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Introduction
This site is a summary of research collection dedicated to cryptocurrency market. Data used in this study is ingested from number of major exchanges. Due to the insufficient maturity of cryptocurrency market, some of the models are still in training stage and some have been successfully deployed. The goal for this research is to identify short/mid/long market direction derived from recurring patterns and traces of market participants significant in this space.
Market Data Providers
List of Market Data Providers
- Bitmex
- Deribit
- Huobi
- Bybit
- CME
- Ftx
Market Data Analysis
Index and Vector
ETH mid term Vector Index Study.
Order Analysis
Projected directions has been classified in three time horizons.
Contract Order Volume and Sequence
Liquidations Sequence on BTC.
Liquidations Sequence on ETH.
Parameters
Price Level Projections
Notification message
Contains target levels with corresponding weights and also include Sequence SFSFT with short term reversal flags.
BTC Short Term Price Projection.
ETH Sequence SFSFT.
Notification message Parameters Breakdown
Below is the list of parameters with interpretation Short Term Target Zone is typically to unfold withing 30 hours from the time of the notification delivery time.
Legend
When change of conditions is detected the new report is forwarded with image to the channel.
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6th year, second grade teacher
Whole Brain Teaching, student led work environment, technology 1:1 program, project based learning, student choice
Yet to be added
BA in Elementary Education (2014) M.Ed in Educational Technology (2019)
Yet to be added
Kindergarten, 1st, 2nd, 3rd, 4th, 5th, 6th, Not Grade Specific
English Language Arts, Spelling, Specialty, Math, Arithmetic, Basic Operations, Graphing, Numbers, Social Studies - History, U.S. History, Arts & Music, Graphic Arts, Math Test Prep, Geography, Religion, Classroom Management, Word Problems, Short Stories, Writing, Reading Strategies, Holidays/Seasonal, Back to School, Christmas/ Chanukah/ Kwanzaa, Mental Math, Halloween, Place Value, For All Subjects, Phonics, Close Reading
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PLEG Planting Day at the Werribee River
Sunday 1st July 9.00 am – 12.00 noon
‘Help us restore the dramatic Werribee River Volcanic Gorge at Pinkerton Flat!’
Pinkerton Landcare and Environment Group (PLEG) will be planting at Pinkerton Flat on the Werribee River, besides Western Water Treatment Plant.
Restoration works are funded by Vision for Werribee Plains (DSE), Western Water and Melbourne Water; at one of the most dramatically scenic sites on the river.
As well as planting beside the river you will have the opportunity to view this section of the Werribee River volcanic gorge. Everyone is welcome.
Find us at: Western Water Treatment Plant, Butlers Road, Strathtulloh.
Melways (2006) Map 221 A9.
Butler's Road leads off Greigs Road at Map 221 3C.
There will be signs along the route into the farm on the day.
- Turn into Butlers Road, through the main WW entrance gate
- Follow Butlers Road through the working farm onto a gravel track, for a distance of about ~ 2 km,
- then take the first right hand turn over an irrigation channel, and past the hay shed on the right,
- The track then passes through a gate, and gradually descends down the escarpment to Pinkerton Flat.
- Park on the river flat.
Morning tea is being provided by Western Water
Daryl Akers
Secretary, Pinkerton Landcare & Environment Group
Mobile: 0438 277 252
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TITLE: In linear algebra when solving an equation is there a reason we solve it in terms of the last variable or is it an arbitrary decision?
QUESTION [0 upvotes]: So if we had a set of linear equations like x + y + z = 5 and x + y + 2z = 10 why do we always parameterize the solution in terms of z? Is it just an arbitrary choice because it's the last variable? And are there situations where it would make more sense to specifically choose a variable to parameterize around?
Like looking through this solution set, every time it's the last variable that the equation is solved around. I figure it's completely arbitrary but I want to make sure.
https://en.wikibooks.org/wiki/Linear_Algebra/Vectors_in_Space/Solutions
REPLY [1 votes]: why do we always parameterize the solution in terms of z? Is it just an arbitrary choice because it's the last variable?
This is not true, presumably it is how it is taught in your region, but there is no particular reason to do so. You could label the variable what you like, e.g if we had a solution to some equations $(11+\frac{z}{4\pi},\frac{z+12\pi}{2\pi},\frac{z}{\pi})$, it is also perfectly valid to say $z=4\pi t$ then we get $(11+t,2t+6,4t)$, which you can adapt however you like to get a nicer form of the solution(purely aesthetic). Since $z$ was arbitrary, $t$ is arbitrary, so these solutions are identical.
And are there situations where it would make more sense to specifically choose a variable to parameterize around?
Again, this is purely notational, you gain no new insight by choosing which variable to parameterise.
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The "n" sounds certainly are not separate phonemes as you say below. The three sounds spelled in these words are as qualitatively different as /d/, /j/, and /g/ as in "daub," "job," and "gob." >/? The tongue always comes off the velum for the ending of "sing" which does make a slight ~g sound. This is the basic sound made for a ~g in quickly flowing speech - a velar pull-off. If the tongue doesn't come off the velum, the air dam is still in place and the nasal is still going. You have zero "~g". Tom Zurinskas, USA - CT20, TN3, NJ33, FL5+ see truespel.com ---------------------------------------- > Date: Tue, 24 Mar 2009 14:09:17 -0400 > From: hfwstahlke at GMAIL.COM > Subject: Re: the three "n" s > To: ADS-L at LISTSERV.UGA.EDU > > ---------------------- Information from the mail header ----------------------- > Sender: American Dialect Society > Poster: Herb Stahlke > Subject: Re: the three "n" s > ------------------------------------------------------------------------------- > > The three sounds spelled in these words are as qualitatively > different as /d/, /j/, and /g/ as in "daub," "job," and "gob." (I > realize that for those of us without the caught/cot merger "daub" has > the vowel of "caught.") The reason they seem not to contrast is that > the contrast is not phonemic before a consonant because nasals are > always homorganic (same place of articulation) to a following > syllable-final consonant. But the differences are physically real. > >/? > > Herb > > On Tue, Mar 24, 2009 at 1:50 PM, Tom Zurinskas wrote: >> ---------------------- Information from the mail header ----------------------- >> Sender: American Dialect Society >> Poster: Tom Zurinskas >> Subject: Re: the three "n" s >> ------------------------------------------------------------------------------- >> >> Perhaps I should say; "three ways of making the sound ~n". I don't think the three n sounds (in winter,finger,danger) are qualitatively different very much, since airflow is rerouted through the nose in each case. The ~n sound is nasal and doesn't matter much what the tongue or mouth does. >> >> I've got a touch of a "g" in "sing", made by the tongue going down off the velum (when saying the word in isolation). But for "sing a song" the "g" becomes fully formed leading the vowel "a". Allophones. >> >> Note: I can't say "sing" without pronouncing the vowel as "long e" ~ee before the "ng" or "nk" in the same syllable. I find this typical of UK or USA English even though dictionaries prescribe short i. >> >> >> Tom Zurinskas, USA - CT20, TN3, NJ33, FL5+ >> see truespel.com >> >> >> >> >> ---------------------------------------- >>> Date: Mon, 23 Mar 2009 19:48:30 -0400 >>> From: paul.johnston at WMICH.EDU >>> Subject: Re: the three "n" s >>> To: ADS-L at LISTSERV.UGA.EDU >>> >>> ---------------------- Information from the mail header ----------------------- >>> Sender: American Dialect Society >>> Poster: Paul Johnston >>> Subject: Re: the three "n" s >>> ------------------------------------------------------------------------------- >>> >>> You mean the n *sounds*--even your description says that. I won't >>> reprise everyone else's argument about the "velar n" being a phoneme >>> on its own. I guess you have a system like English West Midlanders >>> do--where engma IS an allophone. For what t's worth, I've heard such >>> systems from other Easterners--including a colleague of mine from >>> Providence. I don't have it, though. Now, are you like Birmingham/ >>> Liverpool or like Stoke-on-Trent? I. e. do you have a /g/ in sings? >>> I bet you don't. >>> >>> Paul >>> On Mar 23, 2009, at 2:35 PM, Tom Zurinskas wrote: >>> >>>> ---------------------- Information from the mail header >>>> ----------------------- >>>> Sender: American Dialect Society >>>> Poster: Tom Zurinskas >>>> Subject: the three "n" s >>>> ---------------------------------------------------------------------- >>>> --------- >>>> >>>> I'd say there are 3 ways of say the sound "n", the most often >>>> spoken sound of USA English. For each, the tongue forms a dam so >>>> air is routed through the nose to make the "n" sound, but tongue >>>> placement differs, as per example in these three words; winter, >>>> finger, and danger. >>>> >>>> 1. the front n - Winter - (tongue tip contacts top gums) >>>> 2. the back n - Finger - (back of tongue contacts the velum) >>>> 3. the middle n - Danger - (whole tongue rises to palate) >>>> >>>> These "n"s are not noticably different in sound, and are allophones >>>> of each other. >>>> >>>> Tom Zurinskas, USA - CT20, TN3, NJ33, FL5+ >>>> see truespel.com >>>> >>>> >>>> >>>> ---------------------------------------- >>>>> Date: Mon, 23 Mar 2009 10:35:36 -0700 >>>>> From: zwicky at STANFORD.EDU >>>>> Subject: act of show >>>>> To: ADS-L at LISTSERV.UGA.EDU >>>>> >>>>> ---------------------- Information from the mail header >>>>> ----------------------- >>>>> Sender: American Dialect Society >>>>> Poster: Arnold Zwicky >>>>> Subject: act of show >>>>> --------------------------------------------------------------------- >>>>> ---------- >>>>> >>>>> from the Palo Alto Daily News of 21 March 2009, "Palo Alto shooting >>>>> suspect still at large", by Diana Samuels, p. 3: >>>>> >>>>> Police said Gil-Fernandez claimed affiliation with the Norteno street >>>>> gang, though that claim may have been mostly an act of show in Palo >>>>> Alto, where the Norteno gang doesn't have much of a presence. >>>>> ..... >>>>> >>>>> that's "an act of show" 'a show, display, pretense'. a few more >>>>> cites: >>>>> >>>>> I remembered my fiancés first time to pay me a visit at home. I >>>>> was listening to Metallica. At that time, he thought I was just >>>>> playing it to impress him. Later as we knew each other better, he >>>>> realized that it was not an act of show but simply my choice of >>>>> music. >>>>> >>>>> >>>>> Gabby,Lynette,Bree,Susan and Eddie are all part of a clan that has >>>>> secrets lies and webs of drama they all live in such a perfect area >>>>> yet knowing all the perfection is just an act of show!!! >>>>> >>>>> reviews.html&flag=1&pg_rev=4 >>>>> >>>>> Col. Gianfranco Cavallo of the Carabinieri paramilitary police said >>>>> the bomb appeared to be an "act of show" rather than an attack >>>>> planned >>>>> to make victims. >>>>>- >>>>> in-downtown-rome-672350.html >>>>> >>>>> Actually it was our state governor, a female democrat (Ks) that >>>>> ordered them sent! I am sure that it was more of an act of show than >>>>> anything else. >>>>>- >>>>> for-mexico-border-violence >>>>> >>>>> there are also hits for "act of show-off" >>>>> >>>>> do you think if I show up with my guitar it won't be weired? won't >>>>> people think of it as an act of show off? >>>>> >>>>> [Yahoo! Indonesia] >>>>> >>>>> Again, McLeod depicts Ganda Singh as a shallow person of >>>>> pretentious nature whose ego was inflated by the mere receipt of a >>>>> book from McLeod and as an act of show-off, he carried it round under >>>>> his arm, with the title prominently displayed for all to see. >>>>> >>>>> >>>>> They dabble in esoteric themes like mysticism, sexual freedom, >>>>> death, the after-life and such things, as an act of show off. >>>>> >>>>> >>>>> unlike the "act of show" examples, these seem to be from various >>>>> asian >>>>> varieties of english. >>>>> >>>>> arnold >>>>> >>>>> ------------------------------------------------------------ >>>>> The American Dialect Society - >>>> _________________________________________________________________ >>>> Internet Explorer 8 Now Available. Faster, safer, easier. >>>> >>>> >>>> ------------------------------------------------------------ >>>> The American Dialect Society - >>> >>> ------------------------------------------------------------ >>> The American Dialect Society - >> _________________________________________________________________ >> Express your personality in color! Preview and select themes for Hotmail®. >> >> >> ------------------------------------------------------------ >> The American Dialect Society - >> > > ------------------------------------------------------------ > The American Dialect Society - _________________________________________________________________ Internet Explorer 8 Now Available. Faster, safer, easier. ------------------------------------------------------------ The American Dialect Society -
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David Maister’s book Managing The Professional Service Firm remains the gold standard text on the subject, some 16 years after it was first published. In fact, the book is a collection of essays and articles that he had written over previous years, stretching as far back as 1982.
The main thing that struck me about re-reading this again recently was how little things have changed in terms of the major issues still impacting professional service firms of all kinds – everyone from lawyers, accountants, consultants and, of course, PR agencies. For example, he cited systemic under delegation as a key problem back in the early 1980s – and nearly 30 years later, it continues to plague the PR business.
As Maister notes in his chapter on the Motivation Crisis: “It is not uncommon to hear comments such as ‘The practice of law [or accounting or PR consulting] is just not as what might turn out to be less money.”
And Maister wrote this in 1985!
There isn’t a chapter in the book that doesn’t have something of key relevance to everyone working in a PR firm today. Chapter 10 on How Client’s Choose is a good example:
“Buying professional services is rarely a comfortable experience,” says Maister. He goes on to list 10 unpleasant emotions associated with the experience (I’ve editorialised slightly from the original):
- I’m feeling insecure, I’m not sure I know how to detect which of the agencies pitching to me is the genius and which is just good. I’ve exhausted my abilities to make a technical distinction.
- I’m feeling threatened. This is my area of responsibility and even though intellectually I know I need outside expertise, emotionally it’s not comfortable to put my affairs in the hands of others
- I’m taking a personal risk. By putting my affairs in the hands of others, I risk losing control
- I’m impatient. I didn’t call in someone at the first sign of symptoms. I’ve been thinking about this for a while.
- I’m worried. By the very fact of suggesting improvements or changes, these people are implying I haven’t been doing it right up until now. Are they on my side?
- I’m exposed. Whoever I hire, I’m going to have to reveal some proprietary secrets – not all of which is flattering. I will have to undress.
- I’m feeling ignorant – and I don’t like it. I don’t know if I’ve got a simple problem or a complex one – do I trust these PR folk to be honest about that?
- I’m skeptical. I’ve been burned by PR agencies before. You get a lot of promises. How do I know whose promises to buy?
- I’m concerned that they either won’t or can are doing or why, who…., ……who? In short, will these people deal with me in the way I want to be dealt with?
If PR clients felt this way 20 years ago, think how they feel now.
Remember, it may be painful to walk in the other person’s shoes. But David Maister’s advice is as true now as it was 30 years ago: “The single most important talent in selling professional services is the ability to understand the purchasing process (not the sales process) from the client’s perspective. The better a professional can learn to think like a client, the easier it will be to do and say the correct things to get hired.”
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Put your trust and faith in the Lord And will be easy what is hard! - Joshua Aaron Guillory -Joshua Aaron Guillory
Published On: August 08th 2018, Wednesday @ 7:02:41 PM
Total Number of Views: 20
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,” he said.
In the study, to be published in the journal Psychological Science and available online, people who took acetaminophen daily for three weeks reported less emotional suffering over time and showed less activity in regions of the brain previously shown to respond to social rejection than those who took the placebo, Webster said.
“Even so,” Webster said, “we don’t want to tell people to go take Tylenol to cope with their personal problems until more research is done.”
The findings have the potential for acetaminophen to be used eventually to treat minor social pains instead of more powerful drugs, Webster said. Acetaminophen may also show promise in curtailing antisocial behavior, Webster said. Because research has found that being rejected triggers aggression, using acetaminophen to alleviate emotional distress could reduce the likelihood of destructive actions, he said.
“The fMRI (functional magnetic resonance imaging) results from our study show that acetaminophen diminished reactivity in regions of the brain that have been linked to emotional processing, which helps regulate aggression,” he said., he said.
.
Because humans have an extended infancy compared with many other animals in which they are unable to defend or feed themselves, developing social connections from an early age was crucial, Webster said. As a result, humans’ social attachment system may have developed by piggybacking onto the physical pain system and becoming an outgrowth of it in order to promote survival, he said.
“Our findings have important implications because social exclusion is such a common part of life,” he said. “People can feel ostracized at work, snubbed by friends, excluded by their partners or slighted in any number of situations.”
Credits
- Writer
- Cathy Keen, ckeen@ufl.edu, 352-392-0186
- Source
- Gregory Webster , gdwebs@ufl.edu, 352-273-2160
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Two Mile Oak Pub
Two Mile Oak Pub
Two Mile Oak Pub
Friendly staff, nothing is to much trouble, great tasty food.
A friendly traditional Devon Pub, with stunning views of the moor from our Courtyard Garden. The restaurant is on two levels. The rear section formerly a barn, was converted in the early 1990’s. Fresh homemade food, made by our in-house Chefs and the menu ranges from Steak and Kidney Suet Pudding (for which we are famous), to Steaks and Seafood Pancakes. Daily Specials often include freshly caught dishes, such as Seabass and Whole Baked Mackerel. With a large car park at the rear of the building and once parked you walk through our lovely Courtyard Garden to enter the Pub.
Telephone: 01803 812 411
Two Mile Oak Inn
Totnes Road, Ipplepen, Newton Abbot TQ12 6DF, Devon TQ12 6DF
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This website has been a work-in-progress since about 1998. It was once a message board, then it became a regular website and now it's a multi-blog. Soon, I will be ridding the internet of the old website and all that will remain is what I store on this bloggity blog. I've been looking through the old site pages to make sure I don't lose anything important and I came across some old reviews I wrote for Pop Matters. Sometimes I regret not focusing myself more on critical writing but at the end of the day, I would rather be a creator than a critic and I have not had the time to be both. Still, I like writing reviews and will continue to do so as time permits. I just like talkin' 'bout stuff.
Here's dem reviews...
Marvel Boy
Midnight Nation
Peter Parker: Spider-Man
Your best pal ever,
Shannon
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LOQ
Elora
Mar Leather
Made in Spain, the Elora sandal is a summer basic featuring a woven kid leather band. A 3/4 inch cylinder beechwood heel. Vachetta leather footbed and leather sole. Fits true to size.
Vachetta leather is untreated vegetable tanned leather and will develop a patina when exposed to sunlight, oils or water.
FINAL SALE
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Highlights.
What You'll Get
Itinerary
This is a typical itinerary for this product
Stop At: Lake Te Anau, Te Anau, Fiordland National Park New Zealand
o Te Anau is known as the Walking Capital of the World. The mountains to the left are the Kepler Mountains, home of the Kepler Track, one of the popular walking tracks in the area. Te Anau is the gateway to Milford Sound and the Fiordland National Park. The town has a permanent population of approximately 2,500 people but this fluctuates greatly in summer when the town gets an influx of fishermen, hunters, outdoor pursuit seekers and holiday makers. It’s a popular holiday destination. We stop here for morning & afternoon breaks and comfort stops.
Duration: 30 minutes
Stop At: Eglinton Valley, Milford Sound New Zealand: Monkey Creek, Milford Sound New Zealand: 10 minutes
Pass By: Homer Tunnel, 94 S.h., Milford Sound New Zealand: Cruise Milford, 1 Milford Sound Hwy, Milford Sound 9679, New Zealand
o.
Duration: 1 hour 45 minutes
Stop At: The Chasm Walk, Milford Road State Highway 94 , Fiordland National Park, Te Anau, Fiordland National Park New Zealand
o Here you will have the opportunity to go for a 15 minute walk around a loop for photos. The coach will drop you off at one end of the board walk and pick you up at the other end in 15 minutes. Please be back at the coach on time. The walk into the Chasm will take you to a magnificent view of crystal clear waters rushing through a natural chasm where the Cleddau River has carved its way through the solid rock.
Duration: 10 minutes
Additional Information
- Confirmation will be received at time of booking
- Vegetarian option is available, please advise at time of booking if required
- Please be reporting at least 15 minutes prior to your departure times
- This experience requires good weather. If it’s canceled due to poor weather, you’ll be offered a different date or a full refund
- This tour/activity will have a maximum of 50 travelers
- Hotel drop-off (selected central Queenstown hotels only)
- Guide
- Transport by luxury air-conditioned coach
- Milford Sound cruise
- Picnic lunch
- Entry/Admission - Cruise Milford
Exclusions
- Sightseeing flight
Departure Point
Athol Street, Athol St, Queenstown 9300, New Zealand
Traveler pickup is offered
Please check pick up times with provider before travel.
Departure Time
6:30 AM
Duration
12 hours
Return Details
Returns to original departure point
Voucher Info
You can present either a paper or an electronic voucher for this activity.
The tour identified in this promotion is made available through Viator. Groupon is not affiliated with or sponsored by the Gray Line.
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>>. Vimeo PRO Share on Tumblr 5 Recommendations Affiliate Marketing Books Try Ning free (14 days trial). Ning makes it really easy to build social networks from scratch.. All popular spintax styles supported Pixel Domain sold at $1750.00 No File Choosen Likes Received:0 Nasrallah: Despite Israel’s efforts, Hezbollah has acquired ‘precision missiles’ US Airways Login (Guest) said On 02 August, 2018 at 6:00 AM How to Run YouTube Ads in 2018 to Make 6 Figures on AUTOPILOT – Case Study Included Official site: IMPORTANT PAGES: ©2018 Cortx. All rights reserved. Search for: Search Spin Rewriter 8.0 - The best article spinner and Rewriter Avast Free Antivirus. Article-Rewriter For Wordpress CV Writing A couple of years ago when Google unleashed the Penguin update, things started to change. 7 months ago Better writing. No matter what you are working on. Avast Free Antivirus March 12, 2018 / Aaron money ← View all trending topics 14. Article Scraper Tool Hales Jobs 10 Factors That Affect Spin Rewriter 9.0's Longevity. | Click for More 10 Factors That Affect Spin Rewriter 9.0's Longevity. | Click Here 10 Factors That Affect Spin Rewriter 9.0's Longevity. | Download Now
Legal | Sitemap
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It’s May and we are happy to announce more fun and exciting upcoming events. Prints on Paper will be back at it with more awesome printmaking workshops and we are very excited to announce we will be hosting “Capturing State” a group exhibition curated by Isaac Zavale and Minenkulu Ngoyi of Prints on Paper. “Capturing […]
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Pineapple Casserole
by Paula Deen
Listen guys, I know she’s not a great person… but I still can’t deny myself this quick and dirty delicious recipe. Though since Jay doesn’t seem to care for it, I can cut the ingredients in half, throw it in a small bread tin, and eat it in shame all by myself.
One note though… don’t buy the crappy pre-shredded cheese, because it won’t melt… go for a good aged sharp cheddar. Also, don’t you dare use a cheap or healthier version of ritz crackers.
Ingredients
-Instructions
-.
(photo via lauras sweet spot because my old instagram didn’t turn out as appetizing)
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Disc Bulge Treatment: What Is The Right Approach For You? or aching pain accompanied with numbness, tingling and muscle weakness. As well as this characteristic signs of a disc injury include abnormal posture (referred to as antalgic posture), altered gait or walking pattern, reduced movement and generalized
- Traditional manual manipulation involves moving particular joints to realign, reposition and therefore reduce pressure upon discs
- Joint mobilisation is a gentle treatment option in cases that are particular acute or severe. Mobilisation involves stretching and opening up joints to improve overall movement and assist tissue healing
- Traction therapy is particularly favoured in many cases of lumbar disc bulges
Massage
While massage will not directly promote healing of your disc it will facilitate surrounding muscle relaxation to assist with reducing pain and allowing better movement.
Anti-Inflammatory Medications
In particular circumstances anti-inflammatory medications may be advised to assist with reducing your pain. Unfortunately such medications do no assist healing, but in the short-term they will assist with making life a little easier!.
Surgery.
WHAT DOES RESEARCH TELL US?
-
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TITLE: Show that $f$ is not uniformly continuous
QUESTION [1 upvotes]: let $f:\mathbb{R}\rightarrow\mathbb{R}$
There exists a real number $c$ and two sequences $x_n, y_n$ such that
$\forall$ $n$: $|x_n-y_n|=c$
$|f(x_n)-f(y_n)|\xrightarrow[n \to \infty]{}\infty$
Prove that $f$ is not uniformly continuous on $\mathbb{R}$
Hi, i have been trying to solve this question but i keep hitting a brick wall.
What i tried to prove:
there exists an $\varepsilon_0$ such that for all $\delta>0$ there
exists $x,y$ such that $|x-y|<\delta$
but: $|f(x)-f(y)|\geq\varepsilon_0$
So i chose $\varepsilon_0=1$, let $\delta>0$.
If $\delta>c$:
There exists $N$ such that for all $n>N$ : $|f(x_n)-f(y_n)|>1$
let $n>N$:
$|x_n-y_n|=c<\delta$
and $|f(x_n)-f(y_n)|>1$
so $f$ is not uniformly continuous
If $\delta\leq c$: this is where i got stuck.
REPLY [0 votes]: Suppose $f$ is uniformly continuous. Let $\delta' > 0$ be as given in the definition of uniform continuity, corresponding to $\varepsilon = 1$. Put $\delta := \delta'/2$.
Define $N$ be the smallest positive integer such that $N\delta \geqslant c$.
Claim. If $|y - x| = c$, then $|f(y) - f(x)| < N$.
Proof. Let $x, y \in \Bbb R$ be arbitrary such that $y = x + c$. Consider the points $$x, x + \delta, \ldots, x + (N - 1)\delta, y.$$
The consecutive points are $< \delta'$ away in distance and thus, applying triangle inequality repeatedly gives
\begin{align}
|f(y) - f(x)| &\leqslant |f(y) - f(x + (N - 1)\delta)| + \cdots + |f(x + \delta) - f(x)| \\
&<1 + \cdots + 1 = N. \qquad \Box
\end{align}
From this, it follows that a sequence like yours cannot exist.
| 94,715
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Facebook and instant messaging are making it even easier than before for school kids to create closed groups and intentionally exclude others, or worse. To help students counter this trend, an educational musical play on how to better understand and include all students is being performed for more than 3500 students in the Greater Victoria, Nanaimo and Salt Spring Island school districts in late April and early May.
The play “Inclusion Not Seclusion” is a co-production of Mountain Dream Productions () and the Mary Winspear Centre () in Sidney. It is performed by students ages 13-15 who integrate song, dance and dialogue to deliver their message in a lively, engaging performance. The musical play helps children in elementary and middle schools better understand why they should try to include all students – whether of different race, religion, culture or ability.
Margaret Watt, artistic director for Mountain Dream Productions, says she decided to create this musical after observing students at her school. “Seeing children interact and seeing new children arrive from other countries, I realized there’s a lot of integration that needs to go on. This applies to students with disabilities too,” she says. “When I talked to my performing arts students about this, we decided to write a play about ‘inclusion not seclusion’ as everyone should be included in our society today.”
Watt has taught musical theatre for many years. Three times a year she offers her performance arts “Triple Threat” program at the Mary Winspear Centre for children and teens ages 6 to 10 and 11 to 17.
The play “Inclusion Not Seclusion” revolves around three students who are new to a school, including one who has autism. At first the “in” group of students ignores or makes fun of them, but gradually attitudes change as they get to see each other’s strengths.
The young performers have discovered that being in the play has changed the way they think. Andrea says, “At first we thought [the nasty character] wouldn’t act like that… it’s way over the top. And then you go to school and see it happening everywhere.” When reflecting on whether the play had changed her perceptions, Mikayla says, “After being part of the play, if you see someone who is struggling or is different, you would want to go out of your way to make them feel accepted or welcome.”
The play is free of charge to the schools thanks to two provincial grants and another from the Capital Regional District. Students in the Greater Victoria area will be bused to the Mary Winspear Centre to see the play while Nanaimo and Salt Spring students will assemble in school gyms. Much appreciated community support has come from Wilson’s Transportation, Coast Capital Savings, Rotary Club of Royal Oak Centennial, Royal Oak Lions Club, the Municipality of Central Saanich and the Inter-Cultural Association of Victoria.
WHEN AND WHERE:
April 23-24 will perform for six schools in Nanaimo.
May 1-3 will perform at the Mary Winspear Centre for students from Greater Victoria schools.
May 8 will perform for four Salt Spring Island elementary schools.
| 354,986
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Thank
William Harry Leopard.
Erin
I can not find them What was address you sent them to? Will you send them to whleopard@juno.com.
Thank
Iam looking for John and Cecelia Lipsey at one time they lived in Illinois Cecelia was born abt 1827 and so was John. I know they had 2 daughters Nancy E, Sarah Ann, and 3 sons.Samuel, William R. and Eli Thomas. and that all childern were born in Carterville, Illinois.
If any one would have any information about this family I would really like to here from you.
You can reach me at zandy@frontiernet.net
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British Iraqi Fighter Surrenders
6:39am UK, Thursday March 27, 2003
A British man fighting for Saddam Hussein has surrendered to the Desert Rats in Iraq.
The man, in his mid-20s, told the Irish Guards he was from Manchester and wanted to return home - to claim benefits, the Guards claimed.
Born in Iraq, the unidentified prisoner, who speaks in a Mancunian accent, is currently being held in a prisoner of war camp in the south of the country.
A source within the Irish Guards said: "He told us he wanted to fight for the Iraqis because he didn't agree with what Britain and America are doing.
"After handing himself over, he began taunting some of the soldiers that he would soon be back in Britain enjoying state benefits even though he had a wish to fight against our people."
It is understood the man, whose family are still in England, joined a militia group through relatives in the Bas
| 256,533
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Aaron Huey is a masthead photographer for National Geographic Adventure and National Geographic Traveler magazines. His stories from Afghanistan, Haiti, Mali, Siberia, Yemen and French Polynesia (to name just a few) on subjects as diverse as the Afghan drug war and the underwater photography of sharks, can be found in The New Yorker, National Geographic and The New York Times.
Huey serves on the board of directors for the nonprofit Blue Earth Alliance. In 2002, he walked 3,349 miles across America with his dog Cosmo (the journey lasted 154 days), and was recently awarded a National Geographic Expedition Council Grant to hitchhike across Siberia.
If you need translations, you can install "Google Translate" extension into your Chrome Browser.
Furthermore, you can change playback rate by installing "Video Speed Controller" extension.
This website is owned and operated by Tokyo English Network.
The developer's blog is here.
| 263,654
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Description
I Time Travel
Book I – The Red Pill
By Sollog
NEWS – THIS BOOK BANNED BY AMAZON. This series of books is so controversial they have been BANNED BY AMAZON most likely due to the order of the NSA.
Sollog details his Amazing Story as Sollog the Time Traveler who became a World Famous Prophet due to his proven abilities to Travel Through Time with only his Amazing Mind…
Sollog’s new series of Time Travel Books are called the AMAZING ADVENTURES OF SOLLOG THE TIME TRAVLER and are also marketed as the I TIME TRAVEL Series.
The Books in this Series are:
Book 1 – The Red Pill
Book 2 – Awakening of Sollog
Book 3 – Early Life of Sollog
Book 4 – My Alien Operation
Book 5 – 911
Books 6 through 10 are planned in the series but not yet titled.Share
Reviews
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| 224,909
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Lifestyle
Tips To Design Safe And Beautiful Fabric Face Masks
The coronavirus pandemic has affected the whole world. Scientists are searching for a vaccine that can help us to stop the pandemic and restore normal life. Healthcare workers need personal protective equipment.
But due to the growing number of cases of coronavirus, they are facing a shortage in the supply of these protective gear. The surgical or N95 masks are in high demand. But, healthcare workers need these highly secure protective gear as compare to common people.
The normal fabric masks are good for normal people. Therefore, we should prepare masks at home and leave surgical and N95 face masks to healthcare workers who are directly dealing with the coronavirus patients.
You can prepare a custom face mask at home by using old clean cloth. Here, we will discuss the simple tips to prepare beautiful and safe face masks at home. But, before that, we should clear one doubt that is whether cloth face masks are effective in preventing you against coronavirus.
Table of Contents
Efficiency Of Cloth Face Masks
First of all, let me clear you one thing that no face mask is completely secure. Even the surgical face mask or N95 face masks do not have a 100% success rate. But, out of all, the surgical or N95 face masks are highly effective.
After the spread of coronavirus, these face masks are in shortage of supply. If we talk about the cloth face mask, they are quite efficient in creating a gap. Well, the cloth face masks are not ideal. But, it is better to have little protection than nothing.
In addition to this, the CDC department is also recommending the common people to wear cloth face masks before stepping outside of your house. CDC is currently recommending civilians wear cloth face coverings when in public.
How To Wear Fabric Face Masks?
According to the CDC, the cloth face covering should fit properly on your face. But, make sure that the face-covering is comfortable to wear. It is also important to prepare a face mask that ensures safety with multiple layers of fabric and secure ear loops.
Your fabric face mask should also allow you to breathe properly without any problem. Also, the fabric face mask should be reusable after washing it.
Tips To DIY Fabric Face Mask At Home
Take Accurate Measurements
It is imperative to take precise measurements before you start designing and preparing process. You can use your old shirt to prepare the face mask. But, make sure that the fabric is clean, and it is in good condition. Make sure that you cut four pieces of cloth into the same shape.
These different cutting will help in creating a multilayered face mask. Also, you should cut the four 12″ x 1″ strips. These strips will be used as the mask ties. If you have elastic strips at your home, then you can use these strips at the place of fabric strips.
Prepare Mask Layers
Once you have obtained the piece of cutting, then the next thing you have to do is forming a layered mask. You should put two pieces of fabric together. While placing two fabric piece together, you should make sure that the patterned side of the face ask should be facing inside.
After that, you should start sewing the clothes together. You should start sewing from the long curved side. Unfold the curve in order to form the one long piece.
Now, take another two pieces and repeat the same procedure. These two next pieces will help in making the double layer face covering. These two pieces will make inside and outside of your face mask.
Start Sewing Other Three Sides
Now, you have two mask pieces that you have just created in the previous. Now, take these two face mask pieces and place them over one another. Always keep in mind that you should keep the patterned face inside. You should start sewing the two fabric pieces together.
You should sew a short edge as well as a large edge. Make sure that you leave the second large edge open. You should turn the mask and bring the right side in an outward direction. Now, it is time to sew the final side of the fabric mask.
Prepare Straps
Your front face cover has been prepared, and you should fold the strips twice. Now, start sewing, and you will have thin straps in your hands. But, you should keep in mind that straps are thin, and you should slowly move the machine in the right direction.
Fix Straps With Face Masks
If the face mask and straps are complete, then you can sew them together. You should sew these straps with the face mask covering. You should sew the four 12” inches strap with the face mask. You should sew the straps to the outward direction.
Customization
Once you have a complete mask in your hand, then you can customize your own face mask in your own way. If you want to print a special design, then take the help of professionals.
>.
Home Decor
Top 5 Pro Cleaning Tips for 2021
_3<<
But we must emphasize that the body behaves as a whole and as a whole also loses weight. Therefore, it is impossible to arrange for the ladies to lose weight only on the buttocks and thighs and have nice big breasts. But don’t hang your head. With effective exercises, you can strengthen problem areas and shape them nicely. Fatty tissue then turns into muscle. How to help ensure that problem areas are no longer problematic?
Table of Contents
Move regularly
At the outset, it must be said that it is always better to do something than nothing at all. So, even if you get off the bus a little earlier in the morning on the way to work and take a small brisk walk to stretch and flow the blood, it’s just a small snippet in motion, but it’s better than being lazy and doing nothing. Even this seemingly small stretch will benefit your physical condition. Any exercise is a better choice than no exercise.
Suppose you want to start your body losing weight, above all you need to move regularly. How about that, when you are very busy with work, you don’t have time, and when you come home in the evening, and you don’t want to do anything anymore?
Exercise as part of a ritual
Exercising at home is the bare minimum of movement you can start with—just a short ten-minute warm-up in the morning. Pay attention to regularity. Only if you don’t miss it for a day makes sense. Exercising only occasionally when you want it is priceless.
How to force yourself? Just make exercise an obligation. Please take it in the same way you have to comb your hair in the morning, brush your teeth, and wash. It’s just part of a morning ritual you can’t go to work without. Then the various excuses will no longer work.
Do the same in the afternoon. Don’t you want to go to aerobics or the gym after work? Don’t go home and bring sports clothes and shoes straight to work. When you come home, you will not want to go out again, and there will always be an activity at home that will become seemingly more important than sports, or it will be a good excuse for not playing sports.
Don’t keep your friends waiting.
It is also advantageous if you do not go to exercise alone, but with a friend, friend or colleague, or a good group formed in the gym or for an aerobics lesson. Weight loss exercises are done with more pleasure, and you can’t keep your friends waiting.
You burn fat all over your body.
When you move, the essential way is. We have previously described in the introduction that you do not lose weight only in certain areas when you start exercising. Fat begins to burn throughout the body. So when you begin intensely with abdominal exercises, you not only strengthen the abdominal muscles, but you also involve other muscles. For example, interval training or HIIT, i.e., very intensive interval training, are good for burning on the abdomen. But you don’t want to focus only on your abdomen, but you would also like some exercises for the buttocks and thighs. Exercises is also important for erectile dysfunction or you can use Fildena 100 or Fildena 150 to improve your potency.
It is best to combine different types of movement.
Of course, the best-combined exercise that takes your body as a whole and the strengthening exercises are varied. Therefore, it is good to alternate between different types of training, such as functional (focusing on the core, etc.), aerobic (aerobics, cycling, jogging, swimming, etc.), fitness (strengthening and stretching exercises with rehabilitation aids, etc.) or strength (strengthening with dumbbells).
How to do it? In practice, this means that you may reserve one day for strength training in the gym, the next day you will practice for a joint aerobics lesson, the third day you will exercise on overalls and the fourth you will go swimming. Your condition will improve in all respects; you will not neglect any body parts.
One-sided training of only certain games can lead to overtraining. If you also feel that you will not lose weight, it may also be because your body has entered a training stereotype. Try some new sports activity, and you will see a shift.
Exercises to burn fat and strengthen the body
Squats with a jump stand with your feet across the width of your shoulders, your toes facing forward, and try to keep your abdomen fired. Inhale with a deep squat, at least 90 ° between the calves and thighs. Then exhale and then jump up and peel off your feet. We help with movement with our hands. From the jump, smoothly squat and continue this way until the number of repetitions is met.
Boxing
Start the exercise in a standing position with your legs apart and set the toes aside. Pull your buttocks and bend your legs slightly at the knees. Put your hands in a fist in front of your chin so that your elbows are glued to your body. With each exhalation, pull one hand alternately (palm still clenched into a fist) forward into space. Box at a faster pace, don’t forget your cloudy belly.
Climber
Perform a support position in which you have your palms on the ground directly under your shoulders and your head in body extensions. Try to keep your abdomen still tucked up (navel sticking to your spine) and not pull your butt up. Alternately pull your knees to your chin at a fast pace.
Englishmen
From a standing position with your feet slightly apart, you move to a squat position, in which you place your palms on the ground in front of you. In this position, you bounce with your feet behind you to the position of the board. You place the body entirely on the ground and then lift yourself again by clicking in a straight line use fildena 120 and vigora 100 for erection. Beware of incorrect bending in the lower back; try to keep the abdominal muscles still fixed. You leave your hands on the ground and jump back into your squats with your feet. From this position, then jump upwards, clapping your hands over your head in the jump.
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There I saw the joy of the music in the eyes
of the little children as much as in the adults' eyes... Their appreciation
was so sincere and great that I suddenly discovered myself
in a different way...
Later on I was invited to teach harp and chamber
music at the Music Academies of Brussels and Beveren.
If you are interested to study harp, please
just call for more information to the secretariat:
If you would like to attend to the music camps,
please contact the organizer Philippe Broeckmeyer for more information
on telephone : 02/507.83.40-44 or write
to him at :
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bookings@amiraharp.com
| 269,417
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Observing Movies On the internet With Wireless Internet
There used to be a time when you could only observe motion pictures although sitting on your couch in the residing space or in a crowded motion picture theatre. Now you can observe them nearly everywhere in your metropolis when you stream them on-line with wireless Web. You can observe films that you have needed to see for many years whilst you happen to be using the bus to perform in the morning or even though you happen to be sitting in a friend’s living area – all you require is the World wide web.
The Net is slowly and gradually shifting the way men and women do every little thing from communicate to watch tv and videos. You employed to have to get in touch with a friend to tell them what you might be performing for meal that night and now you can just speedily ship them an e-mail. You utilised to have to purchase tickets to videos at the movie theatre but now you can observe them on your computer with an Net relationship. The very best component is that acquiring on-line is even obtaining simpler. You can get on the web with WiMax from practically everywhere in your metropolis relatively than being confined to your desk or home.
Not only can you watch all the videos you could image from your computer, but you can do it while you’re out and about in the town relatively than sitting down in your residence or office. You don’t have to be sitting at a laptop desk to stream your preferred films when you have cellular wimax. You can get on the web and view your favorite videos in the morning or afternoon although you’re driving community transportation back again and forth to and from work. You can even get on the web and look at movies even though you happen to be sitting down in a organization partner’s office waiting for your important meeting to start off. Observing a motion picture isn’t going to have to be a huge affair that demands arranging and good timing due to the fact you can do it anyplace that you have World wide web – and now you can have that wi-fi Web connection anywhere in your metropolis!
You won’t have to be concerned about stopping by the motion picture rental keep on the way home from operate when you can just download or stream the film you want to watch on the internet. You will not even have to go to the film theatres and invest funds when you commit just a minor money to get WiMax technological innovation connecting you to the World wide web at all moments and in all spots. You may, nonetheless, want to make investments in some headphones so the other men and women on the bus or in the place of work can not listen to the movie you happen to be observing. ดูหนัง ‘ll just consider you happen to be doing work on an critical company proposal rather than observing “Dumb and Dumber” on your pc.
You are going to in no way have to established aside unique time to watch motion pictures yet again when you can observe them for the duration of your cost-free time all in excess of the metropolis. All you need to do is get cellular wimax and understand about a handful of on the internet internet sites that you can stream or download motion pictures on and you’ll be watching videos all in excess of the town!August 14, 2020
Recent Comments
| 77,026
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4802
April 27, 2012
We're back with another look at the often bumpy Road to the White House. American Politics expert Dr. Jennifer Victor is our special guest -- she'll talk with the rest of the 4802 crew about politics, campaigning, legislating, lobbying and more. As the road to the White House appears to have only two travelers remaining -- we'll have an inside look at what it takes to get to the finish line -- LIVE at 7:30 - tonight!
| 166,639
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TITLE: Linear dependency of nilpotent matrices
QUESTION [1 upvotes]: I would like to prove that four $2\times 2$ nilpotent matrices are always linearly dependent, using the Cayley-Hamilton theorem or the minimal polynomial in some way.
I think I have proved the statement using a "brute force" method, wherein I just squared every possible $2\times 2$ matrix (there's $16$ different kinds) to see if it vanished. I concluded that the only nilpotent $2\times 2$ matrices are $$\begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix}$$ and $$\begin{bmatrix} 0 & 0 \\ 1 & 0 \end{bmatrix}$$ and any other nilpotent $2\times 2$ matrix is a multiple of one of these. So actually, any three nilpotent matrices are always linearly dependent.
I would like to construct something more "sophisticated". I am also not quite sure my "brute force" approach is $100\%$ correct, anyway!
REPLY [4 votes]: Notice that the trace of any nilpotent matrix is $0$. Using that the map
$$\operatorname{tr}:\mathcal M_2(\Bbb F)\to \Bbb F$$
is a linear form then $\ker \operatorname{tr}$ is an hyperplan of $\mathcal M_2(\Bbb F)$ containing the set of nilpotent matrices. Hence $4$ nilpotent matrices doesn't form a basis of $\mathcal M_2(\Bbb F)$ hence they are linearly dependent.
REPLY [2 votes]: For any nilpotent 2x2 matrix $P$ we have $P^2=0$.
Now consider the space $\Bbb R[X]/(X^2)$. It has dimension two and it isomorphic to the space of nilpotents 2x2 matrices.
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\begin{document}
\title{\large\bf Proof of a conjecture related to divisibility properties of binomial coefficients\footnote{This work was supported by the National
Natural Science Foundation of China, Grant No. 11371195. }}
\date{}
\author{Quan-Hui Yang\footnote{Email: yangquanhui01@163.com.}\\
\small School of Mathematics and Statistics,
\\ \small Nanjing University of Information Science and Technology,\\
\small Nanjing 210044, P. R. China}
\maketitle
\vskip 3mm
\begin{abstract} Let $a,b$ and $n$ be positive integers with $a>b$. In this note,
we prove that
$$(2bn+1)(2bn+3){2bn \choose bn}\bigg|3(a-b)(3a-b){2an \choose an}{an\choose bn}.$$ This
confirms a recent conjecture of Amdeberhan and Moll.
{\it 2010 Mathematics Subject Classifications:} 11B65, 05A10
{\it Keywords:} binomial coefficients, $p$-adic order, divisibility properties
\end{abstract}
\section{Introduction}
Let $\mathbb{Z}$ denote the set of all integers. In 2009, Bober \cite{bober} determined all
cases such that
$$\frac{(a_1n)!\cdots (a_kn)!}{(b_1n)!\cdots (b_{k+1}n)!}\in \mathbb{Z},$$
where $a_s\not=b_t$ for all $s,t$, $\sum{a_s}=\sum{b_t}$ and $\gcd(a_1,\ldots,a_k,b_1,\ldots,b_{k+1})=1$.
Recently, Z.-W. Sun \cite{sun12,sun13} studied divisibility properties of binomial
coefficients and obtained some interesting results. For example,
$$2(2n+1){2n \choose n}\bigg| {6n \choose 3n}{3n \choose n},$$
$$(10n+1){3n \choose n}\bigg| {15n \choose 5n}{5n-1 \choose n-1}.$$
Later, Guo and Krattenthaler (see \cite{guo,{guokra}}) got some similar divisibility results.
For related results, one can refer to \cite{Calkin}-\cite{Fine} and \cite{Guo07}-\cite{Razpet}.
Let $$S_n=\frac{{6n \choose 3n} {3n \choose n}}{2(2n+1){2n \choose n}}\quad \text{and} \quad
t_n=\frac{{15n \choose 5n}{5n-1 \choose n-1}}{(10n+1){3n \choose n}}.$$
In \cite{guonew}, Guo proved the following Sun's conjectures.
\noindent{\bf Theorem A.} (See \cite[Conjecture 3(i)]{sun12}.){\em~ Let $n$ be a positive integer. Then
$$3S_n\equiv 0 \pmod {2n+3}.$$}
\noindent{\bf Theorem B.} (See \cite[Conjecture 1.3]{sun13}.){\em ~ Let $n$ be a positive integer. Then
$$21t_n\equiv 0 \pmod {10n+3}.$$}
Some other results are also proved.
T. Amdeberhan and V. H. Moll proposed the following
conjecture (See Guo \cite[Conjecture 7.1]{guonew}).
\begin{conjecture}\label{conj} Let $a,b$ and $n$ be positive integers with $a>b$. Then
$$(2bn+1)(2bn+3){2bn \choose bn}\bigg|3(a-b)(3a-b){2an \choose an}{an\choose bn}.$$
\end{conjecture}
In this note, we give the proof of this conjecture.
\begin{theorem}\label{thm1} Conjecture \ref{conj} is true.
\end{theorem}
\begin{remark} Let $a=3$ and $b=1$. Then Theorem A follows from Theorem \ref{thm1} immediately.
\end{remark}
\section{Proofs}
For an integer $n$ and a prime $p$, we write $p^{k}\|n$ if $p^k|n$ and $p^{k+1}\nmid n$.
We use $\nu_p(n)$ to denote such integer $k$. It is well known that
\begin{equation}\label{eq1}\nu_p{(n!)}=\sum_{i=1}^{\infty}\left\lfloor \frac {n}{p^i}\right\rfloor,
\end{equation}
where $\left\lfloor x\right\rfloor$ denotes the greatest integer not exceeding $x$.
Before the proof of Theorem \ref{thm1}, we give the following lemma.
\begin{lemma}\label{lem1} Let $x$ and $y$ be two real numbers. Then
$\left\lfloor 2x\right\rfloor +\left\lfloor y\right\rfloor \ge \left\lfloor x\right\rfloor +\left\lfloor x-y\right\rfloor +\left\lfloor 2y\right\rfloor.$
\end{lemma}
\begin{proof} Noting that $2x+y=x+(x-y)+2y$, we only need to prove that
$\{2x\}+\{y\}\le \{x\}+\{x-y\}+\{2y\}$, where $\{z\}$ denotes the fractional part of $z$.
We can prove it by comparing $\{x\}$ and $\{y\}$ with 1/2. We leave the proof to the reader.
\end{proof}
\begin{proof}[Proof of Theorem \ref{thm1}] Let
$$T(a,b,n):={2an \choose an}{an \choose bn}\bigg/{2bn \choose bn}=\frac{(2an)!(bn)!}{(an)!(an-bn)!(2bn)!}.$$
By \eqref{eq1}, for any prime $p$, we have
$$\nu_p(T(a,b,n))=\sum_{i=1}^{\infty}\left( \left\lfloor \frac{2an}{p^i}\right\rfloor
+\left\lfloor\frac{bn}{p^i}\right\rfloor-\left\lfloor\frac{an}{p^i}\right\rfloor-\left\lfloor\frac{an-bn}
{p^i}\right\rfloor-\left\lfloor\frac{2bn}{p^i}\right\rfloor\right).$$
By Lemma \ref{lem1}, it follows that each term of $\nu_p(T(a,b,n))$ is nonnegative.
Hence $\nu_p(T(a,b,n))\ge 0$. Therefore, $T(a,b,n)\in \mathbb{Z}$.
Since $\gcd(2bn+1,2bn+3)=1$, it suffices to prove that
$$2bn+1|3(a-b)(3a-b)T(a,b,n)$$ and $$2bn+3|3(a-b)(3a-b)T(a,b,n).$$
Now we first prove the latter part.
Suppose that $p^{\alpha}\| 2bn+3$ with $\alpha\ge 1$. Then we shall prove
\begin{equation}\label{eq5}p^{\alpha}| 3(a-b)(3a-b)T(a,b,n).\end{equation}
Let $p^{\beta}\|a-b$ and $p^{\gamma}\|3a-b$
with $\beta\ge 0$ and $\gamma\ge 0$. Write $\tau=\max\{\beta,\gamma\}$.
If $\alpha \le \tau$,
then \eqref{eq5} clearly holds. Now we assume $\alpha > \tau$.
Suppose that $p\ge 5$.
Next we prove
$$\left\lfloor \frac{2an}{p^i}\right\rfloor+\left\lfloor \frac{bn}{p^i}\right\rfloor-\left\lfloor\frac{an}{p^i}\right\rfloor-\left\lfloor\frac{an-bn}
{p^i}\right\rfloor-\left\lfloor\frac{2bn}{p^i}\right\rfloor=1$$
for $i=\tau+1,\tau+2,\ldots,\alpha$. Noting that $p|2bn+3$ and $p\ge 5$, we have $\gcd(p,n)=1$. Otherwise,
we have $p|3$, a contradiction.
By $p^{\alpha}\| 2bn+3$, it follows that $2bn \equiv p^{\alpha}-3 \pmod {p^{\alpha}}$
and $bn\equiv (p^{\alpha}-3)/2 \pmod {p^{\alpha}}$.
Take an integer $i\in \{\tau+1,\tau+2,\ldots,\alpha\}$. Then
$2bn \equiv p^{i}-3 \pmod {p^{i}}$
and $bn\equiv (p^{i}-3)/2 \pmod {p^{i}}$.
Now we divide into several cases according to the value of $an \pmod {p^i}$.
{\bf Case 1.} $an\equiv t \pmod {p^i}$ with $0\le t<(p^i-3)/2$. It follows that
$2an\equiv 2t \pmod {p^i}$ and $0\le 2t< p^i-3$. We also have $$an-bn\equiv t-(p^i-3)/2+p^i
\pmod {p^i},$$ where $0\le t-(p^i-3)/2+p^i<p^i$.
Hence
\begin{eqnarray*}&&\left\lfloor \frac{2an}{p^i}\right\rfloor+\left\lfloor \frac{bn}{p^i}\right\rfloor-\left\lfloor\frac{an}{p^i}\right\rfloor-\left\lfloor\frac{an-bn}
{p^i}\right\rfloor-\left\lfloor\frac{2bn}{p^i}\right\rfloor\\
&=&\frac{2an-2t}{p^i}+\frac{bn-(p^i-3)/2}{p^i}-\frac{an-t}{p^i}\\
&&-
\left(\frac{an-bn-(t-(p^i-3)/2+p^i)}{p^i}\right)-\frac{2bn-(p^i-3)}{p^i}\\
&=&1.
\end{eqnarray*}
{\bf Case 2.} $an\equiv ({p^i}-3)/2 \pmod {p^i}$. Clearly, $an-bn\equiv 0 \pmod {p^i}$. Since
$\gcd(p,n)=1$, we have $p^i|a-b$. However, $p^{\beta}\|a-b$ and $\beta\le \tau<i$, a contradiction.
{\bf Case 3.} $an\equiv ({p^i}-1)/2 \pmod {p^i}$. It follows that
$$3an-bn\equiv \frac {3({p^i}-1)}{2}-\frac{({p^i}-3)}{2}\equiv 0 \pmod {p^i}.$$
By $\gcd(p,n)=1$, we have $p^i|3a-b$. This contradicts $p^{\gamma}\|3a-b$ and $\gamma<i$.
{\bf Case 4.} $an\equiv t \pmod {p^i}$ with $(p^i+1)/2\le t<p^i$. It follows that
$$2an\equiv 2t-p^i \pmod {p^i}, \quad 0\le 2t-p^i< p^i.$$ We also have $$an-bn\equiv t-(p^i-3)/2
\pmod {p^i},\quad 0\le t-(p^i-3)/2<p^i.$$
Hence
\begin{eqnarray*}&&\left\lfloor \frac{2an}{p^i}\right\rfloor+\left\lfloor \frac{bn}{p^i}\right\rfloor-\left\lfloor\frac{an}{p^i}\right\rfloor-\left\lfloor\frac{an-bn}
{p^i}\right\rfloor-\left\lfloor\frac{2bn}{p^i}\right\rfloor\\
&=&\frac{2an-(2t-p^i)}{p^i}+\frac{bn-(p^i-3)/2}{p^i}-\frac{an-t}{p^i}\\
&&-
\left(\frac{an-bn-(t-(p^i-3)/2)}{p^i}\right)-\frac{2bn-(p^i-3)}{p^i}\\
&=&1.
\end{eqnarray*}
Therefore, $\nu_p(T(a,b,n))\ge \alpha-\tau$, and then $$\nu_p(3(a-b)(3a-b)T(a,b,n))\ge \alpha.$$
That is, \eqref{eq5} holds.
Now we deal with the case $p=3$. If $9|n$, then $3|2bn+3$ and $9\nmid 2bn+3$. It follows that $\alpha=1$,
and then \eqref{eq5} clearly holds.
If $9\nmid n$, then we can follow the proof of the case $p\ge 5$ above. In Case 2, by $an-bn\equiv 0 \pmod {3^i}$,
we have $3^{i-1}|a-b$. In Case 3, we have $3^{i-1}|3a-b$. So, if $i\ge \tau+2$, then $i-1\ge \tau+1$.
It is a contradiction in both cases. Hence
$$\left\lfloor \frac{2an}{3^i}\right\rfloor+\left\lfloor \frac{bn}{3^i}\right\rfloor-\left\lfloor\frac{an}{3^i}\right\rfloor-\left\lfloor\frac{an-bn}
{3^i}\right\rfloor-\left\lfloor\frac{2bn}{3^i}\right\rfloor=1$$
for $i=\tau+2,\tau+3,\ldots,\alpha$.
It follows that $\nu_3(T(a,b,n))\ge \alpha-\tau-1$, and then $$\nu_3(3(a-b)(3a-b)T(a,b,n))\ge \alpha.$$
That is, \eqref{eq5} also holds.
Hence, $2bn+3|3(a-b)(3a-b)T(a,b,n)$.
The proof of $2bn+1|3(a-b)(3a-b)T(a,b,n)$ is very similar. We omit it here.
This completes the proof of Theorem \ref{thm1}.
\end{proof}
| 33,832
|
Humans’ Body Immunity and Natural Pigments
Abstract
The emerging pandemic of Corona Virus Disease (COVID-19) has attracted special concerns regarding the case of very contagious viral infection. Besides the profound waiting toward vaccine development, public awareness was also addressed into the effort of increasing humans’ body immunity. Nowadays, the people are encouraged to consume adequate amount of vitamin C and increase the consumption of fruits and vegetables. In addition, various kinds of food supplements are offered and claimed to give us protection and prevention against viral infection by boosting our immune system. However, the communities are actually lack of proper information regarding the true mechanism of immune system. How is the infection developed in our body? How does the immune system work? Could it be modulated by consuming the bioactive compounds? Here, the stages of infection, the immune system, as well as the role of natural pigments to support the humans’ body immunity will be briefly discussed.
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References
Goldsmith, J. R., and Sartor, R. B., The role of diet on intestinal microbiota metabolism: downstream impacts on host immune function and health, and theraupetic implications, Journal of Gastroenterology 2014, 49, 785-798. doi: 10.1007/s00535-014-0953-z.
Singh, J., and Basu, P. S., Non-nutritive bioactive compounds in pulses and their impact on human health: An overview, Food and Nutrition Sciences, 2012, 3, 1664-1672.
Chandrasekara, A., and Shahidi, F., Herbal beverages: Bioactive compounds and their role in disease risk reduction – A review, Journal of Traditional and Complementary Medicine, 2018, 8, 451-458.
Abbas, A. K., Lichtman, A. H., and Pillai, S., Basic Immunology E-Book: Functions and Disorders of the Immune System. Elsevier Health Sciences, 2019.
Khomich, O. A., Kochetkov, S. N., Bartosch, B., and Ivanov, A. V., Redox biology of respiratory viral infections, Viruses, 2018, 10, 392. doi: 10.3390/v10080392.
Macedo, A. C., Faria, A. O. V., and Ghezzi, P., Boosting the immune system, from science to myth: Analysis the infosphere with Google, Frontiers in Medicine, 2019, 6, 165. doi: 10.3389/fmed.2019.00165.
Nishida, Y., Yamashita, E., and Miki, W., Quenching activities of common hydrophilic and lipophilic antioxidants against singlet oxygen using chemiluminescence detection system, Carotenoid Science, 2007, 11 (6), 16-20.
Havaux, M., Carotenoids as membrane stabilizers in chloroplasts, Trends in Plant Science, 1998, 3 (4), 147-151.
Tan, C., Xue, J., Lou, X., et al., Liposomes as delivery systems for carotenoids: Comparative studies of loading ability, storage stability and in vitro release, Food and Fuction, 2014, 5, 1232-1240. doi: 10.1039/c3fo60498e.
Hosseini, B., Berthon, B. S., Wark, P., and Wood, L. G. Effects of fruit and vegetable consumption on risk of asthma, wheezing and immune responses: A systematic review and meta-analysis, Nutrients, 2017, 9, 341. doi: 10.3990/nu9040341.
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\begin{document}
\title{On the direct product of partial Burnside rings}
\author[Masahiro Wakatake]{Masahiro Wakatake}
\address[Masahiro Wakatake]{Department of Mathematics, Kindai University, 3-4-1 Kowakae, Higashiosaka City, Osaka 577-8502, Japan}
\keywords{Burnside ring ; Coxeter group ; unit group ; parabolic subgroups}
\maketitle
\begin{abstract}
In this paper, we describe the structure of the direct product of partial Burnside rings of relative to the collection of a finite group. In particular, we
t the unit group of the partial Burnside ring relative to the set of all parabolic subgroups of a reducible finite Coxeter group is isomorphic to the direct product of unit groups of the partial Burnside ring of irreducible Coxeter group.
\end{abstract}
\maketitle
\section{Introduction}
Let $G$ be a finite group. The unit group of the Burnside ring of $G$ has been studied (see \cite{Bo07}, \cite{ma82}, \cite{Di79}, \cite{yo90b}).
Yoshida introduced a theory of generalized Burnside ring (GBR for short) with respect to a family of subgroups of $G$ (see \cite{yo90a}). In 2015, the study of the unit group of the GBR relative to a collection of $G$ was started by Idei and Oda (see \cite{io15}). Also it is discussed by \cite{oty16} and \cite{oty17}.
This paper presents some results of the unit group of the GBR of a reducible finite Coxeter group.
The Burnside ring of a finite group $G$ is defined to be the Grothendieck ring of the semi-ring generated by isomorphism classes of finite (left) $G$-sets where the addition and multiplication are given by disjoint unions and Cartesian products. Denote by $\B{G}$ the Burnside ring of $G$. If a family $\D$ of subgroups of $G$ contains $G$ and it is closed under taking conjugation and intersection, then $\D$ is called a collection of $G$. We call the Grothendieck ring of the semi-ring generated by isomorphism classes of finite (left) $G$-sets whose stabilizers of any element lie in a collection $\D$ of $G$ a partial Burnside ring (PBR for shot) relative to $\D$ of $G$. We denote the PBR relative to a collection $\D$ of $G$ by $\B{G,\D}$. Let $W$ be a finite Coxeter group with Coxeter system $(W,S)$. Then a subgroup $P$ of $W$ is called parabolic subgroup if there exists $J\subseteq S$ and $g\in W$ such that $P=g^{-1}\braket{J}g$. If $W$ is a finite Coxeter group, then the set of all parabolic subgroups of $W$ is a collection of $W$ (see \cite{so76}). Denote by $\P_W$ the set of all parabolic subgroups of a Coxeter group $W$.
The paper is organized as follows. In Section 2, we recall the basic definitions and results from GBR and PBR. In Section 3, we consider the relationship between the direct product of PBR, tensor product of PBR, and PBR of direct product. Section 4 determines the structure of the unit group of the PBR relative to the set of all parabolic subgroups of a reducible finite Coxeter group.
For a unitary ring $R$, write $R^\times$ for the unit group of $R$. For the PBR of a finite Coxeter group relative to the set of all parabolic subgroups, the following lemma is well known.
\vspace{5pt}
\noindent{\bf Lemma 4.2 (\cite{so66}).}
Let $W$ be a finite Coxeter group with Coxeter system $(W,S)$. Then the element
\[
\varepsilon_W:=\sum_{J\subseteq S}(-1)^{|J|}[W/\braket{J}]
\]
lies in $\B{W,\P_W}^\times$.
\vspace{5pt}
We call $\varepsilon_W$ in Lemma $\ref{Signunit}$ a {\em sign unit} of $\B{W,\P}$.
The cardinality of the unit group of the PBR relative to $\P_W$ of a finite Coxeter group $W$ is $4$ if $W$ is a irreducible Coxeter group with type ${\rm A}, {\rm B}, {\rm D}, {\rm E}_6, {\rm E}_7$, or ${\rm E}_8$ (see \cite{io15},\cite{oty17}).
The main result of the paper is the following:
\vspace{5pt}
\noindent{\bf Theorem 4.3.}\emph{ Let $W$ be a finite group and let $(W,S)$ be a reducible Coxeter system with $\ell$-irreducible Coxeter system $(W_t,S_t)$ for $t=1,\cdots,\ell$, namely $W=\prod_{i=1}^{\ell}W_i$ and $S=\bigsqcup_{i=1}^\ell S_i$
. Then the following holds:
\begin{itemize}
\item[$(1)$] $|\B{W,\P_W}^\times| = \frac{1}{2^{\ell-1}}\prod_{i=1}^\ell|\B{W_i,\P_{W_i}}^\times|$.
\item[$(2)$] If $\varepsilon_W$ is the sign unit of $\B{W,S}$, then $\varepsilon_W=\prod_{i=1}^{\ell}f_i(\varepsilon_{W_i})$ .
\end{itemize}
}
\vspace{5pt}
Theorem $\ref{theorem}$ is special case of key result (Lemma $\ref{メイン定理B}$), and the result is extension of \cite{ow}.
\section{Preliminaries}
\subsection{Notation}
Let $G$ be a finite group. Denote by $\Sub{G}$ the set of subgroups of $G$. For a family $\D$ of subgroups of $G$ with closed under $G$-conjugation, write $\D^c$ for the set of the conjugacy classes of $\D$. Denote by $(H)$ the set of all $G$-conjugate subgroups of a subgroup $H$ of $G$. Denote by $[X]$ the isomorphism class of finite $G$-set $X$. For a family $\D$ of subgroups of $G$, a $G$-set $X$ is called a $(G,\D)$-set if the stabilizer of any element of $X$ lies in $\D$. If $X$ is a finite set, write $|X|$ for the cardinality of $X$. Denote by $\bigsqcup$ the disjoint union of sets.
\subsection{Partial Burnside rings}
Let $G$ be a finite group. Then the Burnside ring $\B{G}$ of $G$ can be regarded as a free abelian group with basis $\set{[G/H]|(H)\in\Sub{G}^c}$. The multiplication in the ring is given by
\[
[G/H]\cdot[G/K]=\sum_{HgK\in H\backslash G/K} [G/(H\cap gKg^{-1})].
\]
For a family $\D$ of subgroups of $G$ with closed under taking conjugation, we put
\[
\B{G,\D}:= \braket{[G/H]\ |\ (H)\in\D^c}_\Z.
\]
\begin{dfn}
Let $G$ be a finite group and let $\D$ be a family of subgroup of $G$. We call $\D$ a {\em collection of $G$} if $\D$ satisfies the following $3$ conditions:
\begin{itemize}
\item $G\in\D$
\item $H,K\in\D\Rightarrow H\cap K\in\D$
\item $H\in\D,g\in G \Rightarrow gHg^{-1}\in \D$
\end{itemize}
\end{dfn}
If $\D$ is a collection of $G$, then $\B{G,\D}$ is a subring of $\B{G}$, hence $\B{G,\D}$ is called a {\em partial Burnside ring} (PBR for short) {\em relative to $\D$ of $G$}.
For $K\leq G$ and a $G$-set $X$, we set
\[
{\rm inv}_K(X):=\{x\in X|kx=x \text{\ for all\ } k\in K\}.
\]
If $\D$ is a collection of $G$, then for each $K\in\D$, the $\Z$-linear map $\varphi^G_K:\B{G,\D}\rightarrow\Z$ which is induced by $[G/H]\mapsto \#{\rm inv}_K(G/H)$ is a ring homomorphism. By \cite[Theorem 3.10]{yo90a}, the map $\varphi^G=(\varphi^G_K) : \B{G,\D}\rightarrow \prod_{(K)\in\D^c}\Z$ is an injective ring homomorphism, where $\varphi^G$ is called the Burnside homomorphism. Hence the unit group of the Burnside ring of a finite group is an elementary abelian $2$-group.
\section{Direct product of PBRs\label{Direct product of PBRs}}
Let $G_t$ be a finite group, and let $\D_t$ be a collection of $G_t$ for each $t\in\{1,\cdots,\ell\}$. Then $\prod_{i=1}^\ell \D_i$ is a collection of $\prod_{i=1}^\ell G_i$.
\begin{lem}\label{メイン命題}
Let $G_t$ be a finite group, and let $\D_t$ be a collection of $G_t$ for $t=1,2$. Then
\[
\B{G_1,\D_1}\otimes_{\Z}\B{G_2, D_2}\simeq \B{G_1\times G_2, \D_1\times D_2}
\]
as rings.
\end{lem}
\begin{proof}
We put $G:=G_1\times G_2$, $\D:=\D_1\times\D_2$.
Let $X_t$ be a $(G_t, \D_t)$-set for $t=1,2$. The correspondence $(X_1, X_2)\mapsto X_1\times X_2$ induces a bilinear map from $\B{G_1,\D_1}\times_{\Z}\B{G_2, \D_2}$ to $\B{G_1\times G_2, \D_1\times \D_2}$, hence there exists an injective module homomorphism
\[
\pi_2 : \B{G_1,\D_1}\otimes_{\Z}\B{G_2, \D_2} \longrightarrow \B{G_1\times G_2, \D_1\times \D_2}.
\]
If $X_i$ is a $(G_i,\D_i)$-set of cardinality $1$, then $X_1\times X_2$ is a $(G,\D)$-set of cardinality $1$. Moreover, for any $(G_1,\D_1)$-set $X_1,X_2'$, $(G_2,\D_2)$-set $X_2,X_2'$, the map
\[
(X_1\times X_1')\times(X_2\times X_2') \longrightarrow (X_1\times X_2)\times (X_1'\times X_2')
\]
defined by
\[
((x_1,x_1'),(x_2,x_2')) \longmapsto ((x_1,x_2),(x_1',x_2'))
\]
is an isomorphism of $G$-sets. This shows the map $\pi_2$ is a ring homomorphism. The partial Burnside ring $\B{G_t,\D_t}$ is a free abelian group with the basis $\{[G_t/H_t]|(H_t)\in\D_t^c\}$, hence the $\B{G_1,\D_1}\otimes_{\Z}\B{G_2, D_2}$ is a free abelian group with basis $\{[G_1/H_1]\otimes[G_2/H_2]|(H_t)\in\D_t^c,\ t=1,2\}$. Moreover, it is easy to see that $\pi_2([G_1/H_1]\otimes[G_2/H_2])=[(G_1\times G_2)/(H_1\times H_2)]$, so the map $\pi_2$ is surjective. Therefore the map $\pi_2$ is a ring isomorphism.
\end{proof}
In the rest of this section, let $G_t$ be a finite group, and let $\D_t$ be a collection of $G_t$ for $t=1,\cdots,\ell$.
Lemma $\ref{メイン命題}$ shows the following.
\begin{cor}\label{メイン定理A}
The map
\[
\pi_\ell : \bigotimes_{i=1}^{\ell}\B{G_i,\D_i}\longrightarrow\B{\prod_{i=1}^{\ell} G_i \prod_{i=1}^{\ell} \D_i}
\]
induced by a bilinear map
$\prod_{i=1}^{\ell}\B{G_i,\D_i}\rightarrow \B{\prod_{i=1}^{\ell}G_i,\prod_{i=1}^{\ell}\D_i}$
given by
$(X_1,\cdots,X_\ell)\rightarrow X_1\times\cdots\times X_\ell$
is a ring isomorphism.
\end{cor}
The following lemma is well known.
\begin{lem}\label{ringhom}
If the map
\[
\rho_\ell : \prod_{i=1}^{\ell}\B{G_i,\D_i}\longrightarrow \bigotimes_{i=1}^{\ell}\B{G_i,\D_i}
\]
is a canonical $\Z$-linear map, then the map $\rho_\ell$ is a surjective ring homomorphism.
\end{lem}
\begin{comment}
\begin{proof}
We put $G:=\prod_{i=1}^{\ell}G_i$, $\D:=\prod_{i=1}^{\ell}\D_i$. If
$X_i$ is a $(G_i,\D_i)$-set of cardinality $1$, $\rho_\ell(\prod_{i=1}^{\ell}X_i)=\bigotimes_{i=1}^{\ell}X_i$ is a $(G, \D)$-set of cardinality $1$. Moreover for any $(x_1,\cdots ,x_\ell), (x_1',\cdots ,x_\ell')\in\prod_{i=1}^{\ell}\B{G_i,\D_i}$,
\begin{eqn*}
\rho_\ell((x_1,\cdots ,x_\ell)\cdot(x_1',\cdots ,x_\ell'))&=&x_1x_1'\otimes\cdots\otimes x_\ell x_\ell'\\
&=&(x_1\otimes\cdots\otimes x_\ell) (x_1'\otimes\cdots\otimes x_\ell')\\
&=&\rho_\ell(x_1,\cdots ,x_\ell)\rho_\ell(x_1',\cdots ,x_\ell'). \end{eqnarray*}
Hence the map $\rho_\ell$ is a ring homomorphism. Obviously, the $\rho$ is surjective, completing the proof.
\end{proof}
\end{comment}
For each $i\in\{1,\cdots,\ell\}$ let $G_i$ be a finite group and let $\D_i$ be a collection of $G_i$. Then for each $j\in\{1,\cdots,\ell\}$ it is easy to see that the ring homomorphism \[
f_j:\B{G_j,\D_j}\longrightarrow\B{\prod_{i=1}^{\ell}G_i,\prod_{i=1}^{\ell}\D_i}
\]
which is induced by $[G_j/H]\longmapsto [\prod_{i=1}^{\ell}G_i/ \widehat{H}_j]$ is injective, where $\widehat{H}_j=G_1\times\cdots\times G_{j-1}\times H\times G_{j+1}\times\cdots\times G_\ell$.
\begin{lem}\label{バーンサイド準同型の分解補題}
If $x_t \in \B{G_t,\D_t}$ and $H_t\in \D_t$ for $t=1,\cdots,\ell$, then
\[
\varphi^G_{\prod_{i=1}^{\ell}H_i}(f_j(x_j))=\varphi^{G_j}_{H_j}(x_j) \quad (j=1,\cdots,\ell),
\]
where $G=\prod_{i=1}^{\ell}G_i$ and $\widehat{H_j}_j=G_1\times\cdots\times G_{j-1}\times H_j\times G_{j+1}\times\cdots\times G_\ell$.
\end{lem}
\begin{proof}
For $K\in\D_j$,
\[
\begin{split}
[G/\widehat{K}_j]=&[(G_1/G_1)\times\cdots\times (G_{j-1}/G_{j-1})\times (G_j/K)\\ &\times (G_{j+1}/G_{j+1})\times\cdots\times (G_\ell/G_\ell)]
\end{split}
\]
in $\B{G,\prod_{i=1}^{\ell}\D_i}$. Hence
\begin{eqnarray*}
\varphi^{G}_{\prod_{i=1}^{\ell}H_i}(f_j([G_j/K]))
&=&\left|{\rm inv}_{\prod_{i=1}^{\ell}H_i}(G/\widehat{K}_j)\right|\\
&=&\varphi^{G_j}_{H_j}([G_j/K_j]).
\end{eqnarray*}
Since $\varphi^G$ and $f_j$ are ring homomorphisms, this completed the proof of the lemma.
\end{proof}
\begin{lem}\label{メイン定理B}
For each $t\in\{1,\cdots,\ell\}$ let $G_t$ be a finite group and let $\D_t$ be a collection of $G_t$. Then,
\[
\left|\prod_{i=1}^{\ell}\B{G_i,\D_i}^\times\right|=2^{\ell-1} |\B{\prod_{i=1}^{\ell} G_i, \prod_{i=1}^{\ell} \D_i}^\times|.
\]
In particular, if $|\B{G_i,\D_i}^\times|=2^{r_i+1}$ and $\B{G_i,\D_i}^\times=\braket{-1_{\B{G_i}},u_1^{(i)},\cdots,u_{r_i}^{(i)}}$ for each $i\in\{1,\cdots,\ell\}$, then
\[
\begin{split}
\B{&\prod_{i=1}^{\ell} G_i, \prod_{i=1}^{\ell} \D_i}^\times\\
&=\left<\{-1_{\B{G}}\}\cup\bigsqcup_{i=1}^\ell\{f_i(u_1^{(i)}),\cdots,f_i(u_{r_i}^{(i)})\}\right>.
\end{split}
\]
\end{lem}
\begin{proof}
Now, $\bigotimes_{i=1}^{\ell}\B{G_i,\D_i}$ is isomorphic to $\B{\prod_{i=1}^{\ell} G_i, \prod_{i=1}^{\ell} \D_i}$ by Corollary $\ref{メイン定理A}$. The map
\[
\rho_\ell^\times : \prod_{i=1}^{\ell}\B{G_i,\D_i}^\times\longrightarrow \left(\bigotimes_{i=1}^{\ell}\B{G_i,\D_i}\right)^\times,
\]
which obtained by $\rho_\ell$ in Lemma $\ref{ringhom}$, is a group homomorphism. It suffices to show that $|\ker\rho_\ell^\times|=2^{\ell-1}$. We put $G:=\prod_{i=1}^{\ell}G_i$, $\widehat{H}_j=G_1\times\cdots\times G_{j-1}\times H\times G_{j+1}\times\cdots\times G_\ell\leq G$ for $H\in\D_j$.
Since if $K_t\in\D_t$ for each $t=1,\cdots,\ell$, then
\[
\begin{split}
\pi_\ell([G_1/K_1]\otimes\cdots\otimes[G_\ell/K_\ell])&\\
=\prod_{i=1}^{\ell}[G/G_1\times\cdots\times G_{i-1}\times &K_i\times G_{i+1}\times\cdots G_\ell],
\end{split}
\]
Lemma $\ref{バーンサイド準同型の分解補題}$ shows that for any $H,H'\in \D_j$ and $(x_1,\cdots,x_\ell)\in\ker\rho_\ell^\times$,
\[
\begin{split}
\varphi^G_{\widehat{H}_j}\circ\pi_\ell(x_1\otimes\cdots&\otimes x_\ell)\\
&=\varphi^G_{\widehat{H}_j}(\prod_{i=1}^{\ell}f_i(x_i))\\
&=\varphi^{G_1}_{G_1}(x_1)\cdots\varphi^{G_j}_{H}(x_j)\cdots \varphi^{G_\ell}_{G_\ell}(x_\ell)\\
&=\varphi^G_{\widehat{H}_j}\circ\pi_\ell(1_{\B{G_1}}\otimes\cdots\otimes1_{\B{G_\ell}})\\
&=1,
\end{split}
\]
and
\[
\begin{split}
\varphi^G_{\widehat{H'}_j}\circ\pi_\ell(x_1\otimes\cdots&\otimes x_\ell)\\
&=\varphi^G_{\widehat{H'}_j}(\prod_{i=1}^{\ell}f_i(x_i))\\
&=\varphi^{G_1}_{G_1}(x_1)\cdots\varphi^{G_j}_{H'}(x_j)\cdots \varphi^{G_\ell}_{G_\ell}(x_\ell)\\
&=\varphi^G_{\widehat{H'}_j}\circ\pi_\ell(1_{\B{G_1}}\otimes\cdots\otimes1_{\B{G_\ell}})\\
&=1,
\end{split}
\]
for $j=1,\cdots,\ell$. Thus we obtain $\varphi^{G_j}_H(x_j)=\varphi^{G_j}_{H'}(x_j)$ and $\varphi^{G_j}_H(x_j)$ has a value of $1$ or $-1$. Furthermore by injectivity of the Burnside homomorphism $\varphi^{G_j}$, $x_j=1_{\B{G_i}}$ or $-1_{\B{G_i}}$ for $j=1,\cdots,\ell$. Therefore
\[
|\ker\rho_\ell^\times|=\sum_{k=0}^{\lfloor\frac{\ell}{2}\rfloor}\binom{\ell}{2k}=2^{\ell-1}.
\]
Next we assume that $|\B{G_i,\D_i}^\times|=2^{r_i+1}$ and $\B{G_i,\D_i}^\times=\braket{-1_{\B{G_i}},u_1^{(i)},\cdots,u_{r_i}^{(i)}}$ for each $i\in\{1,\cdots,\ell\}$. We put
\[
L:=\{-1_{\B{G}}\}\cup\bigsqcup_{i=1}^\ell\{f_i(u_1^{(i)}),\cdots,f_i(u_{r_i}^{(i)})\}.
\]
Since for $H\in\D_j, K\in\D_k (j<k)$,
\[
[G/\widehat{H}_j]\cdot[G/\widehat{K}_k]=[G/G_1\times\cdots\times H\times\cdots\times K\times\cdots G_\ell]
\]
and $f_j([G/H_j])=[G/\widehat{H}_j]$, it is easy to see that
\[
f_j(u^{(j)}_k)\notin\braket{L\backslash f_j(u^{(j)}_k)}
\]
for any $j,k\in\{1,\cdots,\ell\}$. Therefore
\[
\begin{split}
|<L>|&=2^{1+r_1+\cdots+r_\ell}\\
&=\frac{1}{2^{\ell-1}}\left|\prod_{i=1}^{\ell}\B{G_i,\D_i}^\times\right|\\
&=|\B{\prod_{i=1}^{\ell} G_i, \prod_{i=1}^{\ell} \D_i}^\times|.
\end{split}
\]
This completes the proof.
\end{proof}
\section{PBR of a reducible finite Coxeter group}
\begin{dfn}
Let $W$ be a finite Coxeter group with Coxeter system $(W,S)$. Then a subgroup $P$ is called parabolic subgroup if there exists $J\subseteq S$ such that $(P)=(\braket{J})$. The $\P_W$ denote the set of all parabolic subgroups of $W$.
\end{dfn}
Let $W$ be a finite Coxeter group. Then the set $\P_W$ of all parabolic subgroups of $W$ becomes a collection (see \cite{so76}). So $\B{W,\P_W}$ is the PBR of $W$ relative to $\P_W$.
For the PBR of a finite Coxeter group relative to the set of all parabolic subgroups, the following lemma is well known.
\begin{lem}[\cite{so66}]\label{Signunit}
Let $W$ be a finite Coxeter group with Coxeter system $(W,S)$. Then the element
\[
\varepsilon_W:=\sum_{J\subseteq S}(-1)^{|J|}[W/\braket{J}]
\]
lies in $\B{W,\P_W}^\times$. Moreover, for $P\in\P_W$ with $(P)=(\braket{J})$,
\[
\varphi^W_P(\varepsilon_W)=(-1)^{|J|}.
\]
\end{lem}
We call $\varepsilon_W$ in Lemma $\ref{Signunit}$ a {\em sign unit} of $\B{W,\P}$.
\begin{thm}\label{theorem}
Let $W$ be a finite group and let $(W,S)$ be a reducible Coxeter system with $\ell$-irreducible Coxeter system $(W_t,S_t)$ for $t=1,\cdots,\ell$, namely $W=\prod_{i=1}^{\ell}W_i$ and $S=\bigsqcup_{i=1}^\ell S_i$. Then the following holds:
\begin{itemize}
\item[$(1)$] $|\B{W,\P_W}^\times| = \frac{1}{2^{\ell-1}}\prod_{i=1}^\ell|\B{W_i,\P_{W_i}}^\times|$, where $\P_{W_i}$ is the set of all parabolic subgroups of $W_i$ for $i=1,\cdots,\ell$.
\item[$(2)$] If $\varepsilon_W$ is the sign unit of $\B{W,S}$, then $\varepsilon_W=\prod_{i=1}^{\ell}f_i(\varepsilon_{W_i})$ .
\end{itemize}
\end{thm}
\begin{proof}
Since $P_W=\prod_{i=1}^{\ell}P_{W_i}$, we obtain $(1)$ by Lemma $\ref{メイン定理B}$. Let $J$ be a subset of $S$ with $J=\bigsqcup_{i=1}^\ell J_i$ and $J_i\subseteq S_i$ for each $i$, then $\braket{J}=\braket{\bigsqcup_{i=1}^\ell J_i}=\prod_{i=1}^\ell\braket{J_i}$. Therefore for $P\in\P_W$ if $(P)=(\braket{J})$, then we have
\begin{eqnarray*}
\varphi^W_P(\prod_{i=1}^{\ell}f_i(\varepsilon_{W_i}))&=&\varphi^W_{\prod_{i=1}^{\ell}\braket{J_i}}(\prod_{i=1}^{\ell}f_i(\varepsilon_{W_i}))\\
&=&\prod_{i=1}^{\ell}\varphi^{W_i}_{\braket{J_i}}(\varepsilon_{W_i})\\
&=&(-1)^{\sum_{i=1}^\ell|J_i|}\\
&=&(-1)^{|\bigsqcup_{i=1}^\ell J_i|}\\
&=&(-1)^{|J|}\\
&=&\varphi^W_P(\varepsilon_W)
\end{eqnarray*}
by Lemma $\ref{バーンサイド準同型の分解補題}$. Therefore we obtain (2) by injectivity of the Burnside homomorphism $\varphi^W$.
\end{proof}
For a unit group of the PBR relative $\P_W$ of a irreducible finite Coxeter group ,there are the following theorems.
\begin{thm}[\cite{io15}]
Let $W=W(A)$ is a Coxeter group of type {\rm A}. Then,
\[
|\B{W,\P_W}^\times| = 4.
\]
\end{thm}
\begin{thm}[\cite{oty17}]
Let $W$ be a irreducible finite Coxeter group. If the type of $W$ is ${\rm B}, {\rm D}, {\rm E}_6, {\rm E}_7$ or ${\rm E}_8$, then
\[
|\B{W,\P_W}^\times| = 4.
\]
\end{thm}
It is easy to see that the following by calculation.
\begin{thm}
Let $W=W({\rm I}_2(m))$ is a Coxeter group of type ${\rm I}_2(m)$. Then,
\[
|\B{W,\P_W}^\times| = 4.
\]
\end{thm}
Also Theorem $\ref{theorem}$ shows the following.
\begin{cor}
Let $W$ be a finite group and let $(W,S)$ be a reducible Coxeter system with $\ell$-irreducible Coxeter system $(W_t,S_t)$ for $t=1,\cdots,\ell$. If $|\B{W_t,\P_{W_t}}^\times|=4$ for any $t\in\{1,\cdots,\ell\}$, then the following holds:
\begin{itemize}
\item[$({\rm i})$] $|\B{W,\P_W}^\times|=2^{\ell+1}$.
\item[$({\rm ii})$] $\B{W,\P_W}^\times=\braket{-1_{\B{W}},f_1(\varepsilon_{W_1}),\cdots,f_\ell(\varepsilon_{W_\ell})}$, where $\varepsilon_{W_i}$ is a sign unit of $\B{W_i,\P_{W_i}}$ for each $i\in\{1,\cdots,\ell\}$.
\end{itemize}
\end{cor}
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\begin{document}
\title{Counterexamples in the Levin-Wen model, group categories, and Turaev unimodality}
\author{Spencer D. Stirling}
\email{stirling@physics.utah.edu}
\affiliation{Department of Physics and Astronomy,
University Of Utah, Salt Lake City, UT 84112, USA}
\affiliation{Department of Mathematics,
University of Utah, Salt Lake City, UT 84112, USA}
\date{\today}
\newcommand{\Vcat}{\mathcal{V}}
\newcommand{\id}{\text{id}}
\newcommand{\grp}{\mathcal{D}}
\newcommand{\trio}{{(\grp,q,c)}}
\newcommand{\triotrunc}{{(\grp,q)}}
\newcommand{\grpcat}{\mathscr{C}_\triotrunc}
\newcommand{\qmodz}{\mathbb{Q}/\mathbb{Z}}
\newcommand{\Hom}{\text{Hom}}
\newcommand{\modfunct}{\mathscr{F}}
\newcommand{\unitobj}{\mathbb{1}}
\newcommand{\idmat}{I}
\newcommand{\cplx}{\mathbb{C}}
\newcommand{\real}{\mathbb{R}}
\newcommand{\integers}{\mathbb{Z}}
\newcommand{\ribv}{\text{Rib}_\Vcat}
\newcommand{\ribi}{\text{Rib}_I}
\newcommand{\progplanar}{\text{ProgPlanar}}
\newcommand{\progthreed}{\text{Prog3D}}
\newcommand{\polarplanar}{\text{PolarPlanar}}
\newcommand{\piv}{\text{piv}}
\newcommand{\qtr}{\text{tr}_q}
\newcommand{\qdim}{\text{dim}_q}
\setcounter{secnumdepth}{1}
\numberwithin{equation}{section}
\theoremstyle{plain}
\newtheorem{theorem}[equation]{Theorem}
\newtheorem{proposition}[equation]{Proposition}
\newtheorem{corollary}[equation]{Corollary}
\newtheorem{lemma}[equation]{Lemma}
\newtheorem{fact}[equation]{Fact}
\newtheorem{conjecture}[equation]{Conjecture}
\theoremstyle{definition}
\newtheorem{definition}[equation]{Definition}
\newtheorem{example}[equation]{Example}
\theoremstyle{remark}
\newtheorem{remark}[equation]{Remark}
\begin{abstract}
We remark on the claim that the string-net model of Levin and Wen \cite{levin_wen}
is a microscopic
Hamiltonian formulation of the Turaev-Viro topological quantum field theory \cite{turaevviro}.
Using simple counterexamples
we indicate where interesting extra structure may be needed in the Levin-Wen model for
this to hold (however we believe that some form of the correspondence is true).
In order to be accessible to the condensed matter community we provide a very brief and
gentle introduction to the relevant concepts in category theory (relying heavily on analogy
with ordinary group representation theory). Likewise, some physical ideas are
briefly surveyed for
the benefit of the more mathematical reader.
The main feature of group categories under consideration is Turaev's unimodality. We pinpoint
where unimodality should fit into the Levin-Wen construction, and show that the simplest example
\cite{levin_wen} fails to be unimodal.
Unimodality is straightforward to compute for group categories, and we provide a complete classification
at the end of the paper.
\end{abstract}
\maketitle
\section{Introduction}
In this note we briefly consider some aspects in the
relationship between the Levin-Wen string-net model \cite{levin_wen}
and the Turaev-Viro/Barrett-Westbury state sum model \cite{turaev,barrettwestbury}.
Although some work towards a correspondence proof has
been provided \cite{rasetti1,rasetti2}, we emphasize an important extra
structure appears necessary in the Levin-Wen model. This structure may have physical implications.
Although the details come from category theory, for the physics community we use simple
analogies with ordinary group representation theory. Also, our discussion of the physics is kept
brief and simple so that interested mathematicians may have access.
The key issue is that $6$j-symbols with \textit{tetrahedral symmetry} play a defining role in both models.
In order to build $6$j-symbols with tetrahedral symmetry,
Turaev's construction uses the \textit{unimodality} condition.
In particular unimodality assures that the ``band-breaking'' maneuver depicted in
Fig~(\ref{fig:turaevbandbreak}) is well-defined.
However, in the Levin-Wen model the analogous maneuver
(see Fig~(\ref{fig:unimodalmove})) is implicitly allowed without restriction. To highlight
how this may be problematic, we show here that the first example computed by Levin-Wen
is, in fact, \textbf{not} unimodal. This implies that the Turaev construction cannot be applied.
This example, along with most of the examples computed by Levin-Wen, are
\textit{group categories} \cite{stirling_thesis}.
At the end we formulate a theorem that clarifies conditions when a group category is unimodal.
Before proceeding into the heart of the examples, we mention the recent work of
Hong \cite{seungmoonhong} that (partially) generalizes the Levin-Wen model to unitary spherical
categories using so-called \textit{mirror conjugate symmetry} rather than tetrahedral
symmetry. It would be interesting to explore the examples considered here in this
more general context.
\section{Levin-Wen model}
Levin-Wen \cite{levin_wen} consider a model on a fixed trivalent graph (with
oriented edges) embedded in $2$d. A
typical configuration is pictured in Fig~(\ref{fig:samplestringnet}) where
each edge is labelled by a \textbf{string type} $j$. There are finitely-many string
types, hence we simply refer to them by number $\{0,1,\ldots,N\}$.
There is a duality $j\mapsto j^*$
that satisfies $()^{**}=()$, and the $0$ label means ``no string'', hence we require
$0^*=0$. We are allowed to reverse the orientation
of an edge if we reverse its label $j\mapsto j^*$.
\begin{figure}[!htb]
\centering
\scalebox{1}{\input{samplestringnet.pspdftex}}
\caption{A sample string-net configuration: a trivalent lattice in $2$-dimensions
with strings ``colored'' by integer string types $i$. In category language the labels are
simple objects $V_i$. The dashed lines indicate that
the string-net continues outside of the picture.}
\label{fig:samplestringnet}
\end{figure}
In order to quantize this Levin and Wen introduce a Hilbert space. It is defined by
promoting every configuration $X$ to be an orthonormal basis vector $\vert X\rangle$. An arbitrary state
in the Hilbert space is a formal linear combinations of these.
Given any string-net configuration $\vert X\rangle$,
an arbitrary state $\vert\Phi\rangle$ has an associated probability amplitude
$\Phi(X):=\langle X|\Phi \rangle$ of being in that particular configuration.
From now on we do not differentiate between states $\vert \Phi\rangle$ and their wavefunctions $\Phi$.
We are interested in a subspace of states $\{\Phi\}$ that encodes the topological information of the phase.
One of their main results \cite{levin_wen} is that $\{\Phi\}$ can be realized as the ground state
subspace of an exactly-soluble Hamiltonian.
Rather than write down an explicit Hamiltonian, it is instructive to review the
strategy that Levin-Wen use to find the topological subspace $\{\Phi\}$. The idea
is based on \textit{renormalization semigroup (RG)} flow. For concreteness suppose that
we start with a (possibly too complicated) Hamiltonian $H$.
Usually $H$ has some parameters that can be adjusted (e.g. coupling constants),
and as we adjust these parameters
we also deform its ground state(s) $\Phi$ (see Fig~(\ref{fig:rgflow})).
Hence we actually have a family $\{(H_\alpha,\Phi_\alpha)\}_\alpha$ of
different (but related) theories where $\alpha$ denotes all of the adjustable parameters.
\begin{figure}[!htb]
\centering
\scalebox{1}{\input{rgflow.pspdftex}}
\caption{Cartoon phase diagram}
\label{fig:rgflow}
\end{figure}
On the other hand, often we do not know the exact Hamiltonian $H$ (nor the true parameters $\alpha$), or if we
do then its complexity is intractable. Fortunately, for a fixed set of parameters $\alpha$
(say in the ``$A$'' region in Fig~(\ref{fig:rgflow}))
the Hamiltonian $H_\alpha$ is usually
approximated by a much simpler \textit{effective} Hamiltonian $H_{A(\alpha)}$
(equipped with its own adjustable parameters $A(\alpha)$ that are in principle
derived from the original parameters $\alpha$).
If we adjust the original parameters $\alpha$ for $H_\alpha$ slightly, then
the effective parameters $A(\alpha)$ adjust slightly.
However, if we make a large adjustment to the parameters $\alpha$ (e.g. $\alpha$ is adjusted
into the ``$B$'' region) then
we cannot expect that the same approximations remain valid.
Instead a different set of approximations may be appropriate, yielding a
different form for the effective Hamiltonian $H_{B(\alpha)}$.
In this way the family of theories $\{(H_\alpha,\Phi_\alpha)\}_\alpha$ is carved into
\textbf{phases}. A phase $A$ is a region of parameters $\alpha$ where the Hamiltonians can all
can be approximated similarly. Although the effective Hamiltonians still depend on adjustable
parameters $A(\alpha)$, the Hamiltonians are of the same \textit{form}.
Assume that we pick a Hamiltonian $H_{A}$ in the $A$ phase (we no longer refer
to the original parameters $\alpha$).
The RG flow procedure involves performing an iterated \textit{scaling out} while averaging the
details. From experience we expect that this preserves the form of the Hamiltonian, but may
adjust the coupling constants, etc. In other words, scaling is just a particular recipe to
flow to a new Hamiltonian $H_{A^\prime}$ that is in the same phase. The new Hamiltonian is
simpler because at each scaling iteration we average out the details.
Eventually scaling produces a system that is minimally simplistic, and further scaling does
not lead to further simplification. This produces a scale-invariant fixed point theory
$(H,\Phi)_{\text{fixed}}$,
i.e. a conformal field theory (CFT).
Since $\Phi_{\text{fixed}}$ is scale independent the
long-range properties of any other groundstate $\Phi_A$ in the phase
are well-approximated by $\Phi_{\text{fixed}}$.
Hence in the following we concentrate only on $\Phi_{\text{fixed}}$,
and we drop the ``fixed'' subscript.
\subsection{6j-symbols}
Since the scale-invariant fixed point is somehow the ``simplest'' system, we
have the best chance of identifying the topological subspace there.
On the other hand, Levin-Wen make the ansatz that the topological phase is completely determined
by the data from a \textbf{modular tensor category} \cite{turaev}.
Combining these ideas,
they propose a set of rules at the fixed point that determines how any state $\Phi$ in the topological subspace
transforms under graphical deformations.
Before describing the rules, it is useful to discuss the origins of the basic data in category
theory. Actually, they really
only need some of the data, starting
with the list of string types $\{0,\ldots,N\}$ described above. In category language the string types are
\textbf{simple objects} $\{V_0,V_1,\ldots,V_N\}$, and they can be viewed as generalized irreducible
group representations. They satisfy fusion rules (generalized Clebsch-Gordan rules)
\begin{equation}
\label{eq:fusionrules}
V_i\otimes V_j = \oplus N^k_{ij}\,V_k\quad N^k_{ij}\in\integers
\end{equation}
Order is important in the generalized tensor product that we encounter in
category theory (i.e. $V_i\otimes V_j\neq V_j\otimes V_i$). From now on
we denote the integers $\{0,\ldots,N\}$ by $I$.
For simplicity, Levin and Wen restrict themselves to fusion rules satisfying
\begin{equation}
N^k_{ij}=0\text{ or }1
\end{equation}
in which case the fusion rules are called \textbf{branching rules}.
The dual representation $(V_i)^*$ of an irrep is also an irrep,
hence we denote it by $V_{i^*}$ with the
convention that $i^{**}=i$. We shall see that this harmless-looking notation is at the
core of the problem, i.e. in a category this rule is subject to some conditions that
are automatically satisfied in ordinary group representation theory.
Given the simple objects and branching rules $\delta_{ijm}:=N^{m^*}_{ij}=0\text{ or }1$,
the necessary data is provided by
a system of tensors
\begin{equation}
(F^{ijm}_{kln},d_i)
\end{equation}
satisfying some consistency equations (there may be many solutions)
\begin{align}
& F^{ijm}_{j^* i^* 0}=\frac{\sqrt{d_m}}{\sqrt{d_i}\sqrt{d_j}}\delta_{ijm}\\
& F^{ijm}_{kln}=F^{lkm^*}_{jin}=F^{jim}_{lkn^*}=
F^{imj}_{k^*nl}\frac{\sqrt{d_m}\sqrt{d_n}}{\sqrt{d_j}\sqrt{d_l}}\nonumber\\
& \sum_{n\in I}F^{mlq}_{kp^*n}F^{jip}_{mns^*}F^{js^*n}_{lkr^*}=F^{jip}_{q^*kr^*}F^{riq^*}_{mls^*}\nonumber
\end{align}
The first line is a normalization condition. The second is the tetrahedral symmetry, and the
third is the Biedenharn-Elliott identity.
If we know the underlying modular tensor category then we can find these tensors explicitly.
$F^{ijm}_{kln}\in\cplx$ is the \textbf{quantum $6j$-symbol} and
each $d_i\in\cplx$ is the \textbf{quantum dimension}
of $V_i$.
\footnote{More appropriately these should be called ``braided $6j$-symbol'' and
``braided dimensions'' since ordinary group representation theory is also ``quantum'', i.e.
bosons and fermions.}
These notions are generalizations of the ordinary $6j$-symbols and vector space dimensions
associated to irreps.
Returning to a wavefunction $\Phi$ in the topological subspace, at the fixed point
Levin-Wen propose transformation rules as in Fig~(\ref{fig:wavefunctionansatz})
(this list is not complete).
Since any graph
can be popped down to nothing using these rules, $\Phi$ is determined uniquely (however if we
embed the graph in a $2$d surface with nontrivial topology then $\Phi$ is not unique).
\begin{figure}[!htb]
\centering
\input{wavefunctionansatz.pspdftex}
\caption{Conditions to determine fixed-point ground state $\Phi$}
\label{fig:wavefunctionansatz}
\end{figure}
\section{Group categories}
Group categories (or \textit{pointed} categories) have been studied by a variety of
authors in a variety of contexts (e.g. \cite{quinn} \cite{frolich_kerler} \cite{joyal_street}).
We recommend \cite{stirling_thesis} for further
details and compatible notation.
\subsection{Braiding}
Ordinary group representation theory is a reasonable prototype to model modular tensor
categories.
\footnote{Actually we neither require nor desire the full structure of modular tensor categories,
hence instead we can consider the simpler structure of \textbf{semisimple ribbon Ab-categories}.
It happens that ordinary group representation theory is a closer prototype for semisimple ribbon
categories anyway.}
Consider a many-body bosonic or fermionic system in $(2+1)$-dimensions.
Each elementary particle is a copy of an irreducible representation $V_i$.
If we have two particles then we have the tensor product (again ordering is
important)
\begin{equation}
V_i\otimes V_j
\end{equation}
(which, of course, can be decomposed using the fusion rules in equation~(\ref{eq:fusionrules})).
For bosons we have an exchange rule
\begin{align}
\text{Perm}:&V_i\otimes V_j\xrightarrow{\sim} V_j\otimes V_i\\
&v\otimes w\mapsto w\otimes v\quad\quad v\in V_i, w\in V_j\nonumber
\end{align}
and for fermions we have a different exchange rule
\begin{align}
\text{Perm}:&V_i\otimes V_j\xrightarrow{\sim} V_j\otimes V_i\\
&v\otimes w\mapsto -w\otimes v\quad\quad v\in V_i, w\in V_j\nonumber
\end{align}
In both cases we have
\begin{equation}
\text{Perm}^2=\idmat\quad\quad\text{(the identity matrix)}
\end{equation}
In this way both bosonic and fermionic systems are considered ``commutative'' since we know what
happens when we exchange the particles $V_i\otimes V_j\xrightarrow{\sim}V_j\otimes V_i$ (and two permutations
is equivalent to doing nothing).
Confusingly, in category language they are both called \textbf{symmetric}
theories (despite the usual physical nomenclature ``antisymmetric'' for fermions).
Categories allow a much more interesting \textit{weakened} form of commutativity: \textbf{braiding}.
Here we have invertible braiding matrices
\begin{equation}
c_{V_i,V_j}:V_i\otimes V_j\xrightarrow{\sim} V_j\otimes V_i
\end{equation}
that do \textit{not} satisfy $c_{V_j,V_i}\circ c_{V_i,V_j}=\idmat$, but rather a more
elaborate set of conditions - the \textbf{hexagon relations}.
For details concerning the hexagon relations
we refer the reader to standard references
\cite{turaev}, \cite{bakalov_kirillov}, \cite{kassel}, \cite{joyal_street}. It turns out that the hexagon
relations come from an obvious picture.
\subsection{Braiding for group categories}
The hexagon relations for group categories are exhaustively solved \cite{stirling_thesis}.
In fact the structure is much richer than will be apparent here.
Every group category $\grpcat$ is constructed from some fundamental data $\triotrunc$. $\grp$ is
any finite abelian group, and $q$ is a function on $\grp$ that returns
a complex phase $\exp(2\pi i q)$. Since $q$ is a phase it is only well-defined up to
the interval $[0,1]$. In fact $q$ takes values in $\qmodz$. The function $q$ must also be a
\textit{pure quadratic form}.
To understand the quadratic form the reader should think about quadratic functions on real numbers
$q(x):=\frac{1}{2}x^2$. We can easily define an induced bilinear function:
$b(x,y):=q(x+y)-q(x)-q(y)=\frac{1}{2}(x+y)^2-\frac{1}{2}(x)^2-\frac{1}{2}(y)^2=xy$.
Likewise, a pure quadratic form is a function such that the induced function
\begin{align}
&b:\grp\otimes\grp\rightarrow\qmodz\\
&b(x,y) := q(x+y)-q(x)-q(y) \pmod{1}\nonumber
\end{align}
is bilinear. The adjective ``pure'' means
\begin{equation}
\label{eq:purity}
q(nx)\equiv n^2 q(x) \pmod{1}\quad\quad n\in\integers
\end{equation}
\begin{example}
\label{ex:z2groupcategory}
The simplest example is when $\grp=\integers_2=\{\hat{0},\hat{1}\}_+$, i.e.
$\hat{1}+\hat{1}=\hat{2}\equiv \hat{0} \pmod{2}$ (we use the hat to differentiate
elements of the group from numbers in $\qmodz$).
Then, since
\begin{equation}
2 b(\hat{1},\hat{1})\equiv b(2\hat{1},\hat{1})\equiv b(\hat{2},\hat{1})
\equiv b(\hat{0},\hat{1})\equiv 0 \pmod{1}
\end{equation}
(bilinearity is used in every step, and $b$ is valued in $\qmodz$) we have
two possibilities for $b$:
\begin{equation}
b(\hat{1},\hat{1}) \equiv 0 \pmod{1}\quad\text{ or }\quad b(\hat{1},\hat{1}) \equiv \frac{1}{2} \pmod{1}
\end{equation}
Because of bilinearity the function $b$ is fully determined by its values on the generator $\hat{1}$
of the group $\grp$. The same statement also holds for the quadratic form $q$ by using the purity
condition in equation~(\ref{eq:purity})).
Suppose we consider the first case, i.e. $b(\hat{1},\hat{1}) \equiv 0 \pmod{1}$. Then there are
two possible pure quadratic forms that produce this $b$:
\begin{equation}
q(\hat{1})\equiv 0 \pmod{1} \quad\text{ or }\quad q(\hat{1})\equiv \frac{1}{2} \pmod{1}
\end{equation}
It turns out that \textit{both} group categories $\grpcat$ (defined below)
constructed from these two quadratic forms
produce the same data $(F^{ijm}_{kln},d_i)$ and hence are the same (from the Levin-Wen
perspective). They both correspond to $\integers_2$-lattice gauge theory.
More interesting examples occur when $b(\hat{1},\hat{1}) \equiv \frac{1}{2} \pmod{1}$.
Then there are
two possible pure quadratic forms that produce this $b$:
\begin{equation}
q(\hat{1})\equiv \frac{1}{4} \pmod{1} \quad\text{ or }\quad q(\hat{1})\equiv \frac{3}{4} \pmod{1}
\end{equation}
The group category $\grpcat$ from the LHS corresponds \cite{stirling_thesis}
to $U(1)$ Chern-Simons at level $2$. The other one corresponds to $U(1)$
Chern-Simons at level $-2$.
On the other hand, both theories produce the same data $(F^{ijm}_{kln},d_i)$ and
hence (from the Levin-Wen perspective) both produce the same \textit{doubled} Chern-Simons theory.
We shall show below that \textbf{neither example is unimodal}.
Since this is the first example considered
by Levin and Wen\cite{levin_wen},
we already encounter the example promised in the introduction.
\end{example}
We continue our construction of the group category $\grpcat$ given the data
$\triotrunc$.
The simple objects (string types) and fusion rules (branching rules) are easy to define.
The list of string types $\{0,\ldots,N\}$ is replaced by $\grp$.
Hence we use string labels like $i,j,k\in I$ interchangeably
with group elements $x,y,z\in\grp$ often here.
For every $x\in\grp$ we
define a simple object (these were denoted $V_i$ above)
\begin{equation}
\cplx_x\quad x\in\grp
\end{equation}
which is a 1-dimensional complex vector space \textit{graded} by the group element $x$.
The fusion rules are also easy to define, and indeed satisfy the
\textit{branching rule} condition $N^z_{xy}=0\text{ or }1$:
\begin{equation}
\cplx_x\otimes \cplx_y = \delta_{z,x+y}\cplx_{z}
\end{equation}
Here $\delta$ is the Kronecker delta.
The braiding is slightly more subtle. First, given the simple fusion rules, it is clear that the
braiding matrix
\begin{align}
c_{x,y}:&\cplx_x\otimes\cplx_y \xrightarrow{\sim} \cplx_y\otimes\cplx_x\\
&\cplx_{x+y}\xrightarrow{\sim}\cplx_{x+y}\nonumber
\end{align}
must be multiplication by a complex number. For group categories this complex
number is a phase:
\begin{align}
c_{x,y}:&\cplx_{x+y}\xrightarrow{\sim}\cplx_{x+y}\\
&v_{x+y}\mapsto \exp(2\pi i s(x,y)) v_{x+y}\nonumber
\end{align}
where $v_{x+y}\in\cplx_{x+y}$ and $s(x,y)\in\qmodz$.
It remains to specify the phase angle $s(x,y)\in\qmodz$.
Every finite abelian group $\grp$ can be decomposed (non-uniquely) as a direct product
of cyclic groups $\integers_{n_1}\times\integers_{n_2}\times\ldots$.
Pick a generator $\hat{1}_s$ for each cyclic group, and denote the order of that
cyclic group by $n_s$.
Then an arbitrary element $x\in\grp$ can be written uniquely (once generators
are picked) as
\begin{equation}
x=\sum_s x_s \hat{1}_s\quad\quad 0\leq x_s \leq n_s
\end{equation}
Pick an arbitrary ordering $\hat{1}_1<\hat{1}_2<\ldots$ on the various generators.
Define for convenience
\begin{align}
&q_s := q(\hat{1}_s)\\
&b_{st} := b(\hat{1}_s,\hat{1}_t)\nonumber
\end{align}
(we refer the reader to \cite{stirling_thesis} for discussion concerning how these results change
for different choices of generators and orderings).
Then if we write arbitrary elements $x,y\in\grp$ in terms of the generators
\begin{align}
&x=\sum_s x_s \hat{1}_s\quad\quad 0\leq x_s \leq n_s\\
&y=\sum_s y_s \hat{1}_s\quad\quad 0\leq y_s \leq n_s\nonumber
\end{align}
then we define the braiding phase angle by
\begin{equation}
\label{eq:braidingphaseangle}
s(x,y) := \sum_{s<t}x_s y_t b_{st} + \sum_s x_s y_s q_s
\end{equation}
\begin{example}
We revisit example~(\ref{ex:z2groupcategory}) in the case
\begin{equation}
b(\hat{1},\hat{1})\equiv\frac{1}{2}\pmod{1}\quad\quad q(\hat{1})\equiv\frac{1}{4}\pmod{1}
\end{equation}
Then it is easy to compute the braiding matrices (here just phases)
\begin{align}
&c_{\hat{0},\hat{0}}=\exp\left(2\pi i \left(0\cdot 0 \cdot \frac{1}{4}\right)\right)=1\label{eq:z2braidings}\\
&c_{\hat{0},\hat{1}}=\exp\left(2\pi i \left(0\cdot 1 \cdot \frac{1}{4}\right)\right)=1\notag\\
&c_{\hat{1},\hat{0}}=\exp\left(2\pi i \left(1\cdot 0 \cdot \frac{1}{4}\right)\right)=1\notag\\
&c_{\hat{1},\hat{1}}=\exp\left(2\pi i \left(1\cdot 1 \cdot \frac{1}{4}\right)\right)=i\notag
\end{align}
\end{example}
\subsection{Twists}
For bosons and fermions in $(3+1)$-dimensions we have the \textit{spin-statistics theorem}.
This relates the effect of exchanging two particles (the exchange statistics when the $\text{Perm}$
operation is applied) to their individual spins.
On the other hand, the spin for an individual particle determines how it is affected under
$3$-dimensional rotations.
Hence we may view the spin-statistics
theorem as a relationship between multi-particle $\text{Perm}$ operations and
single-particle rotation operations.
In $(2+1)$-dimensions we might imagine a similar story, except here the analogue of ``spin''
must describe what happens to an irrep (simple object) under $\text{SO}(2)$ rotations. This is the
\textbf{twist} matrix
\begin{equation}
\theta_i:V_i\xrightarrow{\sim}V_i
\end{equation}
Furthermore, we have seen that in $(2+1)$-dimensions $\text{Perm}$ can be generalized to
braiding $c_{V_i,V_j}$.
In categories with \textit{both} braiding and twist, it is reasonable to assume that a compatibility
relationship exists analogous to the spin-statistics theorem.
Indeed there
is a necessary compatibility between braiding and twists
(with an easy geometric interpretation in terms of ribbons, see e.g. \cite{stirling_wu_bcqm})
called \textbf{balancing}.
Balancing gives certain restrictions on twists (much like fermions have half-integer spin).
In the spirit of this paper we merely give a formula for the twist matrices for group categories.
Given a simple object $\cplx_x$ where $x\in\grp$
the twist matrix is just multiplication by a phase
\begin{align}
\theta_x:&\cplx_x\xrightarrow{\sim}\cplx_x\\
&v_x\mapsto \exp(2\pi i q(x))v_x\quad\quad v_x\in\cplx_x\nonumber
\end{align}
\begin{example}
We revisit the same example~(\ref{ex:z2groupcategory}) in the case
\begin{equation}
b(\hat{1},\hat{1})\equiv\frac{1}{2}\pmod{1}\quad\quad q(\hat{1})\equiv\frac{1}{4}\pmod{1}
\end{equation}
Then it is easy to compute the twist matrices (here just phases)
\begin{align}
&\theta_{\hat{0}}=\exp\left(2\pi i (0)\right)=1\label{eq:z2twists}\\
&\theta_{\hat{1}}=\exp\left(2\pi i \left(\frac{1}{4}\right)\right)=i\notag
\end{align}
\end{example}
\subsection{Duality and ribbon categories}
Duality is a fundamental property of
representation theory. Given an irreducible representation $V_i$ we can
always consider the dual representation $(V_i)^*$, which we dangerously
denoted $V_{i^*}$ above.
In category theory \textbf{rigidity} is the appropriate generalization, however the details can be
rather different than those in ordinary group representation theory. For now, each simple object (string type)
$V_i$ has another simple object associated to it: its \textit{right} dual $(V_i)^*$.
Furthermore, we have a categorical notion of pair creation and annihilation. These
are the \textbf{birth} and \textbf{death} matrices
\begin{align}
&b_{V_i}:\unitobj\rightarrow V_i\otimes (V_i)^*\\
&d_{V_i}:(V_i)^*\otimes V_i\rightarrow \unitobj\nonumber
\end{align}
($\unitobj$ is interpreted as the vacuum).
Notice that the ordering of the objects is exactly opposite for the birth and death matrices,
hence we \textit{cannot} simply birth a pair and then annihilate it without performing
some intermediate moves (such as braiding and twisting).
\footnote{This is how the \textbf{quantum dimension} $d_i$ of a simple object $V_i$
is computed, for example.}
In a category with braiding, twists, \textit{and} rigidity we may desire some compatibility
between all three structures (we already mentioned the ``balancing'' condition between
braiding and twists for example).
Again in the spirit of this paper, rather than discuss this point further we
mention that a \textbf{ribbon category} is a category with braiding, twists, and
rigidity such that all three structures interact appropriately.
Even for group categories rigidity is slightly subtle (again see \cite{stirling_thesis}).
However, for this paper it suffices to merely define when objects are dual to
each other. The reader can guess that each simple object $\cplx_x$ has a right dual
\begin{equation}
\cplx_{-x}
\end{equation}
\section{Unimodality}
We have specified enough information about $\grpcat$
in order to decide when a given group category is unimodal
(we provide a classification theorem below).
Before doing this, however, let us motivate why unimodality is important.
In the Levin-Wen model each \textit{directed} edge is labelled
by a string type $i$
(simple object $V_i$). Because of the dangerous identification $(V_i)^*=V_{i^*}$
we can always
use simplified labels such as $i$ or $j$ or $m^*$, and we may always perform
a ``string-breaking'' move such as in figure~(\ref{fig:unimodalmove}).
\begin{figure}[!htb]
\begin{center}
\centering
\input{unimodalmove.pspdftex}
\caption{String-breaking maneuver is implicit in Levin-Wen model, but non-trivial in
Turaev's construction.}
\label{fig:unimodalmove}
\end{center}
\end{figure}
On the other hand, in Turaev's formulation \cite{turaev} there are non-trivial invertible
matrices
\begin{equation}
w_{i^*}:V_{i^*}\xrightarrow{\sim}(V_i)^*
\end{equation}
that form part of the defining structure of the category. They
also provide a ``band-breaking'' maneuver as in figure~(\ref{fig:turaevbandbreak}), however
the procedure may not be self-consistent. It turns out that an inconsistency
may arise when simple objects are self-dual (i.e. when $V_i=V_{i^*}$),
and hence we must enforce the extra unimodality condition.
\begin{figure}[!htb]
\centering
\input{turaevbandbreak.pspdftex}
\caption{Band-breaking maneuver in Turaev's construction.}
\label{fig:turaevbandbreak}
\end{figure}
Rather than go into details, we give the unimodality condition for group categories:
suppose $\cplx_x$ is self-dual
\begin{equation}
\label{eq:unimodalcondition}
\cplx_x = \cplx_{-x}
\end{equation}
(in other words $x=-x\in\grp$). Then unimodality is a condition on the twist and
braiding:
\begin{equation}
\theta_x\cdot c_{x,x}\overset{?}{=}1\quad\quad\text{whenever }x=-x
\end{equation}
If this is true for all $x=-x$ then the category is \textbf{unimodal}.
\subsection{Counterexample}
Again we revisit the case $\grp=\integers_2$ in example~(\ref{ex:z2groupcategory})
(however \textit{not} $\integers_2$-lattice gauge theory). Given the braiding and twist
tables provided in equations~(\ref{eq:z2braidings}) and (\ref{eq:z2twists}) we compute
that for the self-dual object $\hat{1}$
\begin{equation}
\theta_{\hat{1}}\cdot c_{\hat{1},\hat{1}}=i\cdot i=-1
\end{equation}
This example is \textbf{not unimodal} and Turaev's machinery does not apply.
However, as already discussed this is $U(1)$ Chern-Simons at level $2$ and is the
first example considered by Levin-Wen \cite{levin_wen}.
Alternatively, we can suspect a discrepancy by examining the quantum dimension
$d_{\hat{1}}$ computed in \cite{levin_wen}. There they assert that
\begin{equation}
d_{\hat{1}}=-1
\end{equation}
On the other hand, it is proven in $\cite{stirling_thesis}$ that for group categories
$d_x=1$ for every simple object $\cplx_x$.
\footnote{Our result also agrees with the quantum dimension computed in the quantum group
$\text{SU}_q(2)$ at level $1$. This is relevant since
rank-level duality asserts that $\text{SU}_q(n)$ at level $k$ is isomorphic to
$\text{SU}_q(k)$ at level $n$ whenever $n,k>1$. If $n=1$ (or $k=1)$ then rank-level duality is
slightly modified:
$\text{U}(1)$ Chern-Simons at level $k$ is isomorphic to $\text{SU}_q(k)$ at level $1$.}
\section{Unimodality and group categories}
It is straightforward to classify conditions when a group category is
unimodal:
\begin{theorem}
Let $\grpcat$ be a group category. Arbitrarily decompose the finite abelian group $\grp$ into
cyclic groups of orders $n_s$ and generators $\hat{1}_s$ ($s$ indexes the cyclic factors).
Pick an arbitrary ordering $\hat{1}_1 < \hat{1}_2 < \ldots$ for the generators.
Then we have the following cases:
\begin{enumerate}
\item If $\grp$ contains a cyclic factor of odd order (i.e. if one of the $n_s$ is odd) then
$\grpcat$ is unimodal.
\item If $\grp$ contains no odd-order cyclic factors then $\grpcat$ is unimodal if and only if
\begin{equation}
\sum_s \frac{1}{2}(n_s)^2 q_s \in\integers
\end{equation}
\end{enumerate}
\end{theorem}
\begin{proof}
Suppose $x=-x$. Decompose $x$ and $-x$ using the generators
\begin{align}
&x=\sum_s x_s \hat{1}_s\quad\quad 0\leq x_s\leq n_s\\
&-x=\sum_s (n_s-x_s) \hat{1}_s\nonumber
\end{align}
The condition that $x=-x$ implies
\begin{equation}
n_s-x_s = x_s\quad\forall s
\end{equation}
which implies that $n_s=2x_s$. In particular this implies that every $n_s$ must be even,
hence if there exists any $n_s$ odd then $x\neq -x$ for every $x\in\grp$.
Now suppose that every $n_s$ is even. Then the braiding and twist formulas imply
\begin{equation}
\theta_x\cdot c_{x,x}=\exp(2\pi i q(x))\exp(2\pi i s(x,x))
\end{equation}
Equation~(\ref{eq:braidingphaseangle}) says
\begin{equation}
s(x,x) := \sum_{s<t}x_s x_t b_{st} + \sum_s (x_s)^2 q_s
\end{equation}
The quadratic form $q(x)=q\left(\sum_s x_s \hat{1}_s\right)$ can be decomposed using successive
iterations of the defining formula
\begin{equation}
q(x+y)-q(x)-q(y)=b(x,y)
\end{equation}
rearranged as $q(x+y)=b(x,y)+q(x)+q(y)$.
For example the first iteration is
\begin{align}
q\left(\sum_s x_s \hat{1}_s\right) &=q\left(x_1 \hat{1}_1 +\sum_{s>1} x_s \hat{1}_s\right)\\
&= b\left(x_1 \hat{1}_1,\sum_{s>1} x_s \hat{1}_s\right)+q(x_1 \hat{1}_1)\\
&\quad +q\left(\sum_{s>1} x_s \hat{1}_s\right)\nonumber\\
&=\sum_{s>1} x_1 x_s b_{1s} + (x_1)^2 q_1 +q\left(\sum_{s>1} x_s \hat{1}_s\right)\nonumber
\end{align}
We used bilinearity of $b$ and the fact that $q$ is pure in the last equality.
Now, concentrating on the last term, we repeat the process. Iterating we finally obtain
\begin{equation}
q(x)=\sum_{s<t}x_s x_t b_{st}+\sum_s (x_s)^2 q_s
\end{equation}
which is precisely the formula for $s(x,x)$.
Hence
\begin{equation}
\theta_x\cdot c_{x,x}=\exp(2\pi i (q(x)+s(x,x))=\exp(2\pi i 2q(x))
\end{equation}
Recall that we are only considering $x\in\grp$ such that $x=-x$. Also recall from
that beginning of the proof that then $n_s=2 x_s$. Hence we substitute
$x_s=\frac{1}{2}n_s$ into the expression for $2q(x)$ and obtain
\begin{align}
2q(x)&=\sum_{s<t}2 \frac{n_s}{2} x_t b_{st}+\sum_s 2(\frac{n_s}{2})^2 q_s\\
&=\sum_{s<t} x_t n_s b_{st}+\sum_s \frac{1}{2}(n_s)^2 q_s\nonumber
\end{align}
But $x_t n_s b_{st}=x_t b(n_s \hat{1}_s,\hat{1}_t)=x_t b(0,\hat{1}_t)\equiv 0 \pmod{1}$, hence the
first sum is always an integer and does not play a role in the exponent.
Viewing the second factor we arrive at the desired result, i.e.
\begin{equation}
\theta_x\cdot c_{x,x}=\exp(2\pi i 2q(x))=1\Leftrightarrow \sum_s \frac{1}{2}(n_s)^2 q_s\in\integers
\end{equation}
\end{proof}
\section{Acknowledgements}
During this work SDS was supported by an FQXi grant under Yong-Shi Wu at the University
of Utah.
\bibliographystyle{apsrev}
\bibliography{levinwengroupcategories}
\end{document}
| 192,112
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TITLE: A question regarding orthogonal parametrizations
QUESTION [1 upvotes]: This passage is part of the solution of an exercise in Differential Geometry. Let $S$ be a surface, and let $X: U \longrightarrow S$ be an orthogonal parametrization. If $N^X = \frac{X_u \wedge X_v}{|X_u \wedge X_v|}$, then
$$
\langle X_{uu}, N^X \rangle \langle X_{vv}, N^X \rangle = \langle X_{uu}, X_{vv} \rangle - \langle X_{uu}^T, X_{vv}^T\rangle, \qquad \qquad \qquad (*)
$$
where $T$ denotes the "part tangent to the surface", according to the book.
The exercise asks the reader to compute the Gaussian curvature for a surface parametrized by an orthogonal parametrization.
My questions are the following:
What does "part tangent to the surface" mean? What is the image of the second partial derivative, say $X_{uu}$? How to prove $(*)$?
Thanks in advance and kind regards
REPLY [2 votes]: I got this with a hint by Rodrigo Dias.
$X_{uu}$ lives in $\Bbb{R}^3$, so we can write it as
$$
X_{uu} = \langle X_{uu}, X_u \rangle X_u + \langle X_{uu}, X_v \rangle X_v + \langle X_{uu}, N^X \rangle N^X,
$$
and analogously for $X_{vv}$:
$$
X_{vv} = \langle X_{vv}, X_u \rangle X_u + \langle X_{vv}, X_v \rangle X_v + \langle X_{vv}, N^X \rangle N^X.
$$
Recall that $\Bbb{R}^3 = T_XS \oplus \langle N^X \rangle$.
From these expressions, $(*)$ easily follows, sufficing to write the inner product $\langle X_{uu}, X_{vv} \rangle$.
| 65,788
|
TITLE: Numbers $n$ such that there is some digit occuring in each power of $n$
QUESTION [5 upvotes]: If a positive integer $n$ is congruent to $0$, $1$, $5$ or $6$ modulo $10$, there is some digit occuring in each of the powers of $n$. If the decimal expansion of $n$ ends in a $0$, $1$, $5$ or $6$ then the same is true for the decimal expansions of $n^k$ ($k \geq 1$).
My question is whether there is any other positive
integer $n$ (not congruent to $0$, $1$, $5$ or $6$ modulo $10$) such
that there is some digit that occurs in all powers of $n$.
A computer search shows that if such an $n$ were to exist, it should have at least $3$ digits. For example, $99$ does not have the desired property as $$99^{18} = 834513761450087614416078625185528201$$
does not contain a $9$. Perhaps a number such as $102$ will work, as $102^k$ starts with a $1$ for small $k$ and has a lot of digits for large $k$, so "probably" one of them will be a $1$. On the other hand, I do see no reason why there couldn't be some large integer $N$ for which $102^N$ does not contain a $1$ at all.
I would be interested to know if the answer on the question formulated above is known, and (if this is not the case) whether you think it will be true or not (of course, motivated by some argument).
REPLY [4 votes]: I succeeded in answering my own question (André's comment put me on the right track).
We can take $n=1249$. Since $1249^2 \equiv 1 \mod 10000$, we have
$$
1249^k \equiv
\begin{cases}
1 &\mod 10000 \mbox{ if $k$ is even}; \\
1249 &\mod 10000 \mbox{ if $k$ is odd.}
\end{cases}
$$
Thus, each power of $1249$ contains $1$ as a digit. It remains to find an integer $n$ such that each power of $n$ contains all digits.
| 152,482
|
TITLE: Pythagorean Triple parametrization
QUESTION [0 upvotes]: For the pythagorean tripple (x,y,z) such that $x^{2} + y^{2} = z^{2}$ , we know that $x,y,z$ can't be all odd, which means either x,y or z must be even.
We can choose y even to get many parametrizations with y even ($a^2 - b^2 , 2ab , a^2 + b+2$ ) , or we can choose x even to get many parametrizations with x even ($2ab , a^2 - b^2, a^2 + b^2$).
That's because if we make y=2ab then we have that
$ y^2 = (2a²)(2b²) = z^2 - x^2 = (x+z)(x-z) $
So from that we get $2a² = x+z$ and $2b² = x-z $
then we have $y=2ab , x = a^2 - b^2$ and $z = a^2 + b^2$
Now, I'm interested in the case where we choose z = 2ab, to get many parametrizations with z even ( particularly the ones where either x,y are both even or both odd ).
But i don't know how to get that parametrization :
$z = 2ab$ ..., then i don't know what else to do.
Thanks a lot in advance
REPLY [1 votes]: You should probably go through primitive Pythagorean triples first, that is, those Pythagorean triples $(x, y, z)$ where $x, y$ are coprime. One can easily show that any Pythagorean triple can be obtained from a suitable primitive Pythagorean one $(x, y, z)$ in the form $ (tx, ty, tz)$, for some $t$.
For the primitive Pythagorean triples, you can easily show that $x, y$ have distinct parity, that is, one of them is odd and the other is even, so $z$ is odd.
| 142,200
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TITLE: Using the limit Definition, compute $\frac{\text d}{\text dx}f(x)$ for $f(x) = x-\frac{3}{x}$
QUESTION [0 upvotes]: I got totally lost in this part of the test. Can someone give me some guidance? How do you calculate the answer?
REPLY [1 votes]: Your work is a little difficult to follow, but it looks to me like your mistake was algebra. The limit you set up was $$\lim_{h \to 0} \frac{(x+h)-\frac{3}{x+h}\left(-x-\frac{3}{x}\right)}{h}$$ which is incorrect. It should be $$\begin{align}\lim_{h \to 0} \frac{(x+h)-\frac{3}{x+h}-\left(x-\frac{3}{x}\right)}{h} = \lim_{h \to 0} \frac{x+h-\frac{3}{x+h}-x+\frac{3}{x}}{h} \\ = \lim_{h \to 0} \frac{h-\frac{3}{x+h}+\frac{3}{x}}{h} \\ = \lim_{h \to 0} \frac{h\frac{x(x+h)}{x(x+h)}-\frac{3x}{x(x+h)}+\frac{3(x+h)}{x(x+h)}}{h} \\ = \lim_{h \to 0} \frac{\frac{hx^2+xh^2+3h}{x(x+h)}}{h} \\ = \lim_{h \to 0} \frac{x^2+xh+3}{x(x+h)} \end{align}$$ Can you take it from here?
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\begin{document}
\title[\textbf{Numbers of periodic orbits hidden at fixed points}]{
\textbf{The numbers of periodic orbits of holomorphic mappings hidden at
fixed points}}
\author{Guang Yuan\medskip\ Zhang }
\address{Department of Mathematical Sciences, Tsinghua University, Beijing
100084, P. R. China. \textit{Email:} \textit{gyzhang@math.tsinghua.edu.cn}}
\begin{abstract}
Let $\Delta ^{2}$ be a ball in the complex vector space $\mathbb{C}^{2}$
centered at the origin, let $f:\Delta ^{2}\rightarrow \mathbb{C}^{2}$ be a
holomorphic mapping$,$ with $f(0)=0$, and let $M$ be a positive integer. If
the origin $0$ is an isolated fixed point of the $M$ th iteration $f^{M}$ of
$f,$ then one can define the number $\mathcal{O}_{M}(f,0)$ of periodic
orbits of $f$ with period $M$ hidden at the fixed point $0$, which has the
meaning: any holomorphic mapping $g:\Delta ^{2}\rightarrow \mathbb{C}^{2}$
sufficiently close to $f$ in a neighborhood of the origin has exactly $
\mathcal{O}_{M}(f,0)$ distinct periodic orbits with period $M$ near the
origin, provided that all fixed points of $g^{M}$ near the origin are all
simple.
It is known that $\mathcal{O}_{M}(f,0)\geq 1$ iff the linear part of $f$ at
the origin has a periodic point of period $M.$ This paper will continue to
study the number $\mathcal{O}_{M}(f,0)$. We are interested in the condition
for the linear part of $f$ at the origin such that $\mathcal{O}_{M}(f,0)\geq
2.$
For a $2\times 2$ matrix $A$ that is arbitrarily given, the goal of this
paper is to give a necessary and sufficient condition for $A$,$\ $such that $
\mathcal{O}_{M}(f,0)\geq 2$ for all holomorphic mappings $f:\Delta
^{2}\rightarrow \mathbb{C}^{2}$ such that $f(0)=0,$ $Df(0)=A$ and that the
origin $0$ is an isolated fixed point of $f^{M}.$
\end{abstract}
\subjclass[2000]{32H50, 37C25}
\thanks{Project 10271063 and 10571009 supported by NSFC}
\maketitle
\section{Introduction and the main Theorem\label{S1}}
We denote by $\mathbb{C}^{n}$ the complex vector space of dimension $n$ and
by $\mathcal{O}(\mathbb{C}^{n},0,0)$ the space of all germs of holomorphic
mappings $f$ from a neighborhood of the origin $0$ in $\mathbb{C}^{n}$ into $
\mathbb{C}^{n}$ such that
\begin{equation*}
f(0)=0\ \mathrm{(}0\mathrm{\ denotes\ the\ origin).}
\end{equation*}
Let
\begin{equation*}
f(z)=\lambda z+\mathrm{higher\ terms},
\end{equation*}
be a germ in $\mathcal{O}(\mathbb{C},0,0)$. Then $0$ is a fixed point of $f$
and for each $m\in \mathbb{N}$ (the set of positive integers)$,$ the $m$-th
iteration $f^{m}$ of $f$ is well defined in a neighborhood of $0$. $f^{m}$
is defined as $f^{1}=f,f^{2}=f\circ f,\dots ,f^{m}=f\circ f^{m-1},$
inductively.
If $\lambda =f^{\prime }(0)$ is a primitive $m$-th root of unity, then it is
well known in the theory of one variable complex dynamics [\ref{Mil}] that
either $f^{m}(z)\equiv z,$ or there exist an $\alpha \in \mathbb{N}$ and a
constant $a\neq 0$ such that
\begin{equation*}
f^{m}(z)=z+az^{\alpha m+1}+\mathrm{higher\ terms},
\end{equation*}
and in the later case, $0$ is an isolated fixed point of $f^{m}$ and one can
split this fixed point into one fixed point and $\alpha $ periodic orbits by
small perturbations. More precisely, there exists a sequence of holomorphic
functions $f_{k}$ converging to $f$ uniformly in a neighborhood of the
origin so that $f_{k}(0)=0$ and each $f_{k}$ has $\alpha $ distinct periodic
orbits of period $m$ converging to the origin, and $\alpha $ is the largest
integer with this property. Note that $f$ itself has no periodic orbit of
period $m$ in a neighborhood of $0.$ The number $\alpha $ then can be
interpreted to be the number of periodic orbits of period $m$ of $f$
"hidden" at $0.$ Some authors call this phenomenon that $f$ has $\alpha $
\emph{virtual} periodic orbits of period $m$ at $0$ (see [\ref{CMY}])$.$
This number $\alpha $ will be denoted by $\mathcal{O}_{m}(f,0).$ Here, a
\emph{periodic orbit} of $f_{k}$ of period $m$ means a set $
E=\{p_{1},p_{2},\dots ,p_{m}\}\subset B$ with cardinality $m$ such that $
f_{k}(p_{1})=p_{2},f_{k}(p_{2})=p_{3},\dots ,f_{k}(p_{m-1})=p_{m}$ and $
f_{k}(p_{m})=p_{1}.$ A point is called a \emph{periodic point of period} $m$
if and only if it is contained in a periodic orbit of period $m.$
In higher dimensional cases, there are similar phenomena, but things are far
more complicated. Now, let $f\in \mathcal{O}(\mathbb{C}^{n},0,0)$ and let $
M\in \mathbb{N}$. Then there is a small ball $B$ centered at $0$ so that the
$M$-th iteration $f^{M}$ of $f$ is well defined in $\overline{B}$. If, in
addition, $0$ is an isolated fixed point of $f^{M}$ then we may make $B$
even smaller so that $0$ is the unique fixed point of $f^{M}$ in $\overline{B
}$ and then define $\mathcal{O}_{M}(f,0)$ to be
\begin{equation*}
\max \left\{ m;
\begin{array}{l}
\text{\textrm{there\ exists\ a\ sequence }}f_{k}\in \mathcal{O}(\mathbb{C}
^{n},0,0)\ \mathrm{uniformly\ converging\ to\ }f\ \text{\textrm{in}}\ \\
B\text{\textrm{\ }}\mathrm{such\ that\ each\ }f_{k}\ \mathrm{has\ }m\mathrm{
\ }\text{\textrm{distinct\ }}\mathrm{periodic\ orbits\ of\ period\ }M\
\mathrm{in\ }B.
\end{array}
\right\}
\end{equation*}
It is clear that in this definition, all periodic orbits of period $M$ of $
f_{k}$ located in $B$ converge to $0$ uniformly as $k\rightarrow \infty .$
Therefore, this definition is independent of $B.$
The number $\mathcal{O}_{M}(f,0)$ is a well defined integer, and the
definition for $\mathcal{O}_{M}(f,0)$ agrees with that in the case $n=1.$ In
next section, we shall give another equivalent definition of the number $
\mathcal{O}_{M}(f,0)$ and give some examples for understanding this number.
The following theorem is proved by the author in [\ref{Zh2}].
\begin{theorem}
\label{Th0.1}Let $f\in \mathcal{O}(\mathbb{C}^{n},0,0)$ and assume that the
origin is an isolated fixed point of $f^{M}.$ Then, $\mathcal{O}
_{M}(f,0)\neq 0\ $if and only if the linear part of $f$ at $0$ has a
periodic point of period $M.$
\end{theorem}
The term \textquotedblleft linear part\textquotedblright\ indicates the
linear mapping $l:\mathbb{C}^{n}\rightarrow \mathbb{C}^{n},$
\begin{equation*}
l(x_{1},\dots ,x_{n})=(\sum_{j=1}^{n}a_{1j}x_{j},\dots
,\sum_{j=1}^{n}a_{nj}x_{j}),
\end{equation*}
where
\begin{equation*}
(a_{ij})=Df(0)=\left. \left( \frac{\partial f_{i}}{\partial x_{j}}\right)
\right\vert _{0}
\end{equation*}
is the Jacobian matrix of $f=(f_{1},\dots ,f_{n})$ at the origin$.$ If $M>1,$
then by Lemma \ref{j7}, the linear part of $f$ at $0$ has a periodic point
of period $M\ $if and only if the following condition holds.
\begin{condition}
\label{1.0} $Df(0)$ has eigenvalues $\lambda _{1},\dots ,\lambda _{s},s\leq
n,$ that are primitive $m_{1}$-th,..., $m_{s}$-th roots of unity,
respectively, such that $M$ is the least common multiple of $m_{1},\dots
,m_{s}.$
\end{condition}
Thus, if $M>1,$ by the above theorem, $\mathcal{O}_{M}(f,0)\geq 1$ if and
only if Condition \ref{1.0} holds. This gives rise to the following problem.
\begin{problem}
\label{P1}Assume that Condition \ref{1.0} holds. Under which additional
condition for $Df(0),\ $one must have $\mathcal{O}_{M}(f,0)\geq 2?$
\end{problem}
In this paper, we study this problem for the case $n=2$, and our goal is to
prove the following theorem.
\begin{theorem}
\label{Th1}Let $M>1$ be a positive integer and let $A$ be a $2$ by $2$
matrix. Then the following two conditions \textrm{(A) and (B) }are
equivalent:
\textrm{(A) }For any holomorphic mapping germ $f\in \mathcal{O}(\mathbb{C}
^{2},0,0)$ such that $Df(0)=A$ and that $0$ is an isolated fixed point of
both $f$ and $f^{M},$
\begin{equation*}
\mathcal{O}_{M}(f,0)\geq 2.
\end{equation*}
\textrm{(B) }The two eigenvalues $\lambda _{1}$ and $\lambda _{2}$ of $A$
are primitive $m_{1}$ th and $m_{2}$ th roots of unity, respectively, and
one of the following conditions holds.
\textrm{(b1) }$A$ is diagonalizable, $m_{1}=m_{2}=M$ and $\lambda
_{1}=\lambda _{2}.$
\textrm{(b2) }$m_{1}=m_{2}=M$ and there exist positive integers $\alpha $
and $\beta $ such that $1<\alpha <M$, $1<\beta <M$ and
\begin{equation*}
\lambda _{1}^{\alpha }=\lambda _{2},\lambda _{2}^{\beta }=\lambda
_{1},\alpha \beta >M+1.
\end{equation*}
\textrm{(b3) }$m_{1}|m_{2},$ $m_{2}=M$, and $\lambda _{2}^{m_{2}/m_{1}}\neq
\lambda _{1}.$
\textrm{(b4) }$M=[m_{1},m_{2}],(m_{1},m_{2})>1$ and $\max \{m_{1},m_{2}\}<M.$
\end{theorem}
Here, $[m_{1},m_{2}]$ denotes the least common multiple and $(m_{1},m_{2})$
denotes the greatest common divisor, of $m_{1}$ and $m_{2},$ and $
m_{1}|m_{2} $ means that $m_{1}$ divides $m_{2}.$
\begin{remark}
\label{r1}Consider condition (b2). If in the theorem $m_{1}=m_{2}=M$ but $
\lambda _{1}\neq \lambda _{2},$ say, $\lambda _{1}$ and $\lambda _{2}$ are
distinct primitive $M$ th roots of unity, then it is easy to show that there
uniquely exist positive integers $\alpha $ and $\beta $ such that $1<\alpha
<M$, $1<\beta <M$ and
\begin{equation}
\lambda _{1}^{\alpha }=\lambda _{2},\lambda _{2}^{\beta }=\lambda _{1}.
\label{j6}
\end{equation}
The proof is left to the reader.
(\ref{j6}) implies that $\lambda _{1}^{\alpha \beta }=\lambda _{1},$ and
then by the property of primitive $M$ th roots of unity, one can see that $
\alpha \beta =kM+1$ for some positive integer $k$. Condition (b2) just
permits $k>1.$
\end{remark}
This paper is arranged as follows. In Section \ref{S2}, we shall give
another equivalent definition of the number $\mathcal{O}_{M}(f,0)$ and give
two examples for understanding this number. Sections \ref{S3}--\ref{PM} are
aimed to prove the main theorem. In Sections \ref{S3} and \ref{norm}, we
shall introduce some known results and prove a few consequences. Then, in
Section \ref{cronin-1}, we shall apply Cronin's theorem to compute zero
orders of some germs in $\mathcal{O}(\mathbb{C}^{2},0,0)$. After these
preparations, the proof of the main theorem will be given in the last two
sections.
\section{Another Definition of $\mathcal{O}_{M}(f,0)$ via Dold's Indices and
Some Examples\label{S2}}
Let $f\in \mathcal{O}(\mathbb{C}^{2},0,0)$. Then each component of $f$ can
be expressed as a power series at the origin. All power series in this paper
will be assumed to be convergent in a neighborhood of the origin.
If $p$ is an isolated zero of $f,$ say, there exists a ball $B$ centered at $
p$ such that $f$ is well defined on $\overline{B}$ and that $p$ is the
unique solution of the equation $f(x)=0$ ($0$ is the origin) in $\overline{B}
.$ Then we can define the \emph{zero order} (or \emph{multiplicity}) of $f$
at $p$ by
\begin{equation*}
\pi _{f}(p)=\#\{x\in B;f(x)=q\},
\end{equation*}
where $q$ is a regular value of $f$ such that $|q|$ is small enough and $\#$
denotes the cardinality. $\pi _{f}(p)$ is well defined (see \cite{LL} or [
\ref{Zh}] for the details).
If the origin $0\ $is an isolated fixed point of $f,$ then $0$ is an
isolated zero of the germ $id-f\in \mathcal{O}(\mathbb{C}^{2},0,0),$ which
puts each $x$ near the origin into $x-f(x),$ and then the \emph{fixed point
index} of $f$ at $0$ is well defined by
\begin{equation*}
\mu _{f}(0)=\pi _{id-f}(0).
\end{equation*}
If $0$ is a fixed point of $f$ such that $id-f$ is regular at $0,$ say, the
Jacobian matrix $Df(0)$ of $f$ at $0$ has no eigenvalue $1,$ then $0$ is
called a \emph{simple} fixed point of $f.$ A simple fixed point of a
holomorphic mapping has index $1$ by the inverse function theorem (see Lemma
\ref{lem2-a}). By the definition, it is clear that
\begin{equation*}
\mu _{f}(0)=\pi _{id-f}(0)=\pi _{f-id}(0).
\end{equation*}
If the origin $0$ is a fixed point of $f,$ then for any $m\in \mathbb{N},$
the $m$-th iteration $f^{m}$ is well defined in a neighborhood $V_{m}$ of $
0. $ If for some $M\in \mathbb{N},$ the origin $0$ is an isolated fixed
point of both $f$ and $f^{M}$, then for each factor $m$ of $M,$ $0$ is an
isolated fixed point of $f^{m}$ as well and the fixed point index $\mu
_{f^{m}}(0)$ of $f^{m}$ at $0$ is well defined. Therefore, we can define the
(local) index as in A. Dold's work [\ref{Do}]:
\begin{equation}
P_{M}(f,0)=\sum_{\tau \subset P(M)}(-1)^{\#\tau }\mu _{f^{M:\tau }}(0),
\label{do-1}
\end{equation}
where $P(M)$ is the set of all primes dividing $M,$ the sum extends over all
subsets $\tau $ of $P(M),$ $\#\tau \ $is the cardinal number of $\tau $ and $
M:\tau =M(\prod_{p\in \tau }p)^{-1}$. Note that the sum includes the term $
\mu _{f^{M}}(0)$ which corresponds to the empty subset $\tau =\emptyset $.
If $M=12=2^{2}\cdot 3,$ for example, then $P(M)=\{2,3\},$ and
\begin{equation*}
P_{12}(f,0)=\mu _{f^{12}}(0)-\mu _{f^{4}}(0)-\mu _{f^{6}}(0)+\mu _{f^{2}}(0).
\end{equation*}
The formula (\ref{do-1}) is known as the M\"{o}bius inversion formula \cite
{Hy} (see [\ref{Zh2}] for more interpretations of Dold's index).
\begin{remark}
(1) By Corollary \ref{do1},
\begin{equation*}
\mathcal{O}_{M}(f,0)=P_{M}(f,0)/M.
\end{equation*}
(2) By the definition, $\mathcal{O}_{1}(f,0)=P_{1}(f,0)=\mu _{f}(0).$
\end{remark}
We denote by $O(1)$ any holomorphic function germ at the origin, which may
be different in different places, even in a single equation; by $o(|z|^{k}),$
any holomorphic function germ $\alpha $ defined at the origin $z=0$ such
that
\begin{equation}
\lim_{z\rightarrow 0}|\alpha (z)|/|z|^{k}=0, \label{1-1}
\end{equation}
which is equivalent to the statement that $\alpha $ can be expressed as a
power series in which the terms of degree $\leq k$ are all zero. Also, the
same notation $o(|z|^{k})$ may denote different function germs in different
places, even in a single equation. When $o(|z|^{k})$ denotes a germ of one
variable function, we just write it to be $o(z^{k}).$ Thus, $o(1)=o(|z|^{0})$
means any holomorphic function vanishing at the origin.
\begin{example}
\label{e1}Let $f\in \mathcal{O}(\mathbb{C}^{2},0,0)$ be given by
\begin{equation*}
(x,y)\rightarrow (\lambda _{1}x+o(x),\lambda _{2}y+o(y)),
\end{equation*}
such that $\lambda _{1}$ is a primitive $m_{1}$-th root of unity and $
\lambda _{2}$ is a primitive $m_{2}$-th root of unity, $m_{1}$ and $m_{2}$
are distinct primes, and that $0$ is an isolated fixed point of the $
m_{1}m_{2}$ th iteration $f^{m_{1}m_{2}}$ of $f.$ Then there exist nonzero
constants $a,b$ and positive integers $\alpha $ and $\beta $ such that
\begin{equation*}
f^{m_{1}}(x,y)^{T}=\left(
\begin{array}{l}
x+ax^{\alpha m_{1}+1}(1+o(1)) \\
\lambda _{2}^{m_{1}}y(1+o(1))
\end{array}
\right) ,
\end{equation*}
\begin{equation*}
f^{m_{2}}(x,y)^{T}=\left(
\begin{array}{l}
\lambda _{1}^{m_{2}}x(1+o(1)) \\
y+by^{\beta m_{2}+1}(1+o(1))
\end{array}
\right) ,
\end{equation*}
and
\begin{equation*}
f^{m_{1}m_{2}}(x,y)^{T}=\left(
\begin{array}{l}
x+am_{2}x^{\alpha m_{1}+1}(1+o(1)) \\
y+bm_{1}x^{\beta m_{2}+1}(1+o(1))
\end{array}
\right) .
\end{equation*}
Thus, considering that $\lambda _{1}^{m_{2}}\neq 1$ and $\lambda
_{2}^{m_{1}}\neq 1$, we have by Cronin's Theorem introduced in Section \ref
{cronin-1} that
\begin{equation*}
\mu _{f}(0)=1,
\end{equation*}
\begin{equation*}
\mu _{f^{m_{1}}}(0)=\alpha m_{1}+1,\ \mu _{f^{m_{2}}}(0)=\beta m_{2}+1,
\end{equation*}
\begin{equation*}
\mu _{f^{m_{1}m_{2}}}(0)=\left( \alpha m_{1}+1\right) (\beta m_{2}+1),
\end{equation*}
and then by the formula (\ref{do-1}),
\begin{equation*}
P_{m_{1}}(f,0)=\mu _{f^{m_{1}}}(0)-\mu _{f}(0)=\alpha m_{1},
\end{equation*}
\begin{equation*}
P_{m_{2}}(f,0)=\mu _{f^{m_{2}}}(0)-\mu _{f}(0)=\beta m_{2},
\end{equation*}
\begin{equation*}
P_{m_{1}m_{2}}(f,0)=\mu _{f^{m_{1}m_{2}}}(0)-\mu _{f^{m_{1}}}(0)-\mu
_{f^{m_{2}}}(0)+1=\alpha \beta m_{1}m_{2},
\end{equation*}
and then, by Corollary \ref{do1},
\begin{equation*}
\mathcal{O}_{m_{1}}(f,0)=\alpha ,\mathcal{O}_{m_{2}}(f,0)=\beta ,\mathcal{O}
_{m_{1}m_{2}}(f,0)=\alpha \beta ,
\end{equation*}
say
\begin{equation}
\mathcal{O}_{m_{1}m_{2}}(f,0)=\mathcal{O}_{m_{1}}(f,0)\mathcal{O}
_{m_{2}}(f,0). \label{j1}
\end{equation}
\end{example}
By this example, one may guess that there is a relation between the numbers $
\mathcal{O}_{m_{1}}(f,0),$ $\mathcal{O}_{m_{2}}(f,0)$ and $\mathcal{O}
_{m_{1}m_{2}}(f,0)\ $similar to the above equality (\ref{j1}). But see the
next example.
\begin{example}
\label{E2}Let $k>1$ be any given positive integer and let $f\in \mathcal{O}(
\mathbb{C}^{2},0,0)$ be given by
\begin{equation*}
\left( f(x,y)\right) ^{T}=\left(
\begin{array}{c}
-x+x^{2k+1}+xy^{3} \\
e^{\frac{2\pi i}{3}}y+x^{2}y+y^{3k+1}
\end{array}
\right) .
\end{equation*}
We show that $\mathcal{O}_{2}(f,0)=\mathcal{O}_{3}(f,0)=k,$ but $\mathcal{O}
_{6}(f,0)=1.$
After a careful computation, we have
\begin{equation}
\left( f^{2}(x,y)\right) ^{T}=\left(
\begin{array}{l}
x-2x^{2k+1}(1+o(1))-2xy^{3}(1+o(1)) \\
e^{\frac{4\pi i}{3}}y(1+o(1))
\end{array}
\right) , \label{ggg1}
\end{equation}
\begin{equation}
\left( f^{3}(x,y)\right) ^{T}=\left(
\begin{array}{l}
-x(1+o(1)) \\
y+3e^{\frac{4\pi i}{3}}x^{2}y(1+o(1))+3e^{\frac{4\pi i}{3}}y^{3k+1}(1+o(1))
\end{array}
\right) , \label{ggg2}
\end{equation}
and
\begin{equation*}
\left( f^{6}(x,y)\right) ^{T}=\left(
\begin{array}{l}
x+xh_{1}(x,y) \\
y+yh_{2}(x,y)
\end{array}
\right) ,
\end{equation*}
with
\begin{equation*}
\begin{array}{l}
h_{1}(x,y)=-6x^{2k}(1+o(1))-6y^{3}(1+o(1)), \\
h_{2}(x,y)=6e^{\frac{4\pi i}{3}}x^{2}(1+o(1))+6e^{\frac{4\pi i}{3}
}y^{3k}(1+o(1)).
\end{array}
\end{equation*}
We first compute $\mu _{f^{2}}(0).$ It equals the zero order of the mapping
\begin{equation*}
F(x,y)=(x,y)-f^{2}(x,y),
\end{equation*}
and then, by (\ref{ggg1}) and Corollary \ref{c2}, we have
\begin{equation*}
\mu _{f^{2}}(0)=\pi _{F}(0)=2k+1.
\end{equation*}
Similarly, by (\ref{ggg2}) and Corollary \ref{c2}, we have
\begin{equation*}
\mu _{f^{3}}(0)=3k+1.
\end{equation*}
On the other hand, $0$ is a simple fixed point of $f,$ and then by Lemma \ref
{lem2-a}, $\mu _{f}(0)=1.$ Therefore, by the formula (\ref{do-1}), we have
\begin{eqnarray*}
P_{2}(f,0) &=&\mu _{f^{2}}(0)-1=2k, \\
P_{3}(f,0) &=&\mu _{f^{3}}(0)-1=3k,
\end{eqnarray*}
and then, by Corollary \ref{do1}, we have
\begin{eqnarray*}
\mathcal{O}_{2}(f,0) &=&P_{2}(f,0)/2=k, \\
\mathcal{O}_{3}(f,0) &=&P_{3}(f,0)/3=k.
\end{eqnarray*}
Next, we show that $\mathcal{O}_{6}(f,0)=1.$ It is clear that $\mu
_{f^{6}}(0)$ equals the zero order of the mapping
\begin{equation*}
id-f^{6}:(x,y)\mapsto -(xh_{1}(x,y),yh_{2}(x,y)),
\end{equation*}
and by Lemma \ref{prod1}, the zero order of $id-f^{6}$ at $0$ is the sum of
the zero orders of the four mappings putting $(x,y)\ $into $(x,y),$ $
(x,h_{2}(x,y)),$ $(h_{1}(x,y),y)$ and $(h_{1}(x,y),h_{2}(x,y)),$ which are $
1,$ $3k$, $2k$ and $6,$ respectively, by Cronin's Theorem and Corollary \ref
{c2}. Thus $\mu _{f^{6}}(0)=5k+7,$ and then, by the formula (\ref{do-1}), we
have
\begin{eqnarray*}
P_{6}(f,0) &=&\mu _{f^{6}}(0)-\mu _{f^{2}}(0)-\mu _{f^{3}}(0)+1 \\
&=&5k+7-2k-1-3k-1+1,
\end{eqnarray*}
and then $P_{6}(f,0)=6,$ and $\mathcal{O}_{6}(f,0)=1.$
\end{example}
\begin{remark}
(1). Assume that $m_{1}$ and $m_{2}$ are two positive integers and let $f\in
\mathcal{O}(\mathbb{C}^{2},0,0)$ such that 0 is an isolated fixed point of $
f^{[m_{1},m_{2}]}$, where $[m_{1},m_{2}]$ denotes the least common multiple
of $m_{1}$ and $m_{2}$. Then, by Theorem \ref{Th0.1}, $\mathcal{O}
_{m_{1}}(f,0)\geq 1$ and $\mathcal{O}_{m_{2}}(f,0)\geq 1$ imply $\mathcal{O}
_{[m_{1},\ m_{2}]}(f,0)\geq 1$.
(2). By the main theorem, when $m_{1}$ and $m_{2}$ satisfy certain condition
(for example, if $m_{1}=6$ and $m_{2}=10),$ $\mathcal{O}_{m_{1}}(f,0)\geq 1$
and $\mathcal{O}_{m_{2}}(f,0)\geq 1$ implies $\mathcal{O}_{[m_{1},\
m_{2}]}(f,0)\geq 2.$
\end{remark}
\section{Some basic results of fixed point indices and zero orders\label{S3}}
In this section we introduce some results for later use. Most of them are
known.
Let $U$ be an open and bounded subset of $\mathbb{C}^{2}$ and let $f:
\overline{U}\rightarrow \mathbb{C}^{2}$ be a holomorphic mapping. If $f$ has
no fixed point on the boundary $\partial U$, then the fixed point set
\textrm{Fix}$(f)$ of $f$ is a compact analytic subset of $U,$ and then it is
finite (see [\ref{Ch}]); and therefore, we can define the global fixed point
index $L(f)$ of $f$ as:
\begin{equation*}
L(f)=\sum_{p\in \mathrm{Fix}(f)}\mu _{f}(p),
\end{equation*}
which is just the number of all fixed points of $f$, counting indices. $L(f)$
is, in fact, the Lefschetz fixed point index of $f$ (see the appendix
section in [\ref{Zh2}] for the details).
For each $m\in \mathbb{N},$ the $m$-th iteration $f^{m}$ of $f$ is
understood to be defined on
\begin{equation*}
K_{m}(f)=\cap _{k=0}^{m-1}f^{-k}(\overline{U})=\{x\in \overline{U}
;f^{k}(x)\in \overline{U}\ \mathrm{for\ all\ }k=1,\dots ,m-1\},
\end{equation*}
which is the largest set where $f^{m}$ is well defined. Since $U$ is
bounded, $K_{m}(f)$ is a compact subset of $\overline{U}$. Here, $f^{0}=id.$
Now, let us introduce the global Dold's index. Let $M\in \mathbb{N}$ and
assume that $f^{M}$ has no fixed point on the boundary $\partial U.$ Then,
for each factor $m$ of $M,$ $f^{m}$ again has no fixed point on $\partial U,$
and then the fixed point set $\mathrm{Fix}(f^{m})$ of $f^{m}$ is a compact
subset of $U$. Thus, there exists an open subset $V_{m}$ of $U$ such that $
\mathrm{Fix}(f^{m})\subset V_{m}\subset \overline{V_{m}}\subset U$ and $
f^{m} $ is well defined on $\overline{V_{m}},$ and thus $L(f^{m}|_{\overline{
V_{m}}})$ is well defined and we write $L(f^{m})=L(f^{m}|_{\overline{V_{m}}
}),$ where $f^{m}|_{\overline{V_{m}}}$ is the restriction of $f^{m}$ to $
\overline{V_{m}}.$ In this way, we can define the global Dold's index (see [
\ref{Do}]) as (\ref{do-1}):
\begin{equation}
P_{M}(f)=\sum_{\tau \subset P(M)}(-1)^{\#\tau }L(f^{M:\tau }). \label{p0p}
\end{equation}
Let $m\in \mathbb{N}$. It is clear that, for any compact subset $K$ of $U$
with $\cup _{j=1}^{m}f^{j}(K)\subset U$, there is a neighborhood $V\subset U$
of $K,$ such that for any holomorphic mapping $g:\overline{U}\rightarrow
\mathbb{C}^{2}$ sufficiently close to $f,$ the iterations $g^{j},j=1,\dots
,m,$ are well defined on $\overline{V}$ and
\begin{equation*}
\max_{X\in \overline{U}}|g(X)-f(X)|\rightarrow 0\Longrightarrow \max_{1\leq
j\leq m}\max_{X\in \overline{V}}|g^{j}(X)-f^{j}(X)|\rightarrow 0.
\end{equation*}
We shall use these facts frequently and tacitly.
We denote by $\Delta ^{2}$ a ball in $\mathbb{C}^{2}$ centered at the origin.
\begin{lemma}[\protect\cite{LL}]
\label{lem2-a}Let $f\in \mathcal{O}(\mathbb{C}^{2},0,0)$ and assume that the
origin is an isolated fixed point$.$ Then
\begin{equation*}
\mu _{f}(0)\geq 1,
\end{equation*}
and the equality holds if and only if $1$ is not an eigenvalue of $Df(0)$.
\end{lemma}
\begin{lemma}[\protect\cite{LL}]
\label{lem2-1}(1) Let $f:\overline{\Delta ^{2}}\rightarrow \mathbb{C}^{2}$
be a holomorphic mapping such that $f$ has no fixed point on the boundary $
\partial \Delta ^{2}.$ Then there exists a $\delta >0$ such that any
holomorphic mapping $g:\overline{\Delta ^{2}}\rightarrow \mathbb{C}^{2}$
with $\max_{x\in \overline{\Delta ^{2}}}|g(x)-f(x)|<\delta $ has finitely
many fixed points in $\Delta ^{2}$ and satisfies
\begin{equation*}
L(g)=\sum_{p\in \mathrm{Fix}(g)}\mu _{g}(p)=\sum_{p\in \mathrm{Fix}(f)}\mu
_{f}(p)=L(f).
\end{equation*}
(2) In particular, if $0$ is the unique fixed point of $f$ in $\overline{
\Delta ^{2}},$ then for any holomorphic mapping $g:\overline{\Delta ^{2}}
\rightarrow \mathbb{C}^{2}$ with $\max_{x\in \overline{\Delta ^{2}}
}|g(x)-f(x)|<\delta ,\ $
\begin{equation*}
\mu _{f}(0)=\sum_{p\in \mathrm{Fix}(g)}\mu _{g}(p),
\end{equation*}
and if in addition all fixed points of $g$ are simple, then
\begin{equation*}
\mu _{f}(0)=\#\mathrm{Fix}(g)=\#\{y\in \Delta ^{2};g(y)=y\}.
\end{equation*}
\end{lemma}
This result is another version of Rouch\'{e}'s theorem which is stated as
follows.
\begin{lemma}[Rouch\'{e}'s theorem \protect\cite{LL}]
Let $f:\overline{\Delta ^{2}}\rightarrow $ $\mathbb{C}^{2}$\ be a
holomorphic mapping such that $f$ has no zero on $\partial \Delta ^{2}.$
Then there exists a $\delta >0$ such that any holomorphic mapping $g:
\overline{\Delta ^{2}}\rightarrow $ $\mathbb{C}^{2}$ with $\max_{x\in
\partial \Delta ^{2}}|g(x)-f(x)|<\delta $ has the same number of zeros in $
\Delta ^{2}$ as $f,$ counting zero orders, say,
\begin{equation*}
\sum_{f(x)=0}\pi _{f}(x)=\sum_{g(x)=0}\pi _{g}(x).
\end{equation*}
\end{lemma}
\begin{corollary}
Let $U$ be a bounded open subset of $\mathbb{C}^{2},$ let $M\in \mathbb{N},$
let $f:\overline{U}\rightarrow \mathbb{C}^{2}$ be a holomorphic mapping and
assume that $f^{M}$ has no fixed point on $\partial U.$ If $f$ has a
periodic point $p\in U$ with period $M$, then any holomorphic mapping $g:
\overline{U}\rightarrow \mathbb{C}^{2}$ that is sufficiently close to $f$
has a periodic point with period $M$ in $U.$
\end{corollary}
\begin{proof}
This follows from Lemmas \ref{lem2-a} and \ref{lem2-1} directly (see [\ref
{Zh2}] for a simple proof).
\end{proof}
\begin{corollary}
\label{adc}Let $U$ be a bounded open subset of $\mathbb{C}^{2},$ let $M\in
\mathbb{N},$ let $f:\overline{U}\rightarrow \mathbb{C}^{2}$ be a holomorphic
mapping and assume that $f^{M}$ has no fixed point on $\partial U.$ If $f$
has $k$ distinct periodic points with period $M$, then any holomorphic
mapping $g:\overline{U}\rightarrow \mathbb{C}^{2}$ that is sufficiently
close to $f$ has at least $k$ distinct periodic points with period $M$ in $
U. $
\end{corollary}
\begin{proof}
This follows from the previous corollary directly.
\end{proof}
\begin{corollary}
\label{ad-2-0}Let $\{f_{n}\}\subset \mathcal{O}(\mathbb{C}^{2},0,0)$ be a
sequence converging to $f\in \mathcal{O}(\mathbb{C}^{2},0,0),$ uniformly in
a neighborhood of the origin. If the origin is an isolated zero of $f$, and
there exists an integer $k$ such that $\pi _{f_{n}}(0)\geq k$ for all $n\in
\mathbb{N},$ then $\pi _{f}(0)\geq k.$
\end{corollary}
\begin{proof}
This follows from Rouch\'{e}'s theorem directly.
\end{proof}
\begin{corollary}[{[\protect\ref{Zh2}]}]
\label{cc2-1}Let $M$ be a positive integer, let $U$ be a bounded open subset
of $\mathbb{C}^{2}$, let $f:\overline{U}\rightarrow \mathbb{C}^{2}$ be a
holomorphic mapping and assume that $f^{M}$ has no fixed point on $\partial
U $. Then:
(i). There exists an open subset $V$ of $U,$ such that $f^{M}$ is well
defined on $\overline{V},$ has no fixed point outside $V,$ and has only
finitely many fixed points in $V$.
(ii). For any holomorphic mapping $g:\overline{U}\rightarrow \mathbb{C}^{2}$
sufficiently close to $f$, $g^{M}$ is well defined on $\overline{V},$ has no
fixed point outside $V$ and has only finitely many fixed points in $V;$ and
furthermore,
\begin{equation*}
L(g^{M})=L(f^{M})\mathrm{,}\text{ \ }P_{M}(g)=P_{M}(f).
\end{equation*}
(iii). In particular, if $p_{0}\in U$ is the unique fixed point of both $f$
and $f^{M}$ in $\overline{U},$ then for any holomorphic mapping $g:\overline{
U}\rightarrow \mathbb{C}^{2}$ sufficiently close to $f$,
\begin{equation*}
L(g^{M})=L(f^{M})=\mu _{f^{M}}(p_{0})\mathrm{,}\text{ }
P_{M}(g)=P_{M}(f)=P_{M}(f,p_{0}).
\end{equation*}
\end{corollary}
\begin{remark}
Under the assumption that $f^{M}$ has no fixed point on $\partial U,$ for
any factor $m$ of $M,$ the conclusions (i)--(iii) remain valid if $M$ is
replaced by $m,$ since $f^{m}$ has no fixed point on $\partial U$ as well.
\end{remark}
\begin{lemma}
\label{lem2-1-2}Let $M$ be a positive integer and let $f:\overline{\Delta
^{2}}\rightarrow \mathbb{C}^{2}$\ be a holomorphic mapping such that $f^{M}$
has no fixed point on $\partial \Delta ^{2}$ and each fixed point of $f^{M}$
is simple. Then, $\mathrm{Fix}(f^{M})$ is finite, and
\textrm{(i) }$L(f^{M})=\#\mathrm{Fix}(f^{M})=\sum_{m|M}P_{m}(f);$
\textrm{(ii) }$P_{M}(f)$ is the cardinal number of the set of periodic
points of $f$ of period $M$;
\textrm{(iii) }$P_{M}(f)/M$ is the number of distinct periodic orbits of $f$
of period $M.$
\end{lemma}
\begin{proof}
(i) and (ii) are proved in [\ref{FL}] (see [\ref{Zh2}] for a very simple
proof), and (iii) follows from (ii).
\end{proof}
A fixed point $p$ of $f$ is called \textit{hyperbolic} if $Df(p)$ has no
eigenvalue of absolute $1.$ If $p$ is a hyperbolic fixed point of $f,$ then
it is a hyperbolic fixed point of all iterations $f^{j},j\in \mathbb{N}.$ A
hyperbolic fixed point is a simple fixed point, and so it has index $1$ by
Lemma \ref{lem2-a}$.$
\begin{lemma}
\label{lem2-1-1}Let $M$ be a positive integer, let $V$ be an open subset of $
\Delta ^{2}$ and let $f:$ $\overline{\Delta ^{2}}\rightarrow \mathbb{C}^{2}$
be a holomorphic mapping such that $f^{M}$ is well defined in $\overline{V}$
and has no fixed point on $\partial V$. Then for any positive number $
\varepsilon >0,$ there exists a holomorphic mapping $f_{\varepsilon }:
\overline{\Delta ^{2}}\rightarrow \mathbb{C}^{2}$ such that
\begin{equation*}
\sup_{X\in \overline{\Delta ^{2}}}|f_{\varepsilon {}}(X)-f(X)|<\varepsilon
\end{equation*}
and all the fixed points of $f_{\varepsilon }^{M}$ located in $\overline{V}$
are hyperbolic.
\end{lemma}
A proof of this result follows from the argument in [\ref{B}]. Another proof
can be found in [\ref{Zh1}].
\begin{corollary}
\label{do1}Let $f:\overline{\Delta ^{2}}\rightarrow \mathbb{C}^{2}$\ be a
holomorphic mapping so that $0$ is the unique fixed point of both $f$ and $
f^{M}$ in $\overline{\Delta ^{2}}.$ Then
\begin{equation}
\mathcal{O}_{M}(f,0)=P_{M}(f,0)/M, \label{j3}
\end{equation}
and there exists a $\delta >0,$ such that any holomorphic mapping $g:
\overline{\Delta ^{2}}\rightarrow \mathbb{C}^{2}$ with $\max_{x\in \overline{
\Delta ^{2}}}|g(x)-f(x)|<\delta $ has exactly $\mathcal{O}_{M}(f,0)$
distinct periodic orbits of period $M$ in $\Delta ^{2},$ provided that all
fixed points of $g^{M}$ are simple.
\end{corollary}
\begin{proof}
By Corollary \ref{cc2-1} (iii), there exists a $\delta >0$ such that, for
any holomorphic mapping $g:\overline{\Delta ^{2}}\rightarrow \mathbb{C}^{2}$
,
\begin{equation}
\max_{x\in \overline{\Delta ^{2}}}|g(x)-f(x)|<\delta \label{d1}
\end{equation}
implies
\begin{equation}
P_{M}(g)=P_{M}(f,0). \label{d2}
\end{equation}
Let $\varepsilon $ be any positive number with $\varepsilon <\delta .$ Then
by Lemma \ref{lem2-1-1}, there exists a holomorphic mapping $g_{1}:\overline{
\Delta ^{2}}\rightarrow \mathbb{C}^{2}$ satisfying (\ref{d1}) for $
\varepsilon $, say,
\begin{equation*}
\max_{x\in \overline{\Delta ^{2}}}|g_{1}(x)-f(x)|<\varepsilon ,
\end{equation*}
such that all fixed points of $g_{1}^{M}$ located in $\Delta ^{2}$ are
simple, and then by (\ref{d2}) and Lemma \ref{lem2-1-2} (iii), $g=g_{1}$ has
exactly $P_{M}(f,0)/M$ distinct periodic orbits of period $M.$ Therefore, by
the definition of the number $\mathcal{O}_{M}(f,0)$ and the arbitrariness of
$\varepsilon ,$ we have
\begin{equation*}
\mathcal{O}_{M}(f,0)\geq P_{M}(f,0)/M.
\end{equation*}
We show that the inequality does not occur.
Otherwise, by the definition of $\mathcal{O}_{M}(f,0),$ there exists a
holomorphic mapping $g_{2}:\overline{\Delta ^{2}}\rightarrow \mathbb{C}^{2}$
satisfying (\ref{d1}), say,
\begin{equation*}
\max_{x\in \overline{\Delta ^{2}}}|g_{2}(x)-f(x)|<\delta ,
\end{equation*}
such that $g_{2}$ has at least $P_{M}(f,0)+M$ distinct periodic points of
period $M$ in $\Delta ^{2}.$
Furthermore, by Corollary \ref{adc} and Lemma \ref{lem2-1-1}, there exists a
holomorphic mapping $g_{3}$ satisfying the following three conditions.
(a) $g_{3}$ is so close to $g_{2}$ that $g_{3}$ satisfies (\ref{d1}), say,
\begin{equation*}
\max_{x\in \overline{\Delta ^{2}}}|g_{3}(x)-f(x)|<\delta .
\end{equation*}
(b) $g_{3}$ is so close to $g_{2}$ that it has at least $P_{M}(f,0)+M$
distinct periodic points of period $M$ in $\Delta ^{2}.$
(c) All fixed points of $g_{3}^{M}$ in $\Delta ^{2}$ are simple.
By (c) and Lemma \ref{lem2-1-2} (ii), $g_{3}$ has exactly $P_{M}(g_{3})$
distinct periodic points of period $M,$ and then by (b), $P_{M}(g_{3})\geq
P_{M}(f,0)+M.$ But $g=g_{3}$ satisfies (\ref{d1}) by (a). Then $g=g_{3}$
satisfies (\ref{d2}), say, $P_{M}(g_{3})=P_{M}(f,0)$. This is a
contradiction, and then (\ref{j3}) is proved.
If $g:\overline{\Delta ^{2}}\rightarrow \mathbb{C}^{2}$ is a holomorphic
mapping that satisfies (\ref{d1}), then it satisfies (\ref{d2}), and then,
by Lemma \ref{lem2-1-2} (iii), it has
\begin{equation*}
P_{M}(g)/M=P_{M}(f,0)/M=\mathcal{O}_{M}(f,0)
\end{equation*}
distinct periodic orbits in $\Delta ^{2}$ of period $M,$ provided that all
fixed points of $g^{M}$ are simple. This completes the proof.
\end{proof}
The following result also follows from the above argument.
\begin{lemma}
\label{lem2-2}Let $k$ and $M$ be positive integers and let $f\in \mathcal{O}(
\mathbb{C}^{2},0,0)$. If $0$ is an isolated fixed point of both $f$ and $
f^{M},$ then $\mathcal{O}_{M}(f,0)\geq k$ if there exists a sequence of
holomorphic mappings $f_{j}\in \mathcal{O}(\mathbb{C}^{2},0,0),$ uniformly
converging to $f$ in a neighborhood of the origin, such that
\begin{equation*}
f_{j}(0)=0\;\mathrm{and\;}\mathcal{O}_{M}(f_{j},0)\geq k.
\end{equation*}
\end{lemma}
\begin{lemma}
\label{j7}Let $L:\mathbb{C}^{n}\rightarrow \mathbb{C}^{n}$ be a linear
mapping and let $M>1\ $be a positive integer. Then $L$ has a periodic point
of period $M$ if and only if $L$ has eigenvalues $\lambda _{1},\dots
,\lambda _{s},s\leq n,$ that are primitive $m_{1}$ th,..., $m_{s}$ th roots
of unity, respectively, such that $M=\left[ m_{1},\dots ,m_{s}\right] .$
\end{lemma}
This is a basic knowledge of elementary linear algebra. Recall that $
[m_{1},\dots ,m_{s}]$ denotes the least common multiple of $m_{1},\dots
,m_{s}$. This Lemma is only used once in this paper (in Section 1). We shall
frequently use its special case with $n=2$ in another version:
\begin{lemma}
\label{lem2-5}Let $L:\mathbb{C}^{2}\rightarrow \mathbb{C}^{2}$ be a linear
mapping and let $M>1\ $be a positive integer. Then $L$ has a periodic point
of period $M$ if and only if one of the following conditions holds.
$\mathrm{(a)}\;$One eigenvalue of $L$\ is a primitive $M$-th root of unity.
$\mathrm{(b)}$ The two eigenvalues of $L$\ are primitive $m_{1}$ th and $
m_{2}$ th roots of unity, respectively, such that $[m_{1},m_{2}]=M.$
\end{lemma}
\begin{lemma}
\label{lem2-3}Let $f\in \mathcal{O}(\mathbb{C}^{2},0,0)$\ and let
\begin{equation*}
\mathfrak{M}_{f}=\{m\in \mathbb{N};\;\mathrm{the\;linear\;part\;of\;}f\;
\mathrm{at}\;0\;\mathrm{has\;periodic\;points\;of\;period\;}m\}.
\end{equation*}
Then,
\textrm{(i). }For each $m\in \mathbb{N}\backslash \mathfrak{M}_{f}$ such
that $0$ is an isolated fixed point of $f^{m},$
\begin{equation*}
P_{m}(f,0)=0;
\end{equation*}
\textrm{(ii). }For each positive integer $M$ such that $0$ is an isolated
fixed point of $f^{M},$
\begin{equation*}
\mu _{f^{M}}(0)=\sum_{\substack{ m\in \mathfrak{M}_{f} \\ m|M}}P_{m}(f,0).
\end{equation*}
\end{lemma}
\begin{proof}
(i) and (ii) are essentially proved in [\ref{CMY}] (see [\ref{Zh2}] for a
simple proof).
\end{proof}
\begin{remark}
In the previous Lemma, the set $\mathfrak{M}_{f}$ contains at most four
numbers, by Lemma \ref{lem2-5}.
\end{remark}
\begin{lemma}
\label{lem2-4}Let $k$ be a positive integer, let $f$ and $h$ be germs in $
\mathcal{O}(\mathbb{C}^{2},0,0)$ such that $0$ is an isolated fixed point of
both $f$ and $f^{k}$ and $\det Dh(0)\neq 0,$ and let $g=h\circ f\circ
h^{-1}. $ Then $0$ is still an isolated fixed point of both $g$ and $g^{k}$,
and the following three equalities hold:
\begin{equation*}
\mu _{f^{k}}(0)=\mu _{g^{k}}(0),
\end{equation*}
\begin{equation*}
P_{k}(f,0)=P_{k}(g,0),
\end{equation*}
\begin{equation*}
\mathcal{O}_{k}(f,0)=\mathcal{O}_{k}(g,0).
\end{equation*}
\end{lemma}
\begin{proof}
The first equality is well known. The second follows from the first equality
and the definition of Dold's indices. The last then follows from the second
and Corollary \ref{do1}.
\end{proof}
The following result is due to M. Shub and D. Sullivan [\ref{SS}]. It is
also proved in [\ref{Zh1}].
\begin{lemma}
\label{SS-1}Let $m>1$ be a positive integer and let $f\in \mathcal{O}(
\mathbb{C}^{2},0,0).$ Assume that the origin is an isolated fixed point of $
f $ and that, for each eigenvalue $\lambda $ of $Df(0),$ either $\lambda =1$
or $\lambda ^{m}\neq 1$. Then the origin is still an isolated fixed point of
$f^{m}$ and
\begin{equation*}
\mu _{f}(0)=\mu _{f^{m}}(0).
\end{equation*}
\end{lemma}
\begin{lemma}[\protect\cite{LL}]
\label{prod-ind}Let $h_{1}$ and $h_{2}$ be germs in $\mathcal{O}(\mathbb{C}
^{2},0,0)$. If $0$ is an isolated zero of both $h_{1}$ and $h_{2},$ then the
zero order of $h_{1}\circ h_{2}$ at $0$ equals the product of the zero
orders of $h_{1}$ and $h_{2}$ at $0,$ say, $\pi _{h_{1}\circ h_{2}}(0)=\pi
_{h_{1}}(0)\pi _{h_{2}}(0).$
\end{lemma}
\begin{lemma}[\protect\cite{LL}]
\label{prod1}Let $f=(f_{1},h)$ and $g=(f_{2},h)$ be two germs in $\mathcal{O}
(\mathbb{C}^{2},0,0).$ If $0$ is an isolated zero of both $f$ and $g,$ then $
0$ is also an isolated zero of $F=(f_{1}f_{2},h)$ and
\begin{equation*}
\pi _{F}(0)=\pi _{f}(0)+\pi _{h}(0).
\end{equation*}
\end{lemma}
\section{Normal Forms and Iterations of Normal Forms\label{norm}\protect}
\bigskip
The following lemma is a basic result in the theory of normal forms (see
\cite{AP}, p. 84--85).
\begin{lemma}
\label{nor}Let $f\in \mathcal{O}(\mathbb{C}^{2},0,0)$ and assume that $
Df(0)=(\lambda _{1},\lambda _{2})$ is a diagonal matrix. Then for any
positive integer $r,$ there exists a polynomial transform
\begin{equation}
(y_{1},y_{2})=H(x_{1},x_{2})=(x_{1},x_{2})+\mathrm{higher\ terms}
\label{tra}
\end{equation}
of coordinates in a neighborhood of the origin such that each component $
g_{j}$ of
\begin{equation*}
g=(g_{1},g_{2})=H^{-1}\circ f\circ H
\end{equation*}
has a power series expansion
\begin{equation}
g_{j}(x_{1},x_{2})=\lambda
_{j}x_{j}+
\sum_{_{i_{1}+i_{2}=2}}^{r}c_{i_{1}i_{2}}^{j}x_{1}^{i_{1}}x_{2}^{i_{2}}+
\mathrm{higher\;terms,\;}j=1,2, \label{tra-d}
\end{equation}
in a neighborhood of the origin, where $i_{1}$ and $i_{2}$ are nonnegative
integers and, for $j=1$ and $2,$
\begin{equation}
c_{i_{1}i_{2}}^{j}\neq 0\ \mathrm{only\ if\ }\lambda _{j}=\lambda
_{1}^{i_{1}}\lambda _{2}^{i_{2}}. \label{tra-d-1}
\end{equation}
\end{lemma}
\begin{corollary}
\label{tra-d0}Let $H$ be any transform given by the previous lemma. Then for
each $k\in \mathbb{N},$ the $k$ th iteration $
g^{k}=(g_{1}^{(k)},g_{2}^{(k)}) $ of the germ $g=(g_{1},g_{2})=H^{-1}\circ
f\circ H$ has an expansion similar to (\ref{tra-d}), more precisely, in a
neighborhood of the origin,
\begin{equation}
g_{j}^{(k)}(x_{1},x_{2})=\lambda
_{j}^{k}x_{j}+
\sum_{_{i_{1}+i_{2}=2}}^{r}C_{i_{1}i_{2}}^{kj}x_{1}^{i_{1}}x_{2}^{i_{2}}+
\mathrm{higher\;terms,\;}j=1,2, \label{tra-d-d}
\end{equation}
where $i_{1}$ and $i_{2}$ are nonnegative integers and, for $j=1$ and $2,$
\begin{equation}
C_{i_{1}i_{2}}^{kj}\neq 0\ \mathrm{only\ if\ }\lambda _{j}=\lambda
_{1}^{i_{1}}\lambda _{2}^{i_{2}}. \label{tra-d-d0}
\end{equation}
\end{corollary}
Before starting the proof, we recall that the notation $o(|z|^{r})$ denotes
any holomorphic function germ whose power series expansion at the origin has
no terms of degrees from $0$ to $r.$ The same notation $o(|z|^{r})$ may
denote different function germs in different places, even in a single
equation.
\begin{proof}
By (\ref{tra-d}), the conclusion holds obviously, except (\ref{tra-d-d0}).
We show that the coefficients $C_{i_{1}i_{2}}^{kj}$ satisfy (\ref{tra-d-d0})
for all $k\in \mathbb{N}$ and $j=1,2.$ This is done by induction on $k.$
Since $g^{1}=(g_{1}^{(1)},g_{2}^{(1)})=(g_{1},g_{2})=g,$ the conclusion
holds for $k=1,$ by the previous lemma. Assume that (\ref{tra-d-d0}) is true
for $k=1,\dots ,l.$ We complete the proof by showing that (\ref{tra-d-d0})
is true for $k=l+1\ $and $j=1,2.$
For $j=1,$ and $k=l+1,$ it is clear by the induction hypothesis that
\begin{eqnarray*}
&&g_{1}^{(l+1)}(x_{1},x_{2}) \\
&=&g_{1}^{(l)}\circ
g(x_{1},x_{2})=g_{1}^{(l)}(g_{1}(x_{1},x_{2}),g_{2}(x_{1},x_{2})) \\
&=&\lambda
_{1}^{l}g_{1}(x_{1},x_{2})+\sum_{_{i_{1}+i_{2}=2}}^{r}C_{i_{1}i_{2}}^{l1}
\left[ g_{1}(x_{1},x_{2})\right] ^{i_{1}}\left[ g_{2}(x_{1},x_{2})\right]
^{i_{2}}+o(|g(x_{1},x_{2})|^{r}),
\end{eqnarray*}
and then, writing $o(|g(x_{1},x_{2})|^{r})=o(|x|^{r})$, we have
\begin{eqnarray}
&&g_{1}^{(l+1)}(x_{1},x_{2}) \label{add} \\
&=&\lambda
_{1}^{l}g_{1}(x_{1},x_{2})+\sum_{_{i_{1}+i_{2}=2}}^{r}C_{i_{1}i_{2}}^{l1}
\left[ g_{1}(x_{1},x_{2})\right] ^{i_{1}}\left[ g_{2}(x_{1},x_{2})\right]
^{i_{2}}+o(|x|^{r}), \notag
\end{eqnarray}
and by the induction hypothesis, for each pair $(i_{1},i_{2})$ in the sum in
(\ref{add}),
\begin{equation}
C_{i_{1}i_{2}}^{l1}\neq 0\;\mathrm{only\ if\ }\lambda _{1}=\lambda
_{1}^{i_{1}}\lambda _{2}^{i_{2}}. \label{add-1}
\end{equation}
By (\ref{tra-d}), for the first part of the right hand side of (\ref{add}),
we have
\begin{equation*}
\lambda _{1}^{l}g_{1}(x_{1},x_{2})=\lambda _{1}^{l+1}x_{1}+\lambda
_{1}^{l}
\sum_{_{i_{1}+i_{2}=2}}^{r}c_{i_{1}i_{2}}^{1}x_{1}^{i_{1}}x_{2}^{i_{2}}+
\mathrm{higher\;terms,}
\end{equation*}
where each $c_{i_{1}i_{2}}^{1}$ satisfies (\ref{tra-d-1}) for $j=1,$ say, $
\lambda _{1}^{l}g_{1}(x_{1},x_{2})$ is already in the form of the right hand
side of (\ref{tra-d-d}), together with condition (\ref{tra-d-d0}). Thus, by (
\ref{add}), to complete the induction for $j=1$, it suffices to show that we
can write the sum
\begin{equation}
\sigma (x_{1},x_{2})=\sum_{_{i_{1}+i_{2}=2}}^{r}C_{i_{1}i_{2}}^{l1}\left[
g_{1}(x_{1},x_{2})\right] ^{i_{1}}\left[ g_{2}(x_{1},x_{2})\right] ^{i_{2}}
\label{4.7}
\end{equation}
in (\ref{add}) to be
\begin{equation}
\sigma (x_{1},x_{2})=\sum_{_{j_{1}+j_{2}=2}}^{r}\mathcal{C}
_{j_{1}j_{2}}^{l+1,1}x_{1}^{j_{1}}x_{2}^{j_{2}}+o(|x|^{r})\mathrm{,}
\label{h5}
\end{equation}
such that for each pair $(j_{1},j_{2})$
\begin{equation}
\mathcal{C}_{j_{1}j_{2}}^{l+1,1}\neq 0\mathrm{\ only\ if\ }\ \lambda
_{1}=\lambda _{1}^{j_{1}}\lambda _{2}^{j_{2}}. \label{h6}
\end{equation}
By (\ref{4.7}) and the expressions of $g_{1}$ and $g_{2}$ in (\ref{tra-d}),
we have
\begin{eqnarray*}
&&\sigma (x_{1},x_{2}) \\
&=&\sum_{_{i_{1}+i_{2}=2}}^{r}C_{i_{1}i_{2}}^{l1}\left[ \lambda
_{1}x_{1}+
\sum_{s_{1}+s_{2}=2}^{r}c_{s_{1}s_{2}}^{1}x_{1}^{s_{1}}x_{2}^{s_{2}}+o(|x|^{r})
\right] ^{i_{1}} \\
&&\times \left[ \lambda
_{2}x_{2}+
\sum_{t_{1}+t_{2}=2}^{r}c_{t_{1}t_{2}}^{2}x_{1}^{t_{1}}x_{2}^{t_{2}}+o(|x|^{r})
\right] ^{i_{2}} \\
&=&\sum_{_{i_{1}+i_{2}=2}}^{r}C_{i_{1}i_{2}}^{l1}\left[ \lambda
_{1}x_{1}+
\sum_{s_{1}+s_{2}=2}^{r}c_{s_{1}s_{2}}^{1}x_{1}^{s_{1}}x_{2}^{s_{2}}\right]
^{i_{1}} \\
&&\times \left[ \lambda
_{2}x_{2}+
\sum_{t_{1}+t_{2}=2}^{r}c_{t_{1}t_{2}}^{2}x_{1}^{t_{1}}x_{2}^{t_{2}}\right]
^{i_{2}}+o(|x|^{r})
\end{eqnarray*}
Thus, $\sigma (x_{1},x_{2})$ is a power series at the origin which is the
sum of terms that is either of the form $cx_{1}^{i_{1}}x_{2}^{i_{2}}$ with $
i_{1}+i_{2}>r,$ or of the form
\begin{eqnarray}
D_{j_{1}j_{2}}x_{1}^{j_{1}}x_{2}^{j_{2}} &=&C_{i_{1}i_{2}}^{l1}\left[ \left(
\lambda _{1}x_{1}\right) ^{l_{1}}\prod_{(s_{1},s_{2})\in E_{1}}\left(
c_{s_{1}s_{2}}^{1}x_{1}^{s_{1}}x_{2}^{s_{2}}\right) ^{l_{s_{1}s_{2}}}\right]
\label{h} \\
&&\times \left[ \left( \lambda _{2}x_{2}\right) ^{l_{2}^{\prime
}}\prod_{(t_{1},t_{2})\in E_{2}}\left(
c_{t_{1}t_{2}}^{2}x_{1}^{t_{1}}x_{2}^{t_{2}}\right) ^{l_{t_{1}t_{2}}^{\prime
}}\right] , \notag
\end{eqnarray}
where $l_{1}$, $l_{2}^{\prime },$ $l_{s_{1}s_{2}}$ and $l_{t_{1}t_{2}}^{
\prime }$ are nonnegative integers, and $E_{1}$ and $E_{2}$ are sets of some
pairs of nonnegative integers such that
\begin{equation}
l_{1}+\sum_{(s_{1},s_{2})\in E_{1}}l_{s_{1}s_{2}}=i_{1},\ \mathrm{and}\
l_{2}^{\prime }+\sum_{(t_{1},t_{2})\in E_{2}}l_{t_{1}t_{2}}^{\prime }=i_{2}
\label{h1}
\end{equation}
\begin{equation}
D_{j_{1}j_{2}}=\lambda _{1}^{l_{1}}\lambda _{2}^{l_{2}^{\prime
}}C_{i_{1}i_{2}}^{l1}\left[ \prod_{(s_{1},s_{2})\in E_{1}}\left(
c_{s_{1}s_{2}}^{1}\right) ^{l_{s_{1}s_{2}}}\right] \left[
\prod_{(t_{1},t_{2})\in E_{2}}\left( c_{t_{1}t_{2}}^{2}\right)
^{l_{t_{1}t_{2}}^{\prime }}\right] , \label{h2}
\end{equation}
\begin{equation}
j_{1}=l_{1}+\sum_{(s_{1},s_{2})\in
E_{1}}s_{1}l_{s_{1}s_{2}}+\sum_{(t_{1},t_{2})\in
E_{2}}t_{1}l_{t_{1}t_{2}}^{\prime }, \label{h3}
\end{equation}
and
\begin{equation}
j_{2}=l_{2}^{\prime }+\sum_{(s_{1},s_{2})\in
E_{1}}s_{2}l_{s_{1}s_{2}}+\sum_{(t_{1},t_{2})\in
E_{2}}t_{2}l_{t_{1}t_{2}}^{\prime }. \label{h4}
\end{equation}
We show that
\begin{equation*}
D_{j_{1}j_{2}}\neq 0\mathrm{\ only\ if\ }\lambda _{1}=\lambda
_{1}^{j_{1}}\lambda _{2}^{j_{2}}.
\end{equation*}
By (\ref{h2}), (\ref{tra-d-1}) and (\ref{add-1}), it is clear that $
D_{j_{1}j_{2}}\neq 0$ only if
\begin{equation*}
\lambda _{1}=\lambda _{1}^{i_{1}}\lambda _{2}^{i_{2}},
\end{equation*}
\begin{equation*}
\lambda _{1}=\lambda _{1}^{s_{1}}\lambda _{2}^{s_{2}}\;\mathrm{for\ all\ }
(s_{1},s_{2})\in E_{1},
\end{equation*}
and
\begin{equation*}
\lambda _{2}=\lambda _{1}^{t_{1}}\lambda _{2}^{t_{2}}\;\mathrm{for\ all\ }
(t_{1},t_{2})\in E_{2}.
\end{equation*}
Therefore, if $D_{j_{1}j_{2}}\neq 0,$ then we have, by (\ref{h1})--(\ref{h4}
), that
\begin{eqnarray*}
\lambda _{1}^{j_{1}}\lambda _{2}^{j_{2}} &=&\lambda
_{1}^{l_{1}+\sum_{(s_{1},s_{2})\in
E_{1}}s_{1}l_{s_{1}s_{2}}+\sum_{(t_{1},t_{2})\in
E_{2}}t_{1}l_{t_{1}t_{2}}^{\prime }} \\
&&\times \lambda _{2}^{l_{2}^{\prime }+\sum_{(s_{1},s_{2})\in
E_{1}}s_{2}l_{s_{1}s_{2}}+\sum_{(t_{1},t_{2})\in
E_{2}}t_{2}l_{t_{1}t_{2}}^{\prime }} \\
&=&\left[ \lambda _{1}^{l_{1}}\prod_{(s_{1},s_{2})\in E_{1}}\left( \lambda
_{1}^{s_{1}}\lambda _{2}^{s_{2}}\right) ^{l_{s_{1}s_{2}}}\right] \left[
\lambda _{2}^{l_{2}^{\prime }}\prod_{(t_{1},t_{2})\in E_{2}}\left( \lambda
_{1}^{t_{1}}\lambda _{2}^{t_{2}}\right) ^{l_{t_{1}t_{2}}^{\prime }}\right] \\
&=&\left[ \lambda _{1}^{l_{1}}\prod_{(s_{1},s_{2})\in E_{1}}\left( \lambda
_{1}\right) ^{l_{s_{1}s_{2}}}\right] \left[ \lambda _{2}^{l_{2}^{\prime
}}\prod_{(t_{1},t_{2})\in E_{2}}(\lambda _{2})^{l_{t_{1}t_{2}}^{\prime }}
\right] \\
&=&\left[ \lambda _{1}^{l_{1}+\sum_{(s_{1},s_{2})\in E_{1}}l_{s_{1}s_{2}}}
\right] \left[ \lambda _{2}^{l_{2}^{\prime }+\sum_{(t_{1},t_{2})\in
E_{2}}l_{t_{1}t_{2}}^{\prime }}\right] \\
&=&\lambda _{1}^{i_{1}}\lambda _{2}^{i_{2}}=\lambda _{1}.
\end{eqnarray*}
In other words, $D_{j_{1}j_{2}}\neq 0$ only if $\lambda _{1}=\lambda
_{1}^{j_{1}}\lambda _{2}^{j_{2}}.$ Thus, we can write $\sigma (x_{1},x_{2})$
to be (\ref{h5}) with (\ref{h6}), and then, we have proved that (\ref
{tra-d-d0}) is true for $k=l+1$ and $j=1.$ For the same reason, (\ref
{tra-d-d0}) is true for $k=l+1$ and $j=2.$ The induction is complete.
\end{proof}
\section{Computing Zero Orders Via Cronin's Theorem\label{cronin-1}}
Since fixed point indices are defined via zero orders, to prove the main
theorem, it is useful to compute the zero orders of some special germs of
holomorphic mappings. The following Cronin's theorem plays an important role
in our computations.
\begin{theorem}[\textbf{Cronin \protect\cite{Cro}}]
Let $f\in \mathcal{O}(\mathbb{C}^{2},0,0)$ be given by
\begin{equation*}
f(x_{1},x_{2})=\left(
P_{m_{1}}(x_{1},x_{2})+o(|x|^{m_{1}}),Q_{m_{2}}(x_{1},x_{2})+o(|x|^{m_{2}}
\right) ),
\end{equation*}
where $x=(x_{1},x_{2}),$ $P_{m_{1}}$ and $Q_{m_{2}}$\ are homogeneous
polynomials of degrees $m_{1}$ and $m_{2},$ respectively,\ in $x_{1}$ and $
x_{2}$. If the origin $0$ is an isolated solution of the system
\begin{equation}
\left\{
\begin{array}{c}
P_{m_{1}}(x_{1},x_{2})=0, \\
Q_{m_{2}}(x_{1},x_{2})=0,
\end{array}
\right. \label{cr}
\end{equation}
then $0$\ is an isolated zero of the germ $f$ with zero order
\begin{equation*}
\pi _{f}(0)=m_{1}m_{2}.
\end{equation*}
If $0$ is an isolated zero of $f$ but is not an isolated solution of the
system (\ref{cr}), then
\begin{equation*}
\pi _{f}(0)>m_{1}m_{2}.
\end{equation*}
\end{theorem}
We now apply Cronin's theorem to some special cases.
\begin{corollary}
\label{c1}Let $g=(g_{1},g_{2})\in \mathcal{O}(\mathbb{C}^{2},0,0)$ be given
by
\begin{equation*}
\left\{
\begin{array}{l}
g_{1}(x_{1},x_{2})=x_{1}^{m_{1}+1}(a_{1}+o(1))+x_{2}^{d}(a_{2}+o(1)), \\
g_{2}(x_{1},x_{2})=x_{1}^{m_{1}}x_{2}(a_{3}+o(1))+x_{2}^{d+1}O(1),
\end{array}
\right.
\end{equation*}
where $d$ and $m_{1}$ are positive integers, and $a_{i}$ are constants.
If the origin is an isolated zero of $g,$ then
\begin{equation*}
\pi _{g}(0)\geq dm_{1}+m_{1}+1.
\end{equation*}
If $a_{1}\neq 0,a_{2}\neq 0$ and $a_{3}\neq 0,$ then the origin is an
isolated zero of $g$ with
\begin{equation*}
\pi _{g}(0)=dm_{1}+m_{1}+1.
\end{equation*}
\end{corollary}
\begin{proof}
Let $h\in \mathcal{O}(\mathbb{C}^{2},0,0)$ be given by
\begin{equation*}
h(y_{1},y_{2})=(y_{1}^{d},y_{2}^{m_{1}+1}).
\end{equation*}
then the two component of $g\circ h=(g_{1}\circ h,g_{2}\circ h)$ have the
expressions
\begin{equation}
\left\{
\begin{array}{l}
g_{1}\circ h(y_{1},y_{2})=a_{1}y_{1}^{d(m_{1}+1)}+a_{2}y_{2}^{d(m_{1}+1)}+
\mathrm{higher\ terms,} \\
g_{2}\circ h(y_{1},y_{2})=a_{3}y_{1}^{dm_{1}}y_{2}^{m_{1}+1}+\mathrm{higher\
terms}.
\end{array}
\right. \label{j14}
\end{equation}
If the origin is an isolated zero of $g,$ then it is also an isolated zero
of $g\circ h,$ and then, by Cronin's theorem,
\begin{equation}
\pi _{g\circ h}(0)\geq d(m_{1}+1)(dm_{1}+m_{1}+1), \label{h7}
\end{equation}
and, considering that $\pi _{h}(0)=d(m_{1}+1),$ by Lemma \ref{prod-ind} we
have
\begin{equation}
\pi _{g}(0)=\pi _{g\circ h}(0)/\pi _{h}(0)\geq dm_{1}+m_{1}+1. \label{h8}
\end{equation}
On the other hand, it is clear that when $a_{1},a_{2}$ and $a_{3}$ are all
nonzero, $0$ is an isolated solution of the system
\begin{equation*}
\left\{
\begin{array}{c}
a_{1}y_{1}^{d(m_{1}+1)}+a_{2}y_{2}^{d(m_{1}+1)}=0, \\
a_{3}y_{1}^{dm_{1}}y_{2}^{m_{1}+1}=0,
\end{array}
\right.
\end{equation*}
and then by (\ref{j14}) and Cronin's theorem, $0$ is an isolated zero of $
g\circ h$ and the equality in (\ref{h7}) holds, and then $0$ is also an
isolated zero of $g$ and the equality in (\ref{h8}) holds again\label{?1}.
This completes the proof.
\end{proof}
\begin{corollary}
\label{c2}Let $g=(g_{1},g_{2})\in \mathcal{O}(\mathbb{C}^{2},0,0)$ be given
by
\begin{equation*}
\left\{
\begin{array}{l}
g_{1}(x_{1},x_{2})=x_{1}^{m}(a+o(1))+x_{2}O(1), \\
g_{2}(x_{1},x_{2})=x_{2}(b+o(1))+x_{1}^{m}o(1),
\end{array}
\right.
\end{equation*}
where $a$ and $b$ are constants.
If the origin is an isolated zero of $g,$ then
\begin{equation*}
\pi _{g}(0)\geq m.
\end{equation*}
If $a\neq 0$ and $b\neq 0$, then the origin is an isolated zero of $g$ with
\begin{equation*}
\pi _{g}(0)=m.
\end{equation*}
\end{corollary}
\begin{proof}
Let $h\in \mathcal{O}(\mathbb{C}^{2},0,0)$ be given by
\begin{equation*}
h(y_{1},y_{2})=(y_{1},y_{2}^{m}).
\end{equation*}
Then the germ $g\circ h=(g_{1}\circ h,g_{2}\circ h)$ has the expression
\begin{equation}
\left\{
\begin{array}{l}
g_{1}\circ h(y_{1},y_{2})=ay_{1}^{m}+a_{0}y_{2}^{m}+\mathrm{higher\ terms,}
\\
g_{2}\circ h(y_{1},y_{2})=by_{2}^{m}+\mathrm{higher\ terms},
\end{array}
\right. \label{j4}
\end{equation}
for some constant $a_{0}.$
If the origin is an isolated zero of $g,$ then it is also an isolated zero
of the germ $g\circ h$, and by (\ref{j4}) and Cronin's theorem,
\begin{equation*}
\pi _{g\circ h}(0)\geq m^{2},
\end{equation*}
and the equality holds if $a\neq 0$ and $b\neq 0;$ and then by Lemma \ref
{prod-ind} and by the fact $\pi _{h}(0)=m,$
\begin{equation*}
\pi _{g}(0)=\pi _{g\circ h}(0)/\pi _{h}(0)\geq m,
\end{equation*}
and the equality holds if $a\neq 0$ and $b\neq 0.$ On the other hand, by (
\ref{j4}) and Cronin's theorem, when $a\neq 0$ and $b\neq 0,$ $0$ is an
isolated zero of $g\circ h,$ and then $0$ is also an isolated zero of $g.$
This completes the proof.
\end{proof}
\begin{corollary}
\label{c3}Let $f=(f_{1},f_{2})\in \mathcal{O}(\mathbb{C}^{2},0,0)$ be given
by
\begin{eqnarray*}
f_{1}(x_{1},x_{2}) &=&x_{1}^{n_{1}}O(1)+x_{2}^{n_{2}}O(1), \\
f_{2}(x_{1},x_{2}) &=&x_{1}^{n_{1}}O(1)+x_{2}^{n_{2}}O(1),
\end{eqnarray*}
where $n_{1}$ and $n_{2}$ are positive integers. Assume that $0$ is an
isolated zero of $f.$ Then
\begin{equation*}
\pi _{f}(0)\geq n_{1}n_{2}.
\end{equation*}
\end{corollary}
\begin{proof}
Consider $h\in \mathcal{O}(\mathbb{C}^{2},0,0)$ given by $
h(y_{1},y_{2})=(y_{1}^{n_{2}},y_{2}^{n_{1}}).$ Then $0$ is an isolated zero
of the germ $f\circ h=(f_{1}\circ h,f_{2}\circ h)$ and
\begin{eqnarray*}
f_{1}\circ h(y_{1},y_{2}) &=&y_{1}^{n_{1}n_{2}}O(1)+y_{2}^{n_{1}n_{2}}O(1),
\\
f_{2}\circ h(y_{1},y_{2}) &=&y_{1}^{n_{1}n_{2}}O(1)+y_{2}^{n_{1}n_{2}}O(1).
\end{eqnarray*}
Thus, by Cronin's theorem we have $\pi _{f\circ h}(0)\geq
n_{1}^{2}n_{2}^{2}, $ and then by the fact $\pi _{h}(0)=n_{1}n_{2}$ and by
Lemma \ref{prod-ind}, we have $\pi _{f}(0)=\pi _{f\circ h}(0)/\pi
_{h}(0)\geq n_{1}n_{2}.$
\end{proof}
\begin{corollary}
\label{c4}Let $f=(f_{1},f_{2})\in \mathcal{O}(\mathbb{C}^{2},0,0)$ be given
by
\begin{eqnarray*}
f_{1}(x_{1},x_{2}) &=&x_{1}^{m_{1}}, \\
f_{2}(x_{1},x_{2}) &=&x_{1}^{m_{1}}O(1)+x_{2}^{rm_{2}}O(1),
\end{eqnarray*}
where $r,m_{1}$ and $m_{2}$ are positive integers. Assume that $0$ is an
isolated zero of $f.$ Then
\begin{equation*}
\pi _{f}(0)\geq rm_{1}m_{2}.
\end{equation*}
\end{corollary}
\begin{proof}
It follows from the previous corollary, by taking $n_{1}=m_{1}$ and $
n_{2}=rm_{2}.$
\end{proof}
\begin{corollary}
\label{c5}Let $f=(f_{1},f_{2})\in \mathcal{O}(\mathbb{C}^{2},0,0)$ be given
by
\begin{eqnarray*}
f_{1}(x_{1},x_{2}) &=&x_{1}^{n_{1}}\left[
x_{1}^{rn_{1}}O(1)+x_{2}^{n_{2}}O(1)\right] , \\
f_{2}(x_{1},x_{2}) &=&x_{2}^{n_{2}}\left[
x_{1}^{n_{1}}O(1)+x_{2}^{rn_{2}}O(1)\right] ,
\end{eqnarray*}
where $r,$ $n_{1}$ and $n_{2}$ are positive integers. Assume that $0$ is an
isolated zero of $f.$ Then
\begin{equation*}
\pi _{f}(0)\geq 2n_{1}n_{2}+2rn_{1}n_{2}.
\end{equation*}
\end{corollary}
\begin{proof}
By Lemma \ref{prod1}, the zero order $\pi _{f}(0)$ equals the sum of the
zero orders of the four germs in $\mathcal{O}(\mathbb{C}^{2},0,0)$ given by
\begin{eqnarray*}
g_{1}(x_{1},x_{2}) &=&(x_{1}^{n_{1}},x_{2}^{n_{2}}), \\
g_{2}(x_{1},x_{2}) &=&(x_{1}^{n_{1}},x_{1}^{n_{1}}O(1)+x_{2}^{rn_{2}}O(1)),
\\
g_{3}(x_{1},x_{2}) &=&(x_{1}^{rn_{1}}O(1)+x_{2}^{n_{2}}O(1),x_{2}^{n_{2}}),
\\
g_{4}(x_{1},x_{2})
&=&(x_{1}^{rn_{1}}O(1)+x_{2}^{n_{2}}O(1),x_{1}^{n_{1}}O(1)+x_{2}^{rn_{2}}O(1)).
\end{eqnarray*}
By Cronin's theorem we have $\pi _{g_{1}}(0)\geq n_{1}n_{2};$ by Corollary
\ref{c4} we have $\pi _{g_{2}}(0)\geq rn_{1}n_{2}$ and $\pi _{g_{3}}(0)\geq
rn_{1}n_{2};$ and by Corollary \ref{c3} we have $\pi _{g_{4}}(0)\geq
n_{1}n_{2}.$ Thus, we have
\begin{equation*}
\pi _{f}(0)\geq 2n_{1}n_{2}+2rn_{1}n_{2}.
\end{equation*}
\end{proof}
\begin{lemma}
\label{ad-2-1}Let $f=(f_{1},f_{2})$ and $g=(g_{1},g_{2})$ be germs in $
\mathcal{O}(\mathbb{C}^{2},0,0)$ and assume that $A=(a_{ij})$ is a $2\times
2 $ matrix whose elements $a_{ij}$ are germs of holomorphic functions at the
origin of $\mathbb{C}^{2},$ with $\det A(0)\neq 0$. If
\begin{equation*}
(f_{1},f_{2})=(g_{1},g_{2})A=(g_{1}a_{11}+g_{2}a_{21},g_{1}a_{12}+g_{2}a_{22}),
\end{equation*}
and the origin is an isolated zero of $g,$ then the origin is also an
isolated zero of $f$ and
\begin{equation*}
\pi _{f}(0)=\pi _{g}(0).
\end{equation*}
\end{lemma}
\begin{proof}
By the assumption, there exists a ball $B$ centered at the origin in $
\mathbb{C}^{2}$ such that the origin is the unique zero of $g$ in $\overline{
B},$ $A$ is well defined on $\overline{B}$ and
\begin{equation}
\det A(x_{1},x_{2})\neq 0,\ \ (x_{1},x_{2})\in B. \label{h9}
\end{equation}
Then there exists a regular value $\varepsilon =(\varepsilon
_{1},\varepsilon _{2})$ of $g,$ which can be chosen close to the origin
arbitrarily, such that $g^{-1}(\varepsilon )\cap B$ contains exactly $\pi
_{g}(0)$ distinct points $(a_{1},b_{1}),\dots ,(a_{\pi _{g}(0)},b_{\pi
_{g}(0)})$ in $B.$ Thus, we have
\begin{equation*}
f(a_{i},b_{i})-(\varepsilon _{1},\varepsilon _{2})A(a_{i},b_{i})=0,i=1,\dots
,\pi _{g}(0).
\end{equation*}
In other words, $f_{\varepsilon }(x_{1},x_{2})=f(x_{1},x_{2})-(\varepsilon
_{1},\varepsilon _{2})A(x_{1},x_{2})$ has $\pi _{g}(0)$ distinct zeros in $
B. $ It is clear that $f_{\varepsilon }(x_{1},x_{2})$ is a germ in $\mathcal{
O}(\mathbb{C}^{2},0,0)$ converging to $f$ uniformly on $\overline{B}$ as $
\varepsilon =(\varepsilon _{1},\varepsilon _{2})\rightarrow 0,$ and on the
other hand, the origin is also the unique zero of $f$ in $\overline{B}$ as
well. Thus, by Rouche's theorem, we have
\begin{equation*}
\pi _{f}(0)\geq \pi _{g}(0).
\end{equation*}
But by (\ref{h9}), the inverse $A^{-1}$ of the matrix $A$ is well defined on
$\overline{B},$ which is again a matrix of holomorphic functions, and $
g=fA^{-1}.$ Thus, for the same reason we have $\pi _{f}(0)\leq \pi _{g}(0).$
This completes the proof.
\end{proof}
\begin{corollary}
\label{ad-1}Let $f=(f_{1},f_{2})\in \mathcal{O}(\mathbb{C}^{2},0,0)$, let $h$
be a holomorphic function germ at the origin, and let $g\in \mathcal{O}(
\mathbb{C}^{2},0,0)$ be given by
\begin{equation*}
g=(f_{1},f_{2}+hf_{1})=(f_{1},f_{2})\left(
\begin{array}{cc}
1 & \ h \\
0 & \ 1
\end{array}
\right) .
\end{equation*}
If the origin is an isolated zero of $f,$ then it is an isolated zero of $g$
and
\begin{equation*}
\pi _{g}(0)=\pi _{f}(0).
\end{equation*}
\end{corollary}
\begin{lemma}
\label{dd4}Let $d>1,n_{1}>1$ and $n_{2}>1$ be positive integers and let $
f=(f_{1},f_{2})\in \mathcal{O}(\mathbb{C}^{2},0,0)$ be given by
\begin{equation}
\left\{
\begin{array}{rl}
f_{1}(x_{1},x_{2})= &
x_{1}^{dn_{1}+1}a_{11}(x_{1},x_{2})+x_{1}^{n_{1}+1}x_{2}^{n_{2}}O(1) \\
& +x_{1}x_{2}^{dn_{2}}O(1)+x_{2}^{2dn_{2}+1}a_{12}(x_{1},x_{2}),\medskip \\
f_{2}(x_{1},x_{2})= &
x_{1}^{dn_{1}}x_{2}O(1)+x_{1}^{n_{1}}x_{2}^{n_{2}+1}O(1) \\
& +x_{2}^{dn_{2}+1}a_{22}(x_{1},x_{2})+x_{1}^{2dn_{1}+1}a_{21}(x_{1},x_{2}),
\end{array}
\right. \label{dd}
\end{equation}
where $a_{ij}=a_{ij}(x_{1},x_{2})$ are holomorphic function germs at the
origin. Assume that the origin is an isolated zero of $f.$ Then
\begin{equation}
\pi _{f}(0)\geq 2dn_{1}n_{2}+dn_{1}+dn_{2}+1. \label{dd1}
\end{equation}
\end{lemma}
\begin{proof}
We first assume
\begin{equation}
a_{11}(0)\neq 0,a_{22}(0)\neq 0. \label{dd3}
\end{equation}
Then $[a_{11}(x_{1},x_{2})]^{-1}$ and $[a_{22}(x_{1},x_{2})]^{-1}$ are also
holomorphic function germs at the origin, and we can reduce the germ $f$
into a simpler germ $h=(h_{1},h_{2})$ of the form
\begin{equation}
\left\{
\begin{array}{c}
h_{1}(x_{1},x_{2})=x_{1}r_{1}(x_{1},x_{2}), \\
h_{2}(x_{1},x_{2})=x_{2}r_{2}(x_{1},x_{2}).
\end{array}
\right. \label{ddd1}
\end{equation}
with
\begin{equation}
\left\{
\begin{array}{c}
r_{1}(x_{1},x_{2})=x_{1}^{dn_{1}}a_{11}(x_{1},x_{2})+x_{1}^{n_{1}}x_{2}^{n_{2}}O(1)+x_{2}^{dn_{2}}O(1),
\\
r_{2}(x_{1},x_{2})=x_{1}^{dn_{1}}O(1)+x_{1}^{n_{1}}x_{2}^{n_{2}}O(1)+x_{2}^{dn_{2}}a_{22}(x_{1},x_{2}),
\end{array}
\right. \label{j9}
\end{equation}
and with
\begin{equation}
\pi _{f}(0)=\pi _{h}(0). \label{j10}
\end{equation}
By Corollary \ref{ad-1}, $\pi _{f}(0)$ equals $\pi _{g}(0)$ for the germ $
g=(g_{1},g_{2})\in \mathcal{O}(\mathbb{C}^{2},0,0)$ that is given by
\begin{eqnarray*}
g_{1}(x_{1},x_{2}) &=&f_{1}(x_{1},x_{2}), \\
g_{2}(x_{1},x_{2})
&=&f_{2}(x_{1},x_{2})-x_{1}^{dn_{1}}[a_{11}(x_{1},x_{2})]^{-1}a_{21}(x_{1},x_{2})f_{1}(x_{1},x_{2}),
\end{eqnarray*}
Since $n_{1}>1,n_{2}>1$ and $d>1,$ by (\ref{dd}) we can write
\begin{equation*}
x_{1}^{dn_{1}}[a_{11}(x_{1},x_{2})]^{-1}a_{21}(x_{1},x_{2})f_{1}(x_{1},x_{2})=x_{1}^{2dn_{1}+1}a_{21}(x_{1},x_{2})+x_{1}^{dn_{1}}x_{2}O(1),
\end{equation*}
and then again by (\ref{dd}) we have
\begin{equation}
g_{2}(x_{1},x_{2})=x_{1}^{dn_{1}}x_{2}O(1)+x_{1}^{n_{1}}x_{2}^{n_{2}+1}O(1)+x_{2}^{dn_{2}+1}a_{22}(x_{1},x_{2}).
\label{j13}
\end{equation}
Again by Corollary \ref{ad-1}, $\pi _{g}(0)$ equals $\pi _{h}(0)$ for the
germ $h=(h_{1},h_{2})$ given by
\begin{eqnarray*}
h_{1}(x_{1},x_{2})
&=&g_{1}(x_{1},x_{2})-x_{2}^{dn_{2}}[a_{22}(x_{1},x_{2})]^{-1}a_{12}(x_{1},x_{2})g_{2}(x_{1},x_{2}),
\\
h_{2}(x_{1},x_{2}) &=&g_{2}(x_{1},x_{2}),
\end{eqnarray*}
and, by (\ref{j13}) and the expression of $g_{1}=f_{1}$ in (\ref{dd}), it is
easy to see that $h_{1}(x_{1},x_{2})$ has the expression
\begin{equation*}
h_{1}(x_{1},x_{2})=x_{1}^{dn_{1}+1}a_{11}(x_{1},x_{2})+x_{1}^{n_{1}+1}x_{2}^{n_{2}}O(1)+x_{1}x_{2}^{dn_{2}}O(1),
\end{equation*}
and then the germ $h=(h_{1},h_{2})=(h_{1},g_{2})$ has the expression (\ref
{ddd1}), such that (\ref{j9}) and (\ref{j10}) hold.
By (\ref{dd3}), repeating the above arguments, we can reduce the germ $
r=(r_{1},r_{2})\in \mathcal{O}(\mathbb{C}^{2},0,0)$ into a further simpler
germ $s=(s_{1},s_{2})$ with the expression
\begin{eqnarray*}
s_{1}(x_{1},x_{2})
&=&x_{1}^{dn_{1}}O(1)+x_{1}^{n_{1}}x_{2}^{n_{2}}O(1)=x_{1}^{n_{1}}\left[
x_{1}^{(d-1)n_{1}}O(1)+x_{2}^{n_{2}}O(1)\right] , \\
s_{2}(x_{1},x_{2})
&=&x_{1}^{n_{1}}x_{2}^{n_{2}}O(1)+x_{2}^{dn_{2}}O(1)=x_{2}^{n_{2}}\left[
x_{1}^{n_{1}}O(1)+x_{2}^{(d-1)n_{2}}O(1)\right] ,
\end{eqnarray*}
such that $\pi _{r}(0)=\pi _{s}(0).$ By Corollary \ref{c5}, we have
\begin{equation*}
\pi _{s}(0)\geq 2n_{1}n_{2}+2(d-1)n_{1}n_{2}=2dn_{1}n_{2},
\end{equation*}
and then
\begin{equation*}
\pi _{r}(0)=\pi _{s}(0)\geq 2dn_{1}n_{2}.
\end{equation*}
On the other hand, by Corollary \ref{c2}, for the germs in $\mathcal{O}(
\mathbb{C}^{2},0,0)$ given by
\begin{eqnarray*}
k_{1}(x_{1},x_{2}) &=&(x_{1},x_{2}), \\
k_{2}(x_{1},x_{2}) &=&(r_{1}(x_{1},x_{2}),x_{2}), \\
k_{3}(x_{1},x_{2}) &=&(x_{1},r_{2}(x_{1},x_{2})),
\end{eqnarray*}
we have
\begin{equation*}
\pi _{k_{1}}(0)=1,\pi _{k_{2}}(0)=dn_{1},\pi _{k_{3}}(0)=dn_{2}.
\end{equation*}
Thus, by Lemma \ref{prod1} and (\ref{ddd1}),
\begin{eqnarray*}
\pi _{h}(0) &=&\pi _{k_{1}}(0)+\pi _{k_{2}}(0)+\pi _{k_{3}}(0)+\pi _{r}(0) \\
&\geq &1+dn_{1}+dn_{2}+2dn_{1}n_{2},
\end{eqnarray*}
which implies (\ref{dd1}), by (\ref{j10}).
If (\ref{dd3}) fails, then we consider the germ $f_{\varepsilon }\in
\mathcal{O}(\mathbb{C}^{2},0,0)$ that is obtain from $f$ by just replacing $
a_{ii}(x_{1},x_{2})$ with $a_{ii}(x_{1},x_{2})+\varepsilon ,i=1,2.$ Then $
f_{\varepsilon }$ converges to $f$ uniformly in a neighborhood of the
origin, $f_{\varepsilon }$ has the form of (\ref{dd}) and for sufficiently
small $\varepsilon ,$ $0$ is an isolated zero of $f_{\varepsilon }$ by Rouch
\'{e}'s Theorem, and $a_{ii}(0,0)+\varepsilon \neq 0$ for $i=1$ and $2.$
Thus, the above arguments are applied to such $f_{\varepsilon }$ if $
\varepsilon $ is small enough. In other word, for sufficiently small $
\varepsilon ,$ we have
\begin{equation*}
\pi _{f_{\varepsilon }}(0)\geq 1+dn_{1}+dn_{2}+2dn_{1}n_{2},
\end{equation*}
and then (\ref{dd1}) follows from Corollary \ref{ad-2-0}.
\end{proof}
By Cronin's theorem, one can prove the following result.
\begin{proposition}
\label{c8}Let $f=(f_{1},f_{2})\in \mathcal{O}(\mathbb{C}^{2},0,0)$ be given
by
\begin{eqnarray*}
f_{1}(x_{1},x_{2}) &=&\lambda
_{1}x_{1}+x_{1}(a_{11}x_{1}^{m_{1}}+a_{12}x_{2}^{m_{2}}), \\
f_{2}(x_{1},x_{2}) &=&\lambda
_{2}x_{2}+x_{2}(a_{21}x_{1}^{m_{1}}+a_{22}x_{2}^{m_{2}}),
\end{eqnarray*}
where $\lambda _{1},\lambda _{2}$ are primitive $m_{1}$-th$\ $and $m_{2}$-th
roots of unity, respectively, $m_{1}$ and $m_{2}$ are positive integers that
are relatively prime.
If $a_{11}\neq 0,$ $a_{22}\neq 0$ and $\det (a_{ij})\neq 0,$ then the origin
is an isolated fixed point of $f^{m_{1}},f^{m_{2}}$ and $f^{m_{1}m_{2}}$ and
the following formulae hold.
\begin{equation*}
\mu _{f^{m_{1}}}(0)=(m_{1}+1),
\end{equation*}
\begin{equation*}
\mu _{f^{m_{2}}}(0)=(m_{2}+1),
\end{equation*}
\begin{equation*}
P_{m_{1}m_{2}}(f,0)=m_{1}m_{2}.
\end{equation*}
\end{proposition}
\begin{proof}
When $m_{1}$ and $m_{2}$ are distinct primes, this is proved in [\ref{Zh2}].
But in general, the proof is exactly the same.
\end{proof}
\section{Proof of the Main Theorem: (B)$\Rightarrow $(A) \label{cronin}}
In this section, we deduce (A) from (B) in the main theorem.
Assume that $M>1$ is an integer and $A$ is a matrix that satisfies (B) and
let $f\ $be a germ in $\mathcal{O}(\mathbb{C}^{2},0,0)$ such that
\begin{equation*}
Df(0)=A,
\end{equation*}
and that the origin is an isolated fixed point of $f^{M}.$ We shall show that
\begin{equation}
\mathcal{O}_{M}(f,0)\geq 2. \label{tar}
\end{equation}
By Lemma \ref{lem2-4} and the assumption in (B), we may assume that
\begin{equation*}
A=\left(
\begin{array}{cc}
\lambda _{1} & 0 \\
0 & \lambda _{2}
\end{array}
\right) ,
\end{equation*}
where $\lambda _{1}$ and $\lambda _{2}$ are primitive $m_{1}$ th and $m_{2}$
th roots of unity, respectively, and one of the following conditions holds.
\textrm{(b1) }$m_{1}=m_{2}=M$ and $\lambda _{1}=\lambda _{2}.$
\textrm{(b2) }$m_{1}=m_{2}=M$ and there exists positive integers $1<\alpha
<M $ and $1<\beta <M$ such that
\begin{equation}
\lambda _{1}^{\beta }=\lambda _{2},\lambda _{2}^{\alpha }=\lambda
_{1},\alpha \beta >M+1. \label{aid}
\end{equation}
\textrm{(b3) }$m_{1}|m_{2},$ $m_{2}=M$, and $\lambda _{2}^{m_{2}/m_{1}}\neq
\lambda _{1}.$
\textrm{(b4) }$M=[m_{1},m_{2}],(m_{1},m_{2})>1$ and $\max \{m_{1},m_{2}\}<M.$
Then, in any case from (b1) to (b4), the origin is a simple fixed point of $
f $, and then by Lemma \ref{lem2-a}, we have
\begin{equation}
\mu _{f}(0)=P_{1}(f,0)=1. \label{all}
\end{equation}
We show that any one of the four conditions from (b1) to (b4) deduces (\ref
{tar}), and divide the proof into four parts.\medskip
\begin{description}
\item[\textbf{Part 1}] \textbf{(b1) }$\Rightarrow $\textbf{\ (\ref{tar}
).\medskip }
\end{description}
\begin{proof}
In case (b1), we may assume $\lambda _{1}=\lambda _{2}=\lambda $ and $
\lambda $ is a primitive $M$ th root of unity. Then,
\begin{equation*}
Df(0)=\left(
\begin{array}{cc}
\lambda & 0 \\
0 & \lambda
\end{array}
\right) .
\end{equation*}
By Lemma \ref{nor}, there exists a polynomial transform $
(y_{1},y_{2})=H(x_{1},x_{2})$ in the form of (\ref{tra}), such that each
component of $g=(g_{1},g_{2})=H^{-1}\circ f\circ H$ has the expression
\begin{equation}
g_{j}(x_{1},x_{2})=\lambda
x_{j}+
\sum_{_{i_{1}+i_{2}=2}}^{M}c_{i_{1}i_{2}}^{j}x_{1}^{i_{1}}x_{2}^{i_{2}}+o(|x|^{M}),j=1,2,
\label{nor1}
\end{equation}
in a neighborhood of the origin, where the sum extends over all $2$-tuples $
(i_{1},i_{2})$ of nonnegative integers with
\begin{equation*}
2\leq i_{1}+i_{2}\leq M+1\ \mathrm{and\ }\lambda =\lambda ^{i_{1}+i_{2}},
\end{equation*}
which implies that $M|(i_{1}+i_{2}-1),$ since $\lambda $ is a primitive $M$
th root of unity. Thus (\ref{nor1}) becomes
\begin{equation*}
g_{j}(x_{1},x_{2})=\lambda x_{j}+o(|x|^{M}),j=1,2,
\end{equation*}
and then the $M$ th iteration $g^{M}=(g_{1}^{(M)},g_{2}^{(M)})$ has the form
of
\begin{equation*}
\begin{array}{c}
g_{1}^{(M)}(x_{1},x_{2})=x_{1}+o(|x|^{M}), \\
g_{2}^{(M)}(x_{1},x_{2})=x_{2}+o(|x|^{M}).
\end{array}
\end{equation*}
Then, by Cronin's theorem and Lemma \ref{lem2-4}, we have
\begin{equation*}
\mu _{f^{M}}(0)=\mu _{g^{M}}(0)=\pi _{g^{M}-id}(0)\geq (M+1)^{2},
\end{equation*}
and then, by (\ref{all}) and by Lemmas \ref{lem2-5} and \ref{lem2-3} (ii),
we have
\begin{equation*}
P_{M}(f,0)=\mu _{f^{M}}(0)-P_{1}(f,0)\geq (M+1)^{2}-1>2M,
\end{equation*}
and then, by Corollary \ref{do1}, $\mathcal{O}_{M}(f,0)>2.$ This completes
the proof.\medskip
\end{proof}
\begin{description}
\item[\textbf{Part 2}] \textbf{(b2) }$\Rightarrow $\textbf{\ (\ref{tar}
).\medskip }
\end{description}
\begin{proof}
We first show that
\begin{equation}
\mu _{f^{M}}(0)>M+1. \label{kkk}
\end{equation}
By Lemma \ref{nor} and Corollary \ref{tra-d0}, there exists a polynomial
transform $(y_{1},y_{2})=H(x_{1},x_{2})$ in the form of (\ref{tra}), such
that the $M$ th iteration $g^{M}=(g_{1}^{(M)},g_{2}^{(M)})$ of the germ $
g=(g_{1},g_{2})=H^{-1}\circ f\circ H$ has the expression
\begin{equation*}
\begin{array}{c}
g_{1}^{(M)}(x_{1},x_{2})=\lambda
_{1}^{M}x_{1}+
\sum_{_{i_{1}+i_{2}=2}}^{2M}C_{i_{1}i_{2}}^{1}x_{1}^{i_{1}}x_{2}^{i_{2}}+o(|x|^{2M}),
\\
g_{2}^{(M)}(x_{1},x_{2})=\lambda
_{2}^{M}x_{2}+
\sum_{_{i_{1}+i_{2}=2}}^{2M}C_{i_{1}i_{2}}^{2}x_{1}^{i_{1}}x_{2}^{i_{2}}+o(|x|^{2M}),
\end{array}
\end{equation*}
in a neighborhood of the origin, where, for $j=1$ and $2,$
\begin{equation}
C_{i_{1}i_{2}}^{j}\neq 0\mathrm{\ only\ if\ }\lambda _{j}=\lambda
_{1}^{i_{1}}\lambda _{2}^{i_{2}}. \label{g1}
\end{equation}
By (b2), both $\lambda _{1}$ and $\lambda _{2}$ are primitive $M$ th roots
of unity. Thus, by (\ref{g1}) and (b2), $C_{i_{1}0}^{1}\neq 0$ only if $
M|(i_{1}-1),$ and $C_{0i_{2}}^{1}\neq 0$ only if $i_{2}\geq \alpha ;$ $
C_{0i_{2}}^{2}\neq 0$ only if $M|(i_{2}-1),$ and $C_{i_{1}0}^{2}\neq 0$ only
if $i_{1}\geq \beta .$ Thus, considering that $\lambda _{1}^{M}=\lambda
_{2}^{M}=1$ and that
\begin{equation*}
o(|x|^{2M})=x_{1}^{M+1}O(1)+x_{2}^{M+1}O(1),
\end{equation*}
we can write
\begin{equation}
\left\{
\begin{array}{c}
g_{1}^{(M)}(x_{1},x_{2})=x_{1}+x_{1}^{M+1}O(1)+O(1)x_{1}
\sum_{i_{1}+i_{2}=1}^{2M}D_{i_{1}i_{2}}^{1}x_{1}^{i_{1}}x_{2}^{i_{2}}+x_{2}^{\alpha }O(1),
\\
g_{2}^{(M)}(x_{1},x_{2})=x_{2}+x_{1}^{\beta
}O(1)+O(1)x_{2}
\sum_{i_{1}+i_{2}=1}^{2M}D_{i_{1}i_{2}}^{2}x_{1}^{i_{1}}x_{2}^{i_{2}}+x_{2}^{M+1}O(1),
\end{array}
\right. \label{nor3}
\end{equation}
in a neighborhood of the origin$,$ where
\begin{equation}
D_{i_{1}i_{2}}^{j}\neq 0\ \mathrm{only\ if\ }1=\lambda _{1}^{i_{1}}\lambda
_{2}^{i_{2}},j=1,2. \label{g2}
\end{equation}
By condition (b2), we may write
\begin{equation*}
M=\beta m+\gamma ,
\end{equation*}
where $m$ and $\gamma $ are positive integers with $\gamma <\beta .$ We
first assume
\begin{equation}
\lambda _{1}=e^{\frac{2\pi i}{\beta m+\gamma }},\lambda _{2}=e^{\frac{2\beta
\pi i}{\beta m+\gamma }}. \label{aid-2}
\end{equation}
Then, for any pair $(i_{1},i_{2})$ of nonnegative integers with $\lambda
_{1}^{i_{1}}\lambda _{2}^{i_{2}}=1$ we have
\begin{equation}
i_{1}+\beta i_{2}=0(\mathrm{mod}M\mathrm{),} \label{g3}
\end{equation}
and then by (\ref{g2}), (\ref{g3}) and (\ref{aid}), putting $
(x_{1},x_{2})=h(z_{1},z_{2})=(z_{1},z_{2}^{\beta }),$ we conclude that the
two components of the germ $(g^{M}-id)\circ h$ have the following
expression.
\begin{eqnarray*}
g_{1}^{(M)}\circ h(z_{1},z_{2})-z_{1}
&=&z_{1}^{M+1}O(1)+O(1)z_{1}
\sum_{i_{1}+i_{2}=2}^{M}D_{i_{1}i_{2}}^{1}z_{1}^{i_{1}}z_{2}^{\beta
i_{2}}+z_{2}^{\alpha \beta }O(1) \\
&=&z_{1}\phi _{M}(z_{1},z_{2})+\mathrm{higher\ terms}, \\
g_{2}^{(M)}\circ h(z_{1},z_{2})-z_{2}^{\beta } &=&z_{1}^{\beta
}O(1)+O(1)z_{2}^{\beta
}\sum_{i_{1}+i_{2}=2}^{M}D_{i_{1}i_{2}}^{2}z_{1}^{i_{1}}z_{2}^{\beta
i_{2}}+z_{2}^{\beta (M+1)}O(1) \\
&=&az_{1}^{\beta }+\mathrm{higher\ terms},
\end{eqnarray*}
where $\phi _{M}$ is a homogeneous polynomial of degree $M,$ in $z_{1}$ and $
z_{2},$ and $a$ is a constant. In other words,
\begin{equation}
\left\{
\begin{array}{l}
g_{1}^{(M)}\circ h(z_{1},z_{2})-z_{1}=z_{1}\phi _{M}(z_{1},z_{2})+\mathrm{
higher\ terms}, \\
g_{2}^{(M)}\circ h(z_{1},z_{2})-z_{2}^{\beta }=az_{1}^{\beta }+\mathrm{
higher\ terms}.
\end{array}
\right. \label{ddd}
\end{equation}
Since we have assumed that the origin is an isolated fixed point of $f^{M},$
it is an isolated fixed point of $g^{M}$ by Lemma \ref{lem2-4}, and then the
origin is an isolated zero of the germ $(g^{M}-id)\circ h\in \mathcal{O}(
\mathbb{C}^{2},0,0),$ which has the expression (\ref{ddd}). Thus, by
Cronin's theorem, the zero order $\pi _{(g^{M}-id)\circ h}(0)$ of the germ $
(g^{M}-id)\circ h$ at the origin is not smaller than $\beta (M+1),$ and
since the origin is not an isolated solution of the system of equations
\begin{eqnarray*}
z_{1}\phi _{M}(z_{1},z_{2}) &=&0, \\
az_{1}^{\beta } &=&0,
\end{eqnarray*}
we have by Cronin's theorem,
\begin{equation*}
\pi _{(g^{M}-id)\circ h}(0)>\beta (M+1).
\end{equation*}
Thus we have, by the fact $\pi _{h}(0)=\beta $ and Lemma \ref{prod-ind},
that
\begin{equation*}
\mu _{g^{M}}(0)=\pi _{g^{M}-id}(0)=\pi _{(g^{M}-id)\circ h}(0)/\pi
_{h}(0)>M+1.
\end{equation*}
Thus, by Lemma \ref{lem2-4}, we have proved (\ref{kkk}) under the assumption
(\ref{aid-2}).
When (\ref{aid-2}) fails, we first show that there exists a positive integer
$d$ such that $\Lambda _{1}=\lambda _{1}^{d}$ and $\Lambda _{2}=\lambda
_{2}^{d}$ has expression (\ref{aid-2}), say
\begin{equation}
\Lambda _{1}=\lambda _{1}^{d}=e^{\frac{2\pi i}{\beta m+\gamma }},\Lambda
_{2}=\lambda _{2}^{d}=e^{\frac{2\beta \pi i}{\beta m+\gamma }}.
\label{aid-1}
\end{equation}
Since $\lambda _{1}$ is a primitive $M$ th root of unity, we may assume $
\lambda _{1}=e^{\frac{2k_{1}\pi i}{M}},$ where $k_{1}$ is a positive integer
with $(k_{1},M)=1.$ Then there exists a positive integer $d$ such that
\begin{equation}
dk_{1}+cM=1, \label{aid1}
\end{equation}
where $c$ is an integer. Then,
\begin{equation}
\Lambda _{1}=\lambda _{1}^{d}=e^{\frac{2dk_{1}\pi i}{M}}=e^{\frac{2\pi i}{M}
}. \label{j5}
\end{equation}
By (\ref{aid1}), one has $(d,M)=1,$ and then
\begin{equation*}
\Lambda _{2}=\lambda _{2}^{d}
\end{equation*}
is still a primitive $M$ th root of unity, since $\lambda _{2}$ is a
primitive $M$ th root of unity.
Now, both $\Lambda _{1}$ and $\Lambda _{2}$ are primitive $M$ th roots of
unity, and by (\ref{aid})$,$ we still have
\begin{eqnarray*}
\Lambda _{1}^{\beta } &=&\lambda _{1}^{d\beta }=\lambda _{2}^{d}=\Lambda
_{2}, \\
\Lambda _{2}^{\alpha } &=&\lambda _{2}^{d\alpha }=\lambda _{1}^{d}=\Lambda
_{1},
\end{eqnarray*}
and then by (\ref{j5}), we have (\ref{aid-1}) (recall that $M=\beta m+\gamma
$).
Let $F=f^{d},$ the $d$ th iteration of $f$. Then $\Lambda _{1}$ and $\Lambda
_{2}$ are the two eigenvalues of $DF(0).$ Thus, the above argument for the
case (\ref{aid-2}) works for $\Lambda _{1}$, $\Lambda _{2}$ and $F,$
provided that $0$ is an isolated fixed point of $F^{M}=f^{dM}.$ On the other
hand, it is clear that the two eigenvalues of $Df^{M}(0)$ are both equal to $
1,$ and therefore, by the assumption that $0$ is an isolated fixed point of $
f^{M}$ and by Lemma \ref{SS-1}, $0$ is an isolated fixed point of $
F^{M}=\left( f^{M}\right) ^{d}.$ Thus, applying the above argument to $
\Lambda _{1},\Lambda _{2}$ and $F,$ we have
\begin{equation*}
\mu _{F^{M}}(0)>M+1.
\end{equation*}
and again by Lemma \ref{SS-1} we have
\begin{equation*}
\mu _{f^{M}}(0)=\mu _{\left( f^{M}\right) ^{d}}(0)=\mu _{F^{M}}(0)>M+1,
\end{equation*}
say, (\ref{kkk}) holds, and then we have proved (\ref{kkk}) completely.
Thus, by (\ref{all}), (\ref{kkk}) and by Lemmas \ref{lem2-5} and \ref{lem2-3}
(ii), we have
\begin{equation*}
P_{M}(f,0)=\mu _{f^{M}}(0)-P_{1}(f,0)>M,
\end{equation*}
and then,
\begin{equation*}
\mathcal{O}_{M}(f,0)=P_{M}(f,0)/M\geq 2,
\end{equation*}
for $\mathcal{O}_{M}(f,0)$ is an integer. This completes the proof.\medskip
\end{proof}
\begin{description}
\item[\textbf{Part 3}] \textbf{(b3) }$\Rightarrow $\textbf{\ (\ref{tar}).}
\medskip
\end{description}
\begin{proof}
In case (b3), there exists an integer $d>1$ such that $M=m_{2}=dm_{1}$. Then
$\lambda _{1}$ is a primitive $m_{1}$ th root of unity and $\lambda _{2}$ is
a primitive $dm_{1}$ th root of unity, and
\begin{equation}
\lambda _{2}^{d}\neq \lambda _{1}. \label{ee}
\end{equation}
We first prove the following two conclusions.
\textrm{(i)} If $j_{1}$ and $j_{2}$ are integers with $\lambda
_{1}^{j_{1}}\lambda _{2}^{j_{2}}=1$, then $d|j_{2}.$
\textrm{(ii)} For any positive integer $d_{1}$ with $\lambda
_{2}^{d_{1}}=\lambda _{1},$
\begin{equation*}
d_{1}\geq 2d.
\end{equation*}
We may assume $\lambda _{1}=e^{\frac{2k_{1}\pi i}{m_{1}}}$ and $\lambda
_{2}=e^{\frac{2k_{2}\pi i}{dm_{1}}},$ where $k_{1}$ and $k_{2}$ are positive
integers with $k_{1}<m_{1}$, $k_{2}<dm_{1}$ and
\begin{equation}
(k_{1},m_{1})=(k_{2},dm_{1})=1. \label{mon}
\end{equation}
If $j_{1}$ and $j_{2}$ are integers with $\lambda _{1}^{j_{1}}\lambda
_{2}^{j_{2}}=1$, then we have
\begin{equation*}
\frac{j_{1}k_{1}}{m_{1}}+\frac{j_{2}k_{2}}{dm_{1}}=0(\mathrm{mod}1),
\end{equation*}
and then
\begin{equation*}
j_{1}k_{1}d+j_{2}k_{2}=0(\mathrm{mod}\left( dm_{1}\right) ).
\end{equation*}
Thus by (\ref{mon}), we have $d|j_{2},$ and (i) is proved. If $\lambda
_{2}^{d_{1}}=\lambda _{1},$ then by (\ref{ee}), $d\neq d_{1},$ and by (i), $
d|d_{1},$ which implies (ii).
Next, we show that there exists a polynomial transform $
(y_{1},y_{2})=H(x_{1},x_{2})$ in the form of (\ref{tra}), such that for each
positive integer $k,$ the $k$ th iteration $g^{k}=(g_{1}^{(k)},g_{2}^{(k)})$
of the germ
\begin{equation}
g=(g_{1},g_{2})=H^{-1}\circ f\circ H \label{dddd}
\end{equation}
has the expression
\begin{equation}
\left\{
\begin{array}{l}
g_{1}^{(k)}(x_{1},x_{2})=\lambda _{1}^{k}x_{1}+x_{1}^{m_{1}+1}(a^{(k)}+o(1))
\\
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \
+x_{1}^{2}x_{2}^{d}O(1)+x_{2}^{2d}O(1),\medskip \\
g_{2}^{(k)}(x_{1},x_{2})=\lambda
_{2}^{k}x_{2}+x_{1}^{m_{1}}x_{2}O(1)+x_{1}x_{2}^{d+1}O(1) \\
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ +x_{2}^{2d+1}O(1)+x_{1}^{dm_{1}+1}o(1),
\end{array}
\right. \label{nor4}
\end{equation}
where $a^{(k)}$ is a constant for each $k.$
By Lemma \ref{nor} and Corollary \ref{tra-d0}, there exists a polynomial
transform $H$ in the form of (\ref{tra}) such that the $k$ th iteration $
g^{k}=(g_{1}^{(k)},g_{2}^{(k)})$ of the germ (\ref{dddd}) has the expression
\begin{equation}
g_{j}^{(k)}(x_{1},x_{2})=\lambda
_{j}^{k}x_{j}+
\sum_{_{i_{1}+i_{2}=2}}^{3m_{1}d}C_{i_{1}i_{2}}^{kj}x_{1}^{i_{1}}x_{2}^{i_{2}}+o(|x|^{3dm_{1}})
\mathrm{,\;}j=1,2, \label{aa1}
\end{equation}
in a neighborhood of the origin, in which $i_{1}$ and $i_{2}$ are
nonnegative integers and for $j=1$ and $2,$
\begin{equation}
C_{i_{1}i_{2}}^{kj}\neq 0\mathrm{\ only\ if\ }\lambda _{j}=\lambda
_{1}^{i_{1}}\lambda _{2}^{i_{2}}. \label{aaa}
\end{equation}
By (\ref{aaa}), for $j=1$ and any pair $(i_{1},i_{2})$ in the sum with $
C_{i_{1}i_{2}}^{k1}\neq 0,$ we have $\lambda _{1}^{i_{1}-1}\lambda
_{2}^{i_{2}}=1,$ and then by the assumption that $\lambda _{1}$ and $\lambda
_{2}$ are primitive $m_{1}$ th and $m_{2}$ th roots of unity (note that $
m_{2}=dm_{1})$, respectively, for such pair $(i_{1},i_{2}),$ the following
conclusions from (1) to (4) hold (note that $i_{1}+i_{2}\geq 2$).
(1) If $i_{1}=0,$ then $i_{2}\geq 2d$ (by (\ref{ee}) and (ii))$;$
(2) If $i_{2}=0,$ then $i_{1}\geq m_{1}+1$;
(3) If $i_{1}=1,$ then $i_{2}\geq dm_{1}\geq 2d;$
(4) If $i_{1}\geq 2$ and $i_{2}\geq 1$, then $i_{2}\geq d$ (by (i))$.$
On the other hand, since $d>1$ and $m_{1}>1,$ it is clear that any term of $
o(|x|^{3dm_{1}})$ has either a factor $x_{1}^{dm_{1}+2},$ or a factor $
x_{2}^{2d+1},$ and then one can write
\begin{equation}
o(|x|^{3dm_{1}})=x_{1}^{dm_{1}+2}O(1)+x_{2}^{2d+1}O(1). \label{aa2}
\end{equation}
Thus, by (1)--(4) we obtain the first equation in (\ref{nor4}).
By (\ref{aaa}), for $j=2$ and any pair $(i_{1},i_{2})$ with $
C_{i_{1}i_{2}}^{k2}\neq 0,$ we have $\lambda _{1}^{i_{1}}\lambda
_{2}^{i_{2}-1}=1,$ and then by the assumption, for such pair $(i_{1},i_{2})$
the following conclusions from (5) to (8) hold (note that $i_{1}+i_{2}\geq 2$
).
(5) $i_{2}\neq 0;$
(6) If $i_{1}=0,$ then $i_{2}\geq dm_{1}+1\geq 2d+1$;
(7) If $i_{2}=1,$ then $i_{1}\geq m_{1};$
(8) If $m_{1}>i_{1}\geq 1$ and $i_{2}\geq 2,\;$then $i_{2}\geq d+1$ (by (i))$
.$
Thus, by (\ref{aa2}), the second equation in (\ref{nor4}) holds, and (\ref
{nor4}) is proved.
Since the origin is an isolated fixed point of $f^{dm_{1}},$ it is an
isolated fixed point of $f^{m_{1}},g^{m_{1}}$ and $g^{dm_{1}}$ as well, by
Lemma \ref{lem2-4}. To complete the proof, we first show (\ref{tar}) under
the assumption that
\begin{equation*}
a^{(1)}\neq 0.
\end{equation*}
Then it is easy to see that the coefficient $a^{(k)}$ in (\ref{nor4})
satisfies
\begin{equation}
a^{(k)}=k\lambda _{1}^{k-1}a^{(1)}\neq 0,k\in \mathbb{N}. \label{nor5}
\end{equation}
For $k=m_{1},$ by (\ref{nor4}) and by the fact $\lambda _{1}^{m_{1}}=1\ $and
$\lambda _{2}^{m_{1}}\neq 1,$ one can write
\begin{eqnarray*}
g_{1}^{(m_{1})}(x_{1},x_{2})-x_{1}
&=&x_{1}^{m_{1}+1}(a^{(m_{1})}+o(1))+x_{2}o(1), \\
g_{2}^{(m_{1})}(x_{1},x_{2})-x_{2} &=&bx_{2}+x_{2}o(1)+x_{1}^{m_{1}+1}o(1).
\end{eqnarray*}
where $b=\lambda _{2}^{m_{1}}-1\neq 0,$ and then by (\ref{nor5}) and
Corollary \ref{c2} we have
\begin{equation}
\mu _{g^{m_{1}}}(0)=\pi _{g^{m_{1}}-id}(0)=m_{1}+1. \label{nor5+1}
\end{equation}
For\label{?2} $k=dm_{1},$ by (\ref{nor4}) we have
\begin{equation}
\left\{
\begin{array}{l}
g_{1}^{(dm_{1})}(x_{1},x_{2})=x_{1}+x_{1}^{m_{1}+1}O(1)+x_{1}^{2}x_{2}^{d}O(1)+x_{2}^{2d}O(1),\medskip
\\
g_{2}^{(dm_{1})}(x_{1},x_{2})=x_{2}+x_{1}^{m_{1}}x_{2}O(1)+x_{1}x_{2}^{d+1}O(1)
\\
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \
+x_{2}^{2d+1}O(1)+x_{1}^{dm_{1}+1}o(1),
\end{array}
\right. \label{c62}
\end{equation}
and then, putting $(x_{1},x_{2})=h(z_{1},z_{2})=(z_{1}^{d},z_{2}^{m_{1}}),$
we conclude that the germ $G=(G_{1},G_{2})=\left( g^{dm_{1}}-id\right) \circ
h$ has the expression
\begin{eqnarray*}
G_{1}(z_{1},z_{2}) &=&\phi (z_{1},z_{2})+\mathrm{higher\ terms,} \\
G_{2}(z_{1},z_{2}) &=&\psi (z_{1},z_{2})+\mathrm{higher\ terms,}
\end{eqnarray*}
where $\phi $ and $\psi $ are homogeneous polynomials of degrees\label{?3} $
dm_{1}+d$ and $dm_{1}+m_{1},$ respectively (note that $d>1$ and $m_{1}>1)$.
Since the origin is an isolated fixed point of $f^{M},$ it is also an
isolated fixed point of $g^{M}$ by Lemma \ref{lem2-4}, and then it is an
isolated zero of $G.$ Therefore, by Cronin's theorem we have
\begin{equation*}
\pi _{G}(0)\geq (dm_{1}+d)(dm_{1}+m_{1})=dm_{1}(m_{1}+1)(d+1),
\end{equation*}
and then, we have $\mu _{g^{dm_{1}}}(0)=\pi _{G}(0)/\pi _{h}(0)\geq
(m_{1}+1)(d+1),$ say,
\begin{equation}
\mu _{g^{dm_{1}}}(0)>dm_{1}+m_{1}+1. \label{c64}
\end{equation}
Thus, we have by (\ref{nor5+1}), (\ref{c64}) and Lemma \ref{lem2-4} that
\begin{equation}
\mu _{f^{m_{1}}}(0)=m_{1}+1\ \mathrm{and\ }\mu
_{f^{dm_{1}}}(0)>dm_{1}+m_{1}+1. \label{j11}
\end{equation}
By Lemma \ref{lem2-5}, each periodic point of the linear part of $f$ at the
origin has period $1,$ $m_{1}$ or $dm_{1}.$ Thus, by Lemma \ref{lem2-3}
(ii), we have
\begin{eqnarray*}
P_{dm_{1}}(f,0) &=&\mu _{f^{dm_{1}}}(0)-P_{m_{1}}(f,0)-P_{1}(f,0), \\
P_{m_{1}}(f,0) &=&\mu _{f^{m_{1}}}(0)-P_{1}(f,0),
\end{eqnarray*}
and then, we have by (\ref{j11}) that
\begin{equation*}
P_{dm_{1}}(f,0)=\mu _{f^{dm_{1}}}(0)-\mu _{f^{m_{1}}}(0)>dm_{1},
\end{equation*}
and then $\mathcal{O}_{dm_{1}}(f,0)=P_{dm_{1}}(f,0)/(dm_{1})>1.$ But $
\mathcal{O}_{dm_{1}}(f,0)$ is an integer, we have $\mathcal{O}
_{dm_{1}}(f,0)\geq 2,$ and we have proved (\ref{tar}) under the assumption $
a^{(1)}\neq 0.$
If $a^{(1)}=0,$ then for $\varepsilon \neq 0$ consider the mapping $
g_{\varepsilon }=(g_{1,\varepsilon },g_{2,\varepsilon })$ given by
\begin{equation}
\left\{
\begin{array}{l}
g_{1,\varepsilon }(x_{1},x_{2})=g_{1}(x_{1},x_{2})+\varepsilon
x_{1}^{m_{1}+1}, \\
g_{2,\varepsilon }(x_{1},x_{2})=g_{2}(x_{1},x_{2}),
\end{array}
\right. \notag
\end{equation}
which is obtain from (\ref{nor4}) with $k=1$, by just replacing $a^{(1)},$
the coefficient of $x_{1}^{m_{1}+1}$ of the power series of $g_{1}^{(1)},$
with $\varepsilon $ (note that $
g=g^{1}=(g_{1},g_{2})=(g_{1}^{(1)},g_{2}^{(1)})$). Then $f_{\varepsilon
}=H\circ g_{\varepsilon }\circ H^{-1}$ converges to $f$ uniformly in a
neighborhood of the origin as $\varepsilon \rightarrow 0,$ and then, $0$ is
also an isolated fixed point of $f_{\varepsilon }^{dm_{1}}$ for sufficiently
small $\varepsilon ,$ by the assumption that $0$ is an isolated fixed point
of $f^{dm_{1}}$ and Lemma \ref{lem2-1}.
On the other hand, $g_{\varepsilon }=H^{-1}\circ f_{\varepsilon }\circ H$ is
in the form of (\ref{aa1}) for $k=1,$ together (\ref{aaa}). Thus, by
Corollary \ref{tra-d0}, the $k$ th iteration $g_{\varepsilon }^{k}$ is still
in the form of (\ref{aa1})$,$ together (\ref{aaa}). Thus, repeating the
process for proving (\ref{nor4}), we can prove that $g_{\varepsilon }^{k}$
is still in the form of (\ref{nor4}), more precisely
\begin{eqnarray*}
g_{1}^{(k)}(x_{1},x_{2}) &=&\lambda
_{1}^{k}x_{1}+x_{1}^{m_{1}+1}(A^{(k)}+o(1)) \\
&&+x_{1}^{2}x_{2}^{d}O(1)+x_{2}^{2d}O(1), \\
g_{2}^{(k)}(x_{1},x_{2}) &=&\lambda
_{2}^{k}x_{2}+x_{1}^{m_{1}}x_{2}O(1)+x_{1}x_{2}^{d+1}O(1) \\
&&+x_{2}^{2d+1}O(1)+x_{1}^{dm_{1}+1}o(1).
\end{eqnarray*}
But here $A^{(1)}=\varepsilon \neq 0.$
Therefore, all the above arguments for the case $a^{(1)}\neq 0$ apply to $
f_{\varepsilon },g_{\varepsilon }=H^{-1}\circ f_{\varepsilon }\circ H$ and $
g_{\varepsilon }^{k},$ and then we have $\mathcal{O}_{dm_{1}}(f_{\varepsilon
},0)\geq 2.$ Hence, by Lemma \ref{lem2-2}, we have (\ref{tar}) in the case $
a^{(1)}=0$ as well. This completes the proof.\medskip
\end{proof}
\begin{description}
\item[\textbf{Part 4}] \textbf{(b4) }$\Rightarrow $\textbf{\ (\ref{tar}).}
\medskip
\end{description}
\begin{proof}
By (b4), there exist positive integers $d>1,$ $n_{1}>1$ and $n_{2}>1$ such
that $n_{1}$ and $n_{2}$ are relatively prime and
\begin{equation*}
m_{1}=dn_{1},\ m_{2}=dn_{2},\ M=dn_{1}n_{2}.
\end{equation*}
Then, the two eigenvalues $\lambda _{1}$ and $\lambda _{2}$ of $Df(0)$ are
primitive $dn_{1}$-th, and $dn_{2}$-th roots of unity, respectively.
We first show that there exists a polynomial transform $H$ in the form of (
\ref{tra}) in a neighborhood of the origin such that for each $k\in \mathbb{N
},$ the $k$ th iteration $g^{k}=(g_{1}^{(k)},g_{2}^{(k)})$ of the germ
\begin{equation}
g=H^{-1}\circ f\circ H=(g_{1},g_{2}) \label{h10}
\end{equation}
has the expression
\begin{equation}
\left\{
\begin{array}{lll}
g_{1}^{(k)}(x_{1},x_{2}) & = & \lambda
_{1}^{k}x_{1}+x_{1}^{dn_{1}+1}(a^{(k)}+o(1))+x_{1}^{n_{1}+1}x_{2}^{n_{2}}O(1)
\\
& & +x_{1}x_{2}^{dn_{2}}O(1)+x_{2}^{2dn_{2}+1}O(1),\medskip \\
g_{2}^{(k)}(x_{1},x_{2}) & = & \lambda
_{2}^{k}x_{2}+x_{1}^{dn_{1}}x_{2}O(1)+x_{1}^{n_{1}}x_{2}^{n_{2}+1}O(1) \\
& & +x_{2}^{dn_{2}+1}(b^{(k)}+o(1))+x_{1}^{2dn_{1}+1}O(1),
\end{array}
\right. \label{tra2}
\end{equation}
where $a^{(k)}$ and $b^{(k)}$ are constants.
By Lemma \ref{nor} and Corollary \ref{tra-d0}, there exists a polynomial
coordinate transform $H$ in the form of (\ref{tra}) in a neighborhood of the
origin, such that each component of the $k$ th iteration $
g^{k}=(g_{1}^{(k)},g_{2}^{(k)})$ of the germ (\ref{h10}) has the expression
\begin{equation}
g_{j}^{(k)}(x_{1},x_{2})=\lambda
_{j}^{k}x_{j}+
\sum_{_{i_{1}+i_{2}=2}}^{4dn_{1}n_{2}}C_{i_{1}i_{2}}^{kj}x_{1}^{i_{1}}x_{2}^{i_{2}}+o(|x|^{4dn_{1}n_{2}}),j=1,2,
\label{ad-4.1.2}
\end{equation}
in a neighborhood of the origin, where for each $j=1,2,$ the sum in (\ref
{ad-4.1.2}) extends over all $2$-tuples $(i_{1},i_{2})$ of nonnegative
integers with
\begin{equation}
2\leq i_{1}+i_{2}\leq 4dn_{1}n_{2} \label{0}
\end{equation}
and
\begin{equation}
\lambda _{j}=\lambda _{1}^{i_{1}}\lambda _{2}^{i_{2}}. \label{1}
\end{equation}
We may write
\begin{equation*}
\lambda _{j}=e^{\frac{2k_{j}\pi i}{dn_{j}}},j=1,2,
\end{equation*}
where $k_{j}$ is a positive integer such that $(k_{j},dn_{j})=1,j=1,2.$
Recall that $(k_{j},dn_{j})$ denotes the largest common divisor of $k_{j}$
and $dn_{j}.$
Now, first assume $j=1$ and let $(i_{1},i_{2})$ be any $2$-tuple that
satisfies (\ref{0}) and (\ref{1}). Then we have
\begin{equation*}
\frac{(i_{1}-1)k_{1}}{dn_{1}}+\frac{i_{2}k_{2}}{dn_{2}}=0(\mathrm{{mod}1),}
\end{equation*}
and then
\begin{equation}
(i_{1}-1)k_{1}n_{2}+i_{2}k_{2}n_{1}=0(\mathrm{mod}\left( dn_{1}n_{2}\right)
). \label{3}
\end{equation}
This implies that $n_{1}|(i_{1}-1)$ and $n_{2}|i_{2},$ since
\begin{equation*}
(k_{1},n_{1})=(k_{2},n_{2})=(n_{1},n_{2})=1.
\end{equation*}
Therefore, for $j=1,$ each nonzero term $
C_{i_{1}i_{2}}^{k1}x_{1}^{i_{1}}x_{2}^{i_{2}}$ in (\ref{ad-4.1.2}) is of
type $x_{1}^{sn_{1}+1}x_{2}^{tn_{2}}O(1),$ in which $s=(i_{1}-1)/n_{1}$ and $
t=i_{2}/n_{2}$ are integers. In particular, when $i_{2}=0,$ by (\ref{3}) we
have $\left( dn_{1}\right) |\left( i_{1}-1\right) ,$ and then $i_{1}\geq
dn_{1}+1$ by (\ref{0}); and when $i_{1}=1,$ by (\ref{3}) we have $\left(
dn_{2}\right) |i_{2},$ and then $i_{2}\geq dn_{2}$ by (\ref{0}). On the
other hand, each term in the power series $o(|x|^{4dn_{1}n_{2}})$ has the
form $cx_{1}^{i_{1}}x_{2}^{i_{2}}$ with $i_{1}+i_{2}>4dn_{1}n_{2}.$ Thus
each term in $o(|x|^{4dn_{1}n_{2}})$ has either a factor $x_{1}^{dn_{1}+2}$
or a factor $x_{2}^{2dn_{2}+1}.$ Therefore, the first equality of (\ref{tra2}
) holds for some constant $a^{(k)}$.
For the same reason, the second equality in (\ref{tra2}) holds for some
constant $b^{(k)}$.
By the assumption, we have $\lambda _{1}^{dn_{1}n_{2}}=\lambda
_{2}^{dn_{1}n_{2}}=1,$ and then, by taking $k=dn_{1}n_{2}$ in (\ref{tra2}),
we have
\begin{eqnarray*}
g_{1}^{(dn_{1}n_{2})}(x_{1},x_{2})-x_{1}
&=&x_{1}^{dn_{1}+1}O(1)+x_{1}^{n_{1}+1}x_{2}^{n_{2}}O(1) \\
&&+x_{1}x_{2}^{dn_{2}}O(1)+x_{2}^{2dn_{2}+1}O(1), \\
g_{2}^{(dn_{1}n_{2})}(x_{1},x_{2})-x_{2}
&=&x_{1}^{dn_{1}}x_{2}O(1)+x_{1}^{n_{1}}x_{2}^{n_{2}+1}O(1) \\
&&+x_{2}^{dn_{2}+1}O(1)+x_{1}^{2dn_{1}+1}O(1).
\end{eqnarray*}
By Lemma \ref{dd4}, the zero order $\pi _{g^{dn_{1}n_{2}}-id}(0)$ of $
g^{dn_{1}n_{2}}-id$ at the origin is at least $1+dn_{1}+dn_{2}+2dn_{1}n_{2},$
in other words
\begin{equation*}
\mu _{g^{dn_{1}n_{2}}}(0)\geq 1+dn_{1}+dn_{2}+2dn_{1}n_{2},
\end{equation*}
which implies by Lemma \ref{lem2-4} that
\begin{equation}
\mu _{f^{dn_{1}n_{2}}}(0)\geq 1+dn_{1}+dn_{2}+2dn_{1}n_{2}. \label{dd5}
\end{equation}
Now, let us first prove (\ref{tar}) under the assumption that, in the
expression (\ref{tra2}), $a^{(1)}\neq 0$ and $b^{(1)}\neq 0.$
Then it is easy to see that
\begin{equation*}
a^{(dn_{1})}=dn_{1}\lambda _{1}^{dn_{1}-1}a^{(1)}\neq 0.
\end{equation*}
Thus, by (\ref{tra2}) and the fact that $\lambda _{1}^{dn_{1}}=1$ and $
\lambda _{2}^{dn_{1}}\neq 1$, we can write
\begin{eqnarray*}
g_{1}^{(dn_{1})}(x_{1},x_{2})-x_{1}
&=&x_{1}^{dn_{1}+1}(a^{(dn_{1})}+o(1))+x_{2}o(1), \\
g_{2}^{(dn_{1})}(x_{1},x_{2})-x_{2} &=&cx_{2}+x_{2}o(1)+x_{1}^{dn_{1}+1}o(1),
\end{eqnarray*}
where $c=\lambda _{2}^{dn_{1}}-1\neq 0,$ and then by Corollary \ref{c2} we
have
\begin{equation*}
\mu _{g^{dn_{1}}}(0)=\pi _{g^{dn_{1}}-id}(0)=dn_{1}+1,
\end{equation*}
and, repeating the above argument, by the assumption $b^{(1)}\neq 0,$ we have
\begin{equation*}
\mu _{g^{dn_{2}}}(0)=dn_{2}+1.
\end{equation*}
Thus by Lemma \ref{lem2-4} we have
\begin{equation}
\mu _{f^{dn_{1}}}(0)=dn_{1}+1,\mu _{f^{dn_{2}}}(0)=dn_{2}+1. \label{ai+2}
\end{equation}
On the other hand, by Lemmas \ref{lem2-3} (ii) and \ref{lem2-5}, we have
\begin{eqnarray*}
\mu _{f^{dn_{1}}}(0) &=&P_{dn_{1}}(f,0)+P_{1}(f,0), \\
\mu _{f^{dn_{2}}}(0) &=&P_{dn_{2}}(f,0)+P_{1}(f,0), \\
\mu _{f^{dn_{1}n_{2}}}(0)
&=&P_{dn_{1}n_{2}}(f,0)+P_{dn_{1}}(f,0)+P_{dn_{2}}(f,0)+P_{1}(f,0).
\end{eqnarray*}
Thus, by (\ref{all}) we have
\begin{equation*}
P_{dn_{1}n_{2}}(f,0)=\mu _{f^{dn_{1}n_{2}}}-\mu _{f^{dn_{1}}}(0)-\mu
_{f^{dn_{2}}}(0)+1.
\end{equation*}
and then, by (\ref{dd5}) and by (\ref{ai+2}), we have $P_{dn_{1}n_{2}}(f,0)
\geq 2dn_{1}n_{2},$ and then (\ref{tar}) holds.
Now, we have proved (\ref{tar}) under the condition $a^{(1)}\neq 0$ and $
b^{(1)}\neq 0.$ In general$,$ we consider $g_{\varepsilon
}=(g_{1,\varepsilon },g_{2,\varepsilon })\in \mathcal{O}(\mathbb{C}^{2},0,0)$
given by
\begin{equation}
\left.
\begin{array}{c}
g_{1,\varepsilon }(x_{1},x_{2})=g_{1}(x_{1},x_{2})+\varepsilon x^{dn_{1}+1},
\\
g_{2,\varepsilon }(x_{1},x_{2})=g_{2}(x_{1},x_{2})+\varepsilon
x_{2}^{dn_{2}+1},
\end{array}
\right. \label{2}
\end{equation}
which is obtained from (\ref{tra2}) with $k=1$, by just replacing the
constants $a^{(1)}$ and $b^{(1)}$ with $\varepsilon $, and consider
\begin{equation*}
f_{\varepsilon }=H\circ g_{\varepsilon }\circ H^{-1}
\end{equation*}
where $H$ is the transform in (\ref{h10}).
Since $f_{\varepsilon }$ uniformly converges to $f$ as $\varepsilon
\rightarrow 0$, for sufficiently small $\varepsilon ,$ the origin is an
isolated fixed point of $f_{\varepsilon }^{dn_{1}n_{2}}$ by Rouch\'{e}'s
theorem and the assumption that $0$ is an isolated fixed point of $
f^{dn_{1}n_{2}}$ (note that $M=dn_{1}n_{2})$. Then, it is clear that for
sufficiently small $\varepsilon ,f_{\varepsilon }$ again satisfies condition
(b4) and $g_{\varepsilon }=H^{-1}\circ f_{\varepsilon }\circ H$ is in the
form of (\ref{ad-4.1.2}) of $k=1,$ together with (\ref{0}) and (\ref{1}),
thus $g_{\varepsilon }^{k}=(g_{1,\varepsilon }^{(k)},g_{2,\varepsilon
}^{(k)})$ is in the form of (\ref{ad-4.1.2}) for all $k\in \mathbb{N}$ by
Corollary \ref{tra-d0}, more precisely,
\begin{equation*}
\begin{array}{lll}
g_{1,\varepsilon }^{(k)}(x_{1},x_{2}) & = & \lambda
_{1}^{k}x_{1}+x_{1}^{dn_{1}+1}(A^{(k)}+o(1))+x_{1}^{n_{1}+1}x_{2}^{n_{2}}O(1)
\\
& & +x_{1}x_{2}^{dn_{2}}O(1)+x_{2}^{2dn_{2}+1}O(1), \\
g_{2,\varepsilon }^{(k)}(x_{1},x_{2}) & = & \lambda
_{2}^{k}x_{2}+x_{1}^{dn_{1}}x_{2}O(1)+x_{1}^{n_{1}}x_{2}^{n_{2}+1}O(1) \\
& & +x_{2}^{dn_{2}+1}(B^{(k)}+o(1))+x_{1}^{2dn_{1}+1}O(1),
\end{array}
\end{equation*}
where $A^{(1)}=B^{(1)}=\varepsilon \neq 0$.
Thus, all the above arguments for the case $a^{(1)}b^{(1)}\neq 0$ apply to $
f_{\varepsilon }$ and $g_{\varepsilon }^{k}=(g_{1,\varepsilon
}^{(k)},g_{2,\varepsilon }^{(k)}),$ and then we have $\mathcal{O}
_{dn_{1}n_{2}}(f_{\varepsilon },0)\geq 2,$ and then by Lemma \ref{lem2-2},
we have $\mathcal{O}_{dn_{1}n_{2}}(f,0)\geq 2,$ say (\ref{tar}) holds again$
. $ This completes the proof.
\end{proof}
\section{Proof of the Main Theorem: (A) $\Rightarrow $ (B)\label{PM}}
Assume that $M>1$ is a positive integer and assume that $A$ is a $2\times 2$
matrix such that the following condition holds.
\textrm{(C)} \emph{For any }$f\in \mathcal{O}(\mathbb{C}^{2},0,0)$\emph{\
such that }$Df(0)=A$\emph{\ and that }$0$\emph{\ is an isolated fixed point
of }$f^{M},$
\begin{equation}
\mathcal{O}_{M}(f,0)\geq 2. \label{p1}
\end{equation}
We show that $A$ satisfies condition \textrm{(B)} in Theorem \ref{Th1}.
We first show that the two eigenvalues $\lambda _{1}$ and $\lambda _{2}$ of $
A$ satisfy
\begin{equation}
\lambda _{1}^{M}=\lambda _{2}^{M}=1. \label{p1+1}
\end{equation}
Assume (\ref{p1+1}) fails. Then we may assume $\lambda _{2}^{M}\neq 1,$ and
then by (C), Theorem \ref{Th0.1} and Lemma \ref{lem2-5}, the other
eigenvalue $\lambda _{1}$ must be a primitive $M$ th root of unity. But then
we shall obtain a contradiction by constructing a germ $f\in \mathcal{O}(
\mathbb{C}^{2},0,0)$ such that $Df(0)=A$ and\ that $0$\ is an isolated fixed
point of $f^{M},$ but (\ref{p1}) fails. That is to say, (C) fails!
Since $\lambda _{1}$ is a primitive $M$ th root of unity and $\lambda
_{2}^{M}\neq 1,$ we have $\lambda _{1}\neq \lambda _{2},$ and then the
matrix $A$ is diagonalizable. Thus, by Lemma \ref{lem2-4}, to construct that
$f,$ we may assume that
\begin{equation*}
A=\left(
\begin{array}{cc}
\lambda _{1} & 0 \\
0 & \lambda _{2}
\end{array}
\right) .
\end{equation*}
We show that the germ $f(x_{1},x_{2})\in \mathcal{O}(\mathbb{C}^{2},0,0)$
given by
\begin{equation*}
f(x_{1},x_{2})=(\lambda _{1}x_{1}+x_{1}^{M+1},\lambda _{2}x_{2})
\end{equation*}
is the desired germ.
Since $\lambda _{1}$ is a primitive $M$ th root of unity and $M>1$, it is
clear that $\lambda _{1}\neq 1$ and that the $M$ th iteration $f^{M}$ of $f$
has the expression
\begin{equation*}
f^{M}(x_{1},x_{2})=(x_{1}+cx_{1}^{M+1}(1+o(1)),\lambda _{2}^{M}x_{2}),
\end{equation*}
where $c=M\lambda _{1}^{M-1}\neq 0.$ Then, by the assumption that $\lambda
_{2}^{M}\neq 1$ and by Cronin's theorem, the origin is an isolated zero of $
f^{M}-id$ and the zero order is
\begin{equation*}
\pi _{f^{M}-id}(0)=M+1,
\end{equation*}
and then the origin is an isolated fixed point of $f^{M}$ with
\begin{equation*}
\mu _{f^{M}}(0)=M+1.
\end{equation*}
On the other hand, it is clear that $0$ is a simple fixed point of $f,$ and
then
\begin{equation*}
\mu _{f}(0)=P_{1}(f,0)=1.
\end{equation*}
Therefore, by Lemmas \ref{lem2-3} (ii) and \ref{lem2-5},
\begin{equation*}
P_{M}(f,0)=\mu _{f^{M}}(0)-P_{1}(f,0)=M,
\end{equation*}
and then $\mathcal{O}_{M}(f,0)=1,$ say, (\ref{p1}) fails$.$ Hence, (\ref
{p1+1}) holds.
By \textrm{(C)}, (\ref{p1+1}), Theorem \ref{Th0.1} and Lemma \ref{lem2-5},
we can conclude that there exists positive integers $m_{1}$ and $m_{2}$ such
that
(D) \emph{The two eigenvalues }$\lambda _{1}$\emph{\ and }$\lambda _{2}$
\emph{\ of }$A$ \emph{are primitive }$m_{1}$\emph{\ th and }$m_{2}$\emph{\
th roots of unity, respectively, and}
\begin{equation}
M=[m_{1},m_{2}]. \label{p1+2}
\end{equation}
Without loss of generality, we assume that
\begin{equation*}
m_{1}\leq m_{2}\leq M.
\end{equation*}
If $m_{1}$ and $m_{2}$ do not satisfy any one of the conditions from (b1) to
(b4), then by (D) and Remark \ref{r1}, one of the following conditions must
be satisfied:
\textrm{(b1)'\ }$m_{1}=m_{2}=M,$ $\lambda _{1}=\lambda _{2}$ and $A$ is not
diagonalizable$.$
\textrm{(b2)'\ }$m_{1}=m_{2}=M$ and there exists positive integers $\alpha $
and $\beta \ $with $1<\alpha <M$ and $1<\beta <M$ such that
\begin{equation}
\lambda _{1}^{\alpha }=\lambda _{2},\lambda _{2}^{\beta }=\lambda
_{1},\alpha \beta =M+1. \label{p2}
\end{equation}
\textrm{(b3)'\ }$m_{2}=M,$ $m_{1}|m_{2},m_{1}<m_{2},$ and $\lambda
_{2}^{m_{2}/m_{1}}=\lambda _{1}.$
\textrm{(b4)'} $m_{1}$ and $m_{2}$ are relatively prime, and $m_{2}>m_{1}>1$.
We show that each condition from (b1)' to (b4)' contradicts condition (C).
This will be done in each case, by constructing a germ $F\in \mathcal{O}(
\mathbb{C}^{2},0,0)$ such that $DF(0)=A$ and $0$ is an isolated fixed point
of $F^{M},$ but $\mathcal{O}_{M}(F,0)=1,$ say, (\ref{p1}) fails. We divide
this process into four parts.
\begin{description}
\item[\textbf{Part 1}] \textbf{(b1)' implies the existences of }$F$\textbf{.}
\end{description}
\begin{proof}
In case \textrm{(b1)'}, by Lemma \ref{lem2-4}, we may assume that
\begin{equation*}
A=\left(
\begin{array}{cc}
\lambda _{1} & 0 \\
1 & \lambda _{1}
\end{array}
\right) .
\end{equation*}
Then, consider the germ $F\in \mathcal{O}(\mathbb{C}^{2},0,0)$ given by
\begin{equation*}
F(x_{1},x_{2})=(\lambda _{1}x_{1}+x_{2}^{M+1},x_{1}+\lambda _{1}x_{2}).
\end{equation*}
Since $\lambda _{1}$ is a primitive $M$ th root of unity and $M>1,$ the
origin is a simple fixed point of $f$, say,
\begin{equation}
P_{1}(F,0)=\mu _{F}(0)=1. \label{pf1}
\end{equation}
It is easy to see by induction that the $M$ th iteration $F^{M}$ of $F$ has
the expression
\begin{equation*}
\left( F^{M}(x_{1},x_{2})\right) ^{T}=\left(
\begin{array}{c}
x_{1}+x_{1}\phi _{M}(x_{1},x_{2})+M\lambda _{1}^{M-1}x_{2}^{M+1}+o(|x|^{M+1})
\\
M\lambda _{1}^{M-1}x_{1}+x_{2}+o(|x|)
\end{array}
\right) ,
\end{equation*}
where $x=(x_{1},x_{2})$ and $\phi _{M}$ is a homogenous polynomial of degree
$M.$ Then $F^{M}(x_{1},x_{2})-(x_{1},x_{2})$ has the expression
\begin{equation*}
\left( F^{M}(x_{1},x_{2})-(x_{1},x_{2})\right) ^{T}=\left(
\begin{array}{c}
x_{1}\phi _{M}(x_{1},x_{2})+M\lambda _{1}^{M-1}x_{2}^{M+1}+o(|x|^{M+1}) \\
M\lambda _{1}^{M-1}x_{1}+o(|x|)
\end{array}
\right)
\end{equation*}
and the origin is an isolated zero of the system
\begin{eqnarray*}
x_{1}\phi _{M}(x_{1},x_{2})+M\lambda _{1}^{M-1}x_{2}^{M+1} &=&0, \\
M\lambda _{1}^{M-1}x_{1} &=&0.
\end{eqnarray*}
Thus, by Cronin's theorem, the zero order of the germ $F^{M}-id$ at the
origin equals $M+1,$ and then
\begin{equation*}
\mu _{F^{M}}(0)=M+1.
\end{equation*}
Thus, by (\ref{pf1}) and by Lemmas \ref{lem2-3} (ii) and \ref{lem2-5},
\begin{equation*}
P_{M}(F,0)=\mu _{F^{M}}(0)-P_{1}(F,0)=M,
\end{equation*}
which implies $\mathcal{O}_{M}(F,0)=1.$
\end{proof}
In any other case, $A$ is diagonalizable, and by Lemma \ref{lem2-4}, we may
assume
\begin{equation*}
A=\left(
\begin{array}{cc}
\lambda _{1} & 0 \\
0 & \lambda _{2}
\end{array}
\right) .
\end{equation*}
\begin{description}
\item[\textbf{Part 2}] \textbf{(b2)'}$\ $\textbf{implies the existence of }$
F $\textbf{.}
\end{description}
\begin{proof}
\textrm{In case (b2)',} consider the germ $F=(F_{1},F_{2})\in \mathcal{O}(
\mathbb{C}^{2},0,0)$ given by
\begin{eqnarray*}
F_{1}(x_{1},x_{2}) &=&\lambda _{1}x_{1}+x_{2}^{\beta }, \\
F_{2}(x_{1},x_{2}) &=&\lambda _{2}x_{2}+x_{1}^{\alpha }.
\end{eqnarray*}
By (\textrm{D}) and (b2)', both $\lambda _{1}$ and $\lambda _{2}$ are
primitive $M$ th roots of unity, and then by (\ref{p2}), it is easy to see
that the $M$ th iteration $F^{M}=(F_{1}^{(M)},F_{2}^{(M)})$ of $F$ has the
expression
\begin{eqnarray*}
F_{1}^{(M)}(x_{1},x_{2}) &=&x_{1}+M\lambda _{1}^{M-1}x_{2}^{\beta }+\mathrm{
higher\ terms,} \\
F_{2}^{(M)}(x_{1},x_{2}) &=&x_{2}+M\lambda _{2}^{M-1}x_{1}^{\alpha }+\mathrm{
higher\ terms;}
\end{eqnarray*}
and then by Cronin's theorem, we have
\begin{equation*}
\mu _{F^{M}}(0)=\alpha \beta =M+1.
\end{equation*}
On the other hand, since $M>1$ and both $\lambda _{1}$ and $\lambda _{2}$
are primitive $M$ th roots of unity$,$ the germ $F$ here still satisfies (
\ref{pf1}). Therefore, by (\ref{pf1}) and by Lemmas \ref{lem2-3} (ii) and
\ref{lem2-5}, we have
\begin{equation*}
P_{M}(F,0)=\mu _{F^{M}}(0)-P_{1}(F,0)=M,
\end{equation*}
and then $\mathcal{O}_{M}(F,0)=1$.
\end{proof}
\begin{description}
\item[\textbf{Part 3}] \textbf{(b3)' implies the existence of }$F$\textbf{.}
\end{description}
\begin{proof}
\textrm{In case (b3)',} we consider the germ $F\in \mathcal{O}(\mathbb{C}
^{2},0,0)$ given by
\begin{equation}
F(x_{1},x_{2})=(\lambda _{1}x_{1}+x_{1}^{m_{1}+1}+x_{2}^{d},\lambda
_{2}x_{2}+x_{1}^{m_{1}}x_{2}), \label{p2+2}
\end{equation}
where
\begin{equation*}
d=m_{2}/m_{1}=M/m_{1}>1
\end{equation*}
is a positive integer.
We first assume $m_{1}=1.$ Then $d=m_{2}=M$ and $\lambda _{1}=\lambda
_{2}^{d}=1,$ and then it is easy to show that the $M$ th iteration $
F^{M}=(F_{1}^{(M)},F_{2}^{(M)})$ has the expression
\begin{equation*}
\left\{
\begin{array}{l}
F_{1}^{(M)}(x_{1},x_{2})=x_{1}+Mx_{1}^{2}+Mx_{2}^{d}+x_{1}^{2}o(1)+x_{2}^{d}o(1),
\\
F_{2}^{(M)}(x_{1},x_{2})=x_{2}+M\lambda
_{2}^{M-1}x_{1}x_{2}+x_{1}x_{2}o(1)+x_{2}^{d+1}O(1).
\end{array}
\right.
\end{equation*}
Then by Corollary \ref{c1}, considering that $d=M,$ we have that the zero
order $\pi _{F^{M}-id}(0)$ of $F^{M}-id$ at the origin equals $d+2=M+2.$
Thus,
\begin{equation*}
\mu _{F^{M}}(0)=\pi _{F^{M}-id}(0)=M+2.
\end{equation*}
On the other hand, since $\lambda _{1}=1$ and $\lambda _{2}\neq 1,$ we can
write
\begin{equation*}
F(x_{1},x_{2})=(x_{1}+x_{1}^{2}+x_{2}o(1),\lambda _{2}x_{2}+x_{2}o(1)),
\end{equation*}
and then by Corollary \ref{c2}, the zero order $\pi _{F-id}(0)$ of $F-id$ at
the origin equals $2,$ and then
\begin{equation*}
\mu _{F}(0)=P_{1}(F,0)=\pi _{F-id}(0)=2.
\end{equation*}
Hence, by Lemmas \ref{lem2-5} and \ref{lem2-3} (ii), we have
\begin{equation*}
P_{M}(F,0)=\mu _{F^{M}}(0)-P_{1}(F,0)=M,
\end{equation*}
and then $\mathcal{O}_{M}(F,0)=1$.
Now, assume $m_{1}>1.$ Then we have $d=m_{2}/m_{1}>1$ and $\lambda
_{2}^{m_{1}}\neq 1,$ and then, the $m_{1}$ th iteration $
F^{m_{1}}=(F_{1}^{(m_{1})},F_{2}^{(m_{1})})$ of $F$ given by (\ref{p2+2})
has the expression
\begin{equation}
\left\{
\begin{array}{l}
F_{1}^{(m_{1})}(x_{1},x_{2})=x_{1}+x_{1}^{m_{1}+1}(c_{1}+o(1))+x_{2}^{d}(c_{1}+o(1)),
\\
F_{2}^{(m_{1})}(x_{1},x_{2})=\lambda _{2}^{m_{1}}x_{2}+m_{1}\lambda
_{2}^{m_{1}-1}x_{1}^{m_{1}}x_{2}(1+o(1))+x_{2}^{d+1}o(1),
\end{array}
\right. \label{p3}
\end{equation}
where $c_{1}=m_{1}\lambda _{1}^{m_{1}-1}\neq 0.$ Thus, by Corollary \ref{c2}
, the zero order of the germ $F^{m_{1}}-id$ at the origin equals $m_{1}+1,$
say
\begin{equation}
\mu _{F^{m_{1}}}(0)=m_{1}+1. \label{p4}
\end{equation}
Now, consider the fixed point index $\mu _{F^{M}}(0)$ of $
F^{M}=(F_{1}^{(M)},F_{2}^{(M)}),$ the $M$ th iteration of $F$. By \textrm{
(b3)'} and\ (\ref{p1+2}), $\lambda _{2}^{m_{1}}$ is a primitive $d$ th root
of unity, and then by (\ref{p3}), it is easy to see that the germ (note that
$M=dm_{1}=m_{2})$
\begin{equation*}
F^{M}(x_{1},x_{2})=(F_{1}^{(M)},F_{2}^{(M)})=(F^{m_{1}})^{d}
\end{equation*}
has the expression
\begin{equation*}
\left\{
\begin{array}{l}
F_{1}^{(M)}(x_{1},x_{2})=x_{1}+dx_{1}^{m_{1}+1}(c_{1}+o(1))+dx_{2}^{d}(c_{1}+o(1)),
\\
F_{2}^{(M)}(x_{1},x_{2})=x_{2}+M\lambda
_{2}^{M-1}x_{1}^{m_{1}}x_{2}(1+o(1))+x_{2}^{d+1}o(1),
\end{array}
\right.
\end{equation*}
and then the germ $F^{M}-id=(F_{1}^{(M)},F_{2}^{(M)})-id$ has the expression
\begin{equation*}
\left\{
\begin{array}{l}
F_{1}^{(M)}(x_{1},x_{2})-x_{1}=x_{1}^{m_{1}+1}(dc_{1}+o(1))+x_{2}^{d}(dc_{1}+o(1)),
\\
F_{2}^{(M)}(x_{1},x_{2})-x_{2}=M\lambda
_{2}^{M-1}x_{1}^{m_{1}}x_{2}(1+o(1))+x_{2}^{d+1}o(1),
\end{array}
\right.
\end{equation*}
and then, by Corollary \ref{c1}, we have (note that $dc_{1}\neq 0$ and $
M\lambda _{2}^{M-1}\neq 0)$
\begin{equation}
\mu _{F^{M}}(0)=\pi _{F^{M}-id}(0)=1+m_{1}+dm_{1}. \label{p5}
\end{equation}
By Lemmas \ref{lem2-5} and \ref{lem2-3} (ii), We have
\begin{eqnarray*}
\mu _{F^{M}}(0) &=&P_{M}(F,0)+P_{m_{1}}(F,0)+P_{1}(F,0), \\
\mu _{F^{m_{1}}}(0) &=&P_{m_{1}}(F,0)+P_{1}(F,0).
\end{eqnarray*}
Thus, by (\ref{p4}) and (\ref{p5}) we have
\begin{equation*}
P_{M}(F,0)=\mu _{F^{M}}(0)-\mu _{F^{m_{1}}}(0)=dm_{1}=M,
\end{equation*}
and then $\mathcal{O}_{M}(F,0)=1$.
\end{proof}
\begin{description}
\item[Part 4] \textbf{(b4)' implies the existence of }$F$\textbf{.}
\end{description}
\begin{proof}
\textrm{In case (b4)',} the existence of $F$ follows from Proposition \ref
{c8}. This completes the proof.
Now, we have prove that any case from (b1)' to (b4)' can not occurs. Thus $A$
must satisfy (B) in the main theorem.
\end{proof}
| 172,349
|
\begin{document}
\title{The Futaki Invariant of Kähler Blowups with Isolated Zeros via Localization}
\author{Luke Cherveny}
\maketitle
\begin{abstract}
We present an analytic proof of the relationship between the Calabi-Futaki invariant for a Kähler manifold relative to a holomorphic vector field with a nondegenerate zero and the corresponding invariant of its blowup at that zero, restricting to the case that zeros on the exceptional divisor are isolated. This extends the results of Li and Shi \cite{lishi:2015} for Kähler surfaces. We also clarify a hypothesis regarding the normal form of the vector field near its zero. An algebro-geometric proof was given by Székelyhidi \cite{szekelyhidi:2015} by reducing the situation to the case of projective manifolds for rational data and using Donaldson-Futaki invariants. Our proof will be an application of degenerate localization.
\end{abstract}
\section{Introduction}
Let $M$ be a compact Kähler manifold with Kähler metric $\omega$. A fundamental question in K\"ahler geometry asks whether the class $[\omega]$ contains a canonical Kähler metric. When the first Chern class $c_1(M)$ is zero, celebrated work of Yau \cite{yau:1978} established the existence of a unique Ricci-flat K\"ahler metric in every K\"ahler class, while in the case of negative first Chern class, Yau \cite{yau:1978} and Aubin \cite{aubin:1976} independently proved existence and uniqueness of a Kähler-Einstein metric in $c_1(M)$. The existence of K\"ahler-Einstein metrics when $c_1(M) > 0$ has recently been addressed by Chen-Donaldson-Sun \cite{cds:2015}.
More generally, one could ask whether a K\"ahler class $\Omega \in H^{1,1}(M,\R)$ contains a constant scalar curvature Kähler (cscK) metric. The question is quite subtle and conjecturally related to the algebro-geometric stability of $M$; see \cite{szekelyhidibook} for a survey and references. One obstruction to the existence of a cscK metric is a generalization due to Calabi \cite{calabi:1985} of Futaki's famous obstruction to K\"ahler-Einstein metrics first defined in \cite{futaki:1983}. For any K\"ahler class $\Omega$, this \emph{Calabi-Futaki invariant} is a certain character on the Lie algebra of holomorphic vector fields $\mathfrak{h}$
\[
\fut(\Omega, \cdot): \mathfrak{h} \rightarrow \C.
\]
\noindent whose vanishing on $\mathfrak{h}$ is necessary for $\Omega$ to support a cscK metric.
One approach to calculating the Calabi-Futaki invariant, at least when $M$ is algebraic, is via the algebraic Donaldson-Futaki invariant \cite{donaldson:2005}. Another method is localization, which will be our approach in this paper. When the zero locus of $X$ is nondegenerate, Tian \cite{tian:1996} gave a complete formula reducing $\fut(\Omega, X)$ to a calculation on $\text{Zero}(X)$. When the zero locus of $X$ is degenerate, localization calculations are quite difficult. See section 2 for details.
In this paper we study the following situation where degenerate localization calculations naturally arise: With $M$ as above, suppose that $p\in M$ is a zero of $X \in \mathfrak{h}$. The blowup $\pi: \tilde M \rightarrow M$ at $p$ then admits a holomorphic lift $\tilde X$ of $X$ as well as a natural K\"ahler class
\[
\tilde \Omega = \pi^*\Omega - \epsilon [E],
\]
\noindent where $E$ is the exceptional divisor and $\epsilon$ is sufficiently small \cite{griffithsharris}. A natural question is, what is the relationship between $\fut(\Omega,X)$ and $\fut(\tilde \Omega, \tilde X)$?
We will limit ourselves to the following assumption on $X$ at $p$:
\begin{quote}
($\star$) The Jordan canonical form of the linearization $DX$ at $p$ does not contain multiple Jordan blocks for the same eigenvalue.
\end{quote}
\noindent Geometrically, ($\star$) means $\tilde X$ does not contain a positive dimensional zero locus.
Our main theorem is
\begin{theorem}\label{maintheorem} Let $\pi: \tilde M \rightarrow M$ be the blow-up of $n$-dimensional K\"ahler manifold $M$ at isolated non-degenerate zeros $\{p_1, \dots, p_k\}$ of a holomorphic vector field $X$ with zero-average holomorphy potential $\theta_X$. When ($\star$) holds, the Futaki invariant for $\tilde M$ with respect to the class $\tilde \Omega = \pi^*\Omega - \sum \epsilon_i [E_i]$ and the natural holomorphic extension $\tilde X$ of $X$ to $\tilde M$ satisfies
\[
\textrm{Fut}_{\tilde M}(\tilde \Omega, \tilde X) = \textrm{Fut}_M(\Omega, X) - \sum_{i = 1}^k n(n-1) \theta_X(p) \epsilon_i^{n-1} + O(\epsilon^n)
\]
\end{theorem}
Li and Shi established the result for K\"ahler surfaces \cite{lishi:2015}. In fact, they addressed when $p$ is not isolated or when the zero locus of $\tilde X$ is $E \cong \C P^1$, which are necessarily nondegenerate situations for surfaces but would be quite formidable more generally. An algebro-geometric proof was given by Székelyhidi \cite{szekelyhidi:2015} by reducing the situation to the case of projective manifolds for rational data and using Donaldson-Futaki invariants, with related results recently appearing in the work of Dervan-Ross \cite{dervanross:2017} and Dyrefelt \cite{dyrefelt:2016}. We give an analytic proof via localization, for which the $n=2$ case of Li and Shi is a special case.
One application of Theorem \ref{maintheorem} is to show the nonexistence of cscK metrics on certain blow-ups:
\begin{corollary}
If a K\"ahler manifold $M$ admits a constant scalar curvature metric $\omega \in \Omega$ and there exists a holomorphic vector field $X$ on $M$ vanishing at $p$ and such that $\theta_X(p) \neq 0$, then the blowup $\tilde M$ at $p$ does not admit cscK metrics in any class $\pi^*\Omega - \epsilon [E]$ for small $\epsilon$.
\end{corollary}
The organization of this paper is as follows. Section 2 provides background concerning the Futaki invariant, summarizes Tian's application of Bott localization to its calculation, and sets up the blowup calculation. Section 3 gives a proof of Theorem \ref{maintheorem} in the special case that the linearization of $X$ at $p$ has a single Jordan block. It relies on Lemma \ref{maxdegenlemma}, which generalizes the main calculation of \cite{lishi:2015}. Section 4 addresses the general case of multiple Jordan blocks with distinct eigenvalues. Section 5 discusses the normal forms of a holomorphic vector field $X$ about a singular point and establishes that for the purposes of Theorem \ref{maintheorem}, the simplifying assumption that $X$ is locally biholomorphic to its linearization as in Section 4 is sufficient. Section 6 is an appendix that contains a proof of Lemma \ref{lemmagk} used to sum localization calculations.
\section{Background}
A reference for much of this section is \cite{tian:1996}; to align conventions, we too define the \kahler form $\omega$ and Ricci form $\ric$ without the usual $\sqrt{-1}$, and let the \kahler class be
\[
\Omega = \left[\frac{\sqrt{-1}}{2\pi}\omega\right] \in H^{1,1}(M,\R)
\]
We first recall the definition of the Calabi-Futaki invariant. Let $(M,\omega)$ be a compact \kahler manifold and $F$ the smooth function uniquely determined by
\[
S(\omega) - \bar S = \Delta F \qquad \qquad \int_M (e^F - 1)\omega^n = 0
\]
\noindent where $S(\omega)$ is scalar curvature, $\Delta$ is the Laplacian with respect to $\omega$, and
\[
\bar S = \frac{\int S(\omega) \Omega^n}{\int \Omega^n} = \frac{n \int c_1(M)\cup \Omega^{n-1}}{\int \Omega^n}
\]
\noindent is the average scalar curvature.
The \emph{Calabi-Futaki invariant} is defined for each K\"ahler class $\Omega$ to be a functional on the Lie algebra of holomorphic vector fields $\mathfrak{h}$
\[
\fut(\Omega, \cdot): \mathfrak{h} \rightarrow \C
\]
\noindent given by
\[
\fut(\Omega, X) = \left( \frac{\sqrt{-1}}{2\pi} \right)^n \int_M X(F) \; \omega^n
\]
Futaki \cite{futaki:1988} and Calabi \cite{calabi:1985} showed $\fut(\Omega, \cdot)$ is a Lie algebra character and that the definition is in fact independent of the choice of metric $\omega$ in its K\"ahler class $\Omega$, justifying the notation and making its vanishing for all $X \in \mathfrak{h}$ necessary for $\Omega$ to contain a cscK metric.
Following Tian \cite{tian:1996}, we now explain the localization of $\fut(\Omega,X)$. For every $X \in \mathfrak{h}$, Hodge theory provides a harmonic $(0,1)$-form $\alpha$ and a smooth function $\theta_X$, unique up to addition of a constant, such that
\begin{equation}\label{hodge}
i_X \omega = \alpha - \bar \partial \theta_X.
\end{equation}
\noindent Equivalently, $\theta_X$ is \emph{holomorphy potential} for $X$\footnote{Or rather, recalling that $\omega$ is defined without a $\sqrt{-1}$ on it, $\sqrt{-1}\theta$ is holomorphy potential in the sense that $f: M \rightarrow \C$ is holomorphy potential for holomorphic vector field $X = g^{i\bar j} (\partial_{\bar j} f) \partial_i$, i.e. $X$ is the $(1,0)$ part of the Riemannian gradient of $f$ (up to a factor of 2).}. By applying $\bar \partial^*$ to both sides of (\ref{hodge}) and using integration by parts,
\begin{align}\label{dividebyn}
\fut(\Omega, X) &= \left( \frac{\sqrt{-1}}{2\pi}\right)^n\int_M (\Delta \theta_X) F \; \omega^n \nonumber\\
&= \left( \frac{\sqrt{-1}}{2\pi}\right)^n\int_M \theta_X \Delta F \; \omega^n \nonumber\\
&= \left( \frac{\sqrt{-1}}{2\pi}\right)^n\int_M \theta_X S \; \omega^n - \bar S \int_M \theta_X \; \omega^n \nonumber\\
&= \left( \frac{\sqrt{-1}}{2\pi}\right)^n\int_M n\theta_X \ric \wedge \omega^{n-1} - \frac{\bar S}{n+1} \left( \frac{\sqrt{-1}}{2\pi}\right)^n \int_M (\theta_X + \omega)^{n+1}
\end{align}
\noindent which expresses the Calabi-Futaki invariant without explicit reference to $F$. This expression (\ref{dividebyn}) also shows we may assume $\alpha = 0$ in (\ref{hodge}) for our purposes.
Let $A = (-\Delta \theta_X + \ric)$ and $B = (\theta_X + \omega)$. By using the identity
\begin{equation}\label{combinatorial}
\sum_{j=0}^l (-1)^j {l \choose j} (l-2j)^k = \begin{cases} 0 \qquad k < l \textrm{ or } k=l+1 \\ 2^l l! \qquad k = l \end{cases}
\end{equation}
\noindent one checks that
\begin{align*}
\sum_{j=0}^n (-1)^j &{n \choose j} \int_M (A+(n-2j)B)^{n+1} - (-A + (n-2j)B)^{n+1} \\
&= \sum_{j=0}^n (-1)^j {n \choose j} \int_M \sum_{k=0}^{n+1} {n+1 \choose k} \left[ A^{n+1-k}(n-2j)^kB^k - (-A)^{n+1-k}(n-2j)^kB^k\right] \\
&= 2^nn! \int_M (n+1)2AB^n \hspace{1.5 in} \textrm{(only $k=n$ is non-zero)} \\
&= 2^{n+1}(n+1)! \int_M \left[ n\theta_X \, \ric \wedge \omega^{n-1} -\Delta \theta_X \, \omega^n \right] \\
&= 2^{n+1}(n+1)! \int_M n\theta_X \, \ric \wedge \omega^{n-1}\\
\end{align*}
\noindent Dividing this expression by $n!$ and substituting into the previous calculation (\ref{dividebyn}) yields
\begin{align}\label{expandedfutaki}
2^{n+1}(n+1)\fut (X,\Omega) =
&\sum_{j=0}^n \frac{(-1)^j }{j!(n-j)!} \left( \frac{\sqrt{-1}}{2\pi}\right)^n\int_M [(-\Delta \theta_X + \ric +(n-2j)(\theta_X + \omega) )^{n+1} \nonumber \\
& \hspace{1.8 in} - (\Delta \theta_X - \ric + (n-2j)(\theta_X + \omega) )^{n+1}] \nonumber \\
&- \bar S \sum_{j=0}^{n+1} (-1)^j \frac{(n+1-2j)^{n+1}}{j!(n+1-j)!} \left( \frac{\sqrt{-1}}{2\pi}\right)^n\int_M(\theta_X + \omega)^{n+1}
\end{align}
\noindent The point of expressing $\fut(X,\Omega)$ in this manner is that, as we will see, each integral (for each fixed $j$) may be realized as a Futaki-Morita type invariant, to which holomorphic localization applies.
\subsection{Holomorphic Localization}
We turn for a moment to a general description of Bott's holomorphic localization and its application to Futaki-Morita integrals. Let $(M,g)$ be a Hermitian manifold and $E$ a holomorphic vector bundle on M with Hermitian metric $h$ and curvature $R(h) \in \Omega^{1,1}(\text{End}(E))$ of its Chern connection. Suppose that there exists smooth $\theta(h) \in \Gamma (\text{End}(E))$ satisfying
\[
i_X R(h) = -\bar \partial \theta_X(h)
\]
Given an invariant polynomial
\[
\phi: \mathfrak{gl}(\text{rank}(E),\C) \rightarrow \C
\]
\noindent of degree $n+k$, the \emph{Futaki-Morita} integral is defined as
\[
f_{\phi}(X) := \int_M \phi \left(\theta_X(h) + \frac{\sqrt{-1}}{2\pi}R(h)\right),
\]
\noindent which turns out to be independent of the chosen metrics \cite{futakimorita:1985}.
Futaki-Morita integrals may generally be computed via holomorphic localization: Define a $(1,0)$ form $\eta$ on $M/\textrm{Zero}(X)$ by
\[
\eta(\cdot) = \frac{g(\cdot, \bar X)}{\|X\|^2}
\]
Bott's transgression argument \cite{bott:1967mich} \cite{griffithsharris} shows
\begin{align}\label{transgression}
f_{\phi}(X)
= -\sum_{\lambda} \lim_{r \rightarrow 0^+} \int_{\partial B_{r}(Z_{\lambda})} \phi(\theta_X(h) + R(h)) \wedge \sum_{k = 0}^{n-1} (-1)^k \eta \wedge (\bar\partial \eta)^k
\end{align}
\noindent where $\{Z_{\lambda}\}$ is the zero locus of $X$ and $B_{r}(Z_{\lambda})$ is any small neighborhood of $Z_{\lambda}$.
We say that $\text{Zero}(X)$ is \emph{nondegenerate} when the endomorphism $L_{\lambda}(X)$ of the normal bundle $N_{\lambda}$ to $Z_{\lambda}$ induced by $X$ is invertible\footnote{One may verify this endomorphism is given by $L_{\lambda}(X)(Y) = (\nabla_YX)^{\perp}$}. The work of Bott \cite{bott:1967jdg} essentially showed that when $\text{Zero}(X)$ is nondegenerate,
\begin{equation}\label{bott}
f_{\phi}(X) = \sum_{\lambda} \int_{Z_{\lambda}} \frac{\phi(\theta_X(h) + \frac{\sqrt{-1}}{2\pi}R(h))}{\det \left(L_{\lambda}(X) + \frac{\sqrt{-1}}{2\pi}K_{\lambda}\right)}
\end{equation}
\noindent where $K_{\lambda}$ is the curvature form of the connection induced on $N_{\lambda}$.
On the other hand, when $\text{Zero}(X)$ consists of isolated degenerate points:
\begin{theorem}[Cherveny \cite{cherveny:2016}]\label{degeneratelocalize}
If the zero locus of $X \in \mathfrak{h}$ is an isolated degenerate zero $p$ such that in local coordinates centered at $p$
\[
z_i^{\alpha_i + 1} = \sum b_{ij} X_j
\]
\noindent for some matrix $B = (b_{ij})$ of holomorphic functions, then
\begin{equation}\label{degeneratelocalization}
f_{\phi}(X) = \frac{1}{\prod \alpha_i !} \cdot \frac{\partial^{|\alpha|} \left( \phi(\theta_X(h)) \det B \right)}{\partial z_1^{\alpha_1} \cdots \partial z_n^{\alpha_n}} \bigg|_{z = 0}
\end{equation}
\noindent If $\text{Zero}(X)$ consists of multiple isolated degenerate zeros, then $f_{\phi}(X)$ is the sum of such contributions.
\end{theorem}
A special case of both (\ref{bott}) and (\ref{degeneratelocalization}) is when $p$ is an isolated nondegenerate zero: As $DX$ is invertible near $p$, take $B = (DX)^{-1}$ and $\alpha_i = 0$, giving
\begin{equation}\label{isolatednondeg}
f_{\phi}(X) = \phi(DX_p) \det B = \frac{\phi(DX_p)}{\det DX_p}
\end{equation}
Localization involving a positive dimensional degenerate zero locus is quite complicated and not understood in the general \kahler setting. The calculations in this paper may be viewed as a step in this direction.
\subsection{Localization of $\fut(X, \Omega)$}
Returning to the localization of $\fut(\Omega, X)$, suppose without loss of generality that $\Omega = c_1(L)$ where $L$ is a positive line bundle. Applying the above Futaki-Morita framework to the bundle $E = K_M^{\pm}\otimes L^{n-2j}$, standard computations yield
\begin{gather*}
R_E = \pm \ric + (n-2j) \omega\\
i_X R_E = \pm i_X \ric + (n-2j)i_X \omega \\
-\bar \partial \theta_E = \pm i_X \ric - (n-2j)\bar \partial \theta_X \\
\theta_E = \mp \Delta \theta_X + (n-2j) \theta_X
\end{gather*}
Take $\phi$ to be the invariant polynomial $\phi(A) = \text{Tr}(A^{n+1})$. The first integral in (\ref{expandedfutaki}) is then recognized to be
\[
f_{\phi, E}(X) = \left( \frac{\sqrt{-1}}{2\pi}\right)^n\int_M (-\Delta \theta_X + \ric +(n-2j)(\theta_X + \omega) )^{n+1}
\]
\noindent for $E = K_M\otimes L^{n-2j}$; the second is
\[
f_{\phi, E}(X) = \left( \frac{\sqrt{-1}}{2\pi}\right)^n\int_M (\Delta \theta_X - \ric +(n-2j)(\theta_X + \omega) )^{n+1}
\]
\noindent for $E = K_M^-\otimes L^{n-2j}$; and the third
\[
f_{\phi, E}(X) = \left( \frac{\sqrt{-1}}{2\pi}\right)^n (n+1-2j)^{n+1} \int_M \left(\theta_X + \omega \right)^{n+1}
\]
\noindent for $E = L^{n+1-2j}$. The Calabi-Futaki invariant is thus fully expressible in terms of Futaki-Morita integral invariants.
Applying localization (\ref{transgression}) to each Futaki-Morita invariant in this expression and using the combinatorial identity (\ref{combinatorial}) again to resolve summations yields
\[
\hspace{-.3 in}\fut(\Omega, X) = \lim_{r \rightarrow 0} \left( \frac{\sqrt{-1}}{2\pi}\right)^n \sum_{\lambda} \int_{\partial B_{r}(Z_{\lambda})} \hspace{-.15 in} \left[ (-\Delta \theta_X + \ric)(\theta_X + \omega)^n + \frac{\bar S (\theta_X + \omega)^{n+1}}{n+1} \right] \wedge \sum_{k=0}^{n-1} (-1)^k \eta \wedge (\bar \partial \eta)^k
\]
When $\text{Zero}(X)$ is nondegenerate, this expression was evaluated by Tian using (\ref{bott}), with explicit cohomological simplifications in the specific cases of nondegenerate isolated zeros, or nondegenerate zero loci on a K\"ahler surface, or the Fano case $\Omega = c_1(M)$. See Theorem 6.3 in \cite{tian:1996}; also p. 31 in \cite{tian:2000}.
\subsection{Blowup Situation}
Our interest will be the blowup scenario where degenerate contributions to localization naturally arise. To align notation with Li-Shi \cite{lishi:2015}, let $\mu = \frac{\bar S}{n}$ and define
\begin{gather*}
I_{Z_{\lambda}} := \sum_{\lambda} \lim_{r \rightarrow 0^+} \left( \frac{\sqrt{-1}}{2\pi}\right)^n \int_{\partial B_{r}(Z_{\lambda})} (\Delta \theta_X - \ric)(\theta_X + \omega)^n \wedge \sum_{k = 0}^{n-1} (-1)^k \eta \wedge (\bar\partial \eta)^k \\
J_{Z_{\lambda}} := - \sum_{\lambda} \lim_{r \rightarrow 0^+} \left( \frac{\sqrt{-1}}{2\pi}\right)^n \int_{\partial B_{r}(Z_{\lambda})} (\theta_X + \omega)^{n+1} \wedge \sum_{k = 0}^{n-1} (-1)^k \eta \wedge (\bar\partial \eta)^k\\
\fut_{Z_{\lambda}}(X, \Omega) := I_{Z_{\lambda}} - \frac{n\mu}{n+1} J_{Z_{\lambda}}
\end{gather*}
\noindent so that
\begin{equation}\label{futakidecomposition}
\fut(\Omega, X) = \sum_{\lambda} \fut_{Z_{\lambda}}(X, \Omega) = \sum_{\lambda} \left(I_{Z_{\lambda}} - \frac{n\mu}{n+1} J_{Z_{\lambda}} \right).
\end{equation}
\noindent Also define the summed contribution
\begin{gather*}
J_M(\Omega, X) := \sum_{\lambda} J_{Z_{\lambda}}(\Omega, X) = \left( \frac{\sqrt{-1}}{2\pi}\right)^n \int_M (\theta_X + \omega)^{n+1}
\end{gather*}
We close by expressing a relationship between local contributions to $\fut(X, \Omega)$ on $M$ and those for $\fut(\tilde X, \tilde \Omega)$ on $\tilde M$, where notation is as in the introduction.
\begin{lemma}[Li-Shi \cite{lishi:2015}, Lemma 3.1]\label{contributionslemma} Let $X\in \mathfrak{h}$ vanish at $p$, $\tilde M$ be the blowup of $M$ at $p$, $\tilde \Omega$ be the \kahler class $\pi^*\Omega - \epsilon c_1(E)$, and $\tilde X$ be the extension of $X$ to $\tilde M$. Define $\delta = \tilde \mu - \mu$. Then
\[
\fut_{\tilde M}(\tilde \Omega, \tilde X) = \fut_M(\Omega, X) - \frac{n \delta}{n+1} J_M(\Omega, X) + \sum_{E} \fut_{\tilde Z_{\lambda}}(\tilde \Omega, \tilde X) + \frac{n\delta}{n+1} J_p(\Omega, X) - \fut_p(\Omega, X)
\]
\end{lemma}
The lemma is a consequence of the localization formula for the Futaki invariant (\ref{futakidecomposition}) and the above definitions after separating the fixed components of $\tilde X$ on $\tilde M$ into those contained in the exceptional divisor and those not. The latter type is in one-to-one correspondence with the fixed components of $X$ on $M$ apart from $p$, and moreover these local Futaki invariants agree after an adjustment to $J_{Z_{\lambda}}$ by $\delta$.
Recall that $\theta_X$ is defined up to addition of a constant. Without loss of generality, we may prove our main theorem under the simplifying assumption that this constant is chosen so $\theta_X$ has average value zero, and consequently $J_M(\Omega, X) = 0$. As $p$ is nondegenerate, the term $\fut_p(\Omega, X)$ is immediate using (\ref{isolatednondeg}):
\begin{equation}\label{futakip}
\fut_p(\Omega, X) = I_p - \frac{n\mu}{n+1} J_p = \frac{\text{Tr}(A)}{\det A} \theta_p^n - \frac{n\mu}{n+1} \frac{\theta_p^{n+1}}{\det A}
\end{equation}
\noindent where we have simplified the notation using $A := DX_p$ and $\theta_p := \theta_X(p)$.
With this choice of $\theta_X$, and in light of Lemma \ref{contributionslemma} and (\ref{futakip}),
\begin{lemma}\label{toprovelemma}
To prove Theorem \ref{maintheorem}, it is sufficient to show
\[
\frac{\text{Tr}(A) \theta_p^n}{\det A} - \frac{n(\mu + \delta)}{n+1} \frac{\theta_p^{n+1}}{\det A} - \sum_{j=1}^m \fut_{q_j}(\tilde \Omega, \tilde X) = n(n-1)\theta_p\epsilon^{n-1} + O(\epsilon^n)
\]
\noindent where $\{q_1, \dots, q_m\}$ are the isolated, possibly degenerate zeros of $\tilde X$ in $E$.
\end{lemma}
\section{Case I: Maximally Degenerate Zero}
In this section we will prove Theorem \ref{maintheorem} when $\tilde X$ has a single isolated zero in the exceptional locus, necessarily of maximal degeneracy. The case of multiple zeros builds on these computations and will be given in the next section.
Let $p \in \text{Zero}(X)$ be the zero at which we will blow up $M$. Choosing coordinates about $p$ such that $DX_p$ is in Jordan form, a maximal degenerate zero on the blowup corresponds to $DX_p$ having a single Jordan block.
To be precise, let $A$ denote the $n \times n$ Jordan matrix
\begin{equation}\label{jordanblock}
A = (A_{ij}) = \begin{bmatrix} a & 1 & 0 & \cdots & 0 \\ 0 & \ddots & \ddots & & \vdots \\ \vdots & & \ddots & \ddots & 0 \\ \vdots & & & \ddots & 1 \\ 0 & \cdots & \cdots & 0 & a \end{bmatrix}
\end{equation}
By Poincar\'e's Theorem \ref{poincare}, if $DX_p = A$ then $X$ is biholomorphically equivalent to its linearization on some neighborhood $U$ of $p$ provided that $a \neq 0$ (the zero is nondegenerate). Therefore, without loss of generality, we may assume $X = \sum X_i \frac{\partial}{\partial z_i}$ is given on $U$ by
\begin{equation}\label{oneblock}
X = \sum_{i=1}^{n-1}(az_i + z_{i+1}) \frac{\partial}{\partial z_i} + az_n \frac{\partial}{\partial z_n}
\end{equation}
Following \cite{lishi:2015}, we now describe $X$'s natural extension $\tilde X$ to the blowup in local coordinates. Let $\tilde U = \pi^{-1}(U)$ be the neighborhood of the exceptional divisor on $\tilde M = \textrm{Bl}_p M$ given by
\[
\tilde U = \{\left((z_1, \dots, z_n),[\eta_1, \dots, \eta_n] \right) | z_i\eta_j = z_j \eta_i\} \subseteq \C^n \times \C P^{n-1}
\]
Cover $\tilde U$ with charts $\tilde U_i = \{ \eta_i \neq 0\}$ having local coordinates
\[
(u_1, \dots, u_n) = \left( \frac{\eta_1}{\eta_i}, \dots, z_i, \dots, \frac{\eta_n}{\eta_i} \right)
\]
\noindent Note that the slice of $\tilde U_i$ with $z_i = 0$ is just the standard cover for $E \cong \C P^{n-1}$.
In these coordinates the holomorphic extension $\tilde X$ of any $X \in \mathfrak{h}$ vanishing at $p$ to the blow-up is given by
\begin{equation}\label{xtildelocalcoords}
\tilde X \big|_{\tilde U_i} = X_i \frac{\partial}{\partial u_i} + \sum_{j \neq i} \frac{1}{u_i}(X_j - u_jX_i) \frac{\partial}{\partial u_j}
\end{equation}
In particular, $\tilde X$ over $\tilde U_1$ for the one block case (\ref{oneblock}) presently being considered is
\begin{equation}\label{oneblockblowup}
\tilde X|_{\tilde U_1} = u_1(a + u_2) \frac{\partial}{\partial u_1} + \sum_{i = 2}^{n-1} \left[ (u_{i+1} - u_i u_2) \frac{\partial}{\partial u_i}\right] + (-u_2 u_n)\frac{\partial}{\partial u_n}.
\end{equation}
\noindent One may verify that the isolated zero at the origin in this chart, which we denote $q$, is indeed the only zero of $\tilde X$ in the exceptional divisor. It is ``maximally degenerate" in the sense that zero is an eigenvalue of $D\tilde X_q$ with algebraic multiplicity $n-1$. This is all geometrically obvious when one considers the action induced by $X$ on lines through $p$, where the zero locus of $\tilde X$ in the exceptional divisor corresponds to eigenspaces for $DX_p$.
We now choose a convenient metric in the class $\tilde \Omega = \pi^*\Omega - \epsilon c_1([E])$ following \cite{griffithsharris} (p. 185). Denote by $B_r \subseteq U$ the ball of radius $r$ centered at $p$. Suppose for simplicity $B_1 \subseteq U$, and let $\tilde B_1 = \pi^{-1}(B_1)$. The fiber of $[E]$ over $\tilde B_1$ at a point is simply
\[
[E]_{(z,\eta)} = \{\lambda(\eta_1, \dots, \eta_n), \lambda \in \C\}.
\]
Denote by $h_1$ the metric on $[E]|_{\tilde B_1}$ given by $\|(\eta_1, \dots, \eta_n)\|^2$. Also let $\sigma \in H^0(\tilde M, \mathcal{O}([E]))$ be the holomorphic section whose zero divisor is $E$ and denote by $h_2$ the metric on $[E]|_{\tilde M/E}$ such that $\|\sigma\|_{h_2} \equiv 1$. Finally choose a partition of unity $\{\rho_1, \rho_2\}$ subordinate to the cover $\{\tilde B_1, \tilde M / \tilde B_{1/2}\}$. The metric
\[
h = \rho_1 h_1 + \rho_2 h_2
\]
\noindent has nonzero curvature only on $\tilde B_1$. Our \kahler form will be
\[
\tilde \omega = \pi^*\omega + \epsilon \partial \bar \partial \log h.
\]
A short calculation shows the holomorphy potential $\theta_{\tilde X}$ for $\tilde X$ relative to this $\tilde \omega$ on $\tilde U_1$ is
\begin{equation}\label{thetablowup}
\theta_{\tilde X} = \pi^*\theta_X - \epsilon \left[ a + \frac{u_2 + \bar u_2 u_3 + \bar u_3 u_4 + \dots + \bar u_{n-1} u_n}{1 + |u_2|^2 + \dots + |u_n|^2} \right]
\end{equation}
For brevity, denote $\theta_X (p)$ by $\theta_p$ and likewise for $\tilde \theta_q := \theta_{\tilde X} (q)$, so that
\begin{equation}\label{thetaq}
\tilde \theta_q = \theta_p - a\epsilon.
\end{equation}
\noindent All derivatives of all orders of $\theta_{\tilde X}$ vanish at $q$ with the exception of
\begin{equation}\label{thetaqprime}
\frac{\partial \theta_{\tilde X}}{\partial u_2}(q) = -\epsilon.
\end{equation}
\noindent Another short calculation shows that
\begin{equation}\label{thetalaplacian}
\Delta_{\tilde \omega} \theta_{\tilde X} = a- (n-1)u_2
\end{equation}
By Theorem \ref{degeneratelocalize}, the local contribution to any Futaki-Morita integral in the present situation (and in particular the Futaki invariant) is given by the residue formula
\begin{equation}\label{localizeoneblock}
\text{Res}_q \phi = \frac{1}{\prod (\alpha_j - 1)!} \frac{\partial^{|\alpha|}(\phi \det B)}{\partial u_1^{\alpha_1 - 1} \partial u_2^{\alpha_2 - 1} \cdots \partial u_n^{\alpha_n - 1}} \bigg|_q
\end{equation}
\noindent where $\alpha_j$ are natural numbers and $B = (b_{ij})$ is an $n \times n$ matrix such that
\begin{equation}\label{hilbert}
u_j^{\alpha_j} = \sum b_{ij} \tilde X_i.
\end{equation}
We now construct $B$ in order to calculate (\ref{localizeoneblock}). Choose $k$ such that $2^k < n \leq 2^{k+1}$. It is straightforward to verify for (\ref{oneblockblowup}) that
\begin{align*}
u_1 &= \left[ \frac{1}{a} - \sum_{i=2}^n \left(\frac{-1}{a}\right)^i u_i \right] \tilde X_1 + \sum_{i = 2}^n\left[- u_1 \left(\frac{-1}{a}\right)^i \right] \tilde X_i \\
u_2^n &= \sum_{i = 2}^n \left[- u_2^{n-i} \right] \tilde X_i \\
\end{align*}
For $j = 3, \dots, n$, one calculates
\[
u_j^{2^{k+1}} = \sum_{l = 2}^{j-1} \left[ -u_2^{(j-1)2^{k+1} - l} + u_2^{(j-l-1)2^{k+1}} \prod_{i=0}^k(u_{l+1}^{2^i} +u_2^{2^i}u_{l}^{2^i}) \right] \tilde X_l + \sum_{l = j}^n -u_2^{(j-1)2^{k+1} - l} \tilde X_l\]
\noindent The idea is to repeatedly factor $u_j^{2^{k+1}} - (u_2u_{j-1})^{2^{k+1}}$ into binomials, one of which is eventually $\tilde X_j$, and insert the above expression for $u_2^n$ in terms of the $\tilde X_i$. These relations contain the information necessary to construct $B$ for (\ref{hilbert}) with parameters $\alpha_1 = 1, \alpha_2 = n, \alpha_j = 2^{k+1}$ for $j = 3, \dots, n$ (these are certainly not minimal $\alpha_i$ for every $n$, but are convenient for a general setup).
With these choices, the determinant of $B$ is found by row reduction to be
\begin{equation}\label{detB}
\det B = (-1)^{n-1}\left[ \frac{1}{a} - \sum_{i=2}^n \left(\frac{-1}{a}\right)^i u_i \right] \prod_{j=3}^{n} \prod_{i=0}^{k}(u_j^{2^i} +u_2^{2^i}u_{j-1}^{2^i})
\end{equation}
Applying (\ref{localizeoneblock}), the residue of interest is
\[
\textrm{Res}_q \phi = \frac{1}{(n-1)![(2^k)!]^{n-2}} \cdot \frac{\partial^{|a|}(\phi \det B)}{(\partial u_2)^{n-1} (\partial u_3)^{2^k} \cdots (\partial u_n)^{2^k}} \bigg|_q
\]
Since $\phi$ is a function of $\theta_{\tilde X}$, which depends only on $u_2$ and its derivatives in our case, all other derivatives must be applied to $\det B$. Doing so, the coefficient of $u_3^{2^k} \dots u_n^{2^k}$ in $\det B$ is found to be
\begin{equation}\label{detB1}
(-1)^{n-1} \sum_{i=0}^{n-1} \frac{(-u_2)^i}{a^{i+1}}
\end{equation}
\noindent so that the residue after taking appropriate derivatives and evaluating at $u_3 = ... = u_n = 0$ is
\begin{align*}
\textrm{Res}_q \phi &= \frac{1}{(n-1)!} \frac{\partial}{(\partial u_2)^{n-1}} \left(\phi \sum_{i=0}^{n-1} (-1)^{n-1} \frac{(-u_2)^i}{a^{i+1}}\right) \bigg|_{q} \\
&= \frac{1}{(n-1)!} \sum_{j=0}^{n-1} \binom{n-1}{j} \cdot \frac{(-1)^{n-1}(-1)^{n-1-j}(n-1-j)!}{a^{n-1-j+1}} \cdot \frac{\partial^j \phi}{(\partial u_2)^j}(0) \\
&= \sum_{j=0}^{n-1} \frac{(-1)^j}{j!a^{n-j}} \frac{\partial^j \phi}{(\partial u_2)^j}(0)
\end{align*}
We have shown
\begin{lemma}\label{maxdegenlemma}
If $\phi$ is an invariant polynomial whose value depends only on $u_2$ in the above situation, and $DX_p$ is a single Jordan block with eigenvalue $a$, then the residue contribution to the Futaki-Morita of the blowup at $p$ at the unique isolated zero $q$ is
\[
\textrm{Res}_q \phi = \sum_{i=0}^{n-1} \frac{(-1)^i}{i!a^{n-i}} \frac{\partial^i \phi}{(\partial u_2)^i}(0)
\]
\end{lemma}
\noindent When $n = 2$ we recover Lemma 3.6 of \cite{lishi:2015}, which is the main calculation of the paper and obtained by brute force calculus.
We can now give a direct proof of Theorem \ref{maintheorem} in the case $DX_p = A$ by verifying the identity in Lemma \ref{toprovelemma}. The term to calculate is $\fut_q(\tilde \Omega, \tilde X) = I_q - \frac{n}{n+1}(\delta + \mu)J_q$.
For $J_q$, apply Lemma \ref{maxdegenlemma} with $\phi = \theta_{\tilde X}^{n+1}$. By (\ref{thetaq}) and (\ref{thetaqprime}),
\[
J_q = \sum_{i=0}^{n-1} \frac{(n+1)!(\theta_p - a\epsilon)^{n+1-i}\epsilon^i}{i!a^{n-i}(n+1-i)!}
\]
\noindent Binomial expansion and interchanging summations yields
\[
J_q = (n+1)! \sum_{j=2}^{n+1} \sum_{i=0}^{n+1-j} \frac{(-1)^{n+1-i-j}}{i!j!(n+1-i-j)!a^{j-1}}\theta^j_p \epsilon^{n+1-j} + O(\epsilon^n).
\]
The coefficient of $\epsilon^{n+1-j}$ is proportional to $\sum_{i=0}^{n+1-j} (-1)^i {n+1-j \choose i}$, which vanishes by symmetry of binomial coefficients unless $j = n+1$. It follows that
\[
J_q = \frac{\theta_p^{n+1}}{a^n} + O(\epsilon^n).
\]
On the other hand, to calculate $I_q$ use $\phi = (-\Delta \theta_{\tilde X})\theta_{\tilde X}^n$ in Lemma \ref{maxdegenlemma}. Again by (\ref{thetaq}), (\ref{thetaqprime}), and (\ref{thetalaplacian}),
\[
I_q = \sum_{i=0}^{n-1} \frac{n!\epsilon^{i-1}}{i!a^{n-i}(n-i)!} \left[ \frac{(n-1)i(\theta_p - a\epsilon)^{n+1-i}}{n-i+1} + a(\theta_p - a\epsilon)^{n-i}\epsilon \right]
\]
\noindent By a similar expansion, the summed second term simplifies to
\[
\frac{\theta_p^n}{a^{n-1}} + O(\epsilon^n)
\]
\noindent while the first simplifies to
\[
\frac{(n-1)\theta_p^n}{a^{n-1}} - n(n-1)\theta_p \epsilon^{n-1} + O(\epsilon^n)
\]
\noindent yielding
\[
I_q = \frac{n\theta_p^n}{a^{n-1}} - n(n-1)\theta_p \epsilon^{n-1} + O(\epsilon^{n})
\]
Putting everything together in Lemma \ref{toprovelemma},
\begin{align*}
\fut_p(X, \Omega) &- \fut_q(\tilde X, \tilde \Omega) - \frac{n \delta}{n+1} J_p \\
&= \left[ \frac{an \theta_p^n}{a^n} - \frac{n(\mu + \delta)}{n+1} \frac{\theta_p^{n+1}}{a^n} \right] - \left[\frac{n\theta_p^n}{a^{n-1}} - n(n-1)\theta_p \epsilon^{n-1} - \frac{n(\mu + \delta)}{n+1}\frac{\theta^{n+1}_p}{a^n} + O(\epsilon^{n}) \right] \\
&= n(n-1)\theta_p \epsilon^{n-1} + O(\epsilon^n)
\end{align*}
\noindent which completes our verification of Theorem \ref{maintheorem} in this special case.
\section{Case II: Multiple Degenerate Zeros}
In this section we complete the proof of Theorem \ref{maintheorem}. Suppose the linearization of $X$ at $p$ now has multiple Jordan blocks. Hypothesis ($\star$) means that each Jordan block corresponds to an isolated degenerate zero in the exceptional divisor, and by a change of coordinate we may assume a particular degenerate zero corresponds to the first block. We extend the computations from Section 3 to calculate the contribution to the Futaki invariant from the first block under the influence of other blocks. The net contribution from all degenerate zeros is then a sum given by symmetrizing that formula, which we evaluate using Lemma \ref{lemmagk} (proved via integration of meromorphic differentials in the appendix).
Suppose coordinates centered at $p \in \text{Zero}(X)$ have been chosen such that the linearization $DX$ of $X$ is in Jordan form at $p$:
\[
DX_p = \begin{bmatrix} A_1 & 0 & \cdots & 0 \\ 0 & A_2 & \ddots & \vdots \\ \vdots & \ddots & \ddots & 0 \\ 0 & \cdots & 0 & A_m \end{bmatrix}
\]
\noindent Each Jordan block $A_i$ is of the form (\ref{jordanblock}) with diagonal entries $a_i$ and dimension $n_i$. By Lemma \ref{normalformthrm}, we may assume $X$ is biholomorphic to its linearization near $p$.
Let $s_j = \sum_{k = 1}^j n_k$, so that $s_m = n$. In the coordinates introduced in Section 3,
\begin{equation}\label{generallift}
\tilde X \big|_{\tilde U_1} = u_1(a_1 + u_2) \frac{\partial}{\partial u_1} + \sum_{j = 1}^m \left[ \sum_{i = s_{j-1} + 1}^{s_j - 1} \left[u_{i + 1} - u_i(u_2 + a_1 - a_j) \right] \frac{\partial}{\partial u_i} - u_{s_j}(u_2 + a_1 - a_j) \frac{\partial}{\partial u_{s_j}} \right]
\end{equation}
\noindent (which of course reduces to the one block formula (\ref{oneblockblowup}) when $m = 1$). The zero at the origin in $\tilde U_1$ will be denoted $q_1$.
Our first task is to construct the appropriate $B$ to apply Theorem \ref{degeneratelocalize}. Let $k$ be the natural such that $2^k < n_1 \leq 2^{k + 1}$. The work of Section 3 constructs a matrix $B_1$ expressing powers of $u_1, \dots, u_{s_1}$ in terms of $\tilde X_1, \dots, \tilde X_{s_1}$. For $1 < j \leq m$, one checks
\begin{equation}\label{xtildesj}
u_{s_j} = \tilde X_{s_j} \frac{\prod_{i=0}^k [(a_j - a_1)^{2^i} + u_2^{2^i}]}{(a_j - a_1)^{2^{k+1}}} + u_2^{2^{k+1}} u_{s_j}
\end{equation}
The idea here is to factor $u_{s_j}(u_2^{2^{k+1}} - (a_{s_j} - a_1)^{2^{k+1}})$. We are then done since $u_2^{2^{k+1}}$ is a known linear combination of $\tilde X_1, \dots, \tilde X_{s_1}$ from Section 3. For $u_{s_{j-1}} < u_i < u_{s_j}$,
\begin{equation}\label{uigeneral}
u_i = \frac{\prod_{i=0}^k [(a_j - a_1)^{2^i} + u_2^{2^i}]}{(a_j - a_1)^{2^{k+1}}} \tilde X_i + u_2^{2^{k+1}}u_i - \frac{\prod_{i=0}^k [(a_j - a_1)^{2^i} + u_2^{2^i}]}{(a_j - a_1)^{2^{k+1}}} u_{i+1}
\end{equation}
\noindent so that, in light of (\ref{xtildesj}), we may recursively solve to obtain
\begin{equation}\label{uigeneralB}
u_i = \frac{\prod_{i=0}^k [(a_j - a_1)^{2^i} + u_2^{2^i}]}{(a_j - a_1)^{2^{k+1}}} \tilde X_i + \text{linear combination of $\{\tilde X_1, \dots, \tilde X_{n_1}, \tilde X_{i+1}, \dots, X_{s_j}\}$}.
\end{equation}
It follows that $B$ has the form
\begin{equation}\label{generalB}
B = \begin{bmatrix} B_1 & 0 & 0 & 0 \\ * & B_2 & 0 & 0 \\ * & 0 & \ddots & 0 \\ * & 0 & 0 & B_m \end{bmatrix}
\end{equation}
\noindent where $B_1$ was constructed in the previous section, and each $B_j$ for $j > 1$ is upper triangular with the entries recoverable from (\ref{xtildesj}) and (\ref{uigeneral}).
We have calculated $\det B_1$ in (\ref{detB}), while for $j > 1$,
\[
\det B_j = \left(\frac{\prod_{i=0}^k [(a_j - a_1)^{2^i} + u_2^{2^i}]}{(a_j - a_1)^{2^{k+1}}}\right)^{n_j}
\]
\noindent Clearly the only derivative that may be applied to $\det B_j$ for $j> 1$ is the $u_2$-derivative. The $i$-th $u_2$-derivative of $\det B_j$ evaluated at $u_2 = 0$ is
\[
\frac{\partial}{(\partial u_2)^i} \det B_j = \frac{(n_j + i -1)!}{(n_j - 1)!(a_j - a_1)^{n_j +i}}
\]
Putting everything into (\ref{localizeoneblock}), the residue of interest
\begin{align}\label{generalresidue}
\textrm{Res}_{q_1} \phi &= \frac{1}{(n_1-1)!} \frac{\partial}{(\partial u_2)^{n_1-1} }\left(\phi \prod_{j=1}^m \det B_j \right) \nonumber\\
&= \frac{1}{(n_1 - 1)!} \sum \binom{n_1 - 1}{i, \mu_1, \dots, \mu_m} \cdot \prod_{j=1}^m \frac{\partial \det B_j}{(\partial u_2)^{\mu_j}} \cdot \frac{\partial \phi}{(\partial u_2)^i} \bigg|_{u = 0} \nonumber \\
&= \sum_{i = 0}^{n_1 - 1} \sum_{\mu } \frac{1}{i!} \left( \prod_{j=1}^m \frac{1}{\mu_j!} \frac{\partial \det B_j}{(\partial u_2)^{\mu_j}} \right) \cdot \frac{\partial \phi}{(\partial u_2)^i} \bigg|_{u = 0} \nonumber \\
&= \sum_{i = 0}^{n_1 - 1} \sum_{\mu } \frac{(-1)^{n_1 +\mu_1 -1}}{i! a_1^{\mu_1 + 1}} \left( \prod_{j=2}^m \frac{\binom{n_j + \mu_j - 1}{\mu_j}}{(a_j - a_1)^{n_j + \mu_j}} \right) \cdot \frac{\partial \phi}{(\partial u_2)^i} \bigg|_{u = 0}
\end{align}
\noindent where $\mu = (\mu_1, \dots, \mu_m)$ runs over all partitions of $n_1 - i - 1$ of length $m$. In the last line we have used (\ref{detB1}). This generalizes Lemma \ref{maxdegenlemma} ($m = 1$ is the lemma).
We now complete the proof of Theorem \ref{maintheorem} by verifying the identity in Lemma \ref{toprovelemma}. The holomorphy potential of $\tilde X$ as in (\ref{generallift}) is
\[
\theta_{\tilde X} = \pi^*\theta_X - \epsilon \left[a_1 + \frac{u_2 + \bar u_2 u_3 + \dots + \bar u_{n-1} u_n + \sum_{j=1}^m \sum_{i=s_{j-1} + 1}^{s_j} (a_j - a_1)|u_i|^2}{1 + |u_2|^2 + \dots + |u_n|^2} \right],
\]
\noindent generalizing (\ref{thetablowup}). As in (\ref{thetaq}) and (\ref{thetaqprime}), $\theta_{\tilde X}$ satisfies
\begin{gather}\label{generalthetavalues}
\begin{split}
\tilde \theta_{q_1} = \theta_p - a_1\epsilon \\
\frac{\partial \theta_{\tilde X}}{\partial u_2}(q_1) = -\epsilon
\end{split}
\end{gather}
\noindent while all other derivative of $\theta_{\tilde X}$ vanish at $q_1$, and the Laplacian generalizing (\ref{thetalaplacian}) is
\begin{equation}\label{generallaplacian}
\Delta_{\tilde \omega} \theta_{\tilde X} = \text{Tr}(A) - (n-1)(u_2 + a_1).
\end{equation}
\noindent With (\ref{generalthetavalues}) in mind, $J_{q_1}$ is calculated by applying (\ref{generalresidue}) to $\phi = \theta^{n+1}_{\tilde X}$:
\begin{align*}
J_{q_1} &= \sum_{i = 0}^{n_1 - 1} \sum_{\mu } \frac{(-1)^{n_1 +\mu_1 -1}}{i! a_1^{\mu_1 + 1}} \left( \prod_{j=2}^m \frac{\binom{n_j + \mu_j - 1}{\mu_j}}{(a_j - a_1)^{n_j + \mu_j}} \right) \cdot \frac{(n+1)! (-\epsilon)^i(\theta_p - a_1 \epsilon)^{n+1-i}}{(n+1-i)!}
\end{align*}
\noindent where $\mu$ is still runs over partitions of $n_j - i - 1$ of length $m$.
Interchanging $1 \leftrightarrow j$ gives the sum over all zeros $\{q_1, \dots, q_m\}$ in the exceptional divisor to be
\begin{align}\label{sumofJresidues}
\sum_j J_{q_j} &= \sum_{j=1}^m \sum_{i = 0}^{n_j - 1} \frac{ (n+1)! (-\epsilon)^{i}(\theta_p - a_j \epsilon)^{n+1-i}}{i!(n+1-i)!}
\sum_{\mu} \frac{(-1)^{n_j +\mu_j -1}}{a_j^{\mu_j + 1}}
\left( \prod_{l \neq j}^m \frac{\binom{n_l + \mu_l - 1}{\mu_l}}{(a_l - a_j)^{n_l + \mu_l}} \right) \nonumber \\
&= \sum_{j=1}^m \sum_{i=0}^{n_j-1} \sum_{k=0}^{n+1-i} \frac{(-1)^{n-k}(n+1)!}{k!i!(n+1-i-k)!} \theta_p^k \epsilon^{n+1-k}\sum_{\mu} \frac{(-1)^{n_j +\mu_j}}{a_j^{\mu_j+i+k-n}} \left( \prod_{l \neq j}^m \frac{\binom{n_l + \mu_l - 1}{\mu_l}}{(a_l - a_j)^{n_l + \mu_l}} \right) \nonumber \\
&= \sum_{k=0}^{n+1} (-1)^{n-k} \binom{n+1}{k} \theta_p^k \epsilon^{n+1-k} G_k
\end{align}
\noindent where
\[
G_k = \sum_{j=1}^m \sum_{i=0}^{n+1-k} \binom{n+1-k}{i} \sum_{\mu} \frac{(-1)^{n_j +\mu_j}}{a_j^{\mu_j+i+k-n}} \left( \prod_{l \neq j}^m \frac{\binom{n_l+ \mu_l - 1}{\mu_l}}{(a_l - a_j)^{n_l + \mu_l}} \right).
\]
\noindent We have simplified the notation by allowing $i$ to run into values that make the partition $\mu$ of $n_j - i - 1$ a partition of a negative number. The term is understood to be zero when this happens.
\begin{lemma}\label{lemmagk}
\[
G_k =
\begin{cases}
\frac{-1}{\det A} & k = n+1\\
0 & 1 < k < n+1\\
(-1)^n & k = 1
\end{cases}
\]
\end{lemma}
\noindent See the appendix in Section \ref{appendix} for a proof via integration of meromorphic differentials. It follows from Lemma \ref{lemmagk} that
\[
\sum_j J_{q_j} = \frac{\theta_p^{n+1}}{\det A} + O(\epsilon^n)
\]
Likewise, to compute the contribution $I_{q_j}$ from each fixed point $q_j$, use $\phi = (-\Delta \theta_{\tilde X})\theta_{\tilde X}^n$ in (\ref{generalresidue}) along with (\ref{generalthetavalues}) and (\ref{generallaplacian}). The contribution is
\begin{align}\label{sumofIresidues}
\sum_j I_{q_j} = \sum_{j=1}^m \sum_{i = 0}^{n_j - 1} & \sum_{\mu } \frac{(-1)^{n_1 +\mu_1 -1}}{i! a_j^{\mu_1 + 1}} \left( \prod_{l \neq j}^m \frac{\binom{n_j + \mu_j - 1}{\mu_j}}{(a_j - a_l)^{n_j + \mu_j}} \right) \\
& \cdot \left[ \frac{-n!(\text{Tr}(A) - (n-1)a_j) (-\epsilon)^i(\theta_p - a_j \epsilon)^{n-i}}{(n-i)!} + \frac{i(n-1)n! (-\epsilon)^{i-1}(\theta_p - a_j \epsilon)^{n-i+1}}{(n-i+1)!} \right] \nonumber
\end{align}
This summation is of the form
\[
\sum_{k=1}^n (-1)^{n-k+1} \binom{n}{k} \theta_p^k \epsilon^{n-k} [G_k' + G_k'' + G_k'''] + O(\epsilon^n)
\]
\noindent where
\begin{align*}
G_k' &= \text{Tr}(A) \sum_{j=1}^m \sum_{i=0}^{n-k} \binom{n-k}{i} \sum_{\mu} \frac{(-1)^{n_j +\mu_j}}{a_j^{\mu_j+i+k-n+1}} \left( \prod_{l \neq j}^m \frac{\binom{n_l+ \mu_l - 1}{\mu_l}}{(a_l - a_j)^{n_l + \mu_l}} \right) \\
G_k'' &= -(n-1) \sum_{j=1}^m \sum_{i=0}^{n-k} \binom{n-k}{i} \sum_{\mu} \frac{(-1)^{n_j +\mu_j}}{a_j^{\mu_j+i+k-n}} \left( \prod_{l \neq j}^m \frac{\binom{n_l+ \mu_l - 1}{\mu_l}}{(a_l - a_j)^{n_l + \mu_l}} \right)\\
G_k''' &= -(n-1) \sum_{j=1}^m \sum_{i=1}^{n-k+1} \binom{n-k}{i-1} \sum_{\mu} \frac{(-1)^{n_j +\mu_j}}{a_j^{\mu_j+i+k-n}} \left( \prod_{l \neq j}^m \frac{\binom{n_l+ \mu_l - 1}{\mu_l}}{(a_l - a_j)^{n_l + \mu_l}} \right)
\end{align*}
\noindent and, as with $G_k$ above, any index $i$ producing a partition of a negative number contributes zero.
Clearly $G'_k = \text{Tr}(A)G_{k+1}$, while for each $1 \leq k \leq n$, combinatorial manipulation shows
\[
G_k'' + G_k''' = -(n-1)G_k.
\]
\noindent By Lemma \ref{lemmagk}, (\ref{sumofIresidues}) evaluates to
\[
\sum_j I_{q_j} = \frac{\text{Tr}(A)}{\det A} \theta^n_p - n(n+1)\theta_p \epsilon^{n-1} + O(\epsilon^n).
\]
Putting these results into Lemma \ref{toprovelemma},
\begin{align*}
&\fut_p (\Omega, X) - \sum_{j=1}^m \left(I_{q_j} - \frac{n \tilde \mu}{n+1} J_{q_j} \right) - \frac{n\delta}{n+1} J_p(\Omega, X)\\
&= \left(\frac{\text{Tr}(A)\theta_p^n}{\det A} - \frac{n(\mu+ \delta)}{n+1} \frac{\theta_p^{n+1}}{\det A} \right) - \left(\frac{\text{Tr}(A)\theta^n_p}{\det A} - n(n+1)\theta_p \epsilon^{n-1} - \frac{n \tilde \mu}{n+1} \frac{\theta_p^{n+1}}{\det A} + O(\epsilon^n) \right) \\
&= n(n-1)\theta_X(p)\epsilon^{n-1} + O(\epsilon^n)
\end{align*}
\noindent which concludes the verification of Theorem \ref{maintheorem}.
\section{Normal Forms at Singularities}
Sections 3 and 4 rely on the assumption that $X$ is holomorphically equivalent to its linearization on a neighborhood of $p$, which is in general not true (not even smoothly). Our main result of this section is that for the purposes of proving Theorem \ref{maintheorem} it is sufficient to assume such a normal form.
We recall a well-known condition originally due to Poincar\'e under which this assumption is true. A vector $\lambda = (\lambda_1, \dots, \lambda_n) \in \C^n$ is called \emph{resonant} if there exists a relation
\[
\lambda_k = \sum m_i \lambda_i
\]
\noindent where $m_i$ are nonnegative integers and $\sum m_i \geq 2$. The vector $\lambda$ is said to \emph{belong to the Poincar\'e domain} if the convex hull of $\lambda_1, \dots, \lambda_n$ in $\C$ does not contain the origin.
\begin{theorem}[Poincar\'e \cite{arnold:1988}]\label{poincare}
If the eigenvalues of the linear part of a holomorphic vector field at a zero belong to the Poincar\'e domain and are nonresonant, then the vector field is biholomorphically equivalent to its linearization on a neighborhood of that zero.
\end{theorem}
This condition applies to the one non-trivial Jordan block needed in Li-Shi \cite{lishi:2015} and is key to simplifying their calculation. It remains valid more generally for the case of a single Jordan block in Section 3, but does not apply in Section 4 in general.
We now show that it is sufficient to prove Theorem \ref{maintheorem} under the assumption $X$ is locally biholomorphic to its linearization. To be precise, let $Y$ be a holomorphic vector field on $M$ with an isolated nondegenerate zero at $p$ satisfying ($\star$), so that in coordinates centered at $p$,
\begin{equation}\label{localX}
Y = \sum_j (A_{ij}z_i + O(z_kz_l) )\frac{\partial}{\partial z_j}
\end{equation}
Let $X$ denote the linearization of $Y$, defined on a neighborhood of $p$:
\[
X = \sum (A_{ij}z_i)\frac{\partial}{\partial z_j}
\]
\noindent The extension $\tilde X$ of $X$ to $\tilde M = Bl_pM$ is defined on a neighborhood of the exceptional divisor (see (\ref{xtildelocalcoords}) for the explicit local formula).
\begin{lemma}\label{normalformthrm}
Let $X, Y, \tilde X$, and $\tilde Y$ be as above. The zero loci of $\tilde X$ and $\tilde Y$ on a neighborhood of the exceptional divisor $E$ agree, and for each isolated zero $q \in E$,
\[
\fut_q(\tilde X, \tilde \Omega) = \fut_q(\tilde Y, \tilde \Omega)
\]
\end{lemma}
\begin{proof}
In coordinates centered at $p$, $Y$ has the form in (\ref{localX}). As $p$ is an isolated and nondegenerate zero of $Y$, it is also isolated and nondegenerate for $X$. From expression (\ref{xtildelocalcoords}) it is clear that any zeros of $\tilde Y$ in a neighborhood of E must be on E (one must have $Y_1 = \dots = Y_n = 0$, and this can only happen on $\pi^{-1}(p)$). Likewise for the zeros of $\tilde X$. But the zeros on the exceptional divisor correspond with the eigenspaces of $DX_p = DY_p$, and so the zero loci agree as claimed. In particular, the zeros of $\tilde Y$ are isolated iff the zeros of $\tilde X$ are isolated.
The local Futaki invariant for $X$ was calculated in Section 4 using (\ref{generalresidue}). We will show that adding in higher order terms to form $Y$ changes $\theta_{\tilde X}, \Delta \theta_{\tilde X}$, and $\det B$ each by $O(u_1)$, so that
\begin{align}\label{normalwanttoshow}
\begin{split}
-(\Delta \theta_{\tilde Y})\theta_{\tilde Y}^n \det C &= -(\Delta \theta_{\tilde X})\theta_{\tilde X}^n \det B + O(u_1) \\
\theta_{\tilde Y}^{n+1} \det C &= \theta_{\tilde X}^{n+1} \det B + O(u_1)
\end{split}
\end{align}
\noindent and moreover still no $u_1$ derivatives will be involved. The lemma then follows, as $I_q$ and $J_q$ are calculated via (\ref{degeneratelocalization}) with no $u_1$-derivatives applied to $-(\Delta \theta_{\tilde Y})\theta_{\tilde Y}^n \det C$ and $\theta_{\tilde Y}^{n+1} \det C$, respectively, followed by evaluation involving $u_1 = 0$.
By assumption,
\[
Y_i = X_i + O(z_kz_l)
\]
\noindent for each $i = 1, \dots, n$. Using the coordinates in Section 3 in which the zero of $\tilde Y$ is at the origin in $\tilde U_1$, by (\ref{xtildelocalcoords})
\[
\tilde Y_1 = \tilde X_1 + O(u_1^2)
\]
\noindent while for $2 \leq i \leq n$,
\begin{align*}
\tilde Y_i &= \frac{1}{u_1}[A_{ij}z_j + O(u_1^2) - u_i(A_{ij}z_j + O(u_1^2))] \\
&= \tilde X_i + O(u_1)
\end{align*}
Straightforward calculation then gives
\begin{align}\label{normallaplacian}
\Delta_{\tilde \omega} \theta_{\tilde Y} &= \sum_i \frac{\partial \tilde Y_i}{\partial u_i} \nonumber \\
&= \frac{\partial}{\partial u_1} [\tilde X_1 + O(u_1^2)] + \sum_{i > 1} \frac{\partial}{\partial u_i} [\tilde X_i + O(u_1)] \nonumber\\
&= \Delta_{\tilde \omega} \theta_{\tilde X} + O(u_1)
\end{align}
Next we consider the holomorphy potential itself. On $U$,
\[
\bar \partial \theta_X = \bar \partial \theta_Y + O(z_l z_k)
\]
\noindent so that
\begin{align}\label{normaltheta}
\bar \partial \theta_{\tilde X} &= \pi^* \bar \partial \theta_X - \bar \partial ( \tilde X \log h) \nonumber \\
\bar \partial \theta_{\tilde X} &= \pi^* \bar \partial (\theta_X + O(z_kz_l)) - \bar \partial ( \tilde Y \log h) + \bar \partial (O(u_1)) \nonumber\\
\theta_{\tilde X} &= \theta_{\tilde Y} + O(u_1)
\end{align}
Lastly, let $B = (b_{ij})$ be the matrix of holomorphic functions near $q$ constructed in Sections 3 and 4 satisfying $u_j^{\alpha_j} = \sum_j b_{ij} \tilde X_j$. We will show that there is similarly a matrix $C = (c_{ij})$ such that $u_j^{\alpha_j} = \sum_j c_{ij} \tilde Y_j$ for the same values of $\alpha_j$ (and in particular $\alpha_1 = 1$), such that
\begin{equation}\label{normaldet}
\det C = \det B.
\end{equation}
Theorem \ref{poincare} does adapt to the first Jordan block of $DY_p$, providing a holomorphic coordinate system in which $Y_1, \dots, Y_{n_1}$ agree with $X_1, \dots, X_{n_1}$, and consequently $\tilde Y_1, \dots, \tilde Y_{n_1}$ agree with $\tilde X_1, \dots, \tilde X_{n_1}$ by (\ref{xtildelocalcoords}). As a result, the matrix $C$ sought begins with a block $C_1$ identical to block $B_1$ in (\ref{generalB}).
For the $u_i$ where $i> n_1$, it is known from (\ref{uigeneralB}) that
\[
u_i = \frac{\prod_{i=0}^k [(a_j - a_1)^{2^i} + u_2^{2^i}]}{(a_j - a_1)^{2^{k+1}}} \tilde X_i + \text{linear combination of }\{\tilde X_1, \dots, \tilde X_{n_1}, \tilde X_{i+1}, \dots, X_{s_j}\}.
\]
\noindent Using that $\tilde Y_j = \tilde X_j + O(u_1)$ and that $u_1 = \sum_{j=1}^{n_1} c_{1j} \tilde Y_j$,
\[
u_i = \frac{\prod_{i=0}^k [(a_j - a_1)^{2^i} + u_2^{2^i}]}{(a_j - a_1)^{2^{k+1}}}\tilde Y_i + \text{linear combination of }\{\tilde Y_1, \dots, \tilde Y_{n_1}, \tilde Y_{i+1}, \dots, Y_{s_j}\}.
\]
\noindent It follows that the diagonal entries for $C_j$ are the same as the diagonal entries for $B_j$ for $j>1$ and that $C$ has the same form of $B$ in (\ref{generalB}), which is sufficient to conclude $\det C = \det B$.
With (\ref{normallaplacian}), (\ref{normaltheta}), and (\ref{normaldet}), we have shown (\ref{normalwanttoshow}) and the proof is complete.
\end{proof}
\section{Appendix: Proof of Lemma \ref{lemmagk}}\label{appendix}
We now prove
\begin{lemma}[Lemma \ref{lemmagk}] Let
\[
G_k = \sum_{j=1}^m \sum_{i=0}^{n+1-k} \binom{n+1-k}{i} \sum_{\mu} \frac{(-1)^{n_j +\mu_j}}{a_j^{\mu_j+i+k-n}} \left( \prod_{l \neq j}^m \frac{\binom{n_l+ \mu_l - 1}{\mu_l}}{(a_l - a_j)^{n_l + \mu_l}} \right)
\]
\noindent Then
\[
G_k =
\begin{cases}
\frac{-1}{\det A} & k = n+1\\
0 & 1 < k < n+1\\
(-1)^n & k = 1
\end{cases}
\]
\end{lemma}
The proof is by integration of meromorphic differentials over the Riemann sphere, where the residues will correspond to contributions to $G_k$. If $k = n+1$, consider
\[
\psi_j = \sum_{\mu \perp n_j - 1} \frac{(-1)^{n_j +\mu_j}}{a_j^{\mu_j + \sum_{l \neq j} \mu_l} \cdot z^{1 - \sum_{l \neq j} \mu_l}(z - a_j)} \left( \prod_{l \neq j}^m \frac{\binom{n_l+ \mu_l - 1}{\mu_l}}{(a_l - z)^{n_l + \mu_l}} \right) \; dz.
\]
There are always poles at $z = a_1, \dots, a_m$. There is also a pole at $z = 0$ when $\sum_{l \neq j} \mu_l = 0$, i.e. the partition is $\mu_j = n_j - 1$ and $\mu_l = 0$ for $l \neq j$. There is never a pole at infinity. It is immediate that
\[
\sum_{j=1}^m \text{Res}_{a_j} \psi_{j} = G_{n+1}
\]
\noindent and
\[
\text{Res}_0 \psi_{j} = \frac{1}{\prod_{l=1}^m a_l^{n_l}} = \frac{1}{\det A},
\]
\noindent while standard power series expansion and a consideration of partition recovery shows
\[
\text{Res}_{a_j} \psi_{l} = \text{Res}_{a_j} \psi_{j}
\]
\noindent for $l \neq j$. As the sum of residues over a closed Riemann surface is zero,
\[
0 = \sum_{j=1}^m \left[\text{Res}_0 \psi_j + \sum_{l=1}^m \text{Res}_{a_l} \psi_j \right] = \frac{m}{\det A} + m\sum_{j=1}^m \text{Res}_{a_j} \psi_j = \frac{m}{\det A} + mG_{n+1}
\]
\noindent and the $k = n+1$ case follows.
For $1 < k < n+1$, instead use
\[
\psi_{j,k} = \sum_{i=0}^{n+1-k}\binom{n+1-k}{i} \sum_{\mu} \frac{(-1)^{n_j +\mu_j}}{a_j^{\mu_j + i + k-n + \sum_{l \neq j} \mu_l} \cdot z^{- \sum_{l \neq j} \mu_l}(z - a_j)} \left( \prod_{l \neq j}^m \frac{\binom{n_l+ \mu_l - 1}{\mu_l}}{(a_l - z)^{n_l + \mu_l}} \right) \; dz
\]
The changed power in the denominator results in poles at $a_1, \dots, a_n$, but never at $z = 0$ or $z = \infty$. It is again immediate that the contribution of interest is
\[
\sum_{j=1}^m \text{Res}_{a_j} \psi_{j,k} = G_k
\]
\noindent while power series expansion for this range of $k$ shows $\psi_{j,k}$ has the nice property
\[
\text{Res}_{a_j} \psi_{l,k} = \text{Res}_{a_j} \psi_{j,k}
\]
\noindent for all $l \neq j$ (this fails when $k=1$). As the sum of residues is zero,
\[
0 = \sum_{j=1}^m \left[\sum_{l=1}^m \text{Res}_{a_l} \psi_{j,k} \right] = m\sum_{j=1}^m \text{Res}_{a_j} \psi_{j,k} = mG_k
\]
\noindent and the second part of the lemma is established.
Finally for the $k= 1$ case, use
\[
\psi_j = \sum_{i=0}^n\binom{n+1-k}{i} \sum_{\mu \perp n_j - i- 1} \frac{(-1)^{n_j +\mu_j}}{z^{1 +i - n + \mu_j}(z - a_j)} \left( \prod_{l \neq j}^m \frac{\binom{n_l+ \mu_l - 1}{\mu_l}}{(a_l - z)^{n_l + \mu_l}} \right) \; dz.
\]
The term of interest is
\[
\sum_{j=1}^m \text{Res}_{a_j} \psi_j = G_1
\]
There is a pole at $z = \infty$ with $\text{Res}_{\infty} \psi_j = (-1)^{n+1}$, and again combinatorial manipulation shows
\[
\text{Res}_{a_j} \psi_j = \text{Res}_{a_j} \psi_l
\]
\noindent for all $l \neq j$. We have
\[
0 = \sum_{j=1}^m \left[\sum_{l=1}^m \text{Res}_{a_l} \psi_j + \text{Res}_{\infty}\psi_j \right] = m(G_1 + (-1)^{n+1})
\]
\noindent and the final $k = 1$ case follows.
\newpage
\bibliography{refs}
\bibliographystyle{acm}
\end{document}
| 70,666
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\begin{document}
\begin{abstract}
Let $X$ be a complex nonsingular projective 3-fold of general type.
We show that there are positive constants $c$, $c'$ and $m_1$ such that $\chi (\omega _X)\geq -c\Vol (X)$ and $P_m(X)\geq c'm^3\Vol (X)$ for all $m\geq m_1$.
\end{abstract}
\maketitle
\pagestyle{myheadings} \markboth{\hfill J. A. Chen and C. D. Hacon \hfill}{\hfill On the geography of threefolds of general
type\hfill}
\section{Introduction and known results}
The birational classification of surfaces of general type is well understood.
For example, it is known that if $X$ is a surface of general type then $|mK_X|$ induces a birational map for all $m\geq 5$.
As a general rule, it is not possible to classify surfaces of general type with given invariants.
In general, the best that one can do is to show that
the invariants of a surface $X$ satisfy certain
inequalities.
A fundamental inequality for the invariants of a minimal surface of general type is the Bogomolov-Miyaoka-Yau inequality $K_X^2\leq 9\chi (\OO _X)$.
It is a natural problem to try and extend the results for surfaces to higher dimensions. There have been many partial results for $3$-folds.
For example, it is shown in \cite{CCM} that if $X$ is a Gorenstein minimal $3$-fold of general type, then $|mK_X|$ induces a birational map for all $m\geq 5$.
In fact the proof is based upon the fact that for such $3$-folds, we have
the Miyaoka-Yau inequality $K_X^3\leq 72\chi (\omega _X)$.
Despite many partial results, the geometry of non-Gorenstein 3-folds of general type has proven to be a very challenging topic.
In a recent paper however, the first author and M. Chen \cite{CC} show the remarkable result that if $X$ is a smooth complex projective $3$-fold of general type,
then $P_{12}\geq 1$, $P_{24}\geq 2$ and $|mK_X|$ induces a birational map for all $m\geq 77$. It is then natural to hope that further precise results
on the geography of $3$-folds of general type may be within reach.
The purpose of this paper is to show that using the methods of \cite{CC}
one can in fact prove an inequality similar to the Miyaoka-Yau inequality
which holds for non-Gorenstein $3$-folds of general type. Namely we show that
\begin{thm}\label{T-1} There exists a constant $c>0$ such that for any minimal $3$-fold of general type with terminal singularities $X$, we have
$$\chi(\omega _X)\geq -c K_X^3.$$\end{thm}
Recall that for any minimal $3$-fold of general type with terminal singularities, we have ${\rm Vol}(X)=K_X^3$.
It should be noted that $\chi(\omega _X)$ may be negative
for $3$-folds of general type.
In fact consider curves $C_1$, $C_2$ and $C_3$ with genus $g_i$ and
involutions $\sigma _i$ such that $C_i/<\sigma _i >\cong \mathbb P ^1$.
Then the $3$-fold $X$ given by a desingularization of the quotient of
$C_1\times C_2\times C_3$ by the ``diagonal'' involution, has $\chi (\omega _X)<0$.
In fact if we let $g_1=g_2=g$, then for fixed $g_3$ and for $g\gg 0$ one has that
$-\chi (\omega _X)=O(g^2)$ and $K_X^3=O(g^2)$. So the inequality of Theorem
\ref{T-1} has the right shape. The constant $c$
that may be computed with the methods of this paper and the results of
\cite{CC} is $c=32\cdot 120^3$. We expect that this is far from optimal
and so we make no effort to determine it explicitly.
We remark that if $\Vol(X)\gg 0$, then using the results of \cite{T}, one
can recover $c=2502$.
We also prove the following result concerning the plurigenera of $X$.
\begin{thm}\label{T-2} There exist constants $c'>0$ and $m_1>0$ such that for any minimal $3$-fold of general type with terminal singularities $X$, we have $$P_m(X)\geq c'm^3K_X^3\qquad {\rm for\ all}\ m\geq m_1.$$
\end{thm}
Once again the values of $c'$ and $m_1$ that may be computed with the methods of this paper are far from optimal so we make no effort to determine their values (note that using the results of \cite{CC}, it follows form the arguments below that $c'=\frac 5{89168}$ and $m_1=112$ suffice).
It would be interesting to: \begin{enumerate}
\item Determine the optimal value of $m_1$ in Theorem \ref{T-2};
\item See if Theorem \ref{T-2} can be recovered by using
the methods of \cite{T};
\item See if Theorems \ref{T-1} and \ref{T-2} hold in higher dimensions.\end{enumerate}
We remark that the proof of the results of this paper is based on the methods of \cite{CC}.
We have chosen to keep the exposition of this paper as self contained and simple as possible. Therefore, we include a proof of all the results
of \cite{CC} that we will use (namely inequalities (1) and (2)).
\section{Some inequalities}
In this section, we will prove the following inequalities: \begin{thm}Let $X$ be a minimal $3$-fold of general type with terminal singularities, then $$P_4+P_5+P_6-3P_2-P_3-P_7 \ge 0.
\eqno(1)$$ and
$$ 2P_5+3P_6+P_8+P_{10}+P_{12} \ge \chi(\OO _X) +10P_2+4 P_3
+P_7+P_{11}+P_{13}+14\sigma _{12} \eqno(2)$$
where $\sigma _{12}$ is a positive integer that will be defined below.\end{thm}
Note that a stronger version of the above inequalities is proved in \cite{CC}.
Here we include a simpler and self contained proof of this (weaker) version of the inequalities of \cite{CC}. The stronger version also follows from
the methods of this paper, however it is not necessary for our purposes so we
have chosen not to include it here.
We consider now $X$ a minimal $3$-fold of general type with
terminal singularities. According to Reid (see last section of
\cite{YPG}), there is a ``basket'' of pairs of integers
$\mathscr{B}(X):=\{(b_i, r_i)\}$ such that the Riemann-Roch formula may be written as
$$\chi(\mathcal{O}_X(mK_X))=\frac{1}{12}m(m-1)(2m-1)K_X^3-(2m-1)\chi(\mathcal
{O}_X)+l(m), $$ where the correction term $l(m)$ is computed
by:
$$l(m):=\sum_{Q_i\in \mathscr{B}(X)}l_{Q_i}(m):=\sum_{Q_i\in \mathscr{B}(X)}\sum_{j=1}^{m-1}
\frac{\overline{jb_i}(r_i-\overline{jb_i})}{2r_i}.$$ Here, we assume that
$b_i$ is
co-prime to $r_i$ and $0<b_i\leq \frac{r_i}{2}$. The ratio $\frac{b_i}{r_i}$ is
called the {\it slope of $(b_i, r_i)$}.
For a basket $B$, we let $\sigma (B):=\sum b_i$ and $\sigma_{12} (B):=\sum_{\frac{b_i}{r_i}\leq \frac 1 {12}} b_i$.
Let
$$\overline{M}^j(b,r):=\frac{\overline{jb}(r-\overline{jb})}{2r}, \quad
{M^j}(b,r):=\frac{{jb}(r-{jb})}{2r},$$
$$\Delta^j({b,r}):=\overline{M}^j(b,r)-M^j(b,r).$$ An easy computation
shows that $\Delta^n({b,r})=ibn-\frac{i^2+i}{2}r$, where $i=\lfloor
\frac{bn}{r} \rfloor$.
We will need the following easy computational lemmas.
\begin{lem} \label{nodiff} Let $b_1r_2-b_2r_1=1$. If $0<n\ne xr_1+yr_2$
for any integers $x,y >0$, then there is no rational number $\frac{b}{n} \in (\frac{b_2}{r_2},\frac{b_1}{r_1})$ and we have
$$\Delta^n({b_1+b_2,r_1+r_2})=\Delta^n({b_1,r_1})+\Delta^n({b_2,r_2}).$$
\end{lem}
\begin{proof} We may assume that $\frac {b_2}{r_2}<\frac{b_1}{r_1}$.
Note also that by our assumptions, we have $n\leq r_1r_2$.
If $\frac{b}{n} \in (\frac{b_2}{r_2},\frac{b_1}{r_1})$, then $n= (br_2-b_2n)r_1+(b_1n-br_1)r_2$ with $br_2-b_2n>0$ and $b_1n-br_1 >0$.
Hence, as $n \ne xr_1+yr_2$ for all $x,y >0$, then there is no rational number $\frac{b}{n} \in (\frac{b_2}{r_2},\frac{b_1}{r_1})$.
Let $i_1:=\lfloor
\frac{b_1n}{r_1} \rfloor$, $i_2:=\lfloor
\frac{b_2n}{r_2} \rfloor$ and $i:=\lfloor
\frac{(b_1+b_2)n}{r_1+r_2} \rfloor$. If $\frac{b_1n}{r_1}$ is not an integer, then $$i_2=\lfloor
\frac{b_2n}{r_2} \rfloor \le i_1=\lfloor
\frac{b_1n}{r_1} \rfloor<\frac{b_1n}{r_1}.$$ If $i_1\ne i_2$
then $i_1>\frac {b_2n}{r_2}$ so that $\frac{i_1}n\in (\frac{b_2}{r_2},\frac{b_1}{r_1})$ which is impossible. Therefore $i_1=i_2=i$ and the statement follows from the equation $\Delta=ibn-\frac{i^2+i}2r$.
If $\frac{b_1n}{r_1}$ is an integer, then one sees that $\Delta ^n(b_1,r_1)
=(i_1-1)b_1n-\frac{(i_1-1)^2+(i_1-1)}2r_1$ so that as $i_2=i_1-1$, the statement follows from the definition of $\Delta$.
\end{proof}
\begin{lem} \label{diff} Suppose that $b_1r_2-b_2r_1=1$ and
suppose that $n=xr_1+yr_2$ for some integers $r_2 \ge x >0$, $r_1 \ge
y >0$, then
$$\Delta^n({b_1+b_2,r_1+r_2})=\Delta^n({b_1,r_1})+\Delta^n({b_2,r_2})-\min\{x,y\},$$
\end{lem}
\begin{proof}
We first remark that the expression $n=xr_1+yr_2$ for some integers $r_2 \ge x >0$, $r_1 \ge
y >0$ is unique.
Let $i=xb_1+yb_2$.\\
An easy computation shows that if $r_1\ne y$, then
$$\lfloor \frac{b_1n}{r_1} \rfloor = \lfloor \frac{xb_1r_1+yb_1r_2 }{r_1} \rfloor =\lfloor \frac{xb_1r_1+yb_2r_1+y }{r_1} \rfloor =i+\lfloor \frac{y }{r_1} \rfloor=i,$$
$$\lfloor \frac{b_2n}{r_2} \rfloor = \lfloor \frac{xb_2r_1+yb_2r_2 }{r_2} \rfloor =\lfloor \frac{xb_1r_2-x+yb_2r_2 }{r_2} \rfloor =i+\lfloor \frac{-x }{r_2} \rfloor=i-1,$$
$$\lfloor \frac{(b_1+b_2)n}{r_1+r_2} \rfloor = \lfloor \frac{xb_1r_1+yb_1r_2+xb_2r_1+yb_2r_2 }{r_1+r_2} \rfloor $$
$$=\lfloor \frac{xb_1r_1+yb_2r_1+y+xb_1r_2-x+yb_2r_2 }{r_1+r_2} \rfloor =i+\lfloor \frac{y-x }{r_1+r_2} \rfloor.$$
If $y \ge x$, then $\lfloor \frac{(b_1+b_2)n}{r_1+r_2} \rfloor=i$. Direct computation gives
$$ \Delta^n({b_1+b_2,r_1+r_2})-\Delta^n({b_1,r_1})-\Delta^n({b_2,r_2}) $$
$$=i(b_1+b_2)n-\frac{i^2+i}{2}(r_1+r_2)-ib_1n + \frac{i^2+i}{2}r_1 -(i-1)b_2n+\frac{i^2-i}{2}r_2$$
$$ = b_2n-ir_2 = b_2(xr_1+yr_2)-(xb_1+yb_2)r_2=x(b_2r_1-b_1r_2)=-x.$$
Note that if $r_1=y$, then $\lfloor \frac{b_1n}{r_1} \rfloor =i+1$.
However, one easily sees that
the above formula is unchanged.
If $y \le x$, the computation is similar.
\end{proof}
\begin{proof}[Proof of inequality (1)]
By direct computation, one finds that the $K_X^3$ and $\chi(\OO _X)$ terms coming from the Riemann-Roch formula cancel.
Inequality (1) is then equivalent to
$$ -3 l(2)-l(3)+l(4)+l(5)+l(6)-l(7) \ge 0.$$
Since $l(m)=\sum_{j=1}^{m-1} \overline{M}^j(B)=\sum_{j=1}^{m} \sum
\overline{M}^j(b_i,r_i)$, we must show the inequality
$$
\overline{\Xi}(B):=-2\overline{M}^1(B)+\overline{M}^2(B)+2\overline{M}^3(B)+\overline{M}^4(B)-\overline{M}^6(B)
\ge 0. \eqno(3)$$
We will show that this holds for any single basket $(b,r)$
and hence for any basket
$B$.
We define $$\Xi(B):=-2{M^1}(B)+{M^2}(B)+2{M^3}(B)+{M^4}(B)-{M^6}(B),$$
$$\Xi\Delta(B):=\overline{\Xi}(B)-\Xi(B)=2{\Delta^3}(B)+{\Delta^4}(B)-{\Delta^6}(B),$$
where we have used the fact that as we assumed that $b/r\leq 1/2$, then
${\Delta^1}(B)={\Delta^2}(B)=0$.
{\bf Step 1.} For any single basket $B=\{(b,r)\}$, we have
$\Xi(B)=2b$.
{\bf Step 2.} For the single basket $B=\{(1,2)\}$, we have $\Delta^3({1,2})=1,\Delta^4({1,2})=2$ and $\Delta^6({1,2})=6$.
Hence
$\Xi\Delta(B)=-2$, and $\overline{\Xi}(B)=0$. A similar computation for
$B=\{(1,3)\}$ or $B=\{(1,4)\}$, yields $\Xi\Delta(B)=-2$ and $\overline{\Xi}(B)=0$. When $B=\{(1,5) \}$, then $\Xi\Delta(B)=-1$
and $\overline{\Xi}(B)=1$.
{\bf Step 3.} When $B=\{(1,r)\}$ with $r \ge 6$, we have $\overline{M}^m(1,r)=M^m(1,r)$ for all $m \le 6$. Hence $\overline \Xi(B)=\Xi(B)=2$.
{\bf Step 4.}
Recall that we are assuming $\frac{b}{r} \le \frac{1}{2}$.
Let $S=\{\frac{1}{r}\}_{r \ge 2}$, $S^{(5)}:=S \cup \{\frac{2}{5}\}$ and
for $n\geq 6$ set
$$S^{(n)}=S^{(n-1)} \cup \{\frac{b}{n}| (b,n)=1,0< \frac{b}{n} \le \frac{1}{2} \}.$$ For any $\frac{b}{n} \in S^{(n)}$, let $[0;a_1,...,a_t]$ be its continued fraction expression. Note that as $\frac bn\leq \frac 12$, then $a_1\geq 2$.
If $t>1$, we may consider the rational number
$\frac{b_1}{r_1}$ with continued fraction expression $[0;a_1,...,a_{t-1}]$.
We have that $nb_1-r_1b = \pm 1$ and $\frac{b_1}{r_1}\leq \frac 12$.
Let $b_2=b-b_1, r_2=n-r_1$. Notice that we also
have $b_1r_2-b_2r_1=\pm 1$ and $\frac{b_2}{r_2}\leq \frac 12$.
Then we have $\frac{b_1}{r_1}, \frac{b_2}{r_2} \in S^{(n-1)}$.
{\bf Step 5.} We proceed by showing by induction on $r$ that inequality (3) holds. By Step 1, this is equivalent to showing that
$\Xi \Delta (b,r)\geq -2b$.
We have seen that the inequality (3) holds for $r \le 4$. For $r=5$, we must
consider the single basket $B=\{(2,5)\}$. Notice that
$\Delta^n({2,5})=\Delta^n(1,2)+\Delta^n({1,3})$, for $n=3,4,6$ by Lemma \ref{nodiff}. We see that
$$\Xi\Delta(2,5)=\Xi\Delta(1,2)+\Xi\Delta(1,3)=-4.$$
By Step 1, we have $\overline{\Xi}(2,5)=0$.
For $r=6$, there are no new baskets to consider.
{\bf Step 6.}
For $r \ge 7$, notice that by Step 4, we may assume that $(b,r)=(b_1,r_1)+(b_2,r_2)$ for some
$r_1,r_2 < r$ and (after possibly switching indices) that $b_1r_2-b_2r_1=1$. By induction hypothesis, we have
$\Xi\Delta(b_i,r_i) \ge -2 b_i$. Using Lemma \ref{nodiff}, it is easy to see that
$$\Delta^m(b,r)=\Delta^m(b_1,r_1)+\Delta^m(b_2,r_2)$$for $m \in \{ 3,4,6\}$.
Hence $$\Xi\Delta(b,r)=\Xi\Delta(b_1,r_1)+\Xi\Delta(b_2,r_2) \ge -2b.$$
This completes the proof.
\end{proof}
\begin{proof}[Proof of inequality (2)]
The proof is similar but the computations are a little bit more involved.
Inequality (2) is equivalent to
$$ -10 l(2)-4 l(3)+2 l(5)+3 l(6)-l(7)+l(8)+l(10)-l(11)+l(12)-l(13) \ge 14\sigma _{12},$$
which in turn is equivalent to
$$
\overline{\Xi}(B) := -9\overline{M}^1(B)+\overline{M}^2(B)+5\overline{M}^3(B)+5\overline{M}^4(B)$$
$$\qquad +3\overline{M}^5(B)
+\overline{M}^7(B)-\overline{M}^{10}(B)-\overline{M}^{12}(B) \ge 14\sigma _{12}(B). \eqno(4)$$
We will show that this holds for any single basket and hence for any
basket $B$.
We define $\Xi(B)$ and $\Xi\Delta(B)$ as in the proof of inequality (1).
{\bf Step 1.} For any single basket $B=\{(b,r)\}$, we have
$\Xi(B)=14b$.
{\bf Step 2.} For a single basket $B=\{(1,r)\}$ with $2\leq r \le 11$,
direct computation gives $\overline{\Xi}(1,r)=0,0,0,2,5,6,8,10,12,13$.
{\bf Step 3.} We claim that if $B=\{(b,r)\}$ with $\frac b r\leq \frac 1{12}$, then
$\overline \Xi(B)=14b$.
When $B=\{(1,r)\}$ with $r \ge 12$, we have $\overline{M}^m(1,r)=M^m(1,r)$ for all $m \le 12$, therefore $\overline \Xi(B)=\Xi(B)=14$.
When $B=\{(b,r)\}$ with $\frac b r<\frac 1{12}$ and $b>1$, as in the proof of
inequality (1) Step 4, we
may write $b=b_1+b_2$ and $r=r_1+r_2$ where $b_i$ and $r_i$ are co-prime,
$\frac{b_1}{r_1}>\frac b r > \frac{b_2}{r_2}$ and $b_1r_2-b_2r_1=1$.
By Lemma \ref{nodiff}, we see that as $r=r_1+r_2>12$, then
$\frac 1 {12}\not\in(\frac{b_2}{r_2},\frac{b_1}{r_1})$.
It follows that $\frac{b_1}{r_1}\leq \frac 1{12}$. The claim now follows by induction. In fact, since $r>12$, by Lemma \ref{nodiff}, we have
$$\Delta ^m(b,r)=\Delta ^m(b_1,r_1)+\Delta ^m(b_2,r_2)$$ for all $1\leq m\leq 12$.
We proceed by showing by induction on $r$ that $\Xi \Delta (b,r)\geq -14b$. By Step 1, this is equivalent to $\overline{\Xi} (b,r)\geq 0$ and hence implies that inequality (4) holds.
{\bf Step 4.} $\overline{\Xi} (b,r)\geq 0$ for all baskets $(b,r)$ with $r\leq 12$.
By Step 1, $\overline{\Xi} (b,r)\geq 0$ for all baskets $(b,r)$ with $r \le 4$. For $r=5$, we must
consider the single basket $B=\{(2,5)\}$. By Lemmas \ref{nodiff} and \ref{diff},
one sees that
$$\Delta^n({2,5})-\Delta^n(1,2)-\Delta^n({1,3})=-1,-1,-2,-2\quad {\rm for}\ n=5,7,10,12$$ respectively and $\Delta^n({2,5})-\Delta^n(1,2)-\Delta^n({1,3})=0$
for $n=1,2,3,4$. It follows that
$$\Xi\Delta(2,5)=\Xi\Delta(1,2)+\Xi\Delta(1,3).$$
By Steps 1 and 2, we have $\overline{\Xi}(2,5)=0$.
For $r=6$, there are no new baskets to consider.
We can similarly compute all single baskets $B\in S^{(12)}-S^{(6)}$. Recall that each single basket $(b,r)$,
can be compared with pairs $(b_1,r_1)$ and $(b_2,r_2)$ as described in Step 4 of the proof of inequality (1).
We have that $\Xi\Delta(b,r)\geq \Xi\Delta(b_1,r_1)+\Xi\Delta(b_2,r_2)$ for all
$B\in S^{(12)}-S^{(6)}$ or more precisely that
$\Xi\Delta(b,r)=\Xi\Delta(b_1,r_1)+\Xi\Delta(b_2,r_2) +1$ if $(b,r)\in \{(3,10),(5,12)\}$ and $\Xi\Delta(b,r)=\Xi\Delta(b_1,r_1)+\Xi\Delta(b_2,r_2)$
otherwise. Therefore $\Xi\Delta(b,r)\geq -14b$ for all baskets $(b,r)$ with $r\leq 12$.
{\bf Step 5.} $\overline{\Xi} (b,r)\geq 0$ for all baskets $(b,r)$ with
$r \ge 13$ and $\frac b r >\frac 1{12}$.
We may assume that $(b,r)=(b_1,r_1)+(b_2,r_2)$ for some
$r_1,r_2 < r$. By induction hypothesis, we have
$\Xi\Delta(b_i,r_i) \ge -14 b_i$. By Lemma \ref{nodiff}, we have
$$\Delta^m(b,r)=\Delta^m(b_1,r_1)+\Delta^m(b_2,r_2),$$
for $m \le 12$. Hence $$\Xi\Delta(b,r)=\Xi\Delta(b_1,r_1)+\Xi\Delta(b_2,r_2) \ge -14b.$$
This completes the proof.
\end{proof}
We will also need the following equality
\begin{lem}\label{l-6} For any minimal $3$-fold of general type with terminal
singularities and basket $B$ we have $\sigma (B)=10\chi (\OO _X)+5P_2(X)-P_3(X)$.
\end{lem}
\begin{proof} The equality follows immediately from the Riemann-Roch formula.
\end{proof}
\section{Main result}
In this section we prove the main result of this paper.
\begin{thm}
Let $X$ be a smooth $3$-fold of general type. Then
\begin{enumerate}
\item There are constants $c'>0$ and $m_1>0$ such that $P_m(X) \ge c' m^3\Vol(X)$ for all $m \ge m_1$.
\item There is a constant $c>0$ such that $\Vol(X) \ge c \chi(\OO_X)$.
\end{enumerate}
\end{thm}
\begin{proof}
We will first prove (1). Consider the Riemann-Roch formula.
If $\chi(\OO_X) \le 0$. Then we get
$$P_m \ge \frac{m(m-1)(2m-1)}{12} \Vol(X) \ge \frac{m^3}{16}\Vol(X)$$ for $m \ge 2$ already.
It remains to consider the case when $\chi(\OO_X) >0$. We will need the following.
\begin{lem} There exist constants $m_0,c_1,c_2 >0$ such that
\begin{enumerate} \item $P_{m_0}\geq 2$,
\item $P_m\geq 2$ for all $m\geq 5m_0+6$,
\item $P_m \ge c_1{m}$ for all $m\geq 12m_0+10$
and \item $P_m \ge \frac{c_2 m}{t} P_t$ for any $m \ge 10m_0+2t+10$.\end{enumerate}\end{lem}
\begin{proof} We will repeatedly use the fact that if $P_s>0$ and $P_t>0$, then $P_{s+t}\geq P_s+P_t-1$ and so for all $s\geq t_0=5m_0+6$ and any $t'>0$ such that $P_{t'}\geq 2$, we have
$$P_s> \lfloor \frac{s-t_0}{t'}\rfloor (P_{t'}-1)\geq \frac{s-t_0-t'+1}
{t'}(P_{t'}-1).$$
(1) If $P_i\leq 1$ for $i\in \{ 5,6,8,10,12 \}$, then by inequality (2), we have
$0< \chi(\OO _X)\leq 8$ and $\sigma _{12}=0$.
Since $\sigma =10\chi(\OO _X)+5P_2-P_3$ (cf. Lemma \ref{l-6}), we have
$\sigma=\sum b_i \leq 85$.
Therefore as $\sigma _{12}=\sum _{\frac {b_i}{r_i}\leq \frac 1 {12}}b_i=0$,
there are only finitely many possible
such baskets of singularities and hence there is an integer $m_0$ such that $P_{m_0}(X)\geq 2$.
We may assume that $120$ divides $m_0$.
(2) If $P_i\ge 2$ for some $i\in \{ 5,6,8,10,12 \}$, then we have $P_{120}(X)\geq 2$.
Therefore, if $\chi (\OO _X)>0$, then
$P_{m_0}(X)\geq 2$. By \cite{MC} we have that $|mK_X|$ is birational for all $m\geq 5m_0+6$.
(3) It follows that for all $m\geq 12m_0+10$ we have
$$P_m> \frac{m-6m_0-5}{m_0}(P_{m_0} -1)\geq \frac {m}{2m_0}.$$
(4) If $P_t=0$, the proposed inequality is trivial. If $P_t=1$, the proposed inequality follows from (3) assuming that $c_2\leq c_1$. We now assume that
$P_t\geq 2$ and hence $P_t-1\geq P_t/2$. We have that for $m\geq 10m_0+2t+10$,
$$P_m> \frac{m-5m_0-t-5}{t}(P_t -1)\geq \frac {m} {4t} P_t.$$
\end{proof}
We have that $2P_5+3P_6+P_8+P_{10}+P_{12} \ge \chi(\OO_X)$.
Hence $$(2\frac{5}{c_2 m} + 3 \frac{6}{c_2 m} +\frac{8}{c_2 m}+\frac{10}{c_2 m} + \frac{12}{c_2 m})P_m=\frac{58}{c_2m} P_m\ge \chi(\OO_X)$$
for any $m\geq 10m_0+34$.
Thus, by the Riemann Roch formula $$ (1+\frac{116}{c_2})P_m \ge P_m+2m \chi(\OO_X) \ge \frac{m^3}{16}\Vol(X).$$
This proves the first inequality.
The second inequality holds trivially if $\chi(\OO_X) \le 0$. Hence we assume that $\chi(\OO_X) >0$.
If $P_5,P_6,P_8,P_{10},P_{12} \le 1$, then $\chi(\OO_X) \le 8$.
Since $|(5m_0+6)K_X|$ is birational, then $\Vol (X)\geq \frac 1 {(5m_0+6)^3}$.
Therefore,
$$ \Vol(X) \ge \frac{1}{ (5m_0+6)^3 } \ge \frac{1}{(5m_0+6)^3 \cdot 8} \chi(\OO_X).$$
In general, we have $P_{120} \ge P_t$ for $t\in \{5,6,8,10,12\}$.
Hence $8P_{120} \ge \chi(\OO_X)$. We may assume that $P_{t}\geq 2$ for some $t\in \{5,6,8,10,12\}$ so that
$|120K_X|$ is birational. Therefore
$120^3\Vol(X) \ge P_{120}-3\geq 1$, hence $$4\cdot 120^3\Vol(X) \ge 120^3\Vol(X)+3 \ge P_{120} \ge \frac{1}{8} \chi(\OO_X).$$
\end{proof}
| 220,580
|
\begin{document}
\title{ Singularity Knots of Minimal Surfaces in $\mathbb{R}^4$}
\author{ Marc Soret \& Marina Ville }
\date{ October 20, 2006}
\maketitle
\begin{abstract}
We study knots in $\mathbb{S}^3$ obtained by the intersection of a
minimal surface in $\mathbb{R}^4$ with a small $3$-sphere centered
at a branch point. We construct examples of new minimal knots. In
particular we show the existence of non-fibered minimal knots. We
show that simple minimal knots are either reversible or fully
amphicheiral; this yields an obstruction for a given knot to be an
iterated knot of a minimal surface. Properties and invariants of
these knots such as the algebraic crossing number of a braid
representative and the Alexander polynomial are studied.
\end{abstract}
\medskip
\section{Introduction}
We wish to understand knots associated to specific singularities. Let us
first recall what is meant by singularity of a minimal surface. Let $D$ be
a disk endowed with a Riemann (complex) structure and let $0$ be its center.
Let $X : D \mapsto \mathbb{R}^4$ be a conformal harmonic mapping. If $
dX(0) =0,$ we say that $0$ is a singularity of $X$. It is then clear that
singularities are isolated and correspond to branch points. Let $p = X(0).$
The topology of the singularity at $p$ is entirely determined by a
(possibly singular) knot or link which is obtained as follows : intersect $M
$ with a small 3-sphere of $\mathbb{R}^4$ around p of radius $\epsilon$.
We obtain a curve $K_\epsilon \subset\mathbb{S}^3(\epsilon)$. If $p$ is the
image of only one singularity then $K_\epsilon$ has only one connected
component for $\epsilon $ small enough. If $X(D)$ is embedded in a
neighborhood of $p$, then $K_\epsilon$ is a smoothly embedded curve and
hence is a knot, isotopic to a fixed knot $K$ and $X(D) \cap B(0,\epsilon)$
is topologically a cone over $K$. For example, a holomorphic complex curve
in $\mathbb{R}^4 = \mathbb{C}^2$ is a special case of minimal surface in $
\mathbb{R}^4$. When $X(D)$ is locally determined by the equation $F(z,w) =0$
, equivalent germs of $F$ at p yield isotopic knots in $\mathbb{S}^3$ (cf.
\cite{BK}).
\subsubsection{ Minimal Knots and iterated minimal knots}
As is usual in this context, minimal singularities are given by an
expansion in terms of $z$ and $\bar z$ and one needs to fix some definitions.
We remind the reader that a map $X:D\mapsto\mathbb{C}^2$ from the unit disk of
$\mathbb{C}$ into $\mathbb{C}^2$ is minimal
if and only if $X$ is harmonic with respect to the induced metric
on $D$. However, since a harmonic map from a surface remains
harmonic if we change conformally the metric, $X$ is a minimal if
$X$ is harmonic with respect to the flat metric on $D$ and if $X$ is
conformal; thus harmonicity means that
\begin{equation*}
\Delta_D X = 4 \partial_z \partial_{\bar z} X =0. \qquad ({\mathcal{H}})
\end{equation*}
Let $ z = x+iy$; then conformality means that
\begin{equation*}
\|\frac{\partial X}{\partial x}\|= \|\frac{\partial X}{\partial y}\|,\ \ \ <
\frac{\partial X}{\partial x},\frac{\partial X}{\partial y}>=0. \qquad ({
\mathcal{C}})
\end{equation*}
It is straightforward to derive many examples of germs of minimal surfaces:
\begin{defn}
Let $D$ be the unit disk in $C$ and let $X:D\longrightarrow \mathbb{R}^4$
and suppose that $X(0)=0$. The origin $0$ is a branch point
for $X$ if and only if there is a holomorphic coordinate in $D$ and
a coordinate system on $\mathbb{R}^4$ such that $X$ writes in a
neighbourhood of $0$
$$z\mapsto \left(Re(z^N)+o(|z|^N),\ Im (z^N)+ o(|z|^N),\ o(|z|^N),\ o(|z|^N)\right).$$
\end{defn}
We now assume that $X$ is injective. We denote by $S_\epsilon$
(resp. $B_\epsilon$) the sphere (resp. ball) in ${\bf R}^4$ centered
at $0$ and of radius $\epsilon$. We put $K_\epsilon=S_\epsilon\cap
X(D)$. If $\epsilon$ is small enough
the following two facts are true\\
1) $(B_\epsilon, X(D))$ is a cone over $(S_\epsilon, K_\epsilon)$\\
2) all the knots $K_\eta=S_\eta\cap X(D)$ with $\eta\leq\epsilon$ are isotopic. The
ensuing knot type is said to be {\it associated} to the singularity of $X$ at $0$.\\
It follows from the implicit function theorem that there exists a real function
$r_\epsilon$ such that $K_\epsilon$ is parametrized by $X(r_\epsilon(\theta)e^{i\theta})$,
with $i\theta$ going through $\mathbb{S}^1$.\\
If $X$ is a holomorphic map to $\mathbb{C}$, all its singularities are
branch points. It is a classical result that
the associated knots are iterated torus knots (cf. \cite{BK} or \cite{M}).\\
Holomorphic curves are a special case of area-minimizing surfaces (Wirtinger inequality);
these in turn are a special case of minimal surfaces (i.e. conformal harmonic mappings
of surfaces). \\
Micallef and White in \cite{MW} have researched branch points of minimal surfaces inside
general Riemannian $4$-manifolds. They have shown
that the area minimizing case is strikingly similar to the holomorphic one: the
associated knots are iterated torus knots. In the course of their investigations of
general minimal
surfaces they showed that
other knot types can occur, for example
\begin{equation}
\begin{array}{c}
D \longrightarrow \mathbb{C}\times \mathbb{C} \\
z \mapsto \left( z^3+ o\left(|z|^3\right), z^4-\bar z^4 +z^5+\bar
z^5+ o\left(
|z|^5\right)\right) \\
\end{array}
\end{equation}
They noticed that the corresponding minimal knot is the square knot (see
fig.3).\newline
Micallef and White left open the following question: can every knot isotopy type can be realized as the
knot of a minimal branch point?\\
We investigate in the present paper a specific class of knots of branched points of
minimal disks. They are given by the following Proposition which is inspired by a remark
in \cite{MW}
\begin{prop}
\label{cle}
Let $N$, $p$ and $q$ be integers, $p>N$, $q>N$ and let $\phi$ be a real number.
Suppose that the following map
$$z\mapsto \left(Re(z^N), Im(z^N), Re(e^{i\phi}z^p), Im(z^q)\right)$$
is injective. We denote by $K(N,p,q,\phi)$ the associated
knot type.\\
Then $K(N,p,q,\phi)$ is associated to a branch point of a minimal disk.
\end{prop}
{\bf Proof}. We identify $\mathbb{R}^4$ to $\mathbb{C}^2$ and we look for a map
$X$ of the form $X:z\mapsto (z^N+\bar{f}(z),g(z)+\bar{h}(z))$ where
$f$, $g$ and $h$ are holomorphic functions. Such a map is harmonic;
it is moreover conformal if and only if it satisfies equations $(\mathcal{H})$.
This translates into
$$f'(z)=-\frac{1}{z^{N-1}}g'(z)h'(z).$$
If $g(z)=z^p+z^q$ and $h(z)=z^p-z^q$, the function $f$ exists and verifies
$$|f(z)|=o(|z|^N).$$
In a neighbourhood of $0$, there exists a
function $\phi(z)$ such that\\
$\phi(z)^N=1+\frac{f(z)}{z^N}$. We put $w=z\phi(z)$; we have $w=z+o(|z|)$.\\
Also $w^N=z^N\phi(z)^N$ and $X$ is of the form
$$X(z)=\left(Re(w^N), Im(w^N), Re[(1+a(w))e^{i\phi}w^p], Im[(1+b(w))w^q]\right)$$
where $a(w)$ and $b(w)$ are $o(1)$. We put
$$X_t(w)=\left(Re(w^N), Im(w^N), Re[(1+ta(w))e^{i\phi}w^p], Im[(1+tb(w))(w^q)]\right).$$
We notice the following obvious
\begin{lem}\label{racine}
Let $w$ and $w'$ be complex numbers, $w\neq w'$, which verify \\
$X_t(w)=X_t(w')$.Then there exists $\nu$ a $N$-th root of $1$, $\nu\neq 1$, such that
$w=\nu w'$
\end{lem}
In particular $w=w'$ . Thus, if we let
$w=r e^{i\theta}$ we will know that $X_t$ is injective if and only the following maps is
injective
$$\tilde{X}_t: D-\{0\}\mapsto \mathbb{R}^4$$
$$\tilde{X}_t: w=re^{i\theta}\mapsto (\cos N\theta, \sin N\theta,
Re[(1+ta(w))e^{i\phi}e^{pi\theta}], Im[(1+tb(w))e^{qi\theta}].$$
The value $X_0(re^{i\theta})$ does not depend on $r$ but only on $\theta$, which runs
through the compact space $\mathbb{S}^1$. We derive therof the existence of a positive real number
$C>0$ such that for every $w, \nu$, with $w\neq 0$, $\nu^N=1$ and $\nu\neq 1$,
$$|X_0(w)-X_0(\nu w)|>C.$$
It follows that for $t\in[0,1]$ and $w$ non zero small enough,
$$|X_t(w)-X_t(\nu w)|>\frac{C}{2}.$$
This, together with Lemma \ref{racine} above, proves that the $X_t$'s are injective on a small disk
around $0$. Moreover for $\epsilon$ small enough, $X_t(D)\cap S_\epsilon$ constitutes
an isotopy between the knots associated to the singularities of $X_0$ and $X$.\\
Note that if $q=p$ (and $\phi=1$ but we will see later that this condition is not
necessary), then $K(N,p,q,\phi)$ is a $(N,p)$ torus knot.\\
We call a knot of the type $K(N,p,q,\phi)$ {\it simple minimal}. The word {\it simple} comes
from the fact that general knots of minimal surfaces can be seen as iterated versions of
the $K(N,q,p,\phi)$'s. We plan to devote another paper to these more general knots. For
the moment we focus on simple minimal knots.
Notice that simple minimal knots are similar in
their expression to Lissajous knots \cite{BH}, \cite{JP}, ( see
also KnotPlot \cite{KP} for a generator of Lissajous knots ), and
is a good help in the understanding of the latter ones. Still their
properties are quite different as we shall see.
Let us give a brief outline of the paper.
In section 1, we give a general description of the properties of simple
minimal knots which are parametrized by three numbers and a phase $
K(N, p, q, \phi)$ ; we then give a geometric interpretation of $N,
p, q$ and description of the braid representation that is naturally
attached to them.
In section 2, the minimal braid representation is studied in more
detail and we show that simple knots are invariant by a change of
phase. Minimal knots will be denoted by $K(N, p, q)$.
In section 3, we study the symmetries of minimal knots. Recall that
there are two natural symmetries among knots that are involutions;
the mirror symmetry $s_m$, (symmetry of a knot with respect to a
orientation reversing symmetry of $\mathbb{S}^3$ ) and the inversion
of a knot $s_i$ which maps a knot to the same knot but with the
reverse orientation. $K$ is invertible if it is invariant by $s_i$
and amphicheiral if it has some invariance with respect to $s_m$
Results of section 2 and easy computations yields
\begin{theorem}
A simple minimal knot is either reversible or fully amphicheiral. More
precisely :
\begin{enumerate}
\item All knots $K(N,p,q)$ are strongly invertible.
\item If $p+q$ is odd, then $(N,p,q) $ is strongly +
amphicheiral:
\item If $N$ is even and $p+q$ is even or if $N$ is odd and $p$ and $q$
even, then $K(N,p,q) $ is periodic of order two. The $S^1$ curve by the involution
invariant has a linking number with $K(N,p,q) $ equal to $N$.
\end{enumerate}
\end{theorem}
By a result of \cite{H} and \cite{Mu}, and similarly to the
Lissajous knots (cf. \cite{BH}), these symmetries yields properties
on the Arf invariant and Alexander polynomial that are described in
the section.
Section 3 leads to section 4 where we show the existence of knots that can
not be realized as simple or iterated knots of minimal singularities. These
are the negative amphichireal or chiral knots.
\begin{theorem}
A negative amphicheiral or chiral knot can not be the knot of a simple
minimal or iterated minimal knot
\end{theorem}
The first candidate in the Rolfsen classification is the knot $8_{17}$ which
is the first negative amphicheiral knot, and $9_{32}$ which is the first
chiral knot.
We do not yet know if these chiral knots can be realized as cable
knots of minimal singular knots. This question will be treated
elsewhere.
Section 6 is devoted to the algebraic crossing number of the natural braid
representation (see Section 1) of simple minimal knots.
We will give an upperbound of this number.\par
With the help of KnotTheory and KnotPlot, we describe in section 7 some
examples of minimal knots with their minimal braid representation and
decomposition into prime knots. This allows us to investigate the fibration
of minimal knots.
Let us first recall a few definitions.
\begin{definition}
K is fibered if $S^3\setminus K$ is fibered over $S^1$ : there is a
differentiable mapping $\phi: \mathbb{S}^3\setminus K \mapsto S^1$
which defines a fiber bundle; the fiber $\phi^{-1}(e^{it})$ is the
interior of a compact orientable differentiable surface with
boundary K.
\end{definition}
( Notice that fibered knots have an algebraic characterization: the
commutator of the knot group $\pi_1(\mathbb{S}^3\setminus K)$ is finitely generated .
\cite{BZ}) Knots of holomorphic curves singularities are always fibered; if
the surface is given locally by the equation $F(z,w)=0$, then the (Milnor)
fibration is simply given by
\begin{equation*}
\phi (z,w) =\frac{F(z,w)}{|F(z,w)|}
\end{equation*}
restricted to a sphere of sufficiently small radius $\mathbb{S}^3(p,\epsilon)$.
Thus all torus knots are fibered.
The fibration yields a monodromy mapping $h$ and a gluing map
$\theta$ where $\mathbb{S}^3\setminus K\equiv \frac{I\times \mathcal{F} }{
(x,0) \sim(\theta (x), 1)}$, where $\mathcal{F}$ is the fiber of the
fibration, ie the Seifert surface spanning $K$.
The monodromy map gives a way to compute the Alexander polynomial, another
knot invariant :
\begin{equation*}
P(x) = det(h^* -x Id)
\end{equation*}
where $h^*: H^1(M, \mathbb{R}) \mapsto H^1(M,\mathbb{R} )$.
In particular if the highest order term of $P$ is different from
$\pm 1$ then the knot is can not be fibered.
We construct some examples of minimal braids with three, four or
five strands and identify them. In particular :
\begin{theorem}
The knot $9_{46}$ (see fig. 13 ) is a prime knot which is not
fibered and still is a minimal singularity knot; the minimal
surface is locally given by
\begin{equation*}
z \mapsto \left( z^4 + o(|z|^4),\ z^{13} + \bar z^{13} + z^{5}- \bar z^5
+ o(|z|^{13})\right).
\end{equation*}
\end{theorem}
We conjecture that the simple knots $K(N,p,q)$ with $N $ odd are
fibered.
We are grateful to Harold Rosenberg for introducing us to the problem of
fibration of minimal singularities. We thank Joan Birman for her suggestions
and Dror bar Natan for his help and diligency.
\section{ Definition of a minimal knot and description of its minimal braid}
\subsubsection{ Germ of a minimal singularity}
At a branch point $p$, a minimal surface $M$ still has a tangent
plane $\mathbf{TM_p}$ which is a real plane in $\mathbb{C}^2.$ The
ambiant space is thus split into $\mathbf{TM_p}$ and the normal
plane $\mathbf{NM_p}$. $M$ is a multi-valued minimal graph over a
subdomain of $\mathbf{TM_p}$ locally given by (proposition \ref{cle})
\begin{equation*}
z \mapsto \left(z^N,\ a(z^p + \bar z^p) + b( z^q -\bar z^q)\right)
\end{equation*}
$a,b \in \mathbb{C}$. Let us intersect $M$ with a 3-cylinder
$S_\epsilon \times \mathbf{NM_p} \subset \mathbf{{TM_p}\times {NM_p}
}$, perpendicular
to $\mathbf{TM_p}$; $S_\epsilon$ is a small circle of radius $\epsilon$ in $\mathbf{TM_p}.$
The intersection a knot parametrized by $\theta$ running along $\mathbb{S}
^1(\epsilon)$ $N$ times. As $\epsilon$ tends to zero, this knot converges
clearly to the equator of the corresponding 3-sphere $\mathbb{S}
^3(0,\epsilon)$ hence $K$ can be equally viewed as a knot of the 3-sphere.
This knot can be expressed in terms of circular functions as follows:
\begin{equation}
\left\{
\begin{array}{c}
K: \mathbb{S}^1(0,\epsilon) \longrightarrow \mathbb{R}^4 \\
\theta \mapsto \left( \cos N\theta ,\ \sin N\theta ,\ \cos
\left(p\theta+\phi_p\right) ,\ \sin \left(q\theta+\phi_q\right) \right)
\end{array}
\right.
\end{equation}
Changing $\theta$ into $\theta+\alpha$, we may arrange
that one of
the two phases is zero. We choose the following parametrization
\begin{equation*}
K(N,p,,q,\phi): \theta \mapsto \left( \cos N\theta , \sin N\theta , \cos
(p\theta+\phi) , \sin (q\theta)\right)
\end{equation*}
We parametrize $\mathbb{S}^1$ by $t\in [0,1]$ where $\theta= 2\pi t $.
\begin{defn}
A simple minimal knot, is
isotopic to the curve given by the one-to-one parametrization
\begin{equation*}
B:[0,1] \longrightarrow \mathbf{Cyl}\subset \mathbb{R}^4
\end{equation*}
\begin{equation*}
t \mapsto \left( \cos 2\pi Nt, \sin 2\pi Nt ,\cos (2\pi p t+\phi) , \sin
2\pi q t \right)
\end{equation*}
We denote $B([0,1])$ by $K(N,p,q,\phi)$
\end{defn}
These knots are similar to another type of knots, known as Lissajous
knots because the coordinates of the knot are circular functions of
possibly unequal frequencies.
\subsubsection{ Lissajous Knots}
It is interesting to note another connection between Minimal knots and
Lissajous knots. Let us first recall how it was defined in (\cite{BH} or
\cite{JP})
\begin{defn}
A Lissajous knot is a curve parametrized one-to-one by
\begin{equation*}
L:[0,1] \longrightarrow \mathbb{R}^3
\end{equation*}
\begin{equation*}
t \mapsto \left( \cos 2\pi Nt, \cos (2\pi p t+\phi_1 ) , \cos (2\pi q t
+\phi_2)\right)
\end{equation*}
\end{defn}
Notice that these knots are parametrized by 5 variables instead of 4 for
simple minimal knots.
\begin{lem}
The projection into a vertical 3-plane of a minimal knot is a Lissajoux
knot. If the minimal knot is of type $K(N,p,q)$ then the corresponding
Lissajous knot has a braid representation of 2N strands induced by the
braid representation of $K(N,p,q)$.
\end{lem}
This is a direct consequence of the definition of a Lissajoux knot.
\subsubsection{ The minimal braid representation of a simple minimal knot}
The graph of the function $B : S_\epsilon \mapsto \mathbb{C}$ defined by $
K(N,p,q,\phi) $, is a braid of N strands : there are N functions defined on
the open circle or $[0,1[$ i.e. $k=0,...,N-1$ such that
\begin{equation*}
B_k : [0,1] \longrightarrow \mathbb{C}
\end{equation*}
\begin{equation*}
B_{k}(t) = \cos\left( \frac{2\pi p}{N}(t+k)+\phi_p\right) +i \sin
\left( \frac{2\pi q}{N}(t+k) \right)
\end{equation*}
\begin{figure}[!t]
\fbox{ \includegraphics[width= 6 cm, height = 6 cm]{tresse344}
\includegraphics[width= 6 cm, height = 6 cm]{diatre344} }
\caption[fig1]{braid and braid diagram of K(3,4,4)}
\end{figure}
Reciprocally we reconstruct the knot $K(N,p,q,\phi )$ by closing the braid :
we connect the kth strand to the (k+1)th strand as follows : $B_{k}(1) =
B_{k+1}(0)$ for $k= 0,...,N-1 $.
We denote this braid by $B(N,p,q,\phi).$ We then project the braid onto any $
\mathbb{R}\subset \mathbb{C}$.
\begin{defn}
A braid diagram is the graph of the functions $\pi_D \circ B $ where $\pi_D$
is the projection of $\mathbb{C}$ onto a line $D$ of $\mathbb{C}$.
\end{defn}
If we choose to project the braid onto the y-component of
$\mathbb{C}$, ($z= x+iy$), we obtain a braid diagram that depends
only on $N$ and $q$. We denote it by $K^\perp (N,q)$. This braid
diagram consists of $N$ graphs of functions
\begin{equation*}
h_k= Im B_{k} : [0,1] \longrightarrow \mathbb{R},
\end{equation*}
$k=0,\cdots, N-1$. Note that all the $K^\perp (N,q)$ are identical to
the braid diagram of the $(N,q)$-torus knot. To regain the braid
from the braid diagram, we need to know which
strand is above or below at each intersection of any two strands $h_k$ and $
h_l$. We will first determine the value of the parameter $t$ for
each crossing and will determine which strand is above at each
crossing.
\subsection{ Crossing locus of the braid diagram $K^\perp (N,q)$}
Let us compute the parameter values $t$ at each intersection
points of the braid diagram $K^\perp (N,q)$.
Choose $k<l$ , $0\leq k\leq N-1$, $t\in [0,1[$; we study the intersection of
the strands $h_k$ et $h_l $ ; the crossings correspond to the different
values $t$ for which
\begin{equation*}
h_k(t) - h_l(t) =0.
\end{equation*}
It turns out that for an even number $N$ of strands
the parameter value $t=0$ corresponds to a
value for which two strands $h_{k_1}$ and $h_{k_2}$ intersect ; we thus
introduce a small positive $\epsilon$ such that $\epsilon$ does not parametrize
an intersection of two strands; we study instead the braid diagram on the new
interval $t\in [\epsilon, 1+\epsilon[$.
\begin{lemma}
\label{encadrement} Let $K(N,p,q,\phi)$ be a simple minimal knot. Its braid
diagram $K^\perp (N,q)$ consists of $N$ strands ; its crossing number is $
q(N-1).$ Furthermore the strings $h_k$ and $h_l$, $k<l,\
k=0,...,N-1$, meet at points $(t_{n,k,l}, h(t_{n,k,l})) $, $t\in
[\epsilon, 1+\epsilon[$ where
\begin{equation*}
t(n,k,l) := \frac{ N(2n+1)}{4q}- \frac{k+l}{2};
\end{equation*}
$n $ is any integer such that
\begin{equation*}
\frac{2q\epsilon}{N}+ \frac{q(k+l)}{N} -\frac{1}{2} \leq n < \frac{2q\epsilon
}{N}+ \frac{q(k+l)}{N} -\frac{1}{2} +\frac{2q}{N}.
\end{equation*}
\end{lemma}
\textbf{Proof.}
\begin{equation*}
h_k(t) - h_l(t) =0
\end{equation*}
\begin{equation*}
\Im B_{k}(\theta) - \Im B_{l}(\theta) = \sin 2\pi \frac{q}{N}(t+k) - \sin
2\pi \frac{q}{N}(t+l)
\end{equation*}
using
\begin{equation*}
\sin a -\sin b = 2\cos (a+b/2)\sin(a-b/2)
\end{equation*}
we obtain
\begin{equation} \label{inter}
\Im B_{k}(\theta) - \Im B_{l}(\theta) = 2 \cos \left(2\pi \frac{q t}{N}+ \pi
\frac{q(k+l) }{N} \right)\sin \pi \frac{q(k-l)}{N}
\end{equation}
It follows from (\ref{inter}), that $t$ parametrizes a double point if the
first cosine factor is zero that is
\begin{equation*}
2\pi \frac{qt}{N}+ \pi \frac{q(k+l)}{N} = \pi\frac{2n+1}{2}
\end{equation*}
or
\begin{equation*}
t = \frac{N(2n+1)}{4q}-\frac{k+l}{2}.
\end{equation*}
Hence the number of intersections of $K^\perp(N,q)$ is given by the number
of integers n such that
\begin{equation*}
\epsilon \leq \frac{ N(2n+1)}{4q}- \frac{k+l}{2} <1 + \epsilon
\end{equation*}
\begin{equation*}
\epsilon +\frac{k+l}{2} \leq \frac{ N(2n+1)}{4q} <1+\frac{k+l}{2}+\epsilon
\end{equation*}
\begin{equation*}
\frac{4q}{N}\epsilon +\frac{2q(k+l)}{N} \leq (2n+1) <\frac{4q}{N}+\frac{
2q(k+l)}{N} +\frac{4q}{N}\epsilon
\end{equation*}
\begin{equation*}
\frac{2q}{N}\epsilon+\frac{q(k+l)}{N}-\frac{1}{2} \leq n <\frac{2q}{N}+\frac{
q(k+l)}{N}-\frac{1}{2} +\frac{2q}{N}\epsilon
\end{equation*}
\hfill $\Box$
\textbf{Example 1: N=2} The integer $n$ that parametrizes the solutions
verifies
\begin{equation*}
q \leq (2n+1) <3q
\end{equation*}
\begin{equation*}
\frac{q-1}{2} \leq n < \frac{q-1}{2}+q
\end{equation*}
It follows that the braid diagram has $q$ crossing points.
\subsubsection{ Regularity of $K(N,p ,q ,\protect\phi)$}
We return to the minimal knot and its naturally associated braid.
The braid may be be singular if some strands intersect : if this is
the case,\\
$B_{k}(t) = B_{l}(t)$ for some k, l. and $t$. To solve this
equation, we need the following lemma complementary to lemma \ref{encadrement}
\begin{lemma}
\label{coordx} the projections $\Re B_k$ and $\Re B_l$ $k<l$ of $
K(N,p,q,2\pi\varphi)$ meet at points
\begin{equation*}
t _{n,k,l} := \frac{mN}{2p}- \frac{N\varphi}{2p}- \frac{k+l}{2}
\end{equation*}
where $m$ is any integer such that
\begin{equation*}
\frac{p}{N}(k+l) + 2\varphi \leq n < \frac{p}{N}(k+l) + 2\varphi + 2 \frac{p
}{N}
\end{equation*}
\end{lemma}
\textbf{Proof.} Let $\phi = 2\pi \varphi$\newline
\begin{equation*}
\cos\left(2\pi\frac{p}{N}\left(t+k\right) +\phi \right) - \cos\left(2\pi
\frac{p}{N}\left(t+l\right) +\phi \right) =0
\end{equation*}
if and only if
\begin{equation*}
\sin \left( 2\pi\frac{p}{N}\left(t+ \frac{k+l}{2}\right)+\phi
\right).\sin\left( 2\pi\frac{p}{N}\left(k-l\right)\right)=0
\end{equation*}
\begin{equation*}
2\pi\frac{p}{N}\left(t+ \frac{k+l}{2}\right)+\phi =\pi m
\end{equation*}
\begin{equation*}
t = \frac{mN}{2p}- \frac{N\varphi}{2p}- \frac{k+l}{2}
\end{equation*}
\begin{equation*}
t\in [\epsilon,1+ \epsilon]
\end{equation*}
iff
\begin{equation*}
\epsilon\frac{2p}{N} +\frac{p}{N}\left(k+l\right) + 2\varphi \leq m < \frac{
p}{N}\left(k+l\right) + 2\varphi + 2 \frac{p}{N} +\epsilon\frac{2p}{N}
\end{equation*}
Hence the simple knot has selfintersection, if there is a $t\in [\epsilon,1+
\epsilon[$ that satisfies the hypothesis of lemma \ref{encadrement} and
lemma \ref{coordx}; we conclude that the braid is singular if there are
integers $m$ and $m^{\prime }$ such that
\begin{equation*}
(2m+1)p-2qm^{\prime }= -4\varphi q
\end{equation*}
On the other hand if $N$ and $p$ or $N$ and $q$ are not coprime.then $
K(N,p,q,\phi)$ is always singular : indeed if $N=aN^{\prime },\
p=ap^{\prime }$, then the substitution $t\mapsto t+ \frac{k}{a}$
doesn't alter the first
3 coordinates of the knot: it suffices to verify that that we may choose $
\epsilon<t<1+\epsilon$ such that the last coordinate has the same
value for $t$ and $t+ \frac{k}{a}$.
\begin{proposition}
\label{regularite} Let $K(N,p,q,\phi)$ a minimal knot; if $N\wedge p =
N\wedge q =1$, then for almost all $\phi$ $K(N,p,q,\phi)$ is regular. $
K(N,p,q,\phi)$ is singular for a finite number of $\phi$; and these
$\phi$'s are all of the form $\phi=2\pi\alpha$, where $\alpha$ is
rational.
\end{proposition}
\textbf{Example.} The knots $K(N,p,p,\phi)$ are the torus knots $(N,p)$. The
knot projects on the first component of $\mathbb{C}^2$ onto a circle and on
the second component of $\mathbb{C}^2$ onto an ellipse if $\phi \not = 0$ or
a segment if $\phi =0$. The ''knot" is clearly singular for its value. We
will show in section 3 that, surprisingly, a minimal knot does not change its
type as the phase varies. This is a striking difference with the Lissajous
knots where a suitable conjugate variation of the two phases may change the knot
(\cite{BH}).
\subsection{The $y$-coordinate of a crossing point}
Let $t(m,k,l)$ be a crossing point between the $k$-th and $l$-th strands as
given by Lemma \ref{encadrement}. The $y$-coordinate of the corresponding
point of the braid, i.e. the height of the braid diagram is given by
\begin{equation} \label{coordy}
h_k(t) = h_l(t) = y(t,k,l) =\sin\frac{2\pi}{N}q(t+k)=(-1)^m \sin\left(\frac{
\pi}{N}q\left(k-l\right)+\frac{\pi}{2}\right)
\end{equation}
\subsection{Sign of the crossing points}
\begin{figure}
\picta
\end{figure}
If $t$ is a crossing point between the $k$-th and $l$-th strand, its sign $
S(t,k,l)$ is given by combining the following two pieces of data:
\begin{enumerate}
\item which strand is in front of the other one
\item which strand goes upwards.
\end{enumerate}
(1) is given by the sign of the difference $Re B_k(t)- Re B_l(t)$, that is
the sign of the difference in the x-coordinates
\begin{equation*}
\cos\frac{2\pi}{N} q(t+k+\phi)-\cos \frac{2\pi}{N}q(t+l+\phi)
\end{equation*}
For (2), we notice that the $k$-th strand is going upwards if the derivative
of the function $x\mapsto\sin \frac{2\pi}{N}qx$ at $t+k$ is positive (in
which case its derivative at $t+l$ is negative). It follows that the sign is
given by the product
\begin{equation*}
S(t,k,l)=-[\cos q(t+k)-\cos q(t+l)][\cos p(t+k+\phi)-\cos p(t+l+\phi)].
\end{equation*}
We remind the reader that a difference of cosines can be written as a
product of sines and derive
\begin{equation*}
S(t,k,l)=-4\sin \frac{2\pi}{N}q(t+\frac{k+l}{2})\sin \frac{2\pi}{N}p(t+\frac{
k+l}{2}+\phi) \sin\frac{\pi}{N}p(k-l)\sin\frac{\pi}{N}q({k-l}).
\end{equation*}
We now suppose that $t$ is of the form $t=t(m,k,l)$ as given by lemma
\ref{coordx}. We derive
\begin{equation*}
\sin\frac{2\pi}{N}q(t+\frac{k+l}{2})=\sin(\frac{\pi}{2}+m\pi)=(-1)^m
\end{equation*}
\begin{equation*}
\sin\frac{2\pi}{N}p(t+\frac{k+l}{2}+\phi)=\sin\frac{2\pi}{N}p\left(\phi+
\frac{N}{4q}\left(1+2m\right)\right).
\end{equation*}
{\bf Notations}. If $x$ is a real number we denote its integral part by $[x]$; if $
n$ is an integer, we denote by $P(n)\in\mathbb{Z}_2$ its congruence
modulo $2 $.\newline With these notations, and with the notations of
lemma (\ref{encadrement}) last computations give the following
lemma
\begin{lemma} \label{csf}{\bf Crossing sign formula}\\
The sign at a crossing point of the braid diagram $K(N,q)^\perp $
associated to the simple knot $K(N,p,q,\phi)$ and parametrized by
$t_{m,k,l}\in [\epsilon, 1+\epsilon[$, where $k<l,\ k=0,...,N-1$,
and
$m \in \mathbb{N} \cap
[ \frac{2q\epsilon}{N}+ \frac{q(k+l)}{N} -\frac{1}{2},
\frac{2q\epsilon
}{N}+ \frac{q(k+l)}{N} -\frac{1}{2} +\frac{2q}{N}[, $\\
is given by
the formula
\begin{equation} \label{formuledusigne}
s(k,l,m,\phi)=1+T(k,l)+P(m)+R(m,\phi)
\end{equation}
with respectively
\begin{equation*}
T(k,l)=P\left(\lbrack q \left(\frac{k-l}{N}\right)\rbrack\right)+
P\left(\lbrack p\left(\frac{k-l}{N}\right)\rbrack\right)
\end{equation*}
(the sum is taken in $\mathbb{Z}_2$)\ and
\begin{equation*}
R(m,\phi) = P\left(\lbrack\frac{2}{N}p(\phi+\frac{N}{4q}(1+2m))\rbrack
\right).
\end{equation*}
\end{lemma}
\subsection{An expression of a minimal braid}
We recall here the definition of the braid groups in terms of generators and
relations.\newline
The braid group $\mathbf{B}_N$ is generated by $N-1$ elements, $
\sigma_1,...,\sigma_{N-1}$ subject to the following relations
\begin{equation*}
\forall i,\ \ \sigma_i\sigma_{i+1}\sigma_i=\sigma_{i+1}\sigma_i\sigma_{i+1}
\end{equation*}
\begin{equation*}
\forall i, j\ \text{with}\ |i-j|\geq 2,\ \ \
\sigma_i\sigma_j=\sigma_j\sigma_i.
\end{equation*}
Since the braid $B(N,p,q,\phi)$ has $(N-1)q$ crossing points, we can write
it as a product of $(N-1)q$ $\sigma_i$'s or $\sigma_i^{-1}$'s. It is
straightforward to derive such an expression; however because of the
above-mentioned relations there are several such expressions and we need to
specify exactly which one we take. In other words we will specify one
representative of a preimage of $B(N,p,q,\phi)$ in the free group $\mathbf{F}
_{N-1}$ on the $N-1$ generators $\sigma_i$'s.\newline
Up to the sign of the crossing points, the braid is the same as for a $(N-1)q
$-torus knot, thus we aim for a representation which differs from the
standard representation of the torus knot only by the signs of the terms.
Each crossing point $C(t,k,l)$ is the data of coordinate $t$ and a pair of
strands $k$ and $l$. We number then as
\begin{equation*}
C_1(t_1,\{k,l\}_1),...,C_{q(N-1)}(t_{q(N-1)},\{k,l\}_{q(N-1)}).
\end{equation*}
We require the following ordering conditions
\begin{enumerate}
\item if $u<v$ then $t_u\leq t_v$
\item the first $\frac{(N-1)N}{2}$ crossing points are ordered by
lexicographical order on $t,k+l,k-l$
\item if $u+\frac{N(N-1)}{2}\leq q(N-1)$, then $\{k,l\}_{u+\frac{N(N-1)}{2}
}=\{k,l\}_u$.
\end{enumerate}
Each crossing point $C_i(t_i,\{k,l\}_i)$ has a sign $\epsilon(i)\in\{-1,1\}$
and a $y$-coordinate $y(C_i)$, (cf. \ref{coordy}). We point out that the
value in (cf. \ref{coordy}) takes $N-1$ different values $y_1>y_2>...>y_{N-1}
$. \newline
We define $n(i)$ by
\begin{equation*}
n(i)=s\ \text{if and only if}\ y(C_i)=y_s.
\end{equation*}
We derive a word in $\mathbf{F}_{N-1}$
\begin{equation*}
b_{N,q,p,\phi}=\prod_{i=1}^{(N-1)q}\sigma_{n(i)}^{\epsilon(i)}.
\end{equation*}
Its image in $\mathbf{B}_N$ is a representative of $B(N,p,q,\phi)$.\newline
Since we will be dealing with words of length $q(N-1)$ we introduce, for an
integer $m$, the sets
\begin{equation*}
\mathbb{N}_{m}=\{n\in\mathbb{N}\ | \ 1\leq n\leq m \}.
\end{equation*}
Thus the word $b_{N,q,p,\phi}$ is given by the data of the maps
\begin{equation*}
n:\mathbb{N}_{q(N-1)}\longrightarrow \mathbb{R}^{N-1},\ \ \ \epsilon:\mathbb{
N}_{q(N-1)}\longrightarrow \mathbb{Z}_2.
\end{equation*}
\section{(Non) dependence of the knot type on the phase $\protect\phi$}
\begin{prop}
\label{phaseinv} Let $N,p,q$ be integers as above and let $\phi$ and $
\phi^{\prime }$ be two elements of $[0,2\pi]$. Then the knots $K(N,p,q,\phi)$
and $K(N,p,q,\phi^{\prime })$ - or $K(N,q,p,\phi)$ and the mirror image of $
K(N,q,p,\phi^{\prime })$- can be represented by conjugate braids.
\end{prop}
\begin{cor}
Up to taking a mirror image, the isotopy type of the knot
$K(N,q,p,\phi)$ does not depend on the phase $\phi$.
\end{cor}
\begin{lem}
\label{reduction} It is enough to prove the Proposition in the case
\begin{equation*}
\phi^{\prime }=\phi+\frac{N}{2}\left(\frac{A}{p}+\frac{B}{q}\right)
\end{equation*}
for two integers $A$ and $B$.
\end{lem}
To prove the lemma we introduce
\begin{defn}
An element $\phi\in[0,2\pi]$ is said to be a \textbf{critical phase} if
there exists a crossing point $t=t(m,k,l)$ so that
\begin{equation*}
\cos \frac{2\pi}{N}p(t+k+\phi)=\cos\frac{2\pi}{N}p(t+l+\phi).
\end{equation*}
\end{defn}
In that case $K(N,p,q,\phi)$ has self-intersection at $t(m,k,l)$. This
translates into the existence of an integer $s$ such that
\begin{equation*}
\frac{2\pi}{N}(t+k+\phi)p=-\frac{2\pi}{N}(t+l+\phi)p+2\pi s
\end{equation*}
hence
\begin{equation} \label{phase}
\phi=\frac{sN}{2p}-\frac{N}{4q}(1+2m)
\end{equation}
The braid $B_{N,q,p,\phi}$ is defined if and only if the number $\phi$ is
not a critical phase.
The following is obvious
\begin{lem}
\label{comp} Let $\phi_1,\phi_2$ be two numbers, $\phi_1<\phi_2$ and suppose
that there is no critical phase in the interval $[\phi_1,\phi_2]$. Then the
braids $B_{N,q,p,\phi_1}$ and $B_{N,q,p,\phi_2}$ are identical.
\end{lem}
We have
\begin{lem}
\label{existenceAB} Let $\phi_1$ and $\phi_2$ be two critical phases. Then
there exist two integers $A$ and $B$ such that
\begin{equation*}
\phi_1-\phi_2=\frac{N}{2}\left(\frac{A}{p}+\frac{B}{q}\right).
\end{equation*}
\end{lem}
\textbf{Proof} We write $\phi_1$ and $\phi_2$ as in (\ref{phase}) in terms
of $s_1$, $m_1$ and $s_2$, $m_2$. We have
\begin{equation*}
\phi_1-\phi_2=(s_1-s_2)\frac{N}{2p}-\frac{N}{2q}(m_1-m_2)
\end{equation*}
and the lemma follows.\newline
We order the critical phases between $0$ and $2\pi$ and denote them
respectively
\begin{equation*}
\phi_0,\ \phi_1,\ \cdots, \phi_M.
\end{equation*}
We assume that $\phi_u<\phi<\phi_{u+1}$ and $\phi_v<\phi^{\prime }<\phi_{v+1}
$ for some $u$, $v$, with $0\leq u,v\leq M$. It follows from the lemma that $
\phi_u-\phi_v=X$ with $X=\frac{N}{2}(\frac{A}{p}+\frac{B}{q})$ for some
integers $A$, $B$. Thus $\phi+X$ belongs to the interval $[\phi_v,
\phi_{v+1}]$ (like $\phi^{\prime }$ does). Hence the braids $
B_{N,q,p,\phi^{\prime }}$ and $B_{N,q,p,\phi+\frac{N}{2}(\frac{A}{p}+\frac{B
}{q})}$ are the same. $\Box$
Going back to the formula for the signs of the crossing points, a
straightforward computation yields
\begin{lem}
\label{formuleaux} Let $N,q,p,\phi$ be as above and let $A$ and $B$
be two integers. Then
\begin{equation*}
R\left(m,\phi+\frac{N}{2}(\frac{A}{p}+\frac{B}{q})\right)=P(A)+ R(m+B,\phi).
\end{equation*}
\end{lem}
In view of \ref{csf}, Lemma \ref{existenceAB} solves the problem if $
B=0$ : $B_{N,q,p,\phi}$ is the same (resp. mirror image) of $B_{N,p,q,\phi+
\frac{AN}{2p}}$ if $A$ is even (resp. odd). Hence Lemma \ref{formuleaux}
will be proven (by induction) once we have shown that $K_{N,q,p,\phi}$ and
the mirror image of $K_{N,q,p,\phi+\frac{N}{2q}}$ can be represented by
conjugate braids. More precisely we will show
\begin{lem}
Let $\Phi:B_N\longrightarrow B_N$ be the involutive isomorphism defined by
\begin{equation*}
\forall i,\ \ 1\leq...\leq N-1,\ \ \Phi(\sigma_i)=\sigma_{N-i}.
\end{equation*}
\begin{enumerate}
\item $\Phi(B_{N,q,p,\phi})$ is a braid which represents $K_{N,q,p,\phi}$.
\item The mirror image of $\Phi(B_{N,q,p,\phi})$ and $B_{N,q,p,\phi+\frac{N}{
2q}}$ are conjugate braids.
\end{enumerate}
\end{lem}
\textbf{Proof} 1. is obvious and we set out to prove 2. \newline We
consider the canonical representative $b(N,p,q,\phi+\frac{N}{2q})$
which we have described in the previous paragraph. Then
\begin{equation*}
b(N,p,q,\phi+\frac{N}{2q})=\prod_{i=1}^{q(N-1)}\sigma_n(i)^{\eta(i)}.
\end{equation*}
We then write a word in $\mathbf{F}_{N-1}$ which represents the mirror image
of $\Phi(B_{N,q,p,\phi})$: we take $b_{N,q,p,\phi}$, change every exponent
into its inverse and replace every $\sigma_i$ by the corresponding $
\sigma_{N-i}$; we derive the word
\begin{equation*}
\tilde{b}_{N,q,p,\phi}=\prod_{i=1}^{(N-1)q}\sigma_{N-n(i)}^{-\epsilon(i)}.
\end{equation*}
Both $b_{N,q,p,\phi+\frac{N}{2q}}$ and $\tilde{b}_{N,q,p,\phi}$ are words of
length $q(N-1)$ in the free group $F_{N-1}$; part 2) of the lemma will
derive from
\begin{slem}
The words $b_{N,q,p,\phi+\frac{N}{2q}}$ and $\tilde{b}_{N,q,p,\phi}$ are a
circular permutation of one another.
\end{slem}
The permutation in question is given by
\begin{equation*}
\gamma:\mathbb{N}_{q(N-1)}\longrightarrow \mathbb{N}_{q(N-1)}
\end{equation*}
\begin{equation*}
(1,2,...,q(N-1))\mapsto \left(1+\frac{N(N-1)}{2},2+\frac{N(N-1)}{2}
,\cdots, \frac{N(N-1)}{2}-1, \frac{N(N-1)}{2}\right).
\end{equation*}
In view of the notations we have chosen, proving the Sublemma means proving
the following two facts for every $i\in\mathbb{N}_{q(N-1)}$\newline
(S1) $n(i)=N-n(\gamma(i)),$\newline
(S2) $\eta(i)=1+\epsilon(\gamma(i)).$\newline
\begin{enumerate}
\item \textbf{First case} $i+\frac{N(N-1)}{2}\leq (N-1)q$.\newline
Conditions 1) and 3)of section 2.4 on the ordering of the crossing
points $C_s(t_s,\{k,l\}_s)$ ensure that $C_i$ and
$C_{i+\frac{N(N-1)}{2}}$
are consecutive (in $[0,1]$) crossing points of the pair of strands $
\{k,l\}_i$, hence
\begin{equation*}
\{k,l\}_{\gamma(i)}=\{k,l\}_i.
\end{equation*}
We also derive that, if $t_i=t(m_i,k_i,l_i)=-\frac{k_i+l_i}{2}+\frac{N}{4q}
(1+2m_i)$, then
\begin{equation*}
t_{i+\frac{N(N-1)}{2}}=-\frac{k_i+l_i}{2}+\frac{N}{4q}\left(1+2(m_i+1)
\right).
\end{equation*}
Thus $t(m_i,k_i,l_i)-t(m_i+1,k_i,l_i)=\frac{N}{2q}$. Hence
\begin{equation*}
y(C_i)+y(C_{i+1})=\sin\frac{2\pi}{N}q(t_i)+\sin\frac{2\pi}{N}q(t_i+\frac{N}{
2q})=0
\end{equation*}
It follows that
\begin{equation*}
n(\gamma(i))=N-n(i)\
\end{equation*}
This proves (S1). \newline
The number $\eta(i)$ gives us the sign of the crossing point $
C_i(t_i,k_i,l_i)$ of the braid $B(N,p,q,\phi+\frac{N}{2q})$. It follows from
\ref{csf} and Lemma \ref{comp} that
\begin{equation*}
\eta(i)=1+T(k_i,l_i)+P(m_i)+R(m_i+1,\phi)=T(k_i,l_i)+P(m_i+1)+R(m_i+1,\phi).
\end{equation*}
We derive from the considerations above that the right-hand side is equal to
$1+\epsilon(\gamma(i))$. This proves (S2).\newline
\item \textbf{Second case}. $i+\frac{N(N-1)}{2}>q(N-1)$\newline
Then $C_i$ is the last crossing point of the strands $\{k,l\}_i$; thus the
reasonning of the previous case does not work word for word; however we will
establish a $1-1$ correspondence between the last $\frac{N(N-1)}{2}$
crossing points of the braid and the first $\frac{N(N-1)}{2}$ ones. We put
\begin{equation*}
t(m_i+1,k_i,l_i)=-\frac{k_i+l_i}{2}+\frac{N}{4q}\left(1+2(m_i+1)\right)>1
\end{equation*}
Please note the slight abuse of notation: as we said, $t(m_i+1,k_i,l_i)$ is
NOT a crossing point for the $k_i$-th and $l_i$-th strands. We have
\begin{equation*}
0<t(m_i+1,k_i,l_i)-1=-\frac{k_i+l_i+2}{2}+\frac{N}{4q}\left(1+2(m_i+1)
\right)<1.
\end{equation*}
We recognize the formula given for a crossing point of the braid. \newline
It is clear that the $t(m_i+1, k_i, l_i)-1$ occur in the same order (in the
interval $[0,1]$) as the $t(m_i, k_i, l_i)$'s.\newline
We need to distinguish two subcases.
\begin{enumerate}
\item \textbf{First subcase of second case}. $k_i<l_i<N-1 $.\newline
Then $t(m_i+1,k_i,l_i)-1$ is a crossing point for the strands $k_i+1,l_i+1$.
It is the first crossing point for that pair of strands and writes in our
notation
\begin{equation*}
t(m_i+1,k_i+1,l_i+1).
\end{equation*}
We have $k_\gamma(i)=k_i+1,\ l_\gamma(i)=l_i+1$ and in terms of $y$
coordinates,
\begin{equation*}
y(C(m_i+1,k_i+1,l_i+1))=(-1)^{m_i+1}\sin[\frac{\pi}{N}q(k_i-l)+\frac{\pi}{2}]
=-y(C(t_i,k_i,l_i)).
\end{equation*}
Thus $(S1)$ is true.\newline
Since $T(k_i+1,l_i+1)=T(k_i,l_i)$, the sign of $t(m_i+1,k_i+1,l_i+1)$ as a
crossing point of the $k_i+1$-th and $l_i+1$-th strands of $B_{N,q,p,\phi}$
is given by
\begin{equation*}
\epsilon(\gamma(i))=1+T(k_i,l_i)+P(m_i+1)+R(m_i+1,\phi)
\end{equation*}
\begin{equation*}
=1+T(k_i,l_i)+1+P(m_i)+R(m_i,\phi+\frac{N}{2q})=1+\eta(i).
\end{equation*}
\item \textbf{Second subcase of second case}. $k_i<l_i=N-1$ .\newline
That is, we are considering a point of the form $t(m_i,N-1, l_i)$. We have
\begin{equation*}
t(m_i+1, N-1, l_i)-1=-\frac{N+l+1}{2}+\frac{N}{4q}\left(1+2(m_i+1)\right)
\end{equation*}
\begin{equation*}
=-\frac{l+1}{2}+\frac{N}{4q}\left(1+2(m_i+1-q)\right).
\end{equation*}
Thus $t(m_i+1 ,N-1 ,l_i)-1$ is a crossing point for the $l_i$-th and $0$-th
strands. According to our notations it writes $t(m_i+1-q,0,l_i+1)$. We have
\begin{equation*}
\{k,l\}_{\gamma(i)}=\{N-1,l_{i+1}\}
\end{equation*}
\begin{equation*}
y(C(m+1-q,0,l+1))=(-1)^{m+1-q}\sin\left(\frac{\pi}{N}q(l+1)+\frac{\pi}{2}
\right)
\end{equation*}
On the other hand,
\begin{equation*}
y(C(m_i,N-1,l_i))=(-1)^m\sin[\frac{\pi}{N}q(N-1-l)+\frac{\pi}{2}]
\end{equation*}
\begin{equation*}
=(-1)^m\sin[(q+1)\pi-\frac{\pi}{N}(l+1)-\frac{\pi}{2}]=y(C(m_i+1-q,0,l_i+1))
\end{equation*}
This proves (S1); to investigate the sign of $t(m_i+1-q,0,l_i+1)$ we
need the following identities
\begin{equation*}
T(N-1-l_i,0)=P(p)+P(q)+T(l_i+1,0)
\end{equation*}
\begin{equation*}
P(m_i+1-q)=P(q)+P(m_i)+1
\end{equation*}
\begin{equation*}
R(m_i+1-q,\phi)=P([\frac{2\pi}{N}(\phi+\frac{N}{4q}(1+2m_i)+\frac{N}{2q}-
\frac{N}{2}])
\end{equation*}
\begin{equation*}
=R(m_i,\phi+\frac{N}{2q})+P(p).
\end{equation*}
Summing all these terms, we see that
\begin{equation*}
s(m_i+1-q,l+1, 0, \phi)=1+s(m_i, N-1, l, \phi+\frac{N}{2q}).
\end{equation*}
which proves $(S2)$ in this case.
\end{enumerate}
\end{enumerate}
This concludes the proof of the lemma and hence, of the Proposition.
\section{Symmetries of the knot}
Let us recall some terminology on knots symmetries : There are two natural
symmetries among knots that are involutions; the mirror symmetry $s_m$,
(symmetry of a knot with respect to a orientation reversing diffeomorphism
of $\mathbb{S}^3$ ) and the inversion of a knot $s_i$ which yields the same
knot but with the reverse orientation. $K$ is \textit{invertible} if it is
invariant by $s_i$ and \textit{amphicheiral} if it has some invariance with
respect to $s_m$; more precisely, each type of invariance is given a name
and we recall them here for clarity's sake.
\begin{enumerate}
\item If $s_i(K) = K$ ( equal means isotopic) only, then $K$
is reversible ;
\item if $s_i(K) = K$ and $s_m(K) = K$ then $K$ is fully amphicheiral;
\item If $s_m(K)=K$ only, then $K$ is (positive)
amphicheiral;
\item if $s_m\circ s_i(K) =K$ only, then
$K$ is negative amphicheiral.
\item If $K$ has none of these symmetries then $K$ is chiral.
\end{enumerate}
The knot is said to be strongly symmetric if it is symmetric with respect to
an ambiant isometry of $\mathbb{S}^3$.
Results of section 2 and easy computations yields
\begin{theorem}
A simple minimal knot is either reversible or fully amphicheiral. More
precisely :
\begin{enumerate}
\item All knots $K(N,p,q)$ are strongly invertible.
\item Furthermore, if $p+q$ is odd, then $(N,p,q) $ is strongly fully
amphicheiral:
\item if $N$ is even and $p+q$ is even or if $N$ is odd and $p$ and $q$
even, then $K(N,p,q) $ is periodic of order two : it is invariant by a
rotation of angle $\pi$ around an axis $S^1$ and the linking number of its
axis with the knot is equal to $N$
\end{enumerate}
\end{theorem}
Similarly to the Lissajous knots (\cite{BH}), these symmetries yields
properties on the Arf invariant and Alexander polynomial that are described
in the section.
We first have
\begin{thm}\label{sym}
Let $N, p, q, \phi$ be as above. Then the knot $K_{N, p, q,\phi}$ is
strongly reversible.
\end{thm}
\textbf{Proof} Let us remind the reader that if $\theta$ is a real
number and $m$ is an integer, we have
\begin{equation*}
\sin m(\theta+\pi)=(-1)^m\sin (m\theta),\ \cos
m(\theta+\pi)=(-1)^m\cos(m\theta).
\end{equation*}
The change of parametrization $t\mapsto t +\frac{1}{2}$ doesn't change the
knot as a whole; it is induced by an ambiant symmetry of $\mathbb{S}^3$. We
derive the following
\begin{lem}
The knots $K(N, p, q, 0)$ and $K(N,p, q,\pi)$ are invariant under the
diffeomorphism $\Phi_{N, p, q,}$ of $\mathbb{S}^3$ defined by
\begin{equation*}
(x,y,z,w)\mapsto( (-1)^N x, (-1)^N y, (-1)^q z, (-1)^p w).
\end{equation*}
\end{lem}
It follows that the knots $K(N, p, q, 0)$ and $K(N, p ,q ,\pi)$ -
which are mirror images of one another - are reversible. This fact,
together with Proposition \ref{phaseinv} above, finishes the proof
of the theorem.\hfill \qed\par
According to the parities of the integers
involved we can derive symmetries of the knot. First notice that
$\Phi_{N, q, p}$ is an involution in all cases.
\begin{enumerate}
\item If $N$ is even, then
\begin{enumerate}
\item if $p+q$ is even, $\Phi_{N, q, p}$ is an orientation preserving
symmetry. From the non degeneracy condition ( $K$ is not singular),
$p$ and $q$ can not be both even, hence $p$ and $q$ are odd; then $
\Phi_{N,q,p}$ is a rotation of $\pi$ around an horizontal
$\mathbb{S}^1$. Hence $K$ is periodic of order two but in a weak
sense since the linking number of the invariant axis of the
rotation with $K$ may be zero.
\item if $p+q$ is odd $\Phi_{N,q,p}$ is orientation reversing. and $K$ is
strongly fully amphichireal
\end{enumerate}
\item If $N$ is odd, then
\begin{enumerate}
\item if $p+q$ is even, $\Phi_{N,q,p}$ is orientation preserving. If $p$ and
$q$ are both even, then $\Phi_{N,q,p}$ is a rotation of $\pi$ around an
vertical $\mathbb{S}^1$. its linking number with $K$ is $N$. Hence $K$ is
periodic of order two. If $p$ and $q$ are both odd, then $\Phi_{N,q,p}$ is
orientation preserving and has no fixed points.
\item if $p+q$ is odd $\Phi_{N,q,p}$ is orientation reversing. and $K$ is
strongly fully amphichireal
\end{enumerate}
\end{enumerate}
\subsection{symmetries of Alexander polynomial}
We can deduce from last considerations and \cite{Mu} that
\begin{cor}
The simple minimal knot $K(N, p, q)$'s Alexander polynomial is a square
modulo two, if $p$ and $q$ are not both odd, and the Arf invariant is then zero.
\end{cor}
\subsection{symmetries of the braid}
Using the braid description of the knots, these symmetries translate into
symmetries of the braid. Let us write a point in the braid as $(t,k)$ we mean
the point $h_k(t)$ i.e. on the $k$-th strand. We distinguish two cases :
\begin{enumerate}
\item \textbf{$N$ is even}. We notice that
\begin{equation}
\frac{2\pi}{N}(x+k)+\pi=\frac{2\pi}{N}(x+k+\frac{N}{2})
\end{equation}
Thus, if $p$ and $q$ have the same (resp. a different) parity, $B_{N,q,p,0}$
is preserved (resp. transformed into its mirror image) by the following
transformation :\newline
$\text{if}\ k\leq\frac{N}{2}, \ \ (t,k)\mapsto (t,k+\frac{N}{2})\ \text{if}\
k\geq\frac{N}{2}, \ \ (t,k)\mapsto (t,k-\frac{N}{2}).$\newline
This symmetry switches the strands, while keeping the first coordinate fixed.
\newline
\item \textbf{$N$ is odd}. We notice
\begin{equation*}
\frac{2\pi}{N}(x+k)+\pi=\frac{2\pi}{N}(x+\frac{1}{2}+k+\frac{N-1}{2}) =\frac{
2\pi}{N}(x-\frac{1}{2}+k+\frac{N+1}{2}) .
\end{equation*}
If $p$ and $q$ have the same (resp. a different) parity, $B_{N,q,p,\phi}$ is
preserved (resp. transformed into its mirror image) by the transformation
\newline
if $k\leq\frac{N-1}{2}$ and $t\leq\frac{1}{2}$ then $(t,k)\mapsto (t+\frac{1
}{2},k+\frac{N-1}{2})$\newline
if $k\leq\frac{N-1}{2}$ and $t\geq\frac{1}{2}$ then $(t,k)\mapsto (t-\frac{1
}{2},k+\frac{N-1}{2})$\newline
if $k\geq\frac{N-1}{2}$ and $t\leq\frac{1}{2}$ then $(t,k)\mapsto (t+\frac{1
}{2},k-\frac{N+1}{2})$\newline
if $k\geq\frac{N-1}{2}$ and $t\geq\frac{1}{2}$ then $(t,k)\mapsto (t-\frac{1
}{2},k-\frac{N+1}{2})$\newline
This symmetry switches the first half of a strand with the second half of
another.
\end{enumerate}
\section{A counterexample}
\begin{figure}[tbp]
\fbox{ \includegraphics[width= 6 cm, height = 6 cm]{noeud817}}
\caption[fig1]{ A knot that can not be simple minimal : $8_{17}$ }
\end{figure}
\begin{theorem}
A negative amphicheiral or chiral knot can not be the knot of a
simple minimal or iterated minimal knot.
\end{theorem}
Notice that the first candidate for a counterexample in the Rolfsen classification is the
knot $8_{17}$ which is the first negative amphicheiral knot. This
knot provides the first example of a prime knot which can not be a
simple knot or iterated simple knot.\\
{\bf Proof.} If the knot is simple, then theorem \ref{sym} shows that
it can not be either chiral or negative amphicheiral.\\
Let us examine
the case of an iterated minimal knot (of a
simple knot) Suppose thus that there is an iterated minimal knot $K$
that bounds a negative amphicheiral (or chiral) knot. Then $K$ is
contained in a tubular neighborhood of the companion knot, which is a
simple minimal knot. The linking number $\kappa$ of the satellite
knot with a meridian of the tube is larger than two since the braid
diagram of the satellite knot is identical to the braid diagram of
an iterated torus knot.
But the Alexander polynomial of $K$ is a multiple of $
P(x^\kappa)$ where $P$ is the Alexander polynomial of $K$ which is
impossible.\hfill \qed
Notice though that it may be possible that these chiral knots are
cable knots of minimal singular knots. This case
requires a further investigation as we indicated in the introduction.
\section{The algebraic crossing number}
\subsection{Definition - Background}
We remind the reader that the \textit{algebraic crossing number} of $
B_{N,q,p,\phi}$ is the sum of the signs (i.e. $+1$ or $-1$) of its
crossing points. We will denote it $e(B_{N,q,p,\phi})$. It is an
invariant of the conjugation class of the braid but it is not an
isotopy invariant of the knot $K_{N,q,p,\phi}$. However braids with
three strands have been thoroughly studied by Birman and Menasco and
we can derive from their work,
\begin{thm}
([B-M]). Let $K$ (resp. $K^{\prime }$) be a knot represented by a $3$-braid $
B$ (resp. $B^{\prime }$). If $K$ and $K^{\prime }$ are isotopic and $K$ is
neither trivial nor a $(2,k)$-link, then $e(B)=e(B^{\prime })$.
\end{thm}
In the present case, where the braid comes from a branched immersion in $4$
-space, its algebraic crossing number can be seen as the number of double
points which concentrate at the branch point. Namely
\begin{thm}
([Vi]) Let $\Sigma$ be a closed Riemann surface without boundary,
let $M$ be an orientable $4$-manifold and let
$f:\Sigma\longrightarrow M$ be an embedding which has one branch
point $p$. Suppose that in a neighbourhood of $p$, $f$ is
parametrized as in proposition \ref{cle}, that it is a topological
embedding in the neighbourhood of $p$ and that the associated knot
is $K(N,p,q,\phi)$. Then the degree of the normal bundle $Nf$ is
\begin{equation*}
[f(\Sigma)].[f(\Sigma)]-e(B_{N,q,p,\phi})
\end{equation*}
where $[f(\Sigma)].[f(\Sigma)]$ denotes the self-intersection number of $
f(\Sigma)$.
\end{thm}
We immediately derive from Proposition \ref{phaseinv} above that
\begin{prop}
Up to sign, the algebraic crossing number $e(B(N,p,q,\phi))$ does not depend
on the phase $\phi$.
\end{prop}
and that
\begin{prop}
Suppose $q$ and $p$ are of different parities. Then
\begin{equation*}
e(B(N,q,p,\phi))=0.
\end{equation*}
\end{prop}
{\bf Remark} A knot defined by a $2$-stranded braid is either trivial or is a $
(2,n)$ torus knot in which case the algebraic crossing number is
$n$.
\subsection{An estimate}
The non-dependence on the phase allows us to prove
\begin{prop}
Let $N,q,p$ as above and suppose that $p$ and $q$ are mutually prime. Then
\begin{equation*}
|e(B(N,q,p))|\leq N^2+N-4.
\end{equation*}
\end{prop}
\textbf{Remark} This estimate does not depend on $q$: notice the sharp
contrast with the case of the $(N,q)$-torus knot (i.e. when $p=q$) and
every crossing number is positive: thus we have $e(B(N,q,q)=q(N-1)$.\newline
\textbf{Proof} If $k$ and $l$ are different integers, $1\leq k, l\leq N-1$,
we denote by $M(k,l)$ the set of integers $m$ such that the $t(m,k,l)$ given
by the formula (*) above is a crossing point of the $k$-th and $l$-th
strands.
\begin{lem}
The following correspondance is a bijection between $M(k,l)$ and \newline
$M(N-1-k,N-1-l)$
\begin{equation*}
t(m,k,l)\mapsto t(2q-m-1,N-1-k,N-1-l).
\end{equation*}
\end{lem}
\textbf{Proof}. We notice that, if $m$ verifies \ref{encadrement} w.r.t. the
integers $k$ and $l$, then $2q-m-1$ verifies
\begin{equation*}
\frac{q}{N}(2N-(k+l)-2)-\frac{1}{2}\leq 2q-m-1\leq \frac{q}{N}(2N-(k+l)-2)-
\frac{1}{2} +\frac{2q}{N}.
\end{equation*}
In other words,
\begin{equation*}
\frac{q}{N}((N-k-1)+(N-l-1))-\frac{1}{2}\leq 2q-m-1\leq \frac{q}{N}
(N-1-k)+(N-1-l))-\frac{1}{2} +\frac{2q}{N}.
\end{equation*}
\hfill $\Box$\\
Next, we go back to the expression of $R(.,\phi)$ above and
see that we can choose a phase $\phi$ such that for every $m$, we have
\begin{equation*}
R(m,\phi)=P([\frac{p}{q}m]).
\end{equation*}
At this point we introduce
\begin{equation*}
\hat{M}=\{m\in M(k,l)\ | \ \frac{p}{q}m\in\mathbb{N}\}.
\end{equation*}
Since $p$ and $q$ are mutually prime, an element $m$ of $\hat{M}$ is of the
form $m=aq$, for some integer $a$.
\begin{lem}
\label{prec} If $m\in\hat{M}\cap M(k,l)$, then\newline
i) $m=q$\newline
ii) $k+l=N-2$ or $k+l=N-1$ or $k+l=N$.\newline
Conversely for every $k,l$ verifying ii), $q\in M(k,l)$.
\end{lem}
\textbf{Proof} The equality $m=aq$ yields
\begin{equation*}
\frac{k+l}{N}-\frac{1}{2q}\leq a\leq \frac{k+l}{N}+\frac{2}{N}-\frac{1}{2q}.
\end{equation*}
Since $1\leq k+l\leq 2N-3$, we derive that $a=1$. Thus $k+l$ satisfies
\begin{equation*}
\frac{k+l}{N}-\frac{1}{2q}\leq 1\leq \frac{k+l}{N}+\frac{2}{N}-\frac{1}{2q}.
\end{equation*}
Since $N>q$ the left hand-side yields $k+l\leq N$ and the right-hand side
yields $k+l\geq N-2$.\newline
We leave it to the reader to check the converse. \hfill $\Box$
\begin{lem}
If $m\neq q$, we have
\begin{equation*}
R(m,k,l,\phi)=R(2q-m,N-k-1,N-l-1,\phi)+1.
\end{equation*}
\end{lem}
\textbf{Proof} Since $\frac{p}{q}m$ is not an integer,
\begin{equation*}
R(2q-m,\phi)=P([2p-\frac{p}{q}m])=1+P([\frac{q}{p}m])=1+R(m,\phi).
\end{equation*}
We denote by $\sigma(k, l)$ the signed number of intersection points of the $
k$-th and $l$-th strands.\newline
We have
\begin{equation*}
\sigma(k,l)+\sigma(N-k-1,N-l-1)=-(-1)^{T(k,l)}\Sigma _{m\in
M(k,l)}(-1)^{P(m)}\left((-1)^{R(m,\phi)}-(-1)^{R(2q-m-1,\phi)}\right).
\end{equation*}
If $k+l$ is not equal to either $N-2$, $N-1$ or $N$, then all $m$'s in the
sum above verify Lemma \ref{prec}. Thus
\begin{equation*}
|\sigma_{k,l}+\sigma(N-k-1,N-l-1)|\leq 2.
\end{equation*}
If $k+l=N$ (resp. $k+l=N-2$), then $(N-k-1)+(N-l-1)=N-2$ (resp. $
(N-k-1)+(N-l-1)=N$). \newline
Both pairs of strands $\{k,l\}$ and $\{N-k-1,N-l-1\}$ contain a crossing
point for which $m=q$ and thus which does not verify Lemma \ref{prec}. Thus
\begin{equation*}
|\sigma_{k,l}+\sigma(N-k-1,N-l-1)|\leq 4.
\end{equation*}
Finally if $k+l=N-1$, then $\{k,l\}=\{N-k-1,N-l-1\}$. We have
\begin{equation*}
2|\sigma(k,l)=|\sigma_{k,l}+\sigma(N-k-1,N-l-1)|\leq 4.
\end{equation*}
In order to put together these estimates, we point out the following:
\newline
\textit{if $A$ is an integer, the number of pairs of strands $\{k,l\}$ such
that $k+l=A$ is $A$.} We derive the estimate
\begin{equation*}
|e(b(N,p,q,\phi)|\leq 2\Sigma_{s=1}^{N-3}s+4(N-2)+2(N-1)=N^2+N-4.
\end{equation*}
\section{ Examples of Minimal Knots}
Most of the examples of minimal knots given here can be described with the
help of the Rolfsen table; i.e. they are knots that are the connected sum of
prime knots with at most 10 minimum number of crosssings. In some cases
however, we used the Hoste-Thistlewaite table provided in KnotTheory. It
gives examples with a minimum crossing number reaching 16. We compute the
Alexander Polynomial and Jones polynomial of the subdescribed minimal knots.
In most cases we used the program KnotTheory to compute these polynomials as
well as to draw the natural braids. Knots (in the Rolfsen table) were drawn
using KnotPlot.
We will use Rolfsen's notation for the alexander polynomial :
\begin{equation*}
[a_0+a_1+a_2+\cdots+ a_n] := x^n.\sum _{i=0}^n a_i\left( x +\frac{1}{x}
\right)^i
\end{equation*}
\subsection{ General Properties}
Beforehand we describe some useful properties of these knots.
As the simple knots are phase independant, the will be labelled as $K(N,p,q)$
.
\begin{lemma}
\label{gen1} If
\begin{equation}
\left\{
\begin{array}{c}
x \equiv p\ \mathrm{mod}\ N \\
x \equiv p \ \mathrm{mod}\ 2q
\end{array}
\right.
\end{equation}
then $K(N,x,q)$ is isotopic to $K(N,p,q)$
\end{lemma}
In particular, there is a $2qN$ periodicity for the appearance of knots with
respect to $p$.
\begin{lemma}
\label{gen2} There is an infinite number of minimal representations of the
trivial knot : for any $N$ and $q$, $K(N,N+q,q)$ is isotopic to the trivial
knot.
\end{lemma}
\textbf{Proof}. Choose any two strands; we then can choose the phase such
that the $n$ labelling the crossings satisfies $0\leq n < [\frac{2q}{N}]$.
The crossing numbers appear then with the following sequence of signs as $
+++++------$ where the numbers of plus and minus differ by one or zero. This
means that the two strands can be deformed leaving the ends fixed such that
their number of crossings is either one or zero without taking into account
the other strands. But this is true for any two pairs of strands. Choose
then strand 1. We can deform all others strands such that strand 1 is on top
except for the strand that reaches the right end and that crosses strand 1
only once. We proceed similarly for all other strands.\hfill$\Box$
\begin{lemma}
\label{gen3} For all $N,p,q$, $K(N,p,q)= K(N,q,p)$
\end{lemma}
\textbf{Proof}. We obtain $K(N,q,p)$ from $K(N,p,q)$ by adding the phase $
\pi/2$ and by interchanging coordinates The invariance of the knot by a
phase change yields the result.\hfill $\Box$.\par
We can check that
\begin{lemma}
\label{gen4} for any $N,q,a\in \mathbb{N}$, $K(N, aq, q)$ are isotopic to
two type of knots : the torus knot $T(N,q)$ and a non trivial knot.
Both have a braid group given by
\begin{equation*}
\left(\sigma_1^{\alpha_1}\circ \dots\circ
\sigma_{N-1}^{\alpha_{N-1}}\right)^N
\end{equation*}
In the case $T(N,q)$, all the $\alpha_i =+1.$
\end{lemma}
\subsection{ Torus Knots}
It is known that Torus knots are not Lissajous knots \cite{BH} ; this is in
complete opposition with simple minimal knots :
\begin{thm}
All torus knot are minimal knots: $T(a,b) $ can be realized as $K(a,b,b)$.
Moreover the regular knots $K(2,p,q )$ is trivial if $p\not = q$ ; else it
is the torus knot $T(q,2)$.
\end{thm}
The first part is direct; the second part is a consequence of last section
on the algebraic number of minimal braids with two strands.
\subsection{ Minimal Knots with an odd number of strands}
We will consider first knots whose minimal braid has three strands. The
braid representative of $K(3,p,q)$ is $\prod_{i=1}^q\sigma_1^{\alpha_i}
\sigma_2^{\beta_i}$, $\alpha_i , \beta_i = \pm 1$ and $\sigma_i$ denotes the
ith crossing ordered from left to right on the braid diagram; strands are numbered top down. If the first two crossings of the braid corresponds
respectively to the crossings of the couple of strands $(1,2)$ and $(1,3)$
then the k-th pairs of crossings correspond to the intersection of the
couple of strands $(\tau^k(1),\tau^k(2))$ and $(\tau^k(1),\tau^k(3))$ where $
\tau$ is the cyclic permutation $(1,2,3)$.
\subsubsection{ K(3,\ .\ , 4)}
The number of crossings is at most $q.(N-1) =8$; hence, if theses knots are
prime, they must be in the Rolfsen table; If they are not prime then they
are connected sums of knots in the Rolfsen table. We describe in this
paragraph all such knots. Only five knots appear periodically as $p$ varies;
From lemma \ref{gen1} there is a global 24-periodicity with respect to p
but there are also other symmetries that are not accounted for.
\begin{enumerate}
\item The torus knot $T(3,4)= K(3,4,4p) = K(3,20,4)\cdots $ This case
generalizes to all $K(N,p,p) $ or $K(N,ap,p)$ for suitable $a$. These knots
are clearly reversible.
\item The square knot $3_1 \# \bar 3_1= K(3,5,4)= K(3,4,5)$ . This is the
first example in the literature of a minimal knot that is not toric ( cf.
[MW]). In a way, this the smallest knot with respect to the lexicographic
ordering on $N,p,q$. It is fully amphicheiral and $P = [-1+1]^2$.
\item The trivial knot $K(3,7,4) =1$. This is knot of type $K(N,N+q,q)$
which are all trivial. Notice that $K(3,k,4) =1$ for $k = 5, 11, 13, 19...$
\item The connected sum of the eight knot and the square knot is \newline
$4_1 \# ( 3_1 \# \bar 3_1) = K(3,8,4)$. ( $p= 8,16...$ ). This is a knot
of type $K(N,aq,q)$ ; it is either the torus knot $T(N,p)$ or a knot whose
braid group is $\left(\sigma_1 \sigma_2^{-1} \right)^4$. This knot is fully
amphicheiral
\item The first non toric and non trivial prime knot, the eight knot $4_1 =
K(3,10,4)= K(3,10,4)=K(3,22,4)...$; it is fully amphicheiral. $P= [-3+1]$.
\end{enumerate}
\begin{figure}[tbp]
\fbox{ \includegraphics[width= 6 cm, height = 6 cm]{TRE354}
\includegraphics[width= 6 cm, height = 6 cm]{noeudcarre} }
\caption[fig1]{ minimal braid and knot of K(3,5,4)}
\end{figure}
\begin{figure}[tbp]
\fbox{ \includegraphics[width= 6 cm, height = 6 cm]{TRE374}
\includegraphics[width= 6 cm, height = 6 cm]{noeudtrivial} }
\caption[fig1]{ One of many minimal representation of a trivial knot:
K(3,7,4)}
\end{figure}
\begin{figure}[tbp]
\fbox{ \includegraphics[width= 6 cm, height = 6 cm]{TRE3104}
\includegraphics[width= 6 cm, height = 6 cm]{noeud41bis} }
\caption[fig1]{ The eight knot realized as K(3,10,4)}
\end{figure}
\subsubsection{ K(3, \ . , 5)}
The number of crossings is at most 10 which still allows us to check in the
Rolfsen table. Only four knots appear periodically as $p$ varies. We just
write down the smallest $p$ for each knot.
\begin{enumerate}
\item The torus knot $T(3,5)$. realized by $K(3,5,5)$.
\item The prime knot $10_{155}=K(3,7,5) $; it is reversible. As in the case
of Lissajous knot the appearance of this knot for the values $3,7,5$
suggests that it is difficult to find a topological caracterization of these
minimal knots. This is even clearer in the case $K(4,11,5)$ described below.
\item The trivial knot $K(3,8,4) =1$.
\item The connected sum $ 6_2\# \bar 6_2 = K(3,10,5)$.
This is a knot of type $K(N,aq,q)$ ; ($= K(3,20,5)$, but $ K(3,25,5)=
K(3,35,5) $ is the torus knot $T(3,5)$ and $K(3,10,5)= K(3,40,5)... $)
\end{enumerate}
\begin{figure}[tbp]
\fbox{ \includegraphics[width= 6 cm, height = 6 cm]{TRE355}
\includegraphics[width= 6 cm, height = 6 cm]{noeudt35} }
\caption[fig1]{ An example of a torus knot : K(3,5,5)}
\end{figure}
\begin{figure}[tbp]
\fbox{ \includegraphics[width= 6 cm, height = 6 cm]{TRE375}
\includegraphics[width= 6 cm, height = 6 cm]{noeud10155} }
\caption[fig1]{ minimal braid of $10_{155} = K(3,7,5)$}
\end{figure}
\begin{figure}[tbp]
\fbox{ \includegraphics[width= 6 cm, height = 6 cm]{TRE3105}
\includegraphics[width= 6 cm, height = 6 cm]{noeud62} }
\caption[fig1]{ minimal braid of $K(3,10,5)= (6_2\#\bar 6_2) $ and knot $6_2$
}
\end{figure}
\subsubsection{ K(3,\ . , 7)}
The number of minimal crossings is at most 14, two of them have a minimal
crossing number of 14 ; to check all cases we need consult the
Hoste-Thistlewaite table. Six different knots appear periodically.
\begin{enumerate}
\item The torus knot $K(3, 3, 7) = T(3,7)$.
\item The knot $K(3,8,7) $ is fully amphicheiral and $P = [1-1+1]^2$.
\item The trivial knot $K(3,10,5) =1$.
\item The prime knot $14 N 27120 = K(3,11,7)$ which is reversible! This is
the first occurence of a non amphicheiral knot with odd numbers , $N,p,q$
which is not a torus. $P=[7-5+3-1]$.
\item The connected sum $K(3, 14,7) $; it is amphicheiral, and $P=
[7-6+4-1]^2$. This is a knot of type $K(N,aq,q)$, hence its braid group is $
\left(\sigma_1^{-1}\sigma_2\right)^7$
\item The prime knot $K(3,19,7) = 14 N 11995$; it is reversible and $
P=[7-5+3-1]$.
\end{enumerate}
\begin{figure}[tbp]
\fbox{ \includegraphics[width= 6 cm, height = 6 cm]{TRE3117}
\includegraphics[width= 6 cm, height = 6 cm]{TRE3147} }
\caption[fig1]{ minimal braids of $K(3,11,7)=14 N 27120$ and $K(3,14,7)$ }
\end{figure}
\begin{figure}[tbp]
\fbox{ \includegraphics[width= 6 cm, height = 6 cm]{TRE3197}
\includegraphics[width= 6 cm, height = 6 cm]{TRE5166} }
\caption[fig1]{ minimal braids of $K(3,19,7)=14 N 11995$ and $K(5,16,6)= 14
N 17954$ }
\end{figure}
With $N=3, q\geq 8$ the identification becomes very difficult since the
number of crossings is a priori bigger than 16! We conclude this paragraph
with an example of a minimal knot with $N=5$ i.e. whose minimal braid
representation has 5 strands.
\subsubsection{ K(5,22,6)}
We will consider knots whose minimal braid has five strands. The braid
representative of $K(5,p,q)$ is $\prod_{i=1}^q\sigma_2^{\alpha_i}\sigma_4^{
\beta_i}\sigma_1^{\gamma_i}\sigma_3^{\delta_i}$, $\alpha_i ,
\beta_i,\gamma_i,\delta_i = \pm 1$. The first four crossings of the braid on
the strands corresponds to the couples \\
$( 2, 3), ( 4, 5) , (1, 3), (2, 5 )$
and the $k$-th quadruple corresponds to the crossings $\tau^k (a,b)$ where
$\tau = (1,3,5,4,2)$.\par
We will only describe an example of prime minimal knot. In each case we need
to check on the braid representations that the minimal crossing number is
less than 16. We try to minimize the number of crossings on the minimal
braid for each knot such that the crossing number is less than 15. We will
number the strands top down according to the order of the strands on the far
left of the braid.
\begin{enumerate}
\item $K(5,22,6)= 7_7$ (fig. 11). Consider the minimal braid of this knot; flip to the
bottom the first ''hill" of the strand 4, we reduce the crossing number by
4; Flip the first valley of strand 2 to the top , we can reduce the crossing
number by 4. Then moving downwards the first hill of strand 3 reduces the
number by 2; hence it suffices to check in the table of knots with less than
14 crossings.
\end{enumerate}
\begin{figure}[tbp]
\fbox{ \includegraphics[width= 6 cm, height = 6 cm]{TRE5226}
\includegraphics[width= 6 cm, height = 6 cm]{noeud77} }
\caption[fig1]{ minimal braid of knot $K(5,22,6)= 7_7$ }
\end{figure}
It seems that the Alexander polynomial of all the examples computed have a
highest order coefficient equal to plus or minus one; It may be that all
these simple minimal knots are fibered when $N$ is odd.
hence we conjecture :
\begin{Conj}
Simple minimal knots with an odd number of strands are fibered.
\end{Conj}
This is not the case when $N$ is even as we shall see now.
\subsection{ Minimal Knots with an even number of strands}
We will only discuss the case $N= 4$, the case $N=2$ has been dealt with in
the computation of its algebraic crossing number in section 4.
\subsubsection{ K(4,\ . 5)}
We will consider first knots whose minimal braid has four strands. The braid
group of $K(4,p,q)$ is $\prod_{i=1}^q\sigma_1^{\alpha_i}\sigma_3^{\beta_i}
\sigma_2^{\gamma_i}$, $\alpha_i , \beta_i,\gamma_i = \pm 1$. If the first
three crossings of the braid corresponds respectively to the crossings of
the couple of strands $(1,2),(3,4),(1,4)$, the $k$-th triple corresponds to
the image of the first three by the permutation $\tau^k$ where $\tau =
(1,2,4,3)$
The number of crossings is at most 15 Six different knots appear
periodically.
\begin{enumerate}
\item The torus knot $T(4,5)$.
\item The sum of two prime knots $K(4,7,5) $. it is reversible and $P =
[17-12+4]$. As the highest coefficient of the Alexander polynomial is 4,
this knot is not fibered; it provides the first occurence of a non-fibered
minimal knot.
\item The trivial knot $K(4,9,5) =1$.
\item The prime knot $15 N 166131 = K(4,11,5)$ is reversible and $P=
[37-28+12-2]$.
\item The non fibered prime knot $9_{46} = K(4,13,5)$, reversible and $P=
[5-2]$. In fact we can reduce easily the crossing number of the minimal braid to 13 which
makes computations easier.
\end{enumerate}
\begin{figure}[tbp]
\label{946} \fbox{ \includegraphics[width= 6 cm, height = 6 cm]{TRE4135}
\includegraphics[width= 6 cm, height = 6 cm]{noeud946} }
\caption[fig1]{ The first non-fibered prime knot : K(4,13,5) }
\end{figure}
\begin{corollary}
Singularity knots of minimal branch points are not necessarily fibered
\end{corollary}
\vskip 2 in
Marc Soret at marc.soret@lmpt.univ-tours.fr\\
{\em Universit\'e F. Rabelais, D\'ep. de Math\'ematiques, 37000 Tours, France}\\
Marina Ville at mville@math.jussieu.fr \\
{\em C.N.R.S. \& Dept of Mathematics,\\
Northeastern University, 360 Huntington Avenue, Boston MA 02115, USA}
\newpage
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The Hinterlands
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The Agorran Hinterlands are a diverse region with geographic features ranging from rolling meadows with the occasional copse here and there, to grasslands dotted with thickets of trees and undergrowth, to full dense forest and rocky foothills. It is situated between the foothills of the Khasmodean Range to the north and the Agoras River to the south. What remains of civilization in the Agorran Hinterlands are a spattering of small villages and a few larger trading centers spread out between the ruins that act as a constant reminder of the once great Kingdom of Agorras, which was the first to fall in the aftermath of the Crownbreaker Wars. The Agorran people are a proud, hearty, and diverse people and members of many races are still proud to call themselves Agorran.
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Yes, all these great things about how nutritious they are. On the other hand, we read the nutrition label on a package of nuts, see the high fat and calorie content, and get a little puzzled. Aren’t nuts supposed to be good for us?
Nuts are a perfect example of why we can’t just look at fat and calories when evaluating the nutrition potential of a food. Nuts are extremely calorie dense! But guess what? They’re also extremely nutrient dense.
Check it out:
- A quarter cup serving of walnuts has almost 95% of your recommended daily value (RDV) of omega-3 fatty acids. They’re also chock full of manganese and copper.
- The same serving of almonds contains about 45% RDV of vitamin E and 20% RDV of vitamin B2.
- Cashews have less fat than most nuts, and 90% of its unsaturated fat is oleic acid (the heart-healthy fat found in olive oil). They also contain zinc, iron, and biotin.
- Pecans contain over 19 vitamins and minerals, including vitamins E and A, folic acid, calcium, magnesium, copper, phosphorus, potassium, several B vitamins, and zinc.
- The forgotten Brazil nut packs tons of nutrients, including the all-important selenium (a powerful antioxidant).
Plus, all nuts have a decent amount of protein and fiber. Because they’re so high in fat (the healthy, unsaturated kind!), they’re also really filling. A few nuts go a long way. To get the nutrients of nuts without adding tons of calories, try crushing them first before sprinkling them on salads, oatmeal, or rice. You can get more of the flavor throughout the dish without upping the calories too much. Nuts make great snacks on their own or as part of trail mix. I try to measure out a serving before I snack, though.
Just remember: If something is packed with nutrients and healthy fat, it’s also going to have a few more calories. That’s nothing to be afraid of!
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Lemon-Roasted Chicken with Spicy Zucchini and Spinach
"Drizzling bright, zesty lemon oil over meat, vegetables, or pasta is an easy flavor boost. My recipe takes an ordinary olive oil and adds zing for a fraction of the cost of a 'gourmet' flavored oil," says Carla Hall.
Total Time:
Prep:
Level: Moderate
Yield: 4 servings (cost per serving $1.47)
Serves: 4
Ingredients
- 4 chicken legs
- 3 tbsp. lemon oil (recipe below)
- 1 tbsp. fresh thyme leaves
- kosher salt
- pepper
- 8 clove garlic
- 4 small zucchini
- 2 c. baby spinach
- ½ tsp. crushed red pepper flakes
Lemon Oil
- 1 lemon
- ½.
From: Woman's Day
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Still moping months after being dumped by her Arizona boyfriend Leo, fifteen-year-old Stargirl, a home-schooled free spirit, writes "the world's longest letter" to Leo, describing her new life in Pennsylvania.
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\begin{document}
\title{ Planck-scale number of nodal domains for toral eigenfunctions}
\begin{abstract}
We study the number of nodal domains in balls shrinking slightly above the Planck scale for \textquotedblleft generic'' toral eigenfunctions. We prove that, up to the natural scaling, the nodal domains count obeys the same asymptotic law as the global number of nodal domains. The proof, on one hand, uses new arithmetic information to refine Bourgain's de-randomisation technique at Planck scale. And on the other hand, it requires a Planck scale version of Yau's conjecture which we believe to be of independent interest.
\end{abstract}
\author{Andrea Sartori}
\address{ Departement of Mathematics, King's College London, Strand, London WC2R 2LS, England, Uk }
\email{andrea.sartori.16@ucl.ac.uk}
\maketitle
\section{Introduction}
\subsection{Laplacian eigenfunctions and nodal domains}
Given a compact Riemannian surface $(M,g)$ without boundary, let $\Delta_g$ be the Laplace-Beltrami operator on $M$. There exists an orthonormal basis for $L^2(M,d\vol)$ consisting of eigenfunctions $\{f_{E_i}\}$
\begin{align}
\Delta_g f_{E_i}+ E_i f_{E_i}=0 \nonumber
\end{align}
with $0=E_1<E_2\leq...$ listed with multiplicity, and $E_i\rightarrow \infty$. The \textit{nodal set} of an eigenfunction $f_E$ is the zero set $Z(f_E):= \{x\in M: f_E(x)=0\}$ and it is the union of smooth curves outside a finite set of points \cite{C}. The connected components of $M\backslash Z(F_E)$ are called \emph{nodal domains} and we denote their number by $\mathcal{N}(f_E)$. The main object of our interest is to count the number of nodal domains of $f_E$.
The celebrated Courant Nodal Domains Theorem \cite{CH} implies that there exists an explicit constant $C>0$ such that
\begin{align}
\mathcal{N}(f_E) \leq C \cdot E. \label{Courant}
\end{align}
Stern \cite{ST} showed that, on some planar domains, there exists a sequence of eigenfunctions such that the eigenvalue grows to infinity, but $\mathcal{N}(f_E)= 2$, see also \cite{LEW} for a similar result on the two dimensional sphere. Jung and Zelditch \cite{JZ} proved that for most eigenfunctions on certain negatively curved manifolds $\mathcal{N}(\cdot)$ tends to infinity with the eigenvalue. Ingremeau \cite{I} also gave examples of eigenfunctions with $\mathcal{N}(\cdot) \rightarrow \infty$ on unbounded negatively-curved manifolds.
\subsection{The Random Wave Model}
For \textquotedblleft generic" eigenfunctions, the Random Wave Model proposed by Berry \cite{B1,B2} together with the breakthrough work of Nazarov and Sodin \cite{NS} assert that there exists a constant $c>0$ such that
\begin{align}
\mathcal{N}(f_E)= c\cdot E (1+o(1)). \label{Bourgain}
\end{align}
Remarkably, Bourgain \cite{BU} proved that there exist sequences of eigenfunction on the standard flat torus $\mathbb{T}^2=\mathbb{R}^2/\mathbb{Z}^2$ such that \eqref{Bourgain} holds. Subsequently, Buckley and Wigman \cite{BW} extended Bourgain's work to \textquotedblleft generic'' toral eigenfunctions.
We study a finer form of \eqref{Bourgain}: let $s>0$ and let $\mathcal{N}_{f_E}(s,z)$ be the number of nodal domains lying entirely inside the geodesic ball of radius $s$ around the point $z\in M$; then the Random Wave Model would also predict that
\begin{align}
\mathcal{N}_{f_E}(s,z)= c \cdot E (\pi s^2) (1+o(1))\label{Shrinking}
\end{align}
uniformly in $z$, provided that $s \cdot E^{1/2} \rightarrow \infty$, i.e. provided that the radius of the ball shrinks slightly above the \textit{Planck-scale}. We prove that \eqref{Shrinking} holds for \textquotedblleft generic \textquotedblright toral eigenfunctions with $s> E^{-1/2+ o(1)}$.
\subsection{Statement of main results}
\label{main results}
Every Laplace eigenfunction on $\mathbb{T}^2$ can be written as
\begin{align}
f(x)=f_E(x)=\sum_{ \substack{ \xi \in \mathbb{Z}^2\\|\xi|^2=E} }a_{\xi}e(\langle x,\xi \rangle) \label{function}
\end{align}
where $\{a_{\xi}\}_{\xi}$ are complex coefficients and $e(\cdot)= e(2\pi i \cdot )$ (This normalisation implies that the eigenvalue is $4\pi E$, but we will make no distinction between $E$ and $4\pi E$). The eigenvalues are integers $E\in S:=\{E \in \mathbb{N}: E= a^2+b^2 , \text{ for some} \hspace{1mm} a,b \in \mathbb{Z}\}$ and their multiplicity, which we denote by $N=N(E)$, is given by the number of lattice points on the circle of radius $\sqrt{E}$. Moreover, we assume that $\bar{a_{\xi}}=a_{-\xi}$, that is $f$ is real-valued, and that $f$ is normalised via
\begin{align}
||f||^2_{L^2(\mathbb{T}^2)}=\sum_{ |\xi|^2=E}|a_{\xi}|^2=1 \label{normalisation} .
\end{align}
Thanks to (\ref{normalisation}), we can regard the set $(a_{\xi})_{\xi}$ as points on an $N$-dimensional complex sphere. Then, L\'{e}vy concentration of measure \cite[Theorem 2.3]{LE} implies that, most $a_{\xi}$ are small, $|a_{\xi}|^2\leq (\log N)^{O(1)}/ N$ say, with probability asymptotic to $1$. Therefore, we say that $f$ is \textit{flat} if, for all $\rho>0$
\begin{align}
&\max_{|\xi|^2=E}|a_{\xi}|^2= o(N^{-1+\rho}) & \text{as} \hspace{3mm}N\rightarrow \infty. \nonumber
\end{align}
Also, via \eqref{normalisation}, we associate to $f$ the probability measure on the unit circle $\mathbb{S}^1= \mathbb{R}/\mathbb{Z}$
\begin{align}
\mu_{f}=\sum_{ |\xi|^2=E} |a_{\xi}|^2\delta_{\xi/\sqrt{E}} \label{spectral measure}
\end{align}
where $\delta_{\xi/\sqrt{E}}$ is the Dirac distribution at the point $\xi/\sqrt{E}$. Finally, we denote by $c_{NS}(\mu_{f})$ the \textit{Nazarov-Sodin} constant relative to the measure $\mu_{f}$. In order to present our main result, we differ the discussion about $c_{NS}(\cdot)$ to Section \ref{Gaussian random fields} below. Our principal result is the following:
\begin{thm}
\label{theorem 3}
There exists a density one subset \footnote{
By a density one subset we mean a set $S'\subset S$ such that $\lim\limits_{X\rightarrow \infty } \frac{\#\{E\leq X: E\in S'\}}{\#\{E\leq X: E\in S\}}=1$. } $S'\subset S$ such that for all $\epsilon>0$ we have
\begin{align}
\mathcal{N}_f(s,z)= c_{NS}(\mu_{f})\pi s^2 E(1+o_{E\rightarrow \infty}(1)) \nonumber
\end{align}
uniformly for $f$ flat, $s> E^{-1/2+ \epsilon}$ and $z\in \mathbb{T}^2$.
\end{thm}
\begin{rem}
Using the main result in \cite{SA}, Theorem \ref{theorem 3} still holds if we take $s$ such that for all $m>0$ we have $s \cdot E^{1/2}/ (\log E)^m \rightarrow \infty$. For the sake of elegance of the presentation, we decided not to include it in this manuscript.
\end{rem}
The sequence $\{\mu_{f}\}$, even in the special case $a_{\xi}=1/\sqrt{N}$ for all $\xi's$, does not have a unique limit point with respect to the weak$^{\star}$ topology on $\mathbb{S}^1$ \cite{CI,KW, SA2}. Thus, in order to obtain an asymptotic behaviour for $N(f_E)$, we have to pass to subsequences. Kurlberg and Wigman proved \cite[Theorem 1.3]{KW2} that if $\mu_{f}$ weak$^{\star}$ converges to some probability measure $\mu$ on $\mathbb{S}^1$, then $c_{NS}(\mu_{f})= c_{NS}(\mu)(1+o(1))$. This implies the following version of Theorem \ref{theorem 3}:
\begin{cor}
Under the assumptions of Theorem \ref{theorem 3}, suppose that $\mu_{f}$ weak$^{\star}$ converges to some probability measure $\mu$ on $\mathbb{S}^1$, then
\begin{align}
\mathcal{N}_f(s,z)= c_{NS}(\mu)\pi s^2 E(1+o(1)). \nonumber
\end{align}
uniformly for $f$ flat, $s> E^{-1/2+ \epsilon}$ and $z\in \mathbb{T}^2$.
\end{cor}
\subsection{Nodal length in shrinking balls}
One of the novel aspects in the proof of Theorem \ref{theorem 3} is the study of the \textit{nodal length}, that is the Hausdorff measure of the nodal set, of toral eigenfunctions in shrinking balls. The main (open) question about the nodal length of Laplace eigenfunctions is the following conjecture of Yau: let $f_E$ be a Laplace eigenfunction with eigenvalue $E$ on a smooth, compact manifold without boundaries $M$, then
\begin{align}
\sqrt{E}\ll_M \mathcal{L}(f_E)= \mathcal{H}^{n-1}\{x\in M: f(x)=0\}\ll_M \sqrt{E} \nonumber
\end{align}
Donnelly and Fefferman \cite{DF} showed that Yau's conjecture holds for real-analytic manifolds. Recently, Logunov and Malinnikova \cite{LM,L1,L2} proved the lower-bound in the smooth case and gave a polynomial upper-bound.
As for the nodal domains count, the Random Waves Model suggests that, for \textquotedblleft generic" Laplace eigenfunctions, a rescaled version of Yau's conjecture should hold at small scales, that is for any $z\in M$
\begin{align}
s\sqrt{E}\ll_M s^{-1}\mathcal{L}_f(s,z):= s^{-1}\mathcal{H}^{n-1}\{x\in B(s,z): f(x)=0\}\ll_M s\sqrt{E}. \nonumber
\end{align}
provided that $s$ shrinks slightly above Plank-scale. We prove the following:
\begin{prop}
\label{claim}
Let $f$ be as in \eqref{function} and let $\epsilon>0$, then
\begin{align}
s\sqrt{E}\ll \mathcal{L}_f(s,z)s^{-1}\ll s\sqrt{E} \nonumber
\end{align}
uniformly for $s> E^{-1/2+\epsilon}$ and $z\in \mathbb{T}^2$.
\end{prop}
One particular aspect of Proposition \ref{claim} is that it holds for \textit{every} toral eigenfunction. This might fail on other surfaces: on the $2$-sphere $\mathbb{S}^2= \{x\in \mathbb{R}^3: ||x||^2=1\}$ one can consider the \textquotedblleft sectoral" harmonic $g(\theta,\phi)=\sin (m\phi) P^m_m(\cos(\theta))$ in spherical-coordinates, where $P^m_m(\cdot)$ is the associated Legendre polynomial. Then $\Delta g= -m(m+1) g$ and the upper-bound in Proposition \ref{claim} fails around the North Pole.
\vspace{2mm}
\paragraph{\textit{Application to Laplace eigenfunctions on the square}} The proof of Proposition \ref{claim} is general enough that it can also address Laplace eigenfunctions on the square $[0,1]^2$ with either Dirichlet or Neumann boundary conditions. The study of the nodal length of \textit{random} Laplace eigenfunctions on the square, known as \textit{boundary adapted Arithmetic Random Waves}, was initiated by Cammarota, Klurman and Wigman \cite{CKW}. A major step in their work is to bound the expectation of the nodal length in squares of side $O(1)/\sqrt{E}$, where $E$ is the eigenvalue. We prove the following:
\begin{prop}
\label{claim2}
Let $\tilde{f}$ be a Laplace eigenfunction on the square $[0,1]^2$ with either Dirichlet or Neumann boundary conditions and let $E$ be its eigenvalue. Then we have
\begin{align}
\mathcal{L}_{\tilde{f}}(s,z)s^{-1}\ll s\sqrt{E} + N \nonumber
\end{align}
uniformly for $s>0$ and $z\in \mathbb{T}^2$, where $N=N(E)$ is as in Section \ref{main results}.
\end{prop}
From Proposition \ref{claim2}, it follows that for any fixed $C>0$ we have
\begin{align}
\mathcal{H}^1\{x\in B(C/\sqrt{E},z): \tilde{f}(x)=0 \}\ll \frac{N}{\sqrt{E}}. \label{1.1}
\end{align}
For \textit{random} Laplace eigenfunctions on the square, Cammarota, Klurman and Wigman \cite[Proposition 2.5]{CKW} showed that the bound $\mathcal{H}^1\{x\in B(C/\sqrt{E},z): \tilde{f}(x)=0 \}\ll N^2/\sqrt{E}$ holds with high probability. So \eqref{1.1} not only refines \cite[Proposition 2.5]{CKW} but it also provides a deterministic results which does not rely on moments estimates.
\subsection{Bourgain's de-randomisation in shrinking sets}
Another novel aspect in the proof of Theorem \ref{theorem 3} is an extension of Bourgain's de-randomisation technique to shrinking sets. Let $f$ be as in \eqref{function} and suppose that $a_{\xi}=1$ for all $\xi's$, moreover let $F_x(y)= f(x+y/\sqrt{E})$ for $y\in [-1/2,1/2]^2$. Bourgain \cite{BU} showed that the assemble $\{F_x\}$, where $x$ is drawn uniformly at random from $\mathbb{T}^2$, approximates the Gaussian field with spectral measure the Lebesgue measure on $\mathbb{S}^1$ (see Section \ref{Gaussian random fields} below for some background on Gaussian fields). We use some new informations about sum of lattice points called \textit{quasi-spectral correlations} to show that this approximations still holds even when $x$ is drawn uniformly at random from $B(s,z)$ for $s>E^{-1/2+o(1)}$ and $z\in \mathbb{T}^2$, Proposition \ref{main prop} below.
Since the proof of Proposition \ref{main prop} is quite technical, to give the reader an idea of how such properties of lattice points are exploited, we show here that $F_x(0)=f(x)$ approximates a standard Gaussian random variable when $x$ is drawn uniformly at random from $B(s,z)$. Via the method of moments, we have to evaluate for $l \in \mathbb{N}$
\begin{align}
\frac{1}{\pi s^2} \int_{B(s,z)} |F_x(0)|^{2l} dx&=\frac{1}{\pi s^2} \sum_{\xi_1,...,\xi_{2l}} \int_{B(s,z)}e(\langle x, \xi_1-\xi_2+...+\xi_{2l-1}-\xi_{2l}\rangle) dx .\nonumber
\end{align}
Separating the terms with $\xi_1 +...-\xi_{2l}=0$, known as\textquotedblleft$2l$-spectral correlations", from the other terms, \textquotedblleft $2l$ spectral quasi-correlations" , we obtain
\begin{align}
\frac{1}{\pi s^2} \int_{B(s,z)} |f(x)|^{2l} dx= \sum_{\xi_1- \xi_2+...-\xi_{2l}=0}1
+O\left( \sum_{|\xi_1- \xi_2+...-\xi_{2l}|>0} \frac{J_1(s|\xi_1- \xi_2+...-\xi_{2l}|)}{s|\xi_1- \xi_2+...-\xi_{2l}|} \right) \label{intro3}
\end{align}
where $J_1(\cdot)$ is the Bessel function of the first kind.
The main contribution to the first term in \eqref{intro3} comes from the diagonal solutions $\xi_1=\xi_2$,..., $\xi_{2l-1}=\xi_{2l}$ and their permutations, these contribute $2l!/(2^l\cdot l!)$. Bombieri and Bourgain \cite{BB} showed that, for \textit{generic} $E\in S$, the \textquotedblleft off-diagonal" solutions have lower order as $N\rightarrow \infty$. Thus, the first term on the right hand side of \eqref{intro3} is asymptotic to $2l!/(2^l\cdot l!)$. We are left to show that the second term on the right hand side of \eqref{intro3} tends to $0$ as $N\rightarrow \infty$. Theorem \ref{semi} below implies that for \textit{generic} $E\in S$, $s|\xi_1- \xi_2+...-\xi_{2l}|\geq E^{o(1)}$. Since Bessel functions decay at infinity, this implies that the second term in \eqref{intro3} tends to $0$, as required.
\subsection{Related results}The main body of results regarding statistics of Laplace eigenfunctions in shrinking sets concern their mass distribution. Let $f_E$ be a Laplace eigenfunction on a surface $M$, then one is interested in finding the smallest $s$ such that $\int_{B(s,z)} |f|^2 d\vol= \pi s^2(1+o_{E\rightarrow \infty}(1))$. The celebrated Quantum Ergodicity Theorem \cite{DV,S,Z} asserts that, if the geodesic flow on M is ergodic, then one can take any \textit{fixed} $s>0$ for a density one subsequence of eigenfunctions. Luo and Sarnak \cite{LS} showed that, on the modular surface, one can take $s>E^{-\alpha}$ for some $\alpha>0$ for a density one subsequence, see also \cite{Y}. Hezari, Rivi\`{e}re \cite{HR} and independently Han \cite{H} proved that, if $M$ has negative sectional curvature, then one can take $s> \log(E)^{-\alpha}$ for some small $\alpha>0$ for a density one subsequence. On $\mathbb{T}^2$ Lester and Rudnick \cite{LR} showed that $s> E^{-1/2+ o(1)}$, again for a density one subsequence.
Granville and Wigman \cite{GW} and subsequently Wigman and Yesha \cite{WY} studied the mass distribution of eigenfunctions on $\mathbb{T}^2$ \emph{at Planck scale} by drawing the centre of the ball randomly uniformly. They showed that, for certain eigenfunctions the mass equidistributes in almost every ball, see also \cite{HU,HU1} for similar work on the modular surface. The author \cite{SA1} classified all limiting mass-distributions \emph{at Planck scale} for \textquotedblleft generic" toral eigenfunctions.
Results regarding the zero set are more modest: Benatar, Marinucci and Wigman \cite{BMW} studied the behaviour of nodal length for \emph{random} toral eigenfunctions at scales $s=E^{-1/2+o(1)}$ and found the asymptotic law of the variance.
To the best of the author's knowledge, our own Theorem \ref{theorem 3} is the only asymptotic result on nodal domains at small scales.
\subsection{Notation}
Let $t \rightarrow \infty$ be some parameter, we say that the quantity $X=X(t)$ and $Y=Y(t)$ satisfy $X\ll Y$ , $X\gg Y$ if there exists some constant $C$, independent of $t$, such that $X\leq C Y$ and $X\geq CY$ respectively. We also write $O(X)$ for some quantity bounded in absolute value by a constant times $X$ and $X=o(Y)$ if $X/Y\rightarrow 0$ as $t\rightarrow \infty$, in particular we denote by $o(1)$ any function that tends to $0$ (arbitrarily slowly) as $x\rightarrow \infty$. We denote by $B(s,z)$ the (open) ball of radius $s$ with centre $z$, by $B(s)$ for the ball centred at $0$ and by $\overline{B}(s)$ the closure of $B(s)$. When the specific radius is unimportant, we simply write the ball as $B$ and $\frac{1}{2}B$ for the concentric ball with half the radius. Finally, we denote by $\Omega$ an abstract probability space where every random object is defined.
\section{Preliminaries}
\subsection{Number theoretic background}
\label{NTpre}
Recall that $S=\{n\in \mathbb{N}: n= a^2+b^2 , \text{ for some} \hspace{1mm} \\ a,b \in \mathbb{Z}\}$. In this section we collect some number theoretic results that will be used to define the set $S'\subset S$ in Theorem \ref{theorem 3}. Let $E\in S$ and write its prime factorisation as $E= \prod_{p\equiv 1 \pmod 4}p ^{\alpha_p}\prod_{q\equiv 3 \pmod 4} q^{2\beta_q}$ where $\alpha_p,\beta_q\in \mathbb{N}$. It follows that $N(E)=4 \prod_{p\equiv 1 \pmod 4}(\alpha_p+1)$. Thus, by the divisor bound, we have
\begin{align}
N(E) \ll \exp \left( \frac{\log E}{\log\log E}\right). \label{divisor bound}
\end{align}
Moreover, by the Erd\"{o}s-Kac Theorem \cite[Theorem 12.3]{E}, for almost all integers (representable as sum of two squares) the number $\#\{ p|E: p\equiv 1 \pmod 4\}\rightarrow \infty$ as $E\rightarrow \infty$. So we also have the following lemma:
\begin{lem}
\label{N infinity}
For a density one subset of $E\in S$, $N(E) \rightarrow \infty$ as $E\rightarrow \infty$.
\end{lem}
To state the next results we need some notation: let $l\in \mathbb{N}$ and $E\in S$, denote by $\mathcal{S}(l,E)$ the number of solutions to
\begin{align}
\xi_1+...+ \xi_l=0 \label{14}
\end{align}
where $\xi_j\in \mathbb{Z}^2$ and $|\xi_j|^2=E$, that is \textit{$l$-spectral correlations}. When $l$ is odd, by congruence obstruction modulo $2$, there are no solutions to \eqref{14}. When $l$ is even, we have the following \cite[Theorem 17]{BB} and \cite[Lemma 4]{BU}:
\begin{thm}[Bombieri-Bourgain]
\label{BB}
Let $B=B(E)$ be an arbitrarily slow growing function of $E$, $l\in \mathbb{N}$ and $0<\gamma<1$. Then, for a density one subset of integers $E\in S$, we have
\begin{align}
\mathcal{S}(2l,E)= \frac{(2l)!}{2^l \cdot l!}N^{l} +O(N^{\gamma l}) \nonumber
\end{align}
uniformly for all $l\leq B$, where the constant implied in the notation is absolute.
\end{thm}
Moreover, provided that is not zero, one can give a quantitative lower bound to the sum in \eqref{14}, see \cite[Theorem 1.4]{BMW} and the refinement \cite[Theorem 1.1]{SA}:
\begin{thm}
\label{semi}
Let $B=B(E)$ be an arbitrarily slow growing function of $E$, $l\in \mathbb{N}$ and $Q=Q(E)$ be a function such that $Q \cdot E^{1/2}/(\log E)^m\rightarrow \infty$ for all $m\geq 0$. Then, for a density one subset of integers $E\in S$, we have
\begin{align}
||\xi_1+...+\xi_l||>Q\nonumber
\end{align}
uniformly for all choices of $\xi_1,...,\xi_l$ and $l\leq B$.
\end{thm}
\textbf{The set $S'$}. We are now ready to define the subset in Theorem \ref{theorem 3}: let $S'$ be the set of $E\in S$ which satisfy the conclusion of Lemma \ref{N infinity}, Theorem \ref{BB} and Theorem \ref{semi}. By the discussion in this section, $S'$ has density one.
\subsection{Gaussian fields background}
\label{Gaussian random fields}
We briefly collect some definitions about Gaussian fields (on $\mathbb{R}^2$). A (real-valued) Gaussian field $F$ is a measurable map $F: \mathbb{R}^2 \times \Omega\rightarrow \mathbb{R}$ for some probability space $\Omega$, such that all finite dimensional distributions $(F(x_1, \cdot),...F(x_n,\cdot))$ are multivariate Gaussian. $F$ is \textit{centred} if $\mathbb{E}[F]=0$ and \textit{stationary} if its law is invariant under translations $x\rightarrow x+\tau$ for $\tau \in \mathbb{R}^2$. The \textit{covariance} function of $F$ is
\begin{align}
\mathbb{E}[F(x)\cdot F(y)]= \mathbb{E}[F(x-y)\cdot F(0)]. \nonumber
\end{align}
Since the covariance is positive definite, by Bochner's theorem, it is the Fourier transform of some measure $\mu$ on the $\mathbb{R}^2$. So we have
\begin{align}
\mathbb{E}[F(x)F(y)]= \int_{\mathbb{R}^2} e\left(\langle x-y, \lambda \rangle\right)d\mu(\lambda). \nonumber
\end{align}
The measure $\mu$ is called the \textit{spectral measure} of $F$ and, since $F$ is real-valued, satisfies $\mu(-I)=\mu(I)$ for any (measurable) subset $I\subset \mathbb{R}^2$. By Kolmogorov theorem, $\mu$ fully determines $F$, so we may simply write $F=F_{\mu}$.
\subsection{Nazarov-Sodin constant}
Nazarov and Sodin \cite{NS} found the asymptotic law of the expected number of nodal domains of a stationary Gaussian field in growing balls around the origin, provided its spectral measure satisfies certain (simple) properties. We state here a simplified and slightly adapted form of their Theorem, see \cite[Proposition 1.1]{KW2}:
\begin{thm}
\label{Nazarov-Sodin}
Let $\mu$ be a probability measure on $\mathbb{S}^1$, invariant by rotation by $\pi$ and let $\mathcal{N}(F_\mu,R)$ be the number of nodal domains of $F_{\mu}$ in a ball of radius $R>0$ centred ad the origin. Then, there exists some constant $c_{NS}(\mu)$ such that
\begin{align}
\mathbb{E}[ \mathcal{N}(F_{\mu},R) ]= c_{NS}(\mu)R^2 + O\left(R \right). \nonumber
\end{align}
Moreover $c_{NS}(\mu)>0$ if $\mu$ does not have any atoms.
\end{thm}
We will need the following version of Theorem \ref{Nazarov-Sodin}, see \cite[Proposition 3.4 and Proposition 3.5]{BW}.
\begin{prop}
\label{stability}
Let $R>1$ and $\mu_{f}$ be as in \eqref{spectral measure}. Then, for any function $\psi$ with $||\psi||_{\mathcal{C}^1}$ sufficiently small in terms of $R$, we have
\begin{align}
\mathbb{E}[\mathcal{N}( F_{\mu_{f}} +\psi,R)]= c_{NS}(\mu_{f})R^2 (1+o(1)) \hspace{8mm} \text{as} \hspace{2mm} R\rightarrow \infty.\nonumber
\end{align}
\end{prop}
We conclude this section mentioning another result concerning the positivity of $c_{NS}(\mu_{f})$. Suppose that $\mu_{f}$ is invariant under $\pi/2$ rotations and reflection on the $X$-axis (i.e. $(x_1,x_2)\rightarrow (x_1,-x_2)$). Among these measures, Kurlberg and Wigman \cite[Theorem 1.5]{KW2} showed that there are only two with vanishing Nazarov-Sodin constant:
\begin{align}
&\nu= \sum_{k=1}^4 \delta_{e^{i\pi k/2}} &\tilde{\nu}= \sum_{k=1}^4 \delta_{e^{i(\pi k/2+\pi/4)}} \nonumber.
\end{align}
\section{Nodal length of toral eigenfunctions in shrinking sets}
\label{semi-loc}
The aim of this section is to prove the Proposition \ref{claim} and Proposition \ref{claim2}. First we show the following consequence of Proposition \ref{claim}:
\begin{prop}
\label{semi-locality}
Let $R>1$, $\epsilon>0$ and let $f$ be as in \eqref{function}. Then, uniformly for $s>E^{-1/2+\epsilon}$ and $z\in \mathbb{T}^2$, we have
\begin{align}
\mathcal{N}_f(s,z)= \frac{E}{R^2}\int_{B(s,z)} \mathcal{N}_f\left(\frac{R}{\sqrt{E}},x\right)dx + O\left(\frac{Es^2}{\sqrt{R}}\right). \nonumber
\end{align}
\end{prop}
\begin{proof}[Proof of Proposition \ref{semi-locality} assuming Proposition \ref{claim}]
Let $L>1$ be some parameter to be chosen later. By Proposition \ref{claim} the nodal length of $f$ in $B(s,z)$ is, up to rescaling, at most $\sqrt{E}s$. It follows that there are at most $Es^2/L$ nodal domains of diameter bigger than $L/Es^2$. Therefore, if we divide $B(s,z)$ into balls of radius $R/\sqrt{E}$, any nodal domain of diameter smaller than $L/Es^2$ intersects at most $O(L^2/R^2)$ balls. We deduce that
\begin{align}
\mathcal{N}_f(s,z)= \frac{E}{R^2}\int_{B(s,z)}\mathcal{N}_f\left(\frac{R}{\sqrt{E}},x\right)dx + O\left(\frac{Es^2}{L}\right) + O\left(\frac{Es^2 L^2}{R^2}\right). \nonumber
\end{align}
The Proposition follows choosing $L= \sqrt{R}$.
\end{proof}
\subsection{Proof of Proposition \ref{claim}, upper bound}
The proof will be carried out through a series of lemmas, the first is a standard tool to count zeros of analytic functions.
\begin{lem}[Jensen's bound]
\label{Jensen's bound}
Let $h$ be a complex analytic function on some ball $B \subset \mathbb{C}$ and let $Z(h,\frac{1}{2}\overline{B})$ be the number of its zeros in $\frac{1}{2}\overline{B}$. Then,
\begin{align}
Z\left(h,\frac{1}{2}\overline{B}\right)\ll \log \frac{\underset{B}{\sup} |h|}{\underset{\frac{1}{2}\overline{B}}{\max}|h|}. \nonumber
\end{align}
\end{lem}
\begin{proof}
Up to translation and rescaling, we may assume that $h$ is defined on the unit ball, which we again denote by $B$. Let $w_1,...,w_n$ be the zeros of $h$ on $\frac{1}{2}\overline{B}$ counted with multiplicity and consider the Blaschke factor $ D(z,\omega_i)= (z-\omega_i)/(1-z\overline{\omega_i})$. Then, we can write $h(z)= \prod_{i}D(z,\omega_i) g(z)$ for some $g$ analytic on $B$ with $\underset{B}{\sup} |h|= \underset{B}{\sup} |g|$. Since $| D(z,\omega_i)| \leq (4/5)$ for $z\in \frac{1}{2}\overline{B}$, letting $Z= Z\left(h,\frac{1}{2}\overline{B}\right)$, we have
\begin{align}
\underset{\frac{1}{2}\overline{B}}{\max}|h|\leq \left(\frac{4}{5}\right)^{Z}\underset{\frac{1}{2}\overline{B}}{\max}|g|\leq \left(\frac{4}{5}\right)^{Z}\underset{B}{\sup}|g|\leq \left(\frac{4}{5}\right)^{Z}\underset{B}{\sup}|h|. \label{9}
\end{align}
The lemma follows taking the logarithm on both sides of \eqref{9}.
\end{proof}
We also need the following well-known formula of Crofton, see for example \cite{F}.
\begin{lem}
\label{Crofton's formula} Let $f$ be as in \eqref{function}, $s>0$ and $z\in \mathbb{T}^2$, moreover let $g(y)=f(z+sy)$ for $y\in B(1)$. Then, uniformly in $s$ and $z$, we have
\begin{align}
\mathcal{L}_f(s,z)s^{-1}=\mathcal{L}(g)\ll \int_{B(1)} \int_{\mathbb{S}^1} Z( g(u+tw)) d\omega du \nonumber
\end{align}
where $ Z( g(u+tw))$ is the number of zeros of $g$ as a function of $t\in [0,1]$.
\end{lem}
Finally, we need the following lemma, see \cite{N,T}:
\begin{lem}[Nazarov-Turan]
\label{NT}
Let $J\in \mathbb{N}$ and let $h(t)=\sum_{i=1}^{J}a_{\xi}e(\xi_i \cdot t)$ for $t\in \mathbb{C}$ and suppose that $\xi_i \in \mathbb{C}$ are distinct. Then, for any $B\subset \mathbb{C}$ and $\Omega\subset B$ a measurable subset, we have
\begin{align}
\underset{t\in B}{\sup} |h|<\left(c\frac{|\Omega|}{|B|}\right)^{J-1} e ^{ \max_i |\xi_i| |B|} \underset{t\in \Omega}{\sup} |h|. \nonumber
\end{align}
for some explicit $c>0$.
\end{lem}
We are finally ready to prove the upper bound in Proposition \ref{claim}.
\begin{proof}[Proof of the upper bound in Proposition \ref{claim}]
Let $g(y)=f(z+sy)$ for $y\in B(1)$ and let $h$ be the extension of $g$ to the complex unit ball . By Lemma \ref{Crofton's formula}, we have
\begin{align}
\mathcal{L}_f(s,z)s^{-1}=\mathcal{L}(g)\ll \int_{B(1)} \int_{\mathbb{S}^1} Z( g(u+tw)) d\omega du. \label{10.1}
\end{align}
By Lemma \ref{Jensen's bound} and Lemma \ref{NT}, we have
\begin{align}
Z( g(u+t\omega))\leq Z(h(u+z\omega))\ll \log \frac{\underset{D}{\sup} |h|}{\underset{\frac{1}{2}\overline{D}}{\max}|h|} \ll N + s\sqrt{E} \leq s\sqrt{E} \label{10}
\end{align}
uniformly in $u$ and $\omega$. The last inequality in \eqref{10} follows by \eqref{divisor bound} and the fact that $s>E^{-1/2+\epsilon}$. The upper bound then follows by \eqref{10.1} and \eqref{10}.
\end{proof}
\subsection{Proof of Proposition \ref{claim}, lower bound}
\label{lower bound}
The proof of the lower bound is standard, but we include it for completeness. We need the following result about the density of the zero set:
\begin{lem}
\label{density}
Let $f$ be as in \eqref{function}. There exists some absolute constant $c>0$ such that, uniformly for all $z\in \mathbb{T}^2$, the ball $B(c/\sqrt{E},z)$ contains a point where $f$ vanishes.
\end{lem}
\begin{proof}
Let $s>0$, and observe that the function $h(x,t)= f(x)e^{\sqrt{E}t}$ is harmonic in $D=B(s,z) \times [-s,s]$. If $f$ does not vanish in $B(s,z)$, then $h$ is positive; so it satisfies Harnack's inequality:
\begin{align}
\sup_D |h|\leq C \inf_D |h| \label{16}
\end{align}
for some absolute constant $C>0$. One the other hand,
\begin{align}
\sup_D |h| \geq \sup_{B(s,z)}|f| \exp (s\sqrt{E})\geq\inf_{B(s,z)}|f| \exp (s\sqrt{E}) \label{17}
\end{align}
The lemma follows combining \eqref{16} and \eqref{17} and choosing $c$ appropriately.
\end{proof}
We are finally ready to prove the lower bound in Proposition \ref{claim}.
\begin{proof}[Proof of the lower bound in Proposition \ref{claim}]
Using Lemma \ref{density}, we can divide $B(s,z)$ in $O(E s^2)$ balls of radius $c/\sqrt{E}$ for some appropriate $c>0$ such that $f$ vanishes at the centre of each ball. Let $B$ be one of these balls, then the Faber-Krahn inequality \cite[Theorem 1.5]{M} says that every nodal domain has inner radius at least $c_1/E^{1/2}$ for some absolute $c_1>0$, so we have
\begin{align}
\mathcal{H}^{1}\{ x\in B: f(x)=0\}\gg E^{-1/2}\label{11}
\end{align}
Since \eqref{11} holds for each of the $O(s^2E)$ balls, the lower bound follows.
\end{proof}
\subsection{Proof of Proposition \ref{claim2}}
As mentioned in the introduction the proof of Proposition \ref{claim2} follows the proof of Proposition \ref{claim}. We now give some of the details
\begin{proof}[Proof of Proposition \ref{claim2}]. Let $\mathcal{E}_E= \mathcal{E}:=\{\xi\in \mathbb{Z}^2: |\xi|^2=E\}$, we define an equivalence relation on $\mathcal{E}$ as follows: let $\xi=(\xi^1,\xi^2),\eta=(\eta^1,\eta^2) \in \mathcal{E}$, then $\xi \sim \eta$ if $\xi^1=\pm \eta^2$ and $\xi^2=\pm \eta^2$. Then the general Laplace eigenfunction with eigenvalue $\pi E$ (we make no distinction between $E$ and $\pi E$) satisfying either Dirichlet or Neumann boundary conditions is
\begin{align}
&\tilde{f}_{\text{Dirichlet}}(x)= \sum_{ \xi\in \mathcal{E}/ \sim} a_{\xi} \sin( \pi \xi^1 x^1)\sin( \pi \xi^2 x^2) \label{Di} \\
&\tilde{f}_{\text{Neuman}}(x)= \sum_{ \xi\in \mathcal{E}/ \sim} b_{\xi} \cos( \pi \xi^1 x^1)\cos( \pi \xi^2 x^2) \label{Neu}
\end{align}
where $x=(x^1,x^2)$. Using the formulas $\sin(a)\sin(b)= 2^{-1}(\cos(a-b)- \cos(a+b))$ and $\cos(a)\cos(b)= 2^{-1}(\cos(a+b)+ \cos(a-b))$, we can rewrite \eqref{Di} and \eqref{Neu} as
\begin{align}
&\tilde{f}_{\text{Dirichlet}}(x)= \sum_{ \xi\in \mathcal{E}/ \sim} \tilde{a}_{\xi} e\left(\langle \xi, x\rangle\right)
&\tilde{f}_{\text{Neuman}}(x)= \sum_{ \xi\in \mathcal{E}/ \sim} \tilde{b}_{\xi} e\left(\langle \xi, x\rangle\right) \nonumber
\end{align}
for some complex coefficients $\tilde{a}, \tilde{b}$. The proof now follows step by step the proof of the upper bound in Proposition \ref{claim}.
\end{proof}
\section{Bourgain's de-randomisation in shrinking sets}
\label{BourgainS}
Let $R>1$ be fixed, and consider the restriction of $f$, as in \eqref{function}, to a small square centred at $x\in \mathbb{T}^2$:
\begin{align}
F_x(y)= f\left( x+ \frac{R}{\sqrt{E}}y\right). \label{F}
\end{align}
for $y\in B(1)$. In this section we are going to show that if we sample $x$ uniformly at random from $B(s,z)$, where $z\in \mathbb{T}^2$ and $s>E^{-1/2+\epsilon}$, then the ensemble $\{F_x\}_{x\in B(s,z)}$ approximates the Gaussian field $F_{\mu_{f}}$. The proofs are based on \cite{BU,BW}; nevertheless, the use of Theorem \ref{semi} is required to control the averaging over $B(s,z)$.
\subsection{Approximating $f$ in small squares}
\label{approximating}
In this section, we construct an auxiliary function $\phi_x(y)$ which approximates $F_x(y)$ for most $x\in \mathbb{T}^2$. We begin with some notation: let $K>1$ be some (large) parameter and divide the circle $\mathbb{S}^1$ into arcs $I_k$, of length $1/2K$ for $k\in\{-K,...,K\}$. Furthermore, let $\delta>0 $ be some (small) parameter and denote by $\mathcal{K}\subset \{-K,...,K \}$ the subset of indices such that if $k\in \mathcal{K}$ then
\begin{align}
\mu_f(I_k)>\delta. \label{41}
\end{align}
Finally, let $\mathcal{E}^k=\mathcal{E}_E^k:=\{ |\xi|^2=E: \xi \in I_k\}$ and let $\zeta_k$ be the mid point of $I_k$.
We are ready to begin the construction, first we re-write $F_x$ as
\begin{align}
F_x(y)= \sum_{k\in \mathcal{K}}\sum_{\xi\in \mathcal{E}^k}a_{\xi}e(\langle \xi, x \rangle)e\left(\left\langle\frac{\xi}{\sqrt{E}},Ry\right\rangle\right) + \sum_{k\not\in \mathcal{K}}\sum_{\xi\in \mathcal{E}^k}a_{\xi}e(\langle \xi, x \rangle)e\left(\left\langle\frac{\xi}{\sqrt{E}},Ry\right\rangle\right). \label{42}
\end{align}
Second we approximate $\xi/\sqrt{E}$ by $\zeta_k$ for all $\xi\in \mathcal{E}^k$, and define the function
\begin{align}
\phi_x(y)&=\sum_{k\in \mathcal{K}}\left(\sum_{\xi\in \mathcal{E}^k}a_{\xi}e(\langle \xi, x \rangle) \right)e(\langle R\zeta^{k},y \rangle)= \sum_{k\in \mathcal{K}}\mu_{f}(I_k)^{1/2}b_k(x)e(\langle R\zeta^{k},y \rangle) \label{phi}
\end{align}
where
\begin{align}
b_k(x)=\frac{1}{\mu_{f}(I_k)^{1/2}}\sum_{\xi\in \mathcal{E}^{(k)}}a_{\xi}e(\langle\xi,x\rangle). \label{bk}
\end{align}
The following lemma shows that $\phi_x(y)$ is a good approximation to $F_x(y)$ for most $x\in \mathbb{T}^2$.
\begin{lem}
\label{first approx}
Let $\epsilon>0$, $R,K,\delta$ be as in Section \ref{approximating}, $F_x$, $\phi_x$ be as in \eqref{F} and \eqref{phi} respectively and $S'$ be defined in Section \ref{NTpre}. Then, for all $E\in S'$ we have
\begin{align}
\frac{1}{\pi s^2}\int_{B(s,z)} || F_x- \phi_x||_{\mathcal{C}^1(B(1))} dx \ll R^{6}K\delta + R^{8}K^{-2} + R^{8}E^{-(1/3)\epsilon} \nonumber
\end{align}
uniformly for all $s>E^{1/2+\epsilon}$ and $z\in \mathbb{T}^2$.
\end{lem}
\begin{proof}
Thanks to the Sobolev embedding Theorem, we bound the $\mathcal{C}^1$ norm by the $H^3$ norm
\begin{align}
\frac{1}{\pi s^2}\int_{B(s,z)} || F_x- \phi_x||_{\mathcal{C}^1} dx \ll \frac{1}{\pi s^2}\int_{B(s,z)} \int_{B(1)} | D^{\alpha} (F_x (y)- \phi_x(y)|^2 dy \label{49}
\end{align}
where $|\alpha|= |(\alpha_1,\alpha_2)| \leq 3$ and $D^{\alpha}= \partial_{\alpha_1}\partial_{\alpha_2}$. First, we estimate the contribution coming from the second term on the right hand side of \eqref{42}. Expanding the square and using the triangular inequality we obtain
\begin{align}
\frac{1}{\pi s^2}\int_{B(s,z)} \int_{B(1)} \left| D^{\alpha}\sum_{k\not\in \mathcal{K}}\sum_{\xi\in \mathcal{E}^k}a_{\xi}e(\langle \xi, x \rangle)e\left(\left\langle\frac{\xi}{\sqrt{E}},Ry\right\rangle\right)\right|^2 dxdy \nonumber \\
\ll \frac{1}{\pi s^2}R^{2|\alpha|} \sum_{k,k'\not\in \mathcal{K}} \sum_{\substack{\xi\in \mathcal{E}^k \\ \xi'\in \mathcal{E}^{k'}}}| a_{\xi}\overline{a_{\xi'}}|\left| \int_{B(s,z)}e(\langle \xi-\xi', x \rangle)dx \right| .\label{43}
\end{align}
Observe that for $a\in \mathbb{R}^2$
\begin{align}
\int_{B(s,z)} e(\langle a,x\rangle) dx = \begin{cases}
\pi s^2 & a=0 \\
\pi s^2 e(\langle a,z\rangle) \frac{J_1(s|a|)}{s|a|} & a\neq 0
\end{cases}. \label{orthogonality}
\end{align}
So we separate the terms with $\xi=\xi'$ from the others on the right hand side of \eqref{43} to obtain
\begin{align}
\eqref{43}\ll R^{2|\alpha|}\sum_{k\not\in \mathcal{K}}\sum_{\xi\in \mathcal{E}^k} |{a_{\xi}}|^2 + R^{2|\alpha|} \sum_{k,k'} \sum_{\xi \neq \xi'} |a_{\xi}a_{\xi'}| \frac{J_1\left( s|\xi-\xi'|\right)}{s|\xi-\xi'|} \label{44}
\end{align}
Since $k\not\in \mathcal{K}$ implies $\sum_{\xi\in \mathcal{E}^k} |{a_{\xi}}|^2= \mu_{f}(I_k)\leq \delta$, the first term on the right hand side of \eqref{44} is bounded by $R^{2|\alpha|} K \delta$. By Theorem \ref{semi} $|\xi-\xi'|\gg E^{1/2-2\epsilon}$ so $s|\xi-\xi'|\gg E^{\epsilon}$, it follows that
\begin{align}
\frac{J_1\left( s|\xi-\xi'|\right)}{s|\xi-\xi'|} \ll E^{(-2/3)\epsilon} \label{45}
\end{align}
where we have used the bound $J_1(T)\ll T^{-1/2}$ valid for all sufficiently large $T$. Using \eqref{45}, estimating trivially $|a_{\xi}|\leq 1$ and bearing in mind \eqref{divisor bound}, we obtain
\begin{align}
\sum_{k,k'} \sum_{\xi \neq \xi'} |a_{\xi}a_{\xi'}| \frac{J_1\left( s|\xi-\xi'|\right)}{s|\xi-\xi'|} \ll E^{(-2/3)\epsilon} \cdot N^2 \ll E^{(-1/3)\epsilon}. \nonumber
\end{align}
All in all, we have shown that
\begin{align}
\eqref{43} \ll R^{2|\alpha|} K \delta + R^{2|\alpha|}E^{-(1/3)\epsilon}. \label{48}
\end{align}
Now we turn our attention to bounding the difference between $\phi_x$ and the first term on the right hand side of \eqref{42}. Expanding the square and using the triangular inequality, we have
\begin{align}
\frac{1}{\pi s^2}\int_{B(s,z)} \int_{B(1)} \left| D^{\alpha}\sum_{k\in \mathcal{K}}\sum_{\xi\in \mathcal{E}^k} a_{\xi}e( \langle\xi,x\rangle)\left( e\left(\left\langle\frac{\xi}{\sqrt{E}},Ry\right\rangle\right) - e(\langle R\zeta_k, y\rangle) \right) \right|^2 dxdy \nonumber \\
\ll \frac{R^{2|\alpha|+2}}{\pi s^2} \sum_{k,k'} \sum_{\substack{\xi\in \mathcal{E}^k \\ \xi'\in \mathcal{E}^{k'}}}|a_{\xi}a_{\xi'}|\left|\frac{\xi}{\sqrt{E}}-\zeta_k\right|\left|\frac{\xi'}{\sqrt{E}}-\zeta_{k'}\right|\left| \int_{B(s,z)}e(\langle \xi-\xi', x \rangle)dx \right|. \label{47}
\end{align}
Similarly to the above, via \eqref{orthogonality} and \eqref{45}, the contribution from the terms with $\xi\neq \xi'$ is at most $R^{2|\alpha|+2} E^{-(1/3)\epsilon} $. The contribution of the terms with $\xi=\xi'$, bearing in mind that $|\xi/\sqrt{E}-\zeta_k|\leq 1/K$, can be bounded by
\begin{align}
R^{2|\alpha|+2}\sum_k \sum_{\xi\in \mathcal{E}^k} |a_{\xi}|^2\left|\frac{\xi}{\sqrt{E}}-\zeta_k\right|^2 \ll \frac{ R^{2|\alpha|+2}}{K^2}\sum_k \mu_f(I_k)\leq \frac{ R^{2|\alpha|+2}}{K^2}.\label{46}
\end{align}
All in all we have,
\begin{align}
\eqref{47} \ll \frac{ R^{2|\alpha|+2}}{K^2} + R^{2|\alpha|+2} E^{-(1/3)\epsilon}. \label{50}
\end{align}
The lemma follows combining \eqref{49}, \eqref{48} and \eqref{50}.
\end{proof}
\subsection{Gaussian moments}
Recall the notation \eqref{bk}, we are going to show that the vector $(b_k)_{k\in \mathcal{K}}$ approximates a Gaussian vector $(c_k)_{k\in \mathcal{K}}$, where $c_k$ are i.i.d. complex standard Gaussian random variables subject to $\overline{c}_k=c_{-k}$. We prove the following quantitative lemma:
\begin{lem}
\label{independence in shrinking}
Let $\epsilon>0$, $b_k$ be as in \eqref{bk} and $\mathcal{K},K,\delta$ be as in Section \ref{approximating}. Moreover let $B$ be some large parameter and fix two sets of positive integers $\{r_k\}_{k\in \mathcal{K}}$ and $\{s_k\}_{k\in \mathcal{K}}$ such that $\sum r_k + s_k \leq B$. Suppose that $E \in S'$, then
\begin{align}
\left|\frac{1}{\pi s^2}\int_{B(s,z)}\prod_{k\in \mathcal{K}}b_k^{r_k}\overline{b}_k^{s_k}dx- \mathbb{E}\left[\prod_{k\in \mathcal{K}}c_k^{r_k}\overline{c}_k^{s_k}\right]\right|= o_{\delta,K,B}(1) \hspace{5mm} \text{as} \hspace{2mm} N\rightarrow \infty \nonumber
\end{align}
uniformly for $f$ flat, $s>E^{-1/2+\epsilon}$ and $z\in \mathbb{T}^2$.
\end{lem}
\begin{proof}
Expanding the product, we have
\begin{align}
\frac{1}{\pi s^2} \int_{B(s,z)}\prod_{k}b_k^{r_k}\overline{b}_k^{s_k}dx= \prod_{k} \mu_{f}(I_k)^{-(r_k+s_k)/2}\sum \left( \prod_{i=1}^{r_k}\prod_{j=1}^{s_k}a_{\xi_{i,k}}\overline{a}_{\xi'_{j,k}}\right) \times \nonumber\\ \times \int_{B(s,z)} e\left( \Bigl\langle \sum_{k,i,j} (\xi_{i,k}- \xi'_{j,k}),x \Bigr\rangle\right)dx \label{formula1}
\end{align}
where the out most sum is over all the choices $\xi_{1,1},...,\xi_{1,r_1}, \xi'_{1,1}...,\xi'_{1,s_1},..., \xi_{k,1},..,.\xi_{k,r_k}, \xi'_{k,1}, \\ ...,\xi'_{1,s_k}$. We split the sum in \eqref{formula1} according to \eqref{orthogonality}: we first consider the contribution from the \emph{constant term} $\sum_{k,i,j}(\xi_{i,k}- \xi'_{j,k})=0$ and then the contribution from the \emph{oscillatory term} $|\sum_{k,i,j}(\xi_{i,k}- \xi'_{j,k})|>0$. Furthermore, we subdivide the constant term into \textquotedblleft diagonal" solutions, namely $\{\xi_{i,k}\}= \{\xi'_{j,k}\}$ for each $k\in \mathcal{K}$, and all the other solutions, which we call \textquotedblleft off-diagonal".
\textbf{Constant term}, \textit{\textquotedblleft diagonal" solutions}.
If $\{\xi_{i,k}\}= \{\xi'_{j,k}\}$, then $r_k= s_k$, so, taking into account the possible rearrangements and by definition of $I_k$ and $\mu_{f}(I_k)$, we have a contribution to the right hand side of \eqref{formula1} of
\begin{align}
g_k:=r_k! \cdot \mu_{f}(I_k)^{-r_k}\sum_{ \{\xi_{i,k}\}= \{\xi'_{j,k}\} } \prod_{i=1}^{r_k} |a_{\xi_{i,k}}|^2 = \mathbb{E}[ |c_k|^{2r_k}]. \nonumber
\end{align}
Multiplying together the contributions from all $k$'s, we obtain
\begin{align}
\prod_{k} g_k= \mathbb{E}\left[\prod_{k}|c_k|^{2 r_k}\right].\label{3.1}
\end{align}
\textbf{Constant term}, \textit{\textquotedblleft off-diagonal" solutions}.
Let $B_1=\sum_{k}r_k+ s_k$. Since $E\in S'$, Theorem \ref{BB} implies that the number of off-diagonal solutions is at most $O(N^{\gamma B_1})$. Since $\mu_{f}(I_k)\geq \delta$ for all $k \in \mathcal{K}$ and $|a_{\xi}|^2\leq N^{-1+o(1)}$, we obtain as $N\rightarrow \infty$
\begin{align}
\left|\sum_{\text{off-diagonal}}\prod_{k} \mu_{f}(I_k)^{r_k+s_k/2}\sum \prod_{k} \prod_{i=1}^{r_k}\prod_{j=1}^{s_k}a_{\xi_{i,k}}\overline{a}_{\xi_{j,k}}\right|\ll_{K} N^{-B_1/2+ \gamma B_1 +o(1)} \delta^{-B_1/2}=o_{\delta, B, K}(1). \label{3.2}
\end{align}
\textbf{Oscillatory term}. If $|\sum_{k,i,j}(\xi_{i,k}- \xi'_{j,k})|>0$, Theorem \ref{semi} implies that $|\sum_{k,i,j}(\xi_{i,k}- \xi'_{j,k})|>E^{1/2-2\epsilon}$, therefore $s|\sum_{k,i,j}(\xi_{i,k}- \xi_{j,k})|>E^{\epsilon}$. So, bearing in mind that $J_1(T) \ll T^{1/2}$ for $T$ sufficiently large, we have
\begin{align}
\frac{J_1(s|\sum_{k,i,j}(\xi_{i,k}- \xi_{j,k})|)}{s|\sum_{k,i,j}(\xi_{i,k}- \xi_{j,k})|}\ll E^{-2/3 \epsilon} .\label{3.5}
\end{align}
Since the maximum number of terms in the outer sum in \eqref{formula1} is $N^{B_1}$, $\mu_{f}(I_k)\geq \delta$ for all $k \in \mathcal{K}$, $|a_{\xi}|\leq N^{-1+o(1)}$ and bearing in mind \eqref{3.5} and \eqref{divisor bound} , we obtain as $N\rightarrow \infty$
\begin{align}
\prod_{k} \mu_{f}(I_k)^{-(r_k+s_k)/2}\sum \left( \prod_{k} \prod_{i=1}^{r_k}\prod_{j=1}^{s_k}a_{\xi_{i,k}}\overline{a}_{\xi_{j,k}}\right) \frac{J_1(s|\sum_{k,i,j}(\xi_{i,k}- \xi_{j,k})|)}{s|\sum_{k,i,j}(\xi_{i,k}- \xi_{j,k})|} \nonumber \\ \ll_{K} N^{B_1/2 +o(1)} \delta^{-B_1/2} E^{-2/3 \epsilon}=o_{\delta,K,B_1}(1). \label{3.3}
\end{align}
The Proposition follows combining \eqref{3.1}, \eqref{3.2}, \eqref{3.3}.
\end{proof}
Lemma \ref{independence in shrinking}, by the method of moments, implies that the vector $(b_k)_{k\in \mathcal{K}}$ converges in distribution to the vector $(c_k)_{k\in \mathcal{K}}$ . We restate this fact in the following convenient way, more details can be found in \cite[Lemma 6.5]{BW} and \cite[Page 9]{BU}, see in particular \cite[Lemma 6.4]{BW} for the fact that the measure induced by the $b_k$'s is absolutely continuous with respect to the Lebesgue measure.
\begin{cor}
\label{tau}
Let $\epsilon>0$ and $\alpha_1,\alpha_2>0$ be given, let $\delta,K,B,$ as in Lemma \ref{independence in shrinking} and $f$ be as in \eqref{function}. Suppose that $E \in S'$ is sufficiently large depending on $\epsilon,\alpha_1,\alpha_2,K,\delta$ and $B$. Then, uniformly for all $f$ flat, $s>E^{-1/2+\epsilon}$ and $z\in \mathbb{T}^2$, there exists a measurable map $\tau: \Omega\rightarrow B(s,z)$ and a subset $\Omega^1\subset \Omega$ with the following properties:
\begin{enumerate}
\item For any measurable $A\subset \Omega^1$, we have $\vol(\tau(A))= \pi s^2\mathbb{P}(A)$.
\item $\mathbb{P}(\Omega^1)>1-\alpha_1$ .
\item For all $\omega\in \Omega^1$, we have $|b_k(\tau(\omega))- c_k(\omega)|\leq \alpha_2$ uniformly for all $k\in \mathcal{K}$.
\end{enumerate}
\end{cor}
\subsection{Discarding $\phi_x$}
Before proving the main result of this section, we need the following lemma:
\begin{lem}[Lemma 4, \cite{SO}]
\label{Sodin}
Let $R>1$ $\alpha_3,\alpha_4>0$, $\{\mu_n\}_{n\in \mathbb{N}}$ be a sequence of probability measures on $\mathbb{S}^1$ such that $\mu_n$ weak$^{\star}$ converges to some probability measure $\mu$. Then, for all $n$ sufficiently large depending on $\alpha_3,\alpha_4$ and $R$, we have
\begin{align}
||F_{\mu_n}- F_{\mu}||_{\mathcal{C}^1(B(R))}\leq \alpha_3 \nonumber
\end{align}
outside an event of probability $\alpha_4$.
\end{lem}
\begin{proof}
We can associate to $\mu$ the Gaussian measure $G$ defined on $\mathbb{R}^2$ as follows: for any open and measurable (with respect to $\mu$) subset $A$ of $\mathbb{R}^2$ we let
\begin{align}
G(A)= N(0,\mu(A))\nonumber
\end{align}
where $N(0,\mu(A))$ is a Gaussian random variable with mean zero and variance $\mu(A)$. Moreover, if $A\cap B= \emptyset$, we require $G(A)$ and $G(B)$ to be independent. We define $G_n$ with respect to $\mu_n$ similarly. Since $\mu$ is compactly supported, we see that $G_n$ weak$^{\star}$ converges to $G$ and, since a normal random variable is square integrable, we obtain $G_n\rightarrow G$ in $L^2(\Omega)$ (recall that $\Omega$ is the common probability space of our random objects). By \cite[Theorem 5.4.2]{AT}, we have the $L^2(\Omega)$ representations
\begin{align}
&F_{\mu_n}(x)= \int_{\mathbb{S}^1}e(\langle x\cdot\lambda\rangle)G_n(d\lambda) &F_{\mu}(x)= \int_{\mathbb{S}^1}e(\langle x\cdot\lambda\rangle)G(d\lambda). \label{13}
\end{align}
Since $\mu$ and $\mu_n$ are compactly supported, we can differentiate under the integral in \eqref{13}; bearing in mind that $G_n$ weak$^{\star}$ converges to $G$, we have $||F_{\mu_n}- F_{\mu}||_{\mathcal{C}^1(B(R))}\rightarrow 0$ as $n\rightarrow \infty$ in $L^2(\Omega)$. This implies the conclusion of the Lemma.
\end{proof}
We are finally ready to state and prove the main result of this section:
\begin{prop}
\label{main prop}
Let $\epsilon>0$, $R>1$ and $\eta_1,\eta_2>0$, $f$ be as in \eqref{function}. Suppose that $E \in S'$ is sufficiently large depending on $\epsilon,\eta_1,\eta_2$ and $R$. Then, uniformly for all $f$ flat, $s>E^{-1/2+\epsilon}$ and $z\in \mathbb{T}^2$, there exists a measurable map $\tau: \Omega\rightarrow B(s,z)$ and a subset $\Omega'\subset \Omega$ with the following properties:
\begin{enumerate}
\item For any measurable $A\subset \Omega$, we have $\vol(\tau(A))= \pi s^2\mathbb{P}(A)$.
\item $\mathbb{P}(\Omega')>1-\eta_1$ .
\item For all $\omega\in \Omega'$, we have $||F_{\tau(\omega)}(y)- F_{\mu_{f}}(Ry,w)||_{ \mathcal{C}^1(B(1))}\leq \eta_2$
\end{enumerate}
\end{prop}
\begin{proof}
Let $\mathcal{K},K,\delta$ be as in Section \ref{approximating} and let $F_K(Ry,\omega):=\sum_{k\in \mathcal{K}}\mu_{f}(I_k)^{1/2} c_k(\omega) e(\langle \zeta_k,Ry\rangle)$. Thanks to Corollary \ref{tau} with $\alpha_1=\eta_1/3$ and $\alpha_2=1/K^2$, there exist $\tau: \Omega\rightarrow B(s,z)$ and $\Omega^1\subset \Omega$ such that:
\begin{itemize}
\item For any measurable $A\subset \Omega'$, we have $\vol(\tau(A))= \pi s^2\mathbb{P}(A)$.
\item $\mathbb{P}(\Omega^1)>1-\eta_1/3$ .
\item For all $\omega\in \Omega'$, we have
\begin{align}
||\phi_{\tau(\omega)}(y) -F_K(Ry,\omega)||_{\mathcal{C}^1}\ll RK \alpha_2= \frac{R}{K}\leq \eta_2/3 \label{main1}
\end{align}
provided $K$ is sufficiently large depending on $R$ and $\eta_2$.
\end{itemize}
\begin{claim}
\label{claim1}
There exists some $\Omega^2\subset \Omega$ with $\mathbb{P}(\Omega^2)>1-\eta_1/3$ such that
\begin{align}
||F_K(Ry,\omega)- F_{\mu_{f}}(Ry,\omega)||_{\mathcal{C}^1}\leq \eta_2/3\label{main2}
\end{align}
for all $K$ sufficiently large depending on $\eta_1,\eta_2$ and $R$.
\end{claim}
To prove the claim, observe that $F_K$ is a Gaussian field with spectral measure
\begin{align}
\mu_K= \sum_{k\in \mathcal{K}} \mu_{f}(I_k) \delta_{\zeta_k}. \nonumber
\end{align}
By definition of $\mu_f$, we have $\sup_{A\subset \mathbb{S}^1} |\mu_{f}(A)- \mu_K(A)| \ll \delta K $. So, taking $\delta<1/K^2$ and $K$ sufficiently large, the claim follows from Lemma \ref{Sodin}.
\vspace{2mm}
Finally, by Lemma \ref{first approx} and Markov's inequality, we have
\begin{align}
|| F_x- \phi_x||_{\mathcal{C}^1} \leq \eta_2/3 \label{main3}
\end{align}
for all $x\in B\subset B(s,z)$, where
\begin{align}
(\pi s^2)^{-1} \vol(B(s,z) \backslash B)\ll \eta_2^{-1}\left( R^{6}K\delta + R^{8}K^{-2} + R^{8}E^{-(1/3)\epsilon} \right) \leq \eta_1/3 . \label{5.1}
\end{align}
for $K$ and $E$ sufficiently large in terms of $R$, $\eta_1$ and $\eta_2$. We briefly summaries our choices of parameters: $R,\eta_1,\eta_2$ are fixed, $\delta<1/K^2$, $K$ is large depending $R,\eta_1,\eta_2$ and $E$ is large depending on $R,\eta_1,\eta_2$ and $K$. We are now ready to conclude the proof. Let $\Omega'= \Omega^1 \cap \Omega^2 \cap \tau^{-1} (B)$, then $\tau$ restricted to $\Omega'$ satisfies $(1)$. By Corollary \ref{tau}, Claim \ref{claim1} and \eqref{5.1}, we also have $\mathbb{P}(\Omega')\geq 1- \eta_1$ so $(2)$ holds. Finally, $(3)$ follows by \eqref{main1}, \eqref{main2} and \eqref{main3}, valid for all $\omega \in \Omega'$. This concludes the proof of the Proposition.
\end{proof}
\section{Concluding the proof}
\label{end}
We are finally ready to prove Theorem \ref{theorem 3}.
\begin{proof}[Proof of Theorem \ref{theorem 3}]
Pick some $\eta_1, \eta_2>0$ to be chosen later, and let $\tau$ and $\Omega'$ be given by Proposition \ref {main prop}. Then, by Proposition \ref{semi-locality}, we have
\begin{align}
\mathcal{N}_f(s,z)&= \frac{E}{R^2}\int_{B(s,z)} \mathcal{N}( F_x)dx + O\left(\frac{Es^2}{\sqrt{R}}\right) \nonumber \\
&= \frac{E}{R^2}\int_{\tau^{-1}(\Omega')} \mathcal{N}( F_x)dx + \frac{E}{R^2} \int_{B(s,z)\backslash \tau^{-1}(\Omega')} \mathcal{N}(F_x)dx + O\left(\frac{Es^2}{\sqrt{R}} \right). \label{53}
\end{align}
By part $(3)$ of Proposition \ref{main prop}, we may write $F_x(y)= F_{\mu_{f}}(Ry) +\psi$ for $x\in \tau^{-1}(\Omega')$ and some function $\psi$ with $||\psi||_{\mathcal{C}^1}\leq \eta_1$. Thus, we can rewrite \eqref{53}, bearing in mind part $(1)$ of Proposition \ref{main prop}, as
\begin{align}
\mathcal{N}_f(s,z)&= \frac{\pi s^2 E}{R^2}\int_{\Omega'} \mathcal{N}( F_{\mu_{f}} +\psi, R )d\omega + \frac{E}{R^2}\int_{B(s,z)\backslash \tau^{-1}(\Omega')} \mathcal{N}(F_x)dx + O\left(\frac{Es^2}{\sqrt{R}}\right) \nonumber \\
&= \frac{ \pi s^2 E}{R^2}\left(\int_{\Omega} \mathcal{N}( F_{\mu_{f}} +\psi, R ) d\omega- \int_{\Omega\backslash \Omega'} \mathcal{N}( F_{\mu_{f}} +\psi, R ) d\omega \right) \nonumber \\
& \hspace{1cm} + \frac{ E}{R^2}\int_{B(s,z)\backslash \tau^{-1}(\Omega')} \mathcal{N}(F_x)dx + O\left(\frac{Es^2}{\sqrt{R}}\right) \label{52}
\end{align}
where in the second equality we set $\psi(y,\omega)=0$ for $\omega \not \in \Omega'$. By the Faber-Krahn inequality, $\mathcal{N}(F_x)\ll R^2$, and since $F_{\mu_{f}} +\psi= F_{\tau(\omega)}$, we have $ \mathcal{N}( F_{\mu_{f}} +\psi ) \ll R^2 $ uniformly for all $\omega \in \Omega'$. For $\omega \not \in \Omega'$, by definition, we have $\Delta F_{\mu_{f}}=-R^2F_{\mu_{f}}$. Thus, again by the Faber-Krahn inequality, $ \mathcal{N}( F_{\mu_{f}} +\psi ) \ll R^2 $ holds uniformly for all $\omega \in \Omega$. Therefore, taking $\eta_2= 1/\sqrt{R}$, the second and third integrals in \eqref{52} are bounded by $O(R^{3/2})$. Thus, \eqref{52} can be re-written as
\begin{align}
\mathcal{N}_f(s,z)=\frac{\pi s^2 E}{R^2}\mathbb{E}[ \mathcal{N}( F_{\mu_{f}} +\psi, R )] +O\left(\frac{Es^2}{\sqrt{R}}\right). \nonumber
\end{align}
Taking $\eta_1$ small enough in terms of $R$ via Proposition \ref{stability} and then taking $R\rightarrow \infty$, we deduce
\begin{align}
\mathcal{N}_f(s,z)= c_{NS}(\mu_{f})\pi s^2 E(1+ o_{R\rightarrow \infty}(1)) . \nonumber
\end{align}
and the Theorem follows by taking $R$ to be an arbitrarily slowly, depending on all the parameters, growing function of $E$.
\end{proof}
\section*{Acknowledgement}
The author would like to thank Igor Wigman for many useful discussions. Alejandro Rivera and Maxime Ingremeau for stimulating conversations and Priya Lakshmi for her comments on an early draft of this article. This work was supported by the Engineering and Physical Sciences Research Council [EP/L015234/1].
The EPSRC Centre for Doctoral Training in Geometry and Number Theory (The London School of Geometry and Number Theory), University College London.
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LAMINATED/BULLET-PROOF CURVED GLASS is available in 2-3-4 or more sheets, and with the insertion of plastic materials such as polycarbonate. Bear in mind that the thickness must always be reasonably proportional to the panel’s size and requisites (accident-prevention, safety, bulletproof).
CHEMICALLY TEMPERED LAMINATED GLASS: Chemical tempering allows to realize almost any curved shape adding reinforcement to the glass sheets and making these glass panels suitable for all those applications where aesthetic and safety must be accompanied by an excellent mechanical strength.
INSULATING CURVED GLASS is produced by assembling two or more curved panels of even different type and thickness with a dry air chamber and then joining the panels together by using a special appropriately sealed spacer. In this way, curved insulating glass panels with exceptional thermal and sound insulation properties are created with all the BOW GLASS aesthetic features to make a product that is truly unique in its kind.
All the types indicated above are also available in reflecting, coloured, sandblasted versions.
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jute in English
- jute ⇄ jute, noun.
1. a strong, glossy fiber used for making coarse fabrics or rope. Jute is obtained from the bark of certain tropical plants.
Ex. Jute from the farms goes into Calcutta factories to make canvas, rope, and rough bags (Junior Scholastenglish
- Jute ⇄ Jute, noun.
a member of a Germanic tribe from Denmark and northern Germany. Some of the Jutes, with the Angles and Saxons, invaded and settled in England in the A.D. 400's and 500's.english
| 52,470
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Monday 20 Oct 2014
By Julia Halperin. Web onlyPublished online: 27 November 2013
One of Ukraine’s richest couples has funded a contemporary art exhibition in Kiev about HIV/Aids. “Where There’s a Will, There’s a Way”, produced by Elena Pinchuk’s Antiaids Foundation and presented at her husband Victor Pinchuk’s Pinchuk Art Centre ahead of World Aids Day on 1 December, includes around 30 works by international and Ukrainian artists including Ai Weiwei and Sergiy Bratkov (through 5 January).
The show takes its name from Damien Hirst’s 2007 pill cabinet sculpture filled with clusters of antiretroviral pills used to treat HIV/Aids. “The works… represent a story of how art communities have been reacting to the Aids epidemics,” Elena Pinchuk says in a statement. The show opens with works made in New York during the 1980s, when American artists such as David Wojnarowicz and Keith Haring struggled with the disease, and closes with work from present-day Ukraine, where Aids is one of the largest epidemics in the country’s history. The show features several new commissions, including a performance by the Ukrainian artist Ilya Chichkan and a series of photographs by Nan Goldin shot on location in Kiev. A large-scale video projection on the art centre’s façade by Tony Oursler, Transmission, 2013, also produced for the exhibition, depicts three figures whispering to one another in English, Ukrainian and Russian. The video aims to capture “this cultural component of HIV, of misinformation surrounding it, [which] can be seen as having viral properties too in this Information age”, the artist says in a statement. Versions of Felix González-Torres untitled 1991 billboard work have been installed throughout the city. The massive photographic piece shows the artist’s empty bed with two rumpled pillows and is often seen as an elegy to his partner, who died from Aids-related complications that
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- Calories 61
- Caloriesfromfat 38%
- Fat 2.6g
- Satfat 0.4g
- Monofat 1.8g
- Polyfat 0.3g
- Protein 1.4g
- Carbohydrate 9g
- Fiber 2.3g
- Cholesterol 0.0mg
- Iron 0.8mg
- Sodium 201mg
- Calcium 28mg
Roasted Tomato-and-Pepper Salad
How to Make It.
| 175,449
|
LonginesMens Master Silver Dial Bracelet Watch L2.755.3.78.7
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Product Description
Expertly crafted from sublime stainless steel, this stunning timepiece consists of a subtly patterned dial with beautiful black swirl inspired numbering. The contrasting blue dial hands provide an eye catching touch, alongside the brands trademark logo branding and automatic labelling positioned below the trio of hands. Finished with a stainless steel bracelet composed of matte and polished links with a push deployment clasp fastening.
| 310,140
|
This week, on As The Golf World Turns…
Tiger Woods’ ex-wife Elin Nordegren is not a fan of her husband’s new relationship with Olympic skier Lindsey Vonn, according to Us Weekly..
Woods and Vonn announced their relationship in a Facebook post in March, and since then Vonn has been a regular at Woods’ golf tournaments as she continues to rehab her knee following a harrowing ski crash in February. The two also attended Met Gala, one of the New York City’s most exclusive social events, in May.
Woods and Nordegren divorced in 2010. Photo: Elin Nordegren at the 2009 Presidents Cup in San Francisco (AP).
| 375,672
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Shallow streams, a great lake, a nationally recognized waterway, small lakes and ponds can all be found in the South Shore. We've put together a list of water trails for beginners complete with rental locations.
Have your own equipment? Looking for a challenge? Check out our list of the top 12 challenging water trails for paddlers in Northwest Indiana.
Northwest Indiana Paddling Association It's recommended to learn the basics before going out on chippy waters, deep canals or in strong currents. The Northwest Indiana Paddling Association is a great place to start - offering paddling events from spring to fall and the Association has additional rentals and loaner boats available at a number of their events.
Paddling Rentals Don't own your own? The South Shore has rentals available!
Wolf Lake Memorial Park in Hammond has canoe, kayak, paddleboat and paddleboard/hour to a paddleboat or itBike for $5/half! Lake George Kayak Rentals offers rentals by the hour or by the day. The lake has access to parks and the city's downtown.
Lake Michigan Paddling - The Indiana Dunes Paddling the waters of Lake Michigan offers a different vantage point than from shore and is a wonderful place to start your paddling adventures. You can launch your kayak from at nearly all of the Lakeshore's beaches. There are plenty of other activities at the Indiana Dunes National Lakeshore including hiking, biking, swimming and fishing.
View a map of the hundreds of miles of water and bike trails in Northwest Indiana HERE. Don't forget to check the weather conditions before you venture out on the waters in the South Shore.
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For those of you who have gardens, there comes a time when everything is going off and you need to find something to do with all of it. Give away some of it, yes, but also consider making things that use a lot garden harvest and that keep a long time. Kim chee is one of those things.
My kim chee recipe is a combination of several, namely, “The Korean Kitchen” by Copeland Marks wth Manjo Kim and Tyler Florence, who I’ve found has excellent tips on international dishes.
A few notes about kim chee: according to “The Korean Kitchen,” kim chee has seven components including a hot or sweet chili taste, saltiness, sweetness, sourness, bitterness, astringency, and any ingredient that will intensify or enhance flavor. The great thing about kim chee is that the salt in it softens all kinds of greens, even ones that have gotten a little tough from late harvesting. In my recipe, I use kale, bok choy, radicchio (Italian chicory), and chard, plus I add garlic chives and green onions, which I also grow. My husband, who lived in Korea while he was in the Air Force, likes the bitter greens because he says it tastes more authentic than the pure cabbage kim chee that we buy in the store.
The other nice thing about kim chee is that you can make many different kinds from greens to cucumber, eggplant, radish, and anything else you’d like to experiment with. I store mine in Ball jars and they keep in the fridge for weeks (it would last several months, but we eat it too quickly). Each night for dinner, we place it on the table like a condiment and it goes surprisingly well with a variety of meals. It’s also terrific for breakfast with rice and a fried egg.
So here’s my kim chee recipe:
Ingredients:
Garden greens, about 1 to 1-1/2 pounds (kale, bok choy, radicchio, chard, etc.), rough chopped into 1-inch pieces (can be larger)
¼ cup salt
½ cup rice vinegar
1 tablespoon sugar
3 tablespoons kim chee base (I use Momoya brand)
1-inch piece of ginger, grated
3 cloves garlic, finely chopped
3 green onions, finely sliced
10 garlic chives, finely sliced
¼ Maui onion or other sweet onion
Pinch of Hawaiian salt (if needed)
Remove the outer leaves from the greens and rough chop into 1-inch pieces (I leave some pieces larger for variety). Put greens into a colander, add salt, and mix well. Place over a bowl and let drain, covered, about two hours.
In another bowl, combine vinegar, sugar, kim chee base, ginger, garlic, green onions, garlic chives, and Maui onion and stir. After the greens have been sitting with the salt for two hours, squeeze them out until you’ve gotten most of the liquid out of them. Rinse the greens for a few minutes, until you’ve gotten a lot of the salt off, and squeeze again until the greens are almost dry. Add the greens to the vinegar mixture and stir to combine. If needed, add a pinch of Hawaiian salt and stir again. Put the mixture into sterile glass jars and pack the greens down. Add enough water to cover. Close the jar and refrigerate for at least four hours.
A little goes a long way so if you make enough jars, you can give some away to family and friends. Ma-ni dŭ-se-yo!
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TITLE: Cohomology of Grassmannian
QUESTION [11 upvotes]: Let $G_r$ the infinite complex Grassmannian manifold. We know that $H^{*}(G_r)=\mathbb{C}[x_{1}, \cdots, x_{n}]$ where $x_i$ are the Chern classes of tautological bundle. But $H^{*}(G_r)$ is also isomorphic to the ring $\mathbb{C}[c_{1}, \cdots, c_{n}]$ where $c_i$ are the symmetric polynomials in $y_i$ where $y_i$ are the variables in $\mathbb{C}[y_1, \cdots, y_n]$. How can I see Chern classes as symmetric polynomials?
REPLY [14 votes]: Recall the splitting principle for complex vector bundles.
Theorem. (Complex Splitting Principle) For all rank $n$ complex vector bundles $p: E \longrightarrow X$, there exists a manifold $Y$ and a map $f: Y \longrightarrow X$ such that
$f^\ast: H^\ast(X) \longrightarrow H^\ast(Y)$ is injective.
$f^\ast E = L_1 \oplus \cdots \oplus L_n$ where the $L_i$'s are complex line bundles.
The splitting principle tells us that for purposes of computation, we may consider a complex vector bundle as a Whitney sum of complex line bundles.
Given such a splitting $f^\ast E = L_1 \oplus \dots \oplus L_n$, write $y_k = c_1(L_k)$. Then by the Whitney product formula and naturality of Chern classes, we have
\begin{align}
f^\ast c(E) & = c(f^\ast E) \\
& = c(L_1 \oplus \cdots \oplus L_n) \\
& = \prod_{k = 1}^n c(L_k) \\
& = \prod_{k = 1}^n (1 + c_1(L_k)) \\
& = \prod_{k = 1}^n (1 + y_k),
\end{align}
which upon expansion shows that $f^\ast c_k(E)$ is the $k^\text{th}$ elementary symmetric polynomial in the $y_i$'s.
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- .
Here at Best Laminate, we offer a wide variety of maple laminate flooring options. Your selection ranges from darker laminates to light laminates, from 1-strip laminates to 3-strip. Maple can come with unique designs that look fresh, exotic, or nature inspired.
Don’t know which maple laminate flooring choice is right for you? Get inspired by a special furniture piece or artwork and design your room around it!
Kronoswiss Swiss Prestige Maple D654PR Laminate Flooring
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Alloc City Scapes Brighton Maple 171313 Laminate Flooring
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Quick-Step Home Sound Blonde Maple SFS031 Laminate Flooring + Attached Pad$1.69 /sf
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| 350,257
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Business of Law, Law Office Management, Tech for Lawyers
How Your Laptop Can Improve Your Work-Life Balance (Part 1)
To achieve more workplace flexibility, use your laptop as your communication "control center."
alphaspirit/shutterstock
You will hear many people complain about how modern technology has made us less efficient, distracts us, or causes us to work even more than before. But here’s the truth: those people are not taking control of their technology and their time. If I ran my small firm the old fashioned way, I would have to commute to my office 5-6 days a week, work in a fluorescent-lit commercial office, get two weeks of vacation time per year, and have pasty white skin that never gets a lick of Vitamin D.
With the help of certain tech tools, I am able to work from wherever I happen to be: on an airplane, in my home office, on the beach, and—my personal favorite—in my pajamas. I also get to work whenever I want. This is very different from “all the time,” which is where most people miss the benefit of technology and turn it into shackles connected to their desk. These products and services give me the flexibility to work in a different time zone, at midnight when I can’t sleep, or early in the morning before anyone gets to the office—and never when the golf course is open or the surf’s up. I can easily work around those times, leaving them free from the “distractions” of work.
If you want to transform technology from an oppressive leash to a vehicle for flexible productivity, spend less time on the phone by maximizing use of your laptop to communicate with clients and colleagues.
Using My Laptop as my Communication “Control Center”
All of my channels of non-urgent communication filter through the laptop: my voicemails, emails, text messages, faxes and even social networking notifications. That way, I can access everything when I want to, not when the phone rings or a fax comes in. I have my laptop with me and I open it when I am ready to work—it’s in my control.
My preferred mode of communication is email, but I also make the most of voicemail and document-sharing tools.
Deliver Voicemail to Your Inbox
Use Google Voice! It’s a free service that will save you hours of time each week. If you have a Google account, you can add Google Voice and easily set up any phone number (your home, cell, or office*) to forward all voicemail to your assigned Google Voice number. Your callers will never even know the difference, but you will love that their voice messages get delivered to your email inbox and/or sent to you via text message. You receive a transcription of the message (albeit a bit rough) that you can see immediately upon looking at the email, and an attached audio file of the voice message itself.
Typically, you can decipher from the transcription what you need to do with the message. Does it require an immediate call back? Can you forward it to your assistant to respond? Is it a sales call you are glad to have missed? If you need to investigate the message more thoroughly, you can click on the audio file and listen to it from your computer.
Aside from giving you a way to retrieve voicemail through your email inbox, Google Voice also makes your voice messages “portable.” You can forward them, reply to them, file them, flag or star them, add them to your to-do list, delete them, or ignore them—all without picking up the phone.
If you receive a voice message that does not require a call back, you can simply email your response, and forward the Google Voice message in your reply for reference.
*Note that some office phone systems will not allow voicemail forwarding. For the most versatile and economic phone service, avoid office phone systems that don’t permit Google Voice. Once you go digital with your phone lines, you can easily integrate Google Voice into your system.
Maintain a Collaborative Status Document
To minimize emails and phone calls between you and your staff or co-workers, create and maintain collaborative documents (I use Google Docs and Google Drive) or message boards where everyone can update their work on certain matters. This serves as a hub where everyone can quickly scan one document to get updated on specific matters.
Also, if you set up a schedule of review for the document—that is, specific times everyone has to check in on the document—you can ask and address questions on the document. This is a lot more focused and less distracting than sending and receiving multiple emails on one subject or going back and forth on the phone.
Next time, we’ll dive into a detailed list of communication tools I use for phone calls, video conferences, faxes, cloud storage, and virtual assistance that help untether me from my brick-and-mortar office.
Sally Morin is managing principal of Sally Morin Law in San Francisco.
Reader Comments
| 158,851
|
- This event has passed.
International Tabletop Day
April 28 @ 10:00 am - 5:00 pm
International Tabletop Day is an international day that is dedicated to gaming and we wanted to join in on the FUN! Warm up those dice and dust off your meeples and join us for ITTD! This is a great opportunity to nerd out and scratch your tabletop itch, or to come see what all of this board game fuss is about in the first place. FREE program for all ages!
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\section{Probability inequalities}
\label{sec:prob}
This section collects several probability inequalities used in the analysis of
\Cref{alg:perm}.
Let $\sigma_i(\vM)$ denote the $i$-th largest singular value of the matrix
$\vM$.
\paragraph{Extreme singular values of Gaussian random matrices.}
\begin{lemma}[Eq.~3.2 in \citep{rudelson2010non}]
\label{thm:gaussian-smallest-singular-values}
Let $\vA$ be an $n \times d$ matrix whose entries are i.i.d.~$\Normal(0,1)$
random variables and $n\geq d$.
For any $\eta \in \intoo{0,1}$,
\begin{align*}
\Pr\del{ \sigma_d(\vA) \leq \frac{\eta}{\sqrt{d}} }
& \ \leq \ \eta
\,.
\end{align*}
\end{lemma}
\begin{lemma}[Theorem II.13 in \citep{davidson2001local}]
\label{thm:gaussian-largest-singular-values}
Let $\vA$ be an $n \times d$ matrix whose entries are i.i.d.~$\Normal(0,1)$
random variables.
For any $\eta \in \intoo{0,1}$,
\begin{align*}
\Pr\del{ \sigma_1(\vA) \geq \sqrt{n} + \sqrt{d} + \sqrt{2\ln(1/\eta)} }
& \ \leq \ \eta
\,.
\end{align*}
\end{lemma}
\paragraph{Tail bounds for Gaussian and $\chi^2$ random variables.}
\begin{lemma}
\label{lem:gaussian-conc}
Let $Z \sim \Normal(0,1)$.
For any $\eta \in \intoo{0,1}$, $\Pr(Z^2 \geq 2\ln(2/\delta)) \leq \eta$.
\end{lemma}
\begin{proof}
This follows from the standard Chernoff bounding method.
\end{proof}
\begin{lemma}[Lemma 1 in \citep{laurent2000adaptive}]
\label{lem:chi2-conc}
Let $W \sim \chi_k^2$.
For any $\eta \in \intoo{0,1}$, $\Pr(W \geq k + 2\sqrt{k\ln(1/\eta)} +
2\ln(1/\eta)) \leq \eta$.
\end{lemma}
\paragraph{Anti-concentration bounds for Gaussian and $\chi^2$ random variables.}
\begin{lemma}
\label{lem:gaussian-anticonc}
Let $Z \sim \Normal(0,1)$.
For any $\eta \in \intoo{0,1}$, $\Pr(Z^2 \leq \pi\eta^2/2) \leq \eta$.
\end{lemma}
\begin{proof}
This follows from direct integration.
\end{proof}
\begin{lemma}[Lemma 9 in \citep{pananjady2016linear}]
\label{lem:chi2-anticonc}
Let $W \sim \chi_k^2$.
For any $\eta \in \intoo{0,1}$, $\Pr(W \leq k\eta^{2/k}/4) \leq \eta$.
\end{lemma}
\begin{lemma}
\label{lem:gaussian-quadratic}
Let $\vx \in \R^d$ be any vector, $\vM \in \R^{n \times n}$ be any matrix, and
$\vA$ a random $n \times d$ matrix of i.i.d.~$\Normal(0,1)$ random variables.
For any $\eta \in \intoo{0,1/2}$,
\begin{align*}
\Pr\del{
\norm[0]{\vA^\T \vM \vA \vx}_2
\ \leq \
\norm{\vM}_F
\cdot \norm{\vx}_2
\cdot \sqrt{\frac{(d-1)\pi}{8n}}
\cdot \eta^{1+1/(d-1)}
}
& \ \leq \ 2\eta
\,.
\end{align*}
\end{lemma}
\begin{proof}
Let $\vu_1 := \vx / \norm{\vx}_2$, and extend to an orthonormal basis $\vu_1,
\vu_2, \dotsc, \vu_d$ for $\R^d$.
Let $\vg_i := \vA\vu_i$ for each $i \in [d]$, so $\vg_1, \vg_2, \dotsc, \vg_d$
are i.i.d.~$\Normal(\v0,\vI_n)$ random vectors.
We first show that
\begin{align}
\Pr\del{
\norm{\vM\vg_1}_2
\ \leq \
\norm{\vM}_F \cdot \sqrt{\frac{\pi}{2n}} \cdot \eta
}
& \ \leq \ \eta
\,.
\label{eq:g1}
\end{align}
To see this, note that the distribution of $\norm{\vM\vg_1}_2^2$ is the same
as that of $\sum_{i=1}^n \sigma_i(\vM)^2 \cdot Z_i^2$, where $Z_1, Z_2,
\dotsc, Z_n$ are i.i.d.~$\Normal(0,1)$ random variables.
Therefore, \Cref{lem:gaussian-anticonc} and the fact $\norm{\vM}_2^2 \geq
\norm{\vM}_F^2 / n$ proves the claim in~\eqref{eq:g1}.
Next, observe that
\begin{align}
\norm[0]{\vA^\T \vM \vA \vx}_2^2
& \ = \
\norm{\vx}_2^2 \cdot
\vu_1^\T \vA^\T \vM^\T \vA
\del[4]{ \sum_{i=1}^d \vu_i \vu_i^\T }
\vA^\T \vM \vA \vu_1
\nonumber \\
& \ = \
\norm{\vx}_2^2 \cdot
\vg_1^\T
\vM^\T
\del[4]{ \sum_{i=1}^d \vg_i \vg_i^\T }
\vM \vg_1
\nonumber \\
& \ \geq \
\norm{\vx}_2^2 \cdot
\sum_{i=2}^d \del{\vg_i^\T\vM\vg_1}^2
\,.
\label{eq:chi2}
\end{align}
Conditional on $\vg_1$, the final right-hand side in~\eqref{eq:chi2} has the
same distribution as $\norm{\vx}_2^2 \cdot \norm{\vM\vg_1}_2^2 \cdot W$, where
$W \sim \chi_{d-1}^2$ is a chi-squared random variable with $d-1$ degrees of
freedom.
Therefore, \Cref{lem:chi2-anticonc} implies
\begin{align*}
\Pr\del{
\norm[0]{\vA^\T \vM \vA \vx}_2
\ \leq \
\norm{\vx}_2 \cdot \norm{\vM\vg_1}_2 \cdot \frac{\sqrt{d-1}}{2} \cdot
\eta^{1/(d-1)}
}
& \ \leq \
\eta
\,.
\end{align*}
Combining this inequality with the inequality from~\eqref{eq:g1} and a union
bound proves the claim.
\end{proof}
\paragraph{Lattice basis size.}
The following \namecref{lem:lattice-size} is used to bound the size of the
lattice basis vectors constructed by \Cref{alg:perm} (via \Cref{alg:subsetsum}).
Recall that there are $n^2+1$ basis vectors; one has length $\sqrt{1 +
\beta^2y_0^2}$, and the remaining $n^2$ have length $\sqrt{1 +
\beta^2c_{i,j}^2}$ for $(i,j) \in [n] \times [n]$.
\begin{lemma}
\label{lem:lattice-size}
For any $\eta \in \intoo{0,1/5}$, with probability at least $1-5\eta$,
\begin{align*}
|y_0|
& \ \leq \
\norm[0]{\bar{\vw}}_2 \sqrt{2\ln(2/\eta)}
\,,
\\
\abs{c_{i,j}}
& \ \leq \
\norm[0]{\bar{\vw}}_2
\cdot \sqrt{2\ln(2n/\eta)}
\cdot \frac{d}{\eta^2}
\cdot \sqrt{d + 2\sqrt{d\ln(n/\eta)} + 2\ln(n/\eta)}
\cdot \sqrt{2\ln(2n/\eta)}
\,,
\quad
(i,j) \in [n] \times [n]
\,.
\end{align*}
\end{lemma}
\begin{proof}
By \Cref{thm:gaussian-smallest-singular-values}, \Cref{lem:gaussian-conc}, and
\Cref{lem:chi2-conc}, with probability at least $1-5\eta$,
\begin{align*}
\norm[1]{(\vX^\T\vX)^{-1}}_2
& \ \leq \ \frac{d}{\eta^2}
\,,
\\
\abs[0]{\vx_0^\T\bar{\vw}}
& \ \leq \
\norm[0]{\bar{\vw}}_2 \sqrt{2\ln(2/\eta)}
\,,
\\
\abs[0]{\vx_{\bar{\pi}(i)}^\T\bar{\vw}}
& \ \leq \
\norm[0]{\bar{\vw}}_2 \sqrt{2\ln(2n/\eta)}
\,,
\quad
i \in [n]
\,,
\\
\abs[0]{\tilde{\vx}_j^\T\vx_0}
& \ \leq \
\norm[0]{\tilde{\vx}_j}_2 \sqrt{2\ln(2n/\eta)}
\,,
\quad j \in [n]
\,,
\\
\norm[0]{\vx_j}_2
& \ \leq \
\sqrt{d + 2\sqrt{d\ln(n/\eta)} + 2\ln(n/\eta)}
\,,
\quad
j \in [n]
\,.
\end{align*}
In this event, we have for each $(i,j) \in [n] \times [n]$,
\begin{align*}
\abs{c_{i,j}}
& \ = \
\abs[0]{\vx_{\bar{\pi}(i)}^\T\bar{\vw}}
\cdot \abs[0]{\tilde{\vx}_j^\T\vx_0}
\\
& \ \leq \
\norm[0]{\bar{\vw}}_2
\cdot \sqrt{2\ln(2n/\eta)}
\cdot \norm[0]{\vX^\dag\ve_j}_2
\cdot \sqrt{2\ln(2n/\eta)}
\\
& \ = \
\norm[0]{\bar{\vw}}_2
\cdot \sqrt{2\ln(2n/\eta)}
\cdot \norm[0]{(\vX^\T\vX)^{-1}\vX^\T\ve_j}_2
\cdot \sqrt{2\ln(2n/\eta)}
\\
& \ \leq \
\norm[0]{\bar{\vw}}_2
\cdot \sqrt{2\ln(2n/\eta)}
\cdot \frac{d}{\eta^2}
\cdot \sqrt{d + 2\sqrt{d\ln(n/\eta)} + 2\ln(n/\eta)}
\cdot \sqrt{2\ln(2n/\eta)}
\,,
\end{align*}
and $\abs[0]{y_0} \leq \norm[0]{\bar{\vw}}_2 \sqrt{2\ln(2/\eta)}$.
\end{proof}
| 216,978
|
Milan
Who cares that people say that Canadian-born designer Philippe Malouin exploited his interns in order to make this rug. People just love rumors. I was at the studio when the girls worked on the rug and they didn’t look unhappy. Philippe even dedicated the rug to one of the girls because she worked on it from the beginning. That’s why it is called Yachiyo.
Last week Philippe presented his unique Yachiyo metal rug which was commissioned by the new Carwan Gallery in Milan during the Design Week. The Beirut-based gallery says that they are the ‘first pop-up gallery promoting limited-edition design in the Middle East.’
The rug was Philippe’s most complex and time-consuming work. About 12 people, including himself, were working on it. It consists of galvanised steel wire and is completely handmade with the Japanese ’12-in-2′ chain mail method.
The pic has more of a fashion kinda flava.
The carpet looks great!
| 396,564
|
/2017
I am sorting out my spare bedroom for my baby daughter whilst she is away. As I began working...
Plasterer / Renderer
Added 06/04/2017
Small room 11'6 x 7'6 requires skimming in preparation for decoration.
Plasterer / Renderer
Added 24/01/2017
I need to replace the ceiling in my Kitchen and have it skimmed (Flat), I need to get plumber...
Plasterer / Renderer
Added 08/12/2016
Ceiling 3.4m x 5.3m Two walls approx 3m x 5m one wall approx 3m x 1.1m plus small section above...
Plasterer / Renderer
Added 08/11/2016
rendering the exterior of a large detached house. scaffold is already in place. 2 story house...
Plasterer / Renderer
Added 07/10/2016
currently have 1980s artex ceilings want then plastered. ceilings are lounge, dining.
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With a vast number of direct-to-user/patient shipments happening today in healthcare, especially as the nation battles the contagious coronavirus by complying with stay-at-home edicts, we are seeing unprecedented activity in last-mile deliveries. What challenges do pharmaceutical companies and pharmacies need to solve? How will this experience affect the future of pharmaceutical logistics?
Valerie Metzker, head of business development for Roadie, offers several ideas in this X-minute interview. A crowdsourced delivery company, Roadie boasts the largest local same-day delivery footprint in America. Its point of differentiation is that it works with consumers, small businesses, and national companies across virtually every industry to provide faster and cheaper scheduled, same-day, and/or urgent delivery. With more than 150,000 verified drivers, Roadie reaches 89% of US households, according to the company.
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$159.95
The Andis Super 2-speed AGC2 clipper is extremely quiet, and great for use on cattle. High speed is over 4,000 strokes per minute. 14' cord. Made in the USA.
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Description
Andis®®
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AGC Super 2-Speed Professional Clipper
FREE EGT UltraEdge Pink Blade 2-pk with purchase of ProClip SMC Ecel Clippers!
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Wahl KM10 Clipper, Turquoise
FREE 7F Blade with purchase of a KM10 2-Spd Clipper
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Oster Turbo A5 Two-Speed Clipper
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| 287,274
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Fiamma Carry-Bike 200DJ 2 (Fiat, Peugeot, Citroen) Bike Rack
Designed for vans with double rear doors. Bike carrier for vans with double rear doors. It is not necessary to drill holes in the vehicle, except for the security screws. The rear door can be opened without removing the bicycles.
Can be combined with the Fiamma Deluxe 6 DJ ladder.
Supplied with:
2 Rail Premium
2 Bike-Block Pro Red (1 and 2), 1 Rack Holder
1 Security Strip
Can be installed on the following vehicles:
200 DJ (X250)
Fiat Ducato >06 2006
Peugeot Boxer >06 2006
Citroen Relay >06 2006)
Weight 9kg
Max Load 9 / 35 kg
200 DJ Ducato (X244) < 06 2006
• Fiat Ducato after 2000 and before 06/2006
• Peugeot Boxer after 2000
• Citroën Jumper/Relay after 2000
Weight 9kg
Max Load 9 / 35 kg
| 401,667
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TITLE: Manometer Equation Error
QUESTION [0 upvotes]: This is the question I was trying to solve.
This was the solution offered by the book.
When I was solving it, I applied the manometer equation at the initial level, but I got a different answer.
I can't figure out where I'm going wrong... any help is appreciated.
REPLY [1 votes]: Two points at the same height have the same pressure only if:
1- The liquid is static. (met)
2- The liquid is continous. (met)
3- Same liquid. (not met)
You see after the initial condition changed by x, the initial level line no longer contains the same liquid on both sides, thus you can not say that at the initial level (after the the liquids dislocated) the pressure is still the same. That's why you need to apply equal pressure at 27.2cm below the surface of the liquid on the left side, as that is where the 3 conditions are met.
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14, 2016
The Ice Warriors
Three stories into Troughton's second season and we're again in a base under siege. It's under siege before the story begins but by nature - in the form of a whacking great glacier - not the monstrous. Indeed, as this story progresses you might even argue that it is a story about two bases under siege.
The humans in their lovely futuristic looking clothes are besieged by the Ice Warriors but only as long as they don't know what kind of engines power the Ice Warriors craft. But the Ice Warriors to are under siege. The humans are using an Ionizer to try and keep the glaciers back. That Ionizer is capable of destroying the Ice Warriors ship if it is used. So for a large chunk of the story, it's a Mexican stand-off.
And I'm afraid we come to a problem: the Ice Warriors themselves. I've read that Mark Gatiss had trouble persuading Steven Moffat to let him bring the Ice Warriors back because to Moffat they were the perfect illustration of the slow, clumsy and incomprehensible Classic Doctor Who monster and on the basis of this story I'd have to agree with him. They look great when standing still but as soon as they start moving they waddle slowly about like a duck inside a turtle shell. Their whispering voices, which actually I like, are - on occasion - difficult to understand. But they were obviously memorable enough on their first appearance to get a follow-up.
My other complaint about them is their so stupid. They move straight to violence without thinking about it. Varga threatens to kill the Doctor pretty much instantly even though he's been banging on about 'trapping' someone for ages, which his excuse for keeping Victoria alive. They're just blundering oafs, which is a shame really but explains the...well...we'll keep that for The Seeds of Death.
The best thing about this story though are the performances. It (almost) goes without saying that Troughton is fabulous. He is. Deborah Watling gets to be scared a lot but she's got that ability to be both scared and brave at the same time. An ability that Barry Letts attributed to Liz Sladen. She's terrified almost all the time but she still tries to do the right thing, which makes her heroic. I also love her costume in this story. Frazer Hines never fails to make Jamie feel real and his chemistry with Troughton and Watling is excellent. This is a TARDIS crew that suffered badly from the archive losses but which now might be in need of some reassessment following the return to the archive of The Enemy of the World and The Web of Fear.
The main cast is given superb support here though from Peter Barkworth as Clent. He's the very definition of a man out of his depth, but then who wouldn't be. He's become reliant on the computer to confirm his decisions and the computer can't handle the mess it is in and so he hovers around the edges of a nervous breakdown without quite having one. I ended up feeling a little sorry for Clent who is challenged by The Doctor, by Penley (Peter Sallis) and then gets told by Varga (Bernard Bresslaw) , the Ice Warrior Commander, that his life is pretty much worthless. I'm not sure anyone would come out of such a testing well.
Peter Sallis also gives an excellent performance as Penley. I think this is the only thing apart from 'Last of the Summer Wine' that I've ever seen Peter Sallis is in and it makes me want to find dig out other things he's done. Perhaps Toby Hadoke can help me? [Pops off to Twitter for a moment.] Penley has abandoned Clent's team and fled to the wilds shacking up with Storr (Angus Lennie), who is an anti-science man. Storr is destined to meet a sticky end. Penley, forced to go back to Clent due to Jamie's injuries, will end up winning the computer v humans argument.
There's a nice little scene at the end between Penley and Clent where their obvious friendship comes through and Penley gives Clent the nudge he needs to reset himself back to 'normal'. It nicely played.
Clent's closest colleague, Miss Garrett (Wendy Gifford) tries throughout to stay committed to the computer and Clent. She defends Clent to Penley and the Doctor. She's dressed, like everyone in this story, in a late-60s version of future fashion. If anything dates this story it is the combination of the fashion. But I'm afraid I developed a little crush on Miss Garrett/Wendy Gifford, which I'm sure you don't need to know.
So to conclude a blog that is in danger of escaping from me The Ice Warriors is a story to watch for the performances. It's plot isn't spectacularly original, it's final episode is a messy little thing but Peter Sallis, Angus Lennie, Peter Barkworth, Wendy Gifford, Frazer Hines, Deborah Watling and most particularly Patrick Troughton make it well-worth watching.
Posted by Tony Cross at 20:43
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TITLE: How to calculate the following integral $I=\int_{0}^{1}\frac{1}{\sqrt{y}}e^{-(x^2+y^2)/2y}dy$?
QUESTION [3 upvotes]: I need to calculate the following integration $$I=\int_{0}^{1}\frac{1}{\sqrt{y}}e^{-(x^2+y^2)/2y}dy$$
I am trying to find some substitution but failed so far. I don't see if there any way to simplify this further. So how to do it? Can anyone give me a hint or a solution? Thanks.
Note
this integral came up when I was trying to solve a probability question involving two dimensional joint pdf. I need to find $f_X(x)$
REPLY [3 votes]: Let $I(x)$ be the integral given by
$$I(x)=\int_0^1 \frac{e^{-(x^2+y^2)/2y}}{\sqrt y}\,dy \tag 1$$
Note that $I(0)=\int_0^1 \frac{e^{-\frac12 y}}{\sqrt{y}}\,dy=\sqrt{2\pi}\text{erf}\left(\frac{\sqrt 2}{2}\right)$.
Since $I(x)$ is an even function of $x$, we may assume without loss of generality that $x> 0$.
We first enforce the substitution $y\to y^2$ in $(1)$ to obtain
$$\begin{align}
I(x)&=2\int_0^1 e^{-(x^2+y^4)/2y^2}\,dy \tag 2
\end{align}$$
Next, enforce the substitution $y=\sqrt{x}y$ to find
$$\begin{align}
I(x)&=2\sqrt{x} \int_0^{1/\sqrt{x}} e^{-\frac12 x\left(y^2+\frac{1}{y^2}\right)}\,dy \\\\
&=2\sqrt{x} e^{x}\int_0^{1/\sqrt{x}} e^{-\frac12 x\left(y+\frac{1}{y}\right)^2}\,dy \\\\
&=2\sqrt{x} e^{x}\int_0^{1/\sqrt{x}} e^{-\frac12 x\left(y+\frac{1}{y}\right)^2}\,\left(\frac12 +\frac1{2y^2}+\frac12-\frac{1}{2y^2}\right)\,dy \\\\
&=\sqrt{x} e^{x}\int_0^{1/\sqrt{x}} e^{-\frac12 x\left(y+\frac{1}{y}\right)^2}\,\left(1-\frac{1}{y^2}\right)\,dy\\\\
&+\sqrt{x} e^{-x}\int_0^{1/\sqrt{x}} e^{-\frac12 x\left(y-\frac{1}{y}\right)^2}\,\left(1+\frac{1}{y^2}\right)\,dy\\\\
&=-\sqrt{x} e^{x}\int_{\sqrt{x}+1/\sqrt{x}}^\infty e^{-\frac12 xu^2}\,du\\\\
&+\sqrt{x} e^{-x}\int_{-\infty}^{-\left(\sqrt{x}-1/\sqrt{x}\right)} e^{-\frac12 xu^2}\,du\\\\
&=-\sqrt{2}e^x\int_{(x+1)/\sqrt{2}}^\infty e^{-t^2}\,dt\\\\
&+\sqrt{2}e^{-x}\int_{-\infty}^{-(x-1)/\sqrt{2}} e^{-t^2}\,dt\\\\
&=\sqrt{\frac{\pi}{2}}e^{-x}\text{erfc}\left(\frac{x-1}{\sqrt{2}}\right)-\sqrt{\frac{\pi}{2}}e^x\text{erfc}\left(\frac{x+1}{\sqrt{2}}\right)
\end{align}$$
which after exploiting the evenness of $I(x)$ (i.e., replace $x$ with $|x|$) agrees with the result posted earlier by @Felix Marin!
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VDH reports 340,197 have had PCR tests, 381,539 have had PCR and antibody tests, 46,905 total cases, 4,884 hospitalizations, 1,428 deaths. RAHD reports Stafford has 738 cases (one new hospitalization), 590 in Spotsylvania, 70 in KG, 172 in Fred., 76 in Caroline.
The Governor has provided guidelines for Phase II, that starts Friday. You can find more information at the link below.
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- Government
- Departments
- Public Health
- Emergency Preparedness
- Mass Notification
Mass Notification
During an emergency event, communication (even between individuals in the same building) can often be the most challenging aspect of a response.
To help streamline this process, EPR is encouraging health providers and long term care facilities to utilize an emergency mass notification platform (presently we are utilizing Everbridge) if they are not already utilizing a communication platform. Everbridge serves as an automated calling tree that can reach staff and/or clients via phone, email and/or text messaging.
If you are interested in learning more about Everbridge and how your organization can benefit from this service, contact Crystalynn Kuntz (701-355-1546) or Kalen Ost (701-355-1556) to schedule an appointment.
Drills
While there is not a monetary cost associated with utilizing Everbridge, Everbridge.
Crystalynn KuntzEmergency Preparedness Regional CoordinatorPhone: 701-355-1540
Kalen OstEmergency Preparedness Information SpecialistPhone: 701-355-1556
Public Health
Physical Address
500 E. Front Ave.
Bismarck, ND 58504
Mailing Address
P.O. Box 5503
Bismarck, ND 58504
Phone: 701-355-1540Fax: 701-221-6883
| 394,865
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This week, I’m giving away two mysteries set in the first half of the twentieth century. Irene Fleming’s The Brink of Fame takes readers into the early days of American cinema. Emily Daggett Weiss thinks the film production company she owns with her husband is thriving, until she arrives in Arizona in 1914 to find he lost the studio in a poker game. Left penniless, Emily gets a job in California. It has one catch. She has to find a missing film star before she can direct her first film.
Or you could win Mignon F. Ballard’s Miss Dimple Disappears, a mystery set in Georgia in 1942. No one at the school in Elderberry, Georgia, suspect anything more than a heart attack when the custodian is found dead, apparently of a heart attack. But, when Miss Dimple Kilpatrick, a fixture at the school for more than forty years, disappears the following day, town residents are shaken. School teacher Charlie Carr and her friend, Annie, decide to investigate, and they uncover danger surprisingly close to home.
1914 or 1942, which do you prefer? You can enter to win both contests, but I need separate entries. Send your entries to me at Lesa.Holstine@gmail.com. The subjects should read, either “Win The Brink of Fame” or “Win Miss Dimple Disappears.” Please include your name and mailing address in the body of the entry. Entries from the U.S. only, please.
The contest will end at 6 p.m. MT on Thursday, Feb. 2. I’ll use a random number generator to pick the winners, and mail the books out on Friday
2 comments:
Both certainly sound interesting, just a bit different from my usual read. It's a good thing I like to branch out every now and then!
Thank you Lesa! You always choose such interesting books! Kat B.
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Charity & Third Sector
The Charity & Third Sector industry provides an indispensable source of support for all areas of life in Scotland.
At Stronachs we act for a number of local and national charities which undertake a broad range of activities in health, social care, welfare support and education.
Charitable organisations face stringent regulation, with continually evolving legislation creating a constant challenge to those companies and trusts established to support those less fortunate.
Whether large or small, local or national, Stronachs’ charity and third sector team will give you the help you need to support the causes that matter to you most. We will work with you to ensure that your charity or group takes the most suitable form for your requirements.
Smaller groups may come together to form an unincorporated association or a charitable trust with its well-known and well established familiarity or others may choose a SCIO (Scottish Charitable Incorporated Organisation) or a company limited by guarantee to enjoy the benefits of limited liability and a clear governance structure. We will help you choose the form which is right for you.
As a full service firm, we have the technical expertise needed to assist voluntary and community groups with all aspects of regulatory and commercial matters.
We understand that in order to support others, your group needs support too and we are committed to providing you with the advice and assistance you need to achieve your aims.
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Phil G. – :
Free shipping puts this product over the edge as one of the best deals currently around for Level III+ armor. Used to recommend AR500 armor, but this stuff is just as good if not better at a much better price.
Review Brandon H. – :
I like the color, it isn’t as drab as most FDE and isn’t as bright as some sands, just about a perfect medium desert color at least for out here in California. The armor itself is pretty typical heavy steel, but it does its job and I’d trust my life to it with all of the certifications these guys have.
Review Kennedy S. – :
The carrier itself is of good quality and I’d expect it’d handle a few seasons of field use, the armor in it will probably last until the end of time, but it is heavy as a result, though it kind of needs to be to have the capability their advertising. Overall a good buy, but not cutting edge by any means, just beefier than other offerings.
Review Jeff T. – :
This stuff can take a beating, I was testing out a few different options for armor for my family and this one took an entire AK magazine with barely any denting on the other side. Impressive to say the least, will definitely be ordering more and recommending to anyone who asks
Review Alejandro D. – :
I like the layout and construction of the vest, makes the plates sit really close to the body and therefore hide the weight of the plates themselves. While themselves are heavy (probably for a good reason), but they don’t feel heavy once you put them on. Honestly 18bs isn’t that much, most of the stuff the guys oversea’s are running top 40 pounds
Review Frank G. – :
Shipped free and fast. Arrived well packed and undamaged. The vest itself is well made and obviously sturdy, should last for years to come if my time in the military showed me anything.
Review Jesse P. – :
Their customer service is top notch, I had to cancel an order because I accidently ordered the wrong color. One simple call and they took care of everything. Two days later the vest I wanted arrived at my door. Great work guys!
Review Jake L. – :
The Best Bulletproof’s NIJ certification makes the whole buying process easier and less stressful. You can buy a vest knowing that it will perform above government standards without having to buy a spare to test out yourself.
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52 OUTSTANDING WEDDING TABLE DECORATIONS SO POPULAR
Table flowers are a detail in the wedding arrangement, and also a highlight in the wedding arrangement; just as the bouquet is the crowning touch of the bride’s style, the table flower is also an important decoration in the wedding. The exquisite and beautiful table flower is a beautiful scenery of the wedding banquet, allowing guests to enjoy the double experience of taste and vision during the meal.
Although flowers are the first choice for people in many materials, they are increasingly personalized with wedding arrangements. The choice of table flowers is not limited to flowers. Different materials and various creative table flowers are arranged at the wedding.
For example: sea wedding shell decoration, country wedding wheat, balloon-themed balloons, forest wedding vines, fairy wedding cartoon toys can be used as table flowers to embellish<<
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Our people
We have a highly experienced management team each with a proven track record in managing contracts, meeting targets and delivering results for our clients.
Our staff are all dedicated and well qualified – over 90% of our advisers are educated to level 4 or above in advice and guidance. Staff are organised into teams offering specialist expertise, including careers information, advice and guidance, work with NEET young people, IAG and CEIAG training, support to employers and training providers, and ICT services.
Our management team
- Gerald, Chief Executive
- Steve, Head of Services (Young People)
- Katren, Head of Business Development
- Sarah, Head of Corporate Services
'Strategic direction and effective internal and external support from the Executive team and wider management group was frequently praised by delivery staff as being fundamental to maintaining the organisation's visibility and credibility across local authority areas.' matrix report, January 2016., and one from the National Children's Centre. Our Chair of the Board is Stephen Boyd, Head of Careers & Employability at the University of Huddersfield.
'The.
'Leadership and direction provided by the Executive Board and middle management team continues to be a key organisational strength ... Staff are very clear on the organisational direction of travel and the goal to achieve exceptional performance. The workforce feels supported and values of autonomy and trust prevail.' matrix Report, January 2016.
'My internal induction was fantastic. It enabled me to truly understand my new role and how it links to the C&K Careers strategic aims'. matrix Report, January 2016.
'The TUPE into C&K Careers was so smooth and supportive. As a team we were immediately at home in an organisation whose whole business is quality IAG'. matrix Report, January 2016.
'People have a respect for and belief in top managers – they believe they are capable and competent.' IiP Review Report, December 2011.
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_3<<
We review the two hottest indie rock releases of the summer — Stars’ The Five Ghosts and The Arcade Fire’s The Suburbs and deem how they stack up to each other and the bands’ past work.
The Silversun Pickups are an indie band that doesn’t sound like your average indie band. They have bits of emocore, classic rock revitalism, and psychedelia blended in. Together they offer up a unique brand of rock and roll.
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Spending some time camping with your dog is one of the most enjoyable things. Are you a dog lover? Do you love camping? If the answers to these questions are yes; then this article is for you. Maybe you have never thought of spending time together outdoors with your dog. Now there is a good reason to try it and guess what? It will be a very enjoyable experience. You only need a few tips and guide to make it happen.
Training Your Dog is a Great Starting Point
If she is trained; you will be ready. The most vital training you need for the dog is a dependable recall. Does your dog respond when you call? Your dog should respond even when there are distractions or other dogs. Trust your skills and train your dog on how to respond promptly when you call. You might need to seek the services of a professional dog trainer if you are not sure.
Some simple commands that you can use to train your dog are explained here.
- “Come” – This command tells your dog to come to you
- “Leave it” – This command tells the dog to stop what she is doing and drop what’s in her mouth. This command is very useful when you go camping. You can use it to command your dog to stop sniffing a snake or dead crow
- “Down” – This command tells your dog to lay down and wait for the next command
- “Okay” – This command informs the dog everything is okay or cool
You’ll want to train the dog to stay in the car until they are invited to come along. This training is important because you’ll enhance the safety of your dog. A dog can follow you when you aren’t aware and get into an accident. Through the training; you’ll get the dog to trust you and obey your commands. Remember the more command your dog knows and responds to; the better for you. The training is a just a process of creating a good relationship with the dog in preparation for camping.
Untrained or un-socialized dogs will draw negative attention almost immediately and may put both you and the dog in harms way.
Remember to Carry Enough Supplies
Water should be the first item on the supplies list. Set aside a plastic water bottle and water bowl for your dog when packing supplies. It’s good to carry separate water for you and the dog; just to ensure there’s enough. Because dogs don’t sweat; they tend to drink a lot of water to cool down. When you break to drink water; give your dog some water too. The other item on the supplies is food. Don’t be tempted to change the dog’s diet. If she feeds on homemade food; you can pack some. But you might want to decrease the portions.
Backpacks are a good choice to carry camping supplies. The choice will depend on the number of days you’ll be camping. You can rely on a dog backpack to carry some supplies because it’ll relieve you of some weight. A healthy dog can carry about 25% of their body weight. Make sure to strap the backpack on safely without hurting the dog or putting on too much excess weight. The best practice is to put most of the weight low and forward; over her front shoulders.
Don’t forget a good tent when planning. A pop up canopy tent is great; although you might want to carry hammock. Remember to consider tent stakes. The right equipment will make camping with your dog smooth and memorable.
Related: Prepping Dog and Cat Food
A First Aid Kit is a Great Idea
Your dog can be injured just like you. In fact; a dog can be at higher risk of being injured than you. A dog isn’t a good a judge of what they can do and might end up with injuries. Fearless dogs are more likely to be injured. Running and scrambling over rocks is fun and adventurous; but an invitation for injuries. The dog’s foot pads can be injured. Don’t assume that your dog’s thorn-proof! A pricking on foot can cause a lot of pain and even bleed. When it comes to the dog’s first aid kit; you’re the expert. Carry the drugs that have been prescribed by your veterinarian. Pain relievers are good to carry along as well.
When packing the first aid kit; consider the place where you’ll be camping. This way; you can prepare sufficiently. In almost any camp ground; there’s a threat of small and large animals. Some animals like moose and bears tend to avoid humans. They may however not avoid curious dogs. Your dog can be at risk of injury if the camping site has these animals. The dog can be kicked; bitten or even infected with a disease by interacting with these animals. Small animals can scratch the nose of your dog if they are curious.
In some places; ticks can be a problem. You’re spending time outdoors with the dog; and you should inspect the dog daily for any ticks. There might also be wood ticks in the camping site. You should consider immunizing the dog for Lyme. These tips will be sufficient to keep your dog safe. But at times; your dog might get a big injury that can’t be managed with the first aid kit. It’ll be a great idea to go back home and seek the help of your veterinarian if the injury is not life-threatening. If it is; seek the closet professional help you can find. Fortunately; you can avoid all the worry by preparing well for the specific terrain and risks in the place where you will be camping.
Embrace Good Campsite Etiquette
While at the campsite practice good campsite etiquette. Be courteous to your friends and fellow campers. Use the commands that your dog has learned to keep the dog away from strangers. It’s your responsibility to properly dispose of the dog’s waste each and every time. To keep the dog safe; keep it close to you all times and don’t leave her unsupervised at the campsite. Even in the vehicle ensure your dog is safe. Knowing how to keep the dog safe and close to you will help to prevent unforeseen risks.
Finally, Enjoy with Your Dog />
Now that you are camping with your dog enjoy to your fullest! Play with your dog; take some good photos and relax. Just stay cool. At night enjoy a campfire and let your dog enjoy too. Combine different tips that you have learned to make the experience memorable.
Related: 7 Crucial Things to Your Pet’s Survival
Originally posted 2019-06-19 00:28:21.
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OPENINGS: Look for The Glass Knife, a swanky new café/pâtisserie, to open this Friday on South Orlando Avenue in Winter Park ... The folks behind the Tamale Co. food truck will open Tamale & Co. Take Out in the former Kalienne's Bakeshop space in Altamonte Springs ... The Patina Restaurant Group will launch three concepts in Disney Springs early next year: Maria & Enzo's, Enzo's Hideaway and Pizza Ponte ... Noodles & Rice will open an outpost in Avalon Park sometime next year ... Grills Restaurant Lakeside hopes to open on the west side of Fairview Shores by the end of the year ... Buda Libre, by Iron Chef America contestant Roberto Trevino, is poised to open any day now on Church Street ... Over in Lake Nona Landing, Urban Hibachi will open their third restaurant later next year ... See's Candies has opened a pop-up "holiday gifting center" in the Mall at Millenia. The center will be open until Dec. 27 ... Frank Anderson, who ran the kitchens at James Beard Award-winning Animal and Son of a Gun in L.A., has been named culinary director for the Rez Grill, opening Dec. 14 at the Seminole Hard Rock Hotel & Casino ... Chianti's Pizza and Pasta will open a 120-seat restaurant in Longwood Village in the space that previously housed Carmela's of Brooklyn.
CLOSINGS: Nova Restaurant in Ivanhoe Village has closed ... The Farmacy has ceased its organic grocery operations inside the North Quarter Market.
NEWS: Growing Orlando, a nonprofit urban farm initiative, needs your help. Last month, thieves stole many of the farming tools and equipment they need to run their South Street Garden. Go to growingorlando.org/donate to make a contribution ... Hawkers Asian Street Fare will move its headquarters and test kitchen to Church Street, and will utilize the glassed-in covered bridge as a mock dining/bar area as well as for special tasting events.
EVENTS: Kevin Fonzo (formerly of K Restaurant) pairs up with Kathleen Blake for the Feast of Fonzo Thursday, Nov. 9, at the Rusty Spoon. Cost for the four-course dinner with wine pairings is $65 ... A Taste of Dr. Phillips, with 22 Orlando restaurants participating, is Sunday, Nov. 12, at the Dellagio Town Center. Cost is $70.
Got restaurant dish? Send tips to dining@orlandoweekly.com
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TITLE: $\alpha x^2+\beta y^2=\gamma$ solvable over $\mathbb Q$ iff $ax^2+by^2=z^2$ solvable over $\mathbb Z$ with coprime $x,y,z$?
QUESTION [3 upvotes]: I want to understand an algorithm from [1] to solve
$$\alpha x^2+\beta y^2=\gamma \text{ over } \mathbb{Q}$$
with $\alpha, \beta, \gamma\in\mathbb{Q}$. As far is I understood the process the following happens: Multiply the equation by the $\gcd$ of the denominators of $\alpha, \beta, \gamma$ to obtain an equation
$$\alpha x^2+\beta y^2=\gamma$$
where $\alpha, \beta, \gamma\in\mathbb Z$. We may furthermore assume that $\alpha$ and $\beta$ are square free, since else we can solve the equation
$$\bar{\alpha}x'^2+\bar{\beta}y'^2=\gamma$$
where $\bar{\alpha}, \bar{\beta}$ are the nonsquare-parts of $\alpha$ and $\beta$ and with new variables $x'=(\tilde{\alpha}x)^2$ and $y'=(\tilde{\alpha}y)^2$ where $\tilde{\alpha}$ and$\tilde{\beta}$ are the square parts of $\alpha$ and $\beta$. So without loss of generalty, we are left with an equation
$$\alpha x^2+\beta y^2=\gamma \text{ over } \mathbb{Q}$$
where $\alpha, \beta, \gamma\in\mathbb Z$ and $\alpha, \beta$ are square free.
But now the magic happens: This equation should be solvable if and only if an equation
$$ax^2+by^2=z^2$$
is solvable over $\mathbb Z$ with coprime $x,y,z$. Unfortunately I don't see how the coefficients $\alpha$, $\beta$ and $\gamma$ should be related to the coefficient $a$ and $b$ and therefore am not able to understand the equivalence of solving these two equations.
I think that there might be a very easy number-theoretic argument that I don't know and of course it would be awesome if there is indeed an elementary argument for this.
As a remark, that might or might not help: A friend gave me the idea that $z$ could have something to do with the square part $\tilde{\gamma}$ of $\gamma$.
The argument I'm referring to is on page 22 in[1], the middle after "Suppose $\mathbb F=\mathbb Q$".
[1] On the complexity of cubic forms
REPLY [2 votes]: The process is homogenization: we switch to integers by introducing a new variable.
Suppose that we are interested in the solvability in rationals $(x,y)$ of
$$\alpha x^2+\beta y^2=\gamma,\tag{$1$}$$
where without loss of generality $\alpha$, $\beta$, and $\gamma$ are integers. Assume $\gamma\ne 0$. There is a rational solution of $(1)$ iff there are integers $u$, $v$, $w$, with $w\ne 0$ such that
$\alpha \left(\frac{u}{w}\right)^2+\beta\left(\frac{v}{w}\right)^2=\gamma$, or equivalently
$$\alpha u^2+\beta v^2=\gamma w^2.\tag{$2$}$$
Equation $(2)$ has an integer solution with $w\ne 0$ iff the equation $(\gamma\alpha)u^2+(\gamma\beta)v^2=\gamma^2w^2$ has such a solution.
Or equivalently, Equation $(2)$ has an integer solution with $w\ne 0$ iff the equation $(\gamma\alpha)u^2+(\gamma\beta)v^2=z^2$ has an integer solution with $z\ne 0$. (The $\gamma$ terms on the left force divisibility of $z$ by $\gamma$ since $\gamma$ has no square-part without loss of generality).
| 45,020
|
TITLE: Cluster point of set $\{a_n \,|\, n \in \mathbb N\}$ is limit point of sequence $(a_n)_n$
QUESTION [1 upvotes]: Let $(a_n)_n$ be a sequence in $\mathbb R$ and $A := \{a_n \,|\, n \in \mathbb N\}$ the set of elements of the sequence. I want to show
If $(a_n)_n$ is injective, then a limit point of $(a_n)_n$ is also a cluster point of $A$.
My definitions are
$x_0$ is a cluster point of a set $A \iff$ for every $\varepsilon \gt 0$ there exists $a \in A$, such that $0 \lt |a - x_0| \lt \varepsilon$.
$x_0$ is a limit point of a sequence $(a_n)_n \iff$ there is a subsequence which converges to $x_0$ or equivalently if for all $\varepsilon \gt 0$ and all $N \in \mathbb N$ there exists $n \ge N$, such that $|a_n - x_0| \lt \varepsilon$.
I don't know how to start. How should I use the injective property?
REPLY [1 votes]: The injective property is needed because otherwise for example for $a_n=(-1)^n$ we have $A=\{-1,1\}$ and $-1$ is a limit point of $a_n$ but not a limit point of $A$...
Let $a$ be a limit point of $a_n$. There is a subsequence $a_{n_k}$ converging to $a$. Let $\epsilon>0$ be given. There is $K_1$ such that for all $k>K_1$ we have:
\begin{align}
|a_{n_k}-a|<\epsilon
\end{align}
Not that we cannot have $a_{n_k}=a$ for all $k>K_1$, because $a_n$ is injective. Let $K_2$ be such that $a_{n_{K_2}}=a$ if there is one. If there is no $K_2$ with that property then we take $K_2=1$. Now take $K=\max\{K_1,K_2 \}$. So for all $k>K$ we have $0<|a_{n_k}-a|<\epsilon$. Since $\epsilon>0$ was arbitrary we can conclude $a$ is a limit point of $A$.
| 147,825
|
\begin{document}
\maketitle
\begin{abstract}
We consider the numerical evaluation of one dimensional projections of
general multivariate stable densities introduced by
\cite{Abdul-Hamid:Nolan:1998}. In their approach higher order
derivatives of one dimensional densities are used, which seem
to be cumbersome in practice. Furthermore there are some
difficulties for even dimensions. In order to overcome these
difficulties we obtain the explicit finite-interval integral
representation of one dimensional projections for all dimensions.
For this purpose we utilize the imaginary part of complex
integration,
whose real part corresponds to the derivative of the one dimensional
inversion formula.
We also give summaries on relations between various
parameterizations of stable multivariate density and its one
dimensional projection.
\end{abstract}
\section{Introduction}
Stable distributions are known as the limiting distributions of the
general central limit theorem and the time invariant distributions for L\'evy
processes. Because of these probabilistically very important properties, many
studies have been carried out on their theoretical
aspects. Comprehensive books have been published
(\cite{Araujo:Gine:1980} or \cite{Christoph:Wolf:1993}).
Furthermore
applications of stable distributions
to heavy-tailed data have been growing over past few decades.
These applications have appeared in many fields like finance, Internet
traffic or physics. Especially in time series, the observations from the
causal stable ARMA or fractional
stable ARIMA model can be considered as a multivariate stable random vector.
We refer to \cite{Uchaikin:Zolotarev:1999},
\cite{Rachev:2003} or \cite{Adler:Feldman:Taqqu:1998} for these applications.
As various multivariate heavy tailed data have become available in these fields,
the treatments of multivariate
stable distributions are needed.
However, the multivariate stable
distributions require handling of their spectral measures which is computationally difficult.
We briefly review some important recent progresses concerning the
multivariate stable distributions.
Concerning the calculation of the integral with respect to the
spectral measure, \cite{Byczkowski:Nolan:Rajput:1993} have
approximated the spectral measure by a discrete spectral
measure which is uniformly close to the original spectral measure.
They also have given several instructive graphs
of the stable densities.
Concerning the approximations,
\cite{Davydov:Nagaev:2002} have outlined possible directions although
their results are still theoretical. For the estimation of the parameters,
see \cite{Marcus:Luis:2003} and \cite{Davydov:Paulauskas:1999} and
for the hypothesis testing see \cite{Mittnik:Rachev:Ruschendorf:1999}.
Although there are many other researches,
accurate calculations of the multivariate stable densities are clearly of
basic importance.
For the calculations of the multivariate stable densities,
the method of \cite{Abdul-Hamid:Nolan:1998}
based on one dimensional projection is very promising.
In the approach of \cite{Abdul-Hamid:Nolan:1998}, the basic ingredient
is a function $g_{\alpha,d}(v,\beta)$ defined by
\begin{equation}
\label{eq:g-alpha-d}
g_{\alpha,d}(v,\beta)
=
\begin{cases}
\frac{1}{(2\pi)^d}\int_0^\infty \cos
\left(vu-\left(\beta\tan\frac{\pi\alpha}{2}\right)
u^\alpha\right)u^{d-1}e^{-u^\alpha}du, & \alpha \neq 1, \\
\frac{1}{(2\pi)^d}\int_0^\infty \cos
\left(vu+\frac{2}{\pi}\beta u\log u\right)u^{d-1}e^{-u}du, & \alpha=1,
\end{cases}
\end{equation}
where $\alpha$ is the characteristic exponent of the stable
distribution, $d$ is the dimension and $\beta$ corresponds to the
skewness parameter.
Evaluating $g_{\alpha,d}(v,\beta)$ using this very definition
is not very practical because the integrand is
infinitely oscillating and changing its sign.
Therefore an alternative evaluation of
$g_{\alpha,d}(v,\beta)$ is desirable.
In this paper, we obtain explicit finite-interval integral
representations of $g_{\alpha,d}(v,\beta)$, which do not require the
derivatives of one dimensional densities.
The integrand in our representation is well-behaved without infinite
oscillations. Furthermore, our integral representation covers
all dimensions and all parameter values, except for an exceptional
case of $\alpha \le 1, \beta =1$ and even dimension, where we need an
additional term which is easily tractable.
The calculation of $g_{\alpha,d}(v,\beta)$ corresponds to
the evaluation of the equation (2.2.18)
of \cite{Zolotarev:1986}, which is stated without an explicit proof.
For odd dimension, the real part of the equation is equivalent to $g_{\alpha,d}(v,\beta)$
and for even dimension, the imaginary part is equivalent to $g_{\alpha,d}(v,\beta)$.
Complete derivations of the equation (2.2.18) of \cite{Zolotarev:1986}
is also useful for the numerical evaluations of the derivatives of one dimensional
symmetric densities investigated in \cite{Matsui:Takemura:2006}.
This paper is organized as follows. For the rest of this section we
prepare preliminary results based on \cite{Abdul-Hamid:Nolan:1998}
and \cite{Nolan:1998}.
In Section \ref{sec:finite-integral} we derive
finite-interval integral representations of one dimensional
projections.
In Section \ref{sec:various-projections} we summarize
relations between
various parameterizations of the multivariate stable density and its
one dimensional projection as propositions.
Some discussions and directions for further studies are given
in Section \ref{sec:procedures-future-work}.
Some proofs are postponed to Section \ref{sec:proofs}.
\subsection{Definitions and preliminary results}
\label{sec:definitions}
Let $\mathbf{X} =(X_1,X_2,\ldots,X_d)$ denote a $d$-dimensional $\alpha$-stable random
vector with the characteristic function
$$
\Phi_{\alpha,d}(\mathbf{t}) = E \exp (i\langle \mathbf{t}, \mathbf{x} \rangle ),
$$
where $\mathbf{t}=(t_1,t_2,\ldots,t_d)$ and $\langle\ ,\ \rangle$ denotes
the usual inner product.
Among several definitions of stable distributions, we give the spectral representation of
$\Phi_{\alpha,d}(\mathbf{t})$.
This requires integration over the unit sphere
$\mathbb{S}^d=\{\mathbf{s} : \Vert\mathbf{s}\Vert=1\}$
in $\mathbb{R}^d$ with respect to a spectral measure
$\Gamma$
(Theorem 2.3.1 of \cite{Samorodnitsky:Taqqu:1994}, Theorem 14.10 of \cite{Sato:1999}).
The spectral definition is given as
\begin{equation}
\label{eq:definition-characteristic-a-representation}
\Phi_{\alpha,d}(\mathbf{t})=
\begin{cases}
\exp\left(-\int_{\mathbb{S}^d} | \langle
\mathbf{t},\mathbf{s} \rangle |^\alpha
\left(1-i\sign(\langle
\mathbf{t}, \mathbf{s} \rangle)
\tan\frac{\pi\alpha}{2}\right)\Gamma(d\mathbf{s})+i\langle
\mathbf{t}, \mathbf{\mu} \rangle
\right), & \alpha \neq 1, \\
\exp\left(-\int_{\mathbb{S}^d} | \langle
\mathbf{t},\mathbf{s} \rangle |
\left(1+i\frac{2}{\pi}\sign(\langle
\mathbf{t}, \mathbf{s} \rangle)\log|\langle
\mathbf{t},\mathbf{s}\rangle| \right)\Gamma(d\mathbf{s})+i\langle
\mathbf{t}, \mathbf{\mu} \rangle
\right), & \alpha=1.
\end{cases}
\end{equation}
The pair $(\Gamma,\mu)$ is unique.
The definition (\ref{eq:definition-characteristic-a-representation})
corresponds to Zolotarev's (A) representation and
we use the notation
$S_{\alpha,d}(\Gamma,\mu)$ for
this definition following \cite{Nolan:1998}.
For another definition, we give a multivariate version of Zolotarev's
(M) parameterization, which is also defined in \cite{Nolan:1998}. We use
notation $S^M_{\alpha,d}(\Gamma,\mu_0)$.
\begin{align}
\label{eq:definition-characteristic-m-representation}
& \Phi_{\alpha,d}(\mathbf{t}) \nonumber \\
& \quad = \begin{cases}
\exp\left(-\int_{\mathbb{S}^d} | \langle
\mathbf{t},\mathbf{s} \rangle |^\alpha
\left(1+i\sign(\langle
\mathbf{t}, \mathbf{s}\rangle) \tan\frac{\pi\alpha}{2}( |\langle
\mathbf{t},\mathbf{s} \rangle|^{1-\alpha}-1)
\right)\Gamma(d\mathbf{s})+i\langle
\mathbf{t}, \mathbf{\mu_0} \rangle
\right),
& \alpha\neq 1, \\
\exp\left(-\int_{\mathbb{S}^d} | \langle
\mathbf{t},\mathbf{s} \rangle |
\left(1+i\frac{2}{\pi}\sign(\langle
\mathbf{t}, \mathbf{s} \rangle)\log|\langle
\mathbf{t},\mathbf{s}\rangle| \right)\Gamma(d\mathbf{s})+i\langle
\mathbf{t}, \mathbf{\mu_0} \rangle
\right),
& \alpha=1.
\end{cases}
\end{align}
In the equation (\ref{eq:definition-characteristic-m-representation}),
we avoid the discontinuity at $\alpha=1$ and the
divergence of the
mode as $\alpha \to 1$ which are observed in the definition
(\ref{eq:definition-characteristic-a-representation})
in the non-symmetric case. The difference of (M)
parameterization from
(A) representation is only
$$
\mu_0=\mu-\tan\frac{\pi\alpha}{2}\int_{\mathbb{S}^d}\mathbf{s}\
\Gamma(d\mathbf{s}), \qquad \alpha\neq 1.
$$
Now consider inverting the characteristic function
$\Phi_{\alpha,d}(\mathbf{t})$ to evaluate the multivariate stable density.
The useful idea of \cite{Nolan:1998} and \cite{Abdul-Hamid:Nolan:1998} is to perform this
inversion with respect to the polar coordinates.
The projection of an $\alpha$-stable random vector onto one dimensional
stable random variable is explained as follows.
Here we follow the arguments of Ex.2.3.4 of \cite{Samorodnitsky:Taqqu:1994}.
For any $\mathbf{t}\in \mathbb{R}^d$ consider the characteristic
function $E\exp(iu\langle \mathbf{t},\mathbf{x}\rangle)$ of
a linear combination $Y=\sum_i t_i X_i$,
i.e.,
\begin{eqnarray*}
&&E\exp(iu\langle \mathbf{t},\mathbf{x}-\mu \rangle) \\
&& \quad
=
\begin{cases}
\exp\left(-|u|^\alpha\left[ \int_{\mathbb{S}^d} | \langle
\mathbf{t},\mathbf{s} \rangle |^\alpha\Gamma(d\mathbf{s})
-i\sign(u)\tan\frac{\pi\alpha}{2}
\int_{\mathbb{S}^d}\sign(\langle
\mathbf{t}, \mathbf{s} \rangle)| \langle
\mathbf{t},\mathbf{s} \rangle |^\alpha
\Gamma(d\mathbf{s})\right] \right),
&
\alpha \neq 1, \\
\exp\left(-|u| \left[\int_{\mathbb{S}^d} | \langle
\mathbf{t},\mathbf{s} \rangle | \Gamma(d\mathbf{s})
+i\frac{2}{\pi}\sign(u)\int_{\mathbb{S}^d} \langle
\mathbf{t},\mathbf{s} \rangle \left(\log|u|+\log|\langle
\mathbf{t},\mathbf{s}\rangle|\right)\Gamma(d\mathbf{s})\right]
\right),
&\alpha = 1.
\end{cases}
\end{eqnarray*}
Then, from the definition below, $E\exp(iu\langle
\mathbf{t},\mathbf{x}\rangle)$ is considered as a characteristic function
of one dimensional stable distribution with the following parameters.
\begin{align}
\label{def-A-sigma}
\sigma(\mathbf{t})&=\left(\int_{\mathbb{S}^d}|\langle \mathbf{t}, \mathbf{s}
\rangle|^\alpha \Gamma (d \mathbf{s}) \right)^{1/\alpha},\\
\label{def-A-beta}
\beta(\mathbf{t})&=(\sigma(\mathbf{t}))^{-\alpha}\int_{\mathbb{S}^d} \sign
\langle \mathbf{t}, \mathbf{s} \rangle\ |\langle \mathbf{t}, \mathbf{s}
\rangle|^\alpha\ \Gamma (d \mathbf{s}), \\
\label{def-A-mu}
\mu(\mathbf{t}) &=
\left\{
\begin{array}{ll}
0,& \alpha \neq 1, \\
-\frac{2}{\pi}\int_{\mathbb{S}^d}\langle \mathbf{t}, \mathbf{s}
\rangle \log| \langle \mathbf{t}, \mathbf{s}
\rangle|\ \Gamma (d \mathbf{s}), & \alpha=1.
\end{array}
\right.
\end{align}
Regarding $u$ as the length and $\mathbf{t}$ as the angle, we
transform the $d$-dimensional rectangular integral into
the polar coordinate integral. In the following theorem of
\cite{Nolan:1998}, multivariate stable densities of (A) representation
are projected onto one dimensional densities in (A)
representation. Since in the original theorem (\cite{Nolan:1998})
there are some trivial typographical errors,
we correct them in the formula (\ref{eq:g-alpha-d}) and the following theorem.
\begin{thm} {\rm (\cite{Nolan:1998})} \quad
\label{thm:A-A-representation}
Let $0<\alpha<2$ and $\mathbf{X} =(X_1,X_2,\ldots,X_d)$ be
an $\alpha$-stable random vector $S_{\alpha,d}(\Gamma,\nu)$ with $d\ge
2$. Then the density $f_{\alpha,d}(\mathbf{x})$ of $\mathbf{X}$ is given as
\begin{equation*}
f_{\alpha,d}(\mathbf{x})=
\left\{
\begin{array}{ll}
\int_{\mathbb{S}^d} g_{\alpha,d}\left(
\displaystyle
\frac{ \langle \mathbf{x}-\nu,\mathbf{s} \rangle
}{\sigma(\mathbf{s})},\beta(\mathbf{s})\right)\left(\sigma(\mathbf{s})\right)^{-d} \ d \mathbf{s},
& \alpha\neq 1,\\
\int_{\mathbb{S}^d}g_{1,\alpha}\left(
\displaystyle
\frac{\langle \mathbf{x}-\nu,\mathbf{s} \rangle -\mu(\mathbf{s})-(2/\pi)\beta(\mathbf{s})\sigma(\mathbf{s})\log\sigma(\mathbf{s})}{\sigma(\mathbf{s})},\beta(\mathbf{s})
\right)\left(\sigma(\mathbf{s})\right)^{-d} \ d \mathbf{s},
& \alpha=1,
\end{array}\right.
\end{equation*}
where $g_{\alpha,d}(v,\beta)$ is given in (\ref{eq:g-alpha-d}).
\end{thm}
\cite{Abdul-Hamid:Nolan:1998} expressed
$g_{\alpha,d}(v,\beta)$ as an integral over a finite interval, utilizing
the integral expressions of one dimensional densities.
They explained the method when $d$ is odd as follows.
{}From Theorem 2.2.3 of \cite{Zolotarev:1986}, the one dimensional density $g_{\alpha,1}(v,\beta)$
can be written as
\begin{equation*}
g_{\alpha,1}(v,\beta)=\int_a^b h_g(\theta;\alpha,\beta,v) d\theta,
\end{equation*}
where $-\pi/2 \le a<b \le \pi/2$ and $h_g$ is a somewhat complicated
but explicit function. Then, for odd $d$, differentiating
$g_{\alpha,1}$ $d-1$ times with respect to $v$ gives the representation
$$
g_{\alpha,d}(v,\beta)= c(d)\int_a^b
\frac{\partial^{d-1}h_g(\theta;\alpha,\beta,v)}{\partial v^{d-1}}d \theta,
$$
where $c(d)$ is some constant.
However, the computation of
$\partial^{d-1}h_g(\theta;\alpha,\beta,v)/\partial v^{d-1}$ seems to
be cumbersome. Furthermore, they did not give the corresponding
expression for even $d$.
In the following, we obtain explicit finite-interval integral
representations of $g_{\alpha,d}(v,\beta)$, which do not require the
derivatives of one dimensional densities.
\section{Finite-interval integral representations of projected
one dimensional functions}
\label{sec:finite-integral}
In this section we obtain the finite-interval integral representations of
$g_{\alpha,d}(v,\beta)$ in (\ref{eq:g-alpha-d}).
It is based on the function
$h^n(x;\alpha,\beta)$ defined below in
(\ref{eq:n-derivative-invesion}), which
corresponds to the $n$-th order derivative of one
dimensional inversion formula including the imaginary part.
We first prepare some notations which are the same as in
\cite{Zolotarev:1986}.
\allowdisplaybreaks{
\begin{align*}
K(\alpha)&=\alpha-1+\sign (1-\alpha), \qquad \alpha\neq 1,\\
\theta&=\beta K(\alpha)/\alpha, \qquad \alpha\neq 1,\\
U_\alpha(\varphi;\theta)&=\left(\frac{\sin\frac{\pi}{2}(\varphi+\theta)}{\cos\frac{\pi}{2}\varphi}\right)^{\alpha/(1-\alpha)}
\frac{\cos\frac{\pi}{2}\{(\alpha-1)\varphi+\alpha\theta\}}{\cos\frac{\pi}{2}\varphi},
\qquad \alpha \neq 1,\\
U_1(\varphi;\beta)&= \frac{\pi}{2} \frac{1+\beta\varphi}{\cos\frac{\pi}{2}}
\exp\left( \frac{\pi}{2} \left(\varphi+\frac{1}{\beta} \right)
\tan\frac{\pi}{2}\varphi\right), \\
r(\varphi) &=
\left\{
\begin{array}{ll}
\left(
\frac{\sin\alpha(\varphi+\pi\theta/2)}{x\cos\varphi}
\right)^{1/(1-\alpha)},
&\alpha \neq 1, \\
\exp(-x/\beta+(\varphi+\pi/(2\beta))\tan \varphi),
&\alpha =1,
\end{array}
\right. \\
\tau &=
\left\{
\begin{array}{ll}
(\alpha/x)^{1/(1-\alpha)},
&\alpha \neq 1, \\
\exp(-x-1),
&\alpha =1,
\end{array}
\right. \\
V_n(\varphi) &= r^n\left(\frac{\pi}{2}\varphi\right)
\left\{ r'\left(\frac{\pi}{2}\varphi
\right)\sin\frac{\pi}{2}(n+1)(\varphi+1)+r\left(\frac{\pi}{2}\varphi\right)\cos\frac{\pi}{2}(n+1)(\varphi+1) \right\},\\
W_n(\varphi) &= r^n\left(\frac{\pi}{2}\varphi\right)
\left\{ r\left(\frac{\pi}{2}\varphi \right)\sin\frac{\pi}{2}(n+1)(\varphi+1)-r'\left(\frac{\pi}{2}\varphi\right)\cos\frac{\pi}{2}(n+1)(\varphi+1) \right\}.
\end{align*}
We now define a function $h^n(x;\alpha,\beta)$ as
\begin{eqnarray}
\label{eq:n-derivative-invesion}
h^n(x;\alpha,\beta) &=& \frac{1}{\pi}\int_0^\infty (iz)^n
\exp\left(izx+\psi(z;\alpha,-\beta)\right)dz,
\end{eqnarray}
where
\begin{eqnarray*}
\psi(z;\alpha,\beta) &=&
\left\{
\begin{array}{ll}
-z^{\alpha}\exp(-i\frac{\pi}{2}\theta\alpha),
&\alpha \neq 1, \\
-\frac{\pi}{2}-i\beta z \log z,
&\alpha =1,
\end{array}
\right.
\end{eqnarray*}
is the logarithm of the characteristic function of
stable distributions which are analytically extended from the semi-axis
$\mathrm{Re}\;z>0$ into the complex plane. Then,
$g_{\alpha,d}(v,\beta)$ in (\ref{eq:g-alpha-d}) is expressed in terms
of $h^n(x;\alpha,\beta)$ as follows. Note that $\beta$ in
$g_{\alpha,d}(v,\beta)$ is not equal to $\beta$ in
$h^n(x;\alpha,\beta)$, because
$\beta$ in $h^n(x;\alpha,\beta)$ is based on one dimensional (B)
parameterization. Therefore, we need the following definitions,
$$
\beta_B = \frac{2}{\pi K(\alpha)} \arctan \left(\beta \tan \frac{\pi \alpha}{2} \right),\quad
\theta_B = \beta_B K(\alpha)/\alpha, \qquad \alpha\neq 1.\\
$$
\begin{proposition}
\label{lem:relation-g-h}
Define
$x=\left(\cos(\pi/2\alpha\theta_B)\right)^{1/\alpha}v$ and
$y=\pi/2v+\beta\log (\pi/2)$. Then a
representation of $g_{\alpha,d}(v,\beta)$ in (\ref{eq:g-alpha-d})
in terms of $h(\cdot\ ;\alpha,\beta)$ is given as follows:
\begin{eqnarray*}
g_{\alpha,d}(v,\beta) &=&
\left\{
\begin{array}{ll}
\mathrm{Re}\
\frac{\left(\cos\left(\pi/2\alpha\theta_B
\right)\right)^{d/\alpha}}{2^d(\pi i)^{d-1}} h^{d-1}(|x|;\alpha,\beta_B^\ast),
& \alpha \neq 1,\\ &
\\
\mathrm{Re}\
\frac{\pi}{4^d i^{d-1}} h^{d-1}(y^\ast;\alpha,|\beta|),
& \alpha=1,
\end{array}
\right.
\end{eqnarray*}
where $\beta_B^\ast=\beta_B\ \sign x$ and $y^\ast=y\ \sign \beta$.
\end{proposition}
Proof of this proposition is given in Section \ref{sec:proofs}.
Although the imaginary number $i$ is present in the above formulas, it
is for notational convenience and we do not have to write
$\mathrm{Re}h(\cdot)$ or $\mathrm{Im}h(\cdot)$ depending on the
dimension $d$.
By Proposition \ref{lem:relation-g-h}, it suffices to express
$h^n(x;\alpha,\beta)$ as a finite-interval integral.
We now give our main theorem.
We mention that the following results are
partially stated in \cite{Zolotarev:1986} for the real case without
proofs and the equation (2.2.34) of \cite{Zolotarev:1986} is not complete.
Therefore we give the complete results including the calculation of
imaginary parts. The full proof is given in Section \ref{sec:proofs}.
\begin{thm}
\label{thm:differential-density}
Let $x>0$ in the case $\alpha \neq 1$ and $\beta>0$ in the case
$\alpha=1$.
Then the finite-interval integral representations of
$h^n(x;\alpha,\beta)$ are as follows. \\
{\rm (a)}\ $(\alpha \neq 1,\beta \neq 1)$ or $(\alpha>1, \beta=1)$
\begin{eqnarray*}
\mathrm{Re}\ h^n(x;\alpha,\beta) &=& \frac{1}{2}\int_{-\theta}^1 \exp
\left(-x^{\alpha/(\alpha-1)}U_\alpha(\varphi;\theta)\right)
V_n(\varphi)d\varphi. \\
\mathrm{Im}\ h^n(x;\alpha,\beta) &=& \frac{1}{2}\int_{-\theta}^1 \exp
\left(-x^{\alpha/(\alpha-1)}U_\alpha(\varphi;\theta)\right)
W_n(\varphi) d\varphi.
\end{eqnarray*}
{\rm (b)}\ $(\alpha=1,\beta\neq 1)$
\begin{eqnarray*}
\mathrm{Re}\ h^n(x;1,\beta) &=& \frac{1}{2}\int_{-1}^1 \exp
\left(-e^{-x/\beta}U_1(\varphi;\beta)\right)
V_n(\varphi)d\varphi. \\
\mathrm{Im}\ h^n(x;1,\beta) &=& \frac{1}{2}\int_{-1}^1 \exp
\left(-e^{-x/\beta}U_1(\varphi;\beta)\right)
W_n(\varphi)d\varphi.
\end{eqnarray*}
{\rm (c)}\ $(\alpha<1,\beta=1)$ \\
$\mathrm{Re}\ h^n(x;\alpha,1)$ is the same as in {\rm (a)}.
\begin{eqnarray*}
\mathrm{Im}\ h^n(x;\alpha,1) &=& \frac{1}{2}\int_{-\theta}^1 \exp
\left(-x^{\alpha/(\alpha-1)}U_\alpha(\varphi;\theta)\right)
W_n(\varphi) d\varphi -\frac{1}{\pi}\int_0^\tau \exp(xr-r^\alpha) r^n dr.
\end{eqnarray*}
{\rm (d)}\ $(\alpha=1,\beta=1)$ \\
$\mathrm{Re}\ h^n(x;1,1)$ is the same as in {\rm (b)}.
\begin{eqnarray*}
\mathrm{Im}\ h^n(x;1,1) &=& \frac{1}{2}\int_{-1}^1 \exp
\left(-e^{-x/\beta}U_1(\varphi;\beta)\right) W_n(\varphi) d\varphi-\frac{1}{\pi}\int_0^\tau \exp\left(xr+r\log r\right)r^n dr.
\end{eqnarray*}
\end{thm}
Interestingly, the case $\beta=1$ is different from other values of
the parameters. As suggested by \cite{Zolotarev:1986}, the representations of
Theorem \ref{thm:differential-density} are essentially different from
the $n$-th order derivative of
the integral representations of densities given in the equation (2.2.18)
of \cite{Zolotarev:1986}.
\section{Relations between representations of
multivariate stable densities and one dimensional projections}
\label{sec:various-projections}
In this section, we present propositions concerning the
projections of a stable random vector onto one dimensional stable random
variable.
The idea of projection is useful for the purpose of the computation of
the multivariate densities.
There exist two representations (A) and (M) for the
multivariate stable distributions, whereas three
representations (A), (B) and (M) are known for general one dimensional
stable distributions. The projection of the multivariate representation
(A) onto one dimensional representation (A) is given in Theorem
\ref{thm:A-A-representation}, which is obtained by \cite{Nolan:1998}.
Although \cite{Nolan:1998} mentioned the projection of (M) onto (M)
as a theorem without proof, his is different from ours. Therefore we describe the projection of
the multivariate representation (M) onto the projected one dimensional representation (M)
with a precise proof. Note that the idea of projection entirely belongs
to \cite{Nolan:1998}. Further we present results for all projections without
proofs by adding the other 4 projections.
Note that for
$\alpha=1$ multivariate representations (A) and (M) and one dimensional
representations (A) and (M) are the same. Therefore we omit
statements
on the projections (A $\to$ M), (M $\to$ A) and (M $\to$ B) for $\alpha=1$.
\subsection{One-dimensional projections in various representations}
Here we present propositions on various projections. Note that the
parameters $\sigma(\mathbf{t})$, $\beta(\mathbf{t})$ and
$\mu(\mathbf{t})$ used in this section have already been defined by (\ref{def-A-sigma}),
(\ref{def-A-beta}) and (\ref{def-A-mu}) in Section \ref{sec:definitions}, respectively.
\begin{proposition}[The projection (M $\to$ M)]
\label{thm:M-M-representation}
Let $0<\alpha<2$ and $\mathbf{X} =(X_1,X_2,\ldots,X_d)$ be an $\alpha$-stable random vector $S_{\alpha,d}^M(\Gamma,\nu)$ with $d\ge 2$. Then,
the density $f_{\alpha,d}^M(\mathbf{x})$ is given as follows.
\begin{eqnarray*}
f_{\alpha,d}^M( \mathbf{x} )
=\int_{\mathbb{S}^d}g_{\alpha,d}^M\left(\frac{ \langle \mathbf{x}-\nu,\mathbf{s} \rangle
-\mu_M^M(\mathbf{s})}{\sigma(\mathbf{s})},\beta(\mathbf{s})\right)
\left(\sigma(\mathbf{s})\right)^{-d} \ d \mathbf{s},
\end{eqnarray*}
where \\
\begin{eqnarray*}
g_{\alpha,d}^M(v,\beta)=\frac{1}{(2\pi)^d}\int_0^\infty \cos \left(vu+
\left(\beta\tan\frac{\pi\alpha}{2}(u-u^\alpha)\right)\right)u^{d-1}e^{-u^\alpha}du
\end{eqnarray*}
and
$$
\mu_M^M(\mathbf{t})=\tan\frac{\pi\alpha}{2}\left(
\beta(\mathbf{t}) \sigma(\mathbf{t})-
\int_{\mathbb{S}^d} \langle \mathbf{t}, \mathbf{s}
\rangle \ \Gamma (d \mathbf{s})\right).
$$ \end{proposition}
When $\alpha=1$, (A) parametrization in Theorem
\ref{thm:A-A-representation} and (M)
parametrization in Proposition \ref{thm:M-M-representation} are the same.
It is easy to confirm $f_{1,d}^M(
\mathbf{x})=f_{1,d}( \mathbf{x})$ and
$g_{1,d}^M(v,\beta)=g_{1,d}(v,\beta)$ only by notational change.
Note that at $\alpha=1$, $\mu_M^M(\mathbf{t})$ becomes
\begin{eqnarray*}
\label{eq:mu-at-alpha=1}
\mu_M^M(\mathbf{t})&=&\frac{2}{\pi}\left(
\beta(\mathbf{t}) \sigma(\mathbf{t})\log \sigma(\mathbf{t})-
\int_{\mathbb{S}^d} \langle \mathbf{t}, \mathbf{s}
\rangle \ \log | \langle \mathbf{t}, \mathbf{s}
\rangle |\Gamma (d \mathbf{s}) \right) \\
&=&
\mu(\mathbf{s})+(2/\pi)\beta(\mathbf{s})\sigma(\mathbf{s})\log\sigma(\mathbf{s}). \nonumber
\end{eqnarray*}
\begin{proposition}[The projection (A $\to$ B)]
\label{thm:A-B-representation}
Under the same conditions and notations as in Theorem
\ref{thm:A-A-representation}, the density $f_{\alpha,d}( \mathbf{x} )$
is given as follows.
\\
\begin{equation*}
f_{\alpha,d}(\mathbf{x})=
\left\{
\begin{array}{ll}
\int_{\mathbb{S}^d}g_{\alpha,d}^B\left(\frac{ \langle \mathbf{x}-\nu,\mathbf{s} \rangle
}{\sigma_B(\mathbf{s})},\beta_B(\mathbf{s})\right)
\left(\sigma_B(\mathbf{s})\right)^{-d} \ d \mathbf{s},
& \alpha\neq 1,\\
\int_{\mathbb{S}^d}g_{1,d}^B\left(\frac{ \langle \mathbf{x}-\nu,\mathbf{s} \rangle
- \mu_B(\mathbf{s})}{\sigma_B(\mathbf{s})},\beta(\mathbf{s})\right)
\left(\sigma_B(\mathbf{s})\right)^{-d}\ d \mathbf{s},
& \alpha=1,
\end{array}\right.
\end{equation*}
where
\begin{equation*}
g_{\alpha,d}^B(v,\beta)
=
\begin{cases}
\frac{1}{(2\pi)^d}\int_0^\infty \cos \left(vu-
u^\alpha \sin \left(\frac{\pi}{2}K(\alpha)\beta \right)
\right)u^{d-1}e^{-u^\alpha\cos\left(\pi/2K(\alpha)\beta_B\right)}du,
& \alpha \neq 1, \\
\frac{1}{(2\pi)^d}\int_0^\infty \cos \left(vu+ \beta u \log u \right)u^{d-1}e^{-\pi/2u}du,
& \alpha=1,
\end{cases}
\end{equation*}
and
\begin{equation*}
\sigma_B(\mathbf{t})
=
\begin{cases}
\frac{ \sigma(\mathbf{t}) }{ \left( \cos\left(
\frac{\pi}{2} K(\alpha) \beta_B(\mathbf{t}) \right)
\right)^{1/\alpha}},
& \alpha \neq 1, \\
\frac{2}{\pi}\sigma(\mathbf{t}),
& \alpha=1,
\end{cases}
\end{equation*}
$$
\beta_B(\mathbf{t}) = \frac{2}{\pi K(\alpha)} \arctan \left(\beta (
\mathbf{t}) \tan
\frac{\pi \alpha}{2} \right),
$$
$$\mu_B(\mathbf{t})=\sigma_B(\mathbf{t})\beta(\mathbf{t})\log \sigma_B(\mathbf{t})-\frac{2}{\pi}\int_{\mathbb{S}^d} \langle \mathbf{t}, \mathbf{s}
\rangle \ \log | \langle \mathbf{t}, \mathbf{s}
\rangle |\Gamma (d \mathbf{s}).$$
\end{proposition}
\begin{proposition}[The projection (A $\to$ M)]
Under the same conditions and relations as in Theorem
\ref{thm:A-A-representation}, the density $f_{\alpha,d}(\mathbf{x})$ is
given as follows.
\begin{eqnarray*}
f_{\alpha,d}( \mathbf{x} )
=\int_{\mathbb{S}^d}g_{\alpha,d}^M\left(\frac{ \langle \mathbf{x}-\nu,\mathbf{s} \rangle
-\mu_M(\mathbf{s})}{\sigma(\mathbf{s})},\beta(\mathbf{s})\right)
\left(\sigma(\mathbf{s})\right)^{-d} \ d \mathbf{s}, \quad
\alpha \neq 1,
\end{eqnarray*}
where
$$\mu_M(\mathbf{t})=\sigma(\mathbf{t})\beta(\mathbf{t})
\tan \frac{\pi\alpha}{2}. $$
\end{proposition}
\begin{proposition}[The projection (M $\to$ A)]
Under the same conditions and relations as in Proposition
\ref{thm:M-M-representation}, the density $f^M_{\alpha,d}(\mathbf{x})$ is
given as follows.
\begin{eqnarray*}
f_{\alpha,d}^M( \mathbf{x} )
=\int_{\mathbb{S}^d}g_{\alpha,d}\left(\frac{ \langle \mathbf{x}-\nu,\mathbf{s} \rangle
-\mu^M_A(\mathbf{s})}{\sigma(\mathbf{s})},\beta(\mathbf{s})\right)
\left(\sigma(\mathbf{s})\right)^{-d} \ d \mathbf{s}, \quad
\alpha \neq 1,
\end{eqnarray*}
where
$$\mu_A^M(\mathbf{t})=-\tan\frac{\pi\alpha}{2} \int_{\mathbb{S}^d} \langle \mathbf{t}, \mathbf{s}
\rangle \ \Gamma (d \mathbf{s}). $$
\end{proposition}
\begin{proposition}[The projection (M $\to$ B)]
Under the same conditions and relations as in Proposition \ref{thm:M-M-representation}, the density $f^M_{\alpha,d}(\mathbf{x})$ is
given as follows.
\begin{eqnarray*}
f_{\alpha,d}^M( \mathbf{x} )
=\int_{\mathbb{S}^d}g^B_{\alpha,d}\left(\frac{ \langle \mathbf{x}-\nu,\mathbf{s} \rangle
-\mu^M_A(\mathbf{s})}{\sigma_B(\mathbf{s})},\beta_B(\mathbf{s})\right)
\left(\sigma_B(\mathbf{s})\right)^{-d} \ d \mathbf{s},\quad
\alpha \neq 1.
\end{eqnarray*}
\end{proposition}
\subsection{Finite-integral representations for other representations}
In \cite{Zolotarev:1986} only the analytic extension of the
characteristic function defined by (B) representation is considered
and Theorem \ref{thm:differential-density} corresponds to (B) representation of stable
distributions. Therefore, we have introduced the parameters $\beta_B$
and $\theta_B$ in the finite integral-representation for (A) of
Proportion \ref{lem:relation-g-h}.
Although it is possible to derive (A) or (M) representation versions of
Theorem \ref{thm:differential-density} utilizing the results in \cite{Nolan:1997},
we have to consider the analytic extension of the characteristic function
defined by (A) or (M) representation which needs very complicated arguments.
Therefore, in this paper, we confine our results to the case of (B)
representation and for (A) and (M) representation, we utilize
Proposition \ref{lem:relation-g-h} and \ref{lem:relation-gM-h}. Note that, accordingly,
for $g_{\alpha,d}^B(v,\beta)$ in Proposition \ref{lem:relation-gB-h}, we do not need
to use the extra parameters like $\beta_B$ or $\theta_B$ other than that
defined in Theorem \ref{thm:differential-density}.
\begin{proposition}
\label{lem:relation-gM-h}
Let $g_{\alpha,d}^M(v,\beta)$ be as in Proposition
\ref{thm:M-M-representation}. Define
$x= \left(\cos\left(\pi/2\alpha\theta_B\right)\right)^{1/\alpha}\left(v+\tan\left(\pi/2\alpha\theta_B\right) \right)$ and $y$ as in
Proposition \ref{lem:relation-g-h}. Then the
representation of $g^M_{\alpha,d}(v,\beta)$ using function
$h(\cdot\ ;\alpha,\beta)$ is as follows.
\begin{eqnarray*}
g^M_{\alpha,d}(v,\beta) &=&
\left\{
\begin{array}{ll}
\mathrm{Re}\
\frac{\left(\cos\left(\pi/2\alpha\theta_B
\right)\right)^{d/\alpha}}{2^d(\pi i)^{d-1}} h^{d-1}(|x|;\alpha,\beta_B^\ast),
& \alpha \neq 1,\\ &
\\
g_{1,d}(v,\beta)
, & \alpha=1,
\end{array}
\right.
\end{eqnarray*}
where $\beta_B^\ast=\beta\ \sign x$.
\end{proposition}
\begin{proposition}
\label{lem:relation-gB-h}
Let $g_{\alpha,d}^B(v,\beta)$ be as in Proposition
\ref{thm:A-B-representation}. Then the representation of $g^B_{\alpha,d}(v,\beta)$ using function
$h(\cdot\ ;\alpha,\beta)$ is as follows.
\begin{eqnarray*}
g^B_{\alpha,d}(v,\beta) &=&
\left\{
\begin{array}{ll}
\mathrm{Re}\
\frac{1}{2^d(\pi i)^{d-1}} h^{d-1}(|v|;\alpha,\beta^\ast),
& \alpha \neq 1,\\ &
\\
\mathrm{Re}\
\frac{1}{2^d (\pi i)^{d-1}} h^{d-1}(v^\ast;1,|\beta|)
, & \alpha =1,
\end{array}
\right.
\end{eqnarray*}
where $\beta^\ast=\beta\ \sign v$ and $v^\ast = v\
\sign \beta$.
\end{proposition}
\section{Some discussions and future works}
\label{sec:procedures-future-work}
In this paper we focused on one dimensional projections $g$ of the general
multivariate stable density. This is only one step in calculating the
density itself.
We need to substitute parameters which are functions on the unit
sphere $\mathbf{s}\in \mathbb{S}^d$ into $g$ and then
we need to integrate over the unit sphere
to evaluate the density itself.
We presented improvements in evaluation of $g$. Improvements
of other steps are also important for the evaluation of the
multivariate density.
By further careful examinations of the integrand of the
function $h^n(x;\alpha,\beta)$, we might find some useful
regularities like one dimensional finite integral representations
stated in Section 3 of \cite{Nolan:1997}.
For our future work we consider showing
various properties like tail dependencies or relations between the spectral
measure and tails. Theoretically the asymptotic estimates of
multivariate stable densities are obtained by \cite{Watanabe:2000} or
\cite{Hiraba:2003}. Relations between these
theoretical results and our representations or numerical results are of our next concern.
Furthermore, we can also consider the expansions or
the asymptotic expansions of functions $g$'s for the boundary values.
Many expansions of densities are found in \cite{Zolotarev:1986}. The method of
the expansions may be directly applied to the projected functions $g$'s.
\section{Proofs}
\label{sec:proofs}
In this section, we give the proofs of Proposition \ref{lem:relation-g-h},
Theorem \ref{thm:differential-density} and Theorem \ref{thm:M-M-representation}.
\subsection{Proof of Proposition \ref{lem:relation-g-h}}
For $\alpha \neq 1$, define
\begin{equation*}
Ig_{\alpha,d}(v,\beta)=\frac{1}{(2\pi)^d} \int_0^\infty \sin
\left(vu-\left(\beta \tan\frac{\pi\alpha}{2}\right) u^\alpha
\right) u^{d-1} e^{-u^\alpha}du.
\end{equation*}
Then by simple notational change we have
\begin{eqnarray*}
g_{\alpha,d}(v,\beta)+iIg_{\alpha,d}(v,\beta)
& =&\frac{1}{(2\pi)^d}\int_0^\infty
e^{iuv-i\left( \beta\tan\frac{\pi\alpha}{2}\right) u^\alpha}u^{d-1}e^{-u^\alpha}
du\\
& =& \frac{1}{2^d (\pi i)^{d-1}}\frac{1}{\pi}\int_0^\infty
(iu)^{d-1}e^{iuv}e^{-u^\alpha
\left\{1+i\tan\left(\frac{\pi}{2}\alpha \theta_B \right) \right \} } du\\
& =& \frac{1}{2^d (\pi i)^{d-1}}\frac{1}{\pi}\int_0^\infty
(iu)^{d-1}e^{iuv}
e^{-u^\alpha/\cos\left(\frac{\pi}{2}\alpha\theta_B\right) e^{i(\pi/2\alpha\theta_B)}}du.
\end{eqnarray*}
Replacing $u$ by
$u=\left(\cos\left(\frac{\pi}{2}\alpha\theta_B\right)\right)^{1/\alpha}t$, we have
\begin{eqnarray*}
&& \frac{1}{2^d(\pi i)^{d-1}}\left(\cos\left(\frac{\pi}{2}\alpha
\theta_B\right)\right)^{\alpha/d}\frac{1}{\pi}\int_0^\infty
(it)^{d-1}
e^{it \left( \cos\left(\frac{\pi}{2}\theta_B\right) \right)^{d/\alpha}v}
e^{-t^\alpha e^{i(\pi/2\alpha\theta_B)}}dt \\
&&\hspace{2mm} = \frac{1}{2^d(\pi i)^{d-1}}\left(\cos\left(\frac{\pi}{2}\alpha
\theta_B\right)\right)^{\alpha/d}\frac{1}{\pi} \int_0^\infty
(it)^{d-1} \exp (itx+\psi(t;\alpha,-\beta_B) ) dt \\
&&\hspace{2mm} = \frac{1}{2^d(\pi i)^{d-1}}\left(\cos\left(\frac{\pi}{2}\alpha
\theta_B\right)\right)^{\alpha/d} h^{d-1}(x;\alpha,\beta_B).
\end{eqnarray*}
Taking the real part of this equation, we obtain the desired result when
$x>0$. When $x<0$, replacing $v$ by $-v$ and $\beta$ by $-\beta$, we can
obtain the result in a similar fashion. Note that $g_{\alpha,d}(-v,-\beta)=g_{\alpha,d}(v,\beta)$.
For $\alpha=1$, define
$$
Ig_{1,d}(v,\beta)=\frac{1}{(2\pi)^d} \int_0^\infty
\sin\left(vu+\frac{2}{\pi}\beta u\log u\right)u^{d-1}e^{-u}du.
$$
What we need is to calculate
$
g_{1,d} (v,\beta)+iIg_{1,d}(v,\beta)
$
and to consider the real part of this equation. Since the proof is similar
to the proof of the case $\alpha\neq 1$,
we omit the rest of the proof for $\alpha = 1$.
\hfill $\Box$
\subsection{Proof of Theorem \ref{thm:differential-density}}
The function $h^0(x;\alpha,\beta)$, which can be regarded as the
inversion formula including the imaginary part, has the following
representation ((2.2.20) of \cite{Zolotarev:1986}):
\begin{equation*}
h^0(x;\alpha,\beta)=\frac{1}{\pi}\int_0^\infty \exp\left(izx+\psi(z;\alpha,-\beta)\right)dz.
\end{equation*}
Differentiating $h^0(x;\alpha,\beta)$ $n$ times with respect to $x$, we have
\begin{equation*}
h^n(x;\alpha,\beta)=\frac{1}{\pi}\int_0^\infty (iz)^n
\exp\left(izx+\psi(z;\alpha,-\beta)\right)dz.
\end{equation*}
For the calculation of $h^n(x;\alpha,\beta)$ we consider the same contour as in \cite{Zolotarev:1986}, which is
\begin{equation*}
\Gamma =\left\{z:\mathrm{Im} \left(izx+\psi(z;\alpha,-\beta)\right)=0,\quad
\frac{\pi}{2}k \le \arg z \le \frac{\pi}{2} \right\},
\end{equation*}
where
\begin{eqnarray*}
k &=&
\left\{
\begin{array}{ccc}
-\theta & \mbox{if}\quad \alpha \neq 1, & \\
-1 & \mbox{if}\quad \alpha =1, &\beta \neq -1, \\
1 & \mbox{if}\quad \alpha \le 1, &\beta = -1.
\end{array}
\right.
\end{eqnarray*}
Since readers can refer to
\cite{Zolotarev:1986} if necessary, the details of the contour are omitted.
{}From Lemma 2.2.3 of \cite{Zolotarev:1986}, we only have to calculate the integration along the contour $\Gamma$.
\medskip
\noindent
(a)\ $(\alpha \neq 1,\beta \neq 1)$ or $(\alpha>1, \beta=1)$. Direct
calculation gives
\begin{eqnarray}
h^n(x;\alpha,\beta)&=&\frac{1}{\pi}\mathrm{Re}\ \int_\Gamma
\exp\left(izx+\psi(z;\alpha,-\beta)\right)(iz)^n dz \nonumber \\
&&+\frac{1}{\pi}\mathrm{Im} \int_\Gamma
\exp\left(izx+\psi(z;\alpha,-\beta)\right)(iz)^n dz \nonumber \\
&=&\frac{1}{\pi}
\int_\Gamma \exp \left\{\mathrm{Re}\ (izx+\psi(z;\alpha,-\beta))\right\}\mathrm{Re}\
\{(iz)^n dz\} \nonumber \\
&&+\frac{1}{\pi}
\int_\Gamma \exp \left\{\mathrm{Re}\ (izx+\psi(z;\alpha,-\beta))\right\}\mathrm{Im}\
\{(iz)^n dz\} \nonumber \\
&=&
\frac{1}{\pi}\int_\Gamma \exp (-W(\varphi))\left(\mathrm{Re}\ \{(iz)^n dz \}+\mathrm{Im}\
\{(iz)^n dz\} \right), \label{eq:h-proof}
\end{eqnarray}
where
\begin{eqnarray*}
W(\varphi) &=&
\left\{
\begin{array}{ll}
x^{\alpha/(\alpha-1)} U_\alpha(2\varphi/\pi;\theta),
&\alpha \neq 1, \\
\exp(-x/\beta)U_1(2\varphi/\pi;\beta),
&\alpha =1.
\end{array}
\right.
\end{eqnarray*}
For details for derivation of $W(\varphi)$, see p.76 of \cite{Zolotarev:1986}.
Replacing $z$ by $z=re^{i\varphi}$ gives
\begin{eqnarray*}
(iz)^n&=&(re^{i(\varphi+\frac{\pi}{2})})^n \\
&=& r^n e^{in(\varphi+\frac{\pi}{2})} \\
&=& r^n (\cos n(\varphi+\frac{\pi}{2})+i\sin n(\varphi+\frac{\pi}{2}) ).
\end{eqnarray*}
Since
\begin{eqnarray*}
d(r\cdot \sin \varphi) &=& r'\sin\varphi \;d\varphi + r\cos\varphi \;d\varphi, \\
d(r\cdot \cos \varphi) &=& r'\cos\varphi \;d\varphi - r\sin\varphi \;d\varphi,
\end{eqnarray*}
$$
dz = r'\cos \varphi \;d\varphi
- r\sin\varphi \;d\varphi +i\{r'\sin \varphi+r\cos \varphi\} \;d\varphi.
$$
Combining the equations above, we have
\begin{eqnarray*}
\mathrm{Re}\ (iz)^n dz &=& r^n \left( r'\sin(n+1)\left(\varphi+\frac{\pi}{2}\right)+
r\cos(n+1)\left(\varphi+\frac{\pi}{2}\right)\right)d\varphi,
\\
\mathrm{Im}\ (iz)^n dz &=& r^n \left( r\sin(n+1)\left(\varphi+\frac{\pi}{2}\right)-
r'\cos(n+1)\left(\varphi+\frac{\pi}{2}\right)\right)d\varphi.
\end{eqnarray*}
{}From these equations and the relations of angles and the contour $\Gamma$
on p.76 of \cite{Zolotarev:1986},
($\ref{eq:h-proof}$) becomes
\begin{eqnarray*}
\mathrm{Re}\ h(x;\alpha,\beta) &=& \frac{1}{\pi}
\int_{-\frac{\pi}{2}\theta}^\frac{\pi}{2}
\exp\left(-x^{\alpha/(\alpha-1)}U_\alpha(2\varphi/\pi;\theta)\right)\\
&&\times r^n \left( r'\sin(n+1)\left(\varphi+\frac{\pi}{2}\right)+
r\cos(n+1)\left(\varphi+\frac{\pi}{2}\right)\right)
d\varphi \\
&=& \frac{1}{2}
\int_{-\theta}^1
\exp\left(-x^{\alpha/(\alpha-1)}U_\alpha(\varphi;\theta)\right)V_n(\varphi)
d\varphi
\end{eqnarray*}
and
\begin{eqnarray*}
\mathrm{Im}\ h(x;\alpha,\beta) &=& \frac{1}{\pi}
\int_{-\frac{\pi}{2}\theta}^\frac{\pi}{2}
\exp\left(-x^{\alpha/(\alpha-1)}U_\alpha(2\varphi/\pi;\theta)\right)\\
&& \times r^n \left( r\sin(n+1)\left(\varphi+\frac{\pi}{2}\right)-
r'\cos(n+1)\left(\varphi+\frac{\pi}{2}\right)\right)d\varphi \\
&=& \frac{1}{2}
\int_{-\theta}^1
\exp\left(-x^{\alpha/(\alpha-1)}U_\alpha(\varphi,\theta)\right)W_n(\varphi)
d\varphi.
\end{eqnarray*}
\medskip\noindent
(b)\ $(\alpha=1,\beta\neq 1)$. \
Since similar results hold in this case, we omit the
poof.
\medskip\noindent
(c)\ $(\alpha<1,\beta=1)$. \ Here we must consider additional contour
$\Gamma^\ast=\{z:\mathrm{Re}\ z=0,-\tau \le \mathrm{Im}\ z \le 0\}$ as
stated in p.76 of \cite{Zolotarev:1986}. The contour
$\Gamma^\ast$ satisfies
$$
\mathrm{Im}\ ((izx)+\psi(z;\alpha,-1))=0
$$
for $z \in \Gamma^\ast$. The integral along $\Gamma^\ast$ becomes
$$
\frac{1}{\pi} \int_{\Gamma^\ast} \exp(izx+\psi(z;\alpha,-1))(iz)^n dz =
\frac{1}{\pi} \int_{\Gamma^\ast} \exp(\mathrm{Re}\ (izx+\psi(z;\alpha,-1)))(iz)^n dz.
$$
Replacing $z$ by $z=r e^{-\frac{\pi}{2}i}$ gives
$$
\frac{1}{\pi}\int_0^\tau \exp\left(\mathrm{Re} (izx+\psi(z;\alpha,-1)) \right)
r^n (-i) dr
= -\frac{i}{\pi}\int_0^\tau \exp\left(-W\left(-\frac{\pi}{2}\right) \right)
r^n dr,
$$
where
\begin{eqnarray*}
W\left(-\frac{\pi}{2}\right) &=&xr \sin\left(-\frac{\pi}{2}\right)+r^\alpha
\cos \alpha \left(-\frac{\pi}{2}+\frac{\pi}{2}\theta \right) \\
&=& -xr+r^\alpha
\end{eqnarray*}
for $\alpha \neq 1$ and
\begin{eqnarray*}
W\left(-\frac{\pi}{2}\right) &=& xr \sin
\left(-\frac{\pi}{2}\right)+r\log r\sin
\left(-\frac{\pi}{2}\right)+\left(-\frac{\pi}{2}+\frac{\pi}{2}r\cos
\left(-\frac{\pi}{2}\right) \right) \\
&=& -xr-r\log r
\end{eqnarray*}
for $\alpha =1$. \hfill $\Box$
\subsection{Proof of Proposition \ref{thm:M-M-representation} }
Without loss of generality, we assume $\nu=0$ throughout the proof.
The characteristic function $\phi_{\alpha,d}^M(t)$ can be written
as
\begin{eqnarray*}
\Phi_{\alpha,d}^M(\mathbf{t})&=&\exp\Bigg( -\int_{\mathbb{S}^d}|\langle \mathbf{t}, \mathbf{s}
\rangle|^\alpha \Gamma (d \mathbf{s}) +
i \tan\frac{\pi\alpha}{2} \int_{\mathbb{S}^d} \sign
\langle \mathbf{t}, \mathbf{s} \rangle\ |\langle \mathbf{t}, \mathbf{s}
\rangle|^\alpha\ \Gamma (d \mathbf{s}) \\
&&\hspace{1cm}
-i\tan\frac{\pi\alpha}{2} \int_{\mathbb{S}^d} \sign
\langle \mathbf{t}, \mathbf{s} \rangle\ | \langle \mathbf{t}, \mathbf{s}
\rangle |\ \Gamma (d \mathbf{s})
\Bigg)\\
&=&
\exp\left(-(\sigma(\mathbf{t}))^\alpha+i(\sigma(\mathbf{t}))^\alpha\tan
\frac{\pi\alpha}{2}\beta(\mathbf{t})
-i\tan\frac{\pi\alpha}{2} \int_{\mathbb{S}^d}\langle \mathbf{t}, \mathbf{s}
\rangle \ \Gamma (d \mathbf{s}) \right)\\
&=&
\exp\left(-(\sigma(\mathbf{t}))^\alpha-i(\sigma(\mathbf{t}))^\alpha\tan
\frac{\pi\alpha}{2}\beta(\mathbf{t})\left((\sigma(\mathbf{t}))^{1-\alpha}-1 \right)+i\mu_M^M(\mathbf{t}) \right) \\
&=&\exp\left( -\psi^M\left(\mathbf{t}
\right) \right).
\end{eqnarray*}
Then, the inversion formula gives
\begin{eqnarray*}
f_{\alpha,d}^M(\mathbf{x})&=&
(2\pi)^{-d}
\int_{\mathbb{R}^d }e^{-i\langle \mathbf{x},\mathbf{t} \rangle
}\exp\left( -\psi^M\left(\mathbf{t}
\right)\right) d \mathbf{t}.
\end{eqnarray*}
Note that for a positive real number $r>0$ and a vector $\mathbf{s}\in
\mathbb{R}^d$, $\mu_M^M(r\mathbf{s})=r\mu_M^M(\mathbf{s})$,
$\sigma(r\mathbf{s})=r\sigma(\mathbf{s})$ and $\beta(r\mathbf{s})=\beta(\mathbf{s})$.
Putting $\mathbf{t}=r\mathbf{s}$ where $r>0$ and $\mathbf{s} \in
\mathbb{S}^d$, we obtain
\begin{eqnarray*}
f_{\alpha,d}^M(\mathbf{x})
&=&(2\pi)^{-d}\int_{\mathbb{S}^d}\int_0^\infty
\exp\Big[-i\langle \mathbf{x},\mathbf{s}\rangle r-(r\sigma(\mathbf{s}))^\alpha-i(r\sigma(\mathbf{s}))^\alpha
\tan\frac{\pi\alpha}{2}\beta(\mathbf{s})\left( (r\sigma(\mathbf{s}))^{1-\alpha}-1 \right)\\
&& \hspace{3.8cm}
+ir\mu_M^M(\mathbf{s})\Big]
r^{d-1} dr d \mathbf{s}.
\end{eqnarray*}
Furthermore, replacing $r$ by $\mu=r \sigma(\mathbf{s})$, we get
\begin{eqnarray*}
f_{\alpha,d}^M(\mathbf{x})&=&(2\pi)^{-d}\int_{\mathbb{S}^d}\int_0^\infty
\exp\bigg[ -i\frac{\langle \mathbf{x},\mathbf{s}\rangle
-\mu_M^M(\mathbf{s})}{\sigma(\mathbf{s})} u
-\left\{u^\alpha+iu^\alpha
\tan\frac{\pi\alpha}{2}\beta(\mathbf{s})\left( u^{1-\alpha}-1 \right)
\right\}\bigg]\\
&& \hspace{4cm}
u^{d-1} (\sigma(\mathbf{s}))^{-d} du d \mathbf{s}.
\end{eqnarray*}
Taking the real part of the integrand, we obtain
\begin{equation*}
f_{\alpha,d}^M(\mathbf{x})=\int_{\mathbb{S}^d}
g_{\alpha,d}^M \left( \frac{ \langle \mathbf{x},\mathbf{s} \rangle
-\mu_M^M(\mathbf{s})}{\sigma(\mathbf{s})},\beta(\mathbf{s})\right)
(\sigma(\mathbf{s}))^{-d}\ d \mathbf{s}.
\end{equation*}
Note that the same conclusion holds if we let $\alpha \to 1$ in the proof due to continuity at $\alpha=1$. The direct but tedious calculation
gives the representation $\mu_M^M(\mathbf{t})$ at $\alpha=1$.
\hfill $\Box$
\bigskip\noindent
{\bf Acknowledgment:} \ The authors are grateful to Prof.\ Makoto
Maejima for very helpful comments.
\bibliographystyle{plainnat}
\bibliography{references}
\end{document}
| 114,042
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This reviews the effects of butyrate on modulating miRNA expression in colorectal cancer. - GreenMedInfo Summary
Epigenetic Regulation of Gene Expression Induced by Butyrate in Colorectal Cancer: Involvement of MicroRNA.
Karen S Bishop
Colorectal cancer (CRC) is the third most common cause of cancer mortality globally. Development of CRC is closely associated with lifestyle, and diet may modulate risk. A Western-style diet is characterised by a high intake of red meat but low consumption of fruit, vegetables, and whole cereals. Such a diet is associated with CRC risks. It has been demonstrated that butyrate, produced by the fermentation of dietary plant fibre, can alter both genetic and epigenetic expressions. MicroRNAs (miRNAs) are small non-coding RNAs that are commonly present in both normal and tumour cells. Aberrant miRNA expression is associated with CRC initiation, progression, and metastasis. In addition, butyrate can modulate cell proliferation, differentiation, apoptosis, and miRNA expression in CRC. In this review, the effects of butyrate on modulating miRNA expression in CRC will be discussed. Furthermore, evidence on the effect of butyrate on CRC risk through reducing oncogenic miRNA expression will be presented.
| 298,729
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On Monday, August 7, 2006 at 11:36 PM EST WWE RAW
August 7, 2006
Memphis, TN
Commentators: Jim Ross & Jerry "The King" Lawler
Reported By: Hunter Golden of WrestleView.com
Vince and Shane were in the back dressed in Elvis outfits. They were recounting the events of last week when Triple H and HBK were bested by the McMahons and Umaga.
The opening credits rolled and here we go.
JR and King put over the rest of tonight’s card.
WWE Women’s Championship
Trish Stratus v. Mickie James
The two locked up in the middle of the ring before Edge and Lita interrupted the proceedings. Edge said he was sick of taking backseats to people because he was WWE Champion. He declared the match over and said he was taking over Raw. He wanted Trish to stay out here. He wanted the world to hear that he was a 2 time WWE Champion. He was sick of getting no respect around here.
He said it wasn’t fair that he had to defend the title against John Cena at Summerslam. He didn’t think it was fair that he had to face him in Boston as well. He said no champion in the history of the company had ever had to put up with as much crap as he does. He showed the Summerslam poster up on the screen and pointed out that DX and Cena were front and center while he was nowhere to be found.
He pointed out Batista being on there before he gets injured again and he said he wasn’t too ticked off about it all. He said he thought for sure he’d be on the cover of WWE magazine coming out tomorrow but believe it or not, he wasn’t. He said he was passed over by someone who hadn’t won a match in a year.
Lita got in Trish’s face and they brawled around the ring before Lita nailed her with a spear. Edge went to land a spear of his own before Carlito came to Trish’s aid. Carlito briefly gained the upper hand before getting speared. Edge and Lita left up the ramp, while Trish and Carlito laid on the mat in pain.
****Commercial****
Intercontinental Champion Johnny Nitro and Melina came out to the announce booth for the next match.
#1 Contender’s Match for the IC Championship
Shelton Benjamin v. Kane
Kane made his return and the look on Shelton’s face was hysterical. As Kane went to pop the corner, Shelton came off the ropes at Kane. Shelton pounded the big man to the corner but found himself whipped to the opposite buckles. Kane caught him for a power slam, but Benjamin squeaked out of the hold and almost got a roll up.
Kane took the offensive who whipped Benjamin repeatedly to the buckles, following up with a clothesline and sidewalk slam respectively. Kane came flying off of the top rope and nailed his flying clothesline before going for his choke slam. Shelton countered out however into a beautiful looking DDT and almost got the pin fall.
Shelton continued to work Kane over and went for his T-bone Suplex, but Kane bullied him to the corner. Kane rushed Benjamin but missed him. Shelton leap-frogged him and came sailing off the top rope only to be caught in the choke slam.
Winner and new #1 Contender: Kane
A promo video aired putting over Summerslam.
****Commercial****
John Cena came out to the ring. He said he needed to be honest with people, Edge, he just didn’t like. He made fun of Edge for being whiney and then said WWE did some market analysis and found out that Edge’s face scares little kids. He said Lita got a job with Seven Eleven “Slurpies for herpes’. He said shame on Edge for whining about not being on the cover of WWE magazine because he was on the cover of “Crying Bitch Illustrated”. He said he didn’t care if Summerslam was in his hometown or not. He said instead of making demands, he said Edge should come out here and try to earn it.
The Coach’s music hit and he came out to the ramp. He said Edge would not be coming out to answer his challenge tonight because he was already in a match. He said Carlito and Trish would be taking on Edge and Lita later tonight. He said it was obvious that Cena was hungry for action. He said he personally found an opponent for him who was always hungry, Viscera.
****Commercial****
John Cena v. Viscera
Cena had yet to gain an advantage in the match before dodging a sit out. He came at Viscera, but got nailed with a Samoan drop. Viscera gathered Cena and whipped him to the buckles. He came rushing in with an avalanche and completely annihilated the former champion. Cena rolled out to the floor to avoid the cover attempt, but big Vis collected him and tossed him back into the ring.
Cena ducked under a clothesline attempt but was caught and planted into the mat with a front slam. Viscera climbed on the back of Cena and laid in with a modified camel clutch. Cena fought to his feet and booted Viscera in the gut. He swatted away at the big man with right hands and some knees to the gut. Cena came flipping over Viscera with a snap mare before calling for the five knuckle shuffle. He came off the ropes but got planted with a black hole-slam like move.
Viscera went for his big splash but Cena rolled out of the way. Cena quickly capitalized and FU'd Viscera for the win.
Winner: John Cena via pin fall
****Commercial****
Vince was in the back giving the cops a little pep talk. They put over how Triple H and HBK had been looking for a fight all day. Shane went to the bathroom and Vince told him to be careful.
Trish was in the hallway with Carlito ranting about how much she hated Lita. Carlito tried to calm her down then she kissed him and walked off.
Shane was looking for HBK and ran into the Highlanders, Candice Michelle, and Torrie. HBK was able to jump Shane from behind and knocked him into the McMahon locker room and into a sea of cops. The police pinned him down as we went to commercial.
****Commercial****
HBK was apologizing to the police officers outside when Triple H came in, asking what was going on. The Police said they didn’t care whether he hit the officer on purpose or not and took him off despite the protests of Triple H.
Carlito & Trish v. Edge & Lita
Carlito caught Edge coming in with a back elbow and then planted him to the mat with a clothesline. Carlito bull-rushed the champion into the corner and began swatting away with right hands before hitting him with a flat-liner.
Edge kicked out of the pin fall attempt and went on the offensive himself. He hit a clothesline and a solid looking body slam and barely missed a cover attempt of his own. Carlito began to fight back with a boot to the gut, back elbow and clothesline combination. Edge tagged in Lita and Carlito tagged in Trish.
Trish went right at Lita hammering away at Lita with some punches and some vicious kicks in the corner. Trish missed a follow up on Lita and was shoved off the buckles by Edge. Lita capitalized, cinching in a hard rear chin lock and eventually it evolved into a sleeper.
Trish tried to elbow out of the hold, but Lita tossed her to the mat by her hair. Lita continued her assault, choking Trish out on the ropes before planting some knees into the former Women’s Champion’s back. Lita went to work on Trish’s back some more before going for some sort of leg lock that got her tossed aside. Carlito came running in and he and his partner his a double drop kick, sending the WWE Champion and his ‘lady-friend’ to the floor.
****Commercial****
Edge had a tight arm bar cinched in on Carlito. Carlito eventually fought out of the hold but was rushed to the corner in time to come flipping off it and barely missing a back cracker. Edge baseball-slide drop kicked Carlito out to the floor, allowing Lita to get a few licks in. Edge jammed Carlito’s back into the ring apron before cinching in a butterfly clutch. Carlito broke the hold with a jaw breaker and a springboard back elbow. Both men were left laying on the mat before Trish and Lita were both tagged in.
Trish hit two clotheslines and a drop kick. She continued to hit some chops before it was broken up but a kick to the gut. Lita went for the twist of fate, but Trish kicked out and went for a move herself, but Edge blocked it. Trish went for a stratusphere and connected. Edge went for a spear on Carlito, but Carlito ducked out of the way in the knick of time only to have Trish get mauled with the move. Carlito chased Edge out of the ring while Lita sneaked over to make the cover for the win.
Winners: Edge and Lita via pin fall
****Commercial****
We learned that Hulk Hogan had a torn meniscus in his right knee, which puts his Summerslam match with Randy Orton in jeopardy. Gresham asked Randy Orton what his feelings were on the situation. Orton said Hogan was trying to get out of the match and that Hogan really did know best, because if he showed up, his legend would get killed.
Ric Flair came out to the ring holding a book. He said for those of us who didn’t know it, he said Memphis got a plug they deserved from John Cena. Flair put over his being born in the city. He said he had spent ten years trying to figure out how to make Mick Foley stand up and be a man and what makes a guy like him ‘tick’.
He said he didn’t wanna be like Mick Foley, dress like him, etc. He wondered how to get him to Summerslam. He said someone came up to him and told him to read his book “Foley is Good”. Flair held up the book, dropped it to the mat, stomped on it twice and gave it two elbow drops. He ripped it up and tossed it into the crowd. He said one page was worth it though. He said deep inside Foley was a fan of Flair’s. He said Foley idolizes “the Naitch”. On page 169, his #1 favorite wrestling match was Funk v. Flair in an “I Quit” match. He said it wasn’t a stunt fest, it was a real “I Quit” match. He said there were one of two reasons he wouldn’t wrestle against Flair 1.) he didn’t know if he could beat him or 2.) he KNEW he couldn’t beat him.
Foley’s music hit and he came out to the ramp. Foley said Flair was absolutely right. He said he idolized Ric Flair, that he was a Nature Boy fan. He said Flair was a better wrestler than he ever was. He said the two hate each other and it was no secret. But he said they also knew that if Foley-Flair is in the ring at Summerslam, the match would be a match for the ages, violent, wild and intense. He said it would put the name Ric Flair right back up on top of the mountain where it belongs. He said, however, he just couldn’t allow that to happen. He said 12 years ago, Flair was in charge of his career and he could have made him, but he stepped on Foley’s dreams by not jobbing to him.
He said how fitting was it that Foley was sitting between Flair and his dreams was Foley. He said Flair didn’t do a damn thing to help him out in 1994 and he would return the favor for him this year. He said unless Flair came back to him with something more than just a match, everyone could kiss his ass.
Flair asked Foley is he could be a man just one day of his life. He asked him if he was going to just walk away from all of his accolades like nothing. He said he was giving him a chance to rewrite history if that greatest hardcore match ever involved him. He said he should wrestle him at Summerslam. Foley said he would not go back on his vow, when Ric was ruining his life. He said for embarrassing him on Raw last week, how about he beats him up? How about he takes Flair on in an “I Quit” match at Summerslam. Flair ‘woo’d’ into the mic and smiled as Foley’s music played.
Foley told him that he wanted him to understand one thing. He said this match will be Flair’s greatest match of them all and he guaranteed it would be Flair’s last. Flair toasted Foley saying ‘may one of us bleed to death in Boston’.
****Commercial****
Randy Orton v. Jerry “The King” Lawler
Orton slapped King in the face before hitting him with a back body drop. Lawler fought back with a slap of his own before Orton raked his eyes and nailed him with a drop kick. Orton clamped in a rear chin lock. Lawler fought out of the hold and hit a scoop slam.
Lawler went for his fist drop but no one was home. Orton went for the RKO but Lawler tossed him aside. Orton hit a low blow, stared King down, and RKO’d him for the win.
Winner: Randy Orton via pin fall
****Commercial****
A video putting over the re-arrival of Jeff Hardy aired.
2006 Diva Search
Jen, Layla, Erica, Marlena, and JT were left standing. Miz pointed out that one of the ladies would be eliminated tonight. Miz did his usual sleazy ‘say the ladies name behind their backs’ before stopping at Erica who was eliminated from the competition. Miz also pointed out that the night was not over and they’d be participating in “Diss the Diva” shortly.
JR ran down the Summerslam card:
WWE Championship—If Edge loses via DQ, he loses the title
Edge v. John Cena
“I Quit Match”
Ric Flair v. Mick Foley
Hulk Hogan v. Randy Orton
D-Generation X v. The McMahons
Now it’s time for “Diss the Diva”. They would have 20 seconds to tell the other why they were better than the other contestants. Jenn talked about being a kitty, and Leyla played the hyper active one, Marlena said something that didn’t make sense, and JT was clearly lost in the chorus of boos reigning in from the crowd. Whatever.
A video putting over the build up to Triple H-Umaga aired.
****Commercial****
Triple H v. Umaga
Umaga came right at the Game, who ducked under and landed a few right hands. Triple H came back at Umaga and was planted with a headbutt. Umaga hammered away at the Game in the corner, before missing a follow up which Triple H was eventually able to work the big Samoan up and over the top rope.
Triple H hit a running clothesline on Umaga, but it didn’t phase the big man who nailed the Game with a spinning wheel kick. Umaga sat out hard on Triple H’s chest before dumping him into the corner upside down and planting him with his flying head butt. Triple H slumped into the corner and was met with the running bump-butt from Umaga.
Umaga went for his flying head butt but no one was home. Triple H fought to his feet and began throwing punches in Umaga’s direction. He hit a face buster on the big man and finally floored him with a clothesline. Shane jumped into the ring and Vince distracted the ref as well. Triple H fought them off and almost pedigreed Umaga, but was back body dropped. Umaga came charging at the Game but got caught in a huge spine buster.
Shane came running in to break up the fun but found himself spine-bustered as well. Triple H went for the cover but Vince pulled the referee out of the ring. Vince jumped in but was met by a HHH kick to the gut. Just as Triple H was going to hit the move, Umaga nailed him with a Samoan spike. The referee was tossed back into the ring and made the three count.
Winner: Umaga via pin fall
After the match, Vince thanked Umaga and loaded up Triple H for a pedigree and nailed the game with his own finishing move. He did some more impersonations of Triple H before leaving the ring triumphantly. RAW THOUGHTSThe Good-- First off, Flair-Foley is feud of the year. Another incredible feud, with perfect reasoning for an “I Quit” match, two guys who’ve revolutionized that style of match. Wow. Great reasoning for the match and the best build of any feud in the WWE in a long, long, long, long time.
I’m actually getting back into the DX-McMahons thing and who would’ve thought all it would take would be two weeks of the McMahons getting the better of DX two weeks in a row. It’s still not totally my cup of Tea, but at least I can deal with it.
Was it me or was the Cena-Viscera match pretty good tonight? The FU was cool, Cena was for the first time in a long time (since the Cabana promo last year with HBK drinking the margarita) Cena was legitimately funny. Hell, even Viscera looked really ‘on’ for Viscera.
Orton-Lawler was kept short as it should have been. Shame about Hogan’s injury. Orton can’t catch a break.
The Carlito-Trish thing is interesting and they killed two birds with one stone there building tension between the faces and restoring a bit of credibility to a WWE Champion who could really use some right now. Just a shame they couldn’t work Cena getting his ass handed to him somewhere in there.
The Bad-- I can’t say there was anything ‘bad’ tonight. I’m as surprised as anyone.
The Ugly-- The “Diva Diss” thing was tearfully bad. Not only did the girls not stick to the program, but two of them, flat out, couldn’t handle a mic for the life of them. A third didn’t know what to do when the boos rang down. Jenn was the only semi-coherent one and should probably win the thing, but Leyla is waaaay more irritating ala Christy Hemme in 04 and will probably win the thing. No more pixels wasted on the Diva search from here on out.
Overall B+– This was an incredibly solid Raw. Minus the Diva search segment, there wasn’t a single bad segment done all night. Everything had purpose and was marginally entertaining. While we certainly didn’t get any match of the year candidates, the ring work was the best all around effort Raw certainly this year and perhaps in the last two or three. Again, no great matches by any means, but even every match had something redeeming about it. If WWE could do this every week and put on two or three passable matches with one or two well done segments, they wouldn’t get half the criticism they get these days.
Quick Results
Mickie v. Trish Stratus ended in a No Contest
Kane def. Shelton Benjamin
John Cena def. Viscera
Edge & Lita def. Carlito & Trish Stratus
Randy Orton def. Jerry Lawler
Umaga def. Triple H
Who’s Hot, Who’s Not?
Biggest pops
1. Ric Flair
2. John Cena
3. Triple H
4. Jerry Lawler
Most Heat
1. Vince & Shane McMahon
2. Mick Foley
3. Randy Orton
4. Umaga
Match of the night: John Cena v. Viscera **
Power RankingsWWE Champion: Edge (Last Week: WWE Champion)—Every word he said in his promo tonight was the truth. That’s sad. He looks strong at least for one week. He’ll go back to being Cena’s punching bag next week.
Intercontinental Champion: Johnny Nitro (Last Week: Intercontinental Champion)—Very interesting Summerslam opponent for Nitro, if Kane is in fact the guy who gets the shot. They’re doing a great job of making Nitro’s title seem important again. Interesting first big title defense for the young champion.
1.Umaga (Last Week: 2)—That’s right. I put him here. He’s defeated John Cena, Shawn Michaels, and Triple H all in one month. Dirty or not, he’s got the wins to back the record. He’s defeated almost every man on the roster. Clearly a #1 contender despite being lightyears away from being a champion, let alone a believable main event act.
2. John Cena (Last Week: 2)—Cena was great this week but Umaga has been on more of a roll and has the win over Cena to break the tie. Cena was just as good on the stick as he was in the ring. Seems to be in coast mode until they give him his title back in two weeks.
3. Randy Orton (Last Week: 3)—It flat out sucks to be Randy Orton. You get pushed to the moon too early, turned face and squashed because you couldn’t carry a load you weren’t ready to carry yet. Then you finally rebuild your career and get into contention to be a top act again and get fried with a suspension that was probobly earned. Then you get the program of a lifetime upon return and the guy’s knee gives out. Ugh.
4. Ric Flair (Last Week: 4)—Ric Flair is the man. I’m looking forward to the Foley match more than any Flair match in the last ten years. Period.
5. Triple H (Last Week: 6)—Gets completely squashed by Umaga tonight. The DX thing just puts a guy who doesn’t work babyface very well anyways in an even weirder place.
6. Shawn Michaels (Last Week: 5)—Barely showed tonight so he stays put.
7. Carlito (Last Week: 8)—Despite coming up on the short end of two segments, Carlito is not only getting a big pop as a babyface, he’s looking like the real deal in the ring as well and is beginning to learn how to use his every expanding move set well.
8. Kane (Last Week: NR)—Kane comes out of nowhere and is now the #1 contender for the IC title, which I think is kind of interesting. I think this feud with Nitro could be interesting if they’re willing to give it a go, but it’d have to move past Summerslam.
9.Mick Foley (Last Week: 9)—Foley doesn’t wrestle and therefore doesn’t get moved up. Unreal promo yet again though.
10. Shelton Benjamin (Last Week: 7)—Squash City for Benji, but he’s showing genuine improvement in the ring these days.
Dropping Out: Matt Striker (Last Week: 10)
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Hotels, flights, car hire & activities - all in one place! Bundle everything you need for your holiday to Gibraltar and save
Reviewed on 15 Nov 2021
Reviewed on 9 Nov 2021
Reviewed on 8 Nov 2021
Discover the southernmost tip of continental Europe at this scenic peninsula, featuring the landmark Rock, breathtaking views and a blend of Spanish and British culture. metres). kilometres),.
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\section{Proof of Lemma~\ref{lemma:swide}}
In this section we prove the key bias reduction lemma that was the core of Theorem~\ref{thm:main}. Our proof will be by induction, just like Claim~\ref{clm:genrw}, so we will need to modify the statement of Lemma~\ref{lemma:swide} so it adheres to an inductive argument.
\subsection{Lemma Statement}
Let $A$ and $B$ be the graphs from Section~\ref{sec:mainthm}. Write $\lambda$ instead of $\lambda_B$ for the expansion of $B$ and recall that $\lambda_A\leq\lambda^2$. Let $f:A\rightarrow\{0,1\}$ be a function such that $\big|\E_a\bigl[(-1)^{f(a)}\bigr]\big|\leq\lambda$. For any $k\geq0$, define $g_k:A\times B\rightarrow\mathbb{R}$ by
\begin{equation}\label{eq:gk}
g_k(a,b)=\E_{(a_0,\dots,a_k)\sim \sw^k(a,b)}\Bigl[(-1)^{f(a_0)\oplus\cdots\oplus f(a_k)}\Bigr].
\end{equation}
Let $\ep_k=\big|\E_{a,b}\bigl[g_k(a,b)\bigr]\big|$ and let $\sigma_k$ be such that $\sigma_k^2+\ep_k^2=\E_{a,b}\bigl[g_k(a,b)^2\bigr]$. We prove the following.
\begin{lemma}[{\bf Implies Lemma~\ref{lemma:swide}}]
\label{lemma:swideinduciton}
Assume the above setup. For all $k\geq0$ \[\ep_k\leq(2\lambda)^{k(1-4/s)};\text{ }\sigma_k\leq (2\lambda)^{(k-2)(1-4/s)}.\]
\end{lemma}
\noindent As mentioned, we prove Lemma~\ref{lemma:swideinduciton} by induction. The following two claims combine to easily prove Lemma~\ref{lemma:swideinduciton}; we will prove them in Sections~\ref{sec:epk} and~\ref{sec:sigmak}.
\begin{clm}[{\bf Base Case.}]
\label{clm:swidebasecase}
Assume the above setup. For all $k=0,1,\dots,s$:
\[\ep_k\leq\frac{1}{2}\cdot(2\lambda)^{k+1
};\text{ }\sigma_k\leq2\cdot(2\lambda)^{k-1}.\]
\end{clm}
\begin{clm}[{\bf Induction Step.}]
\label{clm:swideinduction}
Assume the above setup. For all $k>s$:
\begin{center}\begin{minipage}{.9\linewidth}\begin{itemize}
\item[$\cdot$ $\ep_k\leq$] $\frac{1}{2}(2\lambda)^s(\ep_{k-s}+3\sigma_{k-s})$;
\item[$\cdot$ $\sigma_k^2\leq$]
$\frac{1}{2}(2\lambda)^{s-2}(\ep_{k-2}+\lambda\sigma_{k-1})\bigl(\ep_{k-s}+(2+\lambda)\sigma_{k-s}\bigr)+\lambda^s\sigma_{k-s}\sigma_{k-1}+\lambda^2\sigma_{k-1}^2$
\end{itemize}\end{minipage}\end{center}
\end{clm}
\begin{proof}[Proof of Lemma~\ref{lemma:swideinduciton}] Claim~\ref{clm:swidebasecase} clearly establishes the base cases since $\frac{1}{2}\cdot(2\lambda)^{k+1}\leq(2\lambda)^{k(1-4/s)}$ and $2\cdot(2\lambda)^{k-1}\leq(2\lambda)^{(k-2)(1-4/s)}$. For the first part of the induction step, we have \begin{eqnarray*}\ep_k &\leq& \frac{1}{2}\cdot(2\lambda)^s\cdot(\ep_{k-s}+3\sigma_{k-s})\leq\frac{1}{2}\cdot(2\lambda)^s\cdot\Bigl[(2\lambda)^{(k-s)(1-4/s)}+3\cdot(2\lambda)^{(k-s-2)(1-4/s)}\Bigr]\\ &=& 8\lambda^4\cdot\Bigl[(2\lambda)^{k(1-4/s)}+3\cdot(2\lambda)^{(k-2)(1-4/s)}\Bigr]\leq2\lambda^2(4\lambda^2+3)\cdot(2\lambda)^{k(1-4/s)}\leq(2\lambda)^{k(1-4/s)}.\end{eqnarray*} The bound $2\lambda^2(4\lambda^2+3)\leq1$ holds because $\lambda\leq1/3$. The second part of the induction step is similar:
\begin{eqnarray*}
\sigma_k^2
&\leq&
\frac{1}{2}\cdot(2\lambda)^{s-2}\cdot(\ep_{k-2}+\lambda\sigma_{k-1})\bigl(\ep_{k-s}+(2+\lambda)\sigma_{k-s}\bigr)+\lambda^s\sigma_{k-s}\sigma_{k-1}+\lambda^2\sigma_{k-1}^2\\
&\leq&
\frac{1}{2}\cdot(2\lambda)^2\cdot\Bigl[(2\lambda)^{(k-2)(1-4/s)}+\lambda(2\lambda)^{(k-3)(1-4/s)}\Bigr]\cdot\Bigl[(2\lambda)^{k(1-4/s)}+(2+\lambda)(2\lambda)^{(k-2)(1-4/s)}\Bigr]+\\
&+&
\lambda^s(2\lambda)^{(k-s-2)(1-4/s)}(2\lambda)^{(k-3)(1-4/s)}+\lambda^2(2\lambda)^{2(k-3)(1-4/s)}\\
&=&
2\lambda^2(2\lambda)^{(2k-2)(1-4/s)}+2\lambda^3(2\lambda)^{(2k-3)(1-4/s)}+(4\lambda^2+2\lambda^3)(2\lambda)^{(2k-4)(1-4/s)}+\\
&+&
(4\lambda^3+2\lambda^4)(2\lambda)^{(2k-5)(1-4/s)}+2^{4-s}\lambda^4(2\lambda)^{(2k-5)(1-4/s)}+\lambda^2(2\lambda)^{(2k-6)(1-4/s)}\\
&\leq&
\biggl[2\lambda^2+2\lambda^3+(4\lambda^2+2\lambda^3)+(2\lambda^2+\lambda^3)+2^{3-s}\lambda^3+\frac{1}{4}\biggr]\cdot(2\lambda)^{(2k-4)(1-4/s)}\leq(2\lambda)^{(2k-4)(1-4/s)},\end{eqnarray*}
where the last bound has used $8\lambda^2+6\lambda^3\leq3/4$ which holds because $\lambda\leq1/4$.
\end{proof}
\subsection{Key Intuition}
\label{sec:intuition}
In this section we zoom in on some of the key steps in the coming proofs in order to give extra explanations and intuitions.
\paragraph{$s-$wide Replacement Product Walks in $A$.} Recall that a random $s-$wide replacement product walk in $A$ (\emph{i.e.}, a random sample from $\sw^k$) is produced as follows:
\begin{enumerate}
\item choose base points $(a,b)\sim A\times B$;
\item generate $(b_1,\dots,b_k)\in B^k$ as follows:
\begin{itemize}
\item[$(i)$] set $b_1=b$;
\item[$(ii)$] for $i\geq2$, draw $b_i\sim N(b_{i-1})$ and set $b_i={\sf shift}(b_i)$, where ${\sf shift}$ cycles the coordinates of an element of $B\simeq[d]^s$, so ${\sf shift}\bigl(b_i[1],\dots,b_i[s]\bigr)=\bigl(b_i[2],\dots,b_i[s],b_i[1]\bigr)$.
\end{itemize}
\item generate and output $(a_0,\dots,a_k)\in A^{k+1}$ as follows:
\begin{itemize}
\item[$(i)$] set $a_0=a$;
\item[$(ii)$] for $i\geq1$, set $a_i=\phi(a_{i-1},\hat b_i)$ where $\hat b_i=b_i[1]\in[d]$ denotes the first coordinate of $b_i\in[d]^s$, and where $\phi$ is the rotation map of $A$.
\end{itemize}
\end{enumerate}
\paragraph{Pseudorandomness.} As mentioned in Section~\ref{sec:prelims}, when $k\leq s$ the distributions $\sw^k$ and $\rw_A^{k+1}$ are identical. That is, a random $k-$step $s-$wide replacement product walk in $A$ is just a random $(k+1)-$step random walk in $A$. The following is an example of how this concept manifests itself in the next section. Let $\ep_k(a)=\E_b\bigl[g_k(a,b)\bigr]$. \[\ep_k(a)=\E_{(a_0,\dots,a_k)\sim\sw^k(a)}\Bigl[(-1)^{f(a_0)\oplus\cdots\oplus f(a_k)}\Bigr]=\E_{(a_0,\dots,a_k)\sim\rw^{k+1}_A}\Bigl[(-1)^{f(a_0)\oplus\cdots\oplus f(a_k)}\Bigr]=h_{k+1}(a),\] whenever $k\leq s$, where $h_{k+1}$ is the function defined and analyzed in Claim~\ref{clm:exprw}.
\paragraph{The Ignore First Step Trick.} This refers to a key step in the proof that for all $k\geq1$, \begin{equation}\label{eq:firststeptrick}\sigma_k^2\leq\E_a\bigl[\ep_{k-1}(a)^2\bigr]+\lambda^2\sigma_{k-1}^2.\end{equation} This bound is useful as it reduces the task of bounding $\sigma_k^2$ to the task of bounding $\E_a\bigl[\ep_{k-1}(a)^2\bigr]$, which will turn out to be much easier. The proof of (\ref{eq:firststeptrick}) requires other ideas as well. Recall from the previous paragraph the definition of $\ep_k(a)$; additionally let $\sigma_k(a)$ be such that $\sigma_k(a)^2+\ep_k(a)^2=\E_b\bigl[g_k(a,b)^2\bigr]$. \begin{eqnarray*} \sigma_k^2 &\leq& \sigma_k^2+\ep_k^2=\E_{a,b}\bigl[g_k(a,b)^2\bigr]=\E_{a,b}\Bigl[\E_{b'\sim N(b)}\bigl[g_{k-1}(a',b')\bigr]^2\Bigr]=\E_{\stackalign{&a\sim A\\b&\sim_{B^2}b'}}\bigl[g_{k-1}(a,b)\cdot g_{k-1}(a,b')\bigr]\\ &\leq&\E_a\bigl[\ep_{k-1}(a)^2\bigr]+\lambda^2\E_a\bigl[\sigma_{k-1}(a)^2\bigr]\leq\E_a\bigl[\ep_{k-1}(a)^2\bigr]+\lambda^2\sigma_{k-1}^2.\end{eqnarray*} The second equation on the first line holds because $g_k(a,b)=(-1)^{f(a)}\cdot\E_{b'\sim N(b)}\bigl[g_{k-1}(a',b')\bigr]$, where $a'=\phi(a,\hat b)$; the first inequality on the second line follows from the expander mixing lemma (Definition~\ref{def:expander}) on $B^2$ (a $\lambda^2-$expander); the final inequality has used $\E_a\bigl[\sigma_{k-1}(a)^2\bigr]\leq\sigma_{k-1}^2$ which holds because \[\E_a\bigl[\sigma_{k-1}(a)^2+\ep_{k-1}(a)^2\bigr]=\E_{a,b}\bigl[g_{k-1}(a,b)^2\bigr]=\sigma_{k-1}^2+\ep_{k-1}^2,\] and $\ep_{k-1}^2\leq\E_a\bigl[\ep_{k-1}(a)^2\bigr]$ (Jensen's inequality). The ignore first step trick is the reasoning behind the final equation on the first line. The observation is that the distribution which draws $(a,b)\sim A\times B$ and $b',b''\sim N(b)$ and outputs $(a',b',b'')$ where $a'=\phi(a,\hat b)$ is identical to the distribution which draws $a'\sim A$ and a random edge $b'\sim_{B^2}b''$ in $B^2$ and outputs $(a',b',b'')$. See Figure~\ref{fig:2} for intuition.
\begin{figure}[h]
\begin{center}
\includegraphics[scale=0.3]{sections/sw2.png}
\caption{``Ignore first step'' trick.}
\vspace{-1.5 em}
\label{fig:2}
\end{center}
\end{figure}
\paragraph{Starting the Replacement Walk in the Middle.} A useful feature of random walks on an undirected $d-$regular graph is that the steps can be generated out of order. Specifically, the vertices in a $k-$step random walk can be generated by choosing $a_i\sim A$ first for any $i\in[k]$ and then drawing two walks $(a_i,a_{i+1},\dots,a_k)\sim\rw_A^{k-i+1}(a_i)$, $(a_i,a_{i-1},\dots,a_1)\sim\rw_A^i(a_i)$ and outputting $(a_1,\dots,a_k)$. Replacement product walks also have this feature, though correctly formulating it requires precision. We will use that the following distribution is identical to $\sw^k$ for any $i\in\{0,1\dots,k-1\}$:
\begin{enumerate}
\item $a_i\sim A$ and a random edge $b_i\sim b_{i+1}$ in $B$; set $b_{i+1}={\sf shift}(b_{i+1})$;
\item generate $(b_1,\dots,b_k)\in B^k$ as follows:
\begin{itemize}
\item[$(i)$] for $j\geq i+2$, draw $b_j\sim N(b_{j-1})$ and set $b_j={\sf shift}(b_j)$;
\item[$(ii)$] for $j\leq i-1$, draw $b_j\sim N(b_{j+1})$ and set $b_j={\sf shift}^{-1}(b_j)$;
\end{itemize}
\item generate and output $(a_0,\dots,a_k)\in A^{k+1}$ as follows:
\begin{itemize}
\item[$(i)$] for $i\geq i+1$, set $a_i=\phi(a_{i-1},\hat b_i)$ where $\hat b_i=b_i[1]\in[d]$ denotes the first coordinate of $b_i\in[d]^s$, and where $\phi$ is the rotation map of $A$;
\item[$(ii)$] for $j\leq i-1$, set $a_j=\phi^{-1}(a_{j+1},\hat b_j)$ where $\phi^{-1}(a,\hat b)=\phi(a,\hat b')$ where $\hat b'$ is the local inverse of $\hat b$.
\end{itemize}
\end{enumerate}
An example of how this is used is the first step of the bound for $\ep_k$ when $k>s$:
\begin{eqnarray*}\ep_k &=& \bigg|\E_{(a_0,\dots,a_k)\sim\sw^k}\Bigl[(-1)^{f(a_s)}\cdot(-1)^{f(a_0)\oplus\cdots\oplus f(a_s)}\cdot(-1)^{f(a_s)\oplus\cdots\oplus f(a_k)}\Bigr]\bigg|\\ &=& \bigg|\E_{\stackalign{a_s&\sim A\\b_s&\sim b_{s+1}}}\Bigl[(-1)^{f(a_s)}\cdot\backwardsvec{g}_s(a_s,b_s)\cdot g_{k-s}(a_s,b_{s+1})\Bigr]\bigg|,\end{eqnarray*} where $\backwardsvec{g}_s(a,b)$ indicates that the repalcement walk is drawn in the ``backwards'' fashion according to Steps 2(ii) and 3(ii) above. Equivalently, $\backwardsvec{g}_s(a,b)$ is the expectation of $(-1)^{f(a_0)\oplus\cdots\oplus f(a_s)}$ over $(a_0,\dots,a_s)\sim\sw^s$ conditioned on $(a_s,b_s)=(a,b)$.
\begin{figure}[h]
\begin{center}
\includegraphics[scale=0.3]{sections/sw3.png}
\caption{Starting the Replacement Walk in the Middle.}
\vspace{-2 em}
\label{fig:3}
\end{center}
\end{figure}
\subsection{Bounding the $\ep_k$ Terms}
\label{sec:epk}
In this section we bound the $\ep_k$ terms in Claims~\ref{clm:swidebasecase} and~\ref{clm:swideinduction}, thereby proving half of each claim. We bound the $\sigma_k$ terms in the next section.
\paragraph{The Base Case.} This follows directly from the pseudorandomness property, and the analysis already done in Section~\ref{sec:techniques} (Claim~\ref{clm:exprw}). Specifically, when $k\leq s$, we have \[\ep_k=\Big|\E_a\bigl[\ep_k(a)\bigr]\Big|=\Big|\E_a\bigl[h_{k+1}(a)\bigr]\Big|\leq\frac{1}{2}\cdot(2\lambda)^{k+1},\] where $\ep_k(a)=h_{k+1}(a)$ by pseudorandomness ($h_{k+1}$ is the function defined in Claim~\ref{clm:exprw}).
\paragraph{The Induction Step.} Fix $k>s$. We have \[\ep_k =\bigg|\E_{\stackalign{a&\sim A\\b&\sim b'}}\Bigl[(-1)^{f(a)}\cdot\backwardsvec{g}_s(a,b)\cdot g_{k-s}(a,b')\Bigr]\bigg|\leq \bigg|\E_{a\sim A}\Bigl[(-1)^{f(a)}\cdot\backwardsvec{\ep}_s(a)\cdot\ep_{k-s}(a)\Bigr]\bigg|+\lambda\sigma_s\sigma_{k-s},\] where the equality holds by starting the replacement walk in the middle, and the inequality is the expander mixing lemma (Definition~\ref{def:expander}) on $B$. We are using the shorthand $\backwardsvec{\ep}_s(a)$ for $\E_b\bigl[\backwardsvec{g}_s(a,b)\bigr]$, and we have used Cauchy-Schwarz to bound the standard deviation terms, just as we did in the computation in the ``ignore first step trick'' paragraph in Section~\ref{sec:intuition}. Specifically, \[\E_a\bigl[\backwardsvec{\sigma}_s(a)\cdot\sigma_{k-s}(a)\bigr]\leq\sqrt{\E_a[\backwardsvec{\sigma}_s(a)^2]}\sqrt{\E_a[\sigma_{k-s}(a)^2]}\leq\backwardsvec{\sigma}_s\sigma_{k-s}=\sigma_s\sigma_{k-s}.\] By pseudorandomness, $(-1)^{f(a)}\cdot\backwardsvec{\ep}_s(a)=(-1)^{f(a)}\cdot h_{s+1}(a)=\E_{a'\sim N(a)}\bigl[h_s(a')\bigr]=\E_{a'\sim N(a)}\bigl[\ep_{s-1}(a')\bigr]$, and so we get the desired bound on $\ep_k$ via the expander mixing lemma on $A$, as follows: \begin{eqnarray*}\ep_k &\leq& \Big|\E_{a\sim a'}\bigl[\ep_{s-1}(a)\cdot\ep_{k-s}(a')\bigr]\Big|+\lambda\sigma_s\sigma_{k-s}\leq\ep_{s-1}\ep_{k-s}+\lambda^2\sigma_{s-1}\sigma_{k-s}+\lambda\sigma_s\sigma_{k-s}\\ &\leq& \frac{1}{2}(2\lambda)^s(\ep_{k-s}+3\sigma_{k-s}).\end{eqnarray*}
\subsection{Bounding the $\sigma_k$ Terms}
\label{sec:sigmak}
\paragraph{The Base Case.} We have already noted that when $1\leq k\leq s$, $\ep_{k-1}(a)=h_k(a)$ by pseudorandomness. Thus, $\E_a\bigl[\ep_{k-1}(a)^2\bigr]=\E_a\bigl[h_k(a)^2\bigr]\leq(2\lambda)^{2k-2}$, by Claim~\ref{clm:exprw}. It follows from the first step trick that $\sigma_k^2\leq(2\lambda)^{2k-2}+\lambda^2\sigma_{k-1}^2$, which implies $\sigma_k\leq(2\lambda)^{k-1}+\lambda\sigma_{k-1}$. Iterating this bound gives \[\sigma_k\leq\lambda^{k-1}\cdot\bigl(2^{k-1}+2^{k-2}+\cdots+2+1\bigr)\leq2\cdot(2\lambda)^{k-1}.\]
\paragraph{The Induction Step.} Fix $k>s$. As mentioned in the ``ignore first step trick'' paragraph in Section~\ref{sec:intuition}, $\sigma_k^2\leq\E_a\bigl[\ep_{k-1}(a)^2\bigr]+\lambda^2\sigma_{k-1}^2$ holds and so it suffices to bound $\E_a\bigl[\ep_{k-1}(a)^2\bigr]$. By starting the replacement walk in the middle, we get \[\E_a\bigl[\ep_{k-1}(a)^2\bigr]=\E_{\stackalign{a_{s-1}&\sim A\\ b_{s-1}&\sim b_s}}\Bigl[(-1)^{f(a_{s-1})}\cdot g_{k-s}(a_{s-1},b_s)\cdot G(a_{s-1},b_{s-1})\Bigr],\] where $G:A\times B\rightarrow\R$ is defined by $G(a,b):=\E_{(a_0,\dots,a_{s-1})}\bigl[(-1)^{f(a_{s-1})\oplus\cdots\oplus f(a_0)}\cdot\ep_{k-1}(a_0)\bigr]$, where the expectation is over $(a_0,\dots,a_{s-1})$ drawn as follows:
\begin{itemize}
\item[$\cdot$] set $b_{s-1}=b$; for $1\leq i\leq s-2$, draw $b_i\sim N(b_{i+1})$ and then set $b_i={\sf shift}^{-1}(b_i)$;
\item[$\cdot$] set $a_{s-1}=a$; for $0\leq i\leq s-2$ set $a_i=\phi^{-1}(a_{i+1},\hat b_{i+1})$.
\end{itemize}
The expander mixing lemma (Definition~\ref{def:expander}) on $B$ gives \[\E_a\bigl[\ep_{k-1}(a)^2\bigr]\leq\E_a\Bigl[(-1)^{f(a)}\cdot\ep_{k-s}(a)\cdot\mu_G(a)\Bigr]+\lambda\sigma_{k-s}\sigma_G,\] where $\mu_G:=\E_{a,b}\bigl[G(a,b)\bigr]$, $\mu_G(a):=\E_b\bigl[G(a,b)\bigr]$ and $\sigma_G$ is such that $\sigma_G^2+\mu_G^2=\E_{a,b}\bigl[G(a,b)^2\bigr]$. By pseudorandomness, $\mu_G(a)=\E_{(a_0,\dots,a_{s-1})\sim\rw^s_A(a)}\bigl[(-1)^{f(a_0)\oplus\cdots\oplus f(a_{s-1})}\cdot\ep_{k-1}(a_{s-1})\bigr]=\hat h_s(a)$, where $\hat h_s:A\rightarrow\R$ is given by $\hat h_s(a)=\E_{(a_1,\dots,a_s)\sim\rw_A^s}\bigl[(-1)^{f(a_1)\oplus\cdots\oplus f(a_s)}\cdot\ep_{k-1}(a_s)\bigr]$. Note this is the function defined in Claim~\ref{clm:genrw}, instantiated with $H(a)=\ep_{k-1}(a)$. We have $(-1)^{f(a)}\cdot\mu_G(a)=\E_{a'\sim N(a)}\bigl[\hat h_{s-1}(a')\bigr]$, and so by the expander mixing lemma on $A$ and Claim~\ref{clm:genrw} we have \begin{eqnarray*}\E_a\bigl[\ep_{k-1}(a)^2\bigr]&\leq&\E_{a\sim a'}\bigl[\ep_{k-s}(a)\cdot\hat h_{s-1}(a')\bigr]+\lambda\sigma_{k-s}\sigma_G\\ &\leq& \ep_{k-s}\cdot2^{s-3}(\lambda^{s-2}\cdot\hat\ep_1+\lambda^{s-1}\hat\sigma_1)+\lambda^2\sigma_{k-s}\cdot2^{s-3}(\lambda^{s-3}\hat\ep_1+\lambda^{s-2}\hat\sigma_1)+\lambda\sigma_{k-s}\sigma_G,\end{eqnarray*} where $\hat\ep_1$ and $\hat\sigma_1$ are the notations from Claim~\ref{clm:genrw}. In our case, $\hat\ep_1=\E_a\bigl[(-1)^{f(a)}\cdot\ep_{k-1}(a)\bigr]=\ep_{k-2}$, and $\hat\sigma_1=\sqrt{\E_a[\ep_{k-1}(a)^2]-\hat\ep_1^2}\leq\sqrt{\E_{a,b}[g_{k-1}(a,b)^2]-\hat\ep_1^2}=\sqrt{\sigma_{k-1}^2+\ep_{k-1}^2-\ep_{k-2}^2}\leq\sigma_{k-1}$. We have used Jensen's inequality and that $\ep_{k-2}\geq\ep_{k-1}$. Using these values and remembering the bound $\sigma_k^2\leq\E_a\bigl[\ep_{k-1}(a)^2\bigr]+\lambda^2\sigma_{k-1}^2$ gives \begin{equation}\label{eq:almostdone}\sigma_k^2\leq\frac{1}{2}(2\lambda)^{s-2}(\ep_{k-2}+\lambda\sigma_{k-1})(\ep_{k-s}+\lambda\sigma_{k-s})+\lambda\sigma_{k-s}\sigma_G+\lambda^2\sigma_{k-1}^2.\end{equation} This is almost the required bound except we still need to simplify $\sigma_G$. For this purpose, let us add a parameter to our notation for $G$, writing $G_{s-1}$ instead of $G$, since it is an expectation over a length $(s-1)$ ``backwards'' replacement walk. For $r\leq s-1$, let $\mu_r:=\E_{a,b}\bigl[G_r(a,b)\bigr]$, let $\mu_r(a):=\E_b\bigl[G_r(a,b)\bigr]$ and $\tau_r$ such that $\tau_r^2+\mu_r^2=\E_{a,b}\bigl[G_r(a,b)^2\bigr]$. We need to bound$\tau_{s-1}$. By the ignore first step trick and expander mixing lemma on $B^2$, \[\tau_{s-1}^2\leq\E_{a,b}\bigl[G_{s-1}(a,b)^2\bigr]=\E_{\stackalign{a&\sim A\\ b&\sim_{B^2} b'}}\Bigl[G_{s-2}(a,b)\cdot G_{s-2}(a,b')\Bigr]\leq\E_a\bigl[\mu_{s-2}(a)^2\bigr]+\lambda^2\tau_{s-2}^2.\] We have already seen that $\mu_{s-2}(a)=\hat h_{s-1}(a)$, and so by Claim~\ref{clm:genrw} and our computation of $\hat\ep_1$ and $\hat\sigma_1$ above, $\tau_{s-1}^2\leq(2\lambda)^{2s-6}(\ep_{k-2}+\lambda\sigma_{k-1})^2+\lambda^2\tau_{s-2}^2$, which implies $\tau_{s-1}\leq(2\lambda)^{s-3}(\ep_{k-2}+\lambda\sigma_{k-1})+\lambda\tau_{s-2}$. Iterating this bound (and using $\tau_0\leq\sigma_{k-1}$) gives \[\tau_{s-1}\leq\lambda^{s-3}(\ep_{k-2}+\lambda\sigma_{k-1})(2^{s-3}+2^{s-4}+\cdots)+\lambda^{s-1}\tau_0\leq2\cdot(2\lambda)^{s-3}(\ep_{k-2}+\lambda\sigma_{k-1})+\lambda^{s-1}\sigma_{k-1}.\] Plugging this into (\ref{eq:almostdone}) gives the desired bound: \[\sigma_k^2\leq\frac{1}{2}(2\lambda)^{s-2}(\ep_{k-2}+\lambda\sigma_{k-1})\bigl(\ep_{k-s}+(2+\lambda)\sigma_{k-s}\bigr)+\lambda^s\sigma_{k-s}\sigma_{k-1}+\lambda^2\sigma_{k-1}^2.\]
\
| 198,820
|
TITLE: Bijections Between Different Branches of the Inverse of a Polynomial of a Single Complex Variable
QUESTION [4 upvotes]: Let $F\left(z\right)$ be a polynomial of degree $d≥2$ with complex coefficients. Let $D$ be an open disk in the complex plane containing no critical points of $F\left(z\right)$.
Let $f\left(z\right)$ denote the inverse of $F\left(z\right)$; since $F\left(z\right)$ is not one-to-one, $f\left(z\right)$
is a multivalued function with d distinct branches: $f_{1}\left(z\right),f_{2}\left(z\right),...,f_{d}\left(z\right)$. Let $m,n$
be two distinct integers in the set $\left\{ 1,2,...,d\right\}$
Then, does there necessarily exist a function:
$T_{m,n}\left(z\right)=a_{m,n}z+b_{m,n}$
(with the constants $a_{m,n}$ and $b_{m,n}$ being complex numbers to be determined) such that:
$T_{m,n}\left(f_{m}\left(z\right)\right)=f_{n}\left(z\right),\textrm{ }\forall z\in D$
This little question has been something of a thorn in my research. I'd like to think that it is true.
Example: Let $F\left(z\right)=\left(\frac{z}{2}-i\right)^{3}-1$
.
$F$
has three inverses:
$f_{m}\left(z\right)=2i+2\omega^{m-1}\left(z+1\right)^{1/3}$
where $\omega=e^{\frac{2\pi i}{3}}$
and $m=1,2,3$.
These inverses are related to one another by the linear function:
$T_{m,n}\left(z\right)=2i+\omega^{m-n}\left(z-2i\right)$
Any thoughts on whether or not this is true? And how one might go about showing it?
REPLY [1 votes]: It's not true for cubics in general. Try
$$ F(X) = X^3 + c_2 X^2 + c_1 X + c_0$$
The roots of $F(X) - z$ are $f_1, f_2, f_3$ with
$$ \eqalign{f_1 + f_2 + f_3 &= -c_2\cr
f_1 f_2 + f_1 f_3 + f_2 f_3 &= c_1\cr
f_1 f_2 f_3 &= -c_0 - z\cr} $$
Note that the first two equations don't involve $z$. We can eliminate $f_3$ from the first two equations, obtaining
$$ f_1^2 + f_1 f_2 + f_2^2 + c_2 (f_1 + f_2) + c_1 = 0$$
But unless this factors into linear factors, you can't have an equation of the form $f_2 = a f_1 + b$. You happened to pick one that factors.
| 123,960
|
600,000 athletes from every level of athletics having experienced the Parisi Speed School, the program has become nationally recognized as the leader in performance enhancement training. In early 2012, there are over 75 Parisi Franchise locations in 26 different states across America..
Redeem A Free Speed Pass
| 315,797
|
TITLE: Poisson distribution Question - Expectation
QUESTION [2 upvotes]: A company takes on an average of $0.1$ new employees per week. Assume that the actual
number of employees taken on in a given week has a Poisson distribution.
Let $Y$ denote the number of weeks that pass without a new employee being taken on.
Calculate the expectation of $Y$. Calculate the probability that, over the course of a year ($50$ working weeks), there is just one week when more than one new employee is taken on.
REPLY [0 votes]: Alex has answered part of the question.
The probability that more than one new employee is hired in a given week is $1$ minus the probability that only $0$ or $1$ new employees are hired. Thus it is
$$
1- \frac{0.1^0 e^{-0.1}}{0!} - \frac{0.1^1 e^{-0.1}}{1!} = 1 - e^{-0.1} - 0.1e^{-0.1}.
$$
Call this number $p$. Then we're looking for the probability of $1$ success in $50$ trials with probability $p$ of success on each trial --- thus a binomial distribution. The probability is
$$
\binom{50}{1} p^1 (1-p)^{50-1}.
$$
| 218,156
|
COP23 report 2nd - 5th Nov
This first report from the BK COP team includes:
- Summary of the meeting
- News of India One being used for publicity by the German Government
- News from Hamburg, Bremmen and Bonn
- The delegation setting up
- News from Conference of Youth
See the report and photos
COP23 - BK Programmes
Seminar: Climate Change, Ethics, Renewable Energy (6 Nov 2017)
Panel: Hope and Strength for a World in Transition (10 Nov 2017)
(
Faltblatt auf Deutsch) (
Leaflet in English)
Panel: Visionary Leadership for a Sustainable Future (14 Nov 2017)
COP23 - UN Climate Change Conference
At the UN Climate Change Conference this year (COP23, from 6 to 17 November) nations of the world will meet to advance the aims and ambitions of the Paris Agreement and achieve progress on its implementation guidelines.
German Environment ministry has chosen India One as part of its campaign "Ready for the Future" for CoP23. Minister Barbara Hendricks officially presented the posters to the public.
Brahma Kumaris getting ready for COP23
The Brahma Kumaris COP23 team is now assembling and getting ready for a busy time at the COP23 event in Bonn, Germany. These are documents prepared to date:
Statement: Empowering_People_for_a_World_
COP23: Calendar of events
COY13: Calendar of events
COP23 - Visionary Leadership for a Sustainable Future
A panel discussion will be held Tuesday 14th November, 7:00 – 21:00 hrs (reception at 6:15pm) including a number of top speakers:
Sister Jayanti, Director, Brahma Kumaris
Anne Barre, Women Engage for a common Future (WECF), Coordinator Gender & Climate Policy, Strategic Partnerships.
Joachim Golo Pilz, Energy Advisor Brahma Kumaris, Head of India-One solar thermal power plant
The moderator will be:
Thomas Bruhn, Potsdam Research Institute for Advanced Sustainability Studies (iass-potsdam.de)
Welcome to an evening where different pioneers share their visions on how a sustainable future society can look like.
See the detailed introduction...
| 404,925
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Authors XI bring the house down at DLF
The stereotypical image of the writer – or any intellectual, for that matter – is of the retiring, unathletic figure. This idea is not a new one, and it must have goaded enough writers into putting down their pens momentarily, to pick up their bats, balls and wicket stumps to form the Authors cricket team in 1891. The team had such luminaries as Arthur Conan Doyle, and PG Wodehouse, to name some. After the First World War, the original team evolved, losing some old players, picking up new ones, and so it has been since.
At present, the torch is being proudly carried by Charlie Campbell, literary agent and captain of the Authors Cricket Club. He has led this team of writers in over 130 games, and they have faced the Rajasthan Royals, the Vatican and the national team of Japan along the way. Their book, The Authors XI: A Season of English Cricket from Hackney to Hambledon, was shortlisted for the Cricket Society MCC Book of the Year Award. Campbell was present, along with team-mates Anthony McGowan, Alex Preston, and Richard Beard in a light-hearted, interactive panel titled “Of Googlies and Chinaman”, moderated by Khademul Islam.
The discussion kicked off with a general discussion about the recent cricket matches between England and Bangladesh, and Alex Preston garnered hoots and cheers when he declared “Never have I ever cheered as hard as I did for a team in a match against England.” To which Anthony McGowan, author of two literary thrillers and a slew of books for young adults, quipped “You’re such a suckup, Alex!” This of course prompted a series of digs about McGowan’s game, with Charlie Campbell and Richard Beard ganging up with Preston to provide anecdotes.
The rest of the panel continued in a similar vein, peppered with hilarious anecdotes about the team’s travels around the world and their matches (and spectacular losses!) against the unlikeliest opponents. One such story worth mentioning was their match in the Vatican. “Two of our players met the Pope, who was minding his own business, expecting us to talk about religion, and suddenly he’s got two cricketers giving him a cricket cap” said Campbell. Apparently, the Authors lost that match. “It was the only game where we were beaten and then forced to pray with our conquerers. That hasn’t happened in a while” he added.
Themes such as sports writing, and how the nature of the cricket – with its nail-biting moments and moments of mind-numbing boredom, is beneficial for the writers, were explored. Many of the questions from the audience centered around sledging, and the answers, mirthful as they were provided some insight on the egos, insecurities and emotions of the writers. There was also some musing on the role of cricket in alleviating political tension, and building communication between teams and nations.
This wonderful, fun-filled session ended on a positive note, with the team expressing their wish to return to Bangladesh for an opportunity to play with our league teams, to explore places like Chittagong and the Sundarbans, and certainly to attend more literary festivals.
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TITLE: Poisson summation formula as a special case of the trace formula
QUESTION [1 upvotes]: For $f \in L^1(\mathbb R)$, the Fourier transform $\hat{f}: \mathbb R \rightarrow \mathbb C$ is defined by
$$\hat{f}(x) = \int_{\mathbb R} f(y) e^{2\pi i xy}dy.$$
The Poisson summation formula asserts that for $f$ smooth and compactly supported,
$$\sum\limits_{n\in \mathbb Z} f(n) = \sum\limits_{n \in \mathbb Z} \hat{f}(n).$$
I've heard that this formula arises as a special case of a trace formula for an integral operator of trace class on the Hilbert space $L^2(\mathbb R/\mathbb Z)$, or that Selberg or Arthur's trace formula is a "nonabelian Poisson summation formula." How can the Poisson summation formula be seen as arising in this way?
REPLY [1 votes]: Generally, if $v_1, v_2, ... $ are an orthonormal basis of a given Hilbert space, and $T$ is a sufficiently nice operator on the given Hilbert space (specifically of "trace-class"), then the trace of $T$ can be defined by the formula
$$\operatorname{Tr}(T) = \sum\limits_{n\in \mathbb Z} \langle Tv_i, v_i \rangle$$
and this is independent of the choice of orthonormal basis. The Poisson summation formula comes from looking at the trace of a certain operator on the Hilbert space $V = L^2(\mathbb R/\mathbb Z)$ in two ways.
For $f$ smooth and compactly supported on $\mathbb R$, we set $K(x,y) = \sum\limits_{n \in \mathbb Z} f(-x+n+y)$, which is a finite sum for all pairs of real numbers $(x,y)$, and lies in the Hilbert space space $L^2(\mathbb R/\mathbb Z \times \mathbb R/\mathbb Z)$.
We define a bounded linear operator $T$ on $V$ by the formula
$$T(\varphi)(x) = \int\limits_{\mathbb R/\mathbb Z} K(x,y) \varphi(y)dy. \tag{$\varphi \in V$}$$
This operator turns out to be of trace class, so we may compute its trace $\operatorname{Tr}(T)$ in two ways.
First way (spectral expansion)
By the Peter-Weyl theorem, the characters $\phi_m(x) = e^{2\pi i mx} : m \in \mathbb Z$ form an orthonormal basis for $V$. Each vector $\phi_m$ is actually an eigenvector for the operator $T$, with eigenvalue $\hat{f}(n)$:
$$T(\phi_m)(x) = \int\limits_{\mathbb R/\mathbb Z} \sum\limits_{n \in \mathbb Z} f(-x+n+y) \phi_m(y)dy = \int\limits_{\mathbb R/\mathbb Z} \sum\limits_{n \in \mathbb Z} f(-x+n+y) \phi_m(n+y)dy = \int\limits_{\mathbb R} f(-x+y)\phi_m(y)dy $$
$$ = \int\limits_{\mathbb R} f(y) \phi_m(x+y)dy = \int\limits_{\mathbb R} f(y) e^{2 \pi i m(x+y)}dy = e^{2\pi i mx} \int\limits_{\mathbb R} f(y) e^{2\pi i my}dy = \hat{f}(m)e^{2\pi i mx}.$$
Therefore, we can compute $\operatorname{Tr}(T)$ as a sum of the eigenvalues $\hat{f}(n)$:
$$\operatorname{Tr}(T) = \sum\limits_{n\in \mathbb Z} \langle T(\phi_n), \phi_n \rangle = \sum\limits_{n\in \mathbb Z} \hat{f}(n) \langle \phi_n, \phi_n \rangle = \sum\limits_{n \in \mathbb Z} \hat{f}(n).$$
Second way (the trace as an integral along the diagonal)
On the other hand, we claim that the trace of $T$ can be calculated by integrating $K(x,y)$ along the diagonal $(x,x)$. Again by the Peter-Weyl theorem, the characters $\phi_{n,m}(x,y) = \phi_n(x) \overline{\phi_m(y)}$ of $(\mathbb R / \mathbb Z) \times(\mathbb R / \mathbb Z)$ form an orthonormal basis for $L^2(\mathbb R / \mathbb Z \times \mathbb R / \mathbb Z)$, and we may therefore write
$$K(x,y) = \sum\limits_{i,j} c_{ij} \phi_{i,j}(x,y)$$
for some complex numbers $c_{ij}$. We may compute
$$\int\limits_{\mathbb R/\mathbb Z} K(x,x)dx = \int\limits_{\mathbb R/\mathbb Z} \sum\limits_{i,j} c_{ij} \phi_{i,j}(x,x) dx = \sum\limits_{i,j} \int\limits_{\mathbb R/\mathbb Z} \phi_i(x) \overline{\phi_j(x)}dx = \sum\limits_{i,j} c_{i,j} \delta_{ij} = \sum\limits_i c_{i,i}$$
and at the same time,
$$\operatorname{Tr}(T) = \sum\limits_{n \in \mathbb Z} \int\limits_{\mathbb R/\mathbb Z} T(\phi_n)(x) \overline{\phi_n(x)}dx = \sum\limits_n \int\limits_{\mathbb R/\mathbb Z} \space \int\limits_{\mathbb R/\mathbb Z} K(x,y) \phi_n(y) \overline{\phi_n(x)} dx = \sum\limits_n \int\limits_{\mathbb R/\mathbb Z} \space \int\limits_{\mathbb R/\mathbb Z} \sum\limits_{i,j} c_{i,j} \phi_i(x) \overline{\phi_j(y)} \phi_n(y) \overline{\phi_n(x)}dydx$$
$$ = \sum\limits_{n,i,j} c_{i,j} \Bigg(\int\limits_{\mathbb R/\mathbb Z} \phi_i(x) \overline{\phi_n(x)}dx \Bigg) \Bigg( \int\limits_{\mathbb R/\mathbb Z} \phi_n(y) \overline{\phi_j(y)}dy \Bigg) = \sum\limits_{n,i,j} c_{i,j} \delta_{i,n} \delta_{n,j} = \sum\limits_{i,i} c_{i,i}.$$
Therefore,
$$\operatorname{Tr}(T) = \int\limits_{\mathbb R/\mathbb Z} K(x,x)dx.$$
Third way (geometric expansion):
Now that we have established that the trace of $T$ can be gotten by integrating $K(x,y)$ along the diagonal, we can compute the trace in one more way:
$$\operatorname{Tr}(T) = \int\limits_{\mathbb R/\mathbb Z} K(x,x) dx = \int\limits_{\mathbb R/\mathbb Z} \sum\limits_{n\in \mathbb Z} f(-x+n+x)dx = \int\limits_{\mathbb R/\mathbb Z} \sum\limits_{n\in \mathbb Z} f(n) dx = \sum\limits_{n \in \mathbb Z}f(n).$$
Thus the Poisson summation formula
$$\sum\limits_{n \in \mathbb Z}\hat{f}(n) = \sum\limits_{n \in \mathbb Z}f(n)$$
holds by looking at the trace of $T$ in two ways, first as a sum of eigenvalues, and second as an integral along the diagonal.
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Aliona Vilani has confirmed she would love to make a return to Strictly Come Dancing this year after sensationally quitting at the end of 2015.
The ballroom professional - who announced her departure and revealed she wanted to start a family - had admitted she is ready to stage a shock U-turn and make a comeback when the programme is back on air in autumn.
Speaking to the Daily Star Sunday, she said: "I won't be dancing with a celebrity in the next series. But hopefully I'll still be involved in the show in other ways.
"I am missing it already. When I decided to leave after the final I cried my eyes out. It was very upsetting for me. It was a really hard decision. I had enjoyed the season too much."
Read more: Strictly winner Aliona slams show for over-working dancers
Aliona - who is the only professional dancer to win the show twice alongside celebrity partners Harry Judd and Jay McGuiness - also admitted she would even consider returning full-time, if she's given days off.
She added: ."
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TITLE: Theorems about functions with uncountable number of discontinuities
QUESTION [4 upvotes]: I have seen a nice number of theorems that start with "suppose that $f$ is continuous function" or with some equivalent claim and then, with only that, or with some additional assumptions some theorem follows.
But I would like to know about theorems that start with the assumptions like "suppose that $f$ is discontinuous function" and then end with some truth about discontinuous functions.
Suppose that we work with real functions of a real variable.
Because the function can be discontinuous in a finite number of points, in countably infinite number of points and in uncountably infinite number of points let us talk here only about functions that have uncountably infinite number of discontinuities.
So the question is:
Can you give me some examples of theorems that start with the assumption that "$f$ is real function of a real variable which has an uncountable number of discontinuities" (and possibly with some other assumptions) and then some fact about such functions follows?
REPLY [4 votes]: If a function $f : (a,b) \to \Bbb R$ has an uncountable number of discontinuities, then only a countable number of them may be jump discontinuities, the others (uncountably many) being essential discontinuities.
If a function $f : (a,b) \to \Bbb R$ has an uncountable number of discontinuities, then it cannot be monotonic.
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