a bit shaken up.
Kiya shook her head. "Paytah is a different sort of man. He's fiercely protective of our heritage and he tends to be a little militant about it."
"He relented a little when I told him I was your guest and was going to be helping with the computer classes."
Kiya sighed softly as she pushed back her chair. The other women seemed to take that as a signal and they got up as well, filling their hands with dishes and then taking them to the kitchen. The guys helped as well and then filtered outside. Bryce was about to follow them when Kiya spoke again.
"Life on the reservation is hard," she began. "Our family is lucky. My husband is a good man who works hard far from home in the oil fields in North Dakota. Others from the reservation do the same, but very often, little of the money they make gets back to their families. Instead, it's spent on alcohol and other things. It's sad but that's the way it is." She stopped rinsing the dishes and turned to Bryce. "Paytah has not had an easy life, and the blame for most of that he places at the feet of white men. I don't think that's fair, because people are people, but that's how Paytah sees it. He wouldn't harm anyone." She transferred the rinsed dishes to the dishwasher. "John should have insisted on going."
"It was okay. I know there are people here who are suspicious of white people. It doesn't really bother me." Bryce thought for a few moments. "I guess I just wasn't expecting it," he said.
Kiya chuckled softly. "Honey, you never expect it."
Ichante hurried into the kitchen. "Uncle Jerry says we're ready to go set up camp," she told him, taking him by the hand.
"Oh, goody," Bryce said softly, and he heard Kiya laugh again as Ichante lead him outside to the van. Mato was already inside and waiting. John and Jerry were talking to a few of the men, but their group broke up when they saw him coming.
"Ready for camping!" Jerry cried enthusiastically. The kids voiced their excitement, but Bryce was much less enthused. All the van doors were closed, and with everyone in their seats, they headed off with excited waves from the kids.
As promised, the camping place wasn't far from Kiya's, and it was beautiful, with a small stream running through a shallow but lush valley. They unloaded the stuff, and the guys put up the tents. Bryce wasn't sure what he was supposed to do, so he mainly carried things from the van in what seemed like a never-ending caravan of equipment. "That's the last of it," Bryce said as he set down the cooler on the grass. They had tents, chairs, a small fire pit, and not much else, but they all seemed pleased. The kids almost immediately took off their shoes and socks to wade in the shallows of the stream while John and Jerry took a moment to sit quietly. They offered Bryce a chair as well.
"This isn't so bad, is it?" Jerry asked, and Bryce grudgingly admitted that the view and quiet were a balm that his spirit needed.
"But at the first clap of thunder or weird sound outside the tent, I'm making for the van," Bryce quipped with a sarcastic smile.
"It's supposed to be nice all weekend, and the bears will leave us alone if we leave them alone," John said, and Bryce jumped to his feet, ready to head to the van.
"You said there weren't any bears," Bryce cried, knowing then he'd been had. "Fine, but I'm still keeping the van on reserve."
The kids joined them, keeping their shoes off, while John built a small fire. As darkness fell, they roasted hot dogs, told stories, and simply talked. A coyote howled in the distance, and Bryce looked all around, but didn't immediately race to the van. "It's okay, Uncle Bryce," Mato said. "They don't like people and they won't come near the fire."
Bryce nodded and listened as the haunting call was picked up by another and then another. "The land has its own music. If you close your eyes and let go, you'll really hear it," John explained.
Bryce wasn't so sure, but he closed his eyes anyway and listened. At first, he heard the coyotes, then the crackle of the fire, followed by the overlapping chirp of grasshoppers. The stream joined with its soothing gurgle, with tiny animals scurrying through the grasses. A bullfrog croaked loudly, and Bryce jumped, nearly toppling his chair, and the children added their laughter to the chorus. He closed his eyes again, leaning back in his chair and listening. The sounds, no longer scary or confusing, worked their way into his mind. As he listened, he heard a rock tumble down from a nearby bluff. No one said anything, and Bryce breathed deeply and calmly.
"It's beautiful," Bryce said as he felt tears well in his eyes. He made no move to wipe them away. Percy would have loved it out here, and Bryce had never been interested in camping. I'm sorry, Bryce thought, and a tiny breeze came up, caressing his cheek. The tears started in earnest as he swore he heard Percy's voice on the wind telling him it was all right and time for him to move on. I'll always love you, but there's another out there who will love you too, it seemed to say. The breeze caressed his cheek once more and then it was gone.
Bryce opened his eyes and saw Jerry and John staring back at him. "You fell asleep," John said as he poked the fire.
"So it was a dream," Bryce whispered as he looked around. He saw the kids through the screen of the tent, already asleep. "How long?"
"About an hour," John answered in a whisper before lifting his gaze to meet his. "What did you see?"
"I didn't see anything, but I heard Percy on the wind," Bryce said and then waited, but there was no wind, the air still and rapidly cooling. "He told me he'd always love me, but it was time for me to move on." Bryce swallowed back tears. "But it was just a dream."
John shook his head slowly. "No. You were given a blessing. The spirits allowed him to come to you and give you a message. Don't dismiss it or think it didn't happen." John placed his stick on the fire, stood up, and stretched his back in varying directions. Then he sat back down again, and no one talked for a while as Bryce sank deep into his own thoughts. Maybe it was time for him to stop grieving.
Bryce stood up and walked toward the van. He pulled out his suitcase and in the darkness changed into a pair of shorts and T-shirt before making his way to the tent he was sharing with the kids. They had insisted he sleep with them, so Bryce climbed into his sleeping bag on the air mattress and closed his eyes.
IN THE morning, Bryce woke to an empty tent and voices outside. Obviously he was the only one not up yet. Reluctantly, he got out of the sleeping bag and stepped into the morning air. Looking at his watch, he groaned. What was everyone doing up at this ungodly hour? Thankfully, John had made coffee and pressed a mug into his hand. "Sorry, I should have warned you that the kids get up at the crack of dawn out here." Even John yawned, and Bryce considered finishing his coffee and going back to bed, caffeine or not.
Bryce yawned again and sipped from his mug. After setting it on the ground, he grabbed his sleeping bag, unzipped it, and wrapped it around him in the morning chill before sitting in one of the chairs. Wrapped in his cocoon of warmth, he sipped coffee and closed his eyes. He felt happier and more content than he had in a very long time. He'd spent a long time asking the universe why Percy had been taken from him, and then he'd had to fight the airline responsible for the driver who'd killed Percy as well as Percy's family, after which he'd simply spent months being angry at everyone. It was a miracle he had any friends left after that. Now he felt quiet inside, a bit empty, but the shouting and rage were gone and it was okay. He could go on with his life. How he was going to do that, he wasn't sure, but he knew it was okay to try.
"You okay?" Jerry asked quietly, and Bryce smiled. Not one of those forced smiles so people would leave him alone, but a genuine smile that reached to his eyes. "I haven't seen that in a while," Jerry told him, patting him on the shoulder. "Who's ready for breakfast?"
A cry went up, and Bryce added his own voice as his stomach rumbled. Jerry made a basic breakfast of eggs and bacon over the fire, and Bryce added the doughnuts he'd bought the day before. But what surprised Bryce was that everything tasted different. It was only eggs and bacon, things he'd had a million times before, but they were the best he'd ever eaten. "Everything tastes better when you cook it over an open fire," Jerry explained, like he'd been reading his mind. "I found that out the first time John took me camping."
Bryce saw Jerry gaze contentedly at his lover, and soon John moved next to him. No words were said, but Bryce could almost feel them pull together. That was what he'd thought he had with Percy. No, he wasn't going there again. Percy was gone. Bryce had loved him very much and Percy would always be with him, but he had to let go. "So what sort of class are we teaching today?" Bryce asked.
"The last time we were here, a number of the business owners asked for help with business software," Jerry said. "Most of them are still doing their books by hand. So we're going to teach them to use a small-business accounting and management software package. I got the software company to donate it. I think there will be six students, and because this will be a little more complex class, we're going to have to give them some extra attention. So I thought we'd each work with two students during the exercises to make sure they really understand what they're doing." Jerry finished his breakfast and threw the paper plate in the fire before carefully placing the other trash in a bag.
"Leave nothing but footprints," John said, and Bryce nodded his understanding. He finished his breakfast and then burrowed back into his cocoon, closing his eyes for a nap. He listened to the kids playing, smiling at their happiness as his mind sort of floated.
"We need to leave soon so we can get things set up," Jerry said, and Bryce reluctantly stood up and placed the sleeping bag in the tent before quickly changing his clothes behind the van. Then he got his kit, shaved, and cleaned up with water John had heated for him. He was ready and waiting by the time Jerry and John got the kids ready to go. All the food was placed in the van, along with the cooler, and they left the campsite. Jerry dropped the kids at their grandmother's, and then they headed to the tribal community center.
It appeared that some of their students were already waiting for them. They got people set up as the other students filtered in. The class was about to start when the last person arrived.
Paytah still wore his hard expression when he looked at him, and Bryce wondered if he always looked like that, but then he turned to John and smiled, his face lighting up. Well, that answered that question.
"Please take a seat," Jerry said. "Bryce will get the software installed for you while we start the class."
Jerry handed him the CD, and Bryce sat in the chair next to Paytah and began working on his computer. He tried not to look at him as he worked, but he couldn't help it. That smile had been luminous. A few times he noticed Paytah glancing at him too, but his expression didn't seem to change. Jerry ran down an overview of the software, and Bryce finished the installation as Jerry was wrapping up. "You should be all set," Bryce explained in a whisper. "Just be sure to password-protect the files."
"Thank you," Paytah said in a gruff voice, and Bryce wondered how anyone could make a pleasantry sound so unpleasant.
"It's no problem," Bryce said and sat back in his chair. Jerry guided them through setting up their business profile and then actually using the software. After each section, he gave them exercises to do. It quickly became obvious to Bryce that Paytah wasn't picking up the concepts as quickly as the others. "Don't let the computer intimidate you."
Paytah turned toward him, and at least some of the hardness was gone from his expression. "You're doing all of this today by hand," Bryce said. "The computer is just a tool to hopefully allow you to get more done in less time."
"But...," Paytah began, but his face hardened again before he turned back to the computer. He picked his way through the rest of the exercise, but it was painful to watch. Bryce knew he could help if Paytah would let him.
"Let's take a break," Jerry said. "We'll start again in ten minutes."
Paytah got up and left the room with most of the other students, and John slid into the chair next to his. "What's going on?"
"He's not keeping up, and I tried to help, but he wouldn't let me."
"It's part of the warrior's attitude. He'll persevere through just about anything, but he won't actually ask or accept help. I can work with him if you like."
"It's okay," Bryce said. John returned to his place and the students returned. Paytah sat back in his seat, and Jerry began the class again. This time, when he saw him struggling, Bryce asked Jerry a question that led to a discussion of the issue Paytah was having. There was definitely more than one way to skin a cat. If he wouldn't accept help directly, then Bryce would do it in a sneakier way.
The class continued for much of the morning, and as noon approached, Jerry wrapped things up. "I hope this was helpful for you." Everyone clapped, including Paytah.
"Who's watching the store?" Bryce asked as Paytah packed up his computer.
"Alowa," he answered.
Bryce wanted to ask what he'd ever done to him, but then John sidled over and he didn't think it was a good idea, so he excused himself and said good-bye to the other students as they filed out.
"I don't understand why white people are teaching us like this," Paytah told John in a whisper.
"You need to let that go," John told him, and Paytah shook his head. "Jerry set up this program and has helped a lot of people here, including the kids. He loves Mato and Ichante." Bryce saw John glance at him and he turned away so it wouldn't look like he was listening.
"Jerry, I understand, but what about him?" Paytah said softly, and Bryce smiled as he folded the chairs to put them away.
"Bryce is a good guy," he heard John say. "And I think the two of you might have more in common than you think." Bryce figured their little conversation was over, but John continued. "You feel Native American rights are very important, I get that—so do I. But we've been persecuted because of invalid prejudices for centuries. Don't you think the way you treated Bryce was a little like the way you don't want people acting around you?"
Bryce turned around and saw Paytah's expression heat up, and he left the room to get out of the line of fire. In the hall, he sat down next to Jerry to wait. "John wasn't happy with the way you were being treated," Jerry explained.
"I got that," Bryce said. "But I can take care of myself, and Paytah was only hurting himself, not me." Bryce relaxed until John came out of the room.
"Let's go get the kids and have fun for the rest of the weekend," John said to Jerry before kissing him and then heading for the door. John said nothing about what they'd talked about, and Bryce didn't let on that he'd heard anything at all. Jerry followed John outside, and Bryce looked at the closed door to the room they'd been using. Slowly he opened it and peered inside. Paytah sat at the place he'd occupied without moving, looking lost and maybe a bit confused, supporting his head with one hand. Bryce wondered if he was all right, but any help from him probably wouldn't be welcomed, so he silently closed the door before Paytah saw him, and then he left the building.
The van ride back to Kiya's was quiet, but that changed as soon as the kids piled in. They were obviously ready for the rest of the outdoor time, and if Bryce were honest with himself, he was too. He'd slept well, and right now nothing seemed to bother him, not even a surly Native American with an attitude problem and a killer smile.
"Is everyone ready to go?" Jerry asked rhetorically, not waiting for an answer. "Wave good-bye to Grandma," he added as he backed down the drive. The kids waved, and their grandmother waved in return.
The drive to the campsite went fast. Everyone got out and unloaded the supplies they'd brought with them. Nothing had disturbed their camp. The kids had eaten at their grandmother's, so Bryce made sandwiches for the three adults before sitting for a while in the late spring sun. "I love how this spot changes through the year," John said as he sat down. "Right now everything is fresh from the spring rains, but in another month or so, the grasses on the slopes will start to brown, and then a month after that, the dryness will get closer and closer to the stream as it loses much of its water, but then fall comes and the heat breaks. Rain often returns and there's another burst of green before winter covers everything."
"I take it you camped here when you were a kid?"
John nodded. "I used to come with my dad sometimes." John turned to Jerry. "Mom said he should be home for a few weeks next month. We should make a point to come while he's here."
"Of course," Jerry said after swallowing the last of his sandwich. "See if you can arrange some workspace for us, and we'll work here for a few weeks."
"That I can probably do," John said before pulling Jerry into a deep kiss.
"Ewwww, kissing," Mato teased, but he quickly turned his attention to wading in the stream, and Bryce watched the kids to give John and Jerry a moment of privacy.
"You're welcome to come with us if you want, Bryce," Jerry said, "or you can stay and use the office at home if you want."
"We won't have to camp for two weeks, will we?" he asked, and both Jerry and John laughed.
"We'll stay in a cabin that one of John's friend's owns. It's a bit rustic, but there's electricity and running water."
The thought of going back to being alone again was not particularly palatable any longer. "I think I can do that," Bryce said with a smile.
The rest of the day was peaceful, quiet, and passed like wildfire. How sitting, talking, and really doing nothing but eating could make the time pass so quickly was beyond Bryce, but before he knew it, they were having dinner as the sun slid behind the valley walls.
John once again built a small fire, and as the evening wore on, the night sounds started. This time Bryce knew what they were and he was ready for them. Closing his eyes, he listened and relaxed. The stream and the crickets played, with bullfrogs joining in. There was definitely a breeze—he could feel it on his skin. Percy, are you there?
Bryce continued to listen as small animals skittered through the grass or blurped into the water. Again he felt a breeze, but there was no voice on it, just the wind adding its own tempo to the melody. Bryce's first thought was to be sad, but then he realized the coyotes were missing; they weren't calling. Bryce kept his eyes closed and strained to hear them, but they weren't singing.
Then, piercing the night, a single howl cut the air, mournful and long, ending for a few seconds before beginning again. Sadness gripped Bryce's heart, not for himself, but for the animal. What could have hurt it so badly it would make that sound?
"Wolf," John said softly, breaking Bryce's illusion. Slowly, he opened his eyes to see the two men get up to put the kids to bed. Jerry returned a few minutes later, sitting back in his chair.
"John is telling the kids stories about wolves," Jerry said. "They love to hear his stories, and it's good for them to know their heritage." Jerry got back up and returned with a couple of blankets, handing one to Bryce. "It's going to be cold tonight."
Bryce took the blanket and wrapped it around his shoulders before returning his gaze to the dancing flames. "I was hoping Percy would come back," Bryce admitted.
"I know," Jerry said softly. "But I think whatever you got to experience was a onetime deal."
"I wish I understood it," Bryce said.
"No, just accept it for the gift that it was." The wolf howled again, and Bryce sat back, warm now with the blanket and the fire. Closing his eyes, he just listened. The loneliness in the wolf's cry tugged at Bryce's heart. He didn't know how he knew that wolves mated for life, maybe John had told him once, but this wolf had lost its mate. Bryce knew that in his heart. The mournfully gut-wrenching sound echoed off the canyon walls, drowning out everything else. I understand how you feel, but you have to move on, Bryce silently told the wolf.
Another cry split the air, louder and closer. Bryce started and opened his eyes, but he didn't see the fire or Jerry. Instead, he saw Paytah's brilliant smile.
Bryce nearly fell over backward. Only Jerry catching him prevented him from falling flat on his back. "Are you okay? Did you stand up too fast?"
"Yeah," Bryce said breathily, looking all around, but he saw nothing out of place. Breathing deeply, Bryce wondered why he'd seen Paytah.
"Sit back down." Jerry guided him to the chair, and Bryce sat. John joined them a little while later, and the three of them talked quietly. But Bryce couldn't get the image of Paytah's smile out of his mind, and even after he went to bed, he could still see it whenever he closed his eyes.
# Chapter 3
"CAN we have ice cream?" Ichante asked as she closed the door to the freezer in the cabin. They'd arrived the day before, and to Bryce's surprise, the cabin was much nicer than he'd expected. Yes, it was rustic-looking and wasn't big on modern appliances, but it had the basics, and that included running water, a real bathroom, and walls between him and the coyotes, wolves, and potential bears. Everything else he could deal with.
"If you like," Bryce answered before the other two adults could say anything. "What kind would you like?"
"They'll be happy with whatever's at the trading post," Jerry said with a wink as he moved the laptop from his lap to the log table in front of the sofa.
"I'll go," Bryce said, closing his laptop and getting up. He was going a bit stir-crazy and could use a chance to stretch his legs. Bryce noticed the expressions that both Jerry and John sent his way. "I'll be fine. I made sure I have plenty of tens before embarking on this little adventure."
Jerry pulled his laptop back onto his lap, and John turned in his chair. "Do you know the way?"
"I paid attention as we came in, so I should be able to find my way," Bryce said, telling John how to get back to the reservation center. John nodded and went back to work while Bryce got ready to leave. "By the way, Jerry, I have those programs done and checked in for the West Houghton change," Bryce said, and he saw Jerry smile and nod. He'd been working on that account for weeks and they'd kept changing their specifications. Bryce had finally gotten them to make up their minds and he hoped it was done. He was tired of working on those programs.
"Great, thanks. I'll check them out and bundle them up," Jerry said, flashing him a smile. "Good job getting them reined in."
Bryce snorted slightly as he pushed open the door and stepped out into the near wilderness just outside the door. Trees and shrubs were a short distance from the building, and the drive seemed like a long tunnel of green out to the main road. According to John, there had been more rain than normal, and the plants seemed to be taking advantage of it. Bryce walked down the steps to the van and got inside, started the engine, and then pulled down the tunnel of green, being careful to stay on the narrow drive. A few branches lightly brushed the sides of the van as he moved forward. When he reached the road, barely more than a two-track, he slowly turned and carefully drove until he reached the main road and headed to "town."
Of course, he got lost and had to stop to ask for directions, but he made it without too much backtracking and parked outside the store. He got out of the truck and made another check that he didn't have any of the "offensive" money before going inside.
Paytah sat behind the counter, hunched over his computer, two fingers slowly typing at the keyboard. "Heeyah," he said softly, grumbling under his breath as he worked. Bryce coughed softly to get his attention.
"I can help you if you'll let me," Bryce said from across the counter. Paytah looked as though he were about to refuse, but then sighed. "I know you don't like me because I'm white, but I can help."
Paytah grumbled again and returned to his typing. Bryce gave up and turned away to begin finding what he'd come for. "This was supposed to save time, but I've spent a month on this and it still isn't working," Paytah groused with frustration. He stopped what he was doing, stood up, and stepped away from the desk. Bryce took that as an invitation and moved around the counter, settling in the chair. He made sure to save all the work before looking through the program. It didn't take him long to see what was happening.
"I think you're doing things a little backwards," Bryce explained gently as he turned around. "I'm Bryce, in case you don't remember."
"I remember," Paytah said, and Bryce felt himself smile for some strange reason. "Paytah."
"I remember too," Bryce said, and for a second he forgot what he was doing as he sort of lost himself in Paytah's eyes. Realizing what he was doing, Bryce blinked and returned his mind to the task at hand. "What I think you're doing is entering the numbers in the wrong place first and the program is fighting you." Bryce pointed to the screen. "You're entering the numbers you think are correct here, right?"
Paytah nodded. "It takes a long time to add everything up."
"Yes. But if you'd enter the individual values here, the system would record them and add them, filling in this entire screen for you, and you wouldn't have to add the invoices, or have the system fight you. They really should make this screen noneditable, because lots of people make this error." Bryce shifted the program to another screen. "Go ahead and enter your invoices on this screen one at a time, and when you're done, we'll check the other screen." Bryce got up and let Paytah sit down.
"Does this look familiar from class?" Bryce asked.
"I couldn't find it again," Paytah said.
"I bookmarked it for you," Bryce explained as he leaned over Paytah, reaching to the mouse. "Just look here and you'll see the Bookkeeping tab. In there is the link; just click on it and the computer will bring you right to it." Bryce demonstrated. Paytah turned to watch Bryce, his long, silky hair trailed over Bryce's arm, and Bryce nearly groaned. "Give it a try while I do the shopping I came in for."
Bryce walked down the aisles, grabbing a few things. He wasn't going to get the ice cream until the last minute, and he wanted to make sure Paytah understood what he needed to do. Bryce watched him a few times, and though he seemed to be working slowly, some of the frustrated rigidness left his body. Once he had everything except the frozen stuff, Bryce walked up to the counter.
"It's better, thank you," Paytah said with an actual smile on his face. "And the numbers match on the other screen."
Bryce nodded. "It's no problem. John's dad is home for a few weeks, so we've moved the entire office to the reservation so John and the kids can be closer to his dad while he's home. If you have questions, you can call," Bryce said. Paytah was about to start ringing up his purchases when a man rushed into the store. He almost skidded to a halt when he saw Bryce.
"Who are you? Oh yeah, you're one of those guys who teaches the computer classes, aren't you?" the man asked.
"Yes," Bryce answered.
"Paytah, it's Chay. I think you need to come," the man said urgently. "He's on the top of Rainbow Bluff and won't come down."
Bryce saw Paytah grow pale and become nervous. "Okay," he finally said. "I'll close the store once I'm done here."
"I can watch things for you if you want," Bryce volunteered and almost instantly wished he hadn't, based on the looks he got from both men. "I worked in a store when I was in college and I know how to run a register." Bryce turned to Paytah. "This way you won't have to close the store." Bryce had originally offered to help, but now his pride was up and he was tempted to tell them both where they could stick it.
"Okay," Paytah said, quickly showing Bryce where everything was.
"Come on, we have to get to Chay," the other man said.
"My brother can wait two seconds, Coyote," Paytah groused back.
"I'll be okay," Bryce told him. "Go help your brother." Bryce shot Paytah a smile, and Paytah looked longingly toward the door. "I'll be fine," Bryce told him, and Paytah rushed out of the store. Bryce turned, and within seconds the truck that had been parked next to the van pulled away from the building and took off down the road in a cloud of dust. Bryce turned back to the store and shook his head. He'd volunteered to do this on a whim, and now he wondered just what in hell he was doing.
The store was completely quiet, so he pulled out his phone and looked at a picture of him and Percy—the wallpaper image. Reminding himself for the millionth time to change it, he called Jerry. "I got a little delayed," Bryce said when Jerry answered. "Tell the kids I'll bring their ice cream as soon as I can."
"Where are you?" Jerry asked, sounding a bit concerned.
"I'm at the store," Bryce answered, and he could hear Jerry repeating things to John. "I got sort of roped into watching it because Paytah was called away to help his brother. It sounded like an emergency or something." Once again he heard Jerry talking to John, and this time John said something back, but Jerry didn't repeat it. "I'll be back as soon as I can," Bryce said and hung up. Shoving the phone in his pocket, he looked around and picked up a magazine off the rack. There was absolutely nothing he wanted to read, so he sat behind the counter with a copy of Field & Stream, reading about big-mouth bass before tossing the magazine aside, crossing his arms over his chest, and waiting for his first customers.
He had to wait a while. Eventually a woman came in, and at first she looked at him like he was from outer space. "What are you doing here?" she asked.
"I'm watching the store for Paytah," Bryce answered her. "I'm a friend of John, I mean Akecheta Black Raven. Paytah got pulled away, so I'm helping." Bryce needed to remember that here John was referred to mostly by his Lakota name, Akecheta.
Her expression changed almost instantly, the suspicion turning to a smile, her plain, middle-aged features brightening and instantly making her look younger. "You're one of the men teaching computers," she said as she placed her purchases on the counter. "My son and daughter took one of your classes at the school." She emptied her basket, and Bryce began to ring her up.
"I'm glad we could help," Bryce told her.
"You've done more than help. You've begun bringing our children into this century," she told him, and Bryce grinned as he finished ringing her up. He told her the amount, and she paid for the groceries, all in tens. Bryce thanked her, and she left the store. Bryce reminded himself to tell Jerry about her. It would make his day that people really thought that much of the program he and John had put together. Unfortunately, once she left, the store was quiet again and Bryce had nothing to do but wait.
Bored, Bryce pulled a few pieces of paper out of the printer to fan himself in the warm store. The fans were on at least, moving the warm air. A few times, Bryce wandered over to the frozen foods and pulled open the door to cool off. It worked until he closed the door again and a few seconds later he was just as warm as he'd been before.
The door opened and a kid entered. He must have been nine or ten, with eyes that widened, big-time, when he saw Bryce. "Hello," he said, and then he continued on. Bryce saw him pick up a basket that appeared nearly as big as he was. Bryce had seen a few kids like this on his visits, hands and face dirty, clothes in need of a good washing and a size or two too big. He was definitely in need of someone to take care of him. Bryce settled behind the counter, expecting the kid to go for the candy. Instead, he walked directly to the aisle that held the few baby items Paytah carried. A can of formula went into the basket, as did a small package of diapers and wipes. The kid continued on and seemed to deliberate over the diaper cream until he put the smallest tube into the basket.
The kid set down the basket and pulled a small piece of paper out of his pocket, checking it before folding it up again. He shoved it back into his pocket, picked up the basket again, and continued on, disappearing for a few seconds off to the side. Bryce craned his neck and could barely see the top of the kid's head by the refrigerator case. He heard the door open and close, and a few minutes later the kid carried the now full hand basket toward the counter. He'd added milk and a small loaf of bread.
As he got closer, Bryce saw how thin he was, with almost hollow eyes and straggly hair. Bryce leaned over the counter and helped him lift the basket. Bryce saw a round protrusion in one of the kid's pants pockets, the corner of a package of lunchmeat poking out through a tear in the pants, and a square one in the other pocket that Bryce figured was a package of cheese slices. His throat constricted as he rang up the kid's purchases. "That'll be $24.79," Bryce said when he saw the total, and the kid held out a handful of wadded fives and ones that he'd carefully fished out of his pockets while trying not to give away what was in them. Bryce completed the sale and closed the register drawer. "May I ask you a question?"
"Yes, sir," the kid said in a small voice, shifting nervously from foot to foot.
"I bought a doughnut earlier and couldn't eat it. Would you like it?" Bryce asked, and the kid widened his eyes in total surprise. Bryce lifted the dome and handed the kid a cinnamon-sugar doughnut. He took it and licked his lips. Bryce put the lid back on the doughnuts and began bagging up the kid's purchases, watching as he downed the doughnut in less than ten seconds. Bryce knew from personal experience just how good Paytah's doughnuts were, but a kid that small eating that fast.... "Here you go," Bryce said, handing the kid the bag.
The kid wiped his lips on the back of his hand. "Thanks, mister," he said as he took the bag. Then he walked quickly toward the door, pushed it open, and hurried outside. Bryce watched him through the window as he set the bag on the ground, then pulled the lunchmeat and cheese out of his pocket. He added them to the bag and then walked across the open area before disappearing down the road.
"Jesus Christ," Bryce said, nearly falling back into Paytah's desk chair. Yes, the kid had stolen from the store, but lunchmeat and cheese? Bryce shook his head wondering just how many kids were waiting at home for that small bit of food. Obviously there was a baby, and Bryce had the suspicion that the kid who'd just left was the primary caregiver. He seemed so careful to get what the baby needed first.
Bryce walked from behind the counter to the refrigerator case and saw where he thought the lunchmeat and cheese had been. He picked up identical packages and brought them to the register, ringing them up along with a doughnut. He paid for them with a ten from his pocket and then took them back. He was just returning to the front when he saw a truck pull up outside. Paytah got out of the passenger seat, helping another man get out. Coyote hurried around from the driver's side and took the man, who looked drunk, and helped him shuffle away from the truck. Paytah opened the door to the store.
"I'll take care of Chay," Bryce heard Coyote say, and then Paytah entered. He closed the door and looked all around, like he'd expected to come back to a pile of rubble, or bare shelves. The store wasn't large, by any means, and he wandered up and down the aisles before returning to the counter.
"A young kid came in here, about so big," Paytah said, indicating the kid's height with his hands.
"Yes," Bryce answered, wondering how he knew.
"Can of his brother's formula is missing," Paytah said as an explanation. He continued wandering but didn't seem to see anything amiss. "Did he take anything?" Paytah knew about the kid and that he stole. Jesus, could I have been any dumber?
Bryce felt completely foolish. He'd felt sorry for the kid, but apparently he did this all the time. Bryce nodded and walked around the counter. "Don't worry—I paid for what he took. He—" Bryce cut himself off and picked up the basket of things he'd filled. He hurried to the freezer to get ice cream, making sure it was frozen solid before taking everything to the register. So he'd been played by a kid with huge eyes and a needy expression. Fuck, he'd even given the kid a doughnut. What a sap he must have looked like.
Paytah rang up his purchases, staring at him like he was trying to figure Bryce out. "Thank you for helping me out," Paytah said when Bryce handed over the money.
Bryce nodded, unable to look the man in the eye. "I can't believe I fell for that...," Bryce said as an apology. "I'll see you." He grabbed his bags, hurrying to the van before he somehow made a bigger fool of himself. He opened the side door and placed the ice cream in the cooler he'd brought, rearranging the ice packs before closing the lid and climbing inside. He started the van and backed away, getting out of there.
Bryce filled his head with recriminations the entire drive back to the cabin. Once he got through the tunnel of vegetation, he parked the van and carried everything inside. The house was quiet, with Jerry and John still working. The kids were playing a game on the floor, and after putting everything away, Bryce sat down on the sofa and lifted the laptop so he could get to work. "What happened?" John asked.
"Don't want to talk about it," Bryce said, looking at both John and Jerry before opening his laptop and going to work.
"Can we have ice cream?" Ichante asked, and Jerry turned to him. Bryce nodded, indicating he'd bought some, and Jerry put his work aside to help the kids while Bryce continued sulking behind his screen. When Jerry returned, he brought a bowl of ice cream to John and set one down for Bryce as well. Not wanting it to melt, Bryce set the laptop aside and let the frozen minty goodness soothe his ego.
"Thank you, Uncle Bryce," both kids said, and he smiled at them. It was a bit hard to sulk when those two were around. They were happy kids who played and laughed. They didn't have to worry about where their next meal came from. And so what if that kid had played him—Bryce had acted on what he'd thought was right and the way he felt.
"You're welcome," Bryce said, and Ichante climbed onto the sofa to sit next to him. She grinned up at him before taking another bite. Mato climbed up as well, sitting on his other side. The kids were exactly the cure he needed. "Is there a place we can go swimming?" Bryce asked John, who shook his head.
"There aren't any lakes, just streams like the one we camped against last time," John explained.
"There's a water park in Rapid City," Jerry offered. "If we got our work done, we could see if Grandma and Grandpa want to join us on a short road trip tomorrow." The kids cheered, and Bryce nodded and smiled.
They finished their ice cream, and the kids returned to their game. Bryce went back to work. His little jaunt into town had cost him enough work time, so he knuckled down and continued working until it was nearly time for dinner. Only then did he close his computer and emerge from the concentrative cocoon of his work. Wandering through the cabin, he realized he was alone, but heard voices outside. Following them, he found John and a man he didn't immediately recognize talking quietly in what sounded like Lakota.
"Bryce, we didn't mean to disturb you," John said, and Paytah turned around. Bryce wasn't sure if he should approach or leave them alone, but then John excused himself.
"After you left, I realized you might misunderstand what happened," Paytah said, stepping closer. "I was not accusing you of anything wrong."
"Little Wamblee, the child that came in the store, he bought formula, diapers, and baby things, right?" Paytah asked, and Bryce nodded. "His mother... well, she gives him money, and sometimes it's not enough. He always buys what his baby brother needs, but there isn't always money left over." Paytah's expression was very serious. "What did he take?"
"Lunchmeat and cheese," Bryce answered, the lump in his throat he'd felt in the store returning.
"He didn't have enough money left to buy food for himself," Paytah told him. "He does that sometimes, and I pay for it." Paytah sighed. "There is much poverty here."
Bryce nodded slowly. He'd seen it many times. "I gave him a doughnut, and he wolfed it down."
"It was probably the first he'd eaten all day, and the lunchmeat and cheese will probably feed him lunch for the entire week," Paytah said, and Bryce turned away, blinking hard as he remembered the skinny body and his hollow eyes.
"Damn it," Bryce swore, his throat rough. "No kid should have to live like that." He rubbed his eyes before turning back to Paytah.
"There are many here who sometimes do not get enough to eat, young and old," Paytah said as if it were a fact of life. The injustice made Bryce's blood boil. He clenched his fists in frustration, but didn't know what to do about it.
"That doesn't make it right," Bryce said.
"No," Paytah said softly. "I need to get back to the store, but I didn't want you to think you had done anything wrong. You helped him, and that was kind." Paytah turned and walked toward the front of the cabin.
"Why doesn't the council do something?" Bryce asked, and Paytah stopped.
"With what?" Paytah asked with a shrug. "They have no resources. There is no money and very little chance of getting any." Anger seared behind Paytah's eyes. Bryce could see the frustration and maybe hatred burning inside him. Bryce stared back, meeting Paytah's steely gaze with his own. "You really do care," Paytah said.
"You're fucking damned right I care," Bryce spat, not meaning to yell at Paytah, but everything that had been building all day seemed to come to the surface all at once. "I don't know where this distrust of people just because they're white comes from with you. Others I've met here don't have it. But, yes, I care that a kid has to steal to get enough food, and that a ten-year-old is at home taking care of the baby. Of course I fucking care!" Bryce's voice seemed to echo through the land. "But I don't know what to do about it."
"Do?" Paytah said quietly. "There isn't anything to do. Things are the way they are, and most people wouldn't cross the street to help someone who doesn't look like them." Paytah turned and walked away. Soon Bryce heard a truck engine. Turning his head to the sky, he seethed and simply voiced his frustration with a scream that left him shaking from head to foot.
"What's going on?" Jerry asked, running around the building with the two kids right behind him.
"Sorry," Bryce said. "Nothing." He was still shaking. "Paytah is just the most frustrating man I have ever met."
"Is that all?" John said, joining them. "You know, if every person who felt that way yelled at the sky, we'd never be able to hear the coyotes or wolves." John chuckled. "He has that effect on people."
"How can he take things like this lying down?" Bryce asked, turning to John and Jerry, who both gave him blank looks.
"You didn't want to talk about it, remember?" John said, and Bryce nodded.
"Well, I want to talk about it now," Bryce said, marching toward the cabin door, leaving the others to follow in his wake. Bryce stopped before going inside, whirling around. "How can anyone so... so... gorgeous be such a pain in the—" He saw the kids. "—backside!" John smirked, and Bryce realized what he'd said. "Not that it matters what he looks like, because it doesn't matter."
John walked past him and pulled open the door. "So what did Paytah do to ruffle your feathers so much? He told me he came here because he felt bad about earlier and he wanted to thank you for helping him out."
"He did that," Bryce said as he flopped down onto the sofa. "There was this kid at the store who stole some food. I felt sorry for him and paid for it myself once he'd left. When Paytah returned, he didn't seem surprised, and I thought I'd been conned. Turns out, according to Paytah, he does this sometimes when he's hungry." Bryce told himself he wouldn't sniffle.
"Did Paytah say the kid's name?"
"Little Wamblee," Bryce said, and John nodded. "Anyway, when I told him something had to be done, he just shrugged and said there was nothing anyone could do." Bryce clenched his fists tight. "He practically made fun of me."
John sat on the edge of the chair across from him. "Bryce, Paytah may be right. There is only so much the tribe can do. Resources are very tight. They do try, but there isn't enough to help everyone."
John," Bryce pleaded, "this kid spends money to take care of the baby. He was so thin, and his eyes... God, his eyes looked hollow and so tired. I gave him a doughnut and he ate it faster than you could imagine. I barely saw it go into his mouth, he ate it so fast." Bryce wanted badly to make John understand. Lifting his eyes from the floor, his gaze met John's, and he realized John did understand. "God...."
"I know, Bryce. We have it really good. Mato and Ichante will never wonder where their next meal will come from. Little Wamblee will probably always wonder if he'll get enough to eat, and most of the time that answer will be no." John sighed. "Little Wamblee was named after my dad and it hurts to see him hungry. When the kids are in school, it's better, because most of them get free lunch and many get breakfast as well. They're guaranteed to get at least one and sometimes two good meals a day."
"But in the summer, there are children making a week's worth of lunches out of a package of pilfered lunchmeat, cheese, a loaf of bread, and a doughnut." Bryce could see Little Wamblee's face in his mind, and he realized he'd been stupid to think the kid had been playing him. That kid was probably hungry all the time. The only thing that made him feel better about this whole thing was that he'd bought the kid's food.
"Hey, because of you, whether he knows it or not, he and his sister will have something to eat this week," John told him, and Bryce held his head in his hands.
"There're two of them?" Bryce said. "What in hell can we do?" John didn't answer, and Bryce let his head fall back against the cushions. "Aren't there any organizations that can help?"
"There are a few groups that have a presence on the reservation, and they do what they can, but the need is greater than they have the resources for," John explained, and Jerry joined them, sitting next to Bryce on the sofa. "We do what we can, Bryce. Jerry and I have tried to give the kids here a better future. Teach them so they have skills that will hopefully open doors for them."
"But what good does that do when they don't have enough to eat now?" Bryce asked. Both Jerry and John nodded, but neither offered a suggestion, not that Bryce had any bright ideas either. But he knew one thing—he definitely owed Paytah an apology. He'd let his frustration get the best of him. After all, Paytah helped when he could, as well, and Bryce had taken his feelings of helplessness out on him.
"I know how you feel," Jerry said as Mato and Ichante hurried into the room.
"Uncle Bryce was yelling and he said naughty words," Mato said as he jumped up on the sofa.
"I know, and I'm sorry," Bryce said.
"Come on," Jerry announced. "It's time to start dinner, and you two are going to help." Jerry ushered the kids toward the kitchen. Bryce stayed where he was. John didn't move either, and once the others were gone, John's serious expression shifted to a wry smile.
"So you think Paytah is gorgeous," John said, and Bryce groaned. He knew he was going to pay for that single word. John chuckled softly. "Freudian slip," he teased. "You know there's nothing wrong with liking someone, and yes, Paytah is a good-looking guy." John smiled, and Bryce tried to smile as well. "I know you still miss and love Percy—you probably always will—but isn't it time to think about letting him go?"
"What's this got to do with him?" Bryce asked defensively.
"Come on," John said, lowering his gaze slightly. "You still carry a picture of him everywhere you go. There's one on your phone still, and I know if I went into your room, there would be one on the nightstand right now. You love him, I know that, but Percy is gone, and he'd want you to move on."
Bryce lowered his gaze. "I know. Percy would hate me for not moving on, but I.... It's not fair that I had him for so short a time. He and I deserved what you and Jerry have."
"I know you did. But if you let yourself move on, you could have that with someone else. Percy is gone, and carrying a torch for him the way you are isn't helping yourself." John settled back in his chair. "I'm not asking you to forget him or to stop loving him, but maybe make room in your heart for someone else too."
"And you think it's Paytah?" Bryce asked, even though he was teasing. There was no way, gorgeous or not, that he was going to let himself fall for the cantankerous man.
"I didn't say that, but may I remind you that you were the one who said he was gorgeous? And you're also the one who lets him get under your skin." John winked at him, and Bryce growled softly, grinding his teeth together. The man was way too observant sometimes.
"He's so exasperating," Bryce said.
John sighed softly. "He hasn't had an easy life, I know that. He and I were never particularly close, because Paytah never let anyone close. His brother Chayton was the only person I ever saw him really close with, and Chay has his own problems."
"They said something about getting him from Rainbow Bluff, whatever that is," Bryce said, and he watched John nod slowly.
"Chay has a problem with alcohol. A lot of people on the reservation do. When Chay drinks, he gets maudlin at first, and then thinks he's invincible. I heard that a few times he thought he could fly and had to be pulled back from the edge. Rainbow Bluff would be a place where Chay might go if he was feeling that way. There are sheer walls with many different layers of rock. It's very beautiful up there. If you were out of your head and thought you could fly, that place would be a very enticing location to give it a try." John looked away, like he was contemplating something. "Poor Paytah. He always seems to have the weight of the world on his shoulders." Then John seemed to shake off what he was thinking. "There are worse people you could like than Paytah, but I really don't know what his deal is."
"Deal?" Bryce asked.
"Yeah. I have never seen him go out with anyone, boy or girl, other than his brother and Coyote. I somehow doubt if Paytah has ever given any thought to who he might like, either."
"How can you say that?" Bryce asked, figuring John must be using some figure of speech.
"Paytah was a year ahead of me in school, and I always remember him being one of those guys who laughed a lot and had a ton of friends. Up until he turned fourteen or fifteen, he was the center of attention. You've seen how he looks. Well, imagine him at fourteen with those eyes and lips, skin that glowed, and a smile that lit up a room. He was gorgeous, and everyone seemed to want to be around him. Then, within a year, something changed. I don't know what, none of us did, but he withdrew. The friends sort of fell away, and he kept to himself. I have no idea what caused it, but a lot of people remarked on it when we were in school, mostly because no one could figure out why." John stood up and wandered to the blackened stone fireplace, cold because there was no need for a fire in this heat. "Paytah and Chay were raised mostly by their father, who ran the store before Paytah. He was an okay man, but he was one of those people who never seemed happy with whatever his kids did." John shook his head slightly. "Anyway, he died a year or so after Paytah got out of high school, and he took over the trading post. Chay helped sometimes at first, but he wanted more, so he left the reservation. Didn't stay away long, though. When Chay came back, he crawled into a bottle and seems to have pretty much stayed there."
"Good God," Bryce said softly. No wonder he seemed so dour. Bryce stared down at his shoes. Now he felt even worse for how he'd treated Paytah. "No wonder he feels like he can't do anything—he's just been trying to hold things together for so long." Bryce lifted his gaze with a soft sigh hopefully only he could hear. "Do you know what happened to him? Did he withdraw when his mother...." He wasn't sure what word to use. "Left?" he said, going for something safe.
John shook his head and wandered back to the chair, flopping back down into it. "No. His mother died when Paytah and Chay were little. They may remember her, but they would be old memories, and I doubt whatever happened to him had anything to do with her." John looked like he was searching his memory. "No, nothing comes to mind. But something definitely happened, and the only person who knows what it is would be Paytah."
Bryce was intrigued, and not just by what John had said. Like it or not, something about Paytah pulled at him and roused his curiosity. He wished he knew why. "Did you ever ask him?"
John widened his eyes. "Are you kidding? I heard that someone tried to talk with him about it once and he yelled at them and didn't allow them into the trading post for a month, and that was after telling them to mind their own business and threatening to get his gun. After that, I doubt anyone ever tried to bring the subject up again, and Paytah continued on the way he was." John became quiet for a while. "He's always been very Native-American-centric, even as a kid. He was exceedingly proud of the tribe and where he came from. But he got more and more vocal about it when he spoke, which wasn't much, and even before his father died, when he was working in the store, he refused to accept twenties. Once his dad died, it became the rule that everyone lived by if you wanted to do business with him, and since it's the only place on the reservation to get the basics, everyone does."
"I see," Bryce said, even though he really didn't.
"Actually, it's a good thing. It started off as quirky, but now every kid in school learns about Andrew Jackson and what he did to our people. There isn't a person here who doesn't know why Paytah doesn't take twenties, and many other people don't either now. It started off as a bit of weirdness on one person's part, but it's spread now and even raised awareness. So it's been a good thing."
"What do you think about it?" Bryce asked John.
"When I'm here I don't use twenties, but at home I do because I live in the real world. But I know plenty about Andrew Jackson and will tell people what he did. And that's all because of Paytah." John smiled, and Bryce did the same. "So you see, it isn't because he doesn't care. Like many of us, the problem is so large that we don't know where to begin, so we all do what we can." John stood up and walked to the large window that provided a great view of the black hills in the distance. "Look around, Bryce—this area was set aside as a reservation because no one else wanted it. There isn't anything around. The Black Hills are sacred to us, but most of the land was taken away because it contained gold. Our beliefs and generation upon generation of history didn't matter. Instead, they gave us this land because they didn't want it. There isn't much water except when it rains in the spring. The lakes that there are often dry up during the heat of the summer. The few streams that don't are used to the best of our ability, but without resources, there isn't much hope." The frustration in John's voice came through loud and clear. "The tribe has thought of turning to casino gambling like so many others have, but the land we have is out of the way. We tried to buy some land closer to Rapid City, but another tribe already has a casino there, and it doesn't make sense for there to be two, so we couldn't get a license."
"So how do people here survive?" Bryce asked. He'd already seen part of it, but he needed to know the entire picture.
"Some on government welfare. I'm sure that's where Little Wamblee's mother gets what little money she has. I know she works when she can, but there isn't much here, and with the kids, she doesn't have the ability to do what others have done—mainly what my dad does, work in the North Dakota oil and gas fields." John shook his head and continued looking out the window. "I wish he didn't have to do that, but it's the only way he, Mom, and the rest of the family can survive."
"Your dad is amazing," Bryce said. "After everything and all the time away, he still laughs and looks happy."
"He's making good money, and I know Mom is keeping tabs on every cent so he won't have to do this forever," John said. "But others don't have that luxury." John continued staring out the window and their conversation trailed off.
"Dinner is almost ready," Ichante said, hurrying excitedly into the room. "I did the cooking," she announced proudly, and John turned away from the window. Rushing across the room, he scooped her up. She squealed and laughed as he swung her around and into a huge hug. Mato came in to see what the ruckus was, and Bryce scooped him up to the same laughter. After what he and John had been talking about, they both probably needed to hear laughter and happiness—Bryce knew he certainly did.
"Let's go see what you made," John said as he carried Ichante into the kitchen.
"Daddy, Jerry called Grammy and Paw-Paw, and they are going to go with us to the water park tomorrow," Mato told John, and Bryce swung him around, to raucous laughter.
"Do you think your grandma and grandpa are going to go down the large water slide?" Bryce asked as he tickled Mato.
"No, but I am," he said, giggling.
"You are, are you?" Bryce teased, throwing Mato over his shoulder and carrying him into the kitchen.
"Yes." Mato laughed. "And so are you," he added.
"We'll see about that," Bryce said as he set Mato back on his feet. "I think I'll stick to the kiddie pools."
"Naw," Mato said as he pulled Bryce toward the table. "You're not a chicken."
"Is your grandpa a chicken?" Bryce teased.
"No, but he and grandma are old," Mato retorted.
"Don't let Grammy hear you say that," John warned as they all got the table ready for dinner. The laughter and joking extended through dinner, with all of them talking, telling jokes—mostly bad ones—and the kids having the time of their lives. Bryce couldn't remember the last time he'd laughed so much and had such a good time.
"What's got you so quiet?" John asked Bryce when the meal was over and the kids had followed Jerry to the other room.
"I don't know. We were laughing and talking all through dinner, and I was just thinking that I bet Paytah probably hasn't had a meal like that in a very long time."
John snorted slightly and then shook his head before turning away. Bryce thought he heard him mutter, "You got it bad," but he could have been wrong.
# Chapter 4
"WE'RE going to the water park!" Mato yelled through the cabin at dang near the crack of dawn. Bryce rolled over and groaned loudly. He knew the kids were excited, but the sun was barely up, and he certainly wasn't ready to get out of bed. He'd closed his eyes again when he heard both John and Jerry shushing the kids and getting them settled down. Of course, that only lasted for so long. Eventually his door opened and Mato rushed in and jumped on the bed.
"It's time to get up." Mato told him, and Bryce wondered which of his so-called friends had put Mato up to this.
"Mato, let Bryce get up and get dressed," John said from behind him, and he had his answer. Sometimes John was a big kid himself when he got excited about something.
"I'll be up in a few minutes," Bryce promised Mato, and he left the room, John closing the door behind them. Bryce got up and put on his robe before checking, and surprise, surprise, the bathroom was free. Bryce hurried and got his things together, but still had to race before Ichante got there first.
He cleaned up and quickly showered before dressing, careful to leave some hot water for the next person. Before leaving the bathroom, he made sure to clean up after himself and then hurried to his room. He finished dressing and joined the others in the living room. "My parents will be here at about ten," John said, and Bryce looked at the clock and groaned.
"It's not even eight," Bryce groused. Jerry came in from the kitchen and handed him a cup of coffee. Bryce sipped from the mug and felt the bitter, hot energy work its way into his veins. "Okay, I'm better now," Bryce said with a smile as he continued sipping. "Why did we have to get up this early?"
Jerry shook his head as he sat on the sofa. "We tried, but the kids have been up for hours already. They're so excited. We've been here working for three days, and I think they're getting a bit stir-crazy just playing games." Jerry lifted his mug.
"What are we having for breakfast?" Bryce asked.
Jerry peered over his mug at John, and Bryce wondered what they were up to. "I think the kids want doughnuts." Jerry sipped, and Bryce knew he was trying to keep from laughing.
"They do, huh?" Bryce asked, shifting his gaze to John, who tossed him the keys to the van and hid his eyes behind his own mug. Bryce caught the keys and stared at the two of them before finishing his coffee. Then he stood up. He placed his mug in the sink and headed to the door. As he pulled it open, he heard Jerry and John snicker. Bryce leaned so only they could see him, flipping both of them the bird. They laughed as Bryce shut the door.
He hurried to the van. After getting inside, he started the engine and headed down the foliage tunnel toward the road. Of course, this time he didn't get lost and reached the store in no time. He walked through the door, then stopped when he saw Paytah sitting behind the counter, working on his computer. "Are you doing okay? Any problems?"
Paytah looked up and glared at Bryce, but said nothing before returning to his work.
"I suppose I deserved that," Bryce said, leaning against the counter. Damn, why was it so hard to say he was sorry? "I shouldn't have taken my frustration out on you yesterday, and I'm sorry I called into question the way you fight for or care about the people here." Paytah stopped keying and turned to him. "I know you do everything you can to help everyone here." Bryce swallowed hard, waiting to see how Paytah would react, but he simply stared at him.
"You hurt me," Paytah said, his eyes blazing with something Bryce couldn't read. He knew it wasn't anger, but he couldn't figure it out.
"I didn't mean to," Bryce said. "I was upset about what happened earlier in the day and I took it out on you. I shouldn't have. I accused you of not helping, but I've been thinking and I can't come up with any answers either."
"Some questions don't have answers, no matter how much we might want them," Paytah said, and his expression softened. "Is that all you came in for?"
"Well," Bryce began, not sure what else to say. "Is your brother okay? He looked in pretty rough shape yesterday."
"Yes," Paytah answered curtly, glancing back to his computer screen. Bryce was tempted to get his doughnuts and leave, but he finally understood the look in Paytah's eyes—loneliness, deep and profound.
"Does that happen often?" Bryce asked, trying to sound sympathetic.
Paytah looked once again at the monitor, and Bryce figured he'd turn away and go back to work without answering. "More than it should," Paytah answered. "He drinks too much." Paytah looked around the empty store. "Many people do. They have nothing, so they drink to forget or to simply pass the time. No liquor is sold on the reservation, so they travel to get it. Either that or they distill it themselves." The sadness in Paytah's eyes hit Bryce hard. "There is nothing here. Jobs are scarce, so people take their monthly government checks and spend them trying to forget."
"That's what John told me, and I'm trying to understand."
"I don't know if you can. I was born here, and hunger, poverty, alcoholism, and even...." Paytah stopped and swallowed. For a second, the fear in his eyes was almost palpable. "I've been around this all my life and it isn't likely to change." Paytah turned away this time.
"I know you're probably right," Bryce said.
"No 'probably' about it," Paytah muttered before swiveling his chair toward Bryce. "What I don't understand is why you care. You're here for a few weeks and then you go home to your comfortable house in a nice neighborhood. You'll be back a few times to spend a few days teaching a class or two and then leave again. You don't have to live here and see the pain and hunger day after day. You don't have to live with children not getting enough to eat and stealing from their brothers in order to eat." Paytah was about ready to explode, Bryce could feel it.
"Do you think that just because I don't live here that I don't care about what's happening here? That Jerry doesn't? That's why he's trying to do something." Bryce looked at the computer screen. "The kids in the tribe can now be a part of the outside world. They can see that there's more than just what happens around them." Bryce took a deep breath. "Yes, I know they can't eat the computers, but hopefully that equipment will allow them to learn more, get a better education, and they can make something of themselves."
Paytah chuckled. "Eat the computers," he said, rolling his eyes slightly, then his expression turned serious again. "You do good work. You help people here. I don't mean to sound like I'm complaining, because I'm not. You're helping to teach people to fish and that is good."
Now Bryce chuckled. "That's what Jerry says too. He and John have gotten the donated pieces of equipment, refurbished them, and over the past few years, gotten enough for the school kids as well as other people in the tribe," Bryce said proudly, because he'd helped with those efforts... and so had Percy, but Bryce pushed that thought away.
"I know, you helped me too, remember?" Paytah said, and Bryce nodded. He did remember the time he'd spent close to Paytah, and he remembered helping him with his computer and the way his hair felt when it brushed against his arm. Bryce had sudden wicked thoughts surrounding Paytah's hair and just what he'd like to do with those long, silky strands. Instead, Bryce nodded and tried to return his mind to why he'd come. He had come here for something other than to talk to Paytah, but dang if he could remember.
"Oh, yeah, I need a dozen doughnuts. We and the Black Ravens are taking the kids to the water park in Rapid City. They're going a little stir-crazy at the cabin."
Paytah bagged up the doughnuts for him, and Bryce passed over a ten. Paytah was handing him the change when the door opened. "Morning, Paytah," a bright-faced woman said as she strode over to where they were standing and then went around the counter, shooing Paytah away. "It's Thursday, remember?" she said, and Paytah nodded. Bryce thought he was going to growl at her, but he walked around to Bryce's side of the counter. "I have things here. Don't worry. Go home and rest or something."
"Alowa, I'm fine," Paytah said but she ignored him.
Bryce figured that now was a good time to leave, so he grabbed his bag and headed to the door. He expected a fight from Paytah, but he followed him outside. Bryce walked to the van and got inside, watching through the windshield as Paytah stood in front of the store, looking this way and that like he wasn't quite sure what to do with himself. Bryce started the van and was about to put it in gear when he banged his hand on the wheel. "I know I'm going to regret this," he said out loud before turning off the engine and getting out of the van again. "Paytah, you'd be welcome to come and spend the day with us. We're going to the water park, and then we'll get dinner somewhere before coming back. We won't be too late because of the kids, but it should be fun."
Paytah looked like he was considering it.
"Go on and have some fun," Alowa said through the open door, closing it behind her. She eyed Paytah through the glass.
"It should be fun," Bryce offered, and Paytah nodded. "I'll wait while you get your things," Bryce said, and Paytah walked to the small house next door. Bryce knew he was definitely going to regret this, but he still couldn't keep his eyes off Paytah's backside, his gaze following him as he moved with a surprising grace.
"It's about time he did something other than sit in this store. I'm Alowa," the woman said as she walked up to the van.
"I'm Bryce," he said as he studied the attractive woman more closely. "I'm here with Akecheta Black Raven."
"I know. The rumor mill works at lightning speed here," Alowa said with a smile. "Thank you for offering to take him with you. I work in the store one day a week, and most of the time he sits around at home doing nothing. I'm supposed to give him a day off, but he rarely does anything." The door to Paytah's house closed with a bang, and Bryce shifted his gaze, watching as Paytah moved closer. When he turned back to Alowa, she had a knowing look on her face, and Bryce instantly schooled his expression. "Have fun, ya grump," Alowa told Paytah, nudging his shoulder, and Paytah grunted softly before getting into the van. She went back into the store. Bryce got in as well, starting the engine and the blessed air-conditioning before beginning the drive back to the cabin.
Paytah was no conversationalist, and that showed on the ride to the cabin. He sat in the passenger seat without saying a word.
"I'm not transporting you to your death or something," Bryce said.
"Sorry," Paytah said, but then he went quiet again. "Alowa has the hearing of a wolf and she can be really pushy."
"You're going to have a good time. The kids will have fun, and as hot as it's going to be, it will be nice to cool off." Bryce felt the prospect of some excitement starting to catch on with him. "I will warn you. The kids have been up since some ungodly hour, and they're wired." Bryce thought about the doughnuts resting on the console. He probably should have specified the ones without sugar, but it was too late now.
"They're good kids," Paytah said. "They are always well-behaved when they come in the store. Akecheta takes good care of them."
"He and Jerry love those kids to death," Bryce said, slowing down cautiously to take a nearly blind corner.
"They do not suffer because they are being raised by two men?" Paytah asked. Bryce glanced over at him, but he didn't see any judgment in Paytah's expression, only curiosity.
"No. Children respond to love, and those two are loved. John and Jerry fought very hard to get those children. You may have heard part of the story." Paytah nodded. "John adopted the children a while ago, and now they don't have any more issues with child services."
"Do they have any... problems because they're a gay couple?" Paytah hesitated as he asked his question.
"Yes, sometimes. It goes with the territory, I guess. But we have good friends, and most of the time, our families come to understand. Not always, but most of the time." Bryce continued driving down the narrow road. "My mother was supportive, but it still took her a while to get used to the idea. Jerry and John had a neighbor who tried to cause trouble, but he ended up being the one in trouble when the police came knocking on his door for making false accusations."
After another quiet minute or two, Bryce asked, "Do people on the reservation know about you?" He thought Paytah was going to choke. Paytah sputtered and then began to cough. Once Paytah got ahold of himself, Bryce began to chuckle slightly.
"Am I...." Paytah gasped, and then coughed again, appearing a little nervous.
"No," Bryce answered, anticipating the rest of Paytah's question. "No one would know to look at you. I got this feeling in my stomach when I met you, and my gut is rarely wrong." Bryce's chuckles slipped away as Paytah settled in the seat again. "I don't think other people in the tribe know unless you've told them. John didn't, but I think your friend Alowa knows, or at least suspects."
Paytah groaned loudly. "That woman has a huge mouth and talks all the time. My business will be all over the tribe before I get back."
"I don't think so. From the expression on her face, she's probably known for a while," Bryce said.
"You told her?" Paytah sounded almost panicked.
"Of course not, but she saw me watching you, and I know she knew," Bryce admitted, desperate to keep Paytah from panicking. The normally still man settled down some, but still fidgeted in his seat. Bryce breathed a sigh of relief when he turned into the long driveway.
"You were watching me?" Paytah asked as they passed through the tunnel of green.
"Duh," Bryce responded as he parked at the cabin. The kids ran out to meet him and pulled open the door.
"You got the doughnuts?" Mato asked, his excitement lasting until he saw Paytah. It was like a switch had turned him off.
"Yes, I have them," Bryce said, lifting out the bag and handing it to Mato, who walked quickly back toward the cabin, looking over his shoulder every few seconds until he disappeared inside. "Come on in. You can leave your things in here," Bryce said to Paytah before closing his door and leading the way inside.
John met them at the door, and he and Paytah clasped hand to wrist in a greeting Bryce had seen a few times before, then John motioned for Paytah to enter. The kids waited in the kitchen, already eating, while Jerry got out plates for the rest of them. "This is a surprise," Jerry said as he got an extra plate. "Would you like some coffee?"
"Yes, please," Paytah answered, and Jerry got him a mug and filled it from the pot. He also filled Bryce's mug for him.
"John's mother and father will be here in about half an hour. They are going to follow us in their car and they want the kids to ride with them," Jerry told them. "Glad you could join us, Paytah, it should be a lot of fun." Jerry smiled and filled his mug before placing the pot back on the coffeemaker. Bryce glanced at Paytah and then at his friends. He hadn't figured they would be upset that he'd asked Paytah, and they seemed to take it in stride.
"You two finish your doughnuts and then say thank you to Uncle Bryce before you get your things together to go swimming. I'll be in to check you have everything in a few minutes," John said, and Mato shoved the last of his doughnut into his mouth, chewing like a huge chipmunk. "Take it easy," John told him, and Mato drank from his glass of milk, then chewed awhile before swallowing.
"Thank you, Uncle Bryce," Mato said.
"Paytah made them," Bryce said, and, grinning, Mato looked up at the stoic man.
"Thank you, they were good," Mato added quickly before hurrying out of the room at a near run. Ichante finished her breakfast and milk and then she too was gone. The sound of their light bickering as they got ready drifted into the kitchen.
John ate his doughnut and then grabbed another one. "I better make sure they get all their things," John said before taking a bite of his second doughnut. "Glad you're coming with us," he added to Paytah before following the kids.
"I need to get my things together too," Bryce said, finishing his breakfast before rushing to his room. He'd brought a small bag with him and he filled it with what he thought he'd need. He was just finishing when the kids' yelling came through his door and Bryce knew their grandparents had arrived. Bryce got the last of his things together and joined the others. Of course, everyone had to eat before they left, but with the coaxing of grandchildren, it was surprising how quickly the older generation could move.
Soon, somehow, they had everyone and everything in the two vehicles. John started the van and got it turned around and then drove down the drive with the other car behind.
"How's business at the store?" John asked once they reached the main road north off the reservation.
"The usual. It doesn't change much," Paytah answered.
"I suppose you have to stay pretty close most of the time," Jerry said, and Paytah agreed, but didn't say much more. "Have you been to the park before?"
"No. I don't get off the reservation much except to pick up supplies and merchandise. I was only able to come today because Alowa is watching the store."
"I haven't seen her in a while. How is she doing?" John asked. "She and I were in the same class."
"She's fine. Pushy as ever," Paytah said, and Bryce had to keep himself from laughing. But he wasn't going to complain. She might have been pushy, but she was the reason Paytah had agreed to come, so he reminded himself to bring her something back. After all, Bryce had the sneaking suspicion that the highlight of the day wasn't going to be the water slides, but the sight of Paytah without a shirt on.
"She is that," John said with a small laugh. "But she was always a great person and fun to be around." John sped up when they reached the highway and continued heading north. "I thought if we had time on the way home, we could stop at Crazy Horse. The kids keep asking to see it again."
That was fine with Bryce, and Paytah nodded as well, but Jerry shook his head.
"They close at five during the week. We'll have to take the kids another day. I know they want to see it, but after we get done at the park, we'll need dinner, and on the way home they'll both probably fall asleep anyway. We can probably go on Saturday, stop there, maybe take the kids to the caves and see the buffalo herd."
Bryce smiled because all that stuff sounded like fun to him.
"Have you seen Crazy Horse lately?" Bryce asked Paytah, who shook his head.
"My dad used to take us to see it once a year. He said it was something we all should be proud of," Paytah said. "I haven't seen it since he died." The conversation in the car quieted after that. Bryce rode and watched the scenery out the window. The highway made travel faster, and in under an hour John pulled the van into the parking lot of the water park. John's parents parked next to them, and the kids were out of the car almost before it stopped.
They all got their things, and Bryce thought John and Jerry looked like pack mules as they carried everything while Mato and Ichante ran ahead, occasionally stopping to peer through the fence into the happy water wonderland beyond. "Can we go in now?" they said in unison.
"Yes," John said as Jerry paid the admission, and they went inside. Bryce went up the window and paid for him and Paytah. He got a stern look in return.
"I invited you to come," Bryce said and turned away, signing the credit card slip, and then they went inside.
The entire place was one screaming child after another, presided over by a huge water slide and filled with pool after pool with all sorts of fun stuff to do. "I found a quieter place over here," Jerry said as he met them and guided them under an awning, where the grandparents were already reclining on loungers. "John took the kids to change."
Bryce grabbed his bag and motioned Paytah toward the changing rooms. He picked up his stuff as well, and they made their way through the throngs of running, playing kids. The changing rooms were private, and Bryce got into his bathing suit, then slipped a T-shirt on. He dug out his flip-flops and placed them on his feet before folding his clothes and placing everything in his bag. He rolled up his towel and looped it over his shoulders before heading back out.
Bryce waited outside the door to the changing area, watching as people came in and out. He felt a bit like a roadway median for a few seconds, and then Paytah came out of the changing room and Bryce forgot about everything else. Bryce had imagined, especially last night, what Paytah would look like out of the baggy clothes he tended to wear, and during the drive, he'd watched and wondered what Paytah would look like in only a bathing suit, but the reality was so much better than what he'd imagined. He wasn't cut or anything, but he was lean, with smooth, flawless, reddish-brown skin and muscles that flowed under his skin rather than bulging. He'd gathered his flowing hair into a ponytail and....
"Is everything all right?" Paytah asked, looking down at himself. "You were staring."
Bryce nodded slowly. "Everything is fine," he said. "You look just fine." Bryce thought he saw Paytah color under his darker skin. He tore his eyes away from Paytah and cleared his throat. "Let's put our things by the chairs," Bryce suggested.
John's parents agreed to watch their stuff. They didn't seem to be ready to get wet. "We will later," Kiya explained. "Right now, we're relaxing." She looked at her husband with a soft smile, and he gently took her hand.
"Uncle Bryce," the kids called as they ran up, dripping water everywhere. "Come with us on the big slide," Mato said, pointing to the top of a wooden tower with stairs. "Please," Mato added, taking his hand.
"Are you big enough?" Bryce asked.
"I can go on the twisty one," he clarified. Bryce glanced at Paytah, who had an indulgent look on his face. Bryce followed Mato, and Paytah got in line behind him, with Ichante quickly joining them. The line went halfway down the stairs, and they found the correct one and waited. Water dripped from the steps above them, but no one seemed to care. It was already hot and getting hotter, so the water felt good.
"Are you an Indian chief?" the kid behind them asked Paytah. He must have been about six years old, and his eyes were as wide as saucers. Paytah didn't seem to know how to answer the kid.
"No, he's not a chief—he's a warrior," Bryce said, and Paytah became very serious, and the little kid stepped back. Then Paytah smiled, and the kid relaxed and smiled back. The line moved, and they began climbing the stairs. The little kid behind them kept watching Paytah and smiling at him whenever Paytah turned around.
Slowly, they climbed higher and higher, the park laid out below them like a water-filled map complete with lakes, a river, islands, boats, and rafts. Bryce wasn't so sure how much he liked this, but he'd promised the kids, so he tried his best not to look down.
"Don't like heights?" Paytah whispered from behind him, and Bryce nodded.
"I'll be okay," he said and continued to look up to where they were going instead of down. They finally reached the top. Bryce looked around, closed his eyes, and clutched Paytah's arm. "Give me a minute."
"Uncle Bryce, look, there's Grammy and Paw-Paw," Ichante called, waving down at her grandparents, who waved back.
"If you want to go down, it's okay," Paytah said, and Bryce shook his head, forcing his attention to getting Mato into the slide. When his turn came, Mato cheered and disappeared into the slide. He emerged at the bottom, jumping out into the water. Ichante was next, and she slid down as well. Then it was Bryce's turn. He got into position and waited until the man told him to go. Bryce closed his eyes and launched himself into the tube.
After a few seconds, his fear subsided and he started having fun. Bryce twisted and turned, the water zooming him downward, and he quickly found that if he arched his back, he went faster, which added to the fun. At the end of the ride, he zoomed out of the tube into a pool of water.
"Let's do it again," the kids said as he stepped out of the pool. Bryce wasn't ready for that yet. Paytah came down next, splashing into the pool. Bryce couldn't stop a smile when Paytah stood up, water cascading down his skin. Damn, it had been a long time since he'd seen anyone hotter than Paytah. "Let's do it again."
"Go see if John or Jerry wants to go down with you," Bryce said, and the kids took off. Thank goodness for parents. "I think I saw something quieter back here," Bryce said, and Paytah led the way. They got in line for the Lazy River, and after picking up rafts, they joined the other adults, who were floating down a curving loop of slow-moving water. Paytah held onto Bryce's raft and they floated together.
"This is nice," Paytah said, and Bryce hummed his agreement, relaxing in the refreshing water. They floated in silence for a while, which didn't surprise Bryce. Paytah seemed to be a man of few words. "Did you have a boyfriend before?"
Bryce lifted his head, almost startled that Paytah had asked him a question. "Yes. His name was Percy. We were together for almost three years and were going to get married, but he died in an accident." Bryce waited for the waves of sadness that always came when he talked about Percy, but nothing happened. "He was a nurse and had gone to New York for some special training. There was an accident at the airport and he was hit by a luggage truck. The driver was drunk on the job. Have you had anyone?"
"Once," Paytah said. "A long time ago, but we were kids." Paytah fell silent again, and Bryce hoped he would continue, but he didn't.
"What happened?" Bryce asked as they passed under one of the light waterfalls.
"We were kids. He was nice and we spent a lot of time together. Eventually we started messing around. I thought there was more to what we were doing than he did, and when we were almost caught once, he backed away. His father was on the council, and after about a year, he went to college and I was left on the reservation." Paytah scoffed lightly. "Once he left, he never came back for other than short visits. He stopped in the store a few times, but acted like nothing happened between us. Last I heard he got married." Paytah shrugged and lay back in his float.
"The spineless bastard," Bryce said softly. "I knew a guy like that when I was in school, too. I think we all meet the same guy at one time or other. Mine is a minister, married, with two kids. I saw him once at one of the bars in Sioux Falls. He saw me, and I made sure he knew I recognized him. After that, he took off, but I wouldn't be surprised to find him hanging around rest area bathrooms or something." Bryce lowered his voice when some kids paddled closer and then went past them, laughing and having a good time. "After that I dated a number of guys, none of them for very long, until I met Percy. He was special," Bryce said, breathing evenly through his mouth.
"It must be hard to lose someone like that," Paytah said.
"It was." Bryce paddled slightly to move away from the wall. "My mother was diagnosed with cancer a short time after Percy's death, so I had plenty to worry about. She's doing okay now, but we both know it's a matter of time. She's in remission, but no one can say how long it will last. Hopefully quite a while, but you never know." Bryce moved toward the edge of the river, positioning them toward the exit. This was too slow and it was giving him way too much time to think and talk about things that would be better talked about while everyone around them wasn't having fun. They got out of the current and headed up the exit, then stacked their floats with the others.
They made their way to one of the larger pools with pads floating on the surface and ropes strung over it. Bryce was trying to figure the best way across when Paytah darted over the floats like he was walking on water. The kids in the pool watched as Paytah did it again. "How'd you do that?" a boy asked with a touch of awe.
"Speed and balance," Paytah said with a smile. Bryce tried it multiple times and fell in the water after two steps. Some of the kids tried as well, but the only one who ever made it across was Paytah. "I'm going to go down the fast slides," Paytah told him after a while, and Bryce nodded. There was no way he was getting back to the top of that platform, so he found Jerry, John, and the kids playing together near where the grandparents watched.
There was an empty lounger in the shade next to Kiya, and Bryce sat down, instantly cooling once he was out of the sun. "You better put on sunblock," Kiya warned, and Bryce fished it out of his bag and slathered it on himself.
"I'm going to change," John's dad said, and his mother handed him the bag.
"I take it you like Paytah," Kiya said once they were alone, or as alone as they were going to get. "He's a nice young man. Had a hard life, though."
"He is nice in an intense way," Bryce agreed, getting a laugh in return
"That's one way to put it," she said with a smile that lasted for a few seconds.
"John told me that he used to be different—happier," Bryce said, and Kiya nodded her agreement. "He said he always figured something happened to him, but he doesn't know what it is," Bryce said, and from the dark look in her eyes, Bryce figured she either knew or had a pretty good idea.
"There were rumors some time ago, but that's all I know," she said as she sat up and adjusted the back of her lounger. "So I don't really know what happened, but... it isn't my story to tell even if I knew for sure. For what it's worth, I agree with John. I believe something did happen, but what it was, I—" She seemed to be searching for the words. "I can't really say, because at the time it was a lot of people taking a few facts and filling in the blanks." She was clearly uncomfortable with the entire subject. "You should ask him." Bryce knew it wouldn't do any good to press, so he finished putting on the sunscreen and then lay back on the lounger, closing his eyes for a few minutes.
"Paytah's about to go down the big slide," Mato said from next to him. Bryce opened his eyes and sat up, shifting so he could see the top of the platform. It was definitely Paytah. Bryce would know those wide shoulders and bearing anywhere now—they were etched on his brain. Bryce stood up and waited as Paytah got into position and then sluiced down the slide. He reached the pool at the bottom in a matter of seconds, sending a wall of water ahead of him as he came to a stop. "I wanna do that," Mato said, "but I'm not big enough."
"You will be," his grandfather said from behind Mato before scooping him off his feet. The youngster giggled and laughed as his grandfather carried him away before jumping into the nearest pool with him.
"Those are lucky kids," Paytah said softly as he walked up, and Bryce shook off the thought that all kids should have that.
They played in the water for another hour, and then Jerry gathered the kids and everyone headed to the snack bar for a small lunch to tide them over until a nicer dinner. John made the kids sit still for a while and then he cut them loose in the park.
"Shall we swim for a while?" Bryce asked Paytah, and they headed for one of the larger pools unadorned with play items. They dove in the deeper end and swam around for a while until the water began to roll around them, the waves becoming larger and larger. They were joined for a while by other people intent on enjoying the waves. Once the timer ran out and the water settled, they found themselves largely alone again. "Do you do many things other than run the store?" Bryce asked. "I mean, I know you don't get out much, but do you have any hobbies?"
"Not so much lately, but when my dad was alive, I used to work with wood. I'd hunt for branches and bits of log that I would carve into small animals. I even made a few pipes and things, but since Dad died, I haven't done much of it. There isn't the time, and I don't feel much like it any longer." Paytah seemed to look beyond the immediate area, like he was looking back through time. "Chay and I used to do a lot of things together when Dad was still around, but once he died, Chay began pulling away, and I had to take over more and more of the duties of the store." Paytah kicked his legs in the water and then settled back against the side of the pool. "I didn't realize at first that Chay had pulled away because he was drinking, but it soon became apparent."
"How bad is it?" Bryce asked.
"He's an alcoholic, without a doubt. He can't seem to make it through a single day without drinking and regularly empties a bottle with the intent of getting drunk." Paytah rested his head against the side of the pool. He looked weary and tired. "I don't know what to do. I know I should try to do something, but it never does any good. Last year he nearly got himself killed when he ran his car off the road. He promised that he was going to give up drinking, and for a few weeks he was the brother I remembered, but it didn't last. I should have known it couldn't. So I really don't have time for hobbies or much of anything other than somehow making enough in the store to keep myself and Chay fed. I only give him enough to buy food, but he always seems to have enough money to buy alcohol. I really don't know where he gets it and I don't think I want to. Someday I know someone is going to come in the store and tell me he's dead." Paytah pushed himself away from the side of the pool, gliding through the water, and Bryce stared at him, wondering how he could take something like that so lightly.
Bryce realized that Paytah probably wasn't taking it lightly, but he just couldn't do anything about it. Deciding that Paytah had probably told him things he rarely spoke about, Bryce smiled and pushed off the side, following Paytah across the pool.
They stayed at the water park for a few more hours and then gathered all their stuff and the kids to get ready to leave. Bryce and Paytah helped Jerry and John pack everything in the van, and once everyone was dry and in the cars, they headed out. It was a little early for dinner, so Jerry drove to the mall and they wandered around for a while. Bryce had promised his mother that he'd bring her something back, so he stopped in one of the gift shops and looked around. Paytah followed him in. After a while, Bryce found Paytah standing in front of a display of tacky totem poles and baskets. They were made in China and looked cheap and as fake as they actually were. "I hate things like this," Paytah said. "My people have a noble history and culture, and people manufacture this crap to sell."
"I know," Bryce said, gently removing the brightly colored peace pipe with fake feathers from Paytah's hand and putting it back on the shelf. "John said there was a place where I could buy some handcrafts there were real, but I haven't been there yet."
Paytah chuckled, turning away from the cheap souvenirs. "You've driven past it every time you came to the store. I'll show you when we get back to the reservation. It's the missionary place. They sell some nice artist-made items, but they're too expensive for most people on the reservation, so mostly they sit there."
Bryce looked around the store again as the ghost of an idea began to form. "Can I help you?" a clerk asked as she approached them.
"No, thank you," Bryce said, heading for the exit. He wanted to get out of this place almost as badly as he figured Paytah did. "Are there many artists on the reservation?" Bryce asked as they hurried away from the tackiness as if it might follow them down the mall if they went too slowly.
"Some. In the past ten years or so, some people have returned to the reservation for the quiet to work. A few sell their work in places like New York or Houston. Why?" Paytah asked, his curiosity obviously piqued.
"I have what might be the start of an idea and I need some advice," Bryce said absently, his mind already turning. "One of those artists might be good to talk to. Could you help me?"
"I'll ask when I see one of them," Paytah said, and Bryce nodded, still mulling things over. Bryce saw Paytah scowl slightly. "Is this some harebrained idea?"
"No," Bryce answered quickly, his defenses going up. "I don't know what it is, but if it's harebrained, I won't pursue it." Bryce walked faster, his anger propelling him down the mall hallway. Then he stopped and whirled around. "If you don't want to help, then say so."
Paytah continued walking closer at his usual pace. "Sorry," he said cautiously.
"Okay," Bryce said with suspicion.
"We once had a guy who got this idea that we could mine the iron that colors part of Rainbow Gorge. He wanted us to destroy a place that has been sacred to our people for hundreds of years. He said we could make a lot of money, and the council actually considered it." Paytah hissed between his teeth. "We get crap like that all the time, and while I'm suspicious, I shouldn't have judged an idea I haven't heard."
"I would never do anything that would hurt anyone," Bryce said, and Paytah seemed to think for a few seconds before nodding slowly. "And besides, if this works, you'll be involved."
"Me?" Paytah said surprised.
"Yes, you," Bryce said, and the more they talked, the more his idea crystalized in his mind. "Let's go find the others. I think I've had enough of this place," Bryce added as they walked past the tacky souvenir store again. They caught up with the others near the food court, where the kids were asking for smoothies.
"We're going to get dinner," John said calmly, lifting a whimpering Mato into his arms.
"They won't have smoothies," he said, using logic as only a seven-year-old can.
"You're tired and you've done a lot today. Grandpa bought you french fries and ice cream. You had hot dogs and whatever else you got your grandma to buy you. I think you'll live without a smoothie."
"But we can't get them at home," Mato said, lifting his head off John's shoulder.
"Yes, you can," Bryce said. "I make great smoothies. When we get back to the reservation, I'll stop at Paytah's store and get the stuff. Tomorrow I'll make you my extra special smoothies," Bryce promised.
"You're sure?" Mato begged.
"Yes, I'm sure," Bryce said, and that seemed to end the smoothie crisis. "We do have a blender in the cabin, right?"
John shook his head.
Twenty minutes later they left the mall, Bryce carrying a blender. That's what he got for opening his big mouth. "Smoothies tomorrow, right, Uncle Bryce," Mato said as he took the bag, hugging the blender box to him as he walked. At the cars, they stowed everything and got everyone inside and seated before driving to the restaurant. John's parents had chosen it, and Bryce had expected a buffet or family restaurant, but from the look of the place, they were all vastly underdressed. They parked and got out, and Bryce looked at the building, wondering if they were in the right place. Other people arrived who weren't dressed much better than they were, so Bryce followed the rest of their party inside.
The softly lit dining room was surprisingly quiet. "I'm so sorry," the hostess said to John's father as she looked over the reservation book. "If you'd like to wait about half an hour, we'll have a table large enough for your party. Otherwise, we have a table for six right now." She was visibly distressed.
"That's fine. John and I can sit at a separate table," Jerry said, and Bryce saw John nudge him in the side. The hostess looked visibly relieved and led them all through the dining room. Somehow, Bryce wasn't quite sure how, but he and Paytah ended up at their own table a slight distance from the others.
"This is nice," Paytah said quietly after they'd sat down. He also looked a little nervous. Bryce nodded slowly and opened the menu, instantly seeing the source of Paytah's worry. From what he'd seen, the food looked really good, but the prices on the menu were probably much more than Paytah was used to.
"It really is," Bryce said, setting aside his menu. "I'm glad I'm here with you." The day was beginning to feel like a date, and he hoped he wasn't the only one who felt that way. His question was answered when he felt Paytah lightly touch his hand. Bryce smiled and turned his palm up, easily sliding their fingers together.
"They're holding hands," Ichante stage-whispered to John.
"Shhh," John said. "You mind your own business." Ichante quieted down, and Bryce turned his attention back to where Paytah's hand touched his. Bryce had expected Paytah's hands to be sort of rough, but they were as smooth as the rest of him. Bryce closed his eyes when Paytah circled his thumb lightly over Bryce's palm.
"Good evening," the waiter said, and Paytah pulled his hand away and picked up the menu. Bryce turned toward the waiter, expecting shock or something worse, but instead saw a slight, knowing smile. "Would you like something from the bar?"
"Just a diet soda," Paytah said, and Bryce made it two. It didn't surprise him that Paytah didn't drink, and while Bryce might ordinarily have had a beer, he certainly didn't need one. The server left and Bryce chuckled softly. "What's so funny?" Paytah whispered.
"You don't need to be self-conscious," Bryce said.
"But the waiter, he saw us," Paytah whispered back.
"Yeah, and he probably wishes he was the one sitting here, getting to hold your hand," Bryce said softly, and Paytah glanced to where the waiter was already returning with their drinks. "Yes, really," Bryce said, easily reading Paytah's expression. "Just relax."
The waiter brought the drinks for the other table as well as theirs and then began to take orders. Paytah looked over the menu, and Bryce could almost see him figuring the prices. "Don't worry, dinner is on me." Paytah shook his head. "You can pay the next time." Bryce flashed a quick smile, and Paytah looked a bit shocked.
"You want there to be a next time?" Paytah asked. "Next time for what?" Paytah seemed a bit confused, and Bryce held his gaze until Paytah's mouth formed an "O." The conversation from their friends filtered over, and Bryce turned toward the other table, seeing them all looking at them.
"What?" he asked, lifting his eyebrows, and they returned to their own conversation. "We seem to be a source of amusement for everyone else."
"I guess so," Paytah said before glaring at John and his parents. They made a show of looking away, and Paytah turned back to Bryce, flashing him a wicked smile. Instantly Paytah's mask seemed to fall away before Bryce's eyes.
"I knew it," Bryce whispered. "I knew that grumpiness of yours was just an act to get people to leave you alone." Bryce leaned back in his chair.
"It's no act," Paytah said. "I am grumpy and grouchy." He scowled, and Bryce began to chuckle. Now that he'd seen below the surface, the scowl didn't seem so menacing; rather, it was almost comical, like one of those theater masks that exaggerated the emotion to an extreme.
"Not deep down," Bryce countered. Their salads arrived, interrupting their discussion. They ate and sometimes joined in the conversation with the other table.
"What do you mean, 'not deep down'?" Paytah asked after the server had cleared their salad plates. "You act like there's another part of me somewhere that I don't know about. I'm just me."
"There is another part of you," Bryce said. "Just like there's a part of me not everybody sees. We're all like that. Most of us try to put our best foot forward when we meet new people because we want them to like us. You act like a grump, but I don't think it's because you want people to hate you, but because you want them not to get too close."
"Thank you for that analysis," Paytah said rather sharply.
"See?" Bryce figured he would press a little further. "I started to see who you are and you pulled away." He reached across the table, taking Paytah's hand lightly. "You don't have to hide from me. I'm not out to hurt you and I don't want anything from you." Bryce could almost see the denial forming in Paytah's mind, but he didn't say anything. Instead, Paytah narrowed his eyebrows and tried to look intimidating.
Mato chose that second to look over from the other table. "Uncle Bryce, why is Paytah making faces at you?" That did it. Paytah broke into a smile, and Bryce knew for sure that he'd been right.
"He's just playing," Bryce said, and Mato put his fingers in his mouth, pulling his cheeks wide as he stuck out his tongue.
"Stop that," John scolded mildly.
"But I wanna make faces too," Mato said, turning back to the table.
"You can make all the faces you want in the car on the way home," John said, glaring at Bryce, who looked away, trying not to laugh. He definitely needed to remember that little ears pick up everything.
"What makes you think I'm different deep down?" Paytah asked, leaning slightly across the table.
"The way you smile when you don't think anyone's looking," Bryce told him. "John said you used to be different when you were in school, happier, more outgoing...."
"I don't want to talk about it," Paytah said flatly.
What everyone had told Bryce was correct. Something had happened. Bryce's curiosity was definitely piqued, but now was not the time to pursue it. Maybe Paytah would never want to talk about it. "Didn't say you had to. I'm just saying that person is still there, and he's really rather handsome and attractive when he makes an appearance." That got him a smile that Paytah tried to hide, but the smile broke through anyway. "You have dimples," Bryce teased.
"I do not."
"Yes, you do," Bryce said. "You definitely have smile dimples." Bryce wondered for a split second if Paytah had dimples in other places. That thought sent an image racing through his mind. Bryce tried to stop it, but it was too late. He was happy he had his napkin on his lap. Paytah definitely had hip dimples; Bryce had seen them peeking out of his bathing suit, and in his mind, as the imaginary Paytah turned around, there were butt dimples as well.
"Sorry," Bryce said as Ichante's voice from the other table broke through his imagination.
"Are you going to go with us when we see Crazy Horse?" Ichante asked.
"Of course. I've never seen it," he answered. "I'm looking forward to it." Bryce was about to ask if they could see Mount Rushmore as well, but then he remembered that was a sore subject. All the Black Hills were considered sacred by many Native tribes. He'd ask about it later. "Papa Jerry says we're also going to go to a cave and see buffalo." She sounded so excited.
"I heard there's also a place nearby where you can go horseback riding," Bryce said, and Ichante turned to Jerry.
"We'll see," Jerry said. The server refilled their glasses and then left before returning with their meals. He set the impressive plates, both in arrangement and portion size, in front of each of them. The conversation died down as everyone began to eat.
"Are there buffalo on the reservation?" Bryce asked Paytah.
"Not really. The herd is largely at Custer State Park."
"Is there land for buffalo?" he asked.
"Yes," Kiya answered. "The southwest portion of the reservation had buffalo many years ago, but they were largely poached off by hunters. Why?"
"Just curious. I thought it would be cool if you had some, that's all. Your ancestors lived off buffalo, so I thought it would be fitting if there were some on the reservation." Bryce turned his attention to his dinner, and the conversation fell off slightly. Bryce's mind ran a mile a minute and he had to try to stop the thoughts that all buzzed through his head at the same time. He didn't have much luck until he concentrated on Paytah. Then his mind cleared and he focused only on him.
"Do you always do that?" Paytah asked. "You seem to get these ideas all of a sudden."
Bryce shrugged. "Sometimes," he answered after swallowing. "My mind sometimes goes into overdrive and tons of ideas come to me. Most of them are crap and I let them go, but sometimes I have a good one. I mean, it would be cool if there were buffalo on the reservation. Cattle would really suck, but I know there are places where buffalo are raised in a semiwild setting. They live on the range, but they're managed like cattle—tracked, cared for, and eventually used for meat. There was buffalo on the menu, and that wasn't from some guy going out to kill an animal somewhere—it was raised on a sort of farm. I was curious if there was land and if anyone had ever thought of reintroducing them on the reservation and using them as a source of money." Bryce took a bite of his beef before turning toward the other table. "You have land, and while I'm not sure if there's enough water now, there must have been at one time if buffalo were indigenous." Bryce shrugged again, but he saw that John's parents and Paytah were listening. "I don't know if anyone has looked into it, but it might be a way to combine your heritage and the land the tribe owns, and allow the community to make some money at it without destroying the land. You'd be bringing back an animal that once lived there."
John's father set down his fork. "Where would you get the animals?" he asked, and John nodded.
"I read online once that the state park manages their herd and needs to keep it at a certain number," Bryce said. "Petition the state to see if you couldn't arrange to have the overflow. You might also be able to petition the federal government for some animals. There are also farms where they raise them and you might be able to buy some animals, but that would be more difficult." Sometimes his ability to retain useless information came in handy.
John's dad looked around the table, clearly interested. "I don't know if anyone has thought of that, but it might be worth bringing up," he said with a smile, and Bryce returned it before turning his attention to Paytah, who was smiling as well.
"Now that was worth it," Bryce said.
"What was?" Paytah asked.
"Seeing you smile," Bryce said, and Paytah turned away. "Don't. I like it when you smile. You're incredibly handsome when you smile." Bryce took Paytah's hand for a few seconds, squeezing it before releasing it so they could eat.
"What other ideas do you have?" John's dad asked, and Bryce shrugged.
"I'll let you know tomorrow," he quipped, and everyone laughed before returning to their meals.
The kids were definitely winding down by the time they were done eating. The server asked if they wanted dessert, but everyone declined. Mato sat on Jerry's lap, half asleep, and Ichante looked about ready to fall asleep in her chair. The adults finished their coffee, and the server brought the check. John's father took it and would brook no argument. "I don't get to spend time with my grandchildren that often," he said, placing his credit card in the holder before the server took it. Thank-yous were said all around from both tables, and once the bill was signed, they got up to leave. The kids were placed in the backseat of their grandparent's car, and Bryce figured they would probably both be asleep minutes after they pulled out of the parking lot. Bryce and Paytah got in the backseat of the van, and he found his eyes closing as they rode back south toward the reservation.
They rode for a while, the roads becoming quieter, traffic thinning until it seemed like it was only them. They entered the reservation and drove to Paytah's, parking outside his house. He got out and said good night, then walked away toward his front door.
"Go on. We'll wait a few minutes," Jerry told him, and Bryce got out of the backseat and hurried to Paytah's closed door. He knocked softly, and when the door opened, Paytah stared at him, obviously puzzled.
"Did I forget something?" Paytah asked.
"Yes, you forgot this," Bryce said. Standing in his tiptoes, he moved closer to Paytah, put an arm around his shoulders and brought their lips together. At first, Paytah didn't move, but after a few seconds, he kissed Bryce back. Paytah tasted of woods and the outdoors, clean and fresh, regardless of how late it was in the day. Bryce broke the kiss, settling back on his feet. "I'll see you soon," he said, and Paytah nodded in the light that came out of the open doorway.
Bryce backed away and then turned to walk back toward the van. He knew they'd most likely had an audience for their kiss, but Bryce didn't care. He barely felt the ground under his feet the entire way back to the van. He pulled open the door and got inside, seeing Paytah still standing in the doorway. Bryce rolled down his window and waved good-bye as John started the engine. Slowly, they backed up, and Bryce continued watching Paytah's doorway until they rounded the corner. Then Bryce turned around and stared straight ahead, lightly licking his lips in the hopes of capturing some last lingering taste of him.
John and Jerry talked quietly about mundane things, like getting the kids in bed, but Bryce ignored them, turning over that kiss in his mind.
# Chapter 5
OVER the course of the next few days, Bryce, John, and Jerry worked almost nonstop. The kids were spending a few days at their grandparents, so everything was quiet at the cabin. Their day off at the water park had been great, but they still had work to do, so they spent some time catching up. Bryce managed to finish a website he'd been developing for a client. He set it up for testing and turned it over to the customer for review almost a week early. Jerry was thrilled. Bryce was thrilled as well. "You're on fire," Jerry told him. "I bet I know what's got you so riled up."
Bryce felt himself blush. "It was just a kiss," he said.
"I know, but you haven't stopped thinking about him, have you?" Jerry asked with a wicked smile. "You know, it's okay," he added seriously. "There's nothing wrong with developing feelings for someone else. Percy would want you to move on."
"I know," Bryce said. "But I'm wondering if I'm ready. I still miss Percy a great deal. Am I being fair to Paytah or anyone if I still feel that way?"
"You'll always miss Percy," John said, and Bryce shifted his gaze to him. "You were too happy and too much in love to stop missing him. He was too important to you. Missing him doesn't mean you aren't ready. I guess the question is, what does your heart want? For a long time, you only wanted to mourn Percy, but I think that time has passed and your heart is ready to move on."
Bryce nodded. "I think so, but what if I'm wrong?"
"Your heart will never lead you wrong," John told him, returning to his computer. Bryce turned to Jerry, who nodded his agreement, and then he too went back to work. "Sometimes you think too much," John added from behind his computer, and Bryce tossed one of the old sofa pillows at John, hitting him on the side of the head. John tossed it back, and then they all got back to work.
"John's parents are bringing the kids back this afternoon, so let's get done what we can. Then tomorrow we'll take them to see Crazy Horse, the caves, and buffalo," Jerry encouraged. "There isn't a lot we need to finish, but let's get it done so we can enjoy ourselves." They all quieted down and the room filled with the soft clicking sounds from keyboards. Bryce's mind sank into the code, and within a few hours, he was deep inside the program, the code singing to him. Jerry and John moved around sometimes, but he barely heard them. Everything was simply clicking into place. It was Bryce's stomach that finally intruded.
Blinking a few times, he looked away from his screen and checked on the time. "There's a sandwich on the counter for you," John said. "You were really into what you were doing, and we didn't want to disturb you." All three of them had experienced times like that and they all knew to respect the zone. Bryce stood up, his legs a bit stiff, and slowly headed for the kitchen. He got a soda from the refrigerator and ate his sandwich quickly before returning to work. He had just a few more hours of work and then he'd be done. Bryce knew Jerry would look it over before sending it out, but everything was running smoothly and he didn't want to stop.
Bryce worked the rest of the afternoon and he indeed finished drafting the program. He tested it and then set his laptop aside. He'd already stared at the screen enough, and he definitely needed a change of scenery.
"We're going to take a ride to town. Do you want to ride along?" Jerry asked.
"God, yes," Bryce said, hurrying to his room to get his shoes. Once he had them on, he went out to the van and got in the backseat.
The now familiar ride didn't take long, and once they reached the reservation center, Bryce hurried to the trading post. Paytah was helping a customer with a special order, so Bryce wandered through the small store, picking up a few items he needed. There were others doing their shopping, so Bryce hung back until Paytah was done. "Hey," Bryce said with a smile as he approached the counter.
"Hey, yourself," Paytah said rather gruffly. "You didn't call and haven't been in."
"Sorry," Bryce said softly. "We've been working for days to get some projects completed. I wasn't ignoring you." Bryce leaned slightly over the counter. "I thought about you a lot, actually."
"You did?" Paytah asked cautiously.
"Yeah. I should have come in to see you, but we were under a deadline and the cabin was quiet, so we all wanted to make the most of it," Bryce explained, wishing now he had called. "The kids come back today, and tomorrow we're doing some of the touristy things. I was wondering if you'd like to come with us. Sort of a second date."
Paytah smiled, but it quickly faded. "I don't think I can," Paytah said. "I don't have anyone to watch the store. I'd like to, though." Bryce nodded slowly. He'd known it was a long shot. "I'll have to see," Paytah added before rummaging around behind the counter. "I put it right...." He came up and handed Bryce a business card. "Running Deer is one of the artists on the reservation. He does a number of things, including pottery and some woodcraft. I wasn't sure if you were still interested, but when he came in yesterday, I got his card, and he said to call—any time after five or six is best."
"Thank you," Bryce said, taking the card. He still wasn't sure his idea would work, but the more he thought about it, the more positive he became. The part he wasn't sure about was how he could really get it started. "I can't stay very long now because I rode in with Jerry and John," Bryce said, "but I promise the next time I see you I'll explain my entire idea."
Paytah was about to answer when the door opened. Bryce turned along with Paytah as an older white man swaggered in like he owned the place. "Hi, Peter, how are things going?" he asked. Paytah nodded without taking his eyes off the man. The door opened again, and a boy of about eleven or twelve hurried up behind the man, a look of awe on his young face. "Go on and pick out whatever ice cream you want." The boy couldn't seem to believe his luck, and the man repeated himself with a smile. The kid rushed through the store to the frozen food case. The silver-haired man walked over to the counter and leaned casually against it. Bryce alternated between watching the kid and the man, wondering who he was. His behavior seemed somehow out of place.
"Mark Grantham," the man said, answering Bryce's unasked question.
"Bryce Morton." Bryce put out his hand, and Paytah nudged his back and got his attention. "What brings you into the store?" Bryce asked before turning to Paytah.
"I'll get your things rung up," Paytah said briskly. "I know you have to meet Jerry and John." Bryce had never seen Paytah move so quickly before.
Mark didn't seem to notice at all. "I have a charitable foundation and I work with a number of the boys down at the tribal school. The foundation tries to help give some of the boys at the school experiences they wouldn't otherwise be able to have. Once a year we take a group of kids to Yellowstone for a week, run sports weeks, and things like that. We also help sponsor some of the food programs at the school to help ensure that these kids get enough to eat."
"That's very noble, but difficult during the summer when they're out of school," Bryce said, but Mark wasn't paying attention. The boy who had come in after him brought a wrapped ice cream sandwich to the counter, and Mark lifted him into his arms.
"Peter, would you add that to the tab?" Mark asked as they headed for the door. Bryce couldn't help following Mark with his eyes. There was something, he wished he knew what it was, that made him want to watch. The door closed, and Bryce continued watching through the window until they passed out of sight.
"He seems like a nice...." Bryce trailed off when he saw Paytah staring nails out the window. He nearly stepped back at the combination of rage and terror in Paytah's eyes. It only lasted a second. He was about to ask what was wrong when Paytah's expression became more normal. "What is it? Who is he?"
Paytah shook his head violently. "You should find John and Jerry," Paytah said, bagging up his groceries. "Can I call you if I find someone to watch the store?" Paytah seemed almost distracted, though he was looking directly at Bryce, who nodded and gave him his number. He also got Paytah's.
"Maybe Alowa can come down. I'd really like it if you could come with us, but I'll understand if you can't," Bryce said, taking the groceries. As he was leaving the store, he turned to take another glance at Paytah, but realized Paytah wasn't paying attention. Wherever he was, it was miles away, and by the pained expression on Paytah's face, it wasn't a pleasant place.
"Bryce," Jerry called as they both walked toward the van. "Is everything okay?"
"I don't know." He had the most unsettling feeling, but he wasn't quite sure why. Bryce shook it off. "I got a few things for dinner as well as the stuff for smoothies for the kids." They really seemed to love them.
"Is Paytah going to be able to come with us tomorrow?" John asked as he joined them, unlocking the door. Bryce placed his groceries on the backseat and then climbed inside.
"He doesn't know if he can get someone to watch the store," Bryce answered, settling on the seat as the others got in. "Do you know who Mark Grantham is?"
"Sure," John said excitedly. "He's a former pro football player. Years ago he set up a foundation. They do a lot on the reservation, help at the schools, take some of the kids on educational trips, have picnics a couple times a year for all the kids in the tribe, stuff like that. Mark's the guiding force behind the charity. You see him on the television occasionally chairing this event or that to raise money. He's been a real godsend to a lot of the kids on the reservation. Why?" John sounded like he too was a bit in awe of the man.
"He was in the store buying one of the kids ice cream. We talked briefly, and I wondered who he was," Bryce explained. He was still puzzled by Paytah's reaction, so he kept it to himself. "He seemed sort of bigger than life. Acted it too." Bryce quieted and sat back as John talked most of the way back to the cabin, mostly about Mark Grantham and all the things he'd done for people on the reservation. Did the guy walk on water?
"He must be quite a guy," Bryce said toward what he hoped was the end of the litany of the wonderful things the man had done. John's parents had already arrived, and once they pulled to a stop, the kids ran over. John scooped Mato into his arms, and Jerry did the same with Ichante, and then they traded before leading them all inside.
"Can we have smoothies?" Mato asked, and Bryce began getting everything out, making a batch of strawberry-raspberry-orange smoothies. The kids delivered the glasses to everyone, and then Bryce cleaned up and put everything away.
"You look puzzled," Kiya said when Bryce joined the others.
"I'm fine," Bryce said, but as he sat down, he wondered why he couldn't stop thinking about Mark Grantham, or more importantly, Paytah's reaction to him.
THE kids were excited about tomorrow. Jerry and John finally got them to bed, and the cabin was quiet. Bryce had been hoping for much of the evening that Paytah would call. His phone had rung once and he'd answered it quickly, hoping it was Paytah, but it had been his mother. Not that he'd been disappointed that she'd called; he'd just been hoping it was someone else. Bryce had talked to his mother for a while. She was still doing well, which was a relief. "Are you trying to get me off the phone?" she asked toward the end of their conversation.
"No, I'm not trying to get you off the phone," Bryce answered.
"Yes, he is," John said from behind him. "He's hoping he'll get a call from a boy," John teased as though they were in junior high.
His mother laughed. "I'm glad you're doing well and starting to move on. Is it serious?"
Bryce moved outside so he could talk more privately. "I'm not sure. He's a... different sort of man, quiet and incredibly intense."
"He sounds interesting. Is he someone you met at the reservation?" she asked expectantly.
"Yes. He runs the trading post on the reservation and...." Bryce faltered. "He's very serious and he seems sad. He smiles around me, though. I think something happened to him some time ago and it hurt him pretty badly, and I think it changed him. No one seems to know what it is, and he won't talk about it." Bryce's mother had been a grief counselor for years, so his next question was on target. "I was hoping you might know a way I can get him to talk about it."
"There are no magic words, and if he doesn't want to talk about it, you can't make him—no one can," she told him, and Bryce figured that was going to be her answer.
"John said Paytah was a happy, normal kind of kid until he was about fourteen. Then he became quiet and withdrew from everything. He said Paytah was like a different person all of a sudden. He and John weren't close at that time, but he told me what he saw. I asked his mother about it, and she said there were rumors in the tribe, but she wouldn't talk about it either. Something happened to him. I know it," Bryce said. "There are times when he forgets and smiles, and I see the person shine through from before he was hurt."
He heard her sharp intake of breath. "Dear, I know you want to help, and this Paytah must be pretty special, but what you've told me could be the result of him being who he is."
"No. John said that people have asked him about what happened, and Paytah nearly punched one guy out. Something did happen." Bryce was sure of that.
"Then it's his secret to share," she told him patiently. "And he might share it with you when he trusts you enough. But you'll have to earn his trust and probably his love before he'll open up. And he may not even then. If he's kept his hurt buried this long, he may keep it forever." That's what Bryce was afraid of. "Be patient." Bryce promised he would. "Now, tell me all about the guy who's captured your interest." She sounded happy, and Bryce told her all about Paytah. After they ended their call, he went back inside and joined the others in the quiet cabin.
John and Jerry were curled together on the sofa, talking quietly. Bryce went right to his room to give them some time alone. Setting his phone on the dresser, he decided to get cleaned up. He got his clean clothes and carried them to the bathroom. Being as quiet as he could so he wouldn't wake the kids, Bryce showered and got ready for bed.
Back in his bedroom, he checked his phone and saw he had a message. He called his voice mail, punched in his code, and heard Paytah's voice. He sounded excited. "Alowa said she would watch the store tomorrow," Paytah said. "Please call me if you still want me to go with you." The message continued in silence, like Paytah wanted to say something else but didn't. Then the message ended and clicked off.
Bryce called Paytah's number right away. "Yes, I want you to come," he said excitedly when Paytah answered. "I've been hoping you'd call, but it was getting late and I was about to give up hope that you would be able to." The words tumbled out of Bryce's mouth.
"I'm glad I can come too," Paytah said. "Alowa was as pushy as ever and said I needed to get out more." Bryce made a mental note to thank her the next time he saw her.
"Should we pick you up at the store? At nine?"
"That would be fine," Paytah answered, and their conversation stalled. There were many things Bryce wanted to ask, things he wanted to talk about, but they all seemed off-limits right now. "I'll see you then."
"Okay," Bryce said, and they ended the call. Bryce put his phone on the dresser, and a soft knock sounded before the door opened and John stepped in. "Paytah can come with us after all. Alowa is going to watch the store," Bryce told him. "Please remind me to pick up something for her as a thank-you."
John smiled. "So you're saying that woman's pushiness is coming in handy?"
Bryce chuckled lightly. "Yeah, I think she's pushing him in my direction." Bryce walked around the bed, closer to where John stood. "Do you think I'm being stupid?"
John tilted his head slightly to the side. "How so?"
"Do you think things could work with Paytah and me?"
"It's too early to worry about things like that. Get to know each other and see where things lead. Enjoy it and don't think too much. Paytah's a good guy, you know that, but he's going to need time. If you call the other day a date, that's probably the first one he's had in years, if at all."
"There's so much pain there," Bryce said. "I saw some of it today while I was in the store. It just lasted a second, but it was there and a bit frightening."
"Do you know what caused it?" John asked, stepping further into the room.
Bryce nodded slowly. "I don't think you'll be happy when I tell you, but I think his pain and anger were centered on Mark Grantham."
"Bryce, you have to be imagining things. Mark Grantham is a great guy. He could spend his time in corporate board rooms or with rich family and friends in Rapid City. Instead, he comes here and spends a great deal of time with the kids here on the reservation." John sat on the edge of Bryce's bed. "Besides, if I remember right, he and Paytah were very close when Paytah was growing up. They went all kinds of places together. I remember how jealous I was that Paytah was one of the guys Mark chose to spend time with. I would have liked to have been able to do some of the stuff they did." John turned so Bryce could fully see his expression. "His foundation tries to help the neediest of the kids on the reservation, in particular, and I didn't qualify, but at the time I wished I had."
"Like I said, John, I knew you weren't going to like my answer, but you asked me what I thought and I told you. The expression on Paytah's face made me shiver, and it was directed at this Mark guy," Bryce said.
"Maybe they had a falling out," John suggested, and Bryce let it go. What he'd seen in Paytah's eyes had not been the result of a falling out. He'd seen terror, and the way Paytah had moved to keep Bryce from touching the other man. There was much more to it than that, but Bryce wasn't going to convince John.
"Maybe," he agreed halfheartedly, and John stood up. "I'll see you in the morning," Bryce added.
"Night," John said, leaving the room and then closing the door behind him. Bryce finished getting ready for bed. He listened to the crickets and small animals as they skittered outside his open window. He tried closing his eyes, but every single time he did, Paytah's face flashed into his mind, contorted in pain, with the anger and terror he'd seen permanently glowing in his eyes. Finally, Bryce fell asleep.
When he woke in the morning, the cabin was still quiet. Bryce got up and used the bathroom first. Then he went into the kitchen and started the coffee and preparations for breakfast. It was definitely his turn to cook, and they had the ingredients, so Bryce whipped up pancake batter. The room filled with the scent of cooking, and one by one, the smell lured the others out of bed. Both Mato and Ichante yawned as they shuffled into the kitchen. Bryce waited while they both climbed onto stools and rubbed the sleep out of their eyes. Bryce got them glasses of juice and began cooking. John and Jerry joined them a short while later, and they looked like hell too. "Bad night?" Bryce asked, and both Jerry and John nodded, with Jerry reaching for the coffeepot.
"Yeah," John said once Jerry had poured him his coffee. "My grandmother would say that my dreams were haunted by bad spirits, and I'd have to agree with her." John turned into the living room and sat on the sofa while Jerry sat with the kids, and Bryce began flipping pancakes.
"Can you do it in the air?" Mato asked, and Bryce flipped the completed pancake into the air and then placed it on Mato's plate. The kids smiled, and Bryce began making a pancake for Ichante, flipping it for her before sliding it onto her plate. The kids began to eat and talk, dispelling the residual gloom that lingered from the night before.
Bryce continued making pancakes. Jerry and John joined the kids at the table, and everyone ate. Laughter quickly filled the kitchen, and gradually even Bryce's memory of the dreams he'd had the night before faded. Bryce then made some pancakes for himself and joined the others at the table, eating heartily.
"I'll clean up," Jerry said as he finished eating and took his plate to the sink. Once the kids were done, John had them clean up and get dressed. Bryce helped Jerry and then got his things together for the day.
It didn't take long to get everyone in the van and then they were on their way. They stopped to pick up Paytah, and Jerry didn't let the kids get out of the van. Paytah was waiting and he climbed in, sitting next to Bryce in the far backseat, and off they went.
They pulled into the drive for the Crazy Horse memorial just after ten, and the kids were bouncing by the time they drove into the parking lot. They burst from the van, jumping up and down, pointing toward the unfinished sculpture. "How long before it's done?" Bryce heard Mato ask Jerry as he took Mato's hand.
"Probably not until you're as old as Paw-Paw," Jerry answered. Bryce climbed out of the van and waited for Paytah. After closing the sliding door, Bryce walked with Paytah toward the visitor center.
"It's something," Bryce said, looking at the huge mountain in the distance with its finished face and the top of the outstretched arm really beginning to take shape, along with the basic shape of the horse's head.
"It's come far since I was last here," Paytah said as they climbed the steps to the visitor center and paid their entrance fee.
"There's a guided tour to the base of the mountain in half an hour," Jerry said, and they all paid the few dollars extra so they could get a closer look. Then they climbed the steps to the visitor center.
"Is that what it will look like?" Ichante asked, pointing to the completed marble sculpture that stood to the side.
"Yes, only much bigger," Paytah answered, and he put his arm around her shoulders and walked her to the front of the deck. They seemed to talk for a while, and Bryce wasn't sure he wanted to interrupt them. Eventually he wandered over anyway and heard the two of them speaking in Lakota, with Ichante giggling and Paytah laughing. God, that was so good to see. Bryce wanted Paytah to laugh like that more often. The usually moody man set Ichante back on her feet, and she wandered to where John, Jerry, and Mato were admiring the model of what the mountain would look like.
"You two were having a good time," Bryce said as Paytah approached.
Paytah nodded slowly without taking his eyes away from the massive sculpture in progress. Bryce had no idea what was going through Paytah's mind at that moment, so he stood quietly next to him. "The last time my dad brought me here, they hadn't finished the face yet. Now that's done, and they're working to finish the arm. I probably won't live to see it done, but even if it's never finished, people will remember."
"They will remember. Your history isn't as lost to the world in general as you think it is," Bryce said softly, and Paytah turned toward Bryce. "When I was a kid, Native Americans were portrayed as drunks, sidekicks, or bad guys in the movies. Now there are movies from the Native perspective that show just how wrong we were. Your people are portrayed as a full, realized people instead of a mere caricature."
"It's not enough," Paytah said.
"Then the only people who can change that is you—the collective you," Bryce said, and Paytah glared at him for a few seconds, but then his expression softened.
"I bet you have some ideas," Paytah said, and Bryce smiled.
"Yes, I do, and I think they might actually work," Bryce said. "I'll tell you all about it when we have some time alone." Bryce turned toward the visitor center, with Paytah following. They joined the others to watch a short video on the history of the project and where the designers expected to go from here. Then they all looked around for a while before leaving to board their open tram to the mountain.
"Good morning," the guide said as they started to board the tram. "It's going to be a great day, and you picked a fun day to visit. We aren't blasting today, so we can go all the way to the mountain this morning." Everyone boarded, and they rode through pine forests to the base of the mountain, looking up the arm toward the face. The guide gave basic facts about the mountain and project before letting everyone off to take pictures.
"Stay with us," Jerry cautioned the kids, and Bryce took Mato's hand while Paytah took Ichante's. They looked around for a few minutes, marveling up at the sculpture that was taking shape.
"At the current rate of work, it's anticipated that work will continue for at least fifty years, maybe longer," the guide said. "The work being done is financed strictly by private donations. Our founder decided that no government money would be taken and therefore there would be no government say or influence on the project. Because of that, the sculpture is taking decades to complete, but it will be true to its original vision."
"There's someone up there," an old lady said, pointing to the arm.
"That's our sculptor. The son or grandson of our founder—it's hard to tell from here. They're already hard at work, and we need to leave so they can continue." The guide herded them all onto the tram and they rode back to the visitor center."
They got off and walked up into the small store area. Bryce was pleased to see it devoid of the usual tourist crap. They wandered around, looking at the displays. The kids talked Jerry into buying a rock. For a dollar, they could buy a rock that had been blasted off the mountain, and the kids wanted one to put in the garden at home. Jerry forked over the buck, and Mato chose a chunk of rock, which John dutifully hauled to the van.
They rode toward Custer State Park, and Jerry paid for the sticker to enter the park, then they slowly drove the road, looking out the windows. For the longest time they saw nothing, and then there they were: a large group of buffalo congregating in and around the road like nothing was happening. They were eating grass and wandering here and there, backing up a stream of cars, and no one really seemed to mind.
"Are those real buffalos?" Mato asked, his face plastered to the window.
"Yes. Those are real buffalos, and they wander all over the park just like the giant herds did a long time ago," John said. Bryce knew that wasn't exactly true. The park did have boundaries, and the herd was kept to a manageable level more by man than by letting nature take its course. But, yes, it was a real buffalo herd. The cars ahead began to move, and then all of a sudden a buffalo turned and ran at one of the cars. The car sped up, and the animal veered off toward one of the calves.
"What happened?" Ichante asked as she leaned to her brother's side to get a better view.
"The car got between a mama and her baby," Bryce said as they slowly inched forward, the road clearing somewhat and the cars moving on as the herd moved off into the grasslands at the side of the road.
"This is cool," Mato said. Bryce reached into his pocket and pulled out his phone, then snapped a few pictures through the glass.
"Yes it is," Paytah echoed as he watched out the window.
"What are you thinking about?" Bryce asked.
"How traditionally our ancestors hunted these animals on horseback with bows and arrows. If you fell off your horse during the hunt, the chances were you would be trampled, but if you were successful, if the tribe was successful, then everyone ate during the winter and there were skins to keep warm and make clothes and shelter," Paytah said. "Nothing was wasted." Bryce swallowed hard. He knew what Paytah was saying.
They continued on, but didn't see any more buffalo. Donkeys wandered up to the car, peering in the windows before moving on. When they reached the end of the drive, Jerry parked the van near a snack-bar area and they got out. The kids got in line for food, and Bryce thought they had the right idea.
"Can we ride horses?" Ichante asked, pointing toward a paddock once they were seated at a picnic table with hot dogs, chips, and water.
John checked his watch. "I don't think we can do the caves and ride horses, so we'll take a vote. Who wants to ride horses?" John asked. Mato and Ichante both raised their hands excitedly. Bryce looked at Paytah, who nodded, so Bryce raised his hand too, and so did Paytah. "Looks like we'll ride horses," John said, looking a bit disappointed.
"Why don't I wander over and see what times they offer riding," Bryce proposed, and he got up, throwing the trash of his lunch in the can as he went. As he got closer, he saw that the times were posted, with the last ride leaving at four. Bryce returned to the group. "It's twelve thirty now, and the last ride is at four. We could sign up, go to the caves for a tour, and then return to ride the horses."
With that settled, they got the kids moving and spent part of the afternoon deep in the earth. Bryce found the caves fascinating, but for the kids it was a little slow, and they definitely got restless. By the time they were ready to leave, Mato had fallen asleep on John's shoulder, and Ichante was dragging. They rode the elevator back to the surface and then went out into the lobby and gift shop. Bryce did his best to ignore the tacky made-in-China souvenirs until he passed the register on the way out.
"Don't you have anything authentic?" a woman was asking a clerk. "Everything is made in China or the Philippines." The clerk shrugged, and the woman stepped away. Bryce turned to Paytah with a smile.
"What's that for?" Paytah asked as Bryce headed for the exit. Outside, he found a bench and sat down to wait for the others.
"You heard that woman—she was looking for something real, something nice, to take home. Something that didn't cost a fortune, but was real, not tourist crap."
"Yeah," Paytah said.
"So that's my idea. You and the tribe make it," Bryce said.
"Are you saying we weave blankets by hand to sell?" Paytah asked as a woman carried a fake "Indian" blanket out of the store.
"No. I'm saying...." Bryce snuffed softly. "Okay, you saw the tacky peace pipes with plastic feathers? Well, what if you and the tribe were to make real pipes based on authentic designs?" Paytah looked skeptical. "You don't carve them by hand, but you make them in an assembly line process. Say you turn a hundred pipe bowls on a lathe and then make a hundred pipe stems. They would be assembled and decorated by hand in traditional, authentic ways. You do the same for dolls, or any craft item that will work."
"But they'll be expensive," Paytah said.
"More expensive than the crap in the tourist stores, yes, but we'll brand it, say 'Sioux Made' or 'Lakota Made'. Each comes with a certificate stating that the item was made on the reservation by Native Americans and the proceeds go to help projects on the reservation. That's why I wanted to talk to some artists. They could design what you make and keep the items simple, but authentic and usable, or at least potentially usable." Bryce paused to watch Paytah's reaction. "The people on the reservation would be paid, bringing in income, and we could set up a website if it gets going to sell the items on the Internet."
"Wouldn't that be charity?" Paytah asked.
"How so? The tribal members would be working and making things to sell. It would bring money into the tribal economy. The items could be sold in gift shops around the area. I'm not saying we go full steam ahead. We would need to get the tribe's blessing, and then we could test it to see how it worked."
"Wouldn't we be exploiting our heritage?" Paytah asked evenly, and Bryce could tell he'd captured his attention.
"No. It would be celebrating it. The crap in these stores that costs five bucks is exploiting your heritage. It isn't close to the craftsmanship and care that went into making each of the items your people used every day. A knife, a pipe, a doll, children's toys—they were all handmade with care and love. I'm saying you should do the same thing on a slightly bigger scale. Not massive, but large enough that it might employ some of the people on the reservation and return money to the tribe to help the families who need it, like Little Wamblee's." For some reason, he could not get that child out of his mind.
"Okay. Let's say I think this is a good idea and we are able to get the council's support. How do we get it started? I know about selling stuff, but not making stuff. You know about computers."
"My idea is to ask the tribe. Get as many people in the tribe invested in the idea as possible. This isn't mine or yours—it has to be theirs. I bet there's someone who understands production. I bet there are people who retired from factories or places like that. See if they'll be interested in helping. The idea isn't to make me money, but to help the tribe so it can help others." Bryce stopped speaking and waited.
Paytah seemed to be thinking things over. "It's a good idea. If we can get something like that going, it would give families work and help the tribe at the same time."
"It's a start. I mean, it isn't going to make anyone rich, but it could help some families and add to the economy on the reservation."
"You're probably right; it could be a start," Paytah said as he looked around. "It seems like there's so little hope there. Many people have given up and stay in misery or are leaving." Paytah shifted his gaze, and Bryce followed it to where Jerry and John were walking toward them with the kids. "Not that I blame John at all."
"Blame me for what?" John asked as they got closer.
"I was just talking about how people leave the reservation to find work," Paytah said. "Bryce has an interesting idea."
"Why am I not surprised?" Jerry asked with a self-satisfied grin. Jerry had taken a chance on both him and John when he'd hired them three years earlier, and he loved it when Bryce and John proved him right.
"We should get moving or we'll be late," John said, and they got back in the van.
An hour later, after signing forms, finding helmets, and getting set up with horses, they started on their trail ride, with one guide in front and another in back. Bryce had a chestnut-brown horse, and right behind him was Paytah on an almost identical model, except Paytah's was definitely bigger. The thought made Bryce blush. They rode through the woods and crossed a small creek. The horses seemed to know exactly where they were going, and Bryce figured they'd probably been taking this route every day for years.
"Can I ask something?" Paytah said from behind him, and Bryce turned. Paytah seemed very contemplative.
"Of course," Bryce answered, turning so he was looking forward again, but he was still listening. They were toward the back of the group, with Jerry, John, and the kids toward the front. Bryce figured that was matchmaker Jerry's doing. Not that it was a bad thing.
"Why do you care so much?" Paytah asked just loud enough for Bryce to hear. "I mean, you aren't a member of the tribe, and in another week or so, you'll go home and back to your own life. If you don't want to, you'll never have to see anyone here again. So why? It isn't because of me, is it?"
"Would it be so bad if it was because of you?" Bryce asked, looking back over his shoulder, and Paytah seemed a little surprised. Bryce turned back around and figured he'd let Paytah chew on that for a while. Sometimes Paytah seemed completely shocked that Bryce could be interested in him. The ride continued, and there were no more questions from behind him. In fact, Paytah didn't talk much at all for the rest of the ride. Bryce enjoyed the quiet as they continued along the path through the wooded area. It was pleasant and incredibly relaxing. All around him people laughed and talked, but at least for a while, Bryce let himself sink a bit into his own mind.
All too soon the ride was over. The kids groaned good-naturedly when they had to get down. Bryce dismounted and an attendant led away his horse. He also turned in his helmet and found himself walking a little stiffly back toward the van. The kids ran ahead, filled with excitement, chattering on about their ride. Jerry unlocked the van, and everyone piled in and soon they were off.
"What did you mean when we were on the trail?" Paytah whispered.
Bryce thought back to their conversation. "What if I was doing this for you? Would that be so bad?" Bryce said, biting his lips to keep from smiling. "I'm not. Well, I'm not only doing it for you."
"Then why?" Paytah asked, and Bryce looked over at the kids and then back to Paytah, who seemed to understand, and they rode quietly back to the reservation. John parked the van outside Paytah's small house. They worked their way out, and Bryce accompanied Paytah to the door. "I had a nice time."
"I did too. Maybe next time it can be just the two of us," Bryce offered, and he received a warm smile in return. Bryce leaned closer, and Paytah met him for a kiss that started soft, but quickly heated. Bryce had to pull away to keep from giving everyone in the van a show. He took a step back, and as Paytah gazed at him, Bryce felt a surge of excitement run through him. God, the things he wanted to do to Paytah, with Paytah, but.... Slowly, Bryce turned away to head back to the van.
"Could you stay for dinner?" Paytah asked. Bryce approached the van, and John lowered the passenger window.
"I'm going to stay with Paytah for dinner," Bryce explained. "He invited me," he added in an excited whisper.
"Sure," John said. "Call us if you need a ride home." John actually winked at him. Bryce rolled his eyes and stepped away. Jerry backed the van out of its spot, and Bryce joined Paytah outside his door. Paytah unlocked the house, and Bryce followed him inside what looked like a time capsule. The furniture, the rugs, the pictures—everything looked like it had come out of the forties.
"The stuff is all a little old," Paytah said a bit dismissively.
"It's amazing," Bryce said, looking all around him. The house had seen a lot of use, but it was obvious that Paytah took care of what he had and maintained the place beautifully. He hadn't changed anything.
"This is how my mother furnished the place, and I haven't had the heart to make a lot of changes. I had to replace the curtains, but I used something close to what Mom had," Paytah said, and Bryce wandered over to a set of rustic-looking shelves lined with gorgeous pottery pieces. "Mom was an artist, and those are some of her pieces. When she died, I kept what she'd saved. She made her own clay from here on the reservation."
"They're the same color as your skin, warm and rich," Bryce said, and Paytah looked away. "You have nothing to be ashamed or frightened of."
"I'm not scared," Paytah countered.
"Sorry, for a second you looked it," Bryce said as he moved closer to where Paytah stood. "I meant it. You have nothing to be ashamed of. You're warm and rich," Bryce whispered, lightly stroking Paytah's smooth cheek.
Paytah stilled Bryce's hand with his own. "I can't," Paytah whispered.
"Can't what? Allow yourself to be happy or cared for?" Bryce asked gently. "You deserve to be cared for and to have someone to be with."
Paytah shook his head and moved away. "I don't."
"Why?" Bryce asked, and Paytah stiffened. "This has something to do with that guy in the store yesterday, doesn't it? That Mark What's-his-name guy?" Bryce moved closer again, this time hugging Paytah until his tight muscles relaxed just a bit. "I'm not going anywhere, and regardless of what you seem to think, you do deserve to be cared for. I don't care what anyone told you or how anyone made you feel, that's a simple fact." God, whoever had hurt this man was going to answer to him. Bryce was going to find out what happened and then rip the nuts off whoever had done this to Paytah. Bryce got angrier by the second until Paytah finally began to hug him back. Then the anger slipped away. "I shouldn't push you, I'm sorry."
"You really care," Paytah said softly enough that Bryce wondered if Paytah was murmuring to himself.
"Of course I do. I don't understand why you would think I wouldn't." There had to be a reason for Paytah's doubt, and Bryce knew now that this Mark guy was at the bottom of it somehow. Yes, he was jumping to conclusions, but the way Paytah stiffened and braced himself every time Bryce mentioned him was a huge giveaway.
"I...," Paytah began.
"Let's have some dinner and talk about something else, something nicer," Bryce suggested, and some of the anxiety slipped from Paytah's posture.
Paytah went into the kitchen, where Bryce joined him, sitting in one of the chairs. "It will be simple," Paytah said, like Bryce expected something fancy.
"Whatever you make will be wonderful," Bryce said, and Paytah looked at him with such doubt that Bryce wanted to scream. What or who beat Paytah down like this? The thought had Bryce's anger rising again.
The kitchen filled with the scent of bacon, and then Bryce saw Paytah cutting fresh tomatoes and tearing leaf lettuce. "Do you have a garden?" Bryce asked as the scent of the tomatoes, which had to be fresh, mixed with the bacon.
"Alowa does, and she brings stuff over all the time. Says I eat too much crap," Paytah answered as he continued working. "She fancies herself as some sort of Indian earth mother." It was the first time Bryce had ever heard Paytah refer to anyone as Indian. "She's a situational vegetarian. When other people are around, she professes to be vegetarian, but when no one's looking I've seen a burger disappear in her presence." Paytah laughed, and Bryce joined him, the last of the residual tension disappeared from the room. "I hope BLTs are okay."
"They're my favorite," Bryce said. "After all, everything's better with bacon."
Paytah toasted bread and then began building hefty sandwiches. He also added a slice of turkey to each one, plated them, and then carried the plates to the small plank table. He returned to the kitchen for drinks, and then they both sat down. Paytah began to eat right away, looking down at his plate. At first, Bryce wondered if he was being ignored, and then it occurred to him that Paytah was used to eating alone, so he just ate.
"Did your dad build the house?" Bryce asked, and Paytah looked up from his food, setting his sandwich on his plate.
"My father's brother originally built it. Uncle Black Wolf died shortly after he finished it, and left it to my dad. He made a lot of the furniture too. After that, we moved in and Mom got her hands on the place. She loved old things and did all this. There are only two bedrooms. Mom and Dad had one; Chay and I had the other. There wasn't a lot of room, but it was always home."
"Where does Chay live now?" Bryce asked after swallowing a small bite.
"He has a place in the hills somewhere," Paytah answered through a sigh. "He used to live here, but I couldn't stand his drunkenness, and he wasn't willing to help himself. Eventually he moved out so we wouldn't fight any longer. Now he figures he's free to live however and do whatever he wants."
"Must be hard to want to help and not be able to," Bryce observed, and Paytah nodded slowly as he picked up his sandwich.
"I know now that there's nothing I or anyone can do for him," Paytah said, pausing with the sandwich in his hands. "That was the hardest realization—that he has to want to change. It has to come from inside him, but he isn't interested. I keep hoping that he will be someday, and I'll be there when he is." Paytah resumed eating. "I thought we were going to talk about something happy."
"Okay," Bryce said mischievously. "Let's talk about your hair."
"My hair?" Paytah asked, setting down his sandwich again in surprise.
"Yeah, it's really sexy," Bryce said, and from the expression on his face, Paytah thought he was kidding. "The other day I was helping you at the computer and it brushed against my hand." Bryce suppressed a shiver. "Like I said, really sexy. How long have you been growing it?" He reached over the table and lightly touched Paytah's midnight-black hair.
"The last time I cut it was when my father died," Paytah said, "and I won't cut it now unless something happens to Chay."
Bryce combed his fingers through Paytah's sensuously silky strands. "I hope you never have to cut it." The hair slowly fell away from Bryce's fingers. "I think it's amazingly beautiful." It took Bryce a few moments to look away, and then he slowly began to eat again. "Do you think I should let my hair grow out like yours?"
"No, I like it like it is," Paytah told him, and then he went back to eating, his expression shy. "I never said those things to anyone before. You know, caring-type things."
"There's nothing wrong with saying how you feel." Bryce smiled softly. "It's nice when you do." Paytah looked down at his food, and Bryce knew it was his way of hiding, not letting Bryce see too much.
"I was eleven years old when I first met Mr. Mark. That was what he let me call him. He was this important man, and I was just a poor kid that he took an interest in." Paytah looked like he was far away. "My dad ran the store then, but we didn't have any extra money. Sort of like now, we made enough to get by on and that was about all. Mr. Mark offered to take me on his camping trips. My dad thought I'd done something really good to be asked and he was so proud. Mostly he was impressed by Mr. Mark the same as everyone else was, just like I was." Paytah pushed his plate back, stood up, and began pacing. "He used to buy me things like ice cream and candy, just like he did when he was in the store the other day." Paytah's voice hitched. "Everyone around this place loves him, they think the sun rises and sets on his ass, but it doesn't!"
Bryce felt a lump form in his throat, slightly afraid of what Paytah was going to tell him. He tried to swallow the lump away but it stayed there.
"Mr. Mark had been some big-shot athlete in college, so at one point he started a sports league, sort of an after-school program where we could learn to play sports like basketball and football. And he chose me to be a part of it." Paytah puffed his chest out slightly. "I knew I was special, then. Mr. Mark liked me, and I would do anything for him." Paytah's chest returned to normal and he shifted his gaze to the floor. "Most of the programs took place in the summer, but some happened after school. As I look back on it, I realize what he was doing, but then... we were kids. It was hot, so we took off our shirts and Mr. Mark told us he was going to teach us how to wrestle."
Bryce closed his eyes. He could already see where this was leading, but he said nothing. Paytah had probably never told anyone about this, or if he had, it had been a long time ago, and Bryce didn't want him to stop now. Paytah needed to say what he wanted to say, he needed to tell someone. Bryce was close to tears that Paytah was opening up to him.
"Mr. Mark would take off his shirt and wrestle with us. We'd laugh and climb all over him, and he'd show us the holds, or at least his version of the holds. Now I know what he was doing was getting us used to being touched in ways we shouldn't have been touched." Paytah continued slowly pacing the room. "When I was about twelve, we got to go camping with him. He'd take two or three of us boys on a weekend camping trip. We'd all sleep in the same tent, but one night it got cold, so Mr. Mark had me climb into his sleeping bag. I didn't know at the time that he was naked. I know now that he got off on us being near him." Paytah turned away, and Bryce shivered before getting up. Without a second thought, he put his arms around Paytah.
"You don't have to go on if you don't want to. I'll understand."
Paytah shook his head. "Those camping trips that I'd looked forward to when I was eleven turned into trips to hell by the time I was fourteen. His touching turned into him asking to be touched, which turned to kissing, and then oral sex. Eventually... he said no one would believe me if I told and that he'd hurt my father, so I let him." Paytah placed his hands over his face. "I was fourteen years old the first time he had sex with me, and it continued for almost a year whenever we were alone—camping, after school, in his car. When I turned fifteen, I told my father that I didn't want to be a part of the group anymore. The only way I could think of to get away was to withdraw, and I guess I pulled away from everyone and everything. Mark tried to lure me back, and he even used his influence with the principal to talk to me about rejoining the group. The principal actually tried to help get me back into that man's group so he could have sex with me some more."
"Do you think he knew?" Bryce asked.
Paytah shrugged. "At the time I wasn't sure, but probably not. Now that I think back, he was probably in as much awe of Mark as everyone else seems to be around here and thought he was trying to help." Paytah was shaking, and Bryce held him. He didn't know what else to do. He wasn't about to tell him a bunch of mindless platitudes. "I still dream about it, and I know he's still doing it. I've seen him. He lures the kids in and then preys on them."
"Did you tell anyone what happened to you?" Bryce asked, and Paytah nodded.
"I couldn't tell my dad. I know I should have, but I was so ashamed and afraid he'd hate me. As an adult I know it sounds stupid, but he had very... definite opinions and was very vocal with them. I finally got up the courage to talk with the counselor at school. He'd always been nice to me and seemed to have noticed the change in me. He'd actually asked if something was wrong. So I made an appointment." Paytah shuddered. "I remember sitting in his office and I told him the whole story, everything. By the time I was done I felt better, figuring at least now someone knew."
"What happened?"
"He looked shocked and horrified at first. Then I told him who was doing these things to me and his whole attitude changed on a dime. He told me what a great guy Mark was and that it couldn't have been him who did this to me. 'Mark would never do that,' he actually said, and then started questioning me as if my dad was the one who was hurting me. I was so horrified and began yelling that it was Mark, but he didn't believe me." Paytah took a deep breath. "I left the counselor's office near tears, but kept myself together and went back to class, trying to forget everything. I don't remember much of the rest of that day, but I made it through and went home."
Bryce's mouth hung open in disbelief. This was the most unbelievable thing he'd ever heard. "Is that counselor still around?"
Paytah shrugged. "I don't know. He could be. He wasn't that old, and working at the school is a good job here on the reservation because the federal government pays most of the costs."
"Son of a bitch," Bryce swore. "He should have listened to you. That was his job." Bryce calmed himself down when he realized Paytah wasn't done. "Is there more?"
Paytah nodded slowly. "The next day I was called down to the principal's office. The counselor was there and so was Mark."
"Jesus Christ," Bryce muttered.
"They all worked to convince me that I'd either imagined the whole thing or that someone else had done these things to me. None of them believed me. Mark had such a grand reputation here that they believed him." Paytah pulled away from Bryce and went back to pacing, faster and more deliberate now. "I didn't know what else to do, so I kept quiet and left the office when they said I could go. All I wanted to do was get out of there." Paytah stopped nearly dead still. "I made it through the rest of the school day, and when I left the building, Mark was waiting for me."
"Jesus Christ," Bryce swore again, more loudly this time. He could feel the tension building in the back of his head and wondered what hell Paytah must be going through.
"He practically kidnapped me into his car, and we sped away. He was yelling and threatening me the entire time. 'No one is ever going to believe you, so you may as well shut your mouth,' he yelled at me. 'You know I can make sure no one comes in your father's store again. I might open another one on the reservation and run you out of business.' He said all this while we were speeding down a gravel road. All I wanted to do was get the hell away from him and never see him again. I honestly thought he might try to kill me, but then I realized he was taking me toward one of the campsites he used—where he took us for sex."
"Oh my God," Bryce gasped and placed his hand over his mouth.
"He slowed down to take one of the corners, so I pushed open the door and jumped out. The car wasn't moving very fast, and I rolled on the ground before taking off as fast as I could cross-country. I knew the reservation pretty good and stayed away from the roads as I made my way home. It took me a long time, but I made it back home and snuck into my room. I didn't want to see anyone. I sort of felt like my life was over." Paytah took a deep breath. "But as days went by, I started to feel better. Mark stayed away, and the few times I saw him, he kept his distance, and that was all I wanted. For a while, he stayed away from the school altogether, and I don't know if that was because he was afraid of me, but I bet the principal and counselor had him keep away to contain the matter. Everything went back to normal, and after I graduated I started seeing Mark around the school again."
"You know what he's doing," Bryce said, and Paytah nodded.
"He's finding other little boys to do what he did to me, I know that. I tried speaking up, but no one believed me," Paytah said.
"I know." Bryce's heart ached for Paytah and he didn't know what to do for him. What was left of their dinners sat on the table, and obviously neither of them was interested in eating any longer. Bryce's stomach didn't seem to be able to make up its mind if it was happy about what he'd already eaten.
"I think you better go," Paytah said. "Jerry and John will be wondering about you."
Bryce pulled out his phone and called Jerry. "Could you pick me up in the morning?" Jerry chuckled. "It's not like that," Bryce said seriously. "I'll call you in the morning and let you know what's happening."
"I take it this is serious," Jerry said.
"Yes," Bryce answered and heard the phone shift.
"Bryce, it's John. Is everything okay?" he asked urgently.
"It will be," he answered. "I'll see you in the morning." Bryce hung up. "Jerry and John won't worry now."
"You're staying?" Paytah asked.
"Yes. I'll sleep on the sofa out here if you want, but I'm not leaving you alone," Bryce told Paytah emphatically.
"I'm not a child," he countered.
"No, you're not. But you've just told and relived one of the most traumatic events of your life and you shouldn't be alone. You aren't alone. I know you've felt that way for a long time, but that isn't the case." Bryce hugged the larger man. "I believe you," Bryce said softly. "Each and every word."
Paytah trembled in Bryce's arms. "You're the only person to tell me that."
"I know," Bryce said. There were so many things he wanted to say, to tell Paytah, but he kept quiet. Now was not the time. Paytah was exhausted and seemed to sway slightly. Gently, Bryce guided Paytah down onto the sofa and then cleared the table, cleaning up as best he could before returning to sit down next to Paytah.
"I'm tired," Paytah said after a long silence, and Bryce got up with him. He followed Paytah to a small bedroom. Paytah walked inside and then stood still, his eyes unfocused and unmoving. Bryce took him by the hand and guided him toward the bed. Carefully, he removed Paytah's shirt and pants, then helped him get into bed. Paytah stared at him, still looking a bit blank. Bryce thought about leaving the room, but instead took off his own shirt and pants and got into bed next to Paytah, holding him close.
"Just go to sleep. I'll be here with you," Bryce said as he stroked Paytah's hair. "You aren't alone, and I believe you." Bryce kept repeating the words until Paytah sighed softly and closed his eyes. Bryce lay awake for a long time, hoping that Paytah would sleep. He did, eventually, and then Bryce finally dozed off from sheer exhaustion.
# Chapter 6
BRYCE woke in the small hours of the morning. He immediately remembered where he was and felt Paytah thrashing in the bed next to him. He settled for a few moments and then began moving again. Bryce lightly stroked Paytah's arm, trying to calm him, but it got worse.
Paytah gasped and sat straight up in the bed, pulling the sheets around his hips. He sucked air loudly, and Bryce saw him look first one way and then the other. "Bryce, is that you?"
"Yes, I'm here," Bryce soothed.
"I thought you were him," Paytah whispered. "For a few seconds I thought I was back with him. But I'm not, am I?"
"No, and you never will be again," Bryce said, sitting up as well. "You're fine and you're safe. He's never going to get to you again. He can't. I'll rip his balls off first." The vehemence of Bryce's response startled even him. "If he looks at you cross-eyed, I'll make sure he walks funny for a very long time, the sick fuck." Bryce forced his anger away. It wasn't going to help Paytah. "Lie back down, it's okay."
Paytah settled back on the bed, but seemed restless. Bryce tugged him close and lightly kissed Paytah's shoulder. Damn, his skin tasted good, and Bryce had to restrain himself.
"What if he's ruined me forever?" Paytah asked. "I told you I did some stuff when I was a kid, and I did. After I thought I'd left Mark behind. It didn't work too well, and I got scared and backed away."
"So you weren't ready. That doesn't mean you're ruined or damaged. It just means you need to take it slow and know the person you're with cares for you. Sex isn't what it was with Mark, not really. That was him being greedy and taking what he wanted no matter what you had to give. That's not caring—that's abuse." Bryce quieted and kissed Paytah's shoulder, then he did it again, moving a little closer to Paytah's neck. Bryce felt Paytah stiffen, so he moved away. "Nothing is going to happen that you don't want to happen." Bryce kissed him lightly again. "Do you want me to stop?" Paytah shook his head. "You have to say what you want."
"No. Don't stop," Paytah whispered into the darkness, and this time Bryce kissed him at the base of his neck. He used his tongue to find that little spot, and Paytah groaned softly. Bryce was willing to bet no one had ever done this with him before.
"Not until you tell me," Bryce whispered before licking up Paytah's skin to just behind his ear. "See, you're not broken. You just need someone to take care and show you what making love really is."
"Making"—Paytah swallowed—"love."
Bryce sucked lightly on Paytah's ear, chuckling softly. He lightly stroked his hand over Paytah's smooth chest, something he'd wanted to do since the first time he'd seen him without his shirt. He wanted to ask if anyone had ever done anything to make Paytah feel good, but he didn't want Paytah to associate this with his past in any way. Paytah seemed to be enjoying Bryce's touches, and Bryce wanted Paytah to associate pleasure and caring with his touch rather than the fear and self-loathing that had been his past.
"Yes, Pay, love. That's what it's called when two people who care for one another give themselves to each other. That's what you've been missing all this time." That's what Mark stole from you, he wanted to say, but kept quiet, his anger at bay.
Bryce shifted on the bed so he could see Paytah's eyes. Then he slowly leaned closer, giving him a chance to back away. He didn't. Paytah held still, and Bryce kissed him. Just like outside the door, the kiss deepened quickly, and Bryce let Paytah take charge. Bryce figured Paytah would need to feel as though he were in control. He felt Paytah hold him closer, but the kiss didn't deepen further, and Bryce could feel Paytah's reticence. That was okay, as far as he was concerned. Bryce liked being held, and Paytah's doing it felt great.
They continued kissing, and Bryce continued exploring Paytah's skin with the lightest touch he could muster. "I love your hair," Bryce said as he carded his fingers through the long, strands, pulling out the rubber bands Paytah had used to keep it together. "I like it loose and flowing."
"Do you have some kind of kinky hair thing?" Paytah teased, and Bryce clamped his eyes closed.
"Maybe," Bryce said, knowing if Paytah could tease, he must be feeling more comfortable. Bryce wriggled his hips against Paytah's and giggled. "I could sure develop some sort of hair fetish as long as it was your hair." Bryce laughed outright and tugged Paytah even closer. Then he kissed him again, going slowly and carefully, but his own body was working against him. He was achingly hard, but he wasn't going to press anything on Paytah. That was what Mark had done, so he tried to put that out of his mind and simply reveled in the feel of Paytah against him.
Then he realized Paytah was as hard as he was. Bryce didn't want to be too aggressive, but Paytah moved his hips and he heard Paytah's breathing shorten and become more ragged. "That's it, Pay, let it all out." Bryce pressed closer, giving Paytah more friction, and he felt it the second he tumbled over the edge. Paytah held his breath and made little sounds as he came. Bryce held Paytah tight, moving his hips slightly, and soon the ground fell away for him and he came as well.
Bryce lay next to Paytah, breathing deeply, with a wet mess in his shorts. The only consolation was that Paytah was in the same predicament. Bryce began to chuckle slightly and carefully worked his shorts off, using them to clean up. He felt Paytah doing the same thing, and once Paytah settled, Bryce moved close once again. "Is this okay?" Bryce asked.
"Yes," Paytah answered, and Bryce closed his eyes. "I'm not going to break, Bryce. You don't have to tiptoe around me."
"I'm not. I just want the memories we make together to be good ones," Bryce said, "and I don't want to rush you."
"You aren't rushing me," Paytah said as he tugged Bryce to him. "I'll admit it feels a bit strange to share a bed with someone. The last time I shared a bed, it was probably with Chay, when we were kids." Paytah grew quiet, but Bryce could tell he wasn't asleep. "This feels nice."
"It does," Bryce said, and he thought about Percy and smiled. He'd just had sex, well, a sexual encounter, with another man for the first time since Percy, and it was okay. In fact, he thought about Percy for a while and how different Paytah was from him. But he knew Percy would have liked Pay. Yes, they were very different, but deep down they had many things in common—the important things, like honesty and a good heart. Bryce knew that was what had attracted him to Percy, and he saw the same things in Pay. Bryce closed his eyes and tried to go to sleep.
"Why are you calling me Pay?" Paytah's voice rumbled with sleep in the darkness.
"I liked it and I wanted a special name for you," Bryce said, trying to stifle a yawn, but failed. "Something only I would call you."
"Did you have a name like that for Percy?"
"Yes. I called him Sweetums. He hated it at first, but he got used to it. I didn't do it in public, but it was my way of telling him how much I cared for him." Bryce rolled over to face Paytah. "I know it sounds a bit stupid, but it made me feel better to call him that because it was how I felt. He was sweet and very special to me, and I could say all that in just one word." Bryce yawned again and rested his head against Paytah's shoulder. "If you don't like it, I won't use it anymore." Bryce really hoped Paytah was okay with it because he really liked it. Paytah didn't say anything, but he held Bryce closer, and Bryce took that as a yes.
BRYCE fell asleep at some point and woke with sun shining through the bedroom windows. The windows were open, and the cool early-morning breeze wafted into the room. Paytah was still asleep, his face relaxed and oh so beautiful. Afraid to move in case he woke him, Bryce turned his head and smiled as Paytah shifted slightly and continued sleeping. Bryce simply watched him, quiet, relaxed, peaceful.
"What you doing?" Paytah mumbled, opening his dark eyes slightly and then closing them again.
"Watching you," Bryce said. "You feeling okay?"
"Uh-huh," Paytah mumbled, opening his eyes again, keeping them open this time. "Sorry I got all blubbery yesterday."
"You didn't," Bryce said, shifting a bit. "I have something to tell you and then ask you. My mother has worked as a grief counselor for a long time and she has been trying to figure out when she could come out to visit me. If she does, would you be willing to talk to her?" Bryce felt Paytah stiffen and he could almost feel the denial coming. "She could help you. I know you're still hurting, and I hope talking to me helped, but Mom might be able to really help you."
"You said you believed me," Paytah whispered.
"I do. I don't want you to talk to her because I think you're lying. I want you to talk to her because you were telling the truth."
"She can't change it," Paytah said.
"No." Bryce shook his head slowly. He wished he could take all his pain away, but he couldn't. "Mom is the best listener I've ever come across. You know how some kids have parents who don't seem to hear them? I never had that problem. Mom always listened, and then she did her best to help. I'm not saying you have to. It's only an offer, and you don't need to give me an answer now." Bryce lightly touched Paytah's cheeks. "You've carried this around with you for a very long time, and if you're ready, there are people who will believe you and try to help you." Bryce widened his eyes when he realized he'd forgotten the most important thing. "What is it you really want?"
"What kind of question is that? What do I really want?" Paytah's vehemence took Bryce by surprise. Paytah sat up quickly, the covers tumbling away from him. "I want all this to go away. I want to have never met that man so I could have my childhood back and be happy the last years I had my father instead of miserable and afraid of everyone and everything." He turned to Bryce, his eyes blazing. "I want that bastard's dick cut off so he can't hurt any other kid the way he hurt me." Paytah clenched his fists. "I want to hurt that fucker so bad he can't walk again! I want...."
Bryce let Paytah rage and vent some of the hurt that must have been building up for years. When he quieted, his eyes still burned, but the rest of him seemed much calmer. Bryce tried again. "I meant, what do you want for you? I understand wanting to hurt Mark—I personally want to kill him—but what do you want for yourself?"
"I don't know," Paytah answered.
"Do you want to go after him? You're an adult now, not a kid, and you have a powerful voice. Do you want to use it? Do you want help so you can deal with what happened to you and maybe move on?" Bryce touched Paytah's arm. "I know you'd really like all this to go away and I don't blame you, but Mark was in the store and...." Bryce couldn't bring himself to vocalize what he was thinking, and he could tell Paytah thought the same thing, because the fire in his eyes blazed again.
"He's hurting other kids," Paytah said. "Innocent kids with no more recourse than I had." Bryce nodded his agreement, and Paytah took a deep breath and held it, then released it slowly.
"You don't have to decide now. Take your time and think about what you want. This is one time when you have to put yourself first and make sure you know what you want to do." They were in an awkward position on the bed, but Bryce held Paytah as best he could before shifting and moving closer. "I believe you, and I'll help any way I can." Bryce's stomach was flipping all over the place. Paytah's choices weren't pleasant or easy. In the end, Bryce figured Paytah would do the right thing, but that had to be his decision because it was his pain. "I'd do anything to take the pain and hurt away."
"But you can't; no one can," Paytah snapped softly.
"Except you," Bryce countered.
"How?" Paytah asked.
Bryce shrugged. "I don't know. But someone like my mother does."
Paytah sighed and didn't move for a long time, but then he said, "Please call her." Paytah shifted on the bed, and Bryce moved closer, practically sitting on Paytah's lap. He kissed him hard, pressing Paytah gently back against the bedding.
"Let me help you forget," Bryce said, and Paytah nodded. Bryce grinned and pushed away the sheet that had covered them. In the morning light, Bryce let his gaze roam down Paytah's entire body. "You're amazing," Bryce said, and Paytah smiled a bit nervously. "Remember, Pay, all you have to say is stop," Bryce reminded, but he hoped to God he never heard that word from him.
Bryce straddled Pay's legs, resting his balls right over Pay's hard cock. Then he leaned forward and stroked Pay's strong chest and shoulders. He wriggled his hips slightly, and Pay hissed softly. "You're evil," Pay told him, and Bryce smiled.
"That's the whole point. I want to make you feel good. I want you to know that I care," Bryce told him, locking his gaze on Pay's deep-brown, penetrating eyes. If Bryce weren't already naked, he would have felt it, like Pay was probing deep into his soul. Reluctantly, he broke their locked gaze and leaned forward to suck lightly on a caramel-colored nipple. Bryce loved the taste of Pay's skin, and he licked and sucked as Pay thrust his chest forward, moaning deeply.
"Bry," Pay panted softly, and Bryce smiled against Pay's skin. He wasn't quite sure if he was smiling because of the moan or Pay's new nickname for him. He quickly decided it was for both and switched to the other nipple, which he sucked a little harder. Pay moaned again and again, and with each one, Bryce gave him more. When he stopped, Pay groaned deeply. He wanted time to explore Pay from head to toe, but he also wanted to impress as many pleasant sensations on Pay's brain as possible, so Bryce slid down Pay's legs, settling between them. With a wicked grin, he looked into Pay's eyes and then licked up his shaft, watching his reaction. It was blissful. That was the only word that came to mind when he saw that look: sheer bliss.
"Has anyone ever done this?" Bryce asked, and Pay shook his head. "Then you're in for a treat." And so was he. Bryce smiled and then opened his mouth, sucking the thick head of Pay's cock into his mouth. Pay groaned under his breath, and Bryce took that as encouragement. Sliding his lips further, he took all he could of Pay's length down his throat. God, the man was big, and Bryce loved it as he bobbed his head, the thick crown sliding over his tongue, filling his mouth with Pay's unique flavor. Bryce loved this. He was talented when it came to cocksucking, and he loved the small sounds Pay made, the groan when he sucked him deep, and the tiny whine when he slid his lips away.
Pay bucked his hips lightly, and Bryce met each movement with more and more suction, encouraging him to take what he wanted, what he needed. For Pay, sex had been about being used, and Bryce wanted more than anything to show him just how mind-blowing it could be. Bobbing his head faster, Bryce sucked harder as he ran his hands up and down Pay's stomach and over his chest, tweaking his nipples lightly between his fingers whenever he passed over them. Pay's breathing became ragged, coming in shallow pants, and Bryce felt him tense. Pay was close, very close, Bryce could feel it. His own desire urged him on, but caution held him back.
Bryce let Pay slip from his lips and wrapped his fingers tightly around Pay's shaft, stroking fast and hard, watching Pay's expression shift from confusion to wonder to ecstasy as he came. Bryce oversaw it all, taking in the total beauty of Pay in the throes of passion. Once his lover stopped quivering, Bryce released him and climbed up Pay's body. Too far gone to stop, Bryce kissed Pay deep and hard, holding him tight as he moved against him, using Pay's come to smooth the way. "You're so amazing," Bryce gasped between kisses, and within minutes his own release was upon him. Bryce cried out, and Pay held him through the shakes. Bryce rested his head against Pay's shoulder, breathing hard, holding Pay tightly in his arms. "I could stay like this forever with you."
Pay hummed an affirmative sound and returned the hug. Bryce eventually dozed off on top of him.
# Chapter 7
BRYCE called John and Jerry once he and Pay got up. They were planning to stop by for doughnuts (Bryce knew that was an excuse), and they agreed to bring his laptop with them. "You don't have to stay with me," Pay said from behind him while he was on the phone.
"Just a minute," Bryce said and moved the phone away. "Do you want me to go? Because I'll understand if you need some time alone." Pay shook his head, his long, loose hair almost shimmering in the light. "Thanks, Jerry," Bryce said into the phone, "I appreciate it." Bryce hung up, and then he and Pay showered and dressed. Bryce wished he'd thought to ask Jerry to bring him some clothes, but he borrowed some from Pay, which he nearly swam in, and they walked over to the store.
Jerry and John showed up with the kids a little while later, and thankfully Jerry had thought to bring him some fresh clothes. Bryce changed in the bathroom. When he came out, the whole group was still huddled around the register, talking.
"Can I talk to you?" John asked, and Bryce nodded before following him outside. "What happened last night?" John asked. Then he clarified, "I don't want details, but I don't think I've seen Paytah laugh like that in a long time." John smirked. "Now I know why Percy was so happy all the time." He jumped back when Bryce took a light swipe at him.
"It wasn't like that," Bryce protested. "Well, it was, but I think he was able to get some things off his chest, and I hope he feels better, about some things, anyway."
"Was it what you thought?" John asked skeptically.
"Yes, but it was much worse than I thought," Bryce said without turning to look in the window. "If Pay ever decides to tell you what happened, you have to believe him, no matter what."
John narrowed his eyes. "Are you telling me what I think you're telling me? That Grantham is involved?"
"Yes, but what happened isn't my story to tell." Bryce shivered slightly at the memory of what Pay had told him. "I will tell you this. Somehow we have got to get that man away from the children on the reservation," Bryce hissed. "I don't care what you have to do, but you must get him away from the kids. Please find out if he's still spending time at the schools." John didn't look convinced, so Bryce played his ace. "If I'm right, would you want Mato anywhere near this man?" John went completely still. "I didn't think so." Bryce turned away and schooled his expression before going back inside.
"Uncle Bryce," Mato called as he rushed over, half crashing into Bryce's legs, holding his doughnut so it didn't get smashed. "Is Paytah your boyfriend now?" he asked in the loudest whisper ever. "I hope so, 'cause he makes great doughnuts."
"You scamp," Bryce said, ruffling Mato's hair.
"Well, is he?" Mato pressed before taking a bite.
Bryce glanced at Paytah, who seemed to be as interested in the answer as Mato was. "Yes, but that doesn't mean you get to mooch doughnuts off of him," Bryce said before adding, "next time go for ice cream." Mato giggle-snorted, watching Paytah, who turned stern for about two seconds before smiling, which sent Mato into renewed peals of laughter. "Go on, you giggle monster," Bryce teased, lifting Mato into his arms before tickling him.
"Come on, Paytah has a store to run and we have things to do too," Jerry called, herding everyone out of the store. Bryce caught John's eye.
"We'll talk back at the cabin," John whispered, and Bryce nodded slightly. The crowd left, and the store quieted. Bryce set up his computer at a small table in one corner and went to work. Not that he got a great deal completed. He found he kept looking over at Pay to see what he was doing.
"You're bad for my concentration," Bryce said with a smile and then turned back to his screen when a customer entered the store. Bryce didn't recognize her, but Paytah talked to the woman briefly and then she began to do her shopping.
"That's Little Wamblee's mother," Paytah said, and Bryce watched as she walked around the store. Bryce noticed the stuff she bought and shook his head. No wonder her kids stole to eat. Yes, she bought stuff for the baby, but she also got a carton of cigarettes and a bunch of magazines. Bryce wanted to say something, but held his tongue. After a few minutes, Little Wamblee joined her, carrying what looked like a six- or seven-month-old.
"Can I hold the baby?" Bryce asked, and Little Wamblee looked at his mother before shuffling closer to Bryce. He took the round-faced boy and smiled at him. Judging from his clothes, the baby was in need of a bath, just like Little Wamblee. "You're a big boy, aren't you?" Bryce asked, tickling the baby's belly, and the baby laughed, opening his eyes and mouth wide, showing his two bottom teeth. "What's his name?"
"We call him Frankie," Little Wamblee supplied, looking nervously from the baby to his mother. "He's a good baby."
Bryce tickled Frankie one more time. "And I'm sure you're a good big brother," Bryce told him, and Little Wamblee smiled. "Since you're letting me hold him, you can get an ice cream out of the cooler."
Little Wamblee smiled excitedly and then glanced at his mother for permission. When she nodded, he ran to the freezer, and Bryce walked a little closer to where Frankie's mother was checking out. "He is a wonderful baby," Bryce told her as Frankie reached for his shirt collar. Bryce continued playing with him, and Frankie grabbed one of Bryce's fingers, taking it toward his mouth.
"Be careful, those two teeth are sharp and he's teething. He'll gnaw you to the bone." Frankie's mother reached into her purse and brought out an old toy, which Frankie grabbed and shoved into his little mouth.
"Are you having any luck with the job?" Paytah asked as he continued working.
"They're giving me a few more hours at the pool in Hot Springs because it's summer, and the diner is steady, thank God, but nothing other than that," she answered. "I'm Hanna," she added.
"Bryce," he replied, returning his attention to Frankie.
Bryce lightly bounced the baby in his arms, Frankie now smiling up at him, and Bryce grinned at the baby and made farty noises that made Frankie laugh. The front door opened and Bryce shifted his attention to look and the smile fell from his face. Frankie began to whimper, and Bryce shifted him onto his shoulder, bouncing slightly while he rubbed his back. Grantham smiled at all of them, and Bryce returned a hard glare. He was tempted to hand Frankie back to his mother, stride up to the smug bastard, and smack that fucking fake smile off his face.
"Morning, Peter," Grantham said, and Bryce sneered.
"His name is Paytah, not Peter," Bryce corrected sharply, and Hanna looked at him curiously. Bryce schooled his expression. "Is there something we can help you with?" Bryce asked, wanting to get him the hell out of there as fast as possible.
"We came in for some ice cream and supplies," Grantham said before moving farther into the store. Bryce was about to ask what the supplies were for when the door opened and three young boys, all with dark hair pulled into ponytails, came into the store. "Get a snack if you're hungry while I get the rest of the food," Mark told them, and they all headed for the freezer, talking animatedly about what they wanted.
"It's okay, Bryce," Paytah whispered just loud enough for him to hear.
"Do you have lighter fluid?" Grantham asked, and Little Wamblee raced back to show Grantham where it was. Bryce cringed, glancing at Hanna, who was looking in Grantham's direction like he was the messiah himself. Still carrying the baby, Bryce wandered over to where Grantham was wrestling a bag of charcoal and a hand basket. "I was going to ask for a hand but I see yours are full," Grantham said to Bryce.
"Yeah," he agreed in a clipped tone, glaring at him. "Make sure your mother doesn't need any help," Bryce told Little Wamblee, and the boy rushed back to where she was finishing up her shopping. Bryce wanted to say something, to tell this asshole that he knew what he'd been doing, but he swallowed the words. He didn't have proof beyond what Paytah had told him. "Where are you camping?" Bryce asked, trying to sound normal.
"Out by the canyons," Grantham answered nebulously. "Come on, boys. Take what you want up to the register because we need to get going."
"Have fun," Bryce said. "But not too much fun," he added in a very low tone. Grantham snapped his head to Bryce, who met his gaze hard and sure. I see you and I know what the hell you've been doing, Bryce thought, refusing to look away. Finally, Grantham glanced at one of the shelves, suddenly interested in cans of beans.
"I'm not someone to mess with," Grantham said between clenched teeth, his former jock personality coming through full force. "Don't go there." His eyes blazed, and Bryce knew this guy had a temper. Bryce also realized that Pay had probably seen that temper and been on the receiving end of it. No wonder he'd been afraid as a kid and had a hard time getting away. This wasn't just an adult threatening a kid, but a man with power who wasn't afraid to use it.
"Neither am I," Bryce said with equal force, watching as Grantham stepped around him and up to the register. The bully wasn't used to people standing up to him, that was for sure. No wonder the asshole stuck to children. Bryce shuddered at the thought.
"Just bill the foundation," Grantham told Paytah as he rang up the purchases. The boys brought up their treats as well, and Paytah rang everything up. Grantham left the store with the three boys trailing behind, and Little Wamblee looking longingly after them. Bryce rolled his eyes at Pay, who looked concerned, confused, and maybe a touch afraid.
"I applied to that foundation through the school for Little Wamblee to join their after-school activities and things, but they said he's too young," Hanna said.
Bryce thanked heaven for small favors on that count, but couldn't help glancing at Pay, who cringed slightly. Little Wamblee ran over and showed what he'd picked out to Pay, then he opened it and began eating, fast, like it was going to melt in three seconds.
"Your mother was a basket weaver wasn't she?" Pay asked Hanna, who nodded. "Did she teach you?"
"Yes, and I used to weave in my spare time, but that was a while ago, and I haven't done anything since Frankie was born. There are still a few of my pieces in the mission store, but I haven't sold many. Why?" she asked as she handed over the money for her purchases.
"Bryce has an idea that we're looking into, and if it gets off the ground, we'll contact you. It may be an opportunity," Pay said, and she lifted her purchases.
"There aren't many of those around here, so whatever you're thinking, if it helps bring money into this place, it will be a godsend," Hanna said.
Little Wamblee had already finished the ice cream bar and dropped his trash in the can. "Mato and Ichante are in town for another few days. Maybe Little Wamblee could come out to the cabin to play one afternoon," Bryce offered.
"Those are Akecheta Black Raven's kids, right?" she asked. Bryce didn't quibble that he was actually their uncle and nodded. "I think he'd like that." She smiled and then left the store. Bryce handed Frankie back to Little Wamblee, who hefted him into his arms and followed behind his mother to an old car that Bryce figured was only still running by the grace of God.
"What am I going to do?" Pay asked almost as soon as the door closed. "I can't let him get anywhere near that boy. But what if no one believes me?"
Bryce wanted to tell him that they would believe him, but he knew better. Everyone seemed to be enamored of Mark. "Then we'll have to make them believe it, but if that's what you want to do, I think you should try telling your friends first."
"Do you think so?" Pay asked, and Bryce nodded vehemently.
"You need to have support," Bryce answered, and Pay became quiet. Bryce remembered that he needed to call his mother and stepped outside before dialing her number. "Hey, Mom."
"Hello to you too," she said.
"You must be feeling good," he said with a grin. He loved that she was feeling better.
"The doctors gave me the okay to travel, and I thought I'd come for a visit."
"That's what I wanted to talk to you about. You know I'm at the reservation with John and Jerry for another week," he explained.
"I can come when you get home," she clarified. "I don't need to rush there this second."
"That's not it. I told you about Paytah when I called last and... he told me some things, and I was wondering if you could come here. I think he needs to talk to you."
"Isn't there someone there he can speak with?" she asked.
"No, Mom, and I doubt he'd talk to just anyone. He's agreed to talk to you because you're my mother," Bryce said. "It's important."
She remained quiet for a long time. "Did he tell you what happened?"
"Yes, and believe me, it hurt him. And there are other children still in danger." Bryce thought about how he could explain it to her. "Do you remember the patient you had when I was sixteen, and after he opened up you didn't sleep for five days?"
"Yeah," she said.
"It's that important," Bryce said, and he heard his mother gasp softly.
"I'll make arrangements to get there. You'll have to find me a place to stay while I'm there." He could already hear her moving around. "Mom, take it easy, we're all here for another week, and I can probably stay after that and work from here," Bryce explained, but he knew his mother. There was no keeping her back if she thought she could help. She'd been that way when he was a kid, and the loss of energy had been the hardest thing about her being ill.
"I need something to do. Having everyone around here treating me like I'm about to break is driving me crazy," she said, but Bryce knew there was more to it than that. His mother spent a lot of time alone—maybe too much time. She'd always been a very social person, but when she got sick, she'd been confined to either the hospital or home.
"Okay, but be careful and don't overdo it. I'll ask John to send you directions from the highway, and you can call and let me know when to expect you." Bryce's worry ramped up. Maybe this wasn't such a good idea.
"Don't mother-hen me. I'll be just fine. I'm not leaving this very second. I'll get packed and make some arrangements. I'll probably leave sometime tomorrow." She paused, and Bryce heard what sounded like dishes clinking. "I'm looking forward to seeing my wayward son."
"I haven't been that wayward," Bryce protested with a smile he knew she couldn't see. "Things have been hard, but I think they're getting better now, for both of us."
"I'm looking forward to meeting this young man," she said. "Now, I need to get things done, but I'll call you when I leave." They said good-bye and disconnected, and Bryce went back into the store.
"My mother is coming," Bryce told Pay, who appeared a bit nervous. "She's going to love you," Bryce added, "and she's one of the nicest people you'll ever meet. She's also tough as nails and has talked with and helped people who have been through extremely difficult experiences."
"But what if she blames me?" Pay asked.
"Oh, fuck," Bryce muttered to himself. "What he did to you is not your fault. You were a kid, and he used his power to hurt you."
"But he always said I led him on," Pay said softly.
"That was his way of making you feel obliged to him. Mom told me once that's what predators do. They make the people they're hurting think they deserve to be hurt, and that's what Mark did to you. I know that, and Mom will too. She's a smart, caring person, and I know she can help you." Bryce walked around the counter. "Last night you said she couldn't change the past, and she can't, but what she can do is help you realize how Mark hurt you so you don't feel like anything that happened was your fault." From the look on his face, Bryce saw Pay didn't believe him. "Let me ask you this: Would you think Little Wamblee led him on if he was to tell you that Mark had hurt him?"
Pay thought for a few seconds. "No. He's an innocent kid."
"So were you when Mark hurt you," Bryce said emphatically. "And he alone is responsible for his actions."
"But what are we going to do to keep the kids safe?" Pay asked urgently.
"I don't know," Bryce said. "I'm not plugged in to the way things are done here. John is," Bryce offered, "and you are...." Bryce let his suggestion hang in the air.
Pay sighed but didn't say more, and Bryce didn't push. It had been hard for Pay to tell Bryce what happened. It had been painful for Bryce to hear. Telling others was not going to be a picnic, and if Pay decided to keep it to himself, then Bryce would have to find another way, but somehow he was going to close off Mark Grantham's access to the reservation's children. He'd thought about going to the police, but he had no proof. No, involving the police was something Pay would have to do. Bryce returned to his computer and his work, trying hard to keep his mind on his task.
BY THE end of the day, Bryce had racked his brain to come up with some sort of plan, but he was no further along than he had been that morning. He had eventually managed to get some work done, but he'd had a devil of a time concentrating on anything other than Pay and that asshole Grantham.
"I'm going to lock up the store. Are you ready to go?" Pay asked, and Bryce began the shutdown process on his laptop.
"Yes. This will just take a minute," Bryce said, waiting for the power to shut down before closing the lid. He unplugged the computer and shoved it and the cords into his bag. After making sure he had everything, he followed Pay outside.
"I'll drive you back to the cabin," Pay said, and Bryce nodded. He wasn't going to force himself on Pay if he wasn't wanted, so Bryce followed Pay to his car and got inside. They rode in near silence down the darkening roads before pulling into the drive. Bryce got out of the car, expecting Pay to pull away, but he followed him instead.
Pay approached the door apprehensively, and Bryce held it open for him. He wasn't sure what Pay had in mind and he wasn't about to ask.
"Uncle Bryce," Mato said as they entered, hurrying over. Ichante was nowhere to be seen at first, but she came out of the bedroom a few minutes later, carrying one of her dolls. "Did you bring doughnuts?" Mato asked.
"Mato!" John scolded sharply as he approached. "That's rude."
"Sorry," he said softly, hiding behind Bryce.
"Come in and sit down," John told Pay. "Can I get you anything? We're about to start dinner. I hope you can stay." John was already heading to the kitchen. "I have soda and water."
"Water, please," Pay said before slowly lowering himself to the sofa. Bryce sat next to him, and John brought a soda and a bottle of water, setting both on coasters on the coffee table. "There's something I want to talk about," Pay said softly.
"I sort of gathered that," John said. "Let's have dinner, and we'll put the kids to bed," he suggested before returning to the kitchen. He and Jerry cooked a quick dinner, and everyone went to sit at the table. Dinner was a bit subdued. The kids were obviously tired and their usual energy seemed absent. Once they were done eating, John got them ready for bed. Bryce helped clean up, and after the dishes were done, everyone gathered in the living room, Bryce sitting next to Pay on the sofa.
Bryce looked to Pay, who looked back to him. "Pay needs your help," Bryce said, addressing his remarks directly to John, but glancing at Pay, who nodded his agreement. Bryce took a deep breath and began. "I'm not going into details, but Pay did with me last night." Bryce looked at Jerry and John. "Pay took part in the activities of Mark Grantham's foundation when he was a kid. John knows who he is."
"I've met him as well," Jerry said. "Personable, but a bit weird."
"Eccentric," John said, and Bryce ignored it.
"The thing is that he used that contact to get close to Pay, and then used that closeness to his advantage." Bryce moved closer to Pay. "He told me all the terrible things that went on, and I believe him."
"What happened?" John asked.
"Mark Grantham used his position with his foundation to get close to Pay, sexually. It started when he was eleven, and Pay was finally able to get away from him when he was about fifteen." Bryce locked gazes with a disbelieving John. "Pay was fourteen," Bryce said, glancing at Pay for confirmation, "the first time he penetrated him." John went whiter than Bryce. "They had an illegal and coerced sexual relationship until Pay managed to get away from him."
"My God," Jerry said.
"Yeah," Bryce continued. "The thing is, as far as we know, he still has access to the school, and if he's done this to Pay, he's done this to other kids and is likely still doing it." Bryce looked toward the closed door to the kids' room and then back at John. "There's no mistake. What Pay told me last night was traumatic and real. It happened, and to make matters worse, he told the counselor at school back then, who did nothing." Bryce took Pay's hand. "Everyone here seems so enamored of Grantham that they've allowed a predator to remain in their midst, one who is probably still preying on the children."
"It's hard to believe," John said without much conviction.
"Of course it is, but when I was a kid, Mom told me that people who might want to hurt me look like everyone else. In cartoons, the bad guys always look like the bad guys. Even in the movies, we know who the bad guys are because, eventually, they look like the monsters they are. That isn't true in real life, and predators like him count on that." Bryce knew it was going to take more to convince John.
Jerry stood up and walked to where Bryce and Pay were sitting and wrapped Pay in a hug. "I believe you," Bryce heard him say. Jerry released him, straightening back up. "If he comes anywhere near our kids, I'll kill him."
Bryce looked to John, who still looked shocked. "He's... damn it," John said. "I believe you, I do," he added. "He's been important to many people on the reservation for years, and he's supposedly helped so many people. Has he been...," John sputtered. "Damn it."
"It's the trust he's built on the reservation that has allowed him to do what he's been doing. If a few kids make noise, they're shut down because of Grantham's reputation. The kid is just a kid like Pay was, and Grantham's done a lot for people, so they believe him and turn a blind eye."
"The counselor thought I was lying, and he told the principal, who called Mark and told him," Pay said.
"Jesus Christ, these are the people who are supposed to be protecting our children," John swore, and Bryce knew they'd truly broken through. "What do we do?"
Everyone looked at everyone else, but no ideas came. "Paytah could go to the council," John suggested, but Paytah vetoed that, and Bryce couldn't blame him.
"They aren't any more likely to believe it than you were at first," Bryce said, "and there are more of them to convince. No, I don't think...." Bryce swallowed. "What if they have already had reports, like Pay telling the counselor at school?" Bryce turned to John and then to Pay. "John told me when I first met you that he remembered you being a happy, outgoing person. He wondered at the time what happened to change you, and now we know. I asked Kiya about it when we were at the water park, and she said she'd heard vague rumors, but either didn't remember or wasn't going to pass on gossip."
"You asked about me?" Pay questioned.
"Of course. You caught my attention." Bryce leaned closer. "It was really the hair," he added with a wink before returning to the subject. "If Kiya might have heard things, then the council may have heard things, or worse, may have been told things and didn't pursue it because it would mean the help and resources Grantham brought to the tribe would stop."
"So where does that leave us?"
"Nowhere," Pay answered, and Bryce could almost feel him turning inward.
"Not necessarily. My mother is coming," Bryce announced. "Pay has agreed to speak with her, and I suspect she may have some recommendations for us. She's dealt with a lot of situations like this, including helping people with the authorities. She'll know what to do." God, he certainly hoped so.
"Laura is a force to be reckoned with, Paytah. She's one of the most understanding, patient people I have ever met, but she has a spine of steel and she knows the system and how to put it to work," Jerry said before turning to Bryce. "When will she be here?"
"In a couple of days, but she'll need someplace to stay," Bryce said.
"No problem. I'll call Mom," John said. "She'll have room for her, and with Dad leaving in a few days, having a guest will help keep her mind off things. She always gets a bit depressed and lonely when Dad leaves, so this might help."
Bryce yawned, and the others followed right behind him. Jerry and John said good night. "Turn out the lights when you're ready," Jerry said as they left the room, closing their door behind them.
"I should go," Pay said and stood up, walking toward the door. Bryce got up as well and followed him. "I'll see you soon?"
Bryce said nothing, moving close and tugging Pay into a kiss. "How about you stay here with me?"
Pay looked around the cabin. "Are you sure... with everyone else here?"
"Yes, I'm sure. We need to be sort of quiet, so the trapeze is out, but I think we can manage to spend the night sleeping next to one another without screaming our lungs out." Bryce shifted slightly, sucking on one of Pay's ears. "Although it's definitely going to be difficult," Bryce added before taking Pay's hand and leading him through the cabin to his room. Bryce turned out the lights, and then they stepped into his small, dark room. When Bryce closed the door, they were in their own, small world.
The clicking of crickets and other sounds of the night drifted in through the open window as Bryce slowly removed his clothing, letting it fall to the floor. He could see almost nothing, but knew exactly where Pay was at any moment. He could feel his heat, and Bryce was drawn to him even in total darkness. He heard Pay moving, and then the bed squeaked slightly. Bryce moved toward the sound and felt hot skin as he slid his leg against Pay's. Then Pay stroked Bryce's belly and around to his hips.
The bed welcomed them as Bryce pressed Pay back onto the mattress. "You're so hot," Pay whispered, and Bryce shivered at his light touch even in the warmth of the summer night. "When I first saw you I wondered if you might be gay, and then when I sort of knew you were I didn't think you could possibly be single," Pay told him. "You're way too cute to be single."
"I'm not sure I like that. You know being cute is the kiss of death," Bryce said, and he felt Pay roll onto his side, tugging him close.
"Not as far as I'm concerned," Pay said, and Bryce rolled over to face him. He wasn't going to argue and instead kissed Pay deeply, pressing their bodies together, lips to lips, chest to chest, hips to hips. Pay moaned softly, and Bryce pressed closer, moving slightly.
Pay moaned again, and Bryce kissed him hard, silencing him even as he increased the intensity of the sensation. "I want you so badly," Bryce whispered. "I can't wait until I can have you inside me." Pay whimpered, and Bryce felt him thrust harder and faster. "I want to be able to make love to you properly and fully, join with you in love and make everything from before pale in comparison."
Bryce felt his own desire rise at his words. He wanted Pay badly, and for so many more reasons than just hopefully wiping away some of Pay's past experiences. He was falling for Pay, and with every stroke, heartbeat, and touch of Pay's skin on his, Bryce fell a little farther. He wasn't sure if that was a good thing, but it was happening. Pay covered Bryce's mouth with his, holding him as tight as Bryce held Pay, each silencing the cries from the other that threatened to split the night. Bryce could feel the same urgency and need that welled inside reflected back at him by Pay. In that moment, they were one, and they moved together in a dance that became more and more urgent with each touch and each snap of their hips, Bryce moving past Pay, who then moved past him.
"Can't last," Bryce gritted through his teeth before kissing Pay hard, pouring all his feeling into the touch of their lips and bodies.
"Let it go," Pay whispered, and Bryce clamped his eyes closed, desperately trying to swallow the cry threatening to erupt. He buried his face in Pay's neck, mouth open as he came, his entire body shaking. On the periphery of the thralls of ecstasy, he realized Pay was coming as well. Bryce held Pay tight, and he felt Pay holding him. They clung together to hold their vocal cork in the bottle until the shakes and pressure passed and Bryce was left with the warmth of afterglow flowing through him.
Neither of them made any effort to move. "I like when you hold me," Bryce said without opening his eyes. He was too content and happy right where he was to move a muscle.
"I like it too. I didn't realize how much I needed it," Pay said, and Bryce moved upward slightly before lowering again as Pay took a deep breath. "How much longer are you staying?" Pay asked.
Bryce lifted his head to try to see Pay's eyes, but it was too dark. "We're supposed to leave at the end of the week. We stayed this long because John's dad had time off from the oil fields and John and Jerry wanted a chance for him to spend time with the kids. I can stay another week after that, but then I have clients I need to meet with back in Sioux Falls." The thought of leaving left a hole in his heart. Up till now, he hadn't given it much thought as far as Pay was concerned. He'd been happy and was enjoying Pay's company, but then things had begun moving pretty fast and now... Bryce swallowed hard as he realized he didn't want to leave.
Bryce felt tension building in Pay, his body coiling beneath him. "So all those ideas you had, all those things, they were just talk. You threw them out there, making big plans, and what? You're going to leave?" Pay shifted, and Bryce rolled a little on the bed as Pay stood up. He snatched a few tissues and quickly wiped his skin.
"Where are you going?" Bryce asked. He heard the zip and snap of denim as Pay's legs slid into his jeans.
"What does it matter? You're going to leave," Pay said.
"It matters," Bryce began, reaching for Pay's arm, "because I don't know any more than you do what's going to happen. Yes, I have to leave because my work is there. We've been able to work from here, but it's been difficult for all of us to get done what we need to." Pay pulled away, and Bryce got out of bed. "I don't know where things are heading between us, but I want to find out." Pay turned around, and Bryce sighed. "Things have happened rather quickly between us." Pay pulled away again. "I'm not saying that's bad, just that we don't need to rush. Yes, I have to go, but I'll definitely be back." Bryce took Paytah by the hand, and relief washed over him when he felt Pay let him tug him back onto the bed.
"What if you go back and you meet someone else?" Pay asked as he settled on the bed again, his denim pants scratching Bryce's legs. "There are going to be lots of guys interested in you." Pay placed his head on Bryce's shoulder. "I can't compete with them."
Bryce stilled as he realized just what he was being told. All those years living with his mother had taught him a few things. "It doesn't matter who's interested in me, because I'm not interested in them," Bryce reassured Pay, lightly stroking his hair as he read between the lines. "Besides, we have work to do, artisans to speak with, and people with traditional craft skills to find." Pay nodded but didn't say anything more. "Come back to bed," Bryce said, and Pay finally moved, removing his pants and then settling in the exact same spot. His mind in high gear, Bryce resumed stroking Pay's hair.
"Your heart's racing," Pay said softly.
"That's because of you," Bryce lied. Well, it wasn't a lie, exactly, Pay did make his heart race, but right now it was anger that had his heart pounding and his blood racing so fast he thought he could hear it. Pay came across as the strong, silent type, but that silence and stern demeanor hid a fragile soul. Bryce knew Pay had been hurt, but what he hadn't taken into account was the deep insecurity the abuse Pay had suffered had left with him. Pay had just given him a brief glimpse into the depths of that pain, and Bryce's heart ached both for Pay and to make the cause of that pain suffer. "I'm not going anywhere for a while, and I certainly won't leave without saying good-bye and telling you when I'll be back."
Pay lifted his head, and there was just enough moonlight shining through the window for Bryce to see the hope tinged with underlying doubt in Pay's eyes. If that look had come from Mato, it would have been accompanied by a, "You promise?" But Pay's questions remained unsaid even if his eyes asked them as clearly as words.
Bryce continued stroking Pay's hair, and the doubt slowly faded from Pay's expression. Bryce knew he could try to reassure Pay with words all he wanted, but true reassurance would only come from his actions. Eventually, Bryce saw Pay close his eyes, and he lay in the bed, staring at the dark light fixture in the middle of the ceiling. Mark Grantham, you are going to pay for what you did to Paytah and everyone else you hurt. Bryce didn't know how or when, which was eminently frustrating and made his vow in his head sound a bit stupid, but somehow he was going to suffer the consequences. Bryce smiled as an image of him as Scarlett O'Hara came to mind, specifically, the scene of her in the field with the sunset behind her. "With God as my witness, I will rip his fucking nuts off!" Bryce knew Scarlett had never said that, but he figured in this case, she just might.
# Chapter 8
"HI, MOM," Bryce said as he pulled open her car door. She got out, and Bryce hugged her. She was thinner, but seemed stronger since he'd last seen her, and she returned his hug with some of her old gusto. "I've missed you," he said, "and I'm glad you were able to come. Sorry Dad couldn't make it too."
"So is he. But he sends his love," she said, hugging Bryce again. "Something's wrong," she said once she took a look at him.
"Not really. I'm just a little worried," Bryce said, glossing over things a bit. Pay had been, well, a bit more distant in the past few days. He hadn't stayed away physically, but Bryce could feel some of Pay's old walls starting to be rebuilt.
"Do you want to tell me about it?" she asked as Bryce stepped away from her. "Does it have to do with the reason I'm here?" Bryce sighed softly. "I'll take that as a yes."
"He's pulling away, and while I understand why, I don't like it," Bryce said, but he purposely didn't tell her what he thought. His mother would talk to Pay and make up her own mind. "Come inside and I'll introduce you." Bryce led his mother to the door of the trading post. He'd given her directions there, figuring it would be much easier to find than the cabin or John's parents' house. He was also anxious for her to meet Pay.
Bryce held the door, and she walked inside, looking around. "This is like a step back in time," she said. "It reminds me of the small store near where I grew up." She looked all around until her gaze fell on Paytah. Bryce watched as her eyes widened. Slowly she walked to where Pay stood a bit stiffly behind the counter. "You must be Paytah. I'm Bryce's mother, Laura Morton." She held out her hand and Paytah shook it. "I love your place here," she told Paytah, and he smiled.
"Thank you, ma'am," Pay said formally.
"Please call me Laura," she said with a smile. "So, you're the man who's got my son's heart all aflutter." She actually winked at Paytah, and Bryce saw him smile back at her and then at him. "Not that I blame him." Laura opened her purse and pulled out a tissue, using it to wipe her eyes. Her eyes watered more since her treatments.
"Bryce said you could help me," Pay said softly.
"I will certainly try."
"Then how do we do this?" Pay asked nervously. Bryce wasn't sure if he was anxious to talk to someone or simply wanted to get things over with.
"Do you need to rest after your trip?" Bryce asked. "I can get you to John's parents' and you can talk tomorrow."
"I'm fine, dear," she told Bryce before turning back to Paytah. "I suggest we go someplace quiet and comfortable for you where we can talk," she said. "I'll listen, and you'll tell me whatever you want to. What's said between us stays between us, and that goes even for Bryce. If you wish to tell him what we talk about, that's up to you. But I will divulge nothing unless you give me permission."
"I'll watch the store for you, Pay," Bryce offered, joining Pay behind the counter. "Don't worry. She's my mother. Let her help you." Pay nodded and walked around the counter. Bryce wanted to pull him back and kiss him just to reassure him, but held back. They'd never discussed public displays of affection.
Pay led the way out of the store, and Bryce watched through the window until they disappeared from view. Bryce hadn't really planned on his mother and Pay talking right away, but he was thankful he rarely went anywhere without his laptop. He pulled the computer out of his bag and set it up next to Paytah's at his desk. The trading post was quiet, so Bryce got to work.
The tinkling of the bell on the door broke his concentration, and he checked the clock. He'd already been working an hour. He looked up, expecting to see his mother and Pay, but instead saw Little Wamblee's face peeking over the counter. "Where's Paytah?" he asked in a rush, and Bryce heard a wail from below the counter.
"He's away for a while," Bryce said. "Is something wrong?"
"Frankie won't stop crying, and Mama's at work," he said, biting his lower lip. "I made up his bottle, but he's not hungry and I just changed him." The boy was two seconds from crying too, now.
Bryce hurried around the counter and saw the red-faced baby in Little Wamblee's arms. Bryce lifted Frankie and carefully placed him on his shoulder. The baby curled his little legs under him and let out another wail. "It's okay, Frankie," Bryce said as he walked the baby around the store. "When did he eat last?"
Little Wamblee looked at the clock on the wall. "Two and a half hours ago," he answered nervously.
"It's okay," Bryce said as he continued walking, jiggling the baby slightly. Frankie calmed and stretched out for a few minutes before crying once again, trying to curl his little legs under him. Bryce patted his back and continued walking. He considered going to Pay's to get his mother. He checked Frankie's forehead, but he didn't seem to have a fever. "He's going to be okay," Bryce said to reassure Little Wamblee, even though he had little idea what he was doing. "Does your mother leave you alone with the baby often?"
"No. The babysitter was supposed to come, but she never showed up," Little Wamblee said.
"It's okay," Bryce said, relieved this wasn't a regular occurrence. "It's okay," he repeated for both of them. Bryce had hoped it was trapped gas or something like that, but Frankie was obviously in pain. He was about to go over to Pay's to get his mother when a familiar car pulled up in front and Kiya got out. Bryce had never been so happy to see anyone in his life.
"What's this?" she asked as soon as she entered the store. She took Frankie and cradled him. "It's colic," she said. "He's gotten this before." She walked him and sang to him softly. Frankie calmed, and then everyone in the store made the dirty-diaper face.
"Get a package of diapers, cream, and wipes from the shelf," Bryce told Little Wamblee, and he hurried away. "Oh, God," Bryce said as the smell permeated everything.
"Actually, that's good. It means his little system is taking care of things again," Kiya said as Bryce held his nose. Little Wamblee returned, and off toward the bathroom they went. The room became quiet, with only the occasional cry from the bathroom. Bryce breathed a sigh of relief. He went to the shelf and looked up the price of the things Little Wamblee had taken, then returned to ring them up and pay for them.
Little Wamblee and Kiya returned with a freshly diapered Frankie in her arms. She carried him to the counter. He looked much more contented now, but still not happy or smiley like he'd been with Bryce a few days earlier. "He needs a bottle and a nap," she said before gently handing Frankie to Little Wamblee. "Do you know the number for your mother at work?"
"They don't like it when we call her," Little Wamblee said, but he gave her the number and she dialed. Like Moses parting the Red Sea, Kiya soon had Hanna on the phone. She explained the situation like a pro.
"I'm going to take Little Wamblee and Frankie home with me. They can spend the day, and you can pick them up when you get done with work." She paused to listen. "It's no problem. Don't worry; I'm glad I can help." They talked briefly, and then she hung up. "Do you have a car seat?" she asked Little Wamblee, and he nodded. "Okay. We'll stop by and get it." She took the baby, and Frankie curled onto her shoulder, already looking sleepy and much happier. "Oh, before I forget, I came in to see if your mother made it okay," she told Bryce.
"She's with Pay now. They've been talking quite a while," Bryce said, and she nodded.
"That's good. Call me when your mother leaves so I can make sure everything is ready," she said.
Bryce gave Little Wamblee a bag for the baby supplies, and they left the store, the youngster swinging the bag back and forth as he walked. A few minutes later, other customers came in, and a few of the older men in the tribe stopped by for their coffee. They then sat outside on one of the benches Pay kept in front. The more time Bryce spent at the trading post, the more he realized this place was the center of the community.
"Do you need anything else?" Bryce asked as he poked his head out of the door. The old men shook their heads in unison, and Bryce went back inside. Customers came and went for the next hour, keeping Bryce remarkably busy. "Where's Paytah?" almost every single one asked.
Bryce always provided the same answer. "He's away for a bit. I'm a friend of his and Akecheta Black Raven's." That seemed to satisfy everyone. He got a call from Kiya, saying Frankie had eventually eaten and fallen asleep, with Little Wamblee watching over him.
Finally, after more than two hours, Pay and his mother walked back into the store. Pay was definitely a bit shell-shocked but he also smiled quite a bit and had surprisingly little tension in his posture. "Thank you, Laura," he said, taking her hand. "You helped a lot.'
"I'll be here for a few days. We can talk again," she said before turning to Bryce. "I'm going to get settled. Can you give me directions?"
"Of course. You're staying with Kiya," Bryce told her. They'd met a number of times at John and Jerry's, so Bryce felt sure his mother would be comfortable. He wished she could have stayed at the cabin, but there wasn't enough room. Bryce gave her directions as well as the phone number in case she got lost, but that wasn't likely. His mother had an uncanny sense of direction. "I'll see you for dinner at the cabin. It's not hard to find from Kiya's." He hugged her, and she kissed his cheek before heading out to her car. Bryce followed her with his eyes and out of curiosity walked to the door, where he saw the old men had all stopped what they were doing to watch as his mother got in her car. She'd always been attractive, and Bryce knew she would have seen them watching.
"How did it go?" Bryce asked as he walked back to where Pay was already behind the counter. "Did she help?"
"Yes," Pay said. "She was able to help me understand that what happened to me wasn't my fault."
"I know it wasn't."
Pay sighed. "I always wondered why he chose me. Did I do something to lead him on or something special to capture his attention? Laura said he probably chose me because I was a vulnerable and caring person who wanted to please him." Pay took a deep breath. "I know I was. I wanted him to notice me, and I wanted to make him happy. But that didn't give him the right to do what he did."
"It didn't then and it doesn't now," Bryce said. "You know he's probably abusing other children right now?"
"I know, and that's what hurts most. If I go after him, then everyone will think I'm a liar, when it's him who's fooling everyone," Pay said sadly. "Laura understood that too. She said she'd think on it for me." Pay groaned softly. "I want all this to go away. I have for so long." He sighed again. "Your mother helped me see that it won't no matter how much I want it to unless I make it go away."
"She said that?" Bryce asked.
"Not in so many words, but I'm tired of being afraid of him and what people will think." Pay's gaze met his, full of the fire Bryce usually only saw when they were making love. "He made me afraid of everyone and everything, including life, and I want it to end."
"Then it will," Bryce said.
"That's what Laura said too," Pay told him. "So did anything happen while I was away?"
Bryce told Pay about Little Wamblee and Frankie. "They're at Kiya's, and Hanna will pick them up after she's done with work."
"We have to help them somehow," Pay said.
"I know," Bryce agreed as Pay picked up the phone.
BRYCE rode with Pay back to the cabin. He'd planned to spend what was left of the day working. Pay had called the artist he'd spoken to a few days earlier, and Bryce had spent much of the afternoon with Running Deer, or Dave, as he preferred to be called. Bryce had explained what he had in mind, and Dave had loved the idea and seemed willing to champion it. "You don't want anything for the idea?" Dave had asked.
"No. The only thing I want is for the profits to go to the tribe to help those in need," Bryce had said. "I got the idea when we were out doing things with the kids, and I don't have the skills to start something like that. I'm a computer guy; it's what I love. You and the other artisans on the reservation know about making and selling your craft, I don't. Now, when it comes to websites and online commerce, I can help."
They talked for quite a while. Dave agreed to come up with some product ideas and was obviously excited. "Anything that helps bring business and hope to the reservation is a good thing," Dave said thoughtfully before he left. They exchanged contact information, and Bryce explained that he was returning to Sioux Falls in a few days, but that he expected to be back on a regular basis. He'd noticed the unhappy expression on Pay's face and wished he could change it.
"I hate that you're leaving," Pay said as they rode toward the cabin.
"I know. I don't want to leave either." Bryce liked being there. "But you know I'll never fit in here." As he said the words, it felt like fingers squeezed his heart. "I'll always be an outsider, no matter what I do. This afternoon, the old men sat in their usual place, and when I asked them if they needed anything they shook their heads and looked away."
"They're old and set in their ways," Pay explained as he drove.
"Pay, I watched the store for you today, and almost everyone who came in asked where you were and looked at me like they expected me to make off with everything in the store. I do like it here, and I'm developing feelings for you that I haven't felt since... well, since Percy. But I can't stay here indefinitely, and I don't think I can live here." Bryce wanted to ask Pay to come back with him, but his brother, the store, and his friends were all on the reservation. Pay pulled through the tunnel of foliage and parked next to the van. He turned off the engine, but didn't open his door.
"What do we do?" Pay asked.
Bryce reached across, taking Pay's hand. "I don't know. But I'm not going to walk away from you. I'll set you up with Skype and we'll talk over the computer." Bryce squeezed Pay's fingers lightly. His heart was engaged again, and Bryce wasn't letting go unless Pay wanted him to. "I don't have all the answers, so the best I can give you is we'll have to see." Bryce had intended to come here with Jerry and John, work, hold a few classes, and have a bit of fun. He hadn't imagined that Pay would capture his heart.
"You will be back?" Pay asked softly.
"I promise," Bryce said before leaning in. He kissed Pay lightly and was about to pull away when Pay cupped his head with his hand, deepening the kiss. Pay pressed for entrance, and Bryce parted his lips. He heard a seat belt open and then snap back before Pay leaned over him, pressing Bryce back against the door. Bryce wrapped his arms around Pay's neck and held on as Pay took possession of his mouth. Bryce whimpered as his pants tightened and his dick throbbed. He thought about asking to go back to Pay's for a while, but it was too late. Pay kissed him deeper, his tongue and lips tugging, reaching to Bryce's soul.
Pay pulled away, and Bryce blinked up at him, taking a deep breath. "What was that for?" Bryce asked. "Whatever it was, I'll do it again."
"Just so you remember your promise," Pay answered seriously. Bryce nodded slowly, knowing he'd probably remember that kiss for the rest of his life. Bryce shivered with excitement, eyes wide, brain switched off as he panted in the now warm car. Pay stared back at him as if to emphasize his point.
"I always remember my promises," Bryce mumbled, and then he jumped when someone tapped on the window behind him. Great, nothing like being caught making out by your mother, Bryce thought as he slowly opened his door.
"You two having fun?" she asked seriously but quickly smiled. "You know you should probably do that when you don't have two kids watching out their windows." Bryce rolled his eyes, and his mother flashed him one of her scolding looks.
"Good God, Mother, those looks don't work anymore," Bryce said, getting out of the car and then closing the door with more force than was necessary.
"Sure they don't," she retorted, carefully walking around to the front. Bryce waited for Pay and took his hand before following her. They easily caught up, and Bryce held the door.
"Uncle Bryce and Uncle Paytah were kissing," Mato said before sticking out his tongue and closing his eyes in the international sign for "yuck."
"Don't knock it till you've tried it," Bryce's mother said.
"Mato, help your sister set the table," Jerry said with a dish of pasta in his hand. "It's great to see you, Laura. We're both so glad you're doing better." Jerry leaned in for a cheek kiss before heading to the table. "Dinner will be ready in five minutes," Jerry told them. "Kids, finish up with the table and then wash up." Silverware clanged a few times, and then Mato ran to the bathroom ahead of his sister.
"They have such energy," his mother commented, and Bryce took her hand.
"You'll get yours back. It will just take a little more time."
"Oh, don't get me wrong," she said to Bryce. "I'm feeling better than I have in a long time. I just wish I had some of their energy." She set down her purse and went into the kitchen, where Bryce heard her greeting John.
A few minutes later, they all gathered around the table. Bryce looked over the sumptuous Italian feast on the table and thought a bottle of red wine would be perfect, but with both John's and Pay's aversion to alcohol, they had ice tea instead. "So how do you find our little backwater?" John asked Laura a little sourly.
"It's quite nice, but I do see what you mean."
"Bryce has plans to turn it into a hub of commerce," Pay said, bumping Bryce's shoulder.
"We're going to try," Bryce said as the dishes were passed around. Jerry and John helped the kids, and Bryce held the heavy bowls for his mother. He explained his craft idea to her. "Dave thinks he can design a number of wooden items, including pipes and drums. He knows Hanna, Little Wamblee's mother, and he's going to work with her to design some simple baskets. He also said that molds could be made for traditional pots that could be hand-painted with simple traditional decorations. We're also going to see about dolls and other traditional toys."
"You know," Laura said, "I bet you could get a grant to help with the start-up costs. There are a number of organizations and government programs designed around the preservation of Native American crafts." She set down her fork. "You'd have to check into the requirements, but that could provide some of the initial cash. You'd need to develop a business plan, of course."
"I figured that, but don't know the first thing about it," Bryce said.
"You could check that Internet you're always using," she told him with a grin.
Pay nudged him slightly. "If you want this to be something for the tribe, then they need to help. I'll put out feelers to see if anyone can help put that together."
"I think Mom has experience writing requests for grants and things. She's the one who helped get the money for the school ten years ago. She can probably help with the business plan too," John said, and Bryce instantly felt better. He might have come up with the idea, but it needed to be the tribe's project to execute and he was thrilled to see that starting to happen.
"Did you bring doughnuts?" Mato asked.
"You know he doesn't have to bring you doughnuts every time he comes to visit," John scolded, and Mato looked down at his plate.
"Sorry," Mato said softly. "Doughnuts are good," he added in his defense.
"Finish your dinner and you can have a fruit pop," John said, and Mato picked up his fork, mollified for the time being.
"We saw you kissing," Ichante told Bryce with a snicker.
"Finish your dinner," John reminded her gently, and she stabbed a few pieces of pasta.
"Grandpa said we're going to get buffalo," Ichante said before popping the pasta into her mouth.
"That's not what he said, exactly," John corrected. "But it seems Bryce's idea is gaining some traction. Dad called this afternoon to say good-bye and he told me that the council met on Monday and the topic of discussion was your suggestion to return buffalo to the reservation. Dad said they loved the idea and were going to look into the possibility and requirements for starting their own herd. They feel the remaining grasslands and meadows are large enough to support a herd of about five hundred head, and it would be like returning some of our history and heritage to the land." John smiled at him, and so did Pay, who bumped him lightly with his shoulder again. Bryce reached under the table, taking Pay's hand.
The conversation around the table continued. When the kids were done, Jerry gave them each a fruit pop and made them sit at the table to eat them. The adults lingered around the table over decaf coffee until it was time to put the kids to bed.
"We have a problem I'm hoping you can help with," Bryce said, addressing his mother once the door was closed to Mato and Ichante's room and everyone else was seated in the living area.
"I sort of assumed that," she said, and Bryce looked to Pay.
"The man who"—Pay swallowed and Bryce took his hand—"hurt me is still on the reservation running programs for children." Pay looked away, and Bryce squeezed his hand. He felt the urge to take over for him, but he stopped himself. Pay had to do this himself. He had to be the one to ask for help. "If he did what he did to me, then he's doing the same thing to other children."
Bryce's mother was quiet for a few seconds. "Does he volunteer or use the school?"
"He did when I was a kid, and that's what scares me," Pay said.
"You could press charges," Laura said. "There would be an investigation, and other boys might come forward."
"And then they might not," Pay said.
"The tribe thinks the sun rises and sets around this guy. It's going to be hard to convince people, and Pay is going to suffer because of it," Bryce said. The last thing he wanted was Pay hurt again. "I want Grantham to suffer, not Pay," he added vehemently.
"I see," she murmured. "Has anyone requested a review of their background checks? Everyone who volunteers at a school has to have a background check. Usually they're done by the FBI and need to be reissued every few years. I think it's three." She was quiet again. "Okay, so this could work," she said to herself a few minutes later. "How about this? Paytah, you make an anonymous call to the FBI from the reservation giving as much detail as you can without giving yourself away—dates, places, everything you feel you can."
"What does that do?" Bryce asked feeling protective.
"It's using the process to our advantage. Paytah makes a report that's credible, and then we get someone like your mother," she said, looking at John, "to request the records on background checks. The background checks are the law, and they have to have current ones. If your mother threatens to make a big deal out of it, they'll scramble to make sure they have them. Grantham's background check will fail because of Paytah's report, and he will immediately lose access to the school. Furthermore, the FBI will investigate or forward the information to local authorities. They'll have to look into the matter, and who knows what they'll turn up? As part of the investigation, you can come forward."
"I don't want the local authorities involved. They're as bad as everyone else," Pay said nervously.
"Tell the FBI that when you call. Explain that's why you're calling them," Laura suggested. "I know this is hard, but you have to take charge of this and see it through. He needs to be punished for what he did to you, and you have to help stop him from hurting other children, you know that."
Pay nodded slowly. Laura stood up. "Thank you for a lovely dinner," she told John and Jerry before turning to Paytah. "This is a good thing you're doing for all the right reasons. This isn't revenge, but justice, and helping spare other children from going through what you did. No matter what anyone thinks, you're doing what's right." Bryce saw her cup Paytah's cheek. "People may give you grief to begin with, but truth will win out, and you'll know you're doing what's right. Wear that like armor if you need to, but don't doubt yourself. You're a lot stronger than you give yourself credit for." She lightly patted Paytah's shoulder. "You may also find you have a lot more friends and supporters than you thought."
Bryce was so proud of his mother at that moment he thought he would burst, and Pay lifted his gaze to meet hers. When Bryce saw strength and determination in Pay's eyes, he knew calling her had been the absolute right thing to do. Bryce stood up and hugged his mother. "Thank you," he whispered as she hugged him back. Bryce released her and his mother said good night.
"Do you need directions?" John asked.
"No, I'm fine," she said as she walked to the door. "Paytah, would you walk me to my car?" She waited, and Paytah got up from the sofa and then the two of them left the cabin together. Bryce wondered what was up and kept his eye on the door.
Pay returned a short while later with a smile on his face. "She gave me a little pep talk," Pay said but didn't go into any additional detail.
"Do you want to go back to your place?" Bryce asked, and Pay shook his head. "You want to stay here?" Bryce asked, and Pay shook his head again.
"I have some things in my truck, if you'll come with me," Pay said, and Bryce got up, taking Pay's hand. "Good night and thank you for everything," Pay said before stopping in the doorway. "I'm going to make that telephone call first thing in the morning. It's time to put an end to this."
"I'll call my mom," John explained.
"Tell Kiya I appreciate her help," Pay said and then continued outside, and Bryce closed the door behind them.
"Where are we going?"
"I... whenever I was confused or upset, I used to go out to nature. Mark took that away from me, and I want to take it back. I want to spend the night outdoors with you. I want us to be together outside." Pay stopped near the car. Before Bryce could answer, Pay pulled him against his body and kissed him hard. Pay roamed his hands down Bryce's back, then cupped his butt in a signal of exactly what Pay wanted. Bryce hummed his affirmative answer, forgetting completely about the outside part of the proposal as Pay plundered his mouth.
Bryce moaned, but thankfully Pay cut it off with a kiss when he slipped his hands into Bryce's jeans and underwear, cradling his bare butt cheeks. "Is this okay?" Pay asked, and Bryce nodded and whined before Pay kissed him again. Outside, inside, none of that mattered right now. Bryce was so turned on he'd agree to being with Pay on top of an ant hill covered in honey.
Pay broke their kiss, and Bryce moaned softly. "Get in the car. It won't take long to get where we're going."
"I think I'm already halfway there," Bryce said and then burst into a fit of giggles. If he didn't know better, he'd have thought he'd been drinking. Pay's kisses must have short-circuited his brain. Stumbling slightly, Bryce walked around the car and got into the passenger seat.
Pay was obviously excited, bouncing a little in the seat as he started the car and carefully turned it around. The foliage tunnel was ominous, all light disappearing as they passed through. Normally, Bryce's imagination would be working overtime, seeing things in the darkness, but he only had eyes for Pay. "It's not far," Pay said, and Bryce sat back, every muscle in his body strung tight and ready to pounce.
Pay kept his word. The drive took less than ten minutes, and Pay turned off the road, the headlights shining off the ground before dropping away to nothing.
"Where are we?" Bryce asked.
"My spot," Pay said. "After my mom died, this was where I came to be alone and think." Pay turned off the engine, but left the lights on. "And when things started to get bad with Mark and I didn't know what I was going to do or how I could get him to stop, I came here. This was where I decided to tell the counselor what was happening to me and where I came after the bastard didn't believe me." Pay opened his door and got out of the car, saying nothing more. Bryce wondered what he was supposed to do.
Pay opened the trunk and rummaged around inside it. Bryce opened his door and carefully got out. He knew the drop off wasn't going to reach up to get him, but he had no intention of getting anywhere near it. Pay slammed the trunk closed and walked back to the front of the car, where he reached through the open window to turn off the lights. "Give it a few minutes and your eyes will adjust," Pay said in the darkness.
Bryce closed his eyes and waited, afraid to move away from the car. Then he opened his eyes and let them adjust, and a whole world opened up to him. The canopy of stars stretched as far as his eyes could see. "Amazing, isn't it?" Pay said from next to him, and Bryce turned.
"Yes," he agreed softly, blown away by all he could see. The landscape around them, hidden in the darkness, made what was visible even more amazing and astounding.
"Come on," Pay said, taking his hand. "Don't move fast." Pay clicked on a flashlight, shining it on the ground and led Bryce away from the car. A shape broke the flat ground, and as they got closer Bryce saw a rustic, three-sided shelter with bark roof and sides. It appeared to be near a small stand of scrub trees, but Bryce couldn't be quite sure in the darkness.
"Did you build this?" Bryce asked as Pay shined the light on the enclosure.
"Yes. They don't usually survive the winter. I build one each year so I can come here to think," Pay said as he set down the bundle in his arms. He handed Bryce the flashlight and spread out blankets. "Come here," Pay said faintly, and Bryce approached. Pay took the flashlight and clicked it off before guiding Bryce onto the bedding. It was surprisingly soft, and Bryce wanted to ask what he was sitting on, but thought better of it. He'd find out in the morning.
"It's beautiful here," Bryce said as Pay joined him.
"Hmmm," Pay said, and then he kissed him, his lips hot, sweet, and demanding. Bryce returned the kiss, letting go of everything around him and giving himself over to Pay, who pressed him back against the bedding. Pay tugged Bryce's shirt out of his pants before pulling it up his torso. Bryce lifted his arms and Pay pulled the shirt off. The cool night air kissed his skin, with just enough chill to raise goose bumps. "Get undressed," Pay whispered, and Bryce pulled off his shoes and opened his pants before shimmying them off his legs. Pay guided him under the blankets and then joined him moments later.
The chill from the air dissipated instantly when Pay's heat surrounded him. "I've never brought anyone else here," Pay said, and Bryce's heart lurched. "Only you. I don't even think Chay knows about this place." Pay kissed him hard, like he had before they left, short-circuiting Bryce's brain before pulling them together.
Pay pressed his smooth skin to Bryce's. Bryce loved that hot smoothness against his skin, Pay's weight pressing on him. Bryce parted his legs, curling them around Pay's, sliding a heel up and down his calf. Bryce returned Pay's hug, taking their kiss deeper and deeper. Pay cupped his head as their lips and tongues dueled. Skin to skin—that was how he loved Pay best. Bryce thrust his hips forward slightly, his cock throbbing in its erotic trap between their bodies.
"I love you," Pay whispered in the darkness, his hot breath tickling over Bryce's lips. Most people only got to hear those words; Bryce got to feel them as well. "I know we haven't known each other that long and I'll understand if it's too soon, but I wanted you to know how I felt." Pay lightly brushed his fingers over Bryce's cheek. "Are you crying?" Pay asked.
"No," Bryce lied, badly, and then gave it up. "I didn't think I'd hear those words again after Percy died." Bryce's voice cracked. Instead of trying to talk, he held Pay's face in his hands, guiding their lips together, pouring his own love, relief, gratitude, amazement, and passion into his kiss, not trusting his voice any longer. Pay mirrored everything Bryce kissed to him, their shared passion building. Bryce arched when Pay touched his hip and gasped, openmouthed, into the kiss when Pay lightly pinched first one nipple and then the other.
"I brought things," Pay whispered.
"Thank God," Bryce sighed, feeling Pay search around them. Bryce could only see Pay's shadow against the stars outside. "I want to feel you inside me," Bryce whispered, stroking Pay's thigh. "I want to be with you." The muscles in Pay's leg tightened, and Bryce shifted so he could try to see him better. "What is it?"
"I've never done that. I want to. I want to be with you that way, but...." Pay paused, and Bryce waited for him. "I don't want to hurt you."
Bryce scooted closer to Pay, soothing him back toward their outdoor nest. "You could never hurt me; you never would. What happened to you wasn't love and it was designed to hurt—to put him in control and to give him power over you. That isn't what's happening here." Bryce slid his hand down Pay's arm, finding his closed fist and its contents. "I want you to make love to me, more than anything." Bryce needed to show Pay that physical closeness, being together and joined, was beautiful and nothing like what he'd experienced. "I've done this before. Just remember to use lots of lube." Bryce kissed Pay hard and then rolled over on the blankets. But he didn't hear or feel anything. "Touch me, Pay," Bryce said, and he felt Pay's hand on his back. "Can you feel that?"
"You jumped," Pay said.
"Yes. You excite me. I want to feel you, want you to touch me," Bryce whispered and felt Pay move closer, a second hand joining the first. Pay stroked down his back and over his butt. Bryce moaned softly, pressing back. "That's it," Bryce encouraged, and Pay increased the pressure on his skin. "Use a finger, Pay, get me ready for you."
Bryce shook with anticipatory excitement. He heard a rip that sounded as loud as an earthquake out here in the near silence. Bryce laughed when a cold drop of lube fell on him, but quickly forgot about it when Pay touched him again. "Is this okay?" Pay asked, lightly touching his opening.
"Yes," Bryce whispered urgently, pressing back into the touch. "More, please. More." Pay made tiny circles and then carefully, with agonizing slowness, slid into Bryce's body.
"I'm not hurting you, am I?" Pay asked as Bryce's muscles went wild, gripping Pay's finger hard.
"No," Bryce gasped. "Deeper," he said, and Pay slid his finger further into him. "Curl it slightly." Pay did, and Bryce thought his head was going to explode. Half the nerves in his body fired all at once.
"Is that good?" Pay asked, stopping all movement.
"It's heaven," Bryce said through gritted teeth as he tried to keep some sort of control. "Do it again," he begged, and Pay repeated the motion.
"Yes!" Bryce cried, his voice echoing around them. He squirmed on the blanket and felt Pay press a second finger next to the first. The stretch was marvelous, and he sighed softly as Pay moved inside him. "That's it, get me ready for you." Pay was definitely bigger than two fingers, but Bryce was getting impatient.
Pay scissored the fingers inside him and then slowly pulled them away. Bryce groaned when the connection between them broke. Bryce heard Pay open a condom package and waited for him to roll it on. Then Pay positioned himself between Bryce's splayed legs. More lube was added to Bryce's opening, and then he felt Pay pressing into him. "That's it," Bryce gasped as his body began to open and Pay entered him for the first time.
Bryce opened his mouth wide in a silent gasp as Pay damned near split him wide open. The man was huge, and Bryce heaved in a deep breath as Pay slowly pressed deeper. Bryce reached for Pay, tapping his leg. Pay stilled, and Bryce caught his breath as he willed his muscles to relax and stretch. It took a little time, but his body eventually adjusted, and Bryce pressed back against Pay, signaling him to move.
Pay slid deeper, and Bryce arched his back as Pay's cock slid along the spot inside him. Stretched wide and as full as he could ever remember being, Bryce heaved a small sigh of relief when he felt Pay's hips hug his butt. "Jesus Christ," Bryce swore under his breath and then shivered when Pay's cock throbbed and jumped inside him.
"Is that bad?" Pay asked.
Bryce shook his head and then remembered Pay couldn't see it. "No. Just full, so full." Bryce swallowed and held still. Pay slowly began to move, pulling out and then back in, small, slow movements that drove Bryce wild. He clutched the blankets, meeting each of Pay's movements with his own. "Damn, you feel good," Bryce moaned.
"You're a furnace," Pay said, continuing to move very slowly.
"I'm not going to break," Bryce told Pay. "You don't need to treat me like I'm going to shatter."
"I won't hurt you," Pay said, and Bryce chuckled for two seconds until the sound shifted to a deep, throaty moan.
"You won't," Bryce said, slamming his butt back against Pay, who gasped in surprise. He desperately needed more. "Athletic is good," Bryce said, and Pay picked up the pace. "Yes! That's it!" Bryce said, his cries coming back to him time and time again.
"Okay?" Pay asked.
"Hold it a second," Bryce said. "I want to kiss you." Pay slowly pulled out, and Bryce shifted to his back, lifting his knees to his chest. Then he guided Pay back inside him and tugged him down into a hard kiss. Pay began to move, and Bryce grunted and moaned. "Harder," he encouraged, and Pay responded. "Yes!" Bryce nearly screamed, and his call came back, joining with others that built louder and louder, higher and higher.
Bryce stroked himself, surrounded by a world of sensation. He could see almost nothing, but the world was rich with sound—Pay's breathing, the slap of skin against skin, his own constant moan that Pay kept wringing from him. And the scents—sweat, musk, the bark that surrounded them. All of it combined, adding to the intensity and intimacy of the moment, because nothing mattered outside the shelter. The entire world. Everything Bryce wanted was right there, right then.
He stroked harder and faster as Pay drove into his body. Pressure quickly built inside him. Stroking with one hand, he used the other to pull Pay into a sloppy kiss. Their bodies vibrated and shook with each thrust, and eventually it seemed as though the entire world shook along with them. "Paytah!" Bryce hollered as his climax built. "God, yes!" Bryce careened over the cliff of his own pleasure, coming and coming, stars dancing in his eyes as he yelled his lungs out.
Through his haze of pleasure, Bryce heard Pay cry and saw him throw his head back. With a mighty thrust, Pay snapped his hips forward, sending a shock wave through Bryce as Pay throbbed and danced deep inside Bryce's body. Pay stayed where he was, groaning softly for a long time before going completely quiet. Bryce tugged Pay on top of him, his lover's long hair flopping against Bryce's shoulder. Bryce stroked Pay's back as he smiled and held him close. "Was that okay?" Pay asked in a whisper.
"That—" Bryce gasped before swallowing to get rid of his dry mouth. Then he cradled Pay's head, guiding their lips together. "That was completely and totally amazing. There was nothing whatsoever just okay about that." Pay kissed him, and Bryce closed his eyes, reveling in the weight and warmth of his lover.
# Chapter 9
BRYCE woke with the sun in his eyes. He turned away, groaning softly as he reached for Pay. His hand fell on empty space, but he wasn't ready to really wake up yet. He and Pay had been up half the night making love, and Bryce felt the residual wonderful soreness in every part of his body. He rolled over with a sigh and cracked his eyes open.
Pay was on the far side of the clearing, already dressed, talking to another man, both of them sitting on the ground. When the other man turned, Bryce saw it was Pay's brother, Chayton. They were talking softly, and Bryce heard their voices on the breeze. He closed his eyes again and tried to sleep, but it wouldn't come. If Pay had been next to him to help keep him warm then maybe, but it wasn't going to happen now. Opening his eyes again, he saw Pay and his brother hug, and it looked like Chayton was crying.
"If only I'd known," Pay said, his words reaching Bryce's ears. "I could have helped you if you'd have told me. We could have helped each other." Bryce saw Chayton look toward him and he closed his eyes again, leaving them in peace.
"Does he know?"
"Yes. Bryce knows what happened to me. He's a good and special man."
"But he's white," Chayton said, and Bryce kept himself from smiling, knowing that wasn't going to fly with Pay. What a difference a few weeks made. "You, the guy who doesn't take twenties or speak to white people is suddenly... screwing one and everything's okay?" The smack of skin on skin almost made Bryce jump.
"Talking about Bryce like that is not, and will never be, okay. Got it?" Pay hissed, and Bryce saw Chayton rub his cheek as he nodded. "You should know better than to judge someone by the color of their skin. We all should. How many times have we been judged for ours? He's got ideas that can help our people here. You heard the council is looking into reintroducing buffalo? That was Bryce's idea. Kiya and Wamblee Black Raven, Akecheta's parents, are championing it, but it was his idea, and he has more, so don't discount him. I love him." Pay paused. "But it's more important than that, because he may turn out to be the best real friend either of us or the tribe has had in a long time."
"You're serious," Chayton said, still rubbing his cheek.
"Yes, I'm serious. I was wrong, and so are you." Pay pointed at Bryce. "He saw I needed some help and he called his mother and got her to drive hundreds of miles because he thought she could help me. She did, and by extension she helped you too." Pay stepped closer to his brother, and Bryce realized it was going to get pretty hard to keep pretending to be asleep very soon. "Because she and Bryce are why we talked today, and why, maybe, if you're man enough, you can crawl out of the bottle you've hidden in for years, grow up, and become a member of this tribe and this family instead of a drain on both."
Bryce began to stretch and move, groaning softly and yawning. He hoped his performance was believable. "You're awake," Pay said to him, and Bryce nodded, but didn't make any effort to actually get up. He wasn't particularly interested in Chayton seeing his bits. Pay turned back to his brother. "It's up to you, man, but I think it's time you lived your life instead of running from it." Chayton glared at Pay for a long time, and Bryce wondered who was going to blink first. Then Chayton lowered his eyes.
"You've done a lot of running yourself," Chayton said.
"Yes, but I think it's time for both of us to stop," Pay said as he walked to where Bryce was lying. "You remember my brother Chay?"
"Yes, how are you?" Bryce asked. "I'd get up but...." Bryce figured there was no need to finish his thought. "I thought he didn't come here," Bryce whispered to Pay.
Chayton howled with laughter from behind Paytah. "I don't. But anyone within miles of you two heard you last night." Chayton continued chuckling. "I'll give you one thing, you got a healthy pair of lungs on you." Chayton walked toward the path out of the clearing. "I'll see you round, bro."
"I'm serious. If you want to talk, Laura is a great listener," Pay said. "And she understands a lot." Chayton paused and then continued walking out of the clearing without saying another word.
"How much did you hear?"
"Enough," Bryce said softly.
"Mark got to Chay too," Pay told him. Bryce had sort of figured something like that must have happened, but the confirmation sent a chill through him. "Isn't anyone safe from that man? Or is it just my family that he'd been intent on ruining?"
"Hey. It's out in the open now. I take it you two talked," Bryce added.
"Yeah, we talked. Thankfully, it seems Coyote somehow got wind of what was happening with Chay, and the two of them were able to stop things from going as far as they did with me, but Mark really messed Chay up."
"Is he gay?" Bryce asked, and Pay shook his head. "I see."
"Yeah. Chay was a teenager interested in girls, and Mark pressured him into things that really hurt him. He said he thought Mark was going to try to rape him when Coyote intervened."
Bryce found his clothes and began getting dressed. "The sad part is that if Chay had gotten help when he was younger, he might not have turned to alcohol." Yet another long-term consequence of Mark's abuse.
"Chay's problems are bigger than that, but I hope he'll get some help," Pay said with a soft sigh. "I'd like to have my brother back, but I don't know if it's possible. The will has to come from him, and I don't know if he has it in himself any longer, but I'm hoping that telling me what happened will open the door for other things."
Bryce finished dressing, then sat back down on the bedding to put on his shoes before getting up again and wandering to where he could see, but without getting too close to the edge. "This is so beautiful," Bryce said as he peered down the canyon to what looked like a thin ribbon of water. "Hard to believe that carved all this," Bryce muttered almost to himself and then stepped back as his legs began to shake slightly.
"When it rains, the creek becomes a torrent," Pay said from behind him as he slipped his arms around Bryce's waist. "I don't want you to go."
"I know," Bryce said, and he felt Pay shake his head.
"I don't want to be alone anymore," Pay whispered. "You opened my eyes and my world to include you and other people. I haven't had that in a long time, and I don't want to go back to being alone." Pay held him closer, and Bryce lightly stroked his arm. "I know you need to go back and I don't mean to make you feel guilty. I just want you here with me."
Bryce snuggled back into Pay's warmth. There wasn't anything Bryce could say that would change the near future. Bryce turned in Pay's embrace, placing his arms around Pay's neck, kissing him gently. "I'm going to miss you too, but I'm not gone yet."
"No," Pay said with definite resignation. "And I have a phone call to make."
BRYCE helped Pay get everything in his trunk, and they rode back to the reservation center. Pay parked outside the store. He sat, hands still on the wheel, without moving for a while. Bryce sat with him quietly and let Pay think. "It's time," Pay finally said. "He took my childhood and part of my life, and it's time I took them back." Pay opened his door and got out. Bryce hung behind watching as Pay strode across the open area to the old pay phone standing near a telephone pole. Bryce saw Pay glance at him. Up till that point, he wasn't sure if Pay wanted to do this alone or not. Bryce got out of the car and stood next to Pay as he fished his wallet out of his pocket and found a scrap of paper. Then Pay inserted the coins in the phone and dialed the number.
Bryce listened, heart in hand, as Pay told the person who answered what he wanted. He was obviously transferred to someone else and then Pay began to talk. He explained why he was calling and then gave details about what he'd endured at the hands of Mark Grantham. Bryce had heard it before, but hearing most of the story a second time tore at his heart. At the point where Pay was about to describe exactly what Mark had done to him, Pay's voice faltered. Bryce took his hand and held it as Pay finished telling his story. Bryce was grateful it was still very early and no one was about. He felt no eyes watching them and saw no one else. "There are other people. I know I'm not the only one," Pay was saying. "No, I really don't want to give my name, not now, but I will aid an investigation." Pay listened to whatever the person on the other side of the line was saying. "The local police will not help. They're all Mark's buddies. That's why I'm calling you," Pay said and then hung up the phone without saying another word.
"You did good, Pay. They'll have to do something now."
"No, they won't. She didn't sound convinced," Pay said.
"It's her job to ask questions and get as much information as she can. And it isn't important if she believed you. The report has been made and they'll trace the call back to the reservation. So now all we can do is wait. Kiya will stir the pot even more today or tomorrow and we'll see what happens. They can't know it was you who made the call, so you're safe."
"Not necessarily," Pay said. "Mark will figure out who called if they begin an investigation. If they give him any details, he'll know, but by then I hope it won't matter so much."
"It doesn't matter now. He can't hurt you any longer, and you were strong enough to put aside your own fears and report what happened." Together they began walking back toward Pay's car.
"I'll take you back to the cabin before I open the store and get to work," Pay said. Bryce was about to open the car door when Pay's brother stepped out of the small road near the store, with Coyote and a boy who looked about seventeen trailing just behind them.
"Did you call?" Chayton asked, and Pay nodded. "Good," he said and held out his hand. "Would you give me the number?"
Bryce caught Chay's eye and then shifted his gaze to the kid, who shuffled his feet and basically looked like he was trying to disappear behind Chayton and Coyote. Now that he was closer, Bryce realized the boy was younger than he thought, probably fourteen or fifteen, with short ragged hair that looked like it had been hacked off with a knife. Bryce glanced at Pay, who looked like he was going to be sick.
The kid tugged on Coyote's shirt, whispering a few words Bryce couldn't hear, and Coyote nodded.
"It's okay, Kangee," Pay said as he moved closer to the boy. "We all understand here." Kangee looked first at Coyote and then at Chayton, both of them nodding slowly. Then he lowered his head and silently began to cry. Bryce could see he was trying to hold it in, but couldn't. "Let's go inside," Pay said, and he turned, striding to the store and then quickly unlocking the door. Everyone followed, and Pay locked the door behind them. "There's nothing to be ashamed of. You didn't do anything wrong."
"He said he'd hurt my grandfather," Kangee said, trying to gain the upper hand in the battle with his emotions and losing. To Bryce's surprise, it was Chayton who gathered Kangee into his arms and hugged him as the wrenching tears came.
"It's okay. He's not going to get to you any longer. No one will. Paytah and I are going to see to that," Chayton said. "I promise."
Bryce wiped his eyes and turned away from them to give Kangee some privacy. "See? You already have supporters and friends. One person, people can try to fight; two, they might discount; but three...." Bryce shook his head. "No way, and there will be more," he whispered.
"That's what I'm afraid of," Pay said as Bryce moved into his arms, comforting Pay as the sound of Kangee's tears filled the room to the very ceiling.
"We're okay here," Coyote said to Pay, and he nodded, leading Bryce out of the store and to his car.
"I'll take you back to the cabin," Pay said quietly.
"Are you sure? I'll stay here with you if you want," Bryce offered.
"No. I need to be alone and have some time to think, and you need to work," Pay said before opening his car door. Bryce wanted to argue, but after everything, maybe they both needed a chance to think for a while. Bryce got in the car, and Pay started the engine before putting the car in gear. They rode in silence to the cabin. Bryce leaned over the seat and gave Pay a light kiss before getting out.
"Will you call if...," Bryce began, and Pay nodded, but Bryce knew he wouldn't call. Bryce was fairly sure this was something Pay felt he needed to deal with on his own. "Will I see you tonight?" Bryce asked.
"It's not supposed to rain," Pay said, and Bryce nodded his understanding.
"I'll wait here until you pick me up," Bryce said and then closed his door and walked around to the front door of the cabin. It was still quiet inside, and Bryce made as little noise as possible as he went to his room and changed clothes, then got his laptop and settled on the sofa to work for a while.
"How did it go last night?" John asked quietly. Bryce looked up from where he'd been immersed in his work.
"Nice," Bryce said, and he felt himself color. "Pay took me to his... outdoor place, and we spent the night under a shelter."
John began to laugh softly. "You've come a long way from the guy who just weeks ago was scared he'd be eaten by bears," he teased.
"Yeah, I guess so. I still don't feel comfortable being outdoors, but Pay was there."
"Exactly," John began as he sat down next to him. "You feel safe with Paytah, and that's how it should be with the person you love." John yawned and then put his bare feet up on the coffee table. "So have you decided what you're going to do about it?" He bumped Bryce's arm. "You know, the whole 'he's here, you're there' thing."
"Don't know. Pay wants me to stay, but...." Bryce paused and set his laptop on the table beside John's legs. "I like it here, but I don't fit in. I know you're going to say that Jerry did, but you've seen the hesitation and sideways glances the same as I have. Jerry is accepted, more like tolerated, by many people because he's with you. When I meet people, I have to tell them I'm a friend of yours and Pay's before they stop looking at me funny. Even then the suspicion stays."
"It would take time, yes, but people would accept you," John said.
"Yes, they might, but then there are the other things. I can't buy a house here because of the land restrictions. It's a reservation and I have to be a member of the tribe or I'd have to buy a house with Pay. I would only be allowed here as long as I was with Pay. If anything happened to that relationship, I'd have to leave. We've only known each other for a few weeks. I love him, John, but... it's too soon to uproot my entire life. I think we need to spend some time together, get to know each other better, date." Bryce motioned with his hands. "I told him I'd come back, and I hope he can come to visit, but I don't think either of us is ready to completely change our lives." Bryce was trying to be realistic even as his heart was telling him something very different.
"I know how you feel, and from a logical perspective, you're probably right. But we can't always live from a logical perspective." John yawned again and tried to cover it. "Look at it from Paytah's perspective. He trusts you, and that's saying a lot. He told you his deepest, most painful secret, and you helped him."
"Shouldn't I have?" Bryce asked defensively.
"Of course you should. But in the same few weeks you've known each other, you've gone from a distrusted stranger to the one person he trusts most. That's saying a lot for anyone, but for Paytah, that's like moving the world. I know you'll be back and you'll do what you say, but for Paytah, it's hard to see that now because he's come to rely on you."
"So what do I do?" Bryce asked. "I don't want to hurt him."
"Just be patient and don't expect him to jump for joy at your decision. Yes, it's probably the right one. Love can hit hard and burn hot but then sputter away, or it can grow over time. You two probably need to find out which you have, but it's most likely hard for Paytah to see that right now," John explained before yawning yet again. "I need coffee."
John got up and went into the kitchen. Soon the scent of fresh coffee filled the room. "By the way, Pay made his call this morning and there are at least two other people," Bryce said, getting up, not willing to mention any names.
"Already," John said, and Bryce nodded.
"Please have your mother do her thing at the school, but I'm not sure how much it's going to be needed. Once this hits, it's going to be huge, and victims are going to come out of the woodwork." He leaned against the counter, waiting for the coffee to finish. "I just wish I could see that son of a bitch's face when all hell breaks loose."
"Me too."
"I'd like to...," Bryce began, but then swallowed his words when Ichante came into the room. "Morning," Bryce said, noticing for the first time what a striking young lady she was becoming. "God, you're growing up so fast," he sighed.
"Tell me about it," John said as he hugged his niece good morning. "Soon she'll be too big for hugs and want to borrow the car."
"Uncle Akecheta," she said, giggling. "I won't ask to borrow the car. I'll ask you to buy me one."
"How old are you? Thirteen going on twenty?" Bryce asked, and Ichante giggled. The little girl was definitely smart and was going to give them all a run for their money when she got older.
"Where's Mr. Paytah?" she asked as John poured her a glass of juice.
"He had to work," Bryce began as Mato barreled into the room, "to make this one's doughnuts." Bryce laughed and hugged Mato to him. These kids, Jerry, and John were family to him just like his mother was. "What are you guys doing today?"
"Going to Paw-Paw's," Ichante said, glaring at her brother. "We're having a Lakota day," she announced. "We're only speaking Lakota once we get to Paw-Paw's," she reiterated to her brother, and Mato looked about as thrilled as if he'd been told about an impending dental appointment.
"Sit down for breakfast," Jerry said as he joined John. "You can speak all the Lakota you want, but Grammy isn't here yet, so English will have to do." Jerry bumped John, and together they cooked up a spread. Bryce was wondering who was going to eat all this when the door opened and both mothers came in.
"I love a man who can cook," Kiya said before being mobbed by the kids. Bryce kissed his mother hello, and everyone greeted everyone else before heading to the table.
"I was wondering what all the food was for," Bryce said, making sure the two women were on the other side of the table, "then the appetites got here, and mystery solved." Both women glared at him before finally breaking into good-natured smiles. Bryce let his mother catch him.
"Look who's talking. Someone's feeding you well—you're getting a belly," she said, patting Bryce's flat stomach before taking a seat.
"So did Paytah make his phone call?" Kiya asked.
"Yes. It went as well as could be expected, and then we got a surprise. Two more in the same situation showed up." Bryce was speaking generally so the children wouldn't know what they were talking about.
"Do you need me to do my thing?" Kiya asked.
"It couldn't hurt. Get Mark out of the school before things heat up," Bryce said, and she nodded before turning to his mother.
"Feel like raising a little hell?" Kiya asked, and Bryce saw his mother smile.
"I'd love to," she answered, popping a bit of fruit into her mouth. Bryce actually sighed. He couldn't quite decide if getting these two together was a good thing or not.
"Okay, but before you two go into battle, I need some advice," Bryce said, looking specifically at Kiya. "I need something traditional that I can give to Pay, something that will let him know I'm thinking of him when I go. I want it to be special and something that will have meaning for him, for both of us."
"Just give him something from your heart," Laura said.
"Let me think about it," Kiya answered.
"Thank you," Bryce said and went back to his breakfast.
Once they were done eating, Kiya and his mother took the kids, and Bryce went to work. The cabin was quiet, and he got tons accomplished. Jerry and John seemed to get a lot done as well. But while Bryce was productive, he kept part of his attention on the phone, hoping that Pay would call. But unfortunately he didn't.
PAY never did call. Thankfully for Bryce, he stopped by the cabin in the late evening. By that time, Bryce had been going crazy wondering what was going on, but Pay had been quiet at the cabin and on the drive out to the shelter. Bryce hadn't pushed, and Pay hadn't talked, at least not in words, but Pay's thoughts had come through in his almost desperate lovemaking that ended with both of them screaming their releases that echoed and melded over the land before dying away. Bryce realized he was alone and opened his eyes, not that he could see much other than stars on the nearly moonless night. "Pay," he said softly when he heard movement nearby.
"Pay, talk to me," Bryce whispered, hearing light footsteps getting closer.
"Nothing to say," Pay said. "I don't want you to leave, but I know you have to and I know I need you to." Pay hesitated near the shelter entrance, and Bryce figured he was taking off his shoes, but he couldn't see well enough to know. Then Pay joined him under the blankets.
"You want me to leave?" Bryce asked, figuring he must have misheard.
"No, but I think I need you to leave. Not forever." Bryce heard Pay roll over, and then he was pulled close to his lover's warmth. "I've been alone a long time and I like having you with me, knowing you're here, and I could get used to this really fast. And that's the problem."
"I know." Bryce rolled over to face Pay, placing a hand over Pay's heart after he'd settled on the bedding. "I was alone after Percy died, and it can feel overwhelming. But I'll be back, and over time we'll figure something out." He had no idea what that was, but a solution would present itself. "I'm hoping you can come visit me too."
"I'd like that. Alowa would probably push the car the entire way if she thought she could get me to spend some time away from the store," Pay said.
"She cares about you," Bryce said. "Everyone seems to care about you. The few times I've watched the store, whenever someone comes in, they ask where you are, almost every single customer. They wouldn't do that if they didn't care." Bryce brought Pay's lips to his. "So you see, you aren't alone, and even when I'm gone, you won't be alone because the entire tribe is there for you, if you'll let them be." Bryce kissed Pay, nibbling lightly on his lower lip. "That's the beauty of having a tribe, a real tribe, one that supports and accepts you no matter what. You have that, and all you need to do is allow them to support you."
Bryce felt Pay nod slightly. "I've felt apart for so long and sort of like I was fighting them."
"But the fight was within," Bryce supplied.
"Yeah, and as long as I fought myself, Mark won," Pay whispered into the darkness. Bryce wasn't sure if the words were for him or the universe, but he closed his eyes and held Pay closer as sleep finally came.
# Chapter 10
"NOT looking forward to going home tomorrow, are you?" John asked, and Bryce looked up from behind his computer, where he'd been pretending to work. He wasn't getting a thing done and, giving up, shut down the machine.
"Yes and no. It'll be good to get home again and be in my own place."
"But you're going to miss Pay," John supplied.
"Very much, and I don't know what we're going to do."
"Take it one step at a time," John suggested. "As long as you're both willing to make the effort."
"I know, I'm just going to miss him," Bryce explained, and they got quiet.
"Did your mother make it home okay?" Jerry asked as he entered the room.
"Yes. She texted me late last night." Bryce huffed slightly. "She was supposed to take it easy and stop along the way, but she pushed herself and made it home." He wasn't sure if he was happy about that, but at least she was home safe. "Have you started packing?"
"Yeah, you?" Jerry said.
"Uh-huh," Bryce hummed. "So what's on for today?"
"The kids are going to my mom's, and we were wondering if you wanted to go in to be with Paytah." John looked over at Jerry, and Bryce got the message loud and clear. They wanted a bit of alone time.
"That would be great. Is Kiya picking up the kids?"
"Yes. She'll be here soon," John said as Kiya walked in the back door.
"Are they ready?" she asked, placing her purse on the counter.
"They're in the bedroom, playing. I'll get them," John said and left the room.
Kiya rummaged in her purse and pulled out a bundle of cloth, handing it to Bryce. "This is what we talked about," she said, pressing the bundle into his hand. "Also, I just heard that there are investigators on the reservation. They showed up late yesterday and have been quietly questioning people. Of course, in a place like this, little is kept quiet for long."
"Thank you," Bryce said urgently and shifted his attention to John, biting his lower lip nervously. This was a good thing, but it was going to be hard on Pay and those he cared about.
"I'll take you to Paytah," John said. He stood up and kissed his mother. "I'll see you before we leave," John added, and they hurried out to the van after Bryce grabbed his computer bag. Bryce had been expecting this and he'd been waiting for something to happen.
"How do you think Grantham will react when he finds out he's under investigation?" Bryce asked, kicking himself that he hadn't thought about this earlier. "I mean, this kind of abuse is all about control, so what will happen when Grantham realizes he doesn't have control any longer?" Bryce's heart pounded in his chest. His imagination was getting the better of him and he was probably overreacting, but all he could think was what if this guy came after Pay.
John pulled up in front of the store, but it was still closed.
"Thanks, I'm going to head to Pay's," Bryce said.
"I'll wait until you're inside," John said, and Bryce headed around to Pay's small house. A dark sedan was parked outside and Bryce walked beside it, peering inside, but seeing nothing unusual.
Bryce knocked on Pay's door and waited. He was about to leave when Pay opened it. "You're okay?" Bryce asked, and Pay nodded before motioning him inside. Instantly on guard as Bryce came in, two men sat on Pay's sofa. They stood as Bryce approached.
"This is my boyfriend, Bryce Morton," Pay told the men. "These are the agents investigating the report of abuse."
"I'm Agent Williams, and this is Agent Donley," the shorter of the two men said. "We're trying to determine if the reported account is accurate."
"They arrived just a few minutes ago," Pay explained.
"The call we received was made from the pay phone just across the way, and we were wondering if you might have seen who made the call. We're sort of at a loss as to where to start other than speaking to the tribal leaders. But with the proximity of the phone, we thought it logical to begin here."
Bryce locked gazes with Pay and nodded. "Yes," Pay began, "I know who made the call." Pay motioned toward the sofa and sat in one of the chairs while Bryce perched on the arm next to him. "It was me."
The agents began making notes. "Would you tell us what happened?" the agent asked, and Pay began recalling the incidents from his childhood while Bryce held his hand. The detail he relayed was astounding and even more eerily gripping than when Pay had first told him.
"Mr. Stillwater, how old were you when you first met Mr. Grantham?"
"Eleven," Pay answered. "He started playing his touching games when I was about twelve, and things progressed to full sex by the time I was fourteen." Pay related his dealings at the school with the counselor and principal as well.
"I want you to get them, as well. They looked the other way instead of protecting the children in their care," Bryce said. "They probably still are."
The agents checked some notes. "We got requests for background checks on a number of school volunteers recently. They're probably covering their asses."
"Mothers can raise a great deal of hell," Bryce commented and then explained how they'd brought the background checks about. "We weren't sure what would happen or how fast, but we need to get this guy out of the school."
"Are you aware of anyone who might have confronted Mr. Grantham?" Agent Donley asked.
"No," Pay answered, and Bryce squeezed his hand. "Not that I'm aware of."
"Do you know of other victims?" Agent Williams questioned.
Pay turned to Bryce, who answered for him. "We can confirm at least two others. One is an adult and the other is currently... fifteen?" Bryce asked, and Pay answered with a nod.
"One of them is my brother," Pay said levelly. "But I can't speak for him or the boy who came to us."
"We need you to give us their names," Agent Williams said, but Pay shook his head.
"I will ask them to contact you, but that's all I can do. Their stories are theirs to tell, and the decision to come forward will be a hard one, but it's their decision to make, just like it was my decision to call you."
The agents stood up. "Can I ask one more question?" Agent Donley asked. "When you made the original call, why did you remain anonymous?"
"I was afraid of what people would think," Pay answered.
"What's changed?" the agent added.
"I guess I realized I could care less what other people think and more about what I think of myself," Pay explained. "I also came to understand, with Bryce's help, that if I didn't do something, Grantham was going to hurt other kids. Let me change that: he is hurting other kids, and that has to stop." Not giving a damn what the agents thought, Bryce leaned closer to Pay as a lump formed in his throat. Pay needed his support. "I stayed quiet and let him hurt me for years, long after I got away from him physically. I pulled away from family and friends because I was ashamed of what happened. I didn't think anyone would believe me because the counselor at school hadn't believed me. I don't want anyone else to hurt like I did." Pay leaned forward. "And I sure as hell don't want Mark Grantham to determine the way I live the rest of my life. So, you investigate, and I'll do my best to get others to come forward." Pay stood up and stepped toward the agents. "I want you to nail Mark's balls to the wall. I never want him anywhere near another kid again. I don't care if he has to live in a prison at the fricking North Pole, just as long as he can't hurt anyone else, ever!"
The agents stood up. Bryce expected skepticism or doubt, but what he saw in both of them was compassion. "We'll do our very best," Agent Williams said. "Please call us if you have any other information you feel you can share."
"I'll ask my brother to call you," Pay said.
"Thank you for all your help," Agent Williams said. "We'll show ourselves out."
Pay watched them go.
"How do you feel?" Bryce asked.
"Surprisingly good," Pay said and then smiled. "I really did something to get rid of Mark for good." Pay sat down and pulled Bryce onto his lap, and Bryce giggled like a schoolgirl until Pay kissed him, and then thoughts of everything else vanished. "I feel alive," Pay said once he broke the kiss, holding Bryce tight.
"Do you want to show me just how alive?" Bryce asked sheepishly.
"I'd love to, but if I don't get over to the store, the old guys are going to think I died or something and they'll come knocking on the door."
Bryce got off Pay's lap and they left the house, heading over to the store together. Sure enough, a few of the men were gathered outside. Pay unlocked the door, and the men wandered inside and made the coffee themselves while Pay began making doughnuts in his small kitchen in the back. Bryce sat at the table in the corner and opened his computer to work for a while.
When Pay carried a tray of fresh doughnuts from the back, the entire store filled with the aroma. The men all bought doughnuts and carried them outside to their bench, where they sat in front of the store with their coffee and food, talking and looking at each other. "Remind me to get off my ass when I want to do nothing but sit all day," Bryce whispered to Pay, who smiled.
"It's what they do," Pay said as if that explained everything. "They're mostly alone and they keep each other company. They'll only be out there until it gets warm and then they'll move on to the tribal center or head home. They do it almost every day."
"I noticed," Bryce said and saved his work before closing the lid on his laptop. "They need something to do."
"What are you cooking up?" Pay asked with a smile.
"Nothing. But if we get the craft company up and running, those guys may be able to help. I bet they remember a lot about your culture and may have ideas of things we can make," Bryce explained, and Pay rolled his eyes. "I know—I'm always thinking."
"It's a good idea," Pay said as he looked out of the window.
"Before I forget and someone else comes in, I have something for you," Bryce said, and he opened one of the pockets of his computer bag and pulled out the cloth Kiya had given him.
The front door banged open, crashing back on its stops. The sound ricocheted off the walls and Bryce nearly dropped what he was holding.
"You stupid little son of a bitch! I'm going to make you pay!" Mark Grantham threatened as the door banged closed behind him. "I had a visit from the FBI this morning and I know this is all your doing." He stalked toward the register, where Pay stood immobile. Hell, Bryce did the same thing, unable to believe anyone would think acting this way or making threats in front of a witness was a good idea. "You redskin piece of shit." Mark glanced for a second at Bryce, and Bryce felt the rage pouring off the other man. "If you think anyone is going to believe you or your little white scrawny piece of ass here, you're sadly mistaken. No one will believe you, and once I'm done, not a single person here will shop in your little store anymore either." Mark reached the counter and leaned over it to get right into Pay's face.
"You know, as a kid," Mark said, his voice soft enough that Bryce could barely hear him, "you were great. I cared for you a great deal, and what we did together was special and you loved every minute of it—the attention, the care. You were my boy. What makes you mad is that you aren't my boy any longer. That's what bothers you and why you made up these filthy lies." For a second there, Bryce had thought Mark was admitting to everything he'd done to Pay, but that was too good to be true. Of course he was going to deny it. A confession would be too easy. "Those kids need me, and I give them something no one else around here can, just like I gave that to you." Pay hadn't moved, and Bryce was beginning to wonder if he was buying this load of crap. "I've helped people here for years. They look at me and see someone who has helped many people on his reservation. I'm rich, strong, and loved. When they look at you, they see quiet, pathetic, and now they'll see a liar."
Bryce seethed with anger, but he wasn't sure if he was angry with Mark or with Pay for just standing there and listening without moving a muscle. How could he just take this? Bryce thought Pay had come a long way in the past few weeks, but maybe that was wishful thinking.
Bryce took a step closer, and Mark turned to him. "Just stay away, pencil neck, you don't want any part of this!" Mark said vehemently between clenched his teeth, bits of spittle flying in Bryce's direction. For a few seconds, Bryce thought he looked a bit like a big dog, and then Mark turned his fury back to Pay. "I should have known you were a useless little liar," Mark said and then backed away slightly, glaring at Pay. "Nothing to say?" Mark asked. "I should have known. You never said much at all. I always thought you might have been stupid, kind of retarded, and maybe I was right. Yeah, the FBI is really going to believe a stupid Indian retard." Mark swung his hand, sending everything on the counter flying, including the displays of gum and candy and the doughnut case, which hit the floor with a crash and the clear plastic shattered, scattering doughnuts and pieces of plastic all over the floor. "Clean that up, and don't think of billing the foundation or me for any of it. I'll just deny the payment."
The door to the store cracked open, and Bryce saw one of the old men peer inside briefly. Then the door closed again, and Mark, who had glanced away at the sound, looked back at Pay. "See? No one gives a damn about you around here, and no one is going to believe a fucking thing you say. So if you want to have any friends left, I suggest you tell the FBI you made a mistake. Maybe then I'll consider forgiving you."
"You forgive him?" Bryce said. "After what you did, you useless, self-obsessed, pile of—" Bryce stepped forward, but Pay put out a hand to stop him.
"No, Bryce, I can handle this," Pay said softly, and Bryce wondered how in hell he could be so calm.
"So, you can talk. I was beginning to think you'd lost your voice along with your ability to think," Mark sneered. Pay slowly moved out from behind the counter.
"I can talk, and I've done plenty of it," Pay said levelly. "I talked to the FBI and I talked to Bryce's mother. I also talked to other friends here on the reservation, and now I'm talking to you." Paytah continued to slowly move closer to Mark. "And do you know what I found when I talked to people? I found others you hurt like you hurt me. A kid you tried to manipulate and use just like you did me. I found someone who managed to get away from you the way I wished I'd been able to but couldn't for so long. You can think what you want and call me a liar, but I'm not. You're a pedophile who uses children and rapes them." Pay moved still closer to Mark. "You stole my childhood and you stole my innocence. But worse than that, you stole my peace of mind and part of who I was, but I'm taking it back. I'm taking all of it back." Pay's voice became stronger. "I'm going to save other kids from going through what I did and I'm going to send you to prison. I don't care who believes me here and who doesn't, because that doesn't fucking matter. Bryce believes me, and so do other people, and so will a jury!" Pay continued moving forward, and Mark took a single step back. "You're a bully, a coward, and a criminal."
Pay slowly closed in on Mark, and Bryce wondered what Pay was going to do, but he stayed back and let him work it out. Bryce was ready to pounce and beat the shit out of Mark if necessary, but Pay's stiff posture and the way he clenched and unclenched his fists told him he wasn't going to have to do anything.
"Pay," Bryce said softly, now thinking it might be best if Pay let Mark go, but he ignored him.
"You like having sex with little boys. You get off on it. Well, that's going to stop. Once it gets around what you've been doing, all the people you've hurt are going to come forward. They'll be coming out of the woodwork. Your FBI clearance to volunteer at the school is going to be denied, and I'll make damned sure everyone knows why. I may be quiet and an 'Indian retard', but I set your ass up and no one is going to be able to stand up for you when I'm done. Newspapers are going to want to know what happened, and I'm going to talk loud and clear. Your face will be spread all over the state. Your family is going to be so proud of you," Pay added sarcastically before taking a deep breath. "Now get out of my store and off my reservation."
"I'm not going anywhere!" Mark growled, stepping closer to Pay.
Bryce saw Pay clench his fist, but everything else was a blur. Before Bryce could move, he saw Pay snap his arm out and then heard the crunch of bone. Blood spattered from Mark's nose. Instantly, Mark grabbed his face, blood spilling from between his fingers.
"I said get the hell out, and for fuck's sake, stop bleeding on my floor. Go bleed in the dirt with the rest of the animals." Bryce rushed to open the door, and Pay shoved Mark through it. Mark lost his footing and fell flat on the ground. "Stay away from our kids, you pervert. You may have fucked me when I was fourteen and scared, but you can't push me around now that I'm an adult," Pay said, loudly enough that anyone within earshot, and there appeared to be quite a few people, could hear him.
Pay came back inside and closed the door, breathing like he'd just run a marathon, his eyes burning with energy. But it didn't last, and Bryce saw Pay slump slightly, and the flames in his eyes faded.
"I'm so proud of you," Bryce said, barely able to speak.
"But I hit him," Pay said, and Bryce smiled.
"He deserved it, and so much more," Bryce said, taking Pay's hand. "Go wash his blood off so I can make sure you're okay." Bryce didn't think Pay had broken the skin on his hand, but he wanted to be sure. "Where's the card from the agents?" Pay handed him his wallet and walked toward the bathroom while Bryce made a phone call. Agent Williams answered. "This is Bryce Morton, I think you should come to the trading post as soon as you can."
"Has something happened?" Agent Williams asked.
"Yes," Bryce said.
"We're on our way," the agent said, and he disconnected the call. Bryce was still holding the bundle wrapped in cloth, and he placed it back in his bag for later. The agents must have been close by because they pulled in front a few minutes later, a cloud of dust following them. They got out of the car as Pay slowly walked back toward the counter.
"I should clean this up," Pay said.
"No, just leave it," Bryce said, turning away from the window.
The agents walked inside and looked over the mess. "What happened?"
"Mark Grantham was here," Bryce said flatly. "He threatened Pay. Didn't confess to what he'd done in so many words, but he threatened Pay, and I guess Pay had had enough because he punched him in the nose and sent him sprawling."
The door opened and two of the old men came inside. "We saw the whole thing," one of the men said, and the other nodded his white-haired head. "Paytah didn't do nothing. Grantham tripped and fell flat on his face. Looked like he maybe broke his nose, though."
The agents looked at one another and then back at Pay without saying a word. "Sounds like reliable witnesses to me. We met with Mr. Grantham this morning and were about to come talk to you again when we got your call. We need corroboration of your story. He, of course, denies everything. We could take him in, but as it is, his lawyer will have him back out as soon as he gets there."
Pay sighed and looked defeated. "Get Chay," Bryce said, and Pay absently pulled his phone out of his pocket and pressed in a number.
"Coyote, it's Paytah. I need Chay, and Kangee if he'll talk," Pay said and then listened. Bryce saw Pay turn pale. "Okay, I'll see you all then." Pay hung up the phone and turned to Bryce. He looked like he was about to cry, and Bryce moved into his arms.
"What happened? Is Chay...?"
"What?" Pay asked. "Oh, God, no," he whispered. "It's Kangee. Coyote said they found him at Rainbow Gorge. He was about to throw himself off the side when they got to him. He's okay, but...." Pay's voice hitched. "They talked him down. Coyote said Chay talked him down," he clarified. "They're on their way. It'll take a while. He's a mess. They're going to drop him at home with his mother."
"Is Kangee the boy you were referring to yesterday?"
"Yes. He's fifteen," Pay said before swallowing hard.
"It's all right," Agent Williams said, and he turned to Agent Donley.
"I'm already on it," the other agent said and pulled out his phone.
"We're going to bring in a counselor. It looks like a lot of people are going to be hurt," Agent Williams said. He asked Pay for more details on what happened with Grantham and took pictures of the mess as well as statements from both of them. Bryce then cleaned up while Pay made more doughnuts and the agents made themselves at home. It appeared that the trading post was going to be Grantham Central for much of the day.
It took awhile, but Chay and Coyote arrived. Bryce stayed out of the agents' way as they talked to Chay. When they were done, Chay looked a bit shell-shocked, but his eyes appeared clear. "I'm sober," Chay told his brother quietly, "have been for days now. You were right. It is time I stopped running and took my life back. I figured if you could do it, then so could I."
"I'm proud of you, bro," Pay said. "It's hard doing what we're doing, but it is the right thing." Bryce turned away and left the two of them to talk. The agents talked together off in the far corner for a while and then they approached Bryce.
"I don't think there's anything more we can do here. Please call us if you need anything," Agent Williams said.
"Are you going to talk to Kangee?" Bryce asked. They didn't answer, but their expressions told Bryce what he needed to know. "I know you'll be as kind and careful as you can with him, but please let him know that what happened wasn't his fault." Bryce looked over and saw Pay still talking with his brother. "Pay said that Mark made him believe for a long time that what happened was his fault. I'm sure he did that to Kangee too."
Agent Williams nodded slowly. "When we talk to him, we'll be gentle, and we'll certainly tell him." Agent Williams turned to leave and then stopped. "That's very common and insidiously mean. It uses the victim's own need to please against them. We'll do our very best to help him any way we can," Agent Williams promised, and Bryce heard the sincerity in his voice. He watched them leave and then wandered to where Pay was talking to his brother. They were both smiling, so he felt it safe to approach.
"Where did they go?" Chay asked.
"They didn't say exactly, but I think they were going to talk to Kangee," Bryce said.
"That's not a good idea," Chay said with concern.
"I don't know. Maybe it's what he needs. Those two have seen a lot and they'll be able to tell him that things aren't his fault. Sometimes we need to hear things from strangers. They also called in a counselor, so they'll be able to get him some help."
"There are more," Chay said. "I don't know who, but there are more, I can feel it." Bryce nodded his agreement.
"I think that maybe it's time for one of you to go to the council. They need to know because they may be able to help other people who haven't come forward yet," Bryce suggested, and the brothers looked at one another. "If you don't want to do it yourselves, then see if Kiya will do it."
"No, I need to tell them," Pay said definitively. "I can't hide behind other people. This is something I have to do."
"Do you want me with you?" Bryce asked.
"No. I'll call the head of the council in a few minutes," Pay said, and Bryce sat back down at the table where his computer was, but he couldn't concentrate. He heard Pay making his phone call and Chay pacing the floor.
Things in the store calmed down for the next few hours. No one came in to threaten Pay, and the agents didn't call or show up with more questions. Pay got back to work, although he was jumpy as hell. Chayton and Coyote stuck around for a while, but they eventually drifted off as well. Customers came in like usual, and Pay handled them, but it was becoming obvious that rumors were flying around the reservation and that people were stopping in to see if they were true.
Bryce had to hand it to Pay—he didn't pull any punches, and when they asked, he told them the truth. Most were shocked, a few didn't seem to believe it, but when Bryce chimed in that there were multiple victims, they changed their tune very quickly. From the expressions on many people's faces, a lot of kids and grandkids were going to be asked some difficult questions that evening.
"Do you think he'll press charges?" Pay asked sometime late in the afternoon.
"If he does, he's got more guts that I'd give him credit for," Bryce said. "No. I think right now, he's trying to figure a way out of this mess, but there isn't one. He could already be in custody. They have your statement as well as Chay's. But it's Kangee's that's the most damaging because it's the most recent." Bryce stepped over to where Pay was looking out one of the store windows. "If you want my advice, I think you should get ready to close up at your normal time. We'll head to the cabin for dinner and then spend the night together before I have to leave." Bryce hated to leave even more than he had when the day started. He'd promised Jerry he'd go back and there were appointments he couldn't postpone, no matter how badly he wanted to.
"It's supposed to rain," Pay said, forlornly.
"Then we can be together at the cabin or at your house," Bryce said. "Unless you'd rather be alone. I'll understand if you need some time." He hoped that wasn't the case, but he wouldn't blame Pay. With everything that had happened today, he could understand if Pay needed some time to himself to think, but it was their last night together for a while and Bryce really wanted to spend it with Pay.
"I was hoping to take you back to the canyon, but the shelter is not good for rain, especially if there's wind too. I'll come to the cabin with you," Pay said. "But no screaming," he added with a wicked grin.
"The screaming is all your fault and you know it," Bryce said, returning Pay's smile before checking his watch. There was only an hour left till closing, and Bryce spent it trying to work, but he got nothing done at all. His mind refused to settle on anything other than Pay, so he eventually gave up and helped Pay around the store. Whenever the door opened, they both glanced to see who it was. Bryce kept expecting the agents to come back, or even the police if Grantham turned out to be monumentally stupid, but it was just customers doing their shopping. Thank God.
At closing, Pay took care of the money and closed everything up before they left the building, locking the door behind them. As they headed over to Pay's car, Bryce's phone rang and then Pay's rang as well. Bryce answered his first.
"Honey, it's Kiya. The word through the grapevine is that Mark Grantham has been arrested." Bryce glanced at Pay, who was also smiling, and Bryce figured he was getting the word from someone else.
"Have you heard anything about Kangee?" Bryce asked. Kiya seemed to know everything about everyone.
"I talked to his mother a little while ago. She says he's going to be okay and they've talked a lot. She wants Mark's head on a platter, but it seems Kangee was most worried about hurting his mother, and once he knew she loved him no matter what, things settled down. She said she's going to find him some help. I also talked to the council president, and he's going to call a special meeting tomorrow. He was shocked off his ass, but agreed to look into bringing in a counselor to help both the kids and the families. This isn't going to be easy for anyone, the tribe the families, the children who were hurt, but we'll do all we can to help."
"The FBI said they were going to bring in counselors as well," Bryce said.
"Yes. But the council realized this will require long-term healing, and they want someone to help with that," Kiya explained.
"Good," Bryce said as he watched Pay continue to smile. That was what really mattered to him. In two weeks, he'd seen the gruff, almost snarly man he'd first met evolve into someone who actually smiled.
"I understand you're leaving tomorrow with Jerry and John. When will you be back?"
"Two weeks. I'm going to teach a class, and Dave said he'd have some possible product designs for me to look at."
"I know. He's already got me looking into possible grant money to help defray start-up costs, and I found someone who might be able to help put together a business plan. So you'll be busy when you visit."
"I appreciate all the help," Bryce said, and Kiya laughed.
"You're the one that's helped. We caught this bastard because of you, and half the tribe is excited about reintroducing buffalo into the range lands on the reservation. And if this craft company idea pans out, you'll have helped add jobs. So it's we who should be thanking you."
"No. I just had the idea," Bryce said modestly.
Kiya humphed softly, and Bryce thought for sure she was going to call him on his modesty. Thankfully she let it go. "You always have a place here. You're family, just like Jerry." Bryce knew Kiya was being nice, and he thanked her before disconnecting the call. He'd like to think he'd made an impression on the people here and that they would come to accept him, but he knew he'd always be an outsider. Eventually, maybe he'd consider living here with Pay. He'd deal with it to be with him once they'd had the time to build a permanent foundation for their relationship.
"He's in jail," Pay said as soon as Bryce hung up the phone.
"Kiya told me," Bryce explained.
"Coyote told me that now the word is out, three more kids have come forward, as well as another two adults." Pay shook slightly. "I knew there would be more, but these are all people I know. Some are customers I see every few days."
"This is going to ripple through the entire reservation. He hurt a lot of people, and everyone here is going to know someone involved. I hope the tribe comes together over this and really stands by you and everyone else who was hurt." Bryce moved closer to Pay and placed an arm around Pay's waist. "I'm here for you and I always will be." Bryce leaned up and was about to kiss Pay when he saw some of the old men walking toward the bench in front of the store for what he presumed was their evening sit and chat. Not really caring who saw or what they thought, Bryce tugged Pay toward him, taking his lips in a deep kiss. Everyone and everything around them faded as soon as their lips touched. Bryce only cared about Pay, and when his kiss was returned and Pay wrapped his arms around Bryce's waist to tug them closer, nothing else mattered.
"Let's go," Pay said once they broke their kiss. "I don't want to kiss you in front of an audience."
"Okay," Bryce agreed, although the rebellious part of him liked the occasional public display of affection. Pay was his and Bryce wanted everyone to know it. "Let's go have some dinner."
"I need to clean up before I can go anywhere," Pay said as they headed to his place.
"Me too. Do you think a bit of water conservation is in order?" Bryce asked while Pay unlocked the door. He turned around, looking at Bryce like he didn't understand, and then he smiled and nodded.
Bryce set his bag on one of the chairs and followed Pay through to the bathroom. Pay turned on the water before pulling off his clothes. Bryce forgot to get undressed as he watched Pay's deep copper skin appear. "Are you going to join me?" Pay asked.
"Yes. I was just watching," Bryce said before pulling Pay closer. "I love to watch you get naked." Bryce kissed Pay, cupping him lightly in his hand, Pay's cock growing hard at his touch. "What I really like is how I can make you do that," Bryce said, his voice a low rumble as he gripped Pay hard, stroking him.
"Get undressed," Pay said as he thrust his hips. Bryce let go and began removing his clothes, with Pay helping. Obviously, he wasn't moving fast enough. Once Bryce was naked, Pay stepped into the shower, and Bryce followed right behind, his gaze locked on Pay's firm, round, perfect butt.
"You have butt dimples," Bryce said as he traced the sides of Pay's butt cheeks with his fingers before pressing himself to Pay's back, nestling his cock perfectly between Pay's butt cheeks. "I love that you have dimples when you smile...." Bryce began to giggle. "Does that mean your butt is smiling for me?"
"Come here, you goof," Pay said as he turned around and then tugged Bryce close, moving them beneath the water. Bryce's giggles subsided when Pay's skin pressed to his, and he forgot about everything other than his lover. Pay reached for the soap and backed away slightly. He worked up a lather and then began caressing Bryce's skin. "Step back," Pay whispered, and Bryce moved out from under the water. "Turn around," Pay said, and Bryce complied.
Pay began at Bryce's shoulders, soaping and massaging them deeply. Then Pay moved down his back and Bryce sighed, closing his eyes. Anticipating that Pay would continue moving lower, he thrust his butt backward, but Pay continued washing his back, and then he began to sing. Bryce didn't understand the words, but that didn't matter. The tone was deep and low in Pay's throat, and the sounds rumbled through Bryce as they filled the small space.
"What is that?" Bryce asked when Pay paused.
"It's a very old love song," Pay answered and began to sing again. He paused in his washing and then started again, this time, finally, caressing Bryce's butt and upper legs. The song became more intense, and Bryce's cock throbbed from both the melody and the way Pay caressed his skin. Pay rubbed him more intensely, parting his cheeks and lightly sliding his soapy fingers over Bryce's opening and then down, between his legs, reaching through to lightly cup his balls and then back.
"Pay," Bryce whined softly, his legs beginning to shake. Bryce held still, and Pay continued his erotic washing while he sang his love song. Pay tapped his hip, and Bryce turned around, his cock jutting forward. Pay got more soap and started washing him again. Bryce's chest was first, followed by his legs, which were shaking so badly Bryce could barely stand.
When Pay wrapped his fingers around Bryce's cock and slowly began to tug, Bryce placed his hands on Pay's shoulders to steady himself. He closed his eyes and let his head loll back as the soapy slickness sent tingles all through him.
Pay stopped rather abruptly, and Bryce opened his eyes in time for Pay to press against him once more, gently pushing him back under the spray. The soap washed away, and the water sluiced over both of them. Then Bryce got the soap and it was his turn. Stepping back from the water, Bryce slicked his hands with mounds of suds, and then Pay stepped from under the water. Bryce touched his skin, washing his shoulders and chest, paying special attention to Pay's nipples. He'd discovered quite early in his explorations of his lover's body that Pay's nipples were sensitive, so each time he came close, Bryce gave them a little tweak, and Paytah would sigh and a shudder would ripple through him. Bryce loved that he could make Pay react with such a simple touch. It was a visual and aural reminder of what Pay felt for him, and Bryce's own love for the man whose rich copper skin, now covered with soap bubbles, flowed beneath his hands. "I love you, Pay," Bryce whispered, a bit overcome, and Pay took him in his arms, pressing and sliding his soapy skin against Bryce's.
"I love you too," Pay said, and Bryce lifted his gaze. "I know you have to leave. You have a job, and so do I. But I know you'll be back."
"Yes, but I hate to leave you when everything seems like it's going to be so hard on you," Bryce said.
"No. You made so many things so much easier, and I feel freer than I have for as long as I can remember." Pay stroked Bryce's cheeks. "Everything is going to be fine, and you'll be back soon." Pay once again moved them under the water. "I have plenty to keep me busy, thanks to you. Chay seems determined to move on with his life, and there will be many kids who are going to need help, so I think I'm going to try to do what I can. I've been through what they're going through now and I survived, so I need to help them do the same."
"But...," Bryce began.
"I think we both need a bit of time to figure out if what's between us is truly lasting. I think it is, but we need to be sure," Pay told him. Bryce wasn't happy to hear his own words come back to him. Pay turned off the water and reached out of the shower, handing him a towel before getting one for himself. They dried themselves, and Pay left the bathroom to get dressed while Bryce gathered up the clothes he'd worn and put them back on, wishing he'd brought fresh things, but he figured he could change once they got to the cabin.
Once dressed, Bryce met Pay in the living room. "I was going to give you this earlier, but we got interrupted." Bryce dug in his bag and pulled out the cloth bundle. "Kiya told me that she wove this bit of cloth when she was a girl and she wanted me to have it to take with me so I would remember everyone here," Bryce said and slowly opened the bundle until he got to the center. "I asked her to help me come up with something I could give you that would be both traditional and special. So I asked Running Deer to help, and he made this." Bryce opened the cloth, and inside was a small wooden carving of a buffalo. The detail was amazing for something so tiny, and on its back were two small gold loops. Bryce lifted out the buffalo and handed it to Pay.
"This is for me?" Pay said, and Bryce nodded, lifting out the two cords he found hidden in one of the folds of the fabric.
"Let me show you." Bryce gently took the buffalo and carefully pulled the two halves apart, so they each had half of the buffalo. "Each has a small pin to hold it together," Bryce said as he fed a cord through the gold loop and handed it to Pay. "You keep one half and I keep the other," Bryce said, feeding the second cord through his half. "Like I said, I wanted something one of a kind and something special. The buffalo was a symbol of your people. Historically, it meant life to them. Without the buffalo, they would have starved as they roamed the plains." Bryce moved closer to Pay. "Without you, I think my soul would starve." Bryce felt his voice begin to falter, but he held steady. "After I lost Percy, I never thought I would feel for another person what I felt for him, but I do with you." Bryce took the cord with its small buffalo suspended from it and carefully placed it around Pay's neck. Then he took his own and handed it to Pay so he could do the same for him. "I love you, Pay," Bryce said as he held still, waiting until Pay had finished.
"I love you too, Bryce," Pay said as he moved back. Bryce touched his neck, feeling the small buffalo that hung at the base of his throat. "Your gift is beautiful, and whenever I see it, I'll always think of you. I know you have to go for now, but with this, I'll always have you near." Pay, too, touched the carved buffalo at his throat. "This is a beautiful and thoughtful gift. Thank you," Pay added before leaning in closer, moving almost in slow motion until their lips touched in a kiss of profound need and hunger that quickly deepened until Bryce was engulfed in Pay's arms, pressed to his strong body, where he wanted to remain always. But that wasn't possible, at least for now, so Bryce reveled in each touch.
"What do you want?" Bryce asked Pay.
"I want to take you back to the cabin and have dinner with your friends, because you need to finish getting ready to leave. Then I want to take you to bed and spend the night giving you something so that when you do go, you'll have part of me to remember, something special in your heart that will last you until you return," Pay said, and then he kissed Bryce hard. Bryce held Pay's shoulders for balance and returned the kiss with everything he had. Then Pay broke the kiss and stepped back before taking Bryce's hand. "Come on, we need to go or Jerry and Akecheta will wonder what happened to us."
"But...," Bryce sputtered.
"We'll have all night," Pay whispered before leaning in to lick the spot behind Bryce's ear. "And I promise you it'll be worth it." Pay licked his skin, and Bryce shivered. Then Pay led him by the hand out of the house and to his car. Bryce got in, glancing over at Pay with a smile every few seconds as they drove out to the cabin. Finally, when he could keep his hands to himself no longer, he reached across and placed his hand lightly on Pay's leg.
When they arrived, they had a welcoming committee. Mato and Ichante hurried out to greet them, excitedly tugging them both in for dinner, which was raucous, joyful, and in a word, wonderful. Bryce packed once they were done. Jerry and John let the kids stay up later than usual so they would sleep on the trip in the morning, and then at nearly ten, with the kids in bed, Bryce took Pay's hand, and after saying good night to John and Jerry, led him to the bedroom and closed the door.
Pay kept his promise, making love to Bryce for most of the night. They were quiet because they had to be, but that only meant Bryce couldn't cry out—not that he didn't want to, over and over again. Eventually, they both fell asleep, only to wake up early so they could make final preparations to leave. Pay helped them load the van, and then John and Jerry got the kids, situating them in the backseat, where they promptly went back to sleep.
"I'll see you in a few weeks," Pay said, and Bryce nodded before throwing himself into Pay's arms. They hugged, and then Pay kissed him. "It won't be forever."
"No, it won't, but it'll seem like forever." Bryce kissed Pay again before hugging him one last time. "I can feel you everywhere," Bryce whispered, and Pay squeezed him hard. Then he let go, and Bryce walked to the van, got in, and quietly closed the door. John turned the van around and then slowly headed down the foliage tunnel. Bryce turned around and waved to Pay one last time before he disappeared from sight.
THE ride home took hours, but they got back to Sioux Falls in the early afternoon. John dropped him off at his house, and they all helped him unload his things before saying good-bye. Bryce waved to the kids and then turned and walked inside, closing the door behind him. Everything was quiet, so quiet he could hear himself breathe. Too quiet. He'd spent the last two weeks with the kids, John and Jerry, and Pay... God, he missed Pay already.
Moping about it wasn't going to help, so Bryce hauled his bags to the laundry room and began filling the washer with a load of dark clothes, placing almost everything he'd brought back into the baskets to be washed. The few clean clothes that were left he carried upstairs to his bedroom to put away.
Bryce pulled opened his dresser drawers, and a picture of Percy he'd put away caught his eye. Pulling out the framed photograph, he stared into the eyes of the man who was to have been his husband. "I found him, Percy. You told me there was someone else for me, and I found him." Bryce traced Percy's face with his finger. "Thank you," Bryce whispered, and then he placed the picture back in the drawer and slowly closed it.
# Epilogue
"BRYCE, we're going to be late," Pay called from the living room, and then Bryce heard his footsteps through the house. "You'll probably need a jacket," Pay added from the doorway.
"That's what I figured," Bryce said, leaning his head toward the bed, where his jacket had been laid out, as he fastened the small carved buffalo around his neck. When it was secured, Bryce lowered his hands to look at Pay. They were both in jeans and heavy shirts. Yes, they were heading for a ceremony, but it was going to be outside in April, so everyone had been instructed to dress appropriately. In the vee of Pay's collar, Bryce saw that he, too, was wearing his buffalo. Bryce grabbed his jacket and hurried to the bathroom to make sure he looked okay before joining Pay back in the bedroom.
"You look fine," Pay said before tugging Bryce in for a kiss. "More than fine, you look edible," Pay growled.
Bryce chuckled deeply. "You didn't get enough last night?"
"No. I haven't seen you in two weeks," Pay said. "Wouldn't matter anyway, I can't seem to get enough of you." A year hadn't mattered one bit in that regard.
"Me too," Bryce moaned as Pay sucked lightly on his ear. "But didn't someone say something about being late?" Bryce asked, and Pay growled but stopped his nibbling. "When we get back," Bryce promised, and Pay grabbed his butt in what Bryce interpreted as his promise of things to come.
Bryce picked his jacket up off the bed again and followed Pay out. As he approached Pay's car, he glanced over at the trading post, its lights off and door closed. Because of the ceremony, Pay had closed the store for the morning. Most people would be at the ceremony anyway, so he wouldn't be losing much business. Bryce slid into the passenger seat and belted himself in, waiting for Pay, who wandered over to the store to do a quick check and then got in the driver's seat.
The drive took a while. The ceremony was being held in a more remote area of the reservation that Bryce wasn't familiar with. As they rode, the roads got rougher and less used, but they were by no means alone. A steady stream of cars, trucks, and vans followed the road, and they joined the convoy as it wove through some of the most picturesque scenery Bryce had seen on the reservation. "How far is it?"
"Not much farther," Pay said, concentrating ahead of him. Eventually the area around the road widened, the trees and scrub fell away, and an area of relatively flat grassland spread out before them. The cars in front pulled off and began to park. Pay followed behind them and parked, and then they got out, with Pay leading Bryce toward a group of tents. "The council decided to make this a festival since everyone was going to be here."
Bryce nodded and looked all around as people in traditional clothing walked past, heading toward an authentic stand of teepees off to one side. "This is amazing," Bryce said, glancing at Pay, who smiled. "You didn't say anything about all this."
"I wanted you to be surprised," Pay told him and led him toward one of the tents. Running Deer, also known as Dave, smiled from in front of his display of carvings and pottery.
"Bryce," Dave said happily as they shook hands. "Did you have a good drive out?"
"Yes. The roads were clear and I'm sort of getting used to it," he said, bumping Pay's hip lightly. "So how are things going?" Bryce asked as he looked through the booth. "Your work keeps getting more and more amazing," A carved buffalo, basically a larger version of the tiny one that hung from his neck, caught his eye. Bryce picked it up and carefully handled it, amazed at the detail in the horns, the rough texture around the buffalo's head, even the snorting expression on the animal's face. The sculpture looked as though it was at a full run. "Would you put this aside for me?" Bryce asked.
"Of course," Dave said with a pleased smile.
"I told you," Pay said to Dave, and the two shared a smirk. "I saw that last week and told him you'd want it as soon as you saw it."
"Uncle Bryce!" He turned around and saw Little Wamblee holding Frankie's hand as the two brothers walked toward him. Little Wamblee let go of Frankie's hand, and he toddled toward Bryce, holding his arms out to be picked up. Bryce swung him into his arms to peals of little-boy giggles.
"You're getting so big," Bryce told the toddler, who pointed to a booth where Hanna was sitting.
"Mama," Frankie said as clear as a bell, pointing toward his mother.
"Uncle Bryce, Mama asked us to bring you over," Little Wamblee said, although he wasn't so little anymore. The youngster was growing like a weed. One thing Bryce noticed was that the hollow-eyed child he'd seen that first time Little Wamblee had come into the store had been replaced with a healthy, joyful boy. Both kids were clean, happy, and wore decent clothes.
"Okay," Bryce said, following Little Wamblee to where he indicated. "Morning, Hanna," Bryce said as they approached, and she smiled up at him from her chair, a partially formed basket in her hands. "How are things going?" Hanna had turned out to be a diamond in the rough. Dave and Pay had brought her in to help with the craft business, and she'd turned into a marketing and sales dynamo.
"Great. The grant Kiya helped us get got us started. We've got eight artisans working part time." She indicated examples of pipes, dolls, baskets, bowls, and a few toys on the table. "So far I have twelve stores in the region that have purchased items for the tourist season, and the website you set up for us is generating traffic and steady sales." She grinned at Bryce. "I even got the national park gift shop to carry our things at Mount Rushmore." Her energy and enthusiasm were amazing. "I'm hoping we'll do well this year." The entire time she was talking, Hanna's hands never stopped their work on the basket. It was simple, but the finished products were painted with traditional designs. The actual items were polished, quite impressive, and each item retailed for between thirty and fifty dollars.
"Are you able to stick to our price points?" Bryce asked, picking up the doll from the table. Frankie reached for it, and Bryce had to lift it out of his grasp.
"Yes. It hasn't been easy, but we're doing it," she explained as she finished, looking down at her hands for the first time, completing the rim of the basket and then setting it aside. "The hardest item is the baskets because of the handwork in them, but they're also one of the best sellers on the website."
Bryce watched her start another basket, pulling the precut strips from the bag beside her. "If it doesn't make sense to sell the baskets in the gift shops, make them a website exclusive. That way the cost isn't as big a factor because you're selling them at full retail rather than wholesale, or you can raise the price if you think it won't hurt sales too badly. Think about it."
"I will," she said happily.
"Come see the teepees," Little Wamblee said, and Bryce let himself be tugged along. Pay had disappeared somewhere into the crowd, but he reappeared at Bryce's side as they approached the conical dwellings. Bryce peered inside and was greeted by many of the men inside sitting around a small fire smoking a pipe, the smoke billowing up and out a small opening in the top. He returned their greeting and talked briefly before moving on.
"Bryce, they're about ready to get started," Pay said. Bryce set Frankie on his feet, and Little Wamblee held his hand as they walked together toward their mother.
"I'll see you later," Little Wamblee called, and Frankie turned and waved so vigorously, he nearly fell over.
"They're great kids," Bryce said softly, unable to get over the difference in the entire family.
"Hanna's happy," Pay said. "She's only working the one job and she's been doing amazing work with the craft company." Pay sighed. "I doubt it will turn a profit this year, but they're making and selling product, which is helping to bring money and hope to the reservation." Pay guided him toward where everyone was gathering near a small, temporary platform.
"Welcome all!" the president of the council said into a microphone, thankfully without any screeching or feedback. "Today is a monumental day for our tribe, because today we celebrate our heritage as well as our future. We have booths displaying and selling traditional crafts, and the ladies have made some incredible food." Bryce's stomach rumbled as a gentle breeze blew the aroma over the crowd. He hadn't realized how hungry he was until now. "We have displays and demonstrations of our culture," the president added, pointing to the teepees and other demonstrations. "This afternoon there will be contests based on traditional skills for all ages." Excited murmurs rose from the crowd, and Bryce looked around at the sea of happy faces as he leaned contentedly against Pay.
"Uncle Bryce," Mato whispered as he came up beside him. "Grammy and Paw-Paw want you." Bryce looked at Pay, who said nothing, and they followed Mato around the crowd and off to the side where Bryce saw Kiya and Wamblee standing together.
The president became somber as he stood in front of the microphone. "I know today is a day of celebration, but we must also note the hardships many of us have suffered in the last months."
Bryce knew he was referring to the business with Mark Grantham. Nearly two dozen children and adults had come forward. The trial had yet to be held, and thankfully the press had kept their distance from the reservation. At first, there had been speculation in the national media that Grantham was innocent, but that had quickly dissipated once it became clear how many people were involved. Mark Grantham's support had evaporated after that, and in addition to criminal charges, the tribe was suing his family's foundation on behalf of all those affected. Of course, that could take years, but in some strange, almost perverse way, what had happened brought the tribe together. It unified them against Grantham, but more importantly, it made them come together and create the best from their situation that they were able. Pay had told him that this fair was just one of the results.
"We will help one another and support one another," the president continued. "We are all each other's brothers and sisters, and what doesn't kill us makes us stronger. We are strong, and we will look out for and help those who need it." He became choked up and paused for a few seconds.
"But today," the president said after clearing his throat, "we are here to dedicate the restoration of a piece of our heritage. Kiya and Wamblee Black Raven came to the council with a proposal last fall to restore buffalo to the reservation. It intrigued us all, and last fall, after petitioning the state, and after demonstrating our ability to care for them, the state donated twenty head from their herd in Custer State Park, with the promise of up to an additional twenty head for the next two years. I'm pleased to report that those twenty head are thriving, and two calves were born this spring. The state has approved the delivery of the second twenty head in the next few months." Everyone clapped and cheered. "This means that these buffalo are the first herd to live on this land, our land, in over a century." Everyone applauded and cheered loudly. The council president waited until the sound faded. "Our plans are to let the herd grow and become strong. It will take time, but it will happen. Then we will manage it and use it to help provide an income for the tribe. As I said, this will take time, but it's already starting." The cheers rose again, and Bryce clapped along with everyone else.
"Now, there's one more story to tell before I shut up and let you all get back to the fun. The idea to bring buffalo back to our land did not originate from someone within our tribe. The man who suggested the idea was also instrumental in getting the state to change its policy of culling their herd and instead donated their excess to us. This man has also contributed to the formation of a company here on the reservation to produce and sell traditionally inspired crafts."
Bryce turned to Pay, who was grinning from ear to ear. "Did you know about this?" Bryce whispered, but Pay, the bastard, only grinned.
"This company has already brought jobs to the reservation, and as it grows, we hope more will be created. So without going on forever"—he paused as people clapped and laughed—"I'd like to announce that the herd of buffalo you see in the distance will be officially known as the Bryce Morton Buffalo Herd." Everyone applauded, and Bryce stood rooted on his spot as all eyes shifted from the council president to him.
"Go on," Pay said, and Bryce walked slowly to the platform, stepping up next to the president of the council. Bryce looked out on the gathered crowd, all of them smiling at him, but Bryce's gaze tended to travel to Pay and stay there. He was the only one he really wanted to see.
"Today we dedicate this buffalo herd in your name. It's because of you that it's here," the president said. Bryce shook the tribal president's hand and then prepared to step down. "I'm not quite finished, young man," he said, and Bryce colored and stepped back. "Now, dedicating what we believe to be the only buffalo herd on Native American land to a man who is not a member of a Native American tribe is not acceptable, and in order to be a full tribal member, you must be born into the tribe, but we hope we've done the next best thing. So, Bryce Morton, we officially adopt you as a member of this tribe. Each and every one of the men and women in front of you are now your brothers and sisters. You've already demonstrated the kind of tribal member you'll be, and I'm personally pleased and honored to welcome you to our family."
Bryce thought he was going to cry right there in front of everyone. He turned and shook the council president's hand before walking off the platform and back onto solid ground. "You all knew," Bryce accused when he reached Pay, Kiya, and Wamblee.
"Yes, we knew," Wamblee said with a huge grin. "But you know if it wasn't for all your ideas, none of this would have happened. No, you deserve every bit of it, and as he said, you are part of this extended family." Wamblee extended his hand, and Bryce shook it. Then Kiya gave him a hug, followed by John and Jerry. Then Bryce turned to Pay.
"I think I proved you wrong," Pay said, and Bryce looked at him, completely confused. "Some months ago, you said that you loved it here, but that you felt you'd never been accepted, that you'd always be an outsider. Well, you have to know you aren't an outsider any longer." Pay took his hand and led him away from the crowds to a grassy spot under a small tree. "You aren't an outsider, not any longer."
"No, I guess not."
"So I want to ask you, formally, to come live here with me. I can turn the second bedroom of the house into an office for you where you can work," Pay said, and Bryce grinned. "What's that for?"
"It seems Jerry is one step ahead of you. He told me a few days ago that he found a small building on the reservation that he was planning to turn into a satellite office. He wanted enough room for two people and enough space that he and John can work there when they spend time here. It seems he's opened another branch, and he wants me to run it for him."
Pay stood quiet until what Bryce had told him sank in. "So are you saying what I think you are?"
"I told Jerry that as long as a certain person—" Bryce paused and moved closer to Pay. "—was willing to put up with me for the next thirty or forty years, I'd be happy to accept his offer." Bryce burst into a smile, and Pay hugged Bryce close, lifting him off his feet.
"Yes," Pay said laughing. Then he set Bryce back on his feet and kissed him hard in full view of half the tribe. "I love you." Bryce truly meant to tell Pay how much he meant to him, but his words were cut off by another kiss, and that was fine—talk was overrated anyhow.
# About the Author
ANDREW GREY grew up in western Michigan with a father who loved to tell stories and a mother who loved to read them. Since then he has lived throughout the country and traveled throughout the world. He has a master's degree from the University of Wisconsin-Milwaukee and works in information systems for a large corporation. Andrew's hobbies include collecting antiques, gardening, and leaving his dirty dishes anywhere but in the sink (particularly when writing). He considers himself blessed with an accepting family, fantastic friends, and the world's most supportive and loving partner. Andrew currently lives in beautiful historic Carlisle, Pennsylvania.
Visit Andrew's website at http://www.andrewgreybooks.com and blog at http://andrewgreybooks.livejournal.com/.
E-mail him at andrewgrey@comcast.net.
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{"url":"https:\/\/www.physicsforums.com\/threads\/finite-element-method-weak-form-to-algebraic-equations.888331\/","text":"# I Finite Element Method: Weak form to Algebraic Equations?\n\n1. Oct 8, 2016\n\n### ramzerimar\n\nOkay, I'm following a series of video lectures on applications of finite element method to engineering, and the tutor started by demonstrating the mathematical background of FEM using a simple heat transfer problem. He derived the governing equation (in just one dimension):\n\n(1) $$k\\frac{d^2 T}{dx^2} + Q = 0$$\n\nWhere K is a constant, T is temperature and Q is the heat generated. The next step was discretizing the domain (in this case, a bar with lenght L). For this, we used the weighed integral form, which is:\n\n(2) $$\\int_{0}^{L}w_e(k\\frac{d^2 T}{dx^2} + Q) = 0$$\n\nWhere w is a arbitrary linear weighting function. I understand that we can't solve (1) by using this discretization, because the temperatures would be discontinuous at the nodes and the second derivative wouldn't be defined, and that's why we integrate it to get only first derivative terms. But I didn't understand the weighting term. Just integrating it wouldn't be enough? What's the purpose of it?\n\nLast edited: Oct 8, 2016\n2. Oct 8, 2016\n\n### Paul Colby\n\nI think the author is backing his way into a variational expression for the equation of motion. The equation written is valid for all reasonable choices of the test function. This allows for a simple matrix form of the equation being modeled that is easy to program.","date":"2017-10-19 18:58:20","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\": 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\": 0.846933126449585, \"perplexity\": 531.8760056366506}, \"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-2017-43\/segments\/1508187823360.1\/warc\/CC-MAIN-20171019175016-20171019195016-00663.warc.gz\"}"} | null | null |
\section{Introduction}
Estimating normals for unstructured point clouds is a fundamental task in 3D computer vision. The estimated normal vectors can be leveraged in various downstream applications, such as surface reconstruction \cite{MichaelKazhdan2006PoissonSR}, registration \cite{pomerleau2015review}, and semantic segmentation \cite{EleonoraGrilli2017AReviewOP}.
Recently, the learning-based methods have achieved promising results by casting the normal estimation task into a regression problem. They resolved this problem by directly learning a mapping from local patches to normal vectors using neural networks. However, these methods fail to observe adequate samples for accurate normal estimation. This leads to poor generalization ability on unseen scenarios such as large-scale scenes.
As a remedy, state-of-the-art methods introduce to predict the normal of a point by fitting a local geometric surface. They learned point-wise weights for neighboring points to fit the surface, and then predict the normal vector from the fitted surface. However, the geometric surface settings (e.g. the constant order of the polynomial surface) are predefined during the training process, which leads to poor fitting results due to the various complexity of the local point patches.
For example, when the geometric surface is more complex than the underlying point patches, it results in overfitting problems. Otherwise, the underfitting leads to over-smoothing details. As a result, the performance of surface fitting methods is dramatically limited in cases with complex topology and geometry.
\begin{figure*}[t]
\centering
\includegraphics*[width=1.0\linewidth]{top.pdf}
\caption{Comparisons between previous learning-based normal estimation approaches and our NeAF. (a) Given a point on the point cloud, we sample its K nearest neighbors as a local patch and output the normal for the point. (b) Existing learning-based normal estimation methods can be roughly divided into (i) regression-based and (ii) fitting-based methods. (i) Regression-based methods directly map the local patch to a 3D vector as the normal. (ii) Surface fitting-based methods estimate the weight for each point in the local patch and compute the normal of the surface fitted with the weighted points. (iii) Our NeAF learns the angle offsets between the random query vectors and the ground truth normal, and outputs the query vector with an angle offset zero as the normal vector. (c) The colors of the shapes indicate normal RMSE errors. The estimated normals can be used for surface reconstruction and significantly affect the quality of the reconstructed surface.}
\label{fig:methods_cpmparison}
\end{figure*}
To resolve these issues, we propose an implicit method to learn the Neural Angle Field (NeAF) of local patches in the spherical coordinate system. Specifically, instead of explicitly predicting the normal from the input point clouds, we design a neural network to implicitly predict the angle offset between the ground truth normal and a randomly sampled query vector. The network observes more diverse samples by learning to model the relationship between vectors from various directions and the target normal, leading to an angle field around the target normal.
With a well-learned angle field, the network will have the ability to predict accurate and robust normal vectors.
At inference time, to predict normals from the learned angle fields, we randomly sample query vectors in a unit spherical space and take the vectors with minimal angle offsets as the predicted normals. To further leverage the prior learned by NeAF, we propose a novel learning scheme for refining the predicted normal vectors. Unlike existing learning-based methods that can only predict the normal via one forward pass, our proposed method can further optimize the predicted normal, which leads to more accurate normal estimation.
Our contributions are summarized as follows.
\begin{itemize}
\item We propose Neural Angle Field (NeAF) for point normal estimation. Unlike previous methods, our implicit function learns an angle field for each point, which implicitly predicts the angle offset of query vectors.
\item Our method can conduct further optimization at inference time to refine the predicted normals by minimizing the angle offsets for more accurate normal estimation.
\item We achieve state-of-the-art results in normal estimation for synthetic data and real scans on widely used benchmarks.
\end{itemize}
\section{Related Work}
\subsection{Normal Estimation}
Traditionally, principal component analysis (PCA) \cite{HoppeHugues1992SurfaceRF} is a simple and efficient method for normal estimation, which is based on constructing a tangent plane from a fixed-scale local neighborhood and analyzes its normal vector. Variants such as truncated Taylor expansion (Jets) \cite{FrdricCazals2003EstimatingDQ}, moving least squares (MLS) \cite{DavidLevin1998TheAP} and multi-scale kernel methods \cite{SamirAroudj2017VisibilityconsistentTS} were proposed to fit more complex local surfaces. However, these methods tend to choose large scales to ensure the robustness of normal estimation. To preserve more details, the approaches \cite{NinaAmenta1998SurfaceRB,QuentinMrigot2011VoronoiBasedCA} were proposed based on analyzing Voronoi cells. In practice, these methods have to tune their parameters to work with different cases, which dramatically limits their application scenarios.
Recently, with the development of deep learning, learning-based methods were proposed for normal estimation. Boulch and Marlet \shortcite{AlexandreBoulch2016DeepLF} proposed to transform the local patches of point clouds into accumulators in a 2D Hough space and estimate normals from the resulting approximate planes. However, the transformation from 3D to 2D resulted in the loss of the structural information contained in the surface. To preserve a complete local shape, Guerrero et al. \shortcite{PaulGuerrero2017PCPNETLL} proposed the PCPNet with a multi-scale point cloud normal estimation architecture using PointNet \cite{CharlesRQi2016PointNetDL}. Based on this method, Zhou et al. \shortcite{JunZhou2019NormalEF} proposed a new scale selection strategy and extra constraints on the feature. In addition, Zhou et al. \shortcite{9693131} proposed to introduce additional feature representations to refine the input initial normal. Li et al. \shortcite{li2022hsurf} used the networks to learn hyper surfaces and obtained an improved performance.
Recent studies have achieved encouraging performance on normal estimation by predicting point-wise weights to select neighboring points softly. Lenssen et al. \shortcite{JanEricLenssen2019DeepIS} proposed to use a graph neural network to iteratively generate point-wise weights, and then estimate the normal by fitting a moving least squares plane. Ben-Shabat et al. \shortcite{ben2020deepfit} used the n-order Jet of weighted points to fit local surfaces. Cao et al. \shortcite{JunjieCao2022LatentTS} used a differentiable random sample consensus module to predict normals from a latent tangent plane representation constructed from the neighboring points. To learn better point-wise weights, Zhang et al. \shortcite{JieZhang2022GeometryGD} proposed to add direct geometric weight guidance which is constructed by distances between points and the tangent plane. Zhu \shortcite{RunsongZhu2021AdaFitRL} proposed to predict an offset to adjust the distribution of clouds and designed a cascaded scale aggregation (CSA) layer adapting to the scale of the local neighborhood, which achieved state-of-the-art results. Li et al. \shortcite{li2022graphfit} proposed to learn graph convolutional feature representation. Although improvements have been made by predicting point-wise weights and fitting surfaces for normal estimation, these methods still suffer from several inherent problems such as the difficulty in determining the form of the fitted surface and the sensitivity to outliers.
\subsection{Neural Implicit Representation in 3D Reconstruction}
Recently, neural implicit representations have gained popularity and are widely used for 3D reconstruction \cite{Zhou2022CAP-UDF, On-SurfacePriors, LPI, PredictableContextPrior, Li2022DCCDIF}. These representations were learned by neural networks to map input 3D coordinates to occupancy probabilities \cite{chen2019learning, mescheder2019occupancy} or distance values \cite{MateuszMichalkiewicz2019ImplicitSR,JeongJoonPark2019DeepSDFLC}, which implicitly represent continuous surfaces of shapes and handle complex shape topology. In addition, Neural Radiance Fields (NeRFs) \cite{mildenhall2020nerf} implicitly encode the geometry and color information of the scene in the network.
Inspired by neural implicit representations, we propose to utilize the network to implicitly represent the normals of points. Unlike existing learning-based normal estimation methods that focus only on the relationship between local neighborhoods and normal vectors, we use the network to observe the difference between random vectors and the ground truth normal. This provides the network with more prior knowledge by learning a neural angle field, which helps to predict more accurate and robust normals.
\section{Method}
In this section, we present NeAF, an implicit normal estimation method for point clouds.
\begin{figure}[t]
\centering
\includegraphics*[width=1.0\linewidth]{overview.pdf}
\caption{Overview of our NeAF. (a) During training, we randomly sample query vectors in a unit spherical space, together with the local patch as the input to the network, and predict the angle offset $\alpha$ between the query vector $\boldsymbol{n}^q$ and the ground truth normal $\boldsymbol{n}^{gt}$. (b) At inference time, we sample $m$ query vectors at different directions, and then select the query vectors with the minimum $l$ angle offsets as the coarse normals. \emph{Match} means to find the query vectors with the same indices as the minimum $l$ angle offsets. \emph{Refine} means coarse normal refinement in a subsequent section.}
\label{fig:overview}
\end{figure}
\subsection{Implicit Angle Functions}
We first define the \emph{implicit angle function} of a given normal vector. The implicit angle function takes an arbitrary query vector in a unit spherical space and a condition $\mathcal{C}$ as input and predicts the angle offset between the query vector and the ground truth vector. The corresponding function is defined as follows,
\begin{equation}
\label{eq:definition of angle function}
f: \mathcal{C} \times \mathbb{R}^3 \to \mathbb{R}.
\end{equation}
Our key idea is to estimate the normal of one point using the implicit angle function that is learned by a neural network. Observing that the normal vector is a local property of a surface, we leverage the local neighborhood of the center point as a condition $\mathcal{C}$ indicating the center point, together with the query vector as the input to feed the network, and the output is the angle offset between the ground truth and the query vector. Since we focus on unoriented normal estimation, the angle offset is either the difference from the positive direction or the negative direction of the ground truth, and its value range is $[0, \frac{\pi}{2}]$. The implicit angle function of point $p_i$ can be represented as:
\begin{equation}
\label{eq:angle function for network}
f_\theta(P_i, \boldsymbol{n}^q) = \alpha, i \in [1,N],
\end{equation}
where $f_\theta$ denotes the network parameterized by $\theta$, $P_i \in \mathbb{R}^{k \times 3}$ is the condition of point $p_i$, it is the set of $k$ nearest neighbors for the center point $p_i$, $N$ is the number of points on the point cloud, $\boldsymbol{n}^q \in \mathbb{R}^3$ is the query vector, and $\alpha \in [0, \frac{\pi}{2}]$ is the angle offset of $\boldsymbol{n}^q$ relative to the target normal.
With this definition, the target normal of $p_i$ is implicitly represented as the zero level set of the implicit angle function, which is denoted as $\boldsymbol{n}^{t}$, i.e. $f_\theta(P_i, \boldsymbol{n}^{t}) = 0$.
\subsection{Training for Implicit Angle Functions}
To learn the parameters $\theta$ of the neural network $f_\theta$, we first randomly sample query vectors $\boldsymbol{n}^q_j, j=1,...M$ on the unit sphere. For a given point $p_i, i=1,...N$, we use the K-Nearest-Neighbors algorithm to sample the nearest neighbors $P_i$ as the local patch centered at $p_i$. Next, the network takes the local patch $P_i$ and query vector $\boldsymbol{n}^q_j$ as inputs and predicts the angle offsets between the ground truth normal $\boldsymbol{n}^{gt}_{i}$ and $\boldsymbol{n}^q_j$. We minimize the difference between the predicted angle offset $f_\theta(P_i, \boldsymbol{n}^q_j)$ and ground truth angle offset $\alpha^{gt}_{ij}$ by the following $L1$ loss,
\begin{equation}
\label{eq:loss}
\mathcal{L}= \frac{1}{NM}\sum_{i \in [1,N]}{\sum_{j\in [1,M]}}|f_\theta(P_i, \boldsymbol{n}^q_j) - \alpha^{gt}_{ij}|,
\end{equation}
and the $\alpha^{gt}_{ij}$ is computed by,
\begin{equation}
\label{eq:compute gt}
\alpha^{gt}_{ij} = arcsin(||\boldsymbol{n}^{gt}_{i}\times \boldsymbol{n}^q_j||),
\end{equation}
where $arcsin(\cdot)$ represents the arcsine function, $||\cdot||$ is the L2-norm, and $\times$ is cross product.
After convergence, the network learns an implicit angle field for each point. In Figure \ref{fig:field visualization}, we visualize the optimizing process of the angle field around a single point, where the well-learned angle field can predict correct angle offsets on the whole spherical surface.
The angle field implicitly represents the target vector as the zero level-set of the learned angle field. At inference time, to predict the target vector from the learned field, we introduce a gradient descent based optimization to explore the prior learned by the angle field. An intuitive implementation is to directly optimize some randomly sampled vectors which cover the whole spherical space until their angle offsets approach zero, but it increases the computational burden to optimize a large number of vectors during inference. To resolve this issue, we propose to first predict a few coarse normals and then optimize these predicted coarse normals by a gradient decent based optimization algorithm.
\begin{figure}[t]
\centering
\includegraphics*[width=1.0\linewidth]{vis_field}
\caption{The visualization of learning an implicit angle field. Each point on the unit sphere represents a query vector, and the color map indicates their angle offsets to the ground truth normal represented as the black line. After convergence, an angle field centered on the ground truth normal is formed.}
\label{fig:field visualization}
\end{figure}
\subsection{Inference of Angle Fields}
\subsubsection{Coarse Normal Prediction.}
To predict coarse normals at inference time, we discretize the continuous unit spherical space into sections, which are the different query directions. Then a certain density of points are sampled from the discrete sections to form a query vector set $S=\{{{\boldsymbol{n}}^q_k}' | k =1, 2, ... , m\}$ to cover $m$ different directions.
If the query vectors in $S$ are infinitely dense to cover the full spherical space, there is a vector $\boldsymbol{n}^o \in S$ that satisfies $f_{\theta}(P', \boldsymbol{n}^o) = 0$, then $\boldsymbol{n}^o$ can be viewed as the approximated direction of ${\boldsymbol{n}^{gt}}'$ .
However, it is impossible to cover the full spherical space during testing since the discrete query vectors approximate the continuous spherical space and it is also time-consuming to infer highly dense query vectors. Therefore, we propose to predict a few coarse normals from the learned angle fields first.
In practice, given a test point $p_i'$, the trained network $f_{\theta}$ takes the local neighborhood $P_i'$ around $p_i'$ and the query vectors in $S$ as input, and predicts the angle offsets $ A = \{\alpha_{ik} | k = 1,2,...,m\}$, that is $f_{\theta}(P'_i, {{\boldsymbol{n}}^q_k}') = \alpha_{ik}$. We select the query vectors $\{ \boldsymbol{n}^c_s|s=1,...,l\}$ from $S$ with the minimum $l$ angle offsets as the predicted coarse normals. The coarse normals will be further refined in the following.
\begin{figure}[t]
\centering
\includegraphics*[width=1.0\linewidth]{self_optimization.pdf}
\caption{Coarse normal refinement. For the coarse normal $\boldsymbol{n}^c_{s},s=1...l$, we use the trained network $f_{\theta}$ with fixed parameters to estimate the angle offset $\alpha_s'$, then we update the vector $\boldsymbol{n}^c_s$ to make $\alpha_s'$ closer to zero. When the further optimization is done, we average all the coarse normals to output the final predicted normal.}
\label{fig:self-optimization}
\end{figure}
\begin{table*}[pt]
\small
\centering
\begin{tabular}{l|cc|cccc|l}
\toprule
\multirow{2}*{Methods} &\multicolumn{2}{c}{Density} &\multicolumn{4}{c|}{Noise} & \multirow{2}{*}{average} \cr \cmidrule{2-7}
& Stripes &Gradients & No noise & Low Noise& Med Noise& High Noise\cr
\midrule
Jets \cite{FrdricCazals2003EstimatingDQ}
&13.39 &13.13& 12.23 &12.84 &18.33 &27.68 & 16.29\cr
PCA \cite{HoppeHugues1992SurfaceRF}
&13.66 &12.81&12.29 &12.87 &18.38 &27.5 & 16.25\cr
HoughCNN \cite{AlexandreBoulch2016DeepLF}
&12.47&11.02&10.23&11.62&22.66&33.39&16.90\cr
PCPNet \cite{PaulGuerrero2017PCPNETLL}
&11.74 &13.42&9.66 &11.46 &18.26 &22.8 & 14.56\cr
Nesti-Net \cite{YizhakBenShabat2018NestiNetNE}
&8.47&9.00&6.99&10.11&17.63&22.28&12.41\cr
IterNet \cite{JanEricLenssen2019DeepIS}
&7.73&7.51&6.72&9.95&17.18&21.96&11.84\cr
DeepFit \cite{ben2020deepfit}
&7.31&7.92&6.51&9.21&16.72&23.12&11.8\cr
Refine-Net \cite{9693131}
&6.61&7.02&6.27&9.18&16.59&22.57&11.37\cr
AdaFit \cite{RunsongZhu2021AdaFitRL}
&6.04&5.90&5.19&\textbf{9.05}&16.44&21.94&10.76\cr
NeAF (Ours)
&\textbf{4.89}&\textbf{4.88}& \textbf{4.20}&9.25&\textbf{16.35}&\textbf{21.74}&\textbf{10.22}\cr
\bottomrule
\end{tabular}
\caption{Comparison of the angle RMSE with the state-of-the-art methods on PCPNet dataset.}
\label{tab:RMSE on PCPNet}
\end{table*}
\subsubsection{Coarse Normal Refinement.}
The analysis of normal prediction by the learned angle fields shows that the random sampling of discrete spherical points cannot perfectly cover the entire continuous spherical space, thus the forward pass prediction cannot make full use of the learned angle fields.
To further improve the accuracy of the predicted normals at inference time, we propose a novel learning scheme for refining the predicted normal vectors by a gradient decent based optimization algorithm. Unlike existing learning-based methods that can only predict normals via a single forward pass during inference, our proposed method can fully leverage the prior learned by NeAF to adjust the predicted normals, and can further optimize the predicted normal at inference time, thus leading to a more accurate estimation.
Specifically, the proposed NeAF represents the relationship between the target normal and an arbitrary vector in the neural network by measuring the angle offset. Thus we can adopt further refinement given the predicted coarse normals by minimizing the angle offsets of these normals.
At inference time, we utilize the network $f_{\theta}$ with fixed parameters to output an angle offset $\alpha'_{is} $ for the coarse normal $\boldsymbol{n}^c_s$, and then update $\boldsymbol{n}^c_s$ to make $\alpha'_{is}$ as close as possible to zero following the formula,
\begin{equation}
\label{eq:optimization}
{\boldsymbol{n}^c_s}' = \arg \min \limits_{\boldsymbol{n}^c_s} \mathcal{L}_1(f_{\theta}(P'_i, \boldsymbol{n}^c_s)-0),
\end{equation}
where $\mathcal{L}_1$ is the $L1$ loss, and $P'_i$ is the local patch for the testing point $p_i'$.
After coarse normal refinement, $\boldsymbol{n}^c_s$ is optimized to approximate the real zero level-set of the learned angle field.
To ensure the robustness of the results, we propose to average these refined normals ${{\boldsymbol{n}}^c_s}', s\in[1,l]$ as the final estimation $\boldsymbol{n}_{pred}$ by the following formula,
\begin{equation}
\label{eq:average coarse normals}
\boldsymbol{n}_{pred} = \frac{1}{l}\sum_{s \in [1,l]}{{\boldsymbol{n}}^c_s}'.
\end{equation}
Since the correct normal vector direction can be easily estimated by several methods \cite{mullen2010signing, wu2015deep, huang2019learning, metzer2021orienting}, we mainly focus on \emph{unoriented normal estimation}. However, the mean of the refined normals described above may lead to a large error due to the uncertain signs. For example, both the vector with an actual angle offset of $\hat{\alpha}$ and the one with an angle offset of $(\pi-\hat{\alpha})$ are with the same angle offset $\hat{\alpha}$, but the mean of the two normals can be far from the ground truth normal by an offset of $\pi/2$. To solve this problem, we normalize the signs of these vectors before averaging them together. We choose one of the refined vectors ${{\boldsymbol{n}}^c_r}'$ as the reference and re-direct the other vectors ${{\boldsymbol{n}}^c_s}', s\in[1,l]$, according to the dot product between ${{\boldsymbol{n}}^c_r}'$ and ${{\boldsymbol{n}}^c_s}'$ below
\begin{equation}
\label{eq: sign}
{{\boldsymbol{n}}^c_s}' = sign ({{{\boldsymbol{n}}^c_r}'}^T {{\boldsymbol{n}}^c_s}'){{\boldsymbol{n}}^c_s}',
\end{equation}
where $sign$ represents the sign function with an output in $\{-1, 1\}$.
\subsection{Implementation Details}
For generating training data, we adopt the method proposed by Muller \shortcite{muller1959note} to randomly and uniformly sample $M=5000$ query vectors in the unit spherical space for training, and the ground truth angle offsets are calculated by Eq. (\ref{eq:compute gt}). As for inference data, we use the same method to sample $m=10000$ query vectors for extracting coarse normals. During training, we randomly select 400 query vectors from the training set as a batch to train the network. At inference time, we select to extract $l=10$ coarse normals and optimize them simultaneously in 5 epochs for coarse normal refinement. We use the same strategy as AdaFit \cite{RunsongZhu2021AdaFitRL} for preprocessing local patches, including centering and normalizing.
We employ an encoder similar to AdaFit to learn features of local patches, which is mainly based on the PointNet \cite{CharlesRQi2016PointNetDL} architecture, and we retain the CSA layer. The local patch size and parameter settings of our networks are also the same as AdaFit. Besides, we adopt a neural network similar to DeepSDF \cite{JeongJoonPark2019DeepSDFLC} to decode the angle offsets, which is composed of 8 fully connected layers with a residual connection.
\section{Experimental Results}
\subsection{Evaluation on synthetic dataset}
\subsubsection{Dataset and metrics.}
For the experiments on synthetic shapes, we adopt the PCPNet dataset provided by Guerrero et al. \shortcite{PaulGuerrero2017PCPNETLL}. We use the same train/test settings and data augmentation strategies. PCPNet samples 100k points on the mesh of each shape to obtain a point cloud. The training set contains 8 shapes, and each shape includes a noise-free point cloud and three point clouds containing Gaussian noise with a standard deviation of 0.125\% (Low), 0.65\% (Med) and 1.2\% (High) of the length of the bounding box diagonal of the shape. In addition to the three noise variants, two additional point clouds with varying densities (Stripes and Gradients) are added to the test set. For training, we use the Adam optimizer with an initial learning rate of $1\times10^{-3}$, and adopt a cosine learning rate decay strategy with warm-up. The model is trained on 2 GTX 1080Ti. In coarse normal refinement, we use the Adam optimizer with an initial learning rate of 0.005. We use the angle root mean square error (RMSE) between the predicted normal and the ground truth normal as a quantification metric, and compute the final result with 5000 points subset sampled by PCPNet.
\subsubsection{Comparisons.}
We make a comparison with traditional normal estimation methods, including PCA \cite{HoppeHugues1992SurfaceRF}, Jets \cite{FrdricCazals2003EstimatingDQ}, and learning-based methods HoughCNN \cite{AlexandreBoulch2016DeepLF}, PCPNet \cite{PaulGuerrero2017PCPNETLL}, Nesti-Net \cite{YizhakBenShabat2018NestiNetNE}, IterNet \cite{JanEricLenssen2019DeepIS}, DeepFit \cite{ben2020deepfit}, Refine-Net \cite{9693131}, AdaFit \cite{RunsongZhu2021AdaFitRL}. A comparison of the angle RMSEs is shown in Table \ref{tab:RMSE on PCPNet}. The results show that our method significantly outperforms both the traditional methods and the state-of-the-art learning-based methods. Especially on the point clouds with density variations, the latest fitting-based method Adafit fails dramatically due to the point sparsity. This leads to a large discrepancy between the explicitly fitted surface and the underlying surface, while our method is not affected by point density and can make an accurate estimation.
We perform a visual comparison with the PCPNet \cite{PaulGuerrero2017PCPNETLL}, DeepFit \cite{ben2020deepfit}, RefineNet \cite{9693131}, and AdaFit \cite{RunsongZhu2021AdaFitRL} in Figure \ref{fig:comparision on PCPNet}, where the numbers represent the angle RMSE between the predicted normals and the real normals. The color of the shape indicates errors, and the closer to yellow the larger the error, the closer to the blue the smaller the error. The results show that our method outperforms others with more details, and can better handle complex regions such as sharp corners and sharp edges.
\begin{figure}[t]
\centering
\includegraphics[width=1.0\linewidth]{compare_PCPNet.pdf}
\caption{Errors of normal estimation on the PCPNet dataset.}
\label{fig:comparision on PCPNet}
\end{figure}
\subsection{Evaluation on real scanned dataset}
We employ the model pretrained on PCPNet dataset to report our results on the scanned dataset.
\subsubsection{SceneNN dataset.} The SceneNN \cite{BinhSonHua2016SceneNNAS} dataset provides indoor scenes in the form of reconstructed meshes. We randomly sample 1 million points on the mesh of the shape to obtain a point cloud, and calculate the normal of each point. We randomly select 40\% of all points to calculate RMSE. The results of the baseline methods are obtained by testing on the dataset using the models provided by \cite{ben2020deepfit} and \cite{RunsongZhu2021AdaFitRL} under the same experimental conditions. The angle RMSE comparison between NeAF, AdaFit \cite{RunsongZhu2021AdaFitRL} and DeepFit \cite{ben2020deepfit} is given in Table \ref{RMSE_error_indoor}, and the qualitative results are shown in Figure \ref{fig:comparision on SceneNN}. The numerical and visual comparisons show that NeAF achieves the best performance.
\begin{figure}[!tb]
\centering
\includegraphics[width=0.9\linewidth]{compare_scenenn.pdf}
\caption{ Error maps of estimated normals on the SceneNN dataset.}
\label{fig:comparision on SceneNN}
\end{figure}
\begin{table}[!tb]
\centering
\begin{tabular}{cccc}
\toprule
& {NeAF} &{AdaFit}&{DeepFit} \cr
\midrule
RMSE & \textbf{9.50}& 10.19& 12.56\cr
\bottomrule
\end{tabular}
\caption{Normal RMSE of NeAF, AdaFit \cite{RunsongZhu2021AdaFitRL} and DeepFit \cite{ben2020deepfit} on the SceneNN dataset.}
\label{RMSE_error_indoor}
\end{table}
\begin{figure}[!tb]
\centering
\includegraphics[width=1.0\linewidth]{compare_semantic3d.pdf}
\caption{Estimated normals on the semantic3D dataset. The point normal vectors are mapped to RGB colors.}
\label{fig:comparision on Semantic3D}
\end{figure}
\subsubsection{Semantic3D dataset.} The Semantic3D \cite{TimoHackel2017Semantic3DnetAN} dataset provides 30 non-overlapping outdoor scenes acquired with the Terrestrial Laser Scanner in the form of point clouds. This dataset does not provide reconstructed meshes, so the ground truth normals are not available, and we mainly report visual comparisons on this dataset. The comparison between NeAF and baseline methods is shown in Figure \ref{fig:comparision on Semantic3D}. The results show that the proposed method can estimate normals at sharp edges more accurately than baseline methods.
\subsection{Surface Reconstruction}
As an important property of local surfaces, normals are often used as the input for 3D surface reconstruction tasks, and accurate normal vectors play a key role in Poisson reconstruction \cite{MichaelKazhdan2006PoissonSR}. We use the estimated normals for surface reconstruction, and the comparison with the baseline methods is shown in Figure \ref{fig: reconstruction}. The results show that the normals estimated by NeAF are more accurate, which helps to reconstruct surfaces with higher accuracy.
\begin{figure}[!tb]
\centering
\includegraphics[width=1.0\linewidth]{reconstruction.pdf}
\caption{The comparison of the Poisson surface reconstruction using the estimated normals from NeAF, DeepFit \cite{ben2020deepfit}, and AdaFit \cite{RunsongZhu2021AdaFitRL}.}
\label{fig: reconstruction}
\end{figure}
\subsection{Ablation Studies}
\begin{table}[tb]
\centering
\begin{tabular}{l|cccc}
\toprule
& No CNP & No CNR & Min & NeAF \cr
\midrule
Stripes & 37.23 & 5.05 & 4.94 & \textbf{4.89} \cr
Gradients & 37.03 & 5.04 & 4.93 & \textbf{4.88} \cr
No Noise & 37.11 & 4.41 & 4.26& \textbf{4.20} \cr
Low Noise & 37.97 & 9.34 & 9.27 & \textbf{9.25} \cr
Med Noise & 40.10 & 16.40 & 16.36 & \textbf{16.35} \cr
High Noise & 41.92 & 21.76 & \textbf{21.73} & 21.74 \cr
\cmidrule{1-5}
Average & 38.56 & 10.33 & 10.25 & \textbf{10.22} \cr
\bottomrule
\end{tabular}
\caption{Effect of framework design.}
\label{ablation_designs}
\end{table}
\subsubsection{Framework design.}
We demonstrate the effectiveness of each design of our framework in Table \ref{ablation_designs}. We first skip predicting the coarse normals and instead directly refine the random vectors the same number of times. As shown by ``No CNP'' , the performance degenerates dramatically. We also predict normals without coarse normal refinement, and the result demonstrates that the coarse normal refinement can effectively improve the accuracy of the predicted normals in all settings as shown by ``No CNR''. We replace averaging coarse normals with selecting the coarse normals with the minimum angle offsets after refinement as shown by ``Min'' and find a drop in performance.
\begin{table}[tb]
\centering
\begin{tabular}{l|cccc}
\toprule
$M$ & 2.5k & 5k & 7.5k & 10k \cr
\midrule
Stripes & 4.94 & \textbf{4.89} & 4.95 & 4.96 \cr
Gradients & 4.98 & \textbf{4.88} & 4.98 & 5.03 \cr
No Noise & 4.29 & \textbf{4.20} & 4.24 & 4.37 \cr
Low Noise & 9.17 & 9.25 & 9.19 & \textbf{9.16} \cr
Med Noise & 16.41 & \textbf{16.35} & 16.49 & 16.41 \cr
High Noise & 21.83 &21.74 & 21.81 & \textbf{21.69} \cr
\cmidrule{1-5}
Average & 10.27 & \textbf{10.22} & 10.28 & 10.27 \cr
\bottomrule
\end{tabular}
\caption{Effect of query vector number $M$.}
\label{ablation_density}
\end{table}
\begin{table}[tb]
\centering
\begin{tabular}{l|cccc c}
\toprule
$l$ & 1 & 5 & 10 & 20 & 50 \cr
\midrule
Stripes & 4.95 & 4.90 & 4.89 & 4.89& 4.89\cr
Gradients & 4.93 & 4.90 & 4.88 & 4.88& 4.88\cr
No Noise & 4.26 & 4.21 & 4.20 & 4.21& 4.21\cr
Low Noise & 9.28 & 9.25 & 9.25 & 9.25 &9.25\cr
Med Noise & 16.37 & 16.35 & 16.35 &16.35 &16.34\cr
High Noise & 21.76 & 21.74 & 21.74 & 21.72& 21.72\cr
\cmidrule{1-6}
Average & 10.26 & 10.23 & \textbf{10.22} &10.22 &10.22\cr
\bottomrule
\end{tabular}
\caption{Effect of coarse normal number $l$.}
\label{ablation_coarse_number}
\end{table}
\subsubsection{Density of query vectors.}
We explore the effect of the sampling number $M$ of query vectors on the angle field learned by $f_{\theta}$. We report the performance of different $M=[2.5k, 5k, 7.5k, 10k]$ in Table \ref{ablation_density}. The query vector set that is too sparse (``2.5k'') cannot provide enough sample vectors for the network to learn the angle field, while the sets that are too dense (``7.5k'', ``10k'') make the implicit angle function more complicated and make it difficult for the network to learn the correct angle offsets. We found ``5k'' is a proper trade-off.
\subsubsection{Number of coarse normals.}
In Table \ref{ablation_coarse_number}, we conduct experiments on the PCPNet dataset to explore how the coarse normal number $l$ affects coarse normal refinement at the inference time. We report the performance of different $l=[1, 5, 10, 20, 50]$, where we find that optimizing a single coarse normal leads to the degradation of results, and the best accuracy is achieved for the first time with 10 coarse normals. Optimizing more coarse normals requires longer time and more memory without improving the accuracy.
\section{Conclusion}
In this paper, we proposed NeAF to estimate point normals implicitly. We randomly sample the query vectors in a unit spherical space, estimate their angle offsets to ground truth normals, and output the query vector with the smallest angle offset as the estimated normal. To fully leverage the prior learned by NeAF, we refine the predicted normal vector by minimizing the estimated angle offset for more accurate normal estimation. NeAF achieves state-of-the-art performance on the synthetic dataset PCPNet and exhibits good generalization on real scans in SceneNN and Semantic3D. Furthermore, the promising results in the surface reconstruction task with normals estimated by NeAF justify our effectiveness in real applications.
\section{Acknowledgments}
The corresponding author is Yu-Shen Liu. This work was supported by National Key R\&D Program of China (2022YFC3800600, 2020YFF0304100), the National Natural Science Foundation of China (62272263, 62072268), and in part by Tsinghua-Kuaishou Institute of Future Media Data.
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Eliseo Angulo (ur. 1902, zm. ?) – boliwijski piłkarz grający na pozycji pomocnika.
Kariera reprezentacyjna
Eliseo Angulo grał w reprezentacji Boliwii w latach dwudziestych. W 1926 uczestniczył w Copa América 1926. Boliwia zajęła na tym turnieju ostatnie, piąte miejsce, a Angulo wystąpił tylko w pierwszym meczu turnieju przeciwko Chile.
Bibliografia
Profil
Mecze w Copa América
Reprezentanci Boliwii w piłce nożnej
Uczestnicy Copa América 1926
Urodzeni w 1902
Zmarli w XX wieku | {
"redpajama_set_name": "RedPajamaWikipedia"
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\section{Introduction}
Let $\mathbf{a}^k=(a_1^k,...,a_k^k) \in {\mathbb R}^k$ and $\mathbf{b}^k=(b_1^k,...,b_k^k) \in {\mathbb R}^k$. By a
Jacobi matrix, we mean a real, symmetric, tridiagonal matrix of the form
\begin{equation}\label{finitejacobi}
J(\mathbf{a}^k,\mathbf{b}^k) = \left(
\begin{array}{ccccc}
a_1^k & b_1^k & 0 & \cdots & 0 \\
b_1^k & a_2^k & b_2^k & \ddots & \vdots \\
0 & b_2^k & a_3^k & \ddots & 0 \\
\vdots & \ddots & \ddots & \ddots & b_{k-1}^k \\
0 & \cdots & 0 & b_{k-1}^k & a_k^k \\
\end{array}
\right).
\end{equation}
Jacobi matrices have a wide range of applications in mathematical sciences. In physics for instance, they
naturally appear in the study of random matrices and discrete Schr\"odinger operators \cite{MR883643}. In
statistics, they provide a useful tool to study stochastic processes such as the birth-death process and random
walks. In classical analysis, they play an important role in the study of orthogonal polynomials \cite{deift96, MR903848}.
The spectral properties of $J(\mathbf{a}^k,\mathbf{b}^k)$ are well-known and can be found in many texts; a good reference is e.g. \cite{deift96}. If $M$ denotes the quantity
\[ M=\sqrt{3} \left[ \max \limits_{1 \leq i \leq k} |a_i^k| + \max \limits_{1 \leq i \leq k} |b_i^k| \right], \]
then one can easily verify that $\sigma(J(\mathbf{a}^k,\mathbf{b}^k))$, the spectrum of $J(\mathbf{a}^k,\mathbf{b}^k)$, consists of $k$ real simple eigenvalues lying inside the interval $[-M,M]$.
Except for few special cases, it is in general impossible to obtain explicit expressions for the eigenvalues of an arbitrary Jacobi matrix. However, an important special case for which the spectrum is explicitly known is the case of tridiagonal Toeplitz matrices. These matrices have the form
\begin{equation}
T_k(a,b) = \left(
\begin{array}{ccccc}
a & b & 0 & \cdots & 0 \\
b & a & b & \ddots & \vdots \\
0 & b & a & \ddots & 0 \\
\vdots & \ddots & \ddots & \ddots & b \\
0 & \cdots & 0 & b & a \\
\end{array}
\right)
\end{equation}
for some $a \in {\mathbb R}$ and $b>0$. The eigenvalues of $T_k(a,b)$ are well-known and are given by
\[ \lambda_j^k = a +2b \cos \left( \frac{j}{k}\pi \right) \qquad \text{ for } j=1,...,k. \]
Using Riemann sums, it is then straightforward to derive the following asymptotic trace formula
\begin{eqnarray*}
\lim_{k \to \infty} \frac{1}{k} \text{Trace}[\phi(T_k(a,b))] & = &
\lim_{k \to \infty} \frac{1}{k} \sum_{j=1}^k \phi \left( a+2b\cos \left( \frac{j}{k} \pi \right) \right) \\
& = & \frac{1}{\pi} \int_0^{\pi} \phi(a+ 2b \cos x) \, dx
\end{eqnarray*}
for any continuous function $\phi$ on the interval $[a-2b,a+2b]$. This trace formula was successfully used to obtain new ones for Jacobi matrices that are small perturbation of a Toeplitz matrix. For instance, this approach can be used to derive Nevai's result \cite{ne79} on the density of the zeros of orthogonal polynomials that belongs to the $M(a,b)$-class.
In this paper, we continue our investigation initiated in our recent work \cite{agbo09}. We start by deriving two asymptotic trace formulas for the moments that extend the ones given in Lemma 2.3 of \cite{agbo09}. In the third section, we derive our main results. Namely, if we assume the bounded sequences $(\mathbf{a}^k)_{k}, (\mathbf{b}^k)_{k}$ to satisfy the small-deviation conditions of Definition \ref{def1} and if $(\mathbf{a}^k,\mathbf{b}^k)_{k}$ is $\mu$-distributed as in the sense of Definition \ref{mu-distributed}, then we have
\begin{equation} \label{main}
\lim_{k \to \infty} \frac{1}{k} \text{Trace}[\phi(J(\mathbf{a}^k,\mathbf{b}^k))] = \frac{1}{\pi} \int_{[0,1]^2} \int_0^\pi \phi(x+2y\cos t) \, dt \, d\mu(x,y)
\end{equation}
for any $\phi \in C[-3,3]$. We also present similar results for unbounded sequences. In addition, we show in the fourth section of the paper that for any given sequence $(\mathbf{a}^k)_k, \ (\mathbf{b}^k)_{k}$ satisfying the small-deviation conditions of Definitions \ref{def1} and \ref{unbounded}, one can always find a subsequence and probability measure $\mu$ for which a trace formula similar to \eqref{main} holds.
\section{Moments}
We start by proving a trace formula for the moments of $J(\mathbf{a}^k,\mathbf{b}^k)$ when the sequences $(\mathbf{a}^k)_k$ and $(\mathbf{b}^k)_k$ are bounded and satisfy the small deviation condition below. We also present results when the sequences are unbounded.
The first thing consists of defining in precise terms what we mean for a sequence $(\mathbf{a}^k)_k$ with $\mathbf{a}^k=(a_1^k,...,a_k^k) \in {\mathbb R}^k$ to satisfy the required small deviation condition.
\begin{defn} \label{def1}
We say that the sequence $(\mathbf{a}^k)_k$ belongs to $\mathcal{S}$ if it satisfies the following two conditions:
\begin{itemize}
\item[(i)] $\sum_{i=1}^{k-1} |a_{i+1}^k-a_{i}^k|= o(k)$ as $k \to \infty$,
\item[(ii)] $0\leq a_i^k \leq 1$ for all $i=1,...,k$.
\end{itemize}
\end{defn}
Note that (ii) holds without loss generality for any bounded sequence. Indeed, if $|a_i^k|<M$ for all $i \leq k$ and $k \in {\mathbb N}$, then the normalized sequence $(\mathbf{\tilde{a}}^k)_k$ with $\tilde{a}_i^k=1/2+a_i^k/2M$ satisfies $(ii)$.
\medskip
\noindent \textbf{Example 1:} Let $(a_k)_k$ be any convergent sequence in $[0,1]$. If we let $a_i^k=a_i$ for $i \leq k$, then it is easy to see that $(\mathbf{a}^k)_k \in \mathcal{S}$.
\medskip
\noindent \textbf{Example 2:} As an example of a sequence $(\mathbf{a}^k)_k \in \mathcal{S}$ that is not necessarily obtained from a convergent sequence, consider a bounded sequence $(\mathbf{a}^k)_k$ that satisfies for any $0<\delta<1$:
$$ \frac{\#\{ 1 \leq j \leq k: |a_{j+1}^k-a_j^k| = {\cal O}(k^{-\delta}) \}}{k} \to 1 \qquad \text{ as } k \to \infty.$$
One can easily verify that $(\mathbf{a}^k)_k$ satisfies condition (i) of Definition \ref{def1}. For instance, any uniformly distributed sequence $(\mathbf{a}^k)_k$ modulo one with discrepancy
$$D_k =\max_{1 \leq i \leq k} \left\{ \left| a_i^k-\frac{i}{k} \right|, \left|a_i^k - \frac{i-1}{k} \right| \right\} = {\cal O}(k^{-\delta})$$
falls into this category \cite{MR0419394}.
\medskip
Our first trace formula is concerned with the moments of Jacobi matrices $J(\mathbf{a}^k,\mathbf{b}^k)$ whose defining sequences $(\mathbf{a}^k)_k$ and $(\mathbf{b}^k)_k$ belong to $\mathcal{S}$.
\begin{prop} \label{prop 1}
Let $(\mathbf{a}^k)_k, (\mathbf{b}^k)_k$ be two sequences in $\mathcal{S}$. For any $n \in {\mathbb N}$, we have
\begin{equation} \label{trace000}
\text{Trace}\left[ J^n(\mathbf{a}^k,\mathbf{b}^k) \right]=\sum_{j=0}^{\lfloor n/2 \rfloor} \frac{n!}{(j!)^2(n-2j)!} \sum_{i=1}^k (a_i^k)^{n-2j} (b_i^k)^{2j} + o(k).
\end{equation}
\end{prop}
\noindent \textit{Proof:} We write $J_k:=J(\mathbf{a}^k,\mathbf{b}^k)$ as the sum of three matrices $$J_k=L_k+D_k+L^T_k$$ where $D_k=\text{diag}(a_1^k,...,a_k^k)$ and $L_k$ is the lower triangular matrix given by
\begin{equation}
L_k = \left(
\begin{array}{ccccc}
0 & 0 & 0 & \cdots & 0 \\
b_1^k & 0 & 0 & \ddots & \vdots \\
0 & b_2^k & 0& \ddots & 0 \\
\vdots & \ddots & \ddots & \ddots & 0 \\
0 & \cdots & 0 & b_{k-1}^k & 0 \\
\end{array}
\right).
\end{equation}
For any $n \in {\mathbb N}$, $J^n_k=(L_k^T+D_k+L_k)^n$ is the sum of $3^n$ matrix monomials of the form $$A_1A_2 \cdots A_n$$ with $A_j$ being either $L_k$, $D_k$ or $L_k^T$. One can easily verify that the monomials having a different number of $L_k$ and $L_k^T$ in their expressions have zero trace. Therefore, it suffices to consider those having the same number of $L_k$ and $L_k^T$.
The assumptions on the sequences $(\mathbf{a}^k)_k$ and $(\mathbf{b}^k)_k$ imply that when we permute any two consecutive matrices in the product $A_1 A_2 \cdots A_n$, the trace of the resulting expression differ from the trace of the original one by $o(k)$. In other words, we have
\[ \frac{1}{k} \text{Trace}[A_1 \cdots A_i A_{i+1} \cdots A_n] = \frac{1}{k} \text{Trace}[A_1 \cdots A_{i+1} A_i \cdots A_n ] + o(1). \]
To see this, let us consider the case where $A_i=L^T_k$ and $A_{i+1}=L_k$; the other cases can be handled in a similar manner. We have that
\begin{eqnarray*}
A_1A_2 \cdots A_n & = & A_1 A_2 \cdots A_{i+1}A_i \cdots A_n \\
& & + A_1 A_2 \cdots A_{i-1} [A_i,A_{i+1}] A_{i+2} \cdots A_n.
\end{eqnarray*}
By an elementary property of the trace, we also have
\begin{multline*}
\text{Trace}[ A_1 A_2 \cdots A_{i-1} [A_i,A_{i+1}] A_{i+2} \cdots A_n] \\= \text{Trace}[ [A_i,A_{i+1}]A_{i+2}\cdots A_n A_1 \cdots A_{i-1} ]
\end{multline*}
The product $A_{i+2} \cdots A_n A_1 \cdots A_{i-1}$ is a matrix whose diagonal elements are bounded, while the commutator matrix $[A_i,A_{i+1}]$ is a diagonal matrix whose elements are $(b_1^k)^2,(b_2^k)^2-(b_1^k)^2,...,(b^k_k)^2-(b_{k-1}^k)^2,(b^k_k)^2$. Under the assumptions that $(\mathbf{a}^k)_k$ and $(\mathbf{b}^k)_k$ belong to $\mathcal{S}$, we easily deduce
\begin{eqnarray*}
\text{Trace}[ [A_i,A_{i+1}] A_{i+2} \cdots A_n A_1 A_2 \cdots A_{i-1} ] & \leq & |b_1^k|^2 + \sum_{j=1}^{k-1} |b_{j+1}^k-b_j^k|^2 + |b_k^k|^2 \\
& = & o(k).
\end{eqnarray*}
Consequently, it suffices to consider terms of the form $(L_kL^T_k)^{2j} D_k^{n-2j}$ for which the trace can easily be computed. Indeed, we have
\[ \text{Trace}[(L_kL^T_k)^{2j} D_k^{n-2j}]=\sum_{i=1}^k (a_i^k)^{n-2j} (b_i^k)^{2j}. \]
Since there are $\frac{n!}{(j!)^2(n-2j)!}$ monomials containing $j$ matrices $L_k$ and $L_k^T$ and $n-2j$ matrices $D_k$, we finally obtain
\begin{eqnarray*}
\lefteqn{\frac{1}{k} \text{Trace}\left[ J^n(\mathbf{a}^k,\mathbf{b}^k) \right]}\\
& = & \frac{1}{k}\sum_{j=0}^{\lfloor n/2 \rfloor} \frac{n!}{(j!)^2(n-2j)!} \text{Trace}[(L_kL^T_k)^{2j} D_k^{n-2j}] + o(1) \\
& = & \frac{1}{k}\sum_{j=0}^{\lfloor n/2 \rfloor} \frac{n!}{(j!)^2(n-2j)!} \sum_{i=1}^k (a_i^k)^{n-2j} (b_i^k)^{2j} + o(1)
\end{eqnarray*}
as desired. $\qed$
\bigskip
The boundedness condition $(ii)$ in Definition \ref{def1} can be removed if we slightly strengthen the first condition. More precisely, we have:
\begin{defn} \label{unbounded}
We say that the sequence $(\mathbf{a}^k)_k$ belongs to $\mathcal{S'}$ if it satisfies the following two conditions: For any $\delta \in (0,1)$,
\begin{itemize}
\item[(i')] $\sum_{i=1}^{k-1} |a_{i+1}^k-a_{i}^k|= {\cal O}(k^{1-\delta})$ as $k \to \infty$,
\item[(ii')] $\max \limits_{1 \leq i \leq k} |a_i^k|= {\cal O}(\log k)$ as $k \to \infty$
\end{itemize}
\end{defn}
It is readily seen that a similar proof as the one of Proposition \ref{prop 1} holds when the sequences $(\mathbf{a}^k)_k $ and $(\mathbf{b}^k)_k$ are both in $\mathcal{S}'$. We state this result as our second trace formula.
\begin{prop} \label{prop 2}
Let $(\mathbf{a}^k)_k, \,(\mathbf{b}^k)_k \in \mathcal{S}'$. For any $n \in {\mathbb N}$, we have
\begin{equation} \label{trace00}
\text{Trace}\left[ J^n(\mathbf{a}^k),(\mathbf{b}^k) \right]= \sum_{j=0}^{\lfloor n/2 \rfloor} \frac{n!}{(j!)^2(n-2j)!} \sum_{i=1}^k (a_i^k)^{n-2j} (b_i^k)^{2j} + o(k).
\end{equation}
\end{prop}
\section{Main trace formulas: The $\mu$-distributed case}
We now derive our main asymptotic trace formulas for Jacobi matrices whose defining sequences $(\mathbf{a}^k)_k$ and $(\mathbf{b}^k)_k$ are distributed according to some probability measure. We consider different cases depending on if the sequences $(\mathbf{a}^k)_k$ and $(\mathbf{b}_k)^k$ are bounded, unbounded or monotone.
\subsection{The bounded case}
Our first results are concerned with compactly supported probability measure $\mu$ on ${\mathbb R}^2$. After normalization, we may assume without loss of generality that the support is contained in $I^2:=[0,1] \times [0,1]$.
\begin{defn} \label{mu-distributed}
We say that a sequence $(\mathbf{a}^k,\mathbf{b}^k)_k$ with $a_j^k$ and $b_j^k$ in $[0,1]$ for all $j=1,...,k$ is $\mu$-distributed if for any continuous function $\psi$ on $[0,1]^2$, one has
\[ \lim_{k \to \infty} \frac{1}{k} \sum_{j=1}^k \psi(a^k_j,b^k_j) = \int_{I^2} \psi(x,y) \ d\mu(x,y). \]
\end{defn}
\noindent \textbf{Example 3:} Consider the sequences $(\mathbf{a}^k)_k$ and $(\mathbf{b}^k)_k$ that are convergent, i.e. $\mathbf{a}^k=(a_1,...,a_k)$ and $\mathbf{b}^k=(b_1,...,b_k)$ with $a_k \to a$ and $b_k \to b$. It is easy to see that $(\mathbf{a}^k,\mathbf{b}^k)_k$ is $\mu$-distributed with $\mu=\delta_a \times \delta_b$, the product of the Dirac measures at $a$ and at $b$.
\medskip
\noindent \textbf{Example 4:} Let $a,b:[0,1] \to [0,1]$ be two continuous functions. We define the sequences $\mathbf{a}^k$ and $\mathbf{b}^k $ by
$$\mathbf{a}^k=(a(1/k),a(2/k),...,a((k-1)/k),a(1))$$
and
$$\mathbf{b}^k=(b(1/k),b(2/k),...,b((k-1)/k),b(1)).$$
It follows that $(\mathbf{a}^k,\mathbf{b}^k)_k$ is $m_f$-distributed where $m_f$ is the probability distribution associated to the random variable $f(x)=(a(x),b(x))$, i.e.
\[ \int_{[0,1]^2} \psi(x,y) \, dm_{f} (x,y) = \int_0^1 \psi(a(x),b(x)) \, dx \]
for any $\psi \in C(I^2)$.
\medskip
In our first theorem of this section, we give a trace formula for Jacobi matrices with bounded sequences that are $\mu$-distributed.
\begin{theo} \label{theo 1}
Let $\mu$ be a probability measure on $I^2$. Suppose that $(\mathbf{a}^k, \mathbf{b}^k)_k$ is a $\mu$-distributed sequence with $(\mathbf{a}^k)_k, \ (\mathbf{b}^k)_k \in \mathcal{S}$. Then, we have
\begin{equation} \label{trace1}
\lim_{k \to \infty} \frac{1}{k} \text{Trace}[\phi(J(\mathbf{a}^k,\mathbf{b}^k))] = \frac{1}{\pi} \int_{I^2} \int_0^\pi \phi(x+2y\cos t) \, dt \, d\mu(x,y)
\end{equation}
for any $\phi \in C[-3,3]$.
\end{theo}
\noindent \textit{Proof:} For any nonnegative integer $n$, Proposition \ref{prop 1} implies
\begin{multline} \label{trace3}
\frac{1}{k} \text{Trace}\left[ J^n(\mathbf{a}^k,\mathbf{b}^k) \right] \\
= \sum_{j=0}^{\lfloor n/2 \rfloor} \frac{n!}{(j!)^2(n-2j)!} \int_{I^2} x^{n-2j}y^{2j} \, d\mu(x,y) + o(1).
\end{multline}
In order to replace the sum $\sum_{j=0}^{\lfloor n/2 \rfloor}$ by the sum over all multi-indices $\alpha=(\alpha_1,\alpha_2,\alpha_3)$ with $|\alpha|=n$, we introduce the function $sinc:{\mathbb R} \to {\mathbb R}$ defined by
\begin{equation*}
\text{sinc}( \xi) = \frac{1}{2} \int_{-1}^1 e^{i\pi t \xi} \, dt.
\end{equation*}
Note that $\text{sinc}(0)=1$ while $\text{sinc}(\pi x)=0$ when $x$ is a non-zero integer. Using this function, \eqref{trace3} becomes
\begin{multline*}
\frac{1}{k} \text{Trace}\left[ J^n(\mathbf{a}^k,\mathbf{b}^k) \right] \\
= \frac{1}{2} \int_{I^2} \int_{-1}^1 \sum_{|\alpha|=n} \binom{n}{\alpha} e^{i\pi t(\alpha_1-\alpha_2)} x^{\alpha_3}y^{\alpha_1+\alpha_2} \, d\mu(x,y) \, dt + o(1).
\end{multline*}
The multinomial theorem and a simple change of variables then yield
\begin{equation} \label{trace5}
\frac{1}{k} \text{Trace}\left[ J^n(\mathbf{a}^k,\mathbf{b}^k) \right]
= \frac{1}{\pi} \int_{I^2} \int_0^\pi (x+2y\cos(t))^n \, dt \, d\mu(x,y) + o(1).
\end{equation}
This establishes the result for the moments of $J(\mathbf{a}^k,\mathbf{b}^k)$. By linearity, the result also holds for polynomials of arbitrary degree. Now let $\phi$ be a continuous function on $[-3,3]$. Note that
$$ \{ x+2y\cos(t): \, (x,y) \in I^2, \, t \in [0,\pi] \} \subseteq [-1,3]$$
and $\sigma(J(\mathbf{a}^k,\mathbf{b}^k)) \subseteq [-2\sqrt{3},2\sqrt{3}]$.
By Weierstrass Approximation Theorem, there is a polynomial $P$ such that $\|\phi-P\|_{\infty} < \epsilon/3$ on $[-3,3]$. In particular, it implies that
\begin{equation} \label{trace7}
\left| \frac{1}{k} \text{Trace}[\phi(J(\mathbf{a}^k,\mathbf{b}^k))] - \frac{1}{k} \text{Trace}[P(J(\mathbf{a}^k,\mathbf{b}^k))] \right| \leq \epsilon/3,
\end{equation}
together with
\begin{multline} \label{trace8}
\left| \frac{1}{\pi} \int_{I^2} \int_0^\pi \phi(x+2y\cos(t)) \, dt \, d\mu(x,y) \right. \\
\left. -\frac{1}{\pi} \int_{I^2} \int_0^\pi P(x+2y\cos(t)) \, dt \, d\mu(x,y) \right| \leq \epsilon/3.
\end{multline}
Finally, \eqref{trace5} shows that we can choose $k$ large enough so that
\begin{equation} \label{trace9}
\left| \frac{1}{k} \text{Trace}\left[ P(J(\mathbf{a}^k,\mathbf{b}^k)) \right]
-\frac{1}{\pi} \int_{I^2} \int_0^\pi P(x+2y\cos(t)) \, dt \, d\mu(x,y) \right| \leq \epsilon/3.
\end{equation}
The conclusion of the theorem follows by combining \eqref{trace7}, \eqref{trace8}, and \eqref{trace9}. $\qed$
\medskip
\noindent \textbf{Example 5:} Let $(\mathbf{a}^k)_k$ and $(\mathbf{b}^k)_k$ be convergent sequences as in Example 3. From Theorem \ref{theo 1}, we easily deduce that
\begin{eqnarray*}
\lim_{k \to \infty} \frac{1}{k} \, \text{Trace}[\phi(J(\mathbf{a}^k,\mathbf{b}^k))] & = & \frac{1}{\pi} \int_0^\pi \phi(a+2b \cos t) \, dt \\
& = & \frac{1}{\pi} \int_{a-2b}^{a+2b} \phi(t) \frac{dt}{\sqrt{(a+2b-u)(u-a+2b)}}.
\end{eqnarray*}
This type of Jacobi matrices naturally arise for orthogonal polynomials that belong to the $M(a,b)$ class introduced by Nevai \cite{ne79}. One can use this result to easily compute the asymptotic distribution of the zeros of many classical orthogonal polynomials such as the Jacobi ($a=1/4$, $b=1/2$), Chebyshev ($a=0$, $b=1/2$), Legendre $(a=0, b=1/2)$ and Gegenbauer $(a=0, b=1/2)$ polynomials.
\medskip
\noindent \textbf{Example 6:} In computing the asymptotic distribution of the zeros of Van Vleck polynomials \cite{bosh08}, one is led to consider a Jacobi matrix of the form
\begin{equation}\label{finitejacobi}
J_k = \left(
\begin{array}{ccccc}
a_1 & b_1 & 0 & \cdots & 0 \\
b_1 & a_2 & b_2 & \ddots & \vdots \\
0 & b_2 & a_3 & \ddots & 0 \\
\vdots & \ddots & \ddots & \ddots & b_{k-1} \\
0 & \cdots & 0 & b_{k-1} & a_k \\
\end{array}
\right).
\end{equation}
where the entries are up to some constants given by
\[ a_j=\frac{(j-1)}{k} \text{ and } b_j=\frac{j\sqrt{1-(j/k)^2}}{k}. \]
It is easy to see that the sequences $(a_k)_k$ and $(b_k)_k$ are both in $\mathcal{S}$, and the sequence $(a_k,b_k)_k$ is $\mu$-distributed with
\[ \mu(x,y) = \frac{\delta_0(y-\sqrt{x(1-x)})}{2\sqrt{x}}. \]
It then follows from Theorem \ref{theo 1} that
\begin{equation*}
\lim_{k \to \infty} \frac{1}{k} \text{Trace}[\phi(J_k)] = \frac{1}{\pi} \int_0^1 \int_0^\pi \phi(x+2\sqrt{x(1-x)} \cos t) \, dt \, \frac{dx}{2 \sqrt{x}}.
\end{equation*}
\medskip
In a recent paper, Kuiljaars and Serra-Cappizano \cite{MR1866252} proved a general result regarding the asymptotic distribution of the eigenvalues of Jacobi matrices. Their results generalize earlier ones obtained by Kuiljaars and Van Assche \cite{kuva99} and Geronimo, Harrell II and Van Assche \cite{MR956176}. We give here a new proof of their result based on our previous trace formula in the generic case $|a_j^k| \leq 1$ and $0< b_j^k \leq 1$.
\begin{cor} (Kuiljaars-Serra Cappizano) Let $(J(\mathbf{a}^k,\mathbf{b}^k))_k$ be a sequence of Jacobi matrices and suppose we can choose two bounded measurable functions $a,b:[0,1] \to [0,1]$ satisfying the conditions: For any $\epsilon>0$,
\begin{equation} \label{ksc cond 1}
|\{ s \in [0,1]: |a_{\lceil sk \rceil}^k - a(s) | \geq \epsilon \}| \to 0 \text{ as } k \to \infty,
\end{equation}
\begin{equation} \label{ksc cond 2}
|\{ s \in [0,1]: |b_{\lceil sk \rceil}^k - b(s) | \geq \epsilon \}| \to 0 \text{ as } k \to \infty.
\end{equation}
Then, we have
\begin{equation*}
\lim_{k \to \infty} \frac{1}{k} \, \text{Trace}[\phi(J(\mathbf{a}^k,\mathbf{b}^k))] = \frac{1}{\pi} \int_0^1 \int_0^\pi \phi(a(x)+2b(x) \cos y) \, dx \, dy
\end{equation*}
for any $\phi \in [-3,3]$.
\end{cor}
\noindent \textit{Proof:} For $k$ large enough, conditions \eqref{ksc cond 1} and \eqref{ksc cond 2} together with Lusin's Theorem imply that we can choose the functions $a$ and $b$ to be continuous on $[0,1]$ except for a set of measure $1/k$ and for which the additional conditions
\[ \#\{j \leq k :|a_j^k-a(j/k)| \geq \epsilon \} =o(k) \]
\[ \#\{j \leq k :|b_j^k-b(j/k)| \geq \epsilon \} =o(k) \]
hold for any given $\epsilon>0$. It easily follows that for any continuous function $\psi$ on $[0,1]^2$,
\begin{equation} \label{trace15}
\lim_{k \to \infty} \frac{1}{k} \sum_{j=1}^k \psi(a_j^k, b_j^k) = \int_0^1 \psi(a(x),b(x)) \, dx.
\end{equation}
If we introduce the map $f:[0,1] \to [0,1]^2$ defined by $f(x)=(a(x),b(x))$, then it follows as in Example 4 that the sequence $(\mathbf{a}^k,\mathbf{b}^k)_k$ is $m_f$-distributed. The conclusion is then an immediate consequence of Theorem \ref{theo 1}. $\qed$
\medskip
In the next result, we derive a simple extension of Example 5. Indeed, we consider sequences $(\mathbf{a}^k)_k$ and $(\mathbf{b}^k)_k$ that have a finite number of accumulation points.
\begin{cor}
Let $(\mathbf{a}^k)_k$ and $(\mathbf{b}^k)_k$ be two sequences in $\mathcal{S}$. Let $\Omega=[\omega_1,...,\omega_n]$ be an $n$ vector for which $0 \leq \omega_i \leq 1$ and
$$|\Omega|:=\sum_{i=1}^n \omega_i=1.$$
Suppose there exists two $n$-tuple $(\alpha_1,...,\alpha_n)$ and $(\beta_1,...,\beta_n)$ such that for any $1\leq p \leq n$, and $\epsilon>0$,
\begin{equation} \label{cond a}
\frac{\#\{ i : |a_i^k-a_p|< \epsilon \text{ and } |b_i^k-b_p| < \epsilon \}}{k} \to \omega_p \ \text{ as } k \to \infty.
\end{equation}
Then, for every continuous function $\phi$ on $[-3,3]$, one has
\begin{equation}
\lim_{k \to \infty} \frac{1}{k} \sum_{j=1}^k \text{Trace}[\phi(J(\mathbf{a}^k,\mathbf{b}^k))] = \frac{1}{\pi} \int_0^\pi \int_{[0,1]^2} \phi(x+2y\cos t) \,d\omega(x,y) \,dt
\end{equation}
where $\omega$ is the probability measure given by the weighted sum of delta measures, i.e.
\begin{equation}
\omega(x,y)= \sum_{j=1}^n \omega_i \, \delta_{a_i}(x) \times \delta_{b_i}(y).
\end{equation}
\end{cor}
\noindent \textit{Proof:} It is readily seen that both sequences satisfy the assumptions of Theorem \ref{theo 1} with $\mu$ given by $\omega$ as defined in the statement of the theorem. $\qed$
\medskip
As a simple consequence of the previous result, we have the following result also due to Kuiljaars and Serra Capizzano \cite{MR1866252} when the sequences $(\mathbf{a}^k)_k$ and $(\mathbf{b}^k)_k$ have a single accumulation point.
\begin{cor} (Kuiljaars-Serra Capizzano) \label{cor ks}
Let $(a_k)_k$ and $(b_k)_k$, $b_k>0$, be two bounded sequences as in Example 1, i.e. $(\mathbf{a}^k)_k$ is the sequence defined by $\mathbf{a}^k=(a_1,...,a_k)$ and similarly for $(\mathbf{b}^k)_k$. Suppose there exist real constant $a$ and $b>0$ such that for every $\epsilon>0$,
\begin{equation}
\#\{j\leq k: |a_j-a| \geq \epsilon \} = o(k) \ \text{ as } k \to \infty,
\end{equation}
and
\begin{equation}
\#\{j\leq k: |b_j-b| \geq \epsilon \} = o(k) \ \text{ as } k \to \infty.
\end{equation}
Then, for every continuous function $\phi$ on $[a-2b,a+2b]$,
\begin{equation}
\lim_{k \to \infty} \frac{1}{k} \sum_{j=1}^k \text{Trace}[\phi(J(\mathbf{a}^k,\mathbf{b}^k))] = \frac{1}{\pi} \int_0^\pi \phi(a+2b\cos x) \,dx.
\end{equation}
\end{cor}
\noindent \textbf{Remark:} Recently, Trench \cite{MR2013470} obtained weaker conditions under which the conclusion of Corollary \ref{cor ks} holds. However, his conditions are of different nature as they are related to the spectrum of $J(a_k,b_k)$ rather than the sequences $(a_k)_k$ and $(b_k)_k$.
\subsection{The unbounded case: Part I}
We now turn our attention to the unbounded case, i.e. we now assume that the sequences $(\mathbf{a}^k)_k$ and $(\mathbf{b}^k)_k$ can take values over the whole real line. Our first result is concerned with sequences that can be contracted in order to fit the bounded case. The next definition is based on the terminology introduced by Van Assche \cite{MR903848}.
\begin{defn} \label{reg}
We say that $r:(0,\infty) \to (0,\infty)$ is a regularly varying function (at infinity) for the sequence $(\mathbf{a}^k,\mathbf{b}^k)_k$ if the normalized sequences
$$ \left( \frac{\mathbf{a}^k}{r(k)} \right)_k, \ \left( \frac{\mathbf{b}^k}{r(k)} \right)_k$$
are in $\mathcal{S}$.
\end{defn}
\medskip
The next result is an immediate consequence of Theorem \ref{theo 1}, so its proof is omitted.
\begin{cor} \label{cor reg 1}
Let $r$ be a regularly varying function for the sequence $(\mathbf{a}^k,\mathbf{b}^k)_k$. Suppose that and $(\mathbf{a}^k/r(k), \, \mathbf{b}^k/r(k))_k$ is $\mu$-distributed for some probability measure $\mu$ on $I^2$. Then,, we have
\[ \lim_{k \to \infty}\frac{1}{k} \text{Trace}\left[\phi\left(J\left( \frac{\mathbf{a}^k}{r(k)}, \,\frac{\mathbf{b}^k}{r(k)} \right) \right) \right] = \frac{1}{\pi} \int_{I^2} \int_0^\pi \phi(x+2y\cos t) \, dt \, d\mu(x,y)\]
for any $\phi \in C[-3,3]$
\end{cor}
\medskip
\noindent \textbf{Example 7:} Using the three-terms recurrence relation for the Hermite polynomials, it is well-known that the zeros $z_1^k,...,z_k^k$ of the $k$th degree Hermite polynomial are the eigenvalues of the tridiagonal matrix
\begin{equation}
H_k=\left(
\begin{array}{ccccc}
0 & 1/2 & 0 & \cdots & 0 \\
1 & 0 & 1/2 & \ddots & \vdots \\
0 & 2 & 0 & \ddots & 0 \\
\vdots & \ddots & \ddots & \ddots & 1/2 \\
0 & \cdots & 0 & k-1 & 0 \\
\end{array}
\right).
\end{equation}
It easy to see that $H_k$ is similar to the Jacobi matrix
\begin{equation}
H_k'=\left(
\begin{array}{ccccc}
0 & 1/\sqrt{2} & 0 & \cdots & 0 \\
1/\sqrt{2} & 0 & 1 & \ddots & \vdots \\
0 & 1 & 0 & \ddots & 0 \\
\vdots & \ddots & \ddots & \ddots & \sqrt{\frac{k-1}{2}} \\
0 & \cdots & 0 & \sqrt{\frac{k-1}{2}} & 0 \\
\end{array}
\right).
\end{equation}
Indeed, $H_k'=D_kH_kD_k^{-1}$ where $D_k=\text{diag}(d_1,...,d_{k})$ is the diagonal matrix whose elements are defined recursively by the relations
\[ d_1=1, \text{ and } d_{i+1}=\frac{d_i}{\sqrt{2i}} \qquad \text{ for } i=1,...,k-1. \]
The regularly varying function is $r(k)=\sqrt{2k}$ and the sequence
\[ \left( \frac{\mathbf{a}^k}{r(k)},\frac{\mathbf{b}^k}{r(k)} \right)_k = \left( \mathbf{0}^k, \left( \sqrt{\frac{j}{k}} \right)_{j=1}^k \right)_k \]
is $\mu$-distributed on $I^2$ with
\[ d\mu(x,y)= 2 y \delta_0(x) \, dx dy. \]
Consequently, the asymptotic distribution of the contracted zeros of Hermite polynomials is given by
\begin{eqnarray*}
\lim_{k \to \infty} \sum_{j=1}^k \phi \left( \frac{z_j^k}{\sqrt{2k}} \right) & = & \frac{1}{\pi} \int_{I^2} \int_0^\pi \phi(x+2y\cos t) dt \, d\mu(x,y) \\
& = & \frac{2}{\pi} \int_0^1 \int_0^\pi \phi(2y\cos t) \, y \, dt \, dy\\
& = & \frac{1}{2\pi} \int_{-2}^2 \phi(x) \sqrt{4-x^2} \, dx
\end{eqnarray*}
for any $\phi \in C[-\sqrt{2},\sqrt{2}]$. We thus obtain another proof of the well-known fact that the asymptotic distribution of the contracted zeros of Hermite polynomials is the Wigner semicircle distribution.
\medskip
\noindent \textbf{Example 8:} One can use a similar approach to derive the well-known fact that the asymptotic distribution of the contracted zeros of Laguerre polynomials is the Marchenko-Pastur distribution. Indeed, the zeros $z_1^k,...,z_k^k$ of the $k$th degree Laguerre polynomial are the eigenvalues of the Jacobi matrix
\begin{equation}
L_k=\left(
\begin{array}{ccccc}
1 & 1 & 0 & \cdots & 0 \\
1 & 3 & 2 & \ddots & \vdots \\
0 & 2 & 5 & \ddots & 0 \\
\vdots & \ddots & \ddots & \ddots & k-1 \\
0 & \cdots & 0 & k-1 & 2k-1 \\
\end{array}
\right).
\end{equation}
Hence, the regularly varying function is $r(k)=k$ and the sequence is
\[ \left( \frac{\mathbf{a}^k}{r(k)},\frac{\mathbf{b}^k}{r(k)} \right)_k = \left( \left(\frac{2j-1}{k}\right)_{j=1}^{k}, \left( \frac{j}{k} \right)_{j=1}^{k} \right)_k \]
is $\mu$-distributed on $I^2$ with
\[ d\mu(x,y)= \delta_0(x-2y) \, dx dy. \]
The asymptotic distribution of the contracted zeros of Laguerre polynomials is therefore given by
\begin{eqnarray*}
\lim_{k \to \infty} \sum_{j=1}^k \phi \left( \frac{z_j^k}{k} \right) & = & \frac{1}{\pi} \int_{I^2} \int_0^{\pi} \phi(x+2y\cos t) \, dt \, d\mu(x,y) \\
& = & \frac{1}{\pi} \int_0^1 \int_0^\pi \phi(2y+2y\cos t) \, dt \, dy\\
& = & \frac{1}{2 \pi} \int_0^4 \phi(x) \ \frac{\sqrt{4x-x^2}}{x} \, dx
\end{eqnarray*}
for any $\phi \in C[0,4]$.
\medskip
More generally, we can use our trace formula in Corollary \ref{cor reg 1} to give a new proof of the following result due to Van Assche.
\begin{theo}
Let $r$ be a regularly varying function with parameter $\alpha \in (0,1)$, i.e. for every $t>0$,
\[ \lim_{x \to \infty} \frac{r(tx)}{r(x)} = t^{\alpha}. \]
Let $(a_k)_k$ and $(b_k)_k$ be two sequences with $a_k \in {\mathbb R}$ and $b_k>0$ satisfying the conditions
\begin{equation} \label{conditions}
\lim_{k \to \infty} \frac{a_k}{r(k)} = a, \qquad \lim_{k \to \infty} \frac{b_k}{r(k)}=b.
\end{equation}
If $J(\mathbf{a}^k,\mathbf{b}^k)$ is the Jacobi matrix with $\mathbf{a}^k=(a_1,...,a_k)$ and $\mathbf{b}^k=(b_1,...,b_k)$, then we have
\begin{equation*}
\lim_{k \to \infty} \frac{1}{k} \text{Trace}\left[\phi\left(J\left( \frac{\mathbf{a}^k}{r(k)}, \,\frac{\mathbf{b}^k}{r(k)} \right) \right) \right] = \frac{1}{\pi}\int_{a-2b}^{a+2b} f(x) \, v_\alpha(x) \, dx.
\end{equation*}
Here $v_\alpha(x)$ denotes the Nevai-Ullman density with parameter $\alpha$ defined as the Mellin convolution
\[ v_\alpha(x)=b_\alpha \ast \omega_{a,b}(x) = \int_0^1 b_\alpha(y) \omega_{a,b}(x/y) \,\frac{dy}{y} \]
with
\[ b_\alpha(y) = \begin{cases}
\alpha y^{\alpha-1} & \text{ if } y \in (0,1),\\
0 & \text{ elsewhere},
\end{cases} \]
and
\[ \omega_{a,b}(y) = \begin{cases}
\frac{1}{\pi} \, \frac{1}{\sqrt{(a+2b-y)(y-a+2b)}} & \text{ if } y \in (a-2b,a+2b).\\
0 & \text{elsewhere.}
\end{cases} \]
\end{theo}
\medskip
\noindent \textit{Proof:} The conditions \eqref{conditions} imposed on the sequence $(a_k)_k$ and $(b_k)_k$ imply that for any $\epsilon>0$,
\[ \frac{ \# \left\{ 1\leq j \leq k: \left| \frac{a_j}{r(k)} - a \left( \frac{j}{k} \right)^{\alpha} \right| < \epsilon, \right\}}{k} \to 1 \]
and
\[ \frac{ \# \left\{ 1\leq j \leq k: \, \left| \frac{b_j}{r(k)} - b \left( \frac{j}{k} \right)^{\alpha} \right| < \epsilon \right\}}{k} \to 1 \]
for $k \to \infty$. In particular, it follows that
\begin{equation*}
\lim_{k \to \infty} \frac{1}{k} \text{Trace}\left[\phi\left(J\left( \frac{\mathbf{a}^k}{r(k)}, \,\frac{\mathbf{b}^k}{r(k)} \right) \right) \right] =
\lim_{k \to \infty} \frac{1}{k} \text{Trace}\left[\phi\left(J\left( a\mathbf{u}^k, \, b\mathbf{u}^k \right) \right) \right]
\end{equation*}
where the sequence $\mathbf{u}^k$ is given by
\[ \mathbf{u}^k=(1/k^\alpha,(2/k)^\alpha,...,((k-1)/k)^\alpha,1). \]
Clearly, $\mathbf{u}^k \in \mathcal{S}$ and by Example 4, the sequence $\left( a\mathbf{u}^k, \, b\mathbf{u}^k \right)_k$ is $\mu$-distributed with $\mu$ satisfying
\[ \int_{I^2} \psi(x,y) \,d\mu(x,y) = \int_0^1 \psi(ax^\alpha,bx^\alpha) \, dx \]
for any $\psi \in C(I^2)$. By Theorem \ref{theo 1}, we deduce that
\begin{eqnarray*}
\lefteqn{\lim_{k \to \infty} \frac{1}{k} \text{Trace}\left[\phi\left(J\left( \frac{\mathbf{a}^k}{r(k)}, \,\frac{\mathbf{b}^k}{r(k)} \right) \right) \right]}\\
& & \qquad \qquad = \frac{1}{\pi} \int_0^1 \int_0^{\pi} \phi(ax^\alpha+2bx^\alpha \cos (\pi t)) \, dt \,dx\\
& & \qquad \qquad = \frac{1}{\pi} \int_0^1 \int_{a-2b}^{a+2b} f(t^\alpha x) \frac{dx \, dt}{\sqrt{(a+2b-x)(x-a+2b)}}
\end{eqnarray*}
Under some simple changes of variable, last is easily seen to be equivalent to the expression given in the statement of theorem. $\qed$
\subsection{The unbounded case: Part II}
Our second result for Jacobi matrices for unbounded sequences is concerned with sequences $(\mathbf{a}^k)_k$ and $(\mathbf{b}^k)_k$ that belong to $\mathcal{S}'$ (see Definition \ref{unbounded}). Before we can go further, we need to reformulate Definition \ref{mu-distributed} for unbounded sequences.
\begin{defn} \label{mu-distributed unb}
We say that a sequence $(\mathbf{a}^k,\mathbf{b}^k)_k$ with $(\mathbf{a}^k)_k$ and $(\mathbf{b}^k)_k$ in $\mathcal{S}'$ is $\mu$-distributed if for any $\psi \in C_b({\mathbb R}^2):=C({\mathbb R}^2) \cap L_\infty({\mathbb R}^2)$, the space of bounded continuous functions on ${\mathbb R}^2$, one has
\[ \lim_{k \to \infty} \frac{1}{k} \sum_{j=1}^k \psi(a^k_j,b^k_j) = \int_{{\mathbb R}^2} \psi(x,y) \ d\mu(x,y) \]
for some probability measure $\mu$ on ${\mathbb R}^2$.
\end{defn}
Recall, a probability measure is said to be tight if for any $\epsilon>0$, one can find a closed bounded rectangle $\mathcal{R}=[\alpha_1,\beta_1] \times [\alpha_2,\beta_2]$ such that
$$\mu({\mathbb R}^2-\mathcal{R}) < \epsilon.$$
\medskip
\begin{theo} \label{theo unbounded}
Let $(\mathbf{a}^k)_k$ and $(\mathbf{b}^k)_k$ be two sequences in $\mathcal{S}'$. If $(\mathbf{a}^k,\mathbf{b}^k)_k$ is $\mu$-distributed for some tight probability measure $\mu$ on ${\mathbb R}^2$, then we have
\begin{equation} \label{unb 0}
\lim_{k \to \infty} \frac{1}{k}\text{Trace}\left[ \phi(J(\mathbf{a}^k,\mathbf{b}^k)) \right]
= \frac{1}{\pi} \int_{{\mathbb R}^2} \int_0^\pi \phi(x+2y \, \cos t) \, d\mu(x,y)
\end{equation}
for any $\phi \in C_b({\mathbb R})$.
\end{theo}
\medskip
\noindent \textit{Proof:} By Proposition \ref{prop 2}, we know that
\begin{equation} \label{unb1}
\frac{1}{k}\text{Trace}\left[ J^n(\mathbf{a}^k,\mathbf{b}^k) \right]
= \frac{1}{k}\sum_{j=0}^{\lfloor n/2 \rfloor} \frac{n!}{(j!)^2(n-2j)!} \sum_{i=1}^k a_i^{n-2j} b_i^{2j} + o(1).
\end{equation}
Since $\mu$ is tight and $(\mathbf{a}^k,\mathbf{b}^k)_k$ is $\mu$-distributed, there exists for any $\epsilon>0$ a closed bounded rectangle $\mathcal{R} \subseteq {\mathbb R}^2$ such that
\begin{equation} \label{unb1.1}
\frac{\#\{ j \leq k: (a_j^k,b_j^k) \notin \mathcal{R} \}}{k} < \epsilon
\end{equation}
for $k$ large enough. We now introduce the smooth cutoff function $\theta:{\mathbb R}^2 \to [0,1]$ defined by
$$\theta(x,y) = \begin{cases}
1 & \text{ if } (x,y) \in \mathcal{R}\\
0 & \text{ if } (x,y) \notin \mathcal{R}_\epsilon
\end{cases}$$
where $\mathcal{R}_\epsilon$ is a compact subset of ${\mathbb R}^2$ with $\mathcal{R} \subseteq \mathcal{R}_\epsilon$ and $m(\mathcal{R}_\epsilon/\mathcal{R})<\epsilon$. Equations \eqref{unb1} and \eqref{unb1.1} yield
\begin{multline} \label{unb2}
\frac{1}{k}\text{Trace}\left[ J^n(\mathbf{a}^k,\mathbf{b}^k) \right] \\
= \frac{1}{k}\sum_{j=0}^{\lfloor n/2 \rfloor} \frac{n!}{(j!)^2(n-2j)!} \sum_{i=1}^k (a_i^k)^{n-2j} (b_i^k)^{2j} \theta(a_i,b_i) + o(1)
\end{multline}
for any nonnegative integer $n$. Moreover, $(\mathbf{a}^k,\mathbf{b}^k)_k$ being $\mu$-distributed implies
\begin{multline} \label{unb3}
\frac{1}{k}\text{Trace}\left[ J^n(\mathbf{a}^k,\mathbf{b}^k) \right] \\
= \sum_{j=0}^{\lfloor n/2 \rfloor} \frac{n!}{(j!)^2(n-2j)!} \int_{{\mathbb R}^2} x^{n-2j} y^{2j} \ \theta(x,y) \, d\mu(x,y)+ o(1).
\end{multline}
We now perform our usual trick, that is we introduce the $sinc$ function in \eqref{unb3} and use the multinomial theorem to obtain
\begin{multline} \label{unb4}
\frac{1}{k}\text{Trace}\left[ J^n(\mathbf{a}^k,\mathbf{b}^k) \right] \\
= \frac{1}{\pi} \int_{{\mathbb R}^2} \int_0^\pi (x+2y \, \cos t)^n \ \theta(x,y) \, d\mu(x,y)+ o(1).
\end{multline}
Note that
$$ K_1:=\{ x+2y \cos t : (x,y) \in \mathcal{R}_\epsilon \text{ and } t \in [0,\pi] \}$$
is a compact subset of ${\mathbb R}$. Moreover, the assumptions on $\mu$ together with \eqref{unb1.1} imply that there exists a compact subset $K_2 \subseteq {\mathbb R}$ such that
$$ \# \{ \lambda : \lambda \in \sigma(J(\mathbf{a}^k,\mathbf{b}^k)), \ \lambda \notin K_2\} =o(k) $$
as gets $k$ arbitrary large. In particular, we have
\begin{equation}
\frac{1}{k}\text{Trace}\left[ J^n(\mathbf{a}^k,\mathbf{b}^k) \right] = \frac{1}{k} \sum_{\lambda \in \sigma(J(\mathbf{a}^k,\mathbf{b}^k)) \cap K_2} \lambda^n +o(1).
\end{equation}
By Weierstrass Approximation Theorem applied tp a closed bounded interval that contains $K_1 \cup K_2$, it is then easy to see that \eqref{unb4} yields
\begin{multline}
\frac{1}{k}\text{Trace}\left[ \phi(J(\mathbf{a}^k,\mathbf{b}^k)) \right] \\
= \frac{1}{\pi} \int_{{\mathbb R}^2} \int_0^\pi \phi(x+2y \, \cos t) \ \theta(x,y) \, d\mu(x,y)+ o(1)
\end{multline}
for any $\phi \in C_b({\mathbb R})$. Furthermore, since the measure $\mu$ is tight, it follows that
\begin{eqnarray} \label{unb5}
\lefteqn{\left| \int_{{\mathbb R}^2} \phi(x+2y \, \cos t) \ (1-\theta(x,y)) \, d\mu(x,y) \right|} \nonumber \\
& & \qquad \leq \left| \int_{{\mathbb R}^2-R_\epsilon} \phi(x+2y \, \cos t) \, d\mu(x,y) \right| + \nonumber\\
& & \qquad \quad \left| \int_{R_\epsilon-R} \phi(x+2y \, \cos t) \ (1-\theta(x,y)) \, d\mu(x,y) \right| \nonumber\\
& & \qquad \leq 2 \|\phi\|_\infty \epsilon.
\end{eqnarray}
The desired trace formula \eqref{unb 0} is then an immediate consequence of \eqref{unb4} and \eqref{unb5}. $\qed$
\medskip
\noindent \textbf{Example 9:} Let $(\mathbf{X}^k)_k$ and $(\mathbf{Y}^k)_k$ be two sequences with $\mathbf{X}^k$ and $\mathbf{Y}^k$ being $k$-vectors whose components are i.i.d. random variables that are normally distributed with mean zero and variance $\sigma^2(k) \to 0$ as $k \to \infty$. First, we note that for all $1 \leq j \leq k$,
\begin{multline} \label{erfc}
\text{Prob}\{ \omega \in {\mathbb R}: |X^k_j(\omega)|> \log k\} \\
= \frac{1}{\sqrt{2}} \, \text{Erfc} \, (\sigma(k) \, \log k )= {\cal O} \left( \frac{e^{-\log^2 k/2\sigma^2(k)}}{k \log k} \right).
\end{multline}
Secondly, Chebyshev's inequality implies that
\begin{eqnarray*}
\lefteqn{\text{Prob} \left\{ \omega \in {\mathbb R}: |X_{j+1}^k(\omega)-X_j^k(\omega)|> \sigma^{1/2}(k) \right\}} \nonumber\\
& & \leq \frac{1}{\sigma(k)} \int_{-\infty}^\infty \left| X_{j+1}^k(\omega)-X_j^k(\omega) \right|^2 \, e^{-\omega^2/2\sigma^2(k)} \, \frac{d\omega \nonumber}{\sqrt{2\pi}} \\
& & \leq \frac{8}{\sigma(k)} \int_{-\infty}^\infty |X_j^k(\omega)|^2 \, e^{-\omega^2/2\sigma^2(k)} \, \frac{d\omega \nonumber}{\sqrt{2\pi}}\\
& & = 8 \sigma(k).
\end{eqnarray*}
From this, we easily deduce that
\begin{equation} \label{cheb}
\frac{1}{k} \sum_{j=1}^k |X_{j+1}^k(\omega)-X_j^k(\omega)| = {\cal O}(\sigma^{1/2}(k))
\end{equation}
for asymptotically almost every (a.a.e.) $\omega \in {\mathbb R}$. In particular, if we assume that $\sigma(k)={\cal O}(k^{-\delta})$ for some $\delta>0$, it then follows from \eqref{erfc} and \eqref{cheb} that $(\mathbf{X}^k(\omega))_k \in \mathcal{S}'$ for a.a.e. $\omega \in {\mathbb R}$. Of course, a similar conclusion also hold for $(\mathbf{Y}^k(\omega))_k$.
Furthermore, under the independence assumption, one can apply the strong Law of Large Numbers to conclude
\begin{eqnarray*}
\frac{1}{k} \sum_{j=1}^k \psi(X_j(\omega),Y_j(\omega))
& = & \frac{1}{2\pi \sigma(k)} \int_{{\mathbb R}^2} \psi(x,y) \, e^{-(x^2+y^2)/2\sigma^2(k)} \, dx \, dy + o(1)\\
& = & \psi(0,0) +o(1)
\end{eqnarray*}
for a.a.e. $\omega \in {\mathbb R}$ and any $\psi \in C_b({\mathbb R}^2)$. In other words, the sequence $(X_k(\omega),Y_k(\omega))_k$ is $\delta_0 \times \delta_0$-distributed in the sense of Definition \ref{mu-distributed unb}. By Theorem \ref{theo unbounded}, we conclude that for a.a.e. $\omega \in {\mathbb R}$,
\begin{equation*}
\lim_{k \to \infty} \frac{1}{k} \text{Trace}[\phi(J(\mathbf{X}^k(\omega),\mathbf{Y}^k(\omega)))] = \phi(0)
\end{equation*}
for any $\phi \in C_b({\mathbb R})$.
\medskip
\subsection{The monotone case}
An obvious, but interesting corollary of Propositions \ref{prop 1} and \ref{prop 2} is obtained by considering sequences that are almost everywhere monotone.
\begin{defn}
We say that $(\mathbf{a}^k)_{k \in {\mathbb N}}$ is a.e. increasing if
\[ \frac{\#\{ 1 \leq i \leq k: a_i^k \leq a_{i+1}^k \}}{k} \to 1 \qquad \text{as } k \to \infty. \]
\end{defn}
\medskip
Of course, a similar condition holds for a.e. decreasing sequences. Clearly, any a.e. monotone sequence that satisfies conditions (ii) or (ii') of Definitions \ref{def1} and \ref{unbounded} respectively will also satisfy conditions (i) or (i') of the same definitions. As a consequence, we obtain the following result.
\begin{prop}
Let $(\mathbf{a}^k)_k, (\mathbf{b}^k)_k$ be two a.e. monotone sequences with either
$$(\mathbf{a}^k)_k, (\mathbf{b}^k)_k \in \mathcal{S}$$
or
$$(\mathbf{a}^k)_k, (\mathbf{b}^k)_k \in \mathcal{S}'.$$
Then, for any $n \in {\mathbb N}$, we have
\begin{equation} \label{trace0}
\text{Trace}\left[ J^n(\mathbf{a}^k,\mathbf{b}^k) \right]=\sum_{j=0}^{\lfloor n/2 \rfloor} \frac{n!}{(j!)^2(n-2j)!} \sum_{i=1}^k (a_i^k)^{n-2j} (b_i^k)^{2j} + o(k).
\end{equation}
\end{prop}
If we assume furthermore that the sequence $(\mathbf{a}^k,\mathbf{b}^k)_k$ is $\mu$-distributed as in the sense of Definitions \ref{mu-distributed} or \ref{mu-distributed unb}, then we easily derive the following two results that are simple consequence of Theorems \ref{theo 1} and \ref{theo unbounded}.
\begin{cor} \label{cor mono 1}
Let $(\mathbf{a}^k)_k$ and $(\mathbf{b}^k)_k$ be two a.e. monotone sequences in $[0,1]$. If $(\mathbf{a}^k,\mathbf{b}^k)_k$ is $\mu$-distributed for some probability measure $\mu$ on $I^2$, then we have
\begin{equation}
\lim_{k \to \infty} \frac{1}{k}\text{Trace}\left[ \phi(J(\mathbf{a}^k,\mathbf{b}^k)) \right]
= \frac{1}{\pi} \int_{I^2} \int_0^\pi \phi(x+2y \, \cos t) \, d\mu(x,y)
\end{equation}
for any $\phi \in C[-3,3]$.
\end{cor}
\begin{cor} \label{cor mono 2}
Let $(\mathbf{a}^k)_k$ and $(\mathbf{b}^k)_k$ be two a.e. monotone sequences in ${\mathbb R}$ that satisfy the estimates
$$ |a_j^k|={\cal O}(\log k) \text{ and } |b_j^k|={\cal O}(\log k) \text{ for all } 1 \leq j \leq k, \ k \in {\mathbb N}.$$
If $(\mathbf{a}^k,\mathbf{b}^k)_k$ is $\mu$-distributed for some tight probability measure $\mu$ on ${\mathbb R}^2$, then we have
\begin{equation}
\lim_{k \to \infty} \frac{1}{k}\text{Trace}\left[ \phi(J(\mathbf{a}^k,\mathbf{b}^k)) \right]
= \frac{1}{\pi} \int_{{\mathbb R}^2} \int_0^\pi \phi(x+2y \, \cos t) \, d\mu(x,y)
\end{equation}
for any $\phi \in C_b({\mathbb R})$.
\end{cor}
\medskip
\noindent \textbf{Example 10:} Let $(\mathbf{X}^k)_k$ and $(\mathbf{Y}^k)_k$ be sequences where
$$\mathbf{X}^k=(X_1^k,...,X_k^k) \text{ and } \mathbf{Y}^k=(Y_1^k,...,Y_k^k)$$
are two $k$ random vectors. We assume for all $1\leq j \leq k$, $k \in {\mathbb N}$, that the $X_j^k$'s and $Y_j^k$'s are i.i.d. random variables of mean zero and variance one on some probability space $(\Omega,P)$. We denote by $\mathbf{X}^{(k)}$ the random vector whose $j$th entry is the $j$th order statistic $X_{(j)}^k$ obtained from $X_1^k,...,X_k^k$, i.e.
$$ \mathbf{X}^{(k)}=(X_{(1)}^k,...,X_{(k)}^k)$$
and similarly for $\mathbf{Y}^{(k)}$. For any $\omega \in \Omega$, we denote by $J_k(\omega)$ the Jacobi matrix given by
\begin{equation}
\mathbf{J}_{(k)}(\omega) = \left(
\begin{array}{ccccc}
X_{(1)}^k(\omega) & Y_{(1)}^k(\omega) & 0 & \cdots & 0 \\
Y_{(1)}^k(\omega) & X_{(2)}^k(\omega) & Y_{(2)}^k(\omega) & \ddots & \vdots \\
0 & Y_{(2)}^k(\omega) & X_{(3)}^k(\omega) & \ddots & 0 \\
\vdots & \ddots & \ddots & \ddots & Y_{(k-1)}^k(\omega) \\
0 & \cdots & 0 & Y_{(k-1)}^k(\omega) & X_{(k)}^k(\omega) \\
\end{array}
\right).
\end{equation}
By definition of order statistics, the sequences $\mathbf{X}^{(k)}$ and $\mathbf{Y}^{(k)}$ are monotone increasing. If we assume that $X_j^k(\omega) \in [0,1]$ and $Y_j^k(\omega) \in (0,1]$, then the strong Law of Large Numbers implies that for a.a.e $\omega \in \Omega$,
\begin{eqnarray*}
\lim_{k \to \infty} \sum_{j=1}^k \psi(X_{(j)}^k(\omega),Y_{(j)}^k(\omega)) = \int_{I^2} \psi(x,y) \ d\mu_{X,Y}(x,y)
\end{eqnarray*}
for any $\psi \in C(I^2)$. Here, we denote by $\mu_{X,Y}$ the joint probability distribution of $X$ and $Y$ where $X$ and $Y$ are two random variables on $\Omega$ that follow the same distribution as $X_j^k$ and $Y_j^k$ respectively. Consequently, we can apply Corollary \ref{cor mono 1} to conclude that
\begin{equation}
\lim_{k \to \infty} \frac{1}{k}\text{Trace}\left[ \phi(\textbf{J}_k(\omega) \right]
= \frac{1}{\pi} \int_{I^2} \int_0^\pi \phi(x+2y \, \cos t) \, d\mu_{X,Y}(x,y)
\end{equation}
for any $\phi \in C[-3,3]$ and a.a.e $\omega \in \Omega$.
Similarly, if we assume that $X_j^k = {\cal O}(\log k)$ and $Y_j^k= {\cal O}(\log k)$, then we can apply Corollary \ref{cor mono 2} to obtain
\begin{equation}
\lim_{k \to \infty} \frac{1}{k}\text{Trace}\left[ \phi(\textbf{J}_k(\omega) \right]
= \frac{1}{\pi} \int_{{\mathbb R}^2} \int_0^\pi \phi(x+2y \, \cos t) \, d\mu_{X,Y}(x,y)
\end{equation}
for any $\phi \in C_b({\mathbb R})$ and a.a.e $\omega \in \Omega$.
\section{Existence Results}
In this section, we would like to address the following question: Given two sequences $(\mathbf{a}^k)_k$ and $(\mathbf{b}^k)_k$ in $\mathcal{S}$ or $\mathcal{S}'$, does there exist a probability measure $\mu$ with support in $I^2$ or a tight measure probability measure $\mu$ with support in ${\mathbb R}^2$ for which the sequence $(\mathbf{a}^k,\mathbf{b}^k)_k$ is $\mu$-distributed? It is easy to construct examples of sequences that are not $\mu$-distributed for any probability measure. For instance, take the sequence $(\mathbf{a}^k,\mathbf{b}^k)_k$ with
\[ (a_j^k,b_j^k) = \begin{cases}
(0,1) & \text{ if } 2 | k \\
(1,1) & \text{ if } 2 \nmid k.
\end{cases} \]
However, as the next result shows one can always find subsequences that are $\mu$-distributed.
\begin{prop} \label{prop ex 1}
Let $(\mathbf{a}^k)_k$ and $(\mathbf{b}^k)_k$ be two sequences in $\mathcal{S}$. Then, there exists a subsequence of positive integers $k_j$ and a probability measure $\mu$ with support in $[0,1]^2$ such that $(\mathbf{a}^{k_j},\mathbf{b}^{k_j})_j$ is $\mu$-distributed.
\end{prop}
\noindent \textit{Proof:} For any $x,y \in {\mathbb R}$, consider the distribution functions $F_k$ defined by
$$F_k(x,y) = \frac{\# \{ j \leq k: a_j^k < x \text{ and } b_j^k<y\}}{k}.$$
By Helly's selection principle (\cite{MR830424}, Theorem 25.9), we can find a subsequence of positive integers $k_j$, $k_j \to \infty$ as $j \to \infty$, such that $F_{k_j}$ converges weakly (in the probability sense) to a distribution function $F$. By Theorem 12.5 in \cite{MR830424}, we can associate to $F_{k_j}$ and $F$ unique probability measures $\mu_{k_j}$ and $\mu$ on $[0,1]^2$ defined by
\[ \mu((a,b) \times (c,d)) =F(b,d)-F(b,c)-F(a,d)+F(a,c) \]
and similarly for $\mu_{k_j}$. Since $\mu_{k_j}$ converges weakly to $\mu$, we can apply Theorem 29.1 in \cite{MR830424} to conclude that $(\mathbf{a}^{k_j},\mathbf{b}^{k_j})$ is $\mu$-distributed as in the sense of Definition \ref{mu-distributed}. $\qed$
\medskip
\begin{prop} \label{prop ex 2}
Let $(\mathbf{a}^k)_k$ and $(\mathbf{b}^k)_k$ be two sequences in $\mathcal{S}'$. Suppose that for any $\epsilon>0$, there exists a bounded rectangle $R$ such that
\begin{equation} \label{tight seq}
\frac{\# \{ 1 \leq j \leq k : |a_j^k| \notin R\}}{k} < \epsilon, \ \text{ and } \ \frac{\# \{ 1 \leq j \leq k : |b_j^k| \notin R\}}{k} < \epsilon
\end{equation}
for $k$ large enough. Then, there exists a subsequence of positive integers $k_j$ and a tight probability measure $\mu$ on ${\mathbb R}^2$ such that $(\mathbf{a}^{k_j},\mathbf{b}^{k_j})_j$ is $\mu$-distributed.
\end{prop}
\noindent \textit{Proof:} The argument is identical as for the proof of Proposition \ref{prop ex 1} with the exception that Theorem 29.1 in \cite{MR830424} must be replaced by Theorem 29.3. $\qed$
\medskip
By combining Proposition \ref{prop ex 1} and Theorem \ref{theo 1}, or Proposition \ref{prop ex 2} together with Theorem \ref{theo unbounded}, we obtain the following two existence results.
\begin{cor} \label{subsequence}
Let $(\mathbf{a}^k)_k$ and $(\mathbf{b}^k)_k$ be two sequences in $\mathcal{S}$. Then, there exists a subsequence of positive integers $k_j$ and a probability measure $\mu$ with support in $I^2$ such that
\begin{equation} \label{trace 1 sub}
\lim_{j \to \infty} \frac{1}{k_j} \text{Trace}[\phi(J(\mathbf{a}^{k_j},\mathbf{b}^{k_j}))] = \frac{1}{\pi}\int_{I^2} \int_0^\pi \phi(x+2y\cos t) \, dt \, d\mu(x,y)
\end{equation}
for any $\phi \in C[-3,3]$.
\end{cor}
\begin{cor}
Let $(\mathbf{a}^k)_k, \ (\mathbf{b}^k)_k \in \mathcal{S}'$ be two sequences satisfying \eqref{tight seq}. Then, there exists a subsequence of positive integers $k_j$ and a probability measure $\mu$ such that
\begin{equation} \label{trace 1 sub unb}
\lim_{j \to \infty} \frac{1}{k_j} \text{Trace}[\phi(J(\mathbf{a}^{k_j},\mathbf{b}^{k_j}))] = \frac{1}{\pi}\int_{{\mathbb R}^2} \int_0^\pi \phi(x+2y\cos t) \, dt \, d\mu(x,y)
\end{equation}
for any $\phi \in C_b({\mathbb R})$.
\end{cor}
\section{Conclusion}
Our results establish new and very general conditions for which one can compute explicitly the asymptotic distribution of the spectrum of Jacobi matrices. Our results are new in the sense that we do not assume any type of convergence for the sequences $(\mathbf{a}^k)_k,(\mathbf{b}^k)_k$ of $J(\mathbf{a}^k,\mathbf{b}^k)$, but we only assume they satisfy a small-deviation condition and they are $\mu$-distributed.
A quick a look at the proofs should convince the readers that similar techniques as the ones developed in this paper can be used to compute an asymptotic trace formula for band matrices. For instance, it is straightforward to extend the proof of Propositions \ref{prop 1} and \ref{prop 2} to matrices with three diagonal bands of the form
\begin{equation}
\left(
\begin{array}{cccccccc}
a_1^k & & & & & b_1^k & & \\
& \ddots & & & & & \ddots & \\
& & \ddots & & & & & b_{k-q}^k\\
c_1^k & & & a_p^k & & & & \\
& \ddots& & & \ddots & & & \\
& & \ddots & & & a_q^k & & \\
& & & \ddots & & & \ddots & \\
& & & & c_{k-p}^k & & & a_k^k
\end{array}
\right).
\end{equation}
where the sequences $(\mathbf{a}^k)_k,(\mathbf{b}^k)_k,(\mathbf{c}^k)_k$ are assumed to satisfy the usual small-deviation conditions. In particular, we would like to use these extensions to derive new results for the asymptotic distribution of difference and Sturm-Liouville operators. We hope to address these points in future research work.
\bibliographystyle{plain}
| {
"redpajama_set_name": "RedPajamaArXiv"
} | 76 |
Nobody Drives in LA — Los Angeles Train Map
Los Angeles was made possible by trains. Rate wars between Southern Pacific and Atchison, Topeka, and Santa Fe railways ignited the first population boom. To this day, the vast, expansive Pacific Electric Railway was the largest interurban electric railway the world has ever known. Its primary purpose was o expand growth in the suburbs — or as they were known then, "toonervilles" — after a popular comic strip, The Toonerville Trolley That Meets All the Trains.
Screen shot of Los Angeles's railways and train-related sites of interest
Metro operates four light rail lines, two subways, and two metro liners. There are currently 86 stations and 158.5 kilometers of rail although more are under construction — the Gold Line Foothill Extension just opened and the Expo Line Extension opens in May. Metrolink maintains 55 stations and their seven lines cover 624 kilometers.
Metrolink, too, is expanding, with the (39 kilometer Perris Valley Extension opening any day now. Los Angeles and Southern California is also served by the national service, Amtrak, national bus services, the ferry service Catalina Express (older than either Metro or Metrolink), two domestic airports, and one international airport (with two more in the region).
Pendersleigh & Sons Cartography's watercolor map of all passenger rail in Southern California
I made this train map for fun and because finding comprehensive rail maps online is surprisingly difficult. Furthermore, most rail maps are designed for clarity — which is nice — but removes a sense of the railways' (and their stations') surroundings. I've also included abandoned rail cars, model railroad clubs, novelty trolleys, &c. I didn't include buses (aside from Metro Liners) here because there are 170 Metro bus lines — not to mention dozens of other transit providers. It would too cluttered (and too much work). I did include the ferries, however, in part to remind people that Los Angeles includes two large islands (Santa Catalina and San Clemente) and also to nudge Angelenos and visitors into thinking differently about public transit actually is.
This map is a joint venture of California Fool's Gold, Nobody Drives in LA, and Pendersleigh & Sons Cartography. It is, like our public transit system, a work in progress. I realize that transit advocates (and opponents) are often passionate and particular so please bear with me and offer additions and corrections with kindness and not blunt, internet outrage. I endeavor to make corrections in a timely manner but I'm not too concerned with exact rail alignments down to the millimeter so if my casual but fairly comprehensive approach disappoints you, I encourage you to make your own map. Thanks! Oh, and if you want to explore a great Pacific Electric map, The Militant's Pacific Electric Archaeology Map, click here).
Eric Brightwell is an adventurer, writer, rambler, explorer, cartographer, and guerrilla gardener who is always seeking writing, speaking, traveling, and art opportunities — or salaried work. He is not interested in writing advertorials, clickbait, listicles, or other 21st century variations of spam. Brightwell's written work has appeared in Amoeblog, diaCRITICS, and KCET Departures. His work has been featured by the American Institute of Architects, the Architecture & Design Museum, the Craft & Folk Art Museum, Form Follows Function, Los Angeles County Store, Skid Row Housing Trust, and 1650 Gallery. Brightwell has been featured in the Los Angeles Times, Huffington Post, Los Angeles Magazine, LAist, Eastsider LA, Boing Boing, Los Angeles, I'm Yours, and on Notebook on Cities and Culture. He has been a guest speaker on KCRW's Which Way, LA? and at Emerson College. Art prints of his maps are available from 1650 Gallery and on other products from Cal31. He is currently writing a book about Los Angeles and you can follow him on Facebook, Instagram, and Twitter.
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\section{Introduction}
One of the major cosmological parameters yet to be determined precisely pertains to
the spatial curvature of the universe. While current observations indicate that the universe
is very nearly flat, they do not yet provide irrefutable evidence as to whether on very large
scales the curvature is positive, negative, or exactly zero (the $k=+1, -1, 0$ Robertson-Walker (RW) models
respectively). The current observations
merely provide evidence of a prediction of inflation; namely, the curvature scale is sufficiently
large such as to appear very nearly flat to an observer. Such a feature is explained by inflation
through a period of accelerated expansion in the early universe that inflates the curvature
scale to very large values. Thus most work on structure formation has assumed
that the universe is exactly flat with $k=0$ which is a good approximation
for the post inflationary epoch. However, it is the period of the early universe where
the curvature can play an important role and thus should not be neglected.
It is also the high energy regime of the early universe where quantum gravity is expected
to be a requirement for a complete description. While no complete and fully accepted quantum
theory of gravity exists, a leading candidate exists which is known as loop quantum gravity (LQG)
\cite{ash10,book, Thiemann:2001yy}. The application of LQG techniques to the cosmological setting,
loop quantum cosmology (LQC),
has so far been restricted to the $k=0,+1$ models (see \cite{Bojowald:2006da} for a review). One of the major successes of the models
of LQC so far is the resolution of the classical singularity predicted in the $k=0,+1$ models \cite{Bojowald:2001xe}
which can result in a repulsive gravitational force at high energies that leads to a big-bounce
of the universe \cite{Ashtekar:2006rx, Ashtekar:2006uz, Ashtekar:BBII, Newkplus1}.
Thus an open question remains as to whether these results hold in the negatively curved $k=-1$
model and whether a loop quantization even exists.
The $k=-1$ model has not been constructed in LQC owing
to technical issues that inhibit a successful quantization. The $k=-1$ model
can be derived as the isotropic limit of the homogeneous Bianchi V model which
lacks a correct Hamiltonian framework \cite{Maccallum:1972vb}. The Hamiltonian
framework is essential to the canonical quantization scheme of both LQG and LQC
and thus this failure presents a roadblock to quantization. Notwithstanding this issue, as we shall
show the $k=-1$ model also leads to subtle features in the choice of dynamical
variables in LQC that require careful attention when attempting a quantization.
In this paper we will show that these issues can be successfully overcome leading
to a loop quantization of the model. We will show that the Hamiltonian framework
can be constructed specifically for the isotropic Bianchi V model and that
the theory can be quantized
incorporating techniques similar to those used in the loop quantization of spherically symmetric models.
The resulting quantum theory is in a form that is similar to the $k=0,+1$ LQC models
and thus shares many of the same features. We show directly that the model
predicts a big-bounce in the backward evolution of the universe sourced
by a massless scalar field. We describe this behavior in terms of
an effective Friedmann equation that is quadratic in the matter energy density.
Furthermore the effective Friedmann equation predicts a vacuum repulsion
in the Planckian curvature regime, whereby a bounce would
be triggered even with vanishing matter density. Finally, we comment
on the inverse volume effects predicted by LQC and show that they
are dependent on the introduction of a scale into the model which
is not determined from the curvature scale or any matter energy scale.
We discuss the phenomenological implications of this.
\section{Classical Framework}
We begin with the classical framework that will
form the basis of the loop quantization for the $k=-1$ model.
Loop quantum gravity (and hence loop quantum cosmology) is based
on a Hamiltonian framework using connection-triad variables as
the gravitational field variables. The goal of this section is to consider
the connection-triad variables which are invariant under the symmetries
of the Bianchi V group (which leads to the $k=-1$ model), and then construct
the Hamiltonian in terms of the reduced variables, and finally show that the equations
of motion derived from the Hamiltonian give back the usual cosmological equations
of motion for the open model.
The starting point for homogeneous cosmological model we consider
are the Bianchi models. The homogeneous metric
is given by
\begin{eqnarray}
ds^2 = - N(t)^2 \;dt^2 + \alpha_{ij}(t)\;\, {}^o\!\omega^i_a \, {}^o\!\omega^j_b \; dx^a dx^b
\end{eqnarray}
where $\alpha_{ij}(t)$ are the dynamical components of the metric,
$N(t)$ is known as the lapse and represents the rescaling freedom of the
time coordinate,
and ${}^o\!\omega_a^i$ are a basis of left-invariant one-forms determined by
the group structure of the Bianchi model being considered.
The left-invariant one-forms satisfy
\begin{equation} \label{MC}
d\,{}^o\!\omega^i = - \frac{1}{2} {C^i}_{jk} \;{}^o\!\omega^j \wedge {}^o\!\omega^k.
\end{equation}
where ${C^i}_{jk}$ are the structure constants of
the isometry group and thus characterize the Bianchi model.
For the open $k=-1$ model, we consider the Bianchi V model
with structure constants that can be taken of the form
\begin{eqnarray} \label{cijk}
{C^i}_{jk} = \delta^i_k \delta_{j 1} - \delta^i_j \delta_{k 1} \,.
\end{eqnarray}
The structure constants satisfy ${C^i}_{ij} \ne 0$ which
in the language of \cite{Ellis:1968vb} implies that the Bianchi V model is class B. This fact
will be important in what we consider later.
In a particular choice of coordinates,
equations (\ref{MC}) can be solved explicitly to give the left-invariant
one-forms as
\begin{eqnarray}
{}^o\!\omega^1 &=& dx \\
{}^o\!\omega^2 &=& e^{-x} dy \\
{}^o\!\omega^3 &=& e^{-x} dz \,.
\end{eqnarray}
where the coordinates $x,y,z $ are valued on the real line representative of the fact
that we are considering the spatially non-compact k=-1 model with topology
homeomorphic to ${\mathbb R}^3$. Thus we have not chosen a particular
compactification of the k=-1 model and work in the usual model
with infinite spatial extent. If we consider the isotropic limit of this
Bianchi model with $\alpha_{ij}(t) = a^2(t) \delta_{ij}$ with $a(t)$ representing
the scale factor, and fix the lapse to be equal to one, then the metric
\begin{eqnarray} \label{cmetric}
ds^2 &=& - dt^2 + a^2 \, \delta_{ij}\; {}^o\!\omega^i_a \, {}^o\!\omega^j_b \; dx^a dx^b \nonumber \\
&=& -dt^2 + a^2 ( dx^2 + e^{-2x} dy^2 + e^{-2x} dz^2)
\end{eqnarray}
can be shown to have constant negative spatial curvature and
hence corresponds to the open $k=-1$ model.
The usual hyperbolic metric in hyperbolic coordinates
$ds^2 = -dt^2 + a^2(d\psi^2 + \sinh^2\psi \big( d\theta^2 + \sin^2\theta d\phi^2))$ can be recovered with the following transformation
\begin{eqnarray}
x &=& - \ln(\cosh\psi -\sinh\psi \cos\theta) \nonumber \\
y &=& \frac{\sin\theta \cos\phi}{\coth\psi - \cos\theta} \nonumber \\
z &=& \frac{\sin\theta \sin\phi}{\coth\psi - \cos\theta} \nonumber \,.
\end{eqnarray}
With the form of the metric (\ref{cmetric}), Einstein's equations
lead to a set of differential equations satisfied by the scale factor $a(t)$ given
by the Friedmann equation
\begin{equation} \label{Fried}
\Big(\frac{\dot{a}}{a}\Big)^2 = \frac{\kappa}{3} \rho_M + \frac{1}{a^2}
\end{equation}
and the acceleration equation
\begin{eqnarray} \label{accel}
\frac{\ddot{a}}{a} = -\frac{\kappa}{6}\Big(\rho_{M} + 3 p_{M}\Big)
\end{eqnarray}
with $\rho_{M}$ and $p_{M}$ being the matter density and pressure respectively.
Here $\kappa=8\pi G$ with $G$ being Newton's constant.
In addition to the left-invariant one-forms, for what follows we
will also need a basis of vector fields ${}^o\!e^a_i$ which also are left-invariant.
The left-invariant
vector fields have commutators which provide a representation of the Lie
algebra under consideration
\begin{eqnarray} \label{coms}
[{}^o\!e_i, {}^o\!e_j] = {C^k}_{ij}\; {}^o\!e_k
\end{eqnarray}
and are also dual to ${}^o\!\omega_a^i$ thus satisfying
\begin{eqnarray}
{}^o\!e^a_i \; {}^o\!\omega_a^j = \delta^j_i \,.
\end{eqnarray}
In the chosen coordinates for the Bianchi V model, ${}^o\!e^a_i$ are
given explicitly as
\begin{eqnarray}
{}^o\!e_1 = \partial_x, \;\;\;\;\;
{}^o\!e_2 = e^{x}\, \partial_y, \;\;\;\;\;
{}^o\!e_3 = e^x \,\partial_z
\end{eqnarray}
whence it is simple to show satisfy equation (\ref{coms}).
The classical framework of loop quantum cosmology (LQC) diverges from the standard
framework in two important ways. The first is that the equations of motion
are derived from a Hamiltonian framework which allows for a canonical quantization
of the theory. Secondly, the variables that form the basis for quantization are not
the usual metric ones (i.e., the scale factor). This framework
follows directly from that used in the full theory of loop quantum gravity
and it is these changes that allow for a rigorous quantization of gravity. The canonical
set of variables consists of an orthonormal triad $E^a_i$ (of density weight one) which encodes the information
of spatial geometry, and an su(2) valued connection $A_a^i$ which is canonically conjugate
to $E^a_i$. The starting point of LQC is to reduce these variables to the symmetry
of the cosmological model. We can use the basis provided by the left-invariant
one-forms and vector fields to accomplish this.
Starting with the triad $E^a_i$, we expand using the basis vector-fields as
\begin{eqnarray} \label{EBV}
E^a_i = \sqrt{\qz}\; \widetilde{p}(t) \; {}^o\!e^a_i
\end{eqnarray}
where $\widetilde{p}(t)$ represents the dynamical component of the triad.
The factor $\sqrt{\qz} = e^{-2x}$ is a density weight provided by the
hyperbolic metric ${}^o\!q_{ab} = {}^o\!\omega_a^i \,{}^o\!\omega_b^i$ which
gives the triad $E^a_i$ its density weight.
$E^a_i$ encodes the spatial geometry in a specific fashion being that it is
related to the spatial three-metric $q_{ab}$ through
\begin{eqnarray}\label{Etom}
E^a_i E^{b i} = |q|\, q^{ab} \,.
\end{eqnarray}
Using this relation, we find that $\widetilde{p}$ is related to the scale factor as
\begin{eqnarray}
|\widetilde{p}| = a^2
\end{eqnarray}
where the absolute value indicates that we are allowing $\widetilde{p}$ to take on positive and
negative values in contrast to the scale factor which is usually assumed to be strictly non-negative. A change in sign of $\widetilde{p}$ corresponds to a change in orientation of
the triad $E^a_i$ leaving the metric $q_{ab}$ invariant.
The first non-triviality of the $k=-1$ arises when we consider a symmetric
connection $A_a^i$. From the $k=0, +1$ models, we expect that an isotropic
connection can be decomposed using the left-invariant one-forms
as $A_a^i = \widetilde{c}(t)\, {}^o\!\omega_a^i$ \cite{Bojowald:2002gz,Bojowald:2003mc} with
$\widetilde{c}$ being the {\em only} dynamical component. In this form, the connection is diagonal
in the basis of left-invariant one-forms.
However, this form must be consistent with the fact
that on the half-shell (after solving Hamilton's equations for $\dot{E}^a_i$), $A_a^i$
is determined from the dynamics of the spatial metric as
\begin{eqnarray}
A_a^i = \gamma K_a^i + \Gamma_a^i
\end{eqnarray}
where $K_a^i$ is the extrinsic curvature, $\gamma$ is known as the Barbero-Immirzi parameter (a real valued ambiguity parameter of loop quantum gravity),
and $\Gamma_a^i$ is the spin-connection.
Upon symmetry reduction, the extrinsic curvature can be shown to be of diagonal
form\footnote{The $\text{sgn}(p)$ arises
because the extrinsic curvature one-form carries the signature
of the triad which is evident from the definition $K_a^i = e^{a i} K_{ab}$ where
$K_{ab}$ is the usual extrinsic curvature of the ADM formulation which does not
carry information about the orientation.}
$K_a^i = \text{sgn}(\widetilde{p}) \dot{a}\, {}^o\!\omega_a^i$
which is consistent with the connection
being diagonal.
However, this is not the case with the spin connection $\Gamma_a^i$.
The formula for the spin connection is given by
\begin{eqnarray}
\label{spinconnection}
\Gamma^i_a = - \frac{1}{2} \epsilon^{ijk} e^b_j\Big( \partial_a e_b^k - \partial_b e_a^k
+e^c_k e_a^l \partial_c e_b^m \delta_{lm} \Big) \,.
\end{eqnarray}
where $e_a^i$ is the physical triad satisfying
\begin{eqnarray}
e_a^i e_b^i = q_{ab} \,.
\end{eqnarray}
The physical triad $e_a^i$ is related to $E^a_i$ through
\begin{eqnarray}
e^a_i = \frac{1}{\sqrt{|q|}} \; E^a_i \,.
\end{eqnarray}
Using the symmetric form of $E^a_i$ (\ref{EBV}) and evaluating
(\ref{spinconnection}), one finds that the spin connection is
given by
\begin{eqnarray}
\Gamma_a^i = \Gamma^i_j \, {}^o\!\omega_a^j
\end{eqnarray}
with
\begin{equation}
\Gamma^i_j = \left( \begin{array}{ccc} 0&0&0\\0&0&-1\\0&1&0 \end{array} \right)
\end{equation}
whence it is clear that the spin connection is non-diagonal, and the assumption that the connection
is diagonal is not consistent. We must therefore take the connection to be of non-diagonal
form
\begin{eqnarray} \label{Atwo}
A_a^i &=& A^i_j(t) \, {}^o\!\omega_a^j \nonumber \\
A^i_j &=& \left( \begin{array}{ccc} \widetilde{c}(t) &0&0\\0&\widetilde{c}(t) &-\widetilde{c}_2(t)\\0&\widetilde{c}_2(t)&\widetilde{c}(t) \end{array} \right) \,.
\end{eqnarray}
In this form, the connection has two dynamical components $\widetilde{c}$ and $\widetilde{c}_2$, where on the half-shell
$\widetilde{c} = \text{sgn}(\widetilde{p}) \gamma \dot{a}$ is determined from the extrinsic curvature and $\widetilde{c}_2 = 1$.
This is in contrast to the $k=0,+1$ models where the connection can safely be
assumed to be diagonal and only has one dynamical component.
With the symmetry reduced connection-triad variables,
the next step is to show that the Hamiltonian formulation leads to the correct classical equations of motion. Yet here another problem
arises. The Bianchi V model is of class B type where, as first shown in
\cite{Maccallum:1972vb}, the ADM Hamiltonian formulation in the general
homogeneous case fails. The main issue is that the equations of motion derived from the symmetry reduced Hamiltonian
do not agree with Einsteins' equations after symmetry reduction. In other words,
the symmetry reduction and Hamiltonian formulation do not commute in the class B models.
While this may seem a fatal issue for the k=-1 model, it was shown in
\cite{Maccallum:1972vb} that the Hamiltonian formulation does not
fail for the isotropic limit of the Bianchi V model which is precisely the case
we are interested. This failure does hinder the extension of the results
presented here to the anisotropic Bianchi V model, but we will now show explicitly that the Hamiltonian
formulation of the isotropic model using the connection-triad variables leads to the
correct equations of motion. Another avenue worth exploration is whether the analysis
of \cite{Maccallum:1972vb} holds in general for the Hamiltonian formulation based on
the connection-triad variables used in loop quantum gravity which is
manifestly different than the ADM Hamiltonian formulation and thus may not
suffer from the same issues. We do not attempt to address this possibility here.
Therefore, our aim now is to plug in the symmetry reduced connection (\ref{Atwo}) and
triad (\ref{EBV}) into the Hamiltonian of the full theory and show that we get back
the correct equations of motion (\ref{Fried}, \ref{accel}). The action
written in terms of the connection-triad variables\footnote{
The action can also be derived from a Legendre transform
of the covariant Holst action written in terms of a
four dimensional so(3,1) connection and a co-tetrad\cite{Holst:1995pc}.} is given as \cite{ash10}
\begin{eqnarray} \label{fullaction}
S_{GR}[E, A, \lambda^i, N^a, N] = \int\!\!dt \int \!\!d^3 x \frac{1}{\kappa \gamma} E^a_i
\mathcal{L}_t A_a^i \nonumber \\
- \left[ \lambda^i G_i + N^a {\mathcal C}_a + N \C_{GR} \right]
\end{eqnarray}
whence the Hamiltonian is a sum of constraints: $G_i$ is the Gauss constraint, ${\mathcal C}_a$ is the
diffeomorphism constraint, and $\C_{GR}$ is the Hamiltonian constraint. The parameters
$\lambda^i, N^a, N$ are Lagrange multipliers which enforce the vanishing of the constraints.
The first term of the action indicates that the connection and triad are
canonically conjugate with Poisson brackets
\begin{eqnarray}
\{ A_a^i(x), E^b_j(y)\} = \kappa \gamma \,\delta^i_j \, \delta_a^b \,\delta(x-y) \,.
\end{eqnarray}
Hamilton's equations
for the connection $A_a^i$ and triad $E^b_j$ can then be shown to be equivalent to
Einstein's equations.
When inserting the symmetry reduced connection and triad into the action, the first issue
we face is that the spatial integration in
the action diverges since we are considering the non-compact $k=-1$ model. This
same issue arises in the non-compact $k=0$ model and would arise in any
cosmological quantization scheme based on a Hamiltonian or action framework.
To overcome this, we choose the follow the technique used in the $k=0$ model
for LQC \cite{Ashtekar:2003hd}; namely, we restrict the spatial integration to a finite sized fiducial cell with
a fixed background volume
\begin{eqnarray}
V_0 = \int\!\! d^3x\, \sqrt{\qz} \;.
\end{eqnarray}
Note that the extent of the fiducial cell is fixed on the manifold or
in other words has fixed comoving coordinates. Thus, as the universe
expands for instance, so would physical size of the fiducial cell. The choice in
the fiducial cell remains a quantum ambiguity and we will be interested in
determining whether the resulting quantum theory makes predictions
dependent on $V_0$. As we now show, the choice in fiducial cell
has no effect classically, but that is not true in the quantum case which we will discuss
later.
Now with the understanding that we are limiting the spatial integrations in the action
to the fiducial cell we can insert
the symmetry reduced connection and triad (\ref{Atwo}, \ref{EBV}).
The canonical term is given by
\begin{eqnarray}
\int\!\!dt \int \!\!d^3 x \,\frac{1}{\kappa \gamma}\, E^a_i\,
\mathcal{L}_t A_a^i = \int\!\!dt\; \frac{3 V_0}{\kappa\gamma}\, \widetilde{p}\; \dot{\widetilde{c}}
\end{eqnarray}
which indicates that $\widetilde{c}$ and $\widetilde{p}$ are canonically conjugate with Poisson brackets
\begin{equation} \label{pbrak}
\{ \widetilde{c}, \widetilde{p} \} = \frac{\kappa \gamma}{3 V_0} \,.
\end{equation}
The Gauss constraint in terms of the reduced variables is given by
\begin{eqnarray}
G_i \equiv \partial_a E^a_i + {\epsilon_{ij}}^{k} A_a^j E^a_k
= \frac{2 V_0}{\kappa \gamma}\, \widetilde{p} \;(\widetilde{c}_2 - 1)\, \delta_{i 1}
\end{eqnarray}
and thus is non-vanishing. This is in contrast to the $k=0,+1$ models
where the Gauss constraint vanishes indicative of the fact that a complete
gauge fixing of the Gauss constraint was performed in those models.
This suggests that we should gauge fix the Gauss constraint by
setting $\widetilde{c}_2$ to be identically equal to one. With this, the Gauss constraint
vanishes and additionally the diffeomorphism constraint $C_a$ can be shown
to vanish.
With this gauge fixing
the connection is now of the form
\begin{eqnarray}\label{Afinal}
A^i_j &=& \left( \begin{array}{ccc} \widetilde{c} &0&0\\0&\widetilde{c} &-1\\0&1&\widetilde{c} \end{array} \right)
\end{eqnarray}
and we are now left with two dynamical phase space variables $\widetilde{p}$ and $\widetilde{c}$
and one surviving constraint, the Hamiltonian constraint.
This is exactly the situation in the $k=0,+1$ models.
The dynamics of the model is now entirely encoded in the Hamiltonian constraint
which is given by
\begin{eqnarray}
\C_{GR} = - \frac{6 V_0}{\gamma^2} \sqrt{|\widetilde{p}|} \,\Big(\widetilde{c}^2 - \gamma^2\Big)
\end{eqnarray}
and the entire gravitational action becomes\footnote{The extra
factor of $2\kappa$ appearing below the lapse $N$ appears because
the Hamiltonian constraint used in previous works of LQC differs from
the Hamiltonian constraint in the full theory given in \cite{ash10} by the factor
of $2 \kappa$. A constant factor multiplying the Hamiltonian constraint
does not affect any physical results.}
\begin{eqnarray}
S_{GR}[\widetilde{p}, \widetilde{c}, N] &=& \int\!\!dt \, \frac{3 V_0}{\kappa \gamma}\,\widetilde{p} \;
\dot{\widetilde{c}} \nonumber \\
&&- \frac{N}{2 \kappa} \left[ - \frac{6 V_0}{\gamma^2} \sqrt{|\widetilde{p}|} \,\Big(\widetilde{c}^2 - \gamma^2\Big) \right]
\end{eqnarray}
whence the total Hamiltonian including matter is given by
\begin{eqnarray} \label{Ham}
{\mathcal H} &=& - \frac{3 V_0 N}{\kappa \gamma^2} \sqrt{|\widetilde{p}|}\, \Big(\widetilde{c}^2 - \gamma^2\Big)
+ {\mathcal H}_M
\end{eqnarray}
with ${\mathcal H}_M$ denoting the matter Hamiltonian.
With the Hamiltonian and Poisson structure we can now derive the
classical equations of motion. We first have Hamilton's equations
$\dot{x} = \{x, {\mathcal H}\}$ for any phase-space variable $x$,
and further the Hamiltonian itself must vanish since it is proportional
to the Hamiltonian constraint.
Starting with Hamilton's equations
for $\widetilde{p}$ we find
\begin{eqnarray}
\dot{\widetilde{p}} = \{\widetilde{p}, {\mathcal H} \} &=& - \frac{\kappa \gamma}{3 V_0} \, \frac{\partial{\mathcal H}}{\partial\widetilde{c}}
\nonumber \\
&=& \frac{2 \sqrt{|\widetilde{p}|}}{ \gamma} \widetilde{c}
\end{eqnarray}
where for the equations of motion we have fixed the lapse $N=1$.
Notice that the factors of $V_0$ cancel appearing both in the numerator of
the Hamiltonian (\ref{Ham}) and in the denominator of the Poisson brackets (\ref{pbrak}).
Furthermore,
we have assumed that the matter Hamiltonian only couples to the spatial geometry
i.e., is only a function of $\widetilde{p}$ and not $\widetilde{c}$. This assumption is true for scalar fields
and perfect fluids, though is not true for fermions for instance which we do not consider
(see \cite{Perez:15} for discussions for the inclusion of fermions in LQG with physical effects
dependent on the Barbero-Immirzi parameter $\gamma$).
Using $|\widetilde{p}| = a^2$ we can write the left-hand side of the Friedmann
equation as
\begin{eqnarray}
H^2 = \Big(\frac{\dot{a}}{a}\Big)^2 = \frac{\widetilde{c}^2}{\gamma^2 |\widetilde{p}| } \,.
\end{eqnarray}
Now we can use the vanishing of the constraint to relate the right-hand side
to the matter density. Using ${\mathcal H}=0$ we find
\begin{eqnarray}
H^2 &=& \frac{\kappa}{3} \frac{{\mathcal H}_M}{V_0 \widetilde{p}^{3/2}} + \frac{1}{|\widetilde{p}|} \nonumber \\
&=& \frac{\kappa}{3} \rho_{M} + \frac{1}{a^2}
\end{eqnarray}
where we have used ${\mathcal H}_M/(V_0 \widetilde{p}^{3/2}) = \rho_{M}$. Thus the reduced Hamiltonian
gives back the correct Friedmann equation. Similarly the acceleration equation
can be derived by considering Hamilton's equation for $\dot{\widetilde{c}}$ once the matter
Hamiltonian is explicitly specified.
This derivation demonstrates explicitly that the Hamiltonian framework presented here
leads to Einsteins equations for the open $k=-1$ model. This also demonstrates
that the equations of motion classically are insensitive to the choice in fiducial
cell $V_0$ which was introduced to regulate the divergent spatial integrals
in the action and resulting Hamiltonian. Furthermore, the Hamiltonian is similar
in form to the $k=0,+1$ models and thus is indicative that a successful
loop quantization is possible.
Before turning to the quantization, we would like to make a closer
connection to the LQC work of the $k=0, +1$ models. There we can
define untilded variables by rescaling $\widetilde{p}$ and $\widetilde{c}$ by a factor
dependent on the fiducial cell as
\begin{eqnarray} \label{untild}
\begin{array}{rcl}
p &\equiv& V_0^{2/3}\, \widetilde{p} \\
c &\equiv& V_0^{1/3} \,\widetilde{c}
\end{array}
\end{eqnarray}
In terms of the untilded variables, the Poisson bracket is now independent of $V_0$
\begin{equation}
\{ c, p \} = \frac{\kappa \gamma}{3 }
\end{equation}
and the Hamiltonian constraint becomes
\begin{equation} \label{Cuntild}
\C_{GR} = - \frac{3 }{\kappa \gamma^2} \sqrt{|p|} \Big(c^2 - V_0^{2/3} \gamma^2 \Big) \,.
\end{equation}
The relation between the rescaled triad and the scale factor is given by
\begin{eqnarray} \label{pofa}
|p| = V_0^{2/3} a^2 \label{ptoaV}
\end{eqnarray}
as well as the half-shell relation
\begin{eqnarray}
c = \text{sgn}(p) V_0^{1/3} \gamma \dot{a}\label{ctoaV} \,.
\end{eqnarray}
We will use the Hamiltonian based on the untilded variables as the starting
point for quantization. Notice that $V_0$ appears explicitly in
the Hamiltonian constraint (\ref{Cuntild}) which is in contrast
to the $k=0$ model where $V_0$ drops out of the constraint.
It will be important in the interpretation of the quantum theory to
understand the physical meaning of the variable $p$. From
the relation (\ref{pofa}), we find
\begin{eqnarray} \label{Vcell}
|p|^{3/2} = V_0 a^3 = V_{\text {cell}}
\end{eqnarray}
which one recognizes as representing the {\em physical} volume
of the fiducial cell. Note that in order to physically measure
the value of $p$, one would need prescribe the size of the fiducial
cell. For instance, if today the fiducial cell is taken to have Planckian physical
volume: $V {\text {cell}} = l_{\mathrm P}^3$, then $p$ is similarly Planckian:
$p = l_{\mathrm P}^2$. This can be so despite the fact that, assuming we live in
an open $k=-1$ universe, the value of the scale factor $a$
is astronomically large. Thus there is no direct correlation between
the value of the scale factor $a$, and the value of $p$. Again the
value of $p$ is highly dependent on the size of the fiducial cell.
Let us conclude this section with a further note about the fiducial cell. Since
we will be interested in whether the quantum predictions are sensitive to
this choice, we would like to know how the classical variables transform
under a change in its size. For instance, let us consider that
the fiducial cell is resized as
\begin{eqnarray}
V_0 \rightarrow V_0' = \xi^3 V_0
\end{eqnarray}
then from their definition (\ref{untild}), the untilded variables
transform as
\begin{eqnarray}
p \rightarrow p' &=& \xi^2 p \label{pscale} \\
c \rightarrow c' &=& \xi c \label{cscale} \,.
\end{eqnarray}
Note that the scale factor $a$ (and therefore $\widetilde{p}$ and $\widetilde{c}$) do not make
reference to the fiducial cell and therefore do not rescale under this change.
The scaling of $p$ can be understood from (\ref{Vcell}) by noting that
the value of $p$ is determined from the physical volume of the cell, and
thus if the cell is enlarged, we expect the value of $p$ to be larger.
The untilded variables therefore do not classically have direct physical meaning as they
can be freely rescaled under this transformation. What is physical
classically are {\em changes} in $p$ and $c$ where for instance
the Hubble rate $H = \frac{1}{2}\big(\frac{\dot{p}}{p}\big)$
is invariant under a resizing of the fiducial cell.
\section{Quantization}
With the classical framework completed, we can now turn to the loop
quantization of the model. To achieve this in the canonical quantization
scheme involves the following steps. First one chooses a set of basic variables
and finds a quantum representation of their algebra in order to construct
what is known as the kinematical Hilbert space. The next step is to construct
an operator corresponding to the Hamiltonian constraint that is self-adjoint in
the kinematical Hilbert space. Lastly, the physical Hilbert space consists of
wavefunctions that are annihilated by the constraint operator and
that have finite norm in a suitable physical inner product (which typically is not
equivalent to the kinematical inner product). One then interprets the theory
by evaluating expectation values of observables on physical wavefunctions.
By considering the $k=-1$ model sourced with a massless scalar field, this
program can be carried out to completion. The construction of
the $k=-1$ model presented here follows closely that of the $k=0$ model presented
in \cite{Ashtekar:BBII}, and thus we will omit many of the technical details and refer the
reader to that article for a complete description.
\subsection{Kinematical Hilbert Space}
To construct the kinematical Hilbert space we first must consider the elementary
variables that will form the basis for quantization. From the full theory
of LQG, one does not take the bare connection $A_a^i$ and triad $E^a_i$
as the basic variables. Rather, in the case of the connection,
one integrates $A_a^i$ along
edges and then exponentiates the quantity leading to a holonomy.
The holonomy variables are then taken as the basic configuration variables.
The momentum variables are fluxes which are constructed by integrating of the triad over a two-surface.
In the cosmological setting, fluxes
are simply proportional to $p$ which therefore forms an elementary variable.
On the other hand, the holonomies amount to exponentials of the connection
$c$ and it is this fact that becomes the departure point of LQC from
previous versions quantum cosmology based on a Schrodinger
type quantization of the Hamiltonian.
Thus let us consider the holonomies in detail.
In the $k=0$ model they consist
of integrating the connection along edges generated by the left-invariant
vector fields and assume the form
$h_i = \cos\big( \frac{\overline{\m} c}{2 }\big) + 2 \sin\big( \frac{\overline{\m} c}{2 }\big)
\tau_i $
where $\overline{\m}$ is equal to the fiducial length of the edge divided by
$V_0^{1/3}$, and
$\tau_i$ are the generators of SU(2) satisfying $\{\tau_i, \tau_j\} = {\epsilon_{ij}}^{k} \tau_k$.
With holonomies of this form, the algebra generated is that of the almost periodic functions
(which look like exponentials of the connection $e^{i \overline{\m} c}$) and the kinematical Hilbert space assumes a simple form\cite{Ashtekar:2003hd}. However, when we consider
holonomies of the connection in the $k=-1$ model considered here, they take on a more
complicated form where for instance the holonomy along the edge generated
by ${}^o\!e^a_2$ is given by
\begin{eqnarray} \label{hA}
&h_2(\overline{\m})& = \cos\frac{\overline{\m}\sqrt{c^2\!+\!V_0^{2/3}}}{2} \nonumber \\
\!\!\!\!\!\!&+&\!\!\!\!\!\!\!\! \frac{2 \left[ c \tau_2 - V_0^{1/3} \tau_3 \right] }{\sqrt{c^2\!+\!V_0^{2/3}}}\sin\frac{\overline{\m}\sqrt{c^2\!+\!V_0^{2/3}}}{2}
\end{eqnarray}
and thus the algebra generated is no longer simply that of the almost
periodic functions. Finding a representation of the algebra would be difficult.
However, we can exploit a technique used in the loop quantization of other
models such as the spherically symmetric models of LQG \cite{Bojowald:2005cb} as well
as the quantization of the Schwarzschild horizon interior \cite{Ashtekar:2005qt}. The complicated
form of the holonomies of the connection arises because of the non-diagonal form
of the connection (\ref{Afinal}). If we consider instead holonomies of the connection
minus the spin-connection (essentially holonomies of the extrinsic curvature) as
done in \cite{Bojowald:2005cb, Ashtekar:2005qt}, then the holonomies are
of a form equivalent to the $k=0,+1$ models
\begin{eqnarray} \label{holon}
h_i = \cos\Big( \frac{\overline{\m} c}{2 }\Big) + 2 \sin\Big( \frac{\overline{\m} c}{2 }\Big)
\tau_i \,.
\end{eqnarray}
where again $c$ refers to the diagonal component of the connection
in (\ref{Afinal}). In the full theory, the extrinsic curvature is not a connection and hence
its holonomies are not defined. However, in the reduced setting we
have performed a complete SU(2) gauge fixing to arrive at symmetric connections
and thus it is possible to regard the extrinsic curvature as a connection. The
resulting quantization will be a slight departure from that predicted by the full
theory and thus care must be taken when interpreting the results. In
section \ref{Discussion}, we will comment on the regime where we expect the differences to occur.
Additionally we shall follow the prescription of \cite{Ashtekar:BBII} leading
to improved dynamics for LQC. Namely, in contrast to the original
literature of LQC, we assume that the parameter $\overline{\m}$ appearing in
the holonomies is a function of $p$ and not a constant. The motivation for
this can be seen as twofold. First let us consider the issue of the fiducial cell
dependence. The quantity $\overline{\m} c$ appears in the holonomies (\ref{holon}) and
we have shown that under a re-sizing of the fiducial cell, the connection $c$ scales
according to equation (\ref{cscale}). Quantum corrections can arise
when $\overline{\m} c$ becomes on the order of one \cite{Willis:thesis} and thus we can generate arbitrarily
large quantum corrections by choosing a larger fiducial cell as long as $\overline{\m}$ is a
fixed constant. However, if $\overline{\m}$ scales as
\begin{eqnarray}
\overline{\m} \propto \frac{1}{\sqrt{|p|}}
\end{eqnarray}
then we find that the quantity $\overline{\m} c$ is invariant under a resizing of the
fiducial cell. A direct result of this in
the $k=0$ model is that the bounce occurs
when the matter energy density is on the order of Planckian \cite{Ashtekar:BBII}
which is to be expected on physical grounds.
On the other hand,
if $\overline{\m}$ is a fixed constant, the bounce can occur even at largely sub-Planckian
densities and even a cosmological
constant can trigger a future recollapse of the universe \cite{Ashtekar:2006rx, Ashtekar:2006uz, Banerjee:2005ga}.
The second motivation for this scaling comes
from the method proposed in \cite{Ashtekar:2003hd} to constrain the value
of $\overline{\m}$ based on using the minimum area eigenvalue of LQG in constructing
the Hamiltonian constraint operator. We will discuss this in more detail when we construct
the constraint operator. For now let us assume that $\overline{\m}$ is given as
\begin{eqnarray} \label{mubar}
\overline{\m} = \sqrt{\frac{\Delta}{|p|}}
\end{eqnarray}
where $\Delta$ is a constant to be fixed later.
The kinematical Hilbert space can then be constructed and a basis is given by eigenstates of the $\wh{p}$ operator
labeled by a real parameter $v$ with eigenvalues
\begin{eqnarray}
\wh{p} \ket{v} = \frac{\kappa \gamma \hbar}{6} \Bigg( \frac{|v|}{K}\Bigg)^{2/3} \ket{v}
\end{eqnarray}
where the constant $K$ is given by
\begin{eqnarray} \label{K}
K = \frac{2}{3} \sqrt{\frac{\kappa \gamma \hbar}{6 \Delta}} \,.
\end{eqnarray}
Similarly the states $\ket{v}$ are eigenstates of the fiducial cell volume operator
\begin{eqnarray}
\wh{V}_{\text{cell}} \, \ket{v} = \Big(\frac{\kappa \gamma \hbar}{6}\Big)^{2/3} \, \frac{|v|}{K} \,\ket{v}
\end{eqnarray}
The parameter $v$ runs over the entire real line, but the spectrum is discrete
in the sense that the states $\ket{v}$ are normalizable satisfying
\begin{eqnarray}
\braket{v'}{v} = \delta_{v' v}
\end{eqnarray}
A general quantum state is a continuous sum over the basis states $\ket{v}$ as well
as any matter degrees of freedom. We will interest ourselves in the inclusion
of a scalar field degree of freedom whence a general quantum state
is given by
\begin{eqnarray}
\ket{\Psi} = \int\! d\phi \sum_{v} \; \Psi(v, \phi) \,\ket{v, \phi}
\end{eqnarray}
with the kinematical inner product between two states given by
\begin{eqnarray} \label{KIP}
\braket{\Psi_1}{\Psi_2}_{\text{kin}} = \int\! d\phi \sum_{v} \,\overline{\Psi}_1(v, \phi)\; \Psi_2(v, \phi)\,.
\end{eqnarray}
A quantum state which lies in the kinematical Hilbert space has
finite kinematical norm which implies
\begin{eqnarray}
\int\! d\phi \sum_{v} \,\overline{\Psi}(v, \phi)\; \Psi(v, \phi) < \infty \,.
\end{eqnarray}
This constitutes the kinematical Hilbert space as well as the action
of the basic flux operator $\wh{p}$. Additional basic operator are required in
the form of holonomy operators which can be built using
the formula (\ref{holon}) and the basic exponential operators
\begin{eqnarray}
\wh{h}_{\pm} = \exp(\mp i \wh{\overline{\m} c}/2) \,.
\end{eqnarray}
The basis $\ket{v}$ has been chosen such that the exponential operators
act simply as shift operators
\begin{eqnarray}
\wh{h}_{\pm} \Psi(v) = \Psi(v \pm 1) \,.
\end{eqnarray}
An important feature of the quantization is that since holonomies
form the basic configuration variables, there is no basic
operator corresponding to the connection $\wh{c}$. In order
to construct such an operator, one has to approximate it using
the basic holonomy operators. An example of this is given
by the Hamiltonian constraint to which we turn now.
\subsection{Quantum Difference Equation}
The next step in quantization is to construct a Hamiltonian constraint
operator that is self-adjoint on the kinematical Hilbert space. The classical
expression for the gravitational part of the constraint is again given
by
\begin{eqnarray}\label{CGR}
\C_{GR} = - \frac{6 }{\gamma^2} \sqrt{|p|} \,\Big(c^2 - \gamma^2 V_0^{2/3} \Big)
\end{eqnarray}
which is equivalent to the $k=0$ model up to the $\gamma^2 V_0^{2/3}$ term in
the parentheses. The main complication in constructing the gravitational part
of the Hamiltonian constraint operator is the lack of an operator for the bare connection.
Thus the $c^2$ term must be quantized using holonomies. Following the results
from the $k=0$ model, the following classical re-expression
\begin{eqnarray}
\C_{GR} = -\frac{4}{\kappa \hbar \gamma^3 \overline{\m}^3} \sum_{ijk}\epsilon^{ijk} \; \mbox{tr} \Big[\big(
h_i h_j h_i^{-1} h_j^{-1} \nonumber \\- 2 \overline{\m}^2 \gamma^2 V_0^{2/3}\tau_i \tau_j \big)
h_k \left\{ h_k^{-1} , V \right\} \Big]
\end{eqnarray}
can be shown to give back the classical expression (\ref{CGR}) in the limit
as $\overline{\m}$ is taken to zero. This expression is now readily quantizable with
the major non-triviality being that we can not take the limit as $\overline{\m}$ goes
to zero as that would require a $\wh{c}$ operator. Thus in the quantum constraint
operator we do not take the limit, instead leaving $\overline{\m}$ to be a finite parameter given
by the expression (\ref{mubar}).
In order to constrain the parameter $\Delta$ in the definition of $\overline{\m}$ (or
equivalently the parameter $K$ in (\ref{K})), in the $k=0$ model
one can connect to the full theory of LQG by shrinking $\overline{\m}$ until the closed loop
spanned by the edges of the holonomies $h_i h_j h_i^{-1} h_j^{-1}$ has the minimum
physical area eigenvalue of LQG.
This fixes the value of $\Delta$ to be equal to the minimum area eigenvalue
of LQG $\Delta = 2\sqrt{3} \pi \gamma l_{\mathrm P}^2$ which implies
that $K$ is given by
$K = \frac{2 \sqrt{2}}{3 \sqrt{3 \sqrt{3}}}
$\cite{Ashtekar:BBII}.
However, in the $k=-1$ model this interpretation does not hold since the edges
do not close. Thus we can not a priori make the same assignment of $K$.
We can however turn to the $k=+1$ model for guidance. There, the quantization
has been performed using holonomies of the extrinsic curvature where the loop similarly
does not close \cite{Bojowald:2003mc,Bojowald:2003xf}. The quantization
of the $k=+1$ involving holonomies of the connection and using a closed
loop for the constraint operator appears in \cite{Newkplus1} and there
the value of $K$ is constrained to the same value as the $k=0$ model using
the same procedure. Furthermore, the quantization using holonomies of the
connection is quantitatively similar to the one using holonomies of the extrinsic
curvature in the $v \gg 1$ regime which can be taken as evidence
that the same value of $K$ should be used in both quantizations.
With this in mind, we will leave this issue open and assume that the parameter $K$ is
on the order of one without explicitly fixing its value.
With the caveats mentioned, the construction of the constraint operator follows that
of the $k=0$ model (see \cite{Ashtekar:BBII} for details). The action on the
the operator is given by
\begin{eqnarray}
\wh{{\mathcal C}}_{GR} \Psi(v) = f_{+}(v) \,\Psi(v+1) + f_0(v)\, \Psi(v) \nonumber \\
+ f_{-}(v) \,\Psi(v - 1)
\end{eqnarray}
with the functions $f$ given by
\begin{eqnarray}
f_{+}(v) &=& \frac{27}{16} \sqrt{\frac{\kappa \gamma \hbar}{6}} \frac{K}{\gamma^2}
\, \big| v+2\big| \; \bigg| \big| v+1 \big| - \big| v+3 \big| \bigg| \\
f_{-}(v) &=& f_{+}(v-4) \\
f_{0}(v) &=& - f_{+}(v) - f_{-}(v) + g(v)
\end{eqnarray}
and the function $g(v)$ representing the modification coming from the $k=-1$ model
given explicitly as
\begin{eqnarray}
g(v) = \frac{3 V_0^{2/3}}{K^{1/3}} \sqrt{\frac{\kappa \gamma \hbar}{6}}
\;|v|^{1/3} \, \bigg| \big| v+1 \big| - \big| v-1 \big| \bigg|
\end{eqnarray}
Thus the contribution from the $k=-1$ model amounts to the addition
of a term $g(v)$ that acts diagonally on the basis states $\ket{v}$.
To discuss dynamics and interpret the difference equation we can add
matter in the form of a massless scalar field as done in \cite{Ashtekar:BBII}. Since
the difference equation will be of similar form the the $k=0$ model most of the results
remain valid. With a massless scalar field, the full constraint is given
by
\begin{eqnarray}
\wh{C} = \wh{C}_{GR} + \kappa \wh{p}^{-3/2} \wh{P}_{\phi}^2
\end{eqnarray}
where $P_{\phi}$ is the canonical momentum to the scalar field.
Since the Hamiltonian is independent of the scalar field $\phi$,
the conjugate momentum $P_{\phi}$ is a constant of motion classically.
The classical Friedmann equation is given by
\begin{eqnarray}
H^2 = \frac{\kappa}{6} \frac{P_{\phi}^2}{V_0^{2} a^6} + \frac{1}{a^2}
\end{eqnarray}
which can be solved explicitly in terms of conformal time $d\eta = a dt$ giving
\begin{eqnarray}
a^2(\eta) = \sqrt{\frac{\kappa P_{\phi}^2}{6 V_0^2}} \sinh(2\eta)
\end{eqnarray}
and similarly the scalar field evolves as
\begin{eqnarray}
\phi(\eta) = \frac{1}{2} \sqrt{\frac{6}{\kappa}} \ln(\tanh\eta) + \phi_0
\end{eqnarray}
Both are monotonic functions and thus can play the role of emergent time. If we choose the
scalar field to play the role of emergent time the evolution of $a$ is given by
\begin{eqnarray}
a^2(\phi) = \sqrt{\frac{\kappa P_{\phi}}{6}}\; \text{csch}\sqrt{2\kappa/3}(\phi-\phi_0)
\end{eqnarray}
For the quantization of the matter part of the constraint,
the operator for $P_{\phi}$ acts
simply as $\wh{P}_{\phi} = -i \hbar\, \partial / \partial\phi$. The
inverse volume operator $\wh{p}^{-3/2}$ requires careful treatment
as the naive inverse of the $\wh{p}$ operator does not lead to a densely
defined self-adjoint operator owing to the fact the the {\em normalizable}
state $\ket{v=0}$ lies in the spectrum of the $\wh{p}$ operator.
Using techniques from the full theory \cite{Thiemann:1996aw,Thiemann:1997rt},
the application to LQC leads to a bounded self-adjoint operator \cite{Bojowald:2002ny}
with eigenvalues given by \cite{Ashtekar:BBII}
\begin{eqnarray}
\wh{p}^{-3/2} \Psi(v) = \Big(\frac{6}{\kappa \gamma \hbar}\Big)^{3/2} B(v) \Psi(v)
\end{eqnarray}
with the function $B(v)$ given by
\begin{eqnarray}\label{B}
B(v) = \Big(\frac{3}{2}\Big)^{3} K |v| \Big| |v+1|^{1/3} - |v-1|^{1/3} \Big|^3
\end{eqnarray}
This inverse volume operator represent one choice among many possible choices
of the types that have been explored in \cite{Bojowald:2002ny}. In particular there is freedom
to use a particular spin $J$ SU(2) representation to define the holonomies.
The operator shown here corresponds to using the fundamental representation
($J=1/2$) in accordance with arguments indicating that the theory should
be quantized using that value\cite{Vandersloot:2005kh,Perez:2005fn}.
The behavior of $B(v)$ changes for $v < 1$ and $v>1$.
For $v < 1$, $B(v)$ behaves polynomially and increases for large values
of $v$ while it vanishes at the singularity $v=0$. For $v > 1$, $B(v)$ approaches
the classical expression $B(v) \approx \Big(\frac{\kappa \gamma \hbar}{6}\Big)^{3/2} p^{-3/2}$.
The physical meaning of $v < 1$ is dependent on the choice of fiducial cell and we
will discuss this in more detail in section \ref{Discussion}.
With the matter constraint operator, the difference equation can be rearranged into
the form
\begin{eqnarray}
\frac{\partial^2 \Psi(v, \phi)}{\partial\phi^2} =
- \wh{\Theta}\Psi(v,\phi) \equiv -( \wh{\Theta}_0 +\wh{\Theta}_{-1}) \Psi(v,\phi)
\end{eqnarray}
with the $\wh{\Theta}_0$ operator equivalent to the $\wh{\Theta}$ operator in the $k=0$ model \cite{Ashtekar:BBII}
\begin{eqnarray}
\wh{\Theta}_0 \Psi(v) &=&- B(v)^{-1} \Big[ C^{+}(v) \Psi(v+4) + C^{0}(v) \Psi(v) \nonumber \\
&&\;\;\;\;\;\;+ \;C^{-}(v) \Psi(v) \Big]\\
C^{+}(v) &=& \frac{3 \kappa K}{64} \, \big| v+2\big| \; \bigg| \big| v+1 \big| - \big| v+3 \big| \bigg| \\
C^{-}(v) &=& C^{+}(v-4) \\
C^{0}(v) &=& - C^{+}(v) - C^{-}(v)\,.
\end{eqnarray}
The $k=-1$ model contributes the $\wh{\Theta}_{-1}$ operator which acts diagonally
on $\Psi(v)$ as
\begin{eqnarray}
\wh{\Theta}_{-1} \;\Psi(v) = -B(v)^{-1} \frac{\kappa \gamma^2 V_0^{2/3}}{12 K^{1/3}}
\;|v|^{1/3} \nonumber \\ \times \bigg| \big| v+1 \big| - \big| v-1 \big| \bigg| \;\Psi(v)
\end{eqnarray}
The combined operator $\wh{\Theta}$ is self-adjoint\footnote{
Technically $\wh{\Theta}$ is self-adjoint on
the Hilbert space $L^2(\R_{\mathrm{Bohr}}, B(v) d\mu_{\text{Bohr}}) $ where
$\R_{\mathrm{Bohr}}$ refers to the almost periodic functions. The extra factor of $B(v)$ is
due to the fact the $\wh{C}_{GR}$ is self-adjoint in the kinematical Hilbert space $L^2(\R_{\mathrm{Bohr}}, d\mu_{\text{Bohr}}) $ and that $\wh{\Theta} \propto
B^{-1}(v) \wh{C}_{GR}$}
but not positive definite. If we restrict ourselves to the positive part of
the spectrum of $\wh{\Theta}$ then the physical inner product can
be constructed in a simple fashion. Namely we restrict to eigenstates
$e_{\omega}(v)$ of the $\wh{\Theta}$ operator: $\wh{\Theta} e_{\omega}(v) = \omega^2 e_k(v)$. Since $\wh{\Theta}$ is self-adjoint and
we are restricting to the positive part of the spectrum, by spectral analysis
we can construct an operator corresponding to the square root
$\sqrt{\wh{\Theta}}$. Solutions to the difference equation then split
into positive and negative frequency solutions satisfying
a
first order Schrodinger like equation
\begin{eqnarray} \label{diffeqs}
\mp i \frac{\partial \Psi(v, \phi)}{\partial\phi} = \sqrt{\wh{\Theta} } \; \Psi(v, \phi)
\end{eqnarray}
In this form, the difference equation is like a standard evolution equation
in terms of the scalar field $\phi$.
We can restrict to the positive frequency solution space when considering
physical wavefunctions whence the physical inner product is given in
analogy with the Schrodinger inner product of quantum mechanics as
\begin{eqnarray}
\braket{\Psi_1}{\Psi_2}_{\text{phys}} = \sum_{v} B(v) \,\overline{\Psi}_1(v, \phi_0)\, \Psi_2(v, \phi_0) \,.
\end{eqnarray}
As in quantum mechanics, the physical inner product can be evaluated at
any `time' $\phi_0$, and the difference equation (\ref{diffeqs})
guarantees that the result is independent of $\phi_0$.
Lastly to interpret the physical wavefunctions requires the evaluation of
expectation values of observables. Technically, we require Dirac observables
which correspond to quantum operators which commute with the constraint
operator so as to lead to unambiguous gauge invariant observables. Following
the $k=0$ model \cite{Ashtekar:BBII} the scalar field momentum
$\wh{P}_{\phi} = -i \hbar\, \partial / \partial\phi$ is an observable whose operator
trivially commutes with the constraint operator. An additional observable
is the value of $v$ at a given instant in `time' $\phi_0$ labeled
$v|_{\phi_0}$. The expectation value of this observable is given by
\begin{eqnarray}
\braketfull{\Psi}{\wh{v}|_{\phi_0}}{\Psi} = \frac{\sum_{v} B(v) \, v \,\overline{\Psi}(v, \phi_0)\, \Psi(v, \phi_0) } {\braket{\Psi}{\Psi}_{\text{phys}}}
\end{eqnarray}
\section{Dynamics}
With the inclusion of the massless scalar field, the resulting dynamics
and interpretation of the theory
can be understood by constructing suitable semi-classical states \cite{Singh:2005xg,Ashtekar:2006uz,Ashtekar:BBII}.
The dynamics of the theory is most easily understood by choosing the scalar field $\phi$ to play the
role of the internal clock. Following the procedure set forth in \cite{Ashtekar:2006uz,Ashtekar:BBII},
eigenfunctions of the $\wh{\Theta}$ operator are calculated and then Fourier
transformed to get physical wavefunction solutions to the quantum difference
equation. Thus given the eigenfunctions $e_{\omega}(v)$ we choose
a Gaussian profile $e^{-(\omega-\omega_*)^2/2\sigma^2}e^{i \omega \phi_*}$
peaked around a large value of the scalar field momentum $P_{\phi} = \hbar \omega_*$
with spread $\sigma$ and peaked around a value of the scalar field $\phi_*$.
Physical wavefunctions are constructed throuh the Fourier transform
\begin{eqnarray} \label{Ftrans}
\Psi(v, \phi) = \int_{-\infty}^{\infty} \!\! d\omega \, e^{-(\omega-\omega_*)^2/2\sigma^2}e^{i \omega \phi_*} e_{\omega}(v)\, e^{i \omega \phi}
\end{eqnarray}
which are thus by construction solutions to the difference equation.
Numerically, the procedure is to first calculate the eigenstates $e_{\omega}(v)$.
Typically, there is a two-fold degeneracy in the eigenstates, and
this is removed by choosing the eigenstate that matches the positive frequency
Wheeler-DeWitt solution\footnote{Wheeler-DeWitt solutions
are eigenstates of the operator $\underline{\wh{\Theta}}$ which is the
continuous differential operator that approximates the difference operator
$\wh{\Theta}$ in the large $v$ limit. See \cite{Ashtekar:2006uz,Ashtekar:BBII} for more details.} for large $v$. Once the eigenstates are calculated, the Fourier transform
(\ref{Ftrans}) is calculated using the Fast Fourier transform
algorithm.
An example of a numerical simulation is shown in figure \ref{3dplot}. The state is initially peaked
around a large value of $v$ and evolves towards the singularity while remaining
sharply peaked. Instead of plunging into the singularity as expected from
the classical dynamics, the state bounces leading to an expanding universe.
The results of the quantum dynamics are qualitatively similar to the $k=0,+1$ models
\cite{Ashtekar:BBII,Newkplus1} and the bounce occurs when the energy density
of the scalar field is Planckian.
\begin{figure}[ht]
\begin{center}
\includegraphics[width=8cm, keepaspectratio]
{3dplot.eps}
\end{center}
\caption{Evolution of semi-classical state initially peaked at a large
value of $v$. The state remains sharply peaked and bounces before
reaching the singularity $v = 0$. After the bounce, the state continues to remain
sharply peaked and leads to an expanding universe. The values of
the numerical parameters used in the simulation were $\omega_* = 700$,
$\sigma = 20$, $V_0 = 1$, and $K=1/2$.}
\label{3dplot}
\end{figure}
The behavior of the dynamics can be understood in terms of an
effective classical description. This amounts to considering an
effective modified Hamiltonian constraint through which effective classical
equations of motion are calculated. Note that by nature this sort of effective
description can not completely encode the predictions from the quantum theory
and care must be taken when applying the effective theory in more general settings.
In particular if the wavefunction becomes non-sharply peaked, then additional modifications
to the dynamics are expected to become appreciable \cite{Bojowald:2006gr}. In the numerical
simulations performed for this work, the wavefunction remains sharply peaked throughout the
evolution, and the effective description provides an accurate description which we show
explicitly now.
The effective Hamiltonian is given by (see \cite{Date:2004zd, Banerjee:2005ga, Vandersloot:2005kh, Willis:thesis, Bojowald:2006gr, Taveras2006} for various discussions
on the issue)
\begin{eqnarray}
{\mathcal H}_{eff} = -\frac{3 \sqrt{|p|}}{\kappa \gamma^2 \overline{\m}^2} \, \sin^2(\overline{\m} c) +\frac{3 \sqrt{|p|} V_0^{2/3}}{\kappa}+ |p|^{3/2}
\rho_{M}
\end{eqnarray}
where again $\overline{\m}$ is a function of $p$ given by
\begin{eqnarray}
\overline{\m} = \sqrt{\frac{\Delta}{|p|}}\, .
\end{eqnarray}
Note that in this effective Hamiltonian, we are implicitly assuming
the $v \gg 1$ limit. In particular, in this limit the $B(v)$ eigenvalues
that would appear in the matter part of the Hamiltonian are approximated
by the classical expression; namely
\begin{eqnarray}
B(v) = \frac{K}{v} + {\mathcal O}(v^{-3})
\end{eqnarray}
and thus the matter density takes on its classical form
\begin{eqnarray} \label{rhomeff}
\rho_{M} = \frac{P_{\phi}^2}{2 p^3} + {\mathcal O}(p^{-9/2})\,.
\end{eqnarray}
In this effective Hamiltonian we are therefore ignoring the inverse volume
corrections to the matter Hamiltonian and will show that this is
a good approximation by comparison with the quantum dynamics.
With this effective Hamiltonian we can derive an effective Friedmann equation. To do
this first we note that the left hand side of the Friedmann equation involving the Hubble
rate squared can be written as
\begin{eqnarray}
H^2 = \Big(\frac{\dot{a}}{a}\Big)^2 = \frac{1}{4}\Big(\frac{\dot{p}}{p}\Big)^2
\end{eqnarray}
which is a simple consequence from the fact that $p \propto a^2$ from equation (\ref{untild}).
The time derivative $\dot{p}$ is calculated from Hamilton's equation $\dot{p} = \{p, {\mathcal H}_{eff}\}$ giving
\begin{eqnarray}
H^2 = \frac{1}{4}\Big(\frac{\dot{p}}{p}\Big)^2 &=& \frac{1}{\gamma^2 \overline{\m}^2 |p|} \sin^2\overline{\m} c \;\cos^2\overline{\m} c \nonumber \\
&=& \frac{1}{\gamma^2 \overline{\m}^2 |p|} \sin^2\overline{\m} c \; (1 - \sin^2\overline{\m} c) \,.
\end{eqnarray}
Lastly we can use the vanishing of the Hamiltonian to relate $\sin^2\overline{\m} c$ to $\rho_{M}$ which gives
\begin{eqnarray}
\sin^2\overline{\m} c = \gamma^2 \overline{\m}^2 V_0^{2/3} +
\frac{\kappa \gamma^2 \overline{\m}^2 |p|}{3} \rho_{M} \,.
\end{eqnarray}
Putting these together and writing in terms of the scale factor $|p| = V_0^{2/3} a^2$
we get for the effective Friedmann equation
\begin{eqnarray} \label{Friedeff}
H^2 = \Big( \frac{\kappa}{3} \rho_{M} + \frac{1}{a^2}\Big) \Big(1 - \frac{\gamma^2 \Delta}{a^2} -
\frac{\kappa \gamma^2 \Delta}{3} \rho_{M} \Big)
\end{eqnarray}
The first term in parentheses is the classical right-hand side of the Friedmann equation
and thus the second term in parentheses represents the quantum modifications. The bounce
can be understood as arising when second term vanishes; namely, when the matter density
reaches a maximum
\begin{eqnarray} \label{rhomax}
\rho_{max} = \frac{3}{\kappa \gamma^2 \Delta} - \frac{3}{\kappa a^2}
\end{eqnarray}
where the first term is precisely the same form as the critical density $\rhoc= \frac{3}{\kappa \gamma^2 \Delta}$ arising in the $k=0$ model
and the second term forms an additional contribution from the $k=-1$ model.
Notice that the actual value of the matter density at the bounce point, depends on the value
of the scale factor at the bounce point. To determine the bounce scale factor $a_c$ and the value
of the bounce energy density for the massless scalar field, we can solve for when the matter density
equals the maximum value
\begin{eqnarray}
\frac{P_{\phi}^2}{2 V_0^2 a_c^6} = \frac{3}{\kappa \gamma \Delta} - \frac{3}{\kappa a_c^2}
\end{eqnarray}
If the scalar field momentum is sufficiently large, then $a_c$ is sufficiently large that the second
term is negligible and we find that the bounce energy density agrees with the form of the $k=0$ critical density
$\rho_{max} \approx \rhoc $. The actual value of $\rhoc$ is dependent
explicitly on the value of $\Delta$ which by (\ref{K}) depends on the value of $K$.
If $K$ is on the order of one, then (\ref{K}) implies that $\Delta$ is on the order
of the Planck length squared, and one finds that $\rhoc$ is on the order
of the Planck density. From the arguments of
\cite{Ashtekar:BBII}, the critical density in the $k=0$ model is valued at $\rhoc = .82 \rho_p$.
It is evident from the effective Friedmann equation (\ref{Friedeff}) and from
the form of the maximum energy density (\ref{rhomax}) that arbitrary matter
with positive energy density will trigger a bounce. Furthermore, the effective
Friedmann equation predicts a minimum scale factor $a_{min}$ that
the open universe can reach. Namely, even in the vacuum energy density
case the right hand side of the effective Friedmann equation is negative
and thus forbidden for values of the scale factor below
\begin{eqnarray}
a_{min} = \gamma \sqrt{\Delta}
\end{eqnarray}
which again is on the order of the Planck length if $\Delta$ is on the order of
the Planck length squared. Thus the open model constructed here predicts
a vacuum repulsion in the high curvature regime.
We can compare the predictions of the effective Friedmann equation
with the quantum dynamics as a method of testing the validity of
the effective theory.
In figure \ref{vmeanplot}, the expectation value of observable $<\wh{v}|_{\phi_0}>$
is plotted along with the spread $<\wh{\Delta v}|_{\phi_0}>$. The solid line is the trajectory predicted
from the effective Friedmann equation (\ref{Friedeff}) which agrees quite well with
the expectation values. We see that the effective Friedmann equation accounts
for the bounce at the right moment and agrees very well in the post bounce
regime. This testifies as to the validity of the effective theory in the
massless scalar field model considered. Furthermore, we have ignored
the inverse volume corrections to the matter part of the effective Hamiltonian
and thus the quantum dynamics are not sensitive to those effects. The reason
for this is that the bounce occurs at a value of $v$ much larger than one. In particular
for the values of the parameters chosen
in figure \ref{vmeanplot}, the bounce value of $v$ is 228.015. In order to probe
the small $v$ regime, one would need a semi-classical state with a small value
of $P_{\phi}$ yet such states behave non semi-classically with large spread and thus
the effective description would not be valid and the quantum state would not
be a good description of our universe..
However, as we mentioned one should
keep in mind that in more complicated models, the effective theory shown
here can in
principle deviate from the quantum dynamics with deviations that may depend
of the quantum state. Thus it is an open issue to
understand better in what regimes the deviations occur and
whether or not the deviations can be accounted for in a more complete
effective picture. An effective theory that takes into
account the quantum degrees of freedom (such as the spread of the wavefunction)
can be found in \cite{Bojowald:2006gr}, and thus merits testing with the
quantum dynamics in more complicated scenarios.
\begin{figure}[ht]
\begin{center}
\includegraphics[width=8cm, keepaspectratio]
{vmeanplot.eps}
\end{center}
\caption{Expectation value (dots) of $v|_{\phi}$ observable
with the error bars representing the dispersion. The expectation
values are approximated well by the predicted values (solid line) from
the effective Friedmann equation (\ref{Friedeff}).}
\label{vmeanplot}
\end{figure}
\section{Discussion}\label{Discussion}
We have shown explicitly that a successful loop quantization of the $k=-1$ model
exists with the correct semi-classical limit. In this quantization the results
of the $k=0,+1$ models are extended and the classical singularity can be resolved
even leading to a big-bounce with a massless scalar field. This is further testament
to robustness of the predictions of LQC.
Several caveats of the model require discussion. First is that our model was constructed
using holonomies of the extrinsic curvature as opposed to holonomies of the connection
as done in the full theory. The reason for using this quantization
is that the holonomies of the connection (an example of which
is given in formula \ref{hA}) are not almost periodic functions thus
rendering a loop quantization difficult.
As stated, this technique has been utilized in the
loop quantization of the spherically symmetric models as well as in the inhomogeneous
cosmological model of \cite{Bojowald:2006qu}. An important question is therefore
what are the implications of the quantization using holonomies of the extrinsic curvature.
We can turn to the closed
$k=+1$ model where both quantizations have been performed, with
holonomies of the extrinsic curvature being used in the earlier work \cite{Bojowald:2003mc} while
holonomies of the connection comprising the quantization in the more recent work of \cite{Newkplus1}. The two quantizations can be shown to be in agreement in
the $v \gg 1$ limit, with the differences restricted to the small volume $v \ll 1$ regime.
We can
understand the reason for this behavior in the following heuristic way. The holonomies of the connection
consist of exponentials of $\overline{\m}$ times the connection; i.e., $\overline{\m} (\gamma K_a^i + \Gamma_a^i)$
where for the non flat models the spin connection components $\Gamma_a^i$ are constant
valued.
Since we are taking $\overline{\m}$ to scale as $p^{-1/2}$ (equivalently $v^{-1/3}$),
then for large values of $v$ the quantity $\overline{\m} \Gamma_a^i$ is guaranteed to be small.
Thus, the difference of holonomies of the connection and extrinsic curvature are expected
to be negligible in the large $v$ limit. This is precisely what is observed in the $k=+1$ model.
Therefore, the results of the $k=+1$ model indicate that for the $k=-1$ quantization
presented here, the results are expected to be valid in the $v \gg 1$ regime. This
does not affect any of the results presented here provided that the semi-classical
state does not approach the $v < 1 $ regime. As we have mentioned,
the bounce occurs at a value of $v \gg 1$ for universes which behave semi-classically.
Furthermore, in the $k=+1$ model similarly the bounce occurs at $v \gg 1$
for universes which reach macroscopic size before recollapsing \cite{Newkplus1}.
Thus we expect that the physical results presented in this paper, such as the quantum bounce, are largely insensitive
to whether the quantization is performed using holonomies of the connection or
extrinsic curvature.
Additionally there is the issue of the dependence of the quantum results
on the size of the fiducial cell. First we can ask if the effective
Friedmann equation (\ref{Friedeff}) is dependent on the fiducial
cell and therefore the prediction of the quantum bounce.
Classical quantities such as the scale factor $a$ and matter energy density
do not make reference to the fiducial cell and thus do not rescale. This
implies that the effective Friedmann equation (\ref{Friedeff}) is invariant
under a change in fiducial cell. Note that the result crucially depends
on the fact that $\overline{\m}$ is not taken to be a constant, but scales
as $p^{-1/2}$. Thus the prediction of the bounce does not make
reference to the fiducial cell.
The same statement can not be made about the inverse volume corrections
appearing in the quantum matter density of the scalar field. The eigenvalues $B(v)$
give back the classical behavior for $v \ll 1$ but in general behave as
\begin{eqnarray}
B(v) \propto \left\{
\begin{array}{lcl}
v^4 &&v \ll 1 \\
v^{-1} &&v \gg 1
\end{array}
\right.
\end{eqnarray}
The parameter $v$ is proportional to the physical volume of the
fiducial cell, and thus must scale if we resize the fiducial cell.
The exact scaling under a resizing of the fiducial cell
$V_0 \rightarrow V_0' = \xi^3 V_0$, is given as
\begin{eqnarray}
v \rightarrow v' = \xi^3 v
\end{eqnarray}
For a given value of the scale factor, a larger fiducial cell implies a larger value
of $v$.
In terms of the scale factor, $v$ is related as
\begin{eqnarray}
v = V_0 K \; \Big( \frac{6}{8\pi \gamma}\Big)^{3/2} \frac{a^3}{l_{\mathrm P}^3}
\end{eqnarray}
which makes evident that the value of $v$ depends explicitly on
the fiducial cell volume $V_0$ for a {\em fixed} value of the scale factor.
If we enlarge the fiducial cell, then the value of $v$ should also
increase which in turn {\em reduces} the effects of the inverse volume
eigenvalues. Vice versa, a smaller fiducial cell implies stronger
inverse volume effects.
Thus, when considering phenomenological applications involving
the inverse volume modifications, one must specify
the scale at which the inverse volume effects are non-negligible. In
other words, the critical scale separating the quantum regime from
the classical regime corresponds to $v = 1$ which in terms
of a critical scale factor $a_*$ gives
\begin{eqnarray}
a_* = \sqrt{\frac{8\pi \gamma}{6}} \frac{l_{\mathrm P}}{K^{1/3}} \; V_0^{-1/3}
\end{eqnarray}
which indicates the explicit dependence on the fiducial cell. Again, a larger
value of $V_0$ implies a smaller $a_*$ which pushes the quantum effects
into the higher curvature regime and vice versa. If the fiducial cell
volume $V_0$ and $K$ are on the order of one, then $a_*$ is on the order
of the Planck length, but note that the critical scale is not necessarily
Planckian.
The issue of the scale dependence of the inverse volume modifications
occurs additionally in the $k=0$ model where again a fiducial cell is required
to quantize the spatially infinite model (see discussions in \cite{kevthesis,Ashtekar:BBII}). The preceding arguments remain valid
for this model and a scale must be introduced. On the other hand, the compact $k=+1$
model does not require a fiducial cell since the spatial integrations do not diverge.
There, inverse volume modifications occur when the physical volume of the entire
universe is Planckian. In other words, the scale at which the quantum effects occur
is provided by physical volume of the universe. For the closed model this
is equivalent to the high curvature Planckian regime.
Since the scale at which the inverse volume effects occur is given by
the physical size of the fiducial cell in the $k=0,-1$ models, an important
issue is to determine what sets the scale in loop quantum cosmology.
The fiducial cell was introduced in order to regulate the infinite spatial
integrations appearing in the action and Hamiltonian and thus
is not expected to be physically relevant. One possibility
is that the scale is provided in an {\em inhomogeneous} treatment
of loop quantum cosmology. An inhomogeneous model of loop quantum
cosmology has been developed in \cite{Bojowald:2006qu} based
on a fixed lattice quantization. In that model, the scale corresponds
to the physical size of the lattice links. Yet, the inhomogeneous
model does not provide a prescription to determine the size of the scale, which
must be specified by hand and is not necessarily tied to matter degrees of freedom or the curvature scale. The naive expectation would be that
the lattice spacing should be Planckian in size, but if the model describes
the current universe then we would expect to see
inverse volume modifications occurring today, a prediction which
is clearly ruled out by observations.
Whatever determines the scale inherent in LQC models, one is faced with
constraining the predictions with observations. As mentioned,
if the scale is too small, then inverse volume corrections
might be predicted in the near past which would alter the Friedmann dynamics and
be observationally detectable. If the lattice links of an inhomogeneous model provide the scale, the links must
be sufficiently larger than the Planck scale in the recent history of the universe, but presumably
not too large to spoil particle physics. If the scale provided by the lattice links
expands with the growing universe (i.e. the lattice links grow with the universe), then ensuring that they are not too large today,
while being not too small in the earlier universe could be challenging and might require
fine tuning. The inhomogeneous model of
of \cite{Bojowald:2006qu} has the behavior that the lattice links expand with
the universe and thus would face this constraint.
However, as mentioned in \cite{Bojowald:2006qu}, one possibility is that
in a more systematically derived inhomogeneous lattice model of loop quantum cosmology, the scale provided by the lattice
links would dynamically change and thus might not grow with the expanding universe.
This type of behavior is mimicked in the homogeneous setting
when $\overline{\m}$ scales as a function of $p^{-1/2}$, a quantization feature which was
first proposed in \cite{Ashtekar:BBII} and has been utilized in
this paper. With this scaling behavior, the holonomy edges defining the Hamiltonian
constraint operator decrease in physical length with the expanding universe.
The results of the improved quantization appear better grounded on a
physical basis and thus this behavior would seem to be a requirement
for constructing inhomogeneous models.
Furthermore,
we have shown that the inverse volume modifications play no important role
in the quantum dynamics for universes which behave semi-classically
since the bounce occurs for $v \gg 1$. Additionally, in the $k=+1$ model, for universes which grow to macroscopic size, again the inverse
volume modifications play no role \cite{Newkplus1}. We these indications,
along with the arguments that the ambiguity parameter $j$ should be
its lowest value $1/2$ \cite{Vandersloot:2005kh,Perez:2005fn},
these results give evidence that the inverse volume modifications may not play a significant role
in the evolution of the universe.
\acknowledgments{The author would like to thank Abhay Ashtekar, Martin Bojowald, and Parampreet Singh for discussions, and to Abhay Ashtekar, Parampreet Singh, and
Tomasz Pawlowski for
helpful comments on the manuscript. This work was supported in part by
the NSF grant PHY-0456913, the Eberly research funds of Penn State,
and the Marie Curie Incoming International Fellowship MIF1-CT-2006-022239.}
| {
"redpajama_set_name": "RedPajamaArXiv"
} | 9,944 |
Modify Old Registers in Signal Handler
===================
This is an example to set old register in signal handler.
It can be used to fix some unexception. such as divide by zero(SIGFPE)、Segmention Fault(SIGSEGV).
Author
---
kXuan <kxuanobj@gmail.com>
License
---
MIT
| {
"redpajama_set_name": "RedPajamaGithub"
} | 2,859 |
2/21/01 Agreed to terms with RHP Rich Garces on a 2001 contract, thus avoiding salary arbitration.
2/16/01 Agreed to terms with RHP Rolando Arrojo on a 2001 contract, thus avoiding salary arbitration.
2/13/01 Agreed to terms with C Jason Varitek on a 2001 contract, thus avoiding salary arbitration.
2/1/01 Agreed to terms with INF Chris Stynes on a 2001 contract, thus avoiding salary arbitration. Invited RHP Rafael Betancourt, RHP Carlos Castillo, RHP Todd Erdos, LHP Luis Arroyo, LHP Casey Fossum, LHP Allen McDill, C Luis Rodriguez, C Angel Encarnacion, C George Williams, INF Jon Shave, OF Izzy Alcantara, OF Yamil Benitez and OF Mike Neill to Major League Spring Training. | {
"redpajama_set_name": "RedPajamaC4"
} | 8,556 |
Canada's online magazine: Politics, entertainment, technology, media, arts, books: backofthebook.ca
Politics, tech, media, culture and more, from a Canadian point-of-view
The Video
You are here: Home / Media / Lockdown
07/04/2008 by backofthebook.ca Leave a Comment
By Frank Moher
The recent Supreme Court of Canada decision in favour of Vancouver broadcaster Rafe Mair was a big step forward for Canadian journalists and their readers. Mair had been sued by a "Christian-values advocate" who thought he'd defamed her, but the Court ruled 9-0 that "an overly solicitous regard for personal reputation" should not "be permitted to 'chill' freewheeling debate on matters of public interest." This brings us a lot closer to the American definition of fair comment (by which just about anything goes, especially when it comes to public figures), and away from the British one, which saw, for example, Conrad Black reflexively suing journalists who'd offended his exquisite sensibilities.
In fact, if it hadn't done so, I'd have had to think harder about the possible legal ramifications of that last sentence. The Supremes' ruling means that comment in our sometimes too deferential country is liable to become livelier, not to mention more fractious.
However, in the way of these things, the Feds were only giving back with one hand part of what they had recently taken away with another.
Bill C-61, the copyright reform bill recently tabled in parliament, is not the spawn-of-hell, civilization-ending piece of legislation Michael Geist and his Geistolytes make out. The Internet has spawned an absurd culture of entitlement, whereby a lot of people have concluded that if they want something to be free, it should be. Try out that theory next time you're in a department or grocery store.
But C-61 is full of flaws, the most pertinent of which, for purposes of this column, is that it makes no provision for fair use of digital content in journalism. In education, yes, but not in journalism. So let's say I wanted to show you (grabbing a DVD off the nearest shelf . . . hmm, let's see . . . yes, this will do) this little scene from The Day After Tomorrow (click on the word "tomorrow") . . .
myvideoplayer.swf
. . . in order to suggest that it's the most effective piece of anti-Bush administration propaganda ever to appear in pop culture. That would traditionally be fair use, in the same way I can quote from a book I'm reviewing.
But under C-61, I couldn't do it. Simply by using decryption software to turn the movie into a file I could edit and post, I've already broken US law, and will one day have broken Canadian law, if C-61 goes through as written.
Nor could I take a clip of Stephen Harper (assuming it arrived with digital locks, or on a TV with digital locks) and add a little smiley-face to it to satirize his "don't worry, be-happy" approach to climate change. I couldn't even take a Coldplay video, change-up the lyrics, ala Weird Al Yankovic, and post it; there's no provision for parody in C-61, despite parody being a long accepted example of fair use.
It wouldn't be the satire itself that would be unlawful; making fun of public figures remains legal, at least until Mr. Bush declares himself emperor of North America. It's the act of using certain software in order to create the parody that would expose me to big fines. If I could figure out a way to do it without cracking the encryption, I'd be okay. Psychokinesis, maybe?
So, the best basis on which to criticize C-61 is not that it protects giant entertainment companies, though it does, but that it's nonsensical. We can't expect governments to tell big business to start giving away its products, but we can ask that its laws be internally consistent. Note, however, that making C-61 internally consistent would also have the effect of continuing to allow sufficiently-motivated individuals to do whatever they wanted with their copy of The Day After Tomorrow, at least until some impregnable encryption technology comes along. In which case everyone — except, as usual, the big entertainment companies — would be happy.
By the way, that little clip from The Day After Tomorrow is indeed a good piece of anti-Bush propaganda, whatwith the infernal Dick Cheney figure telling the doofus George Bush character what to do. Have a look, if you haven't already. And don't worry; if C-61 passes as is, it'll be my ass, not yours.
Filed Under: Media Tagged With: Bill C-61, Canada, Canadian politics, internet, journalism, law, new media, online media, Rafe Mair
Fort McMurray: Shopping time!
By Brady Tighe We're now officially in the aftermath phase of the northern Alberta wildfire crisis. The fire is long gone, and everyone with a home to return to is back in its … [Read More...]
Electoral reform: Hashtag fresh thinking
By Alison@Creekside The most interesting and innovative idea to come out of the first meeting of the all-party Special Committee on Electoral Reform, or ERRE, was Nathan Cullen's suggestion, … [Read More...]
The Trudeau gush fest is getting old
By Jim Henshaw There have been several bewildered as well as angry accounts coming out of the USA lately about how little media time has been spent covering the Democratic Presidential Primary … [Read More...]
My friend, Rick, at the Pride Parade
By Frank Moher On this dreadful day, I don't want to write about the shootings in Orlando. I want to write about my friend, Rick. Rick lives just outside of Nanaimo, a city of about 80,000, … [Read More...]
Triumph of the drama nerds
By Frank Moher Two drama nerds have recently moved into high profile positions. Before I name them (or perhaps you've already guessed who they are; or perhaps you'd like to scroll down and look at … [Read More...]
From "Our Rape Blog": Shooting the Moon
Originally published on Our Rape Blog, the author's account of the aftermath of a violent sexual assault. By Mary Fraughton Have you ever played Hearts? It's a card game. For our purposes, … [Read More...]
The video: Lelu Island: "They will come."
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{"url":"https:\/\/research.nu.edu.kz\/en\/publications\/femtosecond-laser-based-inscription-technique-for-post-fiber-brag","text":"# Femtosecond-Laser-Based Inscription Technique for Post-Fiber-Bragg Grating Inscription in an Extrinsic Fabry-Perot Interferometer Pressure Sensor\n\nSven Poeggel, Dineshbabu Duraibabu, Amedee Lacraz, Kyriacos Kalli, Daniele Tosi, Gabriel Leen, Elfed Lewis\n\nResearch output: Contribution to journalArticle\n\n8 Citations (Scopus)\n\n### Abstract\n\nIn this paper, a novel fiber Bragg grating inscription technique based on a femtosecond laser is presented. The grating was inscribed in close proximity to the tip of an extrinsic Fabry-Perot interferometer (EFPI)-based optical fiber pressure sensor. This therefore represents an optical fiber pressure and temperature sensor (OFPTS) for simultaneous pressure and temperature measurement for use in exactly the same physical location. The temperature measurement can also be used to compensate thermal drift in the EFPI sensor. The Bragg wavelength can be tailored precisely to any given wavelength in the optical spectrum and the degree of reflection can be adjusted to suit the FPI spectrum. The OFPTS has a diameter of 200 $\\mu \\text{m}$ and is fully biocompatible. Furthermore, the sensor shows a high stability after grating inscription, of better than 0.5% in 20 min. The small size and high stability makes the sensor especially interesting for volume restricted areas, like blood vessels or the brain.\n\nOriginal language English 7109831 3396-3402 7 IEEE Sensors Journal 16 10 https:\/\/doi.org\/10.1109\/JSEN.2015.2434772 Published - May 15 2016\n\n### Fingerprint\n\nFabry-Perot interferometers\nPressure sensors\nFiber Bragg gratings\npressure sensors\nUltrashort pulses\nBragg gratings\nOptical fibers\noptical fibers\ntemperature sensors\nTemperature sensors\nTemperature measurement\nfibers\ntemperature measurement\nsensors\nSensors\ngratings\nlasers\nWavelength\nblood vessels\nBlood vessels\n\n### Keywords\n\n\u2022 Fabry Perot interferometer\n\u2022 Femtosecond laser\n\u2022 Fibre Bragg grating\n\u2022 OFPTS\n\u2022 Optical fibre sensor\n\n### ASJC Scopus subject areas\n\n\u2022 Electrical and Electronic Engineering\n\u2022 Instrumentation\n\n### Cite this\n\nFemtosecond-Laser-Based Inscription Technique for Post-Fiber-Bragg Grating Inscription in an Extrinsic Fabry-Perot Interferometer Pressure Sensor. \/ Poeggel, Sven; Duraibabu, Dineshbabu; Lacraz, Amedee; Kalli, Kyriacos; Tosi, Daniele; Leen, Gabriel; Lewis, Elfed.\n\nIn: IEEE Sensors Journal, Vol. 16, No. 10, 7109831, 15.05.2016, p. 3396-3402.\n\nResearch output: Contribution to journalArticle\n\nPoeggel, Sven ; Duraibabu, Dineshbabu ; Lacraz, Amedee ; Kalli, Kyriacos ; Tosi, Daniele ; Leen, Gabriel ; Lewis, Elfed. \/ Femtosecond-Laser-Based Inscription Technique for Post-Fiber-Bragg Grating Inscription in an Extrinsic Fabry-Perot Interferometer Pressure Sensor. In: IEEE Sensors Journal. 2016 ; Vol. 16, No. 10. pp. 3396-3402.\n@article{6f2c6b1d45f044eb9a131986d7f483d0,\ntitle = \"Femtosecond-Laser-Based Inscription Technique for Post-Fiber-Bragg Grating Inscription in an Extrinsic Fabry-Perot Interferometer Pressure Sensor\",\nabstract = \"In this paper, a novel fiber Bragg grating inscription technique based on a femtosecond laser is presented. The grating was inscribed in close proximity to the tip of an extrinsic Fabry-Perot interferometer (EFPI)-based optical fiber pressure sensor. This therefore represents an optical fiber pressure and temperature sensor (OFPTS) for simultaneous pressure and temperature measurement for use in exactly the same physical location. The temperature measurement can also be used to compensate thermal drift in the EFPI sensor. The Bragg wavelength can be tailored precisely to any given wavelength in the optical spectrum and the degree of reflection can be adjusted to suit the FPI spectrum. The OFPTS has a diameter of 200 $\\mu \\text{m}$ and is fully biocompatible. Furthermore, the sensor shows a high stability after grating inscription, of better than 0.5{\\%} in 20 min. The small size and high stability makes the sensor especially interesting for volume restricted areas, like blood vessels or the brain.\",\nkeywords = \"Fabry Perot interferometer, Femtosecond laser, Fibre Bragg grating, OFPTS, Optical fibre sensor\",\nauthor = \"Sven Poeggel and Dineshbabu Duraibabu and Amedee Lacraz and Kyriacos Kalli and Daniele Tosi and Gabriel Leen and Elfed Lewis\",\nyear = \"2016\",\nmonth = \"5\",\nday = \"15\",\ndoi = \"10.1109\/JSEN.2015.2434772\",\nlanguage = \"English\",\nvolume = \"16\",\npages = \"3396--3402\",\njournal = \"IEEE Sensors Journal\",\nissn = \"1530-437X\",\npublisher = \"Institute of Electrical and Electronics Engineers Inc.\",\nnumber = \"10\",\n\n}\n\nTY - JOUR\n\nT1 - Femtosecond-Laser-Based Inscription Technique for Post-Fiber-Bragg Grating Inscription in an Extrinsic Fabry-Perot Interferometer Pressure Sensor\n\nAU - Poeggel, Sven\n\nAU - Duraibabu, Dineshbabu\n\nAU - Lacraz, Amedee\n\nAU - Kalli, Kyriacos\n\nAU - Tosi, Daniele\n\nAU - Leen, Gabriel\n\nAU - Lewis, Elfed\n\nPY - 2016\/5\/15\n\nY1 - 2016\/5\/15\n\nN2 - In this paper, a novel fiber Bragg grating inscription technique based on a femtosecond laser is presented. The grating was inscribed in close proximity to the tip of an extrinsic Fabry-Perot interferometer (EFPI)-based optical fiber pressure sensor. This therefore represents an optical fiber pressure and temperature sensor (OFPTS) for simultaneous pressure and temperature measurement for use in exactly the same physical location. The temperature measurement can also be used to compensate thermal drift in the EFPI sensor. The Bragg wavelength can be tailored precisely to any given wavelength in the optical spectrum and the degree of reflection can be adjusted to suit the FPI spectrum. The OFPTS has a diameter of 200 $\\mu \\text{m}$ and is fully biocompatible. Furthermore, the sensor shows a high stability after grating inscription, of better than 0.5% in 20 min. The small size and high stability makes the sensor especially interesting for volume restricted areas, like blood vessels or the brain.\n\nAB - In this paper, a novel fiber Bragg grating inscription technique based on a femtosecond laser is presented. The grating was inscribed in close proximity to the tip of an extrinsic Fabry-Perot interferometer (EFPI)-based optical fiber pressure sensor. This therefore represents an optical fiber pressure and temperature sensor (OFPTS) for simultaneous pressure and temperature measurement for use in exactly the same physical location. The temperature measurement can also be used to compensate thermal drift in the EFPI sensor. The Bragg wavelength can be tailored precisely to any given wavelength in the optical spectrum and the degree of reflection can be adjusted to suit the FPI spectrum. The OFPTS has a diameter of 200 $\\mu \\text{m}$ and is fully biocompatible. Furthermore, the sensor shows a high stability after grating inscription, of better than 0.5% in 20 min. The small size and high stability makes the sensor especially interesting for volume restricted areas, like blood vessels or the brain.\n\nKW - Fabry Perot interferometer\n\nKW - Femtosecond laser\n\nKW - Fibre Bragg grating\n\nKW - OFPTS\n\nKW - Optical fibre sensor\n\nUR - http:\/\/www.scopus.com\/inward\/record.url?scp=84963930663&partnerID=8YFLogxK\n\nUR - http:\/\/www.scopus.com\/inward\/citedby.url?scp=84963930663&partnerID=8YFLogxK\n\nU2 - 10.1109\/JSEN.2015.2434772\n\nDO - 10.1109\/JSEN.2015.2434772\n\nM3 - Article\n\nVL - 16\n\nSP - 3396\n\nEP - 3402\n\nJO - IEEE Sensors Journal\n\nJF - IEEE Sensors Journal\n\nSN - 1530-437X\n\nIS - 10\n\nM1 - 7109831\n\nER -","date":"2019-10-15 04:59:19","metadata":"{\"extraction_info\": {\"found_math\": true, \"script_math_tex\": 0, \"script_math_asciimath\": 0, \"math_annotations\": 0, \"math_alttext\": 0, \"mathml\": 0, \"mathjax_tag\": 0, \"mathjax_inline_tex\": 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\": 0.35018786787986755, \"perplexity\": 6653.7416406063485}, \"config\": {\"markdown_headings\": true, \"markdown_code\": true, \"boilerplate_config\": {\"ratio_threshold\": 0.3, \"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\/1570986655864.19\/warc\/CC-MAIN-20191015032537-20191015060037-00316.warc.gz\"}"} | null | null |
Q: How do I control which Excel chart axes pair I plot a data series to? I have three data series (x-y) I want to plot as scatter graphs in Microsoft Excel.
I want the first series to be plotted using the bottom x-axis and left y-axis.
I want the second series to be plotted using the top x-axis and the left y-axis (again).
I want the third series to be plotted using the bottom x-axis and the right y-axis.
So I need four different axes on the four sides of the chart area. Excel seemingly can place primary and secondary horizontal and vertical axes okay but I can't seem to control which combination of these my data series are plotted to - only whether they are "primary" or "secondary" which isn't enough detail.
Any help would be greatly appreciated!
Screenscrab of chart area with the three data series present. The blue series is correctly plotted against bottom X and left Y but the orange and red series are plotted against the wrong y-axis
| {
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Avatar: Generations reemerges with new release window, preregistration
By Tomas Franzese January 10, 2023
Free-to-play Avatar: The Last Airbender mobile RPG Avatar: Generations has reemerged under a new publisher, as the game now has an early 2023 release window and is available for pre-registration on Android.
Avatar: Generations is a turned-based RPG like Marvel Strike Force or Star Wars: Galaxy of Heroes that lets fans play through the stories of various Team Avatars. At launch, the focus is on Avatar: The Last Airbender's story, although the press release for this announcement teased that Avatar: Generations will eventually get updates that feature The Legend of Korra stories and characters, as well as never-before-seen Avatars.
Avatar Generations - Official Gameplay Trailer - Coming Soon to Mobile!
As players progress through the game, they will collect various heroes and companions based on characters from the franchise, and then upgrade those characters and build a squad that's best fit to complete a given level. Those who pre-register for the game will receive hero Avatar Aang and companion Appa free of charge. Although it's not the AAA console game Avatar: The Last Airbender fans are yearning for, Avatar: Generations is shaping up to be the most thorough Avatar game yet.
Interestingly, this announcement also confirms that Avatar: Generations publisher is now Crystal Dynamics — Eidos Entertainment, a subsidiary of Embracer Group. Avatar: Generations was initially announced by Square Enix last year, but it appears that the Square Enix London Mobile team that was handling the game was part of Square Enix's western studio sale to Embracer Group last summer.
Even though Embracer Group shut down Onoma, a mobile studio it acquired from Square Enix, it still seems interested in Avatar: Generations, likely due to its existing relationship with Nickelodeon and the fact that Avatar: The Last Airbender is a well-known IP. Avatar: Generations is developed by Navigator Games.
Avatar: Generations will launch on iOS and Android in early 2023.
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What is Embracer Group? Gaming's new megapower, explained
Avatar: Frontiers of Pandora: release date, trailers, and more
Before Your Eyes devs explain why Netflix works as a gaming platform
Discord is making its Android app more like iOS, and in a good way
iOS 16 lets you pair Nintendo Switch controllers to your iPhone | {
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My favorite St. Nicholas story — and probably the one that Santa Claus has its roots in — is how he went secretly to the home of the three daughters of a poor man at night and left them three bags of gold through the window so that they would have dowries and be able to marry. He didn't rail against the dowry system; he didn't get off on his ideological correctness, like those anti-tipping assholes in New York who leave their waiter a little card explaining that tipping in the restaurant industry is exploitative, drafting the hapless kid into their cause by depriving him of income and not leaving him anything except the little card; he goes off flush with Anglo self-rightouness; the waiter goes home broke. St. Nick simply gave three poor sisters what they needed so that they could survive in the world. And we can talk ideology and exploitation later.
← What a great guy…. | {
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Q: ¿Como cambiar un dato tipo fecha a tipo de dato numérico en SAS? Tengo un codigo que me trae la fecha actual
dt = date()
format dt YYMMN.;
Eso lo que me arroja en este formato es: 202002
El date me da un numero de: 21963
Lo que necesito es poder tener el 202002 en numérico, no me sirve el 21963 porque necesito comparar ese 202002.
A: Te comparto una forma de obtener lo que necesitas desde una consulta SQL.
Ya con eso lo puedes adaptar a tu base de datos para que lo reciba tu código.
SELECT CONVERT(INT,SUBSTRING(CONVERT(VARCHAR, SYSDATETIME(),112),1,6))
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{"url":"https:\/\/www.groundai.com\/project\/non-asymptotic-analysis-of-an-optimal-algorithm-for-network-constrained-averaging-with-noisy-links\/","text":"Non-asymptotic analysis of an optimal algorithm for network-constrained averaging with noisy links\n\n# Non-asymptotic analysis of an optimal algorithm for network-constrained averaging with noisy links\n\n## Abstract\n\nThe problem of network-constrained averaging is to compute the average of a set of values distributed throughout a graph using an algorithm that can pass messages only along graph edges. We study this problem in the noisy setting, in which the communication along each link is modeled by an additive white Gaussian noise channel. We propose a two-phase decentralized algorithm, and we use stochastic approximation methods in conjunction with the spectral graph theory to provide concrete (non-asymptotic) bounds on the mean-squared error. Having found such bounds, we analyze how the number of iterations required to achieve mean-squared error scales as a function of the graph topology and the number of nodes . Previous work provided guarantees with the number of iterations scaling inversely with the second smallest eigenvalue of the Laplacian. This paper gives an algorithm that reduces this graph dependence to the graph diameter, which is the best scaling possible.\n\n## I Introduction\n\nThe problem of network-constrained averaging is to compute the average of a set of numbers distributed throughout a network, using an algorithm that is allowed to pass messages only along edges of the graph. Motivating applications include sensor networks, in which individual motes have limited memory and communication ability, and massive databases and server farms, in which memory constraints preclude storing all data at a central location. In typical applications, the average might represent a statistical estimate of some physical quantity (e.g., temperature, pressure etc.), or an intermediate quantity in a more complex algorithm (e.g., for distributed optimization). There is now an extensive literature on network-averaging, consensus problems, as well as distributed optimization and estimation (e.g., see the papers\u00a0[7, 12, 10, 30, 20, 3, 4, 8, 23, 22]). The bulk of the earlier work has focused on the noiseless variant, in which communication between nodes in the graph is assumed to be noiseless. A more recent line of work has studied versions of the problem with noisy communication links (e.g., see the papers\u00a0[18, 15, 27, 2, 29, 19, 24] and references therein).\n\nThe focus of this paper is a noisy version of network-constrained averaging in which inter-node communication is modeled by an additive white Gaussian noise (AWGN) channel. Given this randomness, any algorithm is necessarily stochastic, and the corresponding sequence of random variables can be analyzed in various ways. The simplest question to ask is whether the algorithm is consistent\u2014that is, does it compute an approximate average or achieve consensus in an asymptotic sense for a given fixed graph? A more refined analysis seeks to provide information about this convergence rate. In this paper, we do so by posing the following question: for a given algorithm, how does number of iterations required to compute the average to within -accuracy scale as a function of the graph topology and number of nodes ? For obvious reasons, we refer to this as the network scaling of an algorithm, and we are interested in finding an algorithm that has near-optimal scaling law.\n\nThe issue of network scaling has been studied by a number of authors in the noiseless setting, in which the communication between nodes is perfect. Of particular relevance here is the work of Benezit et al.\u00a0[5], who in the case of perfect communication, provided a scheme that has essentially optimal message scaling law for random geometric graphs. A portion of the method proposed in this paper is inspired by their scheme, albeit with suitable extensions to multiple paths that are essential in the noisy setting. The issue of network scaling has also been studied in the noisy setting; in particular, past work by Rajagopal and Wainwright\u00a0[27] analyzed a damped version of the usual consensus updates, and provided scalings of the iteration number as a function of the graph topology and size. However, our new algorithm has much better scaling than the method\u00a0[27].\n\nThe main contributions of this paper are the development of a novel two-phase algorithm for network-constrained averaging with noise, and establishing the near-optimality of its network scaling. At a high level, the outer phase of our algorithm produces a sequence of iterates based on a recursive linear update with decaying step size, as in stochastic approximation methods. The system matrix in this update is a time-varying and random quantity, whose structure is determined by the updates within the inner phase. These inner rounds are based on establishing multiple paths between pairs of nodes, and averaging along them simultaneously. By combining a careful analysis of the spectral properties of this random matrix with stochastic approximation theory, we prove that this two-phase algorithm computes a -accurate version of the average using a number of iterations that grows with the graph diameter (up to logarithmic factors).1 As we discuss in more detail following the statement of our main result, this result is optimal up to logarithmic factors, meaning that no algorithm can be substantially better in terms of network scaling.\n\nThe remainder of this paper is organized as follows. We begin in Section\u00a0II with background and formulation of the problem. In Section\u00a0III, we describe our algorithm, and state various theoretical guarantees on its performance. We then provide the proof of our main result in Section\u00a0IV. Section\u00a0V is devoted to some simulation results that confirm the sharpness of our theoretical predictions. We conclude the paper in Section\u00a0VI.\n\nNotation: For the reader\u2019s convenience, we collect here some notation used throughout the paper. The notation means that there exists some constant and such for all , whereas means that for all . The notation means that and . Given a symmetric matrix , we denote its ordered sequence of eigenvalues by and also its -operator norm by . Finally we use to denote the Euclidean inner product.\n\n## Ii Background and problem set-up\n\nWe begin in this section by introducing necessary background and setting up the problem more precisely.\n\n### Ii-a Network-constrained averaging\n\nConsider a collection of numbers. In statistical settings, these numbers would be modeled as identically distributed (i.i.d.) draws from an unknown distribution with mean . In a centralized setting, a standard estimator for the mean is the sample average . When all of the data can be aggregated at a central location, then computation of is straightforward. In this paper, we consider the network-constrained version of this estimation problem, modeled by an undirected graph that consists of a vertex set , and a collection of edges joining pairs of vertices. For , we view each measurement as associated with vertex . (For instance, in the context of sensor networks, each vertex would contain a mote and collect observations of the environment.) The edge structure of the graph enforces communication constraints on the processing: in particular, the presence of edge indicates that it is possible for sensors and to exchange information via a noisy communication channel. Conversely, sensor pairs that are not joined by an edge are not permitted to communicate directly.2 Every node has a synchronized internal clock, and acts at discrete times . For any given pair of sensors , we assume that the message sent from to is perturbed by an independent identically distributed variate. Although this additive white Gaussian noise (AWGN) model is more realistic than a noiseless model, it is conceivable (as pointed out by one of the reviewers) that other stochastic channel models might be more suitable for certain types of sensor networks, and we leave this exploration for future research.\n\nGiven this set-up, of interest to us are stochastic algorithms that generate sequences of iterates contained within , and we require that the algorithm be graph-respecting, meaning that in each iteration, it is allowed to send at most one message for each direction of every edge . At time , we measure the distance between and the desired average via the average (per node) mean-squared error, given by\n\n MSE(\u03b8(t)) :=1nn\u2211i=1E[(\u03b8i(t)\u2212\\makebox[0.0pt][l]\u03b8)2]. (1)\n\nIn this paper, our goal is for every node to compute the average up to an error tolerance . In addition, we require almost sure consensus among nodes, meaning\n\n P[\u03b8i(t)=\u03b8j(t)\u2200i,j=1,2,\u22ef,n]\u21921as\u00a0t\u2192\u221e.\n\nOur primary goal is in characterizing the rate of convergence as a function of the graph topology and the number of nodes, to which we refer as the network-scaling function of the algorithm. More precisely, in order to study this network scaling, we consider sequences of graphs indexed by the number of nodes . For any given algorithm (defined for each graph ) and a fixed tolerance parameter , our goal is to determine bounds on the quantity\n\n TG(n;\u03b4) :=inf{t=1,2,\u2026\u2223MSE(\u03b8(t))\u2264\u03b4}. (2)\n\nNote that is a stopping time, given by the smallest number of iterations required to obtain mean-squared error less than on a graph of type with nodes.\n\n### Ii-B Graph topologies\n\nOf course, the question that we have posed will depend on the graph type, and this paper analyzes three types of graphs, as shown in Figure\u00a01. The first two graphs have regular topologies: the single cycle graph in panel (a) is degree two-regular, and the two-dimensional grid graph in panel (b) is degree four-regular. In addition, we also analyze an important class of random graphs with irregular topology, namely the class of random geometric graphs. As illustrated in Figure\u00a01(c), a random geometric graph (RGG) in the plane is formed according by placing nodes uniformly at random in the unit square , and the connecting two nodes if their Euclidean distance is less than some radius . It is known that an RGG will be connected with high probability as long as ; see Penrose\u00a0[26] for discussion of this and other properties of random geometric graphs.\n\nA key graph-theoretic parameter relevant to our analysis is the graph diameter, denoted by . The path distance between any pair of nodes is the length of the shortest path joining them in the graph, and by definition, the graph diameter is the maximum path distance taken over all node pairs in the graph. It is straightforward to see that for the single cycle graph, and that for the two-dimensional grid. For a random geometric graph with radius chosen to ensure connectivity, it is known that .\n\nFinally, in order to simplify the routing problem explained later, we divide the unit square into subregions (squares) of side length in case of grid, and for some constant , of side length in case of RGG. We assume that each node knows its location and is aware of the center of these subregions namely , where for the regular grid, and for the RGG. As a convention, we assume that is the left bottom square, to which we refer to as the first square. By construction, in a regular grid, each square will contain one and only one node which is located at the center of the square. From known properties of RGGs\u00a0[26, 17], each of the given subregions will contain at least one node with high probability (w.h.p.). Moreover, an RGG is regular w.h.p, meaning that each square contains nodes (see Lemma 1 in the paper\u00a0[12]). Accordingly, in the remainder of the paper, we assume without loss of generality that any given RGG is regular. Note that by construction, the transmission radius is selected so that each node in each square is connected to every other node in four adjacent squares.\n\n## Iii Algorithm and its properties\n\nIn this section we state our main result which is followed by a detailed description of the proposed algorithm.\n\n### Iii-a Theoretical guarantees\n\nOur main result guarantees the existence of a graph-respecting algorithm with desirable properties. Recall the definition of the graph respecting scheme, as well as the definition of our AWGN channel model given in Section\u00a0II. In the following statement, the quantity denotes a universal constant, independent of , , and .\n\n###### Theorem 1.\n\nFor the communication model in which each link is an AWGN channel with variance , there is a graph-respecting algorithm such that:\n\n1. Nodes almost surely reach a consensus. More precisely, we have\n\n \u03b8(t)\\lx@stackrela.s.\u27f6\u02dc\u03b8\u21921as\u00a0t\u2192\u221e, (3)\n\nfor some .\n\n2. After iterations, the algorithm satisfy the following bounds on the :\n\n1. For fixed tolerance sufficiently small, we have after\n\n Tcyc(n;\u03b4)\u2264c0nmax{1\u03b4log1\u03b4,MSE(\u03b8(0))\u03c32\u03b42}\n\niterations for a single cycle graph.\n\n2. For fixed tolerance sufficiently small, we have after\n\n Tgrid(n;\u03b4)\u2264c0\u221anmax{1\u03b4log1\u03b4,MSE(\u03b8(0))\u03c32\u03b42}\n\niterations for the regular grid in two dimensions.\n\n3. Assume that , for some fixed sufficiently small. Then we have after\n\n TRGG(n;\u03b4)\u2264c0\u221an(logn)3max{1\u02dc\u03b4log(logn)2\u02dc\u03b4,MSE(\u03b8(0))\u03c32\u02dc\u03b42}\n\niterations for a regular random geometric graph.\n\nHere is some constant independent of , , and , whose value may change from line to line.\n\nRemarks: A few comments are in order regarding the interpretation of this result. First, it is worth mentioning that the quality of the different links does not have to be the same. Similar arguments apply to the case where noises have different variances. Second, although nodes almost surely reach a consensus, as guaranteed in part (a), this consensus value is not necessarily the same as the sample mean . The choice of is intentional to emphasize this point. However, as guaranteed by part (b), this consensus value is within distance of the actual sample mean. Since the sample mean itself represents a noisy estimate of some underlying population quantity, there is little point to computing it to arbitrary accuracy. Third, it is worthwhile comparing part (b) with previous results on network scaling in the noisy setting. Rajagopal and Wainwright\u00a0[27] analyzed a simple set of damped updates, and showed that for the single cycle, and that for the two-dimensional grid. By comparison, the algorithm proposed here and our analysis thereof has removed factors of and from this scaling.\n\n### Iii-B Optimality of the results\n\nAs we now discuss, the scalings in Theorem\u00a01 are optimal for the cases of cycle and grid and near-optimal (up to logarithmic factor) for the case of RGG. In an adversarial setting, any algorithm needs at least iterations, where denotes the graph diameter, in order to approximate the average; otherwise, some node will fail to have any information from some subset of other nodes (and their values can be set in a worst-case manner). Theorem\u00a01 provides upper bounds on the number of iterations that, at most, are within logarithmic factors of the diameter, and hence are also within logarithmic factors of the optimal latency scaling law. For the graphs given here, the scalings are also optimal in a non-adversarial setting, in which are modeled as chosen i.i.d. from some distribution. Indeed, for a given node , and positive integer , we let denote the depth neighborhood of , meaning the set of nodes that are connected to by a path of length at most . We then define the graph spreading function . Note that the function is non-decreasing, so that we may define its inverse function . As some examples:\n\n\u2022 for a cycle on nodes, we have , and hence .\n\n\u2022 for a -grid in two dimensions, we have the upper bound , and hence the lower bound .\n\n\u2022 for a random geometric graph (RGG), we have the upper bound , which implies the lower bound\n\nAfter steps, a given node can gather the information of at most nodes. For the average based on nodes to be comparable to , we require that , and hence the iteration number should be at least . For the three graphs considered here, this leads to the same conclusion, namely that iterations are required. We note also that using information-theoretic techniques, Ayaso et al.\u00a0[1] proved a lower bound on the number of iterations for a general graph in terms of the Cheeger constant [9]. For the graphs considered here, the Cheeger constant is of the order of the diameter.\n\n### Iii-C Description of algorithm\n\nWe now describe the algorithm that achieves the bounds stated in Theorem\u00a01. At the highest level, the algorithm can be divided into two types of phases: an inner phase, and an outer phase. The outer phase produces a sequence of iterates , where is the outer time scale parameter. By design of the algorithm, each update of the outer parameters requires a total of message-passing rounds (these rounds corresponding to the inner phase), where in each round the algorithm can pass at most two messages per edge (one for each direction). To put everything in a nutshell, the algorithm is based on establishing multiple routes, averaging along them in an inner phase and updating the estimates based on the noisy version of averages along routes in an outer phase. Consequently, if we use the estimate , then in the language of Theorem\u00a01, it corresponds to rounds of message-passing. Our goal is to establish upper bounds on that guarantee the MSE is . Figure 2 illustrates the basic operations of the algorithm.\n\n#### Outer phase\n\nIn the outer phase, we produce a sequence of iterates according to the recursive update\n\n \u03b8(\u03c4+1)=\u03b8(\u03c4)\u2212\u03f5(\u03c4){L(\u03c4)\u03b8(\u03c4)+v(\u03c4)}. (4)\n\nHere is a sequence of positive decreasing stepsizes. For a given precision, , we set . For each , the quantity is a random matrix, whose structure is determined by the inner phase, and is an additive Gaussian term, whose structure is also determined in the inner phase. As will become clear in the sequel, even though and are dependent, they are both independent of . Moreover, given , the random vector is Gaussian with bounded variance.\n\n#### Inner phase\n\nThe inner phase is the core of the algorithm and it involves a number of steps, as we describe here. We use to index the iterations within any inner phase, and use to denote the sequence of inner iterates within . For the inner phase corresponding to outer update from , the inner phase takes the initialization , and then reduces as output to the outer iteration. In more detail, the inner phase can be broken down into three steps, which we now describe in detail.\n\nStep 1, deciding the averaging direction The first step is to choose a direction in which to perform averaging. In a single cycle graph, since left and right are viewed as the same, there is only one choice, and hence nothing to be decided. In contrast, the grid or RGG graphs require a decision-making phase, which proceeds as follows. One node in the first (bottom left) square, wakes up and chooses uniformly at random to send in the horizontal or vertical direction. We code this decision using the random variable , where (respectively ) represents the horizontal (respectively vertical) direction. To simplify matters, we assume in the remainder of this description that the averaging direction is horizontal, with the modifications required for vertical averaging being standard.\n\nStep 2, choosing the head nodes This step applies only to the grid and RGG graphs. Given our assumption that the node in the first square has chosen the horizontal direction, it then passes a token message to a randomly selected node in the above adjacent square. The purpose of this token is to determine which node (referred to as the head node) should be involved in establishing the route passing through the given square. After receiving the token, the receiving node passes it to another randomly selected node in the above adjacent square and so on. Note that in the special case of grid, there is only one node in each square, and so no choices are required within squares. After rounds, one node in each square () receives the token, as illustrated in Figure\u00a03. Note that again in a single cycle graph, there is nothing to be decided, since the direction and head nodes are all determined.\n\nStep 3, establishing routes and averaging In this phase, each of head nodes establishes a horizontal path, and then perform averaging along the path, as illustrated in Figure\u00a03(b). This part of algorithm involves three substeps, which we now describe in detail.\n\n\u2022 For , each head node selects a node uniformly at random (u.a.r.) from within the right adjacent square, and passes to it the quantity . Given the Gaussian noise model, node then receives the quantity\n\n \u02dc\u03b31j(1) =\u03b31j(1)+v1j,where\u00a0v1j\u223cN(0,\u03c32),\n\nand then updates its own local variable as . We then iterate this same procedure\u2014that is, node selects another u.a.r. from its right adjacent square, and passes the message . Overall, at round of this update procedure, we have\n\n \u03b3(i+1)j(i+1) =\u03b3(i+1)j(i)+\u02dc\u03b3ij(i),\n\nwhere , and . At the end of round , node can compute a noisy version of the average along the path , in particular via the rescaled quantity\n\n \u03b7j:=\u03b3mj(m)m=1mm\u2211i=1\u03b8sij(t)+vjj=1,2,\u22ef,m.\n\nHere the variable , since the noise variables associated with different edges are independent.\n\n\u2022 At this point, for each , each node which has the noisy version, , of the path average along route ; can share this information with other nodes in the path by sending back to the head node. A naive way to do this is as follows: node makes copies of \u2014namely, , \u2014and starts transmitting one copy at a time back to the head node. Nodes along the path simply forward what they receive, so that after time steps, node receives noisy copies of the average, where . Averaging the copies, node can compute the quantity\n\n \u03b3ij(3m\u2212i\u22121) :=1mm\u2211l=1\u02dc\u03b7(l)ij=1mm\u2211l=1\u03b8slj(\u03c4)+wij,\n\nwhere . Since the noise on different links and different time steps are independent Gaussian random variables, we have , with\n\n \u03c32i =1m\u03c32+(1\u2212im)\u03c32=(1\u2212(i\u22121)m)\u03c32\u2264\u03c32.\n\nTherefore, at the end of rounds, for each , all nodes have the average of the estimates in the path that is perturbed by Gaussian noise with variance at most . Since , we have .\n\n\u2022 At the end of the inner phase , nodes that were involved in a path use their estimate of the average along the path to update , while estimate of the nodes that were not involved in any route remain the same. A given node on a path updates its estimate via\n\n \u03b8sij(\u03c4+1)={1\u2212\u03f5\u2032(\u03c4)}\u03b8sij(\u03c4)+\u03f5\u2032(\u03c4)\u03b3ij(M), (5)\n\nwhere . On the other hand, using to denote the Euclidean inner product, we have , where is the averaging vector of the route with the entries for , and zero otherwise. Combining the scalar updates\u00a0(5) yields the matrix-form update\n\n \u03b8(\u03c4+1) =\u03b8(\u03c4)\u2212\u03f5\u2032(\u03c4){(I\u2212W(\u03c4))\u03b8(\u03c4)+v\u2032(\u03c4)}, (6)\n\nwhere the matrix is a random averaging matrix induced by the choice of routes and the random directions . The noise vector is additive noise. Note that for any given time, the noise at different nodes are correlated via the matrix , but for different time instants , the noise vectors and are independent. Moreover, from our earlier arguments, we have the upper bound .\n\n## Iv Proof of Theorem\u00a01\n\nWe now turn to the proof of Theorem\u00a01. At a high-level, the structure of the argument consists of decomposing the vector into a sum of two terms: a component within the consensus subspace (meaning all values of the vector are identical), and a component in the orthogonal complement. Using this decomposition, the mean-squared error splits into a sum of two terms and we use standard techniques to bound them. As will be shown, these bounds depend on the parameter , noise variance, the initial MSE, and finally the (inverse) spectral gap of the update matrix. The final step is to lower bound the spectral gap of our update matrix.\n\n### Iv-a Setting up the proof\n\nRecalling the averaging matrix from the update\u00a0(6), we define the Laplacian matrix . We then define the average matrix , where the expectation is taken place over the randomness due to the choice of routes;3 in a similar way, we define the associated (average) Laplacian . Finally, we define the rescaled quantities\n\n \u03f5(\u03c4):=\u03bb2(\\makebox[0.0pt][l]S)\u03f5\u2032(\u03c4),L(\u03c4):=1\u03bb2(\\makebox[0.0pt][l]S)S(\u03c4),andv(\u03c4):=1\u03bb2(\\makebox[0.0pt][l]S)v\u2032(\u03c4), (7)\n\nwhere we recall that denotes the second smallest eigenvalue of a symmetric matrix. In terms of these rescaled quantities, our algorithm has the form\n\n \u03b8(\u03c4+1)=\u03b8(\u03c4)\u2212\u03f5(\u03c4)[L(\u03c4)\u03b8(\u03c4)+v(\u03c4)], (8)\n\nas stated previously in the update equation\u00a0(4). Moreover, by construction, we have where . We also, for theoretical convenience, set\n\n \u03f5\u2032(\u03c4)=1\u03bb2(\u00afS)(\u03c4+1\u03b4), (9)\n\nor equivalently for .\n\nWe first claim that the matrix is symmetric and (doubly) stochastic. The symmetry follows from the fact that different routes do not collide, whereas the matrix is stochastic because every row of (depending on whether the node corresponding to that row participates in a route or not) either represents an averaging along a route or is the corresponding row of the identity matrix. Consequently, we can interpret as the transition matrix of a reversible Markov chain. It is an irreducible Markov chain, because within any updating round, there is a positive chance of averaging nodes that are in the same column or row, which implies that the associated Markov chain can transition from one state to any other in at most two steps. Moreover, the stationary distribution of the chain is uniform (i.e., ).\n\nWe now use these properties to simplify our study of the sequence generated by the update equation\u00a0(8). Since is real and symmetric, it has the eigenvalue decomposition , where is a unitary matrix (that is, ). Moreover, we have , where is the eigenvalue corresponding to the eigenvector , for . Since , the eigenvalues of and are related via\n\n = =\n\nSince the largest eigenvalue of an irreducible Markov chain is one (with multiplicity one)\u00a0[16], we have , or equivalently\n\nwith . Moreover, we have , so that the first eigenvector corresponds to the eigenvalue . Let denote the matrix obtained from by deleting its first column, . Since the smallest eigenvalue of is zero, we may write , where , , and . With this notation, our analysis is based on the decomposition\n\n \u03b8(\u03c4) =\u03b1(\u03c4)\u21921\u221an+\u02dcU\u03b2(\u03c4), (10)\n\nwhere we have defined and . Since for all , from the decomposition\u00a0(10) and the form of the updates\u00a0(8), we have the following recursions,\n\n \u03b1(\u03c4+1) =\u03b1(\u03c4)\u2212\u03f5(\u03c4)\u21921T\u221anv(\u03c4),and (11)\n \u03b2(\u03c4+1) =\u03b2(\u03c4)\u2212\u03f5(\u03c4)(L\u2013\u2013(\u03c4)\u03b2(\u03c4)+\u02dcUTv(\u03c4)). (12)\n\nHere is an matrix defined by the relation\n\n UTL(\u03c4)U =[0\u21920T\u21920L\u2013\u2013(\u03c4)]n\u00d7n.\n\n### Iv-B Main steps\n\nAs we show, part (a) of the theorem requires some intermediate results of the proof of part (b). Accordingly, we defer it to the end of the section. With this set-up, we now state the two main technical lemmas that form the core of Theorem\u00a01. Our first lemma concerns the behavior of the component sequences and which evolve according to equations (11) and (12) respectively.\n\n###### Lemma 2.\n\nGiven the random sequence generated by the update equation (4), we have\n\n MSE(\u03b8(\u03c4)) =1nvar(\u03b1(\u03c4))\ue152\ue154\ue154\ue154\ue154\ue154\ue154\ue151\ue150\ue154\ue154\ue154\ue154\ue154\ue154\ue153e1(\u03c4)+1nE[\u2225\u03b2(\u03c4)\u222522]\ue152\ue154\ue154\ue154\ue154\ue154\ue154\ue154\ue151\ue150\ue154\ue154\ue154\ue154\ue154\ue154\ue154\ue153e2(\u03c4). (13)\n\nFurthermore, and satisfy the following bounds:\n\n1. For each iteration , we have\n\n e1(\u03c4) \u2264\u03c32\u03b4[\u03bb2(\u00afS)]2. (14)\n2. Moreover, for each iteration we have\n\n e2(\u03c4) \u2264\u03c32[\u03bb2(\u00afS)]2log(\u03c4+1\u03b4\u22121)\u03c4+1\u03b4\u22121+e2(0)1\u03b4\u22121\u03c4+1\u03b4\u22121, (15)\n\nFrom Lemma\u00a02, we conclude that in order to guarantee an bound on the MSE, it suffices to take such that\n\n 1\u03b4\u22121\u03c4+1\u03b4\u22121\u2264\u03c32\u03b4e2(0)[\u03bb2(\u00afS)]2,andlog(\u03c4+1\u03b4\u22121)\u03c4+1\u03b4\u22121\u2264\u03b4.\n\nNote that the first inequality is satisfied when . Moreover, doing a little bit of algebra, one can see that is sufficient to satisfy the second inequality. Accordingly, we take\n\n \u03c4=max{2\u03b4log1\u03b4,e2(0)[\u03bb2(\u00afS)]2\u03c32\u03b42}\n\nouter iterations.\n\nThe last part of the proof is to bound the second smallest eigenvalue of the Laplacian matrix . The following lemma, which we prove in Section\u00a0IV-D to follow, addresses this issue. Recall that denotes the second smallest eigenvalue of a matrix.\n\n###### Lemma 3.\n\nThe averaged matrix that arises from our protocol has the following properties:\n\n1. For a cycle and a regular grid we have , and\n\n2. for a random geometric graph, we have , with high probability.\n\nIt is important to note that the averaged matrix is not the same as the graph Laplacian that would arise from standard averaging on these graphs. Rather, as a consequence of establishing many paths and averaging along them in each inner phase, our protocol ensures that the matrix behaves essentially like the graph Laplacian for the fully connected graph.\n\nAs established previously, each outer step requires iterations. Therefore, we have shown that it is sufficient to take a total of\n\n T=O(Dnmax{2\u03b4log1\u03b4,e2(0)[\u03bb2(\u00afS)]2\u03c32\u03b42})\n\ntransmissions per edge in order to guarantee a bound on the MSE. As we will see in the next section, assuming that the initial values are fixed, we have , hence . The claims in Theorem\u00a01 then follow by standard calculations of the diameters of the various graphs and the result of the Lemma 3.\n\nIt remains to prove the two technical results, Lemma\u00a02 and\u00a03, and we do so in the following sections.\n\n### Iv-C Proof of Lemma\u00a02\n\nWe begin by observing that\n\n E[(\u03b8(\u03c4)\u2212\u00af\u03b8\u21921)(\u03b8(\u03c4)\u2212\u00af\u03b8\u21921)T] =F1+F2+F3,\n\nwhere , the second term is given by , and\n\n F3:=E[(\u03b1(\u03c4)\u2212\u221an\u00af\u03b8)\u21921\u221an\u03b2(\u03c4)T\u02dcUT]+E[(\u03b1(\u03c4)\u2212\u221an\u00af\u03b8)\u02dcU\u03b2(\u03c4)\u21921T\u221an].\n\nSince has orthonormal columns, all orthogonal to the all one vector (), it follows that , and .\n\nIt remains to compute . Unwrapping the recursion\u00a0(11) and using the fact that initialization implies yields\n\n \u03b1(\u03c4) (16)\n\nfor all . Since , , are zero mean random vectors, from equation (16) we conclude that 4 and accordingly, . Recalling the definition of the MSE (1) and combining the pieces yields the claim (13).\n\n(a) From equation (16), it is clear that each is Gaussian with mean . It remains to bound the variance. Using the i.i.d. nature of the sequence , we have\n\n var(\u03b1(\u03c4)) =E[(\u03c4\u22121\u2211l=0\u03f5(l)\u27e8\u21921\u221an,v(l)\u27e9)2] =\u03c4\u22121\u2211l=0\u03f5(l)2n\u27e8\u21921,C\u21921\u27e9 =\u03c4\u22121\u2211l=0\u03f5\u2032(l)2\u27e8\u21921,C\u2032\u21921\u27e9n,\n\nwhere we have recalled the rescaled quantities\u00a0(7). Recalling the fact that and using the Cauchy-Schwarz inequality, we have . Hence, we obtain\n\n var(\u03b1(\u03c4)) \u2264n\u03c32\u03c4\u22121\u2211l=0\u03f5\u2032(l)2 =n\u03c32[\u03bb2(\u00afS)]2\u03c4\u22121\u2211l=01(1\u03b4+l)2 \u2264n\u03c32[\u03bb2(\u00afS)]2\u222b\u221e1\u03b41x2dx=n\u03c32\u03b4[\u03bb2(\u00afS)]2;\n\nfrom which rescaling by establishes the bound\u00a0(14).\n\n(b) Defining , the update equation (12) can be written as\n\n \u03b2(\u03c4+1)=\u03b2(\u03c4)\u2212\u03f5(\u03c4)H(\u03b2(\u03c4),v(\u03c4)),\n\nfor . In order to upper bound , defined in (13), we need to control . Doing some algebra yields\n\n e2(\u03c4+1)\u2212e2(\u03c4) =1nE[\u27e8\u03b2(\u03c4+1)\u2212\u03b2(\u03c4),\u03b2(\u03c4+1)+\u03b2(\u03c4)\u27e9] =1nE[\u27e8\u2212\u03f5(\u03c4)H(\u03b2(\u03c4,v(\u03c4))),\u2212\u03f5(\u03c4)H(\u03b2(\u03c4,v(\u03c4)","date":"2020-10-27 07:16:13","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\": 0.8846599459648132, \"perplexity\": 322.84338344347293}, \"config\": {\"markdown_headings\": true, \"markdown_code\": true, \"boilerplate_config\": {\"ratio_threshold\": 0.18, \"absolute_threshold\": 10, \"end_threshold\": 15, \"enable\": true}, \"remove_buttons\": true, \"remove_image_figures\": true, \"remove_link_clusters\": true, \"table_config\": {\"min_rows\": 2, \"min_cols\": 3, \"format\": \"plain\"}, \"remove_chinese\": true, \"remove_edit_buttons\": true, \"extract_latex\": true}, \"warc_path\": \"s3:\/\/commoncrawl\/crawl-data\/CC-MAIN-2020-45\/segments\/1603107893402.83\/warc\/CC-MAIN-20201027052750-20201027082750-00531.warc.gz\"}"} | null | null |
\section{Introduction}
\label{sec_intro}
Rigorous information on the QCD phase diagram, schematically depicted
in Fig.~\ref{fig_phasedia}, is scarce. At finite temperature ($T$) and
vanishing quark chemical potential ($\mu_q=0$), lattice QCD (lQCD)
computations predict the existence of a rapid cross-over transition
from hadronic matter to a Quark-Gluon Plasma at a (pseudo-) critical
temperature of $T_c=(180\pm20)$~MeV~\cite{Cheng:2007jq,Aoki:2006br}.
\begin{figure}[!b]
\begin{center}
\includegraphics[width=0.60\textwidth,angle=-90]{phase2.eps}
\end{center}
\caption[]{Schematic diagram of the QCD phase structure in the $\mu_q$-$T$
plane, as characterized by the lowest dimension quark condensates, i.e.,
the chiral quark-antiquark and the diquark one (Cooper pairs). The former
is believed to prevail in the low-$T$ and -$\mu_q$ hadronic world while
the latter occurs at the Fermi surface of cold dense quark matter. Key to
investigating the phase structure is the identification of the relevant
excitations in the respective phases, as indicated in the figure. The
``data'' points are extracted from hadro-chemical analysis of the final
state observed in heavy-ion collisions~\cite{BraunMunzinger:2003zd,
Becattini:2003wp}.
}
\label{fig_phasedia}
\end{figure}
Heavy-ion collisions at SPS and RHIC energies have shown that the produced
medium exhibits a high degree of thermalization and that the achieved
temperatures are on the order of $T_c$ and above. This provides a firm
basis for studying QCD matter in the laboratory. At finite $\mu_q$ and
vanishing $T$, there are compelling arguments that the high-density limit
of QCD is in a color-superconducting phase (attractive quark-quark
interaction plus Cooper theorem). Since this ground state, characterized
by a non-vanishing diquark condensate ($\langle qq\rangle \ne 0$), is
qualitatively different from the chirally broken QCD vacuum
($\langle \bar{q}q\rangle \ne 0$), it is natural to expect a first-order
transition at some intermediate chemical potential, $\mu_q^c\simeq400$~MeV.
Finally, heavy atomic nuclei have been long used to quantify the
properties and excitatins of the finite-density ground state at
$\mu_q = (m_N-E_B)/3 \simeq 310$~MeV, indicated by the square box in
Fig.~\ref{fig_phasedia}.
Heavy-ion experiments performed over a wide range of collision energies
can, in principle, cover a large region of the phase diagram, cf.~the
``data'' points in Fig.~\ref{fig_phasedia}. In particular, as indicated
above, one ought to be able to determine phase changes, down to
temperatures of about 100~MeV where the critical chemical potential may
be around $\mu_q^c\simeq400$~MeV (it is, however, questionable
whether heavy-ion experiments can reach into the color-superconducting
phases, unless the zero-temperature quark pairing gap is well above
100~MeV~\cite{Rapp:1999qa}; even if so, equilibration appears to be
unlikely. More likely could be the production of the so-called
quarkyonic phase~\cite{McLerran:2007qj} which may extend to higher $T$).
Of special interest is the occurrence of a critical second-order
endpoint, whose existence is suggested by the cross-over transition
found in finite-$T$ lQCD and the putative 1.~order transition at $T=0$.
Suitable observables to study the QCD phase structure in heavy-ion
collisions (HICs) may be roughly divided into two categories:
(1) bulk observables driven by the equation of state (EoS), including
collective flow patterns of the most abundant particles ($\pi$, $K$, $p$)
or fluctuation observables in $p_T$ spectra and hadrochemistry;
(2) ``microscopic'' probes associated with specific quantum-number
channels, e.g., the vector current coupling to photons and dileptons.
In some instances the physical origin of type (1) and (2) observables is
closely related. E.g., electric-charge fluctuations are governed by the
electromagnetic (EM) susceptibility, $\chi_{\rm em}$, which can be
expressed as the screening limit of the static EM correlation
function, $\langle Q^2 \rangle - \langle Q\rangle^2
= \chi_{\rm em} = \Pi_{\rm em}(q_0=0,q\to 0)$, while photon and
dilepton spectra are directly proportional to the imaginary part
of $\Pi_{\rm em}$, as discussed below.
In this paper we focus on microscopic probes as realized via dileptons,
charm and charmonia. Corresponding observables are often associated with
``hard probes'', due to a large momentum transfer associated with their
initial production (e.g., $|q^2| \ge 4m_c^2 \simeq6$~GeV$^2$). We will
argue, however, that all of the above 3 probes can provide valuable
information on relatively ``soft'' modes in the medium, at the
temperature scale or below. For dileptons (Sec.~\ref{sec_dilep}), this
relates to the in-medium $\rho$-meson spectral function as reflected in
thermal radiation at low invariant masses ($M\sim {\cal O}(T)\le m_\rho$).
In the open-charm sector (Sec.~\ref{sec_charm}), one can study mechanisms
of thermalization or, more generally, transport properties within a
diffusion equation via modifications of charmed hadron $p_T$ spectra and
elliptic flow (governed by elastic scattering at typical momentum
transfers $|q|\sim {\cal O}(gT)$). Finally, for charmonia
(Sec.~\ref{sec_charmonium}), the key to understanding their in-medium
bound-state properties lies in color-screening effects (at the Debye
scale $m_D\sim{\cal O}(gT)$) as well as inelastic dissociation
reactions (at the binding-energy scale). In each sector, based on current
insights, we will try to identify promising directions of future
investigations specific to the situation of finite $\mu_q$ and/or the
putative critical point.
Concluding remarks are collected in Sec.~\ref{sec_concl}.
\section{Low-Mass Dileptons: $\rho$-Meson Spectroscopy}
\label{sec_dilep}
The basic quantity to calculate thermal emission spectra of
EM radiation from hot and dense matter is the retarded correlation
function of the hadronic EM current, $j^\mu_{\rm em}$,
\begin{equation}
\Pi_{\rm em}^{\mu \nu}(M,q;\mu_B,T) = -i \int d^4x \ e^{iq\cdot x} \
\Theta(x_0) \ \langle[j_{\rm em}^\mu(x), j_{\rm em}^\nu(0)]\rangle_T \ .
\end{equation}
Its imaginary part (EM spectral function) directly figures into the
differential production rates of dileptons ($l^+l^-$) and photons
($\gamma$),
\begin{eqnarray}
\frac{dN_{ll}}{d^4xd^4q} &=& -\frac{\alpha_{\rm em}^2}{\pi^3} \
\frac{L(M)}{M^2} \ f^B(q_0;T) \ {\rm Im}~\Pi_{\rm em} (M,q;\mu_B,T)
\label{Rll}
\\
q_0\frac{dN_{\gamma}}{d^4xd^3q} &=& -\frac{\alpha_{\rm em}}{\pi^2} \
f^B(q_0;T) \ {\rm Im}~\Pi_{\rm em}(M=0,q;\mu_B,T) \ ,
\label{Rgam}
\end{eqnarray}
respectively ($f^B$ is the thermal Bose distribution and $L(M)$ a
final-state lepton phase-space factor relevant close to the dilepton
threshold). In the vacuum, the low-mass regime ($M\le 1$~GeV) of
$\Pi_{\rm em}$ is essentially saturated by the light vector mesons
$\rho$, $\omega$ and $\phi$. Within the vector-dominance model (VDM)
the EM spectral function is directly proportional to the vector-meson
spectral functions,
\begin{equation}
{\rm Im}~\Pi_{\rm em} \sim [ {\rm Im}~D_\rho +
\frac{1}{9} {\rm Im}~D_\omega + \frac{2}{9} {\rm Im}~D_\phi ] \ .
\label{vdm}
\end{equation}
Thus, if VDM remains valid in the medium (see, e.g.,
Ref.~\cite{Harada:2003jx} for an alternative scheme), low-mass dilepton
spectra mostly probe in-medium modifications of the $\rho$ meson, which
have been studied rather extensively in the literature, see, e.g.,
Refs.~\cite{Rapp:2009yu,Leupold:2009kz} for recent reviews.
It turns out that low-mass thermal EM radiation in HICs dominantly
emanates from the hadronic phases of the collisions, even at RHIC
energies~\cite{David:2006sr}. It is therefore in order to study
hadronic medium effects on the $\rho$ propagator,
\begin{equation}
D_\rho(M,q;\mu_B,T) =
[ M^2-{m_\rho^{(0)}}^2-\Sigma_{\rho\pi\pi}-\Sigma_{\rho B}-
\Sigma_{\rho M} ]^{-1} \ ,
\end{equation}
encoded in selfenergy insertions, $\Sigma_\rho$, induced by interactions
with particles in the heat bath. These may be classified as
(a) medium modifications of the pion cloud, $\Sigma_{\rho\pi\pi}$,
due to pion rescattering (most notably on baryons) and thermal Bose
enhancement~\cite{Urban:1999im};
(b) direct $\rho$-baryon couplings~\cite{Friman:1997tc}, e.g.,
$\rho+N\to \Delta, N(1520), N(1720)$, etc.;
(c) direct interactions of the $\rho$ with mesons, e.g.,
$\rho+\pi\to \omega,a_1,...$ or $\rho+K\to K_1,...$ etc.~\cite{Rapp:1999qu}.
The interactions are usually modeled by effective hadronic Lagrangians
which satisfy basic constraints from EM gauge invariance and (mostly
for pions) chiral symmetry. The free parameters (coupling constants and
formfactor cutoffs to account for finite-size effects) can be
constrained empirically by partial decay rates (e.g.,
$a_1\to \pi\rho, \pi\gamma$) or, more comprehensively,
scattering data (e.g., $\pi N\to \rho N$ or photo-absorption cross
sections).
\begin{figure}[!t]
\begin{minipage}{0.5\linewidth}
\includegraphics[width=0.78\textwidth,angle=-90]{Arho69PbT300c.eps}
\end{minipage}
\hspace{0.3cm}
\begin{minipage}{0.5\linewidth}
\vspace{-0.1cm}
\includegraphics[width=0.95\textwidth]{drdm2-Inx160.eps}
\end{minipage}
\vspace{0.1cm}
\caption[]{Left panel: $\rho$-meson spectral function in hot and dense
hadronic matter at fixed $\mu_B=3\mu_q=330$~MeV (corresponding to
baryon densities $\varrho_B/\varrho_0=0.1,0.7,2.6$ at $T=120,150,180$~MeV,
respectively) and 3-momentum $q=0.3$~GeV~\cite{Rapp:1999us}.
Right panel: 3-momentum
integrated thermal dielectron rates in the isovector channel using the
vacuum (dotted line) and full in-medium (solid line) $\rho$ spectral
function; the long-dashed, dash-dotted and short-dashed curves only
include in-medium selfenergies due to either the in-medium pion cloud
($\Sigma_{\rho\pi\pi}$) or direct $\rho$-baryon interactions
($\Sigma_{\rho B}$) or direct $\rho$-meson interactions
($\Sigma_{\rho M}$), respectively (the latter 2 include the free pion
cloud as well).}
\label{fig_arho}
\end{figure}
The left panel of Fig.~\ref{fig_arho} shows an in-medium $\rho$ spectral
functions~\cite{Rapp:1999us} including all of the above components under
conditions roughly resembling HICs at the SPS. A strong broadening with
increasing matter density and temperature occurs, melting the resonance
when extrapolated toward the phase boundary region.
The large low-mass enhancement becomes much more apparent in the
dilepton production rate, due to the Bose factor and photon propagator
in eq.~(\ref{Rll}), cf.~right panel of Fig.~\ref{fig_arho}. At
$M\simeq0.4$~GeV, an order of magnitude enhancement over the rate
from free $\pi\pi$ annihilation is predicted. Also note that the
divergence of the rate for $M\to 0$ is required to produce a finite
photon production rate, eq.~(\ref{Rgam}). Plotting the rate in terms
of the 3 individual selfenergy contributions as introduced above, one
clearly recognizes the prevalence of the baryon-driven medium effects.
This may seem surprising since at SPS energies the observed
pion-to-baryon ratio is about 5:1. However, in the interacting medium,
most of the pions are ``stored'' in excited (meson and baryon)
resonances; e.g., at ($\mu_B$,$T$)=(240,160)~MeV, the total baryon
density, $\varrho_B\simeq0.8\varrho_0$ is quite comparable to the direct
pion density, $\varrho_\pi\simeq0.9\varrho_0$ ($\varrho_0=0.16~$fm$^{-3}$).
An application of the $\rho$ spectral function to recent NA60 low-mass
dimuon data in In-In collisions at SPS~\cite{Arnaldi:2006jq} is shown
in the left panel of Fig.~\ref{fig_dndm}.
\begin{figure}[!b]
\begin{minipage}{0.5\linewidth}
\includegraphics[width=0.95\textwidth]{dndm-semi2.eps}
\end{minipage}
\begin{minipage}{0.5\linewidth}
\includegraphics[width=0.95\textwidth]{Gam-emsf-SPS.eps}
\end{minipage}
\caption[]{Left panel: dimuon invariant-mass spectra (including experimental
acceptance) in semicentral In-In collisions at SPS energies ($E_{\rm lab}=
158$~AGeV); NA60 data~\cite{Arnaldi:2006jq} are compared to theoretical
calculations based on the in-medium $\rho$ spectral function shown
in Fig.~\ref{fig_arho}.
Right panel: in-medium $\rho$ width as a function of temperature along
isentropic trajectories in the phase diagram starting from chemical
equilibrium at $(T,\mu_B)=(175,230)$~MeV (dots: fixed hadrochemistry
including chemical potentials for pions, kaons, etc.; squares: chemical
equilibrium; triangles: without baryonic medium effects).}
\label{fig_dndm}
\end{figure}
Upon convoluting the EM emission rates over an expanding fireball, the
excess radiation is well described by the predicted in-medium $\rho$
line shape (QGP and primordial contributions are
subleading)~\cite{vanHees:2007th}, see
Refs.~\cite{Dusling:2007kh,Ruppert:2007cr,Bratkovskaya:2008bf} for
alternative calculations. This is also true for the excess dielectron
data reported for central Pb-Au by CERES/NA45~\cite{Adamova:2006nu}.
One can quantify the $\rho$ broadening by an approximate average width
which amounts to
$\bar{\Gamma}_\rho^{\rm med}\simeq$~(350-400)~MeV~$\simeq
3~\Gamma_\rho^{\rm vac}$, realized at a representative temperature
of $\bar{T}\simeq$~150-160~MeV, cf.~right panel of Fig.~\ref{fig_dndm}.
This inevitably implies that, toward $T_c$, the $\rho$'s in-medium width
becomes comparable to its mass, i.e., the resonance has indeed melted.
The absolute yield of the excess radiation is quite sensitive to the
total fireball lifetime, enabling a remarkably accurate measurement
of the fireball lifetime for semicentral In-In collision,
$\tau_{\rm FB}\simeq (6.5\pm1)$~fm/$c$. This tool could become
invaluable for detecting significant lifetime changes when approaching
the critical point and/or moving into the first-order regime with an
extended mixed phase (of course, it only works if the in-medium
spectral shape is under sufficient theoretical control).
The NA60 collaboration has recently taken another step forward by
fully correcting their data for experimental
acceptance~\cite{Arnaldi:2008er,Arnaldi:2008fw}. Upon integrating over
transverse momentum, the resulting invariant-mass spectra,
$dN_{\mu\mu}/dMdy$, do justice to the notion of
Lorentz-{\em invariance}, i.e., transverse flow effects have been
eliminated. Thus, one is essentially looking at the (average) emission
rate from the medium, multiplied by the emitting 4-volume, cf.~left
panels in Fig.~\ref{fig_rate}.
\begin{figure}[!t]
\begin{minipage}{0.5\linewidth}
\vspace{-1.2cm}
\includegraphics[width=0.92\textwidth]{dndm-NA60acor-eosd.eps}
\vspace{-0.2cm}
\includegraphics[width=0.92\textwidth]{dndm-NA60acor02-eosd.eps}
\end{minipage}
\hspace{-0.4cm}
\begin{minipage}{0.5\linewidth}
\includegraphics[width=1.00\textwidth]{drdm-mu15-150-180.eps}
\end{minipage}
\caption[]{Left panel: acceptance-corrected dimuon invariant-mass spectra
in semicentral In(158~AGeV)-In~\cite{Arnaldi:2008er,Arnaldi:2008fw} for
transverse pair momenta $q_t=0.2$-2.4~GeV (upper left) and
$q_t=0$-0.2~GeV (lower left).
Right panel: 3-momentum integrated dimuon thermal emission rates in
the isovector ($\rho$) channel at a baryon chemical potential
representative for SPS energies ($\mu_B=330$~MeV)~\cite{Rapp:1999us}.}
\label{fig_rate}
\end{figure}
This provokes a direct comparison to the theoretical input rates based
on Ref.~\cite{Rapp:1999us}, augmented by the muon phase-space factor,
$L(M)$, shown in the right panel of Fig.~\ref{fig_rate} for two
temperatures. The resemblance of the in-medium hadronic rates and the
NA60 spectra is rather astonishing, both in slope and shape. The former
can, in principle, serve as a true thermometer, i.e. free from blue-shift
contamination due to transverse flow. Essential to these
arguments is the prevalence of thermal radiation in the excess spectra
which is borne out of (i) the theoretical calculations, and
(ii) the complete lack of polarization in the measured angular
distribution of the muon pairs~\cite{Arnaldi:2008gp}. The good overall
agreement of theory and data furthermore corroborates that VDM
stays intact even close to the phase boundary (the data also indicate
that the $\phi$ does not radiate dileptons in the hadronic
phase but decouples earlier~\cite{Adamova:2005jr,Arnaldi:2009wr}; this
is, in fact, consistent with its relatively soft $p_T$ spectra).
The importance of baryon-driven medium effects in the interpretation
of the SPS low-mass dilepton data naturally calls for studies at
lower collisions energies where even larger baryon compression
may be achieved.
\begin{figure}[!t]
\begin{minipage}{0.5\linewidth}
\includegraphics[width=0.8\textwidth,angle=-90]{ArhoM69PbT140.ps}
\end{minipage}
\begin{minipage}{0.5\linewidth}
\includegraphics[width=0.8\textwidth,angle=-90]{ArhoM69PbT170.ps}
\end{minipage}
\caption[]{$\rho$-meson spectral functions, weighted by a factor of
inverse mass as figuring into the 3-momentum integrated dilepton rate,
at two temperatures and for baryon densities representative for full SPS
energy (red lines) and CBM energies (blue lines).}
\label{fig_arhom}
\end{figure}
Hadronic many-body calculations identify the mass regime around
$M\simeq0.2$~GeV as the most sensitive one in the $\rho$ spectral
function, with up to a factor of $\sim$2 enhancement under conditions
expected at the Compressed Baryonic Matter (CBM) experiment relative
to full SPS energy. A first glimpse at such an effect may have been
seen by CERES/NA45 in a 40~AGeV Pb-Au run~\cite{Adamova:2002kf}.
A direct way to study baryon effects is provided by cold
nuclear matter, i.e., in atomic nuclei. The advantage over heavy-ion
experiments obviously lies in the essentially static matter environment,
which, however, is limited by nuclear saturation density ($\varrho_0$)
and also exhibits significant spatial gradients. Nevertheless, the
predicted medium effects on the $\rho$ spectral function are appreciable
even at half saturation density, at least at small 3-momentum relative
to the nuclear rest frame, see left panel of
Fig.~\ref{fig_cold}~\cite{Rapp:1999us,Riek:2008ct}.
\begin{figure}[!t]
\begin{minipage}{0.5\linewidth}
\includegraphics[width=1.0\textwidth]{Arhoc05q.eps}
\end{minipage}
\begin{minipage}{0.5\linewidth}
\vspace{-0.3cm}
\hspace{1.4cm}
\includegraphics[width=0.7\textwidth]{dia-rho-phot-pro.eps}
\end{minipage}
\caption[]{Left panel: in-medium $\rho$ spectral function in cold
nuclear matter at half saturation density for various 3-momenta (for
$q$>0, transverse and longitudinal modes split).
Right panel: elementary amplitude for $\rho$ photo-production on the
nucleon~\cite{Riek:2008ct}.
}
\label{fig_cold}
\end{figure}
As in HICs, the dilepton final state is the cleanest way to probe medium
effects. The initial state in nuclear production experiments is, however,
rather different: the $\rho$ has to be created by an external
excitation (cf.~right panel of Fig.~\ref{fig_cold}), as compared to an
approximately thermal medium in HICs.
Thus, a good knowledge of the production process is mandatory, which
can be tested with proton targets. Two additional complications arise:
(a) an in-medium broadening leads to a reduction in the dilepton to
hadronic branching ratio, thus reducing the signal (in HICs the $\rho$
is ``continuously" regenerated in the interacting medium);
(b) to provide the mass of the $\rho$ a rather energetic incoming
particle is needed which usually implies that the $\rho$ carries
significant 3-momentum; this enhances surface and/or escape effects
thus reducing the in-medium signal as well.
The latter point is presumably best dealt with using a photon beam
where all the incoming energy can be converted into mass.
Photo-production of dileptons has recently been studied by the CLAS
collaboration at Jefferson Lab (JLab) using a variety of nuclear
targets~\cite{Wood:2008ee}, with an incoming photon spectrum ranging
over $E_\gamma\simeq$~(1-3.5)~GeV.
\begin{figure}[!t]
\begin{minipage}{0.5\linewidth}
\includegraphics[width=0.93\textwidth]{clas-fe-spec.eps}
\end{minipage}
\begin{minipage}{0.5\linewidth}
\includegraphics[width=1.0\textwidth]{dndm-clas-Fe.eps}
\end{minipage}
\caption[]{Dilepton spectra measured in nuclear photo-production
by CLAS at JLAB~\cite{Wood:2008ee}, including transport model
calculations~\cite{Muhlich:2002tu} for the decays of
light vector mesons $\rho$, $\omega$ and $\phi$. Right panel:
CLAS ``excess spectra" compared to calculations~\cite{Riek:2008ct}
using the same $\rho$ spectral function as for the NA60 data in
Figs.~\ref{fig_dndm}, \ref{fig_rate}.}
\label{fig_clas}
\end{figure}
The left panel of Fig.~\ref{fig_clas} shows the dilepton signal from
iron targets, compared to Boltzmann transport
calculations~\cite{Muhlich:2002tu} for $\rho$, $\omega$ and $\phi$
decays. Since the $\omega$ and $\phi$ peaks are essentially unaffected
by the medium
(and well concentrated in mass), their contribution can be subtracted
from the spectrum (much like for the NA60 data) leading to an ``excess''
signal as shown by the data in the right panel of Fig.~\ref{fig_clas}.
Breit-Wigner fits have been applied resulting in a moderate $\rho$
broadening to $\Gamma_\rho^{\rm med}\simeq220$\,MeV~\cite{Wood:2008ee}.
The $\rho$ spectral function used in the NA60 context has been applied
to the CLAS experiment in combination with a realistic elementary
production amplitude and a somewhat schematic modeling of the spatial
propagation, accounting for the production kinematics and nuclear
density profile~\cite{Riek:2008ct}. The CLAS data are reasonably well
described; it turns out that for the Fe target the typical densities
and 3-momenta probed in the spectral
function are $\bar{\varrho}\simeq0.5\varrho_0$ and $\bar{q}\simeq2$~GeV.
The latter are the main reason for moderating the medium effects
in the current CLAS data (recall left panel of Fig.~\ref{fig_cold}).
However, at high momenta additional medium effects not included in the
spectral function of Ref.~\cite{Rapp:1999us} may occur, see, e.g.,
the discussion in Ref.~\cite{Rapp:1999ej}.
\section{Open Charm and Transport}
\label{sec_charm}
The masses of charm and bottom quarks (and hadrons) are much larger than
the typical temperatures realized in heavy-ion collisions at SPS and
RHIC, $m_{Q}\gg T$ ($Q$=$b$,$c$). Furthermore, a typical momentum transfer
from the heat bath, $|q^2|\simeq T^2$, is parametrically smaller than the
thermal momentum of $c$ and $b$ quarks, $p^2\sim 3m_{Q}T \gg |q^2|$.
Thus a diffusion approximation to the Boltzmann equation becomes
applicable leading to a Fokker-Planck
equation~\cite{Svetitsky:1987gq,vanHees:2004gq,Moore:2004tg,Mustafa:2004dr},
\begin{equation}
\frac{\partial f_Q}{\partial t} = \gamma \frac{\partial (pf_Q)}{\partial p}
+ D \frac{\partial^2 f_Q}{\partial p^2} \ ,
\end{equation}
for the heavy-quark (HQ) phase-space distribution, $f_Q$. The scattering
rate, $\gamma$, and momentum diffusion coefficient, $D$, are related
via the Einstein relation, $T=D/(\gamma m_Q)$.
Applications to RHIC data revealed that perturbative QCD (pQCD) elastic
scattering is insufficient to generate the observed elliptic flow, even
with a strong coupling constant as large as $\alpha_s=0.4$, supporting
the notion of a strongly coupled QGP (sQGP). At strong coupling, diagrams
with large contributions have to be resummed, possibly leading to the
appearance of collective modes (bound states or resonances). In this spirit,
the effective resonance model for $c$- and $b$-quark scattering through
in-medium $D$ and $B$ meson has been introduced~\cite{vanHees:2004gq}.
The pertinent $s$-channel
diagrams are displayed in the left panels of Fig.~\ref{fig_graph}. An
approximately 4-fold decrease of the HQ thermalization time,
$\tau_Q=\gamma^{-1}$, has been found relative to pQCD.
\begin{figure}[!t]
\begin{minipage}{0.4\linewidth}
\begin{center}
\includegraphics[width=0.65\textwidth]{Dreso-graph.eps}
\includegraphics[width=0.9\textwidth]{Dself-graph.eps}
\end{center}
\end{minipage}
\begin{minipage}{0.6\linewidth}
\includegraphics[width=0.95\textwidth]{brueckner.eps}
\end{minipage}
\caption[]{Left panels: $c$-quark scattering off light antiquarks via
$s$-channel $D$-mesons (upper panel) and pertinent $D$-meson
selfenergy (lower panel) within the effective resonance model in the
QGP~\cite{vanHees:2004gq}. Right panels: selfconsistent Brueckner scheme
for heavy quarks in the QGP based on an in-medium heavy-light $T$-matrix
with interaction potential $V$ (upper panel) and pertinent HQ
selfenergy (lower panel)~\cite{vanHees:2007me}.}
\label{fig_graph}
\end{figure}
When implemented into relativistic Langevin simulations within an
expanding fireball for Au-Au collisions at RHIC~\cite{vanHees:2005wb},
the recent data on suppression and elliptic flow of semileptonic decay
electrons are fairly well described~\cite{Adare:2006nq,Abelev:2006db},
cf.~left panel of Fig.~\ref{fig_elec} (see also
Refs.~\cite{Zhang:2005ni,Gossiaux:2008jv,Akamatsu:2008ge}).
Heavy-light quark coalescence in the hadronization
process at $T_c$ plays a significant role in increasing {\em both}
$R_{AA}$ and $v_2$.
\begin{figure}[!t]
\begin{minipage}{0.5\linewidth}
\vspace{-0.5cm}
\includegraphics[width=0.95\textwidth]{RAA-v2-elec-phenix.eps}
\end{minipage}
\begin{minipage}{0.5\linewidth}
\includegraphics[width=0.95\textwidth]{RAA-v2-elec-tmat.eps}
\end{minipage}
\caption[]{RHIC data~\cite{Adare:2006nq,Abelev:2006db,Hornback:2008ur}
for the nuclear modification factor (upper panels) and elliptic flow (lower
panels) of HQ decay electrons in
Au-Au collisions, compared to theory.
Left panels: relativistic Langevin simulations using upscaled pQCD
elastic scattering cross sections (short-dashed and dash-dotted
line)~\cite{Moore:2004tg} or effective resonances+pQCD elastic
scattering (bands)~\cite{vanHees:2005wb}, and radiative energy-loss
calculations (long-dashed lines)~\cite{Armesto:2005mz}.
Right panels: heavy-light quark
$T$-matrix+pQCD elastic interactions~\cite{vanHees:2007me} using
internal energies from quenched (solid line)
and $N_f$=2 (dash-dotted line) thermal lQCD~\cite{Kaczmarek:2003ph};
the dashed lines are obtained if hadronization via heavy-light quark
coalescence at $T_c$ is switched off.
}
\label{fig_elec}
\end{figure}
One may ask whether resonant HQ interactions in the QGP can be
understood more microscopically. This question has been studied
using a heavy-light quark $T$-matrix equation~\cite{vanHees:2007me},
\begin{equation}
T_{qQ} = V_{qQ} + V_{qQ} \, G_{qQ} \, T_{qQ}
\end{equation}
diagrammatically depicted in the upper right panel of Fig.~\ref{fig_graph}.
The key input is the driving kernel (potential), $V_{qQ}$, which
has been assumed to coincide with the static heavy-quark internal
energy computed in thermal lQCD, augmented by relativistic corrections
due to color-magnetic current-current interactions. In addition
to the color-singlet channel, the $Q$-$\bar{q}$ and $Q$-$q$ interactions
in the color-octet, -antitriplet and -sextet channels have been estimated
(using Casimir scaling). It turns out that the interaction
strength is concentrated in the attractive singlet and antitriplet
channels, supporting Feshbach-like meson and diquark resonances close to
the $q$-$Q$ threshold up to temperatures of $\sim$1.7\,$T_c$ and
$\sim$1.4\,$T_c$, respectively. Compared to the resonance model, the HQ
interaction in the $T$-matrix approach of similar strength close
to $T_c$, but weakens at higher $T$ due to color-screening in the
potential which dissolves the resonances.
An open problem remains the nonperturbative treatment of HQ-gluon
interactions. The application of the $T$-matrix approach to RHIC
electron data looks promising (right panels in Fig.~\ref{fig_elec}),
but significant uncertainties remain which currently inhibit definite
conclusions about the microscopic origin of HQ diffusion.
What effects can be expected for charm-quark diffusion at finite
$\mu_q$, relevant for CBM energies and a RHIC energy scan?
The effective resonance model suggests that, in a quark-dominated
environment, anticharm quarks ($\bar{c}+q\to \bar{D}$) interact more
frequently than charm quarks ($c+\bar{q}\to D$). However, in the
$T$-matrix approach, scattering via (anti-) diquarks, $cq$ and
$\bar{c}\bar{q}$, is equally important, thus washing out the
asymmetry. Moreover, in the hadronic phase, the baryon excess
favors $D$-meson scattering via $\Lambda_c N^{-1}$ excitations
over its antiparticle conjugate.
A promising possibility could be the development of the $\sigma$ soft
mode close to the critical point, which is particularly pronounced
in the spacelike regime~\cite{Fujii:2003bz}. {\em If} charm quarks
couple to the $\sigma$, their $t$-channel exchange cross section with
light quarks (and hadrons) could be ``critically" enhanced leaving
observable traces in $D$-meson $p_T$ spectra and $v_2$ (see
Ref.~\cite{Zhuang:1995uf} for a related study in the light-quark
sector).
\section{Charmonia: Screening and Dissociation}
\label{sec_charmonium}
The dissolution temperature of charmonia in medium largely depends
on two mechanisms: color-screening of the inter-quark force and
inelastic dissociation reactions. While the
former largely governs the $Q$-$\bar{Q}$ binding energy, $\varepsilon_B$
(via spacelike gluon exchange), the latter determines the inelastic
width, $\Gamma_\psi^{\rm inel}$, of the bound-state (via dissociation
reactions with timelike (on-shell) partons in the heat bath).
Within a schematic pole ansatz, both mechanisms figure into the
charmonium spectral function as
\begin{equation}
\sigma_\psi(\omega) \sim {\rm Im}~D_\psi(\omega) \sim
{\rm Im}~[\omega-2m_c^* +\varepsilon_B + i\,\Gamma_\psi/2 ]^{-1}
\label{sig-psi}
\end{equation}
(a pole ansatz applies to a well-defined bound-state/resonance).
Note the dependence on the in-medium $c$-quark mass, $m_c^*$, and
that the total width, $\Gamma_\psi$, includes contributions from elastic
scattering as well. However, as we will see below, binding energies and
dissociation reactions mutually influence each other.
Recent years have seen a revival of potential models to evaluate
in-medium quarkonium
properties~\cite{Mocsy:2007yj,Cabrera:2006wh,Alberico:2006vw,Wong:2006bx,Laine:2007qy,Brambilla:2008cx}.
\begin{figure}[!t]
\begin{minipage}{0.5\linewidth}
\includegraphics[width=1.0\textwidth]{etac-MP08.eps}
\end{minipage}
\begin{minipage}{0.5\linewidth}
\includegraphics[width=0.9\textwidth]{ImG-etac-CR.eps}
\includegraphics[width=0.9\textwidth]{RG-etac-CR.eps}
\end{minipage}
\caption[]{$S$-wave $c$-$\bar{c}$ spectral functions and Euclidean-time
correlators (normalized with a ``reconstructed" correlator, $R_G(\tau)\equiv
G(\tau)/G_{\rm rec}(\tau)$) using a HQ potential close to the lQCD free
energy (left panels)~\cite{Mocsy:2007yj} or corresponding to the
lQCD internal energy (right panels)~\cite{Cabrera:2006wh}.}
\label{fig_corr}
\end{figure}
This was largely spurred by the hope that the input potential can be
taken in a model-independent way from the HQ free energy computed in
thermal lQCD, and that resulting spectral functions can be discriminated
via comparisons to Euclidean correlation functions independently
computed in lQCD. There is, however, an ongoing controversy as to
which quantity to identify with a potential (free vs. internal energy),
and concerning the gauge variance of the projection on the color-singlet
channel~\cite{Philipsen:2008qx}. More quantitatively, the current
situation is illustrated in Fig.~\ref{fig_corr}.
Roughly speaking, when the singlet free energy, $F_{Q\bar{Q}}^1(r,T)$,
is used as potential, ground-state charmonia dissolve at temperatures
below 1.5\,$T_c$ (left panel)~\cite{Mocsy:2007yj}. On the other hand,
with the internal energy,
$U_{Q\bar{Q}}^1 =F_{Q\bar{Q}}^1(r,T) +T S_{Q\bar{Q}}^1(r,T)$
($S_{Q\bar{Q}}$: entropy), a $J/\psi$ peak in the
spectral function can survive up to 2.5-3\,$T_c$ (upper right
panel)~\cite{Cabrera:2006wh}.
The spectral functions have been used to calculate temporal Euclidean
correlation functions,
\begin{equation}
G_\psi(\tau,p;T)=
\int\limits_0^\infty {d\omega} \ \sigma_\psi(\omega,p;T) \
\frac{\cosh[(\omega(\tau-1/2T)]}{\sinh[\omega/2T]} \ .
\label{G-tau}
\end{equation}
which are usually normalized to a ``reconstructed" correlator,
$G_{\rm rec}$, computed with a vacuum spectral function (but
identical finite-temperature kernel in eq.~(\ref{G-tau})).
Surprisingly, both weak and strong-binding scenarios for the
spectral function result in correlator ratios which are around
one and depend weakly on temperature (compatible with lQCD
results~\cite{Datta:2003ww,Jakovac:2006sf,Aarts:2007pk}), cf.~inset
in the left panel and lower right panel of Fig.~\ref{fig_corr}.
The reason for this ``redundancy" is the underlying effective quark
mass, which is calculated from the asymptotic value of the HQ
free or internal energy, $m_c^*=m_c^0+\Delta m_c$ with
$2\Delta m_c = F(r\to\infty,T)$ or $U(r\to\infty,T)$.
For the free energy, $\Delta m_c$ is substantially reduced from its
vacuum value, thus lowering the in-medium $c\bar{c}$ threshold
considerably; this compensates for the lack of a bound state in
providing low-energy strength in the spectral function to ensure
a stable correlator ratio. For the internal energy, $\Delta m_c$
is significantly larger which, together with a stronger binding,
leads to an essentially stable $J/\psi$ peak position and thus a
roughly stable $T$-dependence of the correlator ratio.
More work is needed to disentangle these two scenarios.
Finite-width effects on the correlators have received rather little
attention thus far, but they seem to further stabilize
the $T$-dependence~\cite{Cabrera:2006wh}.
If a reliable understanding of quarkonium correlators in the QGP
at $\mu_q=0$ in terms of potential models can been established,
it might serve as a benchmark to extrapolate, e.g., color-screening
effects to finite $\mu_q$.
\begin{figure}[!t]
\begin{minipage}{0.5\linewidth}
\includegraphics[width=0.95\textwidth]{gam-psi-qgp.eps}
\end{minipage}
\begin{minipage}{0.5\linewidth}
\includegraphics[width=0.95\textwidth]{gam-psi-muq.eps}
\end{minipage}
\caption[]{$J/\psi$ dissociation rate (=inelastic width) in a QGP as a
function of $T$ (at $\mu_q$=0, left panel) and $\mu_q$ (at $T$=180~MeV,
right panel) for gluo-dissociation ($g+J/\psi\to c+\bar{c}$) and
quasifree destruction ($p+J/\psi\to p+c+\bar{c}$ with $p=q,\bar{q},g$)
with vacuum and in-medium reduced $J/\psi$ binding energy.}
\label{fig_gam-psi}
\end{figure}
The dissociation width of heavy quarkonia in medium is pivotal
for a quantitative description of suppression (and
regeneration!) reactions in HICs. The prevalent dissociation
mechanism in the QGP depends on the binding energy of the bound
state~\cite{Kharzeev:1995ju,Grandchamp:2001pf,Park:2007zza,Zhao:2007hh},
cf.~left panel of Fig.~\ref{fig_gam-psi}.
For large binding, $\varepsilon_B\sim 3T$, thermal-gluon absorption is
most efficient, $g+J/\psi\to c+\bar{c}$ (and formally the process to
leading-order in $\alpha_s$). For small binding, $\varepsilon_B < T$,
the phase space for gluo-dissociation shrinks leading to a decrease
in its rate with $T$. Thus, ``quasi-free'' dissociation,
$p+J/\psi\to p+c+\bar{c}$, albeit formally of higher order in
$\alpha_s$, takes over. Note that the quasi-free process can be
induced by both gluons and anti-/quarks. This has consequences at
finite $\mu_q$, in that quasifree dissociation is additionally enhanced
over gluo-dissociation due to an increasing abundance of thermal
quarks, cf.~right panel of Fig.~\ref{fig_gam-psi}~\cite{Zhao:2009}.
\section{Conclusions}
\label{sec_concl}
Instead of a formal summary of this paper, let us reiterate what we
find the most promising perspectives regarding the finite-$\mu_q$
dependence of dileptons and charm/onia ath this point.
For low-mass dileptons, we have identified the mass region around
$M\simeq0.2$~GeV as the most sensitive one for baryon-driven medium
effects. In addition, with a good knowledge of the in-medium spectral
shape of the EM correlator, the dilepton yield can finally be used for
an accurate determination of the fireball lifetime in HICs, which might
be useful in detecting (the onset of) an extended quark-hadron mixed
phase. For open charm, a ``critical" enhancement of $\sigma$ exchange
in $c$-quark or $D$-meson scattering in the vicinity of the critical
point may occur, potentially affecting transport properties in an
observable way ($p_t$ spectra and elliptic flow). For charmonia, a
hope is to establish the validity of finite-$T$ potential models and
extrapolate them to finite $\mu_q$, augmented by microscopic
calculations of dissociation rates.
These developments are particularly exciting in view of future tests
in several heavy-ion programs around the world.
\section*{Acknowledgments}
It is a pleasure to thank D.~Cabrera, V.~Greco, M.~Mannarelli, F.~Riek,
H.~van Hees, J.~Wambach and X.~Zhao for their collaboration on various
aspects of the presented topics, and H.~van Hees for a careful reading
of the ms.
This work has been supported by a U.S. National Science Foundation
CAREER award under grant no. PHY-0449489 and by the A.-v.-Humboldt
Foundation (Germany) through a Bessel award.
| {
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Q: Inserting ethernet frames of a particular ethertype in ahead of TCP/IP frames in netdev_queue We have developed an Application Specific Integrated Circuit for power line communications. The chip has an ethernet interface. If the ASIC receives an ethernet frame containing TCP/IP or ARP payload (ethertypes 0x0800 IPv4, 0x0806 ARP and 0x86DD IPv6), it simply forwards the frame onto the power line and does the same in the other direction. We call such frames data frames.
If the ASIC receives an ethernet frame of a specific ethertype (we use 0x88b5 which is allocated for experimental//public use on local networks), it consumes this frame itself. These frames contain configuration settings for the ASIC. We call these configuration frames.
The chip is connected to an Ethernet LAN on one side and to power line on the other end. So it basically bridges the two network. The ASIC requires throttling of the data frames passing through it. This is due to the fact that the speeds over power line are 100 times less than the 100 Mbps Ethernet and also because the number of data frames that the ASIC can handle per second are limited.
We use raw sockets to form the configuration frames and send it via ethernet to the ASIC. Is there a way in which whenever configuration frame (0x88b5), it is queued in front of all the pending data frames (ethertypes 0x0800, 0x0806, 0x86dd) in the netdev_queue?
Can this be done via some supporting functionality implemented using hacks & hooks in a kernel module?
We came across a similar question (although improperly tagged) here: Setting up priority of packets that are being transmitted over the network
| {
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Черний Виталий Николаевич (; 10 февраля 1971 год, Киев) — украинский профессиональный баскетболист и тренер.
Карьера игрока
Виталий Николаевич Черний родился в Киеве. Карьеру игрока начал в 1992 году в ЦСКА Киев, в котором отыграл пять лет. Следующим клубом Черния был мариупольский «Азовбаскет», которому он помог пробиться в Суперлигу. В 1999 году вернулся в ЦСКА Киев, а ещё через два года перешёл в «Сумыхимпром», где и закончил карьеру игрока.
Наивысшими его достижениями были серебряная и бронзовая медаль чемпионата Украины с ЦСКА в 1993 и 2000 годах.
Тренер юниоров и ассистент
В 2004 году новый главный тренер «Сумыхимпром» известный литовский специалист Римантас Эндрияйтис предложил Чернию стать его ассистентом. В этом качестве Виталий Николаевич проработал три года.
В 2007 году стал тренером дубля БК Киев. В том же году был помощником временно исполняющего обязанности главного тренера национальной сборной Украины Валентина Берестнева.
В октябре того же года был назначен главным тренером кадетской сборной Украины (до 16 лет) и помощником главного тренера юниорской сборной Украины(до 18 лет). Черний руководил кадетской сборной на чемпионате Европы 2008 года среди кадетов, где его подопечные, победив Венгрию,Израиль и Латвию, заняли в итоге девятое место. На чемпионате Европы среди юниоров руководимая Олегом Рубаном и Чернием команда заняла 12 место и с трудом, но осталась в Дивизионе А.
Главный тренер
В середине сезона 2008—2009 БК «Киев» постиг тяжелый финансовый кризис. Клуб был вынужден расстаться по ходу сезона со всеми легионерами и в результате чемпионат доигрывала команда, состоящая из исключительно украинских игроков, немалую долю которых составляли молодые игроки, выращенные в дубле Виталием Чернием. Неожиданный успех «украинизированного» состава, сумевшего занять четвёртое место, подвиг руководство клуба назначить главным тренером украинского специалиста — всего второй раз в истории. И этим специалистом стал тренер дубля Виталий Черний.
В сезоне 2009—2010 БК «Киев», состоящий из воспитанников дубля Черния и нескольких опытных украинских баскетболистов удивил всю Украину, заняв четвёртое место в регулярном сезоне Суперлиги и проиграв лишь в полуфинале плей-офф. Искрометная, бесшабашная игра местных воспитанников приглянулась болельщикам и Виталий Черний был признан открытием тренерского цеха Украины. Черний работал в БК «Киев» до конца сезона 2013/ 2014. В нём он вывел команду в плей-офф Суперлиги с 8 места, но проиграл в первом раунде будущему финалисту — «Химику».
8 июля 2014 года Виталий Черний был назначен новым главным тренером киевского «Будивельника». С 2018 по 2020 годы был главным тренером МБК «Николаев». С 2020 по 2021 год был главным тренером баскетбольного клуба «Прометей».
Сборная Украины
В апреле 2010 года Виталий Черний был избран главным тренером национальной сборной Украины по баскетболу. Это решение было принято Федерацией Баскетбола Украины единогласно.
В начале августа сборная Украины примет участие в отборочном турнире чемпионата Европы-2011 года. Соперниками подопечных Черния станут Босния и Герцеговина, Македония, Венгрия и Великобритания.
Черний-тренер
Черний считается жестким тренером, ставящим во главу угла дисциплину, послушание и выполнение указаний тренера. Вместе с тем он энергичен, умеет создать хорошую атмосферу в команде, умеет сплотить коллектив и настроить его на достижения. Он умеет увлечь за собой игроков, добиться от них максимальной самоотдачи в каждом матче. БК Киев под его руководством слывет молодой, но очень цепкой, неуступчивой командой, которая любит играть в быстрый и интересный баскетбол и всегда настраивается бороться до последнего.
Черний также уделяет немало внимания тактической подготовке игроков, основательно готовится к каждому матчу и внимательно изучает соперника. На тренировках тактике уделяется немало времени.
7 марта 2015 года выиграл свой первый крупный трофей в качестве главного тренера — обыграв со счетом 73:60 БК «Днепр», его команда одержала победу в Кубке Украины.
Примечания
Ссылки
Виталий Черний — тренер БК «Киев»
Интервью с Виталием Чернием
Новый тренер баскетбольной сборной умеет работать с молодыми
Профиль на сайте клуба БК «Киев»
Баскетболисты Украины
Игроки БК СКА Киев
Игроки БК «Азовмаш»
Баскетбольные тренеры Украины
Тренеры БК «Литкабелис»
Тренеры БК «Киев»
Тренеры БК «Будивельник»
Тренеры БК «Николаев»
Тренеры БК «Прометей» | {
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Stirling Residences is a joint venture by two of the leading real estate groups in Singapore; Logan Property Holding Company Limited and Nanshan Group.
Founded in 1996, Logan Property Holdings Company Limited ("Logan Property" or "the Group", stock code:3380.HK) is an integrated property developer that focuses on residential property development in the PRC, mainly the Guangdong-Hong Kong-Macao Greater Bay Area, with its products; they mostly target first-time homebuyers and people that make upgrades to buildings.
It was listed on the main board of the Stock Exchange of Hong Kong in 2013. Logan Poperty has Grade-A qualifications in the development of properties, general contracting of construction and engineering design, and is known as one of the best property developers in the PRC in terms of overall strength.
In 2017, the group jumped to 295th in Fortune China's Top 500 Enterprises. In 2018, the Group was ranked 26th in the China Top 100 Real Estate Developers and 4th in the Top 10 Most Profitable Companies, with its brand and overall strength being highly recognized by the market.
They have a building concept called "Building a Better Life", and adhering to it, Logan Property has so far developed more than 110 residential projects and provided quality life services to over 600,000 people.
More than 80% of Logan Property's land bank is located in the Guangdong-Hong Kong-Macao Greater Bay Area, which provides a solid basis for growing in the future.
Founded in 1979, Nanshan Group has developed into a large scale private joint stock enterprise listed in the Top 500 Chinese Enterprises at present after 40 years of work, forming a multi-industry development led by aluminum industry, textiles and garments, finance, aviation, real estate, health, education, tourism, etc.
The most distinctive feature of Nanshan's industrial development is highlighting the main business with chain operation; developing steadily to become stronger and more practical.
Besides the leading industries, Nanshan Group has such supporting enterprises as construction and installation as well as green food manufacturers related to the life of residents and employees, including peanut oil, flourmill, purified water, eggs and meat products.
With the expanding industry scale, Nanshan Group gradually forms a strategic layout centering on Longkou, spreading over the country and aiming towards the world.
At present, it has established branches and offices in Beijing, Tianjin, Qingdao, Yantai, Hainan and Hong Kong, which has also established branches in many countries such as Australia, America, Germany, Italy, Singapore and Indonesia to move towards the international market and participate in the global competition.
The capability of independent innovation is the key to the success of the industrial development of Nanshan.
While developing the economy, Nanshan Group never forgets to fulfill its social responsibilities and successively drives the villagers of dozens of surrounding villages to realize common prosperity, villagers changed to citizens and villages changed to cities.
On top of that, Nanshan Group also establishes charitable mutual fund and education scholarship to help worker families and poor students in trouble, and actively participates in social public welfare career such as road construction, environmental protection, afforestation, disaster relief and donation, making active contributions to boosting the new urbanization construction, promoting the social economy and social development and other aspects.
It has been granted honorary titles such as "National Civilized Village", "National Advanced Primary Party Organization", "National Small Town Construction Demonstration Area", "Meritorious Enterprise during 30 Years of Reform and Opening up in Shandong Province" and "The Most Responsible Brand in Shandong Province". | {
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Michael Anthony Scott (* 13. März 1986 in Indianapolis, Indiana), genannt Mike Scott, ist ein US-amerikanischer Basketballspieler. Nach dem Ende seines Studiums in seinem Heimatland spielte Scott als Profi in Europa, darunter eine Spielzeit in der Basketball-Bundesliga 2010/11 bei der BG 74 aus Göttingen. Anschließend spielte Scott zwei Jahre in der Košarkaška liga Srbije, unterbrochen nur von einem kurzen Engagement bei Spirou BC Charleroi in der Wallonie. Für die Saison 2013/14 ging Scott dann erneut in den französischsprachigen Raum und unterschrieb einen Vertrag beim Erstliga-Rückkehrer EB Pau-Lacq-Orthez im Südwesten Frankreichs.
Karriere
Scott wechselte zum Studium aus seiner Geburtsstadt an die Kent State University in Kent (Ohio), wo er für das Hochschulteam Golden Flashes in der Mid-American Conference der NCAA Division I spielte. Die Golden Flashes gewannen 2006 und 2008 die Meisterschaft dieser Conference, verloren jedoch jeweils ihr Erstrundenspiel in der landesweiten Endrunde der NCAA.
2008 begann Scott seine professionelle Karriere beim türkischen Zweitligisten aus Trabzon, der nach einem ersten Platz in der Hauptrunde jedoch in der Aufstiegsrunde den Sprung in die höchste Spielklasse verpasste. In der Spielzeit 2009/10 war Scott dann für den ungarischen Erstligisten aus Körmend aktiv, der am Ende der Spielzeit einen dritten Platz belegte. Scott selbst war für das All-Star Game der ungarischen Liga in dieser Spielzeit nominiert. Zur Basketball-Bundesliga 2010/11 bekam Scott einen Vertrag beim deutschen EuroChallenge-Gewinner BG 74 Göttingen. Im Eurocup 2010/11 erreichte Scott mit der Göttinger Mannschaft die K.-o.-Spiele im Viertelfinale, in denen man gegen Benetton Treviso ausschied. In der Zwischenrunde der 16 besten Mannschaften in diesem europäischen Vereinswettbewerb wurde Scott als Most Valuable Player des ersten Spieltags ausgezeichnet. Vor Beginn der Play-offs um die deutsche Meisterschaft löste der Verein jedoch den Vertrag mit Scott.
Zur Saison 2011/12 wechselte Scott nach Serbien, wo er für KK Radnički aus Kragujevac in der höchsten serbischen Spielklasse Košarkaška liga Srbije (KLS) und der supranationalen adriatischen Liga (ABA) spielte. Für Radnički, die am Ende der Spielzeit in der ABA den achten Platz belegten, war Scott zweitbester Scorer mit 17,6 Punkten und bester Rebounder mit 6,6 Abprallern pro Spiel. In der Effektivitätswertung erreichte damit den zweiten Platz und war nur ein wenig schlechter als sein Mannschaftskamerad David Simon. In der nationalen Meisterschaft schied Scott mit Radnički in der Halbfinalserie der Play-offs gegen den späteren Vizemeister Roter Stern Belgrad aus. Zur Spielzeit 2012/13 unterschrieb Scott zunächst einen Vertrag beim belgischen Vizemeister Spirou BC aus Charleroi, wechselte aber bereits Ende Oktober 2012 nach drei Meisterschaftsspielen zurück nach Serbien, wo er fortan für Roter Stern spielt. Mit Roter Stern reichte es zu einem serbischen Pokalsieg und zur Vizemeisterschaft hinter KK Partizan Belgrad in der KLS und der ABA. Für die Saison 2013/14 unterschrieb Scott dann einen Vertrag beim traditionsreichen Erstliga-Rückkehrer Élan Béarnais aus Pau in der französischen LNB Pro A.
Weblinks
Mike Scott – Kent State Golden Flashes – Statistiken aus Collegespielzeiten auf statsheet.com (englisch)
Einzelnachweise
Basketballspieler (Vereinigte Staaten)
US-Amerikaner
Geboren 1986
Mann
Basketballspieler (KK Roter Stern Belgrad)
Basketballspieler (Spirou BC Charleroi)
Basketballspieler (BG 74 Göttingen) | {
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{"url":"https:\/\/lavelle.chem.ucla.edu\/forum\/viewtopic.php?f=123&t=55879","text":"## 5G1 true\/false\n\n$PV=nRT$\n\n705121606\nPosts: 68\nJoined: Wed Sep 18, 2019 12:17 am\n\n### 5G1 true\/false\n\nThe question asks to determine if the statement is true or false and explain why. In part c it states \"If one starts with higher pressure of reactant, the equilibrium constant will be larger.\" I thought this was true because if there is higher pressure of reactant it will shift toward the products. Since the equilibrium constant is equal to concentration of products over concentration of reactants, the equilibrium constant would be larger. Can someone explain why the statement is false.\n\n205405339\nPosts: 77\nJoined: Thu Jul 11, 2019 12:16 am\n\n### Re: 5G1 true\/false\n\nInstead of looking at K as Kc look at it as Kp. Kp= (partial pressure of products) \/ (partial pressure of reactants)\naccording to this equation, an increase in the partial pressure of reactants would lead to a decrease in the equilibrium constant instead of an increase in the equilibrium constant, which is what the question is asking. Therefore, it would be false since K would be decreasing instead of increasing\n\nWendy 1E\nPosts: 111\nJoined: Sat Aug 17, 2019 12:17 am\nBeen\u00a0upvoted: 2 times\n\n### Re: 5G1 true\/false\n\nYour reasoning that \"if there is higher pressure of reactant, it will shift towards product\" is correct. However, this does not mean that K will change. Here, only Q is changing because initially Q<K. K is not changing. This is because the temperature is constant. Only when the temperature changes (we will probably learn this in future lectures), will K then change. The pressure of the reactants and products don't affect the equilibrium constant.\n\nWendy 1E\nPosts: 111\nJoined: Sat Aug 17, 2019 12:17 am\nBeen\u00a0upvoted: 2 times\n\n### Re: 5G1 true\/false\n\n205405339 wrote:Instead of looking at K as Kc look at it as Kp. Kp= (partial pressure of products) \/ (partial pressure of reactants)\naccording to this equation, an increase in the partial pressure of reactants would lead to a decrease in the equilibrium constant instead of an increase in the equilibrium constant, which is what the question is asking. Therefore, it would be false since K would be decreasing instead of increasing\n\nThis is not correct because when you increase the pressure of the reactants, the pressure of the products will also increase. Therefore, the ratio would still be the same and K will not be affected. K will only change if the temperature changes.\n\n205150314\nPosts: 106\nJoined: Wed Feb 20, 2019 12:16 am\n\n### Re: 5G1 true\/false\n\n705121606 wrote:The question asks to determine if the statement is true or false and explain why. In part c it states \"If one starts with higher pressure of reactant, the equilibrium constant will be larger.\" I thought this was true because if there is higher pressure of reactant it will shift toward the products. Since the equilibrium constant is equal to concentration of products over concentration of reactants, the equilibrium constant would be larger. Can someone explain why the statement is false.\n\nIt is false because the equilibrium constant is well constant, so even if you start with a higher pressure of reactant the reaction will create more products and in the end lead to the same value of K\n\nJunwei Sun 4I\nPosts: 125\nJoined: Wed Oct 02, 2019 12:16 am\n\n### Re: 5G1 true\/false\n\nThis statement is false since even if you start with a higher pressure of reactants, this would only result in production of more products and in the end the equilibrium constant remains the same.The value of K is not affected by the addition of products and reactants as long as the temperature is the same.\n\nRuth Glauber 1C\nPosts: 100\nJoined: Wed Sep 18, 2019 12:20 am\n\n### Re: 5G1 true\/false\n\nIt's false because when you increase the pressure of the reactants, the pressure of the products will also increase.\n\nrohun2H\nPosts: 100\nJoined: Wed Sep 18, 2019 12:19 am\n\n### Re: 5G1 true\/false\n\nQ would be larger but ultimately the reaction would reach equilibrium and K remains the same.\n\nTyler Angtuaco 1G\nPosts: 130\nJoined: Wed Sep 11, 2019 12:16 am\n\n### Re: 5G1 true\/false\n\nA higher initial amount of reactant would produce more product. The equilibrium values of the reactant and product will also be higher and the numerator and denominator of the equilibrium constant ratio will be proportional to those of a reaction with less reactant if both reactions occur under the same conditions and stoichiometric coefficients are calculated. Keep in mind that the initial values and equilibrium values are not necessarily proportional to those produced from a reaction with less reactant. So, when calculating the equilibrium constant, you might have a higher numerator (due to the larger equilibrium amount of product), and a higher denominator (due to the larger equilibrium amount of reactant), but the ratio will simplify to the same number for a reaction with a smaller amount of reactant as long as both reactions occur under the same conditions, i.e. temperature. The terminology of the statement also points to why it would be false, since it seems to assume higher initial pressures always result in a higher equilibrium constant due to their inclusion of the word, \"will.\"","date":"2020-06-07 10:29:54","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\": 0, \"codecogs_latex\": 1, \"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\": 0.7658523321151733, \"perplexity\": 808.1950770622989}, \"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\/1590348526471.98\/warc\/CC-MAIN-20200607075929-20200607105929-00053.warc.gz\"}"} | null | null |
This is a list of hotels in the Caribbean.
Geographical note: The Caribbean is a region that consists of the Caribbean Sea, its islands (some surrounded by the Caribbean Sea and some bordering both the Caribbean Sea and the North Atlantic Ocean), and the surrounding coasts. The region is southeast of the Gulf of Mexico and the North American mainland, east of Central America, and north of South America.
Hotels in the Caribbean
Anguilla
Cap Juluca Hotel, Cap Juluca
Carimar Beach Club, Mead's Bay, The Valley
CuisinArt Resort and Spa, Rendezvous Beach
Ku Resort, Shoal Bay
Malliouhana Hotel, Mead's Bay, The Valley
The Viceroy Anguilla, Mead's Bay, The Valley
Aruba
Amsterdam Manor Beach Resort
Aruba Bucuti Beach Resort
Aruba Marriott Resort
Aruba Millenium Resort
La Quinta Beach Resort
Occidental Grand Aruba
Radisson Aruba Resort, Casino & Spa
Renaissance Aruba Resort and Casino
Ritz-Carlton Aruba – opened on November 22, 2013
Tierra el Sol Aruba Resort
Barbados
The Fairmont Royal Pavilion in St. James
The Sandy Lane in St. James
Curaçao
Dreams Curaçao Resort
Plaza Hotel Curaçao
Dominica
Garraway Hotel
Puerto Rico
Saint Lucia
Anse Chastanet resort
See also
List of Caribbean-related topics
Lists of hotels
Tourism in the Caribbean
References
Caribbean
Hotels | {
"redpajama_set_name": "RedPajamaWikipedia"
} | 5,790 |
Dallas ICE Agent, 9 Other Ice Agents Sue Obama Administration
By Jack Fink August 23, 2012 at 9:40 pm
Filed Under:administration, agents, ice, lawsuit, Obama
DALLAS (CBSDFW.COM) – Ten federal immigration agents, including an agent here in Dallas, are suing the Obama administration.
They have filed their lawsuit in Dallas over a controversial plan not to deport certain illegal immigrants. The lawsuit accuses Homeland Security Secretary Janet Napolitano of violating federal law.
The agents claim Napolitano did not have the authority this past June to stop deporting illegal immigrants who came to this country as children, and who meet certain conditions.
Last week, in Dallas, and across the country, thousands of young illegal immigrants lined up at consulate offices so they could apply for a two year work permit to avoid being deported. The Pew Hispanic Center estimates 1.7 million people may be covered under this new directive.
In their lawsuit, the agents claim if they follow the new procedure, they will violate their oath of office and federal law. They want a federal judge to intervene.
A Homeland Security Department spokesman says the department is using its discretion properly.
The former U.S. attorney in Dallas, Richard Roper, says Napolitano could make a compelling argument in court.
"We can't go after every illegal alien in the country. We have to allocate and deploy our scarce resources in a certain way to try to address that, and I'm making that call and I have that authority."
Roper believes a federal judge will not intervene in the case.
It is important to note that one of the attorneys representing the ICE agents, Kris Kobach, is an informal adviser to Republican presidential candidate Mitt Romney. We called Kobach, and the Dallas ICE agent who is suing, but we haven't heard back.
Texas Judge Preparing For 'Civil War' If Obama Re-Elected
Employer Allegedly Tells Worker: Get An Abortion Or Get Fired
Fan Sues Dallas Cowboys For Burned Buttocks
Teacher Gets 5 Years For Having Sex With Multiple Students
KRLD Restaurant Week
Jack Fink
More from Jack Fink
Gov. Greg Abbott Says He's Offended By Vetting Of Texas National Guard In DC | {
"redpajama_set_name": "RedPajamaCommonCrawl"
} | 7,798 |
\section{Flowchart}
\section{Preliminaries}
\subsection{Notation}
Let us first review the definition of $\calO(\cdot)$, and generalize the notation to contain relative constants $\theta$, as well as introducing a new notation representing constant functions that we know exactly the order as well as the coefficient in front of the largest order term.
\begin{defn}
\label{defn: Big O notation}
Let $f$ and $g$ both be real valued function, and suppose $g(x)$ is strictly positive for any $x$ large enough. Then
\begin{enumerate}
\item $f(x) = \calO(g(x))$ if and only if $\exists x_0$, $\abs{f(x)} \le Mg(x)$ for any $x \ge x_0$.
\item $f(x) = \logO(g(x))$ if and only if $\exists x_0$ and $\exists k \in \mathbb{Z}$, $\abs{f(x)} \le Mg(x)\log^k(g(x))$ for any $x \ge x_0$.
\begin{comment}
\item $f(x) = \Omega(g(x))$ if and only if $\exists x_0$, $\abs{f(x)} \ge Mg(x)$ for any $x \ge x_0$.
\item $f(x) = \Omega(\theta;g(x))$ if and only if $\exists x_0$, $\abs{f(x)} \ge M(\theta)g(x)$ for any $x \ge x_0(\theta)$.
\item $f(x) = \Theta(g(x))$ if and only if $f(x) = \calO(g(x))$ and $f(x) = \Omega(g(x))$.
\item $f(x) = \Theta(\theta;g(x))$ if and only if $f(x) = \calO(\theta;g(x))$ and $f(x) = \Omega(\theta;g(x))$.
\end{comment}
\item $f(x) = \calO(g(x))$ is a fixed function with regard to $x$ such that $\exists C > 0$, and $\lim_{x\to\infty}\abs{f(x)/g(x)} = C$
\item $f(x) = \calO(\theta;g(x))$ is a fixed function with regard to $x$ such that $\exists C(\theta) > 0$, and $\lim_{x\to\infty}\abs{f(x)/g(x)} = C(\theta)$
\item
For a set of random variables $X_n$ and a corresponding set of constants $a_n$, the notation
\[X_n = o_p(a_n).\]
means that the set of values $X_n/a_n$ converges to zero in probability as $n$ approaches an appropriate limit. Equivalently, $X_n = o_p(a_n)$ can be written as $X_n/a_n = o_p(1)$, where $X_n = o_p(1)$ is defined as
\[X_n \convP 0.\]
\item
For a set of random variables $X_n$ and $Y_n$, where $Y_n$ is almost surely non-zero, the notation
\[X_n = o(Y_n) \as \]
means that
\begin{equation*}
X_n/Y_n \asConv 0
.\end{equation*}
\item
The notation
\[X_n = \calO_p(a_n).\]
means that the set of values $X_n/a_n$ is stochastically bounded. That is, for any $\epsilon > 0$, there exists a finite $M > 0$ and a finite $N > 0$ such that,
\[\P(\abs{X_n/a_n} > M) < \epsilon, \forall n > N.\]
\item
\label{itm: big O as defn}
\[X_n = \calO(a_n) \as\]
if for almost every $\omega \in \Omega$, there exists a number $C(\omega)$ such that $\abs{X_n(\omega)} \le C(\omega)a_n$. In other words, $X_n = \calO(a_n) \as$ if there exists a random variable $C$ such that $\abs{X_n} \le Ca_n \as$ Equivalently,
\begin{equation*}
X_n = \calO(a_n) \as \equivalent \limsup_{n \to \infty} \frac{\abs{X_n}}{a_n} < \infty \as
\end{equation*}
\item
The notation
\[X_n = \logO_p(a_n).\]
means that the set of values $X_n/a_n$ is stochastically bounded up to a constant order of $log(a_n)$. That is, for any $\epsilon > 0$, there exists a finite $M > 0$ , a finite $k \in \mathbb{Z}$, and a finite $N > 0$ such that,
\[\P(\abs{X_n/\log^k(a_n)a_n} > M) < \epsilon, \forall n > N.\]
\end{enumerate}
\textbf{All these definitions can be generalized to vectors or matrices with entry-wise definition.} Without extra specification, all norms $\norm{\cdot}$ (for both vectors and matrices) are meant to be $L_2$ norm $\norm{\cdot}_2$, i.e., operator-2 norm for the matrix.
\end{defn}
Some relationships between these notations are worth keeping in mind: (see Eq.(7) and Eq.(8) in \citet{janson2011probability})
\begin{equation}
\label{eq:small o as to p}
X_n = o(a_n) \as \Longrightarrow X_n = o_p(a_n)
.\end{equation}
\begin{equation}
\label{eq:big o as to p}
X_n = O(a_n) \as \Longrightarrow X_n = O_p(a_n)
.\end{equation}
To carefully track down the constant chosen manually, when we state order bounds like $\calO(\theta;g(x))$, $\theta$ should not contain variables such as $\delta$ which are set fixed when we prove high probability bounds but could be varying later, but could contain global constants such as $A$, $B$, $K$, $P$, $Q$, $R$, dimension $\inputdim$, $\statedim$ and $C_x$, $C_u$, $\tau$, $\beta$ that are fixed throughout the whole algorithm.
In order to differentiate $\calO(\cdot)$ from fixed constants, we denote $\calO(\theta)$ as constant terms which could be potentially varying and only related with $\theta$. That means for the same $\calO(\theta)$ symbol in two different places, they can be different constants. One special symbol is $\calO(1)$ which represents constant that does not rely on any parameters.
\subsection{Extending results to $\beta=1$}\label{sec:app_ext_beta1}
Although the main text only considered vanishing exploration noise (i.e., $\beta<1$), for completeness (and because it is straightforward to do so) we will also consider the case of $\beta=1$ and $\alpha\le 0$ for all of our results.
\subsection{Proof dependency tree}
In order to make the proof more readable and easier to understand, we put the proof outlines first and summarize most useful middle steps by lemmas. These lemmas' proofs often involve more technical details and is deferred to later parts in the appendix. While this may help readers have better understanding in the high level ideas behind the long proof, we realize that it may also cause loops in the proof structure. Thus, we provide a tree (\cref{fig: Proof dependency tree}) which describes the exact proof dependency structure to make sure that there is no circular argument. In \cref{fig: Proof dependency tree}, all conclusions lies in a perfect tree graph except for the loop marked in red between \cref{lemma: Hi prob bounds in theorem 2} and \cref{prop:one_epoch_estimate_withMyalg}. This is not a contradiction because the proof of \cref{prop:one_epoch_estimate_withMyalg} only relies on a subset of conclusions in \cref{lemma: Hi prob bounds in theorem 2}: \cref{eq:bound on eta_p,eq: stochastic bound on B eta p plus varepsilon p}, which do not require \cref{prop:one_epoch_estimate_withMyalg} to hold. Some of the proofs relies on \cref{eq: sum eta_t}, which is not included in the graph but still
self-consistent (does not rely on other results in the paper).
\clearpage
\begin{figure}[p]
\makebox[\linewidth]{
\includegraphics[width=1\linewidth]{fig/LQR_proof_V4.pdf}
}
\caption{Proof dependency tree}
\label{fig: Proof dependency tree}
\end{figure}
\clearpage
\section{The proof of \cref{thm:main tool}}
\label{The proof of thm:main tool}
\begin{thm*}
\cref{alg:myAlg} applied to a system described by \cref{eq:LinearModel} under \cref{asm:InitialStableCondition} satisfies
\begin{equation}
\label{eq: more general CLT condition 1}
D_t^{-1}
\sum_{i=0}^{t-1}
\begin{bmatrix}
x_i\\
u_i\\
\end{bmatrix}
\begin{bmatrix}
x_i\\
u_i\\
\end{bmatrix}^\top
(D_t^\top)^{-1} \convP I_{n+d}
.\end{equation}
\end{thm*}
\subsection{Proof Outline}
\begin{proof}
Let us first examine the Gram matrix $\sum_{i=0}^{t-1} \begin{bmatrix} x_i\\ u_i\\ \end{bmatrix} \begin{bmatrix} x_i\\ u_i\\ \end{bmatrix}^\top $. Denote
\begin{equation}
\label{eq: defn Mt}
\begin{split}
M_t :=& \sum_{i=1}^{t-1}x_ix_i^\top/t^\beta\log^{\alpha}(t) \\
\end{split}
,\end{equation}
and
\begin{equation}
\label{eq: defn Delta t}
\begin{split}
\Delta_t :=& \sum_{i=1}^{t-1}u_ix_i^\top/t^\beta\log^{\alpha}(t) - K M_t \\
=& \sum_{i=1}^{t-1}((\Kh_t -K)x_i + \eta_i)x_i^\top/t^\beta\log^{\alpha}(t)
.\end{split}
\end{equation}
We will show that
\begin{equation*}
\sum_{i=0}^{t-1}u_iu_i^\top/t^{\beta}\log^{\alpha}(t) =
KM_tK ^\top + \Delta_t K^\top + K\Delta_t^\top + \frac{\tau^2}\beta I_d + o_p(1)
,\end{equation*}
and thus we can write our Gram matrix as
\begin{equation*}
\begin{split}
\sum_{i=1}^{t-1} \begin{bmatrix} x_i\\ u_i\\ \end{bmatrix} \begin{bmatrix} x_i\\ u_i\\ \end{bmatrix}^\top /t^{\beta}\log^{\alpha}(t) =&
\left[
\begin{array}{cc}
\sum_{i=0}^{t-1}x_ix_i^\top & \sum_{i=1}^{t-1}x_iu_i^\top \\
\sum_{i=0}^{t-1}u_ix_i^\top & \sum_{i=1}^{t-1}u_iu_i^\top \\
\end{array}
\right]/t^{\beta}\log^{\alpha}(t) \\
= &
\left[
\begin{array}{cc}
M_t & M_tK ^\top + \Delta_t^\top \\
KM_t+ \Delta_t & KM_tK ^\top + \Delta_t K^\top + K\Delta_t^\top + \frac{\tau^2}\beta I_d \\
\end{array}
\right] + o_p(1) \\
=&
\left[
\begin{array}{cc}
I_n & 0\\
K & I_d\\
\end{array}
\right]\left[
\begin{array}{cc}
M_t & \Delta_t^\top\\
\Delta_t & \frac{\tau^2}\beta I_d\\
\end{array}
\right]\left[
\begin{array}{cc}
I_n & K^\top \\
0 & I_d\\
\end{array}
\right] + o_p(1) .
\end{split}
\end{equation*}
Therefore, in order to satisfy
\[D_t^{-1}\sum_{i=0}^{t-1} \begin{bmatrix} x_i\\ u_i\\ \end{bmatrix} \begin{bmatrix} x_i\\ u_i\\ \end{bmatrix}^\top (D_t^\top)^{-1} \convP I_{n+d}, \]
we can pick $D_t^{-1} :=
\left[
\begin{array}{cc}
C_t^{-1/2} & 0\\
0 & \sqrt{\frac{\beta}{\tau^2}} I_d\\
\end{array}
\right]
\left[
\begin{array}{cc}
I_n & 0\\
-K & I_d\\
\end{array}
\right]/t^{\beta/2}\log^{\alpha/2}(t) $. $C_t$ is a deterministic matrix which satisfies $C_t^{-1/2}M_t^{1/2} \convP I_\statedim$ and $C_t^{-1/2}\Delta_t = o_p(1)$ (we will give $C_t$'s exact expression in \cref{eq: Ct definition}). With this choice of $D_t^{-1}$, we have
\begin{equation*}
\begin{split}
& D_t^{-1}\sum_{i=0}^{t-1} \begin{bmatrix} x_i\\ u_i\\ \end{bmatrix} \begin{bmatrix} x_i\\ u_i\\ \end{bmatrix}^\top (D_t^\top)^{-1} \\
=& \left[
\begin{array}{cc}
C_t^{-1/2} & 0\\
0 & \sqrt{\frac{\beta}{\tau^2}} I_d\\
\end{array}
\right]
\left[
\begin{array}{cc}
I_n & 0\\
-K & I_d\\
\end{array}
\right]
\left(\sum_{i=0}^{t-1} \begin{bmatrix} x_i\\ u_i\\ \end{bmatrix} \begin{bmatrix} x_i\\ u_i\\ \end{bmatrix}^\top /t^{\beta}\log^\alpha(t)
\right)
\left[
\begin{array}{cc}
I_n & -K^\top\\
0 & I_d\\
\end{array}
\right]
\left[
\begin{array}{cc}
C_t^{-1/2} & 0\\
0 & \sqrt{\frac{\beta}{\tau^2}} I_d\\
\end{array}
\right] \\
=&
\left[
\begin{array}{cc}
C_t^{-1/2} & 0\\
0 & \sqrt{\frac{\beta}{\tau^2}} I_d\\
\end{array}
\right]
\left[
\begin{array}{cc}
I_n & 0\\
-K & I_d\\
\end{array}
\right]
\left(
\left[
\begin{array}{cc}
I_n & 0\\
K & I_d\\
\end{array}
\right]\left[
\begin{array}{cc}
M_t & \Delta_t^\top\\
\Delta_t & \frac{\tau^2}\beta I_d\\
\end{array}
\right]\left[
\begin{array}{cc}
I_n & K^\top \\
0 & I_d\\
\end{array}
\right]
+ o_p(1)
\right)\\
& \quad \cdot
\left[
\begin{array}{cc}
I_n & -K^\top\\
0 & I_d\\
\end{array}
\right]
\left[
\begin{array}{cc}
C_t^{-1/2} & 0\\
0 & \sqrt{\frac{\beta}{\tau^2}} I_d\\
\end{array}
\right] \\
=&
\left[
\begin{array}{cc}
C_t^{-1/2} & 0\\
0 & \sqrt{\frac{\beta}{\tau^2}} I_d\\
\end{array}
\right]
\left[
\begin{array}{cc}
I_n & 0\\
-K & I_d\\
\end{array}
\right]
\left[
\begin{array}{cc}
I_n & 0\\
K & I_d\\
\end{array}
\right]\left[
\begin{array}{cc}
M_t & \Delta_t^\top\\
\Delta_t & \frac{\tau^2}\beta I_d\\
\end{array}
\right]\left[
\begin{array}{cc}
I_n & K^\top \\
0 & I_d\\
\end{array}
\right] \\
& \quad \cdot
\left[
\begin{array}{cc}
I_n & -K^\top\\
0 & I_d\\
\end{array}
\right]
\left[
\begin{array}{cc}
C_t^{-1/2} & 0\\
0 & \sqrt{\frac{\beta}{\tau^2}} I_d\\
\end{array}
\right] + o_p(1) \quad \text{(we can move $o_p(1)$ outside because $C_t^{-1/2} \to 0$)}\\
=&
\left[
\begin{array}{cc}
C_t^{-1/2} & 0\\
0 & \sqrt{\frac{\beta}{\tau^2}} I_d\\
\end{array}
\right]
\left[
\begin{array}{cc}
M_t & \Delta_t^\top\\
\Delta_t & \frac{\tau^2}\beta I_d\\
\end{array}
\right]
\left[
\begin{array}{cc}
C_t^{-1/2} & 0\\
0 & \sqrt{\frac{\beta}{\tau^2}} I_d\\
\end{array}
\right] + o_p(1) \\
=&
\left[
\begin{array}{cc}
C_t^{-1/2}M_tC_t^{-1/2} & \sqrt{\frac{\beta}{\tau^2}} C_t^{-1/2}\Delta_t^\top \\
\sqrt{\frac{\beta}{\tau^2}} \Delta_t C_t^{-1/2} & I_d\\
\end{array}
\right]
+ o_p(1) \\
=& I_{n+d} + o_p(1)
.\end{split}
\end{equation*}
\paragraph{Components needing further explanation}
In the final step of the above derivation there are still several points that remains unclear, namely
\begin{itemize}
\item $ \sum_{i=0}^{t-1}u_iu_i^\top/t^{\beta}\log^{\alpha}(t) =
KM_tK ^\top + \Delta_t K^\top + K\Delta_t^\top + \frac{\tau^2}\beta I_d + o_p(1)$,
\item $C_t^{-1/2}M_t^{1/2} \convP I_\statedim$, and
\item $C_t^{-1/2}\Delta_t = o_p(1)$.
\end{itemize}
As we will see, the order of $\Delta_t$ is decided by the convergence rate of $\Kh_t -K$. Because of that, the first step in our proof is to identify the convergence rate of $\Kh_t -K$. Then we will prove the three remaining points in The proof of \cref{eq: more general CLT condition 1}.
To summarize,
our proof can be mainly separated into two big steps:
\begin{enumerate}
\item Identify the convergence rate of $\Kh_t -K$. (see \cref{first part of proving main thm})
\item Prove \cref{eq: more general CLT condition 1} holds:
\[D_t^{-1}\sum_{i=0}^{t-1} \begin{bmatrix} x_i\\ u_i\\ \end{bmatrix} \begin{bmatrix} x_i\\ u_i\\ \end{bmatrix}^\top (D_t^\top)^{-1} \convP I_{n+d}.\]
\begin{itemize}
\item Summarize uniform high probability bound for some random variables, which will serve as basic tools for later proof. (see \cref{tools main thm})
\item Prove $C_t^{-1/2}M_t^{1/2} \convP I_\statedim$. (see \cref{2nd part of proving main thm})
\item Prove $C_t^{-1/2}\Delta_t = o_p(1)$. (see \cref{3rd part of proving main thm})
\item Prove $\sum_{i=0}^{t-1}u_iu_i^\top/t^\beta =
KM_tK ^\top + \Delta_t K^\top + K\Delta_t^\top + \frac{\tau^2}\beta I_d + o_p(1)$. (see \cref{4th part of proving main thm})
\end{itemize}
\end{enumerate}
%
Now we will examine these steps in order.
\end{proof}
\subsection{Convergence rate of $\Kh_t - K$}
\label{first part of proving main thm}
As said in the previous part, the main purpose of this section is to derive the convergence rate of $\Kh_t - K$, which is one crucial step in our proof. Denote the stabilizing controller computed by Line \ref{line:ols} \cref{alg:myAlg} as $\Kt_{t+1}$, i.e.,
\begin{align*}
\Kt_{t+1} =
\begin{cases}
\text{Solve DARE \cref{eq:ControllerK,eq:riccati} with } A = \Ah_{t}, B = \Bh_t, &\text{for }(\Ah_{t}, \Bh_{t}) \text{ stabilizable}\\
K_0, &\text{for }(\Ah_{t}, \Bh_{t}) \text{ not stabilizable}\\
\end{cases}
.
\end{align*}
By Line \ref{line:check} \cref{alg:myAlg}, $\Kh_{t+1}$ can be written as:
\begin{align*}
\Kh_{t+1} =
\begin{cases}
K_0, &\text{when }\norm{x_{t}} > C_x\log(t) \text{ or } \norm{\Kh_t} > C_K\\
\Kt_{t+1}, &\text{otherwise}\\
\end{cases}
.
\end{align*}
In particular, the proof can be separated into three parts:
\begin{enumerate}
\item Derive the convergence rate of $\Ah_t$ and $\Bh_t$.
\item Show that $\Kt_{t+1}$ enjoy the same convergence rate as $\Ah_t$ and $\Bh_t$.
\item Show that $\Kh_{t+1}$ is only different from $\Kt_{t+1}$ finitely often, and as a result, $\Kh_{t+1}$ also enjoy the same convergence rate as $\Ah_t$ and $\Bh_t$.
\end{enumerate}
Correspondingly we have the following three propositions:
\begin{prop}[Similar to Proposition C.1 in \citet{dean2018regret}]
\label{prop:one_epoch_estimate}
Let $x_0 \in \R^{\statedim}$ be any initial state. Assume \cref{asm:InitialStableCondition} is satisfied. When applying $\cref{alg:myAlg}$,
\begin{equation*}
\max\left\{ \norm{\Ah_t - A}, \norm{\Bh_t - B}\right\} =
\calO(t^{-\frac{\beta}{2}} \log^{\frac{-\alpha + 1}{2}}(t)) \as
\end{equation*}
\end{prop}
The proof of \cref{prop:one_epoch_estimate} can be found in \cref{section: The proof of one_epoch_estimate}.
\begin{prop}
\label{prop:one_epoch_estimate_withK}
Let $x_0 \in \R^{\statedim}$ be any initial state. Assume \cref{asm:InitialStableCondition} is satisfied. When applying $\cref{alg:myAlg}$,
\begin{equation*}
\max\left\{ \norm{\Ah_t - A}, \norm{\Bh_t - B}, \norm{\Kt_{t+1} - K}\right\}
= \calO(t^{-\frac{\beta}{2}} \log^{\frac{-\alpha + 1}{2}}(t)) \as
\end{equation*}
\end{prop}
The proof of \cref{prop:one_epoch_estimate_withK} can be found in \cref{section: The proof of one_epoch_estimate_withK}.
\begin{prop}
\label{prop:one_epoch_estimate_withMyalg}
Let $x_0 \in \R^{\statedim}$ be any initial state. Assume \cref{asm:InitialStableCondition} is satisfied. When applying \cref{alg:myAlg},
\begin{equation}
\label{eq:uniform high probability bound for Kh}
\max\left\{ \norm{\Ah_t - A}, \norm{\Bh_t - B}, \norm{\Kh_{t+1} - K}\right\}
= \calO(t^{-\frac{\beta}{2}} \log^{\frac{-\alpha + 1}{2}}(t)) \as
\end{equation}
\end{prop}
The proof of \cref{prop:one_epoch_estimate_withMyalg} can be found in \cref{section: The proof of one_epoch_estimate_withMyalg}.
\cref{prop:one_epoch_estimate,prop:one_epoch_estimate_withK,prop:one_epoch_estimate_withMyalg} all hold additionally for a version of \cref{alg:myAlg} that only updates logarithmically often; see \cref{sec: The proof of Propositions}. The takeaway from this section is the uniform bound for $\norm{\Kh_{t+1} - K}$ \cref{eq:uniform high probability bound for Kh}, which is the only property of $\Kh_t$ we need for the rest of the proof.
\subsection{Proving \cref{eq: more general CLT condition 1}}
\subsubsection{Uniform Bounds}
\label{tools main thm}
In this section we will show several basic uniform bounds that will be used frequently in the later The proof of \cref{thm:main tool}.
\begin{lemma}
\label{lemma: Hi prob bounds in theorem 2}
~
\begin{itemize}
\item
\begin{equation}
\label{eq:bound on eta_p}
\norm{\varepsilon_t}, \norm{\eta_t} = \calO(\log^{1/2}(t)) \as
\end{equation}
\item
\begin{equation}
\label{eq: stochastic bound on B eta p plus varepsilon p}
\norm{B\eta_t+\varepsilon_t} = \calO(\log^{1/2}(t)) \as
\end{equation}
\end{itemize}
Assume \cref{eq:uniform high probability bound for Kh}, then:
\begin{itemize}
\item
\begin{equation}
\label{eq: stochastic bound delta_t}
\norm{\delta_t} = \norm{\Kh_{t} - K} =
\calO(t^{-\frac{\beta}{2}} \log^{\frac{-\alpha + 1}{2}}(t)
) \as
\end{equation}
\item
For $t > q$,
\begin{equation}
\label{eq: stochastic bound Lhat product}
\norm{(L + B\delta_{t-1}) \cdots (L + B\delta_{q})} =
\calO(\rhoL^{t-q}) \as
\end{equation}
\item
\begin{equation}
\label{eq: bound norm xi by log t}
\norm{x_t}, \norm{u_t} =
\calO(\log^{1/2}(t)) \as
\end{equation}
\end{itemize}
where $\delta_t := \Kh_{t} - K$, $L := A+BK$, and $\rhoL := \frac{2 + \rho(L)}{3}$. \textbf{Additionally, when $t=0, 1$ all these terms are bounded by $\calO(1) \as$}
\end{lemma}
The proof can be found in \cref{The proof of lemma: Hi prob bounds in theorem 2}. Following \cref{defn: Big O notation} \cref{itm: big O as defn}, \cref{lemma: Hi prob bounds in theorem 2} presents uniform upper bounds for $t \ge 0$.
We will see that all states $x_t$ and actions $u_t$ can be expressed in recursive summations, which can be bounded easily if we have uniform upper bound for each of their components.
Let us briefly explain why these orders makes sense.
\begin{itemize}
\item The first two inequalities come from the tail bound for standard Gaussian random variables, whose maximum scales as $\log^{1/2}(t)$.
\item The third inequality \cref{eq: stochastic bound delta_t} directly follows from \cref{eq:uniform high probability bound for Kh}.
\item The fourth inequality \cref{eq: stochastic bound Lhat product} holds with exponential decay because the $L$ has spectual radius $<1$ and by \cref{eq: stochastic bound delta_t} $\delta_t$ is shrinking to $0$.
\item The fifth inequality \cref{eq: bound norm xi by log t} holds because the system is stabilizable and the effect of previous states and actions are exponentially decaying, leaving the main factor in the norm to come from the recent system noises. By the first two inequalities $\norm{x_t}$ is uniformly bounded by $\log^{1/2}(t)$ scale.
\end{itemize}
\subsubsection{Showing $C_t^{-1/2}M_t^{1/2} \convP I_n$}
\label{2nd part of proving main thm}
We wish to show that $M_t = \sum_{i=0}^{t-1}x_ix_i^\top/t^\beta\log^\alpha(t) =
C_t (1 + o_p(1))$, where
\begin{equation}
\label{eq: Ct definition}
C_t = \log^{-\alpha}(t)
t^{1-\beta} \sum_{p=0}^{\infty}L ^{p}(L ^{p})^\top \sigma^2 + \frac{\tau^2}{\beta}\sum_{q=0 }^{\infty}L ^{q}BB^\top (L ^{q})^\top
\end{equation}
Recall the system definition \cref{eq:LinearModel}:
\[x_{t+1} = A x_t + B u_t + \varepsilon_t.\]
and the input \cref{eq:Myinput}
\[u_t = \Kh_tx_t + \eta_t.\]
Recursively applying these two equations produces the following formula for $x_t$ in terms of $x_0$, $\{\varepsilon_p\}_{p=0}^{t-1}$, and $\{\eta_p\}_{p=0}^{t-1}$.
\begin{lemma}
\label{lemma: StateExpansion}
For any $t \ge 1$,
\begin{equation}
\label{eq: StateExpansion}
x_{t} = \sum_{p=0}^{t-1}(A+B \Kh_{t-1})\cdots(A+B \Kh_{p+1})(B \eta_p+\varepsilon_p) +(A+B \Kh_{t-1})\cdots(A+B K_0)x_0
,\end{equation}
and
\begin{equation*}
u_{t} = \sum_{p=0}^{t-1}\Kh_t(A+B \Kh_{t-1})\cdots(A+B \Kh_{p+1})(B \eta_p+\varepsilon_p) +\Kh_t(A+B \Kh_{t-1})\cdots(A+B K_0)x_0 + \eta_t
.\end{equation*}
Here when $p = t-1$, we define the product $(A+B \Kh_{t-1})\cdots(A+B \Kh_{p+1}) := I_n$.
\end{lemma}
The proof can be found in \cref{The proof of lemma: StateExpansion}. As a result, we can rewrite $\sum_{i=0}^{t-1}x_ix_i^\top$ into a summation in terms of $\{\varepsilon_i, \eta_i\}_{i=0}^{t-1}$. First consider the terms without $x_0$.
\begin{align*}
\sum_{i=1}^{t-1}\sum_{p=0}^{i-1}\sum_{q=0}^{i-1}\left[(A+B \Kh_{i-1})\cdots(A+B \Kh_{p+1})(B \eta_p+\varepsilon_p)\right]
\left[(A+B \Kh_{i-1})\cdots(A+B \Kh_{q+1})(B \eta_q+\varepsilon_q)\right]^\top
.\end{align*}
This whole expression can be separated into four components with the following bounds:
\begin{lemma}
\label{lemma: four components xtxt}
Assume \cref{eq:uniform high probability bound for Kh}, then:
\begin{enumerate}
\item
\begin{align*}
&\sum_{i=1}^{t-1}\sum_{p=0}^{i-1}\sum_{q=0}^{i-1}\left[(A+B K)^{i-p-1}\right](B \eta_p+\varepsilon_p)(B \eta_q+\varepsilon_q)^\top
\left[(A+B K)^{i-q-1}\right]^\top \\
&\hspace{5cm}= t^\beta \log^\alpha(t) (C_t +o_p(1))
.\end{align*}
\item
\begin{equation*}
\begin{aligned}
\sum_{i=1}^{t-1}\sum_{p=0}^{i-1}\sum_{q=0}^{i-1}\left[(A+B \Kh_{i-1})\cdots(A+B \Kh_{p+1}) - (A+B K)^{i-p-1}\right]&\\
\cdot (B \eta_p+\varepsilon_p)(B \eta_q+\varepsilon_q)^\top
\left[(A+B K)^{i-q-1}\right]^\top & = \calO(t^{1-\beta/2}\log^{\frac{-\alpha + 3}{2}}(t)) \as
\end{aligned}
\end{equation*}
\item
\begin{align*}
&\sum_{i=1}^{t-1}\sum_{p=0}^{i-1}\sum_{q=0}^{i-1}\left[ (A+B K)^{i-p-1}\right](B \eta_p+\varepsilon_p)(B \eta_q+\varepsilon_q)^\top
\left[(A+B \Kh_{i-1})\cdots(A+B \Kh_{q+1}) - (A+B K)^{i-q-1}\right]^\top \\
&
\hspace{8cm} = \calO(t^{1-\beta/2}\log^{\frac{-\alpha + 3}{2}}(t)) \as
\end{align*}
\item
\begin{align*}
&\sum_{i=1}^{t-1}\sum_{p=0}^{i-1}\sum_{q=0}^{i-1}\left[(A+B \Kh_{i-1})\cdots(A+B \Kh_{p+1}) - (A+B K)^{i-p-1}\right]
(B \eta_p+\varepsilon_p)(B \eta_q+\varepsilon_q)^\top \\
& \hspace{3cm} \cdot
\left[(A+B \Kh_{i-1})\cdots(A+B \Kh_{q+1}) - (A+B K)^{i-q-1}\right]^\top = \calO(t^{1-\beta/2}\log^{\frac{-\alpha + 3}{2}}(t)) \as
\end{align*}
\end{enumerate}
\end{lemma}
The proof can be found in \cref{The proof of lemma: four components xtxt}.
It remains to consider the remaining terms with $x_0$, which is relatively straight-forward, since the effect of the initial state is exponentially decaying when $t \to \infty$.
\begin{lemma}
\label{lemma: Term with starting point xtxt}
Assume \cref{eq:uniform high probability bound for Kh}, then
\begin{enumerate}
\item $\sum_{i=0}^{t-1}\left[ (A+B \Kh_{i-1})\cdots(A+B K_{0})x_0\right]
\left[\sum_{q=0}^{i-1}(A+B \Kh_{i-1})\cdots(A+B \Kh_{q+1})(B\eta_q+\varepsilon_q)\right]^T = \logO(1) \as$
\item $\sum_{i=0}^{t-1}\left[ (A+B \Kh_{i-1})\cdots(A+B K_{0})x_0\right]
\left[ (A+B \Kh_{i-1})\cdots(A+B K_{0})x_0\right]^T = \calO(1) \as$
\end{enumerate}
\end{lemma}
The proof can be found in \cref{The proof of lemma: Term with starting point xtxt}. As mentioned in \cref{eq:big o as to p}, $\calO \as$ notation is stronger than $\calO_p$ notation.
Summing up all the results in \cref{lemma: four components xtxt} and \cref{lemma: Term with starting point xtxt} we can finally conclude that
\[
\sum_{i=0}^{t-1} x_ix_i^\top =
t^\beta \log^\alpha(t) (C_t +o_p(1)) + \calO_p(t^{1-\beta/2}\log^{\frac{-\alpha + 3}{2}}(t))
.\]
Thus
\begin{equation}
\label{eq:Cov xx}
M_t = \sum_{i=0}^{t-1}x_ix_i^\top/t^\beta\log^\alpha(t) = C_t +o_p(1) + \calO_p(t^{1-3\beta/2}\log^{\frac{-3\alpha + 3}{2}}(t))
,\end{equation}
where $C_t$ is defined in \cref{eq: Ct definition}
This is already very close to our objective $C_t^{-1/2}M_t^{1/2} \convP I_n$, but we still need to show that $C_t$ is an invertible matrix. $C_t$ is already a positive semi-definite (PSD) matrix because it is a weighted summation of PSD matrices $L^p(L ^{p})^\top$ and $L ^{q}BB^\top (L ^{q})^\top$. The only thing we need to ensure is that $C_t$ is a full rank matrix. And that is indeed true because the $p=0$ term is the identity matrix, and adding more PSD matrices $L^p(L ^{p})^\top$ and $L ^{q}BB^\top (L ^{q})^\top$ will not change its positive definite nature. Following \cref{eq: Ct definition},
we have (because $\beta < 1$ or $\beta = 1 $ and $\alpha \le 0$)
\begin{equation}
\label{eq: Ct order}
C_t =
\log^{-\alpha}(t)t^{1-\beta}
\sum_{p=0}^{\infty}L ^{p}\left(\sigma^2I_n + 1_{\{\beta=1,\alpha=0\}}\tau^2 BB^\top\right)(L ^{p})^\top (I_n + o(1))
.\end{equation}
Thus
\begin{equation}
\label{eq: Ct inverse order}
C_t^{-1}
= t^{\beta-1}\log^\alpha(t)
\left( \sum_{p=0}^{\infty}L ^{p}\left(\sigma^2I_n + 1_{\{\beta=1,\alpha=0\}}\tau^2 BB^\top\right)(L ^{p})^\top
\right)^{-1}
(I_n + o(1))
= \calO(t^{\beta-1}\log^\alpha(t))
.\end{equation}
Noticing that
\[\calO(t^{\beta-1}\log^\alpha(t)) \calO_p(t^{1-3\beta/2}\log^{\frac{-3\alpha + 3}{2}}(t)) = \calO_p(t^{-\beta/2}\log^{\frac{-\alpha + 3}{2}}(t)) = o_p(1),\]
we have from \cref{eq:Cov xx}
\begin{equation}
\label{eq: Ct -1 Mt convP I_n}
C_t^{-1}M_t \convP I_n
.\end{equation}
With the help of the following lemma we conclude that $ C_t^{-1/2}M_t^{1/2} \convP I_n$.
\begin{lemma}
\label{lemma:AtBtConvergeIp}
Assume we have two matrix sequences $\{A_t\}_{t=1}^\infty$ and $\{B_t\}_{t=1}^\infty$, where $A_t$ and $B_t$ are $p \times p$ positive definite matrices, then
\[A_t^{2}B_t^2 \convP I_p.\]
iff
\[A_tB_t \convP I_p.\]
\end{lemma}
The proof can be found in \cref{The proof of lemma:AtBtConvergeIp} (Thanks for the help from Haoyi Yang and Yue Li in proving this lemma).
\subsubsection{Proving $C_t^{-1/2}\Delta_t = o_p(1)$}
\label{3rd part of proving main thm}
Recall the definition of $\Delta_t$ from \cref{eq: defn Delta t}:
\begin{equation*}
\Delta_t := \left(\sum_{i=0}^{t-1}(\Kh_i-K )x_ix_i^\top + \sum_{i=0}^{t-1}\eta_ix_i^\top\right)
/t^\beta \log^\alpha(t)
.\end{equation*}
The order of $\Delta_t$ depends on the order of its two components:
\begin{lemma}
\label{lemma: three parts u_tx_t}
Assume \cref{eq:uniform high probability bound for Kh}, then
\begin{enumerate}
\item $
\sum_{i=0}^{t-1}(\Kh_i-K )x_ix_i^\top =
\calO(t^{1-\beta/2}\log^{\frac{-\alpha + 3}{2}}(t)) \as
$
\item $ \sum_{i=0}^{t-1}\eta_ix_i^\top =
o\left(t^{\beta/2}\log^{\frac{\alpha+3}{2}}(t) \right) \as $
\end{enumerate}
\end{lemma}
The proof can be found in \cref{The proof of lemma: three parts u_tx_t}.
The first term has larger order than the second term when
$1/2 \le \beta < 1$ or $\beta = 1$ and $\alpha \le 0$.
As a result, we have
\begin{align}
\label{eq: Delta t order}
\begin{split}
\Delta_t &= \calO(t^{1-3\beta/2}\log^{\frac{-3\alpha + 3}{2}}(t)) \as \quad \text{(when } \beta \in [1/2, 1)\text{)}
\end{split}
\end{align}
Observe from \cref{eq: Ct inverse order}:
\begin{equation*}
C_t^{-1} = \calO(t^{\beta-1}\log^\alpha(t))
.\end{equation*}
Then when $\beta > 1/2$ or $\beta = 1/2, \alpha > 3/2$
\begin{align*}
C_t^{-1/2}\Delta_t =& \calO(t^{-1/2+\beta/2}\log^{\alpha/2}(t)t^{1-3\beta/2}\log^{\frac{-3\alpha + 3}{2}}(t)) \\
=& \calO(t^{1/2-\beta}\log^{\frac{-2\alpha + 3}{2}}(t)) \\
=& o(1) \as
\end{align*}
\subsubsection{Proving $\sum_{i=0}^{t-1}u_iu_i^\top/
t^\beta \log^\alpha(t)
=
KM_tK ^\top + \Delta_t K^\top + K\Delta_t^\top + \frac{\tau^2}\beta I_d + o_p(1)$}
\label{4th part of proving main thm}
Finally we need to check
\[\sum_{i=0}^{t-1}u_iu_i^\top = \sum_{i=0}^{t-1}((K +\delta_i)x_i+\eta_i)((K +\delta_i)x_i+\eta_i)^\top ,\]
where $\delta_i = \Kh_i - K$.
There are six different kinds of terms in the above equation, namely $\sum_{i=0}^{t-1} Kx_ix_i^TK^\top $,
$\sum_{i=0}^{t-1} Kx_ix_i^\top \delta_i^\top $ and $\sum_{i=0}^{t-1} \delta_ix_ix_i^TK^\top $,
$\sum_{i=0}^{t-1} Kx_i\eta_i^\top $ and $\sum_{i=0}^{t-1} \eta_ix_i^TK^\top $,
$\sum_{i=0}^{t-1} \delta_ix_ix_i^\top \delta_i^\top $,
$\sum_{i=0}^{t-1} \delta_ix_i\eta_i^\top $ and $\sum_{i=0}^{t-1} \eta_ix_i^\top \delta_i^\top$,
and $\sum_{i=0}^{t-1} \eta_i\eta_i^\top $. The first three terms can be written as
\[\sum_{i=0}^{t-1} Kx_ix_i^TK^\top/t^\beta \log^\alpha(t) = K M_t K^\top,\]
and
\[\left(\sum_{i=0}^{t-1} Kx_ix_i^\top \delta_i^\top + \sum_{i=0}^{t-1} \delta_ix_ix_i^TK^\top +
\sum_{i=0}^{t-1} Kx_i\eta_i^\top + \sum_{i=0}^{t-1} \eta_ix_i^TK^\top \right)/t^\beta \log^\alpha(t) = K \Delta_t^T + \Delta_t K^T.\]
The remaining terms can be summarized by
\begin{lemma}
\label{lemma: six parts u_tu_t}
Assume \cref{eq:uniform high probability bound for Kh}, then
\begin{enumerate}
\item $\sum_{i=0}^{t-1} \delta_ix_ix_i^\top \delta_i^\top = \calO(t^{1-\beta}\log^{-\alpha+2}(t)) \as$
\item $\sum_{i=0}^{t-1}\delta_ix_i\eta_i^\top = (\sum_{i=0}^{t-1} \eta_ix_i^\top \delta_i^\top)^\top =
o\left(\log^{2}(t)\right) \as$
\item $\sum_{i=0}^{t-1}\eta_i\eta_i^\top = t^\beta\frac{\tau^2}{\beta}\log^\alpha(t)(I_d + o_p(1)) $
\end{enumerate}
\end{lemma}
The proof of \cref{lemma: six parts u_tu_t} can be found in \cref{The proof of lemma: six parts u_tu_t}.
Combining all parts in \cref{lemma: six parts u_tu_t} we have when $\beta > 1/2$ or $\beta = 1/2, \alpha > 1$, the third item dominates the other two. To sum up, we have
\begin{equation}
\label{eq: uu decomp}
\sum_{i=0}^{t-1}u_iu_i^\top/t^\beta\log^\alpha(t) =
KM_tK ^\top + \Delta_t K^\top + K\Delta_t^\top + \frac{\tau^2}\beta I_d + o_p(1)
.\end{equation}
\paragraph{Summary}
Now we have completed all missing proof pieces in the proof of \cref{eq: more general CLT condition 1}, which finishes The proof of \cref{thm:main tool}.
\section{The proof of \cref{thm:regret}}
\label{The proof of thm:regret}
\begin{thm*}
The average regret of the controller $U$ defined by \cref{alg:myAlg} applied through time horizon $T$ to a system described by \cref{eq:LinearModel} under \cref{asm:InitialStableCondition} satisfies, as $T \to \infty$,
\begin{equation*}
\frac{\mathcal{R}(U,T)}{\tau^2\beta^{-1} \Tr(B^\top PB +R)T^{\beta-1}\log^\alpha(T)} \convP 1,
\end{equation*}
with $\beta = 1/2$ therefore achieving the optimal rate \citep{simchowitz2020naive} of
$\mathcal{R}(U,T) = \logO_p(T^{-1/2})$.
\end{thm*}
\subsection{Proof Outline}
\begin{proof}
We are interested in the cost
\[\sum_{t=1}^{T} x_t^\top Qx_t + u_t^\top Ru_t \quad \text{with $u_t = \Kh_t x_t + \eta_t$}.\]
Recall the \cref{eq: StateExpansion} from \cref{lemma: StateExpansion} that
\[ x_{t} = \sum_{p=0}^{t-1}(A+B \Kh_{t-1})\cdots(A+B \Kh_{p+1})(B \eta_p+\varepsilon_p) +(A+B \Kh_{t-1})\cdots(A+B K_0)x_0.\]
Notice that the state $x_t$ has the same expression as if the system had noise $\tilde{\varepsilon}_t = B\eta_t + \varepsilon_t$ and controller $\tilde{u}_t = \Kh_t x_t$. We wish to switch to the new system because there are some existing tools with controls in the form of $\tilde{u}_t = \Kh_t x_t$.
We will first show in \cref{subsection: Cost difference induced by transformation} that the difference between the original cost and transformed cost is
\begin{equation*}
\sum_{t=1}^{T} u_t^\top Ru_t - \tilde{u}_t^\top R\tilde{u}_t = \frac{\tau^2}{\beta}T^{\beta}\log^\alpha(T)\Tr(R)(1+o_p(1))
,\end{equation*}
and then prove in \cref{subsection: Cost of transferred system} the new system cost is
\begin{equation*}
\sum_{t=1}^{T} x_t^\top Qx_t + \tilde{u}_t^\top R\tilde{u}_t = T \sigma^2\Tr (P)
+ \frac{\tau^2}{\beta} T^\beta \log^\alpha(T) \Tr(B^\top P B) (1+o_p(1))
.\end{equation*}
Combining the above two equations, we conclude that
\begin{align*}
\mathcal{J}(U,T)
&= \frac{1}{T}
\left[\sum_{t=1}^{T} x_t^\top Qx_t + u_t^\top Ru_t\right] \\
&= \sigma^2\Tr (P) + \tau^2\beta^{-1} \Tr(B^\top PB +R)T^{\beta-1}\log^\alpha(T) (1+ o_p(1))
.\end{align*}
Based on similar analysis we prove in \cref{subsection: Optimal average cost} that
\begin{align*}
\mathcal{J}(U^*,T)
&= \sigma^2\Tr (P) + \calO_p(T^{-1/2}\log(T))
.\end{align*}
Recall that we choose
$\beta \in [1/2,1]$,
and $\alpha>3/2$ when $\beta=1/2$, which means $T^{\beta-1}\log^\alpha(T)$ is of larger order than $T^{-1/2}\log(T)$. Finally we finish the proof with
\begin{align*}
\mathcal{R}(U,T) &= \mathcal{J}(U,T)- \mathcal{J}(U^*,T) \\
&= \tau^2\beta^{-1} \Tr(B^\top PB +R)T^{\beta-1}\log^\alpha(T) (1+ o_p(1))
.\end{align*}
\end{proof}
\subsection{Cost difference induced by transformation}
\label{subsection: Cost difference induced by transformation}
The difference is expressed as
\begin{equation*}
\begin{split}
\sum_{t=1}^{T} u_t^\top Ru_t - \tilde{u}_t^\top R\tilde{u}_t
=& \sum_{t=1}^{T}(\Kh_t x_t + \eta_t)^\top R (\Kh_t x_t + \eta_t) - \sum_{t=1}^{T} (\Kh_t x_t)^\top R (\Kh_t x_t) \\
=& 2\sum_{t=1}^{T} (\Kh_t x_t)^\top R \eta_t + \sum_{t=1}^{T} \eta_t^\top R \eta_t
.\end{split}
\end{equation*}
We show in \cref{eq: (Ktxt)T R etaT} that
\begin{equation*}
\sum_{t=1}^{T} (\Kh_t x_t)^\top R \eta_t = o\left(T^{\beta/2}\log^{\frac{\alpha+3}{2}}(T) \right) \as
,\end{equation*}
which is a direct corollary of \cref{lemma: three parts u_tx_t}.
Next we consider the order of $\sum_{t=1}^{T} \eta_t^\top R \eta_t$.
Since $\eta_t \sim \calN(0, \tau^2 t^{-1+\beta}\log^\alpha(t)I_d)$,
\begin{equation*}
\begin{split}
\E \sum_{t=1}^{T} \eta_t^\top R \eta_t
&= \sum_{t=1}^{T} \Tr(\E \eta_t\eta_t^\top R ) \\
&= \sum_{t=1}^{T} \tau^2 t^{-1+\beta}\log^\alpha(t) \Tr(R) \\
& \quad \text{(see the proof in \cref{eq: sum eta_t})} \\
&= \tau^2\frac{T^{\beta}}{\beta}\log^\alpha(T)\Tr(R)(1+o(1))
.\end{split}
\end{equation*}
While the variance of $\sum_{t=1}^{T} \eta_t^\top R \eta_t$ is $\calO(\sum_{t=1}^{T} t^{-2+2\beta}\log^{2\alpha}(t)) = \calO(T^{-1+2\beta}\log^{2\alpha}(T))$, which means the standard error $\calO(T^{-1/2+\beta}\log^{\alpha}(T))$ is of lower order than the expectation. Thus
\begin{equation*}
\sum_{t=1}^{T} \eta_t^\top R \eta_t
= \tau^2\frac{T^{\beta}}{\beta}\log^\alpha(T)\Tr(R)(1+o_p(1))
.\end{equation*}
As a conclusion, the error caused by this transformation is of order $\logO_p(T^\beta)$, and the dominating term is $\sum_{t=1}^{T} \eta_t^\top R \eta_t$.
\begin{equation}
\label{eq: regret first part}
\sum_{t=1}^{T} u_t^\top Ru_t - \tilde{u}_t^\top R\tilde{u}_t =
\tau^2\frac{T^{\beta}}{\beta}\log^\alpha(T)\Tr(R)(1+o_p(1))
.\end{equation}
\subsection{Cost of transformed system}
\label{subsection: Cost of transferred system}
Next we proceed as if our system was $x_t$ with system noise $\tilde\varepsilon_t = B \eta_t+\varepsilon_t$ and controller $\tilde{u}_t = \Kh_t x_t$.
The key idea of the following proof is from Appendix C of \citet{fazel2018global}.
We are interested in the cost
\[ \sum_{t=1}^{T} x_t^\top Qx_t + \tilde{u}_t^\top R\tilde{u}_t \quad \text{with $\tilde{u}_t = \Kh_t x_t $},\]
which can be written as
\begin{equation}
\label{eq: regret decomposition}
\begin{split}
\sum_{t=1}^{T} x_t^\top Qx_t + \tilde{u}_t^\top R\tilde{u}_t
=& \sum_{t=1}^{T} x_t^\top Qx_t + (\Kh_t x_t)^\top R\Kh_t x_t \\
=& \sum_{t=1}^{T} x_t^\top (Q + \Kh_t^\top R \Kh_t)x_t \\
=& \sum_{t=1}^{T} \left[x_t^\top (Q + \Kh_t^\top R \Kh_t)x_t + x_{t+1}^\top P x_{t+1} - x_{t}^\top P x_{t}
\right]
+ x_1^\top Px_1 - x_{T+1}^\top P x_{T+1} \\
=& \sum_{t=1}^{T} \left[x_t^\top (Q + \Kh_t^\top R \Kh_t)x_t + ((A+B\Kh_t)x_t +\tilde\varepsilon_{t})^\top P ((A+B\Kh_t)x_t +\tilde\varepsilon_{t}) - x_{t}^\top P x_{t}\right] \\
& \quad + \logO_p(1) \quad \text{(by \cref{lemma: Hi prob bounds in theorem 2}) } \\
=& \sum_{t=1}^{T} \Big[ x_t^\top (Q + \Kh_t^\top R \Kh_t)x_t + x_t^\top (A+B\Kh_t)^\top P (A+B\Kh_t)x_t - x_{t}^\top P x_{t} \\
& \quad+ 2 \tilde\varepsilon_{t}^\top P (A+B\Kh_t)x_t + \tilde\varepsilon_{t}^\top P \tilde\varepsilon_{t}
\Big]
+\logO_p(1)
.\end{split}
\end{equation}
We constructed the specific form of the first term on purpose. The following lemma translates the first term into a quadratic term with respect to $\Kh_t - K$.
\begin{lemma}
\label{lem: useful lemma from fazel}
For any $\Kh$ with suitable dimension,
\begin{equation*}
\begin{split}
&x^\top (Q + \Kh^\top R \Kh)x + x^\top (A+B\Kh)^\top P (A+B\Kh)x - x^\top P x \\
& \qquad = x^\top (\Kh-K)^\top( R + B^\top P B) (\Kh-K)x
.\end{split}
\end{equation*}
\end{lemma}
The proof can be found in \cref{The proof of lem: useful lemma from fazel}.
As a result
\begin{equation*}
\begin{split}
\sum_{t=1}^{T} x_t^\top Qx_t + \tilde{u}_t^\top R\tilde{u}_t
=& \sum_{t=1}^{T} x_t^\top (\Kh_t-K)^\top( R + B^\top P B) (\Kh_t-K)x_t \\
&\quad + 2 \tilde\varepsilon_{t}^\top P (A+B\Kh_t)x_t + \tilde\varepsilon_{t}^\top P \tilde\varepsilon_{t}
+\logO_p(1)
.\end{split}
\end{equation*}
Now we have three terms, and we will examine them in order.
\begin{enumerate}
\item The first term we consider is $\sum_{t=1}^{T} x_t^\top (\Kh_t-K)^\top( R + B^\top P B) (\Kh_t-K)x_t$. Recall from \cref{lemma: Hi prob bounds in theorem 2} that
\begin{equation*}
\norm{x_t}, \norm{u_t} = \calO(\log^{1/2}(t)) \as
\end{equation*}
and
\begin{equation*}
\norm{\Kh_{t} - K} = \calO(t^{-\frac{\beta}{2}} \log^{\frac{-\alpha + 1}{2}}(t)) \as
\end{equation*}
As a result
\begin{equation*}
\begin{split}
&\sum_{t=1}^{T} x_t^\top (\Kh_t-K)^\top( R + B^\top P B) (\Kh_t-K)x_t \\
\le & \sum_{t=1}^{T} \norm{x_t}^2 \norm{\Kh_t-K}^2 \norm{ R + B^\top P B} \\
= & \sum_{t=1}^{T} \calO(\log(t)) \calO(t^{-\beta} \log^{-\alpha + 1}(t))
\as \\
= & \calO(T^{1-\beta}\log^{-\alpha + 2}(T)) \as
\qquad (\text{by \cref{eq: sum eta_t}})
\end{split}
\end{equation*}
\item The second term we consider is $\sum_{t=1}^{T} \tilde\varepsilon_{t}^\top P (A+B\Kh_t)x_t$. Similar as before, we notice that $\tilde\varepsilon_{t} = \varepsilon_{t} + B\eta_t \independent (A+B\Kh_t)x_t$. Then
\begin{equation*}
\E \sum_{t=1}^{T} \tilde\varepsilon_{t}^\top P (A+B\Kh_t)x_t = 0
.\end{equation*}
Next consider
\begin{equation*}
\begin{split}
&\E (\sum_{t=1}^{T} \tilde\varepsilon_{t}^\top P (A+B\Kh_t)x_t)^2 \\
=&\sum_{t=1}^{T} \E (\tilde\varepsilon_{t}^\top P (A+B\Kh_t)x_t)^2 \\
\le & \sum_{t=1}^{T} \E \norm{\tilde\varepsilon_{t}}^2 \norm{P}^2 \norm{(A+B\Kh_t)}^2 \norm{x_t}^2 \\
& \text{ ($\norm{\Kh_t} \le C_K$ based on \cref{alg:myAlg} design)} \\
\le & \sum_{t=1}^{T} \norm{P}^2 (\norm{A} + \norm{B} C_K)^2 \E \norm{\tilde\varepsilon_{t}}^2 \E \norm{x_t}^2 \\
= & \calO(1) \E \sum_{t=1}^{T} \norm{x_t}^2 \\
& \text{ (because of \cref{lem:bound_covariance} } \E \sum_{t=1}^{T} \norm{x_t}^2 = \calO(T\log^2(T) )) \\
= & \calO(T\log^2(T))
.\end{split}
\end{equation*}
Thus
\begin{equation}
\label{eq: regret 2nd item}
\sum_{t=1}^{T} \tilde\varepsilon_{t}^\top P (A+B\Kh_t)x_t = \calO_p(T^{1/2}\log(T))
.\end{equation}
\item The third term we consider is $\sum_{t=1}^{T} \tilde\varepsilon_{t}^\top P \tilde\varepsilon_{t}$. The expectation is
\begin{equation*}
\begin{split}
&\E \sum_{t=1}^{T} \tilde\varepsilon_{t}^\top P \tilde\varepsilon_{t} \\
=& \sum_{t=1}^{T} \Tr (P \E \tilde\varepsilon_{t} \tilde\varepsilon_{t}^\top ) \\
=& \sum_{t=1}^{T} \Tr (P (\sigma^2 I_n + \tau^2 t^{\beta-1}\log^\alpha(t) B B^\top )) \\
=& T \sigma^2\Tr (P) + \frac{\tau^2}{\beta} T^\beta \log^\alpha(T) \Tr(B^\top P B) (1+o(1))
\quad
\text{(By \cref{eq: sum eta_t})}
.\end{split}
\end{equation*}
On the other hand, the variance is the sum of variances for each single summand with total order $\calO(T)$. As a result, when $\beta > 1/2$ or $\beta = 1/2, \alpha > 0$
\begin{equation}
\label{eq: regret 3rd item}
\sum_{t=1}^{T} \tilde\varepsilon_{t}^\top P \tilde\varepsilon_{t}
= T \sigma^2\Tr (P) + \frac{\tau^2}{\beta} T^\beta \log^\alpha(T) \Tr(B^\top P B) (1+o_p(1))
.\end{equation}
\end{enumerate}
Summing up all three parts we have: when $\beta > 1/2$, or $\beta = 1/2, \alpha > 1$,
\begin{equation}
\sum_{t=1}^{T} x_t^\top Qx_t + \tilde{u}_t^\top R\tilde{u}_t = T \sigma^2\Tr (P)
+ \frac{\tau^2}{\beta} T^\beta \log^\alpha(T) \Tr(B^\top P B) (1+o_p(1))
.\end{equation}
Taking the transformation part into consideration (\cref{eq: regret first part}):
\begin{equation*}
\sum_{t=1}^{T} u_t^\top Ru_t - \tilde{u}_t^\top R\tilde{u}_t =
\tau^2\frac{T^{\beta}}{\beta}\log^\alpha(T)\Tr(R)(1+o_p(1))
.\end{equation*}
Finally we have when $\beta > 1/2$, or $\beta = 1/2, \alpha > 1$
\begin{align*}
\mathcal{J}(U,T)
&= \frac{1}{T}
\left[\sum_{t=1}^{T} x_t^\top Qx_t + u_t^\top Ru_t\right] \\
&= \sigma^2\Tr (P) + \tau^2\beta^{-1} \Tr(B^\top PB +R)T^{\beta-1}\log^\alpha(T) (1+ o_p(1))
.\end{align*}
Finally we only need to prove that the optimal average cost can be expressed as:
\begin{align*}
\mathcal{J}(U^*,T)
&= \sigma^2\Tr (P) + \calO_p(T^{-1/2}\log(T))
.\end{align*}
\subsection{Optimal average cost}
\label{subsection: Optimal average cost}
Denote the states and actions following policy $U^*(H_t) = Kx_t$ as $x'_t$ and $u'_t$. Following \cref{eq: regret decomposition} we know that
\begin{equation*}
\begin{split}
&\sum_{t=1}^{T} (x'_t)^\top Qx'_t + (u'_t)^\top Ru'_t \\
=& \sum_{t=1}^{T} \Big[ (x'_t)^\top (Q + K^\top R K)x'_t + (x'_t)^\top (A+BK)^\top P (A+BK)x'_t - (x'_t)^\top P x'_{t} \\
&+ 2 \varepsilon_{t}^\top P (A+B\Kh_t)x_t + \varepsilon_{t}^\top P \varepsilon_{t}
\Big]
+\logO_p(1)
.\end{split}
\end{equation*}
Following \cref{lem: useful lemma from fazel}, since our $\Kh$ is exactly $K$:
\begin{equation*}
(x'_t)^\top (Q + K^\top R K)x'_t + (x'_t)^\top (A+BK)^\top P (A+BK)x'_t - (x'_t)^\top P x'_{t} = 0
\end{equation*}
The remaining terms can be considered in exactly same way as \cref{eq: regret 2nd item} and \cref{eq: regret 3rd item}, which turn out to be:
\begin{equation*}
\sum_{t=1}^{T} \tilde\varepsilon_{t}^\top P (A+B\Kh_t)x_t = \calO_p(T^{1/2}\log(T))
,\end{equation*}
and
\begin{equation*}
\sum_{t=1}^{T} \varepsilon_{t}^\top P \varepsilon_{t}
= T \sigma^2\Tr (P) + \calO_p(T^{1/2})
.\end{equation*}
Finally we arrive at the conclusion that
\begin{align*}
\mathcal{J}(U^*,T)
&= \frac1T \left(\sum_{t=1}^{T} (x'_t)^\top Qx'_t + (u'_t)^\top Ru'_t\right) \\
&= \frac1T \left(\calO_p(T^{1/2}\log(T)) + T \sigma^2\Tr (P) + \calO_p(T^{1/2})\right) \\
&= \sigma^2\Tr (P) + \calO_p(T^{-1/2}\log(T))
.\end{align*}
\section{The proof of \cref{thm:main CLT}}
\label{The proof of thm:main CLT}
\begin{thm*}
\cref{alg:myAlg} applied to a system described by \cref{eq:LinearModel} under \cref{asm:InitialStableCondition} satisfies, as $t \to \infty$,
\begin{equation*}
\vvector \left[
\begin{bmatrix}
\Ah_t - A,\Bh_t- B
\end{bmatrix} D_t\right] \convD
\calN(0, \sigma^2 I_{\statedim(n+d)} )
.\end{equation*}
\end{thm*}
\begin{proof}
One can find the definition of $D_t$ in \cref{eq:D_t Definition}.
The proof heavily relies on the following theorems from \citet{anderson1992asymptotic}. For better understanding, we directly state those theorems with the same notation as our paper.
\begin{theorem}[Theorems 1 and 3 in \citet{anderson1992asymptotic}]
\label{thm: more general CLT}
Let $\left\{x_i, u_i, \varepsilon_i \right\}$, $i=0,1 \cdots$, be a sequence of random vectors described by \cref{eq:LinearModel} under \cref{asm:InitialStableCondition}, and let $\left\{\calF_i\right\}$ be an increasing sequence of $\sigma$-fields such that $\left\{x_i, u_i\right\}$ is $\calF_{i-1}$ measureable and $\varepsilon_i$ is $\calF_i$ measurable. Let the matrix $D_t$ be a deterministic matrix such that
\begin{equation}
\label{eq: more general CLT condition 1 old}
D_t^{-1}\sum_{i=0}^{t-1} \begin{bmatrix} x_i\\ u_i\\ \end{bmatrix} \begin{bmatrix} x_i\\ u_i\\ \end{bmatrix}^\top (D_t^\top)^{-1} \convP C
,\end{equation}
where $C$ is a constant matrix, and
\begin{equation}
\label{eq: more general CLT condition 2}
\max_{1 \le i \le t}
\begin{bmatrix} x_i\\ u_i\\ \end{bmatrix}^\top
(D_tD_t^\top)^{-1}
\begin{bmatrix} x_i\\ u_i\\ \end{bmatrix} \convP 0
.\end{equation}
Suppose further that $\E(\varepsilon_i|\calF_{i-1}) = 0 $ a.s., $\E(\varepsilon_i\varepsilon_i^\top|\calF_{i-1}) = \Sigma_i$ a.s.,
\begin{equation}
\label{eq: trivial condition 1}
\sum_{i=0}^{t-1} \left[\Sigma_i \otimes D_t^{-1}\begin{bmatrix} x_i\\ u_i\\ \end{bmatrix} \begin{bmatrix} x_i\\ u_i\\ \end{bmatrix}^\top (D_t^\top)^{-1}\right] \convP \Sigma \otimes C,
\end{equation}
where $\Sigma$ is a constant positive semi-definite matrix and
\begin{equation}
\label{eq: trivial condition 2}
\sup_{i \ge 1} \E \left[ \varepsilon_i^\top \varepsilon_i \bm{1}_{\varepsilon_i^\top \varepsilon_i > a} | \calF_{i-1} \right] \convP 0
,\end{equation}
as $a \to \infty$. Then
\begin{equation}
\vvector \left[
\begin{bmatrix}
\Ah_t - A,\Bh_t- B
\end{bmatrix} D_t\right] \convD \calN(0, C^{-1} \otimes \Sigma)
.\end{equation}
\end{theorem}
As we have seen in \cref{alg:myAlg} the controller $\Kh_t$ is fully determined by $\{x_i, u_i\}_{i=0}^{t-1}$. Pick
\[\calF_{t-1} =\sigma(\{x_i, u_i, \eta_i\}_{i=0}^{t}, \{\varepsilon_i\}_{i=0}^{t-1}).\]
Now we verified the design vector $\begin{bmatrix}
x_t \\
u_t
\end{bmatrix}$ at stage $t$ is $\calF_{t-1}$ measurable. Since $\varepsilon_t \iid \calN(0, \sigma ^2 I_\inputdim)$, we know that $\varepsilon_t \independent \mathcal{F}_{t-1}$, and $\left\{\varepsilon_t \right\}$ is a martingale difference sequence with respect to an increasing sequence of $\sigma$-fields $\left\{\mathcal{F}_t \right\}$.
\cref{eq: trivial condition 1} holds by the fact that all variances $\Sigma_i = \sigma^2 I_\inputdim$ and \cref{eq: more general CLT condition 1 old}. For \cref{eq: trivial condition 2}, notice that we can remove the $\sup$ since every term has the same value, so the conclusion follows from a standard property of Gaussian distributions.
Actually, \cref{eq: more general CLT condition 1 old} is already shown in \cref{thm:main tool}.
\cref{eq: more general CLT condition 2} requires less effort to prove as we defined $D_t$ by
\begin{equation}
\label{eq: D_t defn}
D_t :=
t^{\beta/2}\log^{\alpha/2}(t)
\left[
\begin{array}{cc}
I_n & 0\\
K & I_d\\
\end{array}
\right]
\left[
\begin{array}{cc}
C_t^{1/2} & 0\\
0 & \sqrt{\frac{\tau^2}{\beta}} I_d\\
\end{array}
\right]
.\end{equation}
As a result, \cref{eq: more general CLT condition 2} is not surprising since $z_t$ should be only of constant order.
\begin{comment}
Finally, we will translate \cref{eq:final Conclusion} into our confidence region at \cref{eq: ellipsoid 2}
\begin{equation*}
\Tr\left[ (\hat{\Theta}_t -\Theta) \left(\sum_{i=0}^{t-1}\begin{bmatrix} x_i\\ u_i\\ \end{bmatrix} \begin{bmatrix} x_i\\ u_i\\ \end{bmatrix}^\top \right) (\hat{\Theta}_t -\Theta)^\top \right]
\convD \chi^2_{n(n+d)}
.\end{equation*}
To summarize,
there are only two remaining parts not proven:
\begin{enumerate}
\item Prove \cref{eq: more general CLT condition 2} (see \cref{5th part of proving main thm}):
\begin{equation*}
\max_{i=1,2\cdots, n} z_i^\top (D_tD_t^\top)^{-1}z_i \convP 0
.\end{equation*}
\item Prove \cref{eq: ellipsoid 2} with Slutsky's theorem (see \cref{subsection: Ellipsoid Confidence Region})
\begin{equation*}
\Tr\left[ (\hat{\Theta}_t -\Theta) \left(\sum_{i=0}^{t-1}\begin{bmatrix} x_i\\ u_i\\ \end{bmatrix} \begin{bmatrix} x_i\\ u_i\\ \end{bmatrix}^\top \right) (\hat{\Theta}_t -\Theta)^\top \right]
\convD \chi^2_{n(n+d)}
.\end{equation*}
\end{enumerate}
Now we will examine them in order.
\end{comment}
\subsection{The proof of \cref{eq: more general CLT condition 2}}
\label{5th part of proving main thm}
Since
\begin{align*}
D_tD_t^\top
= t^{\beta}\log^{\alpha}(t)
\left[
\begin{array}{cc}
I_n & 0\\
K & I_d\\
\end{array}
\right]
\left[
\begin{array}{cc}
C_t & 0\\
0 & \frac{\tau^2}{\beta} I_d\\
\end{array}
\right]
\left[
\begin{array}{cc}
I_n & K^\top\\
0 & I_d\\
\end{array}
\right],
\end{align*}
we have
\begin{align}
\label{eq: DtDt -1 order}
\begin{split}
(D_tD_t^\top)^{-1}
=& t^{-\beta}\log^{-\alpha}(t)
\left[
\begin{array}{cc}
I_n & -K^\top\\
0 & I_d\\
\end{array}
\right]
\left[
\begin{array}{cc}
C_t^{-1} & 0\\
0 & \frac{\tau^2}{\beta} I_d\\
\end{array}
\right]
\left[
\begin{array}{cc}
I_n & 0\\
-K & I_d\\
\end{array}
\right]
\\
=& \calO(t^{-\beta}\log^{-\alpha}(t)) \quad \text{(by \cref{eq: Ct inverse order})}.
\end{split}
\end{align}
Recall that \cref{eq: more general CLT condition 2} is
\begin{equation*}
\max_{1\le i \le t}
\begin{bmatrix} x_i\\ u_i\\ \end{bmatrix}^\top
(D_tD_t^\top)^{-1}
\begin{bmatrix} x_i\\ u_i\\ \end{bmatrix} \convP 0
.\end{equation*}
It suffices to show
\[t^{-\beta/2}\log^{-\alpha/2}(t)\max_{1\le i \le t}\norm{x_i} \convP 0 \;\text{ and }\;t^{-\beta/2}\log^{-\alpha/2}(t)\max_{1\le i \le t}\norm{u_i} \convP 0.\]
Actually we already shown in \cref{lemma: Hi prob bounds in theorem 2} that
\begin{equation*}
\norm{x_t}, \norm{u_t} = \calO(\log^{1/2}(t)) \as \end{equation*}
This is a uniform bound over $t$, thus a direct corollary is
\begin{equation*}
\max_{1\le i \le t}\norm{x_i} , \max_{1\le i \le t}\norm{u_i} = \calO(\log^{1/2}(t)) \as
\end{equation*}
That immediately implies
\[t^{-\beta/2}\max_{1\le i \le t}\norm{x_i} \asConv 0 \;\text{ and }\; t^{-\beta/2}\max_{1\le i \le t}\norm{u_i} \asConv 0.\]
\end{proof}
\section{The proof of \cref{thm:prediction CLT parametric}}
\label{The proof of thm:prediction CLT parametric}
Here we state and prove a generalization of \cref{thm:prediction CLT parametric} that allows for the case when $\beta=1$ and $\alpha\le 0$.
\begin{thm*}
\cref{alg:myAlg} applied to a system described by \cref{eq:LinearModel} under \cref{asm:InitialStableCondition} satisfies, as $t \to \infty$,
\begin{align}
\left(
x_t^\top
\left( \sum_{p=0}^{\infty}(A+BK) ^{p}
\left(I_n + 1_{\{\beta=1,\alpha=0\}}\frac{\tau^2}{\sigma^2}BB^\top\right)
\left((A+BK) ^{p}\right)^\top\right)^{-1}x_t
+
\beta \sigma^2
\lnorm{w_t}^2
\right)^{-1/2} \nonumber\\
\hspace{-2mm}
\cdot\, t^{1/2} \left((\Ah_t - A)x_t + (\Bh_t- B)u_t\right)
\convD \calN(0,I_n).
\end{align}
\end{thm*}
\begin{proof}
We can generalize the input noise $\eta_t$ to $\xi_t$ which is any random vector independent of the data before $t$: $\{\varepsilon_i, \eta_i\}_{i=0}^{t-1}$. Hereafter, $u_t = \Kh_t x_t + \xi_t$ (but $u_i$ for $i < t$ is still $\Kh_i x_i + \eta_i$).
The proof will proceed by showing that $(\Ah_t, \Bh_t)$ acts as if it were independent of $(x_t, u_t)$, and then effectively conditioning on $(x_t, u_t)$ and using $(\Ah_t, \Bh_t)$'s asymptotic distribution from \cref{thm:main CLT}.
Define $\rhoL := \frac{2 + \rho(L)}{3}$ as in \cref{lemma: Hi prob bounds in theorem 2}.
Define replacements of $x_t$ and $u_t$ which are independent of $\Ah_{t-\myfloor{-\frac{\log(t)}{\log(\rho_L)}}}$ and $\Bh_{t-\myfloor{-\frac{\log(t)}{\log(\rho_L)}}}$:
\begin{equation}
\label{eq: tilde x t definition}
\tilde{x}_t := \sum_{p=t-\myfloor{-\frac{\log(t)}{\log(\rho_L)}}}^{t-1}(A+BK)^{t-p-1} ( B\eta_p + \varepsilon_p)
,\end{equation}
and
\begin{equation}
\label{eq: tilde u t definition}
\tilde{u}_t := K\tilde{x}_t + \xi_t= K\sum_{p=t-\myfloor{-\frac{\log(t)}{\log(\rho_L)}}}^{t-1}(A+BK)^{t-p-1} ( B\eta_p + \varepsilon_p) + \xi_t
.\end{equation}
We can show that the difference between $\tilde{x}_t, \tilde{u}_t$ and $x_t, u_t$ is very small:
\begin{lemma}
\label{lem: difference between real and substitutes x u}
\begin{equation*}
x_t = \tilde{x}_t + O(t^{-\frac{\beta}{2}} \log^{\frac{-\alpha + 2}{2}}(t)) \as
\end{equation*}
\begin{equation*}
u_t = \tilde{u}_t + O(t^{-\frac{\beta}{2}} \log^{\frac{-\alpha + 2}{2}}(t)) \as
\end{equation*}
\end{lemma}
The proof can be found in \cref{The proof of lem: difference between real and substitutes x u}. At the same time, the difference between $\Ah_{t-\myfloor{-\frac{\log(t)}{\log(\rho_L)}}}$, $\Bh_{t-\myfloor{-\frac{\log(t)}{\log(\rho_L)}}}$ and $\Ah_t, \Bh_t$ is also small:
\begin{lemma}
\label{lem: difference between real and substitutes A B}
\begin{equation*}
\Ah_{t} = \Ah_{t-\myfloor{-\frac{\log(t)}{\log(\rho_L)}}} + \calO_p(t^{-\beta}\log^{-\alpha+3/2}(t))
.\end{equation*}
\begin{equation*}
\Bh_{t} = \Bh_{t-\myfloor{-\frac{\log(t)}{\log(\rho_L)}}} + \calO_p(t^{-\beta}\log^{-\alpha+3/2}(t))
.\end{equation*}
\end{lemma}
The proof can be found in \cref{The proof of lem: difference between real and substitutes A B}.
These substitutions are very close to our original concern, and they have the good independence property:
\begin{equation*}
\left(\Ah_{t-\myfloor{-\frac{\log(t)}{\log(\rho_L)}}} - A, \Bh_{t-\myfloor{-\frac{\log(t)}{\log(\rho_L)}}}- B\right) \independent (\tilde{x}_t, \tilde{u}_t)
.\end{equation*}
This is because $\Ah_{t-\myfloor{-\frac{\log(t)}{\log(\rho_L)}}}$ and $\Bh_{t-\myfloor{-\frac{\log(t)}{\log(\rho_L)}}}$ are only functions of the system up to time $t-\myfloor{-\frac{\log(t)}{\log(\rho_L)}}-1$, while $\tilde{x}_t$ and $\tilde{u}_t$ are independent with event before time $t-\myfloor{-\frac{\log(t)}{\log(\rho_L)}}$ by definitions in \cref{eq: tilde x t definition,eq: tilde u t definition}.
Our initial target is to identify the distribution of $(\Ah_t - A)x_t + (\Bh_t- B)u_t$. We will start from its substitution
\begin{equation*}
\begin{split}
&(\Ah_{t-\myfloor{-\frac{\log(t)}{\log(\rho_L)}}} - A)\tilde{x}_t + (\Bh_{t-\myfloor{-\frac{\log(t)}{\log(\rho_L)}}}- B)\tilde{u}_t \\
=& ((\Ah_{t-\myfloor{-\frac{\log(t)}{\log(\rho_L)}}} - A) + K (\Bh_{t-\myfloor{-\frac{\log(t)}{\log(\rho_L)}}}- B))\tilde{x}_t + (\Bh_{t-\myfloor{-\frac{\log(t)}{\log(\rho_L)}}}- B)\xi_t
.\end{split}
\end{equation*}
Because of this independence after substitution, the first term is independent with the second term, and their asymptotic distribution can be described by \cref{eq: fast slow rate CLT}.
\begin{lemma}
\label{lem: CLT of substitution}
For any $\xi_t$ independent of the data before $t$: $\{\varepsilon_i, \eta_i\}_{i=0}^{t-1}$:
\begin{align*}
&\left(
\tilde{x}_t^\top
\left( \sum_{p=0}^{\infty}L ^{p}\left(I_n + 1_{\{\beta=1,\alpha=0\}}\frac{\tau^2}{\sigma^2}BB^\top\right)(L ^{p})^\top\right)^{-1}
\tilde{x}_t
+
\frac{\beta \sigma^2}{\tau^2}
t^{1-\beta} \log^{-\alpha}(t)
\lnorm{\xi_t}^2
\right)^{-1/2} \\
&
\hspace{2cm} \cdot t^{1/2} \left[(\Ah_{t-\myfloor{-\frac{\log(t)}{\log(\rho_L)}}} - A)\tilde{x}_t +
(\Bh_{t-\myfloor{-\frac{\log(t)}{\log(\rho_L)}}}- B)(K\tilde{x}_t+\xi_t)\right]
\convD
\calN(0,I_n).
\end{align*}
\end{lemma}
The proof of \cref{lem: CLT of substitution} can be found in \cref{The proof of lem: CLT of substitution}.
With the help of \cref{lem: difference between real and substitutes x u} and \cref{lem: difference between real and substitutes A B}
, we can change all the replacements back to the original form:
\begin{lemma}
\label{lem: CLT original}
For any $\xi_t$ independent of the data before $t$: $\{\varepsilon_i, \eta_i\}_{i=0}^{t-1}$,
\begin{align*}
&\left(
x_t^\top
\left( \sum_{p=0}^{\infty}L ^{p}\left(I_n + 1_{\{\beta=1,\alpha=0\}}\frac{\tau^2}{\sigma^2}BB^\top\right)(L ^{p})^\top\right)^{-1}
x_t
+
\frac{\beta \sigma^2}{\tau^2}
t^{1-\beta} \log^{-\alpha}(t)
\lnorm{\xi_t}^2
\right)^{-1/2} \\
&
\hspace{2cm} \cdot t^{1/2} \left[(\Ah_{t} - A)x_t +
(\Bh_{t}- B)(\Kh_tx_t+\xi_t)\right]
\convD
\calN(0,I_n)
.\end{align*}
\end{lemma}
The proof of \cref{lem: CLT original} can be found in
\cref{The proof of lem: CLT original}.
Since $\eta_t$ is independent with$\{\varepsilon_i, \eta_i\}_{i=0}^{t-1}$, which satisfies the condition of $\xi_t$, we can restate the result with $\eta_t$ replaced by $\xi_t$:
\begin{align*}
&\left(
x_t^\top
\left( \sum_{p=0}^{\infty}L ^{p}\left(I_n + 1_{\{\beta=1,\alpha=0\}}\frac{\tau^2}{\sigma^2}BB^\top\right)(L ^{p})^\top\right)^{-1}
x_t
+
\frac{\beta \sigma^2}{\tau^2}
t^{1-\beta} \log^{-\alpha}(t)
\lnorm{\eta_t}^2
\right)^{-1/2} \\
&
\hspace{2cm} \cdot t^{1/2} \left[(\Ah_{t} - A)x_t +
(\Bh_{t}- B)(\Kh_tx_t+\eta_t)\right]
\convD
\calN(0,I_n)
.\end{align*}
Finally, we have the desired conclusion using $\eta_t = \tau\sqrt{t^{\beta-1}\log^\alpha(t)}\,w_t$:
\begin{align*}
\left(
x_t^\top
\left( \sum_{p=0}^{\infty}L ^{p}\left(I_n + 1_{\{\beta=1,\alpha=0\}}\frac{\tau^2}{\sigma^2}BB^\top\right)(L ^{p})^\top\right)^{-1}
x_t
+
\beta \sigma^2
\lnorm{w_t}^2
\right)^{-1/2} \\
\cdot \; t^{1/2} \left((\Ah_t - A)x_t + (\Bh_t- B)u_t\right)
\convD
\calN(0,I_n)
.\end{align*}
\end{proof}
\section{The proof of Corollaries}
\subsection{The proof of \cref{corr: K CLT parametric}}
\label{subsection: The proof of K CLT parametric}
\begin{corr*}
Assume $A+BK$ is full rank.
\cref{alg:myAlg} applied to a system described by \cref{eq:LinearModel} under \cref{asm:InitialStableCondition} satisfies
\begin{equation*}
\sqrt{\frac{\tau^2}{\sigma^2\beta}}t^{\beta/2} \log^{\alpha/2}(t)
\left(\left(\frac{dK}{d[A,B]}\right)
\left(\begin{bmatrix}
-K^\top \\
I_d
\end{bmatrix}
\otimes
I_n
\right)
\right)^{-1}
\vvector\left(\Kh_t - K\right)
\convD
\calN(0, I_{nd})
.\end{equation*}
\end{corr*}
\begin{proof}
Before we prove this result, we should first examine that the matrix $\left(\frac{dK}{d[A,B]}\right)
\left(\begin{bmatrix}
-K^\top \\
I_d
\end{bmatrix}
\otimes
I_n
\right)$ is indeed invertible. Since$\left(\begin{bmatrix}
-K^\top \\
I_d
\end{bmatrix}
\otimes
I_n
\right)$ has an identity matrix component $I_{dn}$, it is sufficient to show that $\frac{dK}{d[A,B]}$ is full rank.
\subsubsection{$\frac{dK}{d[A,B]}$ is full rank}
We can ignore the effect of $K_0$ and consider $\Kh_t$ to be the same as certainty equivalent controller $\Kt_t$ which is directly calculated by plugging $\Ah_{t-1}, \Bh_{t-1}$ into DARE \cref{eq:ControllerK,eq:riccati}. This is because $\Kh_t = K_0$ only happens finitely often and thus does not affect asymptotic properties; see \cref{section: The proof of one_epoch_estimate_withMyalg}.
Before we start, we need to define how we solve $\frac{dK}{d[A,B]} \in \mathbb{R}^{nd \times n(n+d)}$ and then prove that $\frac{dK}{d[A,B]}$ is indeed a full rank matrix.
Lemmas 3.1 and B.1 from \citet{simchowitz2020naive} gives the relationship between the derivatives of $K, P, A, B$:
\begin{equation}
\label{eq: dK}
dK = -(R+B^\top P B)^{-1}(dB^\top P (A+BK) + B^\top P (dA +dB K) + B^\top dP (A+BK))
,\end{equation}
where $dP$ can be solved from
\begin{equation}
\label{eq: dP}
(A+BK)^\top dP (A+BK) -dP + (dA +dBK)^\top P (A+BK) + (A+BK)^\top P (dA +dBK) = 0
.\end{equation}
Now we can solve $\frac{dK}{d[A,B]}$ by \cref{eq: dK} and \cref{eq: dP}. Denote the kernel space of the derivative matrix $\frac{dK}{d[A,B]}$ as $\mathcal{S}$. It suffices to show that $\mathcal{S}$'s dimension is $n(n+d) - nd = n^2$, which implies $\frac{dK}{d[A,B]}$ is full rank with rank $nd$. The equivalent definition of kernel space $\mathcal{S}$ is the small perturbation $\vvector[dA, dB]$ such that $K$ does not change ($dK = 0$):
\begin{equation*}
dK = \frac{dK}{d[A,B]}\vvector(dA, dB) = 0.
\end{equation*}
Any vector in kernel space $\mathcal{S}$ can be considered as $\vvector[dA, dB]$ which satisfies $dK = 0$ in \cref{eq: dK}, and that means:
\begin{equation}
\label{eq: kernel space eq1}
dB^\top P (A+BK) + B^\top P (dA +dB K) + B^\top dP (A+BK) = 0
.\end{equation}
On the other hand, \cref{eq: dP} describes a linear recursive relationship between $dP$ and $dA +dB K$, so that we can solve $dP$ with the infinite summation:
\begin{align*}
dP =& (A+BK)^\top dP (A+BK) + (dA +dBK)^\top P (A+BK) + (A+BK)^\top P (dA +dBK) \\
=& ((A+BK)^\top)^2 dP (A+BK)^2 \\
&+
(A+BK)^\top\left((dA +dBK)^\top P (A+BK) + (A+BK)^\top P (dA +dBK)\right) (A+BK) \\
&+
(dA +dBK)^\top P (A+BK) + (A+BK)^\top P (dA +dBK)\\
& \text{(recursively plugging in the first equation)} \\
=&
\sum_{i=0}^\infty
((A+BK)^\top)^i
\left((dA +dBK)^\top P (A+BK) + (A+BK)^\top P (dA +dBK)\right)
(A+BK)^i
.\end{align*}
Also recall that $A+BK$ is assumed to be full rank matrix, and we can show that $P$ is also full rank; see \cref{section: The proof of one_epoch_estimate_withK}. Thus we can explicitly solve $dB$ from \cref{eq: kernel space eq1} as a linear equation with regard to $dA + dB K$:
\begin{equation*}
dB^\top = -(P (A+BK))^{-1}(B^\top P (dA +dB K) + B^\top dP (A+BK) )
.\end{equation*}
This tells us the kernel space $\mathcal{S}$ is the image of a function of its linear subspace $dA +dB K \in \mathbb{R}^{n^2}$, which means $dim(\mathcal{S}) \le n^2$. Notice by kernel space definition its dimension should be at least $dim(\mathcal{S}) \ge n(n+d) - nd = n^2$, where the equality is achieved when $\frac{dK}{d[A,B]}$ has full rank $nd$. Combining these two equations we have $dim(\mathcal{S}) = n^2$. Finally we arrived at the desired conclusion that dimension of $\frac{dK}{d[A,B]} \in \mathbb{R}^{nd \times n(n+d)}$'s kernel space $\mathcal{S}$ is exactly $n^2$, which means $\frac{dK}{d[A,B]}$ is full rank.
Next we describe the rest of the proof:
\subsubsection{Proof by the Delta method}
By Taylor expansion and the consistency of $[\Ah_t, \Bh_t]$ (see \cref{prop:one_epoch_estimate}), we have
\begin{equation*}
\vvector\left(\Kh_t - K\right)
= \left(\frac{dK}{d[A,B]}\right) \vvector\left[\Ah_t- A, \Bh_t-B\right]
(1 + o_p(1))
.\end{equation*}
From \cref{remark: fast convergence rate} we know
\begin{equation*}
\Ah_t- A = (\Bh_t-B)(-K) (1+o_p(1))
.\end{equation*}
Then
\begin{equation*}
\vvector\left(\Kh_t - K\right)
= \left(\frac{dK}{d[A,B]}\right) \vvector\left((\Bh_t-B)
\begin{bmatrix}
-K, & I_d
\end{bmatrix}\right)
(1 + o_p(1))
.\end{equation*}
which can be written as
\begin{equation*}
\vvector\left(\Kh_t - K\right)
= \left(\frac{dK}{d[A,B]}\right) \left(\begin{bmatrix}
-K^\top \\
I_d
\end{bmatrix}
\otimes
I_n
\right)
\vvector\left(\Bh_t-B\right)
(1 + o_p(1))
.\end{equation*}
By \cref{eq: fast slow rate CLT},
\begin{equation*}
\sqrt{\frac{\tau^2}{\sigma^2\beta}}t^{\beta/2} \log^{\alpha/2}(t)\vvector\left(\Bh_t-B\right) \convD \calN(0, I_{nd})
.\end{equation*}
Combining the above two equations, finally we have
\begin{equation*}
\sqrt{\frac{\tau^2}{\sigma^2\beta}}t^{\beta/2} \log^{\alpha/2}(t)
\vvector\left(\Kh_t - K\right)
\convD \left(\frac{dK}{d[A,B]}\right)
\left(\begin{bmatrix}
-K^\top \\
I_d
\end{bmatrix}
\otimes
I_n
\right)
\calN(0, I_{nd})
.\end{equation*}
From the fact that $\frac{dK}{d[A,B]}$ is full rank and that $\left(\begin{bmatrix}
-K^\top \\
I_d
\end{bmatrix}
\otimes
I_n
\right)$ has an identity matrix component $I_{dn}$, we can take matrix inverse and get
\begin{equation*}
\sqrt{\frac{\tau^2}{\sigma^2\beta}}t^{\beta/2} \log^{\alpha/2}(t)
\left(\left(\frac{dK}{d[A,B]}\right)
\left(\begin{bmatrix}
-K^\top \\
I_d
\end{bmatrix}
\otimes
I_n
\right)
\right)^{-1}
\vvector\left(\Kh_t - K\right)
\convD
\calN(0, I_{nd})
.\end{equation*}
\end{proof}
\subsection{The proof of \cref{corr:regret}}
\label{The proof of corr:regret}
\begin{corr*}
The average regret of the controller $U$ defined by \cref{alg:myAlg} applied through time horizon $T$ to a system described by \cref{eq:LinearModel} under \cref{asm:InitialStableCondition} satisfies, as $t \to \infty$ and $T \to \infty$,
\begin{equation}
\label{eq:regret my alg observable 2}
\frac{\mathcal{R}(U,T)}{\tau^2\beta^{-1} \Tr(\Bh_{t}^\top \Ph_t \Bh_{t} +R)T^{\beta-1}\log^\alpha(T)} \convP 1
.\end{equation}
\end{corr*}
\begin{proof}
This is a direct corollary from \cref{thm:regret}, which states
\begin{equation*}
\frac{\mathcal{R}(U,T)}{\tau^2\beta^{-1} \Tr(B^\top PB +R)T^{\beta-1}\log^\alpha(T)} \convP 1,
\end{equation*}
and
from \cref{prop:one_epoch_estimate} and \cref{cor: Ph controlled with Ah and Bh} which implies the consistency of $\Bh_t$ and $\Ph_t$. By Slutsky's theorem we can replace the parameters $B$ and $P$ in \cref{eq:regret my alg observable 2} with $\Bh_t$ and $\Ph_t$.
\end{proof}
\subsection{The proof of \cref{thm:main}}
\label{The proof of thm:main}
\begin{corr*}
\cref{alg:myAlg} applied to a system described by \cref{eq:LinearModel} under \cref{asm:InitialStableCondition} satisfies
\begin{equation*}
\Tr\left(
\begin{bmatrix}
\Ah_t - A,\Bh_t- B
\end{bmatrix}
\sum_{i=0}^{t-1}
\begin{bmatrix}
x_i\\
u_i\\
\end{bmatrix}
\begin{bmatrix}
x_i\\
u_i\\
\end{bmatrix}^\top
\begin{bmatrix}
\Ah_t - A,\Bh_t- B
\end{bmatrix} ^\top
\right)
\convD \sigma^2\chi^2_{n(n+d)}
.\end{equation*}
\end{corr*}
\begin{proof}
For notational simplicity denote
$\hat{\Theta}_t := \begin{bmatrix}
\Ah_t, \Bh_t
\end{bmatrix}$
and
$\Theta := \begin{bmatrix}
A, B
\end{bmatrix}$
. By \cref{thm:main CLT} we know
\begin{equation}
\label{eq: thm3 result}
\vvector \left((\hat{\Theta}_t -\Theta)D_t \right) \convD
\calN(0, \sigma^2 I_{\statedim(n+d)} )
.\end{equation}
Potentially we can derive an ellipsoid "confidence region" with the above formula by
\begin{equation}
\label{eq: ellipsoid 1}
\Tr\left( (\hat{\Theta}_t -\Theta) \left(D_tD_t^\top\right) (\hat{\Theta}_t -\Theta)^\top \right)
\convD \sigma^2 \chi^2_{n(n+d)}
.\end{equation}
However, since a true confidence region should not require any knowledge on oracle parameters, we need to replace $D_tD_t^\top$ with some observable expression, which turns out to be:
\begin{equation*}
\Tr\left( (\hat{\Theta}_t -\Theta) \left(\sum_{i=0}^{t-1}\begin{bmatrix} x_i\\ u_i\\ \end{bmatrix} \begin{bmatrix} x_i\\ u_i\\ \end{bmatrix}^\top \right) (\hat{\Theta}_t -\Theta)^\top \right)
\convD \sigma^2 \chi^2_{n(n+d)}
.\end{equation*}
Next we will explain why it is valid to replace $D_tD_t^\top$ by $\sum_{i=0}^{t-1}\begin{bmatrix} x_i\\ u_i\\ \end{bmatrix} \begin{bmatrix} x_i\\ u_i\\ \end{bmatrix}^\top$. We know from \cref{eq: ellipsoid 1} that
\begin{equation*}
\Tr\left( (\hat{\Theta}_t -\Theta) \left(D_t I_{n+d} D_t^\top\right) (\hat{\Theta}_t -\Theta)^\top \right)
\convD \sigma^2 \chi^2_{n(n+d)}
,\end{equation*}
and we can replace $I_{n+d}$ by $D_t^{-1}
\sum_{i=0}^{t-1}
\begin{bmatrix}
x_i\\
u_i\\
\end{bmatrix}
\begin{bmatrix}
x_i\\
u_i\\
\end{bmatrix}^\top
(D_t^\top)^{-1}
+ o_p(1)$ thanks to \cref{thm:main tool}. As a result,
\begin{equation*}
\Tr\left( (\hat{\Theta}_t -\Theta)D_t \left(D_t^{-1}
\sum_{i=0}^{t-1}
\begin{bmatrix}
x_i\\
u_i\\
\end{bmatrix}
\begin{bmatrix}
x_i\\
u_i\\
\end{bmatrix}^\top
(D_t^\top)^{-1}
+ o_p(1)\right) D_t^\top(\hat{\Theta}_t -\Theta)^\top \right)
\convD \sigma^2 \chi^2_{n(n+d)}
.\end{equation*}
By \cref{eq: thm3 result},
$\vvector \left((\hat{\Theta}_t -\Theta)D_t\right)$ is of constant order, and thus the $o_p(1)$ can be ignored. Finally, we have
\begin{equation*}
\Tr\left( (\hat{\Theta}_t -\Theta) \left(\sum_{i=0}^{t-1}\begin{bmatrix} x_i\\ u_i\\ \end{bmatrix} \begin{bmatrix} x_i\\ u_i\\ \end{bmatrix}^\top \right) (\hat{\Theta}_t -\Theta)^\top \right)
\convD \sigma^2 \chi^2_{n(n+d)}
.\end{equation*}
\end{proof}
\subsection{The proof of \cref{corr: K confidence region}}
\label{subsection: The proof of K confidence region}
\begin{corr*}
\cref{alg:myAlg} applied to a system described by \cref{eq:LinearModel} under \cref{asm:InitialStableCondition} satisfies
\begin{equation}
\vvector(
\Kh_t - K
) ^\top
\left(
\left(\frac{dK}{d[A, B]}\right)_t
\left(
\sum_{i=0}^{t-1}
\begin{bmatrix}
x_i\\
u_i\\
\end{bmatrix}
\begin{bmatrix}
x_i\\
u_i\\
\end{bmatrix}^\top
\otimes I_n\right)^{-1}
\left(\frac{dK}{d[A, B]}\right)_t^\top
\right)^{-1}
\vvector(
\Kh_t - K
)
\convD \sigma^2\chi^2_{nd}
,\end{equation}
where $\left(\frac{dK}{d[A, B]}\right)_t \in \mathbb{R}^{nd \times n(n+d)}$ is defined as $\frac{dK}{d[A, B]}$ evaluated at $\Ah_{t-1}, \Bh_{t-1}$.
\end{corr*}
\begin{proof}
Again, let us denote $\hat{\Theta}_t := \begin{bmatrix}
\Ah_t, \Bh_t
\end{bmatrix}$
and
$\Theta := \begin{bmatrix}
A, B
\end{bmatrix}$. Starting from \cref{thm:main CLT}
\begin{equation*}
\vvector \left((\hat{\Theta}_t -\Theta)D_t\right) \convD
\calN(0, \sigma^2 I_{\statedim(n+d)} )
,\end{equation*}
we need to transfer $D_t$ to its observable version in terms of the Gram matrix. More specifically, we need to find another matrix $E_t$ which is observable and satisfies:
\begin{itemize}
\item $D_t^{-1}E_t \convP I_{n+d}$ because we want to use Slutsky's theorem.
\item $E_tE_t^\top = \sum_{i=0}^{t-1}
\begin{bmatrix}
x_i\\
u_i\\
\end{bmatrix}
\begin{bmatrix}
x_i\\
u_i\\
\end{bmatrix}^\top$ because
$D_t^{-1}
\sum_{i=0}^{t-1}
\begin{bmatrix}
x_i\\
u_i\\
\end{bmatrix}
\begin{bmatrix}
x_i\\
u_i\\
\end{bmatrix}^\top
(D_t^\top)^{-1} \convP I_{n+d}$.
\end{itemize}
For now let us assume we have already found such matrix $E_t$, and thus we can replace $D_t$ with $E_t$:
\begin{equation*}
\vvector \left(
(\hat\Theta_t - \Theta)E_t
\right) \convD
\calN(0, \sigma^2 I_{\statedim(n+d)} )
.\end{equation*}
That is:
\begin{equation*}
(E_t^\top \otimes I_n) \vvector \left(
\hat\Theta_t - \Theta
\right) \convD
\calN(0, \sigma^2 I_{\statedim(n+d)} )
.\end{equation*}
Further denote $F_t := E_t^\top \otimes I_n$, and then
\begin{equation}
\label{eq: F_t CLT}
F_t \vvector \left(
\hat\Theta_t - \Theta
\right) \convD
\calN(0, \sigma^2 I_{\statedim(n+d)} )
.\end{equation}
By Taylor expansion and the consistency of $\hat\Theta_t$ (see \cref{prop:one_epoch_estimate}), we have
\begin{equation*}
\vvector\left(\Kh_t - K\right)
= \left(\frac{dK}{d\Theta}\right)_t \vvector\left(\hat\Theta_t - \Theta\right)
(1 + o_p(1))
.\end{equation*}
Since we will prove $D_t^{-1}E_t \convP I_{n+d}$ in \cref{Finding a valid E_t}, $E_t$ is asymptotically invertible, which means we can take inverse of $F_t = E_t^\top \otimes I_n$ in asymptotic equations:
\begin{equation*}
\vvector\left(\Kh_t - K\right)
= \left(\frac{dK}{d\Theta}\right)_t (F_t)^{-1}
F_t \vvector\left(\hat\Theta_t - \Theta\right)
(1 + o_p(1))
.\end{equation*}
We have already shown in \cref{subsection: The proof of K CLT parametric} that $\frac{dK}{d\Theta}$ is full rank, in the same way we can prove that $\left(\frac{dK}{d\Theta}\right)_t$ is almost surely full rank (the only difference is that we replaced $A, B$ with $\Ah_{t-1}, \Bh_{t-1}$). Recall the QR decomposition, we can re-express $\left(\frac{dK}{d\Theta}\right)_t(F_t)^{-1}$ as $\left(\frac{dK}{d\Theta}\right)_t(F_t)^{-1} = Q_tU_t$, where $Q_t \in \mathbb{R}^{nd \times nd}$ is an invertible matrix, and $U_t \in \mathbb{R}^{nd \times n(n+d)}$ satisfies $U_tU_t^\top = I_{nd}$. This implies that
\begin{equation*}
\vvector\left(\Kh_t - K\right)
= Q_tU_tF_t \vvector\left(\hat\Theta_t - \Theta\right)
(1 + o(1)) \as
\end{equation*}
From this and \cref{eq: F_t CLT} we know
\begin{equation*}
Q_t^{-1}\vvector\left(\Kh_t - K\right)
= U_tF_t \vvector\left(\hat\Theta_t - \Theta\right)
(1 + o(1))
\convD \calN(0, \sigma^2 I_{nd} )
.\end{equation*}
That is,
\begin{equation*}
\vvector\left(\Kh_t - K\right)^\top (Q_t^\top)^{-1}Q_t^{-1}\vvector\left(\Kh_t - K\right)
\convD \sigma^2\chi^2_{nd}
.\end{equation*}
\begin{equation*}
\vvector\left(\Kh_t - K\right)^\top (Q_t U_tU_t^\top Q_t^\top)^{-1}\vvector\left(\Kh_t - K\right)
\convD \sigma^2\chi^2_{nd}
.\end{equation*}
Recall that $\left(\frac{dK}{d\Theta}\right)_t(F_t)^{-1} = Q_tU_t$, and thus
\begin{equation*}
\vvector\left(\Kh_t - K\right)^\top
\left(
\left(\frac{dK}{d\Theta}\right)_t
(F_t^\top F_t)^{-1}
\left(\frac{dK}{d\Theta}\right)_t^\top
\right)^{-1}\vvector\left(\Kh_t - K\right)
\convD \sigma^2\chi^2_{nd}
.\end{equation*}
By definition
\begin{align*}
F_t^\top F_t =& (E_t^\top \otimes I_n)^\top(E_t^\top \otimes I_n) \\
=&(E_t \otimes I_n)(E_t^\top \otimes I_n) \\
=& E_t E_t^\top \otimes I_n \\
=&
\sum_{i=0}^{t-1}
\begin{bmatrix}
x_i\\
u_i\\
\end{bmatrix}
\begin{bmatrix}
x_i\\
u_i\\
\end{bmatrix}^\top \otimes I_n.
\end{align*}
Finally we can say
\begin{equation*}
\vvector\begin{bmatrix}
\Kh_t - K
\end{bmatrix} ^\top
\left(
\left(\frac{dK}{d\Theta}\right)_t
\left(
\sum_{i=0}^{t-1}
\begin{bmatrix}
x_i\\
u_i\\
\end{bmatrix}
\begin{bmatrix}
x_i\\
u_i\\
\end{bmatrix}^\top \otimes I_n\right)^{-1}
\left(\frac{dK}{d\Theta}\right)_t^\top
\right)^{-1}
\vvector\begin{bmatrix}
\Kh_t - K
\end{bmatrix}
\convD \sigma^2\chi^2_{nd}
.\end{equation*}
The only remaining task is to find a valid $E_t$ which satisfies $D_t^{-1}E_t \convP I_{n+d}$ and $E_tE_t^\top = \sum_{i=0}^{t-1}
\begin{bmatrix}
x_i\\
u_i\\
\end{bmatrix}
\begin{bmatrix}
x_i\\
u_i\\
\end{bmatrix}^\top$.
Although we already have \cref{thm:main tool}, $E_t = \left(\sum_{i=0}^{t-1}
\begin{bmatrix}
x_i\\
u_i\\
\end{bmatrix}
\begin{bmatrix}
x_i\\
u_i\\
\end{bmatrix}^\top\right)^{1/2}$ is still not necessarily a valid choice, because we can only show $D_t^{-1}\left(\sum_{i=0}^{t-1}
\begin{bmatrix}
x_i\\
u_i\\
\end{bmatrix}
\begin{bmatrix}
x_i\\
u_i\\
\end{bmatrix}^\top\right)^{1/2}$ is asymptotically an orthogonal matrix, but not identity matrix.
\subsubsection{Finding a valid $E_t$}
\label{Finding a valid E_t}
Recall \cref{eq: uu decomp} that
\begin{equation*}
\sum_{i=0}^{t-1}u_iu_i^\top/t^\beta\log^{\alpha}(t) =
KM_tK ^\top + \Delta_t K^\top + K\Delta_t^\top + \frac{\tau^2}\beta I_d + o_p(1)
.\end{equation*}
Now denote
\begin{equation}
\label{eq: definition Delta u}
\Delta_u := \sum_{i=0}^{t-1}u_iu_i^\top/ t^\beta\log^{\alpha}(t)
- \left(KM_tK ^\top + \Delta_t K^\top + K\Delta_t^\top\right) = \frac{\tau^2}\beta I_d + o_p(1)
,\end{equation}
which is asymptotically proportional to the identity matrix, and is also symmetric.
Recall that $D_t$ is defined as
\[D_t :=
t^{\beta/2}\log^{\alpha/2}(t)
\left[
\begin{array}{cc}
I_n & 0\\
K & I_d\\
\end{array}
\right]
\left[
\begin{array}{cc}
C_t^{1/2} & 0\\
0 & \sqrt{\frac{\tau^2}{\beta}} I_d\\
\end{array}
\right].\]
We will verify that the following construction of $E_t$ is a valid choice:
\[E_t :=
t^{\beta/2}\log^{\alpha/2}(t)
\left[
\begin{array}{cc}
I_n & 0\\
K & I_d\\
\end{array}
\right]
\left[
\begin{array}{cc}
(M_t - \Delta_t^\top \Delta_u^{-1}\Delta_t )^{1/2} &
\Delta_t^\top \Delta_u^{-1/2}\\
0 & \Delta_u^{1/2}\\
\end{array}
\right].\]
We shall examine the two conditions $D_t^{-1}E_t \convP I_{n+d}$ and $E_tE_t^\top = \sum_{i=0}^{t-1}
\begin{bmatrix}
x_i\\
u_i\\
\end{bmatrix}
\begin{bmatrix}
x_i\\
u_i\\
\end{bmatrix}^\top$ in order.
~\paragraph{Proving $D_t^{-1}E_t \convP I_{n+d}$}
It suffices to show:
\begin{equation*}
\left[
\begin{array}{cc}
C_t^{-1/2} & 0\\
0 & \sqrt{\frac{\beta}{\tau^2}} I_d\\
\end{array}
\right]
\left[
\begin{array}{cc}
(M_t - \Delta_t^\top \Delta_u^{-1}\Delta_t )^{1/2} &
\Delta_t^\top\Delta_u^{-1/2} \\
0 & \Delta_u^{1/2}\\
\end{array}
\right] \convP I_{n+d}
.\end{equation*}
\cref{eq: Ct inverse order,eq: Ct -1 Mt convP I_n,eq: Delta t order,eq: definition Delta u} states that
\begin{itemize}
\item $C_t^{-1} = \calO(t^{\beta-1}\log^\alpha(t))$
\item $M_t = C_t (1+o_p(1))$
\item $\Delta_t = \calO_p(t^{1-3\beta/2}\log^{\frac{-3\alpha + 3}{2}}(t))$
\item $\Delta_u = \frac{\tau^2}\beta I_d + o_p(1)$
\end{itemize}
With these facts, $C_t^{-1/2}\Delta_t^\top\Delta_u^{-1/2} = \calO_p(t^{1/2-\beta}\log^{\frac{-2\alpha + 3}{2}}(t)) \convP 0$ and $\sqrt{\frac{\tau^2}{\beta}} I_d \Delta_u^{1/2} \convP I_d$ are immediate.
It only remains to show that $C_t^{-1/2} (M_t - \Delta_t^\top \Delta_u^{-1}\Delta_t )^{1/2} \convP I_n$.
Notice
\begin{equation*}
C_t^{-1/2} (M_t - \Delta_t^\top \Delta_u^{-1}\Delta_t )^{1/2}
= (C_t^{-1}M_t - C_t^{-1}\Delta_t^\top \Delta_u^{-1}\Delta_t)^{1/2}
,\end{equation*}
and \cref{eq: Ct -1 Mt convP I_n} shows that $C_t^{-1}M_t \convP I_n$. It only remains to show
\begin{equation*}
C_t^{-1}\Delta_t^\top \Delta_u^{-1}\Delta_t \convP 0
,\end{equation*}
which is true because when $\beta > 1/2$ or $\beta = 1/2$ and $\alpha > 3/2$:
\begin{align*}
&C_t^{-1}\Delta_t^\top \Delta_u^{-1}\Delta_t \\
=& \calO(t^{\beta-1}\log^\alpha(t))\calO_p(t^{1-3\beta/2}\log^{\frac{-3\alpha + 3}{2}}(t))
\calO_p(1)
\calO_p(t^{1-3\beta/2}\log^{\frac{-3\alpha + 3}{2}}(t)) \\
=& \calO_p(t^{-2\beta+1}\log^{-2\alpha+3}(t)) \\
=& o_p(1)
.\end{align*}
~\paragraph{Proving $E_tE_t^\top = \sum_{i=0}^{t-1}
\begin{bmatrix}
x_i\\
u_i\\
\end{bmatrix}
\begin{bmatrix}
x_i\\
u_i\\
\end{bmatrix}^\top$}
\begin{align*}
E_tE_t^\top =&
t^{\beta}\log^{\alpha}(t)
\left[
\begin{array}{cc}
I_n & 0\\
K & I_d\\
\end{array}
\right]
\left[
\begin{array}{cc}
(M_t - \Delta_t^\top \Delta_u^{-1}\Delta_t )^{1/2} &
\Delta_t^\top\Delta_u^{-1/2} \\
0 & \Delta_u^{1/2}\\
\end{array}
\right] \\
& \hspace{1cm} \cdot \left[
\begin{array}{cc}
(M_t - \Delta_t^\top \Delta_u^{-1}\Delta_t )^{1/2} &
0\\
\Delta_u^{-1/2}\Delta_t & \Delta_u^{1/2}\\
\end{array}
\right]
\left[
\begin{array}{cc}
I_n & K^\top\\
0 & I_d\\
\end{array}
\right]\\
=& t^{\beta}\log^{\alpha}(t)
\left[
\begin{array}{cc}
I_n & 0\\
K & I_d\\
\end{array}
\right]\left[
\begin{array}{cc}
M_t & \Delta_t^\top\\
\Delta_t & \Delta_u\\
\end{array}
\right]\left[
\begin{array}{cc}
I_n & K^\top \\
0 & I_d\\
\end{array}
\right] \\
=&
t^{\beta}\log^{\alpha}(t)
\left[
\begin{array}{cc}
M_t & M_tK ^\top + \Delta_t^\top \\
KM_t+ \Delta_t & KM_tK ^\top + \Delta_t K^\top + K\Delta_t^\top + \Delta_u \\
\end{array}
\right] \\
&\text{(By Definitions \cref{eq: defn Mt,eq: defn Delta t,eq: definition Delta u}}) \\
=& \sum_{i=0}^{t-1} \begin{bmatrix} x_i\\ u_i\\ \end{bmatrix} \begin{bmatrix} x_i\\ u_i\\ \end{bmatrix}^\top
.\end{align*}
We will re-use the following equation later:
\begin{equation}
\label{eq: Gram matrix my symbol}
\sum_{i=0}^{t-1} \begin{bmatrix} x_i\\ u_i\\ \end{bmatrix} \begin{bmatrix} x_i\\ u_i\\ \end{bmatrix}^\top =
t^{\beta}\log^{\alpha}(t)
\left[
\begin{array}{cc}
I_n & 0\\
K & I_d\\
\end{array}
\right]\left[
\begin{array}{cc}
M_t & \Delta_t^\top\\
\Delta_t & \Delta_u\\
\end{array}
\right]\left[
\begin{array}{cc}
I_n & K^\top \\
0 & I_d\\
\end{array}
\right]
\end{equation}
\end{proof}
\subsection{The proof of \cref{thm: prediction CLT}}
\label{The proof of thm: prediction CLT}
\begin{corr*}
\cref{alg:myAlg} applied to a system described by \cref{eq:LinearModel} under \cref{asm:InitialStableCondition} satisfies:
\begin{align*}
\begin{split}
\left(
\sigma^2
\begin{bmatrix}
x_t \\
u_{t}
\end{bmatrix}^\top
\left(
\sum_{i=0}^{t-1}
\begin{bmatrix}
x_i\\
u_i\\
\end{bmatrix}
\begin{bmatrix}
x_i\\
u_i\\
\end{bmatrix}
^\top
\right)^{-1}
\begin{bmatrix}
x_t \\
u_{t}
\end{bmatrix}
\right)^{-1/2} \left((\Ah_{t} - A)x_t +
(\Bh_{t}- B)u_{t}\right)
\convD \calN(0, I_n)
.\end{split}
\end{align*}
where $u_t = \Kh_t x_t + \xi_t$ for any $\xi_t$ independent of the data before $t$: $\{\varepsilon_i, \eta_i\}_{i=0}^{t-1}$.
\end{corr*}
\begin{proof}
This one final lemma connects \cref{lem: CLT original} to our desired conclusion by changing the parametric expression to the observable one:
\begin{lemma}
\label{lem: variance equivalence}
For any $\xi_t$ independent of the data before $t$: $\{\varepsilon_i, \eta_i\}_{i=0}^{t-1}$,
\begin{align*}
\left(
x_t^\top
\left( \sum_{p=0}^{\infty}L ^{p}\left(I_n + 1_{\{\beta=1,\alpha=0\}}\frac{\tau^2}{\sigma^2}BB^\top\right)(L ^{p})^\top\right)^{-1}
x_t
+
\frac{\beta \sigma^2}{\tau^2}
t^{1-\beta} \log^{-\alpha}(t)
\lnorm{\xi_t}^2
\right)^{-1/2} \\
\cdot \;t^{1/2}
\left(
\sigma^2
\begin{bmatrix}
x_t \\
u_{t}
\end{bmatrix}^\top
\left(
\sum_{i=0}^{t-1}
\begin{bmatrix}
x_i\\
u_i\\
\end{bmatrix}
\begin{bmatrix}
x_i\\
u_i\\
\end{bmatrix}
^\top
\right)^{-1}
\begin{bmatrix}
x_t \\
u_{t}
\end{bmatrix}
\right)^{1/2} \convP 1
.\end{align*}
\end{lemma}
The proof of \cref{lem: variance equivalence} can be found in \cref{The proof of lem: variance equivalence}. Finally, we can say
\begin{align*}
\left(
\sigma^2
\begin{bmatrix}
x_t \\
u_{t}
\end{bmatrix}^\top
\left(
\sum_{i=0}^{t-1}
\begin{bmatrix}
x_i\\
u_i\\
\end{bmatrix}
\begin{bmatrix}
x_i\\
u_i\\
\end{bmatrix}
^\top
\right)^{-1}
\begin{bmatrix}
x_t \\
u_{t}
\end{bmatrix}
\right)^{-1/2}
\left((\Ah_{t} - A)x_t +
(\Bh_{t}- B)u_t\right)
\convD \calN(0, I_n)
.\end{align*}
\end{proof}
\section{The proof of Propositions}
\label{sec: The proof of Propositions}
\subsection{The proof of \cref{prop:one_epoch_estimate}}
\label{section: The proof of one_epoch_estimate}
\begin{prop*}[Similar to Proposition C.1 in \citet{dean2018regret}]
Let $x_0 \in \R^{\statedim}$ be any initial state. Assume \cref{asm:InitialStableCondition} is satisfied. When applying $\cref{alg:myAlg}$,
\begin{equation*}
\max\left\{ \norm{\Ah_t - A}, \norm{\Bh_t - B}\right\} =
\calO(t^{-\frac{\beta}{2}} \log^{\frac{-\alpha + 1}{2}}(t)) \as
\end{equation*}
\end{prop*}
\subsubsection{Proof Outline}
\begin{proof}
We shall see that all the properties we derived in this section only require the safety condition \cref{alg:myAlg} Line \ref{line:check} without any other requirement on the controller $\Kh_t$, and thus also apply to \cref{alg:myAlg} with logarithmic updates; see \cref{remark: Logarithmically-update estimates}.
According to \cref{alg:myAlg} Line \ref{line:check}, we keep our controller $\Kh_t$ bounded $\norm{\Kh_t} \le C_K$, which means the next state can not be too far from the previous state. At the same time, whenever the state is too large ($\norm{x_t} > C_x\log(t)$), it is tuned down by safe controller $K_0$. Overall speaking, the state $x_t$ is always controlled with at most $\log(t)$ growth. We will see in \cref{lemma:lwm} that when state growth is controlled, we have a decent bound on $\Ah_t, \Bh_t$.
In other words, as long as we still run \cref{alg:myAlg} Line \ref{line:check} at every time step, which is enough to "control" the system by itself, any $\Ah_t, \Bh_t$ generated with Line \ref{line:ols} satisfies
\begin{equation*}
\max\left\{ \norm{\Ah_t - A}, \norm{\Bh_t - B}\right\} =
\calO(t^{-\frac{\beta}{2}} \log^{\frac{-\alpha + 1}{2}}(t)) \as
\end{equation*}
regardless of the estimation result before time $t$.
\cref{lemma:lwm} follows from a result by \citet{simchowitz2018learning} on the estimation of linear response time-series. We present that result in the context of our problem. Let $\Theta := [A, B]$, and define $z_t :=
\begin{bmatrix}
x_t \\
u_t
\end{bmatrix}$. Then, the OLS estimator \cref{eq: AhBh estimator} is
\begin{align}
\label{eq:ols_M}
(\Ah_T, \Bh_T) = \hat{\Theta}_T \in \arg \min_{\Theta} \sum_{t = 0}^{T - 1} \frac{1}{2}\ltwonorm{x_{t + 1} - \Theta z_t}^2.
\end{align}
We know that the accuracy of the OLS estimator is related to the covariance structure of the predictors, which are $\{z_t\}_{t = 0}^{T}$ in our context. To capture such covariance structure, we need the following definiton:
\begin{defn}[BMSB condition]
\label{def:BMSB condition}
The $\{\calF_t\}_{t \geq 0}$-adapted process $\{z_t\}_{t = 0}^T$ is said to satisfy the $(k, \nu, \xi)$-\emph{block martingale small-ball} (BMSB) condition if for any $0 \le j \le T-k$ and $v \in \mathcal{S}^{\statedim + \inputdim-1} := \{x \in \mathbb{R}^{\statedim + \inputdim}: \norm{x} = 1\}$, one has that
\begin{align*}
\frac{1}{k}\sum_{i = 1}^{k} \P\left( |\langle v, z_{j + i}\rangle| \geq \nu {|\calF_j}\right) \geq \xi \as
\end{align*}
\end{defn}
This condition is used for characterizing the size of the minimum eigenvalue of the matrix $\sum_{t = 0}^{T - 1} z_t z_t^\top$.
A larger $\nu$ guarantees a larger lower bound of the minimum eigenvalue. In the context of our problem the result by \citet{simchowitz2018learning} translates as follows.
\begin{comment}
\begin{theorem}[Theorem C.2 in \citet{dean2018regret}, the proof I believe should follow Theorem 2.4 in \citet{simchowitz2018learning}]
\label{lemma:lwm}
\fcmargin{}{The author didn't prove this, and actually I found this result very different from the main theorem in \citet{simchowitz2018learning}, may consider reprove this later}
There exists uniform constant $\calO(1) > 0$, such that fixing $\varepsilon, \delta \in (0,1)$, for every $T$, $k$, $\nu$, and $\xi$ such that $\{z_t\}_{t = 0}^T$ satisfies the $(k, \nu, \xi)$-BMSB and
\begin{equation}
\label{eq:condition of T for AB bound}
\left\lfloor \frac{T}{k} \right\rfloor \ge \calO(1) \frac{\statedim + \inputdim}{\xi^2} \log \left(1 + \frac{\sum_{t = 0}^{T - 1} \Tr (\E z_t z_t^\T)}{k \lfloor T / k \rfloor \xi^2 \nu^2 \delta} \right)
.\end{equation}
the estimate $\hat{\Theta}$ defined in Eq.~\ref{eq:ols_M} satisfies the following statistical rate
\begin{equation}
\label{eq:hiprob AB bound}
\P \left(\ltwonorm{\hat{\Theta}_T - M} > \frac{\calO(1) \sigma }{\xi \nu} \sqrt{\frac{\statedim + \inputdim}{k \lfloor T / k \rfloor} \log \left(1 + \frac{\sum_{t = 0}^{T - 1} \Tr (\E z_t z_t^\T)}{k \lfloor T / k \rfloor \xi^2 \nu^2 \delta} \right) }\right) \leq \delta.
.\end{equation}
\end{theorem}
\end{comment}
\begin{lemma}[A slightly different version of Theorem C.2 in \citet{dean2018regret}]
\label{lemma:lwm}
For $\delta \in (0,\frac{(n+d)\xi^2}{2}]$, for every $T$, $k$, $\nu$, and $\xi$ such that $\{z_t\}_{t = 0}^T$ satisfies the $(k, \nu, \xi)$-BMSB and
\begin{equation}
\label{eq:condition of T for AB bound}
T/k \ge \frac{10(n+d)}{\xi^2} \log\left(\frac{100(n+d)\sum_{t = 0}^{T-1}\Tr(\E z_t z_t^\top)}{T \nu^{2}\xi^2\delta^{1 + \frac1{n+d}}} \right)
.\end{equation}
the estimate $\hat{\Theta}_T$ defined in \cref{eq:ols_M} satisfies the following statistical rate
\begin{equation}
\label{eq:hiprob AB bound}
\P\left(\lnorm{\hat{\Theta}_T-\Theta} >\frac{90\sigma}{\xi\nu}\sqrt{\frac{n+d}{T}\left(1 + \log\left(\frac{10(n+d)\sum_{t = 0}^{T-1}\Tr(\E z_t z_t^\top)}{T\delta^{1 + \frac1{n+d}} \nu^{2}\xi} \right) \right)} \right) \le 3\delta.
\end{equation}
\end{lemma}
The proof of \cref{lemma:lwm} can be found in \cref{The proof of lemma:lwm}.
We will show that $\sum_{t=1}^T \Tr(\E z_t z_t^\top)$ grows linearly with $T$ (ignoring logarithmic terms), which means in \cref{eq:condition of T for AB bound} the LHS grows faster than the RHS, and is thus always satisfied if $T$ is large enough. \cref{lemma:lwm} is saying that for any $T$ larger than some constant, we can control the $L_2$ norm of the system parameter estimate $\hat{\Theta}_T$, which implies we can control the $L_2$ norm of both $\Ah_T$ and $\Bh_T$.
Still there is one more gap from our \cref{prop:one_epoch_estimate}, which requires \emph{uniform} control on $\Ah_T$ and $\Bh_T$. Fortunately, we have the blessing that this high-probability bound is in the log scale w.r.t $\delta$. Because of that, we can choose a series of decaying $\delta_T = 1/T^2$ for each different estimate $\hat{\Theta}_T$, so that $\sum_{T=C}^\infty 1/T^2 \le 1/C$ and we can achieve a uniform high probability bound on $\Ah_T$ and $\Bh_T$ for all $T > C$, which directly leads to the desired conclusion once we plug in appropriate values for $k$, $\nu$, and $\xi$:
\begin{equation*}
\max\left\{ \norm{\Ah_t - A}, \norm{\Bh_t - B}\right\} =
\calO(t^{-\frac{\beta}{2}} \log^{\frac{-\alpha + 1}{2}}(t)) \as
\end{equation*}
To sum up, there are three main steps in our proof of \cref{prop:one_epoch_estimate}:
\begin{itemize}
\item Verify $\{z_t\}_{t = 0}^T$ satisfies the $(k, \nu, \xi)$-BMSB condition in our setting.
\item Replace $\Tr(\E z_t z_t^\top)$ in \cref{lemma:lwm} by an explicit upper bound in terms of $T$.
\item Prove a uniform high probability bound for $\Ah_T$ and $\Bh_T$ by choosing with $\delta_T = 1/T^2$ with \cref{lemma:lwm}.
\end{itemize}
\end{proof}
\subsubsection{Verifying $\{z_t\}_{t = 0}^T$ satisfies the $(k, \nu, \xi)$-BMSB condition}
In order to apply \cref{lemma:lwm}, we need to find $k$, $\nu$, and $\xi$ such that $\{z_t\}_{t = 0}^T$ satisfies the
$(k, \nu, \xi)$-BMSB condition.
\begin{lemma}[Similar to Lemma C.3 in \citet{dean2018regret}]
\label{lem:bmsb}
If we assume \cref{asm:InitialStableCondition}, then apply \cref{alg:myAlg}, the process $\{z_t\}_{t \geq 0}^T $ satisfies the
$(k, \nu, \xi)$-BMSB condition for
\begin{align*}
(k, \nu, \xi) = \left(1, \sqrt{\sigma_{\eta,T}^2\min\left(\frac{1}{2}, \frac{\sigma ^2}{2 \sigma ^2 C_K^2 + \tau^2} \right)} , \frac{3}{10}\right),
\end{align*}
where $\sigma_{\eta,T}^2 = \tau^2 T^{\beta-1}\log^\alpha(T)$.
\end{lemma}
See \cref{The proof of lem:bmsb} for the proof of \cref{lem:bmsb}.
\subsubsection{Upper bound of $\Tr(\E z_t z_t^\top)$ in terms of $T$}
The benefit of a non-random upper bound of $\Tr(\E z_t z_t^\top)$ w.r.t $T$ is two-fold.
\begin{itemize}
\item We can know exactly how large our $T$ should be for \cref{eq:condition of T for AB bound} to hold.
\item Furthermore, we can also substitute the upper bound in to \cref{eq:hiprob AB bound}.
\end{itemize}
\cref{lem:bound_covariance} shows that we have an upper bound of $\Tr(\E z_t z_t^\top)$ that is $\logO(T)$.
\begin{lemma}[Similar to Lemma C.4 in \citet{dean2018regret}]
\label{lem:bound_covariance}
If we assume \cref{asm:InitialStableCondition}, then apply \cref{alg:myAlg}, the process $\{z_t\}_{t \geq 0}^T $ satisfies
\begin{equation}
\begin{split}
\label{eq:lemma3Conclusion}
\sum_{t = 0}^{T - 1} \Tr \left( \E z_t z_t^\T\right) = \calO(T\log^2(T))
\end{split}
.\end{equation}
\end{lemma}
See \cref{The proof of lem:bound_covariance} for the proof of \cref{lem:bound_covariance}.
\subsubsection{Uniform upper bound for $\max\left\{ \norm{\Ah_t - A}, \norm{\Bh_t - B}\right\}$}
With \cref{lem:bmsb} and \cref{lem:bound_covariance} in hand, we can translate \cref{lemma:lwm} into our problem setting. Fixing $\delta \in (0,\frac{(n+d)\xi^2}{2}]$, we already proved by \cref{lem:bmsb} that the process $z_t = \begin{bmatrix}
x_t \\
u_t
\end{bmatrix}$
satisfies the
\begin{align}
\label{eq:BMSB condition}
(k, \nu, \xi) = \left(1, \sqrt{\sigma_{\eta,T}^2\min\left(\frac{1}{2}, \frac{\sigma ^2}{2 \sigma ^2 C_K^2 + \tau^2} \right)} , \frac{3}{10}\right)\text{ BMSB condition.}
\end{align}
\begin{comment}
If we choose $T$ such that
\begin{align*}
\left\lfloor \frac{T}{k} \right\rfloor &\ge \calO(1) \frac{\statedim + \inputdim}{\xi^2} \log \left(1 + \frac{ \logC(A, B, K_0, C_K, C_x, \tau^2, \sigma^2;T)}{k \lfloor T / k \rfloor \xi^2 \nu^2 \delta} \right), \\
&\ge
\calO(1)\frac{\statedim + \inputdim}{\xi^2} \log \left(1 + \frac{\sum_{t = 0}^{T - 1} \Tr (\E z_t z_t^\T)}{\lfloor T \rfloor \xi^2 \nu^2 \delta} \right),
.\end{align*}
Then we can apply our \cref{lemma:lwm}, and if we further pick $\delta$ to be $1/T^2$, then
\begin{align*}
&\calO(1) \frac{\statedim + \inputdim}{\xi^2} \log \left(1 + \frac{ \logC(A, B, K_0, C_K, C_x, \tau^2, \sigma^2;T)}{k \lfloor T / k \rfloor \xi^2 \nu^2 \delta} \right),
\\
=& \calO(n+d)\log \left(1 + \frac{ \logC(A, B, K_0, C_K, C_x, \tau^2, \sigma^2;T)}{\lfloor T \rfloor \sigma_{\eta,T}^2 1/T^2} \right) \qquad \xi = \frac{3}{10}, k = 1 \text{ are fixed constants} \\
=& \calO(n+d)\log \left(\tau^{-2} \logC(A, B, K_0, C_K, C_x, \tau^2, \sigma^2;T^2)T^{1-\beta} \right) \\
\ge& \Clog(A, B, K_0, C_K, C_x, \tau^2, \sigma^2, \beta;T)
.\end{align*}
\end{comment}
If we choose $\delta = \frac1{3T^2}$ and $T$ such that \cref{eq:condition of T for AB bound}
holds with $(k,\nu,\xi)$ in \cref{eq:BMSB condition}, we can apply \cref{lemma:lwm}. By \cref{eq:lemma3Conclusion}, we only need $T$ to satisfy
\begin{align*}
T/k
\ge& \frac{10(n+d)}{\xi^2} \log\left(\frac{100(n+d)\logO(T)}{T \nu^{2}\xi^2\delta^{1 + \frac1{n+d}}} \right)
\\
=& \calO(1)\log \left(\frac{ \logO(T)}{T \sigma_{\eta,T}^2 \frac{\sigma ^2}{2 \sigma ^2 C_K^2 + \tau^2} T^{-2(1 + \frac1{n+d})}} \right) \qquad (\xi = \frac{3}{10} \text{ is fixed constant}) \\
=& \logO(1) \qquad \text{(Recall that } \sigma_{\eta,T}^2 = T^{\beta-1}\log^\alpha(T) )
.\end{align*}
Since $T$ is growing faster than $\logO(1)$, the above condition is essentially saying that our $T$ should be larger than some constant $\calO(1)$. Suppose that is the case, then following \cref{lemma:lwm} and \cref{lem:bound_covariance}, the estimate $\hat{\Theta}_T$ defined in \cref{eq:ols_M} satisfies the following statistical rate
\begin{align*}
& \P\left(\lnorm{\hat{\Theta}_T-\Theta} >\frac{90\sigma}{\xi\nu}\sqrt{\frac{n+d}{T}\left(1 + \log\left(\frac{10(n+d)\logO(T)}{T\delta^{1 + \frac1{n+d}} \nu^{2}\xi} \right) \right)} \right) \\
\leq& \P\left(\lnorm{\hat{\Theta}_T-\Theta} >\frac{90\sigma}{\xi\nu}\sqrt{\frac{n+d}{T}\left(1 + \log\left(\frac{10(n+d)\sum_{t = 1}^T\Tr(\E z_t z_t^\top)}{T\delta^{1 + \frac1{n+d}} \nu^{2}\xi} \right) \right)} \right) \\
\le& 3\delta.
\end{align*}
Notice that $\hat{\Theta}_T = [\Ah_T, \Bh_T]$, and we know that $\max\left\{ \substack{\norm{\Ah_T - A},\\ \norm{\Bh_T - B}}\right\} \le \ltwonorm{\hat{\Theta}_T - \Theta}$. That is to say
\[\P\left(\max\left\{ \substack{\norm{\Ah_T - A},\\ \norm{\Bh_T - B}}\right\} >\frac{90\sigma}{\xi\nu}\sqrt{\frac{n+d}{T}\left(1 + \log\left(\frac{10(n+d)\logO(1)}{T\delta^{1 + \frac1{n+d}} \nu^{2}\xi} \right) \right)} \right) \le 3\delta.\]
Next we substitute $k= 1$, $\xi = \frac{3}{10}$, $\nu = \sqrt{\sigma_{\eta,T}^2\min\left(\frac{1}{2}, \frac{\sigma ^2}{2 \sigma ^2 C_K^2 + \tau^2} \right)}$, and $\delta = \frac1{3T^2}$ into the previous equation
\begin{align*}
&\P \left(\max\left\{ \substack{\norm{\Ah_T - A},\\ \norm{\Bh_T - B}}\right\} > \frac{\calO(1) }{ \sqrt{\sigma_{\eta,T}^2}} \sqrt{\frac{n+d}{T}\left(1 + \log\left(\frac{\logO(T)}{T (3T^{-2})^{1 + \frac1{n+d}} \sigma_{\eta,T}^2} \right) \right)}\right) \le \frac1{T^2}
.\end{align*}
By merging all constant parameters in to the $\calO$ style expression, and noticing that $\sigma_{\eta,T}^2 = \tau^2T^{\beta-1}\log^{\alpha}(T)$, where $\beta \in [1/2,1)$, we have for any $T > \calO(1)$:
\[\P \left(\max\left\{ \substack{\norm{\Ah_T - A},\\ \norm{\Bh_T - B}}\right\} > \calO( T^{\frac{1-\beta}{2}}\log^{-\alpha/2}(T)) \sqrt{\frac{n+d}{T}\calO(\log(T))}\right) \le \frac1{T^2},\]
which implies
\[ \qquad \P \left(\max\left\{ \substack{\norm{\Ah_T - A},\\ \norm{\Bh_T - B}}\right\} > \calO( T^{-\frac{\beta}{2}}
\log^{\frac{-\alpha+1}{2}}(T))
\right) \le \frac1{T^2}.\]
Notice that
\begin{align*}
\sum_{T=C+1}^{\infty} \frac{1}{T^2}
\le \sum_{T=C+1}^{\infty} \frac{1}{T(T-1)}
\le \sum_{T=C+1}^{\infty} \frac{1}{T-1} - \frac{1}{T}
= \frac{1}{C}
.\end{align*}
Therefore we can derive a uniform confidence bound on the estimation error of parameters $\Ah_t$ and $\Bh_t$: For any integer $C > \calO(1)$:
\begin{align*}
&\P \left(\exists t > C , \;s.t. \;\max\left\{ \substack{\norm{\Ah_t - A},\\ \norm{\Bh_t - B}}\right\} > \calO( T^{-\frac{\beta}{2}}
\log^{\frac{-\alpha+1}{2}}(T)) \right) \\
\le& \sum_{t=C+1}^\infty \P \left( \max\left\{ \substack{\norm{\Ah_t - A},\\ \norm{\Bh_t - B}}\right\} > \calO( T^{-\frac{\beta}{2}}
\log^{\frac{-\alpha+1}{2}}(T)) \right) \\
\le& \frac{1}{C}
.\end{align*}
Notice that this is a uniform upper bound for all $t > C$. Recall \cref{defn: Big O notation} \cref{itm: big O as defn}, where we define $X_n = \calO(a_n) \as$ as:
for almost every $\omega \in \Omega$, there exists a number $C(\omega)$ such that $\abs{X_n(\omega)} \le C(\omega)a_n$, where $\Omega$ denotes the sample space of $\{X_n\}_n$. The previous equation is telling us the union of such event $\omega$ happens with at least probability $1-1/C$, and by taking $C \to \infty$ that is exactly the definition of $\calO(a_n) \as$, and thus:
\begin{equation*}
\max\left\{ \norm{\Ah_t - A}, \norm{\Bh_t - B}\right\} =
\calO(t^{-\frac{\beta}{2}} \log^{\frac{-\alpha + 1}{2}}(t)) \as
\end{equation*}
The same bound holds for logarithmic updates. The reason is that for time $t$, the closest estimation update will always be within $t/c$ time steps of $t$, which does not change the order:
\begin{equation*}
\calO((t/c)^{-\frac{\beta}{2}} \log^{\frac{-\alpha + 1}{2}}(t/c))
= \calO(t^{-\frac{\beta}{2}} \log^{\frac{-\alpha + 1}{2}}(t))
.\end{equation*}
\begin{comment}
Now we wish to enlarge the condition $\exists t > C$ to $\exists t \in \mathbb{N}$. We only need to ensure that in the first $C$ steps, the estimator can not exceed our bound. To do that, we need to multiply our bound by some constant $F$. That is
\[F\logC(t^{-\beta/2}) > U_{AB} \text{ for any } t \in {1,2, \cdots, C}.\]
which is
\[F\calO(1)\Clog(t)t^{-\beta/2} > U_{AB} \text{ for any } t \in {1,2, \cdots, C}.\]
Recall our poly-log is always equipped with positive coefficients, thus when $x \ge 1$, we can always assume $\Clog(x) > 1$ by changing $\Clog(x)$ to $\Clog(x) + 2$, thus we only need to ensure
\[Ft^{-\beta/2} \ge U_{AB} \text{ for any } t \in {1,2, \cdots, C}.\]
To satisfy this, simply let $K := U_{AB}C^{\beta/2}$. And thus
\begin{equation*}
\P \left(\exists t \in \mathbb{N} , \;s.t. \;\max\left\{ \substack{\norm{\Ah_t - A},\\ \norm{\Bh_t - B}}\right\} > U_{AB}C^{\beta/2}\logC(t^{-\beta/2}) \right) \le \frac{1}{C}
.\end{equation*}
\end{comment}
\subsection{The proof of \cref{prop:one_epoch_estimate_withK}}
\label{section: The proof of one_epoch_estimate_withK}
\begin{prop*}
Let $x_0 \in \R^{\statedim}$ be any initial state. Assume \cref{asm:InitialStableCondition} is satisfied. When applying $\cref{alg:myAlg}$,
\begin{equation*}
\max\left\{ \norm{\Ah_t - A}, \norm{\Bh_t - B}, \norm{\Kt_{t+1} - K}\right\}
= \calO(t^{-\frac{\beta}{2}} \log^{\frac{-\alpha + 1}{2}}(t)) \as
\end{equation*}
\end{prop*}
\subsubsection{Proof Outline}
\begin{proof}
When the problem parameters $(A, B, Q, R)$ are known the optimal policy is given by linear feedback, $u_t = K x_t$, where $K = - (R + B^\top P B)^{-1}B^\top P A$ and $P$ is the (positive definite) solution to the discrete Riccati equation
\begin{align}
\label{eq:discrete_riccati}
P = A^\top P A - A^\top P B (R + B^\top P B)^{-1} B^\top P A + Q
.\end{align}
In the following context any time we mention $\Ph_t$ and $\Kt_{t+1}$, we are refering to the corresponding certainty equivalent responses.
\begin{align*}
\Ph_t = \Ah_t^\top \Ph_t \Ah_t - \Ah_t^\top \Ph_t \Bh_t (R + \Bh_t^\top \Ph_t \Bh_t)^{-1} \Bh_t^\top \Ph_t \Ah_t + Q
.\end{align*}
\begin{align*}
\Kt_{t+1} = - (R + \Bh_t^\top \Ph_t \Bh_t)^{-1}\Bh_t^\top \Ph_t \Ah_t
.\end{align*}
Since we already controlled the estimation error of $\Ah_t-A$ and $\Bh_t-B$, one natural thing to ask is that, if we have control over $\Ah_t-A$ and $\Bh_t-B$, do we have control over $\Kt_{t+1}-K$? This can be achieved by two steps:
\begin{enumerate}
\item Show that we can control $\Kt_{t+1}$ once $\Ah_t$, $\Bh_t$, and $\Ph_t$ are controlled.
\item Show that we can control $\Ph_t$ once $\Ah_t$ and $\Bh_t$ are controlled.
\end{enumerate}
\end{proof}
\subsubsection{Show that we can control $\Kt_{t+1}$ once $\Ah_t$, $\Bh_t$, and $\Ph_t$ are controlled}
This is already stated by Proposition 1 in \citet{mania2019certainty}.
Denote the quantity
$$\Gamma_1 := 1 + \max\{\norm{A}, \norm{B}, \norm{P}, \norm{K}\}.$$
\begin{prop}[Proposition 1 in \citet{mania2019certainty}]
\label{prop:stability_perturb}
Let $\epsilon > 0$ such
that $\norm{\Ah - A} \leq \epsilon$ and $\norm{\Bh - B} \leq \epsilon$. Also, let $\norm{\Phat - P} \leq \epsilon_P$ such that $\epsilon_P \geq \epsilon$. Assume $\smin(R) \ge 1$ we have
\begin{align*}
\norm{\Kh - K} \leq 7 \Gamma_1^3 \, \epsilon_P .
\end{align*}
\end{prop}
The $\smin(R)$ represents the minimum eigenvalue of $R$.
we can discard the constraint of $\smin(R) \ge 1$ by the following observation. If we replace our $Q$ and $R$ by $Q/\smin(R)$ and $R/\smin(R)$, then the corresponding solution $P$ for \cref{eq:discrete_riccati} will be $P/\smin(R)$. Notice that changing $Q$ and $R$ by the same proportion does not change the LQR problem. With that being said, our LS estimator $\Ah_t$, $\Bh_t$, and the nominal controller $\Kt_t$ will remain the same. By this transformation the minimum eigenvalue condition is satisfied, and we only need to control
\[\norm{\Phat - P}/\smin(R) \le \epsilon_P\]
such that $\epsilon_P \geq \epsilon$, and we will have $\norm{\Kh - K} \leq 7 \Gamma_2^3 \, \epsilon_P$, where $\Gamma_2 := 1 + \max\{\norm{A}, \norm{B}, \norm{P}/\smin(R), \norm{K}\}$. Here we can replace this denominator $\smin(R)$ by any constant smaller than $\smin(R)$, and the whole story would still work. Since later we will also require $\smin(P) \ge 1$, we can choose the shared denominator to be $\min\{\smin(R), \smin(P)\}$. To sum up we have the following corollary of \cref{prop:stability_perturb}.
\begin{corr}
\label{corr:stability_perturb_tmp}
Let $\epsilon > 0$ such
that $\norm{\Ah - A} \leq \epsilon$ and $\norm{\Bh - B} \leq \epsilon$. Also, let $\norm{\Phat - P} \le \min\{\smin(R), \smin(P)\}\epsilon_P$ such that $\epsilon_P \geq \epsilon$. Then we have
\begin{align*}
\norm{\Kh - K} \leq 7 \Gamma_3^3 \, \epsilon_P .
\end{align*}
where $\Gamma_3 := 1 + \max\{\norm{A}, \norm{B}, \norm{P}/\min\{\smin(R), \smin(P)\}, \norm{K}\}$.
\end{corr}
Now we only need to prove that $\norm{\Phat - P} = \calO(\epsilon)$ given $\norm{\Ah - A} \leq \epsilon$ and $\norm{\Bh - B} \leq \epsilon$.
\subsubsection{Show that we can control $\Ph_t$ once $\Ah_t$ and $\Bh_t$ are controlled}
Consider a general square matrix $M$. In order to quantify the decay rate of $\norm{M^k}$, we define
\begin{align*}
\transient{M}{ \rho} := \sup \left\{\norm{M^k} \rho^{-k} \colon k\geq 0 \right\}.
\end{align*}
In other words, $\transient{M}{\rho}$ is the smallest value such that $\norm{M^k} \leq \transient{M}{\rho} \rho^k$ for all $k \geq 0$.
We note that $\transient{M}{\rho}$ might be infinite, depending on the value of $\rho$, and it is always greater than or equal to one. If $\rho$ is larger than $\rho(M)$, we are guaranteed to have a finite $\transient{M}{\rho}$ (this is a consequence of Gelfand's formula). In particular, if $M$ is a stable matrix, we can choose $\rho < 1$ such that $\transient{M}{\rho}$ is finite. Also, we note that $\transient{M}{\rho}$ is a decreasing function of $\rho$; if $\rho \geq \norm{M}$, we have $\transient{M}{\rho} = 1$.
Recall that $L := A + BK$.
The following proposition that upper bounds $\norm{\Ph - P}$ holds in a more general LQG setting where the matrix $Q$ is unknown:
\begin{prop}[Proposition 2 in \citet{mania2019certainty}]
\label{prop:93}
Let $\gamma \geq \rho(L)$ and also let $\epsilon$ be such that $\norm{\Ah - A}$, $\norm{\Bh - B}$, and $\norm{\Qh - Q}$ are at most $\epsilon$. Let $\norm{\cdot}_+ = \norm{\cdot} + 1$. We assume that $R \succ 0$, $(A, B)$ is stabilizable, $(Q^{1/2}, A)$ observable, and $\smin(P) \geq 1$.
\begin{align*}
\norm{\Phat - P} \leq \calO(1)\, \epsilon \, \frac{\transient{L}{\gamma}^2}{1 - \gamma^2} \norm{A}_+^2 \norm{P}_+^2 \norm{B}_+ \norm{R^{-1}}_+,
\end{align*}
as long as
\begin{align*}
\epsilon \leq \calO(1) \frac{(1 - \gamma^2)^2}{\transient{L}{\gamma}^4} \norm{A}_+^{-2} \norm{P}_+^{-2} \norm{B}_+^{-3} \norm{R^{-1}}_+^{-2} \min \left\{ \norm{L}_+^{-2}, \norm{P}_+^{-1} \right\}.
\end{align*}
\end{prop}
Here $\calO(1)$ are pure constants without dependence of any other parameters.
We already assumed in \cref{asm:InitialStableCondition} that $(A,B)$ stabilizable, but we have not defined `observable' yet. An equivalent statement of observable can be found here.
\begin{lemma}[Lemma 2.1 in \citep{payne1973discrete}]
The pair $(C,A)$ is observable if and only if $Ax = \lambda x$, $Cx = 0$ imply $x = 0$
\label{lem:observable}
\end{lemma}
Since we already assumed $Q$ is positive definite, $Qx = 0$ imply $x = 0$, and thus $(Q^{1/2}, A)$ is observable. In the LQAC setting we know $Q$ exactly, so we can remove the estimation bound condition on $Q$.
Now we can restate \cref{prop:93} in the LQAC setting:
\begin{corr}
\label{cor: Ph controlled with Ah and Bh}
Let $\epsilon$ such that $\norm{\Ah_t - A}$, and $\norm{\Bh_t - B}$ are at most $\epsilon$. Let $\norm{\cdot}_+ = \norm{\cdot} + 1$. We assume that $R \succ 0$, $(A, B)$ is stabilizable, and $\smin(P) \geq 1$.
\begin{align*}
\norm{\Phat_t - P} \leq \calO(1)\, \epsilon \, \frac{\transient{L}{\rho(L)}^2}{1 - \rho(L)^2} \norm{A}_+^2 \norm{P}_+^2 \norm{B}_+ \norm{R^{-1}}_+ = \calO(\epsilon)
.\end{align*}
as long as
\begin{align*}
\epsilon \leq \calO(1) \frac{(1 - \rho(L)^2)^2}{\transient{L}{\rho(L)}^4} \norm{A}_+^{-2} \norm{P}_+^{-2} \norm{B}_+^{-3} \norm{R^{-1}}_+^{-2} \min \left\{ \norm{L}_+^{-2}, \norm{P}_+^{-1} \right\} = \calO(1).
\end{align*}
\end{corr}
Here, the upper bound condition on $\epsilon$ is to ensure that $\Ah_t, \Bh_t$ is stabilizable, so that $\Ph_t$ is well defined.
Furthermore, following the paragraph after Proposition 2 in \citet{mania2019certainty}, the assumption $\smin(P) \geq 1$ can be made without loss of generality when the other assumptions are satisfied. The reason is that, when $R \succ 0$ and $(Q^{1/2}, A)$ observable, the value function matrix $P$ is guaranteed to be positive definite.
Similar to how we got \cref{corr:stability_perturb_tmp}, by replacing $Q$, $R$ and $P$ with $Q/\min\{\smin(R), \smin(P)\}$, $R/\min\{\smin(R), \smin(P)\}$ and $P/\min\{\smin(R), \smin(P)\}$, we can remove the constraint $\smin(P) \geq 1$.
\begin{corr}
\label{cor:prop93}
Suppose $\norm{\Ah_t - A} \le \epsilon$ and $\norm{\Bh_t - B} \le \epsilon$. Let $\norm{\cdot}_+ = \norm{\cdot} + 1$. We assume that $R \succ 0$ and $(A, B)$ is stabilizable.
\begin{align*}
\norm{\Phat_t - P}
\leq& \min\{\smin(R), \smin(P)\}\calO(1)\, \epsilon \, \frac{\transient{L}{\rho(L)}^2}{1 - \rho(L)^2} \norm{A}_+^2 \lnorm{\frac{P}{\min\{\smin(R), \smin(P)\}}}_+^2 \norm{B}_+ \lnorm{\left(\frac{R}{\min\{\smin(R), \smin(P)\}}\right)^{-1}}_+ \\
=& \calO(\epsilon)
.\end{align*}
as long as
\begin{align*}
\epsilon
\leq& \calO(1) \frac{(1 - \rho(L)^2)^2}{\transient{L}{\rho(L)}^4} \norm{A}_+^{-2} \lnorm{\frac{P}{\min\{\smin(R), \smin(P)\}}}_+^{-2} \\
&\norm{B}_+^{-3} \lnorm{\left(\frac{R}{\min\{\smin(R), \smin(P)\}}\right)^{-1}}_+^{-2} \min \left\{ \norm{L}_+^{-2}, \lnorm{\frac{P}{\min\{\smin(R), \smin(P)\}}}_+^{-1} \right\} \\
=& \calO(1).
\end{align*}
\end{corr}
\subsubsection{Combining the two results together}
With \cref{corr:stability_perturb_tmp} and \cref{cor:prop93} the following corollary is straightforward.
\begin{corr}
\label{cor:stability_perturb}
Let $\epsilon > 0$ such
that $\epsilon \le \calO(1)$, $\norm{\Ah_t - A} \leq \epsilon$ and $\norm{\Bh_t - B} \leq \epsilon$. Then, we have
\begin{align*}
\norm{\Kt_{t+1} - K} \leq 7 \Gamma^3 \, \epsilon_P = \calO(\epsilon) .
\end{align*}
Here $\Gamma := 1 + \max\{\norm{A}, \norm{B}, \norm{P}/\min\{\smin(R), \smin(P)\}, \norm{K}\}$.
\end{corr}
\begin{proof}
With \cref{cor:prop93} we can find $\epsilon_P$ such that $\norm{\Phat_t - P} = \calO(\epsilon)$. Thus, the condition of \cref{corr:stability_perturb_tmp} is satisfied.
\end{proof}
\subsubsection{Concluding the proof of \cref{prop:one_epoch_estimate_withK}}
\begin{proof}
With \cref{prop:one_epoch_estimate} and \cref{cor:stability_perturb}, it is straightforward to give a new corollary with uniform control on all $\norm{\Ah_t - A}$, $\norm{\Bh_t - B}$, and $\norm{\Kt_{t+1} - K}$. Recall that we already proved the high probability bound in \cref{prop:one_epoch_estimate} that
\begin{equation*}
\max\left\{ \norm{\Ah_t - A}, \norm{\Bh_t - B}\right\} =
\calO(t^{-\frac{\beta}{2}} \log^{\frac{-\alpha + 1}{2}}(t)) \as
\end{equation*}
Basically, to satisfy the constraint in \cref{cor:stability_perturb}, we only need our bound (named $\epsilon$) in \cref{prop:one_epoch_estimate} to satisfy
\[\epsilon = \calO(t^{-\frac{\beta}{2}} \log^{\frac{-\alpha + 1}{2}}(t)) \le \calO(1) \as\]
which is always true when $t$ is large enough. (This also ensures $\Ah_t, \Bh_t$ to be stabilizable so that $K_0$ is only used finitely many times.)
That means,
\begin{equation*}
\norm{\Kt_{t+1} - K} = \calO(\epsilon) = \calO(t^{-\frac{\beta}{2}} \log^{\frac{-\alpha + 1}{2}}(t)) \as
\end{equation*}
Finally, we can say
\begin{equation*}
\max\left\{ \norm{\Ah_t - A}, \norm{\Bh_t - B}, \norm{\Kt_{t+1} - K}\right\}
= \calO(t^{-\frac{\beta}{2}} \log^{\frac{-\alpha + 1}{2}}(t)) \as
\end{equation*}
\end{proof}
\subsection{The proof of \cref{prop:one_epoch_estimate_withMyalg}}
\label{section: The proof of one_epoch_estimate_withMyalg}
\begin{prop*}
Let $x_0 \in \R^{\statedim}$ be any initial state. Assume \cref{asm:InitialStableCondition} is satisfied. When applying \cref{alg:myAlg}
\begin{equation*}
\max\left\{ \norm{\Ah_t - A}, \norm{\Bh_t - B}, \norm{\Kh_{t+1} - K}\right\} = \calO(t^{-\frac{\beta}{2}} \log^{\frac{-\alpha + 1}{2}}(t)) \as
\end{equation*}
\end{prop*}
\begin{proof}
Going thorough the whole \cref{alg:myAlg},
there are two conditions that might cause the difference between $\Kh_{t}$ and $\Kt_{t}$:
\begin{enumerate}
\item $\norm{\Kt_{t}} > C_K$, and
\item $\norm{x_{t}} > C_{x,t} = C_x\log(t)$.
\end{enumerate}
Our objective is to show that, with probability $1$, $\Kh_{t} \neq \Kt_{t}$ will happen only finitely often.
\paragraph{The first case $\norm{\Kt_{t}} > C_K$}
The first case is when $\norm{\Kt_{t}} > C_K$, this will not happen infinitely often. The first case
$\norm{\Kt_{t}} > C_K$ can only happen when
\begin{equation}
\label{eq: first Kh Kt diff condition}
\norm{\Kt_{t} - K} \ge \norm{\Kt_{t}} - \norm{K} > C_K - \norm{K}
.\end{equation}
By \cref{prop:one_epoch_estimate_withK}, we know that $\norm{\Kt_{t} - K}$ is exponentially decaying:
\begin{equation*}
\max\left\{\norm{\Kt_{t} - K}\right\} = \calO(t^{-\frac{\beta}{2}} \log^{\frac{-\alpha + 1}{2}}(t)) \as
\end{equation*}
As a result, \cref{eq: first Kh Kt diff condition} will hold only finitely many times, \as
\paragraph{The second case $\norm{x_{t}} > C_{x,t} = C_x\log(t)$}
To examine how often this would happen, we need to dig into more details of the decomposition of $\norm{x_t}$. Recall the previously derived formula from \cref{lemma: StateExpansion}:
\[x_{t} = \sum_{p=0}^{t-1}(A+B \Kh_{t-1})\cdots(A+B \Kh_{p+1})(B \eta_p+\varepsilon_p) +(A+B \Kh_{t-1})\cdots(A+B \Kh_0)x_0.\]
We hope to get an upper bound for $\norm{x_{t}}$. Apparently the main difficulty here is to bound the norm of $(A+B \Kh_{t-1})\cdots(A+B \Kh_{p+1})$. The following lemma serves as a key.
\begin{lemma}
\label{lem:productBound}
Suppose we have a constant square matrix $M$ with spectral radius $\rho(M) < 1$, and a sequence of uniformly bounded random variables $\{\delta_t\}_{t=0}^\infty$, satisfying $\norm{\delta_t} \asConv 0$.
Denote the constant $\rho_M:= \frac{2 + \rho(M)}{3} < 1$.
Then we have, for any $t, q \in \mathbb{N}$, $t > q$:
\[\norm{(M + \delta_{t-1}) \cdots (M + \delta_{q})} = \calO(\rhoM^{t-q}) \as\]
And as a direct corollary
\[\norm{M^{t-q}} = \calO(\rhoM^{t-q}) .\]
\end{lemma}
The proof can be found in \cref{The proof of lem:productBound}.
Notice that by our \cref{alg:myAlg}, $\norm{\Kh_{t}} \le C_K$ always holds, thus there exists a uniform upper bound on $\norm{B\delta_t} := \norm{B (\Kh_{t} - K)} \le \norm{B}(C_K + \norm{K})$.
Now we can separate the whole $\norm{(A+B \Kh_{t-1})\cdots(A+B \Kh_{p+1})}$ into two parts. If we denote $\rho_0 := \max(\frac{2+\rho(A+B K_{0})}{3}, \frac{2+\rho(A+BK)}{3})$, then with \cref{lem:productBound}, we can simultaneously bound both parts.
\begin{enumerate}
\item The first part contains the $A+B \Kh_{k}$ where $\Kh_k = K_0$, this part of product is denoted as $I_1$. In this part, $A+B \Kh_{k} = A+B K_0$. Suppose this part has $p_1$ same items, by \cref{{lem:productBound}} we know $I_1 \le \calO(\rho_0^{p_1}) \as$
\item The second part contains the $(A+B \Kh_{k})$ where $\Kh_k = \Kt_k$ to be our true certainty equivalent controller, this part of the product is denoted as $I_2$. If we denote $\delta_k := (\Kh_{k} - K)$, then, in this part, $(A+B \Kh_{k}) = (A+B K + B\delta_k)$. Remember our conclusion in \cref{prop:one_epoch_estimate_withK} that $\norm{\Kt_k - K } \asConv 0$, thus $\norm{\delta_k} \asConv 0$, assuming this part has $p_2$ items, then since $\norm{B\delta_k} \le \norm{B}(C_K + \norm{K})$, by \cref{lem:productBound}
\[I_2 \le \calO(\rho_0^{p_2}) \as\]
\end{enumerate}
We know $p_1 + p_2 = t - p -1$. Combining these two parts we have
\begin{align*}
\norm{(A+B \Kh_{t-1})\cdots(A+B \Kh_{p+1})} \le \calO(\rho_0^{t-p}) \as
\end{align*}
Finally we have the bound on $x_t$:
\begin{align*}
\norm{x_t} =& \lnorm{\sum_{p=0}^{t-1}(A+B \Kh_{t-1})\cdots(A+B \Kh_{p+1})(B \eta_p+\varepsilon_p) +(A+B \Kh_{t-1})\cdots(A+B K_0)x_0} \\
\le& \left(\sum_{p=0}^{t-1}\lnorm{(A+B \Kh_{t-1})\cdots(A+B \Kh_{p+1})}\lnorm{B \eta_p+\varepsilon_p} +\lnorm{(A+B \Kh_{t-1})\cdots(A+B K_0)}\lnorm{x_0}\right) \\
\le& \sum_{p=0}^{t-1}\calO(\rho_0^{t-p})\lnorm{B \eta_p+\varepsilon_p} +\calO(\rho_0^{t})\lnorm{x_0} \as \\
\end{align*}
Then
\begin{equation*}
\norm{x_t} = \calO\left(\sum_{p=0}^{t-1}\rho_0^{t-p}\lnorm{B \eta_p+\varepsilon_p} +\rho_0^{t}\lnorm{x_0}\right) \as
\end{equation*}
By Gaussian tail bounds (see \cref{lemma: Hi prob bounds in theorem 2}), we know that
\begin{equation*}
\norm{B\eta_t+\varepsilon_t} = \calO(\log^{1/2}(t)) \as
\end{equation*}
Then
\begin{equation*}
\norm{x_t} = \calO\left(\sum_{p=0}^{t-1}\rho_0^{t-p} \log^{1/2}(t)\right) +o(1)\as
\end{equation*}
Because $\rho_0^{t-p}$ is geometric sequence,
\begin{equation*}
\norm{x_t} \le \calO(\log^{1/2}(t)) \as
\end{equation*}
Thus for almost any $\omega \in \Omega$, $\norm{x_{t}} > C_{x,t} = C_x\log(t)$ will happen only finitely many times.
Finally, because two conditions $\norm{\Kt_{t}} > C_K$ and $\norm{x_{t}} > C_{x,t} = C_x\log(t)$ will happen only finitely many times, $\Kh_t$ and $\Kt_t$ eventually are the same. Following \cref{prop:one_epoch_estimate_withK},
\begin{equation*}
\begin{split}
\max\left\{ \norm{\Ah_t - A}, \norm{\Bh_t - B}, \norm{\Kh_{t+1} - K}\right\} = \calO(t^{-\frac{\beta}{2}} \log^{\frac{-\alpha + 1}{2}}(t)) \as
\end{split}
\end{equation*}
\end{proof}
\section{The proof of lemmas}
\subsection{Lemmas in \cref{The proof of thm:main tool}}
\subsubsection{The proof of \cref{lemma: Hi prob bounds in theorem 2}}
\label{The proof of lemma: Hi prob bounds in theorem 2}
\begin{lemma*}
~
\begin{itemize}
\item
\begin{equation}
\label{eq:bound on eta_p 2}
\norm{\varepsilon_t}, \norm{\eta_t} = \calO(\log^{1/2}(t)) \as
\end{equation}
\item
\begin{equation}
\label{eq: stochastic bound on B eta p plus varepsilon p 2}
\norm{B\eta_t+\varepsilon_t} = \calO(\log^{1/2}(t)) \as
\end{equation}
\end{itemize}
Assume \cref{eq:uniform high probability bound for Kh}, then:
\begin{itemize}
\item
\begin{equation}
\label{eq: stochastic bound delta_t 2}
\norm{\delta_t} = \norm{\Kh_{t} - K} =
\calO(t^{-\frac{\beta}{2}} \log^{\frac{-\alpha + 1}{2}}(t) ) \as
\end{equation}
\item
\begin{equation}
\label{eq: stochastic bound Lhat product 2}
\norm{(L + B\delta_{t-1}) \cdots (L + B\delta_{q})} =
\calO(\rhoL^{t-q}) \as
\end{equation}
\item
\begin{equation}
\label{eq: bound norm xi by log t 2}
\norm{x_t}, \norm{u_t} =
\calO(\log^{1/2}(t)) \as
\end{equation}
\end{itemize}
where $\delta_t := \Kh_{t} - K$, $L := A+BK$, and $\rhoL := \frac{2 + \rho(L)}{3}$. \textbf{Additionally, when $t=0, 1$ all these terms are bounded by $\calO(1) \as$}
\end{lemma*}
\begin{proof}
Outline:
~\paragraph{The proof of \cref{eq:bound on eta_p 2} and \cref{eq: stochastic bound on B eta p plus varepsilon p 2}}
The following lemma give the proof that \cref{eq:bound on eta_p 2} and \cref{eq: stochastic bound on B eta p plus varepsilon p 2} holds with probability at least $1-\delta$, which can be shown by the tail bound for i.i.d Gaussian random variables.
\begin{lemma}
\label{lemma: stochastic bound on B eta p plus varepsilon p}
For the noise $\eta_t \iid \calN(0, \tau^2 t^{1-\beta}\log^\alpha(t))$ and $\varepsilon_t \iid \calN(0, \sigma^2)$, we have that for any $\delta \in (0,1)$, with probability $1-\delta$, the following two equations holds for any $t \ge 1 $:
\begin{equation*}
\norm{\varepsilon_t}, \norm{\eta_t}, \norm{B\eta_t+\varepsilon_t} \le \calO(1) \log^{1/2}(t^2/\delta)
.\end{equation*}
\end{lemma}
We will prove \cref{lemma: stochastic bound on B eta p plus varepsilon p} shortly.
By \cref{defn: Big O notation} \cref{itm: big O as defn}, this implies
\begin{equation*}
\norm{\varepsilon_t}, \norm{\eta_t} \le \calO(\log^{1/2}(t)) \as
\end{equation*}
and
\begin{equation*}
\norm{B\eta_t+\varepsilon_t} \le \calO(\log^{1/2}(t)) \as
\end{equation*}
~\paragraph{The proof of \cref{eq: stochastic bound delta_t 2} and \cref{eq: stochastic bound Lhat product 2}}
\cref{eq: stochastic bound delta_t 2} directly follows from \cref{eq:uniform high probability bound for Kh}.
\cref{eq: stochastic bound Lhat product 2} follows from \cref{lem:productBound} given that we have $\delta_t \asConv 0$ from \cref{prop:one_epoch_estimate_withMyalg}:
\[ \norm{(L + B\delta_{t-1}) \cdots (L + B\delta_{q})} \le \calO(\rhoL^{t-q}) \as\]
~\paragraph{The proof of \cref{eq: bound norm xi by log t 2}}
Finally we need to prove \cref{eq: bound norm xi by log t 2} that
\[ \norm{x_t}, \norm{u_t} = \calO(\log^{1/2}(t)) \as \]
With the fact from \cref{lemma: StateExpansion} that
\begin{align*}
x_{t} =& \sum_{p=0}^{t-1}(A+B \Kh_{t-1})\cdots(A+B \Kh_{p+1})(B \eta_p+\varepsilon_p) +(A+B \Kh_{t-1})\cdots(A+B K_0)x_0 \\
=& \sum_{p=0}^{t-1}(L + B\delta_{t-1}) \cdots (L + B\delta_{p+1})(B \eta_p+\varepsilon_p) +(L + B\delta_{t-1}) \cdots (L + B\delta_{0})x_0
,\end{align*}
combined with the conclusion of \cref{eq: stochastic bound Lhat product 2} and \cref{eq: stochastic bound on B eta p plus varepsilon p 2}, we derive a norm bound on $x_t$:
\begin{align*}
\norm{x_{t}}
\le& \sum_{p=0}^{t-1}\norm{(L + B\delta_{t-1}) \cdots (L + B\delta_{p+1})}\norm{B \eta_p+\varepsilon_p} + \norm{(L + B\delta_{t-1}) \cdots (L + B\delta_{0})}\norm{x_0} \as \\
=& \sum_{p=0}^{t-1}\calO(\rhoL^{t-p})\norm{B \eta_p+\varepsilon_p} + \calO(\rhoL^{t})\norm{x_0} \as\\
=& \sum_{p=0}^{t-1}\calO(\rhoL^{t-p})\calO(\log^{1/2}(p)) + o(1) \as \\
\le& \sum_{p=0}^{t-1}\calO(\rhoL^{t-p})\calO(\log^{1/2}(t)) + o(1) \as\\
=& \calO(\log^{1/2}(t)) \as
\end{align*}
Recall that we have already shown \cref{eq:bound on eta_p 2}:
\[\norm{\eta_t} = \calO(\log^{1/2}(t)) \as \]
That means
\begin{align*}
&\norm{u_i} \\
= &\norm{(A+B\Kh_i)x_i + \eta_i} \\
\le& (\norm{A}+\norm{B}C_K)\norm{x_i} + \norm{\eta_i} \\
= & \calO(\log^{1/2}(t)) \as
\end{align*}
\end{proof}
~\paragraph{The proof of \cref{lemma: stochastic bound on B eta p plus varepsilon p}}
\label{The proof of lemma: stochastic bound on B eta p plus varepsilon p}
\begin{proof}
For any Gaussian variable $X \sim \calN(0, \sigma^2)$,
\[\P(X > t\sigma) \le e^{-t^2/2}, \]
and
\[\P(X^2 > t^2\sigma^2) = 2\P(X > t\sigma) \le 2e^{-t^2/2}.\]
For any multivariate normal vector sequence $X_t \sim \calN(0, \sigma^2I_n)$,
\begin{equation*}
\P(\norm{X_t}^2 > nt\sigma^2) = \P\left(\sum_{i=1}^n X_{t,i}^2 > nt\sigma^2\right) \le \sum_{i=1}^n \P(X_i^2 > t\sigma^2)\le 2ne^{-t/2}
.\end{equation*}
That means for any constant $c > 0$,
\begin{equation*}
\P(\norm{X_t}^2 >n2\log(ct^2/\delta)\sigma^2) \le 2ne^{-2\log(ct^2/\delta)/2} = \frac{2n\delta}{ct^2}.
\end{equation*}
We can sum up all choices of $t$ to get a uniform bound.
A well known equation states that $\sum_{t=1}^\infty 1/t^2 = \frac{\pi^2}6$. Then
\begin{equation*}
\P(\exists t \ge 1: \,\norm{X_t}^2 >2n\sigma^2\log(ct^2/\delta))
\le \sum_{t=1}^\infty \frac{2n\delta}{ct^2}
.\end{equation*}
We can choose $c = \frac{1}{2n}\sum_{t=1}^\infty 1/t^2 = \frac{\pi^2}{6\cdot 2n}$, so that
\begin{equation*}
\P(\exists t \ge 1: \,\norm{X_t}^2 >2n\sigma^2\log(ct^2/\delta))
\le \delta
.\end{equation*}
That is to say, with probability at least $1-\delta$, we have for any $t \ge 1$,
\begin{equation*}
\norm{X_t} \le \calO(1) \log^{1/2}(ct^2/\delta) = \calO(1) (\log(t^2/\delta) + \log(c))^{1/2}
.\end{equation*}
Since $\log(c)$ can be dominated by $\log(t^2/\delta)$, the above equation can simply be written as
\begin{equation*}
\norm{X_t} \le \calO(1) \log^{1/2}(t^2/\delta)
.\end{equation*}
This bound holds for $\varepsilon_t$ which has constant variance and is also true for $\eta_t$ which has shrinking variance. Thus, with probability at least $1-\delta$:
\begin{equation*}
\norm{\varepsilon_t}, \norm{\eta_t} \le \calO(1)
\log^{1/2}(t^2/\delta)
.\end{equation*}
Consider the fact that $\norm{B\eta_t+\varepsilon_t} \le \norm{B}\norm{\eta_t} + \norm{\varepsilon_t}$, which means $\norm{B\eta_t+\varepsilon_t}$ can still be bounded by:
\begin{equation*}
\norm{B\eta_t+\varepsilon_t} \le \calO(1) \log^{1/2}(t^2/\delta)
.\end{equation*}
\end{proof}
\subsubsection{The proof of \cref{lemma: StateExpansion}}
\label{The proof of lemma: StateExpansion}
\begin{lemma*}
\begin{equation*}
x_{t} = \sum_{p=0}^{t-1}(A+B \Kh_{t-1})\cdots(A+B \Kh_{p+1})(B \eta_p+\varepsilon_p) +(A+B \Kh_{t-1})\cdots(A+B K_0)x_0
.\end{equation*}
\begin{equation*}
u_{t} = \sum_{p=0}^{t-1}\Kh_t(A+B \Kh_{t-1})\cdots(A+B \Kh_{p+1})(B \eta_p+\varepsilon_p) +\Kh_t(A+B \Kh_{t-1})\cdots(A+B K_0)x_0 + \eta_t
.\end{equation*}
Here when $p = t-1$, $(A+B \Kh_{t-1})\cdots(A+B \Kh_{p+1}) := I_n$.
\end{lemma*}
\begin{proof}
Consider the following relationship:
\[u_t = \Kh_tx_t + \eta_t.\]
\begin{align*}
x_t =& A x_{t-1} + B u_{t-1} + \varepsilon_{t-1} \\
=& A x_{t-1} + B (\Kh_{t-1}x_{t-1} + \eta_{t-1}) + \varepsilon_{t-1} \\
=& (A+B \Kh_{t-1})x_{t-1} + B \eta_{t-1} + \varepsilon_{t-1}
.\end{align*}
Iteratively do this calculation to the end:
\begin{equation*}
x_{t} = \sum_{p=0}^{t-1}(A+B \Kh_{t-1})\cdots(A+B \Kh_{p+1})(B \eta_p+\varepsilon_p) +(A+B \Kh_{t-1})\cdots(A+B K_0)x_0
.\end{equation*}
\begin{equation*}
u_{t} = \sum_{p=0}^{t-1}\Kh_t(A+B \Kh_{t-1})\cdots(A+B \Kh_{p+1})(B \eta_p+\varepsilon_p) +\Kh_t(A+B \Kh_{t-1})\cdots(A+B K_0)x_0 + \eta_t
.\end{equation*}
\end{proof}
\subsubsection{The proof of \cref{lemma: four components xtxt}}
\label{The proof of lemma: four components xtxt}
\begin{lemma*}
Assume \cref{eq:uniform high probability bound for Kh}, then
\begin{enumerate}
\item
\begin{align*}
&\sum_{i=1}^{t-1}\sum_{p=0}^{i-1}\sum_{q=0}^{i-1}\left[(A+B K)^{i-p-1}\right](B \eta_p+\varepsilon_p)(B \eta_q+\varepsilon_q)^\top
\left[(A+B K)^{i-q-1}\right]^\top \\
& \qquad= t\sum_{p=0}^{\infty}L ^{p}(L ^{p})^\top \sigma^2 + t^\beta\frac{\tau^2}{\beta}\log^\alpha(t)(1+o_p(1))\sum_{q=0 }^{\infty}L ^{q}BB^\top [L ^{q}]^\top \\
& \qquad = t^\beta \log^\alpha(t) (C_t +o_p(1))
.\end{align*}
\item
\begin{equation*}
\begin{aligned}
&\sum_{i=1}^{t-1}\sum_{p=0}^{i-1}\sum_{q=0}^{i-1}\left[(A+B \Kh_{i-1})\cdots(A+B \Kh_{p+1}) - (A+B K)^{i-p-1}\right]\\
&\qquad \cdot(B \eta_p+\varepsilon_p)(B \eta_q+\varepsilon_q)^\top
\left[(A+B K)^{i-q-1}\right]^\top = \calO_p(t^{1-\beta/2}\log^{\frac{-\alpha + 3}{2}}(t))
.\end{aligned}
\end{equation*}
\item
\begin{align*}
&\sum_{i=1}^{t-1}\sum_{p=0}^{i-1}\sum_{q=0}^{i-1}\left[ (A+B K)^{i-p-1}\right](B \eta_p+\varepsilon_p)(B \eta_q+\varepsilon_q)^\top \\
&\qquad \cdot
\left[(A+B \Kh_{t-1})\cdots(A+B \Kh_{q+1}) - (A+B K)^{i-q-1}\right]^\top = \calO_p(t^{1-\beta/2}\log^{\frac{-\alpha + 3}{2}}(t))
.\end{align*}
\item
\begin{align*}
&\sum_{i=1}^{t-1}\sum_{p=0}^{i-1}\sum_{q=0}^{i-1}\left[(A+B \Kh_{t-1})\cdots(A+B \Kh_{p+1}) - (A+B K)^{i-p-1}\right] \\
&\qquad \cdot (B \eta_p+\varepsilon_p)(B \eta_q+\varepsilon_q)^\top \left[(A+B \Kh_{t-1})\cdots(A+B \Kh_{q+1}) - (A+B K)^{i-q-1}\right]^\top = \calO_p(t^{1-\beta/2}\log^{\frac{-\alpha + 3}{2}}(t))
.\end{align*}
\end{enumerate}
\end{lemma*}
\begin{proof}
The first step is to show the order of 2nd, 3rd and 4th part because they follow by the same method, especially the second part is just a transpose of the third part. Then we can focus on analyzing the first part, which is replacing all controllers $\Kh_t$ by optimal controller $K$.
\paragraph{Second Part}
With \cref{lemma: Hi prob bounds in theorem 2} in hand, now we are in good shape to start our proof with the second part showing
\begin{align*}
&\sum_{i=1}^{t-1}\sum_{p=0}^{i-1}\sum_{q=0}^{i-1}\left[(A+B \Kh_{i-1})\cdots(A+B \Kh_{p+1}) - (A+B K)^{i-p-1}\right]\\
&\qquad (B \eta_p+\varepsilon_p)(B \eta_q+\varepsilon_q)^\top
\left[(A+B K)^{i-q-1}\right]^\top = \calO_p(t^{1-\beta/2}\log^{\frac{-\alpha + 3}{2}}(t))
.\end{align*}
Since we have already shown the uniform bound of $(B \eta_p+\varepsilon_p)(B \eta_q+\varepsilon_q)^\top $ in \cref{lemma: Hi prob bounds in theorem 2}, and that $ \left[(A+B K)^{i-q-1}\right]^\top$ has an exponential decay rate, the main difficulty in bounding the second part is to give a tight bound on $\left[(A+B \Kh_{i-1})\cdots(A+B \Kh_{p+1}) - (A+B K)^{i-p-1}\right]$.
Recall the conclusion of \cref{lemma: Hi prob bounds in theorem 2}:
\begin{equation}
\label{eq:L star product bound}
\norm{(L + B\delta_{i-1}) \cdots (L + B\delta_{p+1})} = \calO(\rho_L ^{i-p}) \as
\end{equation}
Thus
\begin{equation}
\label{eq: difference Khat product and K product}
\begin{split}
&\norm{(A+B \Kh_{i-1})\cdots(A+B \Kh_{p+1}) - (A+B K)^{i-p-1}} \\
= &\norm{(L + B\delta_{i-1}) \cdots (L + B\delta_{p+1}) - L^{i-p-1}} \\
\le & \norm{B\delta_{i-1} (L + B\delta_{i-2}) \cdots (L + B\delta_{p+1})} +
\norm{LB\delta_{i-2} (L + B\delta_{i-3}) \cdots (L + B\delta_{p+1})} +
\cdots
\norm{L^{i-p-2}B\delta_{p+1}} \\
&(\text{For example, } (L + B\delta_3)(L + B\delta_2)(L + B\delta_1)- L^3 = \delta_3(L + B\delta_2)(L + B\delta_1) + L \delta_2 (L + B\delta_1) + L^2 B\delta_1 )\\
\le & \norm{B\delta_{i-1}} \norm{(L + B\delta_{i-2}) \cdots (L + B\delta_{p+1})} +
\norm{B\delta_{i-2}} \norm{L(L + B\delta_{i-3}) \cdots (L + B\delta_{p+1})} +
\cdots
\norm{B\delta_{p+1}}\norm{L^{i-p-2}} \\
\le & \calO(\rho_L ^{i-p})(\norm{\delta_{i-1}}+ \cdots + \norm{\delta_{p+1}}) \as \qquad (\text{using \cref{eq:L star product bound}})
\end{split}
\end{equation}
Now the L2 norm of the second term can be bounded as
\begin{equation}
\label{eq:second part step1}
\begin{split}
&\Bigg\|\sum_{i=1}^{t-1}\sum_{p=0}^{i-1}\sum_{q=0}^{i-1}\left[(A+B \Kh_{i-1})\cdots(A+B \Kh_{p+1}) - (A+B K)^{i-p-1}\right]\\
& \qquad \cdot (B \eta_p+\varepsilon_p)(B \eta_q+\varepsilon_q)^\top
\left[(A+B K)^{i-q-1}\right]^\top \Bigg\|\\
& \qquad \le \sum_{i=1}^{t-1}\sum_{p=0}^{i-1}\sum_{q=0}^{i-1}
\lnorm{\left[(A+B \Kh_{i-1})\cdots(A+B \Kh_{p+1}) - (A+B K)^{i-p-1}\right]}\\
& \qquad \qquad \cdot\lnorm{(B \eta_p+\varepsilon_p)(B \eta_q+\varepsilon_q)^\top }
\lnorm{\left[(A+B K)^{i-q-1}\right]^\top }\\
& \qquad \le \sum_{i=1}^{t-1}\sum_{p=0 }^{i-1}\sum_{q=0 }^{i-1}\calO(\rho_L^{i-p})(\norm{\delta_{i-1}}+ \cdots + \norm{\delta_{p+1}})\calO(\rho_L^{i-q}) \norm{(B\eta_p+\varepsilon_p)(B\eta_q+\varepsilon_q)^\top } \as\\
& \qquad \le \sum_{i=1}^{t-1}\sum_{p=0 }^{i-1}\sum_{q=0 }^{i-1}\calO(\rho_L^{i-p})(\norm{\delta_{i-1}}+ \cdots + \norm{\delta_{p+1}})\calO(\rho_L^{i-q}) (\norm{B\eta_p+\varepsilon_p}^2+\norm{B\eta_q+\varepsilon_q}^2) \as
\end{split}
\end{equation}
At first glance it seems like there is no way this would generate the desired bound, because the $\norm{\delta_{i-1}}+ \cdots + \norm{\delta_{p+1}}$ term could diverge when $i$ is large. However, thanks to the exponentially decaying term $\calO(\rho_L^{i-p})$, we can avoid this by changing the order of summation:
\begin{equation}
\label{eq:part2 basic tool}
\begin{split}
\sum_{p=0 }^{i-1}\calO(\rho_L^{i-p})(\norm{\delta_{i-1}}+ \cdots + \norm{\delta_{p+1}})
= &\sum_{p=0 }^{i-1}\calO(\rho_L^{i-p})\sum_{j=p+1}^{i-1}\norm{\delta_j} \\
= &\sum_{p=0 }^{i-1}\sum_{j=p+1}^{i-1}\calO(\rho_L^{i-p})\norm{\delta_j} \\
= &\sum_{j=1 }^{i-1}\sum_{p=0}^{j-1}\calO(\rho_L^{i-p})\norm{\delta_j} \qquad \text{(exchange the order of summation)} \\
= &\sum_{j=1 }^{i-1}\norm{\delta_j}\sum_{p=0}^{j-1}\calO(\rho_L^{i-p}) \\
= &\sum_{j=1 }^{i-1}\norm{\delta_j}\calO(\rho_L^{i-j}) \\
\end{split}
\end{equation}
The final form is almost the same as the beginning, except that the summation of $\delta_i$ disappears.
Restart from \cref{eq:second part step1}, and remember to use \cref{eq:part2 basic tool}
(\textbf{Additionally, when $p=0, 1$, $\calO(\log(p))$ is meant to be $\calO(1) \as$}):
\begin{align}
\label{eq: second part final step}
&\Bigg\|\sum_{i=1}^{t-1}\sum_{p=0}^{i-1}\sum_{q=0}^{i-1}\left[(A+B \Kh_{i-1})\cdots(A+B \Kh_{p+1}) - (A+B K)^{i-p-1}\right] \nonumber\\
&\qquad \cdot(B \eta_p+\varepsilon_p)(B \eta_q+\varepsilon_q)^\top
\left[(A+B K)^{i-q-1}\right]^\top \Bigg\| \nonumber\\
& \qquad \le \sum_{i=1}^{t-1}\sum_{p=0 }^{i-1}\sum_{q=0 }^{i-1}\calO(\rho_L^{i-p})(\norm{\delta_{i-1}}+ \cdots + \norm{\delta_{p+1}})\calO(\rho_L^{i-q}) (\norm{B\eta_p+\varepsilon_p}^2+\norm{B\eta_q+\varepsilon_q}^2) \as \quad \text{(by \cref{lemma: Hi prob bounds in theorem 2})} \nonumber\\
& \qquad \le \sum_{i=1}^{t-1}\sum_{p=0 }^{i-1}\sum_{q=0 }^{i-1}\calO(\rho_L^{i-p})(\norm{\delta_{i-1}}+ \cdots + \norm{\delta_{p+1}})\calO(\rho_L^{i-q}) (\calO(\log(p)) + \calO(\log(q))) \as \nonumber\\
& \qquad \le \sum_{i=1}^{t-1}\sum_{p=0 }^{i-1}\sum_{q=0 }^{i-1}\calO(\rho_L^{i-p})(\norm{\delta_{i-1}}+ \cdots + \norm{\delta_{p+1}})\calO(\rho_L^{i-q}) \calO(\log(t)) \as \nonumber\\
& \qquad = \calO(\log(t))\sum_{i=1}^{t-1}\sum_{p=0 }^{i-1}\calO(\rho_L^{i-p})(\norm{\delta_{i-1}}+ \cdots + \norm{\delta_{p+1}}) \as \nonumber\\
& \qquad =\calO(\log(t)) \sum_{i=1}^{t-1}\sum_{j=1 }^{i-1}\norm{\delta_j}\calO(\rho_L^{i-j}) \as \nonumber\\
& \qquad = \calO(\log(t))\sum_{j=1}^{t-1}\norm{\delta_{j}}\sum_{i=j+1}^{t-1}\calO(\rho_L^{i-j}) \as \qquad \text{(by \cref{eq:part2 basic tool})} \nonumber\\
& \qquad = \calO(\log(t))\sum_{j=1}^{t-1}\norm{\delta_{j}} \as \nonumber\\
& \qquad = \calO(\log(t))\left(\sum_{j=1}^{t-1} \calO(j^{-\frac{\beta}{2}} \log^{\frac{-\alpha + 1}{2}}(j)) \right) \as \qquad \text{(by \cref{eq: sum eta_t})} \nonumber\\
& \qquad =
\calO(\log(t)) \calO(t^{1-\frac{\beta}{2}}\log^{\frac{-\alpha + 1}{2}}(t)) \as \nonumber\\
& \qquad =
\calO(t^{1-\beta/2}\log^{\frac{-\alpha + 3}{2}}(t)) \as
\end{align}
We know that for any matrix $A$, $\norm{A} \le \norm{A}_F \le \sqrt{r} \norm{A}$, where $r$ is the rank of matrix $A$. Thus \cref{eq: second part final step} implies an upper bound on the Frobenius norm, and the Frobenius norm implies entry-wise upper bound:
\begin{align*}
&\sum_{i=1}^{t-1}\sum_{p=0}^{i-1}\sum_{q=0}^{i-1}\left[(A+B \Kh_{i-1})\cdots(A+B \Kh_{p+1}) - (A+B K)^{i-p-1}\right]\\
&\qquad \cdot(B \eta_p+\varepsilon_p)(B \eta_q+\varepsilon_q)^\top
\left[(A+B K)^{i-q-1}\right]^\top = \calO(t^{1-\beta/2}\log^{\frac{-\alpha + 3}{2}}(t)) \as
\end{align*}
\paragraph{Third Part}
This part is the transpose of the second part, thus shares the same result with the second part.
\paragraph{Fourth Part}
We wish to show that
\begin{align*}
&\Bigg\|\sum_{i=1}^{t-1}\sum_{p=0}^{i-1}\sum_{q=0}^{i-1}\left[(A+B \Kh_{i-1})\cdots(A+B \Kh_{p+1}) - (A+B K)^{i-p-1}\right]\\
&\qquad \cdot(B \eta_p+\varepsilon_p)(B \eta_q+\varepsilon_q)^\top
\left[(A+B \Kh_{i-1})\cdots(A+B \Kh_{q+1}) - (A+B K)^{i-q-1}\right]^\top \Bigg\|\\
& \qquad =\calO_p(t^{1-\beta/2}\log^{\frac{-\alpha + 3}{2}}(t))
.\end{align*}
By \cref{lem:productBound} we have
\[
\norm{(A+B \Kh_{i-1})\cdots(A+B \Kh_{q+1})} = \calO(\rho_L ^{i-q}) \as,\]
and
\[
\norm{(A+B K)^{i-q-1}} = \calO(\rho_L ^{i-q}) \as\]
Thus,
\begin{align*}
(A+B \Kh_{i-1})\cdots(A+B \Kh_{q+1}) - (A+B K)^{i-q-1} = \calO(\rho_L ^{i-q}) \as
\end{align*}
Combining this with \cref{eq: difference Khat product and K product},
\begin{align*}
&\norm{\sum_{i=1}^{t-1}\sum_{p=0}^{i-1}\sum_{q=0}^{i-1}\left[(A+B \Kh_{i-1})\cdots(A+B \Kh_{p+1}) - (A+B K)^{i-p-1}\right]\\
&\qquad \cdot (B \eta_p+\varepsilon_p)(B \eta_q+\varepsilon_q)^\top
\left[(A+B \Kh_{i-1})\cdots(A+B \Kh_{q+1}) - (A+B K)^{i-q-1}\right]^\top }\\
& \qquad \le \sum_{i=1}^{t-1}\sum_{p=0 }^{i-1}\sum_{q=0 }^{i-1}\calO(\rho_L^{i-p})(\norm{\delta_{i-1}}+ \cdots + \norm{\delta_{p+1}}) \norm{(B\eta_p+\varepsilon_p)(B\eta_q+\varepsilon_q)^\top }\calO(\rho_L^{i-q}) \as\\
& \qquad \le \sum_{i=1}^{t-1}\sum_{p=0 }^{i-1}\sum_{q=0 }^{i-1}\calO(\rho_L^{i-p})(\norm{\delta_{i-1}}+ \cdots + \norm{\delta_{p+1}})\calO(\rho_L^{i-q}) (\norm{B\eta_p+\varepsilon_p}^2+\norm{B\eta_q+\varepsilon_q}^2) \as
,\end{align*}
which is exactly the same as the final line of \cref{eq:second part step1}. Then following the same proof procedure as in the second part we can get the same order as in the second part: $\calO(t^{1-\beta/2}\log^{\frac{-\alpha + 3}{2}}(t)) \as$
\paragraph{Summarize second, third, and fourth parts}
To sum up, all three parts are bounded by the same order $\calO(t^{1-\beta/2}\log^{\frac{-\alpha + 3}{2}}(t)) \as$
\paragraph{First Part}
\begin{comment}
Since we do not have any changing constants $C$, $C'$ and $\delta$ in this part, we can safely take expectation on the whole probability space.
\end{comment}
It remains to show
\begin{align*}
&\sum_{i=1}^{t-1}\sum_{p=0}^{i-1}\sum_{q=0}^{i-1}\left[(A+B K)^{i-p-1}\right](B \eta_p+\varepsilon_p)(B \eta_q+\varepsilon_q)^\top
\left[(A+B K)^{i-q-1}\right]^\top \\
& \qquad = t\sum_{p=0}^{\infty}L ^{p}(L ^{p})^\top \sigma^2 + t^\beta\frac{\tau^2}{\beta}\log^\alpha(t)\sum_{q=0 }^{\infty}L ^{q}BB^\top [L ^{q}]^\top (I_n+o_p(1))
.\end{align*}
Recall $L = A+BK$. We divide the left hand side into two separate parts:
\begin{itemize}
\item The part where $p \neq q$. We will show this part is dominated by the $p = q$ part and is only of order $\calO_p(t^{1/2})$.
\[G_t \defineas \sum_{i=1}^{t-1}\sum_{p\neq q }^{i-1}L ^{i-p-1}(B\eta_p+\varepsilon_p)(B\eta_q+\varepsilon_q)^\top [L ^{i-q-1}]^\top = \calO_p(t^{1/2}).\]
\item The part where $p = q$. We will show that
\begin{align*}
&\sum_{i=1}^{t-1}\sum_{p = q =0}^{i-1}L ^{i-p-1}(B\eta_p+\varepsilon_p)(B\eta_q+\varepsilon_q)^\top [L ^{i-q-1}]^\top \\
& \qquad = t\sum_{p=0}^{\infty}L ^{p}(L ^{p})^\top \sigma^2 + t^\beta\frac{\tau^2}{\beta}\log^\alpha(t)\sum_{q=0 }^{\infty}L ^{q}BB^\top [L ^{q}]^\top (I_n+o_p(1)).
\end{align*}
\end{itemize}
Let us first consider the part where $p \neq q$.
We will show the order of $G_t$ by considering its expectation and variance. Since $G_t$ is a summation of cross terms and $\E (B\eta_p+\varepsilon_p)= 0 $, $\E(G_t) = 0$.
Now it remains to consider the variance
\begin{equation*}
\begin{split}
\E(\norm{G_t}_F^2) =& \E(\Tr(G_t^2)) \\
=& \E \Bigg(\Tr\Bigg(\sum_{p\neq q }^{t-1}\sum_{i=p \lor q+1}^{t-1}\sum_{j=p \lor q+1}^{t-1}L ^{i-p-1}(B\eta_p+\varepsilon_p)(B\eta_q+\varepsilon_q)^\top [L ^{i-q-1}]^\top \\
& \qquad \cdot L ^{j-p-1}(B\eta_p+\varepsilon_p)(B\eta_q+\varepsilon_q)^\top [L ^{j-q-1}]^\top \Bigg) \Bigg)\\
& +\E \Bigg(\Tr\Bigg(\sum_{p\neq q }^{t-1}\sum_{i=p \lor q+1}^{t-1}\sum_{j=p \lor q+1}^{t-1}L ^{i-p-1}(B\eta_p+\varepsilon_p)(B\eta_q+\varepsilon_q)^\top [L ^{i-q-1}]^\top \\
& \qquad \cdot L ^{j-q-1}(B\eta_q+\varepsilon_q)(B\eta_p+\varepsilon_p)^\top [L ^{j-p-1}]^\top \Bigg) \Bigg)\\
&\qquad \text{(terms with odd power go away in expectation) }
.\end{split}
\end{equation*}
It is sufficient to consider the first term in the previous expression, and the other term can be analyzed in exactly the same way. Notice the following relationship on any square matrix $A$ with dimension $n$
\[\mathbf{Tr}^2(A) \le n\norm{A}_F^2 \le n \cdot n\norm{A}^2.\]
That is
\begin{equation*}
\Tr(A) \le n\norm{A}
.\end{equation*}
Then
\begin{align*}
& \E \Bigg(\Tr\Bigg(\sum_{p\neq q }^{t-1}\sum_{i=p \lor q+1}^{t-1}\sum_{j=p \lor q+1}^{t-1}L ^{i-p-1}(B\eta_p+\varepsilon_p)(B\eta_q+\varepsilon_q)^\top [L ^{i-q-1}]^\top \\
& \qquad \cdot L ^{j-p-1}(B\eta_p+\varepsilon_p)(B\eta_q+\varepsilon_q)^\top [L ^{j-q-1}]^\top \Bigg) \Bigg) \\
& \qquad \le n\E \Bigg\|\sum_{p\neq q }^{t-1}\sum_{i=p \lor q+1}^{t-1}\sum_{j=p \lor q+1}^{t-1}L ^{i-p-1}(B\eta_p+\varepsilon_p)(B\eta_q+\varepsilon_q)^\top [L ^{i-q-1}]^\top \\
& \qquad \qquad \cdot L ^{j-p-1}(B\eta_p+\varepsilon_p)(B\eta_q+\varepsilon_q)^\top [L ^{j-q-1}]^\top \Bigg\|\\
& \qquad \le \E\sum_{p\neq q }^{t-1}\sum_{i=p \lor q+1}^{t-1}\sum_{j=p \lor q+1}^{t-1} \calO(\rho_L^{i-p}\rho_L^{i-q})\norm{B\eta_p+\varepsilon_p}_2^2\norm{B\eta_q+\varepsilon_q}_2^2O(\rho_L^{j-p}\rho_L^{j-q})
\qquad \text{(by \cref{lem:productBound})} \\
& \qquad = \calO\left( \sum_{p\neq q }^{t-1}\sum_{i=p \lor q+1}^{t-1}\sum_{j=p \lor q+1}^{t-1}\rho_L^{2i-p-q}\rho_L^{2j-p-q} \right) \\
& \qquad = \calO\left(\sum_{p > q}^{t-1}\rho_L^{2(p-q)} \right) \quad \text{(WLOG consider the part where } p>q \text{)} \\
& \qquad = \calO\left(\sum_{q=0}^{t-1}1 \right) \\
& \qquad = \calO(t)
.\end{align*}
Thus the entry-wise standard error of $G_t$ is of order $\calO(t^{1/2})$. Combining this with the fact that $\E G_t = 0$, we have
\begin{equation}
\label{eq: Gt order}
G_t \defineas \sum_{i=1}^{t-1}\sum_{p\neq q }^{i-1}L ^{i-p-1}(B\eta_p+\varepsilon_p)(B\eta_q+\varepsilon_q)^\top [L ^{i-q-1}]^\top = \calO_p(t^{1/2}).
\end{equation}
and it remains to consider
\[R \defineas \sum_{i=1}^{t-1}\sum_{p=0 }^{i-1}L ^{i-p-1}(B\eta_p+\varepsilon_p)(B\eta_p+\varepsilon_p)^\top [L ^{i-p-1}]^\top .\]
Consider the expectation of $R$: $\E(R) = R_0$, where
\[R_0 \defineas \sum_{i=0}^{t-1}\sum_{p=0 }^{i-1}L ^{i-p-1}(p^{\beta-1}\log^\alpha(p)BB^\top \tau^2 + I_n\sigma^2)[L ^{i-p-1}]^\top
.\]
Let us first show $R - R_0 = \calO_p(t ^{1/2})$, and after that we only need to consider $R_0$, which is the dominating term. We know that $B\eta_p+\varepsilon_p$ has a finite fourth moment, so the sum of the variances of each element of $R-R_0$ can be written as
\begin{align*}
\E \norm{R-R_0}_F^2
=&\E(\Tr((R-R_0)^2)) \\
\le& \E(\Tr (\sum_{p=0 }^{t-1}\sum_{i=p+1}^{t-1}\sum_{j=p +1}^{t-1}L ^{i-p-1}[(B\eta_p+\varepsilon_p)(B\eta_p+\varepsilon_p)^\top -(p^{\beta-1}\log^\alpha(p)BB^\top \tau^2 + I_m\sigma^2)]\\
&\qquad \cdot [L ^{i-p-1}]^\top L ^{j-p-1}[(B\eta_p+\varepsilon_p)(B\eta_p+\varepsilon_p)^\top -(p^{\beta-1}\log^\alpha(p)BB^\top \tau^2 + I_m\sigma^2)][L ^{j-p-1}]^\top )) \\
\le& n\E \norm{\sum_{p=0 }^{t-1}\sum_{i=p+1}^{t-1}\sum_{j=p +1}^{t-1}L ^{i-p-1}[(B\eta_p+\varepsilon_p)(B\eta_p+\varepsilon_p)^\top -(p^{\beta-1}\log^\alpha(p)BB^\top \tau^2 + I_m\sigma^2)]\\
&\qquad \cdot [L ^{i-p-1}]^\top L ^{j-p-1}[(B\eta_p+\varepsilon_p)(B\eta_p+\varepsilon_p)^\top -(p^{\beta-1}\log^\alpha(p)BB^\top \tau^2 + I_m\sigma^2)][L ^{j-p-1}]^\top }\\
\le& \calO(\E\sum_{p=0 }^{t-1}\sum_{i=p +1}^{t-1}\sum_{j=p +1}^{t-1} \rho_L^{2i-2p}\norm{(B\eta_p+\varepsilon_p)(B\eta_p+\varepsilon_p)^\top -(p^{\beta-1}\log^\alpha(p)BB^\top \tau^2 + I_m\sigma^2)}^2\rho_L^{2j-2p} )\\
=& \calO( \sum_{p=0}^{t-1}\E\norm{(B\eta_p+\varepsilon_p)(B\eta_p+\varepsilon_p)^\top -(p^{\beta-1}\log^\alpha(p)BB^\top \tau^2 + I_m\sigma^2)}^2 ) \\
=& \calO(t)
.\end{align*}
Thus $R - R_0 = \calO_p(t ^{1/2})$. Now we only need to focus on:
\[R_0 = \sum_{i=1}^{t-1}\sum_{p=0 }^{i-1}L ^{i-p-1}(p^{\beta-1}\log^\alpha(p)BB^\top \tau^2 + I_m\sigma^2)[L ^{i-p-1}]^\top .\]
Again, when $p=0,1$, $p^{\beta-1}\log^\alpha(p)$ should be considered as 1.
Let us start from the identity matrix part $\sum_{i=1}^{t-1}\sum_{p=0 }^{i-1}L ^{i-p-1}I_m\sigma^2[L ^{i-p-1}]^\top $.
\begin{equation*}
\begin{split}
\sum_{i=1}^{t-1}\sum_{p=0 }^{i-1}L ^{i-p-1}[L ^{i-p-1}]^\top = &\sum_{i=1}^{t-1}\sum_{q=0 }^{i-1}L ^{q}[L ^{q}]^\top \\
= & \sum_{i=1}^{t-1}(\sum_{p=0 }^{\infty}L ^{p}(L ^{p})^\top - \sum_{q=i}^{\infty}L ^{q}[L ^{q}]^\top ) \\
=& t\sum_{p=0}^{\infty}L ^{p}(L ^{p})^\top - \sum_{i=1}^{t-1}\sum_{q=i}^{\infty}L ^{q}[L ^{q}]^\top
.\end{split}
\end{equation*}
Notice
\begin{equation*}
\begin{split}
\norm{\sum_{i=1}^{t-1}\sum_{q=i}^{\infty}L ^{q}[L ^{q}]^\top } \le & \sum_{i=1}^{t-1}\sum_{q=i}^{\infty}\calO(\rho_L^{2q}) \\
= & \sum_{i=1}^{t-1}\calO(\rho_L^{2i}) \\
= & \calO(1)
.\end{split}
\end{equation*}
Thus
\begin{equation*}
\sum_{i=1}^{t-1}\sum_{p=0 }^{i-1}L ^{i-p-1}[L ^{i-p-1}]^\top = t\sum_{p=0}^{\infty}L ^{p}(L ^{p})^\top + \calO(1)
.\end{equation*}
On the other hand (\textbf{when $p=0, 1$, $p^{\beta-1}\log^\alpha(p)$ is meant to be $1$}),
\begin{equation}
\label{eq: noname_1}
\begin{split}
& \sum_{i=1}^{t-1}\sum_{p=0 }^{i-1}L ^{i-p-1}p^{\beta-1}\log^\alpha(p)BB^\top [L ^{i-p-1}]^\top \\
& \qquad = \sum_{p=0}^{t-2}\sum_{i=p+1}^{t-1}L ^{i-p-1}p^{\beta-1}\log^\alpha(p)BB^\top [L ^{i-p-1}]^\top \\
& \qquad = \sum_{p=0}^{t-2}p^{\beta-1}\log^\alpha(p)\sum_{q=0 }^{t-p-2}L ^{q}BB^\top [L ^{q}]^\top \\
& \qquad = \sum_{p=0}^{t-2}p^{\beta-1}\log^\alpha(p)\left(\sum_{q=0 }^{\infty}L ^{q}BB^\top [L ^{q}]^\top - \sum_{q=t-p-1 }^{\infty}L ^{q}BB^\top [L ^{q}]^\top \right) \\
& \qquad = \sum_{p=0}^{t-2}p^{\beta-1}\log^\alpha(p)\left(\sum_{q=0 }^{\infty}L ^{q}BB^\top [L ^{q}]^\top - \calO(\rhoL ^{2(t-p-1)})\right) \\
& \qquad = \sum_{p=0}^{t-2}p^{\beta-1}\log^\alpha(p)\sum_{q=0 }^{\infty}L ^{q}BB^\top [L ^{q}]^\top + \sum_{p=0}^{t-2}p^{\beta-1}\log^\alpha(p)\calO\left(\rhoL ^{2(t-p-1)}\right)\\
& \qquad \le \sum_{p=0}^{t-2}p^{\beta-1}\log^\alpha(p)\sum_{q=0 }^{\infty}L ^{q}BB^\top [L ^{q}]^\top + \sum_{p=0}^{t-2}\calO(1)\calO\left(\rhoL ^{2(t-p-1)}\right)\\
& \qquad = \sum_{p=0}^{t-2}p^{\beta-1}\log^\alpha(p)\sum_{q=0 }^{\infty}L ^{q}BB^\top [L ^{q}]^\top + \calO(1)
.\end{split}
\end{equation}
Now it remains to calculate $\sum_{p=0}^{t-2}p^{\beta-1}\log^\alpha(p)$.
Let us consider a more general case $\sum_{p=0}^{t}p^{\gamma}\log^\alpha(p)$ where $\gamma > -1$ and $\alpha$ is any real number. It is clear that this summation goes to infinity when $t \to \infty$.
Recall the Stolz--Cesàro theorem:
\begin{theorem}[Stolz--Cesàro]
\label{thm: Stolz-Cesaro theorem}
Let $\{a_t\}_{t\ge 1}$ and $\{b_t\}_{t\ge1}$ be two sequences of real numbers. Assume that $\{b_t\}_{t\ge1}$ is a strictly monotone and divergent sequence and the following limit exists:
\begin{equation*}
\lim_{t \to \infty} \frac{a_{t+1}-a_t}{b_{t+1}-b_t} = l
\end{equation*}
Then, the limit
\begin{equation*}
\lim_{t \to \infty} \frac{a_t}{b_t} = l
\end{equation*}
\end{theorem}
In \cref{thm: Stolz-Cesaro theorem}, we choose $a_t$ and $b_t$ to be $\sum_{p=0}^{t}p^{\gamma}\log^\alpha(p) $ and $t^{\gamma+1}\log^\alpha(t)$, respectively.
\begin{align*}
\lim_{t\to\infty}\frac{a_{t} - a_{t-1}}{b_t - b_{t-1}} =& \lim_{t\to\infty}\frac{t^{\gamma}\log^\alpha(t)}{t^{\gamma+1}\log^\alpha(t) - (t-1)^{\gamma+1}\log^\alpha(t-1)} \\
=& \lim_{t\to\infty}\frac{1}{t - (\frac{t-1}t)^{\gamma}(t-1) \left(\frac{\log(t-1)}{\log(t)}\right)^\alpha} \\
=& \lim_{t\to\infty}\frac{1/t}{1 - (1-\frac{1}t)^{\gamma+1} \left(1+\frac{\log(t-1)-\log(t)}{\log(t)}\right)^\alpha} \\
=& \lim_{t\to\infty}\frac{1/t}{1 - (1 - \frac{\gamma+1}t + o(\frac1t)) \left(1+\frac{-\frac1t + o(\frac1t)}{\log(t)}\right)^\alpha} \\
=& \lim_{t\to\infty}\frac{1/t}{1 - (1 - \frac{\gamma+1}t + o(\frac1t)) \left(1+o(\frac1t)\right)^\alpha} \\
=& \lim_{t\to\infty}\frac{1/t}{1 - (1 - \frac{\gamma+1}t + o(\frac1t)) e^{\alpha\log\left(1+o(\frac1t)\right)}} \\
=& \lim_{t\to\infty}\frac{1/t}{1 - (1 - \frac{\gamma+1}t + o(\frac1t)) e^{\alpha o(\frac1t)}} \\
=& \lim_{t\to\infty}\frac{1/t}{1 - (1 - \frac{\gamma+1}t + o(\frac1t)) \left(1+o(\frac\alpha{t})\right)} \\
=& \lim_{t\to\infty}\frac{1/t}{\frac{\gamma+1}t + o(\frac1t) } \\
=& \frac{1}{\gamma+1}
.\end{align*}
By \cref{thm: Stolz-Cesaro theorem}, we know
\begin{align*}
\lim_{t\to\infty}\frac{a_{t} }{b_t } = \lim_{t\to\infty}\frac{\sum_{p=0}^{t}p^{\gamma}\log^\alpha(p)}{t^{\gamma+1}\log^\alpha(t)} = \frac{1}{\gamma+1}
\end{align*}
That is to say, for any $\gamma > -1$:
\begin{align}
\label{eq: sum eta_t}
\begin{split}
\sum_{p=0}^{t}p^{\gamma}\log^\alpha(p)
& = \frac{1}{\gamma+1} t^{\gamma+1}\log^\alpha(t) (1+o(1))
.\end{split}
\end{align}
Following \cref{eq: noname_1,eq: sum eta_t},
\begin{equation*}
\begin{split}
&\sum_{i=1}^{t-1}\sum_{p=0 }^{i-1}L ^{i-p-1}p^{\beta-1}\log^\alpha(p)BB^\top [L ^{i-p-1}]^\top \\
=& \sum_{p=0}^{t}p^{\beta-1}\log^\alpha(p)\sum_{q=0 }^{\infty}L ^{q}BB^\top [L ^{q}]^\top + \calO(1) \\
= & \frac{t^\beta}\beta \log^\alpha(t)(1+o(1))\sum_{q=0 }^{\infty}L ^{q}BB^\top [L ^{q}]^\top + \calO(1) \\
= & \frac{t^\beta}\beta \log^\alpha(t)\sum_{q=0 }^{\infty}L ^{q}BB^\top [L ^{q}]^\top (I_n+o(1))
.\end{split}
\end{equation*}
To sum up,
\[R_0 = t\sum_{p=0}^{\infty}L ^{p}(L ^{p})^\top \sigma^2 + \frac{t^\beta}\beta \log^\alpha(t)\sum_{q=0 }^{\infty}L ^{q}BB^\top [L ^{q}]^\top (I_n+o(1)) .\]
Recall that $R - R_0 = \calO_p(t^{1/2})$, so
\begin{align*}
R &= \sum_{i=1}^{t-1}\sum_{p=0 }^{i-1}L ^{i-p-1}(B\eta_p+\varepsilon_p)(B\eta_p+\varepsilon_p)^\top [L ^{i-p-1}]^\top \\
&=
t\sum_{p=0}^{\infty}L ^{p}(L ^{p})^\top \sigma^2 + t^\beta\frac{\tau^2}{\beta}\log^\alpha(t)\sum_{q=0 }^{\infty}L ^{q}BB^\top [L ^{q}]^\top (I_n+o_p(1)).
\end{align*}
Recall \cref{eq: Gt order}:
\[\sum_{i=1}^{t-1}\sum_{p\neq q }^{i-1}L ^{i-p-1}(B\eta_p+\varepsilon_p)(B\eta_q+\varepsilon_q)^\top [L ^{i-q-1}]^\top = \calO_p(t^{1/2}).\]
Finally we proved the order of the first part:
\begin{align*}
&\sum_{i=1}^{t-1}\sum_{p=0}^{i-1}\sum_{q=0}^{i-1}\left[(A+B K)^{i-p-1}\right](B \eta_p+\varepsilon_p)(B \eta_q+\varepsilon_q)^\top
\left[(A+B K)^{i-q-1}\right]^\top \\
& \qquad = \sum_{i=1}^{t-1}\sum_{p=0 }^{i-1}\sum_{q=0 }^{i-1}L ^{i-p-1}(B\eta_p+\varepsilon_p)(B\eta_q+\varepsilon_q)^\top [L ^{i-q-1}]^\top \\
& \qquad= t\sum_{p=0}^{\infty}L ^{p}(L ^{p})^\top \sigma^2 + t^\beta\frac{\tau^2}{\beta}\log^\alpha(t)\sum_{q=0 }^{\infty}L ^{q}BB^\top [L ^{q}]^\top (I_n+o_p(1)) \\
& \qquad= t^\beta \log^\alpha(t) (C_t +o_p(1)) \qquad \text{(by $C_t$ definition \cref{eq: Ct definition})}
.\end{align*}
\end{proof}
\subsubsection{The proof of \cref{lemma: Term with starting point xtxt}}
\label{The proof of lemma: Term with starting point xtxt}
\begin{lemma*}
Assume \cref{eq:uniform high probability bound for Kh}, then
\begin{enumerate}
\item $\sum_{i=0}^{t-1}\left[ (A+B \Kh_{i-1})\cdots(A+B K_{0})x_0\right]
\left[\sum_{q=0}^{i-1}(A+B \Kh_{i-1})\cdots(A+B \Kh_{q+1})(B\eta_q+\varepsilon_q)\right]^T = \logO(1) \as$
\item $\sum_{i=0}^{t-1}\left[ (A+B \Kh_{i-1})\cdots(A+B K_{0})x_0\right]
\left[ (A+B \Kh_{i-1})\cdots(A+B K_{0})x_0\right]^T = \calO(1) \as$
\end{enumerate}
\end{lemma*}
\begin{proof}
This can be proved using a similar technique as in \cref{The proof of lemma: four components xtxt}. \textbf{Recall that when $q=0, 1$, $\log^\alpha(q)$ is taken to be $1$.}
\begin{align*}
&\lnorm{\sum_{i=1}^{t-1}\sum_{q=0}^{i-1} (A+B \Kh_{i-1})\cdots(A+B K_{0})x_0(B\eta_q+\varepsilon_q)^\top
\left[(A+B \Kh_{i-1})\cdots(A+B \Kh_{q+1})\right]^\top} \\
& \qquad \le \sum_{i=1}^{t-1}\sum_{q=0}^{i-1} \norm{(L + B\delta_{t-1}) \cdots (L + B\delta_{0})}\norm{x_0}\norm{B\eta_q+\varepsilon_q}
\norm{(L + B\delta_{t-1}) \cdots (L + B\delta_{q+1})}^\top \\
& \qquad \le \sum_{i=1}^{t-1}\sum_{q=0}^{i-1} \calO(\rho_L^i)\norm{x_0}\norm{B\eta_q+\varepsilon_q} \calO(\rho_L^{i-q}) \as \qquad \text{(by \cref{lemma: Hi prob bounds in theorem 2})} \\
& \qquad \le \sum_{i=1}^{t-1}\sum_{q=0}^{i-1} \calO(\rho_L^{2i-q})\calO(1)\calO(\log^{1/2}(q)) \as \quad \text{(by \cref{lemma: Hi prob bounds in theorem 2})}\\
& \qquad \le \sum_{i=1}^{t-1}\sum_{q=0}^{i-1} \calO(\rho_L^{2i-q})\logO(1) \as \\
& \qquad = \sum_{i=1}^{t-1} \calO(\rho_L^{i})\logO(1) \as \\
& \qquad \le \logO(1) \as
\end{align*}
Also,
\begin{align*}
&\lnorm{\sum_{i=1}^{t-1}\left[ (A+B \Kh_{i-1})\cdots(A+B K_{0})x_0\right]
\left[ (A+B \Kh_{i-1})\cdots(A+B K_{0})x_0\right]^T} \\
& \qquad \le \sum_{i=1}^{t-1} \calO(\rho_L^i)\norm{x_0}^2 \calO(\rho_L^{i}) \as \qquad \text{(by \cref{lemma: Hi prob bounds in theorem 2})}\\
& \qquad \le \sum_{i=1}^{t-1} \calO(\rho_L^{2i}) \as \\
& \qquad \le \calO(1) \as
\end{align*}
\end{proof}
\subsubsection{The proof of \cref{lemma:AtBtConvergeIp}}
\label{The proof of lemma:AtBtConvergeIp}
\begin{lemma*}
Assume we have two matrix sequences $\{A_t\}_{t=1}^\infty$ and $\{B_t\}_{t=1}^\infty$, where $A_t$ and $B_t$ are $p \times p$ positive definite matrices, and
\[A_t^{2}B_t^2 \convP I_p.\]
Then
\[A_tB_t \convP I_p.\]
\end{lemma*}
\begin{proof}
The basic idea is to utilize the equivalence of entry-wise convergence and F-norm convergence and the fact that the F-norm is invariant under orthogonal transformation. We know that positive definite matrices can be diagonalized by orthogonal transformation, and these diagonal matrices are easier to deal with.
Starting from our only equation
\[A_t^{2}B_t^2 \convP I_p.\]
Entry-wise convergence implies F-norm convergence:
\[\norm{A_t^{2}B_t^2 - I_p}_F \convP 0.\]
By the positive definiteness of $A_t$ and $B_t$, we can assume they have the diagnolization $A_t = U_{At}\Lambda_{At}U_{At}^\top$ and $B_t = U_{Bt}\Lambda_{Bt}U_{Bt}^\top$, where $\Lambda_{At}$ and $\Lambda_{Bt}$ are diagonal matrices with diagonal values $\lambda_{Ai, t}$ and $\lambda_{Bi, t}$ ($i=1,2, \cdots, p$), and $U_{At}$ and $U_{Bt}$ are orthogonal matrices. With this transformation, we have
\[\norm{U_{At}\Lambda_{At}^2U_{At}^\top U_{Bt}\Lambda_{Bt}^2U_{Bt}^\top - I_p}_F \convP 0.\]
Since orthogonal transformation does not affect F-norm, on RHS inside the F-norm, we can multiply $U_{At}^\top$ on the left and $U_{Bt}$ and on the right and get
\[\norm{\Lambda_{At}^2U_{At}^\top U_{Bt}\Lambda_{Bt}^2 - U_{At}^\top U_{Bt}}_F \convP 0.\]
Because F-norm convergence to zero is equivalent to entry-wise convergence to zero,
\[\Lambda_{At}^2U_{At}^\top U_{Bt}\Lambda_{Bt}^2 - U_{At}^\top U_{Bt} \convP 0.\]
Denote $T_t := U_{At}^\top U_{Bt}$, then
\[\Lambda_{At}^2T_t\Lambda_{Bt}^2 - T_t\convP 0.\]
If we consider the $ij$th element of the above equation:
\[\lambda_{Ai, t}^2 T_{ij} \lambda_{Bj, t}^2 - T_{ij} \convP 0,\]
which is
\[(\lambda_{Ai, t}\lambda_{Bj, t}-1)(\lambda_{Ai, t}\lambda_{Bj, t}+1) T_{ij} \convP 0.\]
Since by positive definiteness we have $\lambda_{Ai, t}, \lambda_{Bj, t} > 0$ , the above equation implies
\[(\lambda_{Ai, t}\lambda_{Bj, t}-1)T_{ij} \convP 0.\]
This holds for every $i, j$ pair. If we write out this equation back to matrix form, we would get
\[\Lambda_{At}T_t\Lambda_{Bt} - T_t\convP 0.\]
By the same trick this is equivalent to the F-norm form
\[\norm{\Lambda_{At}T_t\Lambda_{Bt} - T_t}_F \convP 0,\]
\[\norm{\Lambda_{At} U_{At}^\top U_{Bt} \Lambda_{Bt} - U_{At}^\top U_{Bt} }_F \convP 0.\]
On RHS inside the F-norm, we can multiply $U_{At}$ on the left and $U_{Bt}^\top$ and on the right and get
\[\norm{U_{At}\Lambda_{At} U_{At}^\top U_{Bt} \Lambda_{Bt} U_{Bt}^T - I_p }_F \convP 0.\]
Plug in our definition $A_t = U_{At}\Lambda_{At}U_{At}^\top$ and $B_t = U_{Bt}\Lambda_{Bt}U_{Bt}^\top$:
\[\norm{A_tB_t - I_p }_F \convP 0.\]
And this implies
\[A_tB_t \convP I_p.\]
\end{proof}
\subsubsection{The proof of \cref{lemma: three parts u_tx_t}}
\label{The proof of lemma: three parts u_tx_t}
\begin{lemma*}
Assume \cref{eq:uniform high probability bound for Kh}, then
\begin{enumerate}
\item $
\sum_{i=0}^{t-1}(\Kh_i-K )x_ix_i^\top =
\calO(t^{1-\beta/2}\log^{\frac{-\alpha + 3}{2}}(t)) \as
$
\item $ \sum_{i=0}^{t-1}\eta_ix_i^\top =
o\left(t^{\beta/2}\log^{\frac{\alpha+3}{2}}(t) \right) \as $
\end{enumerate}
\end{lemma*}
\begin{proof}
~\paragraph{First part $\sum_{i=0}^{t-1}(\Kh_i-K )x_ix_i^\top$}
By \cref{lemma: Hi prob bounds in theorem 2} we have a uniform bound for $\delta_i = \Kh_i-K$ and $x_i$. We can derive the result in the first part by directly plugging in the bound for $\norm{\delta_i} $ and $\norm{x_i}$.
By \cref{lemma: Hi prob bounds in theorem 2}
\begin{equation*}
\begin{aligned}
&\norm{x_i} \le \calO(\log^{1/2}(t)) \as \\
\end{aligned}
\end{equation*}
Thus
\begin{equation*}
\begin{split}
\lnorm{\sum_{i=0}^{t-1}(\Kh_i-K )x_ix_i^\top }
=& \sum_{i=0}^{t-1}\norm{\delta_i}\norm{x_ix_i^\top } \\
\le& \calO(\log(t))\sum_{i=0}^{t-1}\norm{\delta_i} \as \qquad \text{(by \cref{lemma: Hi prob bounds in theorem 2}} )\\
\le& \calO(\log(t))\sum_{i=0}^{t-1}\calO(i^{-\beta/2}\log^{\frac{-\alpha+1}{2}}(i)) \as \qquad \text{(by \cref{lemma: Hi prob bounds in theorem 2}} )\\
\le& \calO(\log(t) t^{1-\beta/2}\log^{\frac{-\alpha + 1}{2}}(t)) \as
\qquad \text{(by \cref{eq: sum eta_t}} )\\
\le& \calO(t^{1-\beta/2}\log^{\frac{-\alpha + 3}{2}}(t)) \as
\end{split}
\end{equation*}
which means (by bounding entry-wise terms by the operator norm)
\[\sum_{i=0}^{t-1}(\Kh_i-K )x_ix_i^\top =\calO(t^{1-\beta/2}\log^{\frac{-\alpha + 3}{2}}(t)) \as\]
~\paragraph{Second Part $\sum_{i=0}^{t-1}\eta_ix_i^\top$}
Following Lemma 2 (iii) from \citet{lai1982least}:
\begin{lemma}
\label{lem: Lai and Wei martingale}
Let $\{\epsilon_n\}$ be a martingale difference sequence with respect to an increasing sequence of $\sigma$-fields $\{\calF_n\}$ such that $\sup_n \E(\varepsilon_n^2|\calF_{n-1}) < \infty$ a.s. Let $v_n$ be an $\calF_{n-1}$-measurable random variable for every $n$. Then
\begin{equation*}
\sum_{i=1}^n v_i\epsilon_i < \infty \text{ a.s. on }
\{\sum_{i=1}^\infty v_i^2 < \infty\}
.\end{equation*}
And for any $\eta > 1/2$
\begin{equation*}
\sum_{i=1}^n v_i\epsilon_i =
o\left(
(\sum_{i=1}^n v_i^2)^{1/2} \log^\eta(\sum_{i=1}^n v_i^2)
\right)
\text{ a.s. on }
\{\sum_{i=1}^\infty v_i^2 = \infty\}
.\end{equation*}
\end{lemma}
As a result, with probability 1
\begin{equation}
\label{eq: Lai and Wei martingale}
\begin{split}
\sum_{i=1}^n v_i\epsilon_i
=&
o\left(
(\sum_{i=1}^n v_i^2)^{1/2} \log(\sum_{i=1}^n v_i^2)
\right)1_{\sum_{i=1}^\infty v_i^2 = \infty} + \calO(1)1_{\sum_{i=1}^\infty v_i^2 < \infty} \as \\
=&
o\left(
(\sum_{i=1}^n v_i^2)^{1/2} \log(\sum_{i=1}^n v_i^2)
\right) + \calO(1) \as
\end{split}
\end{equation}
We can apply \cref{lem: Lai and Wei martingale} to our context by noticing
\begin{equation*}
\sum_{i=0}^{t-1}\eta_ix_i^\top
= \sum_{i=0}^{t-1}\eta_ii^{\frac{1-\beta}2}\log^{-\alpha/2}(i)(i^{\frac{\beta-1}2}\log^{\alpha/2}(i)x_i^\top)
.\end{equation*}
Here we normalized all $\eta_i$ to have a fixed normal distribution $\eta_ii^{\frac{1-\beta}2}\log^{-\alpha/2}(i) \sim \calN(0, \tau^2I_d)$. Apply \cref{eq: Lai and Wei martingale} entry-wise, where $v_i$ corresponds to a fixed entry of $i^{\frac{\beta-1}2}\log^{\alpha/2}(i)x_i^\top$ and $\epsilon_i$ corresponds to a fixed entry of $\eta_ii^{\frac{1-\beta}2}\log^{-\alpha/2}(i)$. $v_i$ is bounded by $i^{\frac{\beta-1}2}\log^{\alpha/2}(i)\norm{x_i}$. Thus
\begin{equation*}
\sum_{i=0}^{t-1}\eta_ix_i^\top = o\left(V_t^{1/2}\log(V_t)\right) + \calO(1) \as,
\end{equation*}
where $V_t := \sum_{i=0}^{t-1} (i^{\frac{\beta-1}2}\log^{\alpha/2}(i)\norm{x_i})^2$. Applying the bounds in \cref{lemma: Hi prob bounds in theorem 2} (\textbf{recall that when $i=0, 1$, $i^{\beta-1}\log^\alpha(i)$ is taken to be $1$}):
\begin{equation*}
\begin{split}
V_t =& \sum_{i=0}^{t-1} (i^{\frac{\beta-1}2}\log^{\alpha/2}(i)\norm{x_i})^2 \\
=&\sum_{i=0}^{t-1}
i^{-1+\beta}\log^{\alpha}(i)
\calO(\log(t)) \as \quad \text{(by \cref{lemma: Hi prob bounds in theorem 2}} )\\
=& \calO(t^{\beta}\log^{\alpha}(t))
\calO(\log(t)) \as \quad \text{(by \cref{eq: sum eta_t}} )\\
=& \calO(t^{\beta}\log^{\alpha+1}(t)) \as
\end{split}
\end{equation*}
Thus,
\begin{align*}
\sum_{i=0}^{t-1}\eta_ix_i^\top =& o\left(V_t^{1/2}\log(V_t)\right) + \calO(1) \\
=& o\left(\calO(t^{\beta}\log^{\alpha+1}(t))^{1/2}\log(\calO(t^{\beta}\log^{\alpha+1}(t)))\right) + \calO(1) \\
=& o\left(\calO(t^{\beta/2}\log^{\frac{\alpha+1}{2}}(t)\log(t))\right) + \calO(1) \\
=& o\left(t^{\beta/2}\log^{\frac{\alpha+3}{2}}(t) \right) \as
\end{align*}
\end{proof}
In exactly the same way, we can show that
\begin{equation}
\label{eq: (Ktxt)T R etaT}
\sum_{i=1}^{t} (\Kh_i x_i)^\top R \eta_i = o\left(t^{\beta/2}\log^{\frac{\alpha+3}{2}}(t) \right) \as
\end{equation}
We first standardize $\eta_i$
\begin{equation*}
\sum_{i=1}^{t} (\Kh_i x_i)^\top R \eta_i =
\sum_{i=0}^{t-1}
(i^{\frac{\beta-1}2}\log^{\alpha/2}(i)(\Kh_i x_i)^\top R)
\eta_ii^{\frac{1-\beta}2}\log^{-\alpha/2}(i)
,\end{equation*}
and then $v_i$ is bounded by
\begin{align*}
&i^{\frac{\beta-1}2}\log^{\alpha/2}(i)\lnorm{(\Kh_i x_i)^\top R} \\
& \qquad \le i^{\frac{\beta-1}2}\log^{\alpha/2}(i)\lnorm{\Kh_i}\norm{R} \norm{x_i} \\
& \qquad \le i^{\frac{\beta-1}2}\log^{\alpha/2}(i)C_K\norm{R} \norm{x_i} \quad \text{(by \cref{alg:myAlg}'s design)}
,\end{align*}
which is different from $v_i$ in $\sum_{i=0}^{t-1}\eta_ix_i^\top$ by a constant factor $C_K\norm{R}$. The rest of the proof is all the same.
\subsubsection{The proof of \cref{lemma: six parts u_tu_t}}
\label{The proof of lemma: six parts u_tu_t}
\begin{lemma*}
Assume \cref{eq:uniform high probability bound for Kh}, then
\begin{enumerate}
\item $\sum_{i=0}^{t-1} \delta_ix_ix_i^\top \delta_i^\top = \calO(t^{1-\beta}\log^{-\alpha+2}(t)) \as$
\item $\sum_{i=0}^{t-1}\delta_ix_i\eta_i^\top = (\sum_{i=0}^{t-1} \eta_ix_i^\top \delta_i^\top)^\top =
o\left(\log^{2}(t)\right) \as$
\item $\sum_{i=0}^{t-1}\eta_i\eta_i^\top = t^\beta\frac{\tau^2}{\beta}\log^\alpha(t)(I_d + o_p(1)) $
\end{enumerate}
\end{lemma*}
\begin{proof}
~\paragraph{First part $\sum_{i=0}^{t-1} \delta_ix_ix_i^\top \delta_i^\top$}
Recall the conclusion from \cref{lemma: Hi prob bounds in theorem 2}: $\norm{x_t} = \calO(\log^{1/2}(t)) \as$
and $\norm{\delta_t} = \calO(t^{-\frac{\beta}{2}} \log^{\frac{-\alpha + 1}{2}}(t)) \as$
\begin{align*}
\lnorm{\sum_{i=1}^{t-1} \delta_ix_ix_i^\top \delta_i^\top }
\le&\sum_{i=1}^{t-1} \norm{\delta_i}^2\norm{x_i}^2 \\
\le& \calO(\log(t))\sum_{i=1}^{t-1}\calO(i^{-\beta}\log^{-\alpha+1}(i)) \as \qquad(\text{by \cref{lemma: Hi prob bounds in theorem 2}})\\
=& \calO(t^{1-\beta}\log^{-\alpha+2}(t))\as \qquad (\text{by \cref{eq: sum eta_t}})
\end{align*}
This implies (by bounding the entries by the operator norm, and including the $i=0$ term as $\calO(1)$):
\[\sum_{i=0}^{t-1} \delta_ix_ix_i^\top \delta_i^\top = \calO(t^{1-\beta}\log^{-\alpha+2}(t)) \as\]
~\paragraph{Second part $\sum_{i=0}^{t-1} \eta_ix_i^\top \delta_i^\top$} The representative of the third term is $\sum_{i=0}^{t-1} \eta_ix_i^\top \delta_i^\top$. The proof idea is similar to that in \cref{lemma: three parts u_tx_t} when we prove the bound for $\sum_{i=0}^{t-1} \eta_ix_i^\top $. Here we have an extra shrinking term $\delta_i$ which makes things easier.
Again, we can apply \cref{lem: Lai and Wei martingale} to our context by noticing
\begin{equation*}
\sum_{i=0}^{t-1}\eta_ix_i^\top \delta_i^\top
= \sum_{i=0}^{t-1}\eta_ii^{\frac{1-\beta}2}\log^{-\alpha/2}(i)(i^{\frac{\beta-1}2}\log^{\alpha/2}(i)x_i^\top\delta_i^\top)
.\end{equation*}
Here we normalized all $\eta_i$ to have a fixed normal distribution. Apply \cref{lem: Lai and Wei martingale} entry-wise, where $v_i$ corresponds to a fixed entry of $i^{\frac{\beta-1}2}\log^{\alpha/2}(i)x_i^\top\delta_i^\top$ and $\epsilon_i$ corresponds to a fixed entry of the normalized $\eta_i$. Our $v_i$ is bounded by $i^{\frac{\beta-1}2}\log^{\alpha/2}(i)\norm{x_i}\norm{\delta_i}$. Thus,
\begin{equation*}
\sum_{i=0}^{t-1}\eta_ix_i^\top\delta_i^\top = o\left(V_t^{1/2}\log(V_t)\right) + \calO(1)
.\end{equation*}
where $V_t := \sum_{i=0}^{t-1} (i^{\frac{\beta-1}2}\log^{\alpha/2}(i)\norm{x_i}\norm{\delta_i})^2$. Apply the high probability bound in \cref{lemma: Hi prob bounds in theorem 2} and we have
\begin{equation*}
\begin{split}
V_t =& \sum_{i=1}^{t-1} (i^{\frac{\beta-1}2}\log^{\alpha/2}(i)\norm{x_i}\norm{\delta_i})^2
\\
=&\sum_{i=1}^{t-1}
i^{-1+\beta}\log^{\alpha}(i)
\calO(\log(t))\calO(t^{-\beta} \log^{-\alpha + 1}(t)) \as
\qquad \text{(by \cref{lemma: Hi prob bounds in theorem 2}} )\\
=& \calO(t^{\beta}\log^{\alpha}(t))
\calO(\log(t))
\calO(t^{-\beta} \log^{-\alpha + 1}(t)) \as
\qquad \text{(by \cref{eq: sum eta_t}} )\\
=& \calO(\log^{2}(t)) \as
\end{split}
\end{equation*}
That is to say, $V_t = \calO(\log^{2}(t)) \as$ (adding the $i=0$ term as $\calO(1)$). Thus,
\begin{align*}
\sum_{i=0}^{t-1}\eta_ix_i^\top\delta_i^\top
=& o\left(V_t^{1/2}\log(V_t)\right) + \calO(1) \as \\
=& o\left(\calO(\log^{2}(t))^{1/2}\log(\calO(\log^{2}(t)))\right) + \calO(1) \as\\
=& o\left(o(\log^{2}(t))\right) + \calO(1) \as\\
=& o\left(\log^{2}(t)\right)\as
\end{align*}
\paragraph{Third part $\sum_{i=0}^{t-1} \eta_i\eta_i^\top$}
By \cref{eq: sum eta_t}:
\[\E (\sum_{i=0}^{t-1} \eta_i\eta_i^\top)
= \sum_{i=0}^{t-1} \tau^2 i^{\beta-1}\log^\alpha(i)I_d
= t^\beta\frac{\tau^2}{\beta}\log^\alpha(t)(I_d + o(1)).\]
With a little abuse of notation we use $\Var(\cdot)$ as entry-wise variance of a matrix. Again, $i=0,1$ terms are meant to be $\calO(1)$.
\begin{align*}
\begin{split}
\Var(\sum_{i=0}^{t-1} \eta_i\eta_i^\top)
=& \sum_{i=0}^{t-1} \Var(\eta_i\eta_i^\top) \\
=& \calO\left(\sum_{i=0}^{t-1} i^{2(\beta-1)}\log^{2\alpha}(i)\right) \\
\le& \calO\left(\sum_{i=0}^{t-1} i^{2(\beta-1)}\log^{2\max\{0,\alpha\}}(i)\right) \\
\le& \calO\left(\sum_{i=0}^{t-1} i^{2(\beta-1)}\log^{2\max\{0,\alpha\}}(t)\right) \\
=&\logO\left( \sum_{i=0}^{t-1} i^{2(\beta-1)}\right) \\
=& \logO(t^{2\beta-1}) .
\end{split}
\end{align*}
When $\beta > 1/2$ the last equation follows by \cref{eq: sum eta_t} and when $\beta = 1/2$ it is summation of harmonic series which is $\logO(1)$.
Thus the standard error is only of order $\logO(t^{\beta-1/2})$, which is smaller than $\E (\sum_{i=0}^{t-1} \eta_i\eta_i^\top) $.
That is to say,
\[\sum_{i=0}^{t-1}\eta_i\eta_i^\top
= t^\beta\frac{\tau^2}{\beta}\log^\alpha(t)(I_d + o_p(1)).\]
\end{proof}
\begin{comment}
\subsubsection{The proof of \cref{lem: matrix decomposition}}
\label{The proof of lem: matrix decomposition}
\begin{lemma*}
For a $p \times p$ positive definite matrix $S$, and fixed constant $\sigma > 0$, we have the following matrix decomposition
\begin{equation}
\label{eq:basic matrix decomposition 2}
\left[
\begin{array}{cc}
S & SK ^\top \\
KS & KSK ^\top + \sigma^2 I_d\\
\end{array}
\right] =
\left[
\begin{array}{cc}
I_n & 0\\
K & I_d\\
\end{array}
\right]\left[
\begin{array}{cc}
S & 0\\
0 & \sigma^2 I_d\\
\end{array}
\right]\left[
\begin{array}{cc}
I_n & K^\top \\
0 & I_d\\
\end{array}
\right]
.\end{equation}
\begin{equation}
\label{eq:basic matrix decomposition inverse 2}
\left[
\begin{array}{cc}
S & SK ^\top \\
KS & KSK ^\top +\sigma^2 I_d\\
\end{array}
\right]^{-1}
=\left[
\begin{array}{cc}
I_n & -K^\top\\
0 & I_d\\
\end{array}
\right]\left[
\begin{array}{cc}
S^{-1} & 0\\
0 & \frac{1}{\sigma^2} I_d\\
\end{array}
\right]\left[
\begin{array}{cc}
I_n & 0 \\
-K & I_d\\
\end{array}
\right]
.\end{equation}
\end{lemma*}
\begin{proof}
This only requires straightforward calculation: The first equation can be verified by
\begin{align*}
&\left[
\begin{array}{cc}
I_n & 0\\
K & I_d\\
\end{array}
\right]\left[
\begin{array}{cc}
S & 0\\
0 & \sigma^2 I_d\\
\end{array}
\right]\left[
\begin{array}{cc}
I_n & K^\top \\
0 & I_d\\
\end{array}
\right] \\
=& \left[
\begin{array}{cc}
S & 0\\
KS & \sigma^2 I_d\\
\end{array}
\right]\left[
\begin{array}{cc}
I_n & K^\top \\
0 & I_d\\
\end{array}
\right] \\
=& \left[
\begin{array}{cc}
S & SK^T\\
KS & KSK^T + \sigma^2 I_d\\
\end{array}
\right]
.\end{align*}
The second equation follows with
\begin{align*}
&\left[
\begin{array}{cc}
S & SK^T\\
KS & KSK^T + \sigma^2 I_d\\
\end{array}
\right]^{-1} \\
=& \left(\left[
\begin{array}{cc}
I_n & 0\\
K & I_d\\
\end{array}
\right]\left[
\begin{array}{cc}
S & 0\\
0 & \sigma^2 I_d\\
\end{array}
\right]\left[
\begin{array}{cc}
I_n & K^\top \\
0 & I_d\\
\end{array}
\right]\right)^{-1} \\
=& \left[
\begin{array}{cc}
I_n & K^\top \\
0 & I_d\\
\end{array}
\right]^{-1}\left[
\begin{array}{cc}
S & 0\\
0 & \sigma^2 I_d\\
\end{array}
\right]^{-1}\left[
\begin{array}{cc}
I_n & 0\\
K & I_d\\
\end{array}
\right]^{-1} \\
=& \left[
\begin{array}{cc}
I_n & -K^\top \\
0 & I_d\\
\end{array}
\right]\left[
\begin{array}{cc}
S^{-1} & 0\\
0 & \sigma^{-2} I_d\\
\end{array}
\right]^{-1}\left[
\begin{array}{cc}
I_n & 0\\
-K & I_d\\
\end{array}
\right]
.\end{align*}
\end{proof}
\end{comment}
\subsection{Lemmas in \cref{The proof of thm:regret}}
\subsubsection{The proof of \cref{lem: useful lemma from fazel}}
\label{The proof of lem: useful lemma from fazel}
\begin{lemma*}
For any $\Kh$ with suitable dimension,
\begin{equation*}
\begin{split}
&x^\top (Q + \Kh^\top R \Kh)x + x^\top (A+B\Kh)^\top P (A+B\Kh)x - x^\top P x \\
& \qquad = x^\top (\Kh-K)^\top( R + B^\top P B) (\Kh-K)x
.\end{split}
\end{equation*}
\end{lemma*}
Recall $P$ is the middle step described by the DARE. It should satisfy \cref{eq:ControllerK}
\begin{equation*}
K = - (R + B^\top P B)^{-1}B^\top P A
.\end{equation*}
As a result,
\begin{equation}
\label{eq: P property 1}
(R +B^\top P B)K + B^\top P A = 0
.\end{equation}
Also it is well known that \citep{Jamieson2018Lecture2}:
\begin{equation}
\label{eq: P property 2}
Q + K^\top R K + (A+BK)^\top P(A+BK) = P
.\end{equation}
Let $\Kh$ be another controller, then we have the following useful equation stated by \cref{lem: useful lemma from fazel}.
\begin{equation*}
\begin{split}
&x^\top (Q + \Kh^\top R \Kh)x + x^\top (A+B\Kh)^\top P (A+B\Kh)x - x^\top P x \\
& \qquad = x^\top (Q + (\Kh-K+K)^\top R (\Kh-K+K))x \\
& \qquad \qquad + x^\top (A+B(\Kh-K)+BK )^\top P (A+B(\Kh-K)+BK)x \\
& \qquad \qquad- x^\top P x \\
& \qquad = x^\top (Q + K^\top RK + (A+BK)^\top P(A+BK)) x \\
& \qquad \qquad + 2x^\top (\Kh-K)^\top ( RK + B^\top P (A+BK))x \\
& \qquad \qquad + x^\top (\Kh-K)^\top( R + B^\top P B) (\Kh-K)x \\
& \qquad \qquad- x^\top P x \\
& \qquad = x^\top (Q + K^\top RK + (A+BK)^\top P(A+BK)) x - x^\top P x\\
& \qquad \qquad + 2x^\top (\Kh-K)^\top ( (R+B^\top P B)K + B^\top P A)x \\
& \qquad \qquad + x^\top (\Kh-K)^\top( R + B^\top P B) (\Kh-K)x \\
& \qquad = x^\top (\Kh-K)^\top( R + B^\top P B) (\Kh-K)x \qquad \text{(by \cref{eq: P property 1,eq: P property 2}})
.\end{split}
\end{equation*}
\subsection{Lemmas in \cref{The proof of thm:prediction CLT parametric}}
\subsubsection{The proof of \cref{lem: difference between real and substitutes x u}}
\label{The proof of lem: difference between real and substitutes x u}
\begin{lemma*}
\begin{equation*}
x_t = \tilde{x}_t + O(t^{-\frac{\beta}{2}} \log^{\frac{-\alpha + 2}{2}}(t)) \as
\end{equation*}
\begin{equation*}
u_t = \tilde{u}_t + O(t^{-\frac{\beta}{2}} \log^{\frac{-\alpha + 2}{2}}(t)) \as
\end{equation*}
where
\begin{equation}
\label{eq: tilde xt defn}
\tilde{x}_t := \sum_{p=t-\myfloor{-\frac{\log(t)}{\log(\rho_L)}}}^{t-1}(A+BK)^{t-p-1} ( B\eta_p + \varepsilon_p)
,\end{equation}
and
\begin{equation*}
\tilde{u}_t := K\tilde{x}_t + \xi_t= K\sum_{p=t-\myfloor{-\frac{\log(t)}{\log(\rho_L)}}}^{t-1}(A+BK)^{t-p-1} ( B\eta_p + \varepsilon_p) + \xi_t
.\end{equation*}
\end{lemma*}
\begin{proof}
Recall \cref{lemma: StateExpansion} states that
\begin{equation*}
x_{t} = \sum_{p=0}^{t-1}(A+B \Kh_{t-1})\cdots(A+B \Kh_{p+1})(B \eta_p+\varepsilon_p) +(A+B \Kh_{t-1})\cdots(A+B K_0)x_0
.\end{equation*}
Similarly, we can rewrite $x_t$ as if starting from time $t - \myfloor{-\frac{\log(t)}{\log(\rho_L)}}$:
\begin{equation}
\label{eq: xt another decomposition}
x_{t} = \sum_{p=t-\myfloor{-\frac{\log(t)}{\log(\rho_L)}}}^{t-1}(A+B \Kh_{t-1})\cdots(A+B \Kh_{p+1})(B \eta_p+\varepsilon_p) + (A+B \Kh_{t-1})\cdots(A+B \Kh_{t-\myfloor{-\frac{\log(t)}{\log(\rho_L)}}})x_{t-\myfloor{-\frac{\log(t)}{\log(\rho_L)}}}.
\end{equation}
By \cref{lemma: Hi prob bounds in theorem 2}, we know
\begin{equation*}
\begin{split}
(A+B \Kh_{t-1})\cdots(A+B \Kh_{t-\myfloor{-\frac{\log(t)}{\log(\rho_L)}}})
\le& \calO(\rhoL^{-\log(t)/\log(\rho_L)}) \as\\
=& \calO(e^{-\log(t)}) \as\\
=& \calO(t^{-1}) \as
\end{split}
\end{equation*}
and
\begin{equation*}
\norm{x_t}, \norm{u_t} \le \calO(\log^{1/2}(t)) \as
\end{equation*}
Thus
\begin{equation*}
(A+B \Kh_{t-1})\cdots(A+B \Kh_{t-\myfloor{-\frac{\log(t)}{\log(\rho_L)}}})x_{t-\myfloor{-\frac{\log(t)}{\log(\rho_L)}}} = \calO(t^{-1}\log^{1/2}(t)) \as
\end{equation*}
Next, comparing \cref{eq: tilde xt defn} with \cref{eq: xt another decomposition}, we still need to bound the difference between $(A+B \Kh_{t-1})\cdots(A+B \Kh_{p+1})$ and $(A+BK)^{t-p-1}$. Again by \cref{lemma: Hi prob bounds in theorem 2},
\begin{equation*}
\begin{split}
&\lnorm{\sum_{p=t-\myfloor{-\frac{\log(t)}{\log(\rho_L)}}}^{t-1}
\left[
(A+B \Kh_{t-1})\cdots(A+B \Kh_{p+1}) - (A+B K)^{t-p-1}
\right]
( B\eta_p + \varepsilon_p)
} \\
& \qquad \le
\sum_{p=t-\myfloor{-\frac{\log(t)}{\log(\rho_L)}}}^{t-1}
\calO(\rho_L ^{t-p})(\norm{\delta_{t-1}}+ \cdots + \norm{\delta_{p+1}})
\calO(\log^{1/2}(t)) \as \qquad \text{(by \cref{eq: difference Khat product and K product,eq:bound on eta_p})}\\
& \qquad =
\sum_{p=t-\myfloor{-\frac{\log(t)}{\log(\rho_L)}} }^{t-1}\norm{\delta_{p+1}}\calO(\rho_L^{t-p})
\calO(\log^{1/2}(t)) \as \qquad \text{(by \cref{eq:part2 basic tool})}\\
& \qquad \le
\calO((t/2)^{-\frac{\beta}{2}} \log^{\frac{-\alpha + 1}{2}}(t/2))
\sum_{p=t-\myfloor{-\frac{\log(t)}{\log(\rho_L)}} }^{t-1}\calO(\rho_L^{t-p})
\calO(\log^{1/2}(t)) \as \\
& \qquad \qquad
\text{(by \cref{eq: stochastic bound delta_t} and that asymptotically $p > t/2$)}
\\
& \qquad =
\calO(t^{-\frac{\beta}{2}} \log^{\frac{-\alpha + 1}{2}}(t))
\calO(\log^{1/2}(t)) \as \\
& \qquad =
\calO(t^{-\frac{\beta}{2}} \log^{\frac{-\alpha + 2}{2}}(t)
) \as
\end{split}
\end{equation*}
This is larger than $\calO(t^{-1}\log^{1/2}(t))$. To summarize,
\begin{equation*}
x_t = \tilde{x}_t +
\calO(t^{-\frac{\beta}{2}} \log^{\frac{-\alpha + 2}{2}}(t)
) \as
\end{equation*}
Since $u_t - \tilde{u}_t = K(x_t - \tilde{x}_t)$,
\begin{equation*}
u_t = \tilde{u}_t + O(t^{-\frac{\beta}{2}} \log^{\frac{-\alpha + 2}{2}}(t)) \as
\end{equation*}
\end{proof}
\subsubsection{The proof of \cref{lem: difference between real and substitutes A B}}
\label{The proof of lem: difference between real and substitutes A B}
\begin{lemma*}
\begin{equation*}
\Ah_{t} = \Ah_{t-\myfloor{-\frac{\log(t)}{\log(\rho_L)}}} + \calO_p(t^{-\beta}\log^{-\alpha+3/2}(t))
.\end{equation*}
\begin{equation*}
\Bh_{t} = \Bh_{t-\myfloor{-\frac{\log(t)}{\log(\rho_L)}}} + \calO_p(t^{-\beta}\log^{-\alpha+3/2}(t))
.\end{equation*}
\end{lemma*}
\begin{proof}
We can bound the distance of neighboring estimators by the following recursive LS formula. Denote $\hat\Theta_{t} := [\Ahat_{t}, \Bhat_{t}]$, $z_i := \begin{bmatrix}
x_i \\
u_i
\end{bmatrix}$, $ H_t := (\sum_{i=0}^{t-1} z_{i}z_i^\top)^{-1}$.
Then the LS estimator \cref{eq: AhBh estimator} is
\begin{equation*}
\hat\Theta_{t} = \sum_{i=0}^{t-1} z_{i+1}z_i^\top(\sum_{i=0}^{t-1} z_{i}z_i^\top)^{-1}
= \sum_{i=0}^{t-1} z_{i+1}z_i^\top H_t
.\end{equation*}
For simplicity, denote $a_t := \myfloor{-\frac{\log(t)}{\log(\rho_L)}}$, then our objective is to bound the difference $ \hat\Theta_{t}-\hat\Theta_{t-a_t}$.
\begin{equation*}
\hat\Theta_{t-a_t} = \sum_{i=0}^{t-a_t-1} z_{i+1}z_i^\top H_{t-a_t}
.\end{equation*}
As a result,
\begin{equation*}
\hat\Theta_{t} = (\hat\Theta_{t-a_t}H_{t-a_t}^{-1} +
\sum_{i=t-a_t}^{t-1}z_{i+1}z_i^\top) H_t
.\end{equation*}
And
\begin{equation}
\label{eq: theta t - theta t-at}
\begin{split}
\hat\Theta_{t}-\hat\Theta_{t-a_t} =& \left(\hat\Theta_{t-a_t}(H_{t-a_t}^{-1}-H_t^{-1}) + \sum_{i=t-a_t}^{t-1}z_{i+1}z_i^\top\right) H_t \\
=& \left(-\hat\Theta_{t-a_t}\left(\sum_{i=t-a_t}^{t-1} z_{i}z_i^\top\right) + \sum_{i=t-a_t}^{t-1}z_{i+1}z_i^\top\right) H_t \\
=& \left(-\hat\Theta_{t-a_t}\left(\sum_{i=t-a_t}^{t-1} z_{i}z_i^\top\right) + \sum_{i=t-a_t}^{t-1}(\Theta z_i + \varepsilon_i)z_i^\top\right) H_t \\
=& (\Theta-\hat\Theta_{t-a_t})\left(\sum_{i=t-a_t}^{t-1} z_{i}z_i^\top\right)H_t + \sum_{i=t-a_t}^{t-1}\varepsilon_{i}z_i^\top H_t
.\end{split}
\end{equation}
Following \cref{eq: Dt -1 Gram Dt -1,eq: DtDt -1 order},
\begin{equation}
\label{eq: Ht order}
H_t = \calO_p(t^{-\beta}\log^{-\alpha}(t))
.\end{equation}
Next will bound the first and second term separately.
~\paragraph{First term $(\Theta-\hat\Theta_{t-a_t})(\sum_{i=t-a_t}^{t-1} z_{i}z_i^\top)H_t$}
By \cref{lemma: Hi prob bounds in theorem 2}, \
\begin{equation*}
z_t = \calO(\log^{1/2}(t)) \as
\end{equation*}
Recall that from \cref{eq:final Conclusion,eq: DtDt -1 order}, $\Theta-\hat\Theta_{t-a_t} = \calO_p(t^{-\beta/2}\log^{-\alpha/2}(t))$.
As a result,
\begin{equation}
\label{eq: tilde estimate part 1}
\begin{split}
(\Theta-\hat\Theta_{t-a_t})(\sum_{i=t-a_t}^{t-1} z_{i}z_i^\top)H_t
=& \calO_p\left(t^{-\beta/2}\log^{-\alpha/2}(t)\right)\calO_p(a_t \log(t) t^{-\beta}\log^{-\alpha}(t)) \\
=& \calO_p(t^{-3\beta/2}\log^{-3\alpha/2+2}(t))
.\end{split}
\end{equation}
We will see that this order is smaller than the second term, so that the second term is dominating.
~\paragraph{Second term $ \sum_{i=t-a_t}^{t-1}\varepsilon_{i}z_i^\top H_t$}
Consider the variance of the $jk$-th element of $ \sum_{i=t-a_t}^{t-1}\varepsilon_{i}z_i^\top$, which is applicable to any choice of $j$ and $k$. Fix $j$, $k$. Define $\calF_{t-1}$ as the filtration which contains every variable except for $\varepsilon_{t-1,j}$. We know that $\varepsilon_{t-1,j} \independent \calF_{t-1}$ and $\varepsilon_{t-1,j} \sim \calN(0, \sigma^2)$.
\begin{align*}
&\Var\left(\sum_{i=t-a_t}^{t-1}\varepsilon_{ij}(z_{i} )_k\right) \\
& \quad =\Var\left(\E\left(\sum_{i=t-a_t}^{t-1}\varepsilon_{ij}(z_{i} )_k \Bigg| \calF_{t-1}\right)\right) + \E\left(\Var\left(\sum_{i=t-a_t}^{t-1}\varepsilon_{ij}(z_{i} )_k \Bigg| \calF_{t-1}\right)\right) \\
& \quad =\Var\left(\sum_{i=t-a_t}^{t-2}\varepsilon_{ij}(z_{i} )_k \right) + \E\left((z_{t-1} )_k^2\sigma^2\right) \\
& \quad = \sigma^2 \sum_{i=t-a_t}^{t-1}\E\left((z_{i} )_k^2\right) \qquad \text{(by recursively conditioning on $\calF_{t-2}, \cdots, \calF_{t-a_t} $)}\\
& \quad \le \sigma^2 \sum_{i=t-a_t}^{t-1}\E\norm{z_{i}}^2 \\
& \quad \le \sigma^2 a_t \calO(\log^2(t)) \qquad \text{(by \cref{eq: E zt 2 final})} \\
& \quad \le \sigma^2 \calO(\log^3(t)) \qquad \left(\text{by } a_t := \myfloor{-\frac{\log(t)}{\log(\rho_L)}}\right)
\end{align*}
Since $\E\left(\sum_{i=t-a_t}^{t-1}\varepsilon_{ij}(z_{i}^\top )_k\right) = 0$, we have $\sum_{i=t-a_t}^{t-1}\varepsilon_{ij}(z_{i}^\top )_k = \calO_p(\log^{3/2}(t))$, which implies
\begin{equation*}
\sum_{i=t-a_t}^{t-1}\varepsilon_{i}z_{i}^\top = \calO_p(\log^{3/2}(t)).
\end{equation*}
By \cref{eq: Ht order},
\begin{equation}
\label{eq: tilde estimate part 2}
\sum_{i=t-a_t}^{t-1}\varepsilon_{i}z_{i}^\top H_t = \calO_p(\log^{3/2}(t))\calO_p(t^{-\beta}\log^{-\alpha}(t))
= \calO_p(t^{-\beta}\log^{-\alpha + 3/2}(t)) .
\end{equation}
This is larger than the first term. Combining \cref{eq: theta t - theta t-at,eq: tilde estimate part 1,eq: tilde estimate part 2} we have
\begin{equation*}
\begin{split}
\hat\Theta_{t}-\hat\Theta_{t-a_t} = \calO_p(t^{-\beta}\log^{-\alpha+3/2}(t))
.\end{split}
\end{equation*}
\end{proof}
\subsubsection{The proof of \cref{lem: CLT of substitution}}
\label{The proof of lem: CLT of substitution}
\begin{lemma*}
For any $\xi_t$ independent of the data before $t$: $\{\varepsilon_i, \eta_i\}_{i=0}^{t-1}$:
\begin{align*}
&\left(
\tilde{x}_t^\top
\left( \sum_{p=0}^{\infty}L ^{p}\left(I_n + 1_{\{\beta=1,\alpha=0\}}\frac{\tau^2}{\sigma^2}BB^\top\right)(L ^{p})^\top\right)^{-1}
\tilde{x}_t
+
\frac{\beta \sigma^2}{\tau^2}
t^{1-\beta} \log^{-\alpha}(t)
\lnorm{\xi_t}^2
\right)^{-1/2} \\
&
\qquad \cdot t^{1/2} \left((\Ah_{t-\myfloor{-\frac{\log(t)}{\log(\rho_L)}}} - A)\tilde{x}_t +
(\Bh_{t-\myfloor{-\frac{\log(t)}{\log(\rho_L)}}}- B)(K\tilde{x}_t+\xi_t)\right)
\convD
\calN(0,I_n).
\end{align*}
\end{lemma*}
\begin{proof}
We will start from finding the conditional distribution of
\begin{equation*}
(\Ah_{t-\myfloor{-\frac{\log(t)}{\log(\rho_L)}}} - A)\tilde{x}_t + (\Bh_{t-\myfloor{-\frac{\log(t)}{\log(\rho_L)}}}- B)\tilde{u}_t \bigg\vert \tilde{x}_t = x, \tilde{u}_t = Kx + \xi
.\end{equation*}
where $x$ and $\xi$ are constants. This should be easy because
$
\Ah_{t-\myfloor{-\frac{\log(t)}{\log(\rho_L)}}}- A, \Bh_{t-\myfloor{-\frac{\log(t)}{\log(\rho_L)}}}-B \independent \tilde{x}_t, \tilde{u}_t
$, which means we can directly apply the asymptotic normality result from \cref{thm:main CLT}. Recall \cref{eq: fast slow rate CLT} that
\begin{equation*}
t^{\beta/2} \log^{\alpha/2}(t)
\vvector \left(
\begin{bmatrix}
\Ah_t - A + (\Bh_t- B)K, &\Bh_t- B
\end{bmatrix}
\left[
\begin{array}{cc}
C_t^{1/2} & 0\\
0 & \sqrt{\frac{\tau^2}{\beta}} I_d\\
\end{array}
\right]
\right) \convD
\calN(0, \sigma^2 I_{\statedim(n+d)} )
,\end{equation*}
where $C_t = t^{1-\beta}\log^{-\alpha}(t)
\sum_{p=0}^{\infty}L ^{p}\left(\sigma^2I_n + 1_{\{\beta=1,\alpha=0\}}\tau^2BB^\top\right)(L ^{p})^\top
(I_n + o_p(1))$ (by \cref{eq: Ct order}). Here there are two different convergence speeds and we need to consider them separately.
More precisely,
\begin{align*}
\vvector \left(
\left[
\begin{array}{cc}
(\Ah_t - A + (\Bh_t- B)K)
t^{\beta/2} \log^{\alpha/2}(t)
C_t^{1/2}\sigma^{-1} &
(\Bh_t- B)t^{\beta/2} \log^{\alpha/2}(t)\sqrt{\frac{\tau^2}{\sigma^2 \beta}} I_d\\
\end{array}
\right]
\right) \\
\convD
\calN(0,
I_{n+d}\otimes I_n)
.\end{align*}
That is to say, for any constant vector $x$ and $\xi_t$ independent of data before $t$, we have
\begin{align*}
&\vvector \Bigg(
\left[
\begin{array}{cc}
(\Ah_t - A + (\Bh_t- B)K)
t^{\beta/2} \log^{\alpha/2}(t)C_t^{1/2}\sigma^{-1}
& (\Bh_t- B)t^{\beta/2} \log^{\alpha/2}(t)\sqrt{\frac{\tau^2}{\sigma^2 \beta}} I_d\\
\end{array}
\right]
\\
&\qquad \cdot \begin{bmatrix}
t^{-\beta/2} \log^{-\alpha/2}(t)C_t^{-1/2}\sigma
x \\
t^{(1-\beta)/2} \log^{-\alpha/2}(t)
\sqrt{\frac{\sigma^2 \beta}{\tau^2}} \xi_t
\end{bmatrix}
\Bigg/
\lnorm{\begin{bmatrix}
t^{-\beta/2} \log^{-\alpha/2}(t)C_t^{-1/2}\sigma
x \\
t^{(1-\beta)/2} \log^{-\alpha/2}(t)
\sqrt{\frac{\sigma^2 \beta}{\tau^2}} \xi_t
\end{bmatrix}}
\Bigg)
\convD
\calN(0, I_n)
.\end{align*}
The above equation holds because we are multiplying independent unit vector to the left hand side, so the result is still a normal distribution. Simplifying the equation:
\begin{align*}
\left(
x^\top
\left( \sum_{p=0}^{\infty}L ^{p}\left(I_n + 1_{\{\beta=1,\alpha=0\}}\frac{\tau^2}{\sigma^2}BB^\top\right)(L ^{p})^\top\right)^{-1}
x
+
\frac{\beta \sigma^2}{\tau^2}
t^{1-\beta} \log^{-\alpha}(t)\norm{\xi_t}^2
\right)^{-1/2} \\
\cdot t^{1/2} \left[(\Ah_{t} - A)x + (\Bh_{t}- B)(Kx+\xi_t)\right] \convD
\calN(0,I_n)
.\end{align*}
We can replace $t$ with $t-\myfloor{-\frac{\log(t)}{\log(\rho_L)}}$:
\begin{align*}
&\left(
x^\top
\left( \sum_{p=0}^{\infty}L ^{p}\left(I_n + 1_{\{\beta=1,\alpha=0\}}\frac{\tau^2}{\sigma^2}BB^\top\right)(L ^{p})^\top\right)^{-1}
x
+
\frac{\beta \sigma^2}{\tau^2}
\left(t-\myfloor{-\frac{\log(t)}{\log(\rho_L)}}\right)^{1-\beta} \log^{-\alpha}\left(t-\myfloor{-\frac{\log(t)}{\log(\rho_L)}}\right)
\lnorm{\xi_t}^2
\right)^{-1/2} \\
& \qquad \cdot
\left(t-\myfloor{-\frac{\log(t)}{\log(\rho_L)}}\right)^{1/2} \left[(\Ah_{t-\myfloor{-\frac{\log(t)}{\log(\rho_L)}}} - A)x +
(\Bh_{t-\myfloor{-\frac{\log(t)}{\log(\rho_L)}}}- B)(Kx+\xi_t)\right] \convD
\calN(0,I_n)
.\end{align*}
Because $\left(t-\myfloor{-\frac{\log(t)}{\log(\rho_L)}}\right)^{1/2}t^{-1/2} \to 1$, we can drop the first three instances of $\myfloor{-\frac{\log(t)}{\log(\rho_L)}}$:
\begin{align*}
&\left(
x^\top
\left( \sum_{p=0}^{\infty}L ^{p}\left(I_n + 1_{\{\beta=1,\alpha=0\}}\frac{\tau^2}{\sigma^2}BB^\top\right)(L ^{p})^\top\right)^{-1}
x
+
\frac{\beta \sigma^2}{\tau^2}
t^{1-\beta} \log^{-\alpha}(t)
\lnorm{\xi_t}^2
\right)^{-1/2} \\
& \qquad \cdot
t^{1/2} \left[(\Ah_{t-\myfloor{-\frac{\log(t)}{\log(\rho_L)}}} - A)x +
(\Bh_{t-\myfloor{-\frac{\log(t)}{\log(\rho_L)}}}- B)(Kx+\xi_t)\right]
\convD
\calN(0,I_n)
.\end{align*}
Here we actually used the fact that for $c_t,a_t,b_t > 0$, when $a_t/b_t \to 1$, then $(c_t+a_t)/(c_t+b_t) \to 1$. This is because
\begin{align*}
\labs{\frac{c_t+a_t}{c_t+b_t} - \frac{a_t}{b_t}}
= \labs{\frac{(b_t-a_t)c_t}{(c_t+b_t)b_t}}
\le \labs{\frac{b_t-a_t}{b_t}}
\to 0.
\end{align*}
In our specific context $c_t$ is the constant $x^\top
\left( \sum_{p=0}^{\infty}L ^{p}\left(I_n + 1_{\{\beta=1,\alpha=0\}}\frac{\tau^2}{\sigma^2}BB^\top\right)(L ^{p})^\top\right)^{-1}
x$.
Since $\tilde{x}_t \independent \Ah_{t-\myfloor{-\frac{\log(t)}{\log(\rho_L)}}} - A, \Bh_{t-\myfloor{-\frac{\log(t)}{\log(\rho_L)}}}- B$, we can replace $x$ with $\tilde{x}_t$ by conditioning on $\tilde{x}_t = x$, replace all $x$ with $\tilde{x}_t$, and finally remove the conditioning since they all converge in distribution to standard normal and $\tilde{x}_t$ asymptotically have same distribution.
\begin{align}
\label{eq: tilde prediction clt}
\begin{split}
&\left(
\tilde{x}_t^\top
\left( \sum_{p=0}^{\infty}L ^{p}\left(I_n + 1_{\{\beta=1,\alpha=0\}}\frac{\tau^2}{\sigma^2}BB^\top\right)(L ^{p})^\top\right)^{-1}
\tilde{x}_t
+
\frac{\beta \sigma^2}{\tau^2}
t^{1-\beta} \log^{-\alpha}(t)
\lnorm{\xi_t}^2
\right)^{-1/2} \\
&
\qquad \cdot t^{1/2} \left((\Ah_{t-\myfloor{-\frac{\log(t)}{\log(\rho_L)}}} - A)\tilde{x}_t +
(\Bh_{t-\myfloor{-\frac{\log(t)}{\log(\rho_L)}}}- B)(K\tilde{x}_t+\xi_t)\right)
\convD
\calN(0,I_n).
\end{split}
\end{align}
\end{proof}
\subsubsection{The proof of \cref{lem: CLT original}}
\label{The proof of lem: CLT original}
\begin{lemma*}
For any $\xi_t$ independent of the data before $t$: $\{\varepsilon_i, \eta_i\}_{i=0}^{t-1}$,
\begin{align*}
&\left(
x_t^\top
\left( \sum_{p=0}^{\infty}L ^{p}\left(I_n + 1_{\{\beta=1,\alpha=0\}}\frac{\tau^2}{\sigma^2}BB^\top\right)(L ^{p})^\top\right)^{-1}
x_t
+
\frac{\beta \sigma^2}{\tau^2}
t^{1-\beta} \log^{-\alpha}(t)
\lnorm{\xi_t}^2
\right)^{-1/2} \\
& \qquad \cdot
t^{1/2} \left[(\Ah_{t} - A)x_t +
(\Bh_{t}- B)(\Kh_tx_t+\xi_t)\right]
\convD
\calN(0,I_n)
.\end{align*}
\end{lemma*}
\begin{proof}
Since we already proved \cref{lem: CLT of substitution}, the only thing we need to do is to replace $\tilde{x}_t$ with $x_t$, $K$ with $\Kh_t$, and $\Ah_{t-\myfloor{-\frac{\log(t)}{\log(\rho_L)}}}$, $\Bh_{t-\myfloor{-\frac{\log(t)}{\log(\rho_L)}}}$ with $ \Ah_t$, $\Bh_t$.
~\paragraph{Replacing $\tilde{x}_t$ with $x_t$}
First, we can replace
$$\left(
\tilde{x}_t^\top
\left( \sum_{p=0}^{\infty}L ^{p}\left(I_n + 1_{\{\beta=1,\alpha=0\}}\frac{\tau^2}{\sigma^2}BB^\top\right)(L ^{p})^\top\right)^{-1}
\tilde{x}_t
+
\frac{\beta \sigma^2}{\tau^2}
t^{1-\beta} \log^{-\alpha}(t)
\lnorm{\xi_t}^2
\right)$$
with
$\left(
x_t^\top
\left( \sum_{p=0}^{\infty}L ^{p}\left(I_n + 1_{\{\beta=1,\alpha=0\}}\frac{\tau^2}{\sigma^2}BB^\top\right)(L ^{p})^\top\right)^{-1}
x_t
+
\frac{\beta \sigma^2}{\tau^2}
t^{1-\beta} \log^{-\alpha}(t)
\lnorm{\xi_t}^2
\right)$ in \cref{eq: tilde prediction clt} because $\tilde{x}_t = x_t + o_p(1)$ by \cref{lem: difference between real and substitutes x u}.
\begin{align*}
\begin{split}
&\left(
x_t^\top
\left( \sum_{p=0}^{\infty}L ^{p}\left(I_n + 1_{\{\beta=1,\alpha=0\}}\frac{\tau^2}{\sigma^2}BB^\top\right)(L ^{p})^\top\right)^{-1}
x_t
+
\frac{\beta \sigma^2}{\tau^2}
t^{1-\beta} \log^{-\alpha}(t)
\lnorm{\xi_t}^2
\right)^{-1/2} \\
&
\qquad \cdot t^{1/2} \left((\Ah_{t-\myfloor{-\frac{\log(t)}{\log(\rho_L)}}} - A)\tilde{x}_t +
(\Bh_{t-\myfloor{-\frac{\log(t)}{\log(\rho_L)}}}- B)(K\tilde{x}_t+\xi_t)\right)
\convD
\calN(0,I_n).
\end{split}
\end{align*}
Since $x_t^\top
\left( \sum_{p=0}^{\infty}L ^{p}\left(I_n + 1_{\{\beta=1,\alpha=0\}}\frac{\tau^2}{\sigma^2}BB^\top\right)(L ^{p})^\top\right)^{-1}
x_t$ is bounded away from $0$ with high probability ($x_t$ has the component $\varepsilon_{t-1}$),
\begin{equation*}
\left(
x_t^\top
\left( \sum_{p=0}^{\infty}L ^{p}\left(I_n + 1_{\{\beta=1,\alpha=0\}}\frac{\tau^2}{\sigma^2}BB^\top\right)(L ^{p})^\top\right)^{-1}
x_t
+
\frac{\beta \sigma^2}{\tau^2}
t^{1-\beta} \log^{-\alpha}(t)
\lnorm{\xi_t}^2
\right)^{-1/2} = \calO_p(1)
.\end{equation*}
By \cref{lem: difference between real and substitutes x u}, $\tilde{x}_t = x_t + O_p(t^{-\frac{\beta}{2}} \log^{\frac{-\alpha + 2}{2}}(t))$.
Recall \cref{prop:one_epoch_estimate_withMyalg} states that $\norm{\Ah_{t-\myfloor{-\frac{\log(t)}{\log(\rho_L)}}}-A}, \norm{\Bh_{t-\myfloor{-\frac{\log(t)}{\log(\rho_L)}}}-B}, \norm{\Kh_{t}-K} = \calO_p(t^{-\beta/2}\log^{\frac{-\alpha + 1}{2}}(t))$.
Thus, the error induced by replacing the remaining $\tilde{x}_t$ with $x_t$ in \cref{eq: tilde prediction clt} is
\begin{equation*}
\calO_p(1)t^{1/2}
\calO_p(t^{-\frac{\beta}{2}} \log^{\frac{-\alpha + 2}{2}}(t))
\calO_p(t^{-\frac\beta2}
\log^{\frac{-\alpha + 1}{2}}(t))
=\calO_p(t^{1/2-\beta} \log^{-\alpha + 3/2}(t))
.\end{equation*}
Under our condition $\beta > 1/2$ or $\beta = 1/2, \alpha > 3/2$, this error is of order $o_p(1)$, which is negligible.
Now we can replace all $\tilde{x}_t$ with $x_t$:
\begin{align*}
&\left(
x_t^\top
\left( \sum_{p=0}^{\infty}L ^{p}\left(I_n + 1_{\{\beta=1,\alpha=0\}}\frac{\tau^2}{\sigma^2}BB^\top\right)(L ^{p})^\top\right)^{-1}
x_t
+
\frac{\beta \sigma^2}{\tau^2}
t^{1-\beta} \log^{-\alpha}(t)
\lnorm{\xi_t}^2
\right)^{-1/2} \\
& \qquad \cdot t^{1/2}
\left[
(\Ah_{t-\myfloor{-\frac{\log(t)}{\log(\rho_L)}}} - A)x_t +
(\Bh_{t-\myfloor{-\frac{\log(t)}{\log(\rho_L)}}}- B)(Kx_t+\xi_t)\right]
\convD
\calN(0,I_n)
.\end{align*}
~\paragraph{Replacing $K$ by $\Kh_t $}
Since $\norm{\Bh_{t-\myfloor{-\frac{\log(t)}{\log(\rho_L)}}}-B}, \norm{\Kh_{t}-K} = \calO_p(t^{-\beta/2}\log^{\frac{-\alpha + 1}{2}}(t))$ (see \cref{prop:one_epoch_estimate_withMyalg}), and $x_t = \calO_p(\log^{1/2}(t))$, the final difference is still of order $\calO_p(t^{1/2-\beta} \log^{-\alpha + 3/2}(t)) = o_p(1)$. Thus
\begin{align*}
&\left(
x_t^\top
\left( \sum_{p=0}^{\infty}L ^{p}\left(I_n + 1_{\{\beta=1,\alpha=0\}}\frac{\tau^2}{\sigma^2}BB^\top\right)(L ^{p})^\top\right)^{-1}
x_t
+
\frac{\beta \sigma^2}{\tau^2}
t^{1-\beta} \log^{-\alpha}(t)
\lnorm{\xi_t}^2
\right)^{-1/2} \\
& \qquad \cdot
t^{1/2}
\left[
(\Ah_{t-\myfloor{-\frac{\log(t)}{\log(\rho_L)}}} - A)x_t +
(\Bh_{t-\myfloor{-\frac{\log(t)}{\log(\rho_L)}}}- B)(\Kh_tx_t+\xi_t)\right]
\convD
\calN(0,I_n)
.\end{align*}
~\paragraph{Replacing $\Ah_{t-\myfloor{-\frac{\log(t)}{\log(\rho_L)}}}$, $\Bh_{t-\myfloor{-\frac{\log(t)}{\log(\rho_L)}}}$ with $ \Ah_t$, $\Bh_t$}
By \cref{lem: difference between real and substitutes A B},
\begin{equation*}
\Ah_{t} - \Ah_{t-\myfloor{-\frac{\log(t)}{\log(\rho_L)}}}, \Bh_{t} - \Bh_{t-\myfloor{-\frac{\log(t)}{\log(\rho_L)}}} =
\calO_p(t^{-\beta}\log^{-\alpha+3/2}(t))
.\end{equation*}
Notice the $x_t$ and $\xi_t$ are multiplied by
$$\left(
x_t^\top
\left( \sum_{p=0}^{\infty}L ^{p}\left(I_n + 1_{\{\beta=1,\alpha=0\}}\frac{\tau^2}{\sigma^2}BB^\top\right)(L ^{p})^\top\right)^{-1}
x_t
+
\frac{\beta \sigma^2}{\tau^2}
t^{1-\beta} \log^{-\alpha}(t)
\lnorm{\xi_t}^2
\right)^{-1/2},$$
thus their order is only $\calO_p(1)$. The difference induced by replacing $\Ah_{t-\myfloor{-\frac{\log(t)}{\log(\rho_L)}}}, \Bh_{t-\myfloor{-\frac{\log(t)}{\log(\rho_L)}}}$ with $\Ah_{t}, \Bh_{t} $ is of order $\calO_p(t^{1/2-\beta}\log^{-\alpha+3/2}(t))$. When $\beta > 1/2$ or $\beta = 1/2, \alpha > 3/2$, this error is of order $o_p(1)$. Finally, after replacement we have
\begin{align*}
&\left(
x_t^\top
\left( \sum_{p=0}^{\infty}L ^{p}\left(I_n + 1_{\{\beta=1,\alpha=0\}}\frac{\tau^2}{\sigma^2}BB^\top\right)(L ^{p})^\top\right)^{-1}
x_t
+
\frac{\beta \sigma^2}{\tau^2}
t^{1-\beta} \log^{-\alpha}(t)
\lnorm{\xi_t}^2
\right)^{-1/2} \\
& \qquad \cdot
t^{1/2} \left[(\Ah_{t} - A)x_t +
(\Bh_{t}- B)(\Kh_tx_t+\xi_t)\right]
\convD
\calN(0,I_n)
.\end{align*}
\end{proof}
\subsubsection{The proof of \cref{lem: variance equivalence}}
\label{The proof of lem: variance equivalence}
\begin{lemma*}
For any $\xi_t$ independent of the data before $t$: $\{\varepsilon_i, \eta_i\}_{i=0}^{t-1}$,
\begin{align*}
\left(
x_t^\top
\left( \sum_{p=0}^{\infty}L ^{p}\left(I_n + 1_{\{\beta=1,\alpha=0\}}\frac{\tau^2}{\sigma^2}BB^\top\right)(L ^{p})^\top\right)^{-1}
x_t
+
\frac{\beta \sigma^2}{\tau^2}
t^{1-\beta} \log^{-\alpha}(t)
\lnorm{\xi_t}^2
\right)^{-1/2} \\
\cdot t^{1/2}
\left(
\sigma^2
\begin{bmatrix}
x_t \\
u_{t}
\end{bmatrix}^\top
\left(
\sum_{i=0}^{t-1}
\begin{bmatrix}
x_i\\
u_i\\
\end{bmatrix}
\begin{bmatrix}
x_i\\
u_i\\
\end{bmatrix}
^\top
\right)^{-1}
\begin{bmatrix}
x_t \\
u_{t}
\end{bmatrix}
\right)^{1/2} \convP 1
.\end{align*}
\end{lemma*}
\begin{proof}
By
$u_t = \Kh_tx_t+\xi_t$, it suffices to show
\begin{align*}
&\left(
x_t^\top
\left( \sum_{p=0}^{\infty}L ^{p}\left(I_n + 1_{\{\beta=1,\alpha=0\}}\frac{\tau^2}{\sigma^2}BB^\top\right)(L ^{p})^\top\right)^{-1}
x_t
+
\frac{\beta \sigma^2}{\tau^2}
t^{1-\beta} \log^{-\alpha}(t)
\lnorm{\xi_t}^2
\right)^{-1} \\
& \qquad \cdot
t^{1/2}
\left(
\sigma^2
\begin{bmatrix}
x_t \\
\Kh_tx_t+\xi_t
\end{bmatrix}^\top
\left(
\sum_{i=0}^{t-1}
\begin{bmatrix}
x_i\\
u_i\\
\end{bmatrix}
\begin{bmatrix}
x_i\\
u_i\\
\end{bmatrix}
^\top
\right)^{-1}
\begin{bmatrix}
x_t \\
\Kh_tx_t+\xi_t
\end{bmatrix}
\right) \convP 1
.\end{align*}
By \cref{eq: Gram matrix my symbol}:
\begin{equation*}
\sum_{i=0}^{t-1}
\begin{bmatrix}
x_i\\
u_i\\
\end{bmatrix}
\begin{bmatrix}
x_i\\
u_i\\
\end{bmatrix}
^\top
/t^\beta \log^{\alpha}(t)
=
\left[
\begin{array}{cc}
I_n & 0\\
K & I_d\\
\end{array}
\right]\left[
\begin{array}{cc}
M_t & \Delta_t^\top\\
\Delta_t & \Delta_u\\
\end{array}
\right]\left[
\begin{array}{cc}
I_n & K^\top \\
0 & I_d\\
\end{array}
\right]
.\end{equation*}
Thus
\begin{align*}
\begin{split}
&\begin{bmatrix}
x_t \\
\Kh_tx_t+\xi_t
\end{bmatrix}^\top
\left(
\sum_{i=0}^{t-1}
\begin{bmatrix}
x_i\\
u_i\\
\end{bmatrix}
\begin{bmatrix}
x_i\\
u_i\\
\end{bmatrix}
^\top
\right)^{-1}
\begin{bmatrix}
x_t \\
\Kh_tx_t+\xi_t
\end{bmatrix} t^\beta \log^{\alpha}(t)
\\
& \qquad =
\begin{bmatrix}
x_t \\
\Kh_tx_t+\xi_t
\end{bmatrix}^\top
\left[
\begin{array}{cc}
I_n & K^\top \\
0 & I_d\\
\end{array}
\right]^{-1}
\left[
\begin{array}{cc}
M_t & \Delta_t^\top\\
\Delta_t & \Delta_u\\
\end{array}
\right]^{-1}
\left[
\begin{array}{cc}
I_n & 0\\
K & I_d\\
\end{array}
\right]^{-1}
\begin{bmatrix}
x_t \\
\Kh_tx_t+\xi_t
\end{bmatrix}\\
& \qquad =
\begin{bmatrix}
x_t \\
\Kh_tx_t+\xi_t
\end{bmatrix}^\top
\left[
\begin{array}{cc}
I_n & -K^\top \\
0 & I_d\\
\end{array}
\right]
\left[
\begin{array}{cc}
(M_t-\Delta_t^\top \Delta_u^{-1} \Delta_t)^{-1}
&
-(M_t-\Delta_t^\top \Delta_u^{-1} \Delta_t)^{-1}\Delta_t^\top \Delta_u^{-1}\\
-((M_t-\Delta_t^\top \Delta_u^{-1} \Delta_t)^{-1}\Delta_t^\top \Delta_u^{-1})^\top & (\Delta_u - \Delta_t M_t^{-1} \Delta_t^\top)^{-1}
\end{array}
\right]\\
& \qquad \qquad \cdot
\left[
\begin{array}{cc}
I_n & 0\\
-K & I_d\\
\end{array}
\right]
\begin{bmatrix}
x_t \\
\Kh_tx_t+\xi_t
\end{bmatrix}
\qquad
\text{(by block matrix inversion)}\\
& \qquad =
\begin{bmatrix}
x_t \\
\Kh_tx_t+\xi_t-Kx_t
\end{bmatrix}^\top
\left[
\begin{array}{cc}
(M_t-\Delta_t^\top \Delta_u^{-1} \Delta_t)^{-1}
&
-(M_t-\Delta_t^\top \Delta_u^{-1} \Delta_t)^{-1}\Delta_t^\top \Delta_u^{-1}\\
-((M_t-\Delta_t^\top \Delta_u^{-1} \Delta_t)^{-1}\Delta_t^\top \Delta_u^{-1})^\top &
(\Delta_u - \Delta_t M_t^{-1} \Delta_t^\top)^{-1}\\
\end{array}
\right] \\
& \qquad \qquad \cdot
\begin{bmatrix}
x_t \\
\Kh_tx_t+\xi_t-Kx_t
\end{bmatrix} \\
& \qquad =
x_t^\top (M_t-\Delta_t^\top \Delta_u^{-1} \Delta_t)^{-1} x_t -
2x_t^\top (M_t-\Delta_t^\top \Delta_u^{-1} \Delta_t)^{-1}\Delta_t^\top \Delta_u^{-1} (\Kh_tx_t+\xi_t-Kx_t) \\
& \qquad \qquad +
(\Kh_tx_t+\xi_t-Kx_t)^\top(\Delta_u - \Delta_t M_t^{-1} \Delta_t^\top)^{-1}(\Kh_tx_t+\xi_t-Kx_t)
.\end{split}
\end{align*}
By \cref{eq:Cov xx}, \cref{eq: Delta t order}, \cref{eq: definition Delta u}:
\begin{align}
\label{eq: Mt Delta orders}
\begin{split}
M_t =&
\log^{-\alpha}(t)t^{1-\beta}
\left(\sum_{p=0}^{\infty}L ^{p}\left(\sigma^2I_n + 1_{\{\beta=1,\alpha=0\}}\tau^2BB^\top\right)(L ^{p})^\top
\right)
(I_n + o(1)) \\
M_t^{-1} =&
\log^{\alpha}(t)t^{-1+\beta} \left(\sum_{p=0}^{\infty}L ^{p}\left(\sigma^2I_n + 1_{\{\beta=1,\alpha=0\}}\tau^2BB^\top\right)(L ^{p})^\top
\right)^{-1}(I_n + o(1)) \\
\Delta_t =& \calO_p(t^{1-3\beta/2}\log^{\frac{-3\alpha + 3}{2}}(t)) \\
\Delta_u =& \frac{\tau^2}\beta (I_d + o_p(1))
.\end{split}
\end{align}
As a result, when $\beta > 1/2$ or $\beta = 1/2, \alpha > 3/2$
\begin{align*}
\Delta_t^\top \Delta_u^{-1} \Delta_t =& \calO_p(t^{2-3\beta}\log^{-3\alpha + 3}(t))
= o_p(t^{1-\beta}\log^{-\alpha}(t))
\\
(M_t - \Delta_t^\top \Delta_u^{-1} \Delta_t)^{-1}
=& M_t^{-1} (I_n-o_p(1))^{-1}= M_t^{-1} (I_n+o_p(1))
\\
\Delta_t M_t^{-1} \Delta_t^\top
=& \calO_p(t^{1-3\beta/2}\log^{\frac{-3\alpha + 3}{2}}(t)) \calO_p(t^{\beta-1} \log^{\alpha}(t)) \calO_p(t^{1-3\beta/2}\log^{\frac{-3\alpha + 3}{2}}(t))\\
=& \calO_p(t^{1-2\beta}\log^{-2\alpha + 3}(t)) = o_p(1)
\\
(\Delta_u - \Delta_t M_t^{-1} \Delta_t^\top)^{-1}
=&\Delta_u^{-1}(I_d+ o_p(1))
.\end{align*}
Notice by \cref{lemma: Hi prob bounds in theorem 2}, $\Kh_t-K = \calO_p(t^{-\frac{\beta}{2}} \log^{\frac{-\alpha + 1}{2}}(t))$. Then
\begin{align*}
&\begin{bmatrix}
x_t \\
\Kh_tx_t+\xi_t
\end{bmatrix}^\top
\left(
\sum_{i=0}^{t-1}
\begin{bmatrix}
x_i\\
u_i\\
\end{bmatrix}
\begin{bmatrix}
x_i\\
u_i\\
\end{bmatrix}
^\top
\right)^{-1}
\begin{bmatrix}
x_t \\
\Kh_tx_t+\xi_t
\end{bmatrix} t^\beta \log^{\alpha}(t)
\\
& \qquad =
x_t^\top (M_t-\Delta_t^\top \Delta_u^{-1} \Delta_t)^{-1} x_t +
2x_t^\top (M_t-\Delta_t^\top \Delta_u^{-1} \Delta_t)^{-1}\Delta_t^\top \Delta_u^{-1} (\Kh_tx_t+\xi_t-Kx_t) \\
& \qquad \qquad +
(\Kh_tx_t+\xi_t-Kx_t)^\top(\Delta_u - \Delta_t M_t^{-1} \Delta_t^\top)^{-1}(\Kh_tx_t+\xi_t-Kx_t) \\
& \qquad =
x_t^\top M_t^{-1}(I_n+o_p(1)) x_t +
2x_t^\top M_t^{-1}(I_n+o_p(1))\calO_p(t^{1-3\beta/2}\log^{\frac{-3\alpha + 3}{2}}(t)) \Delta_u^{-1} (\calO_p(t^{-\frac{\beta}{2}} \log^{\frac{-\alpha + 1}{2}}(t))x_t+\xi_t) \\
& \qquad \qquad +
(\calO_p(t^{-\frac{\beta}{2}} \log^{\frac{-\alpha + 1}{2}}(t))x_t+\xi_t)^\top
\Delta_u^{-1}(I_d+ o_p(1))
(\calO_p(t^{-\frac{\beta}{2}} \log^{\frac{-\alpha + 1}{2}}(t))x_t+\xi_t)
.\end{align*}
~\paragraph{Quadratic terms of $x_t$}
Let us first consider all those quadratic terms of $x_t$:
\begin{itemize}
\item $ x_t^\top M_t^{-1}(I_n+o_p(1)) x_t.$
\item
\begin{align*}
&2x_t^\top M_t^{-1}(I_n+o_p(1))\calO_p(t^{1-3\beta/2}\log^{\frac{-3\alpha + 3}{2}}(t)) \Delta_u^{-1} \calO_p(t^{-\frac{\beta}{2}} \log^{\frac{-\alpha + 1}{2}}(t))x_t \\
& \qquad = 2x_t^\top M_t^{-1}(I_n+o_p(1))\calO_p(t^{1-2\beta}\log^{\frac{-4\alpha + 4}{2}}(t)) x_t \\
& \qquad = x_t^\top M_t^{-1}o_p(1) x_t
.\end{align*}
\item
\begin{align*}
&x_t^\top \calO_p\left(t^{-\frac{\beta}{2}} \log^{\frac{-\alpha + 1}{2}}(t)\right) \Delta_u^{-1}(I_d+ o_p(1)) \calO_p\left(t^{-\frac{\beta}{2}} \log^{\frac{-\alpha + 1}{2}}(t)\right)x_t \\
& \qquad = x_t^\top \calO_p\left(t^{-\beta} \log^{-\alpha + 1}(t)\right) x_t \\
& \qquad = x_t^\top M_t^{-1} t^{1-\beta}\log^{-\alpha}(t)\calO_p\left(t^{-\beta} \log^{-\alpha + 1}(t)\right) x_t \qquad \text{(by \cref{eq: Mt Delta orders})}\\
& \qquad = x_t^\top M_t^{-1} \calO_p\left(t^{1-2\beta} \log^{-2\alpha + 1}(t)\right) x_t \\
& \qquad = x_t^\top M_t^{-1} o_p(1) x_t
.\end{align*}
\end{itemize}
Thus the later two items are dominated by the first term, and the quadratic terms of $x_t$ can be summarized by $x_t^\top M_t^{-1}(I_n+o_p(1)) x_t = x_t^\top M_t^{-1} x_t(1+o_p(1))$.
~\paragraph{Quadratic terms of $\xi_t$} That is already in a simple single item form, so we just keep it as $\xi_t^\top \Delta_u^{-1} (I_d+o_p(1))\xi_t = \xi_t^\top \Delta_u^{-1}\xi_t (1+o_p(1))$.
~\paragraph{Cross terms between $x_t$ and $\xi_t$}
Finally consider the cross terms of $x_t$ and $\xi_t$:
\begin{align*}
& 2x_t^\top M_t^{-1}(I_n+o_p(1))\calO_p(t^{1-3\beta/2}\log^{\frac{-3\alpha + 3}{2}}(t)) \Delta_u^{-1} \xi_t
+
2\calO_p(t^{-\frac{\beta}{2}} \log^{\frac{-\alpha + 1}{2}}(t))x_t^\top
\Delta_u^{-1}(I_d+ o_p(1)) \xi_t\\
& \qquad = 2x_t^\top \calO_p(t^{-\frac{\beta}{2}} \log^{\frac{-\alpha + 3}{2}}(t)) \xi_t
+ 2x_t^\top \calO_p(t^{-\frac{\beta}{2}} \log^{\frac{-\alpha + 1}{2}}(t)) \xi_t \qquad \text{(by \cref{eq: Mt Delta orders})}\\
& \qquad = x_t^\top \calO_p(t^{-\frac{\beta}{2}}\log^{\frac{-\alpha + 3}{2}}(t)) \xi_t \\
& \qquad = x_t^\top o_p(t^{\frac{\beta-1}{2}}\log^{\frac{\alpha}{2}}(t)) \xi_t \qquad \text{(because $\beta > 1/2$ or $\beta = 1/2$ and $\alpha > 3/2$)}\\
& \qquad = x_t^\top M_t^{-1/2}o_p(1) \Delta_u^{-1/2}\xi_t \qquad \text{(by \cref{eq: Mt Delta orders})}\\
& \qquad \le o_p(1) \norm{x_t^\top M_t^{-1/2}} \norm{\Delta_u^{-1/2}\xi_t }
\\
& \qquad \le o_p(1) \left(x_t^\top M_t^{-1} x_t + \xi_t^\top \Delta_u^{-1} \xi_t \right)
,\end{align*}
which is dominated by the quadratic part.
To sum up, we have
\begin{equation*}
\begin{split}
&\begin{bmatrix}
x_t \\
\Kh_tx_t+\xi_t
\end{bmatrix}^\top
\begin{bmatrix}
\sum_{i=0}^{t-1}x_ix_i^\top & \sum_{i=1}^{t-1}x_iu_i^\top \\
\sum_{i=0}^{t-1}u_ix_i^\top & \sum_{i=1}^{t-1}u_iu_i^\top \\
\end{bmatrix}^{-1}
\begin{bmatrix}
x_t \\
\Kh_tx_t+\xi_t
\end{bmatrix} t^\beta \log^{\alpha}(t)
\\
& \qquad =
(x_t^\top M_t^{-1} x_t +
\xi_t^\top\Delta_u ^{-1}\xi_t)(1+o_p(1)) \\
& \qquad =
\left(x_t^\top \log^{\alpha}(t)t^{-1+\beta} \left(\sum_{p=0}^{\infty}L ^{p}\left(\sigma^2I_n + 1_{\{\beta=1,\alpha=0\}}\tau^2BB^\top\right)(L ^{p})^\top
\right)^{-1} x_t
+
\xi_t^\top \frac{\beta}{\tau^2}\xi_t\right)(1+o_p(1))
.\end{split}
\end{equation*}
In other words
\begin{equation*}
\begin{split}
&
t\sigma^2
\begin{bmatrix}
x_t \\
\Kh_tx_t+\xi_t
\end{bmatrix}^\top
\begin{bmatrix}
\sum_{i=0}^{t-1}x_ix_i^\top & \sum_{i=1}^{t-1}x_iu_i^\top \\
\sum_{i=0}^{t-1}u_ix_i^\top & \sum_{i=1}^{t-1}u_iu_i^\top \\
\end{bmatrix}^{-1}
\begin{bmatrix}
x_t \\
\Kh_tx_t+\xi_t
\end{bmatrix}\\
& \qquad =
\left( x_t^\top
\left( \sum_{p=0}^{\infty}L ^{p}\left(I_n + 1_{\{\beta=1,\alpha=0\}}\frac{\tau^2}{\sigma^2}BB^\top\right)(L ^{p})^\top\right)^{-1}
x_t
+
\frac{\beta \sigma^2}{\tau^2}
t^{1-\beta} \log^{-\alpha}(t)
\lnorm{\xi_t}^2\right)(1+o_p(1))
.\end{split}
\end{equation*}
\end{proof}
\begin{comment}
\begin{lemma}
\label{lem: conditional weak convergence}
For any continuous random variable sequence $M_t$. If $\tilde{x}_t = x_t + o(1)$, $\tilde{u}_t = \Kh_tx_t+\xi_t + o(1)$. Denote the cumulative distribution function of $M_t \big| x_t, \Kh_tx_t+\xi_t$ as $F_t(p_1|p_2, p_3)$, and the cumulative distribution function of $M_t \big| \tilde{x}_t, \tilde{u}_t$ as $\tilde{F}_t(p_1|p_2, p_3)$, then if $F_t$ and $\tilde{F}_t$ are all continuous functions,
\begin{equation*}
\lim_{t \to \infty} F_t(p_1|p_2, p_3) - \tilde{F}_t(p_1|p_2, p_3) = 0
.\end{equation*}
\begin{proof}
For simplicity let us consider the case with only conditioning on $x_t$. One could easily generalize it to conditioning on both $x_t$ and $\Kh_tx_t+\xi_t$. We want to show
\begin{equation*}
\lim_{t \to \infty} \lim_{\delta \to 0}
\frac{
\P \left(
M_t > p_1, x_t \in (p_2, p_2 + \delta)
\right)}
{\P \left(
x_t \in (p_2, p_2 + \delta)
\right)}
-
\frac{
\P \left(
M_t > p_1, \tilde{x}_t \in (p_2, p_2 + \delta)
\right)}
{\P \left(
\tilde{x}_t \in (p_2, p_2 + \delta)
\right)}
= 0
.\end{equation*}
Given the condition that $\tilde{x}_t = x_t + o(1)$, this would immediately be true if we can exchange the two limitations. We can take advantage of Moore-Osgood theorem, which says we can indeed switch $\lim_{t \to \infty}$, $\lim_{\delta \to 0}$ if
\begin{itemize}
\item
\begin{equation*}
\lim_{t \to \infty}
\frac{
\P \left(
M_t > p_1, x_t \in (p_2, p_2 + \delta)
\right)}
{\P \left(
x_t \in (p_2, p_2 + \delta)
\right)}
-
\frac{
\P \left(
M_t > p_1, \tilde{x}_t \in (p_2, p_2 + \delta)
\right)}
{\P \left(
\tilde{x}_t \in (p_2, p_2 + \delta)
\right)}
= 0
.\end{equation*}
holds uniformly on $\delta \in \mathbb{R}$
\item
\begin{equation*}
\lim_{\delta \to 0}
\frac{
\P \left(
M_t > p_1, x_t \in (p_2, p_2 + \delta)
\right)}
{\P \left(
x_t \in (p_2, p_2 + \delta)
\right)}
-
\frac{
\P \left(
M_t > p_1, \tilde{x}_t \in (p_2, p_2 + \delta)
\right)}
{\P \left(
\tilde{x}_t \in (p_2, p_2 + \delta)
\right)}
.\end{equation*}
exists point-wise over $t \in \mathbb{N}$
\end{itemize}
The second condition is easy to verify since
\begin{equation*}
\begin{split}
&\lim_{\delta \to 0}
\frac{
\P \left(
M_t > p_1, x_t \in (p_2, p_2 + \delta)
\right)}
{\P \left(
x_t \in (p_2, p_2 + \delta)
\right)}
-
\frac{
\P \left(
M_t > p_1, \tilde{x}_t \in (p_2, p_2 + \delta)
\right)}
{\P \left(
\tilde{x}_t \in (p_2, p_2 + \delta)
\right)} \\
& \qquad =
\P \left(
M_t > p_1 \big| x_t = p_2
\right)
-
\P \left(
M_t > p_1 \big| \tilde{x}_t = p_2
\right)
\end{split}
.\end{equation*}
Now it only remains to verify the first condition. Denote $A_t := \{M_t > p_1, x_t \in (p_2, p_2 + \delta)\}$, $B_t := \{x_t \in (p_2, p_2 + \delta)\}$, $\tilde{A}_t := \{M_t > p_1, \tilde{x}_t \in (p_2, p_2 + \delta)\}$ and $\tilde{B}_t := \{\tilde{x}_t \in (p_2, p_2 + \delta)\}$. Then we are considering
\begin{equation*}
\begin{split}
&\eabs{\frac{\P(A_t)}{\P(B_t)} - \frac{\P(\tilde{A}_t)}{\P(\tilde{B}_t)} }\\
& \qquad =\frac{\eabs{\P(A_t)\P(\tilde{B}_t) - \P(\tilde{A}_t)\P(B_t)}}
{\P(B_t)\P(\tilde{B}_t)} \\
& \qquad = \frac{\eabs{\P(A_t)\P(\tilde{B}_t) -\P(B_t) + \P(B_t)) - (\P(\tilde{A}_t)-\P(A_t) + \P(A_t))\P(B_t)}}
{\P(B_t)\P(\tilde{B}_t)} \\
& \qquad = \frac{\P(A_t)\eabs{\P(\tilde{B}_t) -\P(B_t)} +
\eabs{\P(\tilde{A}_t)-\P(A_t)}\P(B_t)}
{\P(B_t)\P(\tilde{B}_t)} \\
\le & \frac{\P(B_t)\eabs{\P(\tilde{B}_t) -\P(B_t)} +
\eabs{\P(\tilde{A}_t)-\P(A_t)}\P(B_t)}
{\P(B_t)\P(\tilde{B}_t)} \\
= & \frac{\eabs{\P(\tilde{B}_t) -\P(B_t)} +
\eabs{\P(\tilde{A}_t)-\P(A_t)}}
{\P(\tilde{B}_t)} \\
\end{split}
.\end{equation*}
Since $x_t \to x:= \sum_{i=0}^{\infty} (A+BK)^i\varepsilon_i$ a.s., $\P(\tilde{B}_t) \to \P(x \in (p_2, p_2 + \delta))$ which is a constant.
\end{proof}
\end{lemma}
\end{comment}
\subsection{Lemmas in \cref{section: The proof of one_epoch_estimate}}
\subsubsection{The proof of \cref{lemma:lwm}}
\label{The proof of lemma:lwm}
\begin{lemma*}[A slightly different version of Theorem C.2 in \citet{dean2018regret}]
Fixing $\delta \in (0,\frac{(n+d)\xi^2}{2}]$, for every $T$, $k$, $\nu$, and $\xi$ such that $\{z_t\}_{t = 0}^T$ satisfies the $(k, \nu, \xi)$-BMSB and
\begin{equation*}
T/k \ge \frac{10(n+d)}{\xi^2} \log\left(\frac{100(n+d)\sum_{t = 1}^T\Tr(\E z_t z_t^\top)}{T \nu^{2}\xi^2\delta^{1 + \frac1{n+d}}} \right)
.\end{equation*}
the estimate $\hat{\Theta}_T$ defined in \cref{eq:ols_M} satisfies the following statistical rate
\begin{equation*}
\P\left[\lnorm{\hat{\Theta}_T-\Theta}_{2} >\frac{90\sigma}{\xi\nu}\sqrt{\frac{n+d}{T}\left(1 + \log\left(\frac{10(n+d)\sum_{t = 1}^T\Tr(\E z_t z_t^\top)}{T\delta^{1 + \frac1{n+d}} \nu^{2}\xi} \right) \right)} \right] \le 3\delta.
\end{equation*}
\end{lemma*}
First let us review the main theorem in \citep{simchowitz2018learning}. \cref{lemma:lwm} is actually a corollary of that.
To capture the excitation behavior observed in the case of linear systems we introduce a general martingale small-ball condition which quantifies the growth of the covariates $X_t$ for vectors (notice that this is different from \cref{def:BMSB condition}).
\begin{defn}[BMSB condition 2]\label{def:bmsb}
Given an $\{\calF_t\}_{t \ge 1}$-adapted random process $\{X_t\}_{t \ge 1}$ taking values in $\R^d$, we say that it satisfies the $(k,\Gamsb,\xi)$-matrix block martingale small-ball (BMSB) condition for $\Gamsb \succ 0$ if, for any $w\in \calS^{d-1}$ and $j \ge 0$, $\frac{1}{k}\sum_{i=1}^{k} \P( |\langle w, X_{j+i}\rangle | \ge \sqrt{w^\top \Gamsb w} | \calF_{j}) \ge \xi \as$
\end{defn}
\begin{theorem}[Theorem 2.4 in \citet{simchowitz2018learning}]
\label{main_thm_simchowitz2018} Fix $\delta \in (0,1)$, $T \in \N$ and $0 \prec \Gamsb \preceq \Gambar$. Then if $\{z_t,x_{t+1}\}_{t \ge 0} \in (\R^{d+n} \times \R^n)^{T}$ is a random sequence such that (a) $x_{t+1} = \Theta z_t + \varepsilon_t$, where $\varepsilon_t | \calF_t$ is $\sigma^2$-sub-Gaussian and mean zero, (b) $z_0,\dots,z_{T-1}$ satisfies the $(k,\Gamsb,\xi)$-small ball condition, and (c) such that $\P[\sum_{t = 0}^{T-1} z_t z_t^\top \npreceq T\Gambar] \le \delta$. Then if
\begin{align*}
T \geq \frac{10k}{\xi^2}\left(\log \left(\frac{1}{\delta}\right) + 2(d+n)\log(10/ \xi) + \log \det (\Gambar \Gamsb^{-1}) \right),
\end{align*}
we have $\hat{\Theta}_T$ defined in \cref{eq:ols_M} satisfies the following statistical rate
\begin{align*}
\P\left[\lnorm{\hat\Theta_T-\Theta} >\frac{90\sigma}{\xi}\sqrt{\frac{n + (n+d)\log \frac{10}{\xi} + \log \det \Gambar \Gamsb^{-1} + \log\left(\frac{1}{\delta}\right)}{T\sigma_{\min}(\Gamsb)}} \right] \le 3\delta.
\end{align*}
\end{theorem}
Now the main task is to translate this theorem to \cref{lemma:lwm}. First we need to derive the (a), (b), (c) three conditions from the assumptions in \cref{lemma:lwm}. Let us check the conditions one by one.
\paragraph{Condition (a)}
\cref{main_thm_simchowitz2018} states the model should be in the form of $x_t = \Theta z_t + \varepsilon_t$, where $\varepsilon_t | \calF_t$ is $\sigma^2$-sub-Gaussian and mean zero.
It is obvious that the system noise satisfy the sub-Gaussian and mean zero condition.
\paragraph{Condition (b)}
$z_1,\dots,z_T$ satisfies the $(k,\Gamsb,\xi)$-small ball condition.
Based on \cref{def:bmsb}, if we pick $\Gamma_{sb} = \nu^2 I_{\statedim + \inputdim}$, then the condition becomes
\begin{equation}
\label{eq: condition b}
\frac{1}{k}\sum_{i=1}^{k} \P( |\langle w, z_{j+i}\rangle| \ge \sqrt{w^\top \Gamsb w} = \nu| \calF_{j}) \ge \xi \as
\end{equation}
Since we already assume $\{z_t\}_{t = 0}^T$ satisfies the $(k, \nu, \xi)$-BMSB (see \cref{def:BMSB condition}) in \cref{lemma:lwm}, \cref{eq: condition b} holds by definition.
\paragraph{Condition (c)}
We need to show that $\P[\sum_{t = 0}^{T-1} z_t z_t^\top \npreceq T\Gambar] \le \delta$ for some choice $\Gambar$. Let us take
\begin{equation}
\label{eq: defn Gambar}
\Gambar = \frac{(n+d)\E\{\sum_{t = 0}^{T-1} z_t z_t^\top\}}{T\delta} \succ 0.
\end{equation}
First we need to show that $\Gambar = \frac{(n+d)\E\{\sum_{t = 0}^{T-1} z_t z_t^\top\}}{T\delta} \succeq \Gamma_{sb}$, and we can prove this from \cref{eq: condition b}:
\[\text{For any $0 \le j \le T-k$,} \quad \frac{1}{k}\sum_{i=1}^{k} \P( |\langle w, z_{j+i}\rangle| \ge \nu| \calF_{j}) \ge \xi.\]
From a high level perspective, this equation allows us to have a lower bound on the minimum eigenvalue of $\E\{\sum_{t = 0}^{T-1} z_t z_t^\top\}$, and then we can choose a $\delta$ small enough so that $\Gambar \succeq \Gamma_{sb} = \nu^2 I_{\statedim + \inputdim}$.
By Markov inequality, for any $0 \le j \le T-k$,
\[\frac{\frac{1}{k}\sum_{i=1}^{k}\E |\langle w, z_{j+i}\rangle|}{\nu} \ge \xi.\]
This is equivalent to
\[\left(\frac{1}{k}\sum_{i=1}^{k}\E |\langle w, z_{j+i}\rangle|\right)^2 \ge \xi^2\nu^2.\]
By Cauchy--Schwarz inequality:
\[\frac{1}{k} \sum_{i=1}^{k}\E |\langle w, z_{j+i}\rangle|^2 \ge \frac{1}{k} \sum_{i=1}^{k}\E^2 |\langle w, z_{j+i}\rangle| \ge \left(\frac{1}{k}\sum_{i=1}^{k}\E |\langle w, z_{j+i}\rangle|\right)^2 \ge \xi^2\nu^2.\]
Thus $\frac{1}{k} \sum_{i=1}^{k}\E |\langle w, z_{jk+i}\rangle|^2 \ge \xi^2\nu^2$. By summing up this inequality with $j = 0, 1, \cdots, \lfloor\frac{T-1}{k}\rfloor - 1$, we have
\[\frac{1}{\lfloor\frac{T-1}{k}\rfloor}\sum_{j=0}^{\lfloor\frac{T-1}{k}\rfloor - 1}\frac{1}{k}\left(\sum_{i=1}^{k}\E |\langle w, z_{jk+i}\rangle|^2\right) \ge \xi^2\nu^2.\]
We can clean up the summation by merging $\sum_j$ and $\sum_i$ into one summation:
\[\frac{1}{k\lfloor\frac{T-1}{k}\rfloor}\sum_{t=1}^{k\lfloor\frac{T-1}{k}\rfloor}\E |\langle w, z_{t}\rangle|^2 \ge \xi^2\nu^2.\]
Recall that $w$ is any vector in $\calS^{d-1}$, so the above equation can be translated into
\begin{align*}
\xi^2\nu^2 \le& \min_{w \in \calS^{d-1}}\frac{1}{k\lfloor\frac{T-1}{k}\rfloor}\sum_{t=1}^{k\lfloor\frac{T-1}{k}\rfloor}\E |\langle w, z_{t}\rangle|^2
\\
=&\; \E\left(\frac{1}{k\lfloor\frac{T-1}{k}\rfloor}\sum_{t=1}^{k\lfloor\frac{T-1}{k}\rfloor} z_{t}z_{t}^T\right)w \\
=& \min_{w \in \calS^{d-1}} w^T \E\left(\frac{1}{k\lfloor\frac{T-1}{k}\rfloor}\sum_{t=1}^{k\lfloor\frac{T-1}{k}\rfloor} z_{t}z_{t}^T\right)w \\
\le&\; \sigma_{\min}\left(\E\left(\sum_{t = 0}^{T-1} z_t z_t^\top\right)/(k\lfloor\frac{T-1}{k}\rfloor)\right)
.\end{align*}
This means
\begin{align*}
\lambdamin{\Gambar} =& \lambdamin{\frac{(n+d)\E\left(\sum_{t = 0}^{T-1} z_t z_t^\top\right)}{T\delta} } \\
=& \lambdamin{\frac{(n+d)\E\left(\sum_{t = 0}^{T-1} z_t z_t^\top/(k\lfloor\frac{T-1}{k}\rfloor)\right)}{T\delta} (k\lfloor\frac{T-1}{k}\rfloor)} \\
\ge& \frac{(n+d)\xi^2\nu^2}{T\delta} k\lfloor\frac{T-1}{k}\rfloor \qquad \\
\ge& \frac{(n+d)\xi^2\nu^2}{T\delta} \frac{T}{2} \qquad \text{(achieved when T is even and $k = T/2$)}\\
=& \frac{(n+d)\xi^2\nu^2}{2\delta}
.\end{align*}
We wish to have $\frac{(n+d)\xi^2\nu^2}{2\delta} \ge \nu^2$ so that $\lambdamin{\Gambar} \ge \nu^2$ and $\Gambar \succeq \Gamma_{sb} = \nu^2 I_{n+d}$. One sufficient condition is
\begin{equation*}
\delta \le \frac{(n+d)\xi^2}{2}
.\end{equation*}
Next we need to show $\P[\sum_{t = 0}^{T-1} z_t z_t^\top \npreceq T\Gambar] \le \delta$. For simplicity denote $Z_T = \sum_{t = 0}^{T-1} z_t z_t^\top$, which is a positive semi-definite matrix.
\begin{align*}
\P[\sum_{t = 0}^{T-1} z_t z_t^\top \npreceq T\Gambar]
=& \P[Z_T \npreceq \frac{\E\{Z_T\}(n+d)}{\delta} ] \qquad \text{(by \cref{eq: defn Gambar})} \\
=& \P[\E^{-1/2}\left(Z_T\right)Z_T\E^{-1/2}\left(Z_T\right) \npreceq \frac{I_{\statedim + \inputdim}(n+d)}{\delta}] \\
=& \P[\lambda_{\max} \{\E^{-1/2}\left(Z_T\right)Z_T\E^{-1/2}\left(Z_T\right)\} \ge \frac{(n+d)}{\delta}] \\
\le& \P[\Tr \{\E^{-1/2}\left(Z_T\right)Z_T\E^{-1/2}\left(Z_T\right)\} \ge \frac{(n+d)}{\delta}] \\
\le& \E[\Tr \{\E^{-1/2}\left(Z_T\right)Z_T\E^{-1/2}\left(Z_T\right)\}] \delta/(n+d) \quad \text{ (by Markov inequality)}\\
=& \Tr [\E \{\E^{-1/2}\left(Z_T\right)Z_T\E^{-1/2}\left(Z_T\right)\}] \delta/(n+d) \\
=& \Tr [I_{n+d}] \delta/(n+d) \\
=& \delta
.\end{align*}
\paragraph{Result}
Now that we verified all conditions of \cref{main_thm_simchowitz2018}, we can now translate the conclusion of \cref{main_thm_simchowitz2018} into our setting. \cref{main_thm_simchowitz2018} requires
\begin{align*}
T \geq& \frac{10k}{\xi^2}\left(\log \left(\frac{1}{\delta}\right) + 2(d+n)\log(10/ \xi) + \log \det (\Gambar \Gamsb^{-1}) \right)
.\end{align*}
First by our choice of $\Gamsb$ and $\Gambar$ we have
\begin{equation}
\label{eq:Gambar Gamsb inv}
\begin{aligned}
\log \det (\Gambar \Gamsb^{-1}) =& \log \det \left(\frac{(n+d)\E\{\sum_{t = 0}^{T-1} z_t z_t^\top\}}{T\delta} \nu^{-2}\right) \\
=& \log\left( \left(\frac{(n+d)}{T\delta \nu^{2}}\right)^{n+d} \det \left(\E\{\sum_{t = 0}^{T-1} z_t z_t^\top\} \right)\right) \\
\le& \log\left(\left(\frac{(n+d)}{T\delta \nu^{2}}\right)^{n+d} \left(\sum_{t = 0}^{T-1}\Tr(\E z_t z_t^\top) \right)^{n+d} \right) \\
=& (n+d)\log\left(\frac{(n+d)}{T\delta \nu^{2}} \sum_{t = 0}^{T-1}\Tr(\E z_t z_t^\top) \right)
.\end{aligned}
\end{equation}
With this in hand, we know that
\begin{align*}
&\frac{10k}{\xi^2}\left(\log \left(\frac{1}{\delta}\right) + 2(d+n)\log(10/ \xi) + \log \det (\Gambar \Gamsb^{-1}) \right) \\
& \qquad \le \frac{10k}{\xi^2}\left(\log \left(\frac{1}{\delta}\right) + 2(d+n)\log(10/ \xi) + (n+d)\log\left(\frac{(n+d)}{T\delta \nu^{2}} \sum_{t = 0}^{T-1}\Tr(\E z_t z_t^\top) \right)\right) \\
& \qquad = \frac{10(n+d)k}{\xi^2}\left(\log \left(\delta^{-\frac1{n+d}}\right) + \log(100/ \xi^2) + \log\left(\frac{(n+d)}{T\delta \nu^{2}} \sum_{t = 0}^{T-1}\Tr(\E z_t z_t^\top) \right)\right) \\
& \qquad = \frac{10(n+d)k}{\xi^2} \log\left(\frac{100(n+d)\sum_{t = 0}^{T-1}\Tr(\E z_t z_t^\top)}{T \nu^{2}\xi^2\delta^{1 + \frac1{n+d}}} \right)
.\end{align*}
Thus one sufficient condition for the requirement in \cref{main_thm_simchowitz2018} is
\begin{align*}
T/k \ge \frac{10(n+d)}{\xi^2} \log\left(\frac{100(n+d)\sum_{t = 0}^{T-1}\Tr(\E z_t z_t^\top)}{T \nu^{2}\xi^2\delta^{1 + \frac1{n+d}}} \right)
.\end{align*}
Finally we need to translate the conclusion of \cref{main_thm_simchowitz2018}:
\begin{align*}
\P\left[\lnorm{\hat\Theta_T-\Theta} >\frac{90\sigma}{\xi}\sqrt{\frac{n + (n+d)\log \frac{10}{\xi} + \log \det \Gambar \Gamsb^{-1} + \log\left(\frac{1}{\delta}\right)}{T\sigma_{\min}(\Gamsb)}} \right] \le 3\delta.
\end{align*}
By \cref{eq:Gambar Gamsb inv} and $\Gamma_{sb} = \nu^2 I_{n+d}$ we have
\begin{align*}
\P\left[\lnorm{\hat\Theta_T-\Theta} >\frac{90\sigma}{\xi}\sqrt{\frac{n + (n+d)\log \frac{10}{\xi} + (n+d)\log\left(\frac{(n+d)}{T\delta \nu^{2}} \sum_{t = 0}^{T-1}\Tr(\E z_t z_t^\top) \right) + \log\left(\frac{1}{\delta}\right)}{T\nu^2}} \right] \le 3\delta.
\end{align*}
Notice that
\begin{align*}
&n + (n+d)\log \frac{10}{\xi} + (n+d)\log\left(\frac{(n+d)}{T\delta \nu^{2}} \sum_{t = 0}^{T-1}\Tr(\E z_t z_t^\top) \right) + \log\left(\frac{1}{\delta}\right) \\
& \qquad \le (n+d)\left(1 + \log \frac{10}{\xi} + \log\left(\frac{(n+d)}{T\delta \nu^{2}} \sum_{t = 0}^{T-1}\Tr(\E z_t z_t^\top) \right) + \log\delta^{-\frac1{n+d}}\right) \\
& \qquad = (n+d)\left(1 + \log\left(\frac{10(n+d)\sum_{t = 0}^{T-1}\Tr(\E z_t z_t^\top)}{T\delta^{1 + \frac1{n+d}} \nu^{2}\xi} \right) \right)
.\end{align*}
Combining this with the previous inequality we have
\begin{align*}
\P\left[\lnorm{\hat\Theta_T-\Theta} >\frac{90\sigma}{\xi\nu}\sqrt{\frac{n+d}{T}\left(1 + \log\left(\frac{10(n+d)\sum_{t = 0}^{T-1}\Tr(\E z_t z_t^\top)}{T\delta^{1 + \frac1{n+d}} \nu^{2}\xi} \right) \right)} \right] \le 3\delta.
\end{align*}
\subsubsection{The proof of \cref{lem:bmsb}}
\label{The proof of lem:bmsb}
\begin{lemma*}[Similar to Lemma C.3 in \citet{dean2018regret}]
If we assume \cref{asm:InitialStableCondition}, then apply \cref{alg:myAlg}, the process $\{z_t\}_{t \geq 0}^T $ satisfies the
$(k, \nu, \xi)$-BMSB condition for
\begin{align*}
(k, \nu, \xi) = \left(1, \sqrt{\sigma_{\eta,T}^2\min\left(\frac{1}{2}, \frac{\sigma ^2}{2 \sigma ^2 C_K^2 + \tau^2} \right)} , \frac{3}{10}\right),
\end{align*}
where $\sigma_{\eta,T}^2 = \tau^2 T^{\beta-1}\log^\alpha(T)$.
\end{lemma*}
\begin{proof}
By \cref{def:BMSB condition} the statement means, for any $v \in \calS^{\statedim + \inputdim}$ and $0 \le t \le T-1$:
\begin{align*}
\P \left(|\langle v, z_{t+1} \rangle | \geq \sqrt{\sigma_{\eta,T}^2\min\left(\frac{1}{2}, \frac{\sigma ^2}{2 \sigma ^2 C_K^2 + \tau^2} \right)} \Bigg| \calF_t \right) \geq 3/10
.\end{align*}
\begin{comment}
Recall \cref{lemma: StateExpansion} states that:
\begin{equation}
\label{eq:StateExpansion 1}
x_{t} = \sum_{p=0}^{t-1}(A+B \Kh_{t-1})\cdots(A+B \Kh_{p+1})(B \eta_p+\varepsilon_p) +(A+B \Kh_{t-1})\cdots(A+B K_0)x_0
.\end{equation}
\begin{equation}
u_{t} = \sum_{p=0}^{t-1}\Kh_t(A+B \Kh_{t-1})\cdots(A+B \Kh_{p+1})(B \eta_p+\varepsilon_p) +\Kh_t(A+B \Kh_{t-1})\cdots(A+B K_0)x_0 + \eta_t
.\end{equation}
\end{comment}
Recall that
\begin{equation*}
x_{t+1} = Ax_t + Bu_t + \varepsilon_t
.\end{equation*}
\begin{equation*}
u_{t+1} = \Kh_{t+1}x_{t+1} + \eta_{t+1} = \Kh_{t+1}(Ax_t + Bu_t + \varepsilon_t) + \eta_{t+1}
.\end{equation*}
Denote the filtration $\calF_t = \sigma(x_0, \eta_0, \varepsilon_0 \ldots, \eta_{t - 1}, \varepsilon_{t - 1}, \eta_t) = \sigma(x_0, u_0, x_1, \cdots, x_t, u_t)$. It is clear that the process $\{z_t\}_{t \geq 0}$ is $\{\calF_t\}_{t \geq 0}$-adapted.
Recall that $\Kh_{t+1}$ is decided by $\Ah_{t}, \Bh_{t}$ in \cref{alg:myAlg}, where our estimator $\Ah_{t}, \Bh_{t}$ is designed to be only dependent on $x_0, u_0, x_1, \cdots, u_{t-1}, x_t$, which means
\[\Kh_{t+1} \in \calF_t = \sigma(x_0, u_0, x_1, \cdots, x_t, u_t).\]
For all $t \geq 1$, denote
\begin{align*}
\xi_{t+1} &:= \Kh_{t+1}(Ax_t + Bu_t) \in \calF_t
.\end{align*}
Now we are ready to prove \cref{lem:bmsb}. We have
\begin{align*}
\begin{bmatrix}
x_{t + 1}\\
u_{t + 1}
\end{bmatrix} = \begin{bmatrix}
A x_{t} + B u_{t} \\
\xi_{t + 1}
\end{bmatrix} +
\begin{bmatrix}
I_\statedim & 0 \\
\Kh_{t+1} & I_\inputdim
\end{bmatrix}\begin{bmatrix}
\varepsilon_t \\
\eta_{t + 1}
\end{bmatrix}
.\end{align*}
Given $\calF_t$,
$\begin{bmatrix}
x_{t + 1}\\
u_{t + 1}
\end{bmatrix}$
only has randomness in
$\begin{bmatrix}
I_\statedim & 0 \\
\Kh_{t+1} & I_\inputdim
\end{bmatrix}\begin{bmatrix}
\varepsilon_t \\
\eta_{t + 1}
\end{bmatrix}$, where
$\begin{bmatrix}
I_\statedim & 0 \\
\Kh_{t+1} & I_\inputdim
\end{bmatrix}$ is fixed given $\calF_t$, and
$\begin{bmatrix}
\varepsilon_t \\
\eta_{t + 1}
\end{bmatrix}$
follows $\calN \left(0,
\begin{bmatrix}
\sigma ^2 I_\statedim & 0\\
0 & \sigma_{\eta,t+1}^2 I_\inputdim
\end{bmatrix} \right)$. That implies
\begin{align*}
\begin{bmatrix}
x_{t + 1}\\
u_{t + 1}
\end{bmatrix} \Bigg| \calF_t \sim&
\calN\left(
\begin{bmatrix}
A x_{t} + B u_{t} \\
\xi_{t + 1}
\end{bmatrix},
\begin{bmatrix}
\sigma ^2 I_\statedim & \sigma ^2 \Kh_{t+1}^\top \\
\sigma ^2 \Kh_{t+1} & \sigma ^2 \Kh_{t+1}\Kh_{t+1}^\top + \sigma_{\eta,t+1}^2 I_\inputdim
\end{bmatrix}
\right)
.\end{align*}
Denote $\mu_{z, t+1}$ and $\Sigma_{z, t+1}$ as the mean and covariance of this multivariate normal distribution. Recall that we denoted $z_{t+1} = \begin{bmatrix}
x_{t+1} \\
u_{t+1}
\end{bmatrix}$.
Let $v \in \calS^{\statedim + \inputdim}$ and then $\langle v, z_{t+1} \rangle\Bigg| \calF_t \sim \calN(\langle v, \mu_{z, t+1} \rangle, v^\top \Sigma_{z, t+1} v)$. Therefore,
\begin{equation}
\label{eq:v zt 3 over 10}
\begin{aligned}
\P \left(|\langle v, z_{t+1} \rangle | \geq \sqrt{\sigma_{\min}(\Sigma_{z, t+1})} \Bigg| \calF_t \right) &\geq \P\left(|\langle v, z_{t+1} \rangle | \geq \sqrt{ v^\top \Sigma_{z, t+1} v} \Bigg| \calF_t \right)\\
&\geq \P\left(|\langle v, z_{t+1} - \mu_{z, t+1} \rangle | \geq \sqrt{ v^\top \Sigma_{z, t+1} v} \Bigg| \calF_t \right) \\
&\geq 3/10.
\end{aligned}
\end{equation}
Here we used the fact that for any $\mu, \sigma^2 \in \R$ and $\omega \sim \calN(0, \sigma^2)$, we have:
\begin{align*}
\P(|\mu + \omega| \geq \sigma) \geq \P(|\omega| \geq \sigma) \geq 3/10.
\end{align*}
Recall in \cref{alg:myAlg}, we force all our controllers $\Kh_t$ to have norm $\norm{\Kh_t} \le C_K$, where $C_K$ is a constant.
Then, by a simple argument based on a Schur complement (Lemma~\ref{lem:lambda_min_block_matrix}):
\begin{align*}
\sigma_{\min}(\Sigma_{z, t+1})
\geq &\sigma_{\eta,t}^2 \min\left(\frac{1}{2}, \frac{\sigma^2}{2 \ltwonorm{\Kh_{t+1} \sigma^2 \Kh_{t+1}^\T} + \sigma_{\eta,t}^2} \right) \\
\geq& \sigma_{\eta,t}^2 \min\left(\frac{1}{2}, \frac{\sigma ^2}{2 \sigma ^2 C_K^2 + \sigma_{\eta,t}^2} \right) \\
\ge& \sigma_{\eta,T}^2\min\left(\frac{1}{2}, \frac{\sigma ^2}{2 \sigma ^2 C_K^2 + \tau^2} \right).
\end{align*}
The desired conclusion directly follows:
\begin{align*}
&\P \left(|\langle v, z_{t+1} \rangle | \geq \sqrt{\sigma_{\eta,T}^2\min\left(\frac{1}{2}, \frac{\sigma ^2}{2 \sigma ^2 C_K^2 + \tau^2} \right)} \Bigg| \calF_t \right) \\
& \qquad \geq
\P \left(|\langle v, z_{t+1} \rangle | \geq \sqrt{\sigma_{\min}(\Sigma_{z, t+1})} \Bigg| \calF_t \right)
\\
& \qquad \geq 3/10 \qquad \text{ by \cref{eq:v zt 3 over 10}}
.\end{align*}
\end{proof}
~\paragraph{Schur complement}
\begin{lemma}[Lemma F.1 in \citet{mania2019certainty}]
\label{lem:lambda_min_block_matrix}
Let $\Sigma$ be a $\statedim \times \statedim$ positive-definite matrix and let $K $ be a real $\inputdim \times \statedim $ matrix.
Then, for any $\sigma_u \in \R$ we have that
\begin{align*}
\sigma_{\min}\left(\bmattwo{\Sigma}{\Sigma K ^\T}{K \Sigma}{K \Sigma K ^\T + \sigma_u^2 I}\right) &\geq
\sigma_u^2 \min\left(\frac{1}{2}, \frac{\sigma_{\min}(\Sigma)}{2 \ltwonorm{K \Sigma K ^\T} + \sigma_u^2} \right) \:.
\end{align*}
\end{lemma}
\subsubsection{The proof of \cref{lem:bound_covariance}}
\label{The proof of lem:bound_covariance}
\begin{lemma*}[Similar to Lemma C.4 in \citet{dean2018regret}]
If we assume \cref{asm:InitialStableCondition}, then apply \cref{alg:myAlg}, the process $\{z_t\}_{t \geq 0}^T $ satisfies
\begin{equation*}
\begin{split}
\sum_{t = 0}^{T - 1} \Tr \left( \E z_t z_t^\T\right) = \calO(T\log^2(T))
\end{split}
.\end{equation*}
\end{lemma*}
\begin{proof}
Now, note that
\[\Tr \left( \E z_t z_t^\T\right) = \E \left( \Tr z_t z_t^\T\right) = \E \norm{z_t}^2 = \E \left(\norm{x_t}^2 + \norm{u_t}^2\right).\]
Since $\norm{u_t} = \norm{\Kh_t x_t + \eta_t} \le \norm{\Kh_t} \norm{ x_t} + \norm{\eta_t} \le C_K\norm{ x_t} + \norm{\eta_t}$, we will show that if we can bound $\norm{ x_t}$, then we can also get a bound for $\norm{u_t}$ in the same order. Next we will focus on deriving the bound for $\norm{ x_t}$.
Define $C_{x,t} := C_x \log(t)$. Since $\rho(A+B K_0) < 1$, there exists some integer $m$ that $\norm{(A+B K_0)^m} < (\frac{\rho(A+B K_0) + 1}2)^m$. Let us denote $\rho := \frac{\rho(A+B K_0) + 1}2 < 1$ just for this \cref{lem:bound_covariance}.
For each $ t > m+1$, one of the following two statement must be true:
\begin{itemize}
\item $\norm{x_{t-i}} > C_{x,t-i}, (i=2,\cdots, m+1)$.
\item $\exists i \in \{2,\cdots, m+1\}$, which satisfies $\norm{x_{t-i}} \le C_{x,t-i}$.
\end{itemize}
We can derive an upper bound for $\norm{x_t}$ in both cases, and thus have an upper bound for every $\norm{x_t}$ by adding up those two bounds in two different cases.
\begin{enumerate}
\item If $\norm{x_{t-i}} > C_{x,t-i}, (i=2,\cdots, m+1)$, recall that if $\norm{x_{t}} > C_{x,t}$, then we assert our controller in the next step to be probing noise: $u_{t+1} = K_0x_{t+1} + \eta_{t+1}$. By assumption we already had $\norm{x_{k}} > C_{x,k}$, for $k= t-m-1, t-m, \cdots, t-2$. That means we have a consecutive $m$ steps of probing noise with $u_{k} = K_0x_{k} + \eta_{k}$, for $k= t-m, t-(m-1), \cdots, t-1$. Now we have
\[x_{k+1} = (A+B K_0) x_{k} + B\eta_{k} + \varepsilon_k , \quad \text{for} \quad k=t-m, t-(m-1), \cdots, t-1 .\]
That is
\[x_{t} = (A+B K_0)^m x_{t-m} + \sum_{k=0}^{m-1}(A+B K_0)^{k}(B\eta_{t-1-k} + \varepsilon_{t-1-k}).\]
which implies
\begin{equation}
\label{eq:lemma3Cond1}
\norm{x_{t}} \le \norm{(A+B K_0)^m} \norm{x_{t-m}} + \sum_{k=0}^{m-1}\norm{(A+B K_0)^{k}}\norm{(B\eta_{t-1-k} + \varepsilon_{t-1-k})}
.\end{equation}
\item If $\exists i \in \{2,\cdots, m+1\}$, which satisfies $\norm{x_{t-i}} \le C_{x,t-i}, (i=2,\cdots, m+1)$, then consider the following relationship
\begin{align*}
x_t &= A x_{t-1} + B u_{t-1} + \varepsilon_{t-1} \\
&= (A+B \Kh_{t-1})x_{t-1} + B\eta_{t-1} + \varepsilon_{t-1}
.\end{align*}
Therefore by our algorithm design that {$\norm{\Kh_t} \le C_K$} for any $t$
\begin{align}
\begin{split}
\label{eq:lemma3Cond2}
\norm{x_t} &\le \norm{A+B \Kh_{t-1}}\norm{x_{t-1}} + \norm{B\eta_{t-1} + \varepsilon_{t-1}} \\
&\le (\norm{A}+\norm{B}\norm{\Kh_{t-1}})\norm{x_{t-1}} + \norm{B\eta_{t-1} + \varepsilon_{t-1}} \\
&\le (\norm{A}+\norm{B}C_K)\norm{x_{t-1}} + \norm{B\eta_{t-1} + \varepsilon_{t-1}} \\
&\le (\norm{A}+\norm{B}C_K)^{i}\norm{x_{t-i}} + \sum_{k=0}^{i-1}(\norm{A}+\norm{B}C_K)^k\norm{B\eta_{t-1-k} + \varepsilon_{t-1-k}} \\
&\le \max\{1, (\norm{A}+\norm{B}C_K)^{m}\}C_{x,t} + \sum_{k=0}^{m-1} (\norm{A}+\norm{B}C_K)^k\norm{B\eta_{t-1-k} + \varepsilon_{t-1-k}} .
\end{split}
\end{align}
\end{enumerate}
By adding up \cref{eq:lemma3Cond1,eq:lemma3Cond2}, we have a bound that is applicable to both cases. Notice our previous assumption that $\norm{(A+BK_0)^m} \le \rho^m$, where $\rho < 1$, further take $\norm{(A+BK_0)^k}$, and $(\norm{A}+\norm{B}C_K)^k$ to be all bounded by a constant $M \ge 1$ for $k = 0, 1, \cdots, m$, which is of order $M = \calO(1)$ (because $m = \calO(1)$). By \cref{eq:lemma3Cond1,eq:lemma3Cond2}
\begin{equation}
\label{eq:lemma3basicEq}
\begin{aligned}
\norm{x_{t}} &\le \norm{(A+BK_0)^m} \norm{x_{t-m}} + \sum_{k=0}^{m-1}\norm{(A+BK_0)^{k}}\norm{B\eta_{t-1-k} + \varepsilon_{t-1-k}} \\
&\qquad + \max\{1, (\norm{A}+\norm{B}C_K)^{m}\} C_{x,t-i} + \sum_{k=0}^{m-1}(\norm{A}+\norm{B}C_K)^k\norm{B\eta_{t-k-1} + \varepsilon_{t-k-1}} \\
&\le \rho^m \norm{x_{t-m}} + M\left(C_{x,t-i} + 2\sum_{k=0}^{m-1} \norm{B\eta_{t-k-1} + \varepsilon_{t-k-1}}\right) \\
&\le \rho^m \norm{x_{t-m}} + M\left(C_{x,t} + 2\sum_{k=0}^{m-1} \norm{B\eta_{t-k-1} + \varepsilon_{t-k-1}}\right)
.\end{aligned}
\end{equation}
\cref{eq:lemma3basicEq} is very promising because it has a shrinking weight on $\norm{x_{t-m}}$. Let us use a simplified notation for the remainder:
\[J_t := M\left(C_{x,t} + 2\sum_{k=0}^{m-1} \norm{B\eta_{t-k-1} + \varepsilon_{t-k-1}} \right).\]
In $\E [J_t^2] $ there are three types of components:
\begin{itemize}
\item $M = \calO(1)$
\item $C_{x,t} = C_x \log(t)$
\item $\E(\sum_{k=0}^{m-1} \norm{B\eta_{t-k-1} + \varepsilon_{t-k-1}})^2 = \calO(1)$.
\end{itemize}
Since
\begin{equation}
\label{eq: E It2}
\E [J_t^2]
\le M^2 \cdot 2(C_{x,t}^2 + 4\E(\sum_{k=0}^{m-1} \norm{B\eta_{t-k-1} + \varepsilon_{t-k-1}})^2)
=
\calO(\log^2(t)),
\end{equation}
we can control $\E\norm{x_t}^2$ by
\begin{equation}
\label{eq: E xt 2}
\begin{split}
\E\norm{x_t}^2 &\le \E\left(\rho^m \norm{x_{t-m}} + J_t\right)^2 \\
&= \rho^{2m} \E\norm{x_{t-m}}^2 + \E J_t^2 + 2\rho^m \E\norm{x_{t-m}}\abs{J_t} \\
&\le \rho^{2m} \E\norm{x_{t-m}}^2 + \E J_t^2 + \frac{1-\rho^{2m}}{2} \E\norm{x_{t-m}}^2 + \frac{2\rho^{2m}}{1-\rho^{2m}}\E J_t^2 \\
&= \frac{1+\rho^{2m}}{2} \E\norm{x_{t-m}}^2 + \frac{1+\rho^{2m}}{1-\rho^{2m}}\E J_t^2 \\
& \qquad \text{ (because } 2ab \le a^2 +b^2 \text{ with } a^2 = \frac{1-\rho^{2m}}{2} \norm{x_{t-m}}^2 \text{ and } b^2 = \frac{2\rho^{2m}}{1-\rho^{2m}} J_t^2).
\end{split}
\end{equation}
By \cref{eq: E It2,eq: E xt 2},
\begin{equation}
\label{eq: E xt 2 mid}
\begin{split}
\E\norm{x_t}^2 &\le
\frac{1+\rho^{2m}}{2} \E\norm{x_{t-m}}^2 + \calO(\log^2(t)) \\
&\le
(\frac{1+\rho^{2m}}{2})^2 \E\norm{x_{t-2m}}^2 + \frac{1+\rho^{2m}}{2}\calO(\log^2(t)) + \calO(\log^2(t))\\
&\le
(\frac{1+\rho^{2m}}{2})^{\lfloor \frac{t}{m}\rfloor} \E\norm{x_{t-m\lfloor \frac{t}{m}\rfloor}}^2 + \sum_{i=0}^{\lfloor \frac{t}{m}\rfloor-1}(\frac{1+\rho^{2m}}{2})^i \calO(\log^2(t)) \\
&\le
\E\norm{x_{t-m\lfloor \frac{t}{m}\rfloor}}^2 + \calO(\log^2(t)) \\
& \qquad \text{(Recall that $\rho < 1$, and thus $\frac{1+\rho^{2m}}{2} < 1$)}.
\end{split}
\end{equation}
Now it only remains to show that $\E\norm{x_{t-m\lfloor \frac{t}{m}\rfloor}}^2$ is bounded by some constant.
Notice that
\begin{align*}
\E\norm{x_t}^2
\le& \E\left((\norm{A}+\norm{B}\norm{\Kh_t})\norm{x_{t-1}} + \norm{B}\norm{\eta_t} + \norm{\varepsilon_t}\right)^2 \\
\le& 3\left((\norm{A}+\norm{B}C_K)^2\E\norm{x_{t-1}}^2 + \norm{B}^2\E\norm{\eta_t}^2 + \norm{\varepsilon_t}^2\right)\\
\le& 3\left((\norm{A}+\norm{B}C_K)^2\E\norm{x_{t-1}}^2 + \norm{B}^2\tau^2 + \sigma^2\right)
.\end{align*}
By iteratively applying this inequality down to $\E\norm{x_0}^2$, we know that for $ t \le m$:
\[\E \norm{x_t}^2 = \calO(1).\]
Thus
following from \cref{eq: E xt 2 mid} we have
\begin{equation*}
\E \norm{x_t}^2 = \calO(\log^2(t)).
\end{equation*}
Since we already controlled the expectation of $\norm{x_t}^2$, it is straightforward to control the expectation of $\norm{u_t}^2$:
\[u_t = \Kh_tx_t + \eta_t.\]
\begin{align*}
\E\norm{u_t}^2 &\le \E\norm{\Kh_tx_t + \eta_t}^2 \\
&\le 2\E(\norm{\Kh_t}^2\norm{x_t}^2 + \norm{\eta_t}^2) \\
&\le 2\E(C_K^2\norm{x_t}^2 + \norm{\eta_t}^2) \\
&\le \calO(\log^2(t))
.\end{align*}
Thus,
\begin{equation}
\label{eq: E zt 2 final}
\E\norm{z_t}^2 = \E\norm{x_t}^2 + \E\norm{u_t}^2 \le \calO(\log^2(t))
\end{equation}
Then we have
\begin{equation*}
\E\sum_{t=0}^{T-1}\norm{x_t}^2, \E\sum_{t=0}^{T-1}\norm{u_t}^2, \E\sum_{t=0}^{T-1}\norm{z_t}^2 \le \calO(T\log^2(T))
.\end{equation*}
\begin{comment}
\begin{align*}
\E z_t z_t^\T &= \begin{bmatrix}
\Phi_x(t, t + 1)\\
\Phi_u(t, t + 1)
\end{bmatrix}
x_0 x_0^\T
\begin{bmatrix}
\Phi_x(t, t + 1)\\
\Phi_u(t, t + 1)
\end{bmatrix}^\T
+ \begin{bmatrix}
0 & 0\\
0 & \sigma_{\eta,t}^2 I_\inputdim
\end{bmatrix}
\\
&\quad + \sum_{k = 0}^{t - 1} \begin{bmatrix}
\Phi_x(t, t - k) \\
\Phi_u(t, t - k)
\end{bmatrix}
(\sigma_{\eta,t}^2 B B^\T + \sigma ^2 I_\statedim)
\begin{bmatrix}
\Phi_x(t, t - k) \\
\Phi_u(t, t - k)
\end{bmatrix}.
.\end{align*}
Since for all $j \geq 1$ we have $\norm{\Phi_x(t, j)} \leq C_x \rho^j$ and $\norm{\Phi_u(t, j)} \leq C_K \rho^j$, we obtain
\begin{align*}
\Tr \E z_t z_t^\T \leq \inputdim \sigma_{\eta,t}^2 + (\statedim C_x^2 + \inputdim C_K^2)\left(\rho^{2t + 2}\ltwonorm{x_0}^2 + (\sigma ^2 + \sigma_{\eta,t}^2 \norm{B}^2) \sum_{k = 1}^{t} \rho^{2k}\right)
.\end{align*}
Therefore, we get that
\begin{align*}
\sum_{t = 0}^{T - 1} \Tr\E z_t z_t^\top \leq \inputdim \sigma_{\eta,t}^2 T + \frac{\rho^2 T}{1 - \rho^2}(\statedim C_x^2 + \inputdim C_K^2)(\sigma ^2 + \sigma_{\eta,t}^2 \norm{B}^2) + \frac{\rho^2}{1 - \rho^2}(\statedim C_x^2 + \inputdim C_K^2)\ltwonorm{x_0}^2,
.\end{align*}
and the conclusion follows by simple algebra.
\end{comment}
\end{proof}
\subsection{Lemma in \cref{section: The proof of one_epoch_estimate_withMyalg}}
\subsubsection{The proof of \cref{lem:productBound}}
\label{The proof of lem:productBound}
\begin{lemma*}
Suppose we have a constant square matrix $M$ with spectral radius $\rho(M) < 1$, and a sequence of uniformly bounded random variables $\{\delta_t\}_{t=0}^\infty$, satisfying $\norm{\delta_t} \asConv 0$.
Denote the constant $\rho_M:= \frac{2 + \rho(M)}{3} < 1$.
Then we have, for any $t, q \in \mathbb{N}$, $t > q$:
\[\norm{(M + \delta_{t-1}) \cdots (M + \delta_{q})} = \calO(\rhoM^{t-q}) \as\]
And as a direct corollary
\[\norm{M^{t-q}} = \calO(\rhoM^{t-q}) .\]
\end{lemma*}
\begin{proof}
Our assumption of stability only says $\rho(M)< 1$, but our analysis prefers similar exponential decay with regard to spectral norm.
First, we need a conversion between spectral radius and spectral norm. Define
\begin{align*}
\transient{M}{ \rho} := \sup \left\{\norm{M^k} \rho^{-k} \colon k\geq 0 \right\}.
\end{align*}
For simplicity, let us denote
\[\tau(M) := \tau\left(M, \frac{1 + \rho(M)}{2}\right).\]
and with Gelfand's Formula
\begin{align*}
\rho (M)=\lim _{{k\to \infty }}\left\|M^{k}\right\|^{{{\frac {1}{k}}}}
.\end{align*}
Thus $\tau(M)$ is finite because $\frac{1 + \rho(M)}{2} > \rho(M)$.
Since $\{\delta_t\}_{t=0}^\infty$ is uniformly bounded, we can assume an upper bound $U_\delta$ for $\norm{M + \delta_i}$. Let us now consider the spectral norm of $(M+\delta_{t-1})\cdots(M+\delta_{q})$.
\begin{equation*}
\begin{split}
\norm{(M+\delta_{t-1})\cdots(M+\delta_{q})}
\le& \sum_{m=0}^{t-q}\norm{M^{t-q-m}} \sum_{q \le k_1 < \cdots < k_m \le t-1} \prod_{j=1}^m \norm{\delta_{k_j}}\\
\le& \sum_{m=0}^{t-q}\tau(M)\left(\frac{1 + \rho(M)}{2}\right)^{t-q-m} \sum_{q \le k_1 < \cdots < k_m \le t-1} \prod_{j=1}^m \norm{\delta_{k_j}}\\
=& \tau(M)\sum_{m=0}^{t-q}\left(\frac{1 + \rho(M)}{2}\right)^{t-q-m} \sum_{q \le k_1 < \cdots < k_m \le t-1} \prod_{j=1}^m \norm{\delta_{k_j}}\\
=& \tau(M)
\left(\frac{1 + \rho(M)}{2}+\norm{\delta_{t-1}}\right)\cdots
\left(\frac{1 + \rho(M)}{2}+\norm{\delta_{q+1}}\right).
\end{split}
\end{equation*}
Since $\norm{\delta_t} \to 0 \as$, for every $\omega$ in the sample space $\Omega$, such that there exists some $T_1(\omega)$, whenever $t > T_1(\omega)$, $\frac{1 + \rho(M)}{2}+\norm{\delta_{t}} < \frac{2 + \rho(M)}{3} < 1$, then
\begin{align*}
\norm{(M+\delta_{t-1})\cdots(M+\delta_{q})}
\le& \tau(M)
\left(\frac{1 + \rho(M)}{2}+\norm{\delta_{t-1}}\right)\cdots
\left(\frac{1 + \rho(M)}{2}+\norm{\delta_{q+1}}\right) \\
\le& \tau(M)
\rho_M^{t-q-T_1(\omega)}(\frac{1 + \rho(M)}{2}+U_\delta)^{T_1(\omega)}
.\end{align*}
Following \cref{defn: Big O notation} \cref{itm: big O as defn}:
\begin{equation*}
\norm{(M+\delta_{t-1})\cdots(M+\delta_{q})}= \calO(\rhoM^{t-q}) \as
\end{equation*}
\end{proof}
\section{Experiment Details}
\label{section: Additional Experiments}
\subsection{Experiment Setting}
\label{subsection: Experiment on Stable System}
\subsubsection{Experiment Setting on Stable System}
\label{subsection: easy setting}
We set
$A = \begin{bmatrix}
0.8 & 0.1 \\
0 & 0.8
\end{bmatrix}$
and
$B = \begin{bmatrix}
0 \\
1
\end{bmatrix}$, with system noise $\sigma = 1$, injected noise baseline $\tau = 1$, $Q = I_2$, $R = 1$ and initial state $x_0 = [0, 0]^\top$. As for the algorithmic hyper-parameters, we set the warning threshold for states $x_t$ at $C_x=1$ (so that $C_{x,t} = \log(t)$), the known stable controller $K_0 = [0,0]$, and the upper bound of the L2-norm for our controller $\Kh_t$ at $C_K = 5$. Note that this is conservative by about a factor of 10, since the true optimal controller in this system is
$K \approx \begin{bmatrix}
-0.10, -0.48
\end{bmatrix}$.
Recall that the choice of these hyper-parameters does not actually affect our theoretical coverage (as long as $C_K>\|K\|$) or regret guarantees, but in practice their values prevent the system from incurring very large regret in the first few time steps. Even for this, they are only needed because we do not assume we are given an initial controller that is very close to $K$;
in contrast, for instance, \citet{dean2018regret} started from a controller fitted with 100 samples of white noise actions. All stable system results are based on 1,000 independent runs of \cref{alg:myAlg} for $T = 10,000$ time steps.
\subsubsection{Experiment Setting on Unstable System}
\label{subsection: hard experiment setting}
The unstable system we simulate is highly unstable, and is largely the same as that in Appendix H of \citet{dean2018regret}.
We set
$A = \begin{bmatrix}
2 & 0 & 0 \\
4 & 2 & 0 \\
0 & 4 & 2
\end{bmatrix}$
and
$B = I_3$, with system noise $\sigma = 1$, injected noise baseline $\tau = 1$, $Q = 10I_3$, $R = I_3$ and initial state $x_0 = [0, 0, 0]^\top$. As for the hyper-parameters, we set the warning threshold for states $x_t$ at $C_x=1$ (so that $C_{x,t} = \log(t)$), and we examined two different choices for the known stabilizing controller: $K_0 =
-\begin{bmatrix}
1.5 & 0 & 0 \\
0 & 1.5 & 0 \\
0 & 0 & 1.5
\end{bmatrix}$ and
$K_0 =
-\begin{bmatrix}
1.5 & 0 & 0 \\
3.5 & 1.5 & 0 \\
0 & 3.5 & 1.5
\end{bmatrix}$. The former choice incurs quite a bit higher regret than the latter, and hence we refer to the former as the `bad' stabilizing controller and to the latter as the `good' stabilizing controller. We set the upper bound of the L2-norm for our controller $\Kh_t$ at the level of $C_K = 1000$. Our choice of $K_0$ is different from the starting point in \citet{dean2018regret}, where they started from a $T=250$ burn in period estimate, and did not report the regret in the first 250 steps.
All unstable system results are based on 1,000 independent runs of \cref{alg:myAlg} for $T = 5,000$ time steps.
\subsection{Experiment on Unstable System}
\label{subsection: Experiment on Unstable System}
In contrast to the stable system simulation summarized in \cref{subsection: Partial experiment result}, in this section we simulate the severely unstable system described in \cref{subsection: hard experiment setting}. In this setting, the specification of $K_0$ is critical due to the costs incurred at the early time steps, an unavoidable consequence of starting from limited information in a system that can rapidly spiral (nearly) out of control.
\subsubsection{Summary of results on unstable system}
We begin with the analogue of \cref{fig:Summary stable system} for the unstable system, given in \cref{fig:Summary unstable system}. The main takeaways are the same as the discussion in \cref{subsection: Partial experiment result}.
\begin{figure}[H]
\centering
\begin{subfigure}{.45\textwidth}
\captionsetup{justification=centering}
\caption{Benefit of stepwise update}
\includegraphics[width = \linewidth]{fig/paper_plot_all_one_half/control_experiment_large_good_start/Compare_Regret.pdf}
\label{fig:Compare Regret one half unstable}
\end{subfigure}
\begin{subfigure}{.45\textwidth}
\caption{Regret Ratio}
\includegraphics[width = \linewidth]{fig/paper_plot_all_one_half/control_experiment_large_good_start/Regret_ratio2.pdf}
\label{fig:regret ratio one half unstable}
\end{subfigure}
\begin{subfigure}{.45\textwidth}
\captionsetup{justification=centering}
\caption{Differing Convergence Rates}
\includegraphics[width = \linewidth]{fig/paper_plot_all_one_half/control_experiment_large_good_start/estimation_rate.pdf}
\label{fig:estimation rate one half unstable}
\end{subfigure}
\begin{subfigure}{.45\textwidth}
\caption{Confidence Region Coverage}
\includegraphics[width = \linewidth]{fig/paper_plot_all_one_half/control_experiment_large_good_start/Coverage.pdf}
\label{fig:Coverage one half unstable}
\end{subfigure}
\begin{subfigure}{.45\textwidth}
\caption{Prediction Region Coverage}
\includegraphics[width = \linewidth]{fig/paper_plot_all_one_half/control_experiment_large_good_start/Safety_Coverage_True.pdf}
\label{fig:Prediction Coverage True one half unstable}
\end{subfigure}
\end{figure}
\begin{figure}[H]
\ContinuedFloat
\caption{Summary of 1000 independent experiments applying \cref{alg:myAlg} with $\beta = 0.5$, $\alpha = 2$, $C_x=1$, $C_K=5$, and $K_0 = -\begin{bmatrix}
1.5 & 0 & 0 \\
3.5 & 1.5 & 0 \\
0 & 3.5 & 1.5
\end{bmatrix}$
to the unstable system described in \cref{subsection: hard experiment setting}. (a) Difference between the regret of \cref{alg:myAlg} using stepwise and logarithmic updates.
(b) The ratio of the empirical regret and our parametric or observable expressions for the regret.
(c) The average Frobenius norm of various estimation errors considered in this paper, with slopes fitted on a log-log scale so that the estimation error is $\logO(t^{\text{slope}})$.
The effect of $\alpha$ was removed from the slopes of $\Kh_t-K$ and $[\Ah_t-A,\Bh_t-B]$ by dividing the error by $\log^{\alpha/2}(t)$.
(d) Coverage of our 95\% confidence regions for $[A,B]$, $K$, and $\E [x_{t+1} \,|\, \{x_i, u_i\}_{i=0}^{t} ] = Ax_t + Bu_t$.
(e) Coverage of our 95\% prediction region for $ x_{t+1} \,|\, \{x_i, u_i\}_{i=0}^{t}$, along with coverage of the naive prediction region given in \cref{eq: prediction region observable simple}.
}
\label{fig:Summary unstable system}
\end{figure}
\subsubsection{Large Regret From Early Time Steps}
\label{subsubsection: Large Regret at Burn in Period}
For the `bad' choice of stabilizing controller $K_0 =
-\begin{bmatrix}
1.5 & 0 & 0 \\
0 & 1.5 & 0 \\
0 & 0 & 1.5
\end{bmatrix}$, we plot the log regret in subplot (a) of \cref{fig:log regret three choices unstable system}.
We observe a rapidly increasing regret in the first roughly 200 time steps, which dominates all the regret in the remaining steps.
We offer a brief explanation why the cost in the early time steps is very large despite assuming knowledge of a stabilizing yet sub-optimal controller $K_0$. Notice $A+BK_0 = \begin{bmatrix}
0.5 & 0 & 0 \\
4 & 0.5 & 0 \\
0 & 4 & 0.5
\end{bmatrix}$. Thus
$(A+BK_0)^2 = \begin{bmatrix}
0.25 & 0 & 0 \\
4 & 0.25 & 0 \\
16 & 4 & 0.25
\end{bmatrix}$,
$(A+BK_0)^3 = \begin{bmatrix}
2^{-3} & 0 & 0 \\
3 & 2^{-3} & 0 \\
24 & 3 & 2^{-3}
\end{bmatrix}$,
$(A+BK_0)^4 = \begin{bmatrix}
2^{-4} & 0 & 0 \\
2 & 2^{-4} & 0 \\
24 & 2 & 2^{-4}
\end{bmatrix}$,
$(A+BK_0)^5 = \begin{bmatrix}
2^{-5} & 0 & 0 \\
1.25 & 2^{-5} & 0 \\
20 & 1.25 & 2^{-5}
\end{bmatrix}$,
$(A+BK_0)^6 = \begin{bmatrix}
2^{-6} & 0 & 0 \\
0.75 & 2^{-6} & 0 \\
15 & 0.75 & 2^{-6}
\end{bmatrix}$. So although we have a controlled system with maximum eigenvalue $0.5$, the power of $(A+BK_0)^k$ can still be very large in the bottom left corner for $k = 2,3,4,5,6$. Because of this, the randomness in the states is enlarged and propagated to several future steps. It turns out that, at the first 200 steps we used this high cost safety policy $K_0$ a lot as we do not have a good estimate of optimal controller $K$, and that is the real reason for this high burn-in period cost. As we will see later, if we change the stabilizing controller $K_0$ to be closer to the optimal $K$, the regret will be much smaller.
\subsubsection{Comparison with Thompson Sampling}
\label{subsubsection: TS unstable system}
For comparison, we implement a straightforward version of Thompson sampling as follows.
Denote $\Theta := [A, B]$. We use a prior of
\begin{equation*}
\vvector[\Theta _{prior}] \sim \calN(\vvector[\Theta], I_{n(n+d)})
.\end{equation*}
Using the Bayesian updating equations and denoting the least-squares estimate of $\Theta$ by $\hat{\Theta}_t = [\Ah_t,\Bh_t]$, the posterior at time $t$ is given by
\begin{equation*}
\vvector[\Theta_t^{TS}] \sim
\calN\left(
\vvector
\left[
\left(
\Theta + \hat\Theta_t\sum_{i=0}^{t-1}\begin{bmatrix} x_i\\ u_i\\ \end{bmatrix} \begin{bmatrix} x_i\\ u_i\\ \end{bmatrix}^\top
\right)
\left(
I_{n+d}+\sum_{i=0}^{t-1}\begin{bmatrix} x_i\\ u_i\\ \end{bmatrix} \begin{bmatrix} x_i\\ u_i\\ \end{bmatrix}^\top
\right)^{-1}
\right],
\left(
I_{n+d}+\sum_{i=0}^{t-1}\begin{bmatrix} x_i\\ u_i\\ \end{bmatrix} \begin{bmatrix} x_i\\ u_i\\ \end{bmatrix}^\top
\right)^{-1}
\otimes
I_n
\right)
.\end{equation*}
At each step, we draw a sample $\Theta_t^{TS}$ from this posterior and use it as the input to the DARE for calculating $\Kh_t$. Since a system is stabilizable if $\text{rank}([A-\lambda I,B])=n$ for any eigenvalue $\lambda$ of $A$ \citep{hautus1970stabilization}, the Gaussian posterior puts probability 1 on stabilizable $\Theta=[A,B]$ and hence defines a unique solution to the DARE with probability 1 as well.
We report the Thompson sampling regret in subplot (b) of \cref{fig:log regret three choices unstable system}, and see that it also suffers from rapidly increasing regret at early time points.f
\subsubsection{Improved Regret When Using `Good' $K_0$}
\label{subsubsection: Start with good choice of K_0}
When we switch from the `bad' stabilizing controller to the `good' one specified in \cref{subsection: hard experiment setting} as
$K_0 = -\begin{bmatrix}
1.5 & 0 & 0 \\
3.5 & 1.5 & 0 \\
0 & 3.5 & 1.5
\end{bmatrix}$, we get that $A+BK_0 = \begin{bmatrix}
0.5 & 0 & 0 \\
0.5 & 0.5 & 0 \\
0 & 0.5 & 0.5
\end{bmatrix}$, which is a much better starting point than the previous $ \begin{bmatrix}
0.5 & 0 & 0 \\
4 & 0.5 & 0 \\
0 & 4 & 0.5
\end{bmatrix}$, and the regret in this setting is indeed much better (see subplot (c) of \cref{fig:log regret three choices unstable system}) and resembles that of the stable system described in \cref{subsection: easy setting}.
\begin{figure}[!htb]
\centering
\includegraphics[width = 0.95\textwidth]{fig/paper_plot_appendix/Regret_unstable_comparison.pdf}
\caption{Regret on the log scale based on 1000 independent experiments on the unstable system for $\beta = 0.5$ and $\alpha = 0$. (a): Bad safety controller $K_0 =
-\begin{bmatrix}
1.5 & 0 & 0 \\
0 & 1.5 & 0 \\
0 & 0 & 1.5
\end{bmatrix}$; (b): Thompson Sampling; (c): Good safety controller $K_0 = -\begin{bmatrix}
1.5 & 0 & 0 \\
3.5 & 1.5 & 0 \\
0 & 3.5 & 1.5
\end{bmatrix}$.
}
\label{fig:log regret three choices unstable system}
\end{figure}
\subsection{Choices of $\beta$ other than $0.5$}
Our simulations consider choices of $\beta$ beyond $0.5$ and even beyond those covered by our theory. In particular, we consider $\beta = 0.1, 0.3, 0.5, 0.7, 0.9$ and observe promising evidence that some of our asymptotic coverage results may generalize to the setting of $\beta < 1/2$.
\subsubsection{Regret}
\label{apdx subsection: regret}
According to \cref{thm:regret} the dominating term for regret should be $ T^\beta \log^\alpha(T) \Tr((B^\top PB +R)\frac{\tau^2}{\beta})$ for any $\beta \in [1/2,1)$ and $\max\{\beta,\,\alpha -1\} > 1/2$, and that indeed matches with our experimental results (see \cref{fig:regret_stable_unstable}). The asymptotic regret expression from \cref{thm:regret} is represented as the black solid curve, which converges to the empirical regret for $\beta >0.5$, but not $\beta <0.5$.
\begin{figure}[!htb]
\centering
\begin{subfigure}{.95\textwidth}
\caption{Stable System}
\includegraphics[width = \linewidth]{fig/paper_plot_appendix/regret_stable_unstable/Regret_log_stable.pdf}
\label{fig:regret_stable_5beta}
\end{subfigure}
\begin{subfigure}{.95\textwidth}
\caption{Unstable System}
\includegraphics[width = \linewidth]{fig/paper_plot_appendix/regret_stable_unstable/Regret_log_unstable.pdf}
\label{fig:regret_unstable_5beta}
\end{subfigure}
\caption{Regret on the log scale based on 1000 independent experiments for $\beta = 0.1, 0.3, 0.5, 0.7, 0.9$ and $\alpha = 0$. (a): stable system; (b): unstable system.}
\label{fig:regret_stable_unstable}
\end{figure}
\subsubsection{Confidence region coverage}
\label{appendix subsection: confidence coverage}
\cref{fig:Coverage_stable_unstable} shows that the finite sample coverage of our confidence regions and prediction region closely matches the asymptotic theory from \cref{thm:main,corr: K confidence region,thm: prediction CLT} for any choice among $\beta = 0.1, 0.3, 0.5, 0.7, 0.9$, with the exception of confidence regions for $K$, which seem to only work for the $\beta\ge 0.5$ covered by our theory.
\begin{figure}[!htb]
\centering
\begin{subfigure}{.95\textwidth}
\caption{Stable System}
\includegraphics[width = \linewidth]{fig/paper_plot_appendix/coverage_stable_unstable/Coverage_stable.pdf}
\label{fig:Coverage_stable_5beta}
\end{subfigure}
\begin{subfigure}{.95\textwidth}
\caption{Unstable System}
\includegraphics[width = \linewidth]{fig/paper_plot_appendix/coverage_stable_unstable/Coverage_unstable.pdf}
\label{fig:Coverage_unstable_5beta}
\end{subfigure}
\caption{Coverage on the log scale based on 1000 independent experiments for $\beta = 0.1, 0.3, 0.5, 0.7, 0.9$ and $\alpha = 0$. (a): stable system; (b): unstable system.}
\label{fig:Coverage_stable_unstable}
\end{figure}
\subsection{Algorithm design}
We now investigate how the details of \cref{alg:myAlg} (the stabilizing controller $K_0$ and the thresholds on $x_t$ and $\|\Kh_t\|$) impact the regret.
\paragraph{The threshold $C_{x,t}$ controls extreme tail behavior}
Although we only trigger the threshold $C_{x,t}$ rarely, without it we can see some extreme behavior with low probability. In particular, when this threshold constraint is removed, we occasionally observe very large regret in early time steps due to the poor estimate $\Kh_t$, which causes instability of the system (see \cref{fig:log regret no Cx} and compare it to the purple line and shaded region in \cref{fig:log regret CK}). The mean value is even higher than the 0.95 quantile curve because of several extremely large regrets induced by the unstable closed-loop system. And compared to when $C_{x,t}$ is used in \cref{fig:log regret CK}, the 0.95 quantile when $C_{x,t}$ is not used is considerably higher, although its median is quite similar to the mean when $C_{x,t}$ is used.
\begin{figure}[!htb]
\centering
\includegraphics[width = 0.95\textwidth]{fig/paper_plot_appendix/LS_experiment_False1/Regret_log_full.pdf}
\caption{Regret on the log scale with no $C_{x,t}$ threshold on $\norm{x_t}$ based on 1000 independent experiments on stable system for $\beta = 0.1, 0.3, 0.5, 0.7, 0.9$ with $C_K = 5$ and $\alpha = 0$.
}
\label{fig:log regret no Cx}
\end{figure}
\paragraph{Stepwise updating improves regret over logarithmic updating}
As our theory provides guarantees for \cref{alg:myAlg} with both stepwise and logarithmic updating, we run experiments to compare the regret of these two choices.
\cref{fig:estimation rate one half,fig:estimation rate one half unstable} show the difference in regret between \cref{alg:myAlg} and the same algorithm but that only updates its estimates of the system parameters logarithmically often, i.e., at times $t=1, 2, 4, 8, \dots$ On average, we see a steady logarithmic increase in regret from switching from stepwise updates to logarithmic frequency.
\paragraph{A stabilizing controller $K_0$ closer to $K$ improves performance}
Although $K_0$ is a stabilizing controller by assumption, bad choices of $K_0$ can still make $(A+BK_0)^k$ large for some finite $k$
(see \cref{subsubsection: Large Regret at Burn in Period} for a concrete example). Thus, unsurprisingly, choosing $K_0$ to be as near as possible to the optimal controller $K$ produces smaller regret, as evidenced by
\cref{fig:log regret three choices unstable system}.
\paragraph{Regret is robust to conservative choices of $C_K$}
To check the sensitivity of the choice of $C_K=5$ in the stable system, we also tried a looser bound $C_K =1000$. We found that the norm of $\Kh_t$ never surpassed the $C_K =1000$ bound. This larger $C_K$ made little difference for settings covered by our theory ($\beta\ge 0.5$), and surprisingly seems to actually improve the regret for smaller $\beta$ (see \cref{fig:log regret CK}).
\begin{figure}[!htb]
\centering
\includegraphics[width = 0.95\textwidth]{fig/paper_plot_appendix/Regret_CK.pdf}
\caption{Regret on the log scale based on 1000 independent experiments on stable system for $\beta = 0.1, 0.3, 0.5, 0.7, 0.9$ with $\alpha = 0$ comparing $C_K = 5$ and $C_K = 1000$.
}
\label{fig:log regret CK}
\end{figure}
\clearpage
\section{Introduction}
\subsection{Problem statement}
Many dynamic systems such as robots,
power grids,
or living cells
can be described at any given time $t$ by a system state $x_t$ that depends on both its previous state $x_{t-1}$ and some internal or external control $u_{t-1}$ that is applied to direct the system to achieve its desired function. Both adaptive control and reinforcement learning address the problem of choosing the controls $u_t$ when the system dynamics, i.e., the relationship between $x_{t+1}$ and $(x_{t},u_{t})$, are \emph{unknown}. But the behavior of the algorithms developed in these fields has been characterized only coarsely, even in the simplest systems, preventing their deployment in high-stakes applications that require precise guarantees on safety and performance.
In this paper we will consider a canonical model for such systems, the discrete-time linear dynamical system:
\begin{equation}
\label{eq:LinearModel}
x_{t+1} = A x_t + B u_t + \varepsilon_t,
\end{equation}
where $x_t \in \mathbb{R}^{n}$ represents the state of the system at time $t$ and starts at some initial state $x_0$, $u_t \in \mathbb{R}^d$ represents the action or control applied at time $t$, $\varepsilon_t \iid \calN(0, \sigma ^2 I_n)$
is the system noise, and $A\in\mathbb{R}^{n\times n}$ and $B\in\mathbb{R}^{n\times d}$ are matrices determining the system's linear dynamics; the fact that they do not depend on $t$ makes this a \emph{time-homogeneous} dynamical model. The states $x_t$ and controls $u_t$ are assumed to have been transformed so that $x_t$ closer to zero represents the system better-performing its function, and $u_t$ closer to zero represents lower control cost/effort. The goal is to find an algorithm $U$ that, at each time $t$, outputs a control $u_t = U(H_t)$ that is computed using the entire thus-far-observed history of the system $H_t = \{x_t, u_{t-1}, x_{t-1},\dots,u_1,x_1,u_0\}$ to maximize the system's function while minimizing control effort.
We formalize this tradeoff by augmenting the linear dynamics \eqref{eq:LinearModel} with the popular \emph{quadratic} cost function, so that
at every time $t$, the system incurs the cost $x_t^\top Qx_t + u_t^\top Ru_t$, for some known positive-definite matrices $Q\in\mathbb{R}^{n\times n}$ and $R\in\mathbb{R}^{d\times d}$.
In order to abstract away finite-sample issues arising from different time horizons $T$, we will focus on the infinite-horizon problem, which seeks to minimize the expected average limiting cost:
\begin{equation}
\label{eq:cost definition}
\mathcal{J}(U) = \lim_{T\rightarrow\infty}
\mathbb{E}\mathcal{J}(U,T),
\hspace{1cm}
\mathcal{J}(U,T) = \frac{1}{T}
\sum_{t=1}^{T} \left(x_t^\top Qx_t + u_t^\top Ru_t\right).
\end{equation}
When the system dynamics $A$ and $B$ are known, the cost-minimizing algorithm is known and called the linear-quadratic regulator (LQR): $U^*(H_t) = Kx_t$, where $K\in\mathbb{R}^{d\times n}$ is the efficiently-computable solution to a system of equations that only depend on $A$, $B$, $Q$, and $R$; we will review the exact expressions for $K$ in Section~\ref{section:review on the LQR problem}. Like the Gaussian linear model in regression and supervised learning, the aforementioned linear-quadratic problem is foundational to control theory because it is conceptually simple yet it provides a remarkably good description for some real-world systems (e.g.,
biological systems \citep{priess2014solutions}, aircraft flight control \citep{choi1999lqr}, or power supply \citep{shabaani2003application}), and insights from its study often translate to innovations and improved understanding in far-more-complex models.
In this paper we consider the case when the system dynamics $A$ and $B$ are unknown, which we call \emph{linear-quadratic adaptive control} (LQAC),
to distinguish it from the LQR setting when $A$ and $B$ are assumed known. Intuitively, one might hope that after enough time observing a system controlled by almost any algorithm, one should be able to estimate $A$ and $B$ (and hence $K$) fairly well and thus be able to apply an algorithm quite close to $U^*$. Indeed the key challenge in LQAC, as in any reinforcement learning problem, is to trade off \emph{exploration} (actions that help estimate $A$ and $B$) with \emph{exploitation} (actions that minimize cost). We will quantify the cost of an LQAC algorithm by its \emph{average regret}:\footnote{Not to be confused with the more-common \emph{cumulative regret}, given by $T\mathcal{R}(U,T)$. Since one is simply $T$ times the other, it makes no mathematical difference which one is considered, but we prefer a regret formulation that does not diverge to infinity.}
\begin{equation*}
\mathcal{R}(U,T) = \mathcal{J}(U,T)-\mathcal{J}(U^*,T).
\end{equation*}
A flurry of recent work has proposed new algorithms for LQAC and studied their regret and estimation error; we review this literature in Section~\ref{subsection: Related Works}. These studies have produced
finite-sample bounds (in terms of the problem parameters) on various performance metrics which capture the rates at which those metrics depend on various values, especially time $T$.
These recent breakthroughs have
advanced the field significantly, but two significant hurdles remain to using their insights to enable reliable, safe, high-performance reinforcement learning.
\begin{itemize}
\item Many of the benefits of a theoretical characterization of the performance of an algorithm (e.g., its regret or estimation error) involve quantifying \emph{differences}, such as the difference in performance between
two algorithms applied to the same system
or the performance difference between applying the same algorithm to two different systems.
But a difference between two rigorous, but loose, bounds that have the same rate can be misleading,
since the difference in the looseness of the bounds can overwhelm the difference in the true performance.
\item
When an expression characterizing an algorithm's performance depends explicitly on the system dynamics (in our case, $A$ and $B$), it
cannot actually be evaluated in practice because the system dynamics are by assumption unknown.
Thus in order to enable certain critical aspects of reinforcement learning such as safety, non-stationarity detection, and generalization to new systems,
there is a pressing need to characterize algorithmic behavior in terms
only of \emph{observable} quantities.
\end{itemize}
\begin{comment}
Consider a longitudinal flight control problem, where we want to keep the aircraft close to a desired state $x_d = (w_d, v_d, \theta_d, q_d)$.
\begin{itemize}
\item $w_d$: velocity of aircraft along body axis
\item $v_d$: velocity of aircraft perpendicular to body axis
\item $\theta_d$: angle between body axis and horizontal
\item $q_d = \dot \theta_t$: angular velocity of aircraft (pitch rate)
\end{itemize}
with input $u_t = (e_t, f_t)$
\begin{itemize}
\item $e_t$: elevator angle
\item $f_t$: thrust
\end{itemize}
Now set our system state as the difference between our current state and the desired state $x_d$. With little abusing of notation we denote such difference as $x_t = (w_t, v_t, \theta_t, q_t)$. We adopt a discretization setting where states are recorded every 1 second. Then our system can be expressed as $x_{t+1} = Ax_t + Bu_t + \varepsilon_t$. When the desired state for Boeing 747 set as level flight, 40,000 feet and 774 ft/sec, the dynamics are
\begin{equation*}
A =
\begin{bmatrix}
.99 & .03 &-.02 & -.32 \\
.01 & .47 &4.7 &.00 \\
.02 & -.06 &.40 &.00 \\
.01 & -.04 &.72 &.99
\end{bmatrix}
, \quad
B =
\begin{bmatrix}
0.01 & 0.99 \\
-3.44 & 1.66 \\
-0.83 & -0.44 \\
-0.47 & -0.25
\end{bmatrix}
\end{equation*}
Since we want to stay as close to the desired state as possible ($x_t$ close to $0$), and in the meantime, paying the least cost ($u_t$ close to $0$), the most natural objective in this scenario is to minimize the weighted summation of $\norm{x_t}^2$ and $\norm{u_t}^2$, which turns out to be our LQR cost function \cref{eq:cost definition}. The weights should be chosen according to user's objective. For example, if one cares about input cost more than tracking the desired state, then he/she could choose a larger $R$.
\end{comment}
\subsection{Our contribution}
This paper presents asymptotically-exact expressions for a number of quantities of interest for a simple LQAC algorithm that achieves the optimal rate of regret. That is, we prove that the performance of the algorithm converges \emph{exactly} to the expressions we present. We have two types of results: asymptotically-exact expressions in terms of non-random system \emph{parameters}, and asymptotically-exact expressions in terms of only \emph{observable} random variables.
\paragraph{Theory for a rate-optimal algorithm with stepwise-update estimates.}
The LQAC algorithm we consider in all of the theory in this paper is very simple and intuitive, using a least-squares estimate of the system dynamics at each time point to estimate the optimal controller $K$ and adding a vanishing exploration noise to that certainty-equivalent control which can be tuned to achieve the optimal rate of regret. All our theory is for a single system trajectory (no independent restarts), and in contrast to existing literature on LQAC we allow our algorithm to update its estimate of the dynamics at \emph{every} time step, although we show our theoretical results can easily be extended to the more common setting of logarithmic updating as well.
\paragraph{Asymptotically-exact expressions characterizing LQAC performance metrics.}
For a number of different performance metrics of interest for the LQAC problem, we provide asymptotically-exact expressions (a) purely in terms of the non-random, unknown system parameters, and (b) purely in terms of the random, observable system history. In particular, we provide both types (a) and (b) of asymptotically-exact expressions for
\begin{itemize}
\item[(i)] the \emph{regret} at any current or future time point,
\item[(ii)] the distribution of the \emph{estimation error} of the least-squares estimate of system dynamics $A$ and $B$, and
\item[(iii)] the distribution of the \emph{prediction error} of the least-squares estimate of a future state.
\end{itemize}
We further use (ii) to derive the estimation error of the least-squares estimate of the optimal controller $K$, and to identify a function of the dynamics, $A+BK$, that can be estimated at a much faster rate than just $A$ or $B$ (although to reiterate, our expressions characterize not just the rates but the exact constants multiplying those rates as well). Our observable expressions for (ii) and (iii) immediately give us asymptotically-exact online confidence regions for the system dynamics (and optimal controller $K$) and prediction regions for a future state, respectively.
\paragraph{Numerical validation of our theory}
We apply our algorithm to both a stable and an unstable simulated system to compare our asymptotic expressions to the performance metrics they characterize, and we find quite good agreement, even at very early time steps.
\subsection{Related work}
\label{subsection: Related Works}
Our study of the asymptotics of the LQAC problem has connections with many works across control theory, machine learning, and statistics, and we defer a more thorough exposition of related work to \cref{subsection: Background}, while here only focusing on the most relevant literature
The LQAC algorithm we consider in this paper falls into the class of algorithms which has been referred to as \emph{certainty equivalent} controllers in the literature. The key idea is to estimate the system dynamics and then apply a control that would be optimal if the estimate were correct. Following this strategy blindly is known to be inconsistent \citep{becker1985adaptive,lai1982iterated}, but a simple fix is to add a vanishing noise term, which was shown by \citet{dean2018regret} to achieve $\logO(T^{-1/3})$ average regret and later by \citet{faradonbeh2018input,faradonbeh2018optimality,mania2019certainty} to achieve $\logO(T^{-1/2})$ average regret. The recent work of \citet{simchowitz2020naive} refined the existing regret bounds and showed $\logO(T^{-1/2})$ to be the \emph{optimal} rate of average regret. To our knowledge, all LQAC algorithms that have been proved to achieve the optimal rate of regret update their estimate of the system dynamics \emph{logarithmically} often,\footnote{The only exception is \citet{abeille2018improved}, whose Thompson sampling algorithm updates its estimates at every step, but their proof only holds for scalar systems ($n=1$).} and their bounds on regret and estimation error hold in finite samples but have conservative constants multiplying the rate.
There is work on \emph{system identification} and in particular on \emph{optimal experimental design} that relates to our characterization of the estimation error of the learned system dynamics. These works focus mainly on minimizing estimation error with little or no consideration for the regret, and hence only consider algorithms with average regret bounded away from zero as this allows the optimal rate of estimation error of $\calO(T^{-1/2})$. For such algorithms (which essentially correspond to our \cref{alg:myAlg} with $\beta=1$), these works do provide asymptotically-exact expressions for the estimation error \citep{LjungSystem,bombois2006least, gerencser2009identification,hjalmarsson2009system,wahlberg2010optimal,6425920,stojanovic2014adaptive,stojanovic2016optimal,7574358}. More recent work provides finite-sample bounds on the estimation error of such algorithms, but with conservative constants multiplying the rate \citep{abbasi2011online,simchowitz2018learning,sarkar2019finite,dean2019sample,oymak2019non,sarkar2019finite,khosravi2020nonlinear,sattar2020non,foster2020learning,zheng2020non,sun2020finite}.
The main distinction between our paper and all these related works is that we consider a stepwise-updating, regret-rate-optimal LQAC algorithm and provide characterizations of the regret, estimation error, and prediction error that are asymptotically-\emph{exact}. To achieve these results, our proofs combine recent finite-sample bounds \citep{dean2018regret,mania2019certainty} with martingale central limit theorems developed in the statistics literature \citep{lai1982least,anderson1992asymptotic}.
\subsection{Preliminaries}
\label{section:review on the LQR problem}
We make the following mild assumption on $A$ and $B$, without which \emph{no} algorithm could even achieve finite average regret.
\begin{assumption}[Stability]
\label{asm:InitialStableCondition}
Assume the system is \emph{stabilizable}, i.e., there exists $K_0$ such that the spectral radius (maximum absolute eigenvalue) of $A+BK_0$ is strictly less than 1.
\end{assumption}
\noindent Under \cref{asm:InitialStableCondition}, there is a unique optimal controller that can be computed from $A$ and $B$, given by the linear feedback controller $u_t = K x_t$, where
\begin{equation}
\label{eq:ControllerK}
K = - (R + B^\top P B)^{-1}B^\top P A .
\end{equation}
Here $P$ is the unique positive definite solution to the discrete algebraic Riccati equation (DARE):
\begin{align}
\label{eq:riccati}
P = A^\top P A - A^\top P B (R + B^\top P B)^{-1} B^\top P A + Q
\end{align}
\section{Algorithm}
\label{section: Algorithm}
The algorithm whose performance we characterize in \cref{section: Main Theorem} is given in \cref{alg:myAlg}. At the end of each step in line~\ref{line:controller}, we apply a plug-in version of the LQR controller, $\Kh_tx_t$, plus added exploration noise
that vanishes asymptotically
with variance
$\tau^2t^{-(1-\beta)}\log^\alpha(t)$.
Larger $\beta$ corresponds to more exploration noise, and we will see that $\beta=1/2$ gives the optimal rate of regret and is the only $\beta$ value for which a nonzero $\alpha$ is needed in our theory.\footnote{$\beta = 1$ and $\alpha=0$ would make the added exploration noise non-vanishing and give the optimal rate of system identification estimation error; see \cref{sec:app_ext_beta1} for the extension of our results to the case of $\beta=1$.}
$\Kh_t$ is taken as the solution to the DARE (\cref{eq:ControllerK,eq:riccati}) with inputs $\Ah_{t-1}, \Bh_{t-1}$ computed in line~\ref{line:ols}.
Line~\ref{line:check} then checks whether the state or controller is too large, and if so, $\Kh_t$ is set to $K_0$, which by assumption stabilizes the system. The cutoffs for `too large' are determined by inputs $C_x$ and $C_K$, with the latter assumed to be greater than $\norm{K}$. We will prove (\cref{prop:one_epoch_estimate_withMyalg}) the cutoffs are only breached, and hence $K_0$ applied, finitely often with probability 1, and none of $K_0$, $C_x$, or $C_K$ appear in any of our expressions characterizing the asymptotic performance of \cref{alg:myAlg}. We note that $\Kh_t$ is computed from $\Ah_{t-1}$ and $\Bh_{t-1}$ as opposed to $\Ah_t$ and $\Bh_t$---we expect this to have little impact on the performance but it is needed for the proof of the key \cref{lem:bmsb}.
Since the algorithm asymptotically always just applies a noisy plug-in version of the LQR controller, it is simple, intuitive, and computationally efficient.\footnote{The least squares estimator can be computed efficiently in a recursive manner \citep{engel2004kernel}.} All our theory and experimental results are exactly based on \cref{alg:myAlg} without any modification, and in particular, we always analyze a single trajectory (no independent restarts) and our estimates of $A$ and $B$ are updated \emph{stepwise}, i.e., at every time step. This last point is a significant departure from existing literature which focuses on \emph{logarithmic} updating. We show in \cref{fig:estimation rate one half,fig:estimation rate one half unstable} that updating stepwise reduces regret compared to updating logarithmically often, but in fact our theory also applies to a logarithmically-updated version of \cref{alg:myAlg}, as made precise in the following remark.
\begin{remark}[Logarithmically-updated estimates]
\label{remark: Logarithmically-update estimates}
All our theoretical results in \cref{section: Main Theorem} also hold when $\Ah_t$ and $\Bh_t$ are only updated $\Theta(\log(t))$ times per $t$ steps. More precisely, assume $\{t_i\}_{i=1}^\infty$ are the times at which $\Kh_{t}$ is updated. As long as there exists a constant $C$ such that $\limsup\limits_{i\rightarrow \infty}\frac{t_{i+1}}{t_i} \le C$, all results in \cref{section: Main Theorem} still hold.
\end{remark}
\begin{center}
\begin{algorithm}[ht]
\caption{Stepwise Noisy Certainty Equivalent Control}
\begin{algorithmic}[1]
\REQUIRE{Initial state $x_0$, stabilizing control matrix $K_0$,
scalars $C_{x} > 0$, $C_{K} > \norm{K}$, $\tau^2 > 0$, $\beta \in [1/2,1)$, and $\alpha>3/2$ when $\beta=1/2$.}
\STATE Let $u_0 = K_0x_0 + \tau w_0$ and $u_1 = K_0x_1 + \tau w_1$, with $w_0,w_1\stackrel{iid}{\sim}\calN(0,I_d)$.
\FOR{$t = 2,3,\dots$}
\STATE Compute
\begin{equation}
\label{eq: AhBh estimator}
(\Ah_{t-1}, \Bh_{t-1}) \in \argmin_{(A', B')} \sum_{k=0}^{t-2} \ltwonorm{x_{k+1} - A' x_k - B' u_k}^2
\end{equation}
and if stabilizable, plug them into the DARE (\cref{eq:ControllerK,eq:riccati}) to compute $\Kh_t$, otherwise set $\Kh_t=K_0$.
\label{line:ols}
\STATE If
$\norm{x_{t}} > C_x\log(t)$ or $\norm{\Kh_t} > C_K$, reset $\Kh_t = K_0$.\label{line:check}
\STATE Let
\begin{equation}\label{eq:Myinput}
u_t = \Kh_tx_t + \eta_t,\hspace{1cm} \eta_t = \tau\sqrt{t^{-(1-\beta)}\log^\alpha(t)}\,w_t,\hspace{1cm}w_t \stackrel{iid}{\sim} \calN(0, I_\inputdim)
\end{equation}\label{line:controller}
\ENDFOR
\end{algorithmic}
\label{alg:myAlg}
\end{algorithm}
\end{center}
\begin{comment}
At the end of every time $T$ our method estimates the parameters $(A, B)$ from whole trajectory up to time $T-1$, i.e. (we can actually choose to make estimate only for some of the time steps, but for now let's assume we make a new estimator at each time $T$)
\begin{align}
\label{eq:app_ols}
(\Ah_t, \Bh_t) = \arg\min_{A', B'} \sum_{k=0}^{t-1} \frac{1}{2}\ltwonorm{x_{k+1} - A' x_k - B' u_k}^2
\end{align}
The purpose of this algorithm is to ensure high probability confidence bounds on the estimation $(\Ah_t, \Bh_t, \Kh_t)$. Algorithm~\ref{alg:myAlg} generates the inputs $u_t$ starting from a feedback
controller $ K_0$ which stabilizes the true system $(A, B)$.
Summarize how we get controller $\Kh_t$ by \cref{alg:myAlg}: when the previous state $x_{t-1}$ is crazily large, we explore more by asserting $\Kh_t = K_0$; when $\norm{\Kh_t} > C_K$, which is known to be too large, we re-normalize $\Kh_t$ by $\Kh_t/\norm{\Kh_t}*C_K$; otherwise we generate $\Kh_t$ by $\Ah_{t-1}$ and $\Bh_{t-1}$ using \cref{eq:ControllerK} and \cref{eq:riccati}.
\begin{remark}
In practice we could choose $C_K$ to be very large i.e. 1,000. We set this threshold only for theoretical guarantees. On the other hand, the constraint $C_{x,t}$ is very important and prevents the system from being unstable in the burn-in period where we don't have enough information for the true system.
\end{remark}
\end{comment}
\section{Theoretical results}
\label{section: Main Theorem}
Almost all of our asymptotic results are based on the following new result which shows that the \emph{Gram matrix}
$
\sum_{i=0}^{t-1}
\begin{bmatrix}
x_i\\
u_i\\
\end{bmatrix}
\begin{bmatrix}
x_i\\
u_i\\
\end{bmatrix}^\top
\in \mathbb{R}^{(n+d) \times (n+d)}$ is asymptotically equal in a certain sense to the deterministic matrix $D_tD_t^\top$, where
\begin{equation}
\label{eq:D_t Definition}
D_t :=
t^{\beta/2}\log^{\alpha/2}(t)
\left[
\begin{array}{cc}
I_n & 0\\
K & I_d\\
\end{array}
\right]
\left[
\begin{array}{cc}
C_t^{1/2} & 0\\
0 & \sqrt{\frac{\tau^2}{\beta}} I_d\\
\end{array}
\right]
,\end{equation}
and
\begin{equation*}
C_t = t^{1-\beta}\log^{-\alpha}(t) \sum_{p=0}^{\infty}(A+BK) ^{p}((A+BK) ^{p})^\top \sigma^2 + \frac{\tau^2}{\beta}\sum_{q=0 }^{\infty}(A+BK) ^{q}BB^\top ((A+BK) ^{q})^\top
.\end{equation*}
\begin{thm}\label{thm:main tool}
\cref{alg:myAlg} applied to a system described by \cref{eq:LinearModel} under \cref{asm:InitialStableCondition} satisfies
\begin{equation}
\label{eq: Dt -1 Gram Dt -1}
D_t^{-1}
\sum_{i=0}^{t-1}
\begin{bmatrix}
x_i\\
u_i\\
\end{bmatrix}
\begin{bmatrix}
x_i\\
u_i\\
\end{bmatrix}^\top
(D_t^\top)^{-1} \convP I_{n+d}
.\end{equation}
\end{thm}
The proof of \cref{thm:main tool} can be found at \cref{The proof of thm:main tool}. The main idea was to first prove \cref{eq: Dt -1 Gram Dt -1} under the simplifying approximation that $\Kh_t = K$, and then to derive novel uniform rate bounds on the estimation error $\Kh_t - K$ by extending existing bounds \citep{mania2019certainty,dean2018regret} to the setting of stepwise update. \cref{thm:main tool} is the key ingredient that will allow us to asymptotically exactly characterize many of the important properties of \cref{alg:myAlg}.
\subsection{Parametric expressions}
\label{subsection: Parametric expressions}
We have three different types of asymptotically-exact expressions characterizing the system performance in terms of only the non-random problem parameters (i.e., the algorithm, system, and cost function parameters): the regret (\cref{sec:regretexpr}), the distribution of the estimation error $[\Ah_t-A,\Bh_t-B]$ (\cref{sec:esterrexpr}), and the distribution of the prediction error $(\Ah_tx_t+\Bh_tu_t) - (Ax_t+Bu_t)$ (\cref{sec:prederrexpr}).
\subsubsection{Asymptotically exact expression for the regret (parametric)}\label{sec:regretexpr}
Our first result in fact does not follow from \cref{thm:main tool} but requires instead a careful decomposition of the regret paired with novel rate bounds.
\begin{thm}
\label{thm:regret}
The average regret of the controller $U$ defined by \cref{alg:myAlg} applied through time horizon $T$ to a system described by \cref{eq:LinearModel} under \cref{asm:InitialStableCondition} satisfies, as $T \to \infty$,
\begin{equation}
\label{eq:regret my alg}
\frac{\mathcal{R}(U,T)}{\tau^2\beta^{-1} \Tr(B^\top PB +R)T^{\beta-1}\log^\alpha(T)} \convP 1,
\end{equation}
with $\beta = 1/2$ therefore achieving the optimal rate \citep{simchowitz2020naive} of
$\mathcal{R}(U,T) = \logO_p(T^{-1/2})$.
\end{thm}
The proof can be found at \cref{The proof of thm:regret}.
To our knowledge, this is the first time an LQAC algorithm's regret has been characterized asymptotically \emph{exactly}, i.e., \cref{eq:regret my alg} not only captures the rate but also the constant multiplying that rate. With an exact expression for the asymptotic regret, a user can understand exactly how the regret of \cref{alg:myAlg} depends on the system parameters, and would be able to compare this expression directly with exact expressions for other algorithms (if they existed).
\subsubsection{Asymptotic distribution of the estimation error (parametric)}\label{sec:esterrexpr}
\cref{thm:main tool} provides the key ingredient in a martingale central limit theorem (CLT) for the estimators $\Ah_t, \Bh_t$ \citep{anderson1992asymptotic},
which gives the exact asymptotic distribution of the estimation error in terms of only the system parameters.
\begin{thm}\label{thm:main CLT}
\cref{alg:myAlg} applied to a system described by \cref{eq:LinearModel} under \cref{asm:InitialStableCondition} satisfies, as $t \to \infty$,
\begin{equation}
\label{eq:final Conclusion}
\vvector \left(
\begin{bmatrix}
\Ah_t - A,\Bh_t- B
\end{bmatrix} D_t\right) \convD
\calN(0, \sigma^2 I_{\statedim(n+d)} )
.\end{equation}
\end{thm}
The proof of \cref{thm:main CLT} can be found at \cref{The proof of thm:main CLT}.
Again, to our knowledge, this is the first time an LQAC algorithm's estimation error has been characterized asymptotically \emph{exactly}
and, similarly, such a result can help a user understand exactly how the distribution of the estimation error of \cref{alg:myAlg} depends on the system parameters.
\begin{remark}[A convergence rate disparity]
\label{remark: fast convergence rate}
Plugging the definition of $D_t$ \cref{eq:D_t Definition} into \cref{eq:final Conclusion} gives different convergence rates for two different parts of $[\Ah_t - A, \Bh_t - B]$. In particular, as $t \to \infty$,
\begin{equation}
\label{eq: fast slow rate CLT}
\vvector \left(
\begin{bmatrix}
t^{\beta/2} \log^{\alpha/2}(t)C_t^{1/2}(\Ah_t - A + (\Bh_t- B)K), & \sqrt{\frac{\tau^2}{\beta}}t^{\beta/2} \log^{\alpha/2}(t)(\Bh_t- B)
\end{bmatrix}
\right) \convD
\calN(0, \sigma^2 I_{\statedim(n+d)} ).
\end{equation}
Thus $\Ah_t - A + (\Bh_t- B)K$ converges at the rate of $\left(t^{\beta/2} \log^{\alpha/2}(t)C_t^{1/2}\right)^{-1} = \calO_p(t^{-1/2})$ for any $\beta$, while $\Bh_t- B$ converges at the slower $\beta$-dependent rate of $\calO_p(t^{-\beta/2}\log^{-\alpha/2}(t))$. The faster convergence rate of $\Ah_t - A + (\Bh_t- B)K$ implies strong dependency between $\Ah_t - A$ and $\Bh_t- B$: $\Ah_t - A \approx -(\Bh_t- B)K$.
\end{remark}
\begin{remark}[Regret-estimation trade-off]
Because of the asymptotic linear relationship $\Ah_t - A \approx -(\Bh_t- B)K$, the estimation error $[\Ah_t - A, \Bh_t - B]$ can be characterized by the asymptotic variance of $\vvector[\Bh_t- B]$: $\frac{\beta\sigma^2}{\tau^2}t^{-\beta}\log^{-\alpha}(t) I_{nd}$. Combining this with \cref{thm:regret} gives the following asymptotic identity that precisely characterizes a fundamental regret-estimation trade-off for \cref{alg:myAlg} with any $\beta$: as $t \to \infty$,
\begin{equation*}
t\mathcal{R}(U,t) \cdot \Cov(\vvector(\Bh_t- B)) \convP \Tr(B^\top PB +R) \sigma^2 I_{nd}
.\end{equation*}
\end{remark}
Because $K$ is a function of $[A, B]$ (and asymptotically, $\Kh_t$ is the same function of $[\Ah_{t-1}, \Bh_{t-1}]$), by the Delta method, we can use its matrix of derivatives $\frac{dK}{d[A, B]} := \frac{d \,\vvector(K)}{d \,\vvector([A, B])} \in \mathbb{R}^{nd \times n(n+d)}$ to translate the asymptotic distribution of $[\Ah_t - A, \Bh_t - B]$ from \cref{thm:main CLT} to the asymptotic distribution of $\Kh_t - K$.
\begin{corr}
\label{corr: K CLT parametric}
Assume $A+BK$ is full rank. Then \cref{alg:myAlg} applied to a system described by \cref{eq:LinearModel} under \cref{asm:InitialStableCondition} satisfies, as $t \to \infty$,
\begin{equation}
\label{eq: K CLT parametric}
\sqrt{\frac{\tau^2}{\sigma^2\beta}}t^{\beta/2} \log^{\alpha/2}(t)
\left(\left(\frac{dK}{d[A,B]}\right)
\left(\begin{bmatrix}
-K^\top \\
I_d
\end{bmatrix}
\otimes
I_n
\right)
\right)^{-1}
\vvector\left(\Kh_t - K\right)
\convD
\calN(0, I_{nd})
.\end{equation}
\end{corr}
The proof of \cref{corr: K CLT parametric} can be found at \cref{subsection: The proof of K CLT parametric}. \cref{eq: K CLT parametric} quantifies the distance from the current control matrix $\Kh_t$ to the optimal control matrix $K$,
and shows implicitly but asymptotically exactly how the distribution of that distance depends on the system dynamics.
\subsubsection{Asymptotic distribution of the prediction error (parametric)}\label{sec:prederrexpr}
If we consider the entire history $\{x_i, u_i\}_{i=0}^{t}$ to be the input of the prediction rule whose goal is to predict the next state $x_{t+1}$, then the optimal (in terms of mean squared error) prediction is given by $\E [x_{t+1} \,|\, \{x_i, u_i\}_{i=0}^{t} ] = Ax_t + Bu_t$, and a natural choice at time $t$ would be to use the least-squares prediction rule given by $\Ah_t x_t + \Bh_t u_t$.
By combining \cref{thm:main CLT}'s asymptotic distribution for $[\Ah_t - A, \Bh_t- B]$ with a careful handling of the asymptotic dependence between $(x_t, u_t)$ and $[\Ah_t - A, \Bh_t- B]$, we can derive the asymptotic distribution of the error $\Ah_tx_t + \Bh_t u_t - (Ax_t + Bu_t)$ of the least-squares prediction rule.
\begin{thm}\label{thm:prediction CLT parametric}
\cref{alg:myAlg} applied to a system described by \cref{eq:LinearModel} under \cref{asm:InitialStableCondition} satisfies, as $t \to \infty$,
\begin{align}
\label{eq:prediction CLT parametric}
\left(
x_t^\top\left( \sum_{p=0}^{\infty}(A+BK) ^{p}
\left((A+BK) ^{p}\right)^\top\right)^{-1}x_t
+
\beta \sigma^2
\lnorm{w_t}^2
\right)^{-1/2}
t^{1/2} \left((\Ah_t - A)x_t + (\Bh_t- B)u_t\right)
\convD \calN(0,I_n).
\end{align}
\end{thm}
The proof of \cref{thm:prediction CLT parametric} can be found at \cref{The proof of thm:prediction CLT parametric}. This expression is parametric in the sense that the first parenthetical only depends on the system parameters and the random variables $x_t$ and $w_t$ that are used by the algorithm in the time step immediately before the prediction is made.
Note that the convergence rate of $\logO_p(t^{-1/2})$ does not depend on $\beta$, as foreshadowed by \cref{remark: fast convergence rate}, but the constant in the convergence does depend on $\beta$.
Thus, \cref{eq:prediction CLT parametric} shows that the optimal asymptotic prediction error is attained at $\beta = 1/2$ ($x_t$'s asymptotic distribution does not depend on $\beta$, so asymptotically the only $\beta$ dependence is in the term $\beta\sigma\|w_t\|^2$), a conclusion we could not have reached had we only considered the rate.
\cref{thm:prediction CLT parametric} can easily be extended to characterize the full prediction error of $x_{t+1} - (\Ah_tx_t + \Bh_t u_t)$ by simply adding $\sigma^2$ to the first parenthetical.
\subsection{Observable expressions}
\label{subsection: Observable expressions}
The previous subsection provides three asymptotically-exact expressions (regret, estimation error, and prediction error) in terms of only the system parameters; in this subsection, we provide three analogous asymptotically exact expressions in terms of only observable random variables.
\subsubsection{Asymptotically exact expression for the regret (observable)}
Define $\Ph_t$ as the plug-in estimator using \cref{eq:riccati}:
\begin{align*}
\Ph_t = \Ah_t^\top \Ph_t \Ah_t - \Ah_t^\top \Ph_t \Bh_t (R + \Bh_t^\top \Ph_t \Bh_t)^{-1} \Bh_t^\top \Ph_t \Ah_t + Q.
\end{align*}
Then by consistency of $\Ah_{t}$ and $\Bh_{t}$ (see \cref{thm:main CLT}), and therefore also $\Ph_t$, the plug-in version of \cref{eq:regret my alg} is an immediate corollary of \cref{thm:regret}.
\begin{corr}
\label{corr:regret}
The average regret of the controller $U$ defined by \cref{alg:myAlg} applied through time horizon $T$ to a system described by \cref{eq:LinearModel} under \cref{asm:InitialStableCondition} satisfies, as $t \to \infty$ and $T \to \infty$,
\begin{equation}
\label{eq:regret my alg observable}
\frac{\mathcal{R}(U,T)}{\tau^2\beta^{-1} \Tr(\Bh_{t}^\top \Ph_t \Bh_{t} +R)T^{\beta-1}\log^\alpha(T)} \convP 1
.\end{equation}
\end{corr}
The proof of \cref{corr:regret} can be found at \cref{The proof of corr:regret}.
Notice when $t \le T$, \cref{corr:regret} tells us that we can consistently estimate the regret at a future time point. Furthermore, the Delta method applied to \cref{thm:main CLT} gives the asymptotic distribution of the denominator in \cref{eq:regret my alg observable}.
\subsubsection{Asymptotic distribution of the estimation error (observable)}
Combining the asymptotic equivalence of Gram matrix and $D_tD_t^\top$ from \cref{thm:main tool}, the asymptotic distribution of the estimation error from \cref{thm:main CLT}, and Slutsky's theorem immediately produces the following very useful corollary.
\begin{corr}\label{thm:main}
\cref{alg:myAlg} applied to a system described by \cref{eq:LinearModel} under \cref{asm:InitialStableCondition} satisfies
\begin{equation*}
\Tr\left(
\begin{bmatrix}
\Ah_t - A,\Bh_t- B
\end{bmatrix}
\sum_{i=0}^{t-1}
\begin{bmatrix}
x_i\\
u_i\\
\end{bmatrix}
\begin{bmatrix}
x_i\\
u_i\\
\end{bmatrix}^\top
\begin{bmatrix}
\Ah_t - A,\Bh_t- B
\end{bmatrix} ^\top
\right)
\convD \sigma^2\chi^2_{n(n+d)}
.\end{equation*}
\end{corr}
The proof of \cref{thm:main} can be found at \cref{The proof of thm:main}. The reason it is useful is it allows us to construct an asymptotically exact ellipsoidal confidence region for the system dynamics $A$ and $B$. In particular, the following confidence region has asymptotic coverage exactly $1-\alpha$ and is entirely and efficiently computable from data observable through time $t$:
\begin{equation}
\label{eq: ellipsoid 2}
\left\{
A, B \,:\,
\sigma^{-2}
\Tr\left(
\begin{bmatrix}
\Ah_t - A,\Bh_t- B
\end{bmatrix}
\sum_{i=0}^{t-1}
\begin{bmatrix}
x_i\\
u_i\\
\end{bmatrix}
\begin{bmatrix}
x_i\\
u_i\\
\end{bmatrix}^\top
\begin{bmatrix}
\Ah_t - A,\Bh_t- B
\end{bmatrix} ^\top
\right)
\le \chi^2_{n(n+d), 1-\alpha}
\right\},
\end{equation}
where $\chi^2_{n(n+d), 1-\alpha}$ is the $1-\alpha$ quantile of a $\chi^2_{n(n+d)}$ random variable. To our knowledge, this is the first asymptotically exact confidence region for the system dynamics in the LQAC problem.
Note the confidence region in \cref{eq: ellipsoid 2} is identical to the confidence region one would compute if the data points $\{x_i, u_i\}_{i=0}^{t-1}$ were i.i.d., but the theory that led us to this result is far more challenging than in the i.i.d. setting.
Analogously to \cref{corr: K CLT parametric}, we can also use the Delta method to derive a confidence region for $K$.
\begin{corr}
\label{corr: K confidence region}
Assume $A+BK$ is full rank. Then \cref{alg:myAlg} applied to a system described by \cref{eq:LinearModel} under \cref{asm:InitialStableCondition} satisfies
\begin{equation*}
\label{eq: K confidence region}
\vvector(
\Kh_t - K
) ^\top
\left(
\left(\frac{dK}{d[A, B]}\right)_t
\left(
\sum_{i=0}^{t-1}
\begin{bmatrix}
x_i\\
u_i\\
\end{bmatrix}
\begin{bmatrix}
x_i\\
u_i\\
\end{bmatrix}^\top
\otimes I_n\right)^{-1}
\left(\frac{dK}{d[A, B]}\right)_t^\top
\right)^{-1}
\vvector(
\Kh_t - K
)
\convD \sigma^2\chi^2_{nd}
,\end{equation*}
where $\left(\frac{dK}{d[A, B]}\right)_t \in \mathbb{R}^{nd \times n(n+d)}$ is defined as $\frac{dK}{d[A, B]}$ evaluated at $\Ah_{t-1}, \Bh_{t-1}$.
\end{corr}
The proof of \cref{corr: K confidence region} can be found at \cref{subsection: The proof of K confidence region}.
\cref{corr: K confidence region} gives the following asymptotically exact ellipsoidal $1-\alpha$ confidence region for $K$:
\begin{equation*}
\left\{
K \,:\,
\sigma^{-2}
\vvector(
\Kh_t - K
) ^\top
\left(
\left(\frac{dK}{d[A, B]}\right)_t
\left(
\sum_{i=0}^{t-1}
\begin{bmatrix}
x_i\\
u_i\\
\end{bmatrix}
\begin{bmatrix}
x_i\\
u_i\\
\end{bmatrix}^\top
\otimes I_n\right)^{-1}
\left(\frac{dK}{d[A, B]}\right)_t^\top
\right)^{-1}
\vvector(
\Kh_t - K
)
\le
\chi^2_{nd, 1-\alpha}
\right\}.
\end{equation*}
\subsubsection{Asymptotic distribution of the prediction error (observable)}
We can obtain an observable expression for the asymptotic distribution of the prediction error as a direct corollary of \cref{thm:main tool,thm:prediction CLT parametric}.
\begin{corr}
\label{thm: prediction CLT}
\cref{alg:myAlg} applied to a system described by \cref{eq:LinearModel} under \cref{asm:InitialStableCondition} satisfies:
\begin{align*}
\begin{split}
\left(
\sigma^2
\begin{bmatrix}
x_t \\
u_{t}
\end{bmatrix}^\top
\left(
\sum_{i=0}^{t-1}
\begin{bmatrix}
x_i\\
u_i\\
\end{bmatrix}
\begin{bmatrix}
x_i\\
u_i\\
\end{bmatrix}
^\top
\right)^{-1}
\begin{bmatrix}
x_t \\
u_{t}
\end{bmatrix}
\right)^{-1/2} \left((\Ah_{t} - A)x_t +
(\Bh_{t}- B)u_{t}\right)
\convD \calN(0, I_n)
.\end{split}
\end{align*}
\end{corr}
The proof can be found in \cref{The proof of thm: prediction CLT}, and is a special case of a more general result that allows the users to choose their own desired input by replacing $u_t = \Kh_t x_t + \eta_t$ with $u_t = \Kh_t x_t + \xi_t$ for any $\xi_t$ constant or independent of the data. Again, \cref{thm: prediction CLT} can easily be extended to characterize the full prediction error of $x_{t+1} - (\Ah_tx_t + \Bh_t u_t)$ by simply adding $\sigma^2$ to the first parenthetical, leading to the following prediction region:
\begin{equation}
\label{eq: prediction region observable}
\left\{
x_{t+1} \,: \,
\sigma^{-2}
\left(
1+
\begin{bmatrix}
x_t \\
u_{t}
\end{bmatrix}^\top
\left(
\sum_{i=0}^{t-1}
\begin{bmatrix}
x_i\\
u_i\\
\end{bmatrix}
\begin{bmatrix}
x_i\\
u_i\\
\end{bmatrix}
^\top
\right)^{-1}
\begin{bmatrix}
x_t \\
u_{t}
\end{bmatrix}
\right)^{-1}
\norm{(\Ah_t x_t + \Bh_t u_t) - x_{t+1}}^2
\le \chi^2_{n, 1-\alpha}
\right\}.
\end{equation}
Having at each time $t$ a computable region with a high probability of containing the next state $x_{t+1}$ is a crucial ingredient in ensuring the \emph{safety} of a learning system, as it both provides a warning about where the system will be next and gives the system the opportunity to change or cancel the control $u_t$ if the prediction region intersects an unsafe part of the state space.
As an additional application of the prediction region \cref{eq: prediction region observable}, since $x_{t+1}$ is observed at the next time step, we can use the agreement between our prediction region and the true $x_{t+1}$ to test certain assumptions about our system. For instance, the hypothesis test which rejects if $x_{t+1}$ does not fall within the prediction region constructed at time $t$ constitutes a asymptotically valid level-$\alpha$ test of our stationary linear dynamics encoded in \cref{eq:LinearModel}. For instance, if we are confident about the linearity of our system but worried that it may be non-stationary, we could use this test to detect whether the dynamics have changed within the first $t+1$ time steps, and more generally, such tests could be strung together to constitute a change detection algorithm \citep{grunwald2019safe,wang2020online}.
Note that the naive prediction region
\begin{equation}
\label{eq: prediction region observable simple}
\left\{
x_{t+1} \,: \,
\sigma^{-2}
\norm{(\Ah_t x_t + \Bh_t u_t) - x_{t+1}}^2
\le \chi^2_{n, 1-\alpha}
\right\}.
\end{equation}
also has asymptotically exact coverage even though it ignores the estimation error in $[\Ah_t,\Bh_t]$. However, our experiments show that our prediction region from \cref{eq: prediction region observable} achieves much better finite-sample coverage by accounting for the estimation error of $[\Ah_t,\Bh_t]$; see \cref{fig:Prediction Coverage True one half}.
\section{Experiments}
\label{section: Experiments}
We verify our algorithm's performance in one stable and one unstable dynamical system.
We focus on comparing the finite sample performance of our algorithm to our theoretical predictions, and defer comparison between our algorithm and other existing algorithms for future work (see \citet{dean2018regret} for a comparison between an algorithm similar to our algorithm except it updates $\Kh_t$ logarithmically often and other algorithms which we will review in \cref{subsection: Background}). In the main text, we will only display the figures with $\beta = 1/2$ and $\alpha = 2$ in the stable system; the remaining figures and details of the experimental setup can be found in \cref{section: Additional Experiments}. \footnote{Source code for reproducing our results can be found at \url{https://github.com/Feicheng-Wang/LQAC_code}.}
\subsection{A representative simulation}
\label{subsection: Partial experiment result}
\cref{fig:Summary stable system} summarizes the results of our experiment with $\beta = 1/2$ and $\alpha = 2$ in a stable system (for the analogous figure in an unstable system see \cref{fig:Summary unstable system}). The main takeaways are:
\begin{itemize}
\item \cref{fig:Compare Regret one half} shows that \cref{alg:myAlg}'s stepwise update leads to lower regret than update logarithmically often, although the difference is small compared with the variability of the regret. The difference is qualitatively similar but quantitatively larger in the unstable system, and the difference can be quite large for poor choices of $K_0$, but pretty robust for choices of $C_K$; see \cref{fig:Compare Regret one half unstable,fig:log regret three choices unstable system,fig:log regret CK}.
\item \cref{fig:regret ratio one half} verifies that the ratio of the true observed regret with either of our regret expressions in \cref{thm:regret} and \cref{corr:regret} is converging to 1. Note that the large confidence band is due to the huge variance in the regret itself. The analogous plots for $\beta \neq 1/2$ and the unstable system can be found at \cref{fig:regret ratio one half unstable,fig:regret_stable_unstable}; larger $\beta$ speeds up the convergence speed.
\item \cref{fig:estimation rate one half} verifies the convergence rate disparity in \cref{remark: fast convergence rate} that $\Ah_t-A$, $\Bh_t-B$, and $\Kh_t -K$ have a slow convergence rate $\logO(t^{-\beta/2})$, while $\Ah_t-A + (\Bh_t-B) K$ has a fast convergence rate $\logO(t^{-1/2})$
; see \cref{fig:estimation rate one half unstable}.
\item
\cref{fig:Coverage one half} shows that, the finite sample coverage of our confidence regions and prediction region closely matches our asymptotic theory in \cref{thm:main,corr: K confidence region,thm: prediction CLT}.
Also \cref{fig:Prediction Coverage True one half} shows that our prediction region \cref{eq: prediction region observable} have better finite sample coverage than the naive region \cref{eq: prediction region observable simple}. In this simulation, the observable expressions have slightly better coverage.
Similar results hold for other choices of $\beta$ and the unstable systems (\cref{fig:Coverage one half unstable,fig:log regret no Cx}).
\end{itemize}
\begin{figure}[H]
\centering
\begin{subfigure}{.45\textwidth}
\captionsetup{justification=centering}
\caption{Benefit of stepwise updates}
\includegraphics[width = \linewidth]{fig/paper_plot_all_one_half/control_experiment_1_CK_5/Compare_Regret.pdf}
\label{fig:Compare Regret one half}
\end{subfigure}
\begin{subfigure}{.45\textwidth}
\caption{Regret Ratio}
\includegraphics[width = \linewidth]{fig/paper_plot_all_one_half/control_experiment_1_CK_5/Regret_ratio2.pdf}
\label{fig:regret ratio one half}
\end{subfigure}
\begin{subfigure}{.45\textwidth}
\captionsetup{justification=centering}
\caption{Differing Convergence Rates}
\includegraphics[width = \linewidth]{fig/paper_plot_all_one_half/control_experiment_1_CK_5/estimation_rate.pdf}
\label{fig:estimation rate one half}
\end{subfigure}
\begin{subfigure}{.45\textwidth}
\caption{Confidence Region Coverage}
\includegraphics[width = \linewidth]{fig/paper_plot_all_one_half/control_experiment_1_CK_5/Coverage.pdf}
\label{fig:Coverage one half}
\end{subfigure}
\begin{subfigure}{.45\textwidth}
\caption{Prediction Region Coverage}
\includegraphics[width = \linewidth]{fig/paper_plot_all_one_half/control_experiment_1_CK_5/Safety_Coverage_True.pdf}
\label{fig:Prediction Coverage True one half}
\end{subfigure}
\caption[]{\textit{(See next page for caption)}}
\end{figure}
\begin{figure}[H]
\ContinuedFloat
\caption{Summary of 1000 independent experiments applying \cref{alg:myAlg} with $\beta = 1/2$, $\alpha = 2$, $C_x = 1$, and $C_K = 5$ on the stable system described in \cref{subsection: easy setting}. (a) Difference between the regret of \cref{alg:myAlg} using stepwise and logarithmic updates.
(b) The ratio of the empirical regret and our parametric or observable expressions for the regret.
(c) The average Frobenius norm of various estimation errors considered in this paper, with slopes fitted on a log-log scale so that the estimation error is $\logO(t^{\text{slope}})$.
The effect of $\alpha$ was removed from the slopes of $\Kh_t-K$ and $[\Ah_t-A,\Bh_t-B]$ by dividing the error by $\log^{\alpha/2}(t)$.
(d) Coverage of our 95\% confidence regions for $[A,B]$, $K$, and $\E [x_{t+1} \,|\, \{x_i, u_i\}_{i=0}^{t} ] = Ax_t + Bu_t$.
(e) Coverage of our 95\% prediction region for $ x_{t+1} \,|\, \{x_i, u_i\}_{i=0}^{t}$, along with coverage of the naive prediction region given in \cref{eq: prediction region observable simple}.
}
\label{fig:Summary stable system}
\end{figure}
\section{Detailed review of related work}
\label{subsection: Background}
The LQAC problem lies at the intersection of adaptive control and reinforcement learning and has drawn considerable attention in the past decade. This line of work differs from much of the work in reinforcement learning that is based on games or other virtual simulators that can be rerun infinitely many times (\citet{vinyals2017starcraft}, \citet{silver2017mastering}) because it is run in one-shot. However, many real-world applications cannot be easily restarted over and over again, and repeating experiments can be prohibitively expensive. Aside from the CE approach taken in this paper and reviewed in \cref{subsection: Related Works}, we classify LQAC algorithms into two broad categories:
\begin{itemize}
\item \textbf{Optimism in the Face of Uncertainty}: This method uses non-convex optimization to repeatedly select a near optimal control (in the regret sense) from a confidence set, achieves the optimal rate of regret \citep{abbasi2011regret,ibrahimi2012efficient,faradonbeh2017finite}.
Later \citet{cohen2019learning} extended this work by replacing non-convex optimization with semi-definite programming and still achieves the optimal regret.
\item \textbf{Thompson Sampling}:
Starting with a prior distribution for the system parameters, one can use Bayes' rule to update a posterior distribution online and can use samples from that posterior to choose controls that balance exploration and exploitation.
The pioneering work \citep{abeille2017thompson} applying this idea to LQAC demonstrated a suboptimal $\logO(T^{-1/3})$ average regret, which is later improved to the optimal rate $\logO(T^{-1/2})$ by \citet{ouyang2017learning,faradonbeh2018optimality}. \citet{abeille2018improved} is the only work which we know of that achieves the optimal rate with stepwise updates, although their proofs only apply in scalar systems (i.e., $n=1$).
\end{itemize}
\paragraph{Logarithmic Regret}
We pause here to clarify that any result achieving logarithmic regret is in a different setting from ours (in our setting, a lower bound of $\logO(T^{-1/2})$ was proven in \citet{simchowitz2020naive}). For example, when the system parameters $A$ and $B$ are known or partially known, a logarithmic rate of regret is achievable due to the extra information in $A$ and $B$ which allows faster estimation of $K$ \citep{foster2020logarithmic,cassel2020logarithmic}. Or, when the states are only partially observed, although the controller receives less information, the optimal controller also has less information, which turns out to allow a logarithmic rate of regret \citep{lale2020logarithmic,tsiamis2020online}. As a final example, when the cost is not an explicit function of the controls $u_t$, a logarithmic rate of regret is achievable using a controller called a self-tuning regulator, which is similar to our certainty equivalent controller
except that it targets a different optimal controller $U^*$ (because the cost function is different) and applies constant size probing steps logarithmically often \citep{lai1986extended,lai1986asymptotically,guo1991astrom,guo1995convergence}.
\paragraph{Sequential Analysis and Time Series}
Establishing asymptotic normality is common in sequential analysis \citep{lai2001sequential} and time series or state space model analysis \citep{kohn1986prediction,pedroni2004panel}, but the focus in these fields is on stationary and Markovian time series (although we assume our system is stabilizable, the data generated by applying our adaptive controller to that system is non-Markovian and non-stationary as the controller depends on the whole history) and on simpler forms of dependence than we consider.
\begin{comment}
Since we are going to use this DARE to solve for our controller $K$, we need to specify the condition when \cref{eq:riccati} has unique positive definite solution. A summary paper can be found here \citep{payne1973discrete}. We use $\rho(A)$ to define the spectral radius of square matrix $A$, which is the largest absolute eigenvalue of matrix $A$.
\begin{defn}[Definition 2.1 in \citep{payne1973discrete}
\label{def:stabilizable}
The pair $(A,B)$ is stabilizable if there exists a matrix $F$ such that $A+BF$ is asymptotically stable. A matrix $A$ is asymptotically stable if $\rho(A) < 1$.
\end{defn}
\begin{defn}[Definition 2.3 in \citep{payne1973discrete}]
\label{def:perfectly observable}
There exists matrix $C$ and $D$ such that
\[C^\top C = Q, D^\top D = R\]
Further define
\[y_k = Cx_k + Du_k\]
Then the pair $(A,B)$ is perfectly observable if $y_k = 0$, $k = 0, 1, \cdots, n-1$ implies $x_0 = 0$.
\end{defn}
Here I cite Theorem 2.1 in \citep{payne1973discrete}.
\begin{thm}[Theorem 2.1 in \citep{payne1973discrete}]
\label{thm:stabilizable iff unique}
Assuming $Q$ and $R$ are positive definite matrix, the pair $(A,B)$ is stabilizable and perfectly observable if and only if there exists a unique positive definite solution $P$ of the DARE, and for which $A+BK$ is asymptotically stable. Here $K = -(R + B^\top PB )^{-1}B^\top PA$.
\end{thm}
That is to say, if we have stabilizable estimates $(\Ah_t, \Bh_t)$ of $(A,B)$, the unique positive definite solution of
\begin{equation}
\label{eq:riccati_hat}
\Ph_t=\Ah_t^{\top}\Ph_t\Ah_t-(\Ah_t^{\top}\Ph_t\Bh_t)\left(R+\Bh_t^{\top}\Ph_t\Bh_t\right)^{{-1}}(\Bh_t^{\top}\Ph_t\Ah_t)+Q.
\end{equation}
should be unique, and then we can define our certainty equivalent controller $\Kh_t$ with
\begin{equation}
\label{eq:controllerGenerator}
\Kh_{t+1}=-(R+\Bh_t^{\top}\Ph_t\Bh_t)^{{-1}}(\Bh_t^{\top}\Ph_t\Ah_t)\,
\end{equation}
\end{comment}
\begin{comment}
getting down to beta=1/2 and alpha=0 by characterizing the O(1/sqrt(n)) terms that we currently can ignore
simplifying the algorithm in the three ways it deviates from the simplest certainty-equivalent controller, or showing that some of these deviations are actually necessary for some aspect(s) of performance
showing some lower-bound on how well it is possible to do at regret and/or system identification, and seeing whether our result matches that lower-bound
getting results that account for the dimensionality of the system, perhaps by allowing the dimension to increase with T (the high-dimensional analogue of what we're doing here)
how to incorporate this to impose a safety constraint using our prediction regions. At the very least this involves dealing with the "overfitting" that occurs when you consider an action but reject it due to its prediction region intersecting an obstacle, and then have to construct another prediction region for the next action you consider and actually have it be valid
dealing with non-stationarity, both in terms of coming up with a formal way to detect it and in terms of changing your controller in the presence of it
getting coverage results that are non-asymptotic and/or "always-valid" or uniform over time
posing and considering solutions to the "safe LQR" problem where you add a penalty for cumulative conditional variability in the next state to the cost function. It seems any rate-optimal solution would require a solution to item 1 above
getting similar results for other algorithms like thompson sampling
getting similar results for LQG
getting similar results for nonlinear systems
extending these results to very similar algorithms, such as allowing I_d in the covariance matrix of \eta_t to be replaced by some other matrix, or even a matrix that depends on the history up to that point (similar to TS?)
Non-stationary system where the changing part is known
\end{comment}
\section{Discussion}
This paper's main contributions are asymptotically exact expressions for the regret and the distributions of the estimation and prediction errors of a stepwise updating noisy certainty equivalent control algorithm in terms of either the system parameters or observable random variables. These results improve the field's understanding of the LQAC problem and open up a number of new research directions:
\begin{comment}
We achieved the following three contributions at the same time with one single algorithm:
\begin{enumerate}
\item The first asymptotic normality result (implies exact inference and prediction) for system parameters estimation. Since the inference requires only the observable history,
This is directly applicable without any prior knowledge on the system parameters.
The importance of these discoveries are three folds:
\begin{enumerate}
\item Safety Guarantee. In autonomous control (i.e. robots, vehicles, quadrotors), it is extremely important to avoid collision. A tight prediction region allow us to evaluate the probability of collision of a given action, a necessary primitive for any safe reinforcement learning algorithm.
\item Non-stationarity detection. In practice, the changing environment can lead to different system dynamics, and we'd like to detect such change as soon as possible — having confidence regions implicitly allows us to perform hypothesis tests of the system parameters. If the null assumption (system does not change) is rejected, we should discard some old observations and refit the system with more recent data.
, so if at one point we thought they had some value but after a while we reject the hypothesis that they still have that value, we need to do something new in our learning algorithm (i.e. discard some obsolete observations).
Transfer learning
\item System identification and transfer learning. Because operating the system can be expensive, it is usually valuable to transfer what we had learned to other similar systems. More precisely, with a tight confidence region, we could make best utilization of the current information to infer our true underlying system, and get a general sense about the uncertainty of our estimators. This could offer a good starting point if we ever want to run similar systems again, and thus, reduce our regret and increase safety guarantee.
If we ever want to use one agent to reinforcement learn a system and apply what it has learned to other systems, then it would be very useful to have accurate uncertainty quantification of the learned system to assess how well those new systems will likely perform.
\end{enumerate}
We'll discuss them in more detail after we introduced our main theorem (\cref{section: Main Theorem}).
\item The first trajectory independent exact expression for the dominating term of regret when $\beta \ge 1/2$. As far as we know, there's still no such exact expression on the regret in existing literature, and only knowing the order of regret is apparently not enough for practitioners. This offers us a prior knowledge of regret before actually running the algorithm, which can be extremely valuable for practitioners with limited budget.
Especially when the system is non-stationary, our algorithm can swiftly detect such changes with the support of our tight confidence region, which is not even possible for algorithm with only $\log(T)$ updates.
\end{enumerate}
\end{comment}
\begin{enumerate}
\item \textbf{Theoretical improvements}. Our simulations support our suspicion that all of our results except for \cref{thm:regret} and \cref{corr:regret} hold under more general version of \cref{alg:myAlg} that allows $\beta = 1/2$ and $\alpha = 0$, the summation in Line \ref{line:ols} to go up to $t-1$, and the removal of Line \ref{line:check}. We expect such extensions to require significantly stronger theoretical machinery, and we hope that future work will prove these extensions and analogues to \cref{thm:regret} and \cref{corr:regret} which account for an expected additional term of order $\calO(T^{-1/2})$.
\item \textbf{Safe reinforcement learning}.
Existing work in safe reinforcement learning relies heavily on prediction regions derived from Bayesian inference \citep{berkenkamp2017safe,koller2018learning}. Our \cref{thm: prediction CLT} provides a tight frequentist asymptotic prediction region that, unlike Bayesian inference, does not assume a prior on the system parameters, providing a potential starting point for new safe reinforcement learning algorithms.
\item \textbf{Non-stationarity reinforcement learning}. As mentioned in the last paragraph of \cref{section: Main Theorem}, our prediction region can be used for change point detection in non-stationary systems. Many existing work designed for reinforcement learning algorithms in the non-stationary environment relies on some form of change point detection, although they focus on discrete state and action spaces \citep{da2006dealing,auer2009near,padakandla2019reinforcement}. Thus, our work may be useful for designing new reinforcement learning algorithms in non-stationary settings with continuous state and action spaces.
\end{enumerate}
\section*{Acknowledgements}
We are grateful to Na Li, Haoyi Yang, and Yue Li for helpful discussions regarding this project.
| {
"redpajama_set_name": "RedPajamaArXiv"
} | 9,332 |
Political Renegades: Gov. Ventura and Willie Nelson challenge two-party stranglehold
Ventura: Patriots must hold government's 'feet to the fire' or else it goes bad
Nelson: Government should have bailed out family farms, not Wall Street
Aaron Dykes
Tuesday, Oct 28, 2008
RELATED: Exclusive photos of Jesse Ventura, Alex Jones and Willie Nelson at the music icon's world headquarters.
IN EXCLUSIVE VIDEO, legendary music icon Willie Nelson meets up with former Governor Jesse Ventura and Alex Jones to discuss the implications of the 2008 Election, the effects of widespread economic downturn and lingering questions about 9/11.
Gov. Ventura hopes to challenge the 'horse-race' voting mentality that has made both parties unresponsive to the people.
"If we break the trend of electing Democrats and Republicans, and start electing independents, we will get out of the problem we're in," Ventura told News 8 Austin, urging people to vote their 'heart' and 'conscience' outside The Backyard, where Willie Nelson performed at the venue's final event.
Jesse Ventura and Alex Jones addressed the need to dissent against the status quo and reminded those watching online that the First Amendment is meant to protect even unpopular speech.The former Minnesota Governor said people cannot be afraid to criticize the government. He insisted that patriots must 'hold government's feet to the fire'– otherwise, bad government will result.
Ventura was a political rarity, as an unexpected independent victor in Minnesota's 1998 Gubernatorial race, a clear threat to the two-party dictatorship that dominates the political landscape.
Willie Nelson lambasted the government's corrupt bailout of Wall Street while the traditional family farm continues to erode away in the wake of economic downturn, allowing corporate farming to further dominate.
Last week, the ardent Farm Aid activist challenged Hank Paulson and Ben Bernanke to bailout the family farm, instead.
"I have written a letter as president of Farm Aid to whoever controls that money that we would like to have $1 billion dollars from that 700….$1 billion dollars is I think all we'll need to turn the economy around if we start out taking care of our family farmers," Nelson told The Alex Jones Show.
Former Gov. Ventura, Author of 'Don't Start the Revolution Without Me' also joined Alex in the car where he explained the problem with the fact that many government agencies simply refuse to discuss the official 9/11 story.
Ventura criticized government agencies for putting out reports and being unwilling to defend them, and simply refusing to answer any questions.
"If they're not hiding something– let me put it this way– they're certainly not cooperative." If government reports are truthful, Ventura challenges, "then why does the government behave in a manner that makes them look guilty of something and then allows conspiracy theorists the foothold to go with that."
Governments, Ventura says, simply issue their reports and take no questions from anyone. And that just can't stand.
Unaccounted government will only lead to greater abuses, which in turn will demand greater resistance on the part of its people. Ventura and Nelson, veteran renegades of the 60s and 70s, say they've seen the history cycle through more than once already.
They only hope people now awake to government abuses can stay awake long enough to make a difference this time around.
Both Gov. Ventura and Willie Nelson have appeared on the Alex Jones Show to discuss their doubts about the official 9/11 story, particularly the apparently inconsistent manner in which the Twin Towers and WTC 7 collapsed after two planes struck the World Trade Center on 9/11. http://www.prisonplanet.com/political-renegades-gov-ventura-and-willie-nelson-challenge-two-party-stranglehold.html | {
"redpajama_set_name": "RedPajamaCommonCrawl"
} | 1,199 |
Category Archives: Sisters of Charity of St. Joseph's
Voices of Students
"Soon scenes may change, soon friends prove untrue
When for I rove, from dear Saint Josephs view
Yet naught can chase, thy image from my mind"
-From "Farewell to St. Joseph's," 1830, by "Remember Maria"
This is part of a yearlong series about the early days of the Sisters of Charity of St. Joseph's commemorating the 200th anniversary of the death of Saint Elizabeth Ann Seton, foundress of the community. In 1850, the Emmitsburg-based Sisters united with the international community of the French Daughters of Charity.
In many archives that gather administrative or corporate records, the voices of the people who used those services can be frustratingly absent. Luckily, the collection of St. Joseph's Academy, the school founded by Mother Seton and the Sisters of Charity of St. Joseph's, features many of the voices of students, including their creations and their writings.
Some of the earliest examples of students' surviving works include their needlework samplers, several of which come from the era when Mother Seton directly ran the Academy before her death in January 1821.
Needlework by Mary Jamison, July 1812, St. Joseph's Academy Student, 1810-1813
While the number of surviving student mementos from Mother Seton's era are limited, they become more voluminous as time goes on until St. Joseph's Academy received a charter as St. Joseph College in 1902. Other markers of the students' work include "premiums," which are what would be called in the modern day certificates of achievement or small awards for top performers in the class in different subjects.
Premium earned by A.C.A. Grace, June 30th, 1825
Book premium with certificate on inner cover, earned by Missouri Kilby, 1836
Later programs listed out all premiums. This one is from 1854
Other materials in the archives contain some more direct creations from students of the Academy. Although we refer to it as "Sr. Joannah Hickey's Journal," this small volume dated 1830 contains writings from a variety of individuals. It includes the complete version of the poem at the start of this post.
"Farewell to St. Joseph's," by "Remember Maria," 1830
The Sister Mary Raphael Smith Scrapbook contains similar pieces, written by Smith herself, other sisters, and students of the Academy. Sister Mary Raphael had been a student at the Academy before becoming a Sister; she later became Directress of the school. In addition to poetry, the scrapbook contains accounts of events that occurred in the Academy between the 1830s and the 1890s.
Accounts of the death of Father Burlando, by Sister Madeleine O'Brien, Mary Huneker, and others
A handful of these additional "Scrapbooks" from the Academy exist across the middle of the 19th century. Other materials address the education provided by the Academy more directly. Katherine McDonough's lecture notes from 1899 show an average day of education in science, geology, and grammar.
The students of the Academy also contributed to a display of their schoolwork for the 1893 World's Fair in Chicago.
Art and Schoolwork displayed at Chicago World's Fair
In addition to a lucky genealogist looking for an Academy student-ancestor who may stumble upon their ancestor's writing or work, the collection provides a valuable tool of the community and its earliest mission in the United States, along with a picture of education during this time period.
St. Joseph's Academy on the true lawn tennis court
Filed under Emmitsburg, Sisters of Charity of St. Joseph's, St. Joseph's Academy
Tagged as St. Joseph's Academy, students
Mother Seton's Successor: Mother Rose White
The Council knew the end was near for their Foundress and Superioress, Mother Seton, when they met on January 2, 1821. The conclusion of the meeting contained the simple declaration: "Sister Rose was elected."
A sketch of Mother Rose White – no other known images of her have been found
Rose White (1784-1841), like Mother Seton, was a widow who a son. Born Rosetta Landry, her husband, Captain Joseph White, disappeared at sea in 1804, and her young daughter passed away not long after. Under the guidance of Father John B. David, P.S.S. – who later briefly became the Superior of the Sisters of Charity of St. Joseph's – Rose turned more and more toward the Church and charity. Her name appears as a member of the board of the Female Human Association, Charity School in Baltimore as early as 1807.
Charitable organization where Mother Rose White served on the board, front cover and page 4. Images courtesy The Catholic University of America Archives and Special Collections
In 1809, Rose was one of the seven women to join Mother Seton at the her school on Paca Street in Baltimore. She was part of the second wagon of women to travel northwest to Central Maryland the next month, along with the boarding pupils and Mother Seton's two boys. Appointed Assistant by Mother Seton, she was a member of the first band of sisters which formed the Sisters of Charity of St. Joseph's.
Front and side view of the first home of the Sisters and orphans on 6th Street in Philadelphia. Date of photos unknown.
Despite Rose having a similar background to her Superioress, Mother Seton's correspondence John Carroll, first Archbishop of Baltimore, shows a sense that Rose was better suited to the religious life than she was. While Mother Seton wished for Father Pierre Babade, P.S.S. to be appointed as the community Superior, Rose and Sister Kitty Mullen preferred Father David or the more practically-minded Father John DuBois. While clearly showing respect for her ability as an organizer, Mother Seton was prepared in the early days of the community for Sister Rose to be appointed Superioress in her place: "Rose's virtues are truly valued by me and by us all, but from the time she knew she was proposed as Mother of this house in my place and that every one in it should prepare themselves for the change (which I was directed myself to inform them by a special letter immediately after my return from Baltimore) her conduct has undergone an intire [sic] change and has been very unfavourable to her happiness and ours." (March 16, 1811).
In 1814, the community began its first ministry outside of Emmitsburg. Father Michael Hurley, pastor at St. Augustine's Church in Philadelphia and member of the board of trustees at Holy Trinity, which administered St. Joseph's Orphan Asylum, invited the Sisters of Charity to care for the orphans there. Sister Rose was appointed local Superior of the three sisters. They took possession of the Asylum on October 6, 1814 with a meager budget of $600 per year. Through Sister Rose's organizing ability and economic management, they were able to gather a cadre of donors and benefactors to extend that money as far as possible. Her abilities at organizing child care and education for orphans led to an invitation to remedy a similar situation at the New York Orphan Asylum in 1817.
Mother Seton left this world on January 4, 1821. In the first Council meeting after her death, it notes that "Sister Cecilia to replace Sr. Rose in N. York." Almost as if realizing that she had mistitled her new Superioress, the final note from that meeting reads, "the candidate Sally Powers petitions for admission to the Novitiate postponed till the arrival of Mother Rose‑‑‑‑‑," referring to her as "Mother" for the first time.
In her six years as Superioress, the community expanded to missions in Baltimore; Frederick MD; and Washington, DC. After she had served the maximum two consecutive three-year terms, Mother Rose went on to lead another orphanage in Brooklyn to prosperity, before being re-elected to Superioress in 1833 for six more years. In this term, the community founded 12 more missions in New York; Baltimore; New Orleans; Conewago, PA; Utica, NY; Richmond (2 missions); Pittsburgh; Philadelphia; Pottsville, PA; Norfolk, VA; Martinsburg, VA; and Vincennes, IN. She also saw the cornerstone laid for the new Deluol Building at the prestigious St. Joseph's Academy in Emmitsburg, which provided space for the students as its enrollment and curriculum expanded.
First page of Mother Rose White's journal
One of the most vital records of the Sisters of Charity of St. Joseph's is Mother Rose White's journal, considered one of the most informative documents about life in the early community. Among the many vivid depictions of the community's hardship, perhaps the event below illustrates their experiences best:
"We became so crowded that it was thought necessary that some of us should come up to the new house, [St. Joseph's House, today the Emmitsburg White House], to sleep. Accordingly, Sister Sally [Thompson], Sister Kitty [Mullen] and Sister Rose [White] were named and for several weeks we slept in one of the unfurnished rooms, and would rise often at two, three and four o'clock and go down to the farm [to the Stone House] thinking it was time for morning prayers, and the ground was rough plowed and often very muddy. Sometimes we would be forced to stay all day at the new house, the rain would be so heavy; one [sister] would go down and bring up something to eat. We had spinning wheels and would keep ourselves employed. While sleeping at the stone house, the snow would drift in; one morning Sister Sally [Thompson] and Sister Rose [White] shoveled out nearly two cart loads of snow in the garret where two of the Sisters were sleeping, and did not discover that their beds were partly covered also with snow until day began to dawn through the cracks of the boards, which were the only fastening for the windows, but happily the Sisters took no cold."
The journal goes on to describe her time in Philadelphia organizing the community for care of some of the neediest. In addition to the journal, surviving documents of Mother Rose in the archives in Emmitsburg include 104 letters addressed to her, 11 from her, 1 copied transcription in her handwriting, and four additional notes, including a simple prayer for Lent.
Mother Rose passed away at St. John's Asylum in Frederick MD, only two years after her final term as Superioress ended.
Filed under Mother Rose White, Sisters of Charity of St. Joseph's
Tagged as Mother Rose White, Mother Seton
The First Mission of Charity
This is part of a yearlong series about the early days of the Sisters of Charity of St. Joseph's commemorating the 200th anniversary of the death of Saint Elizabeth Ann Seton, foundress of the community. In 1850, Emmitsburg-based Sisters united with the global community of the French Daughters of Charity.
After Mother Seton and her companions left Baltimore in June 1809, the small group formed the Sisters of Charity of St. Joseph's in the village of Emmitsburg in northern Frederick County on July 31, 1809. They began to enact their mission of service to those living in poverty and began with their nearby neighbors.
Dated February 5 and addressed to Mrs. Seton, two women, simply named "Cecilia and Catherine" wrote "an account of the first Mission of Charity."
This mission was in the tradition of Saints Vincent de Paul and Louise de Marillac, whose rules for community life the Sisters adapted for the American situation. When visiting the poor, sisters provided nursing care and resources to those in difficult situations. Catherine and Cecilia evidently travelled to a home in the vicinity of Emmitsburg "after some difficulty on the road about eggs."
The family they ministered to on this day was sick, likely from one of the waterborne disease that routinely swept through Western Maryland in the early 19th century. This brief letter described what the sisters observed about the family's situation. The spellings and grammar are kept as written (if you can imagine where the periods go in modern standardized English, it becomes easier to understand):
We found enough to do at first & even now but all the sick are much better 2 of them are now setting up it was yesterday the oldest girl is about though not well she eat but once since her Mothers death until we came. She has eat a tolerable breakfast & was going to wash the bed cloathes in truth they are very dirty. I think it would be much to the comfort of the one who is obliged to stay in bed if we could put something clean on her. She is also getting better & better ever since we came, however we forbid the young girls to wash there is also 2 young men their brothers in & out all the time & perhaps you will not think it necessary to send Sisters for the night as they do not set up now at all.
They note that a doctor has not had the chance to visit yet, but they seem aware that they have done what they could to improve health and comfort for the family going through a difficult time.
The Catherine of the report may be either Sister Catherine Mullan or Catherine Seton, Mother Seton's nine-year old daughter who travelled with her from New York and lived with the community. After her mother's death, Catherine lived with her brother William and travelled around Europe before joining the Sisters of Mercy of New York in 1846.
Cecilia could refer to either Sister Cecilia Seton, Mother Seton's sister-in-law who was one of the first to join the community, or Sister Cecilia O'Conway. However, other correspondence of Cecilia Seton shows a very different handwriting. It shows far more similarity to Cecilia O'Conway's handwriting, although not definitively so.
The authors recognized this event as the First Mission of Charity undertaken by the new community!
Filed under Sisters of Charity of St. Joseph's
Tagged as Emmitsburg, St. Elizabeth Ann Seton | {
"redpajama_set_name": "RedPajamaCommonCrawl"
} | 8,444 |
\section{Introduction}
R\,Coronae Borealis stars (RCBs) are F- and G-type supergiants with
carbon rich and hydrogen deficient atmospheres. In these stars, the measured hydrogen-to-helium ratio
by number is $\le$ 10$^{-2}$ that stands in contrast to a solar-type star's H/He of 10. The surface abundances of
hydrogen poor stars are measured relative to helium, the most abundant element in their atmospheres.
The next most abundant element in their atmospheres is carbon followed by nitrogen and oxygen, and the
rest are the trace elements including hydrogen. Spectroscopic determination of carbon-to-helium ratio is not
possible from the observed optical spectra of RCBs $-$ photospheric neutral helium lines are not expected for
their effective temperatures and neutral carbon lines are roughly of the same strength across the range in
their effective temperature unlike the lines from other elements, for example, iron. Hence, a carbon-to-helium
ratio of 1.0\% by number is assumed to derive the surface abundaces of RCBs. This assumption comes from the
extreme helium stars (EHes), seemingly the close relatives of RCBs, having C/He ratios of about 1.0\%.
Of about 128 known Galactic RCBs \citep{tisserand2020} including the 11 new RCBs from
Palomar Gattini IR (PGIR) survey \citep{karambelkar2021}, abundance analyses are available for only
about 22 $-$ a very small sample indeed \citep{asplund2000,raonlambert2008,hema2017} to draw conclusions on their
origin and evolution as a group.
In this paper, the dependence of the derived surface abundances on the
adopted model atmosphere's C/He ratios are discussed. Here, we report the revised surface abundances
based on the derived carbon abundance, that is the C/He ratio, from the observed C$_2$ bands in an RCB's
spectrum \citep{hema2012,hema2017}. These new abundances are further compared with the abundances of EHes, and are
then discussed in the light of their formation scenarios.
\section{Abundances}
\subsection{Fundamentals}
In an observed star's spectrum, the strength of an absorption line is always defined as the depth of the line
with respect to the continnum. Thus, the strength of an absorption line (weak) of an element X is controlled
by the line absorption coefficient relative to the continuum absorption coefficient. In the case of a
normal (H-rich) star's optical spectrum with effective temperature in the range similar to RCBs, the continuum
and the line absorption coefficients are proportional to the number density of H and X atoms, respectively.
Then, the number density ratio X/H is what dictates the line strength and is normally expressed as the
abundance of an element X. X/H can be further stated in terms of the mass fraction Z(X) by assuming
He/H ratio of 0.1 that comes from hot normal stars and is an unobservable quantity for cool stars.
In the optical, theory and observations of RCBs' atmospheres suggest that photoionization of C\,{\sc i} is the dominant source of
continuum opacity in their line-forming layers \citep{asplund1997a}. The near constant equivalent widths of weak C\,{\sc i} lines
observed in their optical spectra from one star to another is notable (see Figure 1 of \citet{raonlambert96}).
Therefore, the measure of abundance of an element X in RCBs is the ratio X/C. However, it is crucial to express
the measured abundance A(X) = X/C in terms of the more fundamental quantity: the mass fraction Z(X). Which calls
for the determination or assumption of the C/He ratio, since helium is expected to be the most abundant element in their
atmospheres. The mass fraction, Z(X) is given as
\begin{equation}
Z(X)=\frac{\mu_X X}{\mu_H H + \mu_{He} He + \mu_C C + ...}
= \frac{\mu_X X}{(\sum\mu_I I)}
\end{equation}
where $\mu_{I}$ is the atomic mass of element I. The denominator which represents summation over all the elements is a
conserved quantity through the different stages of nuclear burning. Assuming helium to be the most abundant ingredient,
the equation 1 in terms of the measured abundance A(X) = X/C is redefined as
\begin{equation}
Z(X)= \frac{\mu_X A(X)}{H/C + 4He/C + 12 + ..\mu_I I/C }
\end{equation}
Due to hydrogen being very poor in these stars, H/C relative to He/C is very very small and can be ignored like other
trace elements from the denominator, then the above equation reduces to,
\begin{equation}
Z(X) \simeq \frac{\mu_X}{4}\frac{C}{He}A(X)
\end{equation}
The C/He is spectroscopically determined for 14 RCBs from their observed C$_2$ bands \citep{hema2012,hema2017}, and also the abundance
of any element X for a hydrogen-deficient star like RCBs, can be directly measured spectroscopically i.e., A(X) = X/C.
The derived abundances are normalised based on the convention that $\log\epsilon$(X) = $\log(X/H)$ + 12.0 to a scale
in which $\log\sum\mu_I\epsilon(I)$ = 12.15, where 12.15 is determined from solar abundances with He/H $\simeq$ 0.1.
Based on this normalisation convention, from equation 3, the helium abundance $\log\epsilon$(He) is about 11.54 for
a C/He $\le$ 0.01. The abundance of an element X can be expressed in terms of mass fraction Z(X) or number fraction X/He, and these
two quantities are related as shown in equation 3. Here, stars' derived abundances are given in $\log\epsilon$(X) and the notation [X]
represents abundance of X in a star relative to that in the Sun or log number relative to solar.
\subsection{RCB stars}
The surface abundances of about 22 RCBs are available in literature as mentioned in Section\,1. These abundances
were derived by adopting a C/He of 1\%, or that is by assuming the carbon abundance $\log\epsilon$(C) = 9.54.
Note that, from equation 3, the derived abundances depend on the adopted C/He ratio but the abundance ratios do not.
For example, $\log\epsilon$(X/Fe) is independent of the adopted C/He ratio. Four RCBs: VZ\,Sgr, V\,CrA, V854\,Cen, and V3795\,Sgr,
are clasified as minority class RCBs that show relatively lower Fe abundances and higher Si/Fe and S/Fe ratios than the
majority class RCBs \citep{asplund2000}.
In this paper, the surface abundances of 14 RCBs (11 majority class and 3 minority class) including their fluorine abundances
\citep{pandeyetal2008,hema2017} are revised using their spectroscopically determined
carbon abundances from observed C$_2$ bands \citep{hema2012,hema2017}. The revised abundances for all the elements
except helium, can be obtained by a simple scaling down of the derived abundances (for an assumed C/He) by a factor,
that is the difference between the assumed and the determined $\log\epsilon$(C). The factor, $\Delta$C, by which the
derived abundance (for an assumed C/He) was scaled down is given in Table 1. Before presenting their revised abundances,
we comment on the two known facts related to their derived carbon abundances.
First, the predicted strengths of the C\,{\sc i} lines are stronger than the observed. That means, the carbon abundance
derived using state-of-the-art H-deficient model atmospheres \citep{asplund1997a} is about 0.6 dex reduced from that
chosen for costructing the model atmosphere. \citet{asplund2000} gave a name the ``carbon problem'' to this mismatch.
A more severe ``carbon problem'' was found for the observed [C\,{\sc i}] lines \citep{pandey2004b}. Resolutions to the
``carbon problem'' were provided by crafting state-of-the-art model atmospheres by hand \citep{asplund2000,pandey2004b}.
\citet{pandey2004b}'s hand crafted model atmospheres bring the predicted strengths of C\,{\sc i} and [C\,{\sc i}] lines
in agreement with the observations. However, this holds for all the carbon abundances, from
$\log\epsilon$(C)$=$8.5 (C/He$=$0.1\%) through $\log\epsilon$(C)$=$10.5 (C/He$=$10\%), adopted for constructing the
model atmosphere. This implies that predicted strengths of C\,{\sc i} and [C\,{\sc i}] lines are insensitive to the
carbon abundance.
Second, the carbon abundances derived from observed C$_2$ bands are independent of the adopted model's carbon abundance
and this is described as the C$_2$ carbon problem \citep{hema2012}. The solution to the so called C$_2$ carbon problem
may lie in the modification of the model atmosphere's temperature structure as shown by \citet{pandey2004b}. However, it is
yet to be shown that real atmospheres have flatter temperature gradients, as suggested by \citet{asplund2000}, than the
present state-of-the-art model atmospheres. Therefore, it cannot be ruled out that the carbon abundances derived from
C$_2$ bands are the real measure of carbon abundances in these stars. In principle, the carbon abundances measured from
C$_2$ Swan bands and that assumed for the model atmosphere can be equated for a particular choice of C/He that varies from
star to star. This removes the carbon problem for C$_2$ bands. Also, as expected for carbon rich stars, the carbon abundance derived
from C$_2$ bands is correlated with the O abundance \citep{hema2012}. (Note that the carbon abundance from C\,{\sc i} lines is not
well correlated with the O abundance). Hence, we adopt the carbon abundance or the C/He ratio derived from
C$_2$ bands \citep{hema2012,hema2017} for revising the RCBs' abundances. The revised abundances, deduced from the above described
procedure, are given in Table 1.
\section{The Galactic positions and orbits}
It is crucial to know whether RCBs and their apparent relatives, the EHes are members of the Milky Way's bulge (the central regions)
or the halo (the outer regions). To confirm their membership, the distances and orbits of these stars have been
determined. The distances to these stars were determined using the Parallax measurements made by the Gaia satellite.
Note that, the Gaia mission measures a star's proper motion and radial velocity (RV) with unprecedented
precision. The RV measurements of several RCBs and EHes also come from \citet{asplund2000,hema2012,pandeyetal2001,jefferyetal1987,pandey1996}
and the best or an average value is adopted based on our judgement. The orbits around the Galaxy have been estimated in combination
with their distances and velocities. The required data for calculating the orbits for the following RCBs: GU\,Sgr, UX\,Ant, R\,CrB,
RS\,Tel, SU\,Tau, V482\,Cyg, FH\,Sct, V2552\,Oph, V532\,Oph, ASAS-RCB-10, VZ\,Sgr, and MV\,Sgr (the hot RCB) were available. For the
following EHes: LSS\,3184, BD$-$9$^{\circ}$4395, LS\,IV+$6^{\circ}$002, LSE\,78, LSS\,4357, V1920\,Cyg, LSS\,99, HD\,124448,
BD+10$^{\circ}$2179, PV\,Tel, FQ\,Aqr, LS\,IV$-$1$^{\circ}$ 002, BD$-$1$^{\circ}$3438, LS\,IV$-$14$^{\circ}$ 109, and LSS\,3378,
data were available for calculating the orbits.
The initial conditions for the computation of Galactic orbits of stars are their presently observed positions and
velocities with respect to the galactocentric reference frame. Adopting the
solar motion (U, V, W)$_{\odot}$ = (10.0, 5.2, 7.2) km s$^{-1}$ from \citet{dehnenbinney1998}, the local standard of
rest (LSR) velocities of stars U$_{LSR}$, V$_{LSR}$, W$_{LSR}$ and their errors $\sigma_{U}$, $\sigma_{V}$, $\sigma_{W}$
are calculated with the method of \citet{johnsonnsoderblom1987}. The LSR velocities are then corrected to the
Galactic standard of rest (GSR) by adopting the LSR rotation velocity of 220 km s$^{-1}$ \citep{kerrnlynden1986} at
the galactocentric distance of the Sun of 8.5 kpc. The celestial positions ($\alpha$, $\delta$, l, b), parallaxes ($\pi$),
and absolute proper motions ($\mu_{\alpha}$\,cos $\delta$, $\mu_{\delta}$) are adopted from the
Gaia Early Data Release 3 \citep{gaiacollab2016b,gaiacollab2020a}, while the RV measurements are taken from the published
sources as mentioned above. Adopting the measured Galactic spatial positions and velocities, we studied the
dynamics of stars under the influence of a multicomponent, static, axisymmetric Galactic gravitational potential model
of \citet{flynnetal1996}. The relevant code adopted here for the integration of stellar orbits was used previously in the
analysis of kinematics and orbits of a large sample of open clusters \citep{wuetal2009,reddyetal2016}. Starting with a
star's current position and velocity components referenced to the Galactic standard of rest, the trajectory of the star was
followed backward in time over a period of 5 Gyr to ensure that each star could complete sufficient galactic orbits so that
the averaged orbital parameters can be determined with fair certainty.
The analysed RCBs with accurate kinematics have tightly bound orbits, placing them in the inner regions of the Milky Way of radius less
than or about 6.0 kpc and z$_{max}$ about 3.0 kpc, the exceptions being UX\,Ant, V482\,Cyg, SU\,Tau, and R\,CrB. In contrast, the analysed EHes
with accurate kinematics have orbits extending beyond the regions of the Milky Way of radius more than 6.0 kpc, the exceptions being
LSS\,4357, LS\,IV$-$1$^{\circ}$ 002, and LS\,IV$-$14$^{\circ}$ 109. Including MV\,Sgr, we note that about five analysed RCBs
with accurate kinematics have tightly bound orbits and z$_{max}$ similar to the most metal-poor, $\log\epsilon$(Fe)$\sim$3.5,
star SMSS\,J181609.62$-$333218.7 whose orbit is entirely contained within the bulge \citep{howesetal2015}. The revised metallicity range for
the analysed RCBs is $\log\epsilon$(Fe)=3.8 through $\log\epsilon$(Fe)=5.8 (see Table 1), and the revised metallicities are fairly consistent
with their estimated orbits and location in the Galaxy.
\section{RCB and EHe stars: surface compositions}
The RCBs' revised surface compostion, based on the carbon abundances derived from the observed C$_2$ bands, provide new evidences
for their formation history. We discuss by comparing the revised surface compositions of RCBs with EHes that are considered to be their
relatives having higher effective temperatures. The surface compostion of EHes including the hot RCB star DY\,Cen are from
\citet{jefferynheb1992,drillingetal1998,jeffery1998,jefferyetal1998,pandeyetal2001,pandeyetal2004a,pandeyetal2006,pandey2006,pandeynreddy2006,pandeynlamb2011,pandeyetal2014,pandeynlamb2017,jeffery2017,bhowmicketal2020}.
The principal objective is to seek out similarities, differences, and trends if any.
\subsection{The C/He ratios}
The carbon abundances for the RCBs are in the range $\log\epsilon$(C)$=$7.7 to $\log\epsilon$(C)$=$8.8 that is the C/He ratios
are in the range 0.02 to 0.2 per cent including the three minority class RCB stars. In contrast, the C/He ratios of the EHes are in the range
0.3 to 1.0 per cent with three exceptions. The three exceptions are V652\,Her, HD\,144941, and GALEX J184559.8$-$413827, having very low
C/He ratios $\sim$ 0.003 per cent. The two hot RCBs, MV\,Sgr \citep{jeffery1988} and DY\,Cen have C/He ratios of 0.02 and 1.0 per cent,
respectively. However, recent study of \citet{jeffery2020} indicates that DY\,Cen's evolutionary history involves a very late thermal
pulse due to its very rapid evolution, non-negligible surface hydrogen and high surface strontium.
The hot RCB MV\,Sgr is not subject to a carbon problem like EHes. It is remarkable to note that MV\,Sgr's C/He ratio lies in the range that
is derived for the majority and minority class RCBs.
\subsection{Iron: the initial metallicity}
The revised iron abundances for the RCBs are in the range $\log\epsilon$(Fe)$=$3.8 to $\log\epsilon$(Fe)$=$5.8 with the minority
class RCB V854\,Cen at lower end and the majority class RCB R\,CrB at the higher end of this range (see Table 1). On the contrary,
the iron abundances for the EHes are in the range $\log\epsilon$(Fe)$=$5.4 to $\log\epsilon$(Fe)$=$7.2. As discussed in Section 3,
EHes are broadly placed in the outer regions of the Milky Way than the RCBs. Indications are that the RCBs' metallicity range is roughly
consistent with the metal poor population contained within the bulge \citep{howesetal2015}.
Hydrogen- and helium-burning products are clearly observed in the atmospheres of RCBs and EHes. Their derived effective temperatures
and surface gravities suggest that they are evolved low mass stars. Hence, no synthesis of $\alpha$- and Fe-peak elements occurs in
the course of their evolution. The $\alpha$- and Fe-peak elements remain unaltered in their atmospheres by providing us the initial
metallicty to these stars; here $\alpha$-peak elements usually refer to $\alpha$-capture elements heavier than neon.
\subsection{Hydrogen}
The hydrogen abundance of majority class RCB stars show no obvious trend with the iron abundance or the metal abundance. However, it is
notable that two minority class RCBs V\,CrA and V854\,Cen with relatively lower Fe abundance have hydrogen abundance higher than the
majority class RCBs. The third minority class RCB VZ\,Sgr is like the majority class RCBs (See Figure 1: bottom left panel). For a comparison,
EHe stars are shown in Figure 1: bottom right panel. Note the relatively higher hydrogen abundances of DY\,Cen and the three very low C/He
EHe stars.
\subsection{The CNO abundances}
The majority class RCBs' carbon abundances are a function of their Fe abundances but this trend is not very evident for the minority
class RCBs (see Figure 1: top left panel). However, abundances for only three minority class RCBs are available and so it is not
worth looking for any trends. In contrast for EHes, Figure 1: top right panel clearly demonstrates that carbon abundances are
independent of their Fe abundances.
The nitrogen abundances of both RCBs and EHes depend on their Fe abundances (see Figure 2: bottom left panel for RCBs and bottom right panel
for EHes). The observed N abundance is the sum of initial CNO as expected. Hence, providing the evidence that helium is produced from
hydrogen-burning CNO cycle that converts most of the initial C and O to N.
If all of initial C and O is coverted to N, then the CNO cycle processed material should be N enriched with C and O depleted. The observed
C and O abundances for both RCBs and EHes are not depeleted suggesting that these are products of helium-burning via triple-$\alpha$ and
$\alpha$-capture on $^{14}$N and $^{12}$C. The three exceptions that show depleted C and O are the very low C/He or C-poor EHe stars
(see Figure 1: top left panel for RCBs and top right panel for EHes, and Figure 2: top left panel for RCBs and top right panel for EHes).
\subsection{Fluorine}
The fluorine abundances of the majority and minority class RCBs suggest a mild trend with their Fe abundances and F/Fe in these stars is highly
enriched when compared to the solar F/Fe ratio. Most of the EHes have enhanced F like RCBs but show no trend with their Fe abundances unlike RCBs.
See Figure 3: bottom left panel for RCBs and bottom right panel for EHes, for F versus Fe trends.
\subsection{Neon}
For EHes, the neon abundances like N abundances are enhanced and show a trend with their Fe abundances except for the five cool EHes whose Ne
abundances are not corrected for the non-LTE effects. Application of the non-LTE effects brings the Ne abundances of cool EHes in line with the
hot EHes \citep{pandeynlamb2011}. The dependence of the Ne abundance with the Fe abundance suggest that the observed Ne in EHes is essentially
$^{22}$Ne and is produced from two successive $\alpha$-captures on $^{14}$N. Note that the revised Ne abundances are available only for two
RCBs. The Ne versus Fe trends are shown in Figure 3: top left panel for RCBs and top right panel for EHes.
\subsection{Sodium to zinc}
The EHe and RCB abundances of sodium, aluminium, magnesium, silicon, sulphur, calcium, titanium, chromium, manganese, nickel, and zinc scale well
with the iron abundances. For RCBs, Al, Mg, Si, S, Ca, Ti, Ni and Zn broadly vary in concert with Fe, as expected. Similar correlations are seen for
EHes with additions of Cr and Mn. This clearly indicates that the EHe as well as RCB star's Fe abundance is the representative of initial metallicity.
Abundances of phosphorus, argon, chromium, and manganese are not available for RCBs, but in EHes it is worth considering the observed abundances of P
and Ar that suggest weak correlations with Fe. The X versus Fe trends are shown in Figures 4, 5, 6, 7, 8, 9, and 10. The left and the right panels
show RCBs and EHes, respectively. Note the variation of minority class RCBs with respect to the majority class RCBs.
\subsection{Yttrium, zirconium, and barium: heavy elements}
For majority class RCBs, the revised abundances of Y and Zr show insignificant enhancements. However, the minority class RCBs show a range from
insignificant to a maximum enhancement of about 2.0 dex in both Y/Fe and Zr/Fe with respect to the solar Y/Fe and Zr/Fe ratios like the EHes.
Barium for majority as well as minority class RCBs shows insignificant enhancement. Note that only three measurements of Ba are available for EHes
showing a range from no to a maximum enhancement of about 1.8 dex in Ba/Fe with respect to the solar Ba/Fe ratio. See Figures 10 and 11: the left
and the right panels show RCBs and EHes, respectively.
\section{Key spectroscopic features}
The key features of RCBs are specifically their high $^{12}$C/$^{13}$C and low $^{16}$O/$^{18}$O ratios with remarkable F overabundances \citep{hema2012,
clayton07,pandeyetal2008}. $^{12}$C/$^{13}$C and $^{16}$O/$^{18}$O ratios are from observed C$_2$ and CO molecular bands in the spectra of RCBs but these
molecular bands are not present in the observed spectra of EHes due to their higher effective temperatures. Nevertheless, fluorine atomic lines are
observed in EHes' as well in RCBs' spectra and provide the star's F abundance. The atmospheres of EHes are overabundant in F like RCBs \citep{pandey2006,
bhowmicketal2020}. The processes involving fluorine production in these stars needs to be explored. For this reason, it will be crucial
to identify any correlations between fluorine and other elements including any relationship among the abundances of other key elements.
\subsection{F versus CNO and Ne}
In RCBs, the F abundances suggest mild to no correlation with C abundances. In contrast, EHes' F abundances are strongly correlated with their C
abundances. See Figure 12: the top left and the top right panels show RCBs and EHes, respectively.
F versus N show significant correlation in RCBs unlike in EHes. Figure 13: the top left and the top right panels clearly exhibit these
trends for RCBs and EHes, respectively. F versus O and Ne trends are also shown in Figure 14: the left and the right panels
are for RCBs and EHes, respectively.
We note that the relationship of F with C and that with N is distinct for RCB and EHe stars. This indicates that, possibly, fluorine is produced from
two different processes operating in these stars.
\subsection{C and N versus O}
C and O abundances suggest linear correlation for both RCBs and EHes. These trends are clearly shown in Figure 12: the bottom left and the bottom right
panels are for RCBs and EHes, respectively. The N versus O abundances are also shown in Figure 13: the bottom left and the bottom right
panels are for RCBs and EHes, respectively.
\section{Double white dwarf mergers and the abundances of key elements}
The RCB and EHe stars' origins and evolutionary connections are not yet understood despite thorough analyses of their spectra. In broad terms,
the chemical compositions suggest a hydrogen-deficient atmosphere now composed of material exposed to both H- and He-burning. Following the elimination
of several proposals, two principal theories emerged: the ``double-degenerate'' (DD) model and the ``final-flash'' (FF) model.
The FF model, refers to a late or final He shell flash in a post-AGB star. In this model \citep{iben1983}, the ignition of the helium shell in a
post-AGB star, say, a cooling white dwarf, results in what is known as a late or very late thermal pulse \citep{herwig2001}. This converts the star
to a H-poor cool luminous star (i.e., an RCB star), which then evolves to hotter effective temperatures at about constant luminosity (i.e., as an EHe star),
and finally to the white dwarf cooling track. The attendant nucleosynthesis during and following the He shell flash shows that a H-poor supergiant may
result with surface composition characteristic of RCBs and EHes. However, the FF model has failed to account for the key features, particularly,
the high $^{12}$C/$^{13}$C ratios, the low $^{16}$O/$^{18}$O ratios and anomalous F overabundances observed in these stars \citep{pandey2006,clayton07,
pandeyetal2008,raonlambert2008,hema2012,hema2017}. A consensus is now emerging for DD model but a small fraction of H-poor stars may be produced from
FF model \citep{pandeynlamb2011} such as the majority RCB XX\,Cam, and possibly the EHe HD\,124448 \citep{bhowmicketal2020} including
V4334\,Sgr (Sakurai's object) \citep{pandeyetal2008}. The surface abundances of V4334\,Sgr \citep{asplund97b} are revised based on the measured carbon
abundance, that is $\log\epsilon$(C) = 9.7 \citep{hema2012b}, from C$_2$ bands as done for RCBs. For comparison, the revised abundances of V4334\,Sgr
for the 1996 October spectrum are given in Table 1 .
The DD scenario, proposed by \citet{webbink84} and \citet{iben84}, involves the merger of an He white dwarf with a more massive C-O white dwarf
following the decay of their orbit. Other mergers may involve two He white dwarfs. It is clear that the merger product will be H-poor since neither
of the white dwarfs contain much hydrogen, and the hydrogen that survives will be mixed substantially with more helium and possibly other material.
Recall the extraordinary $^{16}$O/$^{18}$O ratios in RCBs and/or the remarkable F overabundances in RCBs as well as in EHes. Neither CO+He
white dwarf binaries nor He+He white dwarf binaries can account for these exceptional abundances without the ensuing nucleosyntheis during
the merger and/or the postmerger phase. Simulations of the merger and postmerger phases with accompanying nucleosynthsis have been attempted for
evolution of a white dwarf merger to the RCB phase \citep{longland11,zhang12a,zhang12b,menon13,menonetal2019,zhangetal2014,lauer2019}.
Simulations of He+He white dwarf mergers are limited to \citet{zhang12a,zhang12b} and appear to be ``in partial agreement" in explaining the observed
abundances of RCBs and EHes. \citet{bhowmicketal2020} note that F is underpredicted by \citet{zhang12b} while minor disagreements between prediction
and observation are found for C, N, O, and Ne. In these mergers, the F is synthesized by
$^{14}$N$(\alpha,\gamma)^{18}$F$(p,\alpha)^{15}$O$(\alpha,\gamma)^{19}$Ne$(\beta^+)^{19}$F.
However, \citet{zhang12a} provide the latitude to account for the C-poor V652\,Her and HD\,144941 \citep{pandeynlamb2017}.
Most recent calculations for a CO+He merger are from \citet{crawford2020}. It is clear that surface composition of the merger products are
a result of mixing during and/or following the merger. In post-merger objects, the temperature of the helium-burning shell at ignition strongly impacts
the yields of CNO-process as well as $\alpha$-capture isotopes. Hence, Crawford et al. investigate the effects of a range of initial He-burning shell
temperatures and include the effects of solar and subsolar metallicities. Their three models that satisfy the maximum criteria in reproducing the
observed surface abundances are SOL8.57, SUB8.48, and SUB8.53; SOL and SUB represent models with metallicity solar and one-tenth of solar, respectively,
and 8.57, 8.48, and 8.53 are the models' initial He-burning shell temperatures (K) in common log values. Crawford et al. were able to identify SUB8.48
as the preferred model that is able to remarkably reproduce abundances closest to those of observed RCBs. SUB8.48 is at 10 per cent of solar
metallicity with an initial He-burning shell temperature of approximately 3.00$\times$10$^8$ K; the other closest model SUB8.53 yields higher C, F, and
Ne with no significant changes in N and O yields (see Figures 1, 2, and 3). The predicted surface Li abundance ($\log\epsilon$(Li)) is between 0.85 and
2.64 for solar models, while the subsolar models have significantly higher Li abundances, between 3.95 and 6.41.
The Crawford et al.'s predicted surface Li can possibly be an upper limit because their reaction network, mesa\_75.net, as noted by them does not include
the crucial reaction $^{7}$Li$(\alpha,\gamma)^{11}$B. However, inclusion of $^{11}$B in the reaction network reduces the predicted surface Li
abundance ($\log\epsilon$(Li)) significantly to about $-$1.5 to 1.0 for one-tenth of solar metallicity white dwarf mergers \citep{munson2021}.
\section{Concluding Remarks}
Predictions of CO+He white dwarf mergers are in good agreement with the observed abundances, in particular with the extraordinary overabundance of F
\citep{menon13,menonetal2019}. \citet{menon13} identify that one source of F is in the He-burning shell where $^{13}$C$(\alpha,n)^{16}$O provides
the neutrons to seed the reaction $^{14}$N$(n,p)^{14}$C$(p,\gamma)^{15}$N$(\alpha,\gamma)^{19}$F. The correlations of F with C and N indicate that
plausibly F production in majority class RCBs is via the reaction channel:
$^{14}$N$(\alpha,\gamma)^{18}$F$(p,\alpha)^{15}$O$(\alpha,\gamma)^{19}$Ne$(\beta^+)^{19}$F; F abundance depends on $^{14}$N and the available
protons (see Section 5.1, and Figures 12 and 13).
While in EHes, the dependence of F with C and N suggests that F is produced via the reaction channel:
$^{14}$N$(n,p)^{14}$C$(p,\gamma)^{15}$N$(\alpha,\gamma)^{19}$F; F abundance depends on the available neutrons but on the other hand, $^{13}$C which
provides neutrons must result from the mixing of protons into a $^{12}$C-rich region, where the $^{12}$C comes from $^{4}$He
(see Section 5.1 and Figures 12 and 13). For example, if F synthesis is via the reaction channel involving neutrons, then the star's F abundance is
expected to be correlated with its $^{12}$C abundance. On the contrary, if the reaction involving $\alpha$-capture on $^{14}$N is producing F, then
the star's F abundance is expected to be correlated with its N abundance. Nonetheless, the trends of F with C, N, O, and Ne show that significant
helium burning after a double white dwarf merger can account for a majority of the observed abundances (see Section 5.1, and Figures 12, 13, and 14).
It is worth noting that the average of observed F abundances is lower by about 1 dex in RCBs than in EHes for stars with common
Fe abundance (see Figure 3).
The revised Fe abundances ($\log\epsilon$(Fe)), that is the metallicity, of RCBs are in the range 3.8 through 5.8. These revised metallicities
are fairly consistent with their estimated orbits and location in the Galaxy. Recent \citep{crawford2020,munson2021}
and the earlier predictions of CO+He white dwarf mergers do not explore the metallicity range lower than $\log\epsilon$(Fe)$=$6.5.
The revised metallicity range observed for RCB stars is lower than $\log\epsilon$(Fe)$=$6.5. Hence, predictions for CO+He white dwarf mergers
are lacking for the revised lower metallicities of RCB stars. However, an extrapolation of Crawford et al.'s
surface abundance preditions to lower metallicities, similar to the revised metallicity range of RCBs, would likely explain the observed carbon
abundances from C$_2$ bands but not all the observed key elemental abundances. The revised C/O ratios that are between 3.0 and 64.0 may possibly
be explained; tweaking of several parameters mainly the initial He-burning shell temperature may likely reproduce the RCBs' revised abundances.
Theoretical studies with a larger parameter space, for example including all the key reactions, needs to be explored to explain the revised
abundances of RCBs, and this would certainly lead to further refinements in the study of white dwarf mergers.
\acknowledgments
We thank the referee for the constructive comments. GP is grateful to Kameswara Rao, David Lambert and Simon Jeffery for introducing him to the topic
of H-poor stars, and for all their help.
\begin{figure}
\epsscale{1.00}
\plotone{C2rcbeheHCFenew.eps}
\caption{$\log\epsilon$(H) vs. $\log\epsilon$(Fe) for RCBs (bottom left panel) and EHes (bottom right panel).
[C] vs. [Fe] for RCBs (top left panel) and EHes (top right panel). The majority and minority class RCBs are
represented by filled triangles and asterisks, respectively. The hot and cool EHes are represented by open squares
and open stars, respectively. C-poor EHes are represented by open triangles. DY\,Cen is represented by open circle.
Filled circles denote the model predictions from \citet{crawford2020} for three Model IDs SOL8.57, SUB8.48,
and SUB8.53, meeting the maximum criteria; details are in the text of Section 6.
[C]$=$[Fe] is denoted by solid line, and [Fe],[C]$=$0,0 represents the Sun.}
\end{figure}
\begin{figure}
\epsscale{1.00}
\plotone{C2rcbeheNOFenew.eps}
\caption{[N] vs. [Fe] for RCBs (bottom left panel) and EHes (bottom right panel).
[O] vs. [Fe] for RCBs (top left panel) and EHes (top right panel). The majority and minority class RCBs are
represented by filled triangles and asterisks, respectively. The hot and cool EHes are represented by open squares
and open stars, respectively. C-poor EHes are represented by open triangles. DY\,Cen is represented by open circle.
Filled circles denote the model predictions from \citet{crawford2020} for three Model IDs SOL8.57, SUB8.48,
and SUB8.53, meeting the maximum criteria; details are in the text of Section 6.
[X]$=$[Fe] are denoted by solid lines, where X represents N, and O. [Fe],[X]$=$0,0 represents the Sun.}
\end{figure}
\begin{figure}
\epsscale{1.00}
\plotone{C2rcbeheFNeFenew.eps}
\caption{[F] vs. [Fe] for RCBs (bottom left panel) and EHes (bottom right panel).
[Ne] vs. [Fe] for RCBs (top left panel) and EHes (top right panel). The majority and minority class RCBs are
represented by filled triangles and asterisks, respectively. The hot and cool EHes are represented by open squares
and open stars, respectively. C-poor EHes are represented by open triangles. DY\,Cen is represented by open circle.
Filled circles denote the model predictions from \citet{crawford2020} for three Model IDs SOL8.57, SUB8.48,
and SUB8.53, meeting the maximum criteria; details are in the text of Section 6.
[X]$=$[Fe] are denoted by solid lines, where X represents F, and Ne. [Fe],[X]$=$0,0 represents the Sun.}
\end{figure}
\begin{figure}
\epsscale{1.00}
\plotone{C2rcbeheNaAlFenew.eps}
\caption{[Na] vs. [Fe] for RCBs (bottom left panel) and EHes (bottom right panel).
[Al] vs. [Fe] for RCBs (top left panel) and EHes (top right panel). The majority and minority class RCBs are
represented by filled triangles and asterisks, respectively. The hot and cool EHes are represented by open squares
and open stars, respectively. C-poor EHes are represented by open triangles. DY\,Cen is represented by open circle.
[X]$=$[Fe] are denoted by solid lines, where X represents Na, and Al. [Fe],[X]$=$0,0 represents the Sun.}
\end{figure}
\begin{figure}
\epsscale{1.00}
\plotone{C2rcbehePMgFenew.eps}
\caption{[P] vs. [Fe] for RCBs (bottom left panel) and EHes (bottom right panel).
[Mg] vs. [Fe] for RCBs (top left panel) and EHes (top right panel). The majority and minority class RCBs are
represented by filled triangles and asterisks, respectively. The hot and cool EHes are represented by open squares
and open stars, respectively. C-poor EHes are represented by open triangles. DY\,Cen is represented by open circle.
[X]$=$[Fe] are denoted by solid lines, where X represents P, and Mg. [Fe],[X]$=$0,0 represents the Sun.}
\end{figure}
\begin{figure}
\epsscale{1.00}
\plotone{C2rcbeheSiSFenew.eps}
\caption{[Si] vs. [Fe] for RCBs (bottom left panel) and EHes (bottom right panel).
[S] vs. [Fe] for RCBs (top left panel) and EHes (top right panel). The majority and minority class RCBs are
represented by filled triangles and asterisks, respectively. The hot and cool EHes are represented by open squares
and open stars, respectively. C-poor EHes are represented by open triangles. DY\,Cen is represented by open circle.
[X]$=$[Fe] are denoted by solid lines, where X represents Si, and S. [Fe],[X]$=$0,0 represents the Sun.}
\end{figure}
\begin{figure}
\epsscale{1.00}
\plotone{C2rcbeheArCaFenew.eps}
\caption{[Ar] vs. [Fe] for RCBs (bottom left panel) and EHes (bottom right panel).
[Ca] vs. [Fe] for RCBs (top left panel) and EHes (top right panel). The majority and minority class RCBs are
represented by filled triangles and asterisks, respectively. The hot and cool EHes are represented by open squares
and open stars, respectively. C-poor EHes are represented by open triangles. DY\,Cen is represented by open circle.
[X]$=$[Fe] are denoted by solid lines, where X represents Ar, and Ca. [Fe],[X]$=$0,0 represents the Sun.}
\end{figure}
\begin{figure}
\epsscale{1.00}
\plotone{C2rcbeheTiCrFenew.eps}
\caption{[Ti] vs. [Fe] for RCBs (bottom left panel) and EHes (bottom right panel).
[Cr] vs. [Fe] for RCBs (top left panel) and EHes (top right panel). The majority and minority class RCBs are
represented by filled triangles and asterisks, respectively. The hot and cool EHes are represented by open squares
and open stars, respectively.
[X]$=$[Fe] are denoted by solid lines, where X represents Ti, and Cr. [Fe],[X]$=$0,0 represents the Sun.}
\end{figure}
\begin{figure}
\epsscale{1.00}
\plotone{C2rcbeheMnNiFenew.eps}
\caption{[Mn] vs. [Fe] for RCBs (bottom left panel) and EHes (bottom right panel).
[Ni] vs. [Fe] for RCBs (top left panel) and EHes (top right panel). The majority and minority class RCBs are
represented by filled triangles and asterisks, respectively. The hot and cool EHes are represented by open squares
and open stars, respectively.
[X]$=$[Fe] are denoted by solid lines, where X represents Mn, and Ni. [Fe],[X]$=$0,0 represents the Sun.}
\end{figure}
\begin{figure}
\epsscale{1.00}
\plotone{C2rcbeheZnYFenew.eps}
\caption{[Zn] vs. [Fe] for RCBs (bottom left panel) and EHes (bottom right panel).
[Y] vs. [Fe] for RCBs (top left panel) and EHes (top right panel). The majority and minority class RCBs are
represented by filled triangles and asterisks, respectively. The hot and cool EHes are represented by open squares
and open stars, respectively.
[X]$=$[Fe] are denoted by solid lines, where X represents Zn, and Y. [Fe],[X]$=$0,0 represents the Sun.}
\end{figure}
\begin{figure}
\epsscale{1.00}
\plotone{C2rcbeheZrBaFenew.eps}
\caption{[Zr] vs. [Fe] for RCBs (bottom left panel) and EHes (bottom right panel).
[Ba] vs. [Fe] for RCBs (top left panel) and EHes (top right panel). The majority and minority class RCBs are
represented by filled triangles and asterisks, respectively. The hot and cool EHes are represented by open squares
and open stars, respectively.
[X]$=$[Fe] are denoted by solid lines, where X represents Zr, and Ba. [Fe],[X]$=$0,0 represents the Sun.}
\end{figure}
\begin{figure}
\epsscale{1.00}
\plotone{C2rcbeheCOCFnew.eps}
\caption{$\log\epsilon$(O) vs. $\log\epsilon$(C) for RCBs (bottom left panel) and EHes (bottom right panel).
$\log\epsilon$(F) vs. $\log\epsilon$(C) for RCBs (top left panel) and EHes (top right panel). The majority and minority
class RCBs are represented by filled triangles and asterisks, respectively. The hot and cool EHes are represented by
open squares and open stars, respectively. C-poor EHes are represented by open triangles. DY\,Cen is represented by
open circle. The solid lines denote the locus of the solar X/C ratios, where X represents O, and F.}
\end{figure}
\begin{figure}
\epsscale{1.00}
\plotone{C2rcbeheNONFnew.eps}
\caption{$\log\epsilon$(O) vs. $\log\epsilon$(N) for RCBs (bottom left panel) and EHes (bottom right panel).
$\log\epsilon$(F) vs. $\log\epsilon$(N) for RCBs (top left panel) and EHes (top right panel). The majority and minority
class RCBs are represented by filled triangles and asterisks, respectively. The hot and cool EHes are represented by
open squares and open stars, respectively. C-poor EHes are represented by open triangles. DY\,Cen is represented by
open circle. The solid lines denote the locus of the solar X/N ratios, where X represents O, and F.}
\end{figure}
\begin{figure}
\epsscale{1.00}
\plotone{C2rcbeheOFNeFnew.eps}
\caption{$\log\epsilon$(F) vs. $\log\epsilon$(O) for RCBs (bottom left panel) and EHes (bottom right panel).
$\log\epsilon$(F) vs. $\log\epsilon$(Ne) for RCBs (top left panel) and EHes (top right panel). The majority and minority
class RCBs are represented by filled triangles and asterisks, respectively. The hot and cool EHes are represented by
open squares and open stars, respectively. C-poor EHes are represented by open triangles. DY\,Cen is represented by
open circle. The solid lines denote the locus of the solar F/X ratios, where X represents O, and Ne.}
\end{figure}
\clearpage
\thispagestyle{empty}
\begin{landscape}
\begin{deluxetable}{lcccccccccccccccc}
\label{t:lines.uves}
\tabletypesize{\scriptsize}
\tablewidth{0pt}
\tablecolumns{17}
\tablecaption{Revised photospheric abundances of RCBs}
\tablehead{
\colhead{Element} & \colhead{GU\,Sgr} & \colhead{UX\,Ant} & \colhead{R\,CrB} &
\colhead{RS\,Tel} & \colhead{SU\,Tau} & \colhead{V482\,Cyg} & \colhead{FH\,Sct} &
\colhead{V2552\,Oph} & \colhead{V532\,Oph} & \colhead{RCB8\tablenotemark{a}} & \colhead{RCB10\tablenotemark{b}} &
\colhead{VZ\,Sgr} & \colhead{V\,CrA} & \colhead{V854\,Cen} & \colhead{FF\tablenotemark{c}} &
\colhead{Sun\tablenotemark{d}}}
\startdata
H & \nodata & 5.7 & 6.2 & 5.3 & 5.9 & 3.6 & 3.8 & 5.3 & 5.3 & 4.6 & 6.1 & 5.5 & 7.6 & 7.7 & 8.2 & 12.0 \\
Li & $\leq -$0.3 & $\leq$1.3 & 2.1 & $\leq$0.2 & 1.1 & $\leq -$0.1 & $\leq -$0.8 & $\leq -$0.4 & $\leq -$0.3 & $\leq -$0.2 & \nodata & $\leq$0.8 & $\leq -$0.2 & $\leq -$0.2 & 3.4 & 1.05 \\
C & 8.1 & 8.3 & 8.8 & 8.3 & 8.0 & 8.3 & 7.7 & 8.1 & 8.2 & 8.3 & 8.2 & 8.8 & 8.4 & 8.3 & 9.7 & 8.43 \\
$\Delta$C\tablenotemark{e} & 1.4 & 1.2 & 0.7 & 1.2 & 1.5 & 1.2 & 1.8 & 1.4 & 1.3 & 1.2 & 1.3 & 0.7 & 1.1 & 1.2 & 0.8 & \nodata \\
N & 7.3 & 7.1 & 7.7 & 7.6 & 7.0 & 7.6 & 6.9 & 7.4 & 7.3 & 7.4 & 6.5 & 6.9 & 6.9 & 5.6 & 8.1 & 7.83 \\
O & 6.8 & 7.6 & 8.3 & 7.1 & 6.9 & 6.9 & 5.9 & 6.6 & 6.7 & 6.9 & 7.4 & 8.0 & 6.7 & 6.7 & 8.6 & 8.69 \\
F & 5.8 & 5.0 & 6.2 & \nodata & 5.5 & 5.4 & 5.4 & 5.3 & 5.3 & 5.7 & 5.5 & 5.7 & 5.4 & 4.5 & $<$5.6 & 4.42 \\
Ne & \nodata & \nodata & \nodata & \nodata & \nodata & \nodata & \nodata & \nodata & 7.3 & \nodata & 7.3 & \nodata & \nodata & \nodata & \nodata & 7.93 \\
Na & 4.6 & 4.6 & 5.4 & 4.8 & 4.2 & 5.1 & 4.3 & 4.6 & 4.9 & 5.2 & 4.6 & 5.1 & 4.6 & 4.2 & 6.0 & 6.24 \\
Mg & 5.5 & \nodata & 5.9 & \nodata & \nodata & \nodata & \nodata & 5.4 & 5.5 & 5.6 & 5.5 & \nodata & 5.5 & 4.0 & 5.7 & 7.60 \\
Al & 4.3 & \nodata & 5.1 & 4.7 & 3.7 & 5.0 & 4.1 & 4.5 & 4.5 & 4.9 & 4.3 & 4.7 & 4.3 & 3.5 & 5.5 & 6.45 \\
Si & 5.8 & 5.7 & 6.5 & 5.9 & 5.2 & 6.0 & 5.3 & 5.4 & 5.7 & 5.9 & 5.4 & 6.6 & 6.5 & 4.8 & 6.7 & 7.51 \\
P & \nodata & \nodata & \nodata & \nodata & \nodata & \nodata & \nodata & \nodata & \nodata & \nodata & \nodata & \nodata & \nodata & \nodata & \nodata & 5.41 \\
S & 5.6 & 5.0 & 6.1 & 5.6 & 5.0 & 5.7 & 5.2 & 5.6 & 5.4 & 5.9 & 5.4 & 6.0 & 6.1 & 4.2 & 6.1 & 7.12 \\
Ca & 4.0 & 4.3 & 4.6 & 4.1 & 3.5 & 4.2 & 3.3 & 3.8 & 3.8 & 4.4 & 3.8 & 4.3 & 4.2 & 2.9 & 4.7 & 6.34 \\
Ti & \nodata & \nodata & \nodata & \nodata & 2.2 & \nodata & \nodata & 2.8 & 2.9 & 3.2 & 2.5 & \nodata & 2.2 & 1.9 & 3.8 & 4.95 \\
Fe & 4.9 & 5.0 & 5.8 & 5.2 & 4.6 & 5.5 & 4.5 & 5.2 & 5.1 & 5.6 & 5.0 & 5.1 & 4.4 & 3.8 & 5.8 & 7.50 \\
Ni & 4.2 & 4.6 & 4.8 & 4.5 & 3.9 & 4.6 & 4.0 & 4.2 & 4.3 & 4.7 & 4.0 & 4.5 & 3.8 & 3.7 & 5.4 & 6.22 \\
Zn & 3.0 & \nodata & \nodata & 3.1 & 2.1 & 3.2 & 2.3 & 2.9 & 3.1 & 3.3 & \nodata & 3.2 & 2.8 & 2.2 & 4.6 & 4.56 \\
Y & 0.6 & 0.3 & 0.8 & 0.7 & $-$0.2 & 1.4 & 0.2 & 1.0 & 0.7 & 1.1 & 0.2 & 2.1 & \nodata & \nodata & 3.4 & 2.21 \\
Zr & \nodata & \nodata & \nodata & \nodata & \nodata & 1.1 & 0.5 & 1.0 & 0.8 & 1.4 & 0.8 & 1.9 & 0.3 & $-$0.1 & 2.7 & 2.58 \\
Ba & $-$0.2 & $-$0.2 & 0.9 & 0.3 & $-$1.2 & 1.4 & $-$0.4 & $-$0.5 & 0.2 & 0.3 & $-$0.5 & 0.7 & $-$0.8 & $-$0.9 & 1.1 & 2.18 \\
\enddata
\tablenotetext{a}{ASAS-RCB-8}
\tablenotetext{b}{ASAS-RCB-10}
\tablenotetext{c}{V4334\,Sgr (Sakurai's object), a FF product, shows no detectable neutral fluorine lines \citep{pandeyetal2008}}
\tablenotetext{d}{\citet{asplund09}}
\tablenotetext{e}{$\Delta$C=C(assumed for the model)$-$C(derived from C$_{2}$ bands). For the RCB stars assumed C/He=1\% or C=9.5 dex, and for FF object assumed C/He=10\% or C$=$10.5 dex.}
\end{deluxetable}
\end{landscape}
\clearpage
| {
"redpajama_set_name": "RedPajamaArXiv"
} | 5,493 |
\section{Introduction}
This article is devoted to the computation of the Malgrange-Galois groupoid of the Painlevé VI family,
\begin{eqnarray*}
u'' =
\frac{1}{2} \left( \frac{1}{u} + \frac{1}{u-1} + \frac{1}{u-x} \right){u'}^2 -\left( \frac{1}{x} + \frac{1}{x-1} + \frac{1}{u-x} \right){u'}
\\
+ \frac{ u(u-1)(u-x)}{x^2(x-1)^2} \left( \frac{c^2}{2} - {\frac{a^2}{2}}\frac{x}{u^2} + {\frac{b^2}{2}}\frac{x-1}{(u-1)^2} + \frac{1-e^2}{2} \frac{x(x-1)}{(u-x)^2} \right);
\end{eqnarray*}
including the parameters $a\,,b\,,c\,,\,e$ as variables. In order to make use of the invariant volume, we consider the Painlevé VI family in its Hamiltonian form, as a vector field in $\mathbb C^3\times\mathbb C^4$:
\begin{align*}
\vec X &= \frac{\partial}{\partial x} + \frac{\partial H}{\partial q} \frac{\partial}{\partial p} -\frac{\partial H}{\partial p}\frac{\partial}{\partial q} \tag{$\rm HP_{VI}$}\\
H &= \frac{1}{x(x-1)} \Big[ p(p-1)(p-x)q^2 - \Big( a(p-1)(p-x) + \\
& b p(p-x) +(e-1)p(p-1)\Big) q +\frac{1}{4}\big( (a+b+e-1)^2 -c^2\big) (p-x) \Big].
\end{align*}
Our main result is that the Malgrange-Galois groupoid of the Painlevé VI family is the algebraic pseudogroup of transformations of $\mathbb C^7$ fixing the values of the parameters (here $\bar\pi : \mathbb C^3 \times \mathbb C^4 \to \mathbb C^4$ stands for the cartesian projection in the parameter space),
inducing a translation on the $x$-axis, preserving the volume form $dp\wedge dq \wedge dx$ and preserving $\vec X$. \\
{\noindent \bf Theorem \ref{th:Final_Mal_groupoid_PVI}. }{\it
The Malgrange-Galois groupoid of Painlev\'e VI equation is given by
\begin{align}
\mathrm{Mal}(\vec X)
&= \Big\{ \phi \mid
\bar\pi \circ \phi = \bar\pi; \;\; \phi^*(dx) = dx; \;\; \phi_*(\vec X)= \vec X;\nonumber
\\
& \qquad \qquad \phi^*(dp\wedge dq) \equiv dp\wedge dq \mod da,db,dc,de,dx
\Big\}.
\end{align}
}
As a consequence we have a functional hyper-transcendence result concerning parameter dependent solutions of PVI. A solution depending analytically on the parameters cannot satisfy any partial differential equation involving derivatives with respect to the independent variable and the parameters, except those equations that are derived from Painlevé VI.\\
{\noindent \bf Corollary \ref{co:par}. }{\it
If $y(x,a,b,c,e)$ is a parameter dependent solution of the sixth Painlev\'e equation then its annihilator in $\mathcal O_{J(\mathbb C^5, \mathbb C)}$ is the $\partial$-ideal generated by the sixth Painlev\'e equation.}\\
Let us us briefly recall the historical development of differential Galois theory. The ideas of E. Galois on algebraic equations were extended by E. Picard \cite{picard1887equations} and E. Vessiot \cite{vessiot1904theorie} to linear differential equations. In his Ph.D. thesis \cite{drach1898essai} J. Drach attempted to define a Galois group for nonlinear differential equations. In his article, the group-like objet is an algebraic pseudogroup. E. Vessiot spent years to correct Drach's mistake. He succeded is \cite{vessiot1946generale} but his work was forgotten. Later, E. Kolchin \cite{kolchin1973} developed Picard-Vessiot theory in the framework of differential field extensions and algebraic groups. To an order $n$ linear differential equation with coefficients in the differential field $\left( \mathbb C(x), \frac{\partial}{\partial x} \right)$ is associated an algebraic subgroup of ${\rm GL}_n(\mathbb C)$. This group measures algebraic relations between a basis of solutions and theirs derivatives.\\
At the turn of the 21st century, H. Umemura \cite{umemura1996differential} and B. Malgrange \cite{malgrange2001} independently defined a group-like objet associated to a (nonlinear) differential equation. In Umemura's work this object is a Lie-Ritt functor, a special case of infinitesimal group. In Malgrange's work the object is an algebraic pseudogroup ($\mathcal D$-groupoid) that we call Malgrange-Galois groupoid. Simultaneously, an important refinement of the Galois group for linear differential equations was introduced by P. Landesman \cite{landesman2008generalized} and by P. J. Cassidy and M. F. Singer \cite{cassidy2005galois}. The ``parameterized'' Galois group of an order $n$ linear differential equation with coefficients\footnote{The extension of parameterized Picard-Vessiot theory that allows coefficients in finitely generated fields can be consulted in \cite{sanz2018differential}.} in $\left(\mathbb C(t,x),\frac{\partial}{\partial x}\right)$ ($t$ is a set of parameters), is an algebraic differential subgroup of
$\left( {\rm GL}_n(\mathbb C(t)), \frac{\partial}{\partial t}\right)$. This ``group'' is a special case of intransitive pseudogroup and it measures the differential relations between a basis of solution when one derives them with respect to $x$ and to $t$. \\
In Malgrange's nonlinear differential Galois theory there is no distinction between parameters and variables. Therefore, in the complex case, the classical and parameterized Picard-Vessiot theory can be submersed in the nonlinear differential Galois theory. The corresponding differential Galois groups are recovered as the diagonal part of the Malgrange-Galois groupoid. Thus, the Malgrange-Galois groupoid of Painlevé VI family is the nonlinear analog of the differential algebraic Galois group in parameterized Picard-Vessiot theory. Nonlinear monodromy of Painlev\'e VI gives elements of Malgrange-Galois groupoid. The nonlinear monodromy seen as a selfmap on $\mathbb C ^2 \times \mathbb C^4$ preserving the projection on the parameters space $\mathbb C^4$ does not satisfy a common algebraic partial differential equation with coefficients in $\left(\mathbb C (p,q,a,b,c,e), \frac{\partial }{\partial p}, \frac{\partial }{\partial q}, \frac{\partial }{\partial a}, \frac{\partial }{\partial b}, \frac{\partial }{\partial c}, \frac{\partial }{\partial e}\right)$ which is not a consequence of the invariance of $dp\wedge dq \wedge da \wedge db \wedge dc \wedge de$. \\
The first attempt to compute the Galois groupoid of a Painlev\'e equation\footnote{For fixed values of parameters.} are done by P. Painlev\'e \cite{painleve1902irreductibilite} and J. Drach \cite{drach1915irreductibilite}, based on the wrong definition of Drach. Computations are now done for all Painlev\'e VI equations \cite{cantat2009dynamics} and for general values of parameters for Painlevé I to V \cite{casaledavy2020}. These works consider fixed values of parameters and compute the Malgrange-Galois groupoid of the corresponding vector field on $\mathbb C^3$. Specifically for the parameters not in the Picard set we have:
\begin{theo}[Cantat-Loray, \cite{cantat2009dynamics}]
For parameters $( a,b,c,e) $ not in Picard parameters set, Malgrange-Galois groupoid is given by:
\begin{align*}
\mathrm{Mal}(\vec X|_{\mathbb C^3 \times \{ ( a,b,c,e)\} } )
&=
\Big\{
\phi : ( \mathbb C^3, *) \overset{\sim}{\longrightarrow} ( \mathbb C^3,\star) \mid \phi^* dx =dx;\,\, \phi_*(\vec X)= \vec X;
\\
& \quad \quad \quad \phi^* dp\wedge dq = dp \wedge dq \mod dx
\Big\}
\end{align*}
In this formula $a,b,c,e$ are fixed parameters. The asterisk and the star stand for arbitrary points in $\mathbb C^3$.
\end{theo}
These groupoids are ``transversally simple''. Therefore the Malgrange-Galois groupoid of the family has to be a $\mathcal D$-groupoid in $\mathbb C^7$ whose restriction to specific values of parameters yields the ``transversally simple groupoid'' of volume preserving transformations. This situation is similar to the linear case treated in \cite{sanz2018differential}, where the computations were carried out by mean of the so-called Cassidy theorem \cite{cassidy1989classification} on the structure of differential subgroups of simple groups. In the nonlinear situation, one has to replace Cassidy theorem by an algebraic version of Kiso-Morimoto theorem \cite{kiso1979local}, and apply the non-existence of isomonodromic transformations to get our main result.
\subsection{Organization of the paper} In section \S\ref{s:Dgroupoids} we recall the notion of algebraic pseudogroup (which from now on will be termed $\mathcal D$-groupoid). Our exposition is a simplified version of that in \cite{malgrange2001, malgrange2010pseudogroupes}. This simplification is justified in Appendix \ref{App:A} (see also \cite{Damien2016specialisation}). In section \S\ref{s:Galois} we introduce the Malgrange-Galois groupoid of a vector field, which is a particular case of the notion of Galois groupoid of a foliation treated in \cite{malgrange2001, malgrange2010pseudogroupes}. In \S\ref{sec:PainleveVI} we put together all the pieces to carry on the computation of the Malgrange-Galois groupoid of Painlev\'e VI family. A key point in the computation is an algebraic version of Kiso-Morimoto theorem whose proof is given in Appendix \ref{s:kiso}.
\subsection{Notation and conventions} Given a complex irreducible smooth algebraic variety $M$ we denote $\mathbb C(M)$ the field of rational functions on $M$, $\mathfrak X_M$ the Lie algebra of rational vector fields on $M$ and $\Omega_M^k$ the space of rational $k$-forms on $M$. Given two smooth complex algebraic varieties $M$ and $N$ we denote by $J_k(M,N)\to M\times N$ the bundle of $k$-jets of germs of holomorphic maps from open subsets of $M$ to $N$. ${\rm Aut}_k(M)\subset J_k(M,M)$ is the Zariski open subset of $k$-jets of local biholomorphisms. The varieties $M$ and $N$ can be pointed, so that $J_k((M,p),(N,q))$ denotes the space of $k$-jets of local holomorphisms with source $p$ and target $q$. For a bundle $E\to M$ we use the notation $J_k(E/M)$ and $J(E/M)$ for the bundle $k$-jets and infinite order jets of sections respectively, $J_k(E/M)$ is a subset of $J_k(M,E)$. In any of the above cases the absence of the subindex $k$ means that we take the projective limit of the jet bundles so that $J(M,N)\to M\times N$ represents the bundle of jets of infinite order (or formal maps) from $M$ to $N$, and so on.
\section{$\mathcal D$-groupoids}\label{s:Dgroupoids}
In this section we develop the basic tools for dealing with $\mathcal D$-groupoids. For more details and complete proofs see \cite{malgrange2001, malgrange2010pseudogroupes, casaledavy2020}. Note that the first analytic definition of $\mathcal D$-groupoid in \cite{malgrange2001} was replaced by an algebraic one in \cite{malgrange2010pseudogroupes}. Here we present an alternative definition (Definition \ref{def:D_groupoid}) with the advantage of not referring to the differential ring sheaf structure of the regular functions on ${\rm Aut}(M)$; the equivalence with Definition 5.2 in \cite{malgrange2010pseudogroupes} is given in appendix. Along this section, $M$ is an affine smooth irreducible complex variety
and $\mathbb C(M)$ is its field of rational functions.
\subsection{Rational groupoids of Gauge transformations}
For the general definitions of grou\-poid, topological groupoid, differentiable groupoid, and Lie groupoid we refer the reader to \cite{mackenzie1987lie, mackenzie2005general}. Let $\pi\colon P\to M$ be a principal bundle modeled over an affine algebraic group $G$. Without loss of generality we assume that $P$ is also an affine and irreducible smooth variety. We consider the groupoid $(s,t)\colon {\rm Iso}(P)\to M\times M$ of $G$-equivariant transformations between the fibers of $\pi$.
$${\rm Iso}(P) = \left\{\sigma \colon P_x\to P_y \,\colon x,y\in M,\,\,\sigma\,\mbox{ is }\,G\mbox{-equivariant}\right\}.$$
It is an algebraic variety and a Lie groupoid locally isomorphic -in the analytic topology- to $M \times M \times G$.
\begin{definition}
A rational groupoid of gauge transformations of $P$ is a subvariety $\mathcal G\subseteq {\rm Iso}(P)$ satisfying:
\begin{itemize}
\item[(a)] There is an open subset $U\subset M$ such that $(s,t)\colon \mathcal G|_U\to U\times U$ is a differentiable groupoid.
\item[(b)] For any open subset $U\subset M$ $\mathcal G|_U$ is Zariski dense in $\mathcal G$.
\end{itemize}
\end{definition}
\begin{remark}
Condition (b) means that $\mathcal G$ is defined on the generic point of $M$. It means that irreducible components of $\mathcal G$ dominate $M$ by the source and by the target projections. If $\mathcal G$ is irreducible and satisfies condition (a) then it satisfies condition (b).
\end{remark}
We give three examples showing different pathologies.
\begin{example}
Let us consider $G = \{e\}$, $S$ a non-trivial closed subset of $M$ with at least two different points, and $U = M\setminus S$. In such case we have ${\rm Iso}(P) = M\times M$. Let us define:
$$\mathcal G = \{(x,x) : x\in M\}\cup (S\times S)$$
It is a subgroupoid of $M\times M$. Moreover $\mathcal G|_{U}$ is a differentiable groupoid, but $\mathcal G|_U$ is not Zariski dense in $\mathcal G$, and therefore $\mathcal G$ is not a rational groupoid.
\end{example}
\begin{example}
Let $M = \mathbb C^n$ $P= M \times \{e\}$ and $P,Q$ two polynomials in $n$ variables such that ${\rm g.c.d.}(P, Q) =1$.
The subvariety
$$
G =\{(x,y) \in M\times M : P(x)Q(y) = Q(x)P(y)\}
$$
is not a subgroupoid but it is a rational subgroupoid as its restriction on the complement of the indeterminacy locus $ \{P=0, Q=0\}$ and the singular locus $\{QdP-PdQ=0\}$ is a differentiable groupoid.
\end{example}
\begin{example}
Let $M = \mathbb C$ and $P = M\times \mathbb C^\ast$. The subvariety of ${\rm Iso}(P) = M\times M \times \mathbb C^\ast$ given by
$$
G = \{(x,y,u) \in {\rm Iso}(P) : yu=x \}
$$
is a subgroupoid but not a differentiable subgroupoid. It is a differentiable subgroupoid in restriction to $U=\mathbb C^\ast\subset M$.
\end{example}
A useful way of defining a groupoid is through its invariants. Let $F\colon P\to V$ be a rational map. We define the groupoid ${\rm Sym}(F)$ of symmetries of $F$ as the Zariski closure of the set:
$$
\left\{\sigma\in {\rm Iso}(P)\colon
\forall p\in P_{s(\sigma)}\cap {\rm dom}(F)\, , \, F(p) = F(\sigma(p))
\right\}
$$
Moreover, for a given subset $\mathbb F\subseteq \mathbb C(P)$ of rational functions we have,
$${\rm Sym}(\mathbb F) = \bigcap_{F\in \mathbb F} {\rm Sym}(F).$$
Reciprocally, given a rational groupoid of Gauge transformations $\mathcal G$ it has an associated field of invariants:
$$
{\rm Inv}(\mathcal G) = \{f\in \mathbb C(P) \colon \forall \sigma\in\mathcal G,\, \forall p\in P_{s(\sigma)}\cap {\rm dom}(f),\, f(p) = f(\sigma(p))\}.
$$
Following \cite[Proposition 2.18]{sanz2018differential}, there is a natural correspondence:
\begin{proposition}[Galois correspondence]\label{pro:Galois_correspondence1}
The assignation $\mathcal G\leadsto {\rm Inv}(\mathcal G)$
is a bijective correspondence (and anti-isomorphism of lattices) between the set of rational groupoids of gauge transformations of $P$ and the set of $G$-invariant subfields of $\mathbb C(P)$ containing $\mathbb C$. Its inverse is given by $\mathbb F \leadsto {\rm Sym}(\mathbb F)$.
\end{proposition}
Let $\vec X$ be a $G$-invariant rational vector field on $P$. Its field of rational first integrals is,
$$\mathbb C(P)^{\vec X} = \{f\in \mathbb C(P) \,\colon\, \vec X \cdot f =0\}.$$
The $G$-invariance of $\vec X$ implies that of the field $\mathbb C(P)^{\vec X}$.
\begin{definition}\label{df:Galois}
The Galois groupoid of the $G$-invariant rational vector field $\vec X$ is,
$${\rm Gal}(\vec X) = {\rm Sym}(\mathbb C(P)^{\vec X}).$$
\end{definition}
\begin{example}
Some examples of Galois groupoids.
\begin{enumerate}
\item Let us consider $M = \mathbb C^2$ with coordinates $x$ and $\alpha$, $G = \mathbb C^*$ with coordinate $u$ and $P = M\times G$. Let us consider $\vec X = \frac{\partial}{\partial x} + \alpha u\frac{\partial}{\partial u}$. Then $\alpha$ is a rational first integral of $\vec X$, moreover the integral curves of $\vec X$ with $\alpha \neq 0$ are Zariski dense in planes with constant $\alpha$. Therefore $\mathbb C(P)^{\vec X} = \mathbb C(\alpha)$ and:
$${\rm Gal}(\vec X) = \{(x_1,\alpha_1,x_2,\alpha_2,\sigma) \colon \alpha_1 = \alpha_2 \} \subset M \times M\times G = {\rm Iso}(P).$$
\item Let us consider $M = \mathbb C$ with coordinate $x$, $G = {\rm GL}_2(\mathbb C)$ with coordinates $u_{11}$, $u_{12}$, $u_{21}$, $u_{22}$ and $P = M\times G$. Let us consider the vector field associated to the fundamental form of Airy equation,
$$\vec X = \frac{\partial}{\partial x} + u_{21}\frac{\partial}{\partial u_{11}} +
u_{22}\frac{\partial}{\partial u_{12}} + xu_{11}\frac{\partial}{\partial u_{21}} +
xu_{12}\frac{\partial}{\partial u_{22}}.$$
It is well known that the only rational invariant of this vector field is the determinant function,
$$\Delta = u_{11}u_{22} - u_{12}u_{21}$$
and therefore,
$${\rm Gal}(\vec X) = {\rm Sym}(\mathbb C(\Delta)) =
\{(x_1,x_2,\sigma)\,\colon\, \sigma\in{\rm GL}_2(\mathbb C),\, {\det}(\sigma) = 1\}.$$
\end{enumerate}
\end{example}
\begin{remark}
Definition \ref{df:Galois} is a particular case of the more general notion of Galois groupoid for a partial $G$-invariant connection given in \cite[Definition 2.21]{blazquez2019differential}. It suffices to consider that $\langle \vec X \rangle$ as a partial $G$-invariant $\langle \pi_*(\vec X ) \rangle$-connection. The reader may check \cite[\S.2]{blazquez2019differential} for a more detailed account of the definition of the Galois groupoid and further examples.
\end{remark}
\subsection{Frame bundles}
Let $m$ be the complex dimension of $M$. Let $\Gamma_k$ be the group of $k$-jets of germs of biholomorphisms of $(\mathbb C^m,0)$. They form a projective systems of affine algebraic groups,
$$
\ldots \to \Gamma_k \to \Gamma_{k-1} \to \ldots \to \Gamma_1 \simeq {\rm GL}_m(\mathbb C) \to \{1\}.
$$
The projective limit $\Gamma = \lim_k \Gamma_k$ exists as a pro-algebraic group: the group of formal automorphims of $(\mathbb C^m,0)$.
\begin{definition}
A $k$-frame in $M$ is the $k$-jet at $0\in \mathbb C^m$ of a germ of biholomorphism from $(\mathbb C^m, 0)$ to $ M$.
\end{definition}
The set of $k$-frames ${\rm R}_kM$ is a variety endowed of a natural projection onto $M$, $j^k_0 \varphi \mapsto \varphi(0)$, and an action of $\Gamma_k$ by composition on the right side that give to ${\rm R}_kM\to M$ the structure of a principal $\Gamma_k$-bundle. The structures for different orders $k$ are compatible, in the sense that the following diagram:
$$
\xymatrix{\Gamma_{k+1} \times {\rm R}_{k+1}M \ar[r]\ar[d] & {\rm R}_{k+1}M\ar[d] \\
\Gamma_{k} \times {\rm R}_{k}M \ar[r] & {\rm R}_{k}M} \quad
\xymatrix{ (j^{k+1}_0\psi , j^{k+1}_0\varphi) \ar[r]\ar[d] & j_0^{k+1}(\psi\circ\varphi)\ar[d] \\
(j^{k}_0\psi , j^{k}_0\varphi) \ar[r] & j_0^k(\psi\circ\varphi)}
$$
is commutative. We have a chain of projections,
$$
\ldots \to {\rm R}_kM \to {\rm R}_{k-1}M \to \ldots \to {\rm R}_1M \simeq {\rm L}({\rm T}M) \to M.
$$
The projective limit ${\rm R}M = \lim_k {\rm R}_kM$ exists as a pro-algebraic bundle and it is principal $\Gamma$-bundle. We have a chain of field extensions:
$$
\mathbb C \subset \mathbb C(M) \subset \mathbb C({\rm R}_1M) \subset \mathbb C({\rm R}_1M) \subset \ldots \subset \mathbb C({\rm R}M) =
\bigcup_{k} \mathbb C({\rm R}_kM)
$$
where elements in $\mathbb C({\rm R}_kM)$ are termed \emph{differential functions} of order $k$. The field $\mathbb C({\rm R}M)$ has an additional structure of a differential field with set of derivations $\Delta = \{\delta_1,\ldots,\delta_m\}$, the set of total derivative operators with respect to the cartesian coordinates $\varepsilon_1,\ldots,\varepsilon_m$ in $\mathbb C^m$. This structure will be called $\Delta$-field. To describe these derivations let us introduce the following functions built from $x_1,\ldots,x_m$ is a transcendence basis of $\mathbb C(M)$ :
$$
\begin{array}{rrcl}
\delta^\alpha x_j :& {\rm R}M &\rightarrow &\mathbb C \\
& \varphi & \mapsto & \partial^\alpha(x_i(\varphi))(0)
\end{array}
$$
Then we will have that the set
$\{\delta^{\alpha}x_j\}_{1\leq |\alpha|\leq k}$ is a transcendence basis and a system of generators of $\mathbb C({\rm R}_kM)$ over $\mathbb C(M)$, thus
$$
\mathbb C(R_kM) = \mathbb C(M)(\{\delta^{\alpha}x_j\,\,\colon\,\,j = 1,\ldots,m\,\, |\alpha|\leq k\}).
$$
The total derivative operators find their formal expression in coordinates as:
$$
\delta_j = \sum_{|\alpha|\geq 0} \sum_{\ell=1}^m (\delta_j\delta^{\alpha}x_\ell)\frac{\partial}{\partial (\delta^{\alpha}x_\ell)}.
$$
There is a geometric mechanism of extension of vector fields (see, for instance \cite[\S 1.8 p. 29]{malgrange2010pseudogroupes} variational equations of higher order) such that any vector field $\vec X$ in $M$ extends to a unique $\Gamma_k$-invariant rational vector field $\vec X^{(k)}$ in ${\rm R}_kM$ with the properties:
\begin{itemize}
\item[(1)] $\vec X^{(k)}$ projects onto $\vec X^{(k-1)}$, and $\vec X^{(0)} = \vec X.$
\item[(2)] The extension commutes with total derivative operators in the following sense:
$$\vec X^{(k+1)} \circ \delta_j = \delta_j \circ \vec X^{(k)}.$$
\end{itemize}
Note that if $\vec X$ is a rational vector field in $M$ then $\vec X^{(k)}$ is a rational vector field in ${\rm R}_kM$.
\begin{example}
Let us consider $M = \mathbb C^m$ with coordinate functions $x_i$. In such case ${\rm R}_k\mathbb C^n$ is an affine space coordinate by functions
$\delta^{\alpha} x_i$ denoted by $x_{i\colon \alpha}$ for short, with
$\alpha\in \mathbb Z_{\geq 0}^m$ and $|\alpha|\leq k$.
The function $x_{i\colon \alpha}$ represents the derivative of the coordinate functions of $\mathbb C^n$ with respect to the parameters $\varepsilon$ in $(\mathbb C^m,0)$ evaluated in $0$; so that $x_{j:0} = x_j$. The tuple
$(x_{j:\alpha}$ corresponds to the $k$-jet at $0$ of the polynomial map:
$$(\mathbb C^m,0) \to \mathbb C^m, \quad (\varepsilon_1,\ldots ,\varepsilon_m) \mapsto\left(\sum_{|\alpha|\leq k} \frac{x_{1:\alpha}}{\alpha!} \varepsilon_1^{\alpha_1}\cdots \varepsilon_m^{\alpha_m}, \ldots,
\sum_{|\alpha|\leq k} \frac{x_{m:\alpha}}{\alpha!} \varepsilon_1^{\alpha_1}\cdots \varepsilon_m^{\alpha_m} \right).$$
Total derivative operators are given by formal series:
$$\delta_i = \sum_{\alpha}\sum_{j=1}^m x_{j:\alpha+\epsilon_i}\frac{\partial}{\partial x_{j:\alpha}}.$$
The $k$-th extension of a vector field $\vec X = \sum f_i\frac{\partial}{\partial x_i}$ is given in coordinates as
$$\vec X^{(k)} = \sum_{i=1}^m \sum_{|\alpha|\leq k} (\delta^\alpha f_i)\frac{\partial}{\partial x_{i:\alpha}}.$$
\end{example}
In the limit we have an extension $\vec X^{(\infty)}$ which can be seen as a derivation of the field $\mathbb C({\rm R}M)$.
The assignation $\vec X\leadsto \vec X^{(\infty)}$ is compatible with the Lie bracket: $[\vec X_1^{(\infty)},\vec X_2^{(\infty)}] =[\vec X_1,\vec X_2]^{(\infty)}$. Let us consider $\mathfrak X_M$ the Lie algebra of rational vector fields on $M$. The field $\mathbb C({\rm R}M)$ has two natural commuting structures of differential field. On one hand, we consider the set $\Delta = \{\delta_1,\ldots,\delta_m\}$ of total derivative operators that endows it with an structure of $\Delta$-fields. On the other hand the Lie algebra $\mathfrak X_M$ acts by derivations in $\mathbb C({\rm R}M)$ endowing it with an structure of $\mathfrak X_M$-field.
\subsection{Groupoid of automorphisms of $M$}
\begin{definition}
\label{df:Aut_kM}
One denotes by $(s,t)\colon \mathrm{Aut}_k(M)\to M\times M$ be the groupoid of $k$-jets of local invertible biholomorphisms of $M$.
\end{definition}
$\mathrm{Aut}_k(M)$ acts by composition on the $k$-th frame bundle:
$$
\mathrm{Aut}_k(M) \times_M {\rm R_k}M \to {\rm R_k}M \quad (j_p^k \sigma, j_0^k\varphi)\mapsto j_0^k(\sigma \circ \varphi).
$$
A direct calculus show that the groupoid $\mathrm{Aut}_k(M)$ is, in fact, the groupoid of gauge transformations of the $k$-th frame bundle, $\mathrm{Aut}_k(M)\cong {\rm Iso}({\rm R}_k M)$. Given $j_p^k\sigma\in {\rm Aut}_k(M)$ with source $p$ and target $q$. Let us denote $(j_p^k\sigma)^{(k)}$ the induced $\Gamma_k$-equivariant map from $({\rm R}_kM)_p$ and $({\rm R}_kM)_q$:
$$
\xymatrix{
(\mathbb C^m, 0) \ar[r]^-{\varphi} \ar[rd]_-{\psi}& (M,p) \ar[d]^-{\sigma} \\ & (M,q)
}\quad (j^k_p\sigma)^{(k)}(j_p^k \varphi) = j_0^k\psi.$$
We also have a projective system:
$$
\mathrm{Aut}_k(M) \to \ldots \to \mathrm{Aut}_k(M) \to {\rm Aut}_0(M) = M\times M
$$
where the projective limit,
$$
{\rm Aut}(M) = \lim_{\leftarrow} {\rm Aut}_k(M)
$$
is the pro-algebraic groupoid of formal non-singular maps. As before, $\mathrm{Aut}(M)\cong {\rm Iso}({\rm R}M)$. Given $\sigma\in {\rm Aut}(M)$ with source $p$ and target $q$ we write $\sigma^{(\infty)}$ for the corresponding $\Gamma$-equivariant map from $({\rm R}M)_p$ to $({\rm R}M)_q$. This map $\sigma^{(\infty)}$ is the limit of a sequence of compatible maps;
$\sigma$ is a sequence $\{j^k_p\sigma_k\}$ with $j_k\sigma_k\in {\rm Aut}_k(M)$ such that $j_p^{\ell}\sigma_k = j_p^\ell\sigma_\ell$ for $\ell\leq k$, in the same way a frame $\varphi$ in $p$ is a sequence $\{j^k_0\varphi_k\}$ with $j_0^{\ell}\varphi_k = j_0^\ell\varphi_\ell$ for $\ell\leq k$. Finally $\sigma^{(\infty)}$ is defined by as the limit of the sequence $\{\sigma_k^{(k)}\}$, that is,
$\sigma^{(\infty)}(\varphi) = \{j^k_0 (\sigma_k\circ \varphi_k) \}$. In the convergent case, when $\sigma = j_p\bar \sigma$ and $\varphi = j_0\bar \varphi$ where $\bar\sigma$ and $\bar\varphi$ are local biholomorphisms we have simply $\sigma^{(\infty)}(\varphi) = j_0(\bar\sigma\circ\bar\varphi)$.
\subsection{$\mathcal D$-groupoids}
By a closed subset in ${\rm Aut}(M)$ we mean a subset that is closed in the initial topology with respect to all projections ${\rm Aut}(M)\to {\rm Aut}_k(M)$. Such closed subset $Z$ is a sequence of Zariski closed subsets $Z = \{Z_k\}_{k\in\mathbb N}$ such that:
\begin{itemize}
\item[(a)] $Z_k\subseteq {\rm Aut}_k(M)$
\item[(b)] $Z_{k+1}$ dominates $Z_k$ by projection.
\end{itemize}
With such definition, a formal jet $j_p\varphi$ is in $Z$ if and only if $j_p^k\varphi\in Z_k$ for all $k$.
\begin{definition}\label{def:D_groupoid}
We say that $\mathcal G$ ($\simeq \{\mathcal G_k\}_{k\in\mathbb N})$ is a $\mathcal D$-groupoid of transformations of $M$ if:
\begin{itemize}
\item[(a)] For all $k$, $\mathcal G_k$ is a rational subgroupoid of ${\rm Aut}_kM$.
\item[(b)] For any $f\in {\rm Inv}(\mathcal G_k)$ and $j = 1,\ldots, m$ we have $\delta_jf\in {\rm Inv}(\mathcal G_{k+1})$.
\end{itemize}
\end{definition}
\begin{remark}
$\mathcal D$-groupoids should be seen as spaces of solutions of certain PDE systems. We say that a local biholomorphisms $\sigma$ between open subset of $M$ in $\mathcal G$ if for all $p$ in its domain of definition $j_p\sigma\in\mathcal G$.
\end{remark}
\begin{example}
Let us consider
$\omega_m = f(x_1,\ldots,x_n)dx_1\wedge\ldots\wedge dx_m$ a rational $m$-form in $\mathbb C^m$. We may consider the subvariety of ${\rm Aut}_1(\mathbb C(M))$ of $1$-jets of biholomorphisms preserving $\omega_m$,
$$\mathcal G_1 = \left\{j_p^1\sigma\in {\rm Aut}_1(\mathbb C^m)|
\sigma^*(\omega_m)(p) = \omega_m(p)\right\}.$$
Its easy to check that $\mathcal G_1$ is a rational subgroupoid of ${\rm Aut}_1(\mathbb C^m)$. If we introduce coordinates,
$$x_i,y_j,y_{i;j} = \frac{\partial y_i}{\partial x_i}$$
in ${\rm Aut}_1(\mathbb C^m)$ then we can compute the equation of $\mathcal G_1$ obtaining:
\begin{equation}\label{eq:exvol}
\mathcal G_1 = \{(x_i,x_j,y_{i;\epsilon_j})\in {\rm Aut}_1(\mathbb C^m)\,\mid\, f(y_1,\ldots,y_m)\det(y_{i;j}) = f(x_1,\ldots,x_m) \}.
\end{equation}
Note that the equation of $\mathcal G_1$ is a first order PDE where $x_1,\ldots,x_m$ are the independent variables and $y_1,\ldots,y_m$ are the unknown functions. In order to derive higher order equations from \eqref{eq:exvol} there are two different ways.
\begin{enumerate}
\item[(a)] The classical way is to differentiate equation \eqref{eq:exvol} with respect to the independent variables (see Appendix \ref{App:A}) to obtain the equations of $\mathcal G_2$, $\mathcal G_3$ and so on.
\item[(b)] The other way is to differentiate the rational invariants determining $\mathcal G_1$ with respect to the infinitesimal parameters $\varepsilon_i$ to obtain the invariants of $\mathcal G_2$, $\mathcal G_3$ and so on. In order to compute the invariants of $\mathcal G_1$ in $\mathbb C({\rm R}\mathbb C^m)$ we need to interpret the $m$-form $\omega_m$ as a rational function in ${\rm R_1}\mathbb C^m$. For each $1$-frame $j_0^1\varphi$ in $p\in M$ with $p$ in the domain of $f$ we have,
$$\omega_m(p) \circ d_0\varphi = \lambda(j_0^1\varphi) d\varepsilon_1\wedge\ldots\wedge d\varepsilon_m.$$
This gives us the rational function $\lambda$ we are looking for $\lambda = \det(x_{i\colon \epsilon_j})f(x_1,\ldots,x_m)$ and $\mathcal G_1 = {\rm Sym}(\mathbb C(\lambda))$.
Therefore we obtain $\mathcal G_k = {\rm Sym}(\mathbb C(\delta^\alpha\lambda)_{0\leq|\alpha|\leq k}).$
\end{enumerate}
In order to give a complete geometric description of the whole $\mathcal D$-groupoid $\mathcal G$ let us recall than an element $\sigma = [j^1_p\sigma_1, j^2_p\sigma_2, j^3_p\sigma_3, \ldots]\in {\rm Aut}(M)$ with source $p$ and target $q$ can be seen as a formal isomorphism from $(\mathbb C^m,p)$ to $(\mathbb C^m,q)$. We see $\sigma^*(\omega_m)$ as a formal $m$-form near $p$:
$$\sigma^*(\omega_m) = [\sigma_1^*(\omega_m)(p), j^1_p\sigma_2^*(\omega_m)(p), j^2_p\sigma_3^*(\omega_m)(p), \ldots ]$$
and then we may give the description of $\mathcal G$ as the $\mathcal D$-groupoid of formal symmetries of $\omega_m$,
that is, its elements are the formal isomorphisms $\sigma$ such that $\sigma^*(\omega_m) = j_p\omega_m$.
Moreover, as a rational $m$-form is completely determined by its infinite order jet at any point of its domain it is usual to identify $\omega_m$ with its jet $j_p\omega_m$ and just write:
$$\mathcal G = \left\{\sigma\in {\rm Aut}(\mathbb C^m)\,|\,
\sigma^*(\omega_m) = \omega_m\right\}.$$
\end{example}
Given a $\mathcal D$-groupoid $\mathcal G$ we define its field of differential invariants as:
$$
{\rm Inv}_\Delta(\mathcal G) = \bigcup_k {\rm Inv}(\mathcal G_k) \, \subseteq \,\, \mathbb C({\rm R}M).
$$
Condition (b) implies that ${\rm Inv}_\Delta(\mathcal G)$ is a $\Delta$-subfield of $\mathbb C({\rm R}M)$. Reciprocally, given a $\Delta$-subfield $\mathbb F$ we can define its $\mathcal D$-groupoid of symmetries ${\rm Sym}_\Delta(\mathbb F) \simeq \{{\rm Sym}_\Delta(\mathbb F)_k\}_{k\in\mathbb N}$ with:
$$
{\rm Sym}_\Delta(\mathbb F)_k = {\rm Sym}(\mathbb F \cap \mathbb C({\rm R}_kM) ).
$$
We have a Galois correspondence between $\mathcal D$-groupoids and their $\Delta$-fields of differential invariants.
\begin{proposition}[$\Delta$-Galois correspondence]\label{pro:Galois_correspondence2}
The assignation $\mathcal G\leadsto {\rm Inv}_\Delta(\mathcal G)$ is a bijective correspondence (and anti-isomorphism of lattices) between the set of $\mathcal D$-groupoids of transformations of $M$ and $\Gamma$-invariant $\Delta$-subfields of $\mathbb C(P)$ containing $\mathbb C$. Its inverse is given by $\mathbb F\leadsto {\rm Sym}_\Delta(\mathbb F)$
\end{proposition}
\begin{proof}
Note that if $\mathbb F$ is a $\Gamma$-invariant $\Delta$-subfield of $\mathbb C({\rm R}M)$ containing $\mathbb C$ then the intersection $\mathbb F\cap \mathbb C({\rm R}_kM)$ is a $\Gamma_k$-invariant subfield of $\mathbb C({\rm R}_kM)$ containing $\mathbb C$. The proposition follows by application of Proposition \ref{pro:Galois_correspondence1}.
\end{proof}
Le $\mathcal G =\{\mathcal G\}_k$ be a $\mathcal D$-groupoid in $M$. By a finiteness theorem of Kolchin \cite[Proposition 14, p 112]{kolchin1973} the $\Delta$-field ${\rm Inv}_{\Delta}(\mathcal G)$ is $\Delta$-finitely generated. There is a minimum $r$ such that ${\rm Inv}_{\Delta}(\mathcal G)$ in $\Delta$-generated by ${\rm Inv}(\mathcal G_r)$. This minimum $r$ is the order of the $\mathcal D$-groupoid $\mathcal G$. Moreover, let $U\subset M$ be a Zariski open subset such that $\mathcal G_r|_U$ is a groupoid, we have that $\mathcal G|_U$ is a groupoid.
\subsection{$\mathcal D$-Lie algebras}
Let us now consider the jet bundle $J({\rm T}M/M)\to M$. Its elements are formal vector fields in $M$
(that is, continuous derivations of the completed local rings $\hat{\mathcal O}_{M,p}$ for $p\in M$)
and thus it is a bundle by Lie algebras with the usual Lie bracket of formal vector fields. Therefore, the space of rational sections
$$
\Gamma_{\rm rat}(J({\rm T}M/M)) = \lim_{\leftarrow} \Gamma_{\rm rat}(J_k({\rm T}M/M))
$$
is a $\mathbb C(M)$-Lie algebra.
\begin{remark}
There is also a Lie algebroid structure for each order $J_k({\rm T}M/M)$, and then a natural Lie algebra structure in each space of sections $\Gamma_{\rm rat}(J_k({\rm T}M/M))$ given by the so-called Spencer bracket. This Lie algebra structure is different from the one we consider, but they coincide along holonomic sections.
\end{remark}
\begin{definition}
A rational linear sub-bundle of $J_k({\rm T}M/M)$ is an irreducible Zariski closed subset $V\subset J_k({\rm T}M/M)$ such that there is a Zariski open subset $U\subseteq M$ such that $V|_U\to U$ is a vector bundle.
\end{definition}
\begin{example}
Order $0$ rational linear sub-bundles of $J_0(TM/M) = TM$ are singular distributions of vector fields $\mathscr F\subset TM$. If $U$ is the complement of the singular set of $\mathscr F$ then $\mathscr F|_U\to U$ is a vector bundle.
\end{example}
As linear bundles are rationally trivial, a rational linear sub-bundle is characterized by its space of rational sections $\Gamma_{\rm rat}(V_k)\subset \Gamma_{\rm rat}(J_k({\rm T}M/M))$. There is a natural bijective correspondence between rational linear sub-bundles of $J_k({\rm T}M/M)$ and $\mathbb C(M)$-subspaces of $\Gamma_{\rm rat}(J_k({\rm T}M/M)).$ In the projective limit $J({\rm T}M/M)$ we consider the initial topology. Thus, a closed subset $V\subset J({\rm T}M/M)$ is a sequence $V = \{V_k\}_{k\in \mathbb N}$ with $V_k$ Zariski closed in $J_k({\rm T}M/M)$ and such that
$V_{k+1}$ dominates $V_{k}$ by projection.
\begin{definition}
A rational linear sub-bundle of $J({\rm T}M/M)$ is an irreducible closed subset $V = \{V_k\}_{k\in \mathbb N}$ such that for all $k$ we have that $V_k$ is a rational linear sub-bundle of $J_k({\rm T}M/M)$.
\end{definition}
Analogously, there is a natural bijective correspondence between rational linear sub-bundles of $J({\rm T}M/M)$ and $\mathbb C(M)$-subspaces of $\Gamma_{\rm rat}(J({\rm T}M/M))$. In what follows we will define the notion of $\mathcal D$-Lie algebra, a certain kind of $\mathbb C(M)$-subspace of \linebreak $\Gamma_{\rm rat}(J({\rm T}M/M))$; but we will apply the notion indistinctly to rational linear sub-bundles of $J({\rm T}M/M)$.
\begin{definition}
The ring of rational differential operator in $M$ is the ring $\mathcal D_M$ of $\mathbb C$-linear endomorphisms of $\mathbb C(M)$ generated by:
\begin{itemize}
\item[(a)] $\mathbb C(M)$ acting by multiplication, that is, each element $h\in\mathbb C(M)$ is seen as an endomorphism $h\colon \mathbb C(M)\to \mathbb C(M)$, $f\mapsto hf$.
\item[(b)] The Lie algebra of rational vector fields $\mathfrak X_M$, that is, rational vector field $\vec X\in\mathfrak X_M$ is seen as a derivation $\vec X\colon \mathbb C(M)\to \mathbb C(M)$, $f\mapsto \vec Xf$.
\end{itemize}
\end{definition}
We set the degree of rational functions equal to $0$ and that of vector fields to be $1$. Thus, $\mathcal D_M$ is a non commutative graded ring, for $h\in\mathbb C(M)$ and $\vec X\in \mathfrak X_M$ we have,
$$f\circ \vec X = f\vec X, \quad \vec X \circ f = \vec Xf + f\vec X.$$
The inclusion $\mathbb C(M)\subset \mathcal D_M$ endows $\mathcal D_M$ with a $\mathbb C(M)$-bimodule structure with a
left ($f\circ\theta\colon g \mapsto f(\theta(g))$) and a right ($\theta\circ f\colon g \mapsto \theta(fg)$) multiplication.
Let $\Omega_M^1 = \mathfrak X_M^*$ be the $\mathbb C(M)$-space of rational $1$-forms in $M$. We define the space of rational differential operators
from $\mathfrak X_M$ to $\mathbb C(M)$ as the tensor product:
$$
{\rm Diff}(\mathfrak X_M,\mathbb C(M)) = \mathcal D_M\tens_{\mathbb C(M)} \Omega_M^1.
$$
Note that the tensor product is constructed by means of the right $\mathbb C(M)$-module structure in $\mathcal D_M$. Thus, ${\rm Diff}(\mathfrak X_M,\mathbb C(M))$ is a left $\mathcal D_M$-module. \\
Differential operators can be seen as rational functions on $J({\rm T}M/M)$ that are linear along fibers of the projection $J({\rm T}M/M)\to M$.
The coupling of a rational differential operator and a formal vector field $\vec X\in J({\rm T}M/M)$ with base point $p$ is given by $(L \otimes \omega)(\vec X) = L(\omega(X))(p)$. Note that, if $p$ is in the domain of $\omega$ and $L$ and $\omega(X)$ and $L(\omega(\vec X))$ are formal functions at $p$. When we couple a rational differential operator with a rational section of $J({\rm T}M/M)$ we obtain a rational function. Moreover we have a duality:
$$
{\rm Diff}(\mathfrak X_M,\mathbb C(M))^* = \Gamma_{\rm rat}(J({\rm T}M/M)).
$$
\begin{definition}\label{def:DLie_alg}
Let $\mathcal L = \{\mathcal L_k\}_{k\in \mathbb N}$ be a rational linear sub-bundle of $J({\rm T}M/M)$. We say that $\mathcal L$ is a $\mathcal D$-Lie algebra of transformations of $M$ if:
\begin{itemize}
\item[(a)] Its space of sections $\Gamma_{\rm rat}(\mathcal L)$ is a Lie subalgebra of $\Gamma_{\rm rat}(J({\rm T}M/M))$.
\item[(b)] Its $\mathcal D_M$-module of vanishing differential operators:
$$
{\rm ann}(\mathcal L) = \{\theta\in {\rm Diff}(\mathfrak X_M,\mathbb C(M))\,\colon \, \forall \vec X \in \mathcal L \,\,\, \theta(\vec X) = 0 \}
$$
is a $\mathcal D_M$-submodule of ${\rm Diff}(\mathfrak X_M,\mathbb C(M))$.
\end{itemize}
\end{definition}
\begin{remark}
$\mathcal D$-Lie algebras should be seen as spaces of solutions of certain linear PDE systems. Let $\mathcal L = \{\mathcal L_k\}_{k\in \mathbb N}$ be a $\mathcal D$-Lie algebra in $M$ seen as a linear sub-bundle of $J({\rm T}M/M)$. We say that a local analytic vector field $\vec X$ is in $L$ if for all $p$ in its domain of definition $j_p\vec X\in\mathcal L$ (or equivalently, for all $p$ and $k$ $j_p^k\vec X\in \mathcal L_k$).
\end{remark}
\begin{remark}
For the sake of simplicity we presented the definition of $\mathcal D$-Lie algebra in terms of $\mathbb C(M)$-vector spaces instead as coherent sheaves of differential operators as it is done in \cite{malgrange2001} \S 3, \cite{malgrange2010pseudogroupes} \S 5.6 or \cite{le2010algebraic} \S 4. However, it is clear that the $\mathbb C(M)$-space of rational differential operators ${\rm ann}(L)$ in Definition \ref{def:DLie_alg} can be seen as a coherent $\mathcal D$-module of differential operators. In particular we have the following results:
\begin{itemize}
\item[(a)] The $\mathcal D$-Lie algebra and its module of vanishing differential operators ${\rm ann}(\mathcal L)$ determine each other.
\item[(b)] There is an open subset $U\subset M$ such that for all $k$ $L_k|_U$ is a linear bundle over $U$ (Proposition 4.1 in \cite{le2010algebraic}).
\end{itemize}
\end{remark}
\subsection{$\mathcal D$-Lie algebra of a $\mathcal D$-groupoid}
Given a $\mathcal D$-groupoid of transformations of $M$ its Lie algebra is usually defined by its infinitesimal generators: vector fields $\vec X$ such their exponential ${\rm exp}(t\vec X)$ is in $\mathcal G$ wherever it is defined.
The following definition, using differential invariants, is easier to use in the algebraic situation than the usual definition. The compatibility of the prolongation of $\vec X$ to $\vec X^{\infty}$ with the flow and the $\Delta$-Galois correspondence \ref{pro:Galois_correspondence2} implies the equivalence of the two point of view.
\begin{definition}
\label{def:lie_D_algebra}
Let $ \mathcal G = \{\mathcal G_k\}_{k\in \mathbb N}$ be a $\mathcal D$-groupoid of transformations of $M$.
The $\mathcal D$-Lie algebra of $\mathcal G$ is
$$
\mathrm{Lie}\, (\mathcal G) = \big\{ \vec X\in J({\rm TM}/M) \,\colon \,\vec X^{(\infty)} {\rm Inv}_\Delta(\mathcal G) = 0 \big\}
$$
\end{definition}
It is clear that as a sequence of rational linear sub-bundles we have ${\rm Lie}(\mathcal G) = \{{\rm Lie}(\mathcal G_k)\}_{k\in \mathbb N}$ where:
$$
{\rm Lie}(\mathcal G_k) = \big\{j_p^k X\in J_k({\rm TM}/M) \,\colon \,X^{(k)} {\rm Inv}(\mathcal G_k) = 0 \big\}
$$
Because $[\vec X_1^{(\infty)}, \vec X_2^{(\infty)}] =[\vec X_1, \vec X_2]^{(\infty)}$, the space of rational sections of ${\rm Lie}(\mathcal G)$ is closed by Lie bracket. Therefore, it is a $\mathcal D$-Lie algebra.
\begin{example}
Let us consider $\mathcal G$ the $\mathcal D$-lie groupoid of formal maps preserving an $m$-form $\omega_m$:
$$\mathcal G = \{\sigma\in {\rm Aut}(M) \,\mid\, \sigma^*(\omega_m) = \omega_m\}$$
its associated $\mathcal D$-Lie algebra is the $\mathcal D$-Lie algebra of formal infinitesimal symmetries of $\omega_m$:
$${\rm Lie}(\mathcal G) = \{\vec X\in J({\rm TM}/M) \,\mid\, {\rm Lie}_{\vec X}\omega_m = 0\}.$$
where, given a formal vector field $\vec X = [\vec X_0(p), j^1_p\vec X_1, j^1_p\vec X_1, \ldots ]$ the Lie derivative of $\omega_m$ is defined as a formal $m$-form near $p$,
$${\rm Lie}_{\vec X}(\omega_m) = [{\rm Lie}_{\vec X_1}(\omega_m)(p), j^1_p{\rm Lie}_{\vec X_2}(\omega_m)(p),
j^2_p{\rm Lie}_{\vec X_3}(\omega_m)(p), \ldots].$$
Moreover, from previous works \cite[Theorem 1.3.2]{casale2004groupoide} we know that transitive $\mathcal D$-groupoids are groupoids of formal symmetries of geometric structures. Their associated $\mathcal D$-Lie algebras are the $\mathcal D$-Lie algebras of formal infinitesimal symmetries of such geometric structures.
\end{example}
\section{Malgrange-Galois $\mathcal D$-groupoid}\label{s:Galois}
\subsection{Malgrange-Galois $\mathcal D$-groupoid of a vector field}
In this section $M$ is an affine smooth algebraic variety of dimension $m$. Let $\vec X$ be a rational vector field on $M$. Let us recall that for all $k$ the $k$-th extension $\vec X^{(k)}$ is a $\Gamma_k$-invariant vector field in $\mathbb C({\rm R}_kM)$ and $\vec X^{(\infty)}$ is a $\Gamma$-invariant derivation of $\mathbb C({\rm R}M)$.
\begin{definition}
The field of rational differential invariants of $\vec X$ is the field of constants of the derivation $\vec X^{(\infty)}$:
$$\mathbb C({\rm R}M)^{\vec X} = \{f\in \mathbb C({\rm R}M) \,\colon\, \vec X^{(\infty)}f = 0\}.$$
\end{definition}
Let us list some elementary self evident properties of $\mathbb C({\rm R}M)^{\vec X}$:
\begin{enumerate}
\item[(a)] As the derivation $\vec X^{\infty}$ commutes with the total derivative operators $\Delta = \{\delta_1,\ldots,\delta_m\}$ it follows that $\mathbb C({\rm R}M)^{\vec X}$ is a $\Delta$-subfield.
\item[(b)] Since for all $k$, $\vec X^{\infty}|_{\mathbb C({\rm R}_kM)} = \vec X^{(k)}$ we have that the field of rational differential invariants of order $\leq k$,
$$\mathbb C({\rm R}_kM)^{\vec X} = \mathbb C({\rm R}_kM)\cap \mathbb C({\rm R}M)^{\vec X}$$
is the field of rational first integrals of the vector field $\vec X^{(k)}$.
\item[(c)] Clearly:
$$\mathbb C({\rm R}M)^{\vec X} = \bigcup_{k} \mathbb C({\rm R}_kM)^{\vec X}.$$
\item[(d)] As $\vec X^{(\infty)}$ is $\Gamma$-invariant the $\Delta$-field of rational differential invariants $\mathbb C({\rm R}M)^{\vec X}$ is $\Gamma$-invariant.
\end{enumerate}
\begin{definition}\label{df:Mal_groupoid}
The Malgrange-Galois $\mathcal D$-groupoid of $\vec X$ is:
$${\rm Mal}(\vec X)= {\rm Sym}_\Delta(\mathbb C({\rm R}M)^{\vec X}),$$ the $\mathcal D$-Lie groupoid corresponding by means of the $\Delta$-Galois correspondence (Proposition \ref{pro:Galois_correspondence2}) to the $\Delta$-field $\mathbb C({\rm R}M)^{\vec X}$.
\end{definition}
The reader may be aware that this definition looks different from that in \cite{malgrange2001, malgrange2010pseudogroupes}, however they are equivalent, as discussed in appendix A. Indeed, the equivalence between three different definitions of ${\rm Mal}(\vec X)$, including Definition \ref{df:Mal_groupoid}
can be found in \cite[Th\'eor\`eme 3.16]{casaledavy2020}.
Computation of Malgrange-Galois $\mathcal D$-groupoid is in general very difficult. In \cite{casale2011}, the reader may found explicit examples written in coordinates such as the computation for linear differential equations or for first order differential equation. An easy case is given in the example \ref{ex:integrable} below.
\begin{remark}
Applying the definition of ${\rm Sym}_\Delta(\mathbb C({\rm R}M)^{\vec X})$ we may write a direct description of the Malgrange-Galois groupoid:
$${\rm Mal}(\vec X) = \left\{\sigma\in {\rm Aut}(M)\,\colon\,
\forall f\in\mathbb C({\rm R}M)^{\vec X}\,\, f \circ \sigma^{(\infty)}= f|_{({\rm} RM)_{s(\sigma)}}
\right\},$$
the Malgrange-Galois groupoid of $\vec X$ is the $\mathcal D$-groupoid of transformations of $M$ that fixes the rational differential invariants of $\vec X$. Remember that $\sigma^{(\infty)}$ is an ismorphism of ${\rm R}M_{s(\sigma)}$ on ${\rm R}M_{t(\sigma)}$.
\end{remark}
\begin{example}{\label{ex:integrable}}
The simplest Malgrange-Galois group is that of a completely integrable vector field. Let us consider $M = \mathbb C^m$
$\vec X = \partial_{x_1}$. In such case we have $\mathbb C(M)^{\vec X} = \mathbb C(x_2,\ldots,x_m)$. Incidentally the extension of $\vec X$ has the same coordinate expression $\vec X^{(\infty)} ={\partial}_{x_1}$ and therefore all the functions $x_{i:\alpha}$ with $|\alpha|>1$ are rational differential invariants of $\vec X$. Elements $\sigma\in {\rm Aut}(\mathbb C^m)$ preserving all the differential functions $x_{i:\alpha}$ is necessarily jets of a translations. Additionally, the invariance of $x_2,\ldots,x_m$ implies that such translation is done along the axis $x_1$. Therefore:
$${\rm Mal}(\vec X) = \{ j_p\tau_\lambda\,\mid p\in \mathbb C^m, \lambda\in \mathbb C\}$$
where $\tau_\lambda \colon (x_1,x_2\ldots,x_n)\to (x_1+\lambda,x_2\ldots,x_n)$ is the translation of magnitude $\lambda$ along the $x_1$ axis.
\end{example}
If is clear that as a sequence of rational subgroupoids of ${\rm Aut}_k(M)$ we have ${\rm Mal}(\vec X) = \{{\rm Mal}_k(\vec X)\}_{k\in\mathbb N}$ where
${\rm Mal}_k(\vec X)$ is the rational subgroupoid of ${\rm Aut}_k(M)$ whose field of invariants is $\mathbb C({\rm R}_kM)^{\vec X}$. Therefore:
$${\rm Mal}_k(\vec X) = {\rm Gal}(\vec X^{(k)}).$$
This ${\rm Mal}_k(\vec X)$ is termed the Malgrange-Galois groupoid of order $k$ of $\vec X$.
\subsection{Specialization theorem}
\label{sec:specialization_theo} Let us examine a parametric version of nonlinear Galois theory and a specialization result stated in \cite[\S4]{casaledavy2020}. We explain the relations between the formalism of \cite{casaledavy2020} and ours and state the theorem referring to \cite{casaledavy2020} for the proof.
Let $\rho\colon M\to S$ be a surjective smooth map of smooth affine varieties with smooth irreducible fibers of dimension $r$. For each $s\in S$ we consider $M_s= \rho^{-1}(\{s\})$. Let $\vec X$ be a rational vector field tangent to $\rho$ \emph{i.e.} such that $d\rho(\vec X)=0$. We assume (by replacing $M$ by an affine open subset if necessary) that $\vec X$ restricts to a rational vector field $\vec X|_{M_s}$ at each fiber $M_s$. The bundle of partial frames of $M$ with respect to $\rho$ of order $k$, is defined as
$$
\mathrm{R}_k(M/S) = \bigcup_{s\in S} {\mathrm R}(M_s).
$$
which turns out to be an affine variety and a principal bundle over $\Gamma'_k = \textrm{Aut}_k(\mathbb C^r ,0)$ with $r = \dim(M/S)$. The groupoid of gauge isomorphisms of ${\rm R}(M/S)$ turn out to be:
$${\rm Aut}_k(M/S) = \bigcup_{s\in S} {\rm Aut}_k(M|_s),$$
which is an algebraic groupoid on $M$ of transformations between fibers of $\rho$.
The rational vector field $\vec X$ can be extended to a $\Gamma_k'$-invariant rational vector field $(\vec X/S)^{(k)}$ in ${\rm R}_k(M/S)$
$$(\vec X/S)^{(k)}(p) = \left (\vec X|_{M_{\rho(p)}}\right)^{(k)}.$$
As in the general theory, we may take the limit in $k$ obtaining the bundle of partial frames of $M$ with respect to $\rho$,
$$
\mathrm{R}(M/S) = \bigcup_{s\in S} {\mathrm R}(M_s).
$$
which turns out to be a principal bundle over $\Gamma_k = \textrm{Aut}(\mathbb C^r ,0)$. The field of rational functions
$\mathbb C({\rm R}(M/S))$ is also endowed with the total derivative operators $\Delta' =\{\delta_1,\ldots,\delta_r\}$ and it is a $\Delta'$-field. The derivation ${(\vec X/S)}^{(\infty)}$ defined as,
$$(\vec X/S)^{(\infty)}f = (\vec X/S)^{(k)}f\quad \mbox{for} \quad f\in \mathbb C({\rm R}_k(M/S)),$$
commutes with $\Delta'$. The groupoid of gauge isomorphisms\footnote{It can be seen as a quotient of the algebraic subgroupoid ${\rm Sym}_{\Delta}(\mathbb C\langle \rho^*\mathbb C(S)\rangle_{\Delta}) \subset {\rm Aut}_k(M)$)
} of ${\rm R}(M/S)$ is,
$${\rm Aut}(M/S) = \bigcup_{s\in S} {\rm Aut}(M_s).$$
The partial Malgrange groupoid is defined for each order,
$$
\mathrm{Mal}(\vec X/S)= \lim_{\leftarrow} \mathrm{Gal}((\vec X/S)^{(k)}).
$$
For each $s\in S$ we have that $\textrm{Aut}( M_s)$ is a Zariski closed subset of $\mathrm{Aut}(M/S)$. Thus, we can speak of the restriction of the partial Galois groupoid to a fibre,
$$
\mathrm{Mal}(\vec X/S)|_{M_s} =
\mathrm{Mal}(\vec X/S) \cap \textrm{Aut}(M_s).
$$
Note that this restriction is $\mathcal D$-groupoid in $M$ for generic values of $s$.
The following is one of the statements in \cite[Th\'eor\`eme 4.8]{casaledavy2020} \footnote{The reader should be aware that the notation in the reference is different. There the Malgrange-Galois groupoid appears as ${\rm Gal}$ and not ${\rm Mal}$.}.
\begin{theorem}
\label{th:specialisation}
For all $s\in S$,
$\mathrm{Mal}(\vec X|_{M_s})\subset \mathrm{Mal}(\vec X/S)|_{M_s}.$
\end{theorem}
\subsection{Projection theorem}
Let $\pi\colon P\to M$ be a $G$-principal bundle. Assume we have a normal algebraic subgroup $K$ of $G$ and write $\overline{G} =G/K$. Let us consider $q\colon P\to \overline{P} = P/K$, $p\mapsto \overline{p}$. Then $\overline P$ is a $\overline{G}$-principal bundle over $M$. The actions commute: $\overline{pg} = \overline{p}\overline{g}$. There is also a projection $q_*$ of groupoids,
$$
\xymatrix{
\mathrm{Iso}\, P \ar[r]^-{q_*} \ar[dr]_{s,t}
& \mathrm{Iso}\, \overline{P}\ar[d]^{s,t}
\\
&M
}
$$
where the isomorphism $\sigma_{p,q}$ (that sends $p\in P$ to $q\in P$) is sent to $\sigma_{\overline{p}, \overline{q}}$.
\begin{theorem}
\label{th:projection_GGal_theorem}
Let $\vec X$ be a rational $G$-invariant vector field in $P$. Then
$$q_*({\rm Gal}(\vec X)) = {\rm Gal}(q_*\vec X).$$
\end{theorem}
\begin{proof}
It is a particular statement of a result in \cite{casale2009preuve}. However it suffices to note that $q^*(\mathbb C(\overline P)^{q_*\vec X}) \subseteq \mathbb C(P)^{\vec X}$.
\end{proof}
Now, as in subsection \S\ref{sec:specialization_theo}, let us consider $\rho\colon M\to S$ with $\dim M = m$ and $\dim M/S =r$. Let $G$ be the subgroup of $\Gamma$ of maps that leave $ \mathbb C^r\subset \mathbb C^m$ invariant, where we identify $\mathbb C^r$ inside $\mathbb C^m$ as $ \mathbb C^r = \{ \varepsilon_{r+1} =\cdots = \varepsilon_{m}=0 \}$. Explicitly we have
$$
G= \Big\{ \phi : (\mathbb C^m,0)\to (\mathbb C^m,0)
\,\colon \,
\frac{\partial \phi_i}{\partial \varepsilon_j} =0 \mbox{ for } i=1,\ldots, r; \quad j=r+1,\ldots, m\Big\}$$
Let us consider $\mathrm{R}^{\rho}_k M$ the set of $k$-jets of biholomorphisms $(\mathbb C^m,0)\to (M,p)$ that send the subspace $\mathbb C^r$ to the fiber $M_{\rho(p)}$. If a frame is in $\mathrm{R}^{\rho}_k M$ then it can be restricted to $(\mathbb C^r,0)$ obtaining a frame of the fiber. By taking projective limit we obtain $ \mathrm{R}^\rho M\subset \mathrm{R}M$ as a $G$-sub-bundle, a reduction of structure group of $\mathrm{R}M$ from $\Gamma$ to $G$.
The gauge groupoid $ \mathrm{Iso}(\mathrm{R}^\rho M )$is identified with $\textrm{Aut}(M)^\rho = {\rm Sym}_\Delta(\rho^*\mathbb C(S))$ the $\mathcal D$-groupoid of formal maps respecting the projection $\rho$. As $d \rho(X)=0$ we have $X^{(\infty)}$ is tangent to $ \mathrm{R}^\rho M$ and $ \mathrm{Mal}(\vec X) \subset\textrm{Aut}(M)^\rho$. The group $G$ acts on $ \mathrm{R}^\rho M$. There is a natural exact sequence,
$$
0 \to K \to G\to \Gamma' \to 0
$$
given by the restriction to $\mathbb C^r$. Here, $K$ is the subgroup of formal maps in $\mathbb C^m$ inducing the identity in $\mathbb C^r$. A frame in $\mathrm{R}^{\rho}M$ can be restricted to $\mathbb C^m$, and therefore we obtain a frame on a fiber of $\rho$. Thus, we have a projection,
$$
\mathrm{R}^\rho M \longrightarrow
\mathrm{R}^\rho(M)/K \simeq
\mathrm{R}(M/S), \quad
\phi \mapsto \phi|_{ \mathbb C^m}.
$$
It is straightforward that $X^{(\infty)}$ is projectable and projects onto $(X/S)^{\infty}$.
By Theorem \ref{th:projection_GGal_theorem} (applied to all finite orders $k$) we get a surjective map,
$$
\mathrm{Mal}(\vec X) \longrightarrow \mathrm{Mal}(\vec X/S).
$$
Let $\sigma\in {\rm Aut}(M)^{\rho}$ and let $s$ be the projection of the source of $\sigma$. The target is $\sigma$ is also in $M_s$ and moreover $\sigma$ restricts to a formal map in ${\rm Aut}(M_s)$. If we fix a fiber $M_s$, taking into account that ${\rm Mal}(\vec X)\subset{\rm Aut}(M)^{\rho}$ then it makes sense to consider the restriction ${\rm Mal}(\vec X)|_{M_s}$ as the restriction to $M_s$ of all elements of ${\rm Mal}(\vec X)$ with source in $M_s$. From the surjectiveness of the above map we obtain $\mathrm{Mal}(\vec X)|_{M_s} = \mathrm{Mal}(\vec X/S)|_{M_s}$. Now, by application of Theorem \ref{th:specialisation} to the right and side of the equality we get:
\begin{corollary}
\label{co:restriction_Mal_fiber}
For all $s\in S$, $ \mathrm{Mal}(\vec X)|_{M_s} \supset \mathrm{Mal}(\vec X|_{M_s})$.
\end{corollary}
\section{Malgrange-Galois groupoid of Painlev\'e VI equation}
\label{sec:PainleveVI}
\subsection{Hamiltonian form for Painlev\'e VI}
Painlev\'e VI equation \index{Painlev\'e VI equation} is a second order differential equation for function $u$ of a complex variable $x$ of the form (see, for example, \cite[ p.119]{iwasaki2013gauss}, and \cite{okamoto1980polynomial})
\begin{align}
\label{eq:Pain_VI}
u'' &=F(x,u,v, a,b,c,e); \quad u'=v
\end{align}
where $F\in \mathbb C(x,u,v,a,b,c,e)$ is
\begin{eqnarray*}
\frac{1}{2} \left( \frac{1}{u} + \frac{1}{u-1} + \frac{1}{u-x} \right){v}^2 -\left( \frac{1}{x} + \frac{1}{x-1} + \frac{1}{u-x} \right){v}
\\
+ \frac{ u(u-1)(u-x)}{x^2(x-1)^2} \left( \frac{c^2}{2} - {\frac{a^2}{2}}\frac{x}{u^2} + {\frac{b^2}{2}}\frac{x-1}{(u-1)^2} + \frac{1-e^2}{2} \frac{x(x-1)}{(u-x)^2} \right).
\end{eqnarray*}
The variables $a$, $b$, $c$, $e$ are the parameters of Painlev\'e VI equation. However, our interest is to consider the role of the parameters in nonlinear differential Galois theory. Therefore, we see Painlev\'e VI equation as the following rational vector field in $\mathbb C^7$,
\begin{align}
\vec Y = \frac{\partial}{\partial x} + v \frac{\partial}{\partial u} + F(x,u,v, a,b,c,e) \frac{\partial}{\partial v}. \tag{$\rm P_{VI}$}
\end{align}
The trajectories of this vector field, parameterized by $x$ are $(x,u(x), u'(x), a,b,c,e)$ for $u$ a solution of $P_{VI}$ with fixed parameters.
Equation Painlev\'e VI \eqref{eq:Pain_VI} admits the following equivalent Hamiltonian form (see \cite[ p.140]{iwasaki2013gauss})
with hamiltonian function
\begin{align*}
H &= \frac{1}{x(x-1)} \Big[ p(p-1)(p-x)q^2 - \Big( a(p-1)(p-x) + b p(p-x) +(e-1)p(p-1)\Big) q \\
&\quad \quad + \frac{1}{4}\big( (a+b+e-1)^2 -c^2\big) (p-x) \Big].
\end{align*}
and hamiltonian vector field
\begin{align}
\label{eq:PainVI_vec_field2}
\vec X = \frac{\partial}{\partial x} + \frac{\partial H}{\partial q} \frac{\partial}{\partial p} -\frac{\partial H}{\partial p}\frac{\partial}{\partial q}.
\tag{$\rm HP_{VI}$}
\end{align}
The dominant finite map $\phi: \mathbb C^7 \to \mathbb C^7$
$$
(x,p,q,a,b,c,e) \mapsto \left(x,p, \frac{\partial H}{\partial q}, \frac{c^2}{2}, \frac{a^2}{2}, \frac{b^2}{2}, \frac{e^2}{2} \right)
$$
sends the vector field $\vec X$ onto the vector field $\vec Y$, giving the equivalence between systems \eqref{eq:Pain_VI} and \eqref{eq:PainVI_vec_field2}, thus a conjugation of their Malgrange-Galois groupoids. Our purpose is to compute the Malgrange-Galois groupoid of vector field $\vec X$, \eqref{eq:PainVI_vec_field2}.
\subsection{Some invariants of the Malgrange-Galois groupoid for Painlev\'e VI}
Along this section let us consider the following diagram of trivial bundles:
$$
\xymatrix{
M=\mathbb C^7 \ar[r]^-{\pi} \ar[dr]_-{\bar\rho} & \mathbb C^5_{x,a,b,c,e}=B \ar[d]^-{\rho} \\
&\mathbb C^4_{a,b,c,e}=S
}
$$
From Definition \ref{df:Mal_groupoid} the Malgrange-Galois groupoid of a vector field $\vec X$ is given by:
$$
\mathrm{Mal} (\vec X) = \left\{ \sigma \in \mathrm{Aut}M \mid \mbox{for all } f\in \mathbb C(\mathrm{R}M)^{\vec X}\!,\;\; f \circ \sigma^{(\infty)} = f|_{({\rm R}M)_{s(\sigma)}}\right\}
$$
\begin{remark}
\label{rmk:frame_invariants}
From the known invariants of $X$ we can obtain some informations of its Malgrange-Galois groupoid. First, the conserved quantities $a,b,c,e$ are by themselves rational differential invariants of order $0$. Therefore, they span a $\Delta$-subfield\footnote{Let us recall that $\Delta = \{\delta_1,\ldots,\delta_7\}$ stands for the system of total derivative operators with respect to $\varepsilon$'s that give to $\mathbb C({\rm R}M)$ the structure of $\Delta$-field.}
$\mathbb C\langle a,b,c,e \rangle_{\Delta}$ of the $\Delta$-field $\mathbb C(\mathrm RM)^{\vec X}$ of rational differential invariants of $\vec X$. The $\mathcal D$-groupoid corresponding to such field is the groupoid of formal maps respecting the projection $\bar\rho$. Therefore,
\begin{align*}
\mathrm{Mal}(\vec X) &\subset \Big\{ \phi \in \mathrm{Aut}(\mathbb C^7) \; \mid
\; \bar\rho\circ \phi = \bar\rho
\Big\}
\end{align*}
Also, from geometric invariants we can obtain information. There is an intrinsic connection between geometric structures and $\mathcal D$-groupoids. As it is shown (\cite{casale2004groupoide} Theorem 1.3.2 p. 10) any transitive $\mathcal D$-groupoid is the groupoid of invariance of a geometric structure. Let us see some examples.
\begin{itemize}
\item We know $\vec Xx =1$ and therefore $\mathrm{Lie}_X(dx) = 0$. Therefore, $1$-form $dx$ can be seen as a rational tensor invariant by $X$, so that
\begin{align*}
\mathrm{Mal}(\vec X) &\subset \Big\{ \phi \in \mathrm{Aut}(\mathbb C^7) \;\mid \; \phi^*(dx) = dx \Big\}
\end{align*}
How can we obtain the $\Delta$-field of rational differential invariants associated to $dx$?
Note that, in any given frame $\varphi$, the pullback
$$\varphi^*(dx)(0) = x_{:\epsilon_1}d_0\varepsilon_1 + \ldots x_{:\epsilon_7}d_0\varepsilon_7$$
is a co-vector in $\mathbb C^m$. Its coordinates $x_{:\epsilon_1}$, $\ldots$, $x_{:\epsilon_7}$ are the differential invariants defining the $\mathcal D$-groupoid ${\rm Sym}_\Delta(\mathbb C\langle x_{:\epsilon_1},\ldots,x_{:\epsilon_7}\rangle_{\Delta})$ of symmetries of $dx$.
\item
A rational vector field $\vec Y$ in $M$ can be seen as a particular case of a geometric structure. In order to associated to this geometric structure as a $\Delta$-field of rational differential invariants we only need to observe that each $1$-frame $j^1_0\varphi$ at $p\in M$ determines a basis $\{A_1(j_0^1\varphi),\ldots,A_7(j_0^1\varphi)\}$ of $T_pM$ where
$A_i(j_0^1\varphi) = d_0\varphi\left(\frac{\partial}{\partial \varepsilon_i} \right)_0.$
The coordinates $Y_i$ of $Y$ in such basis can be seen as rational functions on ${\rm R}_1M$,
$$Y(p) = \sum_{i=1}^7 Y_i(j_0^1\varphi)A_1(j_0^1\varphi).$$
The $\mathcal D$-groupoid of symmetries of $\vec Y$ is then ${\rm Sym}_\Delta(\mathbb C\langle Y_1,\ldots Y_7\rangle_\Delta)$. In particular we have $[\vec X, \vec X] = 0$, so that the Malgrange-Galois groupoid of $\vec X$ is included in the $\mathcal D$-groupoid of symmetries of $\vec X$:
\begin{align*}
\mathrm{Mal}(\vec X) &\subset \Big\{ \phi \in \mathrm{Aut}(\mathbb C^7) \;\mid \;\phi_*(\vec X) = \vec X \Big\}.
\end{align*}
\item The vector field $\vec X$, when restricted to specific values of the parameters, is a non-autonomous Hamiltonian with respect to the form $dq\wedge dp$. It follows that $\mathrm{Lie}_{\vec X} (dq\wedge dp) = d i_X (dp\wedge dq) = d(\frac{\partial H}{\partial q}dq + \frac{\partial H}{\partial p}dp)$ and then $$ {\rm Lie}_{\vec X}(dp\wedge dq)\equiv 0\ \mathrm{mod}\ dx, da,db,dc,de.$$ This means that the rank $2$ bundle $\mathrm{ker}(d(\pi\circ\rho))$ is endowed with an $\vec X$-invariant volume form, this can also seen as a geometric structure in $\mathbb C^7$, defined by the class of $dp\wedge pq$ modulo $\pi^*\Omega_{B}^1$, and yields the following restriction,
\begin{align*}
\mathrm{Mal}(\vec X) &\subset \Big\{ \phi \in \mathrm{Aut}(\mathbb C^7) \;\mid \; \phi^* (dq\wedge dp) \equiv dq\wedge dp \mod \pi^*\Omega^1_B
\Big\}.
\end{align*}
\end{itemize}
\end{remark}
Summarizing Remark \ref{rmk:frame_invariants}, we have the following restrictions of the Malgrange-Galois groupoid of $X$,
\begin{align}
\label{eq:Mal_subset}
\mathrm{Mal}(\vec X) &\subset \Big\{ \phi \in \mathrm{Aut}(\mathbb C^7) \;\mid\;\phi_*\vec X=\vec X; \quad \bar\rho\circ\phi = \bar\rho;\nonumber
\\
& \quad \quad \quad \phi^* dq\wedge dp \equiv dq\wedge dp \mod \pi^*\Omega^1_B;
\quad \phi^* dx =dx
\Big\}
\end{align}
Malgrange-Galois groupoid of Painlev\'e VI equation, with fixed values of parameters, has been found by Cantat--Loray. In what follows it is necessary to distinguish the set \emph{Picard parameters},
the following subset of $\mathbb C^4$:
$$
\Big\{
(a,b,c,e)\in (\textstyle{\frac{1}{2}}+\mathbb Z^4)
\Big\}
\cup
\Big\{
(a,b,c,e)\in \mathbb Z^4 \mid a+b+c+e \mbox{ is even }
\Big\}
$$
\begin{proposition}[ Theorem 6.1 in \cite{cantat2009dynamics}]
\label{pr:picar_param}
For parameters $( a,b,c,e) $ not in Picard parameter set, Malgrange-Galois groupoid is given by:
\begin{align*}
\mathrm{Mal}(\vec X|_{\mathbb C^3 \times \{ ( a,b,c,e)\} } )
&=
\Big\{
\phi : ( \mathbb C^3, *) \overset{\sim}{\longrightarrow} ( \mathbb C^3,\star) \mid \phi^* dx =dx;\,\, \phi_*(\vec X)= \vec X;
\\
& \quad \quad \quad \phi^* dp\wedge dq = dp \wedge dq \mod dx
\Big\}
\end{align*}
In this formula $a,b,c,e$ are fixed parameters. The asterisk and the star stand for arbitrary points in $\mathbb C^3$.
\end{proposition}
As a direct consequence of Theorem \ref{th:specialisation} we have the following:
\begin{proposition}
\label{pr:inclusion_picard_param}
There is an inclusion $\mathrm{Mal} (\vec X\mid_{ \mathbb C^3 \times \{ (a,b,c,e)\}})
\subset
\mathrm{Mal}(\vec X)\mid_{ \mathbb C^3 \times \{ (a,b,c,e)\}}$.
\end{proposition}
\subsection{Transversal part of $\textrm{Mal}(\vec X)$}
Let us consider $\mathrm{Sym}(\mathbb C\langle x \rangle_\Delta)$, the $\mathcal D$-groupoid of formal diffeomorphisms $\phi : (\mathbb C^7,* ) \to (\mathbb C^7, \star) $ that leave invariant the $x$ coordinate, \emph{i.e.} such that $x\circ \phi = x$. We want to prove the equality in Equation (\ref{eq:Mal_subset}). To do this we look at $\mathrm{Mal}(\vec X) \cap \mathrm{Sym}(\mathbb C\langle x \rangle_\Delta)$. We have that
\begin{align}
\label{eq:Mal_cap_inv(x)}
\mathrm{Mal}(\vec X) \cap \mathrm{Sym}(\mathbb C\langle x \rangle_\Delta) &\subset \Big\{ \phi: (\mathbb C^7,*) \to ( \mathbb C^7, \star) \mid \phi^*\vec X = \vec X; \quad \pi\circ \phi = \pi;\nonumber
\\
& \quad \quad \quad
\quad \quad \quad
\quad \quad \quad
\phi^* (dq\wedge dp) = dq\wedge dp \quad\mathrm{mod }\,\, \pi^*\Omega^1_{B}
\Big\}
\end{align}
The plan of the remaining proof is to obtain equality in Equation \eqref{eq:Mal_cap_inv(x)}. To see this we restrict our attention to the $\mathcal D$--Lie algebra
$$\mathcal L = \mathrm{Lie }\big(\mathrm{Mal}(\vec X)\cap \mathrm{Sym}(\mathbb C\langle x \rangle_\Delta)\big).$$
This Lie algebra is seen to be of the kind described in Kiso--morimoto Theorem \ref{th:kiso-mori} so we apply it. By these means we get the desired equality but for $\mathrm{Lie}\big( \mathrm{Mal}(\vec X)\cap \mathrm{Sym}(\mathbb C\langle x \rangle_\Delta)\big)$ instead of $\mathrm{Mal}(\vec X)\cap \mathrm{Sym}_\Delta(\mathbb C\langle x \rangle_\Delta)$. The final steps will be to ``get rid'' of the terms $\mathrm{Lie}$ and $\mathrm{Sym}(\mathbb C\langle x \rangle_\Delta)$.
\begin{proposition}
The $\mathcal D$--Lie algebra
$\mathcal L$
and $\pi\colon \mathbb C^7\to\mathbb C^5$ are under the hypothesis of Theorem \ref{th:kiso-mori}, namely: first, that $\mathcal L$ is tangent to $\pi$ and second, that
there exists a form $\omega\in \Omega^m_M$ such that for all $s\in S$, $\omega|_{M_s}$ is a non--identically zero $m$--form on $M_s$ satisfying
$$
\label{condition on mathcal L in Malgrange chapter}
\mathcal L|_{M_s}
=\big\{
v \mbox{ vector fields on } M_s \colon
\mathrm{Lie}_v\, \omega|_{M_s} =0
\big\}.$$
\end{proposition}
\begin{proof}
First, $\mathcal L$ is tangent to the fibers of $\pi : \mathbb C^7 \to \mathbb C^5$. Second, it preserves the volume form $dq\wedge dp$ on the fibers. Let us compute the restriction of $\mathcal L$ to a fiber $M_s$ with $s = (x_0,a,b,c,e)$. Let us assume that $(a,b,c,e)$ is not in the Picard parameter set. Then we by Proposition n \ref{pr:picar_param}
we have that ${\rm Mal}(\vec X|_{\bar\rho^{-1}(a,b,c,e)})$ is $\mathcal D$-groupoid of symmetries of $\vec X|_{\bar\rho^{-1}(a,b,c,e)}$,
$dx$ and the class of $dq\wedge dp$ mod $dx$. By corollary \ref{co:restriction_Mal_fiber} it coincides with
${\rm Mal}(\vec X)|_{\bar\rho^{-1}(a,b,c,e)}$.
Taking the intersection with ${\rm Sym}_\Delta(\mathbb C\langle x \rangle_\Delta)$ we obtain that the restriction of
${\rm Mal}(\vec X)\cap {\rm Sym}_\Delta(\mathbb C\langle x \rangle_\Delta)$ to $\bar\rho^{-1}(a,b,c,e)$
is the the $\mathcal D$-groupoid of symmetries of $\vec X|_{\bar\rho^{-1}(a,b,c,e)}$, $x$, and the class of $dq\wedge dp$ mod $dx$. It follows that
$\mathcal L|_{\bar\rho^{-1}(a,b,c,e)}$ is the
$\mathcal D$-algebroid of infinitesimal symmetries of $\vec X|_{\bar\rho^{-1}(a,b,c,e)}$, $x$ and the class of $dq\wedge dp$ mod $dx$. When we fix the value of $x = x_0$ then we obtain that
$\mathcal L|_{M_s}$ is the $\mathcal D$-Lie algebroid of infinitesimal symmetries of $dq\wedge dp$.
\end{proof}
Therefore, from
Kiso--Morimoto Theorem \ref{th:kiso-mori}, we have that:
\begin{enumerate}
\item[(a)] There exists a foliation $ \mathcal F$ on $ \mathbb C^5$,
\item[(b)] there exists $ \mathcal H$ on $ \mathbb C^7$ such that $ d\pi( \mathcal H) = \mathcal F$, ${\rm rank}(\mathcal H) = {\rm rank}(\mathcal F)$,
\item[(c)]
$\mathcal L$ is the $\mathcal D$-Lie algebra of vector fields tangent to the fibers of $\pi$, preserving the volume form $dp\wedge dq$, and preserving the $\mathcal F$-connection\footnote{Let us recall that if $M\to B$ is a bundle and $\mathcal F$ is a foliation in $B$, then a $\mathcal F$-connecion is a linear sub-bundle of ${\rm T}M$ of the same rank that $\mathcal F$ and that projects onto $\mathcal F$, see \cite{malgrange2010pseudogroupes}.}
$\mathcal H$.
\begin{align*}
\mathcal L & = \Big\{ v\in J({\rm T}M/M) \mid d\pi(v) =0; \;\; \mathrm{Lie}_v \omega = 0\mod \pi^* \Omega^1_S;
\\
& \qquad \qquad \forall \vec Y \text{ tangent to }\mathcal H \quad [v,\vec Y]\ \text{ is tangent to } {\mathcal H}
\Big\}
\end{align*}
\end{enumerate}
In particular, the vector field $\vec X$ is tangent to $\mathcal H$. Therefore $\pi_*(\vec X) = \frac{\partial}{\partial x}$ is tangent to $\mathcal F$. Thus, the foliation $\mathcal F$ is $\rho$-projectable on a foliation of $\mathbb C^4$ of the same codimension. Let us denote by $\overline{\mathcal F}$ this projection.
\subsection{Affine Weyl group}
Let us note that if $\phi$ is a birational automorphism of $M$ and $\vec X$ is a rational vector field on $M$, then the pullback by the extension, $\phi^{(\infty)}\colon \mathrm{R}M \to \mathrm{R}M$ sends rational differential invariants of $\phi_*(\vec X)$ to rational differential invariants of $\vec X$. If follows clearly that $\phi$ induces an isomorphism between $\mathrm{Mal}(\vec X)$ and $\mathrm{Mal}(\phi_*(\vec X))$. In particular, if $\phi$ is a \emph{discrete birational symmetry} \index{Discrete birational symmetry} of $\vec X$, that is, a birational automorphism of $M$ such that $\phi_*(\vec X) = \vec X$, then $\phi$ leaves $\mathrm{Mal}(\vec X)$ invariant. That is,
$$
j_p\sigma \in \mathrm{Mal}(\vec X) \Longleftrightarrow j_{\phi(p)}(\phi\circ \sigma \circ \phi^{-1})\in \mathrm{Mal}(\vec X).
$$
\begin{lemma}
Foliation $\overline{\mathcal F}$ in $\mathbb C^4$ is regular.
\end{lemma}
\begin{proof}
Equation \eqref{eq:PainVI_vec_field2} admits a discrete group $\widetilde{W}$ of symmetries, known as Backl\"und transformations, isomorphic to the extended affine Weyl group \index{Affine Weyl group} $D_4^{(1)}$. All elements of this group are birational transformations of $\mathbb C^7$. In particular this group contains a subgroup $G_4$ isomorphic to $\mathbb Z^
4$ with the following characteristics:
\begin{itemize}
\item[(a)] The action of $G_4$ in $\mathbb C^7$ is projectable by $\bar\rho$ to the action of translations with integer displacements in $\mathbb C^4$.
\item[(b)] The function $x$ is invariant by the action.
\end{itemize}
This group of translations is listed in \cite[p.6]{noumi2002new}.\\
The action of $ G_4$ preserves $\mathrm{Mal}(\vec X)$. The function $x$ is an invariant for this action, then $ \mathrm{Sym}(\mathbb C\langle x \rangle_\Delta)$ is also preserved. It follows that $G_4$ leaves $ \mathcal L$ invariant, and its defining equations, in particular, for $\phi \in G_4$, $\phi_*( \mathcal H) = \mathcal H$.
Denote by $\bar \phi$ the projection of $\phi$ as a birational transformation of $ \mathbb C^4$ and observe $\bar \phi$ is a translation. Then $\bar \phi$ sends $\overline{ \mathcal F}$ into $\overline {\mathcal F}$, \emph{i.e.} $d \phi( \overline {\mathcal F})=\overline {\mathcal F}$. As the set of singularities of $\overline {\mathcal F}$ is a proper Zariski closed of $\mathbb C^4$, invariant under translation by $\mathbb Z^4$, it must be empty.
\end{proof}
A result of Iwasaki, \cite[Theorem 1.3]{iwasaki2008finite} implies the following:
\begin{lemma}
The solutions to $P_{VI}$ with finite monodromy are algebraic.
\end{lemma}
This allows us to see:
\begin{lemma}
$\overline{\mathcal F}$ is a foliation by points.
\end{lemma}
\begin{proof}
In \cite{lisovyy2014algebraic} a list of all possible algebraic solutions of Painlev\'e VI equation is presented. These appear at special values of the parameters. The solution numbered as 45 in \cite[p.52]{lisovyy2014algebraic} is algebraic with 72 branches, and happens at the parameter $\theta = (1/12,1/12,1/12,11/12)$.
If the dimension of $ \overline{\mathcal F}$ were greater than $0$, we could find a path on a leaf of $ \overline{\mathcal F}$ along which we could extend the given solution to a solution at another parameter, with also $72$ branches. By Iwasaki, this solution is algebraic. But we know the original solution is unique, all other algebraic solutions have less than $72$ branches. We conclude that the foliation cannot have dimension greater than 1 so it must be a foliation by points.
\end{proof}
Therefore $\mathcal F = \left\langle
\frac{\partial}{\partial x}
\right\rangle$ and $\mathcal H = \left\langle \vec X\right\rangle$. Note that, for a vector field $v$ tangent to $\bar\rho$, as $X$ is transversal to $\bar\rho$, it is equivalent to say that $\mathrm{Lie}_v\mathcal H \subset \mathcal H$ or $[v,X]=0$. This completes the proof of the following result.
\begin{proposition}
\label{prop:first_description_of_mathcal_L}
\begin{align*}
\mathcal L
&= \Big\{ \vec{v} \text{ vector field on } \mathbb C^7 \;\mid\;
\vec v\cdot x=\vec v\cdot a=\vec v \cdot b=\vec v\cdot c =\vec v\cdot e =0;
\;\; [\vec v,X]=0; \\
& \qquad \qquad \mathrm{Lie}_{\vec v} (dp\wedge dq) =0 \mod \pi^\ast \Omega_B^1
\Big\}
\end{align*}
\end{proposition}
By integration of the $\mathcal D$-Lie algebra $\mathcal L$ we obtain:
\begin{proposition}
\label{pr:description_of_Mal_cap_Inv}
\begin{align}
\label{eq:description_of_Mal_cap_Inv}
\mathrm{Mal}(\vec X) \cap \mathrm{Sym}(\mathbb C\langle x \rangle_\Delta)
&= \Big\{ \phi \mid
\pi \circ \phi = \pi; \;\; \phi_*(\vec X)=\vec X;\nonumber
\\
& \qquad \qquad \phi^*(dp\wedge dq) \equiv dp\wedge dq \mod \pi^\ast \Omega_B^1
\Big\}
\end{align}
\end{proposition}
\begin{proof}
The Lie algebra $ \mathcal L$ is determined by $ \mathcal L_1$, its first order part. Let us call $\mathcal G$ the right hand side of (\ref{eq:description_of_Mal_cap_Inv})$; \mathcal G$ is determined by $ \mathcal G_1$, its first order part.The Malgrange-Galois groupoid $ \mathrm{Mal}(\vec X)$ is determined by $ \mathrm{Mal}_1(\vec X)$ too. It can be proved that $ \mathcal G_1$ is connected with respect to source and target. Then $ \mathcal G_1$ is the least Lie subgroupoid of $ \mathrm{Aut}_1( \mathbb C^7)$ such that $ \mathrm{Lie}( \mathcal G_1) = \mathcal L_1$. By (\ref{eq:Mal_cap_inv(x)}) we know that $ \mathrm{Mal}_1(X) \subset \mathcal G_1$. As $ \mathrm{Lie}( \mathrm{Mal}(\vec X)) = \mathcal L_1$ then $ \mathrm{Mal}_1(X) = \mathcal G_1$. We conclude that $ \mathrm{Mal}(\vec X) = \mathcal G$.
\end{proof}
\begin{theorem}
\label{th:Final_Mal_groupoid_PVI}
The Malgrange-Galois groupoid of Painlev\'e VI equation is given by
\begin{align}
\label{eq:Mal_final_description}
\mathrm{Mal}(\vec X)
&= \Big\{ \phi \mid
\bar\rho \circ \phi = \bar\rho; \;\; \phi^*(dx) = dx; \;\; \phi_*(\vec X)=\vec X;\nonumber
\\
& \qquad \qquad \phi^*(dp\wedge dq) \equiv dp\wedge dq \mod\pi^\ast \Omega_B^1
\Big\}
\end{align}
\end{theorem}
\begin{proof}
We already have pointed in (\ref{eq:Mal_subset}) that $ \mathrm{Mal}(\vec X)$ is contained in the right side set of Equation \eqref{eq:Mal_final_description}. Let us prove the remaining inclusion. Fix an integer $k$ large enough and let $\psi\colon (\mathbb C^7,z_0)\to
(\mathbb C^7,z_1)$ be a map that satisfy equations in the right hand side of \eqref{eq:Mal_final_description}. We are going to show that $j_{z_0}^k\psi$ is in $ \mathrm{Mal}_k(\vec X)$. Let $ \varepsilon = x(z_1) -x(z_0)$. Without loss of generality, assume that $ \exp( - \varepsilon \vec X)$ is defined in a neighborhood such that if $z_2 = \exp( - \varepsilon \vec X)(z_1)$ then all the points $z_0,z_1$ and $z_2$ are inside an open set where $ \mathrm{Mal}(\vec X)$ is effectively a groupoid. Let the map $\phi\colon (\mathbb C^7,z_0)\to (\mathbb C^7,z_2)$ be defined by $\phi = \mathrm{exp}(-\varepsilon \vec X)\circ \psi$. Observe that $\phi$ respects $\bar \rho$, $dx$, $dp\wedge dq|_{ \mathbb C^2_{p,q}}$, and the field $\vec X$. Moreover, since $\phi^*(dx) = dx$ we have $\phi^*(x) = x + \lambda$ with $\lambda$ a constant, indeed $x(\phi(z_1)) = x(z_1)$ so that $\lambda = 0$ and $\phi^*(x) = x$. It follows that $\phi$ preserves $\pi$. By Proposition \ref{pr:description_of_Mal_cap_Inv} $j_{z_0}^k \phi$ is in $ \mathrm{Mal}_k(X)$. Then $j_{z_0}^k \psi = j_{z_2}^k \exp ( \varepsilon X) \circ j_{z_0}^k\phi$ is in $ \mathrm{Mal}_k (X)$.
\end{proof}
\begin{corollary}\label{co:par}
If $y(x,a,b,c,e)$ is a parameter dependent solution of the sixth Painlev\'e equation then its annihilator in $\mathcal O_{J(\mathbb C^5, \mathbb C)}$ is the $\partial$-ideal generated by the sixth Painlev\'e equation.
\end{corollary}
Using the Hamiltonian formulation, we want to prove that the Zariski closure of a parameters dependant solution $p(x,a,b,c,e) ; q(x,a,b,c,e)$ in $J(\mathbb C^5, \mathbb C^2)$ is $V$ : the subvariety defined by the differential ideal generated by $\partial_x p - \frac{\partial H}{\partial q}$ and $\partial_x q + \frac{\partial H}{\partial p}$.\\
\begin{proof}
To prove this, one studies the action of $\mathrm{Mal}(\vec X)$ on $\partial_x^{tot}$. This action is well defined because the elements of $\mathrm{Mal}(\vec X)$ are fiber preserving transformations of $\mathbb C^5 \times \mathbb C^2 \to \mathbb C^5$. Let $W\subset V$ be the Zariski closure of a parameter dependant solution.\\
The proof follows from the next two lemmas.
\begin{lemma}\label{lm:W}
$\mathrm{Mal}(\vec X)$ preserves $W$
\end{lemma}
\begin{remark} The restriction of $\partial_x^{tot}$ on $J_0(V) = \mathbb C^7$ is the vector field $\vec X$. As $V$ or $W$ are define by $\partial_x^{tot}$-ideals, the extension of $\vec X$ on $J(\mathbb C^5, \mathbb C^2)$ is tangent to $V$ and to $W$. The stabilizer of $W$ is a $\mathcal D$-groupoid containing the flow of $\vec X$: it must contain $\mathrm{Mal}(\vec X)$.
\end{remark}
\begin{lemma}\label{lm:WV}
$\mathrm{Mal}(\vec X)$ acts transitively on $V$ over the parameter space.
\end{lemma}
This means that fibers $\mathrm{Mal}(\vec X)_{a,b,c,e}$ act transitively on fibers $V_{a,b,c,e}$.
\begin{proof}
Fix parameters $(a_0,b_0,c_0,e_0)\in \mathbb C^4$.
Let $(p_1(x,a,b,c,e), q_1(x,a,b,c,e))$ be a germ of solution on $(\mathbb C, x_1) \times (\mathbb C^4, (a_0,b_0,c_0,e_0))$ and $(p_2(x,a,b,c,e), q_2(x,a,b,c,e))$ be a germ of solution on $(\mathbb C, x_2) \times (\mathbb C^4, (a_0,b_0,c_0,e_0))$. One can find family of translations parameterized by $(a,b,c,e)$ : $\mathbb C^2 \times \mathbb C^4 \to \mathbb C^2 \times \mathbb C^4$ sending $(p_1(x_1,a,b,c,e), q_1(x_1,a,b,c,e))$ on $(p_2(x_2,a,b,c,e), q_2(x_2,a,b,c,e))$. Using trajectories of $\vec X$, one can extent this map to neighborhoods of fibers $\{x=x_1\}$ and $\{x=x_2\}$ above translation in the variable $x$ as a map preserving $\vec X$.
As Painlev\'e equation preserves $dp\wedge dq\wedge dx \ \mod da, db , dc, de$, this extension also. It satisfies all conditions and belongs to $\mathrm{Mal}(\vec X)_{a,b,c,e}$
\end{proof}
Finally, by Lemmas \ref{lm:W} and \ref{lm:WV} we have $V = W$ and thus the graph of the parameter dependent solution in $J(\mathbb C^5,\mathbb C^2)$ is Zariski dense in the variety defined by the radical $\partial$-ideal generated by the sixth Painlevé equation.
\end{proof}
| {
"redpajama_set_name": "RedPajamaArXiv"
} | 3,131 |
package com.sonymobile.seeder;
/**
* Interface definition of a bandwidth estimator.
*/
public interface BandwidthEstimator {
/**
* Get the currently estimated bandwidth.
*
* @return the estimated bandwidth in bytes/s
*/
public long getEstimatedBandwidth();
/**
* Called when a data transfer is started
*/
public void onDataTransferStarted();
/**
* Called when a data transfer is stoppped
*/
public void onDataTransferEnded();
/**
* Called to indicate that data have been transferred.
*
* @param byteCount The number of bytes transferred.
*/
public void onDataTransferred(long byteCount);
}
| {
"redpajama_set_name": "RedPajamaGithub"
} | 2,945 |
Rick Goldman is a construction and real estate development executive whose award-winning career spans skyscrapers, billion-dollar retail projects and luxury residences. His career began in 1990 as a project manager, where he collaborated with legendary architects on millions of square feet.
Rick supervised the development and construction of flagship stores for transformational brands like Crate & Barrel and Pottery Barn—on some of the most expensive streets in the world. In the retail market, his work translated groundbreaking concepts into beautiful, visionary spaces. His portfolio includes many iconic developments in Los Angeles, Albuquerque and the Las Vegas Strip.
In 2001, Rick co-founded Golden Spike Development, a real estate firm focused on innovative urban infill and re-use projects. The firm's vision for style and sustainability encompasses residential (condo and multifamily) projects in San Diego, Milwaukee and Albuquerque—where he developed New Mexico's first and only LEED Platinum-certified luxury building.
Rick lives in Chicago with his wife, and still insists on mowing his own lawn. His filmmaker daughter is considering a short film on his obsession. Rick holds a Bachelor of Science degree in Mechanical Engineering granted by the University of Colorado in 1988.
Ike Hong is a construction management executive with more than 30 years' experience solving complicated problems for real estate developers, architects and government agencies. His unrivaled cost estimating and project management skill has supported projects like the Great Lakes Naval Training Center, the Sears Merchandise Group World Headquarters and Chicago's Adler Planetarium.
Ike founded Pacific Construction Services in 1993 and quickly scaled the firm from his kitchen table to 20 employees. PCS successfully led ground-up and retrofit construction projects throughout the Chicago region—on time and under budget. In 2005, he co-founded Golden Spike Development, a real estate development company focused on innovative urban in-fill projects.
In addition to his professional achievements, Ike has held board positions at Asian Human Services, Rush Northshore Hospital and NorthShore University Health System. He currently serves on the boards at The Night Ministry and the Mosaic House Ministry in Chicago.
Ike holds a Bachelor of Science degree in Civil Engineering granted by Purdue University in 1986.
Chris is vice president and partner at PCS, with more than a decade of experience at the firm. He has years of project management experience and a geography degree from the University of Kansas. In Chris' free time he enjoys spending time at the park with his black lab, Sharpie, watching Chicago Sports, and spending time with his family, especially his darling niece and nephew.
Lou is a partner and the chief financial officer at PCS, having previously served as controller and project accountant. Lou has his BS in accounting from the University of Illinois. In Lou's spare time he roots for the Cubs, spends time with his family and two sons, and is an avid poker player.
Dan brings more than 12 years of construction experience (in superintendent and project manager roles) to the firm. Dan has an associate degree in drafting and design, as well as a bachelors in Construction Management from ITT Technical Institute.
In Dan's free time he enjoys reading, riding his motorcycle, snowboarding and wakeboarding.
Mark is a 3rd generation carpenter working in construction for more than 30 years. He is an Eastern Illinois University graduate and is an Alderman in Countryside, IL. His free time is spent with family and friends.
Jack has over 30 years of experience in the construction industry and started as an engineering assistant (Combat Engineer Squadron) in the Air Force. Jack then became an estate manager, overseeing all aspects of a large estate. In Jack's free time he restores cars and builds motorcycles and brews beer.
Jeffrey has over 30 years of experience in the construction industry. He completed a Master of Science Degree from the Illinois Institute of Technology and has successfully estimated and administered more than 300 awarded governmental projects valued at over 400 million dollars. Jeffrey is a sports enthusiast and is a basketball team member qualifying for the 2019 National Senior Olympics. Also, he and his wife love to spend time with their three children and two grandchildren.
Michael has 25 years of experience as a field superintendent. Prior to becoming a superintendent he was a carpenter and foreman. In his free time, he enjoys fishing, camping, and spending time with his lovely wife and grandchildren.
Natalie is the Director of Social Media and Website Management, and assists Project Managers in the office. In her free time, she studies International wines and loves to oil paint.
Paige focuses on estimating and bidding with PCS. Paige graduated from Saint Xavier University with her Bachelor of Arts in Sociology. She enjoys cooking and traveling in her free time.
Justin Pathmann is a construction management executive with 15 years of industry experience. Justin arrives at PCS from his post as president of Pathmann Construction Management—recently acquired by PCS—with a portfolio of commercial and residential construction projects throughout the Chicago region.
Justin has led an elite team of architects, project managers, site superintendents and sub-contractors to success in diverse markets. His approach prioritizes old-world craftsmanship, attention to detail and construction innovations—as well as meeting budgetary requirements and time constraints.
That standard of excellence continues at PCS. In his spare time, Justin enjoys rock climbing, cooking and coaching youth sports. Justin graduated from Southern Illinois University at Edwardsville, with a bachelor's degree in Construction Management.
Stephanie supports project managers, as well as assisting is the Estimating Department. Prior to working at PCS, Stephanie was the event coordinator for Marriott Hotels. In her free time, she enjoys drinking wine and spending time with her family.
Chad has a strong foundation of experience with 13 years in the architectural and construction industries. He is responsible for managing field personnel and staffing needs, as well as the coordination and scheduling of subcontractors, materials, and on-site work for each project.
Working closely with architects, engineers, and owners to ensure projects are completed on time. Chad has a Bachelors in Architecture from Ball State University. In his spare time, Chad enjoys spending time with his family and friends, and watching both of his sons play sports.
Ken quickly rose through the ranks from assistant project manager to project manager. Ken has his BS in Applied Engineering with concentration in construction management from Eastern Illinois University. When Ken isn't at work, he enjoys restoring classic cars, playing guitar, barbecuing and watching football.
Vanessa is the senior project accountant at PCS. She has 20 years of construction accounting experience. In her free time, Vanessa volunteers for Palos Stars, a youth football and cheerleading organization. She enjoys spending quality time with her son and family as well as watching football, and drag racing. | {
"redpajama_set_name": "RedPajamaC4"
} | 7,523 |
As the matter of the fact, all you should know about umrah packages is something that there are many umrah packages that are offered by the many travel agencies operating in the different parts of the world. It would be important to note that the thousands of the Muslims believers belonging to the diverse areas of the global village tend to pay the visit to the city of the Makkah, the city of Saudi Arabia with the ultimate aim of performing the Umrah to be able to meet the rituals in the single year. Due to the expense involved in the umrah, it may turn out to be very difficult for the Muslims to perform the lesser pilgrimage in the best possible manner. It must be witnessed by you that the massive range of the pilgrims tend to have only one chance to perform the umrah in the life. Henceforth, it has to be the recommended to you to be well aware of all the travel requirements of the Umrah so that you can reduce the financial expense easily along with making the journey of Umrah the best one in your life. In case you are the one who is not known about the travel and the umrah driven needs and the requirements, then you would end up facing the many problems pertaining to the arrangements of the travel.
As the matter of the fact, one ought to be aware about the 3 and 4 umrah packages and the 5 umrah packages that would help you reduce the cost of the Umrah along with lessen the burden of the pilgrim in the shape of the travel requirements and the needs related to the performing of the Umrah. It has been mentioned that there are wide range of the aspects of the Umrah that you need to take into account in the form of the travel guide, transportation's services, the hoteling and the accommodation facilities, and the most importantly, food. It ought to be mentioned that if you are the one who is not familiar with the aspects and the important parts of the umrah packages, then you may end up confronting the challenging journey along with it being the difficult for the Muslim believer and the near and dear ones. The best part is when you can choose the appropriate 3, 4, or the 5 umrah packages that can help you meet the needs and the requirements in the best possible manner. In case you are worried about the performing of the Umrah owing to the huge expenditure or the hectic travel requirements, then you should go for the best umrah packages that are provided by the travel agencies and the travel agents. These packages would be suitable for the wide range of the pilgrims having different needs in the form of the food, umrah visas, transportation and the other things that come under the umbrella of Umrah package. | {
"redpajama_set_name": "RedPajamaC4"
} | 5,221 |
\section{Introduction}
\subsection{Research Problem Scoping}
The past decade witnessed artificial intelligence's growing power in almost all walks of life. From visual objects to natural languages, the target data for AI models to learn has become more abstract, complex, and structured \cite{2016Building}. With this trend, the idea of training models to comprehend code, a highly abstract form of language with rich compositional structures, naturally emerged in AI research. Learning and representing the semantics, rules, and structures in programs are regarded as a significant indicator of building more robust AI \cite{allamanis2018survey}.
\vspace{0.5em}
\noindent The recent years have also brought about more scientific findings on the process of software engineering, as it is an indispensable factor for high-tech companies' profits. In an array of research work, the roles of code in software engineering were highlighted. For engineers, reading code and searching for related information occupy about 50\% of their working time \cite{maalej2014comprehension}. On the long timeline of software development, a considerable amount of time and efforts will be spent on code repair, performance optimization, and requirements tracking\cite{Amy2016coop}. The cost of code-related tasks stimulated the increasing need for automatic code comprehension in the industry.
\vspace{0.5em}
\noindent As academic proceedings merge with real-world needs, a practical issue is the introduction of AI-based code comprehension into industrial settings. The software source code is usually large-scale and complicated, posing unprecedented challenges for current AI models: How to represent and process the source code of large industrial projects? How to evaluate the difficulty and performance of real-world code comprehension? How to deal with the complexity of software source code? Our work starts from the long-standing Linux kernel source code, towards relevant solutions by dataset collection, representation construction, and network analysis.
\subsection{The Loss of Current AI-based Code Comprehension}
\noindent On the basis of above problems, the related progress of AI technology has been motivating us to dig deeper. Aware of the analogy between natural languages and programming languages, many researchers recently proposed to tackle the code comprehension problems by natural language processing (NLP) approaches \cite{allamanis2018survey,sun2014empirical}. Through encoding methods or deep learning models for embedding, code snippets were transformed into distributed vector representations and fed into models or frameworks that are previously designed for NLP tasks \cite{chen2019literature}. The performances of such pipelines on code processing tasks were impressive. With vector representations, the encoder-decoder framework could help provide feedback for students' coding assignments \cite{piech2015learning}; transformer-based models achieved state-of-the-art performance in the summarization tasks of Python and Java code \cite{ahmad2020transformer}; the dual learning framework was powerful in both describing code snippets by natural language and generating code snippets from natural language descriptions \cite{wei2019code}.
\vspace{0.5em}
\noindent However, NLP technology's success in processing code snippets does not mean these models are fully capable of code comprehension. While treating code as natural language brings convenience for data processing, it leads to the overdependence on relatively shallow semantic information, such as the naming of variables and functions in code. It also results in the neglect of critical structural information, including recursive calls, control structures, and data flow \cite{nguyen2013statistical, allamanis2015bimodal, ben2018neural}. Admittedly, this methodology can achieve satisfactory results when dealing with programming languages centered on scripts and glued by a bunch of commands, such as Python and SQL \cite{yao2018staqc}. But it still struggles to comprehend more abstract languages, including C and C++, in which complex program structures and call relations are of more importance. As these programming languages are the foundations of applications related to operating systems, databases, compilers, and so on, the targeted comprehension is an increasingly pressing issue for many high-tech companies.
\vspace{0.5em}
\noindent Digging into this phenomenon, we discovered that the capacity and limitations of current AI-based code comprehension are mainly shaped by the code datasets used in model training:
\begin{itemize}
\item From the aspect of data resources, many of the datasets were collected from websites containing open-source code, such as Github and StackOverflow \cite{wong2015clocom, yao2018staqc}. Through web information extraction, massive code data and related annotations were included in a dataset, providing adequate labeled materials for training AI models. However, the code samples in such datasets are usually short, simple, and mutually independent snippets. It is hard to generalize the model trained on these datasets to our industrial software code, which is characterized by long-range dependencies and complex structures, as well as interleaved relations between different code blocks.
\item From the aspect of data preprocessing, the characters related to the syntax of programming languages were excluded from the code snippets in datasets \cite{sun2014empirical}. For example, when preprocessing Python code, indentations in the code were utterly removed \cite{ahmad2020transformer}. While such a data preprocessing method simplifies the subsequent modeling procedure, it discards the structural information within the code, "flattening" the compositional structures into token sequences.
\end{itemize}
\vspace{0.5em}
\noindent The constraints from data hindered the development of higher-level automatic systems for software engineering tasks based on complex code comprehension, including requirement tracking \cite{liou2011toward}, code repair \cite{devlin2017semantic}, and error location \cite{lam2017bug}. If datasets closer to real-world software source code are created, we will directly approach possible solutions to the complexity problem in code comprehension and software engineering.
\subsection{Linux Kernel Source Code as Data Resource}
When it comes to dataset construction on code with industrial strength, Linux kernel source code stands out as a perfect choice. Its repository scale has shown a considerable potential to serve as the data resource for a set of software engineering tasks. According to our statistics, there are currently 29079 C files in the open-source repository of Linux kernel (version 5.8.11) and 15 to 16 functions in each file. The characteristics of the repository are also promising for data collection:
\begin{itemize}
\item Firstly, it is a \emph{self-contained} repository without dependence on any external knowledge resources, as an advantageous feature for efficient data management and knowledge mining.
\item It also possesses a large amount of \emph{labeled} code data. The abundant comments in the source code have provided various labels, including function summaries, input/output descriptions, and indications of important steps within functions.
\item Moreover, the code blocks in Linux kernel repository are \emph{interrelated}, linking with each other by a series of call relations. This feature serves as the prerequisite for exploring complexity problems on source code.
\end{itemize}
\vspace{0.5em}
\noindent With the above considerations, we performed a systematical dataset collection work on the latest open-source repository of Linux kernel. Through our work, we hope to provide a data foundation for tackling complex code structures in large-scale software code, and powering software engineering by reliable code comprehension in real-world settings.
\subsection{Thinking Complexity in Software Engineering}
The phenomena of complexity are common and long-existing in the process of software engineering, including design, cooperation, management, and so on \cite{Amy2016coop}. As an intuitive solution, complex network theories were extensively applied to represent these problems \cite{wen2007software}. For example, the related work has modeled the collaboration of software engineering \cite{myers2003software}, the execution of software\cite{cai2009software}, and the relations between software packages \cite{zheng2008analyzing} as complex networks, providing tremendous insights and convenience for the industry's software development efforts.
\begin{figure}
\centering
\includegraphics[width=0.90\linewidth]{Fig1-2.png}
\nocaptionrule
\caption{\textbf{An example sample in our present dataset}}
\label{fig1}
\end{figure}
\vspace{0.5em}
\noindent However, these network representations were mainly targeted at the high-level processes of software engineering, without naturally extending them to the modeling of source code, which is the most fundamental component for software and the origin of almost all the complexity problems in software engineering. Moreover, the nodes of complex networks in these previous work were abstract concepts manually set by researchers and experts, leading to difficulties in building complex networks directly from data.
\vspace{0.5em}
\noindent Under such a background, we conducted further analysis of complexity problems on the collected dataset. Especially, we applied the complex network theory to software source code and modeled code tokens as network nodes for the first time, with the aspiration to call for the rethinking of complexity in software engineering.
\subsection{ Contributions of the Present Work}
Our work will contribute to related areas in mainly two aspects:
\begin{itemize}
\item We proposed the first code \emph{dataset} mined from large-scale and industrial-strength source code. The dataset contains a massive amount of code data with complex relations, supporting an array of real-world software engineering tasks including code comprehension.
\item By extracting code tokens and constructing relational networks, we pioneeringly incorporated complex network theories into software code representations, and created a novel and implementable \emph{methodology} to measure and address the complexity of source code in industrial settings.
\end{itemize}
\begin{figure*}
\centering
\includegraphics[width=0.8\linewidth]{Fig_2.png}
\vspace{0.4em}
\nocaptionrule
\caption{\textbf{The dataset construction pipeline on Linux Kernel Code Repository}}
\label{fig2}
\end{figure*}
\section{Building Dataset on Linux Kernel Repository}
The dataset construction work on Linux Kernel source code involved two integral parts: automatic data collection and fit-for-purpose data partition (Figure 2 illustrates this framework).
\vspace{0.5em}
\noindent We extracted the comments and functions separately from the source code, and combined them to constitute samples in the dataset (see Figure 1 for example). For comment extraction, we applied the Comment Parser toolkit \cite{commentparser2015}, a Python module designed to extract comments from source code, to the code files of Linux kernel repository. For function extraction, we implemented a pattern matching algorithm to help recognize C functions' beginning and ending positions in code files. Syntax-related characters were reserved in the code data of each extracted functions for the further building of structural representations.
\vspace{0.5em}
\noindent Upon the information extraction on 12 critical folders in the repository, we constructed our LiKe-CoCo (\underline{Li}nux \underline{Ke}rnel \underline{Co}de \underline{Co}mprehension) Dataset, the scale of it amounts to nearly 140k C function samples. Table 1 shows the comprehensive statistics of LiKe-CoCo Dataset.
\vspace{0.5em}
\noindent According to the information contained in each data sample, data cleaning and partition were further performed. We established four subsets of the LiKe-CoCo Dataset to support various software engineering and code comprehension tasks:
\begin{itemize}
\item The \textbf{LiKe-CoCo(files)} subset was
designed for the mining of relations between functions in the range of a file, as well as the overall summarization of a code file. Each sample is a list of all the functions and corresponding comments within a C code file.
\item The \textbf{LiKe-CoCo(gold)} subset was established for multiple code comprehension tasks including function summarization, input/output tracking, and description generation. The samples are C functions with explicit and clearly-written comments before the function beginnings, indicating the summary, input/output description, and detailed description of the function. We view them as gold-quality labeled data.
\item The \textbf{LiKe-CoCo(steps)} subset was collected for the learning of internal processes within long function bodies, as well as the relations between these steps and the overall function. Each of the samples contains an extracted C function, a function summary, and some short comments inside the function body, serving as signs of internal steps towards function implementation.
\item The \textbf{LiKe-CoCo(sumry)} subset was built for the supervised learning of general function summarization tasks. There are only two integral components in a data sample: a C function and a function summary extracted from comments before the function beginning.
\end{itemize}
\section{Model Source Code as Complex Network}
After dataset construction, we tokenized the C code within each sample. Relation parsing, network analysis, and knowledge management were implemented on the extracted tokens to address the complexity of code.
\subsection{Nodes of Network: Token Extraction}
\begin{table}
\centering
\nocaptionrule
\caption{\textbf{The statistics of the overall dataset and four subsets} \\ (including the average length of code (in lines) in the dataset and the amount of samples from 12 folders of Linux kernel repository).}
\vspace{0.5em}
\begin{tabular}{lrrrrr}
\toprule
& \multicolumn{1}{l}{Overall} & \multicolumn{1}{l}{Files} & \multicolumn{1}{l}{Gold} & \multicolumn{1}{l}{Steps} & \multicolumn{1}{l}{Sumry} \\
\midrule
Avg Length & 23.04 & 23.04 & 27.67 & 44.32 & 28.64 \\
Sample Num & 136428 & 8527 & 9617 & 16209 & 36686 \\
\textit{from folders:} & & & & & \\
arch & 47940 & 4681 & 1505 & 4326 & 9683 \\
block & 1936 & 70 & 349 & 248 & 615 \\
crypto & 2004 & 150 & 73 & 168 & 352 \\
certs & 16 & 4 & 2 & 2 & 10 \\
fs & 29073 & 1277 & 2287 & 4851 & 10233 \\
ipc & 306 & 11 & 33 & 42 & 78 \\
kernel & 12168 & 353 & 1302 & 1693 & 4103 \\
lib & 3269 & 341 & 520 & 284 & 927 \\
mm & 4727 & 116 & 531 & 873 & 1775 \\
net & 32184 & 1392 & 2196 & 3248 & 7532 \\
security & 2494 & 125 & 808 & 452 & 1318 \\
virt & 311 & 7 & 11 & 22 & 60 \\
\bottomrule
\end{tabular}%
\label{tab:addlabel}%
\end{table}%
The code tokenization was conducted through Pygments, a generic syntax highlighting engine for source code \cite{Pygments2006}. Each C function in the dataset was transformed into a set of tokens, paired with token labels that explain corresponding tokens' roles. The extracted tokens could be categorized into four types:
\begin{itemize}
\item \textbf{Keywords} of C programming language, including keywords of control structures, such as "if", "for", "continue" and "break", as well as types of variables, such as "int" and "struct". The keyword tokens could help highlight the information flow in programs \cite{hendrix2002effectiveness}, playing a critical role in recognizing the \emph{structures} of code blocks.
\item \textbf{Names}, including function names, class names and variable names. The naming of variables and functions is an indispensable component in comprehending code \emph{semantics}, as the meaning of these names will directly influence our initial understanding of programs \cite{allamanis2015suggesting}.
\item \textbf{Operators}, such as "=", "+", "!", and "\&". In addition to the name tokens, operator tokens also contribute to the \emph{semantics} of code blocks.
\item \textbf{Punctuations}, including "(", ")", "\{", "\}", and ";". Punctuation tokens also help accentuate the \emph{structures} and relations within code tokens.
\end{itemize}
\begin{table*}
\centering
\nocaptionrule
\caption{\textbf{The statistical analysis on tokenized code and network-based representations}}
(including the average amount of various code tokens, as well as the mean value of maximum DC and MD in constructed networks)
\vspace{0.5em}
\begin{tabular}{lrrrrrrrr}
\toprule
& \multicolumn{1}{l}{Token Num} & \multicolumn{1}{l}{Keyw Num} & \multicolumn{1}{l}{Name Num} & \multicolumn{1}{l}{Punc Num} & \multicolumn{1}{l}{Operator Num} & \multicolumn{1}{l}{Node Num} & \multicolumn{1}{l}{Max DC} & \multicolumn{1}{l}{Mean Dist} \\
\midrule
Overall Dataset & 157.58 & 13.45 & 50.56 & 55.29 & 31.87 & 183.43 & 0.23 & 4.11 \\
\textit{from folders:} & & & & & & & & \\
arch & 134.8 & 12.11 & 41.34 & 47.47 & 25.29 & 156.02 & 0.24 & 3.99 \\
block & 136.44 & 11.58 & 44.3 & 47.2 & 29.07 & 158.06 & 0.23 & 4.15 \\
crypto & 179.48 & 15.53 & 57.78 & 62.78 & 36.11 & 207.3 & 0.24 & 4.31 \\
certs & 120.5 & 9 & 37.5 & 47 & 18 & 140 & 0.26 & 4.19 \\
fs & 217.85 & 16.43 & 70.27 & 75.7 & 45.71 & 256.36 & 0.2 & 4.35 \\
ipc & 139.91 & 11.21 & 46.39 & 47.09 & 32.7 & 159.91 & 0.21 & 4.02 \\
kernel & 110.74 & 10.19 & 35.8 & 39.61 & 22.39 & 128.73 & 0.26 & 3.94 \\
lib & 125.17 & 12.92 & 38.42 & 44.57 & 23.89 & 145.03 & 0.26 & 4.11 \\
mm & 140.11 & 13.09 & 47.46 & 53.88 & 26.23 & 168.46 & 0.23 & 4.18 \\
net & 155.79 & 13.28 & 51.05 & 54.29 & 31.91 & 180.02 & 0.22 & 4.07 \\
security & 147.09 & 14.43 & 46.36 & 50.93 & 29.62 & 169.49 & 0.24 & 3.99 \\
virt & 181.55 & 17.45 & 58.36 & 62.45 & 40.73 & 208.91 & 0.18 & 4.43 \\
\bottomrule
\end{tabular}%
\label{tab:addlabel}%
\end{table*}%
\vspace{0.5em}
\noindent The detailed statistics of code tokens are included in Table 2. There are about 150 code tokens in each data sample, much more than the number of code tokens in the datasets\cite{wan2018improving, hu2018summarizing} for NLP-based code comprehension methods (nearly 50 tokens in each Python sample and 120 tokens in each Java sample). Punctuation, name, operator, and keyword tokens respectively occupy nearly 30\%, 30\%, 20\%, and 9\% of the total token amount. Taking these high ratios into account, the neglect of either structural or semantic information contained in the four types of tokens will lead to a consequential loss of comprehensive code understanding.
\subsection{Edges within Network: Relation Parsing}
Taking the extracted tokens as nodes, we constructed network-based representations to represent the structure of each C function. The established structure networks are undirected and unweighted, with edges indicating the compositional relations between nodes.
\vspace{0.5em}
\noindent To identify the network structure within a sequence of code tokens, we incorporated some interpretive nodes, such as "if statement" nodes and "while statement" nodes, to help segment the code tokens and describe the relations between nodes. When parsing code tokens, the relation parser assigned the corresponding interpretive node to a group of tokens within a code block based on its contained keywords and punctuations. The code token group was then recursively decomposed into lower-level groups of tokens, with more detailed interpretive nodes assigned to each group. Figure 3 has shown a clear example of this process.
\vspace{0.5em}
\noindent In contrast to the previous methods based on function call graph \cite{bohnet2006visual} and AST trees \cite{chilowicz2009syntax} to abstractly represent source code, this representation method will clearly reflect the internal structures of C functions, including control structures and long-term dependencies, without oversimplifying or discarding the rich semantics of code tokens. It provides us with a novel approach to leverage both structural and semantic information in code comprehension.
\subsection{Measure the Network: Complexity Evaluation}
\begin{figure}
\centering
\includegraphics[width=0.9\linewidth]{Fig_3.png}
\nocaptionrule
\caption{\textbf{An example of the network-based representations} \\ (The green nodes refer to interpretive nodes, while the blue nodes are representive of code tokens.)}
\label{fig3}
\end{figure}
\vspace{0.5em}
\noindent The large amount of nodes in network representations (see Table 2 for the statistics on network structures) has shown the complexity of source code in data samples to some degree. To systematically measure the structural complexity, we further computed several indicators \cite{lu2009analysis}:
\begin{itemize}
\item To quantify the \textbf{density} of network representations, we focus on the \emph{degree} of nodes in structural graphs. For a complex network with $n$ nodes, the degree $k_i$ of a node $i$ refers to the number of edges related to it. A node's importance can be further evaluated through degree centrality (DC) by formula (1). As the degree centrality of nodes increases, the network becomes denser. Therefore, we measured the density of structure networks by the maximum DC in each sample.
\begin{equation}
DC(i) = \frac{k_i}{n-1} \label{eq:DC}
\end{equation}
\item To evaluate the \textbf{scale} of structural networks, we considered the \emph{distances} between pairs of nodes. For a connected network with $n$ nodes, the distance $d_{ij}$ between node $i$ and node $j$ refers to the length of the shortest path between the two nodes. The mean distance (MD) computed by formula (2) indicates the sparsity between nodes. As the mean distance increases, the scale of the network becomes larger. To calculate this indicator, We implemented the Floyd algorithm\cite{wei2010optimized}, a classical algorithm in finding shortest paths, to compute the shortest distance between node pairs.
\begin{equation}
MD=\frac{\sum_{i \neq j}d_{ij}}{n(n-1)}\label{eq:MD}
\end{equation}
\end{itemize}
\noindent Table 2 has shown the results of our computation. The degree centrality of nodes in each network exceeded the value of 0.2, which means there are some nodes associated with more than 1/5 of all the possible edges, reflecting the high density of relations within structure networks. The mean distances between nodes are also noteworthy. According to the algorithm output, there were averagely more than four edges in the shortest path between each pair of nodes in a network. Such a phenomenon indicates the sparsity and large scale of structure networks.
\subsection{Bridge Networks: Knowledge Base Settings}
While the network representations expressed the internal structure of C functions, a domain-specific knowledge base of Linux Kernel could further help us understand the token relations in the whole repository scope, thus bridging the networks established for individual data samples.
\vspace{0.5em}
\noindent The knowledge base mainly focused on name tokens. The occurrences of same name nodes in multiple networks were seen as the indications of network connection for complex networks and the signs of call relations in software source code. For each name token extracted from Linux kernel source code, we collected its meaning from related comments and the co-occurrence relations with other name tokens. The name, meaning, and relations were then reserved into a database as the domain-specific knowledge of Linux kernel. There currently are 50016 items in the knowledge base, and 9029 of them incorporated detailed descriptions from comments. The average number of co-occurrence relations associated with a name token is 85.31, reflecting the complex token connections within the whole Linux kernel repository.
\vspace{0.5em}
\noindent The established knowledge base will also largely facilitate the knowledge management of Linux kernel source code, thus improving the performance and interpretability of automatic code comprehension. Besides the efficient searching of token explanations, the knowledge base also assists us in the tracking of interleaved relations between tokens, leading to a deeper understanding of token semantics from related tokens, instead of the shallow interpretation from only the naming of tokens.
\section{Discussions}
Starting from code comprehension, we established the first code dataset collected from large-scale software repositories. Detailed analysis of code structures underlined the complexity of token relations in source code. The constructed network representations and knowledge base further help to evaluate and resolve the complexity in understanding code. Our findings and methodologies are promising for future research work on code comprehension, and highly meaningful to the overall process of software engineering.
\subsection{The Reach and Application of Our Work}
\subsubsection{Code Comprehension}
From the perspective of \textbf{Machine Code Comprehension}, the network-based representations of source code have brought great convenience and opportunities for future analysis and application based on the well-developed complex network theory. By analyzing the structure networks, downstream models will quickly recognize the quality, functionality, schema, and composition of source code. For example:
\begin{itemize}
\item The detection of circuits in the constructed network \cite{chen2017robustness} could serve as a kind of evaluation for code robustness, as more circuits in the network means more redundant paths between code tokens.
\item The identification of similar network structures \cite{zhang2018measure} will help the classification of coding schemas, thus facilitating the recognition of code functionalities.
\item The analysis of network modularity \cite{newman2004fast, fortunato2007resolution} will assist in measuring how well the network will be partitioned into a group of sub-networks, providing a possible approach to tackle the complexity of structure networks by disintegration.
\end{itemize}
\vspace{0.5em}
\noindent
From the perspective of \textbf{Human Code Comprehension}, the data collection pipeline on source code will help people understand software code more efficiently. By building datasets and knowledge base on software code repositories, we will better discover and organize the knowledge contained in source code. For example, our dataset could help engineers and students systematically learn about operating systems and Linux kernel. Further applications of this pipeline to high-tech companies' leading products will also enable the establishment of internal datasets and knowledge bases for targeted code comprehension, largely alleviating engineers' working burden.
\subsubsection{More Than Code Comprehension}
\vspace{0.5em}
\noindent From the perspective of \textbf{Software Engineering Process}, our ideas and methods on complexity will facilitate almost all the components of software engineering, including the process of design, coding, debugging, testing, and sustaining, as well as the team management and communication in collaborative engineering \cite{Amy2016coop}. For example:
\begin{itemize}
\item The analysis of code complexity on different versions of software source code will outline software evolution more clearly, bringing convenience for performance optimization, testing, and sustaining.
\item The network-based representations could help the design of software API for different groups of people, such as customers and engineers, by emphasizing different network nodes.
\item The complexity measurement and network representations on code will also facilitate the work assignment of a developers' team and the knowledge sharing between engineers, especially between experts and novices.
\end{itemize}
\subsection{Future Directions}
Our work will strengthen the research on incorporating structural information into the code comprehension process. Several previous works mainly focused on the performance improvement of neural network-based models by including code structure representations \cite{mou2016convolutional,ben2018neural,ahmad2020transformer}. We here highlighted the prospect of conducting code comprehension tasks directly on structural representations such as code structure networks, and exploiting neural network models' strength in processing big data to adjust parameters for the representations. Combining knowledge-based representations and data-driven learning will lead to more controllable, interpretable, and widely applicable comprehensions of source code.
\vspace{0.5em}
\noindent Our work will also inspire the modeling of the software engineering process based on complex network theory. We extended the application of complex networks to software source code for the first time and revealed its power in representing and quantifying code complexity for comprehension. Future research work needs to build more extensive networks on the whole source code repository, enable the network representations to learn from data, and simplify the network complexity for further computations and applications.
\section{Conclusion}
In the present work, we systematically collected a large-scale code dataset from the open-source Linux Kernel repository to support complex code comprehension and further software engineering tasks in industrial settings. Token extraction, network representations, and database construction strengthened the interpretation, abstraction, and application of code structures. The complex network-based analysis on code structure networks provided us with a novel approach to measure the complexity of code for comprehension tasks. Together, we call for further exploration of real-world code comprehension on software source code, comprehensive evaluation of industrial code complexity, and in-depth code understanding that combined both structural and semantic information.
| {
"redpajama_set_name": "RedPajamaArXiv"
} | 621 |
package com.sap.dcode.agency.ui.offline;
import java.util.List;
import android.view.LayoutInflater;
import android.view.View;
import android.view.ViewGroup;
import android.widget.BaseAdapter;
import android.widget.ListView;
import android.widget.TextView;
import com.sap.dcode.agency.R;
import com.sap.dcode.agency.types.OfflineError;
public class OfflineErrorListAdapter extends BaseAdapter {
final private LayoutInflater inflater;
final private OfflineErrorListFragment fragment;
final private List<OfflineError> errors;
final private ListView myList;
public OfflineErrorListAdapter(OfflineErrorListFragment fragment, ListView myList, LayoutInflater inflater, List<OfflineError> errors) {
this.inflater = inflater;
this.fragment = fragment;
this.myList = myList;
this.errors = errors;
}
@Override
public int getCount() {
return (errors!=null)?errors.size():0;
}
@Override
public Object getItem(int position) {
return (errors!=null)?errors.get(position):null;
}
@Override
public long getItemId(int position) {
return 0;
}
@Override
public View getView(int position, View view, ViewGroup parent) {
if (view == null) {
view = inflater.inflate(R.layout.error_list_item, parent, false);
}
OfflineError errorArchive = errors.get(position);
((TextView) view.findViewById(R.id.error_method)).setText(errorArchive.getRequestMethod());
((TextView) view.findViewById(R.id.error_http_code)).setText(errorArchive.getHttpStatusCode());
((TextView) view.findViewById(R.id.request_url)).setText(errorArchive.getRequestURL());
view.setOnClickListener(new ErrorListClickListener(errors.get(position)));
view.setOnLongClickListener(new ErrorListLongClickListener(position));
return view;
}
/**
* Delete an item from the list.
*
* @param i the index of the item to delete
*/
public void deleteItem(int i) {
myList.setItemChecked(i, false); // un-select the item that is being deleted
errors.remove(i);
notifyDataSetChanged();
}
/**
* This listener handles clicks on a travel agency in the list.
*/
private class ErrorListClickListener implements View.OnClickListener {
final private OfflineError item;
private ErrorListClickListener(OfflineError item) {
this.item = item;
}
@Override
public void onClick(final View view) {
fragment.onAgencySelected(item);
}
}
/**
* This listener handles long clicks on a travel agency in the list. Long clicking
* on an agency makes it available for deleting.
*/
private class ErrorListLongClickListener implements View.OnLongClickListener {
final private int i;
private ErrorListLongClickListener(int i) {
this.i = i;
}
@Override
public boolean onLongClick(View view) {
myList.setItemChecked(i, true);
return true;
}
}
}
| {
"redpajama_set_name": "RedPajamaGithub"
} | 1,318 |
Monabanq est une banque en ligne française fondée en 2006 et basée à Villeneuve-d'Ascq, dans le nord de la France en métropole lilloise.
Monabanq est une filiale du groupe Cofidis Participations, lui-même actionnaire de l'enseigne de Crédit du même nom. Son actionnaire de référence est le Crédit mutuel Alliance fédérale (79,99 %). Elle est également adossée au Groupe Argosyn Entreprise qui regroupe les anciennes activités de type commerce en ligne BtoB et les activités financières précédemment détenues par le Groupe 3SI (Groupe 3 Suisses International) détenant 20,01 % de l'entité.
Historique
Covefi, société établie en France, naît en 1987. Elle devient Banque Covefi en 1992, et développe peu à peu le concept de banque à distance. En 1997, elle ajoute le mot Banque à son logo avec le lancement du compte courant.
Le , Banque Covefi devient Monabanq, se présentant comme « la banque nouvelle génération ». Elle lance, en 2011, une gamme de moyens de paiement pour les clients en difficulté, et en 2012, une gamme assurances et prévoyance. En 2014, elle devient la banque en ligne à accepter l'ouverture d'un compte courant en ligne sans aucune condition de revenus. En , Monabanq déclare avoir 189 collaborateurs, 372 millions d'euros d'encours d'épargne brut et 30 millions d'euros de financements.
Depuis le , Monabanq est la première banque en ligne à proposer un dispositif adapté à ses clients et prospects touchés par une déficience auditive, ce service d'accès à Monabanq via la plateforme DEAFI ou l'application mobile Deafiline est gratuit. Fin 2018, Monabanq est élu, pour la deuxième année consécutive, Service client de l'année 2019 dans la catégorie banque en ligne.
Les produits
De par son appartenance au groupe Crédit Mutuel / CIC , Monabanq permet à ses clients d'utiliser les guichets automatiques bancaires de ces deux banques pour des opérations de base (consultations, virements, éditions de RIB, dépôts d'espèces et de chèques etc.). Son service client (conseillers basés à Villeneuve-d'Ascq) est accessible en semaine de 8h à 21h et le samedi de 8h à 16h.
Monabanq propose une gamme complète de produits bancaires :
Comptes courants particulier / auto-entrepreneur
Épargne réglementée : livret A, livret de développement durable
Épargne non réglementée : livret épargne, livret Croissance, compte à terme fixe, compte à terme progressif
Offre jeune : compte épargne jeune, livret jeune, compte chèque jeune
Placement : assurance-vie, plan d'épargne en actions, compte-titres, OPC
Assurances : Santé, Habitation, Auto, Décès, Perte d'emploi
Crédits : Immobilier, Auto, Moto, Travaux, Crédit renouvelable.
Notes et références
Annexes
Articles connexes
Banque en ligne
Liens externes
Banque ayant son siège en France
Entreprise fondée en 1987
Banque en ligne
Crédit mutuel CIC | {
"redpajama_set_name": "RedPajamaWikipedia"
} | 4,279 |
require 'spec_helper'
RSpec.describe Projects::TransferService do
include GitHelpers
let(:user) { create(:user) }
let(:group) { create(:group) }
let(:project) { create(:project, :repository, :legacy_storage, namespace: user.namespace) }
subject(:execute_transfer) { described_class.new(project, user).execute(group) }
context 'with npm packages' do
before do
group.add_owner(user)
end
subject(:transfer_service) { described_class.new(project, user) }
let!(:package) { create(:npm_package, project: project) }
context 'with a root namespace change' do
it 'does not allow the transfer' do
expect(transfer_service.execute(group)).to be false
expect(project.errors[:new_namespace]).to include("Root namespace can't be updated if project has NPM packages")
end
end
context 'without a root namespace change' do
let(:root) { create(:group) }
let(:group) { create(:group, parent: root) }
let(:other_group) { create(:group, parent: root) }
let(:project) { create(:project, :repository, namespace: group) }
before do
other_group.add_owner(user)
end
it 'does allow the transfer' do
expect(transfer_service.execute(other_group)).to be true
expect(project.errors[:new_namespace]).to be_empty
end
end
end
context 'namespace -> namespace' do
before do
allow_next_instance_of(Gitlab::UploadsTransfer) do |service|
allow(service).to receive(:move_project).and_return(true)
end
allow_next_instance_of(Gitlab::PagesTransfer) do |service|
allow(service).to receive(:move_project).and_return(true)
end
group.add_owner(user)
end
it 'updates the namespace' do
transfer_result = execute_transfer
expect(transfer_result).to be_truthy
expect(project.namespace).to eq(group)
end
end
context 'when transfer succeeds' do
before do
group.add_owner(user)
end
it 'sends notifications' do
expect_any_instance_of(NotificationService).to receive(:project_was_moved)
execute_transfer
end
it 'invalidates the user\'s personal_project_count cache' do
expect(user).to receive(:invalidate_personal_projects_count)
execute_transfer
end
it 'executes system hooks' do
expect_next_instance_of(described_class) do |service|
expect(service).to receive(:execute_system_hooks)
end
execute_transfer
end
it 'moves the disk path', :aggregate_failures do
old_path = project.repository.disk_path
old_full_path = project.repository.full_path
execute_transfer
project.reload_repository!
expect(project.repository.disk_path).not_to eq(old_path)
expect(project.repository.full_path).not_to eq(old_full_path)
expect(project.disk_path).not_to eq(old_path)
expect(project.disk_path).to start_with(group.path)
end
it 'updates project full path in .git/config' do
execute_transfer
expect(rugged_config['gitlab.fullpath']).to eq "#{group.full_path}/#{project.path}"
end
it 'updates storage location' do
execute_transfer
expect(project.project_repository).to have_attributes(
disk_path: "#{group.full_path}/#{project.path}",
shard_name: project.repository_storage
)
end
end
context 'when transfer fails' do
let!(:original_path) { project_path(project) }
def attempt_project_transfer(&block)
expect do
execute_transfer
end.to raise_error(ActiveRecord::ActiveRecordError)
end
before do
group.add_owner(user)
expect_any_instance_of(Labels::TransferService).to receive(:execute).and_raise(ActiveRecord::StatementInvalid, "PG ERROR")
end
def project_path(project)
Gitlab::GitalyClient::StorageSettings.allow_disk_access do
project.repository.path_to_repo
end
end
def current_path
project_path(project)
end
it 'rolls back repo location' do
attempt_project_transfer
expect(project.repository.raw.exists?).to be(true)
expect(original_path).to eq current_path
end
it 'rolls back project full path in .git/config' do
attempt_project_transfer
expect(rugged_config['gitlab.fullpath']).to eq project.full_path
end
it "doesn't send move notifications" do
expect_any_instance_of(NotificationService).not_to receive(:project_was_moved)
attempt_project_transfer
end
it "doesn't run system hooks" do
attempt_project_transfer do |service|
expect(service).not_to receive(:execute_system_hooks)
end
end
it 'does not update storage location' do
create(:project_repository, project: project)
attempt_project_transfer
expect(project.project_repository).to have_attributes(
disk_path: project.disk_path,
shard_name: project.repository_storage
)
end
end
context 'namespace -> no namespace' do
let(:group) { nil }
it 'does not allow the project transfer' do
transfer_result = execute_transfer
expect(transfer_result).to eq false
expect(project.namespace).to eq(user.namespace)
expect(project.errors.messages[:new_namespace].first).to eq 'Please select a new namespace for your project.'
end
end
context 'disallow transferring of project with tags' do
let(:container_repository) { create(:container_repository) }
before do
stub_container_registry_config(enabled: true)
stub_container_registry_tags(repository: :any, tags: ['tag'])
project.container_repositories << container_repository
end
it 'does not allow the project transfer' do
expect(execute_transfer).to eq false
end
end
context 'namespace -> not allowed namespace' do
it 'does not allow the project transfer' do
transfer_result = execute_transfer
expect(transfer_result).to eq false
expect(project.namespace).to eq(user.namespace)
end
end
context 'namespace which contains orphan repository with same projects path name' do
let(:fake_repo_path) { File.join(TestEnv.repos_path, group.full_path, "#{project.path}.git") }
before do
group.add_owner(user)
TestEnv.create_bare_repository(fake_repo_path)
end
after do
FileUtils.rm_rf(fake_repo_path)
end
it 'does not allow the project transfer' do
transfer_result = execute_transfer
expect(transfer_result).to eq false
expect(project.namespace).to eq(user.namespace)
expect(project.errors[:new_namespace]).to include('Cannot move project')
end
end
context 'target namespace containing the same project name' do
before do
group.add_owner(user)
create(:project, name: project.name, group: group, path: 'other')
end
it 'does not allow the project transfer' do
transfer_result = execute_transfer
expect(transfer_result).to eq false
expect(project.namespace).to eq(user.namespace)
expect(project.errors[:new_namespace]).to include('Project with same name or path in target namespace already exists')
end
end
context 'target namespace containing the same project path' do
before do
group.add_owner(user)
create(:project, name: 'other-name', path: project.path, group: group)
end
it 'does not allow the project transfer' do
transfer_result = execute_transfer
expect(transfer_result).to eq false
expect(project.namespace).to eq(user.namespace)
expect(project.errors[:new_namespace]).to include('Project with same name or path in target namespace already exists')
end
end
context 'target namespace allows developers to create projects' do
let(:group) { create(:group, project_creation_level: ::Gitlab::Access::DEVELOPER_MAINTAINER_PROJECT_ACCESS) }
context 'the user is a member of the target namespace with developer permissions' do
before do
group.add_developer(user)
end
it 'does not allow project transfer to the target namespace' do
transfer_result = execute_transfer
expect(transfer_result).to eq false
expect(project.namespace).to eq(user.namespace)
expect(project.errors[:new_namespace]).to include('Transfer failed, please contact an admin.')
end
end
end
context 'visibility level' do
let(:group) { create(:group, :internal) }
before do
group.add_owner(user)
end
context 'when namespace visibility level < project visibility level' do
let(:project) { create(:project, :public, :repository, namespace: user.namespace) }
before do
execute_transfer
end
it { expect(project.visibility_level).to eq(group.visibility_level) }
end
context 'when namespace visibility level > project visibility level' do
let(:project) { create(:project, :private, :repository, namespace: user.namespace) }
before do
execute_transfer
end
it { expect(project.visibility_level).to eq(Gitlab::VisibilityLevel::PRIVATE) }
end
end
context 'missing group labels applied to issues or merge requests' do
it 'delegates transfer to Labels::TransferService' do
group.add_owner(user)
expect_next_instance_of(Labels::TransferService, user, project.group, project) do |labels_transfer_service|
expect(labels_transfer_service).to receive(:execute).once.and_call_original
end
execute_transfer
end
end
context 'missing group milestones applied to issues or merge requests' do
it 'delegates transfer to Milestones::TransferService' do
group.add_owner(user)
expect_next_instance_of(Milestones::TransferService, user, project.group, project) do |milestones_transfer_service|
expect(milestones_transfer_service).to receive(:execute).once.and_call_original
end
execute_transfer
end
end
context 'when hashed storage in use' do
let!(:project) { create(:project, :repository, namespace: user.namespace) }
let!(:old_disk_path) { project.repository.disk_path }
before do
group.add_owner(user)
end
it 'does not move the disk path', :aggregate_failures do
new_full_path = "#{group.full_path}/#{project.path}"
execute_transfer
project.reload_repository!
expect(project.repository).to have_attributes(
disk_path: old_disk_path,
full_path: new_full_path
)
expect(project.disk_path).to eq(old_disk_path)
end
it 'does not move the disk path when the transfer fails', :aggregate_failures do
old_full_path = project.full_path
expect_next_instance_of(described_class) do |service|
allow(service).to receive(:execute_system_hooks).and_raise('foo')
end
expect { execute_transfer }.to raise_error('foo')
project.reload_repository!
expect(project.repository).to have_attributes(
disk_path: old_disk_path,
full_path: old_full_path
)
expect(project.disk_path).to eq(old_disk_path)
end
end
describe 'refreshing project authorizations' do
let(:group) { create(:group) }
let(:owner) { project.namespace.owner }
let(:group_member) { create(:user) }
before do
group.add_user(owner, GroupMember::MAINTAINER)
group.add_user(group_member, GroupMember::DEVELOPER)
end
it 'refreshes the permissions of the old and new namespace' do
execute_transfer
expect(group_member.authorized_projects).to include(project)
expect(owner.authorized_projects).to include(project)
end
it 'only schedules a single job for every user' do
expect_next_instance_of(UserProjectAccessChangedService, [owner.id, group_member.id]) do |service|
expect(service).to receive(:execute).once.and_call_original
end
execute_transfer
end
end
describe 'transferring a design repository' do
subject { described_class.new(project, user) }
before do
group.add_owner(user)
end
def design_repository
project.design_repository
end
it 'does not create a design repository' do
expect(subject.execute(group)).to be true
project.clear_memoization(:design_repository)
expect(design_repository.exists?).to be false
end
describe 'when the project has a design repository' do
let(:project_repo_path) { "#{project.path}#{::Gitlab::GlRepository::DESIGN.path_suffix}" }
let(:old_full_path) { "#{user.namespace.full_path}/#{project_repo_path}" }
let(:new_full_path) { "#{group.full_path}/#{project_repo_path}" }
context 'with legacy storage' do
let(:project) { create(:project, :repository, :legacy_storage, :design_repo, namespace: user.namespace) }
it 'moves the repository' do
expect(subject.execute(group)).to be true
project.clear_memoization(:design_repository)
expect(design_repository).to have_attributes(
disk_path: new_full_path,
full_path: new_full_path
)
end
it 'does not move the repository when an error occurs', :aggregate_failures do
allow(subject).to receive(:execute_system_hooks).and_raise('foo')
expect { subject.execute(group) }.to raise_error('foo')
project.clear_memoization(:design_repository)
expect(design_repository).to have_attributes(
disk_path: old_full_path,
full_path: old_full_path
)
end
end
context 'with hashed storage' do
let(:project) { create(:project, :repository, namespace: user.namespace) }
it 'does not move the repository' do
old_disk_path = design_repository.disk_path
expect(subject.execute(group)).to be true
project.clear_memoization(:design_repository)
expect(design_repository).to have_attributes(
disk_path: old_disk_path,
full_path: new_full_path
)
end
it 'does not move the repository when an error occurs' do
old_disk_path = design_repository.disk_path
allow(subject).to receive(:execute_system_hooks).and_raise('foo')
expect { subject.execute(group) }.to raise_error('foo')
project.clear_memoization(:design_repository)
expect(design_repository).to have_attributes(
disk_path: old_disk_path,
full_path: old_full_path
)
end
end
end
end
def rugged_config
rugged_repo(project.repository).config
end
end
| {
"redpajama_set_name": "RedPajamaGithub"
} | 8,222 |
popover (n.)
also pop-over, "light cake," 1859, from pop (v.) + over (adv.). Perhaps so called because it swells over the rim of the tin when baked.
Entries linking to popover
pop (v.)
mid-15c., "to strike so as to cause to make a short, quick sound;" intransitive sense "make a short, quick sound" is from 1570s; imitative. Of eyes, "to protrude" (as if about to burst), from 1670s. Sense of "to appear or to put with a quick, sudden motion" (often with up, off, in, etc.) is recorded from mid-15c. Baseball sense of "to hit a ball high in the air" is from 1867. To pop the question is from 1725, specific sense of "propose marriage" is from 1826. Related: Popped; popping.
over (prep., adv.)
Old English ofer "beyond; above, in place or position higher than; upon; in; across, past; more than; on high," from Proto-Germanic *uberi (source also of Old Saxon obar, Old Frisian over, Old Norse yfir, Old High German ubar, German über, Gothic ufar "over, above"), from PIE root *uper "over."
As an adjective from Old English uffera. The senses of "past, done, finished; through the whole extent, from beginning to end" are attested from late 14c. The sense of "so as to cover the whole surface" is from c. 1400. Meaning "leaning forward and down" is from 1540s. The meaning "recovered from" is from 1929. In radio communication, it is used to indicate the speaker has finished speaking (1926).
Above expresses greater elevation, but not necessarily in or near a perpendicular direction; over expresses perpendicularity or something near it: thus, one cloud may be above another, without being over it. Over often implies motion or extension where above would not; hence the difference in sense of the flying of a bird over or above a house, the hanging of a branch over or above a wall. In such uses over seems to represent greater nearness. [Century Dictionary]
Phrase over and above (mid-15c.) is pleonastic, for emphasis. Adjective phrase over-the-counter is attested from 1875, originally of stocks and shares. To be (someone) all over "be exactly what one expects of (someone)" is by 1721.
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<a href="https://www.etymonline.com/word/popover">Etymology of popover by etymonline</a>
Harper, D. (n.d.). Etymology of popover. Online Etymology Dictionary. Retrieved $(datetime), from https://www.etymonline.com/word/popover
Harper Douglas, "Etymology of popover," Online Etymology Dictionary, accessed $(datetime), https://www.etymonline.com/word/popover.
Harper, Douglas. "Etymology of popover." Online Etymology Dictionary, https://www.etymonline.com/word/popover. Accessed $(datetimeMla).
D. Harper. "Etymology of popover." Online Etymology Dictionary. https://www.etymonline.com/word/popover (accessed $(datetime)).
Dictionary entries near popover
popinjay
popish
popliteal
poppycock | {
"redpajama_set_name": "RedPajamaCommonCrawl"
} | 9,686 |
[](https://travis-ci.org/kyleladd/travis-demonstration)
## Installation
###### Install dependencies with Composer
```
composer install
```
###### Run PHPUnit tests
```
vendor/phpunit/phpunit/phpunit tests
``` | {
"redpajama_set_name": "RedPajamaGithub"
} | 6,435 |
{"url":"https:\/\/gateoverflow.in\/2058\/gate2014-3-24","text":"7.1k views\n\nA bit-stuffing based framing protocol uses an $\\text{8-bit}$ delimiter pattern of $01111110.$\nIf the output bit-string after stuffing is $01111100101,$ then the input bit-string is:\n\n1. $0111110100$\n2. $0111110101$\n3. $0111111101$\n4. $0111111111$\n\nedited | 7.1k views\n0\nDelimiter pattern=$01111110$.From this pattern it is clear that we are going to insert $0$ in input string.It means no. of $1$ in input string and output string(after stuffing 0) should be same.So we can directly say ans $b$\n\n$011111$ *one zero emitted here* $0101$\n\nCorrect Answer: $B$\n\nedited\n0\n+3\nIn bit stuffing we insert 0 after 5 consecutive occurrence of 1.\n0\nSo what is the use of delimiter pattern here ???\n+5\nSee the given delimiter patter also supports the same thing. If in input string 01111110 will be present then it will be wrongly interpreted as delimiter, so to avoid that we are inserting 0 after five consecutive occurrences of 1s.\n+3\nhow about inserting a 1 after 6 occurrences of 1?\n+3\nExplain Clearly Use of Delimeter and why we are using 5 bits instead of 6 bits? Whatever it is please explain in detail. Thanks.\n+34\nWhy a $0$ is stuffed after five $1s$ and not $6$?\n\nThe purpose of bit stuffing is to distinguish between delimiter pattern occurring at start and end of the frame from the delimiter pattern occurring in the data part.\n\nSuppose if our frame is as following,\n\n$\\overbrace{01111110}^\\text{start of frame delimiter}$ \u00a0\u00a0\u00a0\u00a0$\\overbrace{0101\\textbf{01111110}10}^\\text{data bits}$ \u00a0\u00a0\u00a0\u00a0$\\overbrace{01111110}^\\text{end of frame delimiter}$\n\nSee those data bits also have the delimiter pattern. We can see stuffing after five consecutive $1s$ is proper since if we stuff a $0$ after six $1s$, it won't make any difference.\n\nAlso, the number of $1s$ are the same before and after bit stuffing. by counting the number of $1s$ in options, we can answer this question.\n+4\n\n@junk_mayavi Do we always stuff a 0, can't we stuff a 1 after data 0111 111 1 0 ??\n\nThe bold 1 is stuffed.\n\n+11\n\n@Rishabh Gupta 2\u00a0a 0 bit causes signal transition, whereas 1 bit causes no change. if there is no transition for a long time in transmitted data, there is a chance that sender and receiver clock may go out of sync. this can be the case here.\u00a0\u00a0run-length limited encoding deals with this issue. again, this should be implementation specific and defined by RFCs. such an example definition can be seen here\n\n+1\nThanks.\n\nSo, if the delimiter pattern was something like 1000 0001, or something where we have more 0s, then what we are going to use for bit stuffing?\n+5\n\n@Rishabh Gupta 2\u00a0may be in that case, it makes sense if they use 1 to stuff. but those will be implementation specific as I said and it will be given anyway in question.\n\n0\nIf 8-bit delimiter pattern was 01111111 , then we would have added pseudo 0 after 0 followed by 6 1's ?\n0\n\nwe see pattern of delimiter till (n-1) bit prefix???why here (n-2) first have considered\n\ni.e we should match 01111110 pattern? how 01111110?\n\nwhat is wrong with it please specify\n\n0\nsame doubt... Plz help\n0\nyes sandy right!!\n0\nYESS!!\n\nhere delimeter is \"01111110\" .....\n\nso rule will be like this ...\n\nat sender :- add 0\u00a0after each occurance of \"011111\" in input data..\n\nat Reciever :- remove 0 for each occurance\u00a0of \"011111\" in output.....\n\nAccording to que Output string is \"01111100101\n\nso input will be \"01111100101\".. (i.e - 0111110101)\n\nby\nB option\nby\n0\nas there is no presence of delimiter pattern thus it will remain unchanged .Am i right?\n\n8-bit delimiter pattern is 01111110.\n\nThe output bit-string after stuffing is 01111100101.\n\nThe above highlighted bit is stuffed bit.\nSo input bit-string must be 0111110101.\n+3\nyou've highlighted the wrong zero\n\nWhen delimiter pattern appears in data, we do stuffing so that data is not interpreted as a delimiter.\n\nWe usually stuff 0 and break the delimiter pattern present in data.\n\nNow we have some data which has a delimiter pattern. (there could be any sequence of 0's and 1's before and after this delimiter pattern)\n\n ... 0 1 1 1 1 1 1 0 ...\n\nWe can break the pattern by stuffing 0 after\n\n\u2022 2 consecutive 1's\n\u2022 3 consecutive 1's\n\u2022 4 consecutive 1's\n\u2022 5 consecutive 1's\n\nWe cleverly choose to stuff a 0\u00a0 only after 5 consecutive 1's because by doing so we reduce the number of 0's to be stuffed and thereby also number of 0's un-stuffed later while reading this frame.\n\n(Ex: If we stuff a 0 after every 2 consecutive 1's, we will have to stuff every time 2 consecutive 1's appear in the data. This would be too many times. And also while reading the frame, every 0 that occurs after 2 consecutive 1's has to be un-stuffed. This is not so clever)\n\n+1 vote\n\n$\\text{Bit-Stuffing}:$ Bits stuffing is the insertion of noninformation bits into data and it is used for synchronization purpose\u00a0example\n\n$A. 0111110100 \\rightarrow 011111\\color{red}{0}0100$\n\n$B. 0111110101 \\rightarrow 011111\\color{red}{0}0101$\n\n$C. 0111111101 \\rightarrow 011111\\color{red}{0}1101$\n\n$C. 0111111111 \\rightarrow 011111\\color{red}{0}1111$\n\nSo Option $B$\n\n\u20131 vote\nI think end delimiter pattern should be 0111110 (five 1's).\nby","date":"2020-07-14 11:27:56","metadata":"{\"extraction_info\": {\"found_math\": true, \"script_math_tex\": 0, \"script_math_asciimath\": 0, \"math_annotations\": 0, \"math_alttext\": 0, \"mathml\": 0, \"mathjax_tag\": 0, \"mathjax_inline_tex\": 1, \"mathjax_display_tex\": 0, \"mathjax_asciimath\": 0, \"img_math\": 0, \"codecogs_latex\": 0, \"wp_latex\": 0, \"mimetex.cgi\": 0, \"\/images\/math\/codecogs\": 0, \"mathtex.cgi\": 0, \"katex\": 0, \"math-container\": 0, \"wp-katex-eq\": 0, \"align\": 0, \"equation\": 0, \"x-ck12\": 0, \"texerror\": 0, \"math_score\": 0.5375684499740601, \"perplexity\": 2206.4618548902667}, \"config\": {\"markdown_headings\": true, \"markdown_code\": false, \"boilerplate_config\": {\"ratio_threshold\": 0.18, \"absolute_threshold\": 20, \"end_threshold\": 15, \"enable\": true}, \"remove_buttons\": true, \"remove_image_figures\": true, \"remove_link_clusters\": true, \"table_config\": {\"min_rows\": 2, \"min_cols\": 3, \"format\": \"plain\"}, \"remove_chinese\": true, \"remove_edit_buttons\": true, \"extract_latex\": true}, \"warc_path\": \"s3:\/\/commoncrawl\/crawl-data\/CC-MAIN-2020-29\/segments\/1593657149819.59\/warc\/CC-MAIN-20200714083206-20200714113206-00448.warc.gz\"}"} | null | null |
\subsection{Growth of \bm{$1T$}\text{-}\bm{$\mathrm{TaS}_2$} single crystals}
Single crystals of $1T$-$\mathrm{TaS}_2${} were grown via chemical vapor transport (CVT) in Iodine (I$_2$). High purity powders of Ta and S were first mixed in a 1:2 stochiometric ratio and then were finely grounded and pelletized. The pellets were sealed in quartz tubes and sintered at 750 $^\circ\mathrm{C}$ for 24 hours. Afterwards, the pellets were grounded and pelletized again, and the same sintering process was repeated to obtain the polycrystalline power samples. The polycrystalline powder was then sealed in a 200 mm long quartz tube with I$_2$ as the transport agent. The tube was placed in a two-zone tube furnace with the hot end at 1000 $^\circ\mathrm{C}$ and the cool end at 900 $^\circ\mathrm{C}$ and kept at this condition for one week before quenched in water. Large thin hexagonal crystals were finally obtained at the cool end of the tube.
\subsection{Electron microscopy}
Crystallographic orientations were identified by selected area electron diffraction (SAED) on JEOL 2010F and convergent beam electron diffraction (CBED) on JEOL 3100R05 transmission electron microscope, operated at 200 keV and 300 keV, respectively. Rotational calibration between real space image and diffraction pattern were performed by aligning shadow-grams---formed by diverging Bragg spots on SAED and defocusing CBED patterns---to the real space image.
\subsection{RA-SHG, tr-SHG, and tr-RA-SHG measurements}
In the RA-SHG measurements with the normal incidence geometry, the reflected SHG intensity is recorded as a function of the azimuthal angle ϕ between the incident electric polarization and the in-plane crystalline axis $\bm a$, with the reflected electric polarization 0$^\circ$ (90$^\circ$) from the incident one for the parallel (crossed) polarization channel. In this experiment, the incident ultrafast light source was of 800 nm wavelength, 50 fs pulse duration and 200 kHz repetition rate, and was focused on to a 30 $\mu$m diameter spot on the sample with a fluence of $\sim$ 0.5 mJ/cm$^2$. The intensity of the reflected SHG was measured with a single photon counting detector. In the tr-SHG measurements, a pump beam is added in the collinear geometry with the probe beam. The ultrafast light source was of 720 nm wavelength, 75 fs pulse duration and 200 kHz repetition rate, and was focused on to a 65 $\mu$m diameter spot on the sample with a tunable fluence from 0.3 to 1.0 mJ/cm$^2$. The probe beam performs the SHG measurement at a selected polarization angle $\varphi=\varphi_S$ as a function of the time delay between the pump and probe pulses. In the tr-RA-SHG measurements, the pump beam remains the same as that in tr-SHG whereas the probe beam performs RA-SHG measurements at every single time delay.
\subsection{Determination of crystallographic axes and broken mirror symmetries in the NCCDW}
The RA-SHG pattern breaking mirror symmetry under the ferro-rotational point group $S_6$ of the NCCDW phase is confirmed by comparing the orientation of RA-SHG pattern with respect to the crystal axes that were determined by SAED and CBED. As shown in FIG. S1, SAED \cite{RN1} together with RA-SHG was taken on a same thin flake of $1T$-$\mathrm{TaS}_2${}. The RA-SHG pattern in FIG. S1(c), taken in the crossed polarization channel, is plotted with respect to the crystal axes $\bm a$ and $\bm a$ that are calibrated through the CBED measurement (FIG. S1(b) right panel). The observation of the RA-SHG pattern rotating away from the crystal axis $\bm a$ by 22.8$^\circ$ (shaded in blue) confirms that the NCCDW phase breaks the three diagonal mirrors at 30$^\circ$o, 90$^\circ$ and 150$^\circ$ from the axis $\bm a$.
\begin{figure}
\includegraphics[width=\columnwidth]{figureS1.jpg}%
\caption{(a) A thin flake of $1T$-$\mathrm{TaS}_2${} placed on a transmission electron microscopy grid. The bright spot is a focused laser beam for performing the RA-SHG measurements in (c). (b) Left: the transmission electron microscopy image of the flake in part. Right: the CBED pattern showing the Bragg disks from the diffracted diverging electrons when the sample is illuminated by a convergent electron beam. Arrows denote the crystal axes. (c) Polar plot of RA-SHG data taken at 290 K with the normal incidence geometry on the (001) plane and in the crossed polarization channel. The solid line is the fit to the simulated RA-SHG functional form $I^{2\omega}_\perp=A\cos^2 3(\varphi-\varphi_0)$. The blue shade represents the rotation of the RA-SHG from the crystal axis $\bm a$.}
\end{figure}
\subsection{RA-SHG measurements on bulk and film $1T$-$\mathrm{TaS}_2${} samples}
RA-SHG measurements are conducted on both bulk and film (less than 100 nm thick) $1T$-$\mathrm{TaS}_2${} samples at room temperature, with fluences of $\sim$ 0.5 and 0.4 mJ/cm$^2$, respectively. Figure S2 shows the RA-SHG patterns taken at the crossed polarization channel, both normalized with respect to incident fluence and integration time. Noting the penetration depth in $1T$-$\mathrm{TaS}_2${} at 800 nm is $\sim$ 28 nm \cite{RN4}, the significant decrease of SHG intensity from bulk to film samples strongly suggests the bulk origin of the SHG signal, and therefore the assigned $S_6$ point group as well as the EQ process generating the SHG.
\begin{figure}
\includegraphics[width=0.8\columnwidth]{figureS2.jpg}%
\caption{Polar plots of RA-SHG data taken on bulk (blue) and thin flake (orange) $1T$-$\mathrm{TaS}_2${} samples at room temperature.}
\end{figure}
\subsection{Simulations of RA-SHG fitting functions}
In the (N)CCDW phase of $1T$-$\mathrm{TaS}_2${}, the system possesses the point group S6 where spatial inversion is preserved. The observed SHG is dominated by the bulk electric quadrupole (EQ) term \cite{RN2}
\begin{equation}\label{EQRadiation}
P^\mathrm{eff}_i (2\omega) = \chi^\mathrm{EQ}_{ijkl}E_j(\omega)\partial_k E_l(\omega)
\end{equation}
where $P^\mathrm{eff}_i (2\omega)$ is the effective radiation source for the SHG, $\chi^\mathrm{EQ}_{ijkl}$ is the second-order EQ optical susceptibility tensor under the $S_6$ point group, $E_{j,l}(\omega)$ is the incident fundamental electric field, and $\partial_k$ turns into the wavevector $q_k$ of the incident beam. The RA-SHG patterns can be fitted to the expression
\begin{equation}\label{SHGInt}
I^{2\omega}\propto |\bm{e}(2\omega)\cdot\bm{P}^\mathrm{eff}(2\omega)|^2
\end{equation}
where $\bm{e}(2\omega)$ is the unit vector denoting the direction of the analyzer collecting the reflected SHG beam. In our crossed polarization channel, the SHG analyzer is maintained perpendicular to the incident fundamental beam $\bm{e}(2\omega)\perp \bm{E}(\omega)$.
In the frame of the crystal, the non-zero independent elements of the $\chi^\mathrm{EQ}_{ijkl}$ tensor are determined by applying symmetries in the $S_6$ point group and SH permutation symmetry between indices $j$ and $l$ to the tensors. This yields only 18 non-vanishing independent tensor elements \cite{RN3}
\begin{gather*}
\chi_{xxxx} = \chi_{yyyy} = \chi_{xxyy} + \chi_{xyyx} + \chi_{xyxy} \\
\chi_{xxyy} = \chi_{yyxx} = \chi_{xyyx} = \chi_{yxxy} \\
\chi_{xyxy} = \chi_{yxyx} \\
\chi_{yyzz} = \chi_{xxzz} = \chi_{yzzy} = \chi_{xzzx} \\
\chi_{zzyy} = \chi_{zzxx} = \chi_{zyyz} = \chi_{zxxz} \\
\chi_{yzyz} = \chi_{xzxz} \\
\chi_{zyzy} = \chi_{zxzx} \\
\chi_{xyzz} = -\chi_{yxzz} = \chi_{xzzy} = -\chi_{yzzx} \\
\chi_{zzxy} = -\chi_{zzyx} = \chi_{zyxz} = -\chi_{zxyz} \\
\chi_{xzyz} = -\chi_{yzxz}
\end{gather*}
\begin{gather*}
\begin{split}
\chi_{xxxy} & = \chi_{xyxx} = -\chi_{yyyx} = -\chi_{yxyy} \\
& = \chi_{yyxy} + \chi_{yxyy} + \chi_{xyyy}
\end{split}\\
\chi_{yyxy} = -\chi_{xxyx} \\
\chi_{xyyy} = -\chi_{yxxx} \\
\begin{split}
& \chi_{yyyz} = -\chi_{yxxz} = -\chi_{xyxz} = -\chi_{xxyz} \\
= & \chi_{yzyy} = -\chi_{xzyx} = -\chi_{xzxy} = -\chi_{yzxx}
\end{split}\\
\begin{split}
& \chi_{xxxz} = -\chi_{xyyz} = -\chi_{yxyz} = -\chi_{yyxz} \\
= & \chi_{xzxx} = -\chi_{xzyy} = -\chi_{yzxy} = -\chi_{yzyx}
\end{split}\\
\chi_{xxzx} = -\chi_{xyzy} = -\chi_{yxzy} = -\chi_{yyzx} \\
\chi_{zxxx} = -\chi_{zxyy} = -\chi_{zyxy} = -\chi_{zyyx} \\
\chi_{yyzy} = -\chi_{yxzx} = -\chi_{xyzx} = -\chi_{xxzy} \\
\chi_{zyyy} = -\chi_{zyxx} = -\chi_{zxyx} = -\chi_{zxxy} \\
\chi_{zzzz}
\end{gather*}
With normal incidence on the (001) surface of the crystal, only two tensor elements survive in Eq. (\ref{SHGInt})
\begin{equation}\label{SP}
\begin{split}
I^{2\omega}_\perp(\varphi) & =\big|\chi_{xxzx}\sin3\varphi + \chi_{yyzy}\cos3\varphi\big|^2 \\
& = \big|\chi_{xxzx}^2 + \chi_{yyzy}^2\big|\cos3\Big[\varphi-\frac{1}{3}\mathrm{atan}\frac{\chi_{xxzx}}{\chi_{yyzy}}\Big]
\end{split}
\end{equation}
All the RA-SHG patterns in the main text have been fitted to this expression to obtain $A=|\chi_{xxzx}^2+\chi_{yyzy}^2|$ and $\varphi_0=\frac{1}{3}\mathrm{atan}\frac{\chi_{xxzx}}{\chi_{yyzy}}$. From Eq. (\ref{SP}) it is noted that any changes in $\chi_{xxzx}$ and/or $\chi_{yyzy}$ will lead to the modification of RA-SHG amplitude $A$ and orientation $\varphi_0$.
\subsection{Time-resolved linear reflectivity measurements on the CCDW in \bm{$1T$}\text{-}\bm{$\mathrm{TaS}_2$}}
A reference time-resolved linear reflectivity measurement was conducted with 720 nm pump and 800 nm probe pulses. FIG. S3(a) and S3(b) show the trace and its corresponding FFT spectrum. The pump fluence for this measurement is set the same at 0.46 mJ/cm$^2$ as that in FIG. 3 in the main text, in which the FFT spectra from fitting RA-SHG amplitude and orientation traces are also reproduced here in FIG. S3(c) and S3(d) for comparison. In FIG. S3(b) we note an overall broad asymmetric mode peaked at around 2.38 THz where the triplet structure is not seen.
\begin{figure*}[h!]
\includegraphics[width=0.8\textwidth]{figureS3.png}%
\caption{(a) The time-resolved linear reflectivity trace taken at 90 K in the CCDW phase of $1T$-$\mathrm{TaS}_2${}. (b) The FFT spectrum of the trace in (a). (c-d) The FFT spectra of the RA-SHG amplitude and orientation traces. Data in (c) and (d) are reproduced from FIG. 3.}
\end{figure*}
| {
"redpajama_set_name": "RedPajamaArXiv"
} | 8,891 |
Steem Hardware Wallet
If you're thinking of purchasing a Ledger Stax to meet your requirements, there are a few important things to know prior to you make your purchase. We'll explain everything you need to know, including how to obtain an offer code, and what features are available on this gadget.
The Ledger Stax is a wallet made of hardware that is designed to assist users in managing the digital accounts they have. It's a credit-card-shaped device that is slightly different from older versions of Ledger Hardware wallets. The new model comes with a curved E Ink screen, a magnet system, Bluetooth connectivity, and the option of a lock screen.
The Ledger Stax supports more than 5,000 cryptocurrencies, including Litecoin, Dogecoin, Ripple, and Cardano. The device also has NFT (Non-fungible token) support. These are specially designed currency that is able to be kept and traded using Ledger Stax.
The Ledger Stax's security features are backed by a Secure Element chip. This is the same chip found in passports and credit cards. The Secure Element chip makes sure the security of your cryptocurrency assets. protected from thieves.
The Ledger Stax comes with Bluetooth for connecting directly to Ledger Live Mobile app. Users can also make use of Ledger Stax's wireless Qi charging function.
Additionally to that, the Ledger Stax comes with magnetic systems that allows it to be stacked alongside other similar devices. It also has offline mode, which means it is powered on even when it is turned off.
Ledger Stax is a brand-new hardware wallet that was designed to assist you in managing your digital assets. The device provides an unprecedented amount of customization as well as a curved touchscreen and safe elements. It is compatible with Mac, Windows, and Ubuntu, and supports more than 500 cryptocurrencies.
Ledger Stax comes with Bluetooth 5.2 as well as a USB-C port as well as a 200mAh battery. It also has embedded magnets that let you stack multiple Stax devices together.
As a cryptocurrency wallet, the Ledger Stax enables you to securely keep and manage your cryptographic keys. It can also be used as a cold storage device, so it's less susceptible to hacking attempts.
The device comes with a custom OS, allowing you to customize it to suit your preferences. You can also choose the image or name you want for the screen that locks. In addition, Ledger Stax is equipped with a reliable Bluetooth antenna.
Ledger Stax Ledger Stax is built to be compliant with the most stringent security standards in the industry. It features the Secure Element chip, which is certified to the CC EAL5+ requirements. It also has the capability of clear signing, which allows users to confirm the authenticity of their transactions and decrease the chance of Phishing attacks.
What is the process?
Ledger Stax is the latest hardware crypto wallet developed by French start-up Ledger. It features a sleek and modern design, complete isolation of keys, and the ability to store up to 5500 tokens.
The device comes with a touch screen and an E-Ink display that can support up to 500 cryptocurrencies. The battery is also included which can last from weeks to months. You can connect it to a smartphone or laptop via Bluetooth.
To improve the user experience, Ledger has designed a customized lock screen that can be customized. In addition the device is equipped with magnets that can bind multiple devices together. This makes it convenient to carry around.
In order to guarantee the security of its customers, Ledger uses a secure design and a chip that is secure, known as The Secure Element. This offers institutional-grade security for digital assets.
Ledger's most important feature is its transparent signature. It allows users to verify the authenticity of transactions without compromising their privacy. This can help you avoid frauds and attacks of phishing.
Ledger also offers a unique feature that allows you to see NFTs directly on the device. For instance, you could view a QR code, an image, or even a label image.
The Ledger Stax can be described as the most recent hardware wallet made by the company that developed the famous Ledger Nano S and Nano X. They are excellent for long-term investment as well as being easy to carry around.
They are incredibly secure and protect your private key. They also have unique security features, for example, an offline mode. If you are not online however, you are able to access your cryptos through the built-in magnet.
The most thrilling characteristics that comes with Ledger Stax is its touchscreen. Ledger Stax is its touchscreen. The curved display of e-ink provides more space for reading transactions and view NFT collections. It is also fully compatible with iOS, Windows, and macOS.
The device is powered by Bluetooth 5.2, which lets it be connected directly to Ledger Live through a phone. You can also create pin codes for quicker access. There's a specific "Magnet Shell" protection case that is available.
For those living outside of within the United States, Ledger products usually ship for free. However, you may have to pay taxes for the item based on your location.
Ledger Stax is a wallet that Ledger Stax is a hardware wallet that combines the physical experience of a gadget with the ease of using an E-ink display. It is designed to make it easy to manage and store your cryptocurrencies.
Ledger Stax is a powerful wallet. Ledger Stax is an extremely powerful wallet that can handle 500+ NFTs, tokens, and cryptocurrencies. You can also choose to customize the screen with photos or your preferred NFT. A curved touchscreen display can help you keep the track of your cryptocurrency portfolio.
The Ledger Stax wallet is also linked directly to Ledger Live platform, which offers you an easy way to check transaction details as well as to purchase as well as sell currency. A hardware wallet offers security on the chain, since your private keys will not be accessible to unauthorised users.
In addition to storing and managing your cryptocurrencies, the Ledger Stax can be used for managing NFT collections on multiple blockchains. This includes Ethereum and Polygon-based NFTs.
The Ledger Stax is equipped with Bluetooth that allows you to connect it to your smartphone or desktop. It can also be used as a screen cover. To help ensure security, it comes with an inbuilt battery. When the device is not in use the battery will last for three months.
The Ledger Stax is a breakthrough consumer device. It comes with a 4 inch touchscreen, magnets, as well as a secure architecture. In comparison to its predecessors it's Stax offers more protection and gives an improved user experience. Actually, it's better than other hardware wallets!
The Ledger Stax holds more than 5 hundred cryptocurrency, including Ethereum or Polygon-based NFTs. It also boasts an impressive 4000mAh battery and a touch-screen interface as well as a sleek magnetic document kit that allows for the highest capacity to stack. However, if you're looking to get your hands on one of the latest wallets, you'll have wait for a couple of months.
It's not a reason to be concerned, though. Ledger Stax is a great product. Ledger Stax has a long way to go. It's several years away from becoming available for sale. However, in the interim you'll have to get your hands on the Ledger wallet application and which is the Ledger Nano S, or the Ledger S. You can pre-order the latter two from the website of the manufacturer.
If using the Ledger Wallet application is the most reliable option however, there's a better method to set up your new crypto-holding machine. This involves downloading the Ledger Wallet application, installing it on your computer, and allowing it to access your cryptocurrency holdings.
The Ledger Stax is a premium hardware wallet developed in collaboration with the former Apple iPod creator Tony Fadell and design firm LAYER. The wallet was designed to be simple to operate, the device features a new E-Ink display which provides a clear and easy-to-read interface. It also features a curved screen and embedded magnets which make it more durable and practical.
Ledger Stax Ledger Stax supports a wide variety of cryptocurrency, including Litecoin, Dash, Dogecoin, Monero, Ethereum, Binance Coin, BTC, and XRP. It's specifically designed to be compatible with mobile devices using Bluetooth and a USB-C connection.
Apart from being able to store and transfer various digital assets, Stax can also store and send a variety of digital assets. Stax is also built to protect non-fungible tokens (NFTs). Because it stores private keys offline, it's more vulnerable to attacks by hackers than other wallets for crypto.
Users can personalize their lock screen. Users can select the preferred image, select an unique name for the device, and check its status when it locks and unlocks. In addition, the Stax is compatible with Qi wireless charging which permits users to charge it from the Qi charger of their choice.
Battery life is impressively good. Ledger Stax's batteries can last for months with a single charge, depending on the amount of time you spend using the device.
Ledger Stax is a new version of the hardware wallet made by Ledger, the company that created its first electronic wallet. It combines the security of Ledger Live and the ease of use of a touchscreen. The latest model is equipped with several options that make it simple to transfer and receive crypto.
The Ledger Stax's display is made up of an E-ink-sized credit card display. It offers a clear comfortable and natural read experience, avoiding the glare of other screens and delivering a higher contrast. Additionally, the screen has a curved touchscreen for users to view and sign their transactions with the press of a single button.
The Ledger Stax also features a rechargeable lithium-ion battery built in. It allows the device to function for months on a single charge. Additionally, it can be charged wirelessly, or using an USB-C cable.
In terms of security, the Ledger Stax comes with the secure element chip to provide an extra layer of security. Furthermore, the device is certified to the CC EAL5+ standard. Lastly, the device comes with Bluetooth as well as a safe USB-C port to connect with the Ledger Live mobile application.
Ledger Stax vs Nano X Vs Nano S Plus, which should you pick? These devices are the best hardware cryptocurrency wallets. They're easy to use and have numerous options. The Nano X, Nano S Plus and Stax are excellent choices for both advanced and beginner users alike.
If you're looking for a safe, reliable, and easy-to-use device then it's the Ledger Stax can be the right way to go. It was designed in the hands of iPod designer Tony Fadell, and has all the features of that of the Nano X, but at a more affordable price. This model comes with a touch screen and rechargeable battery which helps it last longer.
Ledger has been an early pioneer in the field of crypto. They have developed a range of hardware wallets and each one comes with a set of specific features to protect private keys.
Through NanoX Nano X, you can transfer cryptocurrencies directly to your phone. This is the ideal solution for people who are on the move. Additionally, it comes with Bluetooth support. It is also possible to connect your phone to Nano X via USB C. But, you'll have to turn on the device by pressing a left button for 3 minutes.
If you're in the market to purchase a crypto wallet you'll be glad to learn that Ledger has recently unveiled its new product, the Ledger Stax. The new hardware wallet blends an e-ink-curved screen and magnets to create an extremely mobile, reliable device.
The Ledger Stax is an equivalent to a credit card size crypto wallet that lets you securely manage and store your digital assets. It comes with a unique minimalist design, is equipped with several other options. Some of the features include an e-ink display and NFC capabilities.
The Ledger Stax is built on an aluminum chassis that has a rounded edge and a soft concave surface. The Stax also comes with an integrated magnet which helps in enhancing its stackability. Additionally, there's an Bluetooth(r) antenna and an efficient Bluetooth(r) battery.
It works with iOS 12 and up and Windows 10 and up. It can connect to your smartphone via Bluetooth and a safe USB-C cable.
It is possible to set up your Ledger Stax by pressing a single button. It comes with a touch-sensitive E Ink display that creates a clear visual interface. It allows you to alter the lockscreen and view your photos.
Ledger Stax can be described as a brand new hardware crypto wallet which features industry-leading security. This wallet is designed to provide users with an easy-to-use experience that allows them to make transactions fast and secure.
The device is constructed with a world-class Secure Element chip. The chip is utilized in passports and credit cards, as well as in payment systems, and many other critical data.
Another major feature of Ledger Stax is its E Ink touchscreen. The curved touchscreen is the first of its kind in the market. It has Bluetooth connectivity and is linked to Ledger Live mobile app. When it isn't in use it is concealed by using magnets.
One of the most important features of Ledger Stax lies in the capability to alter your lockscreen. Users can include their preferred photos and NFT. The users can also sign up for through the Infinity Pass to get their free NFT and utility.
Ledger Stax is designed to appeal to those who are passionate regarding their online assets. They want to save them from theft and invest in their assets from loss. While this hardware wallet has an expensive price compared to some of its rivals but it's an excellent choice for serious cryptocurrency investors.
Ledger Stax is latest hardware wallet made by French company Ledger. It is a touchscreen device that features an E Ink feature. It has other interesting features as well.
The Ledger Stax features a curved 3.7 inch touchscreen. It is therefore easier to use. Users can also personalize the screen. Magnets allow users to connect the device to a phone.
The Ledger Stax is also capable of generating QR codes. It can be used to make cashless payments. It can also be used on mobile devices that are cord-free.
Ledger Stax Ledger Stax is compatible to iOS 13+ and Android 9+. It comes with certified chip that is CC EAL5+ security system. However, the cost is expensive. At $279US, it's nearly twice as costly as Nano X. Nano X.
One of the main selling points can be the screen. The display is constructed of E Ink which is curved. A further feature that is important is the capability to store NFTs, or non-fungible tokens.
It is compatible with 5,000 cryptos, including Ethereum, Polygon, and Nano X. It is also compatible with a suite of DeFi protocols.
The Ledger Stax is a new hardware wallet that is designed to help users keep track of their assets digitally. It is designed to look similar to a credit card, the Stax provides users the ease of carrying a crypto wallet, and still maintaining security. It comes with a magnetic case, an LCD touchscreen, and a powerful 200 mAh battery it is Stax is compact enough to fit comfortably in the palm of your hand.
The device comes with a 3.7 inch E Ink curved touchscreen. This display makes it easy to see transactions and read messages. Also, it supports Bluetooth 5.2, a feature that will help you to connect to your mobile.
Stax is a lock screen that Stax is compatible for iOS 10+, Android 9+ and Ubuntu. Users can also personalize the lock screen. They can monitor the state of lock and unlock on their device and keep track of the percentage of the battery. The integrated Ledger Live app allows users to manage their portfolio of over 500 tokens and cryptocurrencies.
The Ledger Stax comes with an e-case which can be used to stack more than one device on top of it. As with it's predecessor, the Nano S Plus, the Stax is made for use at home.
If you have been using Ledger Nano for some time it could be time for you to consider the next-generation crypto device that is the Ledger Stax. This new model is part of an extensive collection of hardware wallets that have become widely used in the Crypto space.
Ledger Nano Stax, like its predecessor, is a device designed for the mobile lifestyle. It's a compact device that can handle a range of cryptocurrency. The device is able to handle up to 5,500 different tokens, including Polygon NFTs as well as Ethereum NFTs.
One of the major selling points for the Ledger Stax is its touchscreen. The screen of the E Ink2 has an angled design which makes it simpler to read.
Another significant aspect of the Ledger Stax is the rechargeable battery. One charge will keep the device operating for months or even weeks. With the Ledger Stax, you'll have the ability to save your digital assets in safe manner.
The Ledger Stax is compatible to work with iOS 13+, and macOS 12and up. It can be connected to your smartphone via Bluetooth or by using the USB-C cable.
Ledger Stax is the latest on the crypto physical wallet marketplace. The sleek, curved, touchscreen wallet is able to securely manage the digital accounts they have. It supports more than 5,000 tokens and coins and is optimized to improve users' experience.
The wallet can be charged using it's Qi wireless charging technology. The wallet also comes with a magnetic locking system. There's a USB Type C port. A 3.7-inch e-ink display lets you see and sign transactions.
The wallet is equipped with the Ledger Live application. Users can also use applications from third parties to increase their selection. Compared with that of the Nano S Plus, the Ledger Stax has a larger screen and more features.
The Stax is a round edge chassis as well as a powerful Bluetooth antenna. It also has an Secure Element chip, the same chip that is used for the Nano X. These features ensure that the device is secure from hackers.
With its large screen and magnet system, Stax offers unrivaled interaction. In contrast to other wallets, it lets you personalize the touch screen by using an NFT image. It also allows you to generate QR codes.
The Ledger Stax is an anticipated Hardware wallet, which is expected to deliver lots. It's designed to increase accessibility, security, and interactivity while maintaining an excellent user experience. If you're considering purchasing one of these, here are a few points to consider.
First of all, it has a curved ink touchscreen. It will be the only one of this kind. It can be used to sign transactions or view NFT collections. Furthermore, it features Bluetooth.
There's also an embedded magnet system which lets you stack several Stax devices. Another cool thing about it is that it can be charged using Qi wireless charging, the same technology that is used in Samsung and Apple.
Additionally, it comes with an internal battery that can last for months or weeks. This is an incredible benefit in the case of an electronic wallet for crypto.
Overall, it's a great device. There are a few downsides to it. For starters, it's quite expensive. If you're prepared to spend money on a hardware wallet, it's worth it.
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Cash App Crypto Wallet | {
"redpajama_set_name": "RedPajamaCommonCrawl"
} | 7,068 |
BREAKING: 1 million gallons of manure dumped in wetland
Spencer-area farmer cited for environmental violations
BREAKING: 1 million gallons of manure dumped in wetland Spencer-area farmer cited for environmental violations Check out this story on fdlreporter.com: http://wdhne.ws/1otjOu1
Bob Dohr, Daily Herald Media Published 9:25 p.m. CT Aug. 12, 2014 | Updated 9:52 a.m. CT Aug. 13, 2014
An unidentified water resource specialist takes water samples of manure runoff at a town of Brighton farm May 6 in this photo provided by the Marathon County Conservation, Planning and Zoning Department. (Photo: Contributed photo )
SPENCER – An estimated 1 million gallons of manure flowed unchecked from a storage tank on a western Marathon County farm for months, running into a nearby wetland and eventually into the Little Eau Pleine River.
The discharge at the 120-cow Willcome dairy farm on Century Road near Spencer was going on for about a year, according to Paul Daigle, land and water program director for the Marathon County Conservation, Planning and Zoning Department.
"Based on air photo evidence, it looked like they stopped pumping sometime in early 2013 and it went on until this May," Daigle said.
The owner of the farm, Patrick Willcome, pleaded no contest Aug. 1 to violating water-pollution regulations and was found guilty and fined $464.10. He did not return calls seeking comment Tuesday.
Daigle said state officials learned of the violation May 2 when an anonymous letter was sent to the Department of Natural Resources indicating the manure storage facility on the town of Brighton farm — a tank designed for about a week's worth of storage — had "not been pumped for months and is running over."
State and county officials visited the farm May 6 to verify the problem and initiate an emergency clean-up, Daigle said.
Workers try to clean up some of the manure at a town of Brighton farm in May after an estimated 1 million gallons of manure flowed out of a storage tank and into a wetland over the course of a year. (Photo: Photo courtesy the Marathon County Conservation, Planning and Zoning Department )
"(The owner) paid for (the clean-up)," Daigle said. "We provided some technical assistance to get some emergency berms up and immediately stop liquid from continuing to run off into the waterways and haul it out in the fields instead, where it should be, so it can be a valuable crop nutrient."
Willcome could not be reached for comment, but Terry Kafka, DNR water resource management specialist,cited "mismanagement" as the reason for the pollution.
"They had two manure spreaders, but both broke," Kafka said. "I do give them credit; they were very receptive and timely to all the requests that we made."
The wet spring didn't help, according to Kafka.
"They were waiting for weather to be cooperative to get out on the field and then apply the manure," Kafka said. "That was also the time of year there were some fields they simply couldn't get to because they were so wet and they'd just get stuck."
Kafka said the manure flowed into an unnamed ditch and wetland, considered state waters, located several hundred yards from the facility. The runoff then entered the Little Eau Pleine River about a mile from the farm.
Kafka said there was no documented fish kill in the river because the manure was diluted by snowmelt.
Daigle said it's difficult to estimate how much pollution reached the ditch or the river.
"What happens in a situation like that is the solids separate out and the liquids keep going," Daigle said. "Some very potent liquids ... made it down into the wetland. I don't want to speculate what made it to the waters of the state."
The waters of the Little Eau Pleine eventually empty into the Wisconsin River.
Kafka said the state's estimate of the amount of manure discharged — using a September-to-May timeframe — is around 600,000 gallons. That's less than the million-gallon figure derived by Daigle, which used a year duration to reach the estimate.
In addition to the citation, Willcome was subject to an enforcement conference that spelled out provisions to ensure the problem wouldn't happen again.
Bob Dohr can be reached at 715-845-0660. Find him on Twitter as @BobDohr1.
Read or Share this story: http://wdhne.ws/1otjOu1 | {
"redpajama_set_name": "RedPajamaCommonCrawl"
} | 5,272 |
Q: Javascript idiom for code structure My javascript gets included multiple times with the script tag, like so:
<script src="code.js></script>
<script src="code.js></script>
<script src="code.js></script>
Right now I have this code inside code.js to make my code run only once without overwriting my namespaces:
if(typeof _ow == "undefined" ){
_ow = {};
// code in here will only run once
_ow.Auth = (function(){
})();
_ow.Template = (function(){
})();
}
Is there any better structure I could use to make my code only run once?
A: var _ow = _ow || { Auth: ... };
if its already defined then it wont be defined again.
A: Are you familiar with Crockford's Javascript Module Pattern?
A slight variation on how to prevent overwriting the namespace:
var _ow;
if(!_ow) _ow = {};
A: While what you are doing will technically work, it is inefficient, because even though your code only gets run once, it does end up being parsed a number of times on some browsers (this is different than downloading a file via the network, and cannot be cached).
It is best to ensure that the script only gets included once. For all the repeated functionality you can expose a function to be called whenever needed.
| {
"redpajama_set_name": "RedPajamaStackExchange"
} | 5,060 |
VIP is a ground-breaking way for young people to learn and create new music. It provides everything they need to create, produce and release their own music including 24/7 access to a complete online recording studio.
We created it to help teachers teach music technology to KS3, GCSE and BTEC levels and to use music technology generally to teach the National Curriculum. It's proven to encourage more young people to get into creating music.
Watch the video or take out a free no-obligation trial to see it for yourself.
We don't ask for any payment details.
At the end of the trial you can choose to order, or simply allow the trial to expire.
Please confirm that you are a teacher in order to proceed. | {
"redpajama_set_name": "RedPajamaC4"
} | 4,604 |
O2 to release flagship HTC mobile phone without charger in eco stance
Back in 2009 the GSM Association were looking to get all the manufacturers to use one type of phone charger by 2012. O2 has been even bolder and announced that the next HTC handset to be released on the UK network will arrive without a charger.
If you're expecting me to start talking about wireless charging then you're on the wrong track. This move is to cut down on all the 'spare' chargers laying around the place.
The network has teamed up with HTC to release the as yet unspecified phone sans charger in the UK soon as per the EU's plans a few years ago to produce standardised chargers with the goal of selling phones without a charger at all.
Inside the box for the mysterious HTC handset owners will still get a USB to Micro-USB cable as is the European standard and HTC will be selling a separate HTC Micro USB charger plug separately but the move will encourage people to use older chargers (any Micro USB charger will connect) or use the USB charger off a laptop or PC.
O2 claims that 70% of customers who buy one of the 30 million new handsets sold in the UK each year already have the relevant charger which results in over 100 million chargers remaining unused across the UK!
By removing the accessory from the pack, O2 and HTC will be able to cut down on packaging and transport costs for the phones as well as boasting their green credentials 😉
The new system will be treated as a trial scheme by O2 and HTC and the companies have stated the handset will be the next "Flagship" smartphone from HTC – a 5-inch version of the One X has been mentioned.
In recent years manufacturers have taken drastic steps to minimise the size and packaging that is included with mobile phones form removing physical manuals, Data Discs and in some cases even providing a specialised USB plug adapter that can be made smaller.
Ronan Dunne, CEO of O2 stated: "I have a simple vision for O2: we want to take chargers out of boxes full stop,"
"We hope that we will be able to pave the way for others to follow us as this has to be a collective effort if we are to achieve the bigger aim of eliminating chargers sold with every new phone in the UK."
Buyers who don't have a suitable charger already will be able to buy one separately from O2 at cost price so the cynics out there may see a monetary benefit for O2 and HTC amongst all the eco-warrior stuff. I just wonder if we will see this reflected in the device pricing. | {
"redpajama_set_name": "RedPajamaCommonCrawl"
} | 1,758 |
'One punch can kill' – Police and medics on the lives devastated by a single blow
Teaming up for their Christmas campaign, police and the NHS warn how a drunken moment of madness can have lifelong consequences for the victim and the attacker.
Christina O'NeillDigital Journalist
"One punch can be devastating. It can disable someone, it can leave them traumatised – and it can potentially kill them."
These are the words of Dr Michael Murray, Director of Medics Against Violence and Consultant Neuro-anaesthetist at Queen Elizabeth University Hospital – speaking as part of Police Scotland's One Punch campaign at the Garage nightclub.
He sees patients coming in with serious head injuries several times a day – amounting to around 500 a year across Scotland.
Last year, six people died as a result of a single punch. Four of the perpetrators were under 18 and on four of those occasions, the culprit was under the influence of alcohol.
Urgent appeal after woman, 48, vanishes in Glasgow city centre
Glaswegians still have lowest life expectancy in Scotland
Police and medics warn that people who have drunk too much on nights out are more likely to become violent - and that a sudden fit of rage could have long-term consequences for the lives for themselves, the victims and their loved ones.
Dr Murray tells Glasgow Live: "The skull is fragile. The brain is a mushy organ and it doesn't take much to bruise it.
"Dropping your brain onto a concrete floor from six feet has enormous consequences. If you imagine you drop a laptop on the floor, it wouldn't like it very much from six feet, but it can be repaired or replaced.
"If you take your brain, which is a very complicated computer, and drop it onto a concrete floor, it's doing roughly around 15 miles per hour by the time it hits the ground – even by falling over. Many people can die within half an hour, or a couple of hours. If you do survive a head injury, you're twice as likely to be disabled."
Assistant Chief Constable Bernard Higgins and Doctor Michael Murray with Niven Rennie from the Violence reduction unit
Dr Murray warns that people who suffer a brain injury as a result of violence are much more likely to develop addictions and get depressed – and that depression is more likely to be resistant to treatment. Tragically, many of these people take their own lives. Even a single concussion can double the chances of suicide in six months.
He added: "Just remember a punch isn't just for Christmas, it's for life – and it could be your life as well. Offer a hand to wish someone 'Merry Christmas', and not any more."
John Black had a promising career with Dundee (Image: SNS)
One heartbreaking case was that of promising Dundee FC youth footballer John Black. The 21-year-old spent 10 days in a coma and was forced to undergo life-saving surgery after he was found bleeding on the ground following an alleged assault last October.
Assistant Chief Constable Bernard Higgins said the Blairgowrie man had been a victim of a single punch.
He said: "He has made a recovery, thankfully, but he can never play football again. If he heads a ball, that could lead to a fatal injury.
"This was a young man who had his whole life ahead of him – he wanted to be a professional footballer, he attained that goal and then it was cruelly snatched away from him because of a moment of madness. He has lost his whole career. That's the impact that can have."
Higgins warns that using your fists can be as fatal as a knife or a gun – and that perpetrators will face a lengthy jail sentence for their actions.
He added: "Alcohol will impair your judgement and if you've had too much to drink, you're going to make bad decisions. We get less inhibited and can become more aggressive.
"What we're saying come out, enjoy yourself, be moderate with your drinking. If you find yourself in a situation where tempers are getting quite agitated just pause, walk away and don't do it. In a split second, you can ruin your life, the person you strike and that of your families."
Key tips to having a safe night out
• Avoid drinking in rounds - it's easy to have too much when trying to keep up with others
• Recognise the signs that it's time to slow down – and drink a soft drink between alcoholic drinks
• Walk away from arguments and confrontation
• If confrontation arises, be responsible and seek out security staff or a Police Officer and make them aware of the issue
• Enjoy your night out but behave responsibly, stick with your friends and look after each other. | {
"redpajama_set_name": "RedPajamaCommonCrawl"
} | 5,221 |
Q: Cannot access cloud storage from www-admin user, but can from normal user with same service account I am trying to write something very basic that copies a file to a storage bucket from a simple CGI script running on a compute engine VM, but I'm failing miserably and my google-fu is weak on this one.
I have the default service account for compute engine set with Storage Admin permissions, and when, as a login user, I gcloud config set account to the service account I can access the buckets. However, if I su to www-data I cannot use gsutil and the service account, or even if I gcloud auth login with my regular admin account.
E.g.:
website:loginuser> gcloud config set account 655644879042-compute@developer.gserviceaccount.com
Updated property [core/account].
website:loginuser> gsutil ls gs://webaccess-bucket-test
gs://webaccess-bucket-test/index.html
website:loginuser> sudo su -l www-data
www-data@website:~$
www-data@website:~$ gcloud config set account 655644879042-compute@developer.gserviceaccount.com
Updated property [core/account].
gsutil ls gs://webaccess-bucket-test
Your "GCE" credentials are invalid. Please run
$ gcloud auth login
OSError: Permission denied.
As mentioned above I've tried using the login user via gcloud auth login and that has the same issue. I can't figure out what the difference is with the www-admin user that means it can't access gsutil. As far as I can tell I set everything the same.
I can't shake the feeling I'm missing something simple and likely obvious.
| {
"redpajama_set_name": "RedPajamaStackExchange"
} | 4,374 |
Happy jazz album vol. II peti je studijski album hrvatske jazz i rock glazbenice Zdenke Kovačiček, kojeg 1994. godine objavljuje diskografska kuća Croatia Records.
Zdenka Kovačiček i trio Vanje Lisaka materijal za album snimaju 2. i 3. siječnja 1993. godine u B.P. klubu i u studiu Radio Zagreba. Kao gosti na snimanje su pozvani poznati zagrebački jazz glazbenici i tamburaški orkestar HRT-a. Album pored standardnih jazz skladbi sadrži i "Dok razmišljam o nama" od Josipe Lisac.
Popis pjesama
"S' wonderful"
"Dok razmišljam o nama"
"Sunny side of the street"
"Summertime"
"Chattanooga choo choo"
"The lady is a tramp"
"Georgia on my mind"
"New York, New York"
"Quell' uomo dei miei sogni"
"Nuages"
"Is you is or is you ain't my baby"
Izvori
Albumi Zdenke Kovačiček
Albumi iz 1994. | {
"redpajama_set_name": "RedPajamaWikipedia"
} | 4,707 |
10/7/2021 October 7, 2021 - 7/4/2022 July 4, 2022
October 7, 2021 @ 8:00 am - October 8, 2021 @ 4:00 pm
Introduction to FFmpeg
Introduction to FFmpeg is a 2-day hands-on workshop in which students will discover the capabilities of this software project that produces libraries and programs for handling multimedia data. Processes will...
October 12, 2021 @ 8:00 am - October 14, 2021 @ 4:00 pm
Digital Video Evidence Recovery
Scottsdale Police Forensic Services 7601 East McKellips Rd Bldg B, Scottsdale, AZ
Summary The Digital Video Evidence Recovery workshop will assist the investigator in recovering video evidence from Digital Video Security Systems. Best practices and guidelines will be discussed as well as...
Introduction to FFmpeg Webinar
701 Kenmore Ave 701 Kenmore Ave, Suite 103, Fredericksburg, VA
Summary Introduction to FFmpeg Webinar is a 1-day workshop in which students will learn the powerful capabilities that exist using the free command-line tool FFmpeg. You will learn how to...
February 8, 2022 @ 8:00 am - February 10, 2022 @ 4:00 pm
Plantation Police Department 451 NW 70th Terrace, Plantation, FL
March 8, 2022 @ 8:00 am - March 10, 2022 @ 4:00 pm
Tacoma Police Department Headquarters 3701 South Pine Street, Tacoma, WA
March 28, 2022 @ 8:00 am - March 30, 2022 @ 4:00 pm
Introduction to Forensic Audio Analysis
St Paul Police Department Richard Rowan Training Facility 600 Lafayette Road North, St Paul, MN
Summary Introduction to Forensic Audio Analysis will introduce the fundamentals of audio, and the processing techniques for forensic audio clarification. This three-day hands-on workshop will cover sound fundamentals, forensic audio...
March 31, 2022 @ 8:00 am - April 1, 2022 @ 4:00 pm
Forensic Audio Enhancement with iZotope
Summary Forensic Audio Enhancement with iZotope is a two-day hands-on workshop that will cover the features, tools, processes, and advanced enhancement techniques of iZotope RX. Students are expected to bring...
April 5, 2022 @ 8:00 am - April 7, 2022 @ 4:00 pm
Tucson Police Department Crime Laboratory 1310 W Miracle Mile, Tucson, AZ
April 25, 2022 @ 8:00 am - April 28, 2022 @ 4:00 pm
Adobe Photoshop for Forensic Video Analysts
Summary The Adobe Photoshop for Forensic Video Analysts workshop is designed to introduce and build upon image processing techniques for video evidence. This thorough workshop is recommended for analysts with...
June 7, 2022 @ 8:00 am - June 9, 2022 @ 5:00 pm
Portland Police Training Center 14912 NE Airport Way, Portland, OR
Adobe Premiere for Forensic Video Analysts
Bensenville Police Department 345 E Green Street, Bensenville, IL
Summary Adobe Premiere for Forensic Video Analysts is a 3-day workshop that assists the investigator in understanding the overall interface of Adobe Premiere as well as the advanced toolset and filters... | {
"redpajama_set_name": "RedPajamaCommonCrawl"
} | 2,133 |
\section{Introduction}
\label{sec:intro}
To prevent climate change, a movement from a high carbon electricity supply to a low-carbon system is required \cite{Kell2020}. Low carbon electricity supply will aid in the decarbonization of the automotive and heating sectors by allowing for low-carbon electricity to be used in place of oil and gas.
Renewable energy costs, such as solar and wind energy, have reduced over the last ten years, making them cost-competitive with fossil fuels. These price drops are projected to continue \cite{IEA2015}. The future cost of operating and capital cost of electricity generation and electricity demand, however, remain uncertain over the long-term future. These uncertainties are risks which investors must analyze while making long-term decisions.
In this paper, we use the deep deterministic policy gradient (DDPG) reinforcement learning algorithm to simulate the behaviour of investors over a 33-year horizon, between 2017 and 2050 using the FTT:Power model \cite{Hunt2016a}. FTT:Power is a global power systems model that uses logistic differential equations to simulate technology switching \cite{Mercure2012}. The model is parameterized and runs from 2007; however, the investment decisions began in 2017. We start in this year due to the prior parameterization of the FTT:Power model with historical data up until this time. We projected until 2050 because this is a common target for governments to reach zero carbon. The environment used was a modified version of the FTT:Power model.
We modified the FTT:Power model to use the DDPG algorithm in place of the logistic differential equations to make investment decisions. In addition, we simulated two countries: the United Kingdom and Ireland. We choose these due to the wealth of prior work on these countries which can be used use for comparison \cite{Hall2016, Hughes2010}. The DDPG algorithm allows us to simulate the decisions made by investors under imperfect information, such as future electricity costs, taxes and demand. This work enabled us to see an optimal, final state electricity generation mix.
Prior work in this domain has tackled the capacity expansion problem. For example, Oliveira \textit{et al.} also use reinforcement learning for the capacity expansion problem~\cite{Oliveira2018}. Whilst Oliveria \textit{et al.} provide detailed calculations of agents for the capacity expansion problem, we reduce this complexity to a series of observations of the environment, to allow for emergent behaviour. Kazempour \textit{et al.} use a mixed-integer linear programming approach to solve the generation investment problem \cite{Kazempour2011}. In contrast our approach removes the requirement for full knowledge of the time-horizon.
Through this work, it is possible to assess whether a low cost, low-carbon electricity mix is viable over the long-term using a deep reinforcement learning investment algorithm, as well as finding what this optimum mix should be. This work enables us to closely match the investment behaviour of rational agents, without knowledge of the future. It can help guide investors on the choice and proportion of technologies to invest in over the long term.
\section{Model and methodology}
\label{sec:methods}
The Future Technology Transformations system for the power sector model (FTT:Power) model represents global power systems based on market competition, induced technological change and natural resource use and depletion \cite{Mercure2012}. This technological change is dependent on previous cumulative investment \cite{Mercure2012}. The model uses a dynamic set of logistic differential equations for competition between technology options.
For this work, we modified the FTT:Power model to use the deep reinforcement learning investment algorithm, DDPG. That is, the DDPG algorithm was used to make the decision on size of investment for each technology. In addition, we reduced the model only to consider the countries of Ireland and the UK. This enables us to iterate through enough episodes for the reinforcement learning to converge to an optimal reward. With more time it would be possible to undertake this optimisation for the entire world.
\subsection*{Reinforcement Learning}
The investment decision-making process can be formulated as a Markov Decision Process (MDP) \cite{puterman2014markov}. In an MDP environment, an agent receives an observation about the state of their environment $s_t$, chooses an action $a_t$ and receives a reward $r_t$ as a consequence of their action and the resultant change on the environment. Solving an MDP consists of maximizing the cumulative reward over the lifetime of the agent.
For our simulation environment, the agent makes continuous investment decisions for each energy technology, in each region and each year, starting from 2017 until 2050. Technology switching is modelled using a pairwise comparison of flows of market shares of different electricity generation capacity. That is, how much capacity flows from one technology to another.
The agent's observation space is a vector consisting of the electricity produced by each technology, total capacity, total \ch{CO2} emissions over the simulation, levelized cost of electricity (LCOE) both with and without taxes, cumulative investment in each technology, investment in new capacity, carrier prices by commodity, fuel costs and carbon costs.
The reward $r$ is defined as:
\begin{equation}
\label{eq:reward_function}
r = -\left(1000\times\ch{CO2}_e + \frac{LCOE}{1000}\right),
\end{equation}
where $\ch{CO2}_e$ is equal to total \ch{CO2} emissions over the simulation. The LCOE is calculated without taxes and the scaling factors are used to place the $LCOE$ and $\ch{CO2}$ on the same scale. The reward was multiplied by -1 due to the reinforcement learning (RL) algorithm maximizing reward and our requirement to reduce both LCOE and \ch{CO2} emissions.
RL approaches have been used to solve MDP through a trial and error based approach \cite{Sutton2015}. Since the paper published by Deep Mind in 2013 \cite{Arulkumaran2017}, RL has been extended to incorporate Deep Reinforcement Learning (DRL). DRL exploits deep neural networks to overcome the problems of memory and computational complexity \cite{Arulkumaran2017}.
We applied the deep deterministic policy gradient (DDPG) DRL algorithm \cite{Hunt2016a} from the Ray RLlib package to act as the investment algorithm \cite{Liang2014}. The DDPG algorithm is made up of an actor and critic network. We designed both of these to have two hidden layers, made up of 400 and 300 units per layer. The training batch size was set to 40,000. We chose these parameters as they were the default implementation in Ray RLlib. We trialled a variety of different configurations for the number of neurons per layer for hyperparameter tuning. To increase the speed of computation, just for the hyperparameter tuning, we reduced the simulation to run from 2007 to 2020. We chose this range as it allows for a change in the electricity mix. However, we found that the approach worked well, irrespective of parameter choice, as shown by Figure \ref{fig:hyperparameter_training}.
\begin{figure}
\centering
\includegraphics[width=0.5\columnwidth]{figures/hyperparameter_plot_params.pdf}
\caption{Training with different hyperparameters, displaying the minimum, mean and maximum rewards per episode. The hyperparameter set [300, 500], for example, refers to two layers for both the actor and critic network, with 300 neurons in the first layers and 500 in the second.}
\label{fig:hyperparameter_training}
\end{figure}
\section{Results}
\label{sec:results}
Our results show that our investment agent can increase its reward over time, as shown in Figure \ref{fig:days_reward_plot}. A total of ${\sim}$400,000 steps were required to see a levelling off in reward. The total time to simulate ${\sim}$400,000 steps was ${\sim}$8 days. We stopped the training and simulation after this time due to diminishing returns and the cost of computation.
Figure \ref{fig:electricity_generated_plot} displays the results of the reinforcement learning algorithm. Before the black vertical line (2017), the investments made are based upon historical data used by FTT:Power. The reinforcement learning algorithm starts to make investments after the black vertical line.
The historical electricity mix before 2017 is based mainly on fossil fuels: coal, combined cycle gas turbine (CCGT) and oil. Additionally, nuclear is a significant component of the electricity mix before 2009. After reinforcement learning optimizes for LCOE and carbon emissions, a rapid change occurs from fossil fuel and nuclear to renewable energy.
This sudden change occurs because the RL algorithm does not take into account the technical and timeframe constraints embedded in the unmodified FTT:Power model. However, although it is likely that whilst the transition speed is unrealistic, the electricity mix found by the reinforcement learning algorithm is likely to be optimal, according to the reward function defined in Equation \ref{eq:reward_function}. We, therefore, show what the future should look like.
\begin{figure}
\centering
\begin{minipage}{.4\textwidth}
\centering
\includegraphics[width=\linewidth]{figures/runtime_steps_plot.pdf}
\caption{Mean, minimum and maximum rewards over run time.}
\label{fig:days_reward_plot}
\end{minipage}%
\begin{minipage}{.4\textwidth}
\centering
\includegraphics[width=\linewidth]{figures/electricity_generated_plot.pdf}
\caption{Electricity mix over time.}
\label{fig:electricity_generated_plot}
\end{minipage}
\end{figure}
The primary source of energy after the reinforcement learning algorithm begins is offshore, followed by onshore, solar photovoltaics (PV) and wave. As can be seen by Figure \ref{fig:emissions_plot}, the carbon emissions reduce significantly at the time that the reinforcement learning algorithm begins to control investments.
This mix of renewable electricity generation across Ireland and the UK allows for demand to be met during the quarterly time periods of the model. The demand scenario is shown in Figure \ref{fig:demand_scenario}, where the demand can be seen to closely match the electricity mix shown by Figure \ref{fig:electricity_generated_plot}.
\begin{figure}
\centering
\begin{minipage}{.4\textwidth}
\centering
\includegraphics[width=\linewidth]{figures/emissions_plot.pdf}
\caption{Carbon emissions.}
\label{fig:emissions_plot}
\end{minipage}%
\begin{minipage}{.4\textwidth}
\centering
\includegraphics[width=\linewidth]{figures/demand_plot.pdf}
\caption{Demand scenario.}
\label{fig:demand_scenario}
\end{minipage}
\end{figure}
\section{Discussion}
A change from a high carbon-emitting electricity grid to a low-carbon system is required. In order to achieve this, investments in electricity generators must be made whilst taking into account future uncertainty. In this paper, we have modelled a central agent which makes investment decisions in an uncertain environment to find an optimal low-cost, low-carbon electricity mix. To achieve this, we used the reinforcement learning algorithm, DDPG. The environment is modelled using FTT:Power.
Through this exercise, we are able to see the optimal electricity mix in the UK and Ireland. We found that a mixture of renewable sources such as wind, solar and wave power would meet demand at quarter year intervals, as well as providing a cost-effective and low-carbon system.
A limitation of this work is the fact that the investment algorithm does not take into account the technical and timeframe constraints of transitions between technologies. It is for this reason that the reinforcement learning algorithm is able to make such a rapid change in 2017. However, we believe that the investment algorithm is able to find a general solution to the problem of investing in a cost-efficient and low-carbon system over a long time horizon. In future work, we would like to model the transition required by incorporating the technical and timeframe constraints for technology switching. This could be undertaken by modifying the reward function to ensure the transition remains within these constraints.
We would like to increase the number of steps of the FTT:Power model to more adequately model the investment behaviour introduced by the reinforcement learning algorithm. A lower number of simulated time steps leads to an overestimation of the supply of renewables and underestimation of storage and dispatchable technologies \cite{Ludig2011}. In addition, an increase in the number of countries modelled would enable us to see a global picture of how different, interdependent regions may evolve in a new climate of a requirement of low-carbon emissions. This would require an exponentially longer runtime for the reinforcement learning algorithm to converge. This is due to the increased number of decisions that the reinforcement learning algorithm would need to make to account for the different countries.
\section{Acknowledgements}
This work was supported by the Engineering and Physical Sci- ences Research Council, Centre for Doctoral Training in Cloud Computing for Big Data [grant number EP/L015358/1].
\bibliographystyle{ieeetr}
\section{Submission of papers to ``Tackling Climate Change with Machine Learning'' at NeurIPS 2020}
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\medskip
\small
[1] Alexander, J.A.\ \& Mozer, M.C.\ (1995) Template-based algorithms for
connectionist rule extraction. In G.\ Tesauro, D.S.\ Touretzky and T.K.\ Leen
(eds.), {\it Advances in Neural Information Processing Systems 7},
pp.\ 609--616. Cambridge, MA: MIT Press.
[2] Bower, J.M.\ \& Beeman, D.\ (1995) {\it The Book of GENESIS: Exploring
Realistic Neural Models with the GEneral NEural SImulation System.} New York:
TELOS/Springer--Verlag.
[3] Hasselmo, M.E., Schnell, E.\ \& Barkai, E.\ (1995) Dynamics of learning and
recall at excitatory recurrent synapses and cholinergic modulation in rat
hippocampal region CA3. {\it Journal of Neuroscience} {\bf 15}(7):5249-5262.
\end{document} | {
"redpajama_set_name": "RedPajamaArXiv"
} | 1,631 |
#ifndef _ASM_TILE_KMAP_TYPES_H
#define _ASM_TILE_KMAP_TYPES_H
/*
* In 32-bit TILE Linux we have to balance the desire to have a lot of
* nested atomic mappings with the fact that large page sizes and many
* processors chew up address space quickly. In a typical
* 64-processor, 64KB-page layout build, making KM_TYPE_NR one larger
* adds 4MB of required address-space. For now we leave KM_TYPE_NR
* set to depth 8.
*/
enum km_type {
KM_TYPE_NR = 8
};
/*
* We provide dummy definitions of all the stray values that used to be
* required for kmap_atomic() and no longer are.
*/
enum {
KM_BOUNCE_READ,
KM_SKB_SUNRPC_DATA,
KM_SKB_DATA_SOFTIRQ,
KM_USER0,
KM_USER1,
KM_BIO_SRC_IRQ,
KM_BIO_DST_IRQ,
KM_PTE0,
KM_PTE1,
KM_IRQ0,
KM_IRQ1,
KM_SOFTIRQ0,
KM_SOFTIRQ1,
KM_SYNC_ICACHE,
KM_SYNC_DCACHE,
KM_UML_USERCOPY,
KM_IRQ_PTE,
KM_NMI,
KM_NMI_PTE,
KM_KDB
};
#endif /* _ASM_TILE_KMAP_TYPES_H */
| {
"redpajama_set_name": "RedPajamaGithub"
} | 7,306 |
Q: Web page as Background Image on VSCode I want vscode to have a background image of a running web page, not a static image, so I can have a dynamic web page as the background for vscode, generated by JavaScript canvas. Is there any way to do that?
| {
"redpajama_set_name": "RedPajamaStackExchange"
} | 3,006 |
\subsection{Aerodynamic constraints verification through an iterative procedure}
blabla
\subsection{Handling reference trajectories with different durations}
In the setting of Section \ref{sec:opti}, the reference trajectories are implicitly assumed to have the same duration $T$ but this may not be the case in practice. For instance, if one is interested in optimising the climb phase for aircraft from a given altitude to a final one, then one may observe duration differences which vary from a few seconds to a few minutes. In such a case, projecting these trajectories onto the same functional basis is not possible and the approach developed in Section \ref{sec:opti} can not be applied.
To deal with this problem, we propose to extend each reference trajectory $y_{R_i} \in L^2\big([0,T_i], \mathbb{R}^D \big)$ to a trajectory $\overline{y}_{R_i}$ over a larger time interval $[0, T]$ defined as follows:
\begin{equation*}
\overline{y}_{R_i}(t) := \left\{
\begin{array}{l}
y_{R_i}(t) \; , \quad \text{if } t \in [0, T_i] \\[2mm]
y_{fin} \; , \quad \text{if } t \in [T_i, T]
\end{array} \right. \; ,
\end{equation*}
where we set $\displaystyle T := \max_{i \in \{1, \dots, I\}} T_i$. In the climb optimisation setting, this consists in concatenating the beginning of an artificial cruise to each reference climb. Thanks to this procedure, it is then possible to project each component of the reference flights on an orthonormal basis of $L^2\big([0,T], \mathbb{R} \big)$. In this setting, we consider then the following optimisation problem:
\begin{equation*}
\overline{y}^\star \in \argmin_{y \in \mathcal{Y}_\mathcal{K}(0,T) \cap \mathcal{B}(y_{init}, y_{fin})} \sum_{i=1}^I \omega_i \int_0^T \Big\| \Lambda^{\frac{1}{2}} \big( y(t) - \overline{y}_{R_i}(t) \big) \Big\|_{\mathbb{R}^D}^2 \, dt + \int_0^T \mathrm{FF}\big(y(t) \big) \, dt \; ,
\end{equation*}
which is equivalent of an optimisation problem on $\mathbb{R}^K$ of the form \eqref{eq:opt_c2} according to \autoref{prop:min_pb_equiv_y_C}.
On the other hand, the duration $T$ to reach the final state $y_{fin}$ from the initial one $y_{init}$ following the trajectory $\overline{y}^\star$ may be too large when compared to the durations $T_i$ of the best reference trajectories. Nevertheless the trajectory $\overline{y}^\star$ may be close to (even reach) the final state $y_{fin}$ before the final time $T$; for instance, this is likely to occur if the duration of the reference trajectories associated with the largest weights are smaller than $T$. We exploit this fact to propose a truncation-adjustment step leading to an optimised trajectory with a more realistic duration in this setting while remaining close to the optimised trajectory $\overline{y}^\star$. We proceed as follows:
\begin{enumerate}
\item \underline{Truncation:}\\
Suppose that the set
\begin{equation*}
\Big\{t \in [0,T] \, \Big| \, \Big\| \Lambda^{\frac{1}{2}} \big( \overline{y}^\star(t) - y_{fin} \big) \Big\|_{\mathbb{R}^D}^2 \leqslant \epsilon \Big\} \; ,
\end{equation*}
where $\epsilon > 0$, is non-empty and take its minimal element $T_\epsilon$ (or infimum). If the error $\epsilon$ is acceptable from an expert point of view, then the truncated trajectory $\overline{y}^\star|_{[0,T_\epsilon]}$ is the output of our optimisation process. Otherwise we consider the two following steps to adjust $\overline{y}^\star|_{[0,T_\epsilon]}$ so that the resulting trajectory satisfies the endpoint constraints.
\item \underline{Projection:}\\
We set $y_\epsilon := \overline{y}^\star|_{[0,T_\epsilon]}$. The truncated trajectory $y_\epsilon$ does not belong necessarily to a finite dimension space. Given $\varepsilon > 0$, we choose $\mathcal{K}_\varepsilon = \big\{K_{\varepsilon,1},\dots,K_{\varepsilon,D} \big\}$ such that the projection $\widetilde{y}_\epsilon$ of $y_\epsilon$ onto $\mathcal{Y}_{\mathcal{K}_\varepsilon}(0,T_\epsilon)$ satisfies
\begin{equation*}
\int_0^{T_\epsilon} \Big\| \Lambda^{\frac{1}{2}} \big( \widetilde{y}_\epsilon(t) - y_\epsilon(t) \big) \Big\|_{\mathbb{R}^D}^2 \, dt \leqslant \varepsilon \; .
\end{equation*}
We define
\begin{equation*}
c_\epsilon := \Phi \, y_\epsilon \in \mathbb{R}^{K_\varepsilon} \qquad , \qquad \widetilde{y}_\epsilon := \Phi|_{\mathcal{Y}_\epsilon(0,T_\epsilon)}^{-1} c_\epsilon \in \mathcal{Y}_{\mathcal{K}_\varepsilon}(0,T_\epsilon) \; ,
\end{equation*}
where $K_\varepsilon := \sum_{d=1}^D (1 + K_{\varepsilon,d})$. Note that the projected trajectory $\widetilde{y}_\epsilon$ is likely not to satisfy the endpoint constraints, \emph{i.e.} $\widetilde{y}_\epsilon \notin \mathcal{B}(y_{init}, y_{fin})$.
\item \underline{Penalised adjustment:}\\
To obtain a final trajectory belonging to $\mathcal{B}(y_{init}, y_{fin})$, we slightly modify the projected trajectory $\widetilde{y}_\epsilon$ so that the endpoint constraints are verified. To do so, we propose to solve the following optimisation problem:
\begin{equation} \label{eq:pen_adj}
y_\eta^\star \in \argmin_{y \in \mathcal{Y}_{\mathcal{K}_\varepsilon}(0,T_\epsilon) \cap \mathcal{B}(y_{init}, y_{fin})} \int_0^{T_\epsilon} \Big\| \Lambda^{\frac{1}{2}} \big( y(t) - \widetilde{y}_\epsilon(t) \big) \Big\|_{\mathbb{R}^D}^2 \, dt + \eta \int_0^{T_\epsilon} \mathrm{FF}\big(y(t) \big) \, dt \; ,
\end{equation}
where $\eta > 0$, which is equivalent to the following one according to \autoref{prop:min_pb_equiv_y_C}:
\begin{equation} \label{eq:pend_adj_c}
c_\eta^\star \in \argmin_{c \in \mathcal{C}(y_{init}, y_{fin})} \Big\| \overline{\Lambda}^{\frac{1}{2}} \big(c - c_\epsilon \big) \Big\|_{\mathbb{R}^K_\varepsilon}^2 + \eta \, \mathrm{TFC}(\Phi|_{\mathcal{Y}_\epsilon(0,T_\epsilon)}^{-1} c) \; .
\end{equation}
Here we choose $\eta > 0$ sufficiently small so that the preceding optimisation problem is strictly convex, namely the cost function is strictly convex.
\end{enumerate}
In the following theorem, we show that the total fuel consumption of $y_\eta^\star$ is necessarily smaller than the one of $\widetilde{y}_\epsilon$. Further we prove that the trajectories $y_\eta^\star$ and $y_\epsilon$ are close if $\eta$ is sufficiently small, if the projected trajectory $\widetilde{y}_\epsilon$ is sufficiently accurate and if the endpoint constraints errors of $\widetilde{y}_\epsilon$ are sufficiently small.
\begin{theorem} \label{thm:stablity_adj}
The solution $y_\eta^\star$ of the optimisation problem \eqref{eq:pen_adj} satisfies the following property:
\begin{equation} \label{eq:TFC_adjust}
\mathrm{TFC}(y_\eta^\star) \leqslant \mathrm{TFC}(\widetilde{y}_\epsilon) - \eta^{-1} \int_0^{T_\epsilon} \Big\| \Lambda^{\frac{1}{2}} \big( y_\eta^\star(t) - \widetilde{y}_\epsilon(t) \big) \Big\|_{\mathbb{R}^D}^2 \, dt \; .
\end{equation}
Assume in addition the matrix $A(0,T_\epsilon) \in \mathbb{R}^{2D \times K_\varepsilon}$ defined in \autoref{prop:linear_const} to be full rank and that there exists $\delta > 0$ such that
\begin{equation*}
\Big\| \Lambda^\frac{1}{2} \big (\widetilde{y}_\epsilon(0) - y_{init} \big) \Big\|_{\mathbb{R}^D}^2 + \Big\| \Lambda^\frac{1}{2} \big (\widetilde{y}_\epsilon(T_\epsilon) - y_{fin} \big) \Big\|_{\mathbb{R}^D}^2 \leqslant \delta \; .
\end{equation*}
Let $\varepsilon > 0$. Then there exists $\eta_\varepsilon > 0$ such that for all $\eta \in (0, \eta_\varepsilon)$ we have
\begin{equation} \label{eq:traj_adjust}
\int_0^{T_\epsilon} \Big\| \Lambda^{\frac{1}{2}} \big( y_\eta^\star(t) - y_\epsilon(t) \big) \Big\|_{\mathbb{R}^D}^2 \, dt \leqslant 4 (\lambda_{max} + 1) \, \varepsilon + \frac{4 \kappa}{\sigma_{min}^2} \, \delta \; ,
\end{equation}
where $\sigma_{min} > 0$ is the smallest singular value of the matrix $A(0,T_\epsilon)$ and
\begin{equation*}
\lambda_{max} := \max_{d \in \{1,\dots, D\}} \lambda^{(d)} \qquad , \qquad \kappa := \frac{\max_{d \in \{1,\dots, D\}} \lambda^{(d)}}{\min_{d \in \{1,\dots, D\}} \lambda^{(d)}} \; .
\end{equation*}
\end{theorem}
\begin{proof}
The trajectory $y_\eta^\star$ is the solution of the minimisation problem \eqref{eq:pen_adj}. Since $\widetilde{y}_\epsilon$ belongs also to $\mathcal{Y}_{\mathcal{K}_\varepsilon}(0,T_\epsilon) \cap \mathcal{B}(y_{init},y_{fin})$, we have
\begin{align*}
& \int_0^{T_\epsilon} \Big\| \Lambda^{\frac{1}{2}} \big( y_\eta^\star(t) - \widetilde{y}_\epsilon(t) \big) \Big\|_{\mathbb{R}^D}^2 \, dt + \eta \int_0^{T_\epsilon} \mathrm{FF}\big(y_\eta^\star(t) \big) \, dt \\
& \hspace{2cm} \leqslant \int_0^{T_\epsilon} \Big\| \Lambda^{\frac{1}{2}} \big( \widetilde{y}_\epsilon(t) - \widetilde{y}_\epsilon(t) \big) \Big\|_{\mathbb{R}^D}^2 \, dt + \eta \int_0^{T_\epsilon} \mathrm{FF}\big(\widetilde{y}_\epsilon(t) \big) \, dt \\
& \hspace{2cm} = \eta \int_0^{T_\epsilon} \mathrm{FF}\big(\widetilde{y}_\epsilon(t) \big) \, dt \; ,
\end{align*}
which leads to inequality \eqref{eq:TFC_adjust}.\\
We prove now the second inequality \eqref{eq:traj_adjust}. First of all, we apply \autoref{lem:continuity_pert} whose hypotheses are satisfied with
\begin{itemize}
\item $A = A(0,T_\epsilon) \in \mathbb{R}^{2D \times K_\varepsilon}$ which is surjective by hypothesis;
\item $b = (y_{init}, y_{fin})^T$;
\item $\displaystyle f(c) = \Big\| \overline{\Lambda}^\frac{1}{2} (c - c_\epsilon) \Big\|_{\mathbb{R}^{K_\varepsilon}}^2$ which is a strictly convex $\mathcal{C}^\infty$-function;
\item $g(c) = \mathrm{TFC}\big( \Phi|_{\mathcal{Y}_\epsilon(0,T_\epsilon)}^{-1} c \big)$ which is a $\mathcal{C}^\infty$-function.
\end{itemize}
This implies the continuity at 0 of the solution $c_\eta^\star$ of the optimisation problem \eqref{eq:pend_adj_c} with respect to $\eta$. Hence there exists $\eta_\varepsilon > 0$ such that
\begin{equation*}
\forall \, \eta \in (0, \eta_\varepsilon) \qquad \big\| c_\eta^\star - c_0^\star \big\|_{\mathbb{R}^{K_\varepsilon}}^2 \leqslant \varepsilon \; .
\end{equation*}
By \autoref{prop:phi_unitary}, it follows
\begin{equation} \label{eq:ineq_1}
\int_0^{T_\epsilon} \Big\| \Lambda^{\frac{1}{2}} \big( y_\eta^\star(t) - y_0^\star(t) \big) \Big\|_{\mathbb{R}^D}^2 \, dt = \Big\| \overline{\Lambda}^\frac{1}{2} \big( c_\eta^\star - c_0^\star \big) \Big\|_{\mathbb{R}^{K_\varepsilon}}^2 \leqslant \lambda_{max} \, \varepsilon \; ,
\end{equation}
for all $\eta \in (0, \eta_\varepsilon)$. Furthermore the hypotheses of \autoref{lem:stability_adj} are satisfied so we have
\begin{equation} \label{eq:ineq_2}
\int_0^{T_\epsilon} \Big\| \Lambda^{\frac{1}{2}} \big( y_0^\star(t) - \widetilde{y}_\epsilon(t) \big) \Big\|_{\mathbb{R}^D}^2 \, dt \leqslant \frac{\kappa(\Lambda)}{\sigma_0\big(A(0,T_\epsilon)\big)^2} \, \delta \; .
\end{equation}
And the set $\mathcal{K}_\varepsilon$ is chosen in such way that
\begin{equation} \label{eq:ineq_3}
\int_0^{T_\epsilon} \Big\| \Lambda^{\frac{1}{2}} \big( \widetilde{y}_\epsilon(t) - y_\epsilon(t) \big) \Big\|_{\mathbb{R}^D}^2 \, dt \leqslant \varepsilon \; .
\end{equation}
By the parallelogram law, one has
\begin{align*}
\int_0^{T_\epsilon} \Big\| \Lambda^{\frac{1}{2}} \big( y_\eta^\star(t) - y_\epsilon(t) \big) \Big\|_{\mathbb{R}^D}^2 \, dt
& \leqslant 4 \int_0^{T_\epsilon} \Big\| \Lambda^{\frac{1}{2}} \big( y_\eta^\star(t) - y_0^\star(t) \big) \Big\|_{\mathbb{R}^D}^2 \, dt \\
& \hspace{1cm} + 4 \int_0^{T_\epsilon} \Big\| \Lambda^{\frac{1}{2}} \big( y_0^\star(t) - \widetilde{y}_\epsilon(t) \big) \Big\|_{\mathbb{R}^D}^2 \, dt \\
& \hspace{2cm} + 4 \int_0^{T_\epsilon} \Big\| \Lambda^{\frac{1}{2}} \big( \widetilde{y}_\epsilon(t) - y_\epsilon(t) \big) \Big\|_{\mathbb{R}^D}^2 \, dt \; .
\end{align*}
Inserting inequalities \eqref{eq:ineq_1}, \eqref{eq:ineq_2} and \eqref{eq:ineq_3} into the preceding one leads to the final result.
\end{proof}
\begin{lemma} \label{lem:stability_adj}
Assume that there exist $\delta > 0$ such that
\begin{equation*}
\Big\| \Lambda^\frac{1}{2} \big (\widetilde{y}_\epsilon(0) - y_{init} \big) \Big\|_{\mathbb{R}^D}^2 + \Big\| \Lambda^\frac{1}{2} \big (\widetilde{y}_\epsilon(T_\epsilon) - y_{fin} \big) \Big\|_{\mathbb{R}^D}^2 \leqslant \delta \; ,
\end{equation*}
and assume that the matrix $A(0,T_\epsilon) \in R^{2D \times K_\varepsilon}$ is full rank. Then the solution of the optimisation problem
\begin{equation} \label{eq:adj}
y_0^\star \in \argmin_{y \in \mathcal{Y}_{\mathcal{K}_\varepsilon}(0,T_\epsilon) \cap \mathcal{B}(y_{init}, y_{fin})} \int_0^{T_\epsilon} \Big\| \Lambda^{\frac{1}{2}} \big( y(t) - \widetilde{y}_\epsilon(t) \big) \Big\|_{\mathbb{R}^D}^2 \, dt
\end{equation}
satisfies
\begin{equation*}
\int_0^{T_\epsilon} \Big\| \Lambda^{\frac{1}{2}} \big( y_0^\star(t) - \widetilde{y}_\epsilon(t) \big) \Big\|_{\mathbb{R}^D}^2 \, dt \leqslant \frac{\kappa}{\sigma_{min}^2} \, \delta \; ,
\end{equation*}
where $\kappa$ and $\sigma_{min}$ are defined in \autoref{thm:stablity_adj}.
\end{lemma}
\begin{proof}
First of all, we consider the following minimisation problem
\begin{equation*}
c_0^\star \in \argmin_{c \in \mathcal{C}(y_{init}, y_{fin})} \Big\| \overline{\Lambda}^{\frac{1}{2}} \big(c - c_\epsilon \big) \Big\|_{\mathbb{R}^{K_\varepsilon}}^2 \; ,
\end{equation*}
which is equivalent to the problem \eqref{eq:adj} following the lines of the proof of \autoref{prop:min_pb_equiv_y_C}.\\
Consider now the singular value decomposition of the matrix $A(0,T_\epsilon)$:
\begin{equation*}
A(0,T_\epsilon) = U \Sigma V^T \; ,
\end{equation*}
where $\Sigma \in \mathbb{R}^{2D \times K_\varepsilon}$ is a rectangular diagonal matrix and $U \in \mathbb{R}^{2D \times 2D}$ and $V \in \mathbb{R}^{K_\varepsilon \times K_\varepsilon}$ are orthogonal matrices. In particular we can write
\begin{equation*}
U \Sigma V^T = U \Big( \Sigma_1 \hspace{3mm} 0_{2D, K_\varepsilon-2D} \Big) \left(
\begin{array}{c}
V_1^T \\[2mm]
V_2^T
\end{array} \right) = U \Sigma_1 V_1^T \; ,
\end{equation*}
where $\Sigma_1 \in \mathbb{R}^{2D \times 2D}$ is a diagonal matrix whose diagonal elements $\{ \sigma_d \}_{d=1}^{2D}$ are the singular values in descending order and $V_1 \in \mathbb{R}^{K_\varepsilon \times 2D}$ is a semi-orthogonal matrix; note that $\Sigma_1$ is non-singular and that $\sigma_d > 0$ for all $d \in \{1,\dots,2D\}$ thanks to the hypothesis $A(0,T_\epsilon)$ is full rank.\\
For all $c \in \mathcal{C}(y_{init}, y_{fin})$, we have the following equivalences:
\begin{align}
A(0,T_\epsilon) c = \left(
\begin{array}{c}
y_{init} \\
y_{fin}
\end{array} \right) \quad
& \Longleftrightarrow \quad U \Sigma_1 V_1^T c = \left(
\begin{array}{c}
y_{init} \\
y_{fin}
\end{array} \right) \nonumber \\
& \Longleftrightarrow \quad V_1^T c = \Sigma_1^{-1} U^T \left(
\begin{array}{c}
y_{init} \\
y_{fin}
\end{array} \right) \; . \label{eq:equiv_end_const}
\end{align}
Let now define the following vector $\widetilde{c} \in \mathbb{R}^{K_\varepsilon}$ through its image $V^T \widetilde{c}$:
\begin{equation*}
\left\{ \begin{array}{l}
V_1^T \widetilde{c} = \Sigma_1^{-1} U^T \left(
\begin{array}{c}
y_{init} \\
y_{fin}
\end{array} \right) \\[4mm]
V_2^T \widetilde{c} = V_2^T c_\epsilon
\end{array} \right. \; .
\end{equation*}
We deduce that $\widetilde{c}$ belongs to $\mathcal{C}(y_{init},y_{fin})$ from the equivalences \eqref{eq:equiv_end_const} and we have
\begin{align}
\Big\| A(0,T_\epsilon) (\widetilde{c} - c_\epsilon) \Big\|_{\mathbb{R}^{2D}}^2
& = \Big\| U \Sigma V^T (\widetilde{c} - c_\epsilon) \Big\|_{\mathbb{R}^{2D}}^2 \nonumber \\
& = \Big\| \Sigma_1 V_1^T (\widetilde{c} - c_\epsilon) \Big\|_{\mathbb{R}^{2D}}^2 \label{eq:U_unit} \\
& = \sum_{d = 1}^{2D} \sigma_d^2 \Big| \big(V_1^T \widetilde{c}\big)_d - \big(V_1^T c_\epsilon\big)_d \Big|^2 \nonumber \\
& \geqslant \sigma_{2D}^2 \Big\| V_1^T (\widetilde{c} - c_\epsilon) \Big\|_{\mathbb{R}^{2D}}^2 \label{eq:min_sing_val} \\
& = \sigma_{2D}^2 \Big\| V^T (\widetilde{c} - c_\epsilon) \Big\|_{\mathbb{R}^{K_\varepsilon}}^2 \label{eq:V_2_zero} \\
& = \sigma_{2D}^2 \big\| \widetilde{c} - c_\epsilon \big\|_{\mathbb{R}^{K_\varepsilon}}^2 \label{eq:V_unit} \; ;
\end{align}
\begin{itemize}
\item \eqref{eq:U_unit}: use the unitary property of $U$ and the equality $U \Sigma V^T = U \Sigma_1 V_1^T$;
\item \eqref{eq:min_sing_val}: use the fact that $\sigma_d \geqslant \sigma_{2D}$ for all $d \in \{1, \dots, 2D \}$;
\item \eqref{eq:V_2_zero}: by the definition of $\widetilde{c}$, we have
\begin{equation*}
\Big\| V^T (\widetilde{c} - c_\epsilon) \Big\|_{\mathbb{R}^{K_\varepsilon}}^2 = \Big\| V_1^T (\widetilde{c} - c_\epsilon) \Big\|_{\mathbb{R}^{2D}}^2 + \Big\| V_2^T (\widetilde{c} - c_\epsilon) \Big\|_{\mathbb{R}^{K_\varepsilon-2D}}^2 = \Big\| V_1^T (\widetilde{c} - c_\epsilon) \Big\|_{\mathbb{R}^{2D}}^2 \; ;
\end{equation*}
\item[\eqref{eq:V_unit}]: use the unitary property of $V$.
\end{itemize}
It follows
\begin{align}
& \Big\| \overline{\Lambda}^\frac{1}{2} (\widetilde{c} - c_\epsilon ) \Big\|_{\mathbb{R}^{K_\varepsilon}}^2 \nonumber \\
& \hspace{2cm} \leqslant \max_{d \in \{1,\dots, D\}} \lambda^{(d)} \big\| \widetilde{c} - c_\epsilon \big\|_{\mathbb{R}^{K_\varepsilon}}^2 \nonumber \\
& \hspace{2cm} \leqslant \frac{\max_{d \in \{1,\dots, D\}} \lambda^{(d)}}{\sigma_{2D}^2} \, \Big\| A(0,T_\epsilon) (\widetilde{c} - c_\epsilon) \Big\|_{\mathbb{R}^{2D}}^2 \label{eq:inv_ineq} \\
& \hspace{2cm} = \frac{\max_{d \in \{1,\dots, D\}} \lambda^{(d)}}{\sigma_{2D}^2} \bigg( \Big\| A(0,T_\epsilon)_1 (\widetilde{c} - c_\epsilon) \Big\|_{\mathbb{R}^D}^2 + \Big\| A(0,T_\epsilon)_2 (\widetilde{c} - c_\epsilon) \Big\|_{\mathbb{R}^D}^2 \bigg) \label{eq:mat_two_parts} \\
& \hspace{2cm} = \frac{\max_{d \in \{1,\dots, D\}} \lambda^{(d)}}{\sigma_{2D}^2} \Big( \big\| y_{init} - \widetilde{y}_\epsilon(0) \big\|_{\mathbb{R}^D}^2 + \big\| y_{fin} - \widetilde{y}_\epsilon(T_\epsilon) \big\|_{\mathbb{R}^D}^2 \Big) \label{eq:recov_init_cond} \\
& \hspace{2cm} \leqslant \frac{\max_{d \in \{1,\dots, D\}} \lambda^{(d)}}{\sigma_{2D}^2 \min_{d \in \{1,\dots, D\}} \lambda^{(d)}} \Big( \Big\| \Lambda^\frac{1}{2} \big (y_{init} - \widetilde{y}_\epsilon(0) \big) \Big\|_{\mathbb{R}^D}^2 + \Big\| \Lambda^\frac{1}{2} \big( y_{fin} - \widetilde{y}_\epsilon(T_\epsilon) \big) \Big\|_{\mathbb{R}^D}^2 \Big) \label{eq:equiv_norms} \\
& \hspace{2cm} \leqslant \frac{\max_{d \in \{1,\dots, D\}} \lambda^{(d)}}{\sigma_{2D}^2 \min_{d \in \{1,\dots, D\}} \lambda^{(d)}} \, \delta \; ; \label{eq:constraints_hyp}
\end{align}
\begin{itemize}
\item \eqref{eq:inv_ineq}: use the inequality
\begin{equation*}
\sigma_{2D}^2 \big\| \widetilde{c} - c_\epsilon \big\|_{\mathbb{R}^{K_\varepsilon}}^2 \leqslant \Big\| A(0,T_\epsilon) (\widetilde{c} - c_\epsilon) \Big\|_{\mathbb{R}^{2D}}^2
\end{equation*}
proved above and the fact that $\sigma_{2D} > 0$;
\item \eqref{eq:mat_two_parts}: the matrices $A(0,T_\epsilon)_1, A(0,T_\epsilon)_2 \in \mathbb{R}^{D \times K_\varepsilon}$ are defined as follows:
\begin{align*}
& \bullet \quad A(0,T_\epsilon)_1 := \left(
\begin{array}{ccccccc}
\varphi_0(0) & \dots & \varphi_{K_{\varepsilon, 1}}(0) & & & & \\
& & & \ddots & & & \\
& & & & \varphi_0(0) & \dots & \varphi_{K_{\varepsilon, D}}(0)
\end{array} \right) \; ; \\
& \bullet \quad A(0,T_\epsilon)_2 := \left(
\begin{array}{ccccccc}
\varphi_0(T_\epsilon) & \dots & \varphi_{K_{\varepsilon, 1}}(T_\epsilon) & & & & \\
& & & \ddots & & & \\
& & & & \varphi_0(T_\epsilon) & \dots & \varphi_{K_{\varepsilon, D}}(T_\epsilon)
\end{array} \right) \; ;
\end{align*}
in particular, they satisfy
\begin{equation*}
A(0,T_\epsilon) = \left(
\begin{array}{c}
A(0,T_\epsilon)_1 \\[2mm]
A(0,T_\epsilon)_2
\end{array} \right) \; ;
\end{equation*}
\item \eqref{eq:recov_init_cond}: by the definitions of the matrices $A(0,T_\epsilon)_1, A(0,T_\epsilon)_2$ and of the trajectory $\widetilde{y}_\epsilon = \Phi|_{\mathcal{Y}_\epsilon(0,T_\epsilon)}^{-1} c_\epsilon$, we have
\begin{equation*}
A(0,T_\epsilon)_1 \, c_\epsilon := \widetilde{y}_\epsilon(0) \qquad , \qquad A(0,T_\epsilon)_2 \, c_\epsilon := \widetilde{y}_\epsilon(T_\epsilon) \; ,
\end{equation*}
and by the fact that the vector $\widetilde{c}$ belongs to $\mathcal{B}(y_{init},y_{fin})$, we obtain
\begin{equation*}
A(0,T_\epsilon)_1 \, \widetilde{c} := y_{init} \qquad , \qquad A(0,T_\epsilon)_2 \, \widetilde{c} := y_{fin} \; ;
\end{equation*}
\item \eqref{eq:equiv_norms}: use the following inequality
\begin{equation*}
\forall x \in \mathbb{R}^D \qquad \min_{d \in \{1,\dots,D\}} \lambda^{(d)} \big\| x \big\|_{\mathbb{R}^D}^2 \leqslant \Big\| \Lambda^{\frac{1}{2}} x \Big\|_{\mathbb{R}^D}^2 \; ;
\end{equation*}
\item \eqref{eq:constraints_hyp}: use the hypothesis
\begin{equation*}
\Big\| \Lambda^\frac{1}{2} \big (\widetilde{y}_\epsilon(0) - y_{init} \big) \Big\|_{\mathbb{R}^D}^2 + \Big\| \Lambda^\frac{1}{2} \big (\widetilde{y}_\epsilon(T_\epsilon) - y_{fin} \big) \Big\|_{\mathbb{R}^D}^2 \leqslant \delta \; .
\end{equation*}
\end{itemize}
By using once again \autoref{prop:phi_unitary} and by using the property that $c_0^\star$ is a minimiser, we obtain finally
\begin{align*}
\int_0^{T_\epsilon} \Big\| \Lambda^{\frac{1}{2}} \big( y_0^\star(t) - \widetilde{y}_\epsilon(t) \big) \Big\|_{\mathbb{R}^D}^2 \, dt
& = \Big\| \overline{\Lambda}^\frac{1}{2} (c_0^\star - c_\epsilon ) \Big\|_{\mathbb{R}^{K_\varepsilon}}^2 \\
& \leqslant \Big\| \overline{\Lambda}^\frac{1}{2} (\widetilde{c} - c_\epsilon ) \Big\|_{\mathbb{R}^{K_\varepsilon}}^2 \\
& \leqslant \frac{\max_{d \in \{1,\dots, D\}} \lambda^{(d)}}{\sigma_{2D}^2 \min_{d \in \{1,\dots, D\}} \lambda^{(d)}} \, \delta \; .
\end{align*}
\end{proof}
\begin{lemma} \label{lem:continuity_pert}
Let $A \in \mathbb{R}^{\widetilde{K} \times K}$ be a full rank matrix with $\widetilde{K} \leqslant K$ and let $b \in \mathbb{R}^{\widetilde{K}}$. Let $f, g :\mathbb{R}^K \longrightarrow \mathbb{R}$ be two $\mathcal{C}^2$-functions such that $f$ is strictly convex and such that the $C^2$-function $h_\eta := f + \eta \, g$ is strictly convex for all $\eta \in \mathcal{N}(0) \subseteq \mathbb{R}$, where $\mathcal{N}(0)$ is an open neighbourhood of $0$. We define the function $x^\star: \mathcal{N}(0) \longrightarrow \mathbb{R}^K$ as
\begin{equation} \label{eq:opti_abst}
x^\star(\eta) := \argmin_{Ax = b} h_\eta(x) \; ,
\end{equation}
which is well-defined. Then the function $x^\star$ is continuous at $0$.
\end{lemma}
\begin{proof}
For all $\eta \in \mathcal{N}(0)$, the optimisation problem \eqref{eq:opti_abst} has a unique solution $x^\star(\eta) \in \mathbb{R}^K$ since the cost function is strictly convex and the constraint is affine. Further the cost function $h_\eta$ is a $\mathcal{C}^2$-function and the matrix $A$ is full rank by hypothesis. By Lagrange multiplier theorem \cite[p. 285]{F2000}, for all $\eta \in \mathcal{N}(0)$, there exists a unique vector $\lambda(\eta) \in \mathbb{R}^{\widetilde{K}}$ such that
\begin{equation} \label{eq:lagrange}
\nabla h_\eta \big( x^\star(\eta) \big) + A^T \lambda(\eta) = 0_{\mathbb{R}^K} \; .
\end{equation}
Define now $u : \mathbb{R}^K \times \mathcal{N}(0) \longrightarrow \mathbb{R}^K$ as follows
\begin{equation*}
u(x, \eta) = \nabla h_\eta \big( x \big) + A^T \lambda(\eta) \; .
\end{equation*}
Then, by equality \eqref{eq:lagrange}, we have
\begin{equation*}
u\big(x^\star(\eta), \eta \big) = 0_{R^K} \; ,
\end{equation*}
and the gradient $\nabla_x u$ of $u$ with respect to its first variable is equal to the Hessian matrix of $h_\eta$. Since $h_\eta$ is supposed to be strictly convex for all $\eta \in \mathcal{N}(0)$, its Hessian is positive definite and hence $\nabla_x u$ is invertible. By the implicit function theorem, we deduce the continuity of $x^\star$ at $0$.
\end{proof}
\subsection{Misc.}
\begin{proposition} \label{prop:phi_unitary}
Let $\| \cdot \|_{\mathbb{R}^K}$ denote the Euclidean norm on $\mathbb{R}^K$ and let $\big\{ \lambda^{(d)} \big\}_{d=1}^D$ be a set of positive real numbers. Define the matrices $\Lambda \in \mathbb{R}^{D \times D}$ and $\overline{\Lambda} \in \mathbb{R}^{K \times K}$ as
\begin{equation*}
\Lambda := \left(
\begin{array}{ccc}
\lambda^{(1)} & & \\
& \ddots & \\
& & \lambda^{(D)}
\end{array} \right) \qquad , \qquad
\overline{\Lambda} := \left(
\begin{array}{ccc}
\lambda^{(1)} \, I_{K_1 + 1} & & \\
& \ddots & \\
& & \lambda^{(D)} I_{K_D + 1}
\end{array} \right)
\end{equation*}
where $I_n$ is the identity matrix of size $n$. Then the operator $\Phi|_{\mathcal{Y}_\mathcal{K}} : \mathcal{Y}_\mathcal{K} \longrightarrow \mathbb{R}^K$ is bijective and satisfies
\begin{equation*}
\forall \, y \in \mathcal{Y}_\mathcal{K} \qquad \int_0^T \Big\| \Lambda^{\frac{1}{2}} \, y(t) \Big\|_{\mathbb{R}^D}^2 \, dt = \Big\| \overline{\Lambda}^\frac{1}{2} \, \Phi|_{\mathcal{Y}_\mathcal{K}} y \Big\|_{\mathbb{R}^K}^2 \; .
\end{equation*}
\end{proposition}
\begin{proof}
Let $y \in \mathcal{Y}_\mathcal{K}$ and let $c := \Phi|_{\mathcal{Y}_\mathcal{K}} y \in \mathbb{R}^{K_d}$. We have by the definition of the matrix $\Lambda$ and by the hypothesis $y \in \mathcal{Y}_\mathcal{K}$,
\begin{equation*}
\int_0^T \Big\| \Lambda^{\frac{1}{2}} \, y(t) \Big\|_{\mathbb{R}^D}^2 \, dt = \sum_{d=1}^D \lambda^{(d)} \int_0^T \Big| y^{(d)}(t) \Big|^2 \, dt = \sum_{d=1}^D \lambda^{(d)} \int_0^T \left| \sum_{k=0}^{K_d} c_k^{(d)} \, \varphi_k(t) \right|^2 dt \; ,
\end{equation*}
and the orthonormal property of the basis $\{\varphi_k\}_{k=0}^{+\infty}$ gives for all $d \in \{1,\dots,D\}$,
\begin{equation*}
\int_0^T \left| \sum_{k=0}^{K_d} c_k^{(d)} \, \varphi_k(t) \right|^2 dt = \sum_{k=0}^{K_d} \Big| c_k^{(d)} \Big|^2 \int_0^T \big| \varphi_k(t) \big|^2 \, dt = \sum_{k=0}^{K_d} \Big| c_k^{(d)} \Big|^2 \; .
\end{equation*}
By using the definition of $\overline{\Lambda}$, we obtain
\begin{equation*}
\int_0^T \Big\| \Lambda^{\frac{1}{2}} \, y(t) \Big\|_{\mathbb{R}^D}^2 \, dt = \sum_{d=1}^D \lambda^{(d)} \sum_{k=0}^{K_d} \Big| c_k^{(d)} \Big|^2 = \Big\| \overline{\Lambda}^\frac{1}{2} \, c \Big\|_{\mathbb{R}^K}^2 \; .
\end{equation*}
As a direct consequence, the operator $\Phi|_{\mathcal{Y}_\mathcal{K}} : \mathcal{Y}_\mathcal{K} \longrightarrow \mathbb{R}^K$ is injective. It is surjective as well since for all $\widetilde{c} = \big( \widetilde{c}_0^{(1)}, \dots, \widetilde{c}_{K_1}^{(1)}, \dots, \; \widetilde{c}_0^{(D)}, \dots, \widetilde{c}_{K_D}^{(D)} \big) \in \mathbb{R}^K$, the trajectory $\widetilde{y} \in \mathcal{Y}_\mathcal{K}$ defined by
\begin{equation*}
\forall \, d \in \{1,\dots,D\} \qquad \widetilde{y}^{(d)} := \sum_{k=0}^{K_d} \widetilde{c}_k^{(d)} \, \varphi_k
\end{equation*}
satisfies $\Phi|_{\mathcal{Y}_\mathcal{K}} \widetilde{y} = \Phi \widetilde{y} = \widetilde{c}$.
\end{proof}
\begin{definition}[Total fuel consumption]
Let $\mathrm{FF}: \mathbb{R}^D \longrightarrow \mathbb{R}_+$ denote the fuel flow. We define the total fuel consumption $\mathrm{TFC}(y) \in \mathbb{R}_+$ of a trajectory $y \in L^2\big([0,T], \mathbb{R}^D\big)$ as follows:
\begin{equation*}
\mathrm{TFC}(y) := \int_0^T \mathrm{FF}\big(y(t) \big) \, dt \; .
\end{equation*}
\end{definition}
\begin{definition} \label{def:phi}
We define the integer $K := \sum_{d = 1}^D K_d$ and the operator $\Phi : L^2\big([0,T], \mathbb{R}^D \big) \longrightarrow \mathbb{R}^K$ as follows:
\begin{equation*}
\Phi y := \Big( c_1^{(1)}, \dots, c_{K_1}^{(1)}, \; c_1^{(2)}, \dots, c_{K_2}^{(2)}, \dots, \; c_1^{(D)}, \dots, c_{K_D}^{(D)} \Big) \; .
\end{equation*}
\end{definition}
\begin{definition}
We define
\begin{enumerate}
\item the projection operator $\Phi : \mathcal{C}\big([0,T], \mathbb{R}^D \big) \longrightarrow \mathbb{R}^K$ as follows:
\begin{equation*}
\Phi y := \Big( c_1^{(1)}, \dots, c_{K_1}^{(1)}, \; c_1^{(2)}, \dots, c_{K_2}^{(2)}, \dots, \; c_1^{(D)}, \dots, c_{K_D}^{(D)} \Big)^T \; ;
\end{equation*}
\item the set $\widetilde{\mathcal{D}}(y_0, y_T) \subset \mathbb{R}^K$ as the set of vectors satisfying the following linear condition:
\begin{equation} \label{eq:endpoint_syst}
A(0,T) \, c = \Gamma \; ,
\end{equation}
where the vector $\Gamma \in \mathbb{R}^{2D}$ is defined as
\begin{equation*}
\Gamma := \Big( y_0^{\ T} \quad y_T^{\ T} \Big)^T \; ,
\end{equation*}
and the matrix $A(0,T) \in \mathbb{R}^{2D \times K}$ is defined as
\begin{equation*}
A(0, T) := \left(
\begin{array}{ccccccc}
\varphi_1(0) & \dots & \varphi_{K_1}(0) & & & & \\
& & & \ddots & & & \\
& & & & \varphi_1(0) & \dots & \varphi_{K_D}(0) \\
\varphi_1(T) & \dots & \varphi_{K_1}(T) & & & & \\
& & & \ddots & & & \\
& & & & \varphi_1(T) & \dots & \varphi_{K_D}(T) \\
\end{array} \right) \; .
\end{equation*}
\end{enumerate}
\end{definition}
\begin{proposition} \label{prop:linear_const}
\begin{enumerate}
\item The restriction of the operator $\Phi$ to the subspace $\mathcal{Y}_\mathcal{K}$, namely $\Phi|_{\mathcal{Y}_\mathcal{K}} : \mathcal{Y}_\mathcal{K} \longrightarrow \mathbb{R}^K$, is bijective.
\item A trajectory $y \in \mathcal{Y}_{\mathcal{K}}$ belongs to $\mathcal{D}(y_0, y_T)$ if and only if $\Phi y$ belongs to $\widetilde{\mathcal{D}}(y_0, y_T)$.
\end{enumerate}
\end{proposition}
\begin{proof}
\begin{enumerate}
\item It is clear that each component $y^{(d)}$ of a trajectory $y \in \mathcal{Y}_\mathcal{K}$ is associated with a unique vector $c^{(d)} \in \mathbb{R}^{K_d}$ given by
\begin{equation*}
\forall \, k = 1, \dots, K_d \qquad c_k^{(d)} = \int_0^T y^{(d)}(t) \, \varphi_{k}(t) \, dt \; .
\end{equation*}
Since the restriction $\Phi|_{\mathcal{Y}_\mathcal{K}}$ is actually the Cartesian product of the bijective functions $\Phi^{(d)} : \text{span} \left\{ \varphi_{k} \right\}_{k = 1}^{K_d} \longrightarrow \mathbb{R}^{K_d}$ defined by
\begin{equation*}
\Phi^{(d)} y^{(d)} := c^{(d)} \; ,
\end{equation*}
it inherits the bijective nature of its components. This proves the first point.
\item Let $y \in \mathcal{Y}_\mathcal{K}$ and let $c := \Phi y \in \mathbb{R}^K$. By the definition of the matrix $A(0,T)$, we have
\begin{align*}
A(0,T) c
& = A(0,T) \Big( c_1^{(1)}, \dots, c_{K_1}^{(1)}, \; c_1^{(2)}, \dots, c_{K_2}^{(2)}, \dots, \; c_1^{(D)}, \dots, c_{K_D}^{(D)} \Big)^T \\
& = \left( \sum_{k=1}^{K_1} c_k^{(1)} \varphi_k(0), \dots, \displaystyle \sum_{k=1}^{K_D} c_k^{(D)} \varphi_k(0), \dots, \sum_{k=1}^{K_1} c_k^{(1)} \varphi_k(T), \dots, \sum_{k=1}^{K_D} c_k^{(D)} \varphi_k(T) \right)^T \\
& = \big( y(0) \quad y(T) \big)^T \; .
\end{align*}
The conclusion follows directly from the preceding relation.
\end{enumerate}
\end{proof}
\begin{remark}
Throughout the paper, we will use interchangeably the two following representations of a vector $c \in \mathbb{R}^K$:
\begin{itemize}
\item $c = (c_1, c_2, \dots, c_K)^T$;
\item $c = \Big( c_1^{(1)}, \dots, c_{K_1}^{(1)}, \; c_1^{(2)}, \dots, c_{K_2}^{(2)}, \dots, \; c_1^{(D)}, \dots, c_{K_D}^{(D)} \Big)^T$.
\end{itemize}
Indeed, for any $k \in \{1, \dots, K\}$, there exists a unique\footnote{Appendix ?} $\big(\widetilde{d}, \widetilde{k}\big)$ such that
\begin{equation*}
k = \widetilde{k} + \sum_{d = 1}^{\widetilde{d}-1} K_d \; ,
\end{equation*}
with $\widetilde{d} \in \{1,\dots,D\}$ and $\widetilde{k} \in \{1, \dots, K_{\widetilde{d}} \}$. In this case, we have $c_k = c_{\widetilde{k}}^{(\widetilde{d})}$.
\end{remark}
\subsection{Interpretation}
\begin{proposition} \label{prop:min_pb_equiv_y_C}
A vector $c^\star \in \mathbb{R}^K$ is solution of the optimisation problem \eqref{eq:opt_c} if and only if the trajectory $y^\star := \Phi|_{\mathcal{Y}_\mathcal{K}}^{-1} c^\star \in \mathcal{Y}_\mathcal{K}$ is solution of the following optimisation problem
\begin{equation*}
y^\star \in \argmin_{y \in \mathcal{Y}_\mathcal{K} \cap \mathcal{B}(y_0, y_T)} \sum_{i=1}^I \omega_i \int_0^T \Big\| \Lambda^{\frac{1}{2}} \big( y(t) - y_{R_i}(t) \big) \Big\|_{\mathbb{R}^D}^2 \, dt + \int_0^T \mathrm{FF}\big(y(t) \big) \, dt \; .
\end{equation*}
\end{proposition}
\begin{proof}
First of all, we define the maps $g_1: \mathbb{R}^K \longrightarrow \mathbb{R}$ as
\begin{equation*}
g_1(c) := \sum_{i=1}^I \omega_i \, \Big\| \overline{\Lambda}^{\frac{1}{2}} \big(c - c_{R_i} \big) \Big\|_{\mathbb{R}^K}^2 + \mathrm{TFC}\big( \Phi|_{\mathcal{Y}_\mathcal{K}}^{-1} c \big) \; ,
\end{equation*}
and $g_2: L^2\big([0,T], \mathbb{R}^d \big) \longrightarrow \mathbb{R}$ as
\begin{equation*}
g_2(y) := \sum_{i=1}^I \omega_i \int_0^T \Big\| \Lambda^{\frac{1}{2}} \big( y(t) - y_{R_i}(t) \big) \Big\|_{\mathbb{R}^D}^2 \, dt + \int_0^T \mathrm{FF}\big(y(t) \big) \, dt \; .
\end{equation*}
Let $c \in \mathbb{R}^K$ and let $y := \Phi|_{\mathcal{Y}_\mathcal{K}}^{-1} c \in \mathcal{Y}_\mathcal{K}$ be its associated trajectory. By the definition of $y$, we clearly have
\begin{equation*}
\mathrm{TFC}\big( \Phi|_{\mathcal{Y}_\mathcal{K}}^{-1} c \big) = \mathrm{TFC}(y) = \int_0^T \mathrm{FF}\big(y(t) \big) \, dt \; .
\end{equation*}
Furthermore Proposition \ref{prop:phi_unitary} implies
\begin{equation*}
\sum_{i=1}^I \omega_i \, \Big\| \overline{\Lambda}^{\frac{1}{2}} \big(c - c_{R_i} \big) \Big\|_{\mathbb{R}^K}^2 = \sum_{i=1}^I \omega_i \int_0^T \Big\| \Lambda^{\frac{1}{2}} \Big( y(t) - y_{R_i}(t) \Big) \Big\|_{\mathbb{R}^D}^2 \, dt \; ,
\end{equation*}
showing that $g_1(c) = g_2\big(y\big)$. Since the set $\mathcal{C}(y_0, y_T)$ is the image of $\mathcal{Y}_\mathcal{K} \cap \mathcal{B}(y_0, y_T)$ under the bijection $\Phi|_{\mathcal{Y}_\mathcal{K}}$ by definition, it follows that minimising the map $g_1$ over the set $\mathcal{C}(y_0, y_T)$ amounts to minimising the function $g_2$ over $\mathcal{Y}_\mathcal{K} \cap \mathcal{B}(y_0, y_T)$.
\end{proof}
\subsection{Quadratic cost for a convex optimisation problem} \label{subsec:quad_model}
\todo{Note ? + Change the position of $\kappa$ and adapt the results}\\
We focus on the case where the function $f$ (defining the cost $F$) is quadratic. In this setting, explicit formulas for the objective functions appearing in the optimisation problems \eqref{eq:opt_proj} and \eqref{eq:opt_proj_2} can be established. In particular this permits to derive sufficient conditions on the parameter $\kappa > 0$ so that the optimisation problems are proved to be equivalent to quadratic programs \citep[Sec. 4.4]{convexopt}, namely the objective functions are convex quadratic. In practice, this allows to make use of efficient convex optimisation libraries to solve numerically the problems.
Throughout this subsection, we suppose that the function $f$ defining the cost $F$ is quadratic, \emph{i.e.}
\begin{equation} \label{eq:quad_f}
f(x) = x^T Q x + w^T x + r = \sum_{d_1, d_2 = 1}^D Q_{d_1 d_2} \, x^{(d_1)} \, x^{(d_2)} + \sum_{d = 1}^D w_d \, x^{(d)} + r \; ,
\end{equation}
where $x \in \mathbb{R}^D$, $Q \in \mathbb{R}^{D \times D}$ is symmetric, $w \in \mathbb{R}^D$ and $r \in \mathbb{R}$. We define now some vectors and matrices which will be used to prove the quadratic programming property.
\begin{definition} \label{def:misc}
We define
\begin{enumerate}
\item the map $\widetilde{\varphi} \in \mathcal{C}\big([0,T], \mathbb{R}^K\big)$ as
\begin{equation*}
\widetilde{\varphi} := \Big( \varphi_1, \dots, \varphi_{K_1}, \; \varphi_1, \; \dots, \varphi_{K_2}, \dots, \; \varphi_1, \dots, \varphi_{K_D} \Big)^T \; ;
\end{equation*}
\item the matrix $\widetilde{Q} \in \mathbb{R}^{K \times K}$ as
\begin{equation*}
\widetilde{Q} := \left(
\begin{array}{ccc}
Q_{11} \, J_{K_1, K_1} & \ldots & Q_{1D} \, J_{K_1, K_D} \\
\vdots & & \vdots \\
Q_{D1} \, J_{K_D, K_1} & \ldots & Q_{DD} \, J_{K_D, K_D}
\end{array} \right) \; ,
\end{equation*}
where $J_{n,m}$ is the all-ones matrix of size $n \times m$;
\item the matrix $\overline{Q} \in \mathbb{R}^{K \times K}$ as
\begin{equation*}
\overline{Q}_{k_1 k_2} := \widetilde{Q}_{k_1 k_2} \int_0^T \widetilde{\varphi}_{k_1}(t) \, \widetilde{\varphi}_{k_2}(t) \, dt \; ;
\end{equation*}
\item the matrix $\overline{Q}_V := V^T \overline{Q} V \in \mathbb{R}^{K \times K}$ which can be written as follows:
\begin{equation*}
\overline{Q}_V = \left( \begin{array}{cc}
\overline{Q}_{V, 11} & \overline{Q}_{V, 12} \\[2mm]
\overline{Q}_{V, 21} & \overline{Q}_{V, 22}
\end{array} \right) \; ,
\end{equation*}
where $\overline{Q}_{V, 11} \in \mathbb{R}^{\sigma \times \sigma}$, $\overline{Q}_{V, 12} \in \mathbb{R}^{\sigma \times (K - \sigma)}$, $\overline{Q}_{V, 21} \in \mathbb{R}^{(K-\sigma) \times \sigma}$ and $\overline{Q}_{V, 22} \in \mathbb{R}^{(K-\sigma) \times (K-\sigma)}$;
\item the vector $\widetilde{w} \in \mathbb{R}^K$ as
\begin{equation*}
\widetilde{w} := \big( w_1 \, J_{1, K_1} \quad \dots \quad w_D \, J_{1, K_D} \big)^T \; ;
\end{equation*}
\item the vector $\overline{w} \in \mathbb{R}^K$ as
\begin{equation*}
\overline{w}_k := \widetilde{w}_k \int_0^T \widetilde{\varphi}_k(t) \, dt \; ;
\end{equation*}
\item the vector $\overline{w}_V := V^T \overline{w} \in \mathbb{R}^K$ which can be written as follows:
\begin{equation*}
\overline{w}_V = \Big( \overline{w}_{V, 1}^{\ T} \quad \overline{w}_{V, 2}^{\ T} \Big)^T
\end{equation*}
where $\overline{w}_{V, 1} \in \mathbb{R}^\sigma$ and $\overline{w}_{V, 2} \in \mathbb{R}^{K-\sigma}$.
\end{enumerate}
\end{definition}
\begin{remark}
The matrix $\widetilde{Q}$ inherits the symmetric property of $Q$. Indeed, since $J_{n, m}^{\ T} = J_{m, n}$ and $Q_{d_1 d_2} = Q_{d_2 d_1}$, we have
\begin{equation*}
\widetilde{Q}^T = \left(
\begin{array}{ccc}
Q_{11} \, J_{K_1, K_1}^{\ T} & \ldots & Q_{D1} \, J_{K_D, K_1}^{\ T} \\
\vdots & & \vdots \\
Q_{1D} \, J_{K_1, K_D}^{\ T} & \ldots & Q_{DD} \, J_{K_D, K_D}^{\ T}
\end{array} \right) = \left(
\begin{array}{ccc}
Q_{11} \, J_{K_1, K_1} & \ldots & Q_{1D} \, J_{K_1, K_D} \\
\vdots & & \vdots \\
Q_{D1} \, J_{K_1, K_D} & \ldots & Q_{DD} \, J_{K_D, K_D}
\end{array} \right) = \widetilde{Q} \; .
\end{equation*}
It follows that the matrices $\overline{Q}$ and $\overline{Q}_V$ are also symmetric.
\end{remark}
The following lemma provides quadratic formulas for the costs $\check{F}: \mathbb{R}^K \longrightarrow \mathbb{R}$ and $\widetilde{F}: \mathbb{R}^\sigma \longrightarrow \mathbb{R}$ in the present setting. They will be used to establish sufficient conditions so that the optimisation problems are convex.
\begin{lemma} \label{lem:ff}
Suppose that the function $f$ is of the form \eqref{eq:quad_f}. Then
\begin{enumerate}
\item we have for all $c \in \mathbb{R}^K$,
\begin{equation*}
\check{F}(c) = c^T \overline{Q} c + \overline{w}^T c + r T \; ;
\end{equation*}
\item we have for all $\widetilde{c}_1 \in \mathbb{R}^\sigma$,
\begin{equation*}
\widetilde{F}(\widetilde{c}_1) = \widetilde{c}_1^T \, \overline{Q}_{V, 11} \, \widetilde{c}_1 + \Big( 2 \, \overline{Q}_{V, 12} \, \widetilde{c}_{2,3} + \overline{w}_{V, 1}^{\ T} \Big)^T \widetilde{c}_1 + \widetilde{c}_{2,3}^{\ T} \, \overline{Q}_{V, 22} \, \widetilde{c}_{2,3} + \overline{w}_{V, 2}^{\ T} \, \widetilde{c}_{2,3} + r T \; ,
\end{equation*}
where $\widetilde{c}_{2,3} := \Big( c_{R_i}^{\ T} V_2 \quad \Gamma^T U \big(S_{A,2}^{-1}\big)^T \Big)^T \in \mathbb{R}^{K - \sigma}$ and $i$ is arbitrarily chosen in $\{1, \dots, I \}$.
\end{enumerate}
\end{lemma}
\begin{proof}
Let $c \in \mathbb{R}^K$ and let $y := \Phi|_{\mathcal{Y}_\mathcal{K}}^{-1} c \in \mathcal{Y}_\mathcal{K}$ be its associated trajectory, which can be represented as follows:
\begin{equation*}
\forall \, d \in \{1, \dots, D\} \qquad y^{(d)} = \sum_{k=1}^{K_d} c_k^{(d)} \, \varphi_k \; .
\end{equation*}
We also remark that each component of the vector
\begin{equation*}
c = \Big( c_1^{(1)}, \dots, c_{K_1}^{(1)}, \; c_1^{(2)}, \dots, c_{K_2}^{(2)}, \dots, \; c_1^{(D)}, \dots, c_{K_D}^{(D)} \Big)^T
\end{equation*}
can be simply described by a single parameter so that we can write $c = (c_1, c_2, \dots, c_K)^T$.
\begin{enumerate}
\item By inserting the preceding representation of $y$ into \eqref{eq:quad_f}, we obtain:
\begin{align*}
f\big( y(t) \big)
& = \sum_{d_1, d_2 = 1}^D \sum_{k_1 = 0}^{K_{d_1}} \sum_{k_2 = 0}^{K_{d_2}} Q_{d_1 d_2} \, c_{k_1}^{(d_1)} c_{k_2}^{(d_2)} \, \varphi_{k_1}(t) \varphi_{k_2}(t) + \sum_{d = 1}^D \sum_{k=0}^{K_d} w_d \, c_k^{(d)} \, \varphi_k(t) + r \; ,
\end{align*}
for all $t \in [0,T]$. The next step of the proof consists in changing the indices of the vectors and matrices. Using the above rewriting of $c$ as well as the matrix $\widetilde{Q}$ and the map $\widetilde{\varphi}$ given in \cref{def:misc} provides the following equality:
\begin{equation*}
\sum_{d_1, d_2 = 1}^D \sum_{k_1 = 0}^{K_{d_1}} \sum_{k_2 = 0}^{K_{d_2}} Q_{d_1 d_2} \, c_{k_1}^{(d_1)} c_{k_2}^{(d_2)} \, \varphi_{k_1}(t) \varphi_{k_2}(t) = \sum_{k_1, k_2 = 1}^{K} \widetilde{Q}_{k_1 k_2} \, c_{k_1} \, c_{k_2} \, \widetilde{\varphi}_{k_1}(t) \widetilde{\varphi}_{k_2}(t) \; .
\end{equation*}
By similar computations, we obtain
\begin{equation*}
\sum_{d = 1}^D \sum_{k=0}^{K_d} w_d \, c_k^{(d)} \, \varphi_k^{(d)}(t) = \sum_{k = 1}^{K} \widetilde{w}_k \, c_k \, \widetilde{\varphi}_k(t) \; ,
\end{equation*}
leading to
\begin{align*}
f\big(y(t) \big) = \sum_{k_1, k_2 = 1}^{K} \widetilde{Q}_{k_1 k_2} \, c_{k_1} \, c_{k_2} \, \widetilde{\varphi}_{k_1}(t) \widetilde{\varphi}_{k_2}(t) + \sum_{k = 1}^{K} \widetilde{w}_k \, c_k \, \widetilde{\varphi}_k(t) + r \; .
\end{align*}
Integrating finally over $[0,T]$ gives
\begin{align*}
\check{F}(c)
& = \int_0^T f\big(y(t)\big) \, dt \\
& = \sum_{k_1, k_2 = 1}^{K} \widetilde{Q}_{k_1 k_2} \int_0^T \widetilde{\varphi}_{k_1}(t) \, \widetilde{\varphi}_{k_2}(t) \, dt \, c_{k_1} \, c_{k_2} + \sum_{k = 1}^{K} \widetilde{w}_k \int_0^T \widetilde{\varphi}_k(t) \, dt \, c_k + r T \\
& = \sum_{k_1, k_2 = 1}^{K} \overline{Q}_{k_1 k_2} \, c_{k_1} \, c_{k_2} + \sum_{k = 1}^{K} \overline{w}_k \, c_k + r T \\
& = c^T \overline{Q} c + \overline{w}^T c + r T \; .
\end{align*}
\item By the definition of $\widetilde{F}$ given in \cref{subsec:modelling}, we have
\begin{equation*}
\widetilde{F}(\widetilde{c}_1) = \check{F}\Big( V \big( \widetilde{c}_1^T \quad \widetilde{c}_{2,3}^{\ T} \big)^T \Big) \; .
\end{equation*}
We use now the result of the preceding point and the definitions of the matrix $\overline{Q}_V$ and the vector $\overline{w}_V$ to obtain
\begin{align*}
\widetilde{F}(\widetilde{c}_1)
& = \big( \widetilde{c}_1^T \quad \widetilde{c}_{2,3}^{\ T} \big) \big( V^T \overline{Q} V \big) \big( \widetilde{c}_1^T \quad \widetilde{c}_{2,3}^{\ T} \big)^T + \big( V^T \overline{w} \big)^T \big( \widetilde{c}_1^T \quad \widetilde{c}_{2,3}^{\ T} \big)^T + r T \\
& = \big( \widetilde{c}_1^T \quad \widetilde{c}_{2,3}^{\ T} \big) \overline{Q}_V \big( \widetilde{c}_1^T \quad \widetilde{c}_{2,3}^{\ T} \big)^T + \overline{w}_V^T \big( \widetilde{c}_1^T \quad \widetilde{c}_{2,3}^{\ T} \big)^T + r T \\
& = \widetilde{c}_1^T \, \overline{Q}_{V, 11} \, \widetilde{c}_1 + \widetilde{c}_1^T \, \overline{Q}_{V, 12} \, \widetilde{c}_{2,3} + \widetilde{c}_{2,3}^{\ T} \, \overline{Q}_{V, 21} \, \widetilde{c}_1 + \widetilde{c}_{2,3}^{\ T} \, \overline{Q}_{V, 22} \, \widetilde{c}_{2,3} \\
& \hspace{1cm} + \overline{w}_{V, 1}^{\ T} \, \widetilde{c}_{1} + \overline{w}_{V, 2}^{\ T} \, \widetilde{c}_{2,3} + r T \; .
\end{align*}
Rearranging the preceding terms and using the fact that $\overline{Q}_V$ is symmetric gives the result.
\end{enumerate}
\end{proof}
\todo{Changer la place de $\kappa$ dans fonction coût ?}\\
In the present setting, the optimisation problem \eqref{eq:opt_proj} is then equivalent to the following quadratic one:
\begin{equation} \label{eq:opt_proj_quad}
\left\{
\begin{array}{l}
\displaystyle \widetilde{c}_1^\star \in \argmin_{\widetilde{c}_1 \in \mathbb{R}^\sigma} \sum_{i=1}^I \omega_i \, \big( \widetilde{c}_1 - \widetilde{c}_{R_i, 1} \big)^T \Lambda_{\Sigma,1}^{-1} \big( \widetilde{c}_1 - \widetilde{c}_{R_i, 1} \big) \\
\hspace{2cm} + \kappa \, \Big( \widetilde{c}_1^T \, \overline{Q}_{V, 11} \, \widetilde{c}_1 + \Big( 2 \, \overline{Q}_{V, 12} \, \widetilde{c}_{2,3} + \overline{w}_{V, 1}^{\ T} \Big)^T \widetilde{c}_1 \Big) \\[2mm]
\widetilde{c}_2 = V_2^T c_{R_i} \\[2mm]
\widetilde{c}_3 = S_{A,2}^{-1} \, U^T \Gamma
\end{array} \; .
\right.
\end{equation}
In the following result, we provide sufficient conditions on the parameter $\kappa > 0$ so that the problem \eqref{eq:opt_proj_quad} is a quadratic program. The proof uses the fact that the symmetric matrix associated with the quadratic objective function is now explicit and given by the sum of two matrices. A perturbation result for matrices is then applied to obtain a bound for $\kappa$ assuring that the symmetric matrix is positive semidefinite.
\begin{theorem} \label{thm:quad_prog}
Let $\rho_1 \geqslant \rho_2 \geqslant \dots \geqslant \rho_\sigma$ and $\lambda_1 \geqslant \lambda_2 \geqslant \dots \geqslant \lambda_K$ be respectively the eigenvalues of the symmetric matrices $\overline{Q}_{V, 11}$ and $\Sigma$.
\begin{enumerate}
\item If $\rho_\sigma \geqslant 0$ then the optimisation problem \eqref{eq:opt_proj_quad} is a quadratic program for any $\kappa > 0$.
\item If $\rho_\sigma < 0$ then the optimisation problem \eqref{eq:opt_proj_quad} is a quadratic program for any $\kappa > 0$ satisfying
\begin{equation*}
\kappa < - \frac{1}{\lambda_1 \, \rho_\sigma} \; .
\end{equation*}
\end{enumerate}
\end{theorem}
\begin{proof}
We first note that all the eigenvalues of the matrix $\Sigma$ are non-negative (because $\Sigma$ is a covariance matrix) and that $\lambda_{\sigma + 1} = \dots = \lambda_K = 0$ (because $\rk \Sigma = \sigma$). In particular, the eigenvalue $\lambda_1$ is positive.\\
Standard calculations show that the symmetric matrix associated with the quadratic objective function of the problem \eqref{eq:opt_proj_quad} is given by
\begin{equation*}
M(\kappa) := \kappa \, \overline{Q}_{V, 11} + \Lambda_{\Sigma, 1}^{-1} \in \mathbb{R}^{\sigma \times \sigma} \; .
\end{equation*}
Let $\mu_1(\kappa) \geqslant \mu_2(\kappa) \geqslant \dots \geqslant \mu_\sigma(\kappa)$ denote the eigenvalues of $M(\kappa)$. Our goal is to prove that $\mu_\sigma(\kappa)$ is non-negative to assure that $M$ is positive semidefinite. Since $M(\kappa)$ can be interpreted as a perturbed version of $\Lambda_{\Sigma, 1}^{-1}$, we can apply Weyl's inequality (see for instance \citet{wang201965}) which implies
\begin{equation*}
\kappa \, \rho_\sigma + \frac{1}{\lambda_1} \leqslant \mu_\sigma(\kappa) \; .
\end{equation*}
We distinguish now the two cases.
\begin{enumerate}
\item In the case where $\rho_\sigma \geqslant 0$ (\emph{i.e.} the matrix $\overline{Q}_{V, 11}$ is positive semidefinite), then $\mu_\sigma(\kappa)$ is positive for any $\kappa > 0$.
\item If $\rho_\sigma < 0$ and $\kappa \leqslant - \frac{1}{\lambda_1 \, \rho_\sigma}$ then $\mu_\sigma(\kappa)$ is non-negative.
\end{enumerate}
\end{proof}
For the sake of completeness, we finish by rewriting the problem \eqref{eq:opt_proj_quad} as a quadratic optimisation problem in $\mathcal{V}_1 \subset \mathbb{R}^K$ with affine constraints function.
\begin{proposition} \label{prop:quad_min_pb}
Suppose that the function $f$ is of the form \eqref{eq:quad_f}. Then the optimisation problem \eqref{eq:opt_proj_quad} is equivalent to the following one:
\begin{equation} \label{eq:quad_min_pb}
c^\star \in \argmin_{c \in \mathcal{V}_1} c^T \Big( \kappa \, \overline{Q} + \Sigma^\dagger \Big) c + \left( \kappa \, \overline{w} - 2 \sum_{i=1}^I \omega_i \, \Sigma^\dagger c_{R_i} \right)^{\hspace{-5pt}T} c \; .
\end{equation}
\end{proposition}
\begin{proof}
As explained above and according to \cref{prop:opt_equiv}, the problem \eqref{eq:opt_proj_quad} is equivalent to the problem \eqref{eq:opt_proj_2} in the present setting. Hence it is sufficient to show that the objective functions $g_1, g_2: \mathbb{R}^K \longrightarrow \mathbb{R}$ defined as
\begin{itemize}
\item $\displaystyle g_1(c) := \sum_{i=1}^I \omega_i \, \big( c - c_{R_i} \big)^T \Sigma^\dagger \big( c - c_{R_i} \big) + \kappa \, \check{F}(c)$ ,
\item $\displaystyle g_2(c) := c^T \Big( \kappa \, \overline{Q} + \Sigma^\dagger \Big) c + \left( \kappa \, \overline{w} - 2 \sum_{i=1}^I \omega_i \, \Sigma^\dagger c_{R_i} \right)^{\hspace{-5pt}T} c$ .
\end{itemize}
have the same minima. Firstly we have by standard calculations,
\begin{align*}
\sum_{i=1}^I \omega_i \, \big( c - c_{R_i} \big)^T \Sigma^\dagger \big( c - c_{R_i} \big)
& = \sum_{i=1}^I \omega_i \, c^T \Sigma^\dagger c - 2 \sum_{i=1}^I \omega_i \, c_{R_i}^{\ T} \Sigma^\dagger c + \sum_{i=1}^I \omega_i \, c_{R_i}^{\ T} \Sigma^\dagger c_{R_i} \\
& = c^T \Sigma^\dagger c - \left( 2 \sum_{i=1}^I \omega_i \, \Sigma^\dagger c_{R_i} \right)^{\hspace{-5pt}T} c + \sum_{i=1}^I \omega_i \, c_{R_i}^{\ T} \Sigma^\dagger c_{R_i} \; ,
\end{align*}
for any $c \in \mathbb{R}^K$, where we have used $\sum_{i=1}^I \omega_i = 1$. Combining now this equality with \cref{lem:ff} implies
\begin{align*}
g_1(c)
& = c^T \Sigma^\dagger c - \left( 2 \sum_{i=1}^I \omega_i \, \Sigma^\dagger c_{R_i} \right)^{\hspace{-5pt}T} c + \sum_{i=1}^I \omega_i \, c_{R_i}^{\ T} \Sigma^\dagger c_{R_i} + \kappa \, \Big( c^T \overline{Q} c + \overline{w}^T c + r T \Big) \\
& = c^T \Big( \kappa \, \overline{Q} + \Sigma^\dagger \Big) c + \left( \kappa \, \overline{w} - 2 \sum_{i=1}^I \omega_i \, \Sigma^\dagger c_{R_i} \right)^{\hspace{-5pt}T} c + \sum_{i=1}^I \omega_i \, c_{R_i}^{\ T} \Sigma^\dagger c_{R_i} + \kappa \, r T \\
& = g_2(c) + \sum_{i=1}^I \omega_i \, c_{R_i}^{\ T} \Sigma^\dagger c_{R_i} + \kappa \, r T \; .
\end{align*}
Since the two last terms of the last right-hand side do not depend on $c$, we deduce that the objective functions $g_1$ and $g_2$ have the same minima.
\end{proof}
\subsection{Modelling} \label{subsec:appli_aero_model}
Here the trajectories are supposed to be in a vertical plane and are defined by the altitude $h$, the Mach number $\mathrm{M}$ and the engines rotational speed $\mathrm{N1}$ (expressed as a percentage of a maximal value). Hence a trajectory $y$ in this setting is a continuous $\mathbb{R}^3$-valued map defined on $[0,T]$, where $T$ is a maximal climb duration fixed by the user. Hence we have
\begin{equation*}
\forall \, t \in [0,T] \qquad y(t) := \big( h(t), \mathrm{M}(t), \mathrm{N1}(t) \big) \; .
\end{equation*}
The quantity to minimise is the total fuel consumption $\mathrm{TFC}: \mathcal{C}\big( [0,T], \mathbb{R}^3 \big) \longrightarrow \mathbb{R}_+$ which is defined via the fuel flow $\mathrm{FF}: \mathbb{R}^3 \longrightarrow \mathbb{R}_+$ as follows\footnote{In the notation of \cref{subsec:quad_model}, $\mathrm{FF}$ and $\mathrm{TFC}$ play respectively the role of $f$ and $F$.}:
\begin{equation*}
\mathrm{TFC}(y) := \int_0^T \mathrm{FF}\big(y(t)\big) \, dt \; .
\end{equation*}
Regarding the endpoints conditions, we require the trajectory to start at the altitude $h_0$ with Mach number $\mathrm{M}_0$ and to end at the altitude $h_T$ with Mach number $\mathrm{M}_T$. In particular, the reference trajectories we use have to verify these conditions.
We consider also additional constraints which are conventional in the aeronautic setting:
\begin{itemize}
\item The rate of climb, \emph{i.e.} the time-derivative of the altitude, has to be upper bounded by a given maximal value $\gamma_{max}$ during the whole climb;
\item The Mach number should not exceed a certain value called the maximum operational Mach ($\mathrm{MMO}$).
\end{itemize}
The final time of the climb is given by $T^\star \in [0,T]$ which is the first time where the aircraft reaches $h_T$ with Mach number $\mathrm{M}_T$.
Finally we mention that the fuel flow model $\mathrm{FF}$ is here estimated. To do so, we exploit the reference trajectories which contain recorded altitude, Mach number, engines power and fuel flow for each second of the flight. Having access to these data, we are in position to fit a statistical model. Following the numerical results in \cite{dewez2020industrywide} which show that polynomials can accurately model aeronautic variables, we consider a polynomial model of degree 2 for the fuel flow. In particular the requirements for the cost function in the current version of \texttt{PyRotor} are fulfilled. The prediction accuracy of the resulting estimated model is assessed in the following subsection.
\subsection{Numerical results}
We present now numerical results based on real flight data for the above aeronautic problem. Here we have access to 2,162 recorded short and medium-haul flights performed by the same narrow-body airliner type, provided by a partner airline. In particular they can not be publicly released for commercial reasons. The data is here recorded by the Quick Access Recorder (QAR).
Before considering the optimisation setting, we estimate a fuel flow model specific to the climb phase and to the considered airliner type. To do so we extract the signals of the four variables of interest (altitude, Mach number, engines rotational speed and fuel flow) and keep the observations from the take-off to the beginning of the cruise without level-off phases. Smoothing splines are then applied to the raw signals to remove the noise. We sample each 5 seconds to reduce the data set size without impacting strongly the accuracy of the resulting models. At the end, we obtain 494,039 observations which are randomly split into training and test sets to fit a polynomial model of degree 2 using the \texttt{scikit-learn} library. The RMSE and MAPE values of this model on the test set are respectively equal to $3.64 \times 10^{-2}$ kg.s$^{-1}$ and 1.73\%.
Regarding the optimisation, we are interested in climb phases from 3,000~ft to 38,000~ft. We mention that we remove lower altitudes because operational procedures constraint heavily the trajectory during the very beginning of the climb. Further the initial and final Mach numbers are required to be equal to 0.3 and 0.78. It is noteworthy that the optimisation solvers used in \texttt{PyRotor} allow linear inequality conditions, permitting to slightly relax the endpoints conditions. Here we tolerate an error of 100~ft for the altitude and an error of 0.01 for the Mach number. The initial and final $\mathrm{N1}$ values are let unconstrained. Finally the $\mathrm{MMO}$ and $\gamma_{max}$ are respectively set to 0.82 and 3,600~ft.min$^{-1}$.
The reference trajectories are given by 48 recorded flights which satisfy the above climb endpoints conditions among the 2,162 available ones. All these selected flights are used to estimate the covariance matrix involved in the optimisation problem. On the other hand, we use only the 5 most fuel-efficient flights in the objective function to focus on a domain containing the most efficient recorded flights. Further the maximal duration $T$ is here fixed to the duration of the longest climb among the 5 most fuel-efficient ones we use.
Legendre polynomials are used as the functional basis spanning the space in which lies the trajectories. Since we consider narrow-body airliners, polynomials are expected to be relevant to describe the slow variations of such aircrafts. Here the dimensions associated with the altitude, the Mach number and the engines power are given respectively by 4, 10 and 6. The reference vectors $c_{R_i}$ are then computed using the formula \eqref{eq:def_c}. At the end, we amount to solving a constrained optimisation problem in a space of dimension 20.
We are then in position to apply the optimisation method developed in \cref{sec:opti} using the \texttt{PyRotor} library. First of all a relevant value for $\nu_{max} > 0$ has to be fixed. In order to propose a realistic optimised climb, we choose a $\nu_{max}$ relatively small so that the optimised climb remains close to the reference ones. In particular, the quadratic objective function in \eqref{eq:opt_proj_3} turns out to be convex for all $\nu \in [0,\nu_{max}]$ permitting to use the quadratic programming solver from \texttt{CVXOPT} software imported in \texttt{PyRotor}. The preprocessing of the reference trajectories and the optimisation steps have been executed 100 times using \texttt{PyRotor} on an Intel Core i7 6 cores running at 2.2~GHz. The mean of the execution time for both steps is equal to 3.76~s with standard deviation 0.11~s, illustrating that the library is time-efficient in this setting.
A plot of the optimised trajectory obtained using \texttt{PyRotor} is given in \cref{fig:opti_climb}. We observe that the optimised trajectory seeks to reach the maximum altitude in the minimum amount of time; this is in accordance with the existing literature (see for instance \citet{dalmau2014} and references therein). In particular, the duration $T^\star$ is equal to 1,033 seconds which is actually slightly shorter than the reference durations. We note also that the optimised Mach number shares a very similar pattern with the references. On the other hand, the optimised engines rotational speed tends to slowly decrease until the cruise regime before reaching the top of climb. This is not the case for the reference engines speed which falls to the cruise regime just after reaching the final altitude. Most of the savings seem to be achieved in these last moments of the climb. At last but not least, the optimised trajectory presents a realistic pattern inherited from the reference trajectories.
For a quantitative comparison, we refer to \cref{tab:savings} which provides statistical information on the fuel savings. The mean savings 16.54\% together with the fact that the optimised trajectory verifies the additional constraints show that these first results are promising, motivating further studies. For instance one could model environmental conditions or take into account Air Traffic Control constraints for more realistic modellings.
\begin{table}
\caption{\label{tab:savings}Statistical description of the fuel savings of the optimised trajectory -- The savings are compared with the 48 recorded flights satisfying the present endpoints and the total consumption of the optimised trajectory is estimated by using the statistical model for the fuel flow - $Q_1$, $Q_2$ and $Q_3$ refer to the first, second and third quartiles.}
\centering
\fbox{%
\begin{tabular*}{\textwidth}{@{\extracolsep{\fill}} l*{6}{c}r}
& Mean & Std & Min & $Q_1$ & $Q_2$ & $Q_3$ & Max \\
\hline
Fuel savings [kg] & 260.38 & 86.21 & 71.79 & 202.40 & 261.87 & 330.32 & 393.73 \\
Percentage [\%] & 16.54 & 4.73 & 5.27 & 13.56 & 16.88 & 20.39 & 23.39
\end{tabular*}}
\end{table}
\begin{figure}
\centering
\includegraphics[width=\textwidth]{figures/altitude_opti.png}
\includegraphics[width=\textwidth]{figures/mach_opti.png}
\includegraphics[width=\textwidth]{figures/n1_opti.png}
\caption{Optimised and reference altitudes, Mach numbers and engines rotational speeds -- The optimised trajectory is represented by the blue curves.}
\label{fig:opti_climb}
\end{figure}
\subsection{Modelling} \label{subsec:appli_phys_nautic}
To model this problem, we suppose without loss of generality that the trajectories are defined on the (time-)interval $[0,1]$ and we let $V: \mathbb{R}^D \longrightarrow \mathbb{R}^D$ denote a vector field. Furthermore the trajectories are assumed here to be continuously differentiable, \emph{i.e.} they belong to $\mathcal{C}^1\big([0,1], \mathbb{R}^D \big)$. The work of $V$ along a trajectory $y \in \mathcal{C}^1\big( [0,1], \mathbb{R}^D \big)$ is defined as
\begin{equation*}
W(y, \dot{y}) := \int_0^1 V\big(y(t)\big)^T \dot{y}(t) \, dt \; ;
\end{equation*}
here $\dot{y}$ denotes the derivative of $y$ with respect to the independent variable $t$. Moreover using Hamilton's principle in Lagrangian mechanics, it can be shown that the trajectory with constant velocity (\emph{i.e.} a straight line travelled at constant speed) is the minimum of the following functional,
\begin{equation*}
J(\dot{y}) = \int_0^1 \big\| \dot{y}(t) \big\|_2^2 \, dt \; ,
\end{equation*}
where the starting and ending points of $y$ are fixed and different. This functional can be then used to control the travelled distance. It follows that minimising the cost function
\begin{equation*}
F_\alpha(y, \dot{y}) := \alpha J(\dot{y}) - W(y, \dot{y}) = \int_0^1 \alpha \big\| \dot{y}(t) \big\|_2^2 - V\big(y(t)\big)^T \dot{y}(t) \, dt \; ,
\end{equation*}
where $\alpha \geqslant 0$ is arbitrarily chosen, is expected to lead to an optimised trajectory reflecting a trade-off between maximising the work and minimising the distance. Further we require the trajectory to stay in the hypercube $[0,1]^D$ and to start and to end respectively at $y_0 \in [0,1]^D$ and $y_1 \in [0,1]^D$.
Now we remark that the above cost function involves the (time-)derivative $\dot{y}$. So one has to derive a formula permitting to compute the derivative of any trajectory $y = \Phi|_{\mathcal{Y}_\mathcal{K}}^{-1} c \in \mathcal{Y}_\mathcal{K}$ from its associated vector $c \in \mathbb{R}^K$, especially to compute $\check{F}(c)$. For instance, this can be easily achieved by assuming that each element of the functional basis is continuously differentiable. Indeed we can differentiate in this case any $y \in \mathcal{Y}_\mathcal{K}$:
\begin{equation*}
\forall \, d = 1, \dots, D \qquad \dot{y}^{(d)} = \sum_{k=1}^{K_d} c_k^{(d)} \dot{\varphi}_k = \left( \frac{d}{dt} \Phi|_{\mathcal{Y}_\mathcal{K}}^{-1} c \right)^{(d)} \; .
\end{equation*}
We deduce then the following formula for $\check{F}(c)$ in the present setting:
\begin{equation*}
\check{F}(c) := F_\alpha\left( \Phi|_{\mathcal{Y}_\mathcal{K}}^{-1} c, \frac{d}{dt} \Phi|_{\mathcal{Y}_\mathcal{K}}^{-1} c \right) \; .
\end{equation*}
Here the vector $c$ contains information on both position and velocity, permitting especially to keep the problem dimension unchanged. To finish, let us remark that it is possible to make the above formula for $\check{F}$ explicit with respect to $c$ in certain settings. For instance it is possible to derive an explicit quadratic formula for $\check{F}(c)$ when the integrand defining $F_\alpha$ is quadratic with respect to $y(t)$ and $\dot{y}(t)$; this formula is implemented in \texttt{PyRotor} and the arguments to obtain it are similar to those proving \cref{prop:quad_min_pb}.
\subsection{Numerical results}
Numerical results based on randomly generated data for the above physical application are presented in this section.
First of all we consider trajectories with two components $y^{(1)}$ and $y^{(2)}$ lying in the square $[0,1]^2$ for the sake of simplicity. We set the starting and ending points as follows:
\begin{equation*}
y^{(1)}(0) = 0.111 \quad , \quad y^{(2)}(0) = 0.926 \quad , \quad y^{(1)}(1) = 0.912 \quad , \quad y^{(2)}(1) = 0.211
\end{equation*}
with a tolerated error $1 \times 10^{-4}$, and the vector field $V: \mathbb{R}^2 \longrightarrow \mathbb{R}^2$ is here defined by
\begin{equation*}
V\Big(x^{(1)}, x^{(2)}\Big) = \Big( 0, x^{(1)} \Big)^T \; .
\end{equation*}
Given the above endpoints and the vector field, we observe that the force modelled by $V$ will be in average a resistance force to the motion. Indeed the force is oriented toward the top of the square while the moving point has to go downward. Further let us note that the integrand of the cost function $F_\alpha$ in the present setting is actually quadratic with respect to $y(t)$ and $\dot{y}(t)$, so that an explicit quadratic formula for $\check{F}(c)$ implemented in \texttt{PyRotor} is available.
Here the reference trajectories are obtained through a random generation process. To do so, we define an arbitrarily trajectory $y_R$ verifying the endpoints conditions and we compute its associated vector $c_R$; Legendre polynomials are once again used and the dimensions of $y^{(1)}$ and $y^{(2)}$ are here set to 4 and 6. Let us note that $y_R$ is designed in such a way that it has a relevant pattern but not the optimal one. Then we construct a set of reference trajectories by adding centered Gaussian noises to $c_R$. It is noteworthy that the noise is generated in such a way that it belongs to the null space of the matrix $A$ describing the endpoints conditions; the resulting noised trajectories satisfy then these conditions. Further the trajectories which go out of the square $[0,1]^2$ are not kept. At the end, we get 122 generated reference trajectories assumed to be realistic in this setting, each of them containing 81 time observations. Among these reference trajectories, we use the 10 most efficient ones with respect to the cost $F_\alpha$.
In the present example, we set a $\nu_{max}$ relatively large to explore a large domain around the reference trajectories. In this case, the objective function of the optimisation problem \eqref{eq:opt_proj_3} may be not convex even if it is still quadratic. So we make use of the generic optimisation solver \texttt{minimize(method='trust-constr')} imported in \texttt{PyRotor}. Regarding the execution time, we have randomly and uniformly generated 100 values in the interval $[0,10]$ for the parameter $\alpha$ and executed \texttt{PyRotor} for each of them. The mean of \texttt{PyRotor} execution time is 0.44~s with standard deviation 0.03~s on an Intel Core i7 6 cores running at 2.2~GHz.
In \cref{fig:opti_work_velocity}, we plot 4 optimised trajectories associated with different values of $\alpha$: 0, 0.35, 1 and 10. As expected the trajectory associated with the largest value of $\alpha$ gives the most straight trajectory while the most curvy one is associated with $\alpha = 0$. In particular, the latter tends to move to the left at the beginning where the force $V$ is the smallest before going to the ending point in a quasi-straightforward way so that the force is perpendicular to the motion. This example illustrates especially that our optimisation approach may lead to optimised trajectories which differ from the reference ones to reduce more the cost.
A quantitative comparison in terms of work gains for different values of $\alpha$ is provided in \cref{tab:max_work}. The results confirm the above observations on the curves and show that a right value for $\alpha$ has to be fixed depending on the setting.
\begin{figure}[t]
\centering
\includegraphics[width=\textwidth]{figures/opti_work_velocity.png}
\caption{Optimised trajectories in the square $[0,1]^2$ for $\alpha \in \{0, 0.35, 1, 10 \}$ -- Optimised and reference trajectories are respectively given by plain and dotted curves -- Coloured dots indicate the power value of the force at different points of the optimised trajectories and the bar shows the scale -- Red arrows represent the pattern of the vector field $V$.}
\label{fig:opti_work_velocity}
\end{figure}
\begin{table}
\caption{\label{tab:max_work}Statistical description of the work gains in percentage for $\alpha \in \{0, 0.35, 1, 10 \}$ -- The values have been computed by using the 122 available reference trajectories -- Negative percentages indicate that no work gains have been obtained -- $Q_1$, $Q_2$ and $Q_3$ refer to the first, second and third quartiles.}
\centering
\fbox{
\begin{tabular*}{\textwidth}{@{\extracolsep{\fill}} l*{6}{c}r}
& Mean & Std & Min & $Q_1$ & $Q_2$ & $Q_3$ & Max \\
\hline
$\alpha = 0$ & 73.43 & 2.36 & 68.63 & 71.90 & 73.25 & 74.67 & 80.69 \\
$\alpha = 0.35$ & 45.88 & 4.81 & 36.09 & 42.75 & 45.49 & 48.39 & 60.66 \\
$\alpha = 1$ & $-6.12$ & 9.43 & $-25.31$ & $-12.26$ & $-6.88$ & $-1.20$ & 22.87 \\
$\alpha = 10$ & $-34.54$ & 11.96 & $-58.87$ & $-42.32$ & $-35.50$ & $-28.30$ & 2.22 \\
\end{tabular*}}
\end{table}
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\begin{document}
\title{An end-to-end data-driven optimisation framework for constrained trajectories}
\author{\textbf{Florent Dewez} \\ [2ex]
Inria, Lille - Nord Europe Research centre, France \\\\
\textbf{Benjamin Guedj} \\ [2ex]
Inria, Lille - Nord Europe Research centre, France \\
\emph{and} Centre for Artificial Intelligence,\\
Department of Computer Science,\\
University College London, United Kingdom \\\\
\textbf{Arthur Talpaert} \\ [2ex]
Inria, Lille - Nord Europe Research centre, France \\\\
\textbf{Vincent Vandewalle} \\ [2ex]
Inria, Lille - Nord Europe Research centre\\
\emph{and} Université de Lille, France\\
}
\date{}
\maketitle
\input{abstract}
\tableofcontents
\section{Introduction} \label{sec:intro}
\input{intro.tex}
\section{An end-to-end optimisation workflow based on observed trajectories} \label{sec:opti}
\input{optimisation.tex}
\section{The Python library Pyrotor} \label{sec:pyrotor}
\input{pyrotor}
\section{Application 1: trajectory optimisation for fuel-efficient aircrafts} \label{sec:appli_aero}
\input{appli_aero}
\section{Application 2: trajectory optimisation to maximise work of a force field} \label{sec:appli_nautic}
\input{appli_nautic}
\section{Conclusion / Discussion}
\input{conclusion}
\section*{Acknowledgements}
\input{acknowledgments}
\subsection{Fuel-efficient trajectories for aircraft} \label{subsec:appli_aero_model}
\todo{Notations for fuel flow and consumption ?}
In this subsection, we consider the aeronautic problem of reducing the total fuel consumption of an aircraft during the climb phase. This example illustrates the key role played by the reference trajectories since we are able to obtain very promising optimised trajectories thanks to a simple modelling involving few constraints; we refer to \cref{sec:appli_aero} for the numerical results. This motivates further studies on this problem using our methodology. For instance one could model environmental conditions or take into account Air Traffic Control constraints for more realistic modellings.
Here trajectories are supposed to be in a vertical plane and are defined by the altitude $h$, the Mach number $\mathrm{M}$ and the engines rotational speed $\mathrm{N1}$ (expressed as a percentage of a maximal value). Hence a trajectory $y$ in this setting is a continuous $\mathbb{R}^3$-valued function defined on $[0,T]$, where $T$ is a maximal climb duration fixed by the user. Hence we have
\begin{equation*}
\forall \, t \in [0,T] \qquad y(t) := \big( h(t), \mathrm{M}(t), \mathrm{N1}(t) \big) \; .
\end{equation*}
The quantity to minimise is the total fuel consumption $\mathrm{TFC}: \mathcal{C}\big( [0,T], \mathbb{R}^3 \big) \longrightarrow \mathbb{R}_+$ which is defined via the fuel flow $\mathrm{FF}: \mathbb{R}^3 \longrightarrow \mathbb{R}_+$ as follows:
\begin{equation*}
\mathrm{TFC}(y) := \int_0^T \mathrm{FF}\big(y(t)\big) \, dt \; .
\end{equation*}
\todo{footnote to refer to preceding notations}
Regarding the endpoints conditions, we require the trajectory to start at the altitude $h_0$ with Mach number $\mathrm{M}_0$ and to end at the altitude $h_T$ with Mach number $\mathrm{M}_T$. In particular, the reference trajectories we use have to verify these conditions.
\todo{itemize}
We consider also two additional constraints which are conventional in the aeronautic setting. The rate of climb, \emph{i.e.} the time-derivative of the altitude, has to be upper bounded by a given maximal value $\gamma_{max}$ during the whole climb and the Mach number should not exceed the maximum operational Mach ($\mathrm{MMO}$). We also consider a third constraint which permits to determine the duration of the climb: if we let denote $T^\star \in [0,T]$ the first time where the aircraft reaches $h_T$, then we require $\mathrm{M}(T^\star)$ to be equal to $\mathrm{M}_T$. Note that this constraint is satisfied by the reference trajectories.
Finally we mention that the fuel flow model $\mathrm{FF}$ is here estimated. To do so, we exploit the reference trajectories which contain recorded altitude, Mach number, engines power and fuel flow for each second of the flight. Having access to these data, we are in position to fit a statistical model. Here we consider a polynomial model of degree 2 to obtain at the end a quadratic program as explained in \cref{subsec:quad_model}. The prediction accuracy of the resulting estimated model is assessed in \cref{sec:appli_aero}.
\subsection{Maximising the work of a physical force along a path} \label{subsec:appli_phys_nautic}
Here we consider the following generic example: given a moving point in a force field, find a trajectory starting and ending at two different given points which maximises the work of the force along the trajectory while minimising the travelled distance. For instance, \todo{motion of point in wind -- nautic}. This second example demonstrates that our generic optimisation approach is flexible enough to take into account derivatives of trajectories and hence to cover dynamics settings.
To model this problem, we suppose without loss of generality that the trajectories are defined on the (time-)interval $[0,1]$ and we let $V: \mathbb{R}^D \longrightarrow \mathbb{R}$ denote a vector field. Furthermore the trajectories are assumed here to be continuously differentiable, \emph{i.e.} they belong to $\mathcal{C}^1\big([0,1], \mathbb{R}^D \big)$. The work of $V$ along a trajectory $y \in \mathcal{C}^1\big( [0,1], \mathbb{R}^D \big)$ is defined as
\begin{equation*}
W(y, \dot{y}) := \int_0^1 V\big(y(t)\big)^T \dot{y}(t) \, dt \; ;
\end{equation*}
here $\dot{y}$ denotes the derivative of $y$ with respect to the independent variable $t$. Moreover using Hamilton's principle in Lagrangian mechanics, it can be shown that the trajectory with constant velocity is the minimum of the following functional,
\begin{equation*}
J(\dot{y}) = \int_0^1 \big\| \dot{y}(t) \big\|_2^2 \, dt \; ,
\end{equation*}
where the starting and ending points of $y$ are fixed and different. This functional can be then used to control the velocity of the moving point. It follows that minimising the cost function
\begin{equation*}
F_\alpha(y, \dot{y}) := \alpha J(\dot{y}) - W(y, \dot{y}) = \int_0^1 \alpha \big\| \dot{y}(t) \big\|_2^2 - V\big(y(t)\big)^T \dot{y}(t) \, dt \; ,
\end{equation*}
where $\alpha \geqslant 0$ is arbitrarily chosen, is expected to lead to an optimised trajectory reflecting a trade-off between maximising the work and minimising the travelled distance. Further we require the trajectory to stay in the hypercube $[0,1]^D$ and to start and to end respectively at $y_0 \in [0,1]^D$ and $y_1 \in [0,1]^D$.
Now we remark that the above cost function involves the (time-)derivative $\dot{y}$. So one has to derive a formula permitting to compute the derivative of any trajectory $y = \Phi|_{\mathcal{Y}_\mathcal{K}}^{-1} c \in \mathcal{Y}_\mathcal{K}$ from its associated vector $c \in \mathbb{R}^K$, especially to compute the cost $\check{F}(c)$ (introduced in \cref{def:cost} and involved in the generic optimisation problem \eqref{eq:opt_proj_2}). For instance, this can be easily achieved by assuming that each element of the functional basis is continuously differentiable. Indeed we can differentiate in this case any $y \in \mathcal{Y}_\mathcal{K}$ as follows:
\begin{equation*}
\forall \, d = 1, \dots, D \qquad \dot{y}^{(d)} = \left( \frac{d}{dt} \Phi|_{\mathcal{Y}_\mathcal{K}}^{-1} c \right)^{(d)} = \sum_{k=1}^{K_d} c_k^{(d)} \dot{\varphi}_k \; .
\end{equation*}
We deduce then the following formula for $\check{F}(c)$ in the present setting:
\begin{equation*}
\check{F}(c) := F_\alpha\left( \Phi|_{\mathcal{Y}_\mathcal{K}}^{-1} c, \frac{d}{dt} \Phi|_{\mathcal{Y}_\mathcal{K}}^{-1} c \right) \; .
\end{equation*}
Here the vector $c$ contains information on both position and velocity, permitting especially to keep the problem dimension unchanged. To finish, let us remark that it is possible to make the above formula for $\check{F}(c)$ explicit with respect to $c$ for certain cost functions $F_\alpha$. For instance this is the case when $F_\alpha$ is quadratic leading to a quadratic function $\check{F}$.
\subsection{Admissible trajectories modelling}
\label{sec:2-1}
We start with definitions.
\begin{definition}[Trajectory]
Let $T > 0$ be a real number and let $D \geqslant 1$ be an integer. Any continuous $\mathbb{R}^D$-valued map $y$ defined on $[0,T]$, \emph{i.e.} $y \in \mathcal{C}\big([0,T], \mathbb{R}^D\big)$, is called a \emph{trajectory} over the time interval $[0,T]$. The $d$-th component of a trajectory $y$ will be denoted by $y^{(d)}$. As such, a trajectory is at least a continuous map on a finite interval.
\end{definition}
When optimising a trajectory with respect to a given criterion, the initial and final states are often constrained, that is to say the optimisation is performed in an affine subspace modelling these endpoints conditions. This subspace is introduced just below.
\begin{definition}[Endpoints conditions] \label{def:endpoints}
Let $y_0, y_T \in \mathbb{R}^D$. We define the set $\mathcal{D}(y_0, y_T) \subset \mathcal{C}\big([0,T], \mathbb{R}^D\big)$ as follows:
\begin{align*}
y \in \mathcal{D}(y_0, y_T) \qquad
& \Longleftrightarrow \qquad \left\{
\begin{array}{l}
y(0) = y_0 \\[2mm]
y(T) = y_T
\end{array} \right. \; .
\end{align*}
\end{definition}
In many applications, the trajectories have to satisfy some additional constraints defined by a set of (nonlinear) functions. For instance these functions may model physical or user-defined constraints. In this paper, this set is not intended to include the dynamics of the system. We define now the set of trajectories verifying such additional constraints.
\begin{definition}[Additional constraints] \label{def:constraints}
For $\ell = 1,\dots,L$, let $g_\ell$ be a real-valued function defined on $\mathbb{R}^D$. We define the set $\mathcal{G} \subset \mathcal{C}\big([0,T], \mathbb{R}^D\big)$ as the set of trajectories over $[0,T]$ satisfying the following $L$ inequality constraints given by the functions $g_\ell$, \emph{i.e.}
\begin{equation*}
y \in \mathcal{G} \qquad \Longleftrightarrow \qquad \forall \, \ell = 1, \dots, L \quad \forall \, t \in [0,T] \qquad g_\ell\big(y(t)\big) \leqslant 0 \; .
\end{equation*}
\end{definition}
To finish we introduce the set of admissible trajectories which satisfy both the endpoints conditions and the additional constraints.
\begin{definition}[Admissible trajectory] \label{def:admissible}
We define the set $\mathcal{A}_\mathcal{G}(y_0, y_T) \subset \mathcal{C}\big([0,T], \mathbb{R}^D\big)$ as follows:
\begin{equation*}
\mathcal{A}_\mathcal{G}(y_0, y_T) := \mathcal{D}(y_0, y_T) \cap \mathcal{G} \; .
\end{equation*}
Any element of $\mathcal{A}_\mathcal{G}(y_0, y_T)$ will be called an \emph{admissible trajectory}.
\end{definition}
\subsection{Projection for a finite-dimensional optimisation problem}
\label{sec:2-2}
In our approach, a theoretical optimisation problem in a finite-dimensional space is desired to reduce the inherent complexity of the problem. This can be achieved by decomposing the trajectories on a finite number of basis functions. While raw signals are unlikely to be described by a small number of parameters, this is not the case for smoothed versions of these signals which capture the important patterns. In particular, given a family of smoothed observed trajectories, one may suppose that there exists a basis such that the projection error on a certain number of basis functions of any trajectory is negligible.
From now on, the trajectories we consider are assumed to belong to a space spanned by a finite number of basis functions. For the sake of simplicity, we assume in addition that all the components of the trajectories can be decomposed on the same basis but with different dimensions. Extension to different bases is straightforward and does not change our findings but burdens the notation.
\begin{definition} \label{def:projected_set}
Let $\{\varphi_k\}_{k=1}^{+\infty}$ be an orthonormal basis of $L^2\big([0,T], \mathbb{R} \big)$ with respect to the inner product
\begin{equation*}
\langle f, g \rangle = \int_0^T f(t) \, g(t) \, dt \; ,
\end{equation*}
such that each $\varphi_k$ is continuous on $[0,T]$ and let $\mathcal{K} := \{K_d\}_{d=1}^D$ be a sequence of integers with $K := \sum_{d = 1}^D K_d$.
We define the space of projected trajectories $\mathcal{Y}_\mathcal{K}(0,T) \subset \mathcal{C}\big([0,T], \mathbb{R}^D\big)$ over $[0,T]$ as
\begin{equation*}
\mathcal{Y}_\mathcal{K}(0,T) := \prod_{d = 1}^D \text{span} \left\{ \varphi_{k} \right\}_{k = 1}^{K_d} \; .
\end{equation*}
If there is no risk of confusion, we write $\mathcal{Y}_\mathcal{K} := \mathcal{Y}_\mathcal{K}(0,T)$ for the sake of readability.
\end{definition}
\begin{remark}
From the above definition, any projected trajectory $y \in \mathcal{Y}_\mathcal{K}$ is associated with a unique vector
\begin{equation*}
c = \Big( c_1^{(1)}, \dots, c_{K_1}^{(1)}, \; c_1^{(2)}, \dots, c_{K_2}^{(2)}, \dots, \; c_1^{(D)}, \dots, c_{K_D}^{(D)} \Big)^T \in \mathbb{R}^K
\end{equation*}
defined by
\begin{equation} \label{eq:def_c}
c_k^{(d)} := \big\langle y^{(d)}, \varphi_k \big\rangle = \int_0^T y^{(d)}(t) \, \varphi_{k}(t) \, dt \; .
\end{equation}
In other words, the vector $c$ is the image of the trajectory $y$ by the projection operator $\Phi : \mathcal{C}\big([0,T], \mathbb{R}^D \big) \longrightarrow \mathbb{R}^K$ defined by $\Phi y := c$, whose restriction $\Phi|_{\mathcal{Y}_\mathcal{K}}$ is bijective (as the Cartesian product of bijective operators). In particular, the spaces $\mathcal{Y}_\mathcal{K}$ and $\mathbb{R}^K$ are isomorphic, \emph{i.e.} $\mathcal{Y}_\mathcal{K} \simeq \mathbb{R}^K$.
\end{remark}
Regarding the endpoints conditions introduced in \cref{def:endpoints}, we prove in the following result that satisfying these conditions is equivalent to satisfying a linear system for a projected trajectory.
\begin{proposition} \label{prop:linear_const}
A trajectory $y \in \mathcal{Y}_{\mathcal{K}}$ belongs to $\mathcal{D}(y_0, y_T)$ if and only if its associated vector $c := \Phi y \in \mathbb{R}^K$ satisfies the linear system
\begin{equation} \label{eq:endpoint_syst}
A(0,T) \, c = \Gamma \; ,
\end{equation}
where the matrix $A(0,T) \in \mathbb{R}^{2D \times K}$ and the vector $\Gamma \in \mathbb{R}^{2D}$ are defined as follows
\begin{equation*}
A(0, T) := \left(
\begin{array}{ccccccc}
\varphi_1(0) & \dots & \varphi_{K_1}(0) & & & & \\
& & & \ddots & & & \\
& & & & \varphi_1(0) & \dots & \varphi_{K_D}(0) \\
\varphi_1(T) & \dots & \varphi_{K_1}(T) & & & & \\
& & & \ddots & & & \\
& & & & \varphi_1(T) & \dots & \varphi_{K_D}(T) \\
\end{array} \right) \quad , \quad
\Gamma := \left(
\begin{array}{c}
y_0\\
y_T
\end{array}
\right) \; .
\end{equation*}
\end{proposition}
\begin{proof}
Let $y \in \mathcal{Y}_\mathcal{K}$ and let $c := \Phi y \in \mathbb{R}^K$. By the definition of the matrix $A(0,T)$, we have
\begin{align*}
A(0,T) \, c
& = A(0,T) \Big( c_1^{(1)}, \dots, c_{K_1}^{(1)}, \; c_1^{(2)}, \dots, c_{K_2}^{(2)}, \dots, \; c_1^{(D)}, \dots, c_{K_D}^{(D)} \Big)^T \\
& = \left( \sum_{k=1}^{K_1} c_k^{(1)} \varphi_k(0), \dots, \displaystyle \sum_{k=1}^{K_D} c_k^{(D)} \varphi_k(0), \dots, \sum_{k=1}^{K_1} c_k^{(1)} \varphi_k(T), \dots, \sum_{k=1}^{K_D} c_k^{(D)} \varphi_k(T) \right)^T \\
& = \left(
\begin{array}{c}
y(0)\\
y(T)
\end{array}
\right) \; .
\end{align*}
The conclusion follows directly from the preceding relation.
\end{proof}
\subsection{Reference trajectories modelling} \label{subsec:ref_traj}
Let us now suppose that we have access to $I$ recorded trajectories $y_{R_1},\ldots,y_{R_I}$, called \emph{reference trajectories}, coming from some experiments. We propose here a statistical modelling for these reference trajectories, permitting especially to exhibit some linear properties. This modelling will permit to take advantage of the information contained in these recorded trajectories when deriving optimisation problems in the next subsection.
These trajectories being recorded, they are in particular admissible and we assume that they belong to the space $\mathcal{Y}_{\mathcal{K}}(0,T)$. As explained previously they may be interpreted as smoothed versions of recorded signals. In particular each reference trajectory $y_{R_i}$ is associated with a unique vector $c_{R_i} \in \mathbb{R}^K$. Moreover we consider each reference trajectory as a noisy observation of a certain admissible and projected trajectory $y_*$. In other words we suppose that there exists a trajectory $y_* \in \mathcal{Y}_{\mathcal{K}} \cap \mathcal{A}_\mathcal{G}(y_0, y_T)$ associated with a vector $c_* \in \mathbb{R}^K$ satisfying
\begin{equation*}
\forall \, i = 1, \dots, I \qquad c_{R_i} = c_* + \varepsilon_i \; .
\end{equation*}
The noise $\varepsilon_i$ is here assumed to be a centered Gaussian whose covariance matrix $\Sigma_i$ is of the form
\begin{equation*}
\Sigma_i = \frac{1}{2 \omega_i} \, \Sigma \; ,
\end{equation*}
where $\Sigma \in \mathbb{R}^{K \times K}$. It is noteworthy that this matrix will not be known in most of the cases but an estimated covariance matrix can be computed on the basis of the reference vectors. The positive real numbers $\omega_i$ are here considered as weights so we require $\sum_{i=1}^I \omega_i = 1 \;$;
each $\omega_i$ plays actually the role of a noise intensity.
Further from the hypothesis that the trajectory $y_*$ and all the reference trajectories $y_{R_i}$ verify the same endpoints conditions, we deduce
\begin{equation*}
A \, c_{R_i} = A \, c_* + A \, \varepsilon_i \qquad \Longleftrightarrow \qquad A \, \varepsilon_i = 0_{\mathbb{R}^{2D}} \qquad \Longleftrightarrow \qquad \varepsilon_i \in \ker A \; ,
\end{equation*}
for all $i = 1, \dots, I$ (we shorten $A(0,T)$ in $A$ when the context is clear). Hence the reference vector $c_*$ satisfies the following $I$ systems:
\begin{equation} \label{eq:model_c}
\left\{ \begin{array}{l}
c_{R_i} = c_* + \varepsilon_i \\[2mm]
\varepsilon_i \sim \mathcal{N}(0_{\mathbb{R}^K}, \Sigma_i) \\[2mm]
\varepsilon_i \in \ker A
\end{array} \right. \; .
\end{equation}
To establish a more explicit system which is equivalent to the preceding one, we require the following preliminary proposition. Here we diagonalise the matrices $\Sigma$ and $A^T A$ by exploiting the fact that the image of the first one is contained in the null space of the other one and vice versa; this is shown in the proof. This property is actually a consequence of the above modelling: the endpoints conditions modelled by $A$ imply linear relations within the components of the vectors, which should be reflected by the covariance matrix $\Sigma$. The following result will be helpful to establish \cref{prop:equiv_syst}.
\begin{proposition} \label{prop:matrix_structure}
We define $\sigma := \rk \Sigma$ and $a := \rk A^T A$. In the setting of system \eqref{eq:model_c}, we have $\sigma + a \leqslant K$ and there exist an orthogonal matrix $V \in \mathbb{R}^{K \times K}$ and two matrices $\Lambda_\Sigma \in \mathbb{R}^{K \times K}$ and $\Lambda_A \in \mathbb{R}^{K \times K}$ of the following form:
\begin{equation*}
\Lambda_\Sigma = \left( \begin{array}{cc}
\Lambda_{\Sigma,1} & 0_{\mathbb{R}^{\sigma \times (K-\sigma)}} \\[2mm]
0_{\mathbb{R}^{(K-\sigma) \times \sigma}} & 0_{\mathbb{R}^{(K-\sigma) \times (K-\sigma)}}
\end{array} \right) \qquad , \qquad
\Lambda_A = \left( \begin{array}{cc}
0_{\mathbb{R}^{(K-a) \times (K-a)}} & 0_{\mathbb{R}^{(K-a) \times a}} \\[2mm]
0_{\mathbb{R}^{a \times (K-a)}} & \Lambda_{A,2}
\end{array} \right) \; ,
\end{equation*}
where $\Lambda_{\Sigma,1} \in \mathbb{R}^{\sigma \times \sigma}$ and $\Lambda_{A,2} \in \mathbb{R}^{a \times a}$ are diagonal matrices with positive elements, such that
\begin{equation*}
\Sigma = V \Lambda_\Sigma V^T \qquad , \qquad A^T A = V \Lambda_A V^T \; .
\end{equation*}
\end{proposition}
\begin{proof}
The starting point of the proof is to remark that we have
\begin{equation} \label{eq:commute_zero}
\Sigma \, A^T A = A^T A \, \Sigma = 0_{\mathbb{R}^{K \times K}} \; .
\end{equation}
Indeed using the hypothesis $\varepsilon_i \in \ker A$ for any $i=1,\dots,I$ gives
\begin{equation*}
\Sigma \, A^T A = 2 \omega_i \, \Sigma_i \, A^T A = 2 \omega_i \, \mathbb{E}(\varepsilon_i \varepsilon_i^T) \, A^T A = 2 \omega_i \, \mathbb{E}\big(\varepsilon_i \, (A \varepsilon_i)^T \big) \, A = 0_{\mathbb{R}^{K \times K}} \; ;
\end{equation*}
similar arguments prove the second equality in \eqref{eq:commute_zero}.
First, we can deduce
\begin{equation} \label{eq:inclusion}
\im \Sigma \subseteq \ker A^T A \; ,
\end{equation}
which leads to $\sigma \leqslant K - a$ by the rank-nullity theorem. Equalities \eqref{eq:commute_zero} show also that $\Sigma$ and $A^T A$ are simultaneously diagonalisable (since they commute) so there exists an orthogonal matrix $V \in \mathbb{R}^{K \times K}$ such that
\begin{equation} \label{eq:diag}
\Sigma = V \Lambda_\Sigma V^T \qquad , \qquad A^T A = V \Lambda_A V^T \; ,
\end{equation}
where $\Lambda_\Sigma \in \mathbb{R}^{K \times K}$ and $\Lambda_A \in \mathbb{R}^{K \times K}$ are diagonal matrices. Permuting if necessary columns of $V$, we can write the matrix $\Lambda_\Sigma$ as follows:
\begin{equation} \label{eq:lambda_sigma}
\Lambda_\Sigma = \left( \begin{array}{cc}
\Lambda_{\Sigma,1} & 0_{\mathbb{R}^{\sigma \times (K-\sigma)}} \\[2mm]
0_{\mathbb{R}^{(K-\sigma) \times \sigma}} & 0_{\mathbb{R}^{(K-\sigma) \times (K-\sigma)}}
\end{array} \right) \; ;
\end{equation}
in other words the $\sigma$ first column vectors of $V$ span the image of $\Sigma$. From the inclusion \eqref{eq:inclusion}, we deduce that these vectors belong to the null space of $A^T A$. Hence the $\sigma$ first diagonal elements of $\Lambda_A$ are equal to zero and, up to a permutation of the $K - \sigma$ last column vectors of $V$, we can write
\begin{equation*}
\Lambda_A = \left( \begin{array}{cc}
0_{\mathbb{R}^{(K-a) \times (K-a)}} & 0_{\mathbb{R}^{(K-a) \times a}} \\[2mm]
0_{\mathbb{R}^{a \times (K-a)}} & \Lambda_{A,2}
\end{array} \right) \; ,
\end{equation*}
which ends the proof.
\end{proof}
\begin{remark}
From equalities \eqref{eq:commute_zero}, we can also deduce
\begin{equation*}
\im A^T A \subseteq \ker \Sigma \; ,
\end{equation*}
showing that $\Sigma$ is singular. Consequently the Gaussian noise $\varepsilon_i$ involved in \eqref{eq:model_c} is degenerate.
\end{remark}
A new formulation of system \eqref{eq:model_c} which makes explicit the constrained and unconstrained parts of a vector satisfying this system is given in the following result. This is achieved by using the preceding result which allows to decompose the space $\mathbb{R}^K$ into three orthogonal subspaces. We prove that the restriction of the noise $\varepsilon_i$ to the first subspace is a non-degenerate Gaussian, showing that this first subspace corresponds to the unconstrained one. The two other subspaces describe affine relations coming from the endpoints conditions and from implicit relations within the vector components. These implicit relations, which may model for instance natural trends, are expected to be contained in the reference vectors $c_{R_i}$ and reflected by the (estimated) covariance matrix $\Sigma$.\\
Prior to this, let us write the matrix $V \in \mathbb{R}^{K \times K}$ introduced in \cref{prop:matrix_structure} as follows:
\begin{equation*}
V = \big( V_1 \quad V_2 \quad V_3 \big) \; ,
\end{equation*}
where $V_1 \in \mathbb{R}^{K \times \sigma}$, $V_2 \in \mathbb{R}^{K \times K-\sigma-a}$ and $V_3 \in \mathbb{R}^{K \times a}$. We emphasise that the column-vectors of the matrices $V_1$ and $V_3$ do not overlap according to the property $\sigma + a \leqslant K$ proved in \cref{prop:matrix_structure}. In particular the matrix $V_2$ has to be considered only in the case $\sigma + a < K$. Further for any $c \in \mathbb{R}^K$, we will use the notations
\begin{equation*}
\widetilde{c} := V^T c \qquad, \qquad \widetilde{c}_\ell := V_\ell^T c \; ,
\end{equation*}
for $\ell=1,2,3$. Finally we consider the singular value decomposition of $A$ coming from the diagonalisation of the symmetric matrix $A^T A$ with $V$:
\begin{equation*}
A = U S_A V^T \; ,
\end{equation*}
where $U \in \mathbb{R}^{2D \times 2D}$ is orthogonal and $S_A \in \mathbb{R}^{2D \times K}$ is a rectangular diagonal matrix of the following form:
\begin{equation} \label{eq:s_a}
S_A = \big( 0_{\mathbb{R}^{2D \times K-2D}} \quad S_{A,2} \big) \; ,
\end{equation}
with $S_{A,2} := \sqrt{\Lambda_{A,2}} \in \mathbb{R}^{2D \times 2D}$.
\begin{proposition} \label{prop:equiv_syst}
Suppose that the matrix $A$ is full rank, \emph{i.e.} $a = 2D$. Then for any $i=1, \dots, I$, system \eqref{eq:model_c} is equivalent to the following one:
\begin{equation} \label{eq:model_c_tilde_2}
\left\{ \begin{array}{l}
\widetilde{c}_{R_i, 1} = \widetilde{c}_{*,1} + \widetilde{\varepsilon}_{i, 1} \\[2mm]
\displaystyle \widetilde{\varepsilon}_{i, 1} \sim \mathcal{N} \left(0_{\mathbb{R}^\sigma}, \frac{1}{2 \omega_i} \, \Lambda_{\Sigma,1} \right) \\[3mm]
\widetilde{c}_{*,2} = V_2^T c_{R_i} \\[2mm]
\widetilde{c}_{*,3} = \, S_{A,2}^{-1} \, U^T \Gamma
\end{array} \right. \; .
\end{equation}
\end{proposition}
\begin{proof}
We first prove that system \eqref{eq:model_c} is equivalent to
\begin{equation} \label{eq:model_c_tilde}
\left\{ \begin{array}{l}
\widetilde{c}_{R_i} = \widetilde{c}_* + \widetilde{\varepsilon}_i \\[2mm]
\displaystyle \widetilde{\varepsilon}_i \sim \mathcal{N} \left(0_{\mathbb{R}^K}, \frac{1}{2 \omega_i} \, \Lambda_\Sigma \right) \\[3mm]
S_A \, \widetilde{c}_* = U^T \Gamma
\end{array} \right. \; .
\end{equation}
The matrix $V$ being orthogonal, it is non-singular and so we have for all $i = 1, \dots, I$,
\begin{equation*}
c_{R_i} = c_* + \varepsilon_i \qquad \Longleftrightarrow \qquad \widetilde{c}_{R_i} = \widetilde{c}_* + \widetilde{\varepsilon}_i \; ,
\end{equation*}
and, since $\Sigma_i = \frac{1}{2 \omega_i} \, \Sigma = \frac{1}{2 \omega_i} \, V \Lambda_\Sigma V^T$, we obtain
\begin{equation*}
\varepsilon_i \sim \mathcal{N}(0_{\mathbb{R}^K}, \Sigma_i) \qquad \Longleftrightarrow \qquad \widetilde{\varepsilon}_i \sim \mathcal{N} \left(0_{\mathbb{R}^K}, \frac{1}{2 \omega_i} \, \Lambda_\Sigma \right) \; .
\end{equation*}
Finally the property $\varepsilon_i \in \ker A$ is equivalent to
\begin{align*}
A \, c_* = \Gamma \qquad
& \Longleftrightarrow \qquad U S_A V^T c_* = \Gamma \\
& \Longleftrightarrow \qquad S_A \, \widetilde{c}_* = U^T \Gamma \; ,
\end{align*}
proving that the systems \eqref{eq:model_c} and \eqref{eq:model_c_tilde} are equivalent.
Now the fact that the $K-\sigma$ last diagonal elements of $\Lambda_\Sigma$ are zero implies that the components $\widetilde{c}_{*,2} \in \mathbb{R}^{K-\sigma-2D}$ and $\widetilde{c}_{*,3} \in \mathbb{R}^{2D}$ are constant. From the first equality of \eqref{eq:model_c_tilde}, we have on one side
\begin{equation*}
\widetilde{c}_{R_i, 2} = \widetilde{c}_{*,2} \qquad \Longleftrightarrow \qquad V_2^T c_{R_i} = \widetilde{c}_{*,2} \; ,
\end{equation*}
for any $i = 1, \dots, I$. On the other side, combining the last relation of the system \eqref{eq:model_c_tilde} with the form of the matrix $S_A$ given in \eqref{eq:s_a} permits to obtain
\begin{align*}
S_A \, \widetilde{c}_* = U^T \Gamma \qquad
& \Longleftrightarrow \qquad S_{A,2} \, \widetilde{c}_{*,3} = U^T \Gamma \nonumber \\
& \Longleftrightarrow \qquad \widetilde{c}_{*,3} = \, S_{A,2}^{-1} \, U^T \Gamma \; ,
\end{align*}
the last equivalence being justified by the hypothesis that the matrix $A$ is full rank (which implies that the diagonal matrix $S_{A,2}$ is non-singular).
\end{proof}
The above decomposition gives us access to non-degenerated density of $\widetilde{c}_{R_i, 1}$ given $\widetilde{c}_{*,1}$ which is later denoted by $u(\widetilde{c}_{R_i, 1}|\widetilde{c}_{*,1})$. In next section we will assume a prior distribution on $\widetilde{c}_{*,1}$ with high density for low values of the cost function $F$.
\subsection{A trajectory optimisation problem via a Maximum A Posteriori approach} \label{subsec:modelling}
Before introducing the Bayesian framework, let first recall that we are interested in minimising a certain cost function $F: \mathcal{C}\big([0,T], \mathbb{R}^D \big) \longrightarrow \mathbb{R}$ over the set of projected and admissible trajectories $\mathcal{Y}_\mathcal{K} \cap \mathcal{A}_\mathcal{G}(y_0, y_T)$. As explained previously, we propose here a methodology leading to a constrained optimisation problem based on the reference trajectories and designed to provide realistic trajectories.
Technically speaking, we seek for the mode of a \emph{posterior} distribution which contains information from the reference trajectories. The aim of this subsection is then to obtain the \emph{posterior} distribution via Bayes's rule, using in particular the precise modelling of the reference trajectories given in \cref{prop:equiv_syst} and defining an accurate prior distribution with high density for low values of the cost function $F$.
To do so, we recall firstly that all the trajectories considered here are assumed to belong to the space $\mathcal{Y}_\mathcal{K}$ which is isomorphic to $\mathbb{R}^K$. So each trajectory is here described by its associated vector in $\mathbb{R}^K$, permitting in particular to define distributions over finite-dimensional spaces. We recall also that the reference trajectories are interpreted as noisy observations of a certain $y_*$ associated with a $c_*$. According to \cref{prop:equiv_syst}, this vector complies with some affine conditions which are described by the following subspace $\mathcal{V}_1$:
\begin{equation} \label{eq:v1}
c \in \mathcal{V}_1 \qquad \Longleftrightarrow \qquad \left\{
\begin{array}{l}
V_2^T c = V_2^T c_{R_i} \\[2mm]
V_3^T c = S_{A,2}^{-1} \, U^T \Gamma
\end{array}
\right. \; .
\end{equation}
Hence a vector $c$ belonging to $\mathcal{V}_1$ is described only through its component $\widetilde{c}_1 := V_1^T c$. In addition we note that the definition of $\mathcal{V}_1$ does not depend actually on the choice of $i$ since $V_2^T c_{R_i}$ has been proved to be constant in \cref{prop:equiv_syst}. Further we emphasise that the matrix $A$ is supposed to be full rank in this case and we have $\mathcal{V}_1 \simeq \mathbb{R}^\sigma$; we recall that $\sigma$ is the rank of the covariance matrix $\Sigma$.
Let us now define the cost function $F$ over the spaces $\mathbb{R}^K$ and $\mathcal{V}_1$. This is necessary to define the \emph{prior} distribution and to establish our optimisation problem.
\begin{definition}[Cost functions] \label{def:cost}
Let $\check{F}: \mathbb{R}^K \longrightarrow \mathbb{R}$ and $\widetilde{F}: \mathbb{R}^\sigma \longrightarrow \mathbb{R}$ be the functions defined by
\begin{itemize}
\item $\displaystyle \check{F}(c) := F \big( \Phi|_{\mathcal{Y}_\mathcal{K}}^{-1} c \big)$ ;
\item $\displaystyle \widetilde{F}(\widetilde{c}_1) := F\Big( \Phi|_{\mathcal{Y}_\mathcal{K}}^{-1} \, V \Big( \widetilde{c}_1^T \quad c_{R_i}^{\ T} V_2 \quad \Gamma^T U \, \big(S_{A,2}^{-1} \big)^T \Big)^T \Big)$ .
\end{itemize}
\end{definition}
\begin{remark}
From the preceding definition, we observe that for any $y \in \mathcal{Y}_\mathcal{K}$ and its associated vector $c \in \mathbb{R}^K$, we have
\begin{equation*}
\check{F}(c) = F \big( \Phi|_{\mathcal{Y}_\mathcal{K}}^{-1} c \big) = F(y) \; .
\end{equation*}
Further for any $c \in \mathcal{V}_1$, we have
\begin{align*}
\check{F}(c) = F \big( \Phi|_{\mathcal{Y}_\mathcal{K}}^{-1} c \big) = F \big( \Phi|_{\mathcal{Y}_\mathcal{K}}^{-1} V \widetilde{c} \big) = F\Big( \Phi|_{\mathcal{Y}_\mathcal{K}}^{-1} \, V \Big( \widetilde{c}_1^T \quad c_{R_i}^{\ T} V_2 \quad \Gamma^T U \, \big(S_{A,2}^{-1} \big)^T \Big)^T \Big) = \widetilde{F}(\widetilde{c}_1) \; .
\end{align*}
We deduce that $\widetilde{F}$ is actually the restriction of $\check{F}$ to the subspace $\mathcal{V}_1$.
\end{remark}
From now on, the trajectory $y_*$ and the associated vector $c_*$ will be considered as random variables and will be denoted by $y$ and $c$. We are interested in the \emph{posterior} distribution
\begin{equation*}
u(\widetilde{c}_1 \, | \, \widetilde{c}_{R_1, 1}, \dots, \widetilde{c}_{R_I, 1}) \; ,
\end{equation*}
which depends only on the free component $\widetilde{c}_1$ of $c \in \mathcal{V}_1$, the two other ones $\widetilde{c}_2$ and $\widetilde{c}_3$ being fixed according to \eqref{eq:v1}. We use Bayes's rule to model the \emph{posterior} via the \emph{prior} and likelihood distributions, leading to
\begin{equation*}
u(\widetilde{c}_1 \, | \, \widetilde{c}_{R_1, 1}, \dots, \widetilde{c}_{R_I, 1}) \propto u(\widetilde{c}_{R_1, 1}, \dots, \widetilde{c}_{R_I, 1} \, | \, \widetilde{c}_1) \, u(\widetilde{c}_1) \; .
\end{equation*}
Assuming now that the vectors $\widetilde{c}_{R_i, 1}$ are independent gives
\begin{equation*}
u(\widetilde{c}_{R_1, 1}, \dots, \widetilde{c}_{R_I, 1} \, | \, \widetilde{c}_1) \, u(\widetilde{c}_1) = \prod_{i=1}^I u(\widetilde{c}_{R_i, 1} \, | \, \widetilde{c}_1) \, u(\widetilde{c}_1) \; .
\end{equation*}
The above likelihood is given by the modelling of the reference trajectories detailed in \cref{prop:equiv_syst}. In this case, we have
\begin{equation*}
u(\widetilde{c}_{R_i, 1} \, | \, \widetilde{c}_1) \propto \exp\Big( -\omega_i \, \big( \widetilde{c}_1 - \widetilde{c}_{R_i, 1} \big)^T \Lambda_{\Sigma,1}^{-1} \big( \widetilde{c}_1 - \widetilde{c}_{R_i, 1} \big) \Big) \; .
\end{equation*}
The prior distribution is obtained by assuming that the most efficient trajectories (with respect to the cost function) are \emph{a priori} the most likely ones:
\begin{equation} \label{eq:model_prior}
u(\widetilde{c}_1) \propto \exp \Big( -\kappa^{-1} \widetilde{F}(\widetilde{c}_1) \Big) \; ,
\end{equation}
where $\kappa > 0$.
Putting everything together and taking the negative of the logarithm gives the following minimisation problem, whose solution is the Maximum \emph{A Posteriori} estimator:
\begin{equation} \label{eq:opt_proj}
\left\{
\begin{array}{l}
\displaystyle \widetilde{c}_1^\star \in \argmin_{\widetilde{c}_1 \in \mathbb{R}^\sigma} \widetilde{F}(\widetilde{c}_1) + \kappa \sum_{i=1}^I \omega_i \, \big( \widetilde{c}_1 - \widetilde{c}_{R_i, 1} \big)^T \Lambda_{\Sigma,1}^{-1} \big( \widetilde{c}_1 - \widetilde{c}_{R_i, 1} \big) \\[2mm]
\widetilde{c}_2 = V_2^T c_{R_i} \\[2mm]
\widetilde{c}_3 = S_{A,2}^{-1} \, U^T \Gamma
\end{array} \; ,
\right.
\end{equation}
where $i$ is arbitrarily chosen in $\{1, \dots, I \}$.
Let us now rewrite the above optimisation problem with respect to the variable $c = V \widetilde{c} \in \mathbb{R}^K$ in order to make it more interpretable.
\begin{proposition} \label{prop:opt_equiv}
The optimisation problem \eqref{eq:opt_proj} is equivalent to the following one:
\begin{equation} \label{eq:opt_proj_2}
c^\star \in \argmin_{c \in \mathcal{V}_1} \check{F}(c) + \kappa \sum_{i=1}^I \omega_i \, \big( c - c_{R_i} \big)^T \Sigma^\dagger \big( c - c_{R_i} \big) \; ,
\end{equation}
where $\Sigma^\dagger \in \mathbb{R}^{K \times K}$ denotes the pseudoinverse of the matrix $\Sigma$.
\end{proposition}
\begin{proof}
From \eqref{eq:lambda_sigma}, we deduce
\begin{align*}
\sum_{i=1}^I \omega_i \, \big( \widetilde{c}_1 - \widetilde{c}_{R_i, 1} \big)^T \Lambda_{\Sigma,1}^{-1} \big( \widetilde{c}_1 - \widetilde{c}_{R_i, 1} \big)
& = \sum_{i=1}^I \omega_i \, \big( \widetilde{c} - \widetilde{c}_{R_i} \big)^T \Lambda_{\Sigma}^\dagger \big( \widetilde{c} - \widetilde{c}_{R_i} \big) \\
& = \sum_{i=1}^I \omega_i \, \big( c - c_{R_i} \big)^T \, V \Lambda_{\Sigma}^\dagger V^T \, \big( c - c_{R_i} \big) \\
& = \sum_{i=1}^I \omega_i \, \big( c - c_{R_i} \big)^T \Sigma^\dagger \big( c - c_{R_i} \big) \; .
\end{align*}
And from the proof of \cref{prop:equiv_syst}, we have
\begin{equation*}
A \, c = \Gamma \qquad \Longleftrightarrow \qquad \widetilde{c}_3 = \, S_{A,2}^{-1} \, U^T \Gamma \; ,
\end{equation*}
proving that $c \in \mathcal{V}_1$.
\end{proof}
To conclude, let us comment on this optimisation problem.
\begin{enumerate}
\item To interpret the optimisation problem \eqref{eq:opt_proj_2} (or equivalently \eqref{eq:opt_proj}) from a geometric point of view, let us consider the following new problem:
\begin{equation} \label{eq:primal}
\begin{array}{ll}
& \displaystyle \min_{\widetilde{c}_1 \in \mathbb{R}^\sigma} \widetilde{F}(\widetilde{c}_1) \\[2mm]
\text{s.t. } & \displaystyle \sum_{i=1}^I \omega_i \, \big( \widetilde{c}_1 - \widetilde{c}_{R_i, 1} \big)^T \Lambda_{\Sigma,1}^{-1} \big( \widetilde{c}_1 - \widetilde{c}_{R_i, 1} \big) \leqslant \widetilde{\kappa}
\end{array}
\end{equation}
where $\lambda \geqslant 0$. Here we suppose that $\widetilde{F}$ is strictly convex and that the problem \eqref{eq:primal} has a solution (which is then unique). By Slater's theorem \citep[Subsec. 5.2.3]{convexopt}, the strong duality holds for the problem \eqref{eq:primal}. It can then be proved that there exists a certain $\lambda^\star \geqslant 0$ such that the solution of \eqref{eq:primal} is the minimiser of the strictly convex function
\begin{equation*}
\widetilde{c}_1 \longmapsto \widetilde{F}(\widetilde{c}_1) + \lambda^\star \sum_{i=1}^I \omega_i \, \big( \widetilde{c}_1 - \widetilde{c}_{R_i, 1} \big)^T \Lambda_{\Sigma,1}^{-1} \big( \widetilde{c}_1 - \widetilde{c}_{R_i, 1} \big) \; ,
\end{equation*}
which is actually the objective function of the optimisation problem \eqref{eq:opt_proj} for $\kappa = \lambda^\star$. Hence the problem \eqref{eq:opt_proj} minimises the cost $\widetilde{F}$ in a ball centered on the weighted average of the reference trajectories. In particular if the reference trajectories are close to an optimal one with respect to $\widetilde{F}$ then one could expect the solution of \eqref{eq:opt_proj} to be equal to this optimal trajectory.
\item Further the optimisation problem \eqref{eq:opt_proj_2} takes into account the endpoints conditions through the subspace $\mathcal{V}_1$ but not the additional constraints. However as explained in the preceding point, the solution is close to realistic trajectories and so is likely to comply with the additional constraints for a well-chosen parameter $\kappa > 0$. We refer to \cref{subsec:iterations} for more details on an iterative method for the tuning of $\kappa$. In particular, a right choice for this parameter is expected to provide an optimised trajectory with a realistic behaviour. This is for instance illustrated in \cref{sec:appli_aero}.
\item Taking into account the linear information from the available data through the covariance matrix $\Sigma$ allows to restrict the search to the subspace $\mathcal{V}_1$ describing these relations. This is of particular interest when implicit relations (modelled by the sub-matrix $V_2$) are revealed by the estimation of $\Sigma$ on the basis of the reference trajectories; in this case, these implicit relations may not be known by the expert.
\item The optimisation problem \eqref{eq:opt_proj_2} has linear constraints and a quadratic penalised term. For instance, if the cost function $\check{F}$ is a convex function then we obtain a convex problem for which efficient algorithms exist.
\end{enumerate}
\subsection{Quadratic cost for a convex optimisation problem} \label{subsec:quad_model}
In this short subsection, we focus on a particular case where the cost function $F$ is defined as the integral of an instantaneous quadratic cost function, \emph{i.e.}
\begin{equation} \label{eq:quad_f}
\forall \, y \in \mathcal{C}\big([0,T], \mathbb{R}^D \big) \qquad F(y) = \int_0^T f(y(t)) \, dt \; ,
\end{equation}
where $f : \mathbb{R}^D \longrightarrow \mathbb{R}$ is quadratic.
Even though such a setting may appear to be restrictive, we emphasise that quadratic models may lead to highly accurate approximations of variables, as it is illustrated in \cref{sec:appli_aero}. For a quadratic instantaneous cost, the associated function $\check{F}: \mathbb{R}^K \longrightarrow \mathbb{R}$ can be proved to be quadratic as well and can be explicitly computed. In the following result, we provide a quadratic optimisation problem equivalent to \eqref{eq:opt_proj_2}.
\begin{proposition} \label{prop:quad_min_pb}
Suppose that the cost function $F$ is of the form \eqref{eq:quad_f} with $f$ quadratic. Then the optimisation problem \eqref{eq:opt_proj_2} is equivalent to the following one:
\begin{equation} \label{eq:quad_min_pb}
c^\star \in \argmin_{c \in \mathcal{V}_1} c^T \Big( \check{Q} + \kappa \, \Sigma^\dagger \Big) c + \left( \check{w} - 2 \kappa \sum_{i=1}^I \omega_i \, \Sigma^\dagger c_{R_i} \right)^{\hspace{-5pt}T} c \; ,
\end{equation}
where $\check{Q} \in \mathbb{R}^{K \times K}$ and $\check{w} \in \mathbb{R}^K$ can be explicitly computed from $f$.
\end{proposition}
\begin{proof}
We defer the proof to the supplementary material.
\end{proof}
In particular this permits to derive sufficient conditions on the parameter $\kappa > 0$ so that the optimisation problem is proved to be equivalent to a quadratic program \citep[Sec. 4.4]{convexopt}, namely the objective function is convex quadratic together with affine constraints. In practice, this allows to make use of efficient optimisation libraries to solve numerically \eqref{eq:quad_min_pb}.
\subsection{Iterative process to comply with additional constraints} \label{subsec:iterations}
As explained in \cref{subsec:modelling}, the trajectory optimisation problem \eqref{eq:opt_proj_2} is constrained by the endpoints conditions and by implicit linear relations revealed by the reference trajectories. Nevertheless the additional constraints introduced in \cref{def:constraints} are not taken into account in this problem. In practice such constraints assure that natural or user-defined features are verified and so a trajectory which does not comply with these constraints may be considered as unrealistic.
Our aim is then to assure that the trajectory $y^\star = \Phi|_{\mathcal{Y}_\mathcal{K}}^{-1} c^\star$, where $c^\star \in \mathcal{V}_1$ is the solution of the optimisation problem \eqref{eq:opt_proj_2}, verifies the additional constraints, \emph{i.e.} belongs to the set $\mathcal{G}$. A first solution would be to add the constraint $\Phi|_{\mathcal{Y}_\mathcal{K}}^{-1} c \in \mathcal{G}$ in the optimisation problem \eqref{eq:opt_proj_2}. However depending on the nature of the constraints functions $g_\ell$, this may lead to nonlinear constraints which could be costly from a numerical point of view. The solution we propose consists rather in exploiting the degree of freedom coming from the parameter $\kappa > 0$ appearing in the problem \eqref{eq:opt_proj_2}.
First of all, let us factorise the problem \eqref{eq:opt_proj_2} by $\kappa$ to obtain the following new one for the sake of presentation:
\begin{equation} \label{eq:opt_proj_3}
c^\star \in \argmin_{c \in \mathcal{V}_1} \nu \, \check{F}(c) + \sum_{i=1}^I \omega_i \, \big( c - c_{R_i} \big)^T \Sigma^\dagger \big( c - c_{R_i} \big) \; ,
\end{equation}
where $\nu := \kappa^{-1}$. On one hand, we observe that the solution of the optimisation problem \eqref{eq:opt_proj_3} for the limit case $\nu = 0$ is given by $\sum_{i=1}^I \omega_i \, c_{R_i}$ which is the average of the reference vectors. In this case, one may expect that the associated average trajectory complies with the constraints but is unlikely to optimise the cost function $F$. On the other hand, for very large $\nu > 0$, the second term of the objective function in \eqref{eq:opt_proj_3} can be considered as negligible compared to the first one. In this case, the cost of the solution will surely be smaller than the costs of the reference trajectories but no guarantee regarding the additional constraints can be established in a general setting.
Given these observations, the task is then to find an appropriate value $\nu^\star > 0$ in order to reach a trade-off between optimising and remaining close to the reference trajectories to comply with the additional constraints. Many methods can be developed to find such a $\nu^\star$ and, among those based on iterative processes, linear or binary search algorithms can be considered. In this case, one has to set firstly a maximal value $\nu_{max}$ so that the solution of \eqref{eq:opt_proj_3} with $\nu_{max}$ is unlikely to satisfy the constraints and to perform secondly the search over the interval $(0,\nu_{max})$. Since the solution for $\nu = 0$ is assumed to be admissible, we expect that the binary search will find a $\nu^\star > 0$ leading to an optimised trajectory belonging to $\mathcal{G}$.
\subsection{Confidence bounds on the integrated cost}
\label{sec:2-7}
In practice the cost function $F$ considered is an estimation of the true cost $F^{\star}$, a random variable which cannot be fully predicted based on $y$. If the distribution $F(y)$ would be known it would be possible to deduce confidence bound on $F^{\star}$. This is for instance possible by considering multivariate functional regression \citep{RHCC2007}.
The simplest case from the estimation point of view is to consider that $F^{\star}$ is the integral of some instantaneous consumption function $f^{\star}$
as in \cref{subsec:quad_model}, and to estimate the parameters of the standard multivariate regression
$$
f^{\star}(y(t)) = f(y(t)) + \varepsilon(t),
$$
where the random noise $\varepsilon(t)$ is assumed to follow a centered Gaussian distribution with variance $\sigma$. In this case $F^{\star}$ can be expressed as the integral of a stochastic process
$$
F^{\star}(y) := \int_{0}^{T} f^{\star}(y(t)) \, dt \; = F(y) + \int_{0}^{T} \varepsilon(t) \, dt \;.
$$
then assuming that $(\varepsilon(t))_{t \in [0,T]}$ independent we obtain
$$
\int_{0}^{T} \varepsilon(t) \, dt \; \sim \mathcal{N}(0,T \sigma^2).
$$
Thus $F^{\star}(y)$ follows a Gaussian distribution centered on $F(y)$ and with variance equals to $T\sigma^2$. This makes it possible to compute confidence bounds on $F^{\star}(y)$. For a confidence level $1-u$, $u\in [0,1]$, a confidence interval for $F^{\star}(y)$ is obtained as
$$ \texttt{CI}^{1-u}(F^{\star}(y)) = F(y) \pm \zeta_{1-\frac{u}{2}}
\sqrt{T}\sigma,$$
where $\zeta_{1-\frac{u}{2}}$ is the quantile of order $1-\frac{u}{2}$ of the standard Gaussian distribution.
The assumption that $f$ and $\sigma^2$ are known is relevant since they are estimated based on a huge amount of training data. The assumption of white Gaussian noise can be seen as unrealistic, however it appears to be the only route to explicit calculus. A more complex strategy could be derived using Gaussian processes, which is beyond the scope of this paper.
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\section{Introduction}\label{intro}
The notion of recollement of triangulated categories was introduced by Beilinson, Bernstein and Deligne in connection with derived categories of sheaves of topological spaces (\cite{BBD}).
One of the authors introduced the notion of stable $t$-structure in a triangulated category
\cite{Mi1}, and studied relations to recollements.
Afterwards this notion was studied by many authors under a lot of names, e.g.
a torsion pair, a semiorthogonal decomposition, Bousfield localization.
\begin{defn} \label{st-tprime}
Let $\mcal{D}$ be a triangulated category with the translation functor $\Sigma$.
A pair $(\mcal{U}, \mcal{V})$ of full subcategories of $\mcal{D}$ is called a {\it stable t-structure} in $\mcal{D}$ provided that
\begin{enumerate}
\rmitem{a} $\mcal{U}=\Sigma\mcal{U}$ and $\mcal{V}=\Sigma\mcal{V}$.
\rmitem{b} $\operatorname{Hom}_{\mcal{D}}(\mcal{U}, \mcal{V}) = 0$.
\rmitem{c} For every $X \in \mcal{D}$, there exists a triangle $U \rightarrow X \rightarrow V \rightarrow
\Sigma U$
with $U \in \mcal{U}$ and $V \in \mcal{V}$.
\end{enumerate}
\end{defn}
In \cite{IKM}, we introduced the notion of polygons of recollements in a triangulated category,
and studied the properties of triangle of recollements in connection with
the homotopy category of unbounded complexes with bounded homologies,
its quotient category by the homotopy category of bounded complexes, and
the stable category of Cohen-Macaulay modules over an Iwanaga-Gorenstein ring.
Moreover, we studied derived categories $\cat{D}_{N}(\mcal{A})$ of $N$-complexes of an
abelian category $\mcal{A}$, and show$\cat{D}_{N}(\mcal{A})$ is a triangle equivalent to
the derived category $\cat{D}(\operatorname{\mathsf{Mor}}_{N-1}(\mcal{A}))$ of ordinary complexes over an abelian category $\operatorname{\mathsf{Mor}}_{N-1}(\mcal{A})$ of $N-1$ sequences of morphisms of $\mcal{A}$ (\cite{IKM2}).
In this article, we study the properties of polygons of recollements in various categories.
\begin{defn}
Let $\mcal{D}$ be a triangulated category, and let $\mcal{U}_{1}, \cdots, \mcal{U}_{n}$
be full triangulated subcategories of $\mcal{D}$.
An $n$-tuple $(\mcal{U}_1, \mcal{U}_2, \cdots , \mcal{U}_n)$ is called
an $n$-gon of recollements in $\mcal{D}$
if $(\mcal{U}_{i}, \mcal{U}_{i+1})$ is a stable t-structure in $\mcal{D}$ ($1 \leq i \leq n$), where
$\mcal{U}_1=\mcal{U}_{n+1}$.
\end{defn}
In Section \ref{t-strecoll}, we recall the notions of stable $t$-structures and recollement
polygons of recollements in a triangulated category.
\begin{prop}[Proposition \ref{p20Apr30}]
Let $\mcal{D}_1$, $\mcal{D}_2$ be triangulated categories.
Let $( \mcal{U}_1 ,\cdots , \mcal{U}_{n})$ and
$( \mcal{V}_1 ,\cdots , \mcal{V}_{n})$
be $n$-gons of recollements in $\mcal{D}_1$ and $\mcal{D}_2$, respectively.
Assume a triangle functor $F:\mcal{D}_{1} \to \mcal{D}_{2}$
sends $( \mcal{U}_1 , \cdots, \mcal{U}_{n} )$ to
$( \mcal{V}_1 , \cdots, \mcal{V}_{n} )$.
If $F\mid _{\mcal{U}_t }$ and $F\mid _{\mcal{U}_{t+1}}$
are triangle equivalences for some $t$, so is $F$.
\end{prop}
In Section \ref{t-strecoll}, we study stable $t$-structures with relation to relations to
contravariantly finite categories and Calabi-Yau triangulated categories.
\begin{prop}[Proposition \ref{n-gon1}]
Let $\mcal{D}$ be an $(m/n)$-Calabi-Yau triangulated category.
For any functorially finite thick subcategory $\mcal{U}_1$ of $\mcal{D}$, we put $\mcal{U}_{i+1}:=\mcal{U}_i^\perp$ for any $i$.
Then we have an $l$-gon $(\mcal{U}_{1}, \cdots , \mcal{U}_{l})$ of recollements in $\mcal{D}$
for some positive divisor $l$ of $2n$.
\end{prop}
In Section \ref{n-gonII}, we constructs polygons of recollements
in the derived derived category of modules over an algebra, and them
in the stable category of Cohen-Mcaulay modules over an Iwanaga-Gorenstein ring.
\begin{thm}[Theorem \ref{n-gon3}]
Let $A$ be a finite dimensional $k$-algebra of finite global dimension such that $A/J_A$ is separable over a field $k$ and
$\cat{D}^{\mrm{b}}(\operatorname{\mathsf{mod}} A)$ is $(m/n)$-Calabi-Yau.
Let $R$ be a coherent $k$-algebra of finite self-injective dimension as both sides.
For any functorially finite thick subcategory $\mcal{U}_1$ of $\cat{D}^{\mrm{b}}(\operatorname{\mathsf{mod}} A)$, we put $\mcal{U}_{i+1}:=\mcal{U}_i^\perp$ for any $i$.
Then there is a positive divisor $l$ of $2n$ such that we have an $l$-gon
$(\mcal{U}_{1}^{R}, \cdots , \mcal{U}_{l}^{R})$ of recollements in $\cat{D}^{\mrm{b}}(\operatorname{\mathsf{mod}} R\ten_k A)$
and an $l$-gon $(Q(\mcal{U}_{1}^{R}), \cdots , Q(\mcal{U}_{l}^{R}))$ of recollements in
$\underline{\operatorname{\mathsf{CM}}}(R\ten_k A)$
\end{thm}
In Section \ref{Tcpx}, we the homotopy category {\small$\cat{K}(\operatorname{\mathsf{Mor}}_{N-1}^{\mrm{sm}}(\mcal{B}))$}
of the category $\operatorname{\mathsf{Mor}}_{N-1}^{\mrm{sm}}(\mcal{B})$ of $N-1$ sequences of split monomorphisms
in an additive category $\mcal{B}$.
\begin{thm}[Theorem \ref{th:ngon@trg01}]
Let $\mcal{B}$ be an additive category.
Then there is a $2N$-gon of recollements in $\cat{K}(\operatorname{\mathsf{Mor}}^{\mrm{sm}}_{N-1}(\mcal{B}))$:
{\small
\[
(\mcal{F}^{[1,N-1]},\mcal{E}^{[2,N-1]}, \mcal{E}^{1}, \mcal{F}^{[1,2]}, \cdots,
\mcal{E}^{s}, \mcal{F}^{[s,s+1]}, \cdots, \mcal{E}^{N-2}, \mcal{F}^{[N-2,N-1]}, \mcal{E}^{N-1}, \mcal{E}^{[1,N-2]})
\]
}
\end{thm}
In Section \ref{TNcpx}, we study the homotopy category $\cat{K}_{N}(\mcal{B})$
of $N$-complexes of objects of an additive category $\mcal{B}$.
\begin{thm}[Corollary \ref{n-cpxgon04}]
We have a recollement of $\cat{K}_{N}(\mcal{B})$:
\[\xymatrix{
\cat{K}_{N-r}(\mcal{B}) \ar@<-1ex>[r]^{i_{s*}}
& \cat{K}_{N}(\mcal{B})
\ar@/_1.5pc/[l]^{i_{s}^{*}} \ar@/^1.5pc/[l]_{i_{s}^{!}} \ar@<-1ex>[r]^{j_{s}^{*}}
& \cat{K}_{r+1}(\mcal{B})
\ar@/_1.5pc/[l]^{j_{s!}} \ar@/^1.5pc/[l]_{j_{s*}}
}\]
\end{thm}
\begin{cor}[Corollary \ref{cpx2N-gon}]
There is a $2N$-gon of recollements in $\cat{K}_{N}(\mcal{B})$:
\[
(\mcal{F}_{1}^{N-2}, \mcal{F}_{0}^{1}, \mcal{F}_{2}^{N-2}, \mcal{F}_{1}^{1},
\cdots, \mcal{F}_{r+1}^{N-2}, \mcal{F}_{r}^{1},
\cdots, \mcal{F}_{N-1}^{N-2}, \mcal{F}_{N-2}^{1}, \mcal{F}_{0}^{N-2}, \mcal{F}_{N-1}^{1})
\]
\end{cor}
In Section \ref{TrieqDN}, we construct a triangle functor
$\underline{F}_{N}: \cat{K}(\operatorname{\mathsf{Mor}}^{\mrm{sm}}_{N-1}(\mcal{B})) \to \cat{K}_{N}(\mcal{B})$
which sends the above $2N$-gon of $\cat{K}_{N}(\mcal{B})$ to
the above $2N$-gon of $\cat{K}_{N}(\mcal{B})$.
Therefore we have the result.
\begin{thm}[Theorem \ref{KNhtp}]
Let $\mcal{B}$ be an additive category, then we have triangle equivalences:
\[
\cat{K}^{\sharp}(\operatorname{\mathsf{Mor}}^{\mrm{sm}}_{N-1}(\mcal{B})) \simeq \cat{K}^{\sharp}_{N}(\mcal{B})
\]
where $\sharp=\text{nothing}, -, +, \mrm{b}$.
\end{thm}
\section{Stable t-structures and recollements}\label{t-strecoll}
We recall the notion of recollements and study their relationship with stable t-structures.
This correspondence enables us to understand (co)localizations and recollements
by way of subcategories instead of quotient categories.
In Proposition \ref{st-t-1second} we see that a recollement corresponds to two consecutive stable t-structures.
First we see that a (co)localization and a stable t-structure
essentially describe the same phenomenon,
using the methods which are similar to ones in recollements \cite{BBD}.
Next we recall the notion of a recollement which consists of a localization and
a colocalization.
\begin{defn} [\cite{BBD}]
We call a diagram
\[\xymatrix{
\mcal{D'} \ar@<-1ex>[r]^{i_{*}}
& \mcal{D}
\ar@/^1.5pc/[l]_{i^{!}} \ar@/_1.5pc/[l]^{i^{*}} \ar@<-1ex>[r]^{j^{*}}
& \mcal{D''}
\ar@/_1.5pc/[l]^{j_{!}} \ar@/^1.5pc/[l]_{j_{*}}
}\]
of triangulated categories and functors a \emph{recollement} if it satisfies the following:
\begin{enumerate}
\item $i_*$, $j_!$, and $j_*$ are fully faithful.
\item $(i^* , i_* )$, $(i_* , i^!)$, $(j_! , j^* )$, and $(j^* , j_* )$ are adjoint pairs.
\item there are canonical embeddings $\operatorname{Im} j_! \hookrightarrow \operatorname{Ker}i^*$, $\operatorname{Im}i_* \hookrightarrow\operatorname{Ker} j^*$, and \\
$\operatorname{Im} j_*\hookrightarrow\operatorname{Ker}i^!$ which are equivalences.
\end{enumerate}
\end{defn}
We remember that a recollement corresponds to a pair of
consecutive stable t-structures.
\begin{prop} [\cite{Mi1}] \label{st-t-1second}
~
\begin{enumerate}
\item Let
\[\xymatrix{
\mcal{D'} \ar@<-1ex>[r]^{i_{*}}
& \mcal{D}
\ar@/^1.5pc/[l]_{i^{!}} \ar@/_1.5pc/[l]^{i^{*}} \ar@<-1ex>[r]^{j^{*}}
& \mcal{D''}
\ar@/_1.5pc/[l]^{j_{!}} \ar@/^1.5pc/[l]_{j_{*}}
}\] be a recollement. Then
$(\mcal{U}, \mcal{V} )$ and $(\mcal{V}, \mcal{W})$
are stable t-structures in $\mcal{D}$
where we put $\mcal{U} = \operatorname{Im}j_{!}$, $\mcal{V} =\operatorname{Im}i_{*}$ and $\mcal{W} =\operatorname{Im}j_{*}$.
\item
Let $(\mcal{U}, \mcal{V})$ and $(\mcal{V}, \mcal{W})$ be stable t-structures in $\mcal{D}$.
Then for the canonical embedding $i_{*}:\mcal{V} \to \mcal{D}$,
there is a recollement
\[\xymatrix{
\mcal{V} \ar@<-1ex>[r]^{i_{*}}
& \mcal{D}
\ar@/_1.5pc/[l]^{i^{*}} \ar@/^1.5pc/[l]_{i^{!}} \ar@<-1ex>[r]^{j^{*}}
& \mcal{D}/\mcal{V}
\ar@/_1.5pc/[l]^{j_{!}} \ar@/^1.5pc/[l]_{j_{*}}
}\]
such that $\operatorname{Im}j_! = \mcal{U}$ and
$\operatorname{Im}j_* =\mcal{W}$.
\end{enumerate}
In each cases, for every object $X$ of ${\mcal D}$ adjunction arrows of adjoints induce triangles
\[ \begin{aligned} i_* i^! X \to X \to j_* j^* X \to \Sigma i_* i^! X , \\
j_! j^* X \to X \to i_* i^* X \to \Sigma j_! j^* X .
\end{aligned} \]
\end{prop}
Thirdly, we introduce the notion of a polygon of recollements.
\begin{defn}
Let $\mcal{D}$ be a triangulated category, and let $\mcal{U}_{1}, \cdots, \mcal{U}_{n}$
be full triangulated subcategories of $\mcal{D}$.
We call $(\mcal{U}_1, \mcal{U}_2, \cdots , \mcal{U}_n)$
an $n$-gon of recollements in $\mcal{D}$
if $(\mcal{U}_{i}, \mcal{U}_{i+1})$ is a stable t-structure in $\mcal{D}$ ($1 \leq i \leq n$), where
$\mcal{U}_1=\mcal{U}_{n+1}$.
\end{defn}
An $n$-gon of recollements results in strong symmetry, and it induces three recollements
as the name suggests.
\begin{prop}[\cite{IKM}]\label{added}
Let $\mcal{D}$ be a triangulated category.
Then $(\mcal{U}_1, \cdots, \mcal{U}_n)$ is an $n$-gon of recollements in $\mcal{D}$ if and only if there is a recollement
\[\xymatrix{
\mcal{U}_t \ar@<-1ex>[r]^{i_{t*}}
& \mcal{D}
\ar@/_1.5pc/[l]^{i_{t}^{*}} \ar@/^1.5pc/[l]_{i_{t}^{!}} \ar@<-1ex>[r]^{j_{t}^{*}}
& \mcal{D}/\mcal{U}_t
\ar@/_1.5pc/[l]^{j_{t!}} \ar@/^1.5pc/[l]_{j_{t*}}
}\]
such that the essential image $\operatorname{Im} j_{t!}$ is $\mcal{U}_{t-1}$, and that
the essential image $\operatorname{Im}j_{t*}$ is $\mcal{U}_{t+1}$
for any $t \!\! \mod \! n$.
In this case all the relevant subcategories $\mcal{U}_t$ and
the quotient categories $\mcal{D}/\mcal{U}_t$ are triangle equivalent.
\end{prop}
Finally we study the case that triangle functors preserve
localizations, colocalizations or recollements, etc.
\begin{defn} Let $\mcal{D}_1$ and $\mcal{D}_2$ be triangulated categories and
let $F:\mathcal{D}_1 \to \mathcal{D}_2$ be a triangle functor.
\begin{enumerate}
\item Let $(\mathcal{U}_n , \mathcal{V}_n)$ be a stable t-structure in $\mathcal{D}_n$ $(n=1,2)$.
We say that \emph{$F$ sends $( \mcal{U}_1 , \mcal{V}_1 )$ to $( \mcal{U}_2 , \mcal{V}_2 )$}
if $F(\mcal{U}_1 )$ is contained in $\mcal{U}_2$ and
$F(\mcal{V}_1 )$ is in $\mcal{V}_2$.
\item Let $( \mcal{U}_{in} , \cdots, \mcal{U}_{in})$ be
an $n$-gon of recollements in $\mathcal{D}_i$ $(i=1,2)$.
We say that \emph{$F$ sends $( \mcal{U}_{1n} , \cdots , \mcal{U}_{1n})$ to
$( \mcal{U}_{2n} , \cdots, \mcal{U}_{2n})$}
if $F(\mcal{U}_{1k} )$ is contained in $\mcal{U}_{2k}$ for any $k$.
\end{enumerate}
\end{defn}
\begin{lem}[\cite{IKM}]\label{17-1Apr30}
If a triangle functor $F:\mcal{D}_{1} \to \mcal{D}_{2}$
sends a stable t-structure $( \mcal{U}_1 , \mcal{V}_1 )$ in $\mcal{D}_1$
to a stable t-structure $( \mcal{U}_2 , \mcal{V}_2 )$ in $\mcal{D}_2$.
Then we have the following:
\begin{enumerate}
\item If $F\mid _{\mcal{U}_1 }$ is full (resp., faithful), then
$\operatorname{Hom} _{\mcal{D}_1}( U, X) \to \operatorname{Hom} _{\mcal{D}_2}( FU, FX)$ is
surjective (resp., injective) for
$U\in {\mcal{U}_1}$ and $X\in {\mcal{D}_1}$.
\item If $F\mid _{\mcal{V}_1 }$ is full (resp., faithful), then
$\operatorname{Hom} _{\mcal{D}_1}( X, V) \to \operatorname{Hom} _{\mcal{D}_2}( FX, FV)$ is
surjective (resp., injective) for
$X\in {\mcal{D}_1}$ and $V\in {\mcal{V}_1}$.
\item If $F$ is full and $F\mid _{\mcal{U}_1}:\mcal{U}_1\to\mcal{U}_2$ and
$F\mid_{\mcal{V}_1}:\mcal{V}_1\to\mcal{V}_2$ are dense,
then $F$ is dense.
\end{enumerate}
\end{lem}
\begin{prop}\label{p20Apr30}
Let $\mcal{D}_1$, $\mcal{D}_2$ be triangulated categories.
Let $( \mcal{U}_1 ,\cdots , \mcal{U}_{n})$ and
$( \mcal{V}_1 ,\cdots , \mcal{V}_{n})$
be $n$-gons of recollements in $\mcal{D}_1$ and $\mcal{D}_2$, respectively.
Assume a triangle functor $F:\mcal{D}_{1} \to \mcal{D}_{2}$
sends $( \mcal{U}_1 , \cdots, \mcal{U}_{n} )$ to
$( \mcal{V}_1 , \cdots, \mcal{V}_{n} )$.
Then the following hold.
\begin{enumerate}
\item In the case that $n$ is odd,
if $F\mid _{\mcal{U}_t }$ is fully faithful (equivalent) for some $t$, so is $F$.
\item In the case that $n$ is even,
if $F\mid _{\mcal{U}_t }$ and $F\mid _{\mcal{U}_{t+1}}$
are fully faithful (equivalent) for some $t$, so is $F$.
\end{enumerate}
\end{prop}
\begin{proof}
By Lemma \ref{17-1Apr30}.
\end{proof}
\section{Contravariantly finite subcategories and Stable $t$-structures}\label{n-gon}
In this section let $k$ be a field and $D:=\operatorname{Hom}_k(-,k)$.
The concept of stable $t$-structures is closely related to functorially finite subcategories \cite{AS}.
If $(\mcal{U},\mcal{V})$ is a stable $t$-structure of $\mcal{D}$, then clearly $\mcal{U}$ (resp., $\mcal{V}$)
is a contravariantly (resp., covariantly) finite subcategory of $\mcal{D}$.
We shall show that a certain converse of this statement holds.
For a full subcategory $\mcal{U}$ of $\mcal{D}$, we put
\begin{eqnarray*}
\mcal{U}^\perp&:=&\{T\in\mcal{D}\ |\ \operatorname{Hom}_{\mcal{D}}(\mcal{U},T)=0\},\\
{}^\perp\mcal{U}&:=&\{T\in\mcal{D}\ |\ \operatorname{Hom}_{\mcal{D}}(T,\mcal{U})=0\}.
\end{eqnarray*}
Recall that an additive category is called \emph{Krull-Schmidt}
if any object is isomorphic to a finite direct sum of objects whose endomorphism rings are local.
\begin{defn}
Let $\mcal{C}$ be an additive category, and $\mcal{U}$ its full subcategory.
For an object $X$ of $\mcal{C}$, a morphism $U_X \xrightarrow{f} X$ with $U_X \in \mcal{U}$ is called
a \emph{right $\mcal{U}$-approximation} if
$\operatorname{Hom}_{\mcal{C}}(U,f):\operatorname{Hom}_{\mcal{C}}(U,U_X) \to \operatorname{Hom}_{\mcal{C}}(U,X)$
is surjective for any $U\in \mcal{U}$.
Moreover, a right $\mcal{U}$-approximation $U_X \xrightarrow{f} X$ is called minimal if $g$ is an isomorphism
whenever $g: U_X \to U_X$ satisfies $f\circ g=f$.
A full subcategory $\mcal{U}$ is called a \emph{contravariantly finite} subcategory if
every object $X$ of $\mcal{C}$ has a right a $\mcal{U}$-approximation.
A left a $\mcal{U}$-approximation and a covariantly finite subcategory are defined dually.
$\mcal{U}$ is called a functorially finite subcategory if it is contravariantly finite and covariantly finite.
\end{defn}
\begin{prop}\label{wakamatsu}
Let $\mcal{D}$ be a Krull-Schmidt triangulated category.
For any contravariantly (resp., covariantly) finite thick subcategory $\mcal{U}$ of $\mcal{D}$,
we have a stable $t$-structure $(\mcal{U},\mcal{U}^\perp)$ (resp., $({}^\perp\mcal{U},\mcal{U})$) in $\mcal{D}$.
\end{prop}
\begin{proof}
This is a consequence of Wakamatsu-type Lemma (see \cite[Prop. 3.6]{IY} for example).
For the convenience of the reader, we give the proof here.
Since $\mcal{D}$ is Krull-Schmidt category and $\mcal{U}$ is closed under direct summands,
for any object $X$ of $\mcal{D}$ there is a triangle
$U_X \xrightarrow{f} X \xrightarrow{g} V \xrightarrow{h} \Sigma U_X$ such that
$U_X \xrightarrow{f} X$ is a minimal right $\mcal{U}$-approximation.
For any morphism $\alpha : U \to V$ with $U \in \mcal{U}$, by octahedral axiom we have a commutative diagram
\[\xymatrix{
\Sigma^{-1}U \ar@{=}[r]\ar[d]^{\Sigma^{-1}\alpha} & \Sigma^{-1}U \ar[d] \\
\Sigma^{-1}V \ar[r]^{-\Sigma^{-1}h} \ar[d]^{\beta} & U_X \ar[r]^{f} \ar[d]^{\gamma}
& X \ar[r]^{g} \ar@{=}[d] & V \ar[d]^{\Sigma\beta} \\
Z \ar[r]^{h'} \ar[d] & M \ar[r]^{f'} \ar[d] & X \ar[r]^{g'} & \Sigma Z \\
U \ar@{=}[r] & U
}\]
where all columns and rows of 4 terms are triangles. Then $M$ belongs to $\mcal{U}$.
Since $f$ is a right $\mcal{U}$-approximation, there is a morphism $\delta:M \to U_X$
such that $f\circ\delta=f'$. Then there is a morphism $\epsilon: Z \to \Sigma^{-1}V$
such that $-\Sigma^{-1}h\circ\epsilon=\delta\circ h'$.
Then $\gamma$ is a split monomorphism because of the minimality of $f$.
Therefore $\beta$ is a split monomorphism, and $\Sigma^{-1}\alpha$ and $\alpha$ are zero.
Hence we have $V \in \mcal{U}^{\perp}$.
\end{proof}
\begin{defn}
Let $\mcal{D}$ be a $k$-linear triangulated category such that \\
$\dim_k\operatorname{Hom}_{\mcal{D}}(X,Y)<\infty$ for any $X,Y\in\mcal{D}$.
An autofunctor $S:\mcal{D}\to\mcal{D}$ is called a \emph{Serre functor} \cite{BK,RV} if there exists a functorial isomorphism
\[\operatorname{Hom}_{\mcal{D}}(X,Y)\simeq D\operatorname{Hom}_{\mcal{D}}(Y, SX)\]
for any $X,Y\in\mcal{D}$.
We say that $\mcal{D}$ is \emph{$(m/n)$-Calabi-Yau} for a positive integer $n$ and an integer $m$ if we have an isomorphism $S^n\simeq\Sigma^m$ of functors
\footnote{We notice that we can not cancel a common divisor of $m$ and $n$.}.
\end{defn}
Immediately we have the following.
\begin{lem}\label{properties of serre}
Let $\mcal{D}$ be a Krull-Schmidt triangulated category with a Serre functor $S$. Then the following hold.
\begin{itemize}
\item[(1)] $\mcal{U}^\perp={}^\perp(S\mcal{U})$ for any subcategory $\mcal{U}$ of $\mcal{D}$.
\item[(2)] If $(\mcal{U},\mcal{V})$ is a stable $t$-structure and $\mcal{U}$ is functorially finite in $\mcal{D}$,
then $(\mcal{V},S\mcal{U})$ is a stable $t$-structure and $\mcal{V}$ is functorially finite in $\mcal{D}$.
\end{itemize}
\end{lem}
\begin{proof}
(1) Immediate from the definition of Serre functor.
(2) We have ${}^\perp(S\mcal{U})=\mcal{U}^\perp=\mcal{V}$ by (1).
Since $S\mcal{U}$ is a covariantly finite subcategory of $\mcal{D}$,
we have that $(\mcal{V},S\mcal{U})$ is a stable $t$-structure by the dual of Proposition \ref{wakamatsu}.
Consequently $\mcal{V}$ is a functorially finite subcategory of $\mcal{D}$ by Proposition \ref{wakamatsu}.
\end{proof}
\begin{prop} \label{n-gon1}
Let $\mcal{D}$ be an $(m/n)$-Calabi-Yau triangulated category.
For any functorially finite thick subcategory $\mcal{U}_1$ of $\mcal{D}$, we put $\mcal{U}_{i+1}:=\mcal{U}_i^\perp$ for any $i$.
Then we have an $l$-gon $(\mcal{U}_{1}, \cdots , \mcal{U}_{l})$ of recollements in $\mcal{D}$
for some positive divisor $l$ of $2n$.
\end{prop}
\begin{proof}
By Propositions \ref{wakamatsu}, we have a stable $t$-structure $(\mcal{U}_1,\mcal{U}_{2})$ in $\mcal{D}$.
Using Lemma \ref{properties of serre}(2) inductively, we have a stable $t$-structure $(\mcal{U}_i,\mcal{U}_{i+1})$ in $\mcal{D}$
such that $\mcal{U}_{i+2}=S\mcal{U}_i$ for any $i$.
We have the statement because of $S^n\simeq \Sigma^m$.
\end{proof}
\section{Constructions of Stable $t$-structures}\label{n-gonII}
In this section, we investigate polygons of recollements in derived categories and in stable module categories.
For a ring $R$, we denote by $\operatorname{\mathsf{Mod}} R$ (resp., $\operatorname{\mathsf{mod}} R$) the category of right
(resp., finitely generated right) $R$-modules, and denote by $\operatorname{\mathsf{Proj}} {R}$ (resp., $\operatorname{\mathsf{Inj}} R$, $\operatorname{\mathsf{proj}} {R}$)
the full subcategory of
$\cat{Mod}R$ consisting of projective (resp., injective, finitely generated projective) modules.
For right (resp., left) $R$-module $M_R$ (resp., ${}_{R}N$), we denote by
$\operatorname{idim}M_R$ (resp., $\operatorname{idim}{}_{R}N$) the injective dimension
of $M_{R}$ (resp., ${}_{R}N$), and by
$\operatorname{pdim}M_R$ (resp., $\operatorname{pdim}{}_{R}N$) the projective dimension
of $M_{R}$ (resp., ${}_{R}N$).
For $\mcal{A}$ be an abelian category and its addtive subcategory $\mcal{B}$.
we denote by $\cat{D}^{\mrm{b}}(\mcal{A})$ (resp., $\cat{K}^{\mrm{b}}(\mcal{B}$)
the derived category (resp., the homotopy category) of bounded complexes of objects of $\mcal{A}$ (resp., $\mcal{B}$).
\begin{defn}
We call a ring $R$ \emph{Iwanaga-Gorenstein} if it is Noetherian
with $\operatorname{idim}_RR<\infty$ and $\operatorname{idim}R_R<\infty$ \cite{Iw}.
We define the category of \emph{Cohen-Macaulay $R$-modules}
\footnote{
In the representation theory of orders \cite{CR, A2, Y}, there is another
notion of Cohen-Macaulay modules which generalizes the classical
notion in commutative ring theory. These two concepts coincide for
Gorenstein orders.
}
and the category of \emph{large Cohen-Macaulay $R$-modules} by
\begin{eqnarray*}
\operatorname{\mathsf{CM}} R&:=&\{X\in\operatorname{\mathsf{mod}} R\ |\ \operatorname{Ext}^i_R(X,R)=0\ (i>0)\},\\
\operatorname{\mathsf{LCM}} R&:=&\{X\in\operatorname{\mathsf{Mod}} R\ |\ \operatorname{Ext}^i_R(X, \operatorname{\mathsf{Proj}} R)=0\ (i>0)\}.
\end{eqnarray*}
\end{defn}
Then $\operatorname{\mathsf{CM}} R$ forms a Frobenius category with the subcategory
$\operatorname{\mathsf{proj}} R$ of projective-injective objects, and the stable
category $\underline{\operatorname{\mathsf{CM}}}R$ forms a triangulated category \cite{H1}.
By \cite{IKM} there exist triangle equivalences
\[
\underline{\operatorname{\mathsf{CM}}} R\simeq\cat{D}^{\mrm{b}}\!(\operatorname{\mathsf{mod}} R)/\cat{K}^{\mrm{b}}(\operatorname{\mathsf{proj}} {R}),\quad
\underline{\operatorname{\mathsf{LCM}}} R\simeq\cat{D}^{\mrm{b}}\!(\operatorname{\mathsf{Mod}} R)/\cat{K}^{\mrm{b}}(\operatorname{\mathsf{Proj}} {R}).
\]
For subcategories $\mcal{U}$ and $\mcal{V}$ of a triangulated category $\mcal{D}$, we put
\[\mcal{U}*\mcal{V}:=\{X\in\mcal{D}\ |\ U \to X \to V \to \Sigma U\
\text{is a triangle in}\ \mcal{D}\ (U\in \mcal{U}, V \in \mcal{V})\}.\]
By octahedral axiom, we have $(\mcal{U}*\mcal{V})*\mcal{W}=\mcal{U}*(\mcal{V}*\mcal{W})$.
\begin{lem}\label{about *}
Let $\mcal{D}$ be a triangulated subcategory, and $\mcal{U}$ and $\mcal{V}$ triangulated (resp., thick) subcategories
of $\mcal{D}$ satisfying $\operatorname{Hom}_{\mcal{D}}(\mcal{U},\mcal{V})=0$.
Then $\mcal{U}*\mcal{V}$ is a triangulated (resp., thick) subcategory of $\mcal{D}$.
\end{lem}
\begin{proof}
We only have to show $(\mcal{U}*\mcal{V})*(\mcal{U}*\mcal{V})\subset\mcal{U}*\mcal{V}$.
Since $\operatorname{Hom}_{\mathcal{D}}(\mcal{U},\Sigma\mcal{V})=0$, we have $\mcal{V}*\mcal{U}=\cat{add}\{\mcal{U},\mcal{V}\}$.
Thus we have $(\mcal{U}*\mcal{V})*(\mcal{U}*\mcal{V})=\mcal{U}*(\mcal{V}*\mcal{U})*\mcal{V}=\mcal{U}*\cat{add}\{\mcal{U},\mcal{V}\}*\mcal{V}\subset(\mcal{U}*\mcal{U})*(\mcal{V}*\mcal{V})=\mcal{U}*\mcal{V}$,
where $\operatorname{\mathsf{add}}\{\mcal{U},\mcal{V}\}$ is the additive subcategory of consisting of finite direct sums of objects of $\mcal{U}$
and $\mcal{V}$.
Thus $\mcal{U}*\mcal{V}$ is a triangulated subcategory of $\mcal{D}$.
If $\mcal{U}$ and $\mcal{V}$ are closed under direct summand, then so is $\mcal{U}*\mcal{V}$ (e.g. \cite[Prop. 2.1]{IY}).
Thus the assertion for thick subcategories follows.
\end{proof}
Let $A$ and $R$ be $k$-algebras.
For a subcategory $\mcal{U}$ of $\cat{D}^{\mrm{b}}(\operatorname{\mathsf{mod}} A)$,
we denote by $\mcal{U}^R$ the thick subcategory of $\cat{D}^{\mrm{b}}(\operatorname{\mathsf{mod}} R\otimes_kA)$ generated by
\[\{L\otimes_kX\ |\ L\in\cat{D}^{\mrm{b}}(\operatorname{\mathsf{mod}} R),\ X\in\mcal{U}\}.\]
The following observation gives us a lot of examples of stable $t$-structures in derived categories.
\begin{prop} \label{n-gon2}
Let $R$ be a $k$-algebra and $A$ a finite dimensional $k$-algebra such that $A/J_A$ is a separable $k$-algebra, where $J_A$ is the Jacobson radical.
For any stable $t$-structure $(\mcal{U},\mcal{V})$ in $\cat{D}^{\mrm{b}}(\operatorname{\mathsf{mod}} A)$, we have a stable $t$-structure $(\mcal{U}^R,\mcal{V}^R)$ in $\cat{D}^{\mrm{b}}(\operatorname{\mathsf{mod}} R\otimes_kA)$.
\end{prop}
\begin{proof}
Let $\mcal{D}:=\cat{D}^{\mrm{b}}(\operatorname{\mathsf{mod}} R\otimes_kA)$.
Since
\[
\operatorname{Hom}_{\mathcal{D}}(L\otimes_kU,M\otimes_kV)=
\operatorname{Hom}_{\cat{D}^{\mrm{b}}(\operatorname{\mathsf{mod}} R)}(L,M)\otimes_k\operatorname{Hom}_{\cat{D}^{\mrm{b}}(\operatorname{\mathsf{mod}} A)}(U,V)
\]
for any $L, M \in \cat{D}^{\mrm{b}}(\operatorname{\mathsf{mod}} R)$ and any $U, V \in \cat{D}^{\mrm{b}}(\operatorname{\mathsf{mod}} A)$,
we have $\operatorname{Hom}_{\mathcal{D}}(\mcal{U}^R,\mcal{V}^R)=0$.
Since $\mcal{U}^R*\mcal{V}^R$ is a thick subcategory of $\mcal{D}$ by Lemma \ref{about *},
we only have to show $\mcal{U}^R*\mcal{V}^R$ contains $\operatorname{\mathsf{mod}} R\otimes_kA$.
Any $R\otimes_kA$-module $M$ is filtered by $R\otimes_kA$-modules $MJ_A^i/MJ_A^{i+1}$ which are semisimple $A$-modules.
We only have to show that any $R\otimes_kA$-module $N$ which is a semisimple $A$-modules belongs to $\mcal{U}^R*\mcal{V}^R$.
Since the map $(A/J_A)\otimes_k(A/J_A)\to A/J_A$, $x\otimes y\mapsto xy$ is a split epimorphism of $A^{\mrm{op}}\otimes_kA$-modules,
we have that the map $N\otimes_k(A/J_A)\to N$, $n\otimes y\mapsto ny$ is a split epimorphism of $R\otimes_kA$-modules.
Since $A/J_A\in\mcal{U}*\mcal{V}$, we have that $N\otimes_k(A/J_A)\in\mcal{U}^R*\mcal{V}^R$. Thus $N\in\mcal{U}^R*\mcal{V}^R$.
\end{proof}
The following result gives a criterion for a stable $t$-structure in the derived category to give a stable $t$-structure in the stable category.
\begin{lem}[\cite{IKM}] \label{st-stK}
Let $\mathcal{D}$ be a triangulated category, $\mathcal{C}$ a thick
subcategory of $\mcal{D}$, and
$Q:\mathcal{D} \to \mathcal{D}/\mathcal{C}$ the canonical quotient \cite{Ne2}.
For a stable $t$-structure $(\mathcal{U}, \mathcal{V})$ in $\mathcal{D}$, the following are equivalent,
where $Q(\mathcal{U})$ is the full subcategory of $\mcal{D}/\mcal{C}$ consisting of objects
$Q(X)$ for $X \in \mcal{E}$.
\begin{enumerate}
\item $(Q(\mathcal{U}), Q(\mathcal{V}))$ is a stable $t$-structure in $\mathcal{D}/\mathcal{C}$.
\item $(\mathcal{U}\cap\mathcal{C}, \mathcal{V}\cap\mathcal{C})$ is a stable $t$-structure in
$\mathcal{C}$.
\end{enumerate}
\end{lem}
The following example provides us a rich source of triangulated categories with Serre functors.
\begin{prop}\label{serre functor for A}
Let $A$ be a finite dimensional $k$-algebra of finite self-injective dimension as both sides.
Then the following hold.
\begin{itemize}
\item[(1)] $\cat{K}^{\mrm{b}}(\operatorname{\mathsf{proj}} {A})$ has a Serre functor $\nu_A:=-{\otimes}^{\boldsymbol{L}}_{A}(DA)$.
\item[(2)] $\cat{K}^{\mrm{b}}(\operatorname{\mathsf{proj}} {A})$ is $(m/n)$-Calabi-Yau if and only if $(DA)^{{\otimes}^{\boldsymbol{L}}_{A}n}\simeq\Sigma^mA$ in \\
$\cat{D}^{\mrm{b}}(\operatorname{\mathsf{mod}} A^{\mrm{op}}\otimes_k A)$.
\end{itemize}
\end{prop}
We have the following main result in this section.
\begin{thm} \label{n-gon3}
Let $A$ be a finite dimensional $k$-algebra of finite global dimension such that $A/J_A$ is separable over $k$ and
$\cat{D}^{\mrm{b}}(\operatorname{\mathsf{mod}} A)$ is $(m/n)$-Calabi-Yau.
Let $R$ be a coherent $k$-algebra of finite self-injective dimension as both sides.
For any functorially finite thick subcategory $\mcal{U}_1$ of $\cat{D}^{\mrm{b}}(\operatorname{\mathsf{mod}} A)$, we put $\mcal{U}_{i+1}:=\mcal{U}_i^\perp$ for any $i$.
Then there is a positive divisor $l$ of $2n$ such that we have an $l$-gon
$(\mcal{U}_{1}^{R}, \cdots , \mcal{U}_{l}^{R})$ of recollements in $\cat{D}^{\mrm{b}}(\operatorname{\mathsf{mod}} R\ten_k A)$
and an $l$-gon $(Q(\mcal{U}_{1}^{R}), \cdots , Q(\mcal{U}_{l}^{R}))$ of recollements in
$\underline{\operatorname{\mathsf{CM}}}(R\ten_k A)$, where
$Q :\cat{D}^{\mrm{b}}(\operatorname{\mathsf{mod}} R\ten_k A)\to \underline{\operatorname{\mathsf{CM}}}(R\ten_k A)$ is the canonical quotient.
\end{thm}
\begin{proof}
According to Propositions \ref{n-gon1} and \ref{n-gon2},
we have an $n$-gon $(\mcal{U}_{1}^{R}, \mcal{U}_{2}^{R}, \cdots , \mcal{U}_{2n}^{R})$
of recollements in $\cat{D}^{\mrm{b}}(\operatorname{\mathsf{mod}} R\otimes_kA)$.
Since $A$ is of finite global dimension, there exists a triangle
\[U_i\to A\to U_{i+1}\to\Sigma U_i\]
with $U_i\in\mcal{U}_i\cap\cat{K}^{\mrm{b}}(\operatorname{\mathsf{proj}} {A})$ and $U_{i+1}\in\mcal{U}_{i+1}\cap\cat{K}^{\mrm{b}}(\operatorname{\mathsf{proj}} {A})$.
Applying $R\ten_k-$, we have a triangle
\[R\ten_kU_i\to R\ten_kA\to R\ten_kU_{i+1}\to\Sigma R\ten_kU_i\]
with $R\ten_kU_i\in\mcal{U}_{i}^{R}\cap\cat{K}^{\mrm{b}}(\operatorname{\mathsf{proj}} {R\ten_kA})$ and $R\ten_kU_{i+1}\in\mcal{U}_{i+1 }^{R}\cap\cat{K}^{\mrm{b}}(\operatorname{\mathsf{proj}} {R\ten_kA})$.
By Lemma \ref{st-stK}, we have a stable $t$-structure $(Q(\mcal{U}_{i}^{R}), Q(\mcal{U}_{i+1 }^{R}))$ in $\underline{\operatorname{\mathsf{CM}}}(R\ten_k A)$.
\end{proof}
We have the following example of recollements by \cite[Cor. 5.11]{Mi1}.
\begin{prop}\label{recollement from idempotent}
Let $A$ be a finite dimensional $k$-algebra, and $e$ an idempotent of $A$.
Assume that $\operatorname{Ext}_{A}^{i}(A/AeA,A/AeA)=0$ ($i>0$), $\operatorname{pdim}{}_A(AeA)<\infty$ and $\operatorname{pdim}(AeA)_A<\infty$.
Then we have a recollement
\[\xymatrix{
\cat{D}^{\mrm{b}}(\operatorname{\mathsf{mod}} A/AeA) \ar@<-1ex>[r]^{i_{e*}}
& \cat{D}^{\mrm{b}}(\operatorname{\mathsf{mod}} A)
\ar@/_1.5pc/[l]^{i_{e}^{*}} \ar@/^1.5pc/[l]_{i_{e}^{!}} \ar@<-1ex>[r]^{j_{e}^{*}}
& \cat{D}^{\mrm{b}}(\operatorname{\mathsf{mod}} eAe)
\ar@/_1.5pc/[l]^{j_{e!}} \ar@/^1.5pc/[l]_{j_{e*}}
}\]
In particuler, $(\operatorname{Im}j_{e!}, \operatorname{Im}i_{e}^{*})$ and
$(\operatorname{Im}i_{e}^{*}, \operatorname{Im}j_{e*})$ are stable $t$-structures in
$\cat{D}^{\mrm{b}}(\operatorname{\mathsf{mod}} A)$.
\end{prop}
\section{Recollement of $\cat{K}(\operatorname{\mathsf{Mor}}_{N-1}^{\mrm{sm}}(\mcal{B}))$}\label{Tcpx}
In this section we study the properties of complexes of the category of $N-1$ sequences
of morphisms in an additive category $\mcal{B}$.
Throughout this section $\mcal{B}$ is an additive category.
Then the category $\cat{C}(\mcal{B})$ of complexes of objects of $\mcal{B}$
is a Frobenius category such that its cconflations are
short exact sequences of which each term is a split exact sequence in $\mcal{B}$.
\begin{defn}\label{smcat}
We define the category $\operatorname{\mathsf{Mor}}^{\mrm{sm}}_{N-1}(\mcal{B})$ (resp., $\operatorname{\mathsf{Mor}}_{N-1}(\mcal{B})$)
of sequences of morphisms in $\mcal{B}$ as follows.
\begin{itemize}
\item An object is a sequence of split monomorphisms (resp., morphisms)
$X: X^{1} \xrightarrow{\alpha_X^1} \cdots \xrightarrow{\alpha_X^{N-2} } X^{N-1}$ in $\mcal{B}$.
\item A morphism from $X$ to $Y$ is an $(N-1)$-tuple $f=(f^1,\cdots , f^{N-1})$ of morphisms $f^i:X^{i} \to Y^{i}$ such that
$f^{i+1}\alpha_X^{i} = \alpha_Y^{i+1} f^{i}$ for $1 \leq i \leq N-2$.
\end{itemize}
\end{defn}
We give technical tools to investigate the homotopy category
$\cat{K}(\operatorname{\mathsf{Mor}}_{N-1}^{\mrm{sm}}(\mcal{B}))$.
\begin{defn}\label{comma01}
For an additive functor $G: \mcal{B} \to \mcal{B}'$ between additive categories,
let $(G\downarrow \mathbf{1}_{\mcal{B}'})$ be a comma category, that is the category of objects
$G(X) \xrightarrow{\alpha} Y$ for $X \in \mcal{A}, Y \in \mcal{B}'$.
We denote by $(G\downarrow^{\mrm{sm}} \mathbf{1}_{\mcal{B}'})$ the subcategory of
$(G\downarrow \mathbf{1}_{\mcal{B}'})$ consisting of objects
$G(X) \xrightarrow{\alpha} Y$, where $\alpha$ are split monomorphisms.
\end{defn}
\begin{exmp}\label{comma02}
For $1 \leq r < N-1$, let $G:\operatorname{\mathsf{Mor}}_{r}(\mcal{B}) \to \operatorname{\mathsf{Mor}}_{N-r-1}(\mcal{B})$
(resp., $G:\operatorname{\mathsf{Mor}}^{\mrm{sm}}_{r}(\mcal{B}) \to \operatorname{\mathsf{Mor}}^{\mrm{sm}}_{N-r-1}(\mcal{B})$)
be an additive functor defined by
\[
G(X^{1} \xrightarrow{\alpha^1} \cdots \xrightarrow{\alpha^{r-1}} X^{r})=Y^{1} \xrightarrow{\beta^1} \cdots \xrightarrow{\beta^{N-r-2}} Y^{N-r-1}
\]
where $Y^{1}=\cdots =Y^{N-r-1}=X^{r}$ and $\beta^{1}=\cdots=\beta^{N-r-2}=1_{X^{r}}$.
Then the category $(G\downarrow \mathbf{1}_{\operatorname{\mathsf{Mor}}_{N-r-1}(\mcal{B})})$
(resp., $(G\downarrow^{\mrm{sm}} \mathbf{1}_{\operatorname{\mathsf{Mor}}_{N-r-1}(\mcal{B})})$)
is equivalent to $\operatorname{\mathsf{Mor}}_{N-1}(\mcal{B})$
(resp., $\operatorname{\mathsf{Mor}}^{\mrm{sm}}_{N-1}(\mcal{B})$).
\end{exmp}
\begin{lem}\label{lem:comma03}
For an additive functor $G: \mcal{B} \to \mcal{B}'$ between additive categories,
the following hold.
\begin{enumerate}
\item Evrey complex of $(G\downarrow^{\mrm{sm}} \mathbf{1}_{\mcal{B}'})$ has the following form:
\[
G(X) \xrightarrow{u_f} C(f)
\]
where $f: Y \to G(X)$ is a morphism of complexes of $\mcal{B}'$,
$C(f)$ is the mapping cone of $f$, and $u_f$ is the canonical morphism.
\item If a complex $X$ of $\mcal{B}$ is homotopically trivial, then
a complex $G(X) \xrightarrow{u_f} C(f)$ of $(G\downarrow^{\mrm{sm}} \mathbf{1}_{\mcal{B}'})$
is isomorphic to $0 \to\Sigma_{\mcal{B}'}Y$ in $\cat{K}(G\downarrow^{\mrm{sm}} \mathbf{1}_{\mcal{B}'})$, where $f: Y \to G(X)$.
\item For a complex $G(X) \xrightarrow{u_f} C(f)$ of $(G\downarrow^{\mrm{sm}} \mathbf{1}_{\mcal{B}'})$,
if $C(f)$ is homotopically trivial, then $G(X) \xrightarrow{u_f} C(f)$ is isomorphic to
$G(X) \xrightarrow{u_1} C(1_{G(X)})$ in in $\cat{K}(G\downarrow^{\mrm{sm}} \mathbf{1}_{\mcal{B}'})$.
\end{enumerate}
\end{lem}
\begin{proof}
(1) It is trivial.\par \noindent
(2) For a morphism $Y \xrightarrow{f} G(X)$ of complexes of $\mcal{B}'$,
we have a triangle
in $\cat{K}(G\downarrow^{\mrm{sm}} \mathbf{1}_{\mcal{B}'})$:
\[\xymatrix{
0 \ar[d] \ar[r] & G(X) \ar@{=}[d] \ar[r]^{1} & G(X) \ar[d]^{u_f} \ar[r] &0 \ar[d]\\
Y \ar[r]^{f} & G(X) \ar[r]^{u_f} & C(f) \ar[r]^{v_f} & \Sigma_{\mcal{B}'}Y
}\]
If $G(X)$ is homotopically trivial, then $G(X) \xrightarrow{1} G(X)$ is $0$ in
$\cat{K}(G\downarrow^{\mrm{sm}} \mathbf{1}_{\mcal{B}'})$, and hence
$G(X) \xrightarrow{u_f} C(f)$ is isomorphic to $0 \to\Sigma_{\mcal{B}'}Y$.\par\noindent
(3) For a morphism $Y \xrightarrow{f} G(X)$ of complexes of $\mcal{B}'$,
we have a morphism between triangles in $\cat{K}(G\downarrow^{\mrm{sm}} \mathbf{1}_{\mcal{B}'})$:
\[
\xymatrix@!0{
& 0\ \ar@{->}[rr]\ar@{->}'[d][dd]
& & G(X)\ \ar@{->}[rr]^{1}\ar@{=}'[d][dd]
& & G(X)\ \ar@{->}[rr]^{1}\ar@{->}'[d]^{u_f}[dd]
& & 0\ \ar@{->}[dd]
\\
0\ \ar@{<-}[ur]\ar@{->}[rr]\ar@{->}[dd]
& & G(X)\ \ar@{->}[rr]^{\quad 1}\ar@{<-}[ur]^{1}\ar@{=}[dd]
& & G(X)\ \ar@{->}[rr]\ar@{<-}[ur]^{1}\ar@{->}[dd]
& & 0\ \ar@{<-}[ur]\ar@{->}[dd]
\\
& Y\ \ar@{->}'[r]^{f}[rr]
& & G(X)\ \ar@{->}'[r]^{u_f}[rr]
& & C(f)\ \ar@{->}'[r]^{v_f}[rr]
& & \Sigma Y\
\\
G(X)\ \ar@{->}[rr]^{1}\ar@{<-}[ur]^{f}
& & G(X)\ \ar@{->}[rr]\ar@{<-}[ur]^{1}
& & C(1_{G(X)})\ \ar@{->}[rr]\ar@{<-}[ur]
& & \Sigma G(X)\ \ar@{<-}[ur]}
\]
If $C(f)$ is homotopically trivial, then $f:Y \to G(X)$ is an isomorphism in $\cat{K}(\mcal{B}')$,
and therefore $(0,f):(0 \to Y) \to (0 \to G(X))$ is an isomorphism in
$\cat{K}(G\downarrow^{\mrm{sm}} \mathbf{1}_{\mcal{B}'})$.
By the above morphism between triangles, $G(X) \xrightarrow{u_f} C(f)$ is isomorphic to
$G(X) \xrightarrow{u_1} C(1_{G(X)})$ in in $\cat{K}(G\downarrow^{\mrm{sm}} \mathbf{1}_{\mcal{B}'})$.
\end{proof}
\begin{defn}\label{adj02}
We define the following functors:
\[\begin{array}{ll}
D_{[s,t]}: \operatorname{\mathsf{Mor}}^{\mrm{sm}}_{N-1}(\mcal{B}) \to \operatorname{\mathsf{Mor}}^{\mrm{sm}}_{t-s+1}(\mcal{B}) &
(1\leq s \leq t \leq N-1) \\
E^{\Uparrow_{r}^{N-1}}:\operatorname{\mathsf{Mor}}_{r}(\mcal{B}) \to \operatorname{\mathsf{Mor}}_{N-1}(\mcal{B}) &
(1\leq r \leq N-2) \\
U_{N-1}:\mcal{B} \to \operatorname{\mathsf{Mor}}_{N-1}(\mcal{B})
\end{array}\]
as follows.
For $X^{1} \xrightarrow{\alpha_X^1} \cdots \xrightarrow{\alpha_X^{N-2} } X^{N-1} \in
\operatorname{\mathsf{Mor}}^{\mrm{sm}}_{N-1}(\mcal{B})$,
\[\begin{aligned}
D_{[s, t]}(X^{1} \xrightarrow{\alpha_X^1} \cdots \xrightarrow{\alpha_X^{N-2} } X^{N-1}) =& \
Y^{1} \xrightarrow{\alpha_Y^1} \cdots \xrightarrow{\alpha_Y^{t-s} } Y^{t-s+1} \\
\end{aligned}\]
where $Y^{i}=X^{i+s-1}, \alpha_Y^i=\alpha_X^{i+s-1}$ $(1 \leq i \leq t-s+1)$.
We denote $D_{[s, s]}$ by $D_{[s]}$.
For $X^{1} \xrightarrow{\alpha_X^1} \cdots \xrightarrow{\alpha_X^{r-1} } X^r \in
\operatorname{\mathsf{Mor}}^{\mrm{sm}}_{r}(\mcal{B})$,
\[\begin{aligned}
E^{\Uparrow_{r}^{N-1}}(X^{1} \xrightarrow{\alpha_X^1} \cdots \xrightarrow{\alpha_X^{r-1} } X^r) =& \
Y^{1} \xrightarrow{\alpha_Y^1} \cdots \xrightarrow{\alpha_Y^{N-2} } Y^{N-1} \\
\end{aligned}\]
{\small
\[
Y^{i}=\begin{cases} 0 \ (1 \leq i < N-r) \\ X^{i-N+r+1} \ (N-r \leq i \leq N-1)\end{cases},
\alpha_{Y}^{i}=\begin{cases} 0 \ (1 \leq i < N-r) \\ \alpha_{X}^{i-N+r+1} \ (N-r \leq i \leq N-1)\end{cases}
\]
}
For $X \in \mcal{B}$,
\[\begin{aligned}
U_{N-1}(X) =& \
Y^{1} \xrightarrow{\alpha_Y^1} \cdots \xrightarrow{\alpha_Y^{N-2} } Y^{N-1} \\
\end{aligned}\]
where $Y^{i}=X, \alpha_Y^i=1_X$ $(1 \leq i \leq N-1)$.
Moreover, we use the same symbols for the corresponding functors
$D_{[s,t]}: \cat{K}(\operatorname{\mathsf{Mor}}^{\mrm{sm}}_{N-1}(\mcal{B})) \to
\cat{K}(\operatorname{\mathsf{Mor}}^{\mrm{sm}}_{t-s+1}(\mcal{B}))$,
$E^{\Uparrow_{r}^{N-1}}:\cat{K}(\operatorname{\mathsf{Mor}}_{r}(\mcal{B})) \to \cat{K}(\operatorname{\mathsf{Mor}}_{N-1}(\mcal{B}))$ and
$U_{N-1}:\cat{K}(\mcal{B}) \to \cat{K}(\operatorname{\mathsf{Mor}}_{N-1}(\mcal{B}))$.
\end{defn}
\begin{defn}\label{ngon@trg01}
We define the following full triangulated subcategories of \\
$\cat{K}(\operatorname{\mathsf{Mor}}^{\mrm{sm}}_{N-1}(\mcal{B}))$:
\[\begin{array}{llll}
\mcal{E}^{[2,N-1]}=\operatorname{Ker}D_{[1]} &
\mcal{E}^{[1,N-2]}=\operatorname{Ker}D_{[N-1]} &
\mcal{E}^{1}=\operatorname{Ker}D_{[2,N-1]} \\
\mcal{E}^{s}=\operatorname{Ker}D_{[1,s-1]}\bigcap\operatorname{Ker}D_{[s+1,N-1]} &
\mcal{E}^{N-1}=\operatorname{Ker}D_{[1,N-2]}
\end{array}\]
For $1\leq s < t \leq N-1$,
$\mcal{F}^{[s,t]}$ is the full triangulated subcategory of $\cat{K}(\operatorname{\mathsf{Mor}}^{\mrm{sm}}_{N-1}(\mcal{B}))$
consisting of objects $X^{1} \xrightarrow{\alpha^1} \cdots \xrightarrow{\alpha^{N-2}} X^{N-1}$
such that $\alpha^{s}=\cdots=\alpha^{t-1}=1$.
\end{defn}
Immediately, we have the following.
\begin{prop}\label{lastpiece}
The following hold.
\begin{enumerate}
\item A functor $E^{\Uparrow_{N-2}^{N-1}}:\cat{K}(\operatorname{\mathsf{Mor}}_{N-2}(\mcal{B})) \to \cat{K}(\operatorname{\mathsf{Mor}}_{N-1}(\mcal{B}))$
induces a triangle equivalence between $\cat{K}(\operatorname{\mathsf{Mor}}_{N-2}(\mcal{B}))$ and $\mcal{E}^{[2,N-1]}$.
\item A functor $U_{N-1}:\cat{K}(\mcal{B}) \to \cat{K}(\operatorname{\mathsf{Mor}}_{N-1}(\mcal{B}))$ induces
a triangle equivalence between $\cat{K}(\mcal{B})$ and $\mcal{F}^{[1,N-1]}$.
\end{enumerate}
\end{prop}
\begin{proof}
(1)
By Lemma \ref{lem:comma03}, every complex of $\mcal{E}^{[2,N-1]}$ is isomorphic to
some complex of the form
\[\xymatrix{
0 \ar[r] & X^2 \ar[r] & \cdots \ar[r] & X^{N-1}
}\]
Then it is easy to see that
a triangle functor $E^{\Uparrow_{N-2}^{N-1}}:\cat{K}(\operatorname{\mathsf{Mor}}_{N-2}(\mcal{B})) \to \mcal{E}^{[2,N-1]}$
is a triangle equivalence.
\par\noindent
(2)
Every complex of $\mcal{F}^{[1,N-1]}$ is of the form
\[\xymatrix{
X^1 \ar@{=}[r] & X^2 \ar@{=}[r] & \cdots \ar@{=}[r] & X^{N-1}
}\]
Then it is easy to see that
a triangle functor $U^{N-1}:\cat{K}(\mcal{B}) \to \mcal{F}^{[1,N-1]}$
is a triangle equivalence.
\end{proof}
\begin{thm}\label{th:ngon@trg01}
Let $\mcal{B}$ be an additive category.
Then a $2N$-tuple of full subcategories
{\small
\[
(\mcal{F}^{[1,N-1]},\mcal{E}^{[2,N-1]}, \mcal{E}^{1}, \mcal{F}^{[1,2]}, \cdots,
\mcal{E}^{s}, \mcal{F}^{[s,s+1]}, \cdots, \mcal{E}^{N-2}, \mcal{F}^{[N-2,N-1]}, \mcal{E}^{N-1}, \mcal{E}^{[1,N-2]})
\]
}
is a $2N$-gon of recollements in $\cat{K}(\operatorname{\mathsf{Mor}}^{\mrm{sm}}_{N-1}(\mcal{B}))$.
\end{thm}
\begin{proof}
First, we prove $\operatorname{Hom}_{\cat{K}(\operatorname{\mathsf{Mor}}^{\mrm{sm}}_{N-1}(\mcal{B}))}(\mcal{X},\mcal{Y})=0$, where $\mcal{X}, \mcal{Y}$
are two successive subcategories of the above $2N$-tuple.
By Example \ref{comma02}, Lemma \ref{lem:comma03} (2),
any complex of $\mcal{E}^{[2,N-1]}$ is isomorphic to
a complex $0 \to X^2 \to \cdots \to X^{N-1}$ in $\cat{K}(\operatorname{\mathsf{Mor}}^{\mrm{sm}}_{N-1}(\mcal{B}))$.
Then $\operatorname{Hom}_{\cat{K}(\operatorname{\mathsf{Mor}}^{\mrm{sm}}_{N-1}(\mcal{B}))}(\mcal{F}^{[1,N-1]},\mcal{E}^{[2,N-1]})=0$ is easy.
By Example \ref{comma02}, Lemma \ref{lem:comma03} (3), for any object of $\mcal{E}^{1}$ there is a complex
$X$ such that it is isomorphic to an object
$
X \to \operatorname{C}(1_X) \to \cdots \to \operatorname{C}(1_X)
$
in $\cat{K}(\operatorname{\mathsf{Mor}}^{\mrm{sm}}_{N-2}(\mcal{B}))$.
Since it is easy to see
{\small
\[
\operatorname{Hom}_{\cat{K}(\operatorname{\mathsf{Mor}}^{\mrm{sm}}_{N-1}(\mcal{B}))}(\mcal{E}^{[2,N-1]}, \mcal{E}^{1})\simeq
\operatorname{Hom}_{\cat{K}(\operatorname{\mathsf{Mor}}^{\mrm{sm}}_{N-2}(\mcal{B}))}(D_{[2,N-1]}(\mcal{E}^{[2,N-1]}), D_{[2,N-1]}(\mcal{E}^{1}))
\]
}
and
\[
D_{[2,N-1]}(X \to \operatorname{C}(1_X) \to \cdots \to \operatorname{C}(1_X))=\operatorname{C}(1_X) \to \cdots \to \operatorname{C}(1_X)
\]
is homotopically $0$ in $\cat{K}(\operatorname{\mathsf{Mor}}^{\mrm{sm}}_{N-2}(\mcal{B}))$,
$\operatorname{Hom}_{\cat{K}(\operatorname{\mathsf{Mor}}^{\mrm{sm}}_{N-1}(\mcal{B}))}(\mcal{E}^{[2,N-1]}, \mcal{E}^{1})=0$.
By Example \ref{comma02}, Lemma \ref{lem:comma03} (2), (3),
we may assume any morphism from $\mcal{E}^{s}$ to $\mcal{F}^{[s,s+1]}$ is of the form
\[\xymatrix{
0 \ar[d] \ar[r] & \cdots \ar[r] & 0 \ar[d] \ar[r] & X^{s} \ar[d] \ar[r] & C(1_X) \ar[d]\ar@{=}[r] & \cdots \ar@{=}[r] & C(1_X) \ar[d] \\
Y^{1} \ar[r] & \cdots \ar[r] & Y^{s-1} \ar[r] & Y^{s} \ar@{=}[r] & Y^{s+1} \ar[r] & \cdots \ar[r] & Y^{N-1}
}\]
It is easy that the above morphism is null homotopic, and then \\
$\operatorname{Hom}_{\cat{K}(\operatorname{\mathsf{Mor}}^{\mrm{sm}}_{N-1}(\mcal{B}))}(\mcal{E}^{s}, \mcal{F}^{[s,s+1]})=0$.
Similarly, we may assume any morphism from $\mcal{F}^{[s,s+1]}$ to $\mcal{E}^{s+1}$ is of the form
\[\xymatrix{
X^{1} \ar[d] \ar[r] & \cdots \ar[r] & X^{s} \ar[d] \ar@{=}[r] & X^{s+1} \ar[d] \ar[r] & X^{s+2} \ar[d] \ar[r] & \cdots \ar[r] & X^{N-1} \ar[d] \\
0 \ar[r] & \cdots \ar[r] & 0 \ar[r] & Y \ar[r] & C(1_Y) \ar@{=}[r] & \cdots \ar@{=}[r] & C(1_Y) \\
}\]
It is easy that the above morphism is null homotopic, and then \\
$\operatorname{Hom}_{\cat{K}(\operatorname{\mathsf{Mor}}^{\mrm{sm}}_{N-1}(\mcal{B}))}(\mcal{F}^{[s,s+1]}, \mcal{E}^{s+1})=0$.
Since any complex of $\mcal{E}^{N-1}$ is isomorphic to $0 \to \cdots \to 0 \to X^{N-1}$, and
any complex of $\mcal{E}^{[1,N-2]}$ is isomorphic to $Y^1 \to \cdots \to Y^{N-2} \to C(1_{Y^{N-1}})$
in $\operatorname{\mathsf{Mor}}^{\mrm{sm}}_{N-1}(\mcal{B})$,
it is easy to see that $\operatorname{Hom}_{\cat{K}(\operatorname{\mathsf{Mor}}^{\mrm{sm}}_{N-1}(\mcal{B}))}(\mcal{E}^{N-1}, \\ \mcal{E}^{[1,N-2]})=0$.
Since we may assume any morphism from $\mcal{E}^{[1,N-2]}$ to $\mcal{F}^{[1,N-1]}$ is of the form
\[\xymatrix{
X^1 \ar[d] \ar[r] & \cdots \ar[r] & X^{N-2} \ar[d] \ar[r] & C(1_{X^{N-1}}) \ar[d] \\
Y^{1} \ar@{=}[r] & \cdots \ar@{=}[r] & Y^{N-2} \ar@{=}[r] & Y^{N-1}
}\]
It is easy to see it is null homotopic, and then
$\operatorname{Hom}_{\cat{K}(\operatorname{\mathsf{Mor}}^{\mrm{sm}}_{N-1}(\mcal{B})}(\mcal{E}^{[1,N-2]}, \mcal{F}^{[1,N-1]})=0$.
\par\noindent
Second, we prove $\cat{K}(\operatorname{\mathsf{Mor}}^{\mrm{sm}}_{N-1}(\mcal{B}))=\mcal{X}*\mcal{Y}$, where $\mcal{X}, \mcal{Y}$
are two successive subcategories of the above $2N$-tuple.
Let $X^{1} \xrightarrow{\alpha^1} X^{2} \xrightarrow{\alpha^2} \cdots \xrightarrow{\alpha^{N-2}} X^{N-1}$ be a complex
of $\cat{K}(\operatorname{\mathsf{Mor}}^{\mrm{sm}}_{N-1}(\mcal{B}))$.
Since we have a short exact sequence of complexes:
\[\xymatrix{
X^1 \ar[d]^{1} \ar@{=}[r] & X^1 \ar[d]^{\alpha^1} \ar@{=}[r] &\cdots \ar@{=}[r]& X^{1} \ar[d]^{\alpha^{N-2}\cdots\alpha^1} \\
X^1 \ar[d] \ar[r]^{\alpha^1} & X^2 \ar[d] \ar[r]^{\alpha^2} &\cdots \ar[r]^{\alpha^{N-2}} & X^{N-1} \ar[d] \\
0 \ar[r] & \operatorname{Cok}\alpha^{1} \ar[r] & \cdots \ar[r] & \operatorname{Cok}\alpha^{N-2}\cdots\alpha^{1}
}\]
we have $\cat{K}(\operatorname{\mathsf{Mor}}^{\mrm{sm}}_{N-1}(\mcal{B}))=\mcal{F}^{[1,N-1]}*\mcal{E}^{[2,N-1]}$.
Since we have a triangle in \\ $\cat{K}(\operatorname{\mathsf{Mor}}^{\mrm{sm}}_{N-1}(\mcal{B}))$:
\[\xymatrix{
0 \ar[d] \ar[r]^{\alpha^1} & X^2 \ar[d]^{1} \ar[r]^{\alpha^2} &\cdots \ar[r]^{\alpha^{N-2}} & X^{N-1} \ar[d]^{1} \\
X^1 \ar[d] \ar[r]^{\alpha^1} & X^2 \ar[d] \ar[r]^{\alpha^2} &\cdots \ar[r]^{\alpha^{N-2}} & X^{N-1} \ar[d] \\
X^1 \ar[d] \ar[r] & \operatorname{C}(1_{X^2}) \ar[d] \ar[r] &\cdots \ar[r] & \operatorname{C}(1_{X^{N-1}}) \ar[d] \\
0 \ar[r]^{\Sigma \alpha^1} & \Sigma X^2 \ar[r]^{\Sigma \alpha^2} &\cdots \ar[r]^{\Sigma \alpha^{N-2}} & \Sigma X^{N-1} \\
}\]
we have $\cat{K}(\operatorname{\mathsf{Mor}}^{\mrm{sm}}_{N-1}(\mcal{B}))=\mcal{E}^{[2,N-1]}*\mcal{E}^{1}$.
Since we have a triangle in $\cat{K}(\operatorname{\mathsf{Mor}}^{\mrm{sm}}_{N-1}(\mcal{B}))$:
{\scriptsize
\[\xymatrix{
X^1 \ar[d]^{1} \ar[r]^{\alpha^1} & \cdots \ar[r]^{\alpha^{s-2}} &
X^{s-1} \ar[d]^{1} \ar[r]^{\alpha^{s-1}} & X^{s} \ar[d]^{\alpha^{s}} \ar[r]^{\alpha^{s}} &
X^{s+1} \ar[d]^{1} \ar[r]^{\alpha^{s+1}} & \cdots \ar[r]^{\alpha^{N-2}} &X^{N-1} \ar[d]^{1} \\
X^1 \ar[d] \ar[r]^{\alpha^1} & \cdots \ar[r]^{\alpha^{s-2}} &
X^{s-1} \ar[d] \ar[r]^{\alpha^{s}\alpha^{s-1}} & X^{s+1} \ar[d] \ar@{=}[r] &
X^{s+1} \ar[d] \ar[r]^{\alpha^{s+1}} & \cdots \ar[r]^{\alpha^{N-2}} &X^{N-1} \ar[d] \\
\operatorname{C}(1_{X^{1}}) \ar[d] \ar[r] & \cdots \ar[r] &
\operatorname{C}(1_{X^{s-1}}) \ar[d] \ar[r] & \operatorname{C}(\alpha^{s}) \ar[d] \ar[r]^{\alpha^{s+1}} &
\operatorname{C}(1_{X^{s+1}}) \ar[d] \ar[r] & \cdots \ar[r] &\operatorname{C}(1_{X^{N-1}}) \ar[d] \\
\Sigma X^1 \ar[r]^{\Sigma \alpha^1} & \cdots \ar[r]^{\Sigma \alpha^{s-2}} &
\Sigma X^{s-1} \ar[r]^{\Sigma \alpha^{s-1}} & \Sigma X^{s} \ar[r]^{\Sigma \alpha^{s}} &
\Sigma X^{s+1} \ar[r]^{\Sigma \alpha^{s+1}} & \cdots \ar[r]^{\Sigma\alpha^{N-2}} & \Sigma X^{N-1} \\
}\]
}
we have $\cat{K}(\operatorname{\mathsf{Mor}}^{\mrm{sm}}_{N-1}(\mcal{B}))=\mcal{E}^{s}*\mcal{F}^{[s,s+1]}$.
Since we have a triangle in $\cat{K}(\operatorname{\mathsf{Mor}}^{\mrm{sm}}_{N-1}(\mcal{B}))$:
{\scriptsize
\[\xymatrix{
X^1 \ar[d]^{1} \ar[r] & \cdots \ar[r]^{\alpha^{s-1}} &
X^{s} \ar[d]^{1} \ar@{=}[r] & X^{s} \ar[d]^{\alpha^{s}} \ar[r]^{\alpha^{s+1}\alpha^{s}} &
X^{s+2} \ar[d]^{1} \ar[r]^{\alpha^{s+2}} & \cdots \ar[r]^{\alpha^{N-2}} &X^{N-1} \ar[d]^{1} \\
X^1 \ar[d] \ar[r] & \cdots \ar[r]^{\alpha^{s-1}} &
X^{s} \ar[d] \ar[r]^{\alpha^{s}} & X^{s+1} \ar[d] \ar[r]^{\alpha^{s+1}} &
X^{s+2} \ar[d] \ar[r]^{\alpha^{s+2}} & \cdots \ar[r]^{\alpha^{N-2}} &X^{N-1} \ar[d] \\
\operatorname{C}(1_{X^{1}}) \ar[d] \ar[r] & \cdots \ar[r] &
\operatorname{C}(1_{X^{s-1}}) \ar[d] \ar[r] & \operatorname{C}(\alpha^{s}) \ar[d] \ar[r]^{\alpha^{s+1}} &
\operatorname{C}(1_{X^{s+1}}) \ar[d] \ar[r] & \cdots \ar[r] &\operatorname{C}(1_{X^{N-1}}) \ar[d] \\
\Sigma X^1 \ar[r]^{\Sigma \alpha^1} & \cdots \ar[r]^{\Sigma \alpha^{s-1}} &
\Sigma X^{s} \ar@{=}[r] & \Sigma X^{s} \ar[r]^{\Sigma \alpha^{s+1}\alpha^{s}} &
\Sigma X^{s+2} \ar[r]^{\Sigma \alpha^{s+2}} & \cdots \ar[r]^{\Sigma\alpha^{N-2}} & \Sigma X^{N-1} \\
}\]
}
we have $\cat{K}(\operatorname{\mathsf{Mor}}^{\mrm{sm}}_{N-1}(\mcal{B}))=\mcal{F}^{[s,s+1]}*\mcal{E}^{s+1}$.
Since we have a triangle in {\small$\cat{K}(\operatorname{\mathsf{Mor}}^{\mrm{sm}}_{N-1}(\mcal{B}))$:}
\[\xymatrix{
0 \ar[d] \ar[r] & \cdots \ar[r] & 0 \ar[d] \ar[r] & X^{N-1} \ar[d]^{1} \\
X^1 \ar[d]^{1} \ar[r]^{\alpha^1} & \cdots \ar[r]^{\alpha^{N-3}} & X^{N-2} \ar[d]^{1} \ar[r]^{\alpha^{N-2}} & X^{N-1} \ar[d] \\
X^1 \ar[d] \ar[r]^{\alpha^1} & \cdots \ar[r]^{\alpha^{N-3}} & X^{N-2} \ar[d] \ar[r] & \operatorname{C}(1_{X^{N-1}}) \ar[d] \\
0 \ar[r] & \cdots \ar[r] & 0 \ar[r] & \Sigma X^{N-1} \\
}\]
we have $\cat{K}(\operatorname{\mathsf{Mor}}^{\mrm{sm}}_{N-1}(\mcal{B}))=\mcal{E}^{N-1}*\mcal{E}^{[1,N-2]}$.
Since we have a triangle in {\small$\cat{K}(\operatorname{\mathsf{Mor}}^{\mrm{sm}}_{N-1}(\mcal{B}))$:}
\[\xymatrix{
X^{1} \ar[d]^{\alpha^{N-2}\cdots\alpha^{1}} \ar[r]^{\alpha^{1}} & \cdots \ar[r]^{\alpha^{N-3}} &
X^{N-2} \ar[d]^{\alpha^{N-2}} \ar[r]^{\alpha^{N-2}} & X^{N-1} \ar[d]^{1} \\
X^{N-1} \ar[d] \ar@{=}[r] & \cdots \ar@{=}[r] &
X^{N-1} \ar[d] \ar@{=}[r] & X^{N-1} \ar[d] \\
\operatorname{C}(\alpha^{N-2}\cdots\alpha^{1}) \ar[d] \ar[r] & \cdots \ar[r] &
\operatorname{C}(\alpha^{N-2}) \ar[d] \ar[r] & \operatorname{C}(1_{X^{N-1}}) \ar[d] \\
\Sigma X^{1} \ar[r]^{\Sigma \alpha^{1}} & \cdots \ar[r]^{\Sigma \alpha^{N-3}} &
\Sigma X^{N-2} \ar[r]^{\Sigma \alpha^{N-2}} & \Sigma X^{N-1}
}\]
we have $\cat{K}(\operatorname{\mathsf{Mor}}^{\mrm{sm}}_{N-1}(\mcal{B}))=\mcal{E}^{N-1}*\mcal{F}^{[1,N-1]}$.
\end{proof}
\section{Recollement of $\cat{K}_{N}(\mcal{B})$}\label{TNcpx}
We fix a positive integer $N\ge2$.
Let $\mcal{B}$ be an additive category.
An \emph{$N$-complex} is a diagram
\[\cdots \xrightarrow{d_X^{i-1}}X^i\xrightarrow{d_X^i}X^{i+1}\xrightarrow{d_X^{i+1}}\cdots\]
with objects $X^i\in\mcal{B}$ and morphisms $d_X^i\in\operatorname{Hom}_{\mcal{B}}(X^i,X^{i+1})$ satisfying
\[d_{X}^{i+N-1}\cdots d_{X}^{i+1}d_{X}^{i}=0\]
for any $i\in\mathbf{Z}$.
A \emph{morphism} between $N$-complexes is a commutative diagram
\[\begin{array}{ccccc}
\cdots \xrightarrow{d_X^{i-1}}&X^i&\xrightarrow{d_X^i}&X^{i+1}&\xrightarrow{d_X^{i+1}}\cdots\\
&\downarrow^{f^i}&&\downarrow^{f^{i+1}}&\\
\cdots \xrightarrow{d_Y^{i-1}}&Y^i&\xrightarrow{d_Y^i}&Y^{i+1}&\xrightarrow{d_Y^{i+1}}\cdots
\end{array}\]
with $f^i\in\operatorname{Hom}_{\mcal{B}}(X^i,Y^i)$ for any $i\in\mathbf{Z}$.
We denote by $\cat{C}_N(\mcal{B})$ the category of $N$-complexes.
A collection $\mcal{S}_N(\mcal{B})$ of conflations is the collection of short exact sequences of $N$-complexes of which each term is a split short exact sequence in $\mcal{B}$.
\begin{prop}[\cite{IKM2}]\label{NcpxFrob}
A category $(\cat{C}_N(\mcal{B}), \mcal{S}_N(\mcal{B}))$ is a Frobenius category.
\end{prop}
\begin{defn}[\cite{IKM2}]\label{cone01}
Let $(X,d)$, $(Y,e)$ be objects and $f: Y \to X$ be a morphism in $\cat{C}_{N}(\mcal{B})$.
Then the mapping cone $C(f)$ of $f$ is given as
{\footnotesize
\[ C(f)^m = X^m \oplus {\displaystyle \coprod_{i=m+1}^{m+N-1}}Y^i,
d_{C(f)}^m = \left( \begin{array}{c|ccccc}
d&f&0&\cdots&\cdots&0\\
\hline
0&0&1&\ddots&&\vdots\\
\vdots&\vdots&\ddots&\ddots&\ddots&\vdots\\
\vdots&0&\cdots&0&1&0\\
0&0&\cdots&\cdots&0&1\\
0&-e^{\{N-1\}}&-e^{\{N-2\}}&\cdots&\cdots&-e\\
\end{array} \right)
\]
\[(\Sigma ^{-1} C(f) )^m \!\!\! = \!\!\! \coprod_{i=m-N+1}^{m-1} X^i \oplus Y^m ,
d_{\Sigma ^{-1} C(f)}^m = \left( \begin{array}{ccccc|c}
-d&1&0&\cdots&\cdots&0\\
-d^2&0&1&\ddots&&\vdots\\
\vdots&\vdots&&\ddots&\ddots&\vdots\\
\vdots&\vdots&\ddots&\ddots&1&0\\
-d^{\{N-1\}}&0&\cdots&\cdots&0&f\\
\hline
0&\cdots&\cdots&\cdots&0&e\\
\end{array} \right).
\]
}
Here $d^{\{N-1\}}$ means the $(N-1)$-power of $d$.
\end{defn}
The above mapping cone induces a morpism between conflations:
\[\xymatrix{
0 \ar[r] & X \ar[d]^{f} \ar[r]^{u_X} & \operatorname{C}(1_X) \ar[d]^{\psi_f} \ar[r]^{v_X} & \Sigma X \ar@{=}[d] \ar[r] & 0 \\
0 \ar[r] & Y \ar[r]^{u_f} & \operatorname{C}(f) \ar[r]^{v_f} & \Sigma X \ar[r] & 0
}\]
Let $I(X)=\operatorname{C}(1_X)$, then $I(X)$ is a projective-injective object in $\cat{C}_{N}(\mcal{B})$.
We call a sequence $Y \xrightarrow{f} X \xrightarrow{u_f} \operatorname{C}(1_X) \xrightarrow{v_f} \Sigma X $ a (distinguished) triangle.
A {morphism} $f:X \to Y$of $N$-complexes
is called \emph{null-homotopic} if there exists $s^i\in\operatorname{Hom}_{\mathcal{B}}(X^i,Y^{i-N+1})$
such that
\[f^i=\sum_{j=1}^{N-1}d_Y^{i-1}\cdots d_Y^{i-N+j}s^{i+j-1}d_X^{i+j-2}\cdots d_X^i\]
for any $i\in\mathbf{Z}$.
We denote by $\cat{K}_N(\mcal{B})$ the homotopy category of $N$-complexes.
\begin{thm}[\cite{IKM2}]\label{NcpxTricat}
A category $\cat{K}_N(\mcal{B})$ is a triangulated category.
\end{thm}
\begin{defn}\label{prolong01}
Let $N$ be an integer greater than 2. For any integer $s$, we define functions
$\iota_{s}^{(N-1)}:\mathbf{Z} \to \mathbf{Z}$, $\rho_{s}^{(N)}:\mathbf{Z} \to \mathbf{Z}$ as follows.
\[\begin{aligned}
\iota_{s}^{(N-1)}(s+i+kN) &=
\begin{cases}\begin{aligned}
&s+k(N-1) &(i=0) \\
&s+i-1+k(N-1) &(0 < i < N)
\end{aligned}\end{cases} \\
\rho_{s}^{(N)}(s+i+k(N-1)) &=
\begin{cases}\begin{aligned}
&s+kN &(i=0) \\
&s+i+1+kN &(0 < i < N-1)
\end{aligned}\end{cases}
\end{aligned}\]
For an $(N-1)$-complex $X=(X^{i}, d_{X}^{i})$, we define a complex $I_{s}^{(N-1)}(X)$ by
\[\begin{aligned}
I_{s}^{(N-1)}(X)^{i} &=X^{\iota_{s}^{(N-1)}(i)}, \\
d_{I_{s}^{(N-1)}(X)}^{i}&=\begin{cases}\begin{aligned}
&d^{\iota_{s}^{(N-1)}(i)} &(\iota_{s}^{(N-1)}(i) < \iota_{s}^{(N-1)}(i+1)) \\
&1 &(\iota_{s}^{(N-1)}(i) = \iota_{s}^{(N-1)}(i+1)) .
\end{aligned}\end{cases}
\end{aligned}\]
For an $N$-complex $Y=(Y^{i}, d_{Y}^{i})$, we define a complex $J_{s}^{(N)}(Y)$ by
\[\begin{aligned}
J_{s}^{(N)}(X)^{i} &=X^{\rho_{s}^{(N)}(i)}, \\
d_{J_{s}^{(N)}(X)}^{i}&=
d^{\rho_{s}^{(N)}(i+1)-1}\cdots d^{\rho_{s}^{(N)}(i)} .
\end{aligned}\]
Then $I_{s}^{(N-1)}: \cat{C}_{N-1}(\mcal{B}) \to \cat{C}_{N}(\mcal{B})$ and
$J_{s}^{(N)}: \cat{C}_{N}(\mcal{B}) \to \cat{C}_{N-1}(\mcal{B})$ are functors.
\end{defn}
\begin{lem}\label{prolong02}
A functor $I_{s}^{(N-1)}: \cat{C}_{N-1}(\mcal{B}) \to \cat{C}_{N}(\mcal{B})$
induces the triangle functor $\underline{I}_{s}^{(N-1)}: \cat{K}_{N-1}(\mcal{B}) \to \cat{K}_{N}(\mcal{B})$.
\end{lem}
\begin{proof}
Let $X \xrightarrow{f} Y$ be a morphism in $\cat{C}_{N-1}(\mcal{B})$.
Consider the mapping cone $C(I_{0}^{(N-1)}(f))$ of $I_{0}^{(N-1)}(X) \xrightarrow{I_{0}^{(N-1)}(f)} I_{0}^{(N-1)}(Y)$, then
we have an exact sequence $0 \to I_{0}^{(N-1)}(Y) \to C(I_{0}^{(N-1)}(f)) \to \Sigma(I_{0}^{(N-1)}(X)) \to 0$
in $\cat{C}_{N-1}(\mcal{B})$.
Let $Z$ be an $N$-complex defined by
{\footnotesize
\[\begin{aligned}
Z^{k}
=\begin{cases}
Y^{\iota^{(N-1)}_{0}(k)}\oplus \coprod_{i=k+1}^{k+N-2}X^{\iota^{(N-1)}_{0}(i)}\oplus X^{\iota^{(N-1)}_{0}(k)}
\ (k \equiv 0 \operatorname{mod} N) \\
Y^{\iota^{(N-1)}_{0}(k)}\oplus \coprod_{i=k+1}^{k+N-1}X^{\iota^{(N-1)}_{0}(i)} \ (k \not\equiv 0 \operatorname{mod} N)
\end{cases} \\
d_Z^{k}
=\begin{cases}
\left(\begin{array}{ccc|c}
1 & \cdots & 0 & 0 \\
\vdots & \ddots & \vdots & \vdots \\
0 & \cdots & 1 & 0 \\\hline
0 & \cdots & 0 & 0
\end{array}\right) \ (k \equiv 0 \operatorname{mod} N) \\
\left(\begin{array}{ccccc|c}
e&f&0&\cdots&0& 0\\
0&0&1&\ddots&\vdots&\vdots\\
\vdots&\vdots&\vdots&\ddots&\vdots&\vdots\\
0&0&0&\cdots&1&\vdots\\
0&-d^{N-1}&d^{N-2}&\cdots&-d&0\\\hline
0&0&0&\cdots&0&0\\
\end{array}\right) \ (k \not\equiv 0 \operatorname{mod} N)
\end{cases} \\
\end{aligned}\]
}
and
$W$ an $N$-complex defined by
{\footnotesize
\[\begin{aligned}
W^{k}
=\begin{cases}
\coprod_{i=k+1}^{k+N-2}X^{\iota^{(N-1)}_{0}(i)}\oplus X^{\iota^{(N-1)}_{0}(k)} \ (k \equiv 0 \operatorname{mod} N) \\
\coprod_{i=k+1}^{k+N-1}X^{\iota^{(N-1)}_{0}(i)} \ (k \not\equiv 0 \operatorname{mod} N)
\end{cases} \\
d_W^{k}
=\begin{cases}
\left(\begin{array}{ccc|c}
1 & \cdots & 0 & 0 \\
\vdots & \ddots & \vdots & \vdots \\
0 & \cdots & 1 & 0 \\\hline
0 & \cdots & 0 & 0
\end{array}\right) \ (k \equiv 0 \operatorname{mod} N) \\
\left(\begin{array}{cccc|c}
0&1&\cdots&0&0\\
\vdots&\vdots&\ddots&\vdots&\vdots\\
0&0&\cdots&1&\vdots\\
-d^{N-1}&d^{N-2}&\cdots&-d&0\\\hline
0&0&\cdots&0&0\\
\end{array}\right) \ (k \not\equiv 0 \operatorname{mod} N)
\end{cases} \\
\end{aligned}\]
}
Let $g: C(I_{0}^{(N-1)}(f)) \to Z$ be a isomorphism defined by
{\tiny
\[
\begin{array}{ll}
g^{k}=
\left(\begin{array}{c|c|ccc}
1 & f & 0 & \cdots & 0 \\\hline
0 & 0 & 1 & \cdots & 0 \\
\vdots & \vdots & 0 & \ddots & \vdots \\
0 & 0 & 0 & \cdots & 1 \\\hline
0 & -1 & 0 & \cdots & 0
\end{array}\right)
\ (k \equiv 0 \operatorname{mod} N), \\
\left(\begin{array}{c|ccc|c}
1 & 0 & \cdots & 0 & 0 \\\hline
0 & 1 & \cdots & 0 & 0 \\
\vdots & \vdots & \ddots & \vdots & \vdots \\
0 & 0 & \cdots & 1 & 0 \\\hline
0 & -d^{\{N-1\}} & \cdots & -d & 1
\end{array}\right)
\ (k \equiv 1 \operatorname{mod} N), &
\left(\begin{array}{c|ccc|c|c}
1 & 0 & \cdots & 0 & 0 & 0 \\\hline
0 & 1 & \cdots & 0 & 0 &0 \\
\vdots & \vdots & \ddots & \vdots & \vdots \\
0 & 0 & \cdots & 1 & 0 & 0 \\\hline
0 & 0 & \cdots &0 & 1 & 1\\\hline
0 & 0 & \cdots &0 & 0& -1
\end{array}\right)
\ (k \equiv 2 \operatorname{mod} N), \\
\left(\begin{array}{ccc|c|c|c}
1 & \cdots & 0 & 0 & 0 & 0 \\
\vdots & \ddots & \vdots & \vdots& \vdots& \vdots \\
0 & \cdots & 1 & 0 & 0 & 0 \\\hline
0 & 0 & \cdots & 1 & 1 & 0 \\\hline
0 & 0 & \cdots &0 & 0 & 1\\\hline
0 & 0 & \cdots &0 &-1& 0
\end{array}\right)
\ (k \equiv 3 \operatorname{mod} N), &
\hspace{54pt} \cdots, \\
\left(\begin{array}{c|c|cccc}
1 & 0 & 0 & 0 & \cdots & 0 \\\hline
0 & 1 & 1 & 0 & \cdots& 0 \\\hline
0 & 0 & 0 & 1 & \cdots & 0 \\
\vdots & \vdots & \vdots & \vdots & \ddots & \vdots \\\hline
0 & 0 & 0 & 0 & \cdots & 1\\\hline
0 & 0 & -1 &0 &\cdots & 0
\end{array}\right)
\ (k \equiv N-1 \operatorname{mod} N) \\
\end{array}
\]
}
Then we have the following isomorphism between short exact sequences:
\[\xymatrix{
0 \ar[r] &I_{0}^{(N-1)}(Y) \ar@{=}[d] \ar[r]^{u} & C(I_{0}^{(N-1)}(f)) \ar[d]_{\wr}^{g} \ar[r]^{v} & \Sigma(I_{0}^{(N-1)}(X))\ar[d]_{\wr}^{g'} \ar[r] & 0\\
0 \ar[r] &I_{0}^{(N-1)}(Y) \ar[r]^{u'} & Z \ar[r]^{v'} & W \ar[r] & 0\\
}\]
where
{\footnotesize
$u'= \begin{pmatrix}
1\cr 0\cr \vdots\cr 0\cr
\end{pmatrix}$,
$v'=
\left(\begin{array}{cccc}
0 & 1 & \cdots & 0 \\
\vdots & \vdots & \ddots & \vdots \\
0 & 0 & \cdots & 1 \end{array}\right)
$}.
Therefore, $C(I_{0}^{(N-1)}(f))\simeq I_{0}^{(N-1)}(C(f))$ and
$\Sigma(I_{0}^{(N-1)}(X))\simeq I_{0}^{(N-1)}(\Sigma(X))$ in $\cat{K}_N(\mcal{B})$.
\end{proof}
\begin{prop}\label{prolong03}
For a functor $J_{s}^{(N)}: \cat{C}_{N}(\mcal{B}) \to \cat{C}_{N-1}(\mcal{B})$,
the following hold.
\begin{enumerate}
\item
$J_{s}^{(N)}$ induces the triangle functor $\underline{J}_{s}^{(N)}: \cat{K}_{N}(\mcal{B}) \to \cat{K}_{N-1}(\mcal{B})$.
\item
$\underline{J}_{s}^{(N)}$ is a right adjoint of $\underline{I}_{s}^{(N-1)}$.
\item
$\underline{I}_{s+1}^{(N-1)}$ is a right adjoint of $\underline{J}_{s}^{(N)}$.
\item
The adjunction arrow $\mathbf{1}\to \underline{J}_{s}^{(N)}\underline{I}_{s}^{(N-1)}$ is an isomorphism.
\item
The adjunction arrow $\underline{J}_{s}^{(N)}\underline{I}_{s+1}^{(N-1)}\to \mathbf{1}$ is an isomorphism.
\end{enumerate}
\end{prop}
\begin{proof}
For $X, Z \in \cat{C}_{N-1}(\mcal{B}), Y \in \cat{C}_{N}(\mcal{B})$,
consider the following morphisms in $\cat{C}_{N}(\mcal{B})$:
\[\xymatrix{
I_s^{(N-1)}(X) : \ar[d] & \cdots \ar[r] & X^{s} \ar@{=}[r] \ar[d]^{f^s} & X^{s} \ar[r]^{d^s} \ar[d]^{d^sf^s} & X^{s+1} \ar[r] \ar[d]^{f^{s+1}} & \cdots \\
Y : \ar[d] : & \cdots \ar[r] & Y^s \ar[r]^{d^s} \ar[d]^{g^s} & Y^{s+1} \ar[r]^{d^{s+1}} \ar[d]^{g^{s+1}} & Y^{s+2} \ar[r] \ar[d]^{g^{s+2}} & \cdots\\
I_{s+1}^{(N-1)}(Z) : & \cdots \ar[r] & Z^s \ar[r]^{d^s} & Z^{s+1} \ar@{=}[r] & Z^{s+1} \ar[r] & \cdots
} \]
Then these morphisms correspond to the following morphisms in $\cat{C}_{N-1}(\mcal{B})$:
\[\xymatrix{
X : \ar[d] & \cdots \ar[r] & X^{s} \ar[r]^{d^s} \ar[d]^{f^s} & X^{s+1} \ar[r] \ar[d]^{f^{s+1}} & \cdots \\
J_{s}^{(N)}(Y) : \ar[d] : & \cdots \ar[r] & Y^s \ar[r]^{d^{s+1}d^s} \ar[d]^{g^s} & Y^{s+2} \ar[r] \ar[d]^{g^{s+2}} & \cdots\\
Z : & \cdots \ar[r] & Z^s \ar[r]^{d^s} & Z^{s+1} \ar[r] & \cdots
} \]
Then it is easy to see that ${J}_{s}^{(N)}$ is a right adjoint of ${I}_{s}^{(N-1)}$, that
${I}_{s+1}^{(N-1)}$ is a right adjoint of ${J}_{s}^{(N)}$, and that
the adjunction arrows $\mathbf{1}\to {J}_{s}^{(N)}{I}_{s}^{(N-1)}$ and
${J}_{s}^{(N)}{I}_{s+1}^{(N-1)}\to \mathbf{1}$ are isomorphisms.
By Lemmas \ref{prolong02} and Proposition \ref{triadjoint02}, we have statements.
\end{proof}
\begin{defn}\label{updown01}
We denote
$\underline{I}_{s}^{\Uparrow_{N-r}^{N}}=
\underline{I}_{s}^{(N-1)}\cdots\underline{I}_{s}^{(N-r)}:
\cat{K}_{N-r}(\mcal{B}) \to \cat{K}_{N}(\mcal{B})$ and
$\underline{J}_{s}^{\Downarrow_{N-r}^{N}}=
\underline{J}_{s}^{(N-r+1)} \cdots \underline{J}_{s}^{(N)}:
\cat{K}_{N}(\mcal{B}) \to \cat{K}_{N-r}(\mcal{B})$.
For $1\leq r < N$, we define the full subcategory of $\cat{K}_{N}(\mcal{B})$
\[
\mcal{F}_{s}^{r}=\{(X^i, d^i) \in \cat{K}_{N}(\mcal{B}) \mid d^{i+k}=1_{X^{i+k}} (0 \leq k < r)
\text{for}\ i\equiv s \mod N\}
\]
\end{defn}
\begin{cor}\label{updown02}
$\underline{I}_{s}^{\Uparrow_{N-r}^{N}}$, $\underline{J}_{s}^{\Downarrow_{N-r}^{N}}$
are triangle functors such that
$\underline{I}_{s}^{\Uparrow_{N-r}^{N}}$ is a left (reps., right) adjoint of
$\underline{J}_{s}^{\Downarrow_{N-r}^{N}}$
(resp., $\underline{J}_{s+1}^{\Downarrow_{N-r}^{N}}$),
and that $\mcal{F}_{s}^{r}=\cat{Im}\underline{I}_{s}^{\Uparrow_{N-r}^{N}}$ is
a full triangulated subcategory of $\cat{K}_{N}(\mcal{B})$.
\end{cor}
\begin{proof}
By Lemma \ref{prolong02}, Proposition \ref{prolong03}.
\end{proof}
\begin{thm}\label{tstNcpx01}
For $1\leq r < N$, $(\mcal{F}_{s}^{r}, \mcal{F}_{r+s+1}^{N-r-1})$ is a stable $t$-structure
in $\cat{K}_{N}(\mcal{B})$.
\end{thm}
\begin{proof}
We may assume $s=0$.
It is easy to see that
$\operatorname{Hom}_{\cat{K}_{N}(\mcal{B})}(\mcal{F}_{0}^{r}, \mcal{F}_{r+1}^{N-r-1})=0$.
Let $X$ be an $N$-complex.
By Proposition \ref{prolong03}, the adjunction arrow induce a triangle
\[
\underline{I}_{0}^{\Uparrow_{N-r}^{N}}
\underline{J}_{0}^{\Downarrow_{N-r}^{N}}(X) \xrightarrow{\varepsilon_{X}} X \xrightarrow{u} \operatorname{C}(\varepsilon_{X}) \xrightarrow{v}
\Sigma\underline{I}_{0}^{\Uparrow_{N-r}^{N}}
\underline{J}_{0}^{\Downarrow_{N-r}^{N}}(X) .
\]
Let $\sigma:=\iota_{0}^{(N-1)}\cdots\iota^{(N-r)}_{0}\rho_{0}^{(N-r+1)}\cdots\rho_{0}^{(N)}$,
and $V$ be an $N$-complex defined by
{\footnotesize
\[\begin{aligned}
V^{k}
&=\begin{cases}
\coprod_{i=k+1}^{k+N-1}X^{\sigma(i)} \
(k\equiv 0, \cdots , r \operatorname{mod} N)\\
X^{k}\oplus \coprod_{i=k+1, i\not\equiv 0 \operatorname{mod} N}^{k+N-1}X^{\sigma(i)} \
\ (k\equiv r+1, \cdots , N-1 \operatorname{mod} N)
\end{cases}
\\
d_{V}^{k}
&=\begin{cases}
\left(\begin{array}{ccc}
1 & \cdots & 0 \\
\vdots & \ddots & \vdots \\
0 & \cdots & 1\end{array}\right)
\ (k \equiv r \operatorname{mod} N) \\
\left(\begin{array}{c|ccc}
0 & 1 & \cdots & 0 \\
\vdots & \vdots & \ddots & \vdots \\
0 & 0 & \cdots & 1 \\\hline
0 & 0 & \cdots & 0
\end{array}\right) \ (\text{othewise}) \\
\end{cases} \\
\end{aligned}\]
}
Let $h: V \to C(\epsilon_{X})$ be a monomorphism defined by
{\scriptsize
\[\begin{array}{ll}
h^{k}=
\left(\begin{array}{cccccc}
0 & \cdots & 0& 0 \\\hline
1 & \cdots & 0& 0 \\
-d & \ddots & \vdots& \vdots \\
\vdots & \ddots & 1&0 \\
0 & \cdots & -d &1
\end{array}\right)
& (k \equiv r \operatorname{mod} N),
\end{array}
\]
\[\begin{array}{lll}
& \hspace{72pt} N-r\\
N-r
&
\left(\begin{array}{cccccccc}
1 & \cdots & 0& 0 & 0 & \cdots & 0 \\
-d & \ddots & \vdots & \vdots & \vdots& \cdots & \vdots \\
\vdots & \ddots & 1& 0 & 0 & \cdots & 0 \\
0 & \cdots & -d &1 & 0 & \cdots & 0 \\
0 & \cdots & 0 &-1&1 & \cdots & 0 \\
0 & \cdots & 0 &\vdots &\ddots & \ddots & \vdots \\
0 & \cdots & 0 &0 &\cdots & -1 &1 \\
0 & \cdots & 0 &0 &\cdots & 0 &-1
\end{array}\right)
& (k \equiv r+1 \operatorname{mod} N)
\end{array}
\]
\[\begin{array}{lll}
& \hspace{72pt} N-r\\
N-r
&
\left(\begin{array}{ccccccc|c}
1 & \cdots & 0& 0 & 0 & \cdots & 0 &0 \\
-d & \ddots & \vdots & \vdots & \vdots& \cdots & \vdots & \vdots \\
\vdots & \ddots & 1& 0 & 0 & \cdots & 0 & 0 \\
0 & \cdots & -d &1 & 0 & \cdots & 0 & 0 \\
0 & \cdots & 0 &-1&1 & \cdots & 0 & 0 \\
0 & \cdots & 0 &\vdots &\ddots & \ddots & \vdots & \vdots \\
0 & \cdots & 0 &0 &\cdots & -1 &1 &0\\
0 & \cdots & 0 &0 &\cdots & 0 &-1&0 \\\hline
0 & \cdots & 0 &0 &\cdots & 0 &0 &1
\end{array}\right)
& (k \equiv r+2 \operatorname{mod} N),\cdots
\end{array}\]
\[\begin{array}{lll}
& \hspace{96pt} r+1\\
r+1
&\left(\begin{array}{cccc|ccccc}
1 & 0 & \cdots & 0 & 0 & \cdots & 0 & 0 \\
-1 & 1 & \cdots& 0 & 0 & \cdots & 0& 0 \\
0 & \ddots & \ddots & \vdots & \vdots & \ddots & \vdots& \vdots \\
\vdots & \ddots & \ddots & 1 & 0 &\cdots & 0& 0 \\
0 & \cdots & 0 & -1 & 0 & \cdots & 0& 0 \\\hline
0 & \cdots & \cdots & 0 & 1 & \cdots & 0& 0 \\
0 & \cdots & \cdots & 0 & -d & \ddots & \vdots& \vdots \\
\vdots & \vdots & \vdots & \vdots & \vdots & \ddots & 1& 0 \\
0 & \cdots & \cdots & 0 & 0 & \cdots & -d & 1
\end{array}\right)
& (k \equiv 0 \operatorname{mod} N)
\end{array}\]
\[\begin{array}{lll}
& \hspace{96pt} r\\
r
&\left(\begin{array}{cccc|cccc}
d & 0 & \cdots & 0 & 0 & \cdots & 0& 0 \\
-1 & 1 & \cdots& 0 & 0 & \cdots & 0& 0 \\
0 & -1 & \ddots & \vdots & \vdots & \ddots & \vdots& \vdots \\
\vdots & \ddots & \ddots & 1 & 0 &\cdots & 0& 0 \\
0 & \cdots & 0 & -1 & 0 & \cdots & 0& 0 \\\hline
0 & \cdots & \cdots & 0 & 1 & \cdots & 0 & 0 \\
0 & \cdots & \cdots & 0 & -d & \ddots & \vdots & \vdots \\
\vdots & \vdots & \vdots & \vdots & \vdots & \ddots & 1 & 0 \\
0 & \cdots & \cdots & 0 & 0 & \cdots & -d & 1
\end{array}\right)
& (k \equiv 1 \operatorname{mod} N), \cdots
\end{array}\]
\[\begin{array}{ll}
\left(\begin{array}{c|cccccc}
d^{r-1} & 0 & \cdots & 0& 0 \\
-1 & 0 & \cdots & 0& 0 \\\hline
0 & 1 & \cdots & 0& 0 \\
0 & -d & \ddots & \vdots& \vdots \\
\vdots & \vdots & \ddots & 1&0 \\
0 & 0 & \cdots & -d &1
\end{array}\right)
& (k \equiv r-1 \operatorname{mod} N)
\end{array}\]
}
Let $p: C(\epsilon_{X}) \to \underline{I}_{r+1}^{\Uparrow_{r+1}^{N}}
\underline{J}_{r}^{\Downarrow_{r+1}^{N}}(X)$ be a epimorphism defined by
{\scriptsize
\[\begin{array}{ll}
p^{k}=
& \left(\begin{array}{cccccc}
1 & 0 & \cdots & 0
\end{array}\right)
\ (k \equiv r \operatorname{mod} N),
\\
& \left(\begin{array}{cccccccc}
d^{N-r-1} & d^{N-r-2}& \cdots & d & 1& \cdots & 1& 1
\end{array}\right)
\ (k \equiv r+1 \operatorname{mod} N),
\\
& \left(\begin{array}{cccccccc}
d^{N-r-2} & d^{N-r-3}& \cdots & d & 1& \cdots & 1& 0
\end{array}\right)
\ (k \equiv r+2 \operatorname{mod} N),
\\
& \hspace{120pt} \vdots \\
& \begin{array}{l}
\hspace{54pt} r+1\\
\left(\begin{array}{cccccccc}
1 & 1& \cdots & 1 & 0& \cdots & 0
\end{array}\right)
\ (k \equiv 0 \operatorname{mod} N),
\end{array}
\\
& \begin{array}{l}
\hspace{60pt} r\\
\left(\begin{array}{cccccccc}
1 & d& \cdots & d & 0& \cdots & 0
\end{array}\right)
\ (k \equiv 1 \operatorname{mod} N),
\end{array}
\\
& \begin{array}{l}
\hspace{60pt} r-1\\
\left(\begin{array}{cccccccc}
1 & d^2 & \cdots & d^2 & 0& \cdots & 0
& \end{array}\right)
\ (k \equiv 2 \operatorname{mod} N),
\end{array}
\\
& \hspace{120pt} \vdots \\
& \left(\begin{array}{cccccccc}
1 & d^{r-1} & 0 \cdots & 0
\end{array}\right)
\ (k \equiv r-1 \operatorname{mod} N).
\\
\end{array}
\]
}
Then we have the following conflations.
\[
0 \to V \xrightarrow{h} \operatorname{C}(\varepsilon_{X}) \xrightarrow{p}
\underline{I}_{r+1}^{\Uparrow_{r+1}^{N}}
\underline{J}_{r}^{\Downarrow_{r+1}^{N}}(X) \to 0
\]
Since $V$ is a projective object in $\cat{C}_{N}(\mcal{B})$, we have an isomorphism
in $\cat{K}_{N}(\mcal{B})$
\[
\operatorname{C}(\varepsilon_{X}) \simeq \underline{I}_{r+1}^{\Uparrow_{r+1}^{N}}
\underline{J}_{r}^{\Downarrow_{r+1}^{N}}(X)
\]
Therefore, we have a triangle $U \to X \to V \to \Sigma X$ such that $U \in \mcal{F}_{s}^{r}$,
$V \in \mcal{F}_{r+s+1}^{N-r-1}$.\end{proof}
\begin{cor}\label{n-cpxgon04}
We have a recollement of $\cat{K}_{N}(\mcal{B})$:
\[\xymatrix{
\cat{K}_{N-r}(\mcal{B}) \ar@<-1ex>[r]^{i_{s*}}
& \cat{K}_{N}(\mcal{B})
\ar@/_1.5pc/[l]^{i_{s}^{*}} \ar@/^1.5pc/[l]_{i_{s}^{!}} \ar@<-1ex>[r]^{j_{s}^{*}}
& \cat{K}_{r+1}(\mcal{B})
\ar@/_1.5pc/[l]^{j_{s!}} \ar@/^1.5pc/[l]_{j_{s*}}
}\]
where
$i_{s}^{*}=\underline{J}_{s-1}^{\Downarrow_{N-r}^{N}}$,
$i_{s*}=\underline{I}_{s}^{\Uparrow_{N-r}^{N}}$,
$i_{s}^{!}=\underline{J}_{s}^{\Downarrow_{N-r}^{N}}$,
$j_{s!}=\underline{I}_{r+s}^{\Uparrow_{r+1}^{N}}$,
$j_{s}^{*}=\underline{J}_{r+s}^{\Downarrow_{r+1}^{N}}$ and
$j_{s!}=\underline{I}_{r+s+1}^{\Downarrow_{r+1}^{N}}$.
\end{cor}
\begin{proof}
By Theorem \ref{tstNcpx01},
$(\mcal{F}_{r+s-N}^{N-r-1}, \mcal{F}_{s}^{r})$ and
$(\mcal{F}_{s}^{r}, \mcal{F}_{r+s+1}^{N-r-1})$ are stable $t$-structures in
$\cat{K}_{N}(\mcal{B})$.
By the proof of Theorem \ref{tstNcpx01}, the adjunction arrows induce
triangles in $\cat{K}(\mcal{B})$.
Hence we have the statement.
\end{proof}
\begin{cor}\label{cpx2N-gon}
For any integer $s$,
\[
(\mcal{F}_{s+1}^{N-2}, \mcal{F}_{s}^{1}, \mcal{F}_{s+2}^{N-2}, \mcal{F}_{s+1}^{1},
\cdots, \mcal{F}_{s+r+1}^{N-2}, \mcal{F}_{s+r}^{1},
\cdots, \mcal{F}_{s+N-1}^{N-2}, \mcal{F}_{s+N-2}^{1}, \mcal{F}_{s}^{N-2}, \mcal{F}_{s+N-1}^{1})
\]
is a $2N$-gon of recollements in $\cat{K}_{N}(\mcal{B})$.
\end{cor}
\section{Triangle equivalence between homotopy categories}\label{TrieqDN}
\[
\mu^{s}_{r}C: X^{s-r+1} \xrightarrow{d^{s-r+1}} \cdots \xrightarrow{d^{s-2}} X^{s-1} \xrightarrow{d^{s-1}} X^{s}
\]
be an $N$-complex satisfying that
$
X^{s-i}=C \ (0 \leq i \leq r-1), \quad d^{s-i}=1_{C} \ (0 < i \leq r-1).
$
\begin{lem}\label{smcatcp}
Let $\mcal{B}$ be an additive category.
Consider $\operatorname{\mathsf{Mor}}^{\mrm{sm}}_{N-1}(\mcal{B})$ as a subcategory
of $\cat{C}_{N}(\mcal{B})$, then the following hold.
\begin{enumerate}
\item
For every object $X$ of $\operatorname{\mathsf{Mor}}^{\mrm{sm}}_{N-1}(\mcal{B})$,
there are objects $C_1, \cdots ,C_{N-1}$ of $\mcal{B}$ such that
$X \simeq \coprod_{i=1}^{N-1}\mu^{N-1}_{i}C_{i}$.
\item
For any $X, Y \in \operatorname{\mathsf{Mor}}^{\mrm{sm}}_{N-1}(\mcal{B})$, we have isomorphisms
\[\begin{aligned}
\operatorname{Hom}_{\operatorname{\mathsf{Mor}}^{\mrm{sm}}_{N-1}(\mcal{B})}(X,Y) &=\operatorname{Hom}_{\cat{C}_{N}(\mcal{B})}(X,Y) \\
&=\operatorname{Hom}_{\cat{K}_{N}(\mcal{B})}(X,Y) .
\end{aligned}\]
\item
For any $X, Y \in \operatorname{\mathsf{Mor}}^{\mrm{sm}}_{N-1}(\mcal{B})$, we have
$\operatorname{Hom}_{\cat{K}_{N}(\mcal{B})}(X,\Sigma^{i}Y) = 0 \ (i\not=0)$.
\end{enumerate}
\end{lem}
\begin{proof}
(1) It is trivial.\par\noindent
(2) It is easy because the term-length of objects of $\operatorname{\mathsf{Mor}}^{\mrm{sm}}_{N-1}(\mcal{B})$ is less than $N$.
\par\noindent
(3) For every $C \in \mcal{B}$, $1 \leq r \leq N-1$ and any integer $i$, the canonical injection and projection induce an conflation in
$\cat{C}(\mcal{B})$
\[\begin{aligned}
0 \to \mu_{r}^{iN+N-1}C \to \mu_{N}^{iN+N-1}C \to \mu_{N-r}^{(i+1)N-r-1}C \to 0 .
\end{aligned}\]
Since $\mu_{N}^{iN+N-1}C$
is a projective-injective objet in
$\cat{C}_{N}(\mcal{B})$,
we have isomorphisms in $\cat{K}_{N}(\mcal{B})$:
\[
\Sigma^{j}\mu_{r}^{N-1}C\simeq\begin{cases}
\mu_{N-r}^{(1-j)N/2-r-1}C \ (j\equiv 1 \operatorname{mod} 2)\\
\mu_{r}^{(2-j)N/2-1}C \ (j\equiv 0 \operatorname{mod} 2)
\end{cases}
\]
For every $C, C' \in \mcal{B}$ and any $1 \leq r,r' \leq N-1$, we have
\[
\operatorname{Hom}_{\cat{K}(\mcal{B})}(\mu_{r}^{N-1}C,\Sigma^{j}\mu_{r'}^{N-1}C') = 0 \ (j\not=0) .
\]
By (1), we have the statement.
\end{proof}
\begin{defn}\label{simpleF}
For every $C \in \mcal{B}$, let
\[
\Xi^{j}\mu_{r}^{N-1}C=\begin{cases}
\mu_{N-r}^{(1-j)N/2+N-r-1}C \ (j\equiv 1 \operatorname{mod} 2)\\
\mu_{r}^{(2-j)N/2-1}C \ (j\equiv 0 \operatorname{mod} 2)
\end{cases}\]
By the proof of Lemma \ref{smcatcp},
for every $M \in \operatorname{\mathsf{Mor}}^{\mrm{sm}}_{N-1}(\mcal{B})$ and any $i \in \mathbb{Z}$, there exist
the projective-injective object $I'(\Xi^iM)$ and
a functorial conflation in $\cat{C}_N(\mcal{B})$
\[
0 \rightarrow \Xi^{i}M \xrightarrow{u'_{\Xi^iM}} I'(\Xi^{i} M) \xrightarrow{v'_{\Xi^iM}} \Xi^{i+1}M \rightarrow 0
\]
that is, every morphism $f: M \to N$ in $\operatorname{\mathsf{Mor}}^{\mrm{sm}}_{N-1}(\mcal{B})$
uniquely determines morphisms $I'(f):I'(M) \to I'(N)$ and $\Xi^if:\Xi^iM \to \Xi^iN$
which have the following commutative diagram:
\[\xymatrix{
0 \ar[r] & \Xi^{i}M \ar[r] \ar[d]^{\Xi^if} & I'(\Xi^{i} M) \ar[r] \ar[d]^{I'(\Xi^if)} & \Xi^{i+1}M \ar[r]
\ar[d]^{\Xi^{i+1}f} & 0 \\
0 \ar[r] & \Xi^{i}N \ar[r] & I'(\Xi^{i} N) \ar[r] & \Xi^{i+1}N \ar[r] & 0
}\]
\end{defn}
For any (ordinary and $N$) complex $X = (X^{i}, d^{i})$, we define
the following truncations:
\[\begin{aligned}
{\tau}_{\leq n}X & : \cdots \rightarrow X^{n-2} \rightarrow X^{n-1}
\rightarrow X^{n} \rightarrow 0 \rightarrow \cdots ,\\
{\tau }_{\geq n}X & : \cdots \rightarrow 0 \rightarrow X^{n}
\rightarrow X^{n+1}\rightarrow X^{n+2}\rightarrow \cdots, \\
{\tau}_{[m,n]}X & : \cdots \rightarrow X^{m} \rightarrow \cdots \rightarrow X^{n} \rightarrow 0 \rightarrow \cdots ,\\
{\tau }_{n}X & : \cdots \rightarrow 0 \rightarrow X^{n} \rightarrow 0 \rightarrow \cdots .
\end{aligned}\]
\begin{lem}\label{prolong}
Let $\mcal{B}$ be an additive category.
There exists an exact functor
$F_{N}: \cat{C}(\operatorname{\mathsf{Mor}}^{\mrm{sm}}_{N-1}(\mcal{B})) \to \cat{C}_{N}(\mcal{B})$
which sends $\operatorname{\mathsf{Mor}}^{\mrm{sm}}_{N-1}(\mcal{B})$ to $\operatorname{\mathsf{Mor}}^{\mrm{sm}}_{N-1}(\mcal{B})$ as a subcategory
of $\cat{K}_{N}(\mcal{B})$
such that $F_{N}$ induces a triangle functor
$\underline{F}_{N}: \cat{K}(\operatorname{\mathsf{Mor}}^{\mrm{sm}}_{N-1}(\mcal{B})) \to \cat{K}_{N}(\mcal{B})$.
\end{lem}
\begin{proof}
Let $T: \cat{C}(\operatorname{\mathsf{Mor}}^{\mrm{sm}}_{N-1}(\mcal{B})) \to \cat{C}(\operatorname{\mathsf{Mor}}^{\mrm{sm}}_{N-1}(\mcal{B}))$
be a translation functor.
We will prove the statement by the following steps:
\par\noindent
Step 1. We have a functor $F_{N}: \coprod_{i}T^{i}\mcal{B} \to \cat{C}_{N}(\mcal{B})$ which preserves split exact sequences.
\par\noindent
For $T^{i} X \in T^{i}\operatorname{\mathsf{Mor}}^{\mrm{sm}}_{N-1}(\mcal{B})$, let $F_{N}(T^{i} X)=\Xi^{i}X$, then
$F_{N}: \coprod_{i}T^{i}\operatorname{\mathsf{Mor}}^{\mrm{sm}}_{N-1}(\mcal{B}) \to \cat{K}_{N}(\mcal{B})$ which preserves split exact sequences.
\par\noindent
Step 2. We have a functor $F_{N}: \cat{C}^{\mrm{b}}(\operatorname{\mathsf{Mor}}^{\mrm{sm}}_{N-1}(\mcal{B})) \to \cat{C}_{N}(\mcal{B})$ which preserves conflations.
\par\noindent
For $i<j$, let $X:X^{i} \xrightarrow{d^i} X^{i+1} \to \cdots \to X^{j} \in \cat{C}^{\mrm{b}}(\operatorname{\mathsf{Mor}}^{\mrm{sm}}_{N-1}(\mcal{B}))$.
By induction on $j-i$, we comstruct a functor $F_{N}$.
For $X$, we have a commutative diagram
in $\cat{C}^{\mrm{b}}(\mcal{B})$:
\[\xymatrix{
0 \ar[r] & T^{-i}X^{i} \ar[r] \ar[d]^{d^i} & I(T^{-i}X^{i}) \ar[r] \ar[d] & T^{-i-1}X^i \ar[r] \ar@{=}[d] & 0 \\
0 \ar[r] & \tau_{\geq i+1}X \ar[r] & \tau_{\geq i}X \ar[r] & T^{-i-1}X^i \ar[r] & 0
}
\]
where all rows are exact. Then we have a commutative diagram
in $\cat{C}_{N}(\mcal{B})$:
\begin{equation}\label{limit01}
\xymatrix@1{
0 \ar[r] & \ar @{} [dr] |{(A)} F_{N}(T^{-i}X^{i}) \ar[r] \ar[d]^{F_{N}(d^i)} & I'(F_{N}(T^{-i}X^{i})) \ar[r] \ar[d]^{\gamma_{i}} & F_{N}(T^{-i-1}X^i) \ar[r] \ar@{=}[d] & 0 \\
0 \ar[r] & F_{N}(\tau_{\geq i+1}X) \ar[r] & C(F_{N}(d^{i})) \ar[r] & F_{N}(T^{-i-1}X^i) \ar[r] & 0
}
\end{equation}
where all rows are conflations. We take $F_{N}(\tau_{\geq i}X)=C(F_{N}(d^i))$.
By successive step, we have a functor $F_{N}: \cat{C}^{\mrm{b}}(\operatorname{\mathsf{Mor}}^{\mrm{sm}}_{N-1}(\mcal{B})) \to \cat{C}_{N}(\mcal{B})$.
For any inflation (resp., deflation) $f:X \to Y$ in $\cat{C}^{\mrm{b}}(\mcal{B})$,
$F_{N}(T^if):F_{N}(T^iX) \to F_{N}(T^iY)$ and $I'(F_{N}(T^if)):I'(F_{N}(T^iX)) \to I'(F_{N}(T^iY))$ are split monomorphism
(resp., epimorphism) by Definition \ref{simpleF}.
Then \[
0 \to F_{N}(T^iX) \to F_{N}(\tau_{\geq i+1}X)\oplus I'(F_{N}(T^{-i}X^i)) \to F_{N}(\tau_{\geq i}X) \to 0
\]
is an inflation.
Let $0 \to X \to Y \to Z \to 0$ be an conflation in
$\cat{C}^{\mrm{b}}(\operatorname{\mathsf{Mor}}^{\mrm{sm}}_{N-1}(\mcal{B}))$.
Consider a small Frobenius subcategory $\mcal{C}$ of
$\cat{C}_{N}(\operatorname{\mathsf{Mor}}^{\mrm{sm}}_{N-1}(\mcal{B}))$ which contains
$F_{N}(T^{i}X), \\ F_{N}(T^{i}Y), F_{N}(T^{i}Z), I'(F_{N}(T^{-i}X), I'(F_{N}(T^{-i}Y), I'(F_{N}(T^{-i}Z),
F_{N}(\tau_{\geq i}X), \\ F_{N}(\tau_{\geq i}Y)$ and $F_{N}(\tau_{\geq i}Z)$ $(i \in \mathbb{Z})$.
Then by the induction on $j-i$, 9-lemma implies that
\[
0 \to F_{N}(\tau_{\geq i}X) \to F_{N}(\tau_{\geq i}Y) \to F_{N}(\tau_{\geq i}Z) \to 0
\]
is a short exact sequence in some abelian category.
According to Proposition \ref{excatemb}, it is a conflation in $\mcal{C}$.
Therefore, $F_{N}$ preserves conflations.
\par\noindent
Step 3. We have a functor $F_{N}: \cat{C}^{-}(\operatorname{\mathsf{Mor}}^{\mrm{sm}}_{N-1}(\mcal{B})) \to \cat{C}_{N}(\mcal{B})$ which preserves
conflations.
\par\noindent
For $X \in \cat{C}^{-}(\mcal{B})$, we have $X =\displaystyle{\lim_{i \to \infty}}\tau_{\geq -i}X$.
By the diagram \ref{limit01}, we have
\[
\tau_{\geq (\left[\frac{i+1}{2}\right])N-1}F_{N}(\tau_{\geq -i}X) =
\tau_{\geq (\left[\frac{i+1}{2}\right])N-1}F_{N}(\tau_{\geq -i-1}X)
\]
Then there exists $\displaystyle{\lim_{i \to \infty}}F_{N}(\tau_{\geq -i}X)$, and
we take $F_{N}(X)=\displaystyle{\lim_{i \to \infty}}F_{N}(\tau_{\geq -i}X)$.
It is not hard to see that $F_{N}$ becomes a functor and
preserves conflations.
\par\noindent
Step 4. We have a functor $F_{N}: \cat{C}(\operatorname{\mathsf{Mor}}^{\mrm{sm}}_{N-1}(\mcal{B})) \to \cat{C}_{N}(\mcal{B})$ which preserves
conflations.
\par\noindent
Let $X:\cdots \to X^{i} \xrightarrow{d^i} X^{i+1} \to \cdots \in \cat{C}(\mcal{B})$.
By induction on $i$, we comstruct a functor $F_{N}$.
For $X$, we have a commutative diagram in $\cat{C}^{-}(\operatorname{\mathsf{Mor}}^{\mrm{sm}}_{N-1}(\mcal{B}))$:
\[\xymatrix{
0 \ar[r] & T^{-i-1}X^{i+1} \ar[r] \ar@{=}[d] & \tau_{\leq i+1}X\ar[r] \ar[d] & \tau_{\leq i}X \ar[r] \ar[d]^{d^i} & 0 \\
0 \ar[r] & T^{-i-1}X^{i+1} \ar[r] & I(T^{-i-1}X^{i+1}) \ar[r] & T^{-i}X^{i+1} \ar[r] & 0
}
\]
where all rows are exact. Then we have a commutative diagram
in $\cat{C}_{N}(\mcal{B})$:
\begin{equation}\label{limit02}
\xymatrix{
0 \ar[r] & F_{N}(T^{-i-1}X^{i+1}) \ar[r] \ar@{=}[d] & \ar @{} [dr] |{(B)}
\Sigma^{-1}\operatorname{C}(F_{N}(d^{i})) \ar[r] \ar[d] &
F_{N}(\tau_{\leq i}X) \ar[r]\ar[d]^{F_{N}(d^i)} & 0 \\
0 \ar[r] & F_{N}(T^{-i-1}X^{i+1}) \ar[r] & I'(T^{-i-1}X^{i+1}) \ar[r] & F_{N}(T^{-i}X^{i+1}) \ar[r] & 0
}
\end{equation}
where all rows are conflations. We take $F_{N}(\tau_{\leq i+1}X)=C(F_{N}(d^i))$.
By the diagram \ref{limit02}, we have
\[
\tau_{\leq \left[\frac{i+2}{2}\right]N-1}F_{N}(\tau_{\leq i}X) =
\tau_{\leq \left[\frac{i+2}{2}\right]N-1}F_{N}(\tau_{\leq i+1}X)
\]
Then there exists $\displaystyle{\lim_{\infty \gets i}}F_{N}(\tau_{\leq i}X)$, and
we take $F_{N}(X)=\displaystyle{\lim_{\infty \gets i}}F_{N}(\tau_{\leq i}X)$.
Since the commutative diagram (B) is a exact square,
it is not hard to see that $F_{N}$ becomes a functor and preserves
conflations.
\par\noindent
Step 5. $F_{N}$
sends projective-injective objects in $\cat{C}(\operatorname{\mathsf{Mor}}^{\mrm{sm}}_{N-1}(\mcal{B}))$
to projective-\\ injective objects in $\cat{C}_{N}(\mcal{B})$.
\par\noindent
Every projective-injective object in $\cat{C}(\operatorname{\mathsf{Mor}}^{\mrm{sm}}_{N-1}(\mcal{B}))$
is a direct summand of some biproduct $\oplus_{i \in \mathbb{Z}}T^i\operatorname{C}(1_{M_i})$
with $M_i \in \operatorname{\mathsf{Mor}}^{\mrm{sm}}_{N-1}(\mcal{B})$.
Since $F_{N}(\operatorname{C}(1_{M_i})) \simeq \operatorname{C}(1_{F_{N}(M_i)})$ is projective-injective
in $\cat{C}_{N}(\mcal{B})$, we have the statement.
\par\noindent
According to Proposition \ref{extr01},
$F_{N}: \cat{C}(\operatorname{\mathsf{Mor}}^{\mrm{sm}}_{N-1}(\mcal{B})) \to \cat{C}_{N}(\mcal{B})$ induces
a triangle functor
$\underline{F}_{N}: \cat{K}(\operatorname{\mathsf{Mor}}^{\mrm{sm}}_{N-1}(\mcal{B})) \to \cat{K}_{N}(\mcal{B})$.
\end{proof}
\begin{lem}\label{send2Ngons00}
The following hold.
\begin{enumerate}
\item Every complex $X$ of $\cat{C}(\operatorname{\mathsf{Mor}}^{\mrm{sm}}_{N-1}(\mcal{B}))$ is of the form
\[\xymatrix{
X_{1} \ar[r]^{\alpha^1}
& \oplus_{i=1}^{2}X_{i} \ar[r]^{\alpha^2}
&\cdots \ar[r]^{\alpha^{N-2}} & \oplus_{i=1}^{N-1}X_{i}
}\]
where $\alpha^i=
\left[\begin{smallmatrix}
1 & \cdots & 0 \\
& \ddots & \\
0 & \cdots & 1 \\
0 & \cdots & 0 \\
\end{smallmatrix}\right]$
and $(\oplus_{i=1}^{r}X_{i}, d_X^i=
\left[\begin{smallmatrix}
d^{i}_{11} & d^{i}_{12} & \cdots & d^{i}_{1r} \\
0 & d^{i}_{22} & \cdots & d^{i}_{2r} \\
\vdots & \vdots & \ddots &\vdots \\
0 & 0 & \cdots & d^{i}_{rr}
\end{smallmatrix}\right])$
is a complex of $\mcal{B}$.
\item For a complex $X$ of $\cat{C}(\operatorname{\mathsf{Mor}}^{\mrm{sm}}_{N-1}(\mcal{B}))$,
the $N$-complex $F_{N}(X)$ is equal to $(Y^j, d_Y^j)$
where
\[
Y^j=\left(\oplus_{r=1}^{k}X_{r}^{2i}\right)\oplus
\left(\oplus_{r=k}^{N-1}X_{r}^{2i-1}\right) \\
\]
\[
d_Y^j=
\begin{cases}
\left[\begin{smallmatrix}
d^{i}_{11} & d^{i}_{12} & \cdots & d^{i}_{1N-1} \\
0 & d^{i}_{22} & \cdots & d^{i}_{2N-1} \\
\vdots & \vdots & \ddots &\vdots \\
0 & 0 & \cdots & d^{i}_{N-1N-1}
\end{smallmatrix}\right]
&\ (j\equiv -1 \mod N),
\\
\left[\begin{matrix}
1 & 0 & \cdots & 0 &d^{i-1}_{1k} & 0 & \cdots & 0 \\
0 & 1 & \ddots & \vdots & d^{i-1}_{2k} & \vdots & & \vdots \\
\vdots & \ddots & \ddots & 0 & \vdots & \vdots & & \vdots \\
\vdots & & \ddots & 1 & d^{i-1}_{k-1k} & \vdots & & \vdots \\
\vdots & & & \ddots & d^{i-1}_{kk} & 0 & & \vdots \\
\vdots & & & & \ddots & 1 & \ddots & \vdots \\
\vdots & & & & & \ddots & \ddots& 0\\
0 & \cdots &\cdots &\cdots &\cdots &\cdots & 0 & 1\\
\end{matrix}\right]
&\ \text{otherwise}
\end{cases}\]
where $i=2\left[\frac{j}{N}\right]$, $0 \leq k \leq N-1$ and $k \equiv j \mod N$.
\item $\underline{F}_{N}(\mcal{E}^{[1, N-2]}) \subset
\mcal{F}_{N-1}^{1}$
\end{enumerate}
\end{lem}
\begin{proof}
(1)
It is trivial.
\par\noindent
(2)
For a complex $X \in \cat{C}^{\mrm{b}}(\operatorname{\mathsf{Mor}}^{\mrm{sm}}_{N-1}(\mcal{B}))$,
we assume $F_{N}(\tau_{\geq 2j}X)$ satisfies the statement.
For $2j=i+1$, we have the following equations in the diagram \ref{limit01}
\[\begin{aligned}
F_{N}(T^{-i}X^{i}) & =\oplus_{r=1}^{N-1}\mu_{N-r}^{N-1+jN}X^{2j-1}_r &
I'(F_{N}(T^{-i}X^{i})) & = \oplus_{r=1}^{N-1}\mu_{N-1}^{N-1+jN}X^{2j-1}_r \\
F_{N}(T^{-i-1}X^{i}) & =\oplus_{r=1}^{N-1}\mu_{r}^{r-1+jN}X^{2j-1}_r \\
\end{aligned}\]
By easy calculations,
$
\gamma_j^{k+jN}: I'(F_{N}(T^{-i}X^{i}))^k \to Y^{k+jN}
$
is equal to
$d_Y^{k+N}$ in the statement for $0 \leq k < N$.
Therefore, we have the statement for $F_{N}(\tau_{\geq i}X)$.
\par\noindent
For $2j=i$, we have the following equations in the Diagram \ref{limit01}
\[\begin{aligned}
F_{N}(T^{-i}X^{i}) & =\oplus_{r=1}^{N-1}\mu_{r}^{r-1+(j+1)N}X^{2j}_r &
I'(F_{N}(T^{-i}X^{i})) & = \oplus_{r=1}^{N-1}\mu_{N-1}^{r-1+(j+1)N}X^{2j}_r\\
F_{N}(T^{-i}X^{i}) & =\oplus_{r=1}^{N-1}\mu_{N-r+1}^{-1+jN}X^{2j}_r \\
\end{aligned}\]
By easy calculations,
$
\gamma_j^{k+jN}: I'(F_{N}(T^{-i}X^{i}))^k \to Y^{k+jN}
$
is a $(k+1)\times k$ matrix
$\left[\begin{smallmatrix}
1 & \cdots & 0 \\
& \ddots & \\
0 & \cdots & 1 \\
0 & \cdots & 0 \\
\end{smallmatrix}\right]$
for $1 \leq k < N-1$, and
$
\gamma_j^{N-1+jN}: I'(F_{N}(T^{-i}X^{i}))^k \to Y^{k+jN}
$
is equal to
$d_Y^{N-1+jN}$ in the statement.
Therefore, we have the statement for $F_{N}(\tau_{\geq i}X)$.
\par\noindent
Similarly, we have the same result for the diagram \ref{limit02}.
\par\noindent
(3) By Lemma \ref{lem:comma03}, any complex of $\mcal{E}^{[1,N-2]}$
is isomorphic to the mapping cone of the following morphism
between complexes:
\[\xymatrix{
X' : \ar[d]_{f} & 0\ar[r] \ar[d] & 0 \ar[r] \ar[d]
&\cdots \ar[r] & 0 \ar[r] \ar[d] & \oplus_{i=1}^{N-1}X_{i} \ar[d]^{1} \\
X: & X_{1} \ar[r]^{\alpha^1}
& \oplus_{i=1}^{2}X_{i} \ar[r]^{\alpha^2}
&\cdots \ar[r]^{\alpha^{N-2}} & \oplus_{i=1}^{N-2}X_{i} \ar[r] & \oplus_{i=1}^{N-1}X_{i}
}\]
Consider a morphism between the Diagram \ref{limit01} for $F_{N}(\tau_{\geq i}X')$
and the Diagram \ref{limit01} for $F_{N}(\tau_{\geq i}X)$.
Then $F_{N}(X')=(Y'^j, d_{Y'}^i)$,
where
\[
Y^j=\begin{cases}
\oplus_{r=1}^{N-1}X_{r}^{2i} &\ (j \equiv -1 \mod N) \\
\oplus_{r=1}^{N-1}X_{r}^{2i-1}&\ \text{otherwise}
\end{cases}
\]
\[
d_Y^j=
\begin{cases}
\left[\begin{smallmatrix}
d^{i-1}_{11} & d^{i-1}_{12} & \cdots & d^{i-1}_{1N-1} \\
0 & d^{i-1}_{22} & \cdots & d^{i-1}_{2N-1} \\
\vdots & \vdots & \ddots &\vdots \\
0 & 0 & \cdots & d^{i-1}_{N-1N-1}
\end{smallmatrix}\right]
&\ (j\equiv -2 \mod N),
\\
\text{identity} &\ \text{otherwise}
\end{cases}
\]
where $i=2\left[\frac{j}{N}\right]$, $0 \leq k \leq N-1$ and $k \equiv j \mod N$.
Moreover, $F_{N}(f): F_{N}(X') \to F_{N}(X)$ is equal to $g:Y' \to Y$
where
\[g^j=
\begin{cases}
\text{identity} &\ (j\equiv 0, -1 \mod N) \\
\left[\begin{matrix}
d_{11}^{i-1} & \cdots & d_{1k}^{i-1} & 0 & \cdots & 0 \\
0 & \ddots & \vdots & \vdots & & \vdots \\
\vdots & \ddots & d^{i-1}_{kk} & 0 & & \vdots \\
\vdots & & \ddots & 1 & \ddots & \vdots \\
\vdots & & & \ddots & \ddots& 0\\
0 &\cdots &\cdots &\cdots &0 & 1 \\
\end{matrix}\right] &\ \text{otherwise}
\end{cases}\]
where $i=2\left[\frac{j}{N}\right]$, $0 < k < N-1$ and $k \equiv j \mod N$.
By the proof of Theorem \ref{tstNcpx01}, we have
$Y'=\underline{I}_{0}^{\Uparrow_{1}^{N}}
\underline{J}_{0}^{\Downarrow_{1}^{N}}(Y)$ and $\epsilon_Y=g$.
Therefore $\operatorname{C}(F_{N}(f)) \simeq \operatorname{C}(g) \in \mcal{F}_{N-1}^{1}$.
\end{proof}
\begin{lem}\label{send2Ngons01}
The following hold for the above triangle functor
$\underline{F}_{N}: \cat{K}(\operatorname{\mathsf{Mor}}^{\mrm{sm}}_{N-1}(\mcal{B})) \\ \to \cat{K}_{N}(\mcal{B})$.
\begin{enumerate}
\item $\underline{F}_{N}(\mcal{E}^{[s, N-1]}\cap\mcal{F}^{[s,N-1]}) \subset
\mcal{F}_{0}^{s-1}\cap\mcal{F}_{s}^{N-s-1}$.
\item $\underline{F}_{N}(\mcal{E}^{s}) \subset \mcal{F}_{s+1}^{N-2}$.
\end{enumerate}
\end{lem}
\begin{proof}
(1)
According to Lemma \ref{lem:comma03}, we may assume any complex $Y$ of
$\mcal{E}^{[s, N-1]}\cap\mcal{F}^{[s,N-1]}$
is of the form
\[
0 \to \cdots \to 0 \rightarrow Y^s \xrightarrow{\alpha^{s}} \cdots \xrightarrow{\alpha^{N-2}} Y^{N-1}
\]
where $X=(X^i, d^i)=Y^s = \cdots =Y^{N-1}$ is a complex of $\mcal{B}$ and
$\alpha^{s}=\cdots =\alpha^{N-2}=1_X$.
Then by Diagrams \ref{limit01}, \ref{limit02},
$\underline{F}_{N}(Y)$ is isomorphic to a complex $(X'^i, d'^i)$
where
\[\begin{array}{ll}
X'^{i} = \begin{cases}
X^{2\left[\frac{i}{N}\right]-1} \ (i\equiv 0, 1, \cdots, s-1 \mod N) \\
X^{2\left[\frac{i}{N}\right]} \ (i\equiv s, s+1, \cdots, N-1 \mod N)
\end{cases}
\\
d'^{i} = \begin{cases}
1_{X^{2\left[\frac{i}{N}\right]}} \ (i\equiv 0, 1, \cdots, s-2 \mod N) \\
d^{2\left[\frac{i}{N}\right]-1} \ (i\equiv s-1 \mod N) \\
1_{X^{2\left[\frac{i}{N}\right]}} \ (i\equiv s, s+1, \cdots, N-2 \mod N) \\
d^{2\left[\frac{i}{N}\right]} \ (i\equiv N-1 \mod N) \\
\end{cases}
\end{array}\]
Therefore $\underline{F}_{N}(\mcal{E}^{[s, N-1]}\cap\mcal{F}^{[s,N-1]}) \subset
\mcal{F}_{0}^{[0,s-1]}\cap\mcal{F}_{s}^{[0,N-s-1]}$.
\par\noindent
(2)
Any complex $Y$ of $\mcal{E}^{s}$ is isomorphic to the mapping cone of a morphism $\iota:Y_1 \to Y_2$
in $\cat{C}(\operatorname{\mathsf{Mor}}^{\mrm{sm}}_{N-1}(\mcal{B}))$:
\[\xymatrix{
Y_1: \ar[d]^{\iota} & 0 \ar[d] \ar[r] & \cdots \ar[r] & 0 \ar[d] \ar[r] & 0 \ar[d] \ar[r] & Y^{s+1} \ar[d]^{1} \ar[r]^{\alpha^{s+1}} & \cdots \ar[r]^{\alpha^{N-2}} & Y^{N-1} \ar[d]^{1} \\
Y_2: & 0 \ar[r] & \cdots \ar[r] & 0 \ar[r] & Y^s \ar[r]^{\alpha^{s}} & Y^{s+1} \ar[r]^{\alpha^{s+1}} & \cdots \ar[r]^{\alpha^{N-2}} & Y^{N-1}
}\]
where $X=(X^i, d^i)=Y^s = \cdots =Y^{N-1}$ is a complex of $\mcal{B}$ and
$\alpha^{s}=\cdots =\alpha^{N-2}=1_X$.
By the construction of $F_{N}$ in Theorem \ref{prolong},
$\underline{F}_{N}(Y_2)=\underline{I}_{0}^{\Uparrow_{N-r}^{N}}
\underline{J}_{0}^{\Downarrow_{N-r}^{N}}(\underline{F}_{N}(Y_1)) \in \mcal{F}_{0}^{s}$ and
$\underline{F}_{N}(\iota)=\underline{I}_{0}^{\Uparrow_{N-r}^{N}}
\underline{J}_{0}^{\Downarrow_{N-r}^{N}}(\underline{F}_{N}(Y_1)) \xrightarrow{\varepsilon_{\underline{F}_{N}(Y_1)}} \underline{F}_{N}(Y_1)$.
By the Proof of Theorem \ref{tstNcpx01},
the mapping cone $\operatorname{C}(\underline{F}_{N}(\iota))$ is isomorphic to a complex $(X'^i, d'^i)$
where
\[\begin{array}{ll}
X'^{i} = \begin{cases}
X^{2\left[\frac{i}{N}\right]-1} \ (i\equiv 0, 1, \cdots, s-1 \mod N) \\
X^{2\left[\frac{i}{N}\right]} \ (i\equiv s \mod N) \\
X^{2\left[\frac{i}{N}\right]+1} \ (i\equiv s+1, \cdots, N-1 \mod N)
\end{cases}
\\
d'^{i} = \begin{cases}
1_{X^{2\left[\frac{i}{N}\right]-1}} \ (i\equiv 0, 1, \cdots, s-2 \mod N) \\
d^{2\left[\frac{i}{N}\right]-1} \ (i\equiv s-1 \mod N) \\
d^{2\left[\frac{i}{N}\right]} \ (i\equiv s \mod N) \\
1_{X^{2\left[\frac{i}{N}\right]}} \ (i\equiv s+1, \cdots, N-1 \mod N) \\
\end{cases}
\end{array}\]
Therefore $\underline{F}_{N}(\mcal{E}^{s}) \subset \mcal{F}_{s+1}^{N-2}$.
\end{proof}
\begin{thm}\label{KNhtp}
Let $\mcal{B}$ be an additive category, then we have triangle equivalences:
\[
\cat{K}^{\sharp}(\operatorname{\mathsf{Mor}}^{\mrm{sm}}_{N-1}(\mcal{B})) \simeq \cat{K}^{\sharp}_{N}(\mcal{B})
\]
where $\sharp=\text{nothing}, -, +, \mrm{b}$.
\end{thm}
\begin{proof}
We prove the statement by the following steps:
\par\noindent
Step 1.
$\underline{F}_{N}: \cat{K}(\operatorname{\mathsf{Mor}}^{\mrm{sm}}_{N-1}(\mcal{B})) \to \cat{K}_{N}(\mcal{B})$
sends \\
{\small $
(\mcal{F}^{[1,N-1]},\mcal{E}^{[2,N-1]}, \mcal{E}^{1}, \mcal{F}^{[1,2]}, \cdots,
\mcal{E}^{s}, \mcal{F}^{[s,s+1]}, \cdots, \mcal{E}^{N-2}, \mcal{F}^{[N-2,N-1]}, \mcal{E}^{N-1}, \mcal{E}^{[1,N-2]})
$}
to \\
$
(\mcal{F}_{1}^{N-2}, \mcal{F}_{0}^{1}, \mcal{F}_{2}^{N-2}, \mcal{F}_{1}^{1},
\cdots, \mcal{F}_{s+1}^{N-2}, \mcal{F}_{s}^{1},
\cdots, \mcal{F}_{N-1}^{N-2}, \mcal{F}_{N-2}^{1}, \mcal{F}_{0}^{N-2}, \mcal{F}_{N-1}^{1}).
$
\par\noindent
According to Lemma \ref{send2Ngons00} (2),
it is easy to see that
\[\begin{array}{lll}
\underline{F}_{N}(\mcal{F}^{[1,N-1]})\subset \mcal{F}_{1}^{N-2}, &\quad
\underline{F}_{N}(\mcal{F}^{[2,N-1]}) \subset \mcal{F}_{0}^{1}, &\quad
\underline{F}_{N}(\mcal{E}^{N-1}) \subset \mcal{F}_{0}^{N-2}, \\
\underline{F}_{N}(\mcal{F}^{[s,s+1]}) \subset \mcal{F}_{s}^{1} & \quad (1\leq s <N-1)
\end{array}\]
By Lemma \ref{send2Ngons01}, we have
$\underline{F}_{N}(\mcal{E}^{s}) \subset \mcal{F}_{s+1}^{N-2}$ $(1 \leq s < N-1)$.
By Lemma \ref{send2Ngons00} (3), we have
$\underline{F}_{N}(\mcal{E}^{[1, N-2]}) \subset \mcal{F}_{N-1}^{1}$.
\par\noindent
Step 2. $\underline{F}_{N}$ induces a triangle equivalence
between $\mcal{F}^{[1,N-1]}$ and $\mcal{F}_{1}^{N-2}$.
\par\noindent
Consider the following diagram:
\[\xymatrix{
\cat{K}(\mcal{B})
\ar[d]_{U_{N-1}}\ar[r]^{id} &
\cat{K}(\mcal{B}) \ar[d]^{\underline{I}_{1}^{\Uparrow_{2}^{N}}} \\
\cat{K}(\operatorname{\mathsf{Mor}}_{N-1}(\mcal{B})) \ar[r]^{\quad \underline{F}_{N}} & \cat{K}_{N}(\mcal{B})
}\]
By Proposition \ref{lastpiece}, Lemma \ref{send2Ngons00},
it is not hard to see that we have a functorial isomorphism
$\underline{F}_{N}\circ U_{N-1} \simeq
\underline{I}_{1}^{\Uparrow_{2}^{N}}$.
By Proposition \ref{lastpiece}, $U_{N-1}$ induces a triangle equivalence between $\cat{K}(\mcal{B})$ and $\mcal{F}^{[1,N-1]}$.
On the other hand, by Corollary \ref{n-cpxgon04}
$\underline{I}_{1}^{\Uparrow_{2}^{N}}$ induces a triangle equivalence between
$\cat{K}_{N-1}(\mcal{B})$ and $\mcal{F}_{1}^{N-2}$.
Therefore $\underline{F}_{N}$ induces a triangle equivalence
between $\mcal{F}^{[1,N-1]}$ and $\mcal{F}_{1}^{N-2}$.
\par\noindent
Step 3. $\underline{F}_{N}$ induces a triangle equivalence
between $\mcal{E}^{[2,N-1]}$ and $\mcal{F}_{0}^{1}$.
\par\noindent
Consider the following diagram:
\[\xymatrix{
\cat{K}(\operatorname{\mathsf{Mor}}_{N-2}(\mcal{B}))
\ar[d]_{E^{\Uparrow_{N-2}^{N-1}}} \ar[r]^{\quad \underline{F}_{N-1}} &
\cat{K}_{N-1}(\mcal{B}) \ar[d]^{\underline{I}_{0}^{\Uparrow_{N-1}^{N}}} \\
\cat{K}(\operatorname{\mathsf{Mor}}_{N-1}(\mcal{B})) \ar[r]^{\quad \underline{F}_{N}} & \cat{K}_{N}(\mcal{B})
}\]
Similarly, we have a functorial isomorphism
$\underline{F}_{N}\circ E^{\Uparrow_{N-2}^{N-1}} \simeq
\underline{I}_{0}^{\Uparrow_{N-1}^{N}}\circ \underline{F}_{N-1}$.
By Proposition \ref{lastpiece}, $E^{\Uparrow_{N-2}^{N-1}}$ induces
a triangle equivalence between $\cat{K}(\mcal{B})$ and $\mcal{F}^{[1,N-1]}$.
On the other hand, by Corollary \ref{n-cpxgon04}
$\underline{I}_{1}^{\Uparrow_{2}^{N}}$ induces a triangle equivalence between
$\cat{K}_{N-1}(\mcal{B})$ and $\mcal{F}_{0}^{1}$.
By the assumption of induction on $N$, $\underline{F}_{N-1}$ is a triangle equivalence.
Therefore $\underline{F}_{N}$ induces a triangle equivalence
between $\mcal{E}^{[2,N-1]}$ and $\mcal{F}_{0}^{1}$.
\par\noindent
According to Proposition \ref{p20Apr30},
$\underline{F}_{N}:\cat{K}(\operatorname{\mathsf{Mor}}^{\mrm{sm}}_{N-1}(\mcal{B})) \to \cat{K}_{N}(\mcal{B})$
is a triangle equivalence.
Moreover, it is easy to see the above proof is available for the case
$\underline{F}^{\sharp}_{N}:\cat{K}^{\sharp}(\operatorname{\mathsf{Mor}}^{\mrm{sm}}_{N-1}(\mcal{B})) \to \cat{K}^{\sharp}_{N}(\mcal{B})$, where $\sharp=-,+, \mrm{b}$.
\end{proof}
In \cite{IKM2}, we studied the derived category of $N$-complexes.
We have results of \cite{IKM2} Corollaries 4.11, 4.12 under the weaker condition.
\begin{lem}\label{qtricat01}
Let $\mcal{D}$ be a triangulated category, $\mcal{C}, \mcal{U}$
full triangulated subcategories of $\mcal{C}$.
We assume that for any $X \in \mcal{D}$ there is a triangle
$C_X \to X \to U_X \to \Sigma C_X$ such that
$C_X \in \mcal{C}$ and $U_X \in \mcal{U}$.
Then we have a triangle equivalence
\[
\mcal{C}/(\mcal{C}\cap\mcal{U}) \simeq \mcal{D}/\mcal{U}.
\]
\end{lem}
\begin{proof}
Let $E: \mcal{C} \to \mcal{D}$ be the canonical embedding,
$Q': \mcal{C} \to \mcal{C}/(\mcal{C}\cap\mcal{U})$,
$Q: \mcal{D} \to \mcal{D}/\mcal{U}$ the canonical quotients.
Then there is a triangle functor $F: \mcal{C}/(\mcal{C}\cap\mcal{U}) \to \mcal{D}/\mcal{U}$
such that $Q\circ E = F\circ Q'$.
By the assumption, $F$ is obviously dense.
Given $X_1, X_2 \in \mcal{C}$, any morphism in $\mcal{D}/\mcal{U}$ is represented by the following in $\mcal{D}$:
\begin{equation}\label{qfmor}
\xymatrix{
X_1 \ar[d]_{s_1} \ar[dr]^{g}\\
F(C_1) & F(C_2)
}
\end{equation}
where $X_1 \xrightarrow{s_1} F(C_1) \rightarrow U_1 \rightarrow \Sigma X_1$ is a triangle in $\mcal{D}$
with $U_1 \in \mcal{U}$.
By the assumption, there is a triangle $C_2 \xrightarrow{s_2} X_1 \rightarrow U_2 \rightarrow \Sigma C_2$
with $U_2 \in \mcal{U}$.
Therefore, $F$ is a full dense functor.
Let $f: C_1 \to C_2$ be a morphism in $\mcal{C}/(\mcal{C}\cap\mcal{U})$
such that $F(f)=0$ in $\mcal{D}_\mcal{U}$.
Then $F(f)$ is represented by the diagram \ref{qfmor}
where $g=0$.
In the above, $gs_2=0$, and then $f=0$ in $\mcal{C}/(\mcal{C}\cap\mcal{U})$.
Hence $F$ is an equivalence.
\end{proof}
We say that $\mcal{A}$ is an Ab3 category provided that
it has any coproduct of objects.
Moreover, $\mcal{A}$ is an Ab4 category provided that
it has any coproduct of objects, and that
the coproduct of monics is monic.
\begin{prop}\label{qtricat02}
Let $\mcal{A}$ be an AB4 category with enough projectives, and $\mcal{P}$ the category of projective objects, $\mcal{P}$ the full subcategory of $\mcal{A}$ consisting of projective objects.
Then the following hold.
\begin{enumerate}
\item We have a triangle equivalence
\[
\cat{K}(\operatorname{\mathsf{Mor}}^{\mrm{sm}}_{N-1}(\mcal{P}))/
\cat{K}^{\phi}(\operatorname{\mathsf{Mor}}^{\mrm{sm}}_{N-1}(\mcal{P}))
\simeq
\cat{D}(\operatorname{\mathsf{Mor}}_{N-1}(\mcal{A})).
\]
\item We have a triangle equivalence
\[
\cat{K}_{N}(\mcal{P})/\cat{K}^{\phi}_{N}(\mcal{P})
\simeq
\cat{D}_{N}(\mcal{A}).
\]
\end{enumerate}
Here $\cat{K}^{\phi}(\operatorname{\mathsf{Mor}}^{\mrm{sm}}_{N-1}(\mcal{P}))$
(resp., $\cat{K}^{\phi}_{N}(\mcal{P})$) is the full subcategory of
Here $\cat{K}(\operatorname{\mathsf{Mor}}^{\mrm{sm}}_{N-1}(\mcal{P}))$
(resp., $\cat{K}_{N}(\mcal{P})$) consisting of
complexes (resp., $N$-complexes) of which all hmologies are null.
\end{prop}
\begin{proof}
(1)
It is easy to see that
$\operatorname{\mathsf{Mor}}^{\mrm{sm}}_{N-1}(\mcal{P})$ is the full subcategory of $\operatorname{\mathsf{Mor}}_{N-1}(\mcal{A})$
consisting of projective objects, and that $\operatorname{\mathsf{Mor}}_{N-1}(\mcal{A})$
is an Ab4 category with enough projectives.
According to \cite{BN}, for any complex $X \in \cat{K}(\operatorname{\mathsf{Mor}}_{N-1}(\mcal{A}))$, there
is a quasi-isomorphism $P \to X$ with $P \in \cat{K}(\operatorname{\mathsf{Mor}}^{\mrm{sm}}_{N-1}(\mcal{P}))$.
By Lemma \ref{qtricat01}, we have
$\cat{K}(\operatorname{\mathsf{Mor}}^{\mrm{sm}}_{N-1}(\mcal{P}))/
\cat{K}^{\phi}(\operatorname{\mathsf{Mor}}^{\mrm{sm}}_{N-1}(\mcal{P}))
\simeq
\cat{D}(\operatorname{\mathsf{Mor}}_{N-1}(\mcal{A}))$.
\par\noindent
(2)
According to \cite{IKM2} Theorem 2.23, for any complex $X \in \cat{K}_{N}(\mcal{A})$, there
is a quasi-isomorphism $P \to X$ with $P \in \cat{K}_{N}(\mcal{P})$.
By Lemma \ref{qtricat01}, we have
$\cat{K}_{N}(\mcal{P})/\cat{K}^{\phi}_{N}(\mcal{P})
\simeq
\cat{D}_{N}(\mcal{A})$.
\end{proof}
\begin{cor}\label{DNAb}
Let $\mcal{A}$ be an abelian category with enough projectives, and $\mcal{P}$ the category of projective objects.
Then we have triangle equivalences
\[\begin{aligned}
\cat{K}^{-}(\operatorname{\mathsf{Mor}}^{\mrm{sm}}_{N-1}(\mcal{P})) & \simeq \cat{K}^{-}_{N}(\mcal{P}),
\cat{K}^{\mrm{b}}(\operatorname{\mathsf{Mor}}^{\mrm{sm}}_{N-1}(\mcal{P})) & \simeq \cat{K}^{\mrm{b}}_{N}(\mcal{P}), \\
\cat{D}^{-}(\operatorname{\mathsf{Mor}}_{N-1}(\mcal{A})) & \simeq \cat{D}^{-}_{N}(\mcal{A}) .
\end{aligned}\]
Moreover if $\mcal{A}$ is an Ab4 category, then
Then we have triangle equivalences:
\[
\cat{D}(\operatorname{\mathsf{Mor}}_{N-1}(\mcal{A})) \simeq \cat{D}_{N}(\mcal{A}).
\]
\end{cor}
\begin{proof}
According to \cite{IKM2} Lemma 4.8, we have a triangle equivalence \\
$\cat{K}^{-}(\operatorname{\mathsf{Mor}}^{\mrm{sm}}_{N-1}(\mcal{P})) \simeq \cat{D}^{-}(\operatorname{\mathsf{Mor}}_{N-1}(\mcal{A}))$.
By \cite{IKM2} Theorem 2.18, we have a triangle equivalence
$\cat{K}^{-}_{N}(\mcal{P})\simeq \cat{D}^{-}_{N}(\mcal{A})$.
By Theorem \ref{KNhtp}, we have the statement.
\end{proof}
\section{Appendix}\label{appendix}
In this section, we give results concerning Frobenius categories.
Let $\mcal{C}$ be an exact category with a collection $\mcal{E}$ of exact sequences in the sense of Quillen \cite{Qu}.
An exact sequence $0 \to X \xrightarrow{f} Y \xrightarrow{g} Z \to 0$ in $\mcal{E}$ is called a conflation, and
$f$ and $g$ are called an inflation and a deflation, respectively.
An additive functor $F: \mcal{C} \to \mcal{C}'$ is called exact if it sends conflations in $\mcal{C}$
to conflations in $\mcal{C}'$.
An exact category $\mcal{C}$ is called a Frobenius category provided that it has enough projectives
and enough injectives, and that any object of $\mcal{C}$ is projective if and only if it is injective.
In this case, the stable category $\underline{C}$ of $\mcal{C}$ by projective objects is a triangulated category
(see \cite{H1}).
\begin{rem}\label{extr00}
For a Frobenius category $\mcal{C}$, we treat the only case that
for any object $X$ of $\mcal{C}$ we can choose conflations
\[\begin{aligned}
0 \rightarrow X \xrightarrow{u_X} I_{\mcal{C}}(X) \xrightarrow{v_X} \Sigma_{\mcal{C}} X \rightarrow 0 \\
0 \rightarrow \Sigma^{-1}_{\mcal{C}}X \xrightarrow{u_{\Sigma^{-1}X}} P_{\mcal{C}}(X) \xrightarrow{v_{\Sigma^{-1}X}} X \rightarrow 0
\end{aligned}\]
where $I_{\mcal{C}}(X)$ and $P_{\mcal{C}}(X)$ are projective-injective objects in $\mcal{C}$.
For a morphism $f:X \to Y$ in $\mcal{C}$, we have a commutative diagram
\[\xymatrix{
0 \ar[r] & X \ar[d]^{f}\ar[r]^{u_X} & I_{\mcal{C}}(X) \ar[d]^{I_f}\ar[r]^{v_X} & \Sigma_{\mcal{C}} X \ar[d]^{\Sigma_f}\ar[r] & 0 \\
0 \ar[r] & Y \ar[r]^{u_X} & I_{\mcal{C}}(Y) \ar[r]^{v_Y} & \Sigma_{\mcal{C}} Y \ar[r] & 0 \\
}\]
It is easy to see that $\Sigma_f$ is uniquely determined in the stable category $\underline{\mcal{C}}$.
Therefore $\underline{C}$ has a suspension functor $\Sigma_{\underline{\mcal{C}}}: \underline{\mcal{C}}
\to \underline{\mcal{C}}$.
\end{rem}
x
\begin{prop}[\cite{Ke1} A.2 Proposition]\label{excatemb}
If $(\mcal{C}, \mcal{E})$ is a small exact category, there is
an equivalence $G : \mcal{C} \to \mcal{M}$ onto a full subcategory $M$ of an abelian category $\mcal{A}$ such
that $\mcal{M}$ is closed under extensions and that $\mcal{E}$ is formed by the collection of sequences
$0 \to X \xrightarrow{f} Y \xrightarrow{g} Z \to 0$ inducing exact sequences in $\mcal{A}$:
\[
0 \to G(X) \xrightarrow{G(f)} G(Y) \xrightarrow{G(g)} G(Z) \to 0
\]
\end{prop}
\begin{prop}\label{extr01}
Let $\mcal{C}, \mcal{C}'$ be Frobenius categories, $F: \mcal{C} \to \mcal{C}'$ an exact functor.
If $F$ sends projective objects in $\mcal{C}$ to projective objents in $\mcal{T}'$, then
it induces the triangle functor $\underline{F} : \underline{\mcal{C}} \to \underline{\mcal{C}}'$.
\end{prop}
\begin{proof}
For $X \in \mcal{C}$, since $F(I_{\mcal{C}}(X)), I_{\mcal{C}'}(F(X))$ are projective-injective
objects in $\mcal{\underline{C}}'$, we have a commutative diagram
\[\xymatrix{
0 \ar[r] & F(X) \ar@{=}[d]\ar[r]^{F(u_X)} & F(I_{\mcal{C}}(X)) \ar[d]^{\theta_X}\ar[r]^{F(v_{X})}
& F(\Sigma_{\mcal{C}} X) \ar[d]^{\eta_X}\ar[r] & 0 \\
0 \ar[r] & F(X) \ar@{=}[d]\ar[r]^{u_{F(X)}} & I_{\mcal{C}'}(F(X)) \ar[d]^{\theta'_X}\ar[r]^{v_{F(X)}} &
\Sigma_{\mcal{C}'}F(X)\ar[d]^{\eta'_X}\ar[r] & 0 \\
0 \ar[r] & F(X) \ar[r]^{F(u_X)} & F(I_{\mcal{C}}(X)) \ar[r]^{F(v_{X})}
& F(\Sigma_{\underline{\mcal{C}}} X) \ar[r] & 0 \\
}\]
There are $\gamma_X: F(\Sigma_{\mcal{C}} X) \to F(I_{\mcal{C}}(X))$
and $\gamma'_X: \Sigma_{\mcal{C}'}F(X) \to I_{\mcal{C}'}(F(X))$ such that
$F(v_X)\gamma_X=1_{F(\Sigma_{\underline{\mcal{C}}} X)}-\eta'_X\eta_X$ and
$v_{F(X)}\gamma'_X=1_{\Sigma_{\mcal{C}'}F(X)}-\eta_X\eta'_X$ in ${\mcal{C}}'$.
Then $\underline{\eta}_X$ is an isomorphism in $\underline{\mcal{C}}'$.
For a morphism $f:X \to Y$ in $\mcal{C}$,
we have the following diagram by the diagrams of the above and Definition \ref{extr00}:
\[
\xymatrix@!0{
& 0\ \ar@{->}[rr]
& & F(X) \ \ar@{->}[rr]^<{F(u_X)}\ar@{=}'[d][dd]
& & F(I(X))\ \ar@{->}[rr]^<{F(v_X)}\ar@{->}'[d][dd]^{\theta_X}
& & F(\Sigma X)\ \ar@{->}[rr]\ar@{->}'[d][dd]^{\eta_X}
& & 0
\\
0\ \ar@{->}[rr]
& & F(Y) \ \ar@{<-}[ur]^(.25){F(f)}\ar@{->}[rr]\ar@{=}[dd]
& & F(I(Y))\ \ar@{->}[rr]\ar@{<-}[ur]^(.25){F(I_f)}\ar@{->}[dd]^<<{\theta_Y}
& & F(\Sigma Y)\ \ar@{->}[rr]\ar@{<-}[ur]^(.25){F(\Sigma_f)}\ar@{->}[dd]^<<{\eta_Y}
& & 0\
\\
& 0\ \ar@{->}'[r][rr]
& & F(X)\ \ar@{->}'[r][rr]
& & I(F(X))\ \ar@{->}'[r][rr]
& & \Sigma F(X)\ \ar@{->}[rr]
& & 0\
\\
0\ \ar@{->}[rr]
& & F(Y)\ \ar@{->}[rr]_>>{u_{F(Y)}}\ar@{<-}[ur]_(.8){F(f)}
& & I(F(Y))\ \ar@{->}[rr]_>{v_{F(Y)}}\ar@{<-}[ur]_(.8){I_{F(f)}}
& & \Sigma F(Y)\ \ar@{->}[rr]\ar@{<-}[ur]_(.8){\Sigma_{F(f)}}
& & 0\ \ar@{}[ur]}
\]
where the diagrams on the top, the bottom, the front and the back surfaces are commutative.
Since
$
(\theta_YF(I_f) -I_{F(f)}\theta_X)F(u_X) =u_{F(Y)}F(f) -I_{F(f)}u_{F(X)} =0
$
there is a morphism $\delta:F(\Sigma X) \to F(I(Y))$ such that
$\theta_YF(I_f) -I_{F(f)}\theta_X=\delta F(v_X)$.
Then we have $\eta_YF(\Sigma_f)-\Sigma_{F(f)}\eta_X=v_{F(Y)}\delta$, and then
$\underline{\eta}_Y\underline{F}(\Sigma(f))=\Sigma(\underline{F}(f)) \underline{\eta}_X$ in $\mcal{C}'$.
Therefore we have a functorial isomorphism
$\underline{\eta}: \underline{F}\circ\Sigma_{\underline{\mcal{C}}} \stackrel{\sim}{\rightarrow}
\Sigma_{\underline{\mcal{C}}'}\circ\underline{F}$.
For a triangle $X \xrightarrow{\underline{f}} Y \xrightarrow{\underline{g}} Z \xrightarrow{\underline{h}} \Sigma X$
in $\underline{\mcal{C}}$,
we may have a morphism between conflations:
\[\xymatrix{
0 \ar[r] & X \ar[d]^{f}\ar[r]^{u_X} & I(X) \ar[d]^{\psi_f}\ar[r]^{v_X} & \Sigma X \ar@{=}[d]\ar[r] & 0 \\
0 \ar[r] & Y \ar[r]^{g} & Z \ar[r]^{h} & \Sigma X \ar[r] & 0
}\]
There are morphisms $\psi_{F(f)}:I_{\mcal{C}'}(F(X)) \rightarrow Z'$ and
$\xymatrix{z:F(Z) \ar@{-->}[r] & Z'}$ such that we have a commutative diagram
\[
\xymatrix@!0{
& 0\ \ar@{->}[rr]
& & F(X) \ \ar@{->}[rr]^<{F(u_X)}\ar@{=}'[d][dd]
& & F(I(X))\ \ar@{->}[rr]^<{F(v_X)}\ar@{->}'[d][dd]^{\theta_X}
& & F(\Sigma X)\ \ar@{->}[rr]\ar@{->}'[d][dd]^{\eta_X}
& & 0
\\
0\ \ar@{->}[rr]
& & F(Y) \ \ar@{<-}[ur]^(.25){F(f)}\ar@{->}[rr]\ar@{=}[dd]
& & F(Z)\ \ar@{->}[rr]\ar@{<-}[ur]^(.25){F(\psi_{f})}\ar@{-->}[dd]^<<{z}
& & F(\Sigma X)\ \ar@{->}[rr]\ar@{=}[ur]\ar@{->}[dd]^<<{\eta_X}
& & 0\
\\
& 0\ \ar@{->}'[r][rr]
& & F(X)\ \ar@{->}'[r][rr]
& & I(F(X))\ \ar@{->}'[r][rr]
& & \Sigma F(X)\ \ar@{->}[rr]
& & 0\
\\
0\ \ar@{->}[rr]
& & F(Y)\ \ar@{->}[rr]\ar@{<-}[ur]_(.8){F(f)}
& & Z' \ \ar@{->}[rr]\ar@{<-}[ur]_(.8){\psi_{F(f)}}
& & \Sigma F(X)\ \ar@{->}[rr]\ar@{=}[ur]
& & 0\ \ar@{}[ur]}
\]
Similarly, there is a morphism $z':Z' \to F(Z)$ such that
we have the above commutative diagram of which all vertical arrows are reversed.
Put $\zeta=z'z+F(\psi_{f})\gamma_XF(h)$, we have a commutative diagram
\[\xymatrix{
0 \ar[r] & F(Y) \ar@{=}[d] \ar[r]^{F(g)} & F(Z) \ar[d]^{\zeta} \ar[r]^{F(h)} & F(\Sigma X) \ar@{=}[d]\ar[r] & 0 \\
0 \ar[r] & F(Y) \ar[r]^{F(g)} & F(Z) \ar[r]^{F(h)} & F(\Sigma X) \ar[r] & 0
}\]
Then $\underline{z}'\underline{z}$ is an isomorphism in $\underline{\mcal{C}}'$.
Similarly $\underline{z}\underline{z}'$ is also an isomorphism in $\underline{\mcal{C}}'$, and
then $\underline{z}$ is an isomorphism in $\underline{\mcal{C}}'$.
Therefore we have a commutative diagram in $\underline{\mcal{C}}'$:
\[\xymatrix{
\underline{F}(X) \ar@{=}[d]\ar[r]^{\underline{F}(\underline{f})} & \underline{F}(Y) \ar@{=}[d]\ar[r]^{\underline{F}(\underline{g})}
& F(Z) \ar[d]^{\underline{z}}\ar[r]^{\underline{F}(\underline{h})} & \underline{F}(\Sigma X) \ar[r]^{\underline{\eta}_X} & \Sigma \underline{F}(X)\ar@{=}[d]\\
\underline{F}(X) \ar[r]^{\underline{F}(\underline{f})} & \underline{F}(Y) \ar[r]^{\underline{g}'} & Z' \ar[rr]^{\underline{h}'} & & \Sigma \underline{F}(X)
}\]
where all vertical arrows are isomorphisms.
Hence $F$ induces a triangle functor $\underline{F}:\underline{\mcal{C}} \to \underline{\mcal{C}}'$.
\end{proof}
\begin{prop}\label{triadjoint02}
Let $\mcal{C}, \mcal{C}'$ be Frobenius categories, $F: \mcal{C} \to \mcal{C}'$, $G: \mcal{C}' \to \mcal{C}$
exact functors such that $F$ is a right adjoint of $G$.
Then $F$ and $G$ induce triangle functors
$\underline{F}: \underline{\mcal{C}} \to \underline{\mcal{C}}'$,
$\underline{G}: \underline{\mcal{C}}' \to \underline{\mcal{C}}$ such that
$\underline{F}$ is a right adjoint of $\underline{G}$.
\end{prop}
\begin{proof}
Let $0 \to A \to B \to C \to 0$ be a conflation in $\mcal{C}$, then
$0 \to F(A) \to F(B) \to F(C) \to 0$ is a conflation in $\mcal{C}'$.
For a projective-injective object $P$ of $\mcal{C}'$,
we have an isomorphism between exact sequences:
\[\xymatrix{
0 \ar[r] & \operatorname{Hom}_{\mcal{C}}(P, F(A)) \ar[d]^{\wr}\ar[r] & \operatorname{Hom}_{\mcal{C}}(P, F(B)) \ar[d]^{\wr}\ar[r]
& \operatorname{Hom}_{\mcal{C}}(P, F(C)) \ar[d]^{\wr}\ar[r] & 0 \\
0 \ar[r] & \operatorname{Hom}_{\mcal{C}}(G(P), A) \ar[r] & \operatorname{Hom}_{\mcal{C}}(G(P), B) \ar[r]
& \operatorname{Hom}_{\mcal{C}}(G(P), C)
}\]
Then $G(P)$ is a projective-injective object in $\mcal{C}$.
Similarly, if $Q$ is a projective-injective object of $\mcal{C}$, then
$F(Q)$ is also a projective-injective object of $\mcal{C}'$.
By Proposition \ref{extr01}, $F$ and $G$ induce the triangle functors
$\underline{F}: \underline{\mcal{C}} \to \underline{\mcal{C}}'$,
$\underline{G}: \underline{\mcal{C}}' \to \underline{\mcal{C}}$.
Given $X \in \mcal{C}'$, consider a conflation
\[
0 \rightarrow X \rightarrow I(X) \rightarrow \Sigma X \rightarrow 0
\]
For any $Y \in \mcal{C}$, we have a isomorphism between exact sequences
\[\xymatrix{
\operatorname{Hom}_{\mcal{C}}(I(X),F(Y)) \ar[d]^{\wr} \ar[r]
& \operatorname{Hom}_{\mcal{C}}(X,F(Y)) \ar[d]^{\wr} \ar[r]
& \operatorname{Hom}_{\underline{\mcal{C}}}(X,F(Y)) \ar[d] \ar[r]
& 0 \\
\operatorname{Hom}_{\mcal{C}}(G(I(X)),Y) \ar[r]
& \operatorname{Hom}_{\mcal{C}}(G(X),Y) \ar[r]
& \operatorname{Hom}_{\underline{\mcal{C}}}(G(X),Y) \ar[r]
& 0 \\
}\]
Then we have an isomorphism
$\operatorname{Hom}_{\underline{\mcal{C}}}(X, F(-)) \simeq \operatorname{Hom}_{\underline{\mcal{C}}}(G(X),-)$.
Similarly we have an isomorphism
$\operatorname{Hom}_{\underline{\mcal{C}}}(-, F(Y)) \simeq \operatorname{Hom}_{\underline{\mcal{C}}}(G(-),Y)$.
\end{proof}
| {
"redpajama_set_name": "RedPajamaArXiv"
} | 1,543 |
Q: Print out on paper is different from print preview in crystal report I have designed a crystal report and used A4 paper size by default. When I run a preview of my crystal report then all the fields' contents show on the correct location on the page but when I make a printout, then all the fields' contents do not print at their proper locations. The printout does not match with my crystal report preview. I am using Crystal Report with Visual Studio 2010. Does anybody have a solution?
A: try these steps :
1, in crystalreportviewer, if you click print, go to flag LAYOUT, and set CENTER TO PAGE
2, in Crystal report application, go to PAGE SETUP (right click on page) and set all the margins to 5
let me know, if it help
nice day
| {
"redpajama_set_name": "RedPajamaStackExchange"
} | 3,948 |
\section{Introduction}
The anomalous magnetic moment of the muon $a_\mu=(g_\mu-2)/2$ is one of the most promising
quantities for
the search of physics beyond the Standard Model of particle physics, since it can be measured
and calculated to a high precision.
The current most precise experimental determination of the anomalous magnetic moment of the muon
has been obtained using polarised muons in a storage ring at
Brookhaven National Laboratory \cite{Bennett:2006fi}
\begin{equation}
a^\textrm{\scriptsize exp}_\mu = 11659209.1 ( 5.4 )( 3.3 ) \times 10^{-10}\,.
\label{eq:amuexp}
\end{equation}
Two upcoming experiments at Fermilab \cite{Venanzoni:2014ixa} and JPARC \cite{Otani:2015jra} plan
to further reduce the uncertainty of the experimental result by a factor of four.
\par
When estimating the Standard Model prediction of $a_\mu$, contributions from the different
fundamental
interactions need to be calculated and a summary of the current most precise results is given in
table \ref{tab:amucontr}. The biggest contribution to $a_\mu$ comes from the electromagnetic
interaction (em) whereas the error is dominated by the QCD contributions,
namely the hadronic vacuum polarisation (HVP) and the hadronic light-by-light scattering (HLBL).
The corresponding diagrams for these processes are shown in figure \ref{fig:hadrdiagrams}.
%
%
%
\par
\begin{table}[h]
\centering
\begin{tabular}{|l|r|r|}
\hline
& $a_\mu\times10^{10}$ & reference\\
\hline\hline
em & $11658471.895(8)$ & \cite{Aoyama:2012wk}\\
weak & $15.36(10)$ & \cite{Gnendiger:2013pva}\\[0.3cm]
HVP & $693.26(2.46)$ & \cite{Keshavarzi:2018mgv}\footnotemark\\
HVP (NLO) & $-9.84(6)$ & \cite{Hagiwara:2011af}\\
HVP (NNLO) & $1.24(1)$ & \cite{Kurz:2014wya} \\
HLBL & $10.5(2.6)$ & \cite{Prades:2009tw} \\
\hline
total & $11659182.4(1)(2.5)^\textrm{\scriptsize HVP}(2.6)^\textrm{\scriptsize HLBL}$ & \\
\hline
\end{tabular}
\caption{Standard Model Contributions to $a_\mu$}
\label{tab:amucontr}
\end{table}
%
\footnotetext[1]{Other recent determinations of the HVP using $R$-ratio data can be found in
\cite{Davier:2019can,Jegerlehner:2017lbd}.}
%
%
%
\par
\begin{figure}[h]
\centering
\includegraphics[scale=1.2]{plots/diagrams/vacuumpolarization}\hspace{1.5cm}
\includegraphics[scale=1.2]{plots/diagrams/g2lightbylight}
\caption{Hadronic contributions to $a_\mu$: The hadronic vacuum polarisation (left) and the
hadronic light-by-light scattering (right).}
\label{fig:hadrdiagrams}
\end{figure}
A comparison of the Standard Model prediction of $a_\mu$ (cf. table \ref{tab:amucontr}) and the
experimental determination reveals a deviation of about $3.5\sigma$, which could potentially hint
to new physics. Further investigation requires both, the experimental as well as
the theoretical uncertainties to be reduced. To match the targeted precision of the upcoming
experiments
($\approx1.5\times 10^{-10}$), this requires know\-ledge of the HVP contribution at the level of
$0.2\%$ accuracy and the HLBL contribution at a level of about $10\%$.
\par
In recent years a lot of effort has been undertaken to calculate the hadronic contributions to the
anomalous magnetic moment of the muon from first principles using Lattice QCD. In the remainder of
this proceedings I will review the
current status of such lattice calculations and discuss the prospects and challenges to match the
precision of the upcoming experiments.
\section{Hadronic Vacuum Polarisation}
Currently the most precise determinations of the HVP contribution to
$a_\mu$ are obtained by using a dispersion relation and experimental data for
the cross section $\sigma(e^+\,e^-\rightarrow
\textnormal{hadrons})$ as an input
\begin{equation}
a_\mu^\textnormal{\scriptsize HVP}}%a_\mu^{hlbl = \left(\frac{\alpha
m_\mu}{3\pi}\right)^2
\int\limits_{m^2_\pi}^\infty \textnormal{d} s\, \frac{R(s) K(s)}{s^2}
\hspace{1.5cm}\textnormal{with}\qquad
R(s)=\frac{\sigma(e^+\,e^-\rightarrow
\textnormal{hadrons},s)}{\sigma(e^+\,e^-\rightarrow
\mu^+\mu^-,s)}
\end{equation}
with an analytically known kernel function $K(s)$. Recent determinations of $a_\mu^\textnormal{\scriptsize HVP}}%a_\mu^{hlbl$ from this
method can be found in
\cite{Keshavarzi:2018mgv,Davier:2019can,Jegerlehner:2017lbd} and have a precision of about $0.5\%$.
However, this approach relies on experimental input and an \textit{ab initio} calculation can be
done using lattice QCD.
\subsection{The HVP from the Lattice}
In the following we will discuss how the HVP contribution to $a_\mu$ can be calculated using
lattice QCD. The hadronic vacuum polarisation tensor
\begin{equation}
\Pi_{\mu\nu}(Q) \equiv
\int\!\textnormal{d}^4 x\,\,
e^{i\,Q\cdot
x}\,\left<j^\gamma_\mu(x)\,\,j^\gamma_\nu(0)\right>
= (Q_\mu Q_\nu -
\delta_{\mu\nu} Q^2)\,\Pi(Q^2)\,,
\end{equation}
is given by the four-dimensional Fourier transform of the
correlation of two electromagnetic currents
\begin{equation}
j^\gamma_\mu = \frac{2}{3}
\overline{u}\gamma_\mu u - \frac{1}{3}
\overline{d}\gamma_\mu d - \frac{1}{3}
\overline{s}\gamma_\mu s + \frac{2}{3}
\overline{c}\gamma_\mu c\,,
\end{equation}
which receive contributions from the different quark flavours multiplied by the respective charge
factors. The contribution to the anomalous magnetic moment can then be determined from the vacuum
polarisation function $\Pi(Q^2)$ as \cite{Blum:2002ii}
\begin{equation}
a_\mu^\textnormal{\scriptsize HVP}}%a_\mu^{hlbl = \left(\frac{\alpha}{\pi}\right)^2
\int\limits_0^\infty \textnormal{d} Q^2\, K(Q^2)\,\hat{\Pi}(Q^2)
\hspace{0.9cm}\textnormal{with}\hspace{0.4cm}\hat{\Pi}(Q^2) =
4\,\pi^2\left[\Pi(Q^2) - \Pi(0)\right]\,.
\label{eq:amusubHVP}
\end{equation}
with an analytically known electromagnetic kernel function $K(Q^2)$. In the last few years it has
become common to calculate the subtracted vacuum polarisation $\hat{\Pi}(Q^2)$ required in
equation~(\ref{eq:amusubHVP}) directly from the vector-vector two-point function $C(t)$ projected
to zero spatial momentum \cite{Bernecker:2011gh,Feng:2013xsa}
\begin{equation}
\hat{\Pi}(Q^2) =
4\pi^2\!\!\!\int\limits_0^\infty\!\!\textnormal{d}
t\,C(t)\!\left[\!\frac{\cos(Qt)\!-\!1}{Q^2}+\frac{1}{2}t^2\right]
\hspace{1cm}\textnormal{with}\qquad
C(t) =
\frac{1}{3}\sum\limits_{k=0}^2 \sum\limits_{\vec{x}}
\left<j^\gamma_k\!(\vec{x},t)j^\gamma_k\!(0)\right>\,.
\end{equation}
Changing the order of integration one can obtain $a_\mu$ directly from $C(t)$ by integrating over
the Euclidean time using the appropriate kernel function $f(t)$
\begin{equation}
\hspace{0.6cm}a_\mu^\textnormal{\scriptsize HVP}}%a_\mu^{hlbl\! = \!\!\!\int\limits_0^\infty\!\!\textnormal{d} t\,f(t) C(t)
\,.
\end{equation}
In isospin symmetric QCD\footnote{We will discuss isospin breaking corrections in section
\ref{subsec:IB}} the vector two-point function can be written in the following flavour decomposition
\begin{equation}
C(t) = \frac{5}{9} C^\ell(t) + \frac{1}{9} C^s(t) + \frac{4}{9} C^c(t) +
C^\textnormal{\scriptsize disc}(t)\,,
\end{equation}
with connected contributions for the light quark $C^\ell$, strange quark $C^s$ and charm $C^c$
quark as well as
a quark-disconnected contribution $C^\textnormal{\scriptsize disc}$. The diagrams corresponding to
the quark-connected and
quark-disconnected Wick contractions are shown in figure \ref{fig:Wick}.
\par
\begin{figure}[h]
\centering
\includegraphics[width=0.7\textwidth]{plots/diagrams/WickcontHVP}
\vspace{-0.3cm}
\caption{Quark-connected (left) and quark-disconnected (right) Wick contraction for the HVP.}
\label{fig:Wick}
\end{figure}
In the following, I will discuss the contributions to $a_\mu^\textnormal{\scriptsize HVP}}%a_\mu^{hlbl$ flavour by flavour, starting with
the
light-quark connected contribution in section~\ref{subsec:light}, followed by the strange- and
charm-quark connected
contributions in section~\ref{subsec:strange}, quark-disconnected contributions in
section~\ref{subsec:disc} and finally isospin breaking corrections in section~\ref{subsec:IB}. A
summary and comparison of the various available results for $a_\mu^\textnormal{\scriptsize HVP}}%a_\mu^{hlbl$ as
well as an outlook for this quantity is given in section \ref{subsec:HVPsummary}.
\subsection{Light-Quark Contribution}
\label{subsec:light}
In this section, I will show results for the light-quark connected contribution to $a_\mu^\textnormal{\scriptsize HVP}}%a_\mu^{hlbl$ and
discuss some of the main challenges for achieving a sub-percent precision calculation, namely the
long distance
noise-to-signal problem, finite volume corrections and accurate scale setting.
\subsubsection{Long-Distance tail of the Vector Correlator}
Figure \ref{fig:lightcorrelator} shows examples for the light-quark vector-vector two-point
function $C(t)$
from the \linebreak HPQCD/Fermilab/MILC \cite{Davies:2019efs} collaboration (left) and the
integration kernel $f(t)\cdot C(t)$
for $a_\mu^\textnormal{\scriptsize HVP}}%a_\mu^{hlbl$ from Mainz \cite{Gerardin:2019rua} (right). Both data sets have been obtained at
physical pion mass.
\par
\begin{figure}[h]
\centering
\includegraphics[width=0.39\textwidth]{plots/papers/fig2_190204223.png}\hspace{1.3cm}
\includegraphics[width=0.49\textwidth]{plots/papers/fig1_190403120.png}
\caption{The vector correlator at physical pion mass: The plot on the left is from
\cite{Davies:2019efs} and shows the two-point function $C(t)$ for two ensembles with the
same parameters and different statistics. The plot on the right is from \cite{Gerardin:2019rua} and
shows the integration kernel $f(t)\cdot C(t)$ for $a_\mu^\textnormal{\scriptsize HVP}}%a_\mu^{hlbl$ with the light-quark contribution shown
in black.}
\label{fig:lightcorrelator}
\end{figure}
As one can see in both plots, the signal in the vector two-point function deteriorates for large
Euclidean times $t$. The statistical error on the raw data can be improved by using noise reduction
techniques such as all-mode-averaging \cite{Blum:2012uh, Bali:2009hu} or low-mode-averaging,
which has been successfully used in \cite{Blum:2018mom, Aubin:2019usy}.
However, the statistical uncertainty for $a_\mu^\textnormal{\scriptsize HVP}}%a_\mu^{hlbl$ after integrating the raw
data of $C(t)$ over $t$ will
still be dominated by the noise from large times. A possible approach to reduce this uncertainty,
is
to replace the correlator by a (multi-) exponential fit for $t>t_c$ (see, e.g.,
\cite{DellaMorte:2017dyu, Davies:2019efs}). A more systematic way to treat the long distance tail
of the correlator is the bounding
method \cite{Borsanyi:2016lpl, Blum:2018mom}, which will be discussed in the following.
\subsubsection{Bounding Method}
The vector correlator can be written using the spectral representation as the sum of exponentials
with positive prefactors
\begin{equation}
C(t) = \sum_n \frac{A^2_n}{2E_n} e^{-E_n t}\hspace{1.2cm}\textnormal{with}\qquad A_n^2>0\,.
\label{eq:specrep}
\end{equation}
The idea of the bounding method is to bound the two-point function $C(t)$ for Euclidean times
larger
than some value $t_c$ from
above and below
\cite{Borsanyi:2016lpl,Blum:2018mom}
\begin{equation}
0< C(t_c)\, e^{-E_{t_c} (t-t_c)} \leq C(t) \leq C(t_c)\, e^{-E_0(t-t_c)}
\hspace{1cm}\textrm{for}\qquad t\geq t_c
\,.
\end{equation}
As a trivial lower bound one can use zero, since the correlator (cf. equation (\ref{eq:specrep}))
is
strictly positive. A more stringent lower bound for $t\geq t_c$ can be obtained by
using the effective mass $E_{t_c}$ at the given $t_c$. Since energy states with higher energies
decay faster, the true correlator $C(t)$ is guaranteed to fall slower than
$C(t_c)\, e^{-E_{t_c} (t-t_c)}$ for $t\geq t_c$. On the other hand, the true correlator is
guaranteed to decay faster than the actual ground state energy $E_0$, and thus $C(t)\leq C(t_c)\,
e^{-E_0(t-t_c)}$ is an upper bound for the correlator. In the vector channel, the ground state
energy $E_0$ is given by the finite volume energy of two pions with one unit of momentum. When
calculating $a_\mu^\textnormal{\scriptsize HVP}}%a_\mu^{hlbl$ one now replaces the correlator for $t\geq t_c$ by the upper and the lower
bound and varies $t_c$. An example of $a_\mu^\textnormal{\scriptsize HVP}}%a_\mu^{hlbl$ plotted against $t_c$ is shown on the left-hand
side of figure \ref{fig:bounding}. As one can see, once $t_c$ is large enough, the result
using the upper and lower bound overlap, and $a_\mu^\textnormal{\scriptsize HVP}}%a_\mu^{hlbl$ can be extracted from this region.
\par
\begin{figure}[h]
\centering
\includegraphics[width=0.98\textwidth]{plots/papers/fig3_190403120.png}
\caption{$a_\mu^\textnormal{\scriptsize HVP}}%a_\mu^{hlbl$ from the bounding method (left) and the improved bounding
method (right) plotted against $t_c$. The plots are taken from \cite{Gerardin:2019rua} and use a
pion mass of $M_\pi\approx200$~MeV.}
\label{fig:bounding}
\end{figure}
\par
The bounding method can be further improved \cite{Gerardin:2019rua,Bruno:2019nzm} if a dedicated
spectroscopy study for the vector channel is available. This requires to calculate two-point
functions using various operators that have overlap with two-pion states. With the
Generalised Eigenvalue Problem (GEVP)~\cite{Blossier:2009kd}, one can extract the energies
$E_n$ and overlap factors $A_n$ for the lowest $N$ states of the spectrum.
Once these energies and overlap factors
have been determined, the long-distance tail of the vector correlator can be reconstructed. An
example at physical pion mass is shown in figure \ref{fig:corrreconstruction}. As one can see, the
more states are used in the reconstruction, the closer the reconstructed data are to the original
correlator (shown in black in figure \ref{fig:corrreconstruction}).
%
%
%
\par
\begin{figure}[h]
\centering
\includegraphics[width=0.48\textwidth]{plots/papers/fig3_191011745.png}
\caption{The integration kernel $f(t)\cdot C(t)$ for $a_\mu^\textnormal{\scriptsize HVP}}%a_\mu^{hlbl$ plotted against $t$. The black
data show the original correlator, the coloured points are reconstructions of the vector two-point
function using one (blue) and up to four (red) states from the GEVP. The figure is taken from
\cite{Bruno:2019nzm}.}
\label{fig:corrreconstruction}
\end{figure}
%
%
%
These reconstructed states can now be used to improve the bounding method as follows. Subtract the
lowest $N$ states from the correlation function and bound the subtracted correlator $\tilde{C}(t)$
\begin{equation}
\tilde{C}(t) = C(t) - \sum_{n=0}^{N-1} \frac{A^2_n}{2E_n} e^{-E_n t}
\hspace{1.9cm}
0\leq \tilde{C}(t_c)\, e^{-E_{t_c} (t-t_c)} \leq \tilde{C}(t) \leq \tilde{C}(t_c)\,
e^{-E_N(t-t_c)}\,.
\end{equation}
Here, the upper bound is obtained using the energy of the $(N+1)$th state, i.e.\ the lightest
state that was not subtracted from the correlator. The right-hand side of figure \ref{fig:bounding}
shows the improved bounding method from \cite{Gerardin:2019rua} with the two lightest states
subtracted. As one
can see in comparison with the unimproved bounding method (left on figure \ref{fig:bounding}) the
upper and the lower bound now overlap at much smaller $t_c$ and $a_\mu^\textnormal{\scriptsize HVP}}%a_\mu^{hlbl$ can be extracted with a
smaller error.
\subsubsection{Finite Volume Effects}
Finite volume (FV) effects in the vector correlator are dominated by the two-pion state and can
thus be expected to be important for large time separations $t$.
Various studies (see, e.g.~\cite{Shintani:2019wai,Aubin:2019usy,Gerardin:2019rua,LehnerLat19}) of
FV effects for lattice calculations of $a_\mu^\textnormal{\scriptsize HVP}}%a_\mu^{hlbl$ suggest that these are of the order of
$\Delta^{\textnormal{\scriptsize FV}}a_\mu^\textnormal{\scriptsize HVP}}%a_\mu^{hlbl \approx 20-30\times10^{-10}$ for typical sizes
of $5-6$~fm of state-of-the-art lattice ensembles used at the
physical point. Thus, it is crucial to carefully study and correct for FV effects when aiming at
percent level precision for the HVP.
\par
A straightforward way to study finite volume effects is using ensembles that
differ only in the volume. Figure \ref{fig:FVShintanietal} shows results from the PACS
collaboration
\cite{Shintani:2019wai} for the $a_\mu^\textnormal{\scriptsize HVP}}%a_\mu^{hlbl$ integrand calculated using two different volumes of
$5.4$~fm and $10.8$~fm at the physical pion mass. One can clearly see, a significant difference
between the data on both ensembles, in particular for larger values of $t$. The authors of
\cite{Shintani:2019wai} found finite volume effects for $a_\mu^\textnormal{\scriptsize HVP}}%a_\mu^{hlbl$ to be about $1.7$ times larger
than what was expected from next-to-leading order (NLO) Chiral Perturbation Theory ($\chi$PT).
\par
\begin{figure}[h]
\centering
\includegraphics[width=0.75\textwidth]{plots/papers/fig3_190200885.png}
\caption{The $a_\mu^\textnormal{\scriptsize HVP}}%a_\mu^{hlbl$ integrand calculated using two different volumes of $5.4$~fm (green
triangles) and $10.8$~fm (blue circles) at physical pion mass. The plot on the right shows a
zoomed-in version of the data on the left. The plot is taken from \cite{Shintani:2019wai}.}
\label{fig:FVShintanietal}
\end{figure}
Finite volume corrections to $a_\mu^\textnormal{\scriptsize HVP}}%a_\mu^{hlbl$ have been determined at next-to-next-to-leading order
\linebreak (NNLO)
in $\chi$PT recently \cite{Bijnens:2017esv,Aubin:2019usy} and the authors of \cite{Aubin:2019usy}
find that additional FV effects from NNLO are about $0.4-0.45$ times the NLO FV corrections.
\par
A systematic way to study and correct for FV effects for the HVP is by writing the long
distance contribution of the vector two-point function in terms of the timelike pion form factor.
In \cite{Meyer:2011um, Francis:2013qna} it was suggested to use the Gounaris-Sakurai
parameterization of the timelike pion form factor to calculate the infinite volume long-distance
vector two-point function and the finite volume equivalent using the Gounaris-Sakurai
parameterization combined with the Lellouch-L\"uscher formalism
\cite{Luscher:1991cf,Lellouch:2000pv}. This approach has been used by Mainz
\cite{Gerardin:2019rua}, ETMC \cite{Giusti:2018mdh} and RBC/UKQCD \cite{LehnerLat19} to study
FV volume corrections and results after correcting for FV effects are found to be consistent when
comparing
ensembles that only differ by volume \cite{Gerardin:2019rua,Giusti:2018mdh}. The plot on the
left-hand side of figure \ref{fig:FVtimelikepff} is from Mainz \cite{Gerardin:2019rua} and shows the
$a_\mu^\textnormal{\scriptsize HVP}}%a_\mu^{hlbl$ integrand for two different
ensembles at the same pion mass $M_\pi=280$~MeV with different volumes. The smaller volume is shown
without (black circles) and with (blue squares) FV correction using the timelike pion form factor.
After the data on the smaller ensemble has been corrected the results for the two different volumes
agree with each other. The plot on the right-hand side of figure \ref{fig:FVtimelikepff} is from
ETMC \cite{Giusti:2018mdh} and shows $a_\mu^\textnormal{\scriptsize HVP}}%a_\mu^{hlbl$ plotted against the pion mass for various ensembles
without (open symbols) and with (closed symbols) FV corrections. In addition to using the timelike
pion form factor to estimate FV effects for large times $t$, the
authors of \cite{Giusti:2018mdh} use perturbative QCD for small $t$ to correct for FV effects. At a
pion mass of around $M_\pi=320$~MeV, where several ensembles are available that only differ in
volume, results are found to be in agreement once FV effects have been corrected for.
\par
\begin{figure}[h]
\centering
\hspace{-0.3cm}
\includegraphics[width=0.52\textwidth,trim = 0mm 0mm 0mm 0mm,
clip=true]{plots/papers/fig5_190403120.png}
\hspace{0.7cm}
\includegraphics[width=0.43\textwidth]{plots/papers/fig13_180800887.png}
\caption{Correcting finite volume effects with the timelike pion form factor. The plot on the
left is taken from \cite{Gerardin:2019rua} and shows the $a_\mu^\textnormal{\scriptsize HVP}}%a_\mu^{hlbl$ integrand for two different
ensembles at the same pion mass $M_\pi=280$~MeV with different volumes. The smaller volume is shown
without (black circles) and with FV correction (blue squares). The plot on the right is taken from
\cite{Giusti:2018mdh} and shows $a_\mu^\textnormal{\scriptsize HVP}}%a_\mu^{hlbl$ vs the pion mass for various ensembles without (open
symbols) and with (closed symbols) FV corrections.}
\label{fig:FVtimelikepff}
\end{figure}
\par
In a recent paper \cite{Hansen:2019rbh}, FV corrections to the HVP have been studied using a
Hamiltonian approach, currently quoting corrections at $\mathcal{O}(e^{-m_\pi L})$, but neglecting
effects of $\mathcal{O}(e^{-\sqrt{2}m_\pi L})$ and higher orders.
\subsubsection{Scale Setting}
Although $a_\mu$ is a dimensionless quantity, it depends on the scale of a given lattice, since the
evaluation of the Kernel function for integrating the HVP requires to input the muon mass in
lattice units. Assuming the scale has been set by some quantity $\Lambda$ with statistical error
$\Delta\Lambda$, using error propagation one can show \cite{DellaMorte:2017dyu}, that the relative
error on $\Lambda$ is going to be enhanced by a factor of $\approx 1.8$ for $a_\mu^\textnormal{\scriptsize HVP}}%a_\mu^{hlbl$
\begin{equation}
\Delta a_\mu^\textnormal{\scriptsize HVP}}%a_\mu^{hlbl = \left|\Lambda\,\, \frac{\textrm{d} a_\mu^\textnormal{\scriptsize HVP}}%a_\mu^{hlbl}{\textrm{d}
\Lambda}\right|\cdot\,\frac{\Delta \Lambda}{\Lambda} = \left|M_\mu\frac{\textrm{d}
a_\mu^\textnormal{\scriptsize HVP}}%a_\mu^{hlbl}{\textrm{d}
M_\mu}\right|\cdot\frac{\Delta \Lambda}{\Lambda} \hspace{2cm}\textnormal{with}\qquad M_\mu =
\frac{m_\mu}{\Lambda}\,.
\end{equation}
Thus, achieving $0.2\%$ accuracy on the HVP requires knowledge of the lattice scale to at least
$0.1\%$. A suitable quantity for high-precision scale setting might be the mass of the
$\Omega$-Baryon. However, whether or not scale setting at $0.1\%$ accuracy is possible in the near
future remains to be investigated.
\subsubsection{Comparison of Results for the Light-Quark Contribution}
\label{subsubsec:lightres}
Figure \ref{fig:complight} shows a comparison of results for the light-quark connected contribution
to the
HVP from various collaborations. The two different panels $N_f=2+1$ and $N_f=2+1+1$ denote the
number of dynamical fermions used in the calculations.
\par
\begin{figure}[h]
\centering
\includegraphics[width=1\textwidth]{plots/comparison/light_comp_standalone}
\vspace{-0.9cm}
\caption{Comparison of lattice results for the light-quark contribution to $a_\mu^\textnormal{\scriptsize HVP}}%a_\mu^{hlbl$. The values
are
taken from CLS Mainz 2019 \cite{Gerardin:2019rua}, PACS-CS 2019 \cite{Shintani:2019wai}, RBC/UKQCD
2018 \cite{Blum:2018mom}, BMW 2018 \cite{Borsanyi:2017zdw}, ETMC 2018 \cite{Giusti:2018mdh},
HPQCD/Fermilab/MILC 2019 \cite{Davies:2019efs}, Aubin et al 2019 \cite{Aubin:2019usy}.}
\label{fig:complight}
\end{figure}
\par
The relative errors on the light-quark contributions of the different results are about $1.3\%$ to
$3.3\%$. For most of the collaborations, the error is dominated by statistics (inner error bar on
the points in figure \ref{fig:complight}). As discussed above, the main challenge in terms of
statistical error is the
growing noise-to-signal ratio for large Euclidean time separations. Thus, a reduction of this error
requires good control of the long-distance tail. A very promising approach for reaching sub-percent
precision on the light quark contribution in the future is the improved bounding method discussed
above. \par
As one can see, there is a slight tension between the smallest and the largest results of
about $2\sigma$. This tension has to be object of further investigations, and in particular needs to
be monitored when the collaborations further reduce the uncertainties in the individual
calculations. A possible approach for
determining the source of potential differences between different groups would be the comparison of
more
intermediate results, e.g. time-moments \cite{Chakraborty:2014mwa} of the vector correlator or
$a_\mu$ calculated from a time window \cite{Blum:2018mom}.
\subsection{Strange- and Charm-Quark Contribution}
\label{subsec:strange}
The connected strange- and charm-quark contributions to the HVP are significantly
smaller than the light-quark contribution and suffer far less from noise-to-signal problems in
the long-distance tail or finite volume corrections. For the charm-quark contribution
discretization effects can
be large and lattice calculations should ideally include at least
three different lattice spacings to reliably extrapolate to the continuum.
\par
The results for the connected strange- and charm-quark contribution to $a_\mu^\textnormal{\scriptsize HVP}}%a_\mu^{hlbl$ from various
collaborations are shown in figure
\ref{fig:compsc} and are in good agreement with each other. The errors correspond to errors of
about $\lesssim0.4\%$ and $\lesssim0.3\%$ on the total HVP for the strange and charm, respectively.
Thus, these contributions are already in a good shape when aiming at sub-percent precision for
$a_\mu^\textnormal{\scriptsize HVP}}%a_\mu^{hlbl$ from a lattice calculation.
\par
\begin{figure}[h]
\centering
\hspace{-0.5cm}
\includegraphics[scale=0.73]{plots/comparison/strange_comp_standalone}
\hspace{-0.05\textwidth}
\includegraphics[scale=0.73,trim = 36mm 0mm 42mm 0mm,
clip=true]{plots/comparison/charm_comp_standalone}
\vspace{-0.4cm}
\caption{Comparison of lattice results for the strange-quark (left) and charm-quark (right)
contributions to $a_\mu^\textnormal{\scriptsize HVP}}%a_\mu^{hlbl$. Results are taken from CLS Mainz 2019 \cite{Gerardin:2019rua}, PACS-CS
2019 \cite{Shintani:2019wai}, RBC/UKQCD
2018 \cite{Blum:2018mom}, BMW 2018 \cite{Borsanyi:2017zdw}, ETMC 2017 \cite{Giusti:2017jof},
HPQCD/Fermilab/MILC 2014 \cite{Chakraborty:2014mwa}.}
\label{fig:compsc}
\end{figure}
\par
\subsection{Quark-Disconnected Contribution}
\label{subsec:disc}
Besides the quark-connected contributions discussed above, the HVP receives a contribution from a
quark-disconnected Wick contraction (cf. right-hand side of figure \ref{fig:Wick}). The calculation
of the respective
disconnected quark-loops requires knowledge of the propagator from all lattice points to all other
lattice points (all-to-all propagator), which has to be determined stochastically and is thus
notoriously noisy.
The combined light- and
strange-quark disconnected contribution is $SU(3)$-flavour suppressed, i.e.\ it would vanish if
$m_s=m_\ell$.
In \cite{Francis:2014hoa} it was shown that a substantial reduction in the statistical error can be
achieved by exactly implementing $SU(3)$ suppression in the lattice calculation
\begin{equation}
C^\textnormal{disc}(t) = \frac{1}{9}
\left<(\Delta^\ell(t)-\Delta^s(t))\cdot(\Delta^\ell(0)-\Delta^s(0))\right>\hspace{1cm}\textrm{with}
\hspace{0.5cm} \Delta_\mu^f(t) = \sum_{\vec{x}} \textnormal{Tr}\left[\gamma_\mu S^f(x,x)\right]
\end{equation}
and stochastically estimate the difference $\Delta^\ell(t)-\Delta^s(t)$ rather than the individual
quark loops. Various stochastic estimators for $\Delta^\ell(t)-\Delta^s(t)$ and further
noise reduction techniques have been proposed, e.g.\ low-mode-averaging and sparsened noise sources
\cite{Blum:2015you}, hierarchical probing \cite{Stathopoulos:2013aci, Gerardin:2019rua} or
frequency-splitting estimators \cite{Giusti:2019kff,HarrisLat19}.
\par
\begin{figure}[h]
\centering
\includegraphics[width=0.80\textwidth]{plots/comparison/disconnected_comp_standalone}
\vspace{-0.7cm}
\caption{Comparison of lattice results for the quark-disconnected contribution to $a_\mu^\textnormal{\scriptsize HVP}}%a_\mu^{hlbl$.
Values
are taken from CLS Mainz 2019 \cite{Gerardin:2019rua}, RBC/UKQCD 2018 \cite{Blum:2018mom}, BMW 2018
\cite{Borsanyi:2017zdw}.}
\label{fig:disccomp}
\end{figure}
Figure \ref{fig:disccomp} shows the published results for the quark-disconnected contribution to
$a_\mu^\textnormal{\scriptsize HVP}}%a_\mu^{hlbl$. The results are in reasonable agreement, although the results from Mainz is about
$2\sigma$ below the two other values. Whether this is due to the chiral extrapolation done in
\cite{Gerardin:2019rua} needs to be investigated by adding data closer or at the physical point in
this calculation.
\par
The errors on the available published results for the quark-disconnected contribution correspond to
errors of about
$0.3-0.7\%$ on the total HVP and thus, are already precise enough for a $1\%$ determination of
$a_\mu^\textnormal{\scriptsize HVP}}%a_\mu^{hlbl$, however, still need to be improved if an error of $<0.2\%$ is targeted.
\par
Work in progress on calculating the quark-disconnected contribution to the hadronic vacuum
polarisation
by the HPQCD/Fermilab/MILC collaboration was presented at this conference
\cite{Davies:2019acq}.
\subsection{Isospin Breaking Corrections}
\label{subsec:IB}
All lattice calculations discussed above have been done in the isospin symmetric limit, where the
up- and the down-quark are treated as being equal. However, in nature, isospin is broken by the
different electromagnetic charges of the up- and the down-quark as well as their different bare
quark
masses. These effects are expected to be of the order of $\mathcal{O}(\alpha)\approx1\%$ and
$\mathcal{O}((m_d-m_u)/\Lambda_\textnormal{\scriptsize QCD})\approx1\%$, respectively. Clearly, a
lattice
calculation
aiming at such precision will need to include these effects. It is important to stress
that the separation of strong isospin breaking and QED effects require to define a renormalisation
prescription. At the same time the definition of the ``physical'' point in a pure QCD simulation
becomes scheme dependent.
Only in full QCD$+$QED with $m_u\neq m_d$ the physical point is unambiguously
defined, e.g.\ by matching a set of hadron masses to their experimental value (see e.g.\
\cite{Aoki:2019cca}).
\subsubsection{Strong Isospin Breaking Correction}
The effect from strong isospin breaking can be taken into account in a lattice calculation by
simply using different input quark masses for the up and the down quark. This has been done for
calculating the strong isospin breaking correction to the HVP by the HPQCD/Fermilab/MILC
collaboration \cite{Chakraborty:2017tqp}. Here,
two different gauge ensembles where used, one with $N_f=2+1+1$ and $m_u\neq m_d$ only in the
valance sector and one with $N_f=1+1+1+1$ taking also strong isospin breaking corrections for the
sea quarks into account. The strong isospin breaking correction is then quoted as the difference
between calculating $a_\mu^\textnormal{\scriptsize HVP}}%a_\mu^{hlbl$ using the average up- and down-quark mass for the light quarks and
using
the up-quark mass for the up and the down-quark mass for the down quark. In
\cite{Chakraborty:2017tqp} the authors find $\delta^{sIB}a_\mu^\textnormal{\scriptsize HVP}}%a_\mu^{hlbl = (7.7\pm3.7)\times10^{-10}$ and
$\delta^{sIB}a_\mu^\textnormal{\scriptsize HVP}}%a_\mu^{hlbl = (9.0\pm2.3)\times10^{-10}$ using the ensemble without and with $m_u\neq m_d$
for the sea quarks, respectively.
\par
A different approach for including strong isospin breaking corrections in a lattice
calculation is by expanding \cite{deDivitiis:2011eh} the path integral in the difference of the
respective quark masses and
their isospin symmetric mass $\hat{m}$
\begin{equation}
\left<O\right>_{m_f\neq \hat{m}_f} = \left<O\right>_{m_f=\hat{m}} +
\Delta m_f \left.\frac{\partial}{\partial
m_f}\left<O\right>\right|_{m_f=\hat{m}} + \mathcal{O}\left(\Delta
m_f^2\right
\label{eq:patintegralmexpansion}
\end{equation}
with $\Delta m_f = m_f - \hat{m}$. At $\mathcal{O}(\Delta m_f)$ one has to calculate contributions
with one
insertion of a scalar current. The corresponding diagrams for the hadronic vacuum polarisation are
shown in figure
\ref{fig:sIBdiagrams}. Diagram $M$ and $O$ are the correction for the valance quarks for the
quark-connected and quark-disconnected HVP, respectively, whereas diagrams $R$ and $R_d$
correspond to sea-quark effects.
\par
\begin{figure}[h]
\centering
\includegraphics[width=0.99\textwidth]{plots/diagrams/scalar_insertion}
\vspace{-0.3cm}
\caption{Mass insertion diagrams for the hadronic vacuum polarisation. The insertion of a scalar
current is denoted by the diamond-shaped vertex.}
\label{fig:sIBdiagrams}
\end{figure}
The expansion of the path integral was used in
\cite{Giusti:2019xct,Blum:2018mom} to calculate the valance effect for the quark-connected HVP
(diagram $M$). The ETMC collaboration finds $\delta^{sIB}a_\mu^\textnormal{\scriptsize HVP}}%a_\mu^{hlbl = (6.0\pm2.3)\times10^{-10}$
\cite{Giusti:2019xct} and RBC/UKQCD obtains $\delta^{sIB}a_\mu^\textnormal{\scriptsize HVP}}%a_\mu^{hlbl =
(10.6\pm4.3\pm6.6)\times10^{-10}$ \cite{Blum:2018mom}.
\subsubsection{QED Correction}
The determination of electromagnetic corrections requires the inclusion of QED when evaluating the
Euclidean
path integral. Since QED is a long range interaction, finite volume (FV) effects for lattice
calculations can be large. Compared to pure QCD, where FV corrections are exponentially suppressed,
in the case of QED finite volume corrections are usually power-law\footnote{A prescription for
infinite volume
QED without power-law finite volume corrections has been recently proposed in \cite{Feng:2018qpx}.}.
For QED$_L$ \cite{Hayakawa:2008an}, which is a commonly used prescription of QED in a finite volume,
all the spatial
zero modes of the photon field are subtracted and finite volume effects for the QED
corrections to the HVP are of $\mathcal{O}(1/(m_\pi L)^3)$
\cite{Bijnens:2019ejw,Hermansson-Truedsson:2019pna,Giusti:2017jof}, and thus negligible within the
required precision for the HVP when using typical lattice sizes with $m_\pi L\gtrsim 4$.
\par
Since the electromagnetic fine structure constant $\alpha$ is small at low
energies, QED can be treated in a perturbative way by expanding the QCD$+$QED path integral in
$e^2$ \cite{deDivitiis:2013xla}
\begin{equation}
\left<O\right>_{\textnormal{\scriptsize QCD}+\textnormal{\scriptsize QED}} =
\left<O\right>_{\textnormal{\scriptsize QCD}} +
\frac{1}{2}\,e^2\left.\frac{\partial^2}{\partial
e^2}\left<O\right>\right|_{e=0} + {\cal{O}}(\alpha^2)\,.
\label{eq:pathintegraleexpansion}
\end{equation}
At $\mathcal{O}(\alpha)$ this requires to calculate diagrams that include one photon propagator.
The respective diagrams for the HVP are shown in figure \ref{fig:QEDdiagrams}. Diagrams $S$ and $V$
are QED corrections to the quark-connected HVP and diagrams $F$ and $D3$ are QED corrections to the
quark-disconnected HVP. All other diagrams correspond to QED effects for the sea quarks.
\par
\begin{figure}[h]
\centering
\includegraphics[width=0.99\textwidth]{plots/diagrams/diagrams_hvp_phys_incldisc}
\vspace{-0.3cm}
\caption{QED correction diagrams for the hadronic vacuum polarisation.}
\label{fig:QEDdiagrams}
\end{figure}
The ETMC \cite{Giusti:2019xct} and RBC/UKQCD collaborations \cite{Blum:2018mom,Gulpers:2018mim} both
have
calculated QED corrections to the quark-connected HVP in the electro-quenched approximation, where
QED corrections for the sea-quarks are not taken into account. Calculating QED corrections at
various masses heavier than physical pion mass and extrapolating to the physical point, ETMC finds
$\delta^\textnormal{\scriptsize QED} a_\mu^\textnormal{\scriptsize HVP}}%a_\mu^{hlbl=1.1(1.0)\times 10^{-10}$
\cite{Giusti:2019xct}. RBC/UKQCD finds $\delta^\textnormal{\scriptsize QED}
a_\mu^\textnormal{\scriptsize HVP}}%a_\mu^{hlbl=5.9(5.7)\times 10^{-10}$ \cite{Blum:2018mom}, consistent with ETMC albeit
with larger error bars, working directly at physical masses at a single lattice spacing. Further
work in progress for QED corrections to the HVP by other collaborations was presented at this
conference \cite{Risch:2019xio, TothLat19}.
\par
The leading QED correction to the quark-disconnected HVP is given by the diagram $F$ shown in
figure \ref{fig:QEDdiagrams}. Other than the same diagram without the photon connecting the two
quark loops (i.e.\ the pure QCD quark-disconnected diagram), this contribution is not
$SU(3)$-flavour suppressed and could thus be important. When calculating this contribution, one is
only interested in the case where, besides the photon line, the quark-loops are in addition
connected by gluons. If no additional gluons connect the quarks, these contributions are
conventionally included in the NLO HVP contribution\footnote{See \cite{Chakraborty:2018iyb} for a
lattice calculation of the NLO HVP contribution to $a_\mu$} to $a_\mu$ (cf. table
\ref{tab:amucontr}) and
thus, need to be subtracted in a lattice calculation to avoid double counting.
RBC/UKQCD calculated this diagram and finds $\delta^{\textnormal{\scriptsize QED, disc}}
a_\mu^\textnormal{\scriptsize HVP}}%a_\mu^{hlbl = -6.9(2.1)(1.4)\times 10^{-10}$ \cite{Blum:2018mom}.
\par
The diagrams in figure \ref{fig:QEDdiagrams} corresponding to sea-quark effects are all at least
either $SU(3)$-flavour or $1/N_c$ suppressed. However, naively, they could still be of the
order of $\approx 33\%$ of the connected contribution. Thus, when aiming at sub-percent precision
for the total HVP contribution to $a_\mu$, these effects will have to be included eventually.
\subsection{Summary - HVP Contribution to $a_\mu$}
\label{subsec:HVPsummary}
Figure \ref{fig:HVPcomp} shows a comparison of results for the total HVP contribution to $a_\mu$.
The upper panel shows determinations using the $R$-ratio data. The coloured points in the panel
labeled ``lattice'' shows the published lattice results from various collaboration. The lowest
point in the plot (``no new physics'') denotes the result when subtracting all other
Standard Model contributions
(as in table~\ref{tab:amucontr}) from the experimental result, i.e.\ the value that $a_\mu^\textnormal{\scriptsize HVP}}%a_\mu^{hlbl$ would
have to take for the Standard Model prediction to be in agreement with experiment. Clearly, at the
current state-of-the-art, lattice calculations are not yet precise enough to distinguish between
the $R$-ratio results and the ``no new physics'' scenario. \par
Furthermore, one can see that the
smallest and largest lattice results disagree at a level of about $2\sigma$. Slight tensions
between
lattice results will have to be subject to further investigation in the future, in particular once
the collaborations reduce the errors, to make sure to achieve consensus between the
various lattice results. A possible approach is the comparison of more
intermediate quantities, e.g. time-moments \cite{Chakraborty:2014mwa} of the vector correlator or
$a_\mu^\textnormal{\scriptsize HVP}}%a_\mu^{hlbl$ calculated from a time window~\cite{Blum:2018mom}.
\par
Finally, the point in the second panel in figure \ref{fig:HVPcomp} shows a result from RBC/UKQCD
\cite{Blum:2018mom} combining lattice and $R$-ratio results using a window method, where the vector
correlator from small and large distances is taken from the $R$-ratio and intermediate distances
from a lattice calculation. The point shown here uses the $R$-ratio data compilation from
``Jegerlehner 2017'' \cite{alphaQED} and clearly shows that it is possible to improve the $R$-ratio
results by
supplementing it with data from a lattice calculation.
\par
\begin{figure}[h]
\centering
\includegraphics[width=0.95\textwidth]{plots/comparison/amu_comp_standalone}
\vspace{-0.7cm}
\caption{Comparison of various determinations of the HVP contribution to the anomalous magnetic
moment of the muon.}
\label{fig:HVPcomp}
\end{figure}
\par
The relative contribution of the various quark-flavours to the total HVP contribution from lattice
calculations is shown by the pie chart on the left-hand side of figure \ref{fig:HVPsize}. Clearly,
the by far biggest contribution comes for the light-quark connected diagram, followed by the
strange-quark contribution.
\par
\begin{figure}[h]
\centering
\includegraphics[scale=1.0]{plots/comparison/hvp_standalone}
\includegraphics[scale=1.0]{plots/comparison/dhvp_standalone}\\
\hspace{1cm}$a^\textnormal{\scriptsize HVP}_\mu$\hspace{7.8cm}$\delta a^\textnormal{\scriptsize
HVP}_\mu$
\caption{The relative size of the various flavour contributions to the HVP is shown on the left.
The plot on the right shows the relative size of the average (statistical + systematic) error on
the various flavour contributions from the available lattice results.}
\label{fig:HVPsize}
\end{figure}
The total errors on the lattice results for the HVP shown in figure \ref{fig:HVPcomp} are all of
the order of $2-3\%$. This is clearly not yet competitive with the $R$-ratio results, which would
require an accuracy of $\lesssim 1\%$, or even the required $0.2\%$ to match the precision of the
upcoming experiments. The relative contribution to the error on the average lattice calculation is
shown in the pie chart on the right-hand side of figure \ref{fig:HVPsize}. The
error on the lattice results is dominated by the error on the light-quark
contribution. The main challenges for reaching sub-percent precision for the light-quark
contributions, that were discussed in
previous sections, are controlling the statistical noise in the long-distance tail, having
careful and reliable estimates of finite volume effects as well as a precise scale setting.
\par
The second biggest contribution to the error on the average lattice calculation (cf. right-hand
side of figure \ref{fig:HVPsize}) comes from isospin breaking corrections (or the systematic error
made by not including those effects).
Given that the first calculations of isospin breaking corrections have only recently become
available, progress and further reduction of statistical error is to be expected in the near
future. However, it is also important to study the effects of including QED for the sea quarks,
since these contributions could potentially be important at the level of sub-percent precision for
the hadronic vacuum polarisation.
\par
For the future, the first goal is to obtain calculations of $a_\mu^\textnormal{\scriptsize HVP}}%a_\mu^{hlbl$ from lattice calculations at
a precision of $\lesssim 1\%$, at which the lattice becomes competitive to $R$-ratio determinations.
Given the recent progress presented at this conference, first results at $1\%$ precision could
be available within the time frame of a year.
\par
Besides the anomalous magnetic moment of the muon, the hadronic vacuum polarisation also enters in
other quantities, like the running of the electromagnetic coupling and the running of the
electroweak mixing angle. Progress in this direction was presented by Mainz at this conference~
\cite{Ce:2019imp}.
\section{Hadronic Light-by-Light Scattering}
The hadronic light-by-light scattering contribution (cf. diagram on the right-hand side of
figure~\ref{fig:hadrdiagrams}) enters the anomalous magnetic moment at order $\alpha^3$. The value
that is
often used for the Standard Model prediction is the so-called ``Glasgow-consensus''
\cite{Prades:2009tw}, which includes model-dependent estimates of various contributions to the
light-by-light scattering, with the largest contribution coming from the $\pi^0$-pole.
Recent work in progress on dispersion relations for the hadronic light-by-light scattering can
be found, e.g., in \cite{Colangelo:2018mtw,Hoferichter:2018dmo,Pauk:2014rfa} and references
therein.
\par
In the following, I will discuss the progress of \textit{ab initio} calculations of the hadronic
light-by-light scattering using lattice QCD.
\subsection{Light-by-Light from the Lattice}
The hadronic part of the light-by-light scattering amplitude is written in terms of the expectation
value of four electromagnetic currents
\begin{equation}
\Pi_{\mu\nu\lambda\rho}(q_1,q_2,q_3) = \int \textnormal{d}^4 x_1\textnormal{d}^4
x_2\textnormal{d}^4 x_3 \,
e^{-i(q_1x_1+q_2x_2+q_3x_3)}
\left<j^\gamma_\mu(x_1)j^\gamma_\nu(x_2)j^\gamma_\lambda(x_3)j^\gamma_\rho(0)\right>\,.
\end{equation}
The fully quark-connected as well as the leading quark-disconnected Wick contraction are shown in
figure \ref{fig:lblwick}. All other possible quark-disconnected contractions are $SU(3)$-flavour
suppressed.
\par
\begin{figure}[h]
\centering
\includegraphics[width=0.3\textwidth]{plots/diagrams/g2lightbylight_con}
\hspace{1cm}
\includegraphics[width=0.3\textwidth]{plots/diagrams/g2lightbylight_leadingdisc}
\vspace{-0.5cm}
\caption{Fully quark-connected (left) and leading quark-disconnected (right) contributions to the
hadronic light-by-light scattering.}
\label{fig:lblwick}
\end{figure}
\par
Currently, two collaborations -- Mainz and RBC/UKQCD -- are actively working on calculating the
hadronic light-by-light scattering contribution $a_\mu^\textnormal{\scriptsize HLBL}$ to the anomalous magnetic moment of the
muon from the lattice. Both collaborations use position space approaches and their methods and
status are described
in sections \ref{subsec:lblmainz} and \ref{subsec:lblrbcukqcd}, respectively.
\subsection{Mainz Status}
\label{subsec:lblmainz}
In the position-space method developed by Mainz \cite{Green:2015mva,Asmussen:2016lse}
the hadronic light-by-light scattering contribution to $a_\mu$ is written as\\[-0.4cm]
\begin{equation}
a_\mu^\textnormal{\scriptsize HLBL} = \frac{m_\mu e^6}{3} \int\!\textnormal{d} x^4\textnormal{d} y^4\,
\overline{\mathcal{L}}_{[\rho,\sigma];\mu\nu\lambda}(x,y)\,\, i
\hat{\Pi}_{\rho;\mu\nu\lambda\sigma}(x,y)\,.
\label{eq:mainzlbl}
\end{equation}
The hadronic part is given by the four-point function\\[-0.3cm]
\begin{equation}
\hat{\Pi}_{\rho;\mu\nu\lambda\sigma} = \int\!\textnormal{d} z^4\,\, iz_\rho
\left<j^\gamma_\mu(x)\,j^\gamma_\nu(y)\,j^\gamma_\lambda(0)\,j^\gamma_\sigma(z)\right>
\end{equation}
where $z$ is the position of the vertex with the external photon, and $x$, $y$ and $0$ the
positions of the quark-photon vertices with the internal photon lines. The QED part of the
light-by-light diagram is given by an electromagnetic
kernel function $\overline{\mathcal{L}}_{[\rho,\sigma];\mu\nu\lambda}(x,y)$, that can be computed
directly in the continuum and infinite volume limit.
\par
\begin{figure}[h]
\centering
\includegraphics[width=0.47\textwidth]{plots/papers/fig3c_191105573.png}
\caption{The quark-connected contribution to $a_\mu^\textnormal{\scriptsize HLBL}$ for different pion masses. The data has
been
integrated up to $|y|_\textnormal{\scriptsize max}$. Plot is taken from \cite{Asmussen:2019act}.}
\label{fig:mainzlbl}
\end{figure}
\par
The current status of calculating $a_\mu^\textnormal{\scriptsize HLBL}$ using this method was presented
at this conference \cite{Asmussen:2019act} and some results for the fully quark-connected
contribution are shown in figure \ref{fig:mainzlbl}.
The plot in figure \ref{fig:mainzlbl} shows $a_\mu^\textnormal{\scriptsize HLBL}$ for three different pion masses at fixed
lattice spacing. The data has been partially integrated up to $|y|_\textnormal{\scriptsize max}$,
such that $a_\mu^\textnormal{\scriptsize HLBL}$ can be extracted from a plateau at large enough values of
$|y|_\textnormal{\scriptsize max}$.
\subsection{RBC/UKQCD Status}
\label{subsec:lblrbcukqcd}
In the position space approach proposed by RBC/UKQCD \cite{Blum:2015gfa} the full hadronic
light-by-light scattering diagram including the muon and photon propagators is treated on the
lattice. To make this calculation feasible, position space sampling is used, where the sum
over the position of two of the quark-photon vertices is sampled stochastically. For the photon
propagators RBC/UKQCD uses either the QED$_L$ \cite{Hayakawa:2008an} prescription leading to
power-law finite volume corrections or the infinite volume formulation (QED$_\infty$) of the
photon propagator as proposed in \cite{Blum:2017cer}.
\par
Figure \ref{fig:lblrbcukqcd} shows results from the recent paper \cite{Blum:2019ugy} using QED$_L$
for the photon propagator. $a_\mu^\textnormal{\scriptsize HLBL}$ obtained on various different gauge ensembles is plotted
against $1/(m_\mu L)^2$ with the box size $L$ for the fully quark-connected diagram (left) and the
leading quark-disconnected diagram (right). The lines in the
plots show the infinite volume and continuum extrapolation with the purple point at $1/(m_\mu
L)^2=0$ showing the result extrapolated to the infinite volume and continuum limit.
\par
\begin{figure}
\centering
\hspace{-1.3cm}
\includegraphics[height=4.6cm]{plots/papers/fig61_191108123.png}
\hspace{0.3cm}
\includegraphics[height=4.4cm]{plots/papers/fig62_191108123.png}
\vspace{-0.2cm}
\caption{Infinite volume and continuum extrapolation of $a_\mu^\textnormal{\scriptsize HLBL}$ using QED$_L$ for the photons for
the fully connected diagram (left) and the leading disconnected diagram (right). In both plots the
purple point at $1/(m_\mu L)^2=0$ shows the result extrapolated
to infinite volume and continuum. The plots are taken from~\cite{Blum:2019ugy}.}
\label{fig:lblrbcukqcd}
\end{figure}
The quark-connected and leading quark-disconnected contribution are found to have opposite signs,
and the result for the sum of both contribution is \cite{Blum:2019ugy}\\[-0.3cm]
\begin{equation}
a_\mu^\textnormal{\scriptsize HLBL}=7.20(3.98)(1.65)\times 10^{-10}\,.
\label{eq:amlblresult}
\end{equation}
extrapolated to the physical point. The result (\ref{eq:amlblresult}) is consistent with the model
estimates used in the Glasgow Concensus (cf. table \ref{tab:amucontr}), albeit with larger
uncertainties.
\par
In addition RBC/UKQCD presented progress \cite{BlumLat19,JinLat19} on calculating $a_\mu^\textnormal{\scriptsize HLBL}$
using QED$_\infty$ at this conference. The long-distance tail of the hadronic light-by-light
scattering is dominated by the $\pi^0$-pole contribution (cf. diagram in figure \ref{fig:pi0pole}).
This can be used to further improve the calculation of $a_\mu^\textnormal{\scriptsize HLBL}$ by supplementing the
long distance by the pion-pole contribution calculated either from a model or directly from the
lattice using the pion transition form factor $\pi^0\rightarrow\gamma\gamma$ and work in progress
in that direction was presented \cite{BlumLat19,JinLat19}.
\begin{figure}[h]
\centering
\includegraphics[width=0.27\textwidth]{plots/diagrams/pi0pole}
\vspace{-0.5cm}
\caption{$\pi^0$ pole contribution to the hadronic light-by-light scattering}
\label{fig:pi0pole}
\end{figure}
\subsection{Summary}
Currently two collaborations are actively working on determining the hadronic light-by-light
scattering contribution to $a_\mu$ from first principles using lattice calculations. RBC/UKQCD has
recently published \cite{Blum:2019ugy} the first result extrapolated to the continuum and infinite
volume limit and Mainz has presented \cite{Asmussen:2019act} promising work in progress using their
position space approach \cite{Green:2015mva,Asmussen:2016lse}.
\par
For the hadronic light-by-light scattering contribution to explain the discrepancy between
experimental measurement and Standard Model prediction of $a_\mu$, one would need a value which is
about three times larger than the number quoted in the Glasgow consensus. However, current lattice
calculations suggest, that this is very unlikely. \par
To match the precision of the upcoming experimental results for $g-2$, the hadronic light-by-light
scattering amplitude needs to be determined to a precision of about $10\%$. A promising proposal to
reduce the statistical noise from the long-distance contribution is to constrain lattice data at
long distances by the pion-pole contribution. A lattice calculation of the pion-pole
contribution requires the
pion transition form factor $\pi^0\rightarrow\gamma\gamma$ (see
\cite{Gerardin:2016cqj,Gerardin:2019vio} for recent calculations from the Mainz collaboration).
\section{Final Remarks}
The persistent discrepancy between the Standard Model prediction (table \ref{tab:amucontr}) and the
experimental result (equation (\ref{eq:amuexp})) for the anomalous magnetic moment of the muon has
triggered a tremendous effort within the lattice community to calculate the hadronic contributions
to $a_\mu$ from first principles. The upcoming experiments at Fermilab \cite{Venanzoni:2014ixa} and
JPARC \cite{Otani:2015jra} aim to further reduce the uncertainty of the experimental result by a
factor of $4$. To match the precision of these upcoming experiments one finally
has to determine the hadronic vacuum polarisation contribution $a_\mu^\textnormal{\scriptsize HVP}}%a_\mu^{hlbl$ to a precision of about
$0.2\%$ and the hadronic light-by-light scattering $a_\mu^\textnormal{\scriptsize HLBL}$ to about $10\%$ accuracy.
\par
In terms of the hadronic vacuum polarisation, which is the leading hadronic contribution to
$a_\mu$, several results from different collaborations with a precision of $2-3\%$ are available
and summarised in figure \ref{fig:HVPcomp}. Recent progress presented at this conference suggests,
that the first results at around $1\%$ precision could be available within the time frame of a
year.
\section*{Acknowledgments}
The author thanks N. Asmussen, C. Aubin, T. Blum, C. DeTar, D. Giusti, C. Lehner, L. Lellouch, A.
Meyer, B. Toth and G. von Hippel for useful discussions and/or providing material
prior to the conference. \par
VG has received funding from the European Research Council (ERC)
under
the
European Union's Horizon 2020 research and innovation programme under grant agreement No 757646.
| {
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\section{Introduction}\vspace{-0.1cm}
The use of unmanned aerial vehicles (UAVs) as flying wireless communication platforms has received significant attention recently \cite{mozaffari2, HouraniModeling,Irem}.
On the one hand, UAVs can be used as wireless relays for improving connectivity of ground wireless devices and extending network coverage. On the other hand, UAVs can act as mobile aerial base stations to provide reliable downlink and uplink communications for ground users, and boost the capacity of wireless networks \cite{Irem} and \cite{Letter}. Compared to the terrestrial base stations, the advantage of using UAV-based aerial base stations is their ability to quickly and easily move. Furthermore, the high altitude of UAVs can enable line-of-sight (LoS) communication links to the ground users.
Due to the adjustable altitude and mobility, UAVs can move towards potential ground users and establish reliable connections with a low transmit power. Hence, they can provide a cost-effective and energy-efficient solution to collect data from ground mobile users that are spread around a geographical area with limited terrestrial infrastructure.
Indeed, UAVs can play a key role in the \emph{Internet of Things (IoT)} which is composed of small, battery-limited devices such as
sensors, and health monitors \cite{dawy,lien,Eragh}. These devices are typically unable to transmit over a long distance due to their energy constraints \cite{lien}. In such IoT scenarios, UAVs can dynamically move towards IoT devices, collect the IoT data, and transmit it to other devices which are out of the communication ranges of the transmitting IoT devices \cite{lien}. In this case, the UAVs play the role of moving aggregators for IoT networks. However, to effectively use UAVs for IoT communications, several challenges must be addressed such as optimal deployment and energy-efficient use of UAVs \cite{mozaffari2}.
In \cite {mozaffari2}, we investigated the optimal deployment and movement of a single UAV for supporting downlink wireless communications. However, this work was restricted to a single UAV and focused on the downlink. The work in \cite{Han} analyzed the optimal trajectory of UAVs to enhance connectivity of ad-hoc networks. Nevertheless, this work did not study the optimal movement of multiple UAVs acting as aerial base stations. The work in \cite{ MozaffariTransport} studied the optimal deployment of UAVs and UAV-users association for static ground users with the goal of meeting users' rate requirements. In \cite{pang}, the authors used UAVs to efficiently collect data and recharge the clusters' head in a wireless sensor network which is partitioned into multiple clusters. However, this work is limited to a static sensor network, and does not investigate the optimal deployment of the UAVs. While the energy efficiency of uplink data transmission in a machine-to-machine (M2M) communication network was investigated in \cite{Nof} and \cite{Tu}, the presence of UAVs was not considered.
In fact, none of the prior studies \cite{mozaffari2, HouraniModeling,Irem}, and \cite{Han, MozaffariTransport, Nof, pang, Tu}, addressed the problem of optimal deployment and mobility of UAVs for enabling reliable and energy-efficient communications for mobile IoT devices.
The main contribution of this paper is to propose a novel approach for deploying multiple, mobile UAVs for energy-efficient uplink data collection from mobile, ground IoT devices. First, to minimize the total transmit power of the IoT devices, we create multiple clusters where each one is served by one of the UAVs. Next, to guarantee energy-efficient communications for the IoT devices in mobile and time-varying networks, we determine the optimal paths of the UAVs by exploiting dynamic clustering and optimal transport theory \cite{villani2003}. Using the proposed method, the total transmit power of the IoT devices required for successful transmissions is minimized by the dynamic movement of the UAVs. In addition, the proposed approach will minimize the total energy needed for the UAVs to effectively move. The results show that, using our proposed framework,
the total transmit power of the devices during the uplink transmissions can be reduced by $56\%$ compared to the fixed Voronoi deployment method. Furthermore, given the optimal trajectories for UAVs, they can serve the mobile IoT devices with a minimum energy consumption.
The rest of this paper is organized as follows. In Section II, we present the system model and problem formulation. Section III presents the optimal devices' clustering approach. In Section IV, we address the mobility of the UAVs using discrete transport theory. In Section V, we provide the simulation results, and Section VI draws some conclusions.\vspace{-0.05cm}
\section{System Model and Problem Formulation}
Consider an IoT system consisting of a set $\mathcal{L}=\{1,2,...,L\}$ of $L$ IoT devices deployed within a geographical area. In this system, a set $\mathcal{K}=\{1,2,...,K\}$ of $K$ UAVs must be deployed to collect the data from the ground devices in the uplink. The locations of device $i\in \mathcal{L}$ and UAV $j\in \mathcal{K}$ are, respectively, given by $(x_i,y_i)$ and $(x_{u,j},y_{u,j},h_j)$ as shown in Figure 1. We assume that devices transmit in the uplink using orthogonal frequency division multiple access (OFDMA) and UAV $j$ can support at most $M_j$ devices simultaneously. Note that, we consider a centralized network in which the locations of devices and UAVs are known to a control center such as a central cloud server. The ground IoT devices can be mobile (e.g. smart cars) and their data availability can be intermittent (e.g. sensors). Therefore, to effectively respond to the network mobility, it is essential that the UAVs optimally move for establishing reliable and energy-efficient communications with the devices. Note that, in our analysis, without loss of generality any mobility model can be accommodated.
For ground-to-air communications, each device will typically have a LoS view towards a specific UAV with a given probability. This LoS probability depends on the environment, location of the device and the UAV, and the elevation angle between the device and the UAV \cite{HouraniModeling}. One suitable expression for the LoS probability is given by \cite{HouraniModeling}:\vspace{-0.12cm}
\begin{equation}\label{PLoS}
{P_{{\rm{LoS}}}} = \frac{1}{{1 + \psi \exp ( - \beta\left[ {\theta - \psi} \right])}},
\end{equation}
where $\psi$ and $\beta$ are constant values which depend on the carrier frequency and type of environment such as rural, urban, or dense urban, and $\theta$ is the elevation angle. Clearly, ${\theta} = \frac{{180}}{\pi } \times {\sin ^{ - 1}}\left( {{\textstyle{{{h_j}} \over { {d_{ij}}}}}} \right)$, where $ {d_{ij}} = \sqrt {(x_i-x_{u,j})^2+(y_i-y_{u,j})^2+h_j^2 }$ is the distance between device $i$ and UAV $j$.
From (\ref{PLoS}), we can observe that by increasing the elevation angle or increasing the UAV altitude, the LoS probability increases. We assume that, the necessary condition for connecting a device to a UAV is to have a LoS probability greater than a threshold ($\epsilon$ close to 1). In other words, ${P_{\text{LoS}}}(\theta ) \ge \varepsilon$, and hence, $\theta \ge P_{\text{LoS}}^{ - 1}(\varepsilon )$ leading to:\vspace{-0.2cm}
\begin{equation}\label{dmin}
d_{ij} \le \frac{h_j}{{\sin \left( {P_{\text{LoS}}^{ - 1}(\varepsilon )} \right)}}.
\end{equation}
Note that (\ref{dmin}) shows the necessary condition for connecting the device to the UAV. Therefore, a device will not be assigned to UAVs which are located at distances greater than $\frac{h_j}{{\sin \left( {P_{\text{LoS}}^{ - 1}(\varepsilon )} \right)}}$.
Now, considering the LoS link, the received signal power at UAV $j$ from device $i$ is given by \cite{HouraniModeling} (in dB):
\begin{equation} \label{Pr}
P_r^{ij} = {P_{t,i}} - 10\alpha \log \left( {\frac{{4\pi {f_c}}}{c}{d_{ij}}} \right) - \eta,
\end{equation}
where $P_{t,i}$ is the transmit power of device $i$ in dB, $f_c$ is the carrier frequency, $\alpha=2$ is the path loss exponent for LoS propagation, $\eta$ is an excessive path loss added to the free space propagation loss, and $c$ is the speed of light.
In our model, the transmit power of the devices must satisfy the minimum signal-to-noise-ratio (SNR) required for a successful decoding at UAVs. For quadrature phase shift keying (QPSK) modulation, the minimum transmit power of device $i$ needed to reach a bit error rate requirement of $\delta$ is:\vspace{-0.2cm}
\begin{equation} \label{Pt_min}
P_t^{ij} = {\left[ {{Q^{ - 1}}\left( \delta \right)} \right]^2}\frac{{{R_b}{N_o}}}{2}{10^{\eta /10}}{\left( {\frac{{4\pi {f_c}{d_{ij}}}}{c}} \right)^2},
\end{equation}
where $Q^{ - 1}(.)$ is the inverse $Q$-function, $N_o$ is the noise power spectral density, and $R_b$ is the transmission bit rate. Note that, to derive (\ref{Pt_min}) using (\ref{Pr}), we use the bit error rate expression for QPSK modulation as ${P_e} = Q(\sqrt{\frac{{2P_r^{ij}}}{{{R_b}{N_o}}}})$ \cite{Goldsmith}.
Our first goal is to optimally move and deploy the UAVs in a way that the total transmit power of devices to reach the minimum SNR requirement for a successful decoding at the UAVs is minimized. In fact, the objective function is:\vspace{-0.2cm}
\begin{align} \label{opt1}
&\min \limits_{{\mathcal{C}_j},{\boldsymbol{\mu_j}}} \sum\limits_{j = 1}^K {\sum\limits_{i \in {\mathcal{C}_j}} {P_t^{ij}} }, \,\,\,j\in \mathcal{K},
\end{align}
where $P_t^{ij}$ is the transmit power of device $i$ to UAV $j$, and $K$ is the number of UAVs. Also, $\mathcal{C}_j$ is the set of devices assigned to UAV $j$, and $\boldsymbol{\mu_j}$ is the 3D location of UAV $j$.
From (\ref{Pt_min}), we can observe that the transmit power is directly proportional to the distance squared. Hence, minimizing the power is equivalent to minimizing the distance squared. Then, using (\ref{dmin}), (\ref{Pt_min}), and (\ref{opt1}), and considering the constraint on the maximum number of devices that can connect to each UAV, our optimization problem can be formulated as:\vspace{-0.2cm}
\begin{align}
&\left\{ {\mathcal{C}_j^*,\boldsymbol{\mu_j^*}} \right\} = \mathop {\arg \min }\limits_{{\mathcal{C}_j},{\boldsymbol{\mu_j}}} \sum\limits_{j = 1}^K {\sum\limits_{i \in {\mathcal{C}_j}} {d_{ij}^2} } \,\,,\,\,j\in \mathcal{K}, \label{opt_main}\\
{\text{s.t.}}\,\,&{\mathcal{C}_j} \cap {\mathcal{C}_m} = \emptyset ,\,\,j \ne m,\,\,\,\, j,m\in \mathcal{K}, \label{cons1} \\
&\sum\limits_{j = 1}^K {|{\mathcal{C}_j}|} = L, \label{cons2}\\\
&{d_{ij}} \le \frac{h_j}{{\sin \left( {P_{\text{LoS}}^{ - 1}(\varepsilon )} \right)}}, \\
&|{\mathcal{C}_j}| \le {M_j},
\end{align}
where $|\mathcal{C}_j|$ is the number of devices assigned to UAV $j$, $L$ is the total number of devices, and $M_j$ is the maximum number of devices that UAV $j$ can support. (\ref{cons1}) and (\ref{cons2}) guarantee that each device connects to only one UAV.
Clearly, we can consider the set of devices which are assigned to a UAV as a cluster, and place the corresponding UAV at the center of the cluster. Placing a UAV at the cluster center ensures that the UAV has a minimum total distance squared to all the cluster members. Hence, to solve problem (\ref{opt_main}), we need to find $L$ clusters, and their centers which effectively correspond to the locations of the UAVs. Note that, in a time varying network in which the location of IoT devices change, the clusters will also change. Consequently, the location of UAVs as the center of the clusters must be updated accordingly. However, moving these UAVs to the center of the clusters should be done with a minimum energy consumption. Therefore, in the mobile scenario, while finding the optimal paths of UAVs, we need to determine which UAV must go to which cluster center. Next, we present the IoT devices' clustering approach for minimizing the total transmit of the IoT devices.
\begin{figure}[!t]
\begin{center}
\vspace{-0.2cm}
\includegraphics[width=8.45cm]{System_model.pdf}
\vspace{-0.4cm}
\caption{ \small System model.\vspace{-0.72cm}}
\label{Nu}
\end{center}
\end{figure}
\section{Clustering IoT Devices}
Our first step is to optimally cluster the devices and deploy the UAVs at the center of the formed clusters so as to minimize the transmit power of the ground IoT devices.
We solve the problem in (\ref{opt_main}) by exploiting the constrained $K$-mean clustering approach \cite{Hoppner}. In the $K$-mean clustering problem, given $L$ points in $\mathds{R}^n$, the goal is to partition the points into $K$ disjoint clusters such that the sum of distances squared of the points to their corresponding cluster center is minimized. Hence, considering (\ref{opt1}) and (\ref{opt_main}), the total transmit power of devices is minimized by placing the UAVs in the center of the optimal clusters. This problem can be solved in two steps using an iterative process. The first step is related to the assignment, and the second step is called update.
\subsection{Assignment Step}
In the assignment step, given the location of the clusters' center (given all $ \boldsymbol{\mu_j}$), we find the optimal clusters for which the total distance squared between the cluster members and their center is minimized. Therefore, in our problem, given the location of the UAVs, we will first determine the optimal assignment of the devices to UAVs which can be written as:\vspace{-0.1cm}
\begin{align} \label{assign}
&\min\limits_{A_{ij}} \sum\limits_{j = 1}^K {\sum\limits_{i = 1}^L {{A_{ij}}||{\boldsymbol{v_i}} - {\mu _j}|{|^2}} }, \\
{\rm{s}}{\rm{.t}}{\rm{.}}&\sum\limits_{i = 1}^L {{A_{ij}}} \le {M_j}, \,\,\,j\in\mathcal{K},\\
&\sum\limits_{j = 1}^K {{A_{ij}}} = 1, \,\,\,i\in\mathcal{L},\\
&{A_{ij}}||{\boldsymbol{v_i}} - {\boldsymbol{\mu_j}}|| \le \frac{h_j}{{\sin \left( {P_{\text{LoS}}^{ - 1}(\varepsilon )} \right)}},\\
&{A_{ij}} \in \{ 0,1\},
\end{align}
where $\boldsymbol{v_i}=(x_i,y_i)$ is the two-dimensional location vector of device $i$, and $A_{ij}$ is equal to 1 if device $i$ is assigned to UAV $j$, otherwise $A_{ij}$ will be equal to 0. The problem presented in (\ref{assign}) is an integer linear programming which is solved by using the cutting plane method \cite{garf}.
\subsection {Update Step}
In the update step, given the clusters obtained in the assignment step, we update the location of the UAVs which is equivalent to updating the center of the clusters. Thus, the update location of UAVs is the solution to the following optimization problem:
\begin{align}\label{update}
&\mathop {\min }\limits_{({x_{u,j}},{y_{u,j}},h_j)} \sum\limits_{i \in {\mathcal{C}_j}} {{{({x_{u,j}} - {x_i})}^2} + {{({y_{u,j}} - {y_i})}^2}}+{h_j}^2 ,\\
&\text{s.t.}\,\,{({x_{u,j}} - {x_i})^2} + {({y_{u,j}} - {y_i})^2} + h_j^2\left( {1 - \frac{1}{{{{\sin }^2}\left( {P_{\text{LoS}}^{ - 1}(\varepsilon )} \right)}}} \right) \le 0, \label{consupdate} \nonumber\\
& \text{for all} \,\,\, i \in {\mathcal{C}_j},\,\, \text{and}\,\, j\in\mathcal{K}.
\end{align}
\begin{theorem}
\normalfont
The solution to (\ref{update}) is $\boldsymbol{{s^*}} = (x_{u,j}^*,y_{u,j}^*,h_j^*) = - \boldsymbol{P{(\lambda )^{ - 1}}Q(\lambda )}$, with the vector $\boldsymbol{\lambda}$ that maximizes the following concave function:
\begin{align}
&\mathop {\max {\rm{ }}}\limits_{\boldsymbol{\lambda}} \frac{1}{2} \boldsymbol{Q{(\lambda )^T} P{(\lambda )^{ - 1}}Q(\lambda )} + r(\boldsymbol{\lambda}),\\
&\textnormal {s.t.} \,\,\,\boldsymbol{\lambda} \ge 0,
\end{align}
where $\boldsymbol{P(\lambda)}= \boldsymbol{P_o} + \sum\limits_i {{\lambda _i}{P_i}}$, $\boldsymbol{Q(\lambda)}=\boldsymbol{Q_o} + \sum\limits_i {{\lambda _i}{Q_i}}$ and $r(\boldsymbol{\lambda})={r_o} + \sum\limits_i {{\lambda _i}{r_i}}$, with $\boldsymbol{P_o}$, $\boldsymbol{Q_o}$, $r_o$, $\boldsymbol{P_i}$, $\boldsymbol{Q_i}$, and $r_i$ given in the proof.
\end{theorem}
\begin{proof}
As we can see from (\ref{update}), the optimization problem is a quadratically constrained quadratic program (QCQP) whose general form is given by:
\begin{align}\label{QCQP}
&\mathop {\min }\limits_{\boldsymbol{s}} \,{\rm{ }}\frac{1}{2}\boldsymbol{{s^T}{P_o}s} + \boldsymbol{Q_o^Ts} + {r_o},\\
&\text{s.t.}\,\,\frac{1}{2}\boldsymbol{{s^T}{P_i}s} + \boldsymbol{Q_i^Ts} + {r_i}\le 0,\,\,\,i = 1,...,|{\mathcal{C}_j}|.
\end{align}
Given (\ref{update}) and (\ref{consupdate}), we have:\vspace{0.2cm}
$\boldsymbol{{P_o}} = \left[ {\begin{array}{*{20}{c}}
{2|{\mathcal{C}_j}|}&0&0\\
0&{2|{\mathcal{C}_j}|}&0\\
0&0&{2|{\mathcal{C}_j}|}
\end{array}} \right]$, $\boldsymbol{{P_i}} = \left[ {\begin{array}{*{20}{c}}
2&0&0\\
0&2&0\\
0&0&\omega
\end{array}} \right]$,\vspace{0.1cm}\\ $\omega = 1 - \frac{1}{{{{\sin }^2}\left( {P_{\text{LoS}}^{ - 1}(\varepsilon )} \right)}}$, $\boldsymbol{Q_o}={\left[ {\begin{array}{*{20}{c}}
{ - 2\sum\limits_{i = 1}^{|{\mathcal{C}_j}|} {{x_i}} }&{ - 2\sum\limits_{i = 1}^{|{\mathcal{C}_j}|} {{y_i}} }&0
\end{array}} \right]^T}$,\vspace{0.2cm}\\ $\boldsymbol{{Q_i}} = {\left[ {\begin{array}{*{20}{c}}
{ - 2{x_i}}&{ - 2{y_i}}&0
\end{array}} \right]^T}$, ${r_o} = \sum\limits_{i = 1}^{|{\mathcal{C}_j}|} {x_i^2} + \sum\limits_{i = 1}^{|{\mathcal{C}_j}|} {y_i^2}$, and \vspace{0.1cm}\\ ${r_i} = x_i^2 + y_i^2$. Note that, $\boldsymbol{P_o}$ and $\boldsymbol{P_i}$ are positive semidefinite matrices, and, hence, the QCQP problem in (\ref{QCQP}) is convex. Now, we write the Lagrange dual function as:\vspace{-0.1cm}
\begin{align}
f(\lambda ) = \mathop \text{inf}\limits_{\boldsymbol s} \biggl[&\frac{1}{2}\boldsymbol{{s^T}{P_o}s} + \boldsymbol{Q_o^Ts} + {r_o}\nonumber \\
&+ \sum\limits_i {{\lambda _i}\left( {\frac{1}{2}\boldsymbol{{s^T}{P_i}s} + \boldsymbol{Q_i^Ts} + {r_i}} \right)}\biggr]\nonumber \\
&= \mathop \text{inf}\limits_{\boldsymbol s} \left[ {\frac{1}{2}\boldsymbol{{s^T}P(\lambda )s} + \boldsymbol{Q{{(\lambda )}^T}s} + r(\boldsymbol{\lambda} )} \right].\nonumber
\end{align}
Clearly, by taking the gradient of the function inside the infimum with respect to $s$, we find $\boldsymbol{{s^*} = - P{(\lambda )^{ - 1}}Q(\lambda )}$. As a result, using $\boldsymbol{{s^*}}$, $f(\boldsymbol{\lambda} ) = \frac{1}{2}\boldsymbol{Q{(\lambda )^T}P{(\lambda )^{ - 1}}Q(\lambda )} + {r(\boldsymbol{\lambda} )}$. Finally, the dual of problem (\ref{QCQP}) or (\ref{update}) will be:\vspace{-0.1cm}
\begin{align}
\text{max}\,\, f(\boldsymbol{\lambda}), \,\,\textnormal {s.t.} \,\,\,\boldsymbol{\lambda} \ge 0,
\end{align}
which proves Theorem 1.
\end{proof}
The assignment and update steps are applied iteratively until there is no change in the location update step. Clearly, at each iteration, the total transmit power is reduced and the objective function is monotonically decreasing. Hence, the solution converges after several iterations.
In summary, for given locations of the ground IoT devices, we determined the optimal locations of the UAVs (cluster centers) for which the transmit power of the IoT devices used for reliable uplink communications is minimized. In a mobile IoT network, the UAVs must update their locations and follow the cluster centers as they evolve due to time-varying dynamics.
Next, we investigate how to optimally move the UAVs to the center of the clusters.
\section{Mobility of UAVs: Optimal Transport Theory}
Here, we find the optimal trajectory of the UAVs to guarantee the reliable uplink transmissions of mobile IoT devices. To move along the optimal trajectories, the UAVs must spend a minimum total energy on their mobility so as to remain operational for a longer time. In the considered mobile ground IoT network, the location of the devices and their availability might change, and hence, the clusters will change. Consequently, the UAVs must frequently update their locations accordingly. Now, given the location of the cluster centers obtained in Section III, and initial locations of the UAVs, we determine which UAV should fly to which cluster center such that the total energy consumed for their mobility is minimized.
In other words, given $\mathcal{I}$ and $\mathcal{J}$, the initial and new sets of UAVs' locations, one needs to find the optimal mapping between these two sets in a way that the energy used for transportations (between two sets) is minimized.
This problem can be modeled using discrete \emph{optimal transport theory} \cite{villani2003}. In its general form, optimal transport theory deals with finding an optimal transportation plan between two sets of points that leads to a minimum transportation cost \cite{villani2003}. These sets can be either discrete or continuous, with arbitrary distributions, and can have a general transportation cost function. The optimal transport theory was originated from the following Monge problem \cite{villani2003}. Given piles of sands and holes with the same volume, find the best move (transport map) to completely fill up the holes with the minimum total transportation cost. In general, this problem does not necessarily have a solution as each point must be mapped to only one location. However, Kantorovich relaxed this problem by using transport plans instead of maps, in which one point can go to multiple points \cite{villani2003}.
In our model, the UAVs need to move from initial locations to the new destinations. The transportation cost for each move is the energy used by each UAV for the mobility. We model this problem based on the discrete Monge-Kantorovich problem as follows \cite{xia}: \vspace{-0.10cm}
\begin{align} \label{transport1}
&\min\limits_{Z_{kl}} \sum\limits_{l \in \mathcal{J}} {\sum\limits_{k \in \mathcal{I}} {{E_{kl}}{Z_{kl}}} }, \\
\text{s.t.}\,&\sum\limits_{l \in \mathcal{J}} {{Z_{kl}}} = {m_k},\\
&\sum\limits_{k \in \mathcal{I}} {{Z_{kl}}} = {m_l},\\
&{Z_{kl}} \in \{ 0,1\},
\end{align}
where $\mathcal{I}$ and $\mathcal{J}$, are the initial and new sets of UAVs' locations. $\boldsymbol{Z}$ is the $\mathcal{|J|\times|I|}$ transportation plan matrix with each element $Z_{kl}$ being 1 if UAV $k$ is assigned to location $l$, and 0 otherwise. $E_{kl}$ is the energy used for moving a UAV from its initial location with index $k \in \mathcal{I}$ to a new location with index $l \in \mathcal{J}$. $m_l$ and $m_k$ are the number of points (UAVs) at the locations with indices $l$ and $k$.
The energy consumption of a UAV moving with a constant speed as a function of distance is given by \cite{di}:\vspace{-0.3cm}
\begin{equation} \label{energy}
E(D,v) = \int\limits_{t = 0}^{t = D/v} {p(v)dt} = \frac{{p(v)}}{v}D,
\end{equation}
where $D$ is the travel distance of the UAV, $v$ is the constant speed, $t$ is the travel time, and $p(v)$ is the power consumption as a function of speed. As we can see from (\ref{energy}), energy consumption for mobility is linearly proportional to the travel distance. Using the Kantorovich-duality, the discrete optimal transport problem in (\ref{transport1}) is equivalent to:\vspace{-0.1cm}
\begin{align}\label{dual}
&\max\limits_{\varphi,\xi} \left[ {\sum\limits_{k \in \mathcal{I}} {\xi (k)} - \sum\limits_{l \in \mathcal{J}} {\varphi (l)} } \right],\\
& \text{s.t.}\,\, \varphi (l)-\xi (k) \le {E_{kl}},
\end{align}
where $\xi :\mathcal{I} \to \mathds{R}$, and $\varphi :\mathcal{J} \to \mathds{R}$ are unknown functions of the maximization problem. The dual problem in (\ref{dual}) is used to solve the primal problem in (\ref{transport1}) by applying the complementary slackness theorem \cite{villani2003}. In this case, the optimal solution including the optimal transport plan between $\mathcal{I}$ and $\mathcal{J}$ is achieved when $\varphi (l)-\xi (k) = {E_{kl}}$ \cite{villani2003}. Here, to find the optimal mapping between initial set of locations and the destination set, we use the revised simplex method \cite{ford}. The result will be the transportation plan ($\boldsymbol{Z}$) that optimally assigns the UAVs to the destinations. Hence, the location of the UAVs are updated according to the new destinations. Subsequently, having the destinations of each UAV at different time instances, we can find the optimal trajectory of the UAVs. As a result, given the optimal paths, the UAVs are able to serve the mobile IoT devices in an energy-efficient way.
\section{Simulation Results and Analysis}
In our simulations, the IoT devices are deployed within a geographical area of size $1.2\,\text{km}\times 1.2 \,\text{km}$. We consider UAV-based communications in an urban environment with $\psi=11.95$, and $\beta=0.14$ at 2\,GHz carrier frequency \cite{HouraniModeling}. Moreover, we use the energy consumption model for UAVs' mobility as $E(D,v) = D\left( {0.95{v^2} - 20.4v + 130} \right)$ \cite{di}. Furthermore, in a time-varying network, to capture the mobility and availability of the ground IoT devices,
we generate the new devices' locations by adding zero mean Gaussian random variables to the initial devices' locations. Table I lists the simulation parameters. Note that,
all statistical results are averaged over a large number of independent runs.
\begin{table}[!t]
\normalsize
\begin{center}
\caption{\small Simulation parameters.}
\label{TableP}
\resizebox{7.2cm}{!}{
\begin{tabular}{|c|c|c|}
\hline
\textbf{Parameter} & \textbf{Description} & \textbf{Value} \\ \hline \hline
$f_c$ & Carrier frequency & 2\,GHz \\ \hline
$v$ & UAV speed & 10\,m/s \\ \hline
$\delta$ & Bit error rate requirement & $10^{-8}$ \\ \hline
$\epsilon$ & $P_\text{LoS}$ requirement & 0.95 \\ \hline
$N_o$ & Noise power spectral density & -170\,dBm/Hz \\ \hline
$R_b$ & Transmission data rate & 200\,Kbps \\ \hline
$B$ & Transmission bandwidth per device & 200\,KHz \\ \hline
$\eta$ & Additional path loss to free space & 5\,dB \\ \hline
\end{tabular}}
\end{center}\vspace{-0.3cm}
\end{table}
Figure \ref{cluster} shows a snapshot of the optimal devices' clustering as well as the optimal UAVs' locations resulting from our proposed approach. In this example, 5 UAVs are used to support 100 IoT devices. We assume that each UAV has a limited number of resource blocks which can be allocated to at most 30 devices. Therefore, we have 5 clusters with a maximum size of 30 devices and 5 cluster centers corresponding to the locations of the UAVs. As we can see from Figure \ref{cluster}, the location of IoT devices significantly impacts the number of devices per cluster and also the optimal locations of the UAVs. In this figure, the minimum and maximum cluster sizes are 3 and 27.
\begin{figure}[!t]
\begin{center}
\vspace{-0.2cm}
\includegraphics[width=7.6cm]{Clusters.eps}
\vspace{-0.4cm}
\caption{ \small Optimal clusters and UAVs' locations in one snapshot.\vspace{-0.7cm}}
\label{cluster}
\end{center}
\end{figure}
Figure \ref{Pt_voronoi} shows the total transmit power of devices versus the number of UAVs averaged over multiple simulation runs. In this figure, the performance of the proposed approach is compared with the fixed Voronoi case which is known to be a typical deployment method for static base stations. Note that, for a fair comparison, we assume that the total number of resource blocks is fixed ($L$), and hence, the number of resources per UAV is $\left\lceil {\frac{L}{K}} \right\rceil$. In other words, the maximum size of each cluster will decrease as the number of UAVs increases. In the Voronoi method, assuming a uniform distribution of devices, we fix the location of UAVs at an altitude of 500\,m, and then, we assign each device to the closest UAV. However, in the proposed clustering algorithm, we find the optimal clusters and deploy the UAVs at the center of the clusters. As shown in Figure \ref{Pt_voronoi}, the proposed method outperforms the classical Voronoi scheme as the UAVs can be placed closer to the devices. As expected, increasing the number of UAVs reduces the total transmit power of IoT devices. For instance, when the number of UAVs increases from 4 to 8, the total transmit power decreases from 77\,mW to 38\,mW for the proposed method, and from 115\,mW to 95\,mW for the Voronoi case. Figure \ref{Pt_voronoi} shows that our approach results in about 56\% reduction in the transmit power of the IoT devices.
\begin{figure}[!t]
\begin{center}
\vspace{-0.2cm}
\includegraphics[width=6.8cm]{Pt_numb_UAVs_MonteCarlo.eps}
\vspace{-0.2cm}
\caption{ \small Average of total transmit power vs. number of UAVs.\vspace{-0.4cm}}
\label{Pt_voronoi}
\end{center}
\end{figure}
Figure \ref{Trajectory} shows the trajectory of one of the UAVs in a mobile IoT scenario derived from optimal transport theory.
Here, we consider 8 UAVs, and 400 devices whose locations are updated at each time by adding a Gaussian random variable with $N(0,50\,\text{m})$ to the previous locations.
Clearly, since the locations of the devices may change over time, the optimal clusters must be updated accordingly. In Figure \ref{Trajectory}, the red dots correspond to the optimal destinations of the UAV at different times. In fact, as the clusters are changing over time, the UAV uses the proposed scheme to optimally move to one of the new cluster centers.
\begin{figure}[!t]
\begin{center}
\vspace{-0.2cm}
\includegraphics[width=7.2cm]{Trajectory.eps}
\vspace{-0.5cm}
\caption{ \small Trajectory of a UAV in a mobile IoT network.\vspace{-0.8cm}}
\label{Trajectory}
\end{center}
\end{figure}
Figure \ref{EnergyConsumption} shows the energy consumed by each UAV during its mobility. In this case, we use 8 UAVs for supporting 400 ground IoT devices. We consider the network at 10 time instances during which the UAVs move at a speed of 10\,\text{m/s} while updating their locations. As shown in Figure \ref{EnergyConsumption}, for the given scenario, the total amount of energy that UAVs use for mobility is around 106\,\text{kJ}. Note that, this is the minimum total energy consumption that can be achieved via the optimal transport of the UAVs. As shown, different UAVs spend different amount of energy on the mobility. Depending on the optimal clustering of devices over time, different UAVs might have different travel distances to the cluster centers. For instance, UAV 1 consumes 1.8 times more energy than UAV 3. Hence, the number of UAVs may also change over time.
\begin{figure}[!t]
\begin{center}
\vspace{-0.5cm}
\includegraphics[width=6.8cm]{EnergyConsumption.eps}
\vspace{-0.4cm}
\caption{ \small Energy consumption of each UAV.\vspace{-0.15cm}}
\label{EnergyConsumption}
\end{center}
\end{figure}
Figure \ref{UAV_loss3} shows the energy consumption per UAV when the number of UAVs changes. Here, we assume that, initially the UAVs are optimally deployed for a given IoT system, however, after a while some of the UAVs ($q$ UAVs) will not be operational due to the lack of battery. Consequently, the number of UAVs decreases and the remaining UAVs must update their locations to maintain the power efficiency of the ground devices.
In Figure \ref{UAV_loss3}, for the average case, we take the average of energy over all possible combinations of removing $q$ UAVs among the total UAVs. However, in the worst-case scenario, we remove the $q$ UAVs whose loss leads to the highest energy consumption for the remaining UAVs.
Clearly, as more UAVs become inoperational, the energy consumption of the functioning UAVs will increase. For example, when the number of lost UAVs increases from 2 to 4, the average energy consumption per UAV increases from 1520\,J to 2510\,J.\vspace{0.09cm}
\begin{figure}[!t]
\begin{center}
\vspace{-0.5cm}
\includegraphics[width=6.6cm]{UAV_loss_MonteCarlo.eps}
\vspace{-0.70cm}
\caption{ \small Energy consumption vs. number of battery depleted UAVs.\vspace{-0.76cm}}
\label{UAV_loss3}
\end{center}
\end{figure}
\section{conclusions}\vspace{-0.007cm}
In this paper, we have proposed a novel framework for efficiently deploying and moving UAVs to collect data from ground IoT devices. In particular, we have determined the optimal clustering of IoT devices as well as the optimal deployment and mobility of the UAVs such that the total transmit power of IoT devices is minimized while meeting a required bit error rate. To perform clustering given the limited capacity of each UAV, we have adopted the constrained size clustering approach.
Furthermore, we have obtained the optimal trajectories that are used by the UAVs to serve the mobile IoT devices with a minimum energy consumption.
The results have shown that by carefully clustering the devices and deploying the UAVs, the total transmit power of devices significantly decreases compared to the classical Voronoi-based deployment. Moreover, we have shown that, by intelligently moving the UAVs, they can remain operational for a longer time while serving the ground devices. \vspace{-0.01cm}
\def1.00{1.00}
\bibliographystyle{IEEEtran}
\vspace{-0.01cm}
| {
"redpajama_set_name": "RedPajamaArXiv"
} | 2,149 |
2018 Cultural Studies Association (CSA) Conference
OPPORTUNITIESFellowships
The Getty Foundation Pre- and Postdoctoral Fellowships 2018
Getty Predoctoral and Postdoctoral Fellowships are intended for emerging scholars to complete work on projects related to the Getty Research Institute's annual research theme. Recipients are in residence at the Getty Research Institute or Getty Villa, where they pursue research projects, complete their dissertations, or expand dissertation for publication. Fellows make use of the Getty collections, join in a weekly meeting devoted to the annual theme, and participate in the intellectual life of the Getty.
Applications for Getty Pre- and Postdoctoral fellowships are welcome from scholars of all nationalities.
Current Getty staff and members of their immediate family are not eligible for Pre- and Postdoctoral fellowships.
Getty Predoctoral Fellowship applicants must have advanced to candidacy by the application deadline and should expect to complete their dissertations during the fellowship period. Successful Predoctoral Fellowship applicants who are awarded their degree after the application deadline but before the fellowship begins, or who receive their doctorate while in residence, automatically become Postdoctoral Fellows. To be eligible to apply for the 2018-2019 scholar year, Postdoctoral Fellowship applicants should not have received their degree earlier than 2013.
Getty Predoctoral Fellows are in residence for nine months from late-September to late-June and receive a stipend of $25,000.
Getty Postdoctoral Fellows are in residence for nine months from late-September to late-June and receive a stipend of $30,000.
Both fellowships also provide a workspace at the Getty Research Institute or the Getty Villa, an apartment in the Getty scholar housing complex, airfare to and from Los Angeles, and makes healthcare options available. These terms apply as of July 2017 and are subject to future changes.
Additional information about the terms of these grants, including healthcare, is available here.
Applicants are notified of the Getty Research Institute's decision approximately six months following the deadline.
All fellowships are awarded on a competitive basis. Applications will be evaluated by the Getty Research Institute based on: (1) the overall quality of the application; (2) how the proposed project bears upon the annual research theme; (3) the applicant's past achievements; and (4) how the project would benefit from the resources at the Getty, including its library and collections.
Applicants are required to complete and submit the online Fellowship application form (which includes uploading a Project Proposal; Selected Bibliography; Doctoral Dissertation Plan or Abstract; Curriculum Vitae; Writing Sample; and Confirmation letter by the deadline.
Two letters of recommendation are also required for this application.
October 2, 2017 (5:00 p.m. PDT).
Visit Official Website for More Information
International FellowshipsopportunitiesPhDScholarships
Center on Conflict and Development (ConDev) Student Media Grant 2018
NASP PhD Scholarship in Political Science in Italy, 2017
Skoltech MSc and PhD Full Tuition Scholarships for International Students in Russia, 2017
Cosmopolis Advanced International MA/PhD Scholarships in Netherlands, 2017 | {
"redpajama_set_name": "RedPajamaCommonCrawl"
} | 6,914 |
VIDEO: MRI Neuro-interventional Suite at Henry Ford Hospital
Magnetic Resonance Imaging (MRI) | May 21, 2019
This is a quick look inside the magnetic resonance imaging (MRI) neuro-interventional suite at Henry Ford Hospital in Detriot, Mich. This neuro-procedure room connects through a door to a traditional operating room and the patient can be transferred using a trolley cot system to move between the two rooms. The MRI is a Philips 1.5T Achieva and the MRI suite includes MRI-safe surgical instruments, furniture and anesthesia system.
Find more Henry Ford Hospital related content.
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Quality Assurance (QA) | February 06, 2013
IBA MagicMaX System Enables One Minute QA Testing
A quality assurance test can be performed within one minute using IBA's MagicMaX QA system for X-ray machines. The system is designed to be simple to use and lightweight.
Radiology Imaging | December 20, 2012
Hitachi Highlights Widest-Bore MRI, New 128-Slice CT
At RSNA 2012, Hitachi featured its Echelon Oval 1.5T MRI system, which features the widest bore on the market at 74 cm, a wide table and the ability to perform non-contrast MR angiography exams. Hitachi also highlighted new features for its Scenaria CT system, which is upgradeable to a 128-slice system, offers new, faster iterative reconstruction software and cardiac imaging packages.
Cardiac Imaging | December 18, 2012
Editor's Choice of Cardiovascular Imaging Technology at RSNA
DAIC Editor Dave Fornell highlights the latest advancements that will impact cardiovascular imaging from the 2012 Radiological Society of North America (RSNA) meeting. RSNA is the largest medical imaging show in the world and most advancements are shown here first.
ITN Editor's Choice of Most Innovative Radiology Technology at RSNA 2012
ITN Editor Dave Fornell highlights his choices for the most innovative radiology technologies and trends at RSNA 2012. Choices include the first wireless ultrasound transducer, noiseless MRI, a 640-slice CT scanner and a printer than creates sculptures from 3-D datasets.
Digital Radiography (DR) | December 10, 2012
Konica Minolta Shows Latest Advances in DR Imaging
At RSNA 2012, Konica Minolta showed its three latest advances in digital radiography (DR) X-ray. The company featured its X70 radiography room, which centers around the Aero DR wireless detector. The room integrates to even locate where the DR detector is located. The company also highlighted its automatic stitching solution, 10 x 12 detector for pediatric and table use, and the Aero Sync solution that wirelessly synchronizes with X-ray generators to eliminate cables.
Quality Assurance (QA) | December 07, 2012
IBA Offers Quality Assurance Solutions for X-Ray, CT, Displays
At RSNA 2012, IBA showcased several quality assurance (QA) solutions for radiology. The MagicMax is an all-in-one QA system for all X-ray systems, including digital radiography (DR), CT and mammography. The Primus L phantom is an all-in-one QA device for all digital X-ray imaging systems. IBA also offers the LXcan to QA diagnostic-quality black and white flat panel monitors, and the LX Chroma to QA color displays.
Computed Tomography (CT) | December 07, 2012
Neurologica Offers Only Truly Portable Whole-Body CT Scanner
Neurologica demonstrated its BodyTom portable whole-body CT scanner during RSNA 2012. The system is the first mobile scanner that can be moved on casters from room to room and is battery powered. The system has an 85 cm gantry and a 60 cm field of view. Unlike traditional CT scanners, the gantry moves over the patient, rather than the patient table being moved through the gantry. This facilitates use in the operating room, where it is not easy to move a patient who may be connected to several devices.
Trends and Analysis of RSNA 2012
Imaging Technology News experts discuss the trends and latest technology they saw on the show floor and in sessions at RSNA 2012. Their discussions include some of the most innovative new devices and software to solve issues facing radiology today.
SwissRay Highlights Latest DR X -Ray Technology
During RSNA 2012, SwissRay featured its new DDR Versa Motion Plus X-ray system. The technologist selects a body part to be images and the X-ray head automatically swings into the proper imaging position. The head includes a touch-screen where information can be entered at the patient bedside. Also featured were the DDR Cruze mobile DR X-Ray system and the DDR Shift retrofit kit that enables conversion of mobile CR systems to wireless DR.
Radiology Imaging | November 12, 2012
Preview of RSNA Highlights New Features for the 2012 Event
Imaging Technology News talks to Mark Watson, executive director of the Radiological Society of North America (RSNA), and Steve Drew, RSNA's executive director for scientific assembly and informatics, about the upcoming RSNA 2012 event, as well as what's ahead for radiologists in 2013.
Digital Radiography (DR) | September 19, 2012
Experience the X-Factor of Carestream DRX Systems
Carestream is changing the DR game and putting you in control of the move to digital.
Carestream DRX-Revolution Mobile X-Ray System On The Move
The Carestream DRX-Revolution is a mobile X-ray system on wheels powered by a wireless DRX detector.
DRX-Evolution Versatility and the X-Factor
See the versatility of the DRX Evolution room with the wireless DRX detector.
Carestream DRX-Transportable / Field Portable System Rugged and Reliable for Military Settings
Carestream's DRX - transportable / field portable X-ray unit is designed and tested for the rigorous conditions of military, disaster and remote locations.
Carestream DRX-Revolution - Convenience is Key for Hamilton General Hospital Technologists
Jeannie Patterson, PSW at Hamilton General Hospital, explains some of the benefits of the Carestream DRX-Revolution mobile X-ray system, including its compact size, easy detector bagging, storage for markers, gloves, bags and wipes, and the swivel image head.
Computed Tomography (CT) | August 28, 2012
Iterative Technique to Eliminate Artifacts from Implants
Dr. Web Stayman of Johns Hopkins University presents an overview of research he presented at the 2012 American Association of Physicists in Medicine (AAPM) annual meeting in Charlotte, N.C. It involves an iterative technique for computed tomography (CT) to better contend with implants to improve image-guided surgery or interventions. The technique takes knowledge about the components and integrates it into the reconstruction to eliminate artifacts.
Mammography | August 28, 2012
A New Technique Utilizes Spectral Mammography to Measure Breast Density
Dr. Sabee Molloi from the School of Medicine at the University of California, Irvine, worked with a team on a study using spectral mammography to develop a quantitative technique to measure volumetric breast density. Their technique also enables a lower dose to be used for a screening mammogram. Two members of the team, Justin Ducote and Huanjun Ding, describe the research, which they presented at the 2012 annual meeting of the American Association of Physicists in Medicine.
Nuclear Imaging | July 26, 2012
A New Name for SNM
Speaking with ITN Editorial Director Helen Kuhl at the SNM annual meeting in Miami Beach, Fla., in June, incoming president Frederic H. Fahey, DSc, explains the reasoning behind the society's name change from Society of Nuclear Medicine to Society of Nuclear Medicine and Molecular Imaging. He also shares highlights of the successful 2012 event.
SNM's Initiatives for the Coming Year
Incoming president Frederic H. Fahey, DSc, describes the primary initiatives the Society of Nuclear Medicine and Molecular Imaging will be undertaking during the coming year, during an interview with ITN Editorial Director Helen Kuhl at the society's annual meeting in June. These include growing global initiatives, including more involvement in developing countries, plus continued education and efforts with regard to radiation dose and dose optimization.
Current Trends in Nuclear Medicine and Molecular Imaging
Frederic H. Fahey, DSc, incoming president of the Society of Nuclear Medicine and Molecular Imaging, shares his views about significant trends in the field, including the emergence of new amyloid imaging agents and other new agents, radionuclide therapy and the ongoing focus on quality and safety.
Mammography | March 30, 2012
Philips Showcases Microdose Mammography at 2012 NCoBC
Philips' new Microdose digital mammography system provides comfort for the patient, efficiency for the physician and department manager, plus 50 percent less dose.
Breast Imaging | March 30, 2012
Highlights of the 2012 NCoBC Meeting: Breast Health Trends
Gary Levine, M.D., program chair/incoming president of the National Consortium of Breast Centers, gives an overview of current trends in technology, diagnosis and treatment of breast cancer, and regulatory activity that will impact women's health.
Women's Health | March 30, 2012
Highlights of the 2012 NCoBC Meeting: Breast Density Discussion
Gary Levine, M.D., program chair/incoming president of the National Consortium of Breast Centers, discusses legislation regarding breast density at the 22nd annual National Interdisciplinary Breast Center Conference (NCoBC), held in Las Vegas in March.
Highlights of the 2012 NCoBC Meeting: Utilizing Social Media
Gary Levine, M.D., program chair/incoming president of the National Consortium of Breast Centers, discusses how breast centers can use social media to educate the public regarding breast health and their services at the 2012 NCoBC meeting, held in Las Vegas in March.
Highlights of the 2012 NCoBC Meeting: IORT
Gary Levine, M.D., program chair/incoming president of the National Consortium of Breast Centers, discusses the emergence of interoperative radiation therapy (IORT) at the 22nd annual National Interdisciplinary Breast Center Conference (NCoBC), held in Las Vegas in March.
Highlights of the 2012 NCoBC Meeting: Politics and Women's Health
Gary Levine, M.D., program chair/incoming president of the National Consortium of Breast Centers, discusses the role of politics on women's health in an election year, during the 2012 National Interdisciplinary Breast Center Conference (NCoBC), held in Las Vegas in March.
Printers | February 14, 2012
Canon - High-Quality DICOM Images Printed on Paper
The Cypher PS print server and Canon imagePRESS C1+ digital printing solution is a cost-effective answer that enables radiology departments to print high-quality digital imaging and communications in medicine (DICOM) images on various types and sizes of paper. This digital printing solution helps to cut the cost of expensive film usage by providing a low-cost, high-quality alternative. This system enables practices to save time and money while expediting the sharing of exam results among medical teams and patients. Not only will this digital printing solution output high-quality DICOM images, but it also enables use as a departmental printer for a variety of documents. For more information: www.usa.canon.com/cypher
Advanced Visualization | February 13, 2012
Advanced Visualization Fly-Through of Zoo Animal CT Scans
The Chicago Zoological Society's (CZS) Brookfield Zoo is the first North American zoo to use 3-D advanced visualization imaging technology. This video shows a video fly-through of reconstructed 3-D computed tomography (CT) images of an aardvark, Humboldt penguin and African crested porcupine. The zoo is using Web-based software from Vizua. For the related story, visit http://www.itnonline.com/article/first-us-zoo-begins-using-3-d-imaging-t...
Radiology Imaging | March 22, 2011
Angio-Suite, Breast Imaging Set the Trends in Imaging
Advanced breast imaging capabilities added elastography to the list, fused MR/CT image data combined with angiography navigation systems to guide percutaneous oncology, and 3.0 Tesla MR debut at the 2009 Radiological Society of North America (RSNA). All these innovations headlined the news at RSNA. To find out where these trends are leading radiology and radiation oncology, Imaging Technology News spoke with The MarkeTech Group's (TMTG) CEO and Founder Dr. Christian Renaudin. In an exclusive interview, Dr. Renaudin analyzes what these key market trends mean to diagnostic imaging. The MarkeTech Group is a CASRO certified international marketing research and consulting firm focused exclusively on medical technology. As a leading ad hoc Voice-of-Customer solution provider in medical imaging, The MarkeTech Group attends the annual RSNA meeting to investigate what new technological innovations in diagnostic imaging manufacturers are displaying on the show room floor. For more information: www.themarketechgroup.com
Main Video Page | {
"redpajama_set_name": "RedPajamaCommonCrawl"
} | 3,520 |
Q: Using the IPAddress Gem As a Custom Field in Mongoid I'm trying to use IPAddress as a custom field for a Mongoid Document by monkey patching the serialization methods to the IPAddress module and I just can't seem to get it...
class Computer
include Mongoid::Document
field :ip_address, type: IPAddress
end
module IPAddress
def mongoize
to_string
end
def self.mongoize(object)
object.to_string
end
def self.demongoize(string)
IPAddress.new string
end
def self.evolve(object)
object.to_string
end
end
Here's what I got right this second... but I've tried lots of other ways and just can't find one that works. Any help would be much appreciated!
| {
"redpajama_set_name": "RedPajamaStackExchange"
} | 6,834 |
Q: how i can do an image slideshow i have this code , and i want to ask how i can use buttons (left ,right ) to slide small images that i imported from a data base like slideshow using jquery
?? thank you for your help
Example for website: http://www.voirfilms.co
$PARAM_hote='localhost';
$PARAM_nom_bd='venteformation';
$PARAM_utilisateur='root';
$PARAM_mot_passe='';
try{
$connexion = new PDO('mysql:host='.$PARAM_hote.';dbname='.$PARAM_nom_bd, $PARAM_utilisateur, $PARAM_mot_passe);
}
catch(Exception $e) {
echo 'Erreur : '.$e->getMessage().'<br />'; echo 'N° : '.$e->getCode();
}
if (mysqli_connect_errno()) {
echo "Echec de la connexion" ;
exit();
}
else
{
$RequetLivre = $connexion->query("select * from livre");
?>
<div id="wrapper" class="col-lg-12">
<div id="page-wrapper" >
<div class="container-fluid">
<?php /**Block ROW 1 *****/?>
<div class="row">
<div class="col-lg-12">
<form name="f_livre" method="POST" action="verification.php">
<fieldset>
<table width="" border="0">
<tr>
<?php
while ($LivreRow = $RequetLivre->fetch(PDO::FETCH_ASSOC))
{
echo "<div class='slideshow'>";
echo"<td>";
echo '<button type="submit" name="idl" value="'.$LivreRow['IDL'].'"><img src="data:image/jpeg;base64,'.base64_encode( $LivreRow['IL'] ).'" width="240" height="300" id="'.$LivreRow['IDL'].'" /></button>';
echo"</td></div>";
}
</tr>
</table>
</fieldset>
</form>
</div>
</div><!-- /#ROW 1 -->
</div>
</div><!-- /#page-wrapper -->
A: I have this code which works which is dynamic slider takes dynamic images and inserts into slider.
<div class="item <?php if($i==1) echo "active"; ?>"> //specifies image active
This above code specifies that 1st image must be active.
<!DOCTYPE html>
<html>
<head>
<meta charset="utf-8">
<meta http-equiv="X-UA-Compatible" content="IE=edge,chrome=1">
<meta name="viewport" content="width=device-width, initial-scale=1">
<title>Dynamic Slider</title>
<link rel="stylesheet" href="http://maxcdn.bootstrapcdn.com/bootstrap/3.3.7/css/bootstrap.min.css">
<script src="https://ajax.googleapis.com/ajax/libs/jquery/1.12.4/jquery.min.js"></script>
<script src="http://maxcdn.bootstrapcdn.com/bootstrap/3.3.7/js/bootstrap.min.js"></script>
<link href="http://www.jqueryscript.net/css/jquerysctipttop.css" rel="stylesheet" type="text/css">
<link rel="stylesheet" href="http://netdna.bootstrapcdn.com/bootstrap/3.3.5/css/bootstrap.min.css">
<script src="https://use.fontawesome.com/0f773d63b5.js"></script>
<style>
/*SLIDER CSS*/
/* Main carousel style */
.carousel {
width: 100%;
}
/* Indicators list style */
.article-slide .carousel-indicators {
bottom: 0;
left: 0;
margin-left: 5px;
width: 100%;
}
/* Indicators list style */
.article-slide .carousel-indicators li {
border: medium none;
border-radius: 0;
float: left;
height: 54px;
margin-bottom: 5px;
margin-left: 0;
margin-right: 5px !important;
margin-top: 0;
width: 100px; /*here will be the size of */
}
/* Indicators images style */
.article-slide .carousel-indicators img {
border: 2px solid #FFFFFF;
float: left;
height: 54px;
left: 0;
width: 100px;
}
/* Indicators active image style */
.article-slide .carousel-indicators .active img {
border: 2px solid #428BCA;
opacity: 0.7;
}
</style>
<script type="text/javascript">
// SLIDER JQUERY
$('.carousel').carousel({
interval: false
});
</script>
</head>
<body>
<!--SECTION ABOUT PLACE-->
<section class="container-fluid">
<div class="carousel slide article-slide carousel_size " id="article-photo-carousel">
<!-- Wrapper for slides -->
<div class="carousel-inner">
<?php
//ESATBLISH CONNECTION
$host = "localhost";
$username_db = "USERNAME";
$password_db = "PASSWORD";
$databaseName = "DATABASE";
$link = new mysqli ($host, $username_db, $password_db, $databaseName);
//check connection
if($link->connect_error){
die ("connection failed : ".$link->connect_error);
}
//DISPLAY IMAGESIN SIDE SLIDER
$id = $_GET['profileId'];
$img_sql = "SELECT * FROM gallery WHERE profileId='$id' LIMIT 5";
$img_res =$link->query($img_sql);
if($img_res){
$i=0;
while($row=$img_res->fetch_assoc()){
$i++;
?>
<div class="item <?php if($i==1) echo "active"; ?>">
<img alt="" title="" src="admin/uploads/<?php echo $row['image'];?>">
</div>
<?php
}
}
?>
</div>
<!-- Left and right controls -->
<a class="left carousel-control" href="#article-photo-carousel" role="button" data-slide="prev">
<span class="glyphicon glyphicon-chevron-left" aria-hidden="true"></span>
<span class="sr-only">Previous</span>
</a>
<a class="right carousel-control" href="#article-photo-carousel" role="button" data-slide="next">
<span class="glyphicon glyphicon-chevron-right" aria-hidden="true"></span>
<span class="sr-only">Next</span>
</a>
</div>
</section>
</body>
</html>
| {
"redpajama_set_name": "RedPajamaStackExchange"
} | 5,277 |
The Nuclear Suppliers Group was formed in reaction to India's test of an atom bomb in 1974, which it had built using imported reactor technology.
Some group members were also considering to demand that exports of technology for uranium enrichment and spent fuel reprocessing should stay prohibited for India, as it can be used not only for fuelling power plants, but also for making bombs.
Currently, nuclear export rules prevent shipments to India, because the country has not signed the Nuclear Non-Proliferation Treaty.
Austria, Ireland, the Netherlands, New Zealand, Switzerland and Scandinavian states are among the countries that are most active in seeking conditions for allowing nuclear trade with India.
An Indian delegation led by Foreign Secretary Shiv Shankar Menon is also in Vienna to lobby Nuclear Suppliers Group members, but will not take part in the deliberations.
Student shot at Maryland school, suspect in custody | {
"redpajama_set_name": "RedPajamaCommonCrawl"
} | 2,922 |
\section{Introduction}
Nonlinear optical properties of graphene have attracted considerable interest in the community. The magnitude of the matrix element of the interaction Hamiltonian describing coupling of massless Dirac electrons to light scales in proportion to $ v_F/\omega \propto \lambda$, i.e.~it grows more rapidly with wavelength $\lambda$ than in conventional materials with parabolic energy dispersion, where it scales roughly as $\sqrt{\lambda}$. This promises a strong nonlinear response at long wavelengths. Unfortunately, graphene is also a centrosymmetric medium for low-energy in-plane excitations, which suppresses second-order nonlinear response in the electric dipole approximation. Therefore, most of the effort was concentrated on the third-order nonlinear processes that are electric dipole-allowed. Recent theoretical proposals and some experiments include third-harmonic generation \cite{kumar2013,hong2013}, four-wave mixing \cite{hendry2010,gu2012,sun2010} and current-induced second-harmonic generation \cite{glazov2014,bykov2012,cheng2014}. In few-layer graphene, second-harmonic generation (SHG) arising from the interactions between layers, which breaks the inversion symmetry, has been observed \cite{dean2009,dean2010}.
The aim of this paper is to show that monolayer graphene does demonstrate quite significant second-order nonlinearity at long wavelengths despite its inversion symmetry. Here and throughout the paper, we will discuss only the 2D (surface) nonlinearity due to in-plane motion of electrons. Like any surface, graphene exhibits anisotropy between in-plane and out-of-plane electron motion. However, the corresponding second-order nonlinearity is very small and we will not discuss it here.
We develop the full quantum-mechanical theory of the in-plane second-order nonlinear response beyond the electric dipole approximation. In this case one has to consider oblique or in-plane propagation of electromagnetic waves. A non-zero in-plane second-order susceptibility $\chi^{(2)}$ of monolayer graphene appears when one includes the dependence of $\chi^{(2)}$ on the in-plane photon wave vectors, i.e. the {\it spatial dispersion}. Physically, this means that the inversion symmetry of graphene is broken by the wave vector direction. The spatial dispersion in momentum space is of course equivalent to the nonlocal response in real space. Spatial dispersion effects turn out to be quite large because of a large magnitude of the electron velocity $v_F$. A non-zero value of the nonlocal $\chi^{(2)}$ has been pointed out before for second-harmonic generation \cite{mikhailovSHG,glazov2011,smirnova2014} (which only included intraband transitions in a free-carrier model), difference-frequency generation \cite{yao2014}, and parametric frequency down-conversion \cite{tokman2016}. The latter two papers developed a quantum theory including both intraband and interband transitions and applied it to the nonlinear generation of surface plasmons. In the recent experiment \cite{constant2016}, evidence for the difference-frequency generation of surface plasmons in graphene was reported. Here we provide a systematic derivation of the second-order nonlinear conductivity tensor, valid for all second-order processes, all frequencies and doping densities, as long as the massless Dirac fermion approximation for a single-particle Hamiltonian is applicable. For graphene, this means the range of frequencies from zero (more precisely, from inverse scattering time) to the near-infrared. Our approach can be applied to any system of massless chiral Dirac fermions, for example surface states in topological insulators such as Bi$_2$Se$_3$. The resulting nonlinear susceptibility tensor satisfies all symmetry and permutation properties, and predicts unusual polarization properties of the nonlinear signal. We also summarize main properties of the linear current as a necessary step in deriving the nonlinear response functions, and present a detailed discussion of its gauge properties and regularization.
\section{Basic equations}
Consider a 2D quantum system which in the absence of external fields can be described by the Dirac Hamiltonian
\begin{equation}
\label{Eq:Hamiltonian}
\hat{H}_0(\hat{\mathbold{p}}) = v_F \hat{\mathbold{\sigma}} \cdot \hat{\mathbold{p}} ,
\end{equation}
where $\hat{\mathbold{p}} = \mathbold{x}_0 \hat{p}_x + \mathbold{y}_0 \hat{p}_y$, $\hat{p}_{x,y} = - i\hbar \frac{\partial}{\partial x, \partial y}$, $\hat{\bm{\sigma}} = \mathbold{x}_0 \hat{\sigma}_x + \mathbold{y}_0 \hat{\sigma}_y$, where $\hat{\sigma}_{x,y}$ are Pauli matrices. The spinor eigenfunctions $\mathbold{\Psi} = \begin{pmatrix} \Psi_1 \\ \Psi_2 \end{pmatrix}$ of the Hamiltonian (\ref{Eq:Hamiltonian}) are
\begin{equation}
\label{Eq:eigen_state}
\mathbold{\Psi}_{\mathbold{k},s}(\mathbold{r}) \equiv \langle \mathbold{r} |\mathbold{k},s \rangle = \frac{e^{i\mathbold{k}\cdot\mathbold{r}}}{\sqrt{2 A}}
\begin{pmatrix}
s \\ e^{i\theta(\mathbold{k})}
\end{pmatrix} ,
\end{equation}
and the eigenenergies are $E = s \hbar v_F k$ where $s = \pm 1$ for conduction and valence bands, respectively; $\mathbold{k} = \mathbold{x}_0 k_x + \mathbold{y}_0 k_y$, $\theta(\mathbold{k})$ is the angle between the electron momentum $\mathbold{k}$ and $x$-axis, and $A$ is the normalization area. This description is valid for carriers in monolayer graphene up to the energies of order 1 eV. For higher energies, quadratic and trigonal warping corrections become non-negligible.
Consider the most general light-matter interaction Hamiltonian utilizing both vector and scalar potentials: $\mathbold{E} = -\nabla \varphi - c^{-1} \dot{\mathbold{A}}$ and $\mathbold{B} = \nabla \times \mathbold{A}$. Following a standard procedure \cite{landau2013quantum, gantmakher1987}, we replace $\hat{\mathbold{p}} \Rightarrow \hat{\mathbold{p}} + \frac{e}{c} \mathbold{A}$ in the unperturbed Hamiltonian $\hat{H}_0(\hat{\mathbold{p}})$ and add the potential energy operator $-e \varphi$ assuming a particle with the charge $-e$. This gives
\begin{equation}
\label{Eq:Hamiltonian_with_int}
\hat{H} = \hat{H}_0 + \hat{H}_{int}^{opt} ,
\hspace{1cm}
\hat{H}_{int}^{opt} = \frac{e v_F}{c} \hat{\mathbold{\sigma}} \cdot \mathbold{A} - e \varphi \cdot \hat{1} ,
\end{equation}
where $\hat{1}$ is a unit 2$\times$2 matrix. The Hamiltonian in Eq.~(\ref{Eq:Hamiltonian_with_int}) leads to the von Neumann equation for the density matrix:
\begin{align}
\label{Eq:density_matrix_eq}
i\hbar \frac{\partial}{\partial t} \rho_{mn}
=
(E_m - E_n) \rho_{mn}
+ \sum_{l} \left[ \left( \hat{H}_{int}^{opt} \right)_{ml} \rho_{ln} - \rho_{ml} \left( \hat{H}_{int}^{opt} \right)_{ln} \right] ,
\end{align}
where $|n\rangle = |\mathbold{k},s\rangle$.
We will consider a monochromatic electromagnetic field in plane of graphene,
\begin{equation}
\label{Eq:optical_field}
\mathbold{E} = \frac{1}{2} \left[ \mathbold{x}_0 E_x(\omega) + \mathbold{y}_0 E_y(\omega) \right] e^{-i\omega t + i q x} + \mathrm{C.C.}
\end{equation}
or its bichromatic combinations. The field component $\mathbold{z}_0 E_z$ can be ignored because neither this field component itself nor the magnetic field it generates can affect the 2D carrier motion. Furthermore, the component of the vector potential $\mathbold{z}_0 A_z$ which generates the z-component of the electric field $\mathbold{z}_0 E_z$ does not enter the Hamiltonian (\ref{Eq:Hamiltonian_with_int}) because $\hat{\mathbold{\sigma}} \cdot \mathbold{z}_0 = 0$. The field described by Eq.~(\ref{Eq:optical_field}) corresponds to the electromagnetic potentials
\begin{align}
\label{Eq:EM_potentials}
\varphi &= \frac{1}{2} \phi(\omega) e^{-i\omega t + i q x} + \mathrm{C.C.}, \nonumber \\
\mathbold{A} &= \frac{1}{2} \left[ \mathbold{x}_0 A_x(\omega) + \mathbold{y}_0 A_y(\omega) \right] e^{-i\omega t + i q x} + \mathrm{C.C.} .
\end{align}
Note that the P-polarized radiation can be defined through both the scalar potential,
\begin{equation}
\label{Eq:scalar_P}
\varphi = \frac{1}{2} \frac{i E_x(\omega)}{q} e^{-i\omega t + i q x} + \mathrm{C.C.} ,
\end{equation}
and the vector potential:
\begin{equation}
\label{Eq:vector_P}
\mathbold{A}_P = \frac{1}{2} \mathbold{x}_0 \frac{c E_x(\omega)}{i\omega} e^{-i\omega t + i q x} + \mathrm{C.C.}
\end{equation}
At the same time, the S-polarized radiation, can be defined only through the vector potential:
\begin{equation}
\label{Eq:vector_S}
\mathbold{A}_S = \frac{1}{2} \mathbold{y}_0 \frac{c E_y(\omega)}{i\omega} e^{-i\omega t + i q x} + \mathrm{C.C.}
\end{equation}
It is convenient to represent the surface current density generated in response to a harmonic field as a sum over spatial harmonics: $\mathbold{j}(\mathbold{r}) = 2^{-1} \sum_{\mathbold{q}} \mathbold{j}^{(q)} e^{i\mathbold{q}\cdot\mathbold{r}} + \mathrm{C.C.}$, where $2^{-1} \mathbold{j}^{(q)} = S^{-1} \int_S \mathbold{j}(\mathbold{r}) e^{-i\mathbold{q}\cdot\mathbold{r}} d^2\mathbold{r}$; the set of in-plane photon wave vectors $\mathbold{q}$ is specified by appropriate conditions on the boundary of a large area $S \gg A$. It is also convenient to choose the area $S$ to be a multiple of the normalization area $A$, so that
\begin{equation}
(2 A)^{-1/2} \int_A \mathbold{\Psi}_n^\ast(\mathbold{r}) \mathbold{\Psi}_m(\mathbold{r}) d^2 r = (2 S)^{-1/2} \int_S \mathbold{\Psi}_n^\ast(\mathbold{r}) \mathbold{\Psi}_m(\mathbold{r}) d^2 r .
\end{equation}
After calculating the matrix elements $\mathbold{j}_{nm}^{(q)}$ of the current density operator and solving independently the master equations (\ref{Eq:density_matrix_eq}), one can calculate the average amplitude of a given current density harmonic, which could be used as a source in Maxwell's equations or to determine the conductivity tensor:
\begin{equation}
\mathbold{j}^{(q)} = \sum_{mn} \mathbold{j}_{nm}^{(q)} \rho_{mn} .
\end{equation}
In order to evaluate $\mathbold{j}_{nm}^{(q)}$ we determine the velocity operator $\hat{\mathbold{v}} = i\hbar^{-1} \left[ \hat{H},\hat{\mathbold{r}} \right]$ and define the current density operator as $\hat{\mathbold{j}} = -e \hat{\mathbold{v}}$:
\begin{equation}
\hat{\mathbold{j}} = -e v_F \hat{\mathbold{\sigma}} .
\end{equation}
Next, we take into account a standard expression for the current density operator in a second-quantized formalism \cite{landau2013quantum}: $\hat{\mathbold{j}}(\mathbold{r}) = \hat{\mathbold{\Psi}}^\dagger \cdot \hat{\mathbold{j}} \cdot \hat{\mathbold{\Psi}}$, where $\hat{\mathbold{\Psi}} = \sum_n \hat{a}_n \mathbold{\Psi}_n(\mathbold{r})$ and $\hat{\mathbold{\Psi}}^\dagger = \sum_m \hat{a}_m^\dagger \mathbold{\Psi}_m^\dagger(\mathbold{r})$ are second-quantized operators, and $\hat{a}_m^\dagger$ and $\hat{a}_n$ are fermion creation and annihilation operators. Treating $\hat{a}_m^\dagger$ and $\hat{a}_n$ as Heisenberg operators and using $\mathbold{j}(\mathbold{r}) = \langle \hat{\mathbold{j}}(\mathbold{r}) \rangle$, $\langle \hat{a}_m^\dagger(t) \hat{a}_n(t) \rangle = \rho_{mn}(t)$, we arrive at $2^{-1} \mathbold{j}^{(q)} = \sum_{mn} \left(e^{-i\mathbold{q}\cdot\mathbold{r}} \hat{\mathbold{j}} \right)_{nm} \rho_{mn}$, which gives
\begin{equation}
2^{-1} \mathbold{j}_{nm}^{(q)} = \langle n | e^{-i\mathbold{q}\cdot\mathbold{r}} \hat{\mathbold{j}} | m \rangle .
\label{Eq:jq_matrix}
\end{equation}
To calculate the matrix elements $\mathbold{j}_{mn}^{(q)}$ and $\left( \hat{H}_{int}^{opt} \right)_{mn}$ we will need the following useful relationships:
\begin{align}
\left(e^{iqx}\right)_{mn} &= \frac{1}{2} \left(s_m s_n + e^{i(\theta_n-\theta_m)}\right) \delta_{\mathbold{k}_m,\mathbold{k}_n+\mathbold{q}} ,
\label{Eq:eiqx_matrix}
\\
\left(\hat{\mathbold{\sigma}} e^{i q x}\right)_{mn} &= \frac{1}{2} \left[ (\mathbold{x}_0 - i\mathbold{y}_0) s_m e^{i\theta_n} + (\mathbold{x}_0 + i\mathbold{y}_0) s_n e^{-i\theta_m} \right] \delta_{\mathbold{k}_m,\mathbold{k}_n+\mathbold{q}} .
\label{Eq:sigma_eiqx_matrix}
\end{align}
The above general equations should allow one to calculate the conductivity in any order with respect to the external optical field. There is however a complication related to the fact that the model described by the effective Hamiltonian Eq.~(\ref{Eq:Hamiltonian}) contains a ''bottomless'' valence band with electrons occupying all states to $k \rightarrow \infty$. Therefore, only the converging integrals make sense:
\begin{equation}
\label{Eq:sum_to_integral}
\sum_{mn} \mathbold{j}_{nm}^{(q)} \rho_{mn} \Rightarrow g \sum_{s s'} \underaccent{\infty'}{\int} \frac{d^2 k'}{4\pi^2} \underaccent{\infty}{\int} \frac{d^2 k}{4\pi^2} \mathbold{j}^{(q)}_{\mathbold{k}'\mathbold{k}s's} \rho_{\mathbold{k}\mathbold{k}'ss'} ,
\end{equation}
where $g$ is the degeneracy factor. Otherwise the optical response could be determined by the electron dispersion far from the Dirac point where the effective Hamiltonian Eq.~(\ref{Eq:Hamiltonian}) is no longer valid. It turns out that the convergence of the linear current depends on the choice of the gauge, whereas for the second-order nonlinear current the integral in Eq.~(\ref{Eq:sum_to_integral}) converges for any gauge. The divergence of the linear response can be regularized as discussed in the next section. In addition, the gauge dependence of the linear response violates gauge invariance, which is a consequence of the fact that the density matrix corresponding to the bottomless Hamiltonian in Eq.~(\ref{Eq:Hamiltonian}) has an infinite trace. In the next section we discuss this issue in more detail.
\section{The linear response of massless Dirac fermions \label{Sec:linear}}
The perturbation expansion of the nonlinear response functions implies that the second-order nonlinear terms depend on the first-order linear response. Therefore, in this section we outline the derivation of the linear current. The nontrivial aspect of this derivation is an apparent violation of gauge invariance and divergence of the linear current. We address these issues in this section and related Appendix sections.
The solution of the density matrix equation (\ref{Eq:density_matrix_eq}) in the linear approximation with respect to the field is
\begin{equation}
\label{Eq:rho_linear}
\rho^{(1)}_{nm}(\omega)_= \frac{1}{2} \frac{\left[ \hat{V}(\omega) e^{i q x} \right]_{nm} (n_m - n_n)}{\hbar\omega - (E_n-E_m)} ,
\end{equation}
where we defined $\hat{H}_{int}^{opt} = 2^{-1} \left[ \hat{V}(\omega) e^{-i\omega t + i q x} + \mathrm{H.C.} \right]$. \\
Here $\hat{V}(\omega) = -e\phi(\omega) \cdot \hat{1} + \frac{e v_F}{c} \left[ \hat{\sigma}_x A_x(\omega) + \hat{\sigma}_y A_y(\omega) \right]$ and $\rho^{(1)}_{nm}(\omega)$ is a complex-valued amplitude of the linear perturbation $\propto e^{-i\omega t}$ of the density matrix. For a monochromatic current $\mathbold{j} = 2^{-1} \mathbold{j}^{(q)}(\omega) e^{-i\omega t + i q x} + \mathrm{C.C.}$ we have
\begin{equation}
\label{Eq:jq_linear}
\mathbold{j}^{(q)}(\omega) = \sum_{mn} \mathbold{j}^{(q)}_{mn} \rho^{(1)}_{nm}(\omega) .
\end{equation}
The expression (\ref{Eq:jq_linear}) is evaluated in Appendix \ref{appendix:linear}. The most straightforward derivation is for a P-polarized field defined through a scalar potential, Eq.~(\ref{Eq:scalar_P}), since in this case the integral (\ref{Eq:sum_to_integral}) converges. If we keep only the terms of the lowest order in $q$ (i.e. the linear terms since $E_x = -i q \phi$), the resulting 2D (surface) conductivity tensor is independent of $q$. In the limit of strong degeneracy or low temperatures, the relevant terms are \\
({\romannumeral 1}) intraband conductivity, which has a Drude-like form:
\begin{equation}
\label{Eq:conductivity_linear_intra}
\sigma^{(intra)}_{xx}(\omega) = \frac{i e^2 v_F k_F}{\pi\hbar(\omega+i\gamma)} ,
\end{equation}
({\romannumeral 2}) and the interband term:
\begin{align}
\label{Eq:conductivity_linear_inter}
\sigma_{xx}^{(inter)}(\omega) = \frac{i e^2}{4\pi\hbar} \ln \left[ \frac{2 v_F k_F - (\omega + i\gamma)}{2 v_F k_F + (\omega + i\gamma)} \right] .
\end{align}
Here $k = k_F$ is Fermi momentum, and we also added the relaxation terms by replacing $\omega \rightarrow \omega + i\gamma$ in Eq.~(\ref{Eq:rho_linear}); in the limit $\gamma \rightarrow +0$ one can obtain from Eq.~(\ref{Eq:conductivity_linear_inter}) the well known result for the interband conductivity \cite{nair2008}: $\mathrm{Re} \sigma^{(inter)}_{xx} = \frac{e^2}{4\hbar} \Theta(\omega - 2 v_F k_F)$, where $\Theta(x)$ is the Heaviside step function.
If we define the optical field with a vector potential, the same calculation will lead to divergent integrals. In this case the finite, and at the same time gauge-invariant, expression for the linear current at frequency $\omega$ can be obtained by subtracting the same current evaluated at zero frequency \cite{falkovsky2007}:
\begin{equation}
\label{Eq:jq_Falkovsky}
\mathbold{j}^{(q)}(\omega) = \sum_{mn} \mathbold{j}^{(q)}_{mn} \left[ \rho^{1,A}_{nm}(\omega) - \rho^{1,A}_{nm}(\omega \rightarrow 0) \right] .
\end{equation}
Here $\rho^{1,A}_{nm}(\omega)$ is Eq.~(\ref{Eq:rho_linear}) with $\phi(\omega) = 0$ in the interaction Hamiltonian. This prescription cancels the divergent term and leads to the Kubo formula for the linear response. In our case Eq.~(\ref{Eq:jq_Falkovsky}) is equal to the sum of Eqs.~(\ref{Eq:conductivity_linear_intra}) and (\ref{Eq:conductivity_linear_inter}) for the diagonal conductivities $\sigma_{yy} = \sigma_{xx}$, and gives $\sigma_{xy} = 0$. The procedure in Eq.~(\ref{Eq:jq_Falkovsky}) can be justified by considering the graphene Hamiltonian with a small quadratic term in the energy dispersion:
\begin{equation}
\label{Eq:dispersion_with_quad}
E = s\hbar v_F k + \epsilon \frac{\hbar^2 k^2}{2} ,
\end{equation}
where $\epsilon$ is a small parameter. Adding this term provides a bottom to the valence band. As shown in Appendix \ref{appendix:current_define}, the linear current for such a system approaches Eq.~(\ref{Eq:jq_Falkovsky}) when $\epsilon \rightarrow 0$.
For a P-polarized field which can be represented through both scalar and vector potentials the renormalization procedure in Eq.~(\ref{Eq:jq_Falkovsky}) is equivalent to the gauge transformation of the density matrix from the A-gauge (\ref{Eq:vector_P}) to the $\varphi$-gauge (\ref{Eq:scalar_P}). Indeed, let the function $\rho^{1,A_P}_{nm}(\omega)$ correspond to the solution of Eq.~(\ref{Eq:rho_linear}) for the field defined in the gauge given by Eq.~(\ref{Eq:vector_P}), whereas the function $\rho^{1,\varphi}_{nm}(\omega)$ correspond to the gauge of Eq.~(\ref{Eq:scalar_P}). Since we just found that the sum $\sum_{mn} \mathbold{j}^{(q)}_{mn} \rho^{(1,\varphi)}_{nm}(\omega)$ is finite, it makes sense to try the transformation $\rho^{1,A_P}_{nm} \Rightarrow \rho^{1,\varphi}_{nm}$. The gauge transformation from $\mathbold{A}$ and $\varphi$ to $\tilde{\mathbold{A}}$ and $\tilde{\varphi}$ corresponds to the unitary transformation of the density matrix (see Appendix \ref{appendix:gauge_transform})
\begin{equation}
\label{Eq:rho_gauge_transformation}
\tilde{\rho}_{nm} = \sum_{qp} \left(e^{-\frac{ief}{\hbar c}}\right)_{nq} \rho_{qp} \left(e^{+\frac{ief}{\hbar c}}\right)_{pm} ,
\end{equation}
where the scalar function $f(t,\mathbold{r})$ determines the gauge transformation of the potentials
\begin{align}
\label{Eq:gauge_transformation}
\tilde{\mathbold{A}} = \mathbold{A} + \nabla f(t,\mathbold{r}),
\hspace{1cm}
\tilde{\varphi} = \varphi - \frac{1}{c} \frac{\partial f(t,\mathbold{r})}{\partial t}.
\end{align}
In particular, the transformation from the vector potential (\ref{Eq:vector_P}) to scalar potential (\ref{Eq:scalar_P}) is
\begin{equation}
\label{Eq:f_Ap2phip}
\nabla f = -\mathbold{A}_P .
\end{equation}
Within the linear approximation with respect to $f$ we obtain from Eq.~(\ref{Eq:rho_gauge_transformation}):
\begin{equation}
\label{Eq:rho1_transfrom_linear}
\rho^{1,A_P}_{nm} \Rightarrow \rho^{1,A_P}_{nm} - \frac{i e}{\hbar c} f_{nm} (n_m - n_n) .
\end{equation}
Next, we will use the general relationship (see e.g. \cite{tokman2009})
\begin{equation}
f_{nm} = \frac{-i\hbar}{E_n - E_m} \left( \frac{\nabla f \cdot \hat{\mathbold{v}} + \hat{\mathbold{v}} \cdot \nabla f}{2} \right)_{nm} ,
\end{equation}
from which we obtain from $\hat{\mathbold{v}} = v_F \hat{\mathbold{\sigma}}$ that
\begin{equation}
\label{Eq:fnm_graphene}
f_{nm} = \frac{-i\hbar v_F \left(\hat{\mathbold{\sigma}} \cdot \nabla f\right)_{nm}}{E_n-E_m} .
\end{equation}
As a result, from Eqs.~(\ref{Eq:rho1_transfrom_linear}), (\ref{Eq:fnm_graphene}) and (\ref{Eq:f_Ap2phip}) one gets
\begin{equation}
\label{Eq:rho1_transfrom_Ap2phip}
\rho^{1,A_P}_{nm}(\omega) \Rightarrow \rho^{1,A_P}_{nm}(\omega) + \frac{e v_F}{c} \frac{ \left[\hat{\sigma}_x A_x(\omega)\right]_{nm} (n_m - n_n)}{E_n - E_m} .
\end{equation}
Taking into account Eq.~(\ref{Eq:rho_linear}), Eq.~(\ref{Eq:rho1_transfrom_Ap2phip}) can be represented as $\rho^{1,A_P}_{nm}(\omega) \Rightarrow \rho^{1,A_P}_{nm}(\omega) - \rho^{1,A_P}_{nm}(\omega \rightarrow 0)$, which is identical to Eq.~(\ref{Eq:jq_Falkovsky}).
The structure of transformation (\ref{Eq:rho_gauge_transformation}) makes it clear why the density matrix with an infinite trace can give rise to the divergent current. Consider the density matrix in the form $\rho_{nm} = \rho_{mm}\delta_{nm} + \xi_{n \neq m}$, where $\xi$ is a small perturbation. The sum $\sum_{mn} \mathbold{j}_{mn} \xi_{nm} $ can converge in a certain gauge even if the trace $\sum_m \rho_{mm} $ diverges. However, the transformation (\ref{Eq:rho_gauge_transformation}) to a different gauge projects the diagonal of the matrix with an infinite trace onto off-diagonal elements, which can lead to the divergence in Eq.~(\ref{Eq:sum_to_integral}). The inverse is also true: the divergence can be eliminated by the transformation (\ref{Eq:rho_gauge_transformation}) as we have just shown above.
It is also clear that the separation of the response into intraband and interband components depends generally on the choice of the gauge since the transformation (\ref{Eq:rho_gauge_transformation}) mixes different contributions. At the same time, a correctly defined current has to be gauge-invariant.
\section{Second-order nonlinear response \label{Sec:2nd_order}}
Now we consider the second-order nonlinear response to the bichromatic field which we will represent through the vector potential in order to describe both P- and S-polarized fields with the same formalism. We will write the in-plane field components at frequencies $\omega_{1,2}$ directed along unit vectors $\bm{\eta}_{1,2}$ as
\begin{align}
\mathbold{A} = \frac{1}{2} \mathbold{\eta}_1 A(\omega_1) e^{i(\mathbold{q}_1\cdot\mathbold{r}_{\|} - \omega_1 t)} + \frac{1}{2} \mathbold{\eta}_2 A(\omega_2) e^{i(\mathbold{q}_2\cdot\mathbold{r}_{\|} - \omega_2 t)} + \mathrm{c.c.}
\end{align}
We need to calculate the perturbation of the density matrix at the sum frequency $\omega_1+\omega_2$. The term quadratic with respect to the field can be written as
\begin{widetext}
\begin{align}
&\phantom{{}={}}\rho_{mn}^{(2)}(\omega_1+\omega_2) = \left(\frac{e}{2c}\right)
\frac{1}{\hbar(\omega_1+\omega_2)-(\epsilon_m-\epsilon_n)} \nonumber \\
&\times \sum_{l\neq m,n} \left[ \left( (\hat{\mathbold{v}} \cdot \mathbold{\eta}_1) e^{i\mathbold{q}_1 \cdot \mathbold{r}} \right)_{ml} A(\omega_1)
\rho_{ln}^{(1)}(\omega_2) - \rho_{ml}^{(1)}(\omega_1) \left( (\hat{\mathbold{v}} \cdot \mathbold{\eta}_2) e^{i\mathbold{q}_2 \cdot \mathbold{r}} \right)_{ln} A(\omega_2) \right] + \left\{ 1 \leftrightarrow 2 \right\} \nonumber \\
&=
\frac{1}{2} \left(\frac{e}{c}\right)^2
\frac{ A(\omega_1) A(\omega_2) }{\hbar(\omega_1+\omega_2)-(\epsilon_m-\epsilon_n)}
\times \sum_{l\neq m,n} \left( (\hat{\mathbold{v}} \cdot \mathbold{\eta}_1) e^{i\mathbold{q}_1 \cdot \mathbold{r}} \right)_{ml}
\left( (\hat{\mathbold{v}} \cdot \mathbold{\eta}_2) e^{i\mathbold{q}_2 \cdot \mathbold{r}} \right)_{ln} \nonumber \\
& \times \left[
\frac{ (\rho_{nn}-\rho_{ll}) }{ \hbar\omega_2 - (\epsilon_l - \epsilon_n) }
- \frac{ (\rho_{ll}-\rho_{mm}) }{ \hbar\omega_1 - (\epsilon_m - \epsilon_l) }
\right]
+ \left\{ 1 \leftrightarrow 2 \right\} .
\end{align}
\end{widetext}
The trace of the corresponding Fourier harmonic of the induced current can be then calculated as
\begin{equation}
\label{Eq:jq_2nd}
\mathbold{j}^{(q_1+q_2)}(\omega_1+\omega_2) = - e \sum_{mn} \left(\hat{\mathbold{v}} e^{-i(\mathbold{q}_1+\mathbold{q}_2)\cdot\mathbold{r}}\right)_{nm} \rho_{mn}^{(2)}(\omega_1+\omega_2) .
\end{equation}
The second-order response at the difference frequency, $\rho^{(2)}_{mn}(\omega_1-\omega_2)$ can be obtained by replacing
\begin{equation}
\omega_2 \Rightarrow -\omega_2, \; \bm{q}_2 \Rightarrow -\bm{q}_2, \; A(\omega_2) \Rightarrow A^*(\omega_2) .
\end{equation}
Next, we transform from summation to integration over $\bm{k}$-states, introduce the corresponding occupation numbers $f(s,\bm{k})$ of the momentum states in each band, apply the momentum conservation in a three-wave mixing process, and take into account spin and valley degeneracy. Note that the integral over the electron momenta converges, as opposed to the linear response calculations where one needs to regularize the integral by either subtracting the contribution at zero frequency or adding a $k^2$ term to the Hamiltonian, as discussed above. The result is
\begin{align}
&\phantom{{}={}} \mathbold{J}^{(2)}(\omega_1+\omega_2) \nonumber \\
&= - \frac{e^3 v_F^3}{16 \pi^2 c^2 \hbar^2} A(\omega_1) A(\omega_2) \sum_{s_m,s_n,s_l} \int d^2\mathbold{k}
\frac{1}
{
(\omega_1+\omega_2)- v_F (s_m | \mathbold{k}+\mathbold{q}_1 | - s_n |\mathbold{k}-\mathbold{q}_2| )
} \nonumber \\
&\times \left[
\frac{ f(s_n,|\mathbold{k}-\mathbold{q}_2|) - f(s_l,|\mathbold{k}|) }
{\omega_2 - v_F (s_l |\mathbold{k}| - s_n |\mathbold{k}-\mathbold{q}_2|)}
-
\frac{ f(s_l,|\mathbold{k}|) - f(s_m,|\mathbold{k}+\mathbold{q}_1|) }
{ \omega_1 - v_F (s_m |\mathbold{k}+\mathbold{q}_1| - s_l |\mathbold{k}|) }
\right] \nonumber \\
&\times \left[ (\eta_{1x} - i\eta_{1y}) s_m e^{i\theta(\mathbold{k})} + (\eta_{1x} + i\eta_{1y}) s_l e^{-i\theta(\mathbold{k}+\mathbold{q}_1)} \right] \nonumber \\
&\times \left[ (\eta_{2x} - i\eta_{2y}) s_l e^{i\theta(\mathbold{k}-\mathbold{q}_2)} + (\eta_{2x} + i\eta_{2y}) s_n e^{-i\theta(\mathbold{k})} \right] \nonumber \\
&\times \left[ (\mathbold{x}_0 + i \mathbold{y}_0) s_m e^{-i\theta(\mathbold{k}-\mathbold{q}_2)} + (\mathbold{x}_0 - i \mathbold{y}_0) s_n e^{i\theta(\mathbold{k}+\mathbold{q}_1)} \right] \nonumber \\
&+ \left\{ 1 \leftrightarrow 2 \right\} .
\label{Eq:J_2ndorder_general}
\end{align}
This equation can be integrated numerically for any given geometry of incident fields and electron distribution. We consider the limit of the Fermi distribution with a strong degeneracy, direct all in-plane photon wave vectors along x-axis, and expand the integrand in Eq.~(\ref{Eq:J_2ndorder_general}) in powers of $q_1, q_2$. The integral over the term of zeroth-order in $q$ vanishes, as expected from symmetry. We will keep the terms linear in $q$. Also we have to evaluate separately the intraband contribution $s_l=s_m=s_n$ and all types of mixed interband-intraband contributions: $s_m = s_n = -s_l$, $s_m = s_l = -s_n$, and $s_n = s_l = -s_m$.
After performing this procedure, we find the following nonzero components of the second-order nonlinear conductivity tensor, while all other components are zero:
\begin{align}
&\phantom{{}={}}\sigma^{(2)}_{xxx}(\omega_1+\omega_2;\omega_1,\omega_2) \nonumber \\
&=
- s(\epsilon_F) \frac{e^3 v_F^2}{2 \pi \hbar^2} \frac{1}{\omega_1^2 \omega_2^2 (\omega_1+\omega_2)} \frac{1}{(\omega_1^2 - 4 v_F^2 k_F^2) (\omega_2^2 - 4 v_F^2 k_F^2) ((\omega_1 + \omega_2)^2 - 4 v_F^2 k_F^2 )} \nonumber \\
&\times \left[ - 4 v_F^4 k_F^4 (q_1 \omega_2^3 (2 \omega_1 + \omega_2) + q_2 \omega_1^3 (\omega_1 + 2 \omega_2) ) + 16 v_F^6 k_F^6 (q_1 \omega_2 (2 \omega_1 + \omega_2) + q_2 \omega_1 (\omega_1 + 2 \omega_2)) \right]
, \label{xxx} \\
&\phantom{{}={}}\sigma^{(2)}_{xyy}(\omega_1+\omega_2;\omega_1,\omega_2) \nonumber \\
&=
- s(\epsilon_F) \frac{e^3 v_F^2}{2 \pi \hbar^2} \frac{1}{\omega_1^2 \omega_2^2 (\omega_1+\omega_2)} \frac{1}
{(\omega_1^2 - 4 v_F^2 k_F^2) (\omega_2^2 - 4 v_F^2 k_F^2) ((\omega_1 + \omega_2)^2 - 4 v_F^2 k_F^2 )} \nonumber \\
&\times \left[ 4(v_F k_F)^2 \omega_1 \omega_2 (\omega_1+\omega_2)^2 (q_1 \omega_2^2 + q_2 \omega_1^2) \right. \nonumber \\
&\phantom{{}={}} + 4 (v_F k_F)^4 (q_1 \omega_2^4 - (6 q_1 + 4 q_2) \omega_1 \omega_2^3 - 8 (q_1+q_2) \omega_1^2 \omega_2^2 - (4 q_1 + 6 q_2) \omega_1^3 \omega_2 + q_2 \omega_1^4 ) \nonumber \\
&\phantom{{}={}} + \left. 16 (v_F k_F)^6 ( q_1 \omega_2 ( 2 \omega_1 - \omega_2) + q_2 \omega_1 ( 2 \omega_2 - \omega_1)) \right] , \label{xyy} \\
&\phantom{{}={}}\sigma^{(2)}_{yxy}(\omega_1+\omega_2;\omega_1,\omega_2) \nonumber \\
&=
- s(\epsilon_F) \frac{e^3 v_F^2}{2 \pi \hbar^2} \frac{1}{\omega_1^2 \omega_2^2 (\omega_1+\omega_2)} \frac{1}{(\omega_1^2 - 4 v_F^2 k_F^2) (\omega_2^2 - 4 v_F^2 k_F^2) ((\omega_1 + \omega_2)^2 - 4 v_F^2 k_F^2 )} \nonumber \\
&\times \left[ 4(v_F k_F)^2 \omega_1^2 \omega_2 (\omega_1+\omega_2)( q_1 \omega_2^2 - q_2 \omega_1 (\omega_1 + 2 \omega_2)) \right. \nonumber \\
&\phantom{{}={}} + 4 (v_F k_F)^4 (q_2 \omega_1 (\omega_1 + 2 \omega_2)^3 - q_1 \omega_2 (4 \omega_1^3 + 4 \omega_1^2 \omega_2 + 2 \omega_1 \omega_2^2 + 3 \omega_2^3)) \nonumber \\
&\phantom{{}={}} + \left. 16 (v_F k_F)^6 ( q_1 \omega_2 (2 \omega_1 + 3 \omega_2) - q_2 \omega_1 (\omega_1+2 \omega_2) ) \right] , \label{yxy} \\
&\phantom{{}={}}\sigma^{(2)}_{yyx}(\omega_1+\omega_2;\omega_1,\omega_2) \nonumber \\
&=
- s(\epsilon_F) \frac{e^3 v_F^2}{2 \pi \hbar^2} \frac{1}{\omega_1^2 \omega_2^2 (\omega_1+\omega_2)} \frac{1}{(\omega_1^2 - 4 v_F^2 k_F^2) (\omega_2^2 - 4 v_F^2 k_F^2) ((\omega_1 + \omega_2)^2 - 4 v_F^2 k_F^2 )} \nonumber \\
&\times \left[ 4(v_F k_F)^2 \omega_1 \omega_2^2 (\omega_1 + \omega_2) (q_2 \omega_1^2 - q_1 \omega_2 (2 \omega_1 + \omega_2)) \right. \nonumber \\
&\phantom{{}={}} + 4 (v_F k_F)^4 (q_1 \omega_2 (2 \omega_1 + \omega_2)^3 - q_2 \omega_1 (3 \omega_1^3 + 2 \omega_1^2 \omega_2 + 4 \omega_1 \omega_2^2 + 4 \omega_2^3)) \nonumber \\
&\phantom{{}={}} + \left. 16 (v_F k_F)^6 ( q_2 \omega_1 (3 \omega_1 + 2 \omega_2) - q_1 \omega_2 (2 \omega_1 + \omega_2) ) \right] . \label{yyx}
\end{align}
Here $s(\epsilon_F) = \pm 1$ depending on whether the Fermi level is in the conduction or valence band. A sketch of the second-order nonlinear process for an obliquely incident light of mixed polarization is shown in Fig.~1. Note that when both pump fields have either S- or P-polarization, the generated nonlinear current has only the $x$-component (along the in-plane direction of propagation of the pumps). When the polarizations are mixed, the $y$-component of the nonlinear current appears due to $yxy$ and $yyx$ components of the nonlinear conductivity (they are different only by permutation of indices 1 and 2 referring to the two pump fields).
\begin{figure}[htb]
\begin{center}
\includegraphics[scale=0.7]{fig1.pdf}
\caption{A sketch of the second order nonlinear current generation in the graphene plane for obliquely incident light. }
\label{Fig:2nd_nonlinear_illustration}
\end{center}
\end{figure}
Apparent ``non-reciprocity'' of the expressions for $\sigma^{(2)}_{yxx} = 0$ (P-in, S-out channel) and $\sigma^{(2)}_{xyy}$ (S-in, P-out channel) has a simple physical explanation: a P-polarized incident field cannot create a current orthogonal to the electric field, whereas an incident S-polarized field creates such a current via the magnetic field component $B_z$ normal to the layer.
The expressions for the nonlinear conductivity tensor that we obtained pass all symmetry and gauge invariance tests. Indeed, one can verify that the value of $\sigma^{(2)}_{xxx}$ agrees with the one derived using scalar potential in the interaction Hamiltonian. Furthermore, after converting the nonlinear conductivity to the nonlinear susceptibility according to
$$ \chi^{(2)}_{ijk}(\omega_1+\omega_2;\omega_1,\omega_2) = \frac{i \sigma_{ijk}^{(2)}(\omega_1+\omega_2;\omega_1,\omega_2)}{\omega_1 + \omega_2 },
$$
one can verify that all components of the nonlinear susceptibility tensor satisfy proper permutation relations; see e.g. Ch.~2.9 in \cite{keldysh}:
\begin{align}
\label{permut}
\chi^{(2)}_{ijk}(\omega_3=\omega_1+\omega_2) = \chi^{(2)}_{jik}(-\omega_1=-\omega_3+\omega_2) = \chi^{(2)}_{kji}(-\omega_2=-\omega_3+\omega_1),
\end{align}
where in-plane wave vectors have to be permuted together with frequencies.
The second-order response goes to zero when the Fermi energy $\epsilon_F$ goes to zero, and has maxima at resonances when one of
the three frequencies involved in three-wave mixing is close to $2 \epsilon_F/\hbar = 2 v_F k_F$. Far from these resonances and for high frequencies or low doping, $2 v_F k_F \ll \omega_1, \omega_2, \omega_1+\omega_2$, expressions for the nonlinear conductivity are greatly simplified (we will give only the expressions for $\sigma^{(2)}_{xxx}$ and $\sigma^{(2)}_{xyy}$ for brevity):
\begin{align}
& \sigma^{(2)}_{xxx} = s(\epsilon_F) \frac{2e^3 v_F^2}{\pi \hbar^2} \frac{v_F^4 k_F^4 \left[ q_1 \omega_2^3 (2\omega_1+ \omega_2) + q_2 \omega_1^3 (2\omega_2 + \omega_1) \right] }{\omega_1^4 \omega_2^4 (\omega_1 + \omega_2)^3} ,
\label{high1} \\
& \sigma^{(2)}_{xyy} = - s(\epsilon_F) \frac{2e^3 v_F^2}{\pi \hbar^2} \frac{v_F^2 k_F^2 \left( q_1 \omega_2^2 + q_2 \omega_1^2\right) }{\omega_1^3 \omega_2^3 (\omega_1 + \omega_2)} .
\label{high2}
\end{align}
An interesting and surprising result contained in these expressions is that the nonlinear frequency conversion of S-polarized
radiation into P-polarized radiation is much more efficient at high frequencies as compared to the P-in, P-out channel: $\sigma^{(2)}_{xyy}/\sigma^{(2)}_{xxx} \propto \frac{\omega^2}{v_F^2 k_F^2} \gg 1$. The dominance of the S-in, P-out channel is due to the chiral nature of Dirac fermion states. In particular, for the second-harmonic generation process $\omega_1 = \omega_2 = \omega$ and $q_1 = q_2 = q$, and the dominant component of the nonlinear conductivity tensor is simply
\begin{align}
& \sigma^{(2)}_{xyy} = - s \frac{2e^3}{\pi \hbar^2} \frac{v_F^4 k_F^2 q}{\omega^5}.
\label{shg}
\end{align}
In the opposite limit of low frequencies or high doping, $2 v_F k_F \gg \omega_1, \omega_2, \omega_1+\omega_2$, we also obtain simplified expressions:
\begin{align}
& \sigma^{(2)}_{xxx} = s \frac{e^3 v_F^2}{4 \pi \hbar^2 \omega_1 \omega_2} \left( \frac{q_1 + q_2}{\omega_1+ \omega_2} + \frac{q_1}{ \omega_1} + \frac{q_2}{ \omega_2} \right) ,
\label{low2-1} \\
& \sigma^{(2)}_{xyy} = - s \frac{e^3 v_F^2}{4 \pi \hbar^2 \omega_1 \omega_2} \left[ \frac{q_1 + q_2}{\omega_1+ \omega_2} + \frac{\omega_1 - \omega_2}{\omega_1 + \omega_2} \left( \frac{q_1}{ \omega_1} + \frac{q_2}{ \omega_2} \right) \right] .
\label{low2-2}
\end{align}
We verified that Eqs.~(\ref{low2-1}) and (\ref{low2-2}) can be derived independently from the single-band kinetic equation, i.e.~in the quasiclassical approximation described in Appendix D. This provides another test of our general expressions, since one should indeed expect that the single-band physics emerges in the limit of a strong doping and low frequencies, when all interband transitions become suppressed by Pauli blocking. Note that although Eqs.~(\ref{low2-1}) and (\ref{low2-2}) do not depend on $k_F$, they are valid only in the high-$k_F$ limit and are completely inapplicable for undoped graphene. In fact, exact expressions (\ref{xxx}) and (\ref{xyy}) give $\sigma^{(2)}_{xxx} = 0$ and $\sigma^{(2)}_{xyy} = 0$ for $k_F = 0$, since in this case the nonlinear currents due to interband and intraband transitions cancel each other. This can be viewed as a manifestation of the electron-hole symmetry in graphene.
The nonlinear conductivity components (\ref{xxx}) and (\ref{xyy}) diverge when one or more of
the three frequencies involved in three-wave mixing is close to $2 \epsilon_F/\hbar = 2 v_F k_F$. Close to resonance with $2 \epsilon_F/\hbar$ one has to include the imaginary part of the frequency which comes from the omitted relaxation term $-\gamma \rho_{mn}$ in the density-matrix equations. This amounts to substituting
$\omega_1 \rightarrow \omega_1 + i \gamma_1$, $\omega_2 \rightarrow \omega_2 + i \gamma_2$, $\omega_1 + \omega_2 \rightarrow \omega_1 + \omega_2 + i \gamma_3$. Note that if we flip the sign of $\omega_2$ to describe the difference frequency generation process, the sign of $+ i \gamma_2$ remains the same, i.e. $\omega_2 \rightarrow -\omega_2 + i\gamma_2$.
Even if dissipation is included, we can still use Eqs.~(\ref{permut}) to derive the components of the nonlinear susceptibility tensor from other components. In order to do that, one needs to use Eqs.~(\ref{permut}) in the absence of dissipation and then add imaginary parts of frequencies. Of course the resulting expressions after adding dissipation won't satisfy the permutation relation Eqs.~(\ref{permut}).
\begin{figure}[htb]
\begin{center}
\includegraphics[scale=0.7]{sigma2_SHG.pdf}
\caption{Nonzero components of the second order nonlinear conductivity tensor for the process of SHG as a function of the fundamental frequency. The pump is incident at 45 degrees. The Fermi energy is 200 meV and all resonances are broadened by the same factor $\gamma$ equal to 5 meV. }
\label{Fig:SHG}
\end{center}
\end{figure}
\begin{figure}[htb]
\begin{center}
\includegraphics[scale=0.7]{sigma2_DFG.pdf}
\caption{Nonzero components of the second order nonlinear conductivity tensor for the process of DFG as a function of one of the pump frequencies ($\omega_2$). Frequency $\omega_1$ is fixed at 400 meV. Both pumps are incident in the $(xz)$-plane at 45 degrees. The Fermi energy is 200 meV and all resonances are broadened by the same factor $\gamma$ equal to 5 meV. }
\label{Fig:DFG}
\end{center}
\end{figure}
\begin{figure}[htb]
\begin{center}
\includegraphics[scale=0.7]{sigma2_SFG.pdf}
\caption{Nonzero components of the second order nonlinear conductivity tensor for the process of SFG as a function of one of the pump frequencies ($\omega_2$). Frequency $\omega_1$ is fixed at 200 meV. Both pumps are incident in the $(xz)$-plane at 45 degrees. The Fermi energy is 200 meV and all resonances are broadened by the same factor $\gamma$ equal to 5 meV. }
\label{Fig:SFG}
\end{center}
\end{figure}
Figures 2-4 illustrate the above properties of the nonlinear conductivity for the processes of the second-harmonic generation (SHG), difference-frequency generation (DFG), and sum-frequency generation (SFG). We used SI units in the figures for easier comparison of the values with known materials. In Fig.~2, absolute values of nonzero components of the nonlinear conductivity tensor for the SHG process $\omega + \omega \Rightarrow 2 \omega$ are plotted as a function of the fundamental frequency $\omega$, assuming that the Fermi energy is 200 meV and all resonances are broadened by the same half-width factor $\gamma$ equal to 5 meV in energy units. The plots for $\sigma^{(2)}_{yxy}$ and $\sigma^{(2)}_{yyx}$ are identical as they should be. There are two prominent resonances at $\hbar \omega = 2 \epsilon_F = 400$ meV and $ 2\hbar \omega = 2 \epsilon_F$. At high frequencies, the $xxx$ component falls off much faster than the $xyy$ component. At low frequencies, both components diverge as $1/\omega^2$. Our treatment, however, becomes invalid in the low-frequency limit $\omega \leq \gamma$ when any of the frequencies becomes lower than the scattering rate; that is why the plots are truncated at $\omega = 20$ meV. The quasi-classical method of the kinetic equation has the same applicability limit.
Figure 3 shows absolute values of the nonzero components of the nonlinear conductivity tensor for the DFG process for the same values of $\epsilon_F$ and $\gamma$, as a function of $\omega_2$. The second frequency $\hbar \omega_1$ is fixed to be 400 meV. The same qualitative behavior is observed: there is a double resonance when both $\omega_1$ and $\omega_2$ are equal to $2 k_F v_F$. Note that there is no divergence at $\omega_1 - \omega_2 \rightarrow 0$ because the same factor $\omega_1 - \omega_2$ appears in the numerator. There is divergence when $\omega_2 \rightarrow 0$ which should be truncated at $\omega_2 \sim \gamma$.
In Fig.~4, the nonzero components of the nonlinear conductivity tensor for the SFG process are shown as a function of $\omega_2$. The second frequency $\hbar \omega_1$ is fixed at 200 meV. As expected, all components show strong resonances when one of the frequencies or their sum is equal to $2 \epsilon_F = 400$ meV.
The magnitude of the nonlinear response generally increases rapidly when one or more of the frequencies is decreased, as is obvious also from analytic expressions. For the DFG process, the magnitude of the nonlinear conductivity components is two orders of magnitude higher as compared to SHG or SFG. As one of the frequencies goes to zero, the treatment becomes invalid, but one could get an order of magnitude estimate of the maximum nonlinear conductivity by putting this frequency equal to $\gamma$. Using the same value of $\gamma = 5$ meV, one gets the nonlinear conductivity for DFG of the order of several m$^2$/(Vs) in the THz range. This is a 2D conductivity. Purely for the sake of comparison with known bulk nonlinear materials, we can convert it to the bulk nonlinear susceptibility dividing by the frequency and the monolayer thickness of 0.3 nm, to arrive at $|\chi^{(2)}_{3D}| \sim 10^{-3}$ m/V. This is a huge value as compared to 1-100 pm/V values for most materials. Of course, only the 2D values of the graphene conductivity or susceptibility enter all physical results such as the intensity of the generated nonlinear signal \cite{yao2014,tokman2016} or the parametric gain \cite{tokman2016}. Still, combination of intrinsically large nonlinear conductivity of graphene and a surface plasmon resonance for the nonlinear signal may lead to quite significant efficiency of the nonlinear processes, as emphasized in the theoretical proposals \cite{yao2014,tokman2016}.
In conclusion, we developed the full quantum-mechanical theory of the in-plane second-order nonlinear response of graphene beyond the electric dipole approximation. We provided a systematic derivation of the second-order nonlinear conductivity tensor, valid for all second-order processes, all frequencies and doping densities, as long as the massless Dirac fermion approximation for a single-particle Hamiltonian is applicable. Our approach can be applied to any system of massless chiral Dirac fermions, for example surface states in topological insulators such as Bi$_2$Se$_3$. We derived useful analytic expressions for the components of the nonlinear conductivity tensor, which satisfy all symmetry and permutation properties, and have a correct quasi-classical limit. We also summarized main features of the linear response, with emphasis on its gauge properties and regularization.
\begin{acknowledgments}
This work has been supported by the Air Force Office for Scientific Research through grants FA9550-15-1-0153 and FA9550-14-1-0376. M.T.~acknowledges support from the Russian Foundation for Basic Research Grant No.~14-22-02034 and thanks I.D.~Tokman for helpful discussions.
\end{acknowledgments}
| {
"redpajama_set_name": "RedPajamaArXiv"
} | 4,771 |
import RecordButton from './record';
export default class BetaFeaturesToggle {
constructor(menu, renderer) {
const features = [new RecordButton(renderer)];
this.elm = menu.menu.querySelector('.menu-beta-features input');
menu.addChangeListener((config) => {
if ('enableBetaFeatures' in config)
this.elm.checked = config.enableBetaFeatures;
else
config.enableBetaFeatures = this.elm.checked;
});
const onChange = (evt) => {
const enabled = this.elm.checked;
menu.submittedConfig.enableBetaFeatures = enabled;
menu.persist();
for (let i = 0; i < features.length; i++) {
const feature = features[i];
if (enabled)
feature.enable();
else
feature.disable();
}
};
this.elm.addEventListener('change', onChange);
// Give the app time to load the initial config
window.setTimeout(onChange, 0);
}
}
| {
"redpajama_set_name": "RedPajamaGithub"
} | 1,576 |
Q: No Hibernate Session bound to thread I'm using Struts + Hibernate + Spring for my project development.
And here is my Spring Context XML file. When I called "sessionFactory.getCurrentSession()" in the beginning of userDao.getXXXX method, the exception whose detail message is "No Hibernate Session bound to thread, and configuration does not allow creation of non-transactional one here" was thrown.
<!-- Hibernate Configuration -->
<bean id="sessionFactory"
class="org.springframework.orm.hibernate3.LocalSessionFactoryBean">
<property name="configLocation">
<value>classpath:hibernate.cfg.xml</value>
</property>
</bean>
<!-- Spring Transaction Manager -->
<bean id="transactionManager"
class="org.springframework.orm.hibernate3.HibernateTransactionManager">
<property name="sessionFactory">
<ref bean="sessionFactory"/>
</property>
</bean>
<!-- Spring Transaction Descriptions -->
<bean id="transactionAttributeSource"
class="org.springframework.transaction.interceptor.MethodMapTransactionAttributeSource">
<property name="methodMap">
<map>
<entry key="com.miaozhen.monitor.service.LoginServiceImpl.*">
<value>PROPAGATION_REQUIRED</value>
</entry>
</map>
</property>
</bean>
<bean id="transactionInterceptor"
class="org.springframework.transaction.interceptor.TransactionInterceptor">
<property name="transactionManager">
<ref bean="transactionManager"/>
</property>
<property name="transactionAttributeSource">
<ref bean="transactionAttributeSource"/>
</property>
</bean>
<bean id="transactionAdvisor"
class="org.springframework.transaction.interceptor.TransactionAttributeSourceAdvisor">
<constructor-arg>
<ref bean="transactionInterceptor"/>
</constructor-arg>
</bean>
<bean id="autoproxy"
class="org.springframework.aop.framework.autoproxy.DefaultAdvisorAutoProxyCreator">
</bean>
<!-- DAO -->
<bean id="userDao"
class="com.miaozhen.dbservice.hibernate.dao.AUserDAO">
<property name="sessionFactory">
<ref local="sessionFactory"/>
</property>
</bean>
<!-- Service Layer -->
<bean id="loginService"
class="com.miaozhen.monitor.service.LoginServiceImpl">
<property name="userDao">
<ref bean="userDao"/>
</property>
</bean>
<!-- Struts Actions for DelegatingActionProxy -->
<bean name="/login"
class="com.miaozhen.monitor.struts.action.LoginAction">
<property name="loginService">
<ref bean="loginService"/>
</property>
</bean>
A: That's quite a complicated configuration, but I suspect that the loginService bean is somehow not being proxied with the transactionAdvisor, even though I can see that's what you're trying to do.
Try making sure that the loginService that's being injected into the controller is in fact a generated proxy object rather than the raw LoginServiceImpl object. A debugger would also be very useful in making sure that the code execution passes through the TransactionInterceptor.
Is there a reason you're doing things this way? There are much easier ways to achieve the same thing, which don't involve creating advisors, autoproxy factories, transaction attribute sources, and so on. For example, using <tx:annotation-driven> and @Transactional make this stuff easy. Maybe it's because your current approach means you should have no references to Spring in your code is what appealed to you?
| {
"redpajama_set_name": "RedPajamaStackExchange"
} | 9,902 |
Situated in the heart of North Kolkata and close to the river Hooghly, Uddipa is a well-designed residential complex with functional spaces.
Launching across three phases, it will aspire to change the skyline of Sinthee More, North Kolkata.
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919 - 1,196 Sq. Ft.
637 - 846 Sq. Ft. | {
"redpajama_set_name": "RedPajamaC4"
} | 4,091 |
Meynau Pieters, död 19 mars 1612, var en nederländsk filantrop. Hon är känd som grundaren av barnhemmet Maagdenhuis i ett före detta kloster i Amsterdam 1584.
Referenser
kvinnor
Födda 1500-talet
Avlidna 1612
Personer i Nederländerna under 1500-talet
Personer i Nederländerna under 1600-talet
Nederländska filantroper | {
"redpajama_set_name": "RedPajamaWikipedia"
} | 2,232 |
In preparation for the 2015 IEEE Power & Energy Society's General Meeting in Denver, which took place July 26-30, Marianna Vaiman will present a tutorial on how to deal with the threat of grid reliability. As the executive vice president and co-founder of V&R Energy, Vaiman presented the following tutorial: "Understanding Cascading Phenomenon: Methodologies and Industry Practice for Analysis of Cascading Failures." Developed by the IEEE Cascading Failure Working Group, the tutorial offers an overview of the cascading phenomenon and explains methods on how to manage system failures. | {
"redpajama_set_name": "RedPajamaC4"
} | 1,684 |
El municipio de Rice (en inglés: Rice Township) es un municipio ubicado en el condado de Sandusky en el estado estadounidense de Ohio. En el año 2010 tenía una población de 1370 habitantes y una densidad poblacional de 22,78 personas por km².
Geografía
El municipio de Rice se encuentra ubicado en las coordenadas . Según la Oficina del Censo de los Estados Unidos, el municipio tiene una superficie total de 60.13 km², de la cual 55,94 km² corresponden a tierra firme y (6,97 %) 4,19 km² es agua.
Demografía
Según el censo de 2010, había 1370 personas residiendo en el municipio de Rice. La densidad de población era de 22,78 hab./km². De los 1370 habitantes, el municipio de Rice estaba compuesto por el 92,26 % blancos, el 1,97 % eran afroamericanos, el 0,36 % eran amerindios, el 0,29 % eran asiáticos, el 3,43 % eran de otras razas y el 1,68 % eran de una mezcla de razas. Del total de la población el 8,32 % eran hispanos o latinos de cualquier raza.
Referencias
Enlaces externos
Municipios de Ohio
Localidades del condado de Sandusky | {
"redpajama_set_name": "RedPajamaWikipedia"
} | 96 |
The Dodgers asked U.S. Bankruptcy Court Judge Kevin Gross to extend their exclusive window to propose a reorganization plan for the team until April 25, according to a court filing late Monday.
The request, if granted, would enable embattled owner Frank McCourt to control the Dodgers through the start of the 2012 season.
Major League Baseball has asked Gross to terminate the Dodgers' exclusivity so the league can propose a reorganization plan of its own — that is, to get the team sold this winter. Gross wrote last month that he intended "a prompt disposition of the key issues" so that the Dodgers could "utilize the approaching off season to prepare for the 2012 season."
The Dodgers said in their filing that they needed the extra time because they have been spurned by MLB and Fox Sports in efforts to hold discussions and build consensus on a reorganization plan.
"In its place, those parties have substituted no-holds-barred litigation," according to the filing.
Fox opposes the centerpiece of McCourt's reorganization strategy — an auction of the Dodgers' television rights that Fox says would violate its current broadcast contract — and the Dodgers say that even seven weeks of mediation with Commissioner Bud Selig proved fruitless.
"Unfortunately, the Commissioner has made clear that ... he will not agree to any plan of reorganization that does not involve a forced sale of the Dodgers," the team said in its filing.
The Dodgers called such extensions "routine" in significant bankruptcy cases. They said the case includes "thousands of creditors" and assets — the baseball team, Dodger Stadium and surrounding land — valued by McCourt and his advisors at "well in excess of $1 billion." | {
"redpajama_set_name": "RedPajamaC4"
} | 7,388 |
const apis = {
}
export default apis | {
"redpajama_set_name": "RedPajamaGithub"
} | 1,691 |
{"url":"https:\/\/stats.stackexchange.com\/questions\/43996\/kernel-logistic-regression-vs-svm\/44002","text":"# Kernel logistic regression vs SVM\n\nAs is known to all, SVM can use kernel method to project data points in higher spaces so that points can be separated by a linear space. But we can also use logistic regression to choose this boundary in the kernel space, so what's the advantages of SVM? Since SVM uses a sparse model in which only those support vectors make contributions when predicting, does this make SVM faster in prediction?\n\nKLRs and SVMs\n\n1. Classification performance is almost identical in both cases.\n2. KLR can provide class probabilities whereas SVM is a deterministic classifier.\n3. KLR has a natural extension to multi-class classification whereas in SVM, there are multiple ways to extend it to multi-class classification (and it is still an area of research whether there is a version which has provably superior qualities over the others).\n4. Surprisingly or unsurprisingly, KLR also has optimal margin properties that the SVMs enjoy (well in the limit at least)!\n\nLooking at the above it almost feels like kernel logistic regression is what you should be using. However, there are certain advantages that SVMs enjoy\n\n1. KLR is computationally more expensive than SVM - $O(N^3)$ vs $O(N^2k)$ where $k$ is the number of support vectors.\n2. The classifier in SVM is designed such that it is defined only in terms of the support vectors, whereas in KLR, the classifier is defined over all the points and not just the support vectors. This allows SVMs to enjoy some natural speed-ups (in terms of efficient code-writing) that is hard to achieve for KLR.\n\u2022 +1 I would just add though that if computational complexity is an issue, it isn't too difficult to construct a sparse kernel logistic regression model by greedily choosing the basis vectors to minimise the regularised loss on the training set, or other approaches. See the papers on the \"Informative Vector Machine\" for example. \u2013\u00a0Dikran Marsupial Nov 20 '12 at 9:55\n\u2022 Also, quite often if you optimise the kernel and regularisation parameters of an SVM you end up with a model where virtually all of the data are support vectors. The sparsity of SVMs is a happy accident, it isn't really a good selling point of the technique as it is generally possible to achieve greater sparsity by other means. \u2013\u00a0Dikran Marsupial Nov 20 '12 at 9:56\n\u2022 @DikranMarsupial Thanks for the pointer to Informative Vector Machine. I know of some works in Sparse KLR but so far I don't think any of them scale well for large datasets. Either way releasing a good implementation of sparse KLR which is user-friendly like libSVM or SVM Light can go a long way in its adoption. Apologies if such implementations already exists, however I am not aware of any.(EDIT: I think you meant \"Import vector machine\" instead of \"Informative vector machine\"?) \u2013\u00a0TenaliRaman Nov 20 '12 at 10:47\n\u2022 If you are ending up with all data points as support vectors, then you are over fitting. This happens with RBF many times. In fact, one of the fundamental thing I have learnt as a user of SVM is to first and foremost check the fraction of points chosen as support vectors. If it is anything more than 30% of the data, I outright reject that model. \u2013\u00a0TenaliRaman Nov 20 '12 at 10:48\n\u2022 It is not correct that all data points being SVs means over-fitting. If the value of C is small, then there is little penalty on the slack variables then you can have a very bland classifier (that makes many errors on the training set) and the margin is so wide that all the data are support vectors. Rejecting non-sparse models is not a good rule of thumb as sometimes the SVM with the best generalisation performance is non-sparse. The number of SVs is an upper bound on the leave-one-out error, but it is often a very lose bound indeed! \u2013\u00a0Dikran Marsupial Nov 20 '12 at 12:25\n\nHere's my take on the issue:\n\nSVMs are a very elegant way to do classification. There's some nice theory, some beautiful math, they generalize well, and they're not too slow either. Try to use them for regression though, and it gets messy.\n\n\u2022 Here's a resource on SVM regression. Notice the extra parameters to twiddle and the in depth discussion about optimization algorithms.\n\nGaussian Process Regression has a lot of the same kernelly math, and it works great for regression. Again, the very elegant, and it's not too slow. Try to use them for classification, and it starts feeling pretty kludgy.\n\n\u2022 Here's a chapter from the GP book on regression.\n\n\u2022 Here's a chapter on classification, for comparison. Notice that you end up with some complicated approximations or an iterative method.\n\nOne nice thing about using GPs for classification, though, is that it gives you a predictive distribution, rather than a simple yes\/no classification.\n\n\u2022 +1 GPs are a good alternative to KLR (although KLR often gives better performance because evidence based model selection can go wrong quite easily if there is model mis-specification) and cross-validation is often preferable. \u2013\u00a0Dikran Marsupial Nov 20 '12 at 9:58\n\nSome conclusions: The classi\ufb01cation performance is very similar. Has limiting optimal margin properties. Provides estimates of the class probabilities. Often these are more useful than the classi\ufb01cations. Generalizes naturally to M-class classi\ufb01cation through kernel multi-logit regression.","date":"2018-10-17 03:04:22","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\": 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\": 0.7226229310035706, \"perplexity\": 822.4292888817761}, \"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-2018-43\/segments\/1539583510969.31\/warc\/CC-MAIN-20181017023458-20181017044958-00258.warc.gz\"}"} | null | null |
package org.wso2.andes.transport.flow.control;
import org.apache.commons.logging.Log;
import org.apache.commons.logging.LogFactory;
import javax.management.*;
import java.lang.management.ManagementFactory;
import java.lang.management.MemoryPoolMXBean;
import java.lang.management.MemoryType;
import java.lang.management.MemoryUsage;
import java.util.List;
public final class MemoryMonitor implements MemoryMonitorMBean, NotificationBroadcaster {
private static Log log = LogFactory.getLog(MemoryMonitor.class);
private NotificationBroadcasterSupport broadcaster = new NotificationBroadcasterSupport();
private Thread task;
public MemoryMonitor(final double recoveryThreshold, final long memoryCheckInterval) {
super();
this.task =
new Thread(new MemoryMonitorTask(recoveryThreshold, broadcaster, memoryCheckInterval));
}
public void start() {
task.start();
}
public void stop() {
task.stop();
}
private static final class MemoryMonitorTask implements Runnable {
private List<MemoryPoolMXBean> pools;
private NotificationBroadcasterSupport broadcaster;
private int notificationSequence;
private double recoveryThreshold;
private long memoryCheckInterval = 20000L;
public MemoryMonitorTask(final double recoveryThreshold,
final NotificationBroadcasterSupport broadcaster,
final long memoryCheckInterval) {
this.pools = ManagementFactory.getMemoryPoolMXBeans();
this.broadcaster = broadcaster;
this.recoveryThreshold = recoveryThreshold;
this.memoryCheckInterval = memoryCheckInterval;
}
@Override
public void run() {
boolean memoryAvailable = false;
while (true) {
if (log.isDebugEnabled()) {
log.debug("Memory Monitor Is Running");
}
for (MemoryPoolMXBean poolMXBean : pools) {
if (MemoryType.HEAP.equals(poolMXBean.getType())) {
if (!poolMXBean.isUsageThresholdSupported()) {
continue;
}
MemoryUsage usage = poolMXBean.getUsage();
long thresholdMemory = (long) Math.floor(usage.getCommitted() * recoveryThreshold);
long availableMemory = (usage.getCommitted() - usage.getUsed());
if (log.isDebugEnabled()) {
log.debug("Maximum Amount Of Memory Available To Be Managed : " + usage.getMax() + " bytes");
log.debug("Memory Recovery Threshold : " + thresholdMemory + " bytes");
log.debug("Available Memory : " + availableMemory + " bytes");
}
if (availableMemory >= thresholdMemory) {
memoryAvailable = true;
break;
}
}
}
if (memoryAvailable) {
if (log.isDebugEnabled()) {
log.debug("Memory Recovery Notification Is Sent");
}
broadcaster.sendNotification(new Notification(
FlowControlConstants.FLOW_CONTROL_MEMORY_THRESHOLD_RECOVERED,
this, ++notificationSequence,
"Flow Control Memory Threshold Recovered"));
}
try {
Thread.sleep(memoryCheckInterval);
} catch (InterruptedException ignored) {
//Do nothing
}
memoryAvailable = false;
}
}
}
@Override
public void addNotificationListener(NotificationListener listener, NotificationFilter filter,
Object handback) throws IllegalArgumentException {
broadcaster.addNotificationListener(listener, filter, handback);
}
@Override
public void removeNotificationListener(
NotificationListener listener) throws ListenerNotFoundException {
broadcaster.removeNotificationListener(listener);
}
@Override
public MBeanNotificationInfo[] getNotificationInfo() {
return new MBeanNotificationInfo[] {
new MBeanNotificationInfo(
new String[]
{ FlowControlConstants.FLOW_CONTROL_MEMORY_THRESHOLD_RECOVERED },
Notification.class.getName(),
"Flow Control Memory Threshold Recovered"
)
};
}
}
| {
"redpajama_set_name": "RedPajamaGithub"
} | 765 |
{"url":"http:\/\/blogs.ams.org\/phdplus\/2012\/09\/11\/to-ph-d-or-not-to-ph-d\/","text":"# To Ph.D. or not to Ph.D.\n\nThe Fall semester, as many of us know, is a popular semester for writing letters of recommendation and advising seniors on their future. As a math professor in a small liberal arts school, I am aware that many (most?) of our majors are not going to go on to graduate school. Certainly, it is not true that math majors can only have careers in academia. There are many options for them and this fact has led to higher enrollments in the mathematical sciences (I think). But this doesn\u2019t mean that we should discourage students from pursuing academic careers, and in this post I will share some of my thoughts on when and how to advise students to pursue higher degrees in mathematics.\n\nI recently read a very good Slate article that I have now shared with a few of my students. The main point I take from this article, and the advice that I have given students in the past, is that if you really enjoy mathematics, thinking about mathematics, doing research in mathematics, then getting a Ph.D. is not a waste of time. In fact, you can only be better off afterwards. The article was in response to other recent articles mentioning how academic jobs are very tough to get and that maybe Ph.D.s in science were not worth the effort. But there are still employment options, essentially the same that you had when you finished college but now you are smarter and more mature to boot. Usually, and the Slate article mentions this, too, you get paid enough to get by, so even though you may not be getting rich you will not be more in debt by the end, either.\u00a0 One thing I make sure to tell my students (and this is something I told myself before accepting to go to Texas for a Ph.D.) is that it is also completely OK to change your mind. Deciding to pursue a Ph.D. is not a prison sentence, and if they realize that they do not enjoy math as much as they thought, or that they don\u2019t really want to spend so much time learning more mathematics, then there is no reason to stay.\n\nOf course, other things besides not enjoying learning math can happen, like not passing quals or prelims or whatever they\u2019re called these days. It is my opinion that people who truly want to stay in graduate school find a way to do so. I have also had friends who have left for a few years, come back, and are very successful. Other friends have switched from one institution to another that was a better fit. This reminds me of a talk Kathryn Leonard gave at MathFest this past August,\u00a0 \u201cI failed, and no one died\u201d. In it, she explains how we need to teach our students (and learn for ourselves) to distinguish between failure and Failure. A Failure is when an airplane pilot fails to land his airplane or a surgeon botches a procedure. These are bad places to fail at what you\u2019re doing. In mathematics, we are constantly faced with little-f-failure. We are working on a research problem that we can\u2019t prove, or we don\u2019t pass our qual, or we get a bad grade on our Real Analysis midterm. These are not huge problems, and no one is going to die. They are also indicative that you need to change something that you\u2019re doing. Either try a new approach to your research problem, figure out how to study for your qual, find a study group for your next midterm. These failures could also indicate other issues, like maybe your problem is much more difficult than you thought (maybe you need to assume GRH!) , and in the other two situations, maybe you realize that you don\u2019t care enough to struggle. This is important, because math is difficult, and if you don\u2019t enjoy it then it is very hard to get through some of these obstacles. This goes for many difficult occupations, by the way.\n\nJust like with \u201cfailure\u201d, \u201csuccess\u201d is also measured in different ways. I think of myself as being successful, because I got the job I wanted to get, which was teaching math and doing research at a good, small liberal arts school, but some people would say that unless you are at a top research institution you really have not made it as a mathematician. I have a friend who recently got his Ph.D. and went on to teach high school, because that is what he wanted to do. Some people were surprised because he was really good and they felt he should be pursuing a more \u201cprestigious\u201d path. But let me ask you who is more successful: the person who loves their job even though some people don\u2019t think too highly of it, or the person who got the job everyone covets but doesn\u2019t enjoy one moment of it? And this gets me again to the point that even though getting an academic job after getting a Ph.D. is becoming more difficult, this is not the only respectable or desirable job option. There are so many things one can do with a degree in mathematics! I always point unsure students (especially the ones that want to explain to their parents why they want to major in math and maybe pursue a higher degree) to this wonderful website by the BYU mathematics department. It is more geared towards undergraduates, but many of these career paths (probably all, actually) are still reasonable pursuits post-Ph.D.\n\nMy advice is probably biased because I enjoyed my own graduate school experience so much. This is why I do try to remind students of the possible obstacles and remind them to give themselves permission not only to fail, but to change their minds. I know of people that did not find the experience as rewarding as I did, and they tend to advise students not to go to graduate school (unless they are extremely gifted and confident). But I ask them to try to think, just like I\u2019m doing, about the possibility of a different experience for their students. At the end of the day, there is no way of knowing what will happen. I saw my fair share of superstars fizzle out during their first few years of graduate school, and some very quiet kids from not well-known schools blow everyone away and become superstars later on.\n\nI am aware that not everyone agrees with me on this topic, and in fact I have had many heated (but friendly) debates with colleagues about this issue. My point of view is perhaps overly optimistic, but hey, it\u2019s my point of view. That said, I do want to encourage everyone to post their own thoughts, experiences, and advice in the comments below. Maybe some of you with more experience (and who have advised graduate students, for example) have a different perspective. We would love to hear what you think.\n\n## 2 comments on \u201cTo Ph.D. or not to Ph.D.\u201d\n\n1. Why is a Ph.D necessary to succeed in research? A person can jump into research without a Ph.D. Why waste so many years of your life working on problems that are of interest to your supervisor and no interest to you? The same time can be spent on problem of your choice.\nThe only reason I see for a Ph.D is that it is easier to get a research\/academic job. Employers\/universities offering academic and research jobs are more interested in licking the Ph.D degree than employing a capable person.\n\nYou may use these HTML tags and attributes: <a href=\"\" title=\"\"> <abbr title=\"\"> <acronym title=\"\"> <b> <blockquote cite=\"\"> <cite> <code> <del datetime=\"\"> <em> <i> <q cite=\"\"> <strike> <strong>","date":"2013-05-19 08:36:06","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\": 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\": 0.33249756693840027, \"perplexity\": 626.2276461925844}, \"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\/1368696400149\/warc\/CC-MAIN-20130516092640-00062-ip-10-60-113-184.ec2.internal.warc.gz\"}"} | null | null |
{"url":"https:\/\/github.com\/pgf-tikz\/pgf\/issues\/689","text":"# Wrong definition of \\pgfutil@translate #689\n\nClosed\nopened this issue Jun 5, 2019 \u00b7 0 comments\n\n2 participants\n\n### rolfn commented Jun 5, 2019\n\n The file pgf\/tex\/generic\/pgf\/utilities\/pgfutil-latex.def contains the following definition: \\ifx\\translate\\@undefined % check if \\translate is available \\def\\pgfutil@translate#1{\\translate{#1}} \\else \\def\\pgfutil@translate#1{#1} \\fi This seems to be wrong. I assume the order should be revers. A related question: The file pgf\/tex\/generic\/pgf\/utilities\/pgfcalendar.code.tex contains a similar definition: \\ifx\\translate\\@undefined \\def\\translate#1{#1} \\fi Are both definitions really necessary? Thanks for your work. Rolf\n\n### hmenke added a commit that referenced this issue Jun 5, 2019\n\n Wrong order in definition of \\translate #689 \n d5aca9f","date":"2019-06-26 10:24:44","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\": 0.65940922498703, \"perplexity\": 11062.072760407433}, \"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-2019-26\/segments\/1560628000266.39\/warc\/CC-MAIN-20190626094111-20190626120111-00150.warc.gz\"}"} | null | null |
\section{Introduction} \label{sec:int}
\subsection{Background and motivation}
Given a sequence $(a_n)_{n \ge 1}$ of the rational integers $\Z$, a prime divisor of a term $a_n$ is called \textit{primitive} if it divides no earlier term.
The sequence is a \textit{divisibility sequence} if $a_m \mid a_n$ whenever $m \mid n$,
and it is a \textit{strong divisibility sequence} if $\gcd(a_m, a_n) = a_d$ with $d = \gcd(m,n)$ for any positive integers $m, n$.
These notions apply to any sequence of a unique factorization domain.
It is a classical and still very active topic in number theory to study primitive prime divisors of an integer sequence.
The classical Zsigmondy theorem \cite{Zsig} in 1892, extending earlier work of Bang \cite{Bang} in the case $b=1$, says that every term beyond the sixth in the sequence
$(a^n - b^n)_{n \ge 1}$ has a primitive prime divisor, where $a,b$ are positive coprime integers.
This theorem was independently rediscovered by Birkhoff and Vandiver \cite{BV}.
Results of this form are often useful in group theory and in the theory of recurrence sequences (see \cite[Section 6.3]{EPSW} for a discussion and references).
In 1913, Carmichael \cite{Car} showed that each term of the Lucas sequence $((a^n - b^n)/(a-b))_{n \ge 1}$ beyond the twelfth has a primitive prime divisor,
where $a, b$ are real algebraic integers such that $a/b$ is not a root of unity, and $a+b$ and $ab$ are coprime integers in $\Z$.
In 1955, Ward \cite{Ward} obtained a similar result for the Lehmer sequence $(s_n)_{n \ge 1}$ with $s_n = (a^n - b^n)/(a-b)$ for odd $n$
and $s_n = (a^n - b^n)/(a^2-b^2)$ for even $n$, where $a,b$ are real, and $(a+b)^2$ and $ab$ are coprime integers in $\Z$.
All these results, including Zsigmondy's theorem, were extended to any number field (that is, $a, b$ do not need to be real) by Schinzel \cite{Sch}
in an effective but not explicit manner (see \cite{PS} for an earlier work),
which was first made explicitly by Stewart \cite{Stewart}.
Furthermore, in 2001, Bilu, Hanrot and Voutier \cite{BHV} listed all the Lucas and Lehmer numbers without primitive prime divisor.
So far, the above classical results have various extensions in different settings.
For example, the extensions to elliptic divisibility sequence \cite{EMW,Sil}, to dynamical sequences \cite{IS,Rice},
to function fields defined over number fields \cite{IMSSS}, to Drinfeld modules \cite{Bam,Quan,ZJ}.
Recently, Flatters and Ward \cite{FW} found an analogue of Zsigmondy's theorem for a polynomial sequence $(f^n - g^n)_{n \ge 1}$,
where $f,g$ are two coprime polynomials in a polynomial ring $K[X]$ ($K$ is a field).
In this paper, we want to establish analogues of Zsigmondy's theorem and the primitive divisor results for the Lucas and Lehmer sequences
in polynomial rings of several variables.
The approach is essentially the same as in \cite{FW}.
It in fact follows the classical one with some modifications needed to avoid terms in the sequence where the Frobenius automorphism precludes primitive divisors.
However, for analogues of polynomial Lucas and Lehmer sequences, it indeed needs some more considerations.
Throughout the paper, let $K$ be a field, and $R = K[X_1, \ldots, X_r]$ the ring of polynomials in varibales $X_1, \ldots, X_r$.
Let $p$ be the characteristic of $K$. Note that $R$ is a unique factorization domain.
Besides, a \textit{prime divisor} of a polynomial $h$ in $R$ means a monic irreducible polynomial in $R$ dividing $h$.
We state the main results in the rest of Section~\ref{sec:int}, and then prove them later on.
\subsection{Main results}
Let $\lambda, \eta$ be non-zero algebraic elements over $R$ such that $\lambda / \eta$ is not a root of unity.
Assume that $(\lambda + \eta)^2$ and $\lambda\eta$ are non-zero coprime polynomials in $R$ which are not both in $K$.
Define the \textit{Lehmer sequence} of $R$:
\begin{equation*}
U_n =
\left\{\begin{array}{ll}
\frac{\lambda^n - \eta^n}{\lambda - \eta} & \textrm{if $n$ is odd,}\\
\\
\frac{\lambda^n - \eta^n}{\lambda^2 - \eta^2} & \textrm{if $n$ is even.}\end{array}\right.
\end{equation*}
We remark that the Lehmer sequence $(U_n)_{n \ge 1}$ satisfies the following recurrence relation over $R$:
$$
U_{n+4} = (\lambda^2 + \eta^2) U_{n+2} - \lambda^2\eta^2 U_n, \quad n = 1, 2, \ldots.
$$
The following two theorems are about the strong divisibility property and the primitive prime divisors of the sequence $(U_n)_{n \ge 1}$, respectively.
\begin{theorem} \label{thm:strong3}
The sequence $(U_n)_{n \ge 1}$ is a strong divisibility sequence.
\end{theorem}
\begin{theorem} \label{thm:primitive3}
Suppose the characteristic $p > 0$ and let $U^\prime$ be the sequence obtained from $(U_n)_{n \ge 1}$ by deleting the terms $U_n$ with $p \mid n$,
then each term of $U^\prime$ beyond the second has a primitive prime divisor.
If $p = 0$, then each term of $(U_n)_{n \ge 1}$ beyond the second has a primitive prime divisor.
\end{theorem}
Applying Theorems~\ref{thm:strong3} and \ref{thm:primitive3},
we can obtain the strong divisibility property and the primitive divisor result for polynomial Lucas sequences.
Let $\alpha, \beta$ be non-zero algebraic elements over $R$ such that the quotient $\alpha/\beta$ is not a root of unity.
Assume that $\alpha + \beta$ and $\alpha \beta$ are coprime polynomials in $R$ which are not both in $K$.
Define the \textit{Lucas sequence} of $R$:
$$
L_n = \frac{\alpha^n - \beta^n}{\alpha - \beta}, \quad n =1,2, \ldots.
$$
We remark that the Lucas sequence $(L_n)_{n \ge 1}$ satisfies the following recurrence relation over $R$:
$$
L_{n+2} = (\alpha + \beta) L_{n+1} - \alpha\beta L_n, \quad n = 1, 2, \ldots.
$$
\begin{theorem} \label{thm:strong2}
The sequence $(L_n)_{n \ge 1}$ is a strong divisibility sequence.
\end{theorem}
\begin{theorem} \label{thm:primitive2}
Suppose the characteristic $p > 0$ and let $L^\prime$ be the sequence obtained from $(L_n)_{n \ge 1}$ by deleting the terms $L_n$ with $p \mid n$,
then each term of $L^\prime$ beyond the second has a primitive prime divisor.
If $p = 0$, then each term of $(L_n)_{n \ge 1}$ beyond the second has a primitive prime divisor.
\end{theorem}
Theorem~\ref{thm:primitive2} in fact implies an analogue of Zsigmondy's theorem in $R$.
Let $f, g$ be non-zero coprime polynomials in $R$ such that $f$ and $g$ are not both in $K$ and the quotient $f/g$ is not a root of unity.
Define the sequence of $R$:
$$
F_n = f^n - g^n, \quad n =1,2, \ldots.
$$
\begin{theorem} \label{thm:strong1}
The sequence $(F_n)_{n \ge 1}$ is a strong divisibility sequence.
\end{theorem}
\begin{theorem} \label{thm:primitive1}
Suppose the characteristic $p > 0$ and let $F^\prime$ be the sequence obtained from $(F_n)_{n \ge 1}$ by deleting the terms $F_n$ with $p \mid n$,
then each term of $F^\prime$ beyond the second has a primitive prime divisor.
If $p = 0$, then each term of $(F_n)_{n \ge 1}$ beyond the second has a primitive prime divisor.
\end{theorem}
For $f, g$ as the above, define the sequence:
$$
S_n = f^n + g^n, \quad n = 1,2, \ldots.
$$
If $p \ne 2$, then $(S_n)_{n \ge 1} \ne (F_n)_{n \ge 1}$.
Note that $F_{2n} = F_n S_n$, which implies that each primitive prime divisor of $F_{2n}$ comes from $S_n$.
Then, the following corollary is a direct consequence of Theorem~\ref{thm:primitive1}.
\begin{corollary} \label{cor:primitive1}
Suppose $p > 0$, and let $S^\prime$ be the sequence obtained from $(S_n)_{n \ge 1}$ by deleting the terms $S_n$ with $p \mid n$,
then each term of $S^\prime$ beyond the second has a primitive prime divisor.
If $p = 0$, then each term of $(S_n)_{n \ge 1}$ beyond the second has a primitive prime divisor.
\end{corollary}
\section{Preliminaries}
Although the sequences we consider are defined over $R$, we prefer to establish some results in a more general setting.
Throughout this section, let $D$ be a unique factorization domain.
Recall that the resultant of two homogeneous polynomials in variables $X$ and $Y$ is defined to the determinant of their Sylvester matrix.
Some basic properties of this resultant are listed in the following lemma; see \cite[Proposition 2.3]{Silverman}.
\begin{lemma} \label{lem:Res}
For two non-constant homogeneous polynomials defined over a field
$$
A(X,Y) = a_0 X^m + a_1 X^{m-1}Y + \ldots + a_m Y^m = a_0 \prod_{i=1}^{m} (X - \alpha_i Y)
$$
and
$$
B(X,Y) = b_0 X^n + b_1 X^{n-1}Y + \ldots + b_n Y^n = b_0 \prod_{j=1}^{n} (X - \beta_j Y),
$$
their resultant is
$$
\Res(A,B) = a_0^n b_0^m \prod_{i=1}^{m} \prod_{j=1}^{n} (\alpha_i - \beta_j) \in \Z[a_0, \ldots, a_m, b_0, \ldots, b_n].
$$
Moreover, there exist $G_1, H_1, G_2, H_2 \in \Z[a_0, \ldots, a_m, b_0, \ldots, b_n][X,Y]$ homogeneous in $X$ and $Y$ such that
\begin{align*}
& G_1A + H_1 B = \Res(A,B)X^{m+n-1}, \\
& G_2A + H_2 B = \Res(A,B)Y^{m+n-1}.
\end{align*}
\end{lemma}
For any integer $n \ge 1$, the $n$-th homogeneous cyclotomic polynomial is defined by
$$
\Phi_n (X,Y) = \prod_{k=1, \, \gcd(k,n)=1}^{n} (X - \zeta_n^k Y) \in \Z[X,Y],
$$
where $\zeta_n$ is a primitive $n$-th root of unity, and we also define the polynomial
$$
P_n(X,Y) = \frac{X^n - Y^n}{X-Y} = \sum_{k=0}^{n-1} X^{n-1-k}Y^k = \prod_{k=1}^{n-1} (X - \zeta_n^k Y).
$$
Then, it is easy to see that
$$
X^n - Y^n = \prod_{d \mid n} \Phi_d (X,Y), \qquad P_n (X,Y) = \prod_{d \mid n, \, d \ge 2} \Phi_d (X,Y).
$$
The following result is \cite[Lemma 2.4]{FW} about the resultant of $P_m(X,Y)$ and $P_n(X,Y)$.
\begin{lemma} \label{lem:Res2}
For any positive coprime integers $m$ and $n$, we have $\Res(P_m, P_n) = \pm 1$.
\end{lemma}
We now want to establish some results about coprime elements in $D$.
First, we prove a general result.
\begin{lemma} \label{lem:coprime}
Let $a, b$ be algebraic elements over $D$.
Assume that $(a+b)^2$ and $ab$ are coprime elements in $D$.
Let $A(X,Y), B(X,Y) \in \Z[X,Y]$ be non-constant homogeneous polynomials with resultant $\Res(A,B) = \pm 1$.
Assume that both $A(a,b)$ and $B(a,b)$ are in $D$.
Then, $A(a,b)$ and $B(a,b)$ are coprime in $D$.
\end{lemma}
\begin{proof}
Let $m = \deg A$ and $n = \deg B$.
By assumption, $a^2 + b^2$ and $ab$ are also coprime in $D$.
Note that for any integer $k \ge 1$, $a^{2k} + b^{2k}$ is in $D$.
Using Lemma~\ref{lem:Res} and noticing $\Res(A,B) = \pm 1$, we obtain that there exist $u_1, w_1, u_2, w_2 \in \Z[a, b]$ such that
$$
u_1 A(a,b) + w_1B(a,b) = a^{2(m+n-1)} + b^{2(m+n-1)} \in D
$$
and
$$
u_2 A(a,b) + w_2 B(a,b) = a^{2(m+n-1)}b^{2(m+n-1)} \in D.
$$
Note that $u_1, w_1, u_2, w_2$ might be not in $D$.
By contradiction, suppose that $A(a,b)$ and $B(a,b)$ are not coprime in $D$.
Then, there is a prime element $\pi \in D$ such that $\pi \mid A(a,b)$ and $\pi \mid B(a,b)$ in $D$.
By the above discussion, we obtain that both
$$
\frac{u_1 A(a,b) + w_1 B(a,b)}{\pi}, \quad \frac{u_2 A(a,b) + w_2 B(a,b)}{\pi}
$$
are in the fraction field of $D$ and integral over $D$
(because $a,b$ are integral over $D$ satisfying the equation $X^4 - (a^2 + b^2)X^2 + a^2b^2 = 0$).
Note that $D$ is an integrally closed domain, so these two quotients are both in $D$.
Hence, we have $\pi \mid a^{2(m+n-1)} + b^{2(m+n-1)}$ and $\pi \mid a^{2(m+n-1)}b^{2(m+n-1)}$ in $D$.
So, $\pi \mid ab$ in $D$.
Let $k = m+n-1$. We have known that $\pi \mid a^{2k} + b^{2k}$ and $\pi \mid ab$ in $D$.
Consider
\begin{align*}
(a^2 + b^2)^k & = a^{2k} + b^{2k} + \sum_{i=1}^{k-1}\binom{k}{i}(a^2)^i (b^2)^{k-i} \\
& = a^{2k} + b^{2k} + a^2b^2 \sum_{i=1}^{k-1}\binom{k}{i}(a^2)^{i-1} (b^2)^{k-1-i}.
\end{align*}
Note that $\sum_{i=1}^{k-1}\binom{k}{i}(a^2)^{i-1} (b^2)^{k-1-i}$ is also in $D$, because it is symmetric in $a^2$ and $b^2$.
Hence, we have $\pi \mid (a^2 + b^2)^{k}$, and then $\pi \mid a^2 + b^2$, so $\pi \mid (a + b)^2$ in $D$ (because $\pi \mid ab$).
This leads to a contradiction with the assumption that $(a + b)^2$ and $ab$ are coprime in $D$.
Therefore, $A(a,b)$ and $B(a,b)$ are coprime in $D$.
\end{proof}
Based on Lemma~\ref{lem:coprime}, we can derive several results about coprime elements in $R$ in the sequel.
\begin{lemma} \label{lem:PmPn2}
Let $a, b$ be two algebraic elements over $D$.
Assume that $a+b$ and $ab$ are coprime elements in $D$.
Then, for any positive coprime integers $m, n$, $P_m(a,b)$ and $P_n(a,b)$ are coprime in $D$.
\end{lemma}
\begin{proof}
Without loss of generality, we can assume $m \ge 2, n \ge 2$.
Clearly, both $P_m(a,b)$ and $P_n(a,b)$ are in $D$. Because both $P_m(X,Y)$ and $P_n(X,Y)$ are symmetric in $X$ and $Y$.
By assumption, $(a+b)^2$ and $ab$ are coprime elements in $D$.
Then, the desired result follows from Lemmas~\ref{lem:Res2} and \ref{lem:coprime}.
\end{proof}
\begin{lemma} \label{lem:PmPn-odd}
Let $a, b$ be defined as in Lemma~\ref{lem:coprime}.
Let $m,n$ be two positive coprime integers such that both $m$ and $n$ are odd.
Then, $P_m(a,b)$ and $P_n(a,b)$ are coprime in $D$.
\end{lemma}
\begin{proof}
Without loss of generality, we can assume $m \ge 3, n \ge 3$.
Since both $(a+b)^2$ and $ab$ are in $D$, we have $a^2 + b^2 \in D$.
Moreover, we have that for any integer $k \ge 1$, $a^{2k} + b^{2k} \in D$.
Note that $P_m(X,Y)$ is homogeneous of even degree $m-1$ and symmetric in $X$ and $Y$.
So, if $X^iY^j$ is a term in $P_m(X,Y)$, then $X^jY^i$ is also a term in $P_m(X,Y)$, and then assuming $i \le j$, we have
$$
a^i b^j + a^j b^i = (ab)^i (a^{j-i} + b^{j-i}) \in D,
$$
where we use the fact that $j - i$ is even (because $i + j = m-1$ is even).
Hence, we have that $P_m(a,b)$ is in $D$.
Similarly, $P_n(a,b)$ is also in $D$.
Now, the desired result follows directly from Lemmas~\ref{lem:Res2} and \ref{lem:coprime}.
\end{proof}
\begin{lemma} \label{lem:PmPn-mix}
Let $a, b$ be defined as in Lemma~\ref{lem:coprime}.
Let $m,n$ be two positive coprime integers such that $m$ is odd and $n$ is even.
Then, $P_m(a,b)$ and $P_n(a,b)/(a + b)$ are coprime in $D$.
\end{lemma}
\begin{proof}
Without loss of generality, we can assume $m \ge 3, n \ge 4$ (because $P_2(a,b)/(a + b)=1$).
Since $m$ is odd, as in the proof of Lemma~\ref{lem:PmPn-odd} $P_m(a,b)$ is in $D$.
For any odd integer $k \ge 1$, note that
$$
\frac{a^k + b^k}{a+b} = a^{k-1} - a^{k-2}b + \ldots - ab^{k-2} + b^{k-1}
$$
is homogeneous of even degree $k-1$ and is symmetric in $a$ and $b$, so it is in $D$.
Hence, for even $n$, since
\begin{align*}
\frac{P_n(a,b)}{a + b} = \frac{a^{n-1} + b^{n-1}}{a+b} + ab \cdot \frac{a^{n-3} + b^{n-3}}{a+b} + \ldots
+ a^{\frac{n-2}{2}}b^{\frac{n-2}{2}} \cdot \frac{a + b}{a+b},
\end{align*}
we have that $P_n(a,b)/(a + b)$ is in $D$.
Denote $T_n(X,Y) = P_n(X,Y)/(X+Y)$, which can be viewed as a polynomial over $\Z$.
Using Lemma~\ref{lem:Res} and applying the same arguments as in the proof of \cite[Lemma 2.4]{FW},
we obtain that the resultant of $P_m(X,Y)$ and $T_n(X,Y)$ is equal to $\pm 1$.
Hence, by Lemma~\ref{lem:coprime}, $P_m(a,b)$ and $T_n(a,b)$ are coprime in $D$.
\end{proof}
\begin{lemma} \label{lem:Pmn}
Let $a, b$ be defined as in Lemma~\ref{lem:coprime}.
Let $m, n$ be two positive integers such that both $m$ and $n$ are odd.
Then, $P_m(a^n,b^n)$ and $(a^n + b^n)/(a + b)$ are coprime in $D$.
\end{lemma}
\begin{proof}
Without loss of generality, we can assume $m \ge 3, n \ge 3$.
As before, since $m$ and $n$ are odd, both $P_m(a^n,b^n)$ and $(a^n + b^n)/(a + b)$ are indeed in $D$.
Define
$$
V_m(X,Y) =P_m(X^n, Y^n), \qquad W_n(X,Y) = \frac{X^n + Y^n}{X+Y}.
$$
Both $V_m$ and $W_n$ can be viewed as polynomials over $\Z$.
So, we first compute their resultant over $\Z$.
Note that
$$
V_m(X,Y) = \prod_{i=1}^{m-1}(X^n - \zeta_m^i Y^n) = \prod_{i=1}^{m-1}\prod_{j=1}^{n} (X - \zeta_n^j \zeta_{mn}^i Y),
$$
and
$$
W_n(X,Y) = \frac{X^n - (-Y)^n}{X+Y} = \prod_{k=1}^{n-1}(X + \zeta_n^k Y).
$$
By Lemma~\ref{lem:Res}, the resultant
$$
\Res(V_m, W_n) = \prod_{i=1}^{m-1}\prod_{j=1}^{n} \prod_{k=1}^{n-1} ( \zeta_{mn}^i \zeta_n^j + \zeta_n^k) \in \Z.
$$
For each factor $\zeta_{mn}^i \zeta_n^j + \zeta_n^k$ in the resultant, we have $\zeta_{mn}^i \zeta_n^j \ne \zeta_n^k$
(because otherwise we would have $\zeta_{mn}^i \zeta_{mn}^{mj} = \zeta_{mn}^{mk}$, and then $m \mid i$, but $1 \le i \le m-1$),
and so $\zeta_{mn}^i \zeta_n^j + \zeta_n^k = \zeta_{m^\prime}(1 -\zeta_2\zeta_{n^\prime})$
for some odd integer $m^\prime$ and odd integer $n^\prime \ge 3$ (noticing both $m,n$ are odd),
and thus it is a unit by \cite[Proposition 2.8]{Was}.
Hence, $\Res(V_m, W_n)$ is a unit in $\Z$, that is, $\Res(V_m, W_n) = \pm 1$.
Therefore, as polynomials over $D$, we also have $\Res(V_m, W_n) = \pm 1$.
Then, by Lemma~\ref{lem:coprime}, $V_m(a,b)$ and $W_n(a,b)$ are coprime in $D$.
\end{proof}
\begin{lemma} \label{lem:abn}
Let $a, b$ be defined as in Lemma~\ref{lem:coprime}.
Then, for any odd integer $n \ge 1$, $(a^n - b^n)/(a-b) $ and $(a+b)^2$ are coprime in $D$.
\end{lemma}
\begin{proof}
Without loss of generality, we fix an odd integer $n \ge 3$.
As before, both $(a^n - b^n)/(a-b) $ and $(a+b)^2$ are in $D$.
As in the proof of Lemma~\ref{lem:Pmn}, we deduce that the resultant of the homogeneous polynomials
$(X^n - Y^n)/(X-Y) $ and $(X+Y)^2$ is equal to $\pm 1$.
Hence, using Lemma~\ref{lem:coprime}, we obtain that $(a^n - b^n)/(a-b) $ and $(a+b)^2$ are coprime in $D$.
\end{proof}
\section{Proofs of Theorems~\ref{thm:strong3} and \ref{thm:primitive3}}
We need to make one more preparation.
Recall that $p$ is the characteristic of the field $K$.
As usual, denote by $v_\pi(h)$ the maximal power to which an irreducible polynomial $\pi$ divides $h \in R$.
Let $M$ be the fraction field of $R$.
By assumption, $M(\lambda)$ is a field extension over $M$ having degree at most four. Note that $\eta \in M(\lambda)$.
For any irreducible polynomial $\pi \in R$, as usual $v_\pi$ induces a valuation of $M$.
It is well-known that the valuation $v_\pi$ in $M$ can be extended to the field $M(\lambda)$; see, for instance, \cite[Theorem 3.1.2]{EP}.
Without confusion, we still denote by $v_\pi$ the corresponding extension of valuation in $M(\lambda)$.
\begin{lemma} \label{lem:vU}
Let $\pi \in R$ be an irreducible polynomial dividing $U_n$ for some $n \ge 3$.
Then, for any $m \ge 1$ with $p \nmid m$ $($including the case $p=0)$,
we have $v_\pi (U_{mn}) = v_\pi (U_n)$.
\end{lemma}
\begin{proof}
First, since $\lambda, \eta$ are both integral over the ring $R$,
we have that $v_\pi(\lambda) \ge 0$ and $v_\pi(\eta) \ge 0$.
Suppose that $v_\pi(\eta) > 0$.
Note that we have either $\lambda^n = \eta^n + (\lambda - \eta)U_n$, or $\lambda^n = \eta^n + (\lambda^2 - \eta^2)U_n$.
Then, since $v_\pi(\eta) > 0$ and $v_\pi(U_n) > 0$, we have $v_\pi(\lambda^n) > 0$.
So, $v_\pi (\lambda) > 0$.
Thus,
$$
v_\pi(\lambda + \eta) >0, \qquad v_\pi(\lambda\eta) > 0,
$$
which contradicts the assumption that $(\lambda + \eta)^2$ and $\lambda\eta$ are coprime in $R$.
Hence, we must have $v_\pi(\eta) = 0$.
Similarly, we must have $v_\pi(\lambda) = 0$.
Assume that $n$ is odd.
Then, $U_n = (\lambda^n - \eta^n)/(\lambda - \eta)$.
So, we have
$$
\lambda^n = \eta^n + (\lambda - \eta)U_n.
$$
Then, we obtain
$$
\lambda^{mn} = \big( \eta^n + (\lambda - \eta)U_n \big)^m
= \eta^{mn} + \sum_{i=1}^{m} \binom{m}{i} (\lambda - \eta)^i U_n^i \eta^{n(m-i)}.
$$
So
$$
\frac{\lambda^{mn} - \eta^{mn}}{\lambda - \eta} = m\eta^{n(m-1)} U_n + \sum_{i=2}^{m} \binom{m}{i} (\lambda - \eta)^{i-1} \eta^{n(m-i)} U_n^i.
$$
Hence, we obtain that for odd $m$
$$
U_{mn} = m\eta^{n(m-1)}U_n + \sum_{i=2}^{m}\binom{m}{i} (\lambda - \eta)^{i-1}\eta^{n(m-i)} U_n^{i},
$$
and for even $m$
$$
(\lambda + \eta)U_{mn} = m\eta^{n(m-1)}U_n + \sum_{i=2}^{m}\binom{m}{i} (\lambda - \eta)^{i-1}\eta^{n(m-i)} U_n^{i}.
$$
We also note that since $n$ is odd and $v_\pi (U_n) > 0$, by Lemma~\ref{lem:abn} we have $v_\pi (\lambda + \eta) = 0$.
Then, the desired result follows.
Finally, assume that $n$ is even.
Then, as the above, for any integer $m \ge 1$ we obtain
$$
U_{mn} = m\eta^{n(m-1)}U_n + \sum_{i=2}^{m}\binom{m}{i} (\lambda^2 - \eta^2)^{i-1}\eta^{n(m-i)} U_n^{i}.
$$
The desired result now follows.
\end{proof}
Now, we are ready to prove the theorems.
\begin{proof}[Proof of Theorem~\ref{thm:strong3}]
Let $d = \gcd(m,n)$.
First, we assume that both $m$ and $n$ are even. Then, $d$ is also even.
By definition, we obtain
$$
U_m = U_d P_{m/d}(\lambda^d, \eta^d), \quad U_n = U_d P_{n/d}(\lambda^d, \eta^d).
$$
By assumption, it is easy to see that $\lambda^d + \eta^d$ and $\lambda^d \eta^d$ are coprime in $R$
(as in the last paragraph of the proof of Lemma~\ref{lem:coprime}).
Hence, by Lemma~\ref{lem:PmPn2}, we know that $P_{m/d}(\lambda^d, \eta^d)$ and $P_{n/d}(\lambda^d, \eta^d)$ are coprime in $R$,
and so we have $\gcd(U_m, U_n) = U_d$ in this case.
Now, we assume that both $m$ and $n$ are odd. Then, $d$ is also odd.
By definition, we have
$$
U_m = U_d P_{m/d}(\lambda^d, \eta^d), \quad U_n = U_d P_{n/d}(\lambda^d, \eta^d).
$$
We also note that $(\lambda^d + \eta^d)^2$ and $\lambda^d \eta^d$ are coprime in $R$.
Then, by Lemma~\ref{lem:PmPn-odd} we know that $P_{m/d}(\lambda^d, \eta^d)$ and $P_{n/d}(\lambda^d, \eta^d)$ are coprime in $R$,
and so we have $\gcd(U_m, U_n) = U_d$.
Finally, when $m$ and $n$ do not have the same parity, without loss of generality, we assume that $m$ is odd and $n$ is even.
Then, $d$ is odd. By definition, we have
$$
U_m = U_d P_{m/d}(\lambda^d, \eta^d),
$$
and
$$
U_n = U_d \cdot \frac{P_{n/d}(\lambda^d, \eta^d)}{\lambda^d + \eta^d} \cdot \frac{\lambda^d + \eta^d}{\lambda + \eta}.
$$
Then, by Lemma~\ref{lem:PmPn-mix} we know that $P_{m/d}(\lambda^d, \eta^d)$ and $P_{n/d}(\lambda^d, \eta^d)/(\lambda^d + \eta^d)$ are coprime in $R$.
Besides, by Lemma~\ref{lem:Pmn} we obtain that $P_{m/d}(\lambda^d, \eta^d)$ and $(\lambda^d + \eta^d)/(\lambda + \eta)$ are coprime in $R$.
Hence, we have $\gcd(U_m, U_n) = U_d$.
This completes the proof.
\end{proof}
\begin{proof}[Proof of Theorem~\ref{thm:primitive3}]
As in \cite{Ward}, we define the sequence $(Q_n)_{n \ge 1}$ of polynomials by $Q_1 = 1, Q_2 = 1$, and
$$
Q_n(X,Y) = \Phi_n(X,Y), \quad n = 3, 4, \ldots.
$$
Then, it is easy to see that for any integer $n \ge 1$ we have
$$
U_n = \prod_{d \mid n} Q_d(\lambda, \eta).
$$
By the M{\"o}bius inversion, we have
$$
Q_n(\lambda, \eta) = \prod_{d\mid n} U_d^{\mu(n/d)}.
$$
So, for any irreducible polynomial $\pi$ in $R$ we have
$$
v_\pi(Q_n(\lambda, \eta)) = \sum_{d \mid n} \mu(n/d) v_\pi(U_d).
$$
Now, assume the characteristic $p > 0$.
suppose that $\pi$ is a prime divisor of $U_n$ which is not primitive, where $p \nmid n$.
Let $m$ be the minimal positive integer such that $\pi \mid U_m$. Automatically, $p \nmid m$.
Then, by Theorem~\ref{thm:strong3} we have $m \mid n$, and by Lemma~\ref{lem:vU}, for any positive integer $k$ with $p \nmid k$
$$
v_\pi(U_{mk}) = v_\pi(U_m).
$$
Hence, if $m < n$, noticing $p \nmid n$ we obtain
\begin{align*}
v_\pi(Q_n(\lambda, \eta)) & = \sum_{d \mid n/m} \mu(n/(dm)) v_\pi(Q_{dm}) \\
& = \sum_{d \mid n/m} \mu(n/(dm)) v_\pi(Q_m) \\
& = v_\pi(Q_m) \sum_{d \mid n/m} \mu(n/(dm)) = 0.
\end{align*}
So, any non-primitive prime divisor of $U_n$ (in the sequence $U^\prime$) does not divide $Q_n(\lambda, \eta)$.
It is easy to see that when $n > 2$, $Q_n(\lambda, \eta) = \Phi_n(\lambda, \eta)$ is non-constant
(because at least one of $\lambda$ and $\eta$ is transcendental over $K$),
and so $Q_n(\lambda, \eta) $ has a prime divisor in $R$.
Thus, when $n > 2$, any prime divisor of $Q_n(\lambda, \eta)$ is primitive,
and so each term in the sequence $U^\prime$ beyond the second has a primitive prime divisor.
The proof for the case $p = 0$ follows exactly the same way.
\end{proof}
\begin{remark} \label{rem:Lehmer}
In the proof of Theorem~\ref{thm:primitive3}, we obtain more: the \textit{primitive part}
(that is, the product of all the primitive prime divisors to their respective powers)
of $U_n$ is $Q_n(\lambda, \eta) = \Phi_n(\lambda,\eta)$, where $n \ge 3$, and $p \nmid n$ if $p > 0$.
\end{remark}
\section{Proofs of Theorems~\ref{thm:strong2} and \ref{thm:primitive2}}
The proofs follow easily from Theorems~\ref{thm:strong3} and \ref{thm:primitive3}.
\begin{proof}[Proof of Theorem~\ref{thm:strong2}]
Fix positive integers $m, n$ with $d= \gcd(m,n)$.
If either both $m,n$ are odd, or both $m,n$ are even,
it follows directly from Theorem~\ref{thm:strong3} that $\gcd(L_m, L_n) = L_d$ (setting $\lambda = \alpha, \eta = \beta$).
Now, without loss of generality, assume that $m$ is even and $n$ is odd. By Theorem~\ref{thm:strong3}, we have
$$
\gcd\Big(\frac{\alpha^m - \beta^m}{\alpha^2 - \beta^2}, \frac{\alpha^n - \beta^n}{\alpha - \beta} \Big)
= \frac{\alpha^d - \beta^d}{\alpha - \beta}.
$$
Using Lemma~\ref{lem:abn} we know that $(\alpha^n - \beta^n)/(\alpha - \beta)$ and $\alpha + \beta$ are coprime in $R$.
Hence, we obtain $\gcd(L_m, L_n) = L_d$.
This completes the proof.
\end{proof}
\begin{proof}[Proof of Theorem~\ref{thm:primitive2}]
Assume that the characteristic $p = 0$.
First, by Lemma~\ref{lem:abn}, we have that for any odd $n \ge 3$, $\Phi_n(\alpha,\beta)$ and $\alpha + \beta$ are coprime in $R$.
Now, fix an even integer $n \ge 4$.
Suppose that there exists an irreducible polynomial, say $\pi$, in $R$ dividing both $\Phi_n(\alpha, \beta)$ and $\Phi_2(\alpha,\beta) = \alpha + \beta$.
This means that the polynomial $X^n - Y^n$, defined over the fraction field of the ring $R$ (mod $\pi$), has a multiple root (that is, $(\alpha, \beta)$).
However, this fraction field has characteristic zero (because it contains the field $K$), which implies that $X^n - Y^n$ is in fact a simple polynomial.
Hence, this leads to a contradiction, and so $\Phi_n(\alpha, \beta)$ and $\alpha + \beta$ are coprime in $R$.
Therefore, by constructions we directly obtain from Remark~\ref{rem:Lehmer} that
the primitive part of $L_n$ is $\Phi_n(\alpha,\beta)$, where $n \ge 3$.
Finally, if the characteristic $p > 0$, then by contruction the above arguments still work
(because in the sequence $L^\prime$ we have deleted those terms $L_n$ with $p \mid n$).
\end{proof}
\begin{remark} \label{rem:Lucas}
In the proof of Theorem~\ref{thm:primitive2}, we obtain more: the primitive part
of $L_n$ is $\Phi_n(\alpha,\beta)$, where $n \ge 3$, and $p \nmid n$ if $p > 0$.
\end{remark}
\section{Proofs of Theorems~\ref{thm:strong1} and \ref{thm:primitive1}}
Clearly, Theorem~\ref{thm:strong1} follows directly from Theorem~\ref{thm:strong2}.
\begin{proof}[Proof of Theorem~\ref{thm:primitive1}]
Assume that the characteristic $p = 0$.
Fix an integer $n \ge 3$.
Taking $\alpha = f$ and $\beta = g$ in Theorem~\ref{thm:primitive2} and noticing Remark~\ref{rem:Lucas},
we know that the primitive part of the term $ (f^n - g^n) / (f-g)$ is $\Phi_n(f,g)$.
As the above, we obtain that $\Phi_n(f,g)$ and $f-g$ are coprime in $R$.
Hence, the primitive part of the term $F_n = f^n - g^n$ is $\Phi_n(f,g)$.
Finally, if the characteristic $p > 0$, then by contruction the above arguments still work
(because in the sequence $F^\prime$ we have deleted those terms $F_n$ with $p \mid n$).
\end{proof}
\begin{remark}
In the proof of Theorem~\ref{thm:primitive1}, we obtain more: the primitive part
of $F_n$ is $\Phi_n(f,g)$, where $n \ge 3$, and $p \nmid n$ if $p > 0$.
\end{remark}
\section{Comments}
In this section, we make some remarks about extending our results to unique factorization domains.
Note that all the lemmas used in proving the strong divisibility property are valid for any unique factorization domain.
So, we have the following result.
\begin{theorem}
The strong divisibility properties in Theorems~\ref{thm:strong3}, \ref{thm:strong2} and \ref{thm:strong1} still hold
when we replace the ring $R$ by a unique factorization domain $D$.
\end{theorem}
In order to extend fully all our results on primitive divisors to a unique factorization domain $D$, we need to assure two properties.
One is about the valuation similar as in Lemma~\ref{lem:vU}.
The other is to assure that $\Phi_n(f,g), \Phi_n(\alpha,\beta)$ and $\Phi_n(\lambda, \eta)$ are all non-zero and non-unit whenever $n \ge 3$.
If $D$ contains a field, then any integer as an element in $D$ is either zero or a unit,
and so the valuation result holds in this case by following the same arguments as Lemma~\ref{lem:vU}.
Hence, in this case, if one can show that $\Phi_n(f,g)$ is non-unit whenever $n > n_0$ for some integer $n_0$,
then one in fact prove the result in Theorem~\ref{thm:primitive1} by replacing ``beyond the second" with ``beyond the $n_0$-th".
Similar things apply to Theorems~\ref{thm:primitive3} and \ref{thm:primitive2}.
We present an example here.
Let $D=K[[X]]$ be the formal power series ring defined over a field $K$ in one variable $X$.
Then, an element $\sum_{n=0}^{\infty}a_n X^n$ in $D$ is a unit if and only if $a_0 \ne 0$.
Let $f$ and $g$ be non-zero, non-unit and coprime in $D$ such that $f/g$ is not a root of unity.
Then, $\Phi_n(f,g)$ is non-zero and non-unit for any $n \ge 1$,
and so Theorem~\ref{thm:primitive1} holds in this case.
In addition, if let $f$ be non-unit and $g$ a unit in $D$, then $\Phi_n(f,g)$ is a unit for any $n \ge 1$,
\section*{Acknowledgement}
The author was partly supported by the Australian Research Council Grant DE190100888.
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