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[The nurse in the decentralization process of the health system].
Study about the decentralization process of the health system in the '80s and '90s in the city of Itabuna-BA. It is aimed at describing the city's health decentralization process, identifying the nurse's insertion/participation in this process. Descriptive/qualitative study of exploratory nature that utilized both the semistructured interview and documental analysis for data collection. The results show that the nurse's insertion in the decentralization process took place according to the changes that occurred at every municipal management, where the nurse acted more effectively as the set of circumstances was established and was influenced by the several conjunctures formed by the implemented policies or implemented in each context of the management of the municipal health system. |
Q:
Does c# have a VB.NET-equivalent for shorthand array declaration like {"string1","string2"}?
In VB.NET, you can instantiate and immediately use an array like this:
Dim b as Boolean = {"string1", "string2"}.Contains("string1")
In c#, however, it appears you have to do this:
bool b = new string[] { "string1", "string2" }.Contains("string1");
Does c# have equivalent shorthand syntax that uses type inference to determine the type of array without it having to be explicitly declared?
A:
Implicitly typed arrays do not have to include their type, provided it can be inferred:
bool b = new [] { "string1", "string2" }.Contains("string1");
A:
It called Implicitly Typed Arrays
You can create an implicitly-typed array in which the type of the
array instance is inferred from the elements specified in the array
initializer. The rules for any implicitly-typed variable also apply to
implicitly-typed arrays.
static void Main()
{
var a = new[] { 1, 10, 100, 1000 }; // int[]
var b = new[] { "hello", null, "world" }; // string[]
}
You can use it also for jagged array.
|
February 14, 2005
The fallback position of the Beltway press on the "Gannon" story has been that it was somehow inappropriate to "go after" his personal life (ie., his homosexuality), as if it were somehow not germane to the discussion of how a non-credentialed "reporter" using a fake name was able to gain access inside the White House. AmericaBlog, which has done some of the most aggressive reporting on the subject, has a detailed post up about some of the more unsavory aspects of Mr. Guckert's "personal life", and why it is relevant to the scandal. It's very explicit, so you've been warned.... |
When he began working on the ending of Mass Effect 2, lead cinematic animator Parrish Ley felt like a fraud. There were so many issues to sort out, he didn’t know where to begin.
By the start of the Suicide Mission, there were hundreds of player choices to account for that resulted in thousands of possible scenarios. As work on the mission progressed, it was like trying to unspool a spider web.
“It was the kind of thing where you think, ‘I don’t know what I’m doing,’” Ley said. “It was a complex piece of branching narrative. We wanted to make sure it felt right for the players who did it, but at the same time, under the hood there were a lot of things running.”
Because they couldn’t say for certain which characters the player would have in their party, how many of them had completed loyalty missions, or which ones might die, even the most basic questions like who would deliver what lines became a nightmare.
Unfortunately, in situations like this, there’s no real Eureka moment or a silver bullet to take the beast down. In reality, there’s a group of people in a room who’ve missed a couple showers and skipped a few meals, working on the problem over and over and over until it’s solved.
“It started out utterly broken to the point where it was totally unplayable and we looked at each other and thought, ‘I don’t know what’s going on. We’re not going to be able to ship this game. This is crazy,’” Ley says. “And we would play it and go back to our desks and work on it. And every day it would get a little bit better, and a little bit better.”
While players remember the emotional moments and intimate scenes, it’s often the hours that went into crafting them that stick with developers. Creating games is an intensely personal experience, says lead level designer Rick Knowles. You help a game develop over months and years, forging it into something you’re proud of.
“As developers, we’re also gamers, and I feel privileged to make games we want to play,” Knowles says. “I think we’d feel we failed if we built a game that we didn’t want to play.”
How you end up can often be miles from where you start out, he says. This was the case with ME3’s multiplayer. The original idea was to have a co-op mode. But the more they tried to hammer out the dents, the worse it felt.
“It took a very long time to settle on a model that we felt worked well. We tried out different layouts, different creatures, different settings,” Knowles says. “We were very conscious from the beginning to not just make a straight-up multiplayer experience. It had to have some narrative context.”
They addressed this by creating objectives, which gave weight and purpose to each mission. And though the stories are smaller and more contained, Knowles and his team built internal narratives into each map that would later inform design decisions. This is why the hazard version of Firebase Dagger features a sandstorm, for instance. It doesn’t just make the map more difficult; it makes sense in the context of downed satellite dishes.
But while you could put a lifetime of hours into developing a game, eventually it has to ship, and developers then must sit back and watch as players take over and experience it.
The day ME2 launched, Kris Schoneberg and her fellow level designers crowded around a computer screen, watching a live stream of an early playthrough. They were eager to see how the player would react to certain plot points, and how they would handle the challenges the team had so lovingly devised.
“They got to the Warden Kuril boss fight in the Jack mission,” Schoneberg says, laughing. “And I watched the guy just die over and over again, and I thought, ‘I’m sorry. I hope you’re having fun.’”
Patrick Weekes, a writer on all three games, had a similar experience on ME2. “I was watching a playthrough of a super-renegade player doing Tali’s loyalty mission. He was about to finish the trial, and after all his renegade decisions, I really didn’t want to watch him break Tali’s heart. Then he pauses the playthrough, goes to the chat box, and says, ‘I don’t know what to do, guys. I don’t wanna hurt Tali.’ Seeing that something I was part of connected with a fan deeply enough to affect his decisions made me realize how special this series really was.”
As time goes on, the series becomes more and more the property of fans, and they begin to develop their own ideas about what makes and defines Mass Effect, says executive producer Casey Hudson. And while this can at times cause resistance toward new ideas, ultimately it’s a sign that the characters and stories the developers created are resonating.
“You can have characters in a story, but that’s kind of different from building memorable characters that transcend the story,” Hudson says. “These characters that we have in the Mass Effect series, people want to takes these characters further out of the story and see them in comics and books and they want to know more about these characters.”
DLC is among the many good ways to tell new stories, says producer Mike Gamble. Here in these smaller, more contained adventures, the team can take more risks and explore ideas that just couldn’t fit into the main game.
It also gives members of the ME team a chance to get back together with old friends, Gamble says—a big part of what made the series great in the first place.
“When we first start to make games, we have everyone from different departments each with an idea of what the game looks like,” he says. “Being able to work together for a long time, you start to develop trust, and those walls go away and you can drive toward a singular purpose.”
While the Citadel DLC was a farewell to Shepard and the crew of the Normandy for fans, inside the studio, the developers said their own goodbyes.
“We were sad when Shepard’s story was over. There was definitely a solemnness in the room,” Schoneberg says. “The last thing I worked on was the party in Citadel, and the party ends with a group photo. I manipulated my save file to show the guests I’d have on my game at home. When that last screen came up with that group photo, I got a little choked up and a little teary eyed, because I realized, ‘Well, this is it.’”
Weekes has a different memory of the party. “I’d always told fans that the one thing we’d never do was a cocktail party, because how could you handle that much conversation? How do you let the players feel like they’re chatting with people and moving around in a natural way with so many possible permutations? And then on Citadel, our lead, Mac Walters, said, ‘I think we need to do a party,’ and sure enough, he and Kris and some of the other writers actually came up with a structure that made it work. It’s just one more time that people on the team proved me wrong and did the impossible to make something really special.”
Though that party in Citadel may mark the end of Shepard’s story, it’s far from the end of Mass Effect. As the fourth title in the ME universe begins its development, the team is carrying forward all the things that made the Shepard’s trilogy so memorable.
“It’s the idea of exploring a vast universe: going out and seeing amazing new things. It’s scale: seeing new planets, new species, and having choices that matter. Because it’s a story, and one that you care about,” Hudson says. “It’s going to continue on, and the things people love about Mass Effect they’ll see even better in the next generation of games.” |
Background
==========
cGMP as a second messenger regulates cardiac contractility and might protect the heart from hypertrophy and failure by acting in distinct subcellular microdomains. However, direct visualization of cGMP in subcellular microdomains of adult cardiomyocytes has been challenging. Little is known also about changes in cardiomyocyte cGMP signalling at an early stage of the disease.
Methods
=======
We used a highly sensitive cytosolic and membrane-targeted Förster resonance energy transfer (FRET)-based biosensors for cGMP and cAMP for real time measurements in freshly isolated sensor-trasngenic adult ventricular cardiomyocytes. Combined with single cell contractility measurements, biochemical techniques and whole-heart recording, the effect of atrial natriuretic peptide (ANP) on cardiomyocyte cGMP/cAMP and contractility in healthy and hypertrophied hearts (after transverse aortic constriction).
Results
=======
Contractility measurements in hypertrophied hearts have unravelled ANP-induced augmentation of catecholamine stimulated increase in force and frequency of contraction which was present only in diseased hearts at the state of early compensated cardiac hypertrophy. Interestingly, this effect was not due to changes in cGMP content, membrane receptor densities or whole-cells phosphodiesterase (PDE) activity. Instead, physical redistribution of the cGMP-stimulated PDE2 and cGMP-inhibited PDE3 between distinct membrane domains led to a change of cAMP compartmentation towards an ANP/cGMP/PDE3-dependent increase of local cAMP levels in a microdomain regulating cardiac contractility \[[@B1]\]. FRET-based cGMP measurements revealed relatively small increase of cGMP upon ANP stimulation \[[@B2]\], stringent compartmentation of ANP/cGMP signals to T-tubular membranes and apparent absence of receptor desensitization in early cardiac hypertrophy \[[@B1]\].
Conclusion
==========
ANP/cGMP signalling can play distinct roles in cardiac disease, including a previously unrecognized contractility augmentation in early hypertrophy which might support heart function upon pressure overload. This can be achieved by PDE2/3 redistribution-dependent changes in cGMP/cAMP compartmentation.
|
News, observations and reader questions about the Sacramento Kings and the NBA.
July 30, 2008
We know the ending. Ron Artest to the Rockets for Bobby Jackson, Donte Greene and a No. 1, with minor pieces to be added for financial purposes, a deal that will become official Aug. 14 when the calendar cooperates and the holdup of a salary-cap technicality expires.
In retrospect, Houston was a very logical Artest destination. I had thought so for the last week or so, but could not nail anything down. The Rockets were one of the few teams that fit the unique set of circumstances that would prompt someone to undertake the challenge of the Tour de Ron, so no great surprise.
It makes so much sense in the end. It's like that backwards episode on "Seinfeld." Start with the outcome and fit the pieces together that led everyone to this point and it becomes easy to see the Rockets at the finish line.
One reason they would not have been an obvious fit, and it's a big one: position need. The Rockets were more than covered at small forward with Shane Battier, a much better complementary fit for the Tracy McGrady / Yao Ming foundation. Battier keeps the ball moving, does not disrupt, and last season was easily rated a better defender.
Otherwise, a match.
The standings. The team that takes on Artest has to be a team pushing to make something happen now. The risk is too great. A young club ramping up for 2010 or '11, a fringe playoff possibility, an owner content to float around No. 8 in either conference and make money -- they don't give up a first-round pick this year (Greene) and next for a guy who could bolt in 11 months as a free agent and is wildly unpredictable no matter the contract status.
The Warriors were a potential exception because Chris Mullin loves Artest and Artest could have played power forward in the Don Nelson smallball world. Maybe a team in the Eastern Conference, knowing anything's possible there. Mostly, though, the Kings' trade partner had to need to push the envelope.
The Rockets won 55 games... and finished third in the Southwest Division and tied for fourth in the Western Conference. Yao played just 55 games and McGrady 66, so Houston would have been better in 2008-09 just by being healthy, but so would the Lakers with Andrew Bynum back and the Hornets with the valuable postseason experience and the addition of free agent James Posey. This was no time for the Rockets to feel content.
The price. Again, no playoff team was going to rip a chunk from its rotation and no playoff wannabe was going to mortgage the future for a free agent-in-waiting with his history. The Rockets spent one player who was on the team last season, and Jackson was there for 26 games, mostly as a reserve. Greene is 20 years old and the No. 28 selection in the draft after one season at Syracuse. He could have spent 2008-09 near the end of the bench. The pick next June projects to be somewhere in the 20s.
The atmosphere. It's one of the reasons the Lakers made so much sense, because of Kobe Bryant and Phil Jackson on board as strong personalities to keep Artest in check, or close to it. The Rockets don't have that. But the Rockets do have someone with a history of reaching Artest.
Rick Adelman coached Artest for all of three-plus months with the Kings in 2006, after Artest came from Indiana in the Peja Stojakovic deal, but it was a good short-term match. Adelman has good feelings for Artest and Artest will arrive in Houston with good feelings for Adelman. There's every chance that will change, but it's a good start, at least.
The Kings. They were not going to do a deal just to get away from Artest and they were not going to take back any more bad contracts. Longer contracts for good young players, yes. Longer, bloated contracts for average old players, no. Given the choice between the latter and holding on to Artest the entire season, they would have held and considered his departure as a free agent in the summer of 2009 as the gain.
The Rockets could offer the proper dollar considerations and the desired Sacramento roster infusion. Jackson is 35 and on the books for $6.49 million, but an expiring contract. No great financial commitment there. Greene is hope for the future. The No. 1 in '09 is hope for the future. |
shopt -s expand_aliases
# from an austin-group report
alias foo="echo 'Error:"
foo bar'
# from some FreeBSD sh tests
v=1
alias a='unalias -a
v=2'
eval a
[ "$v" = 2 ] && echo ok 1
v=1
alias a='unalias a
v=2'
eval a
[ "$v" = 2 ] && echo ok 2
# make sure command doesn't ever reset anything even if it's made a keyword
unalias -a
alias command=command
alias true='echo bad'
eval 'command true'
unalias -a
alias alias0=command
alias true='echo bad'
eval 'alias0 true'
# make sure null aliases are ok
unalias -a
alias nullalias=''
alias foo='echo '
foo nullalias text
unalias foo
# aliases shouldn't be expanded in quoted strings even when the previous word
# is an alias whose expansion ends in a space
alias foo="echo 'whoops: "
foo nullalias'
unalias -a
# recursive alias definitions
alias echo=echo
eval echo foo
alias echo='echo a'
echo
echo b
eval echo b
echo $(eval echo b)
unalias -a
# alias expansion when in a command position after redirections
alias e=echo
eval '</dev/null e ok 3'
eval 'a=true e ok 4'
alias comment=#
comment
alias long_comment='# for x in '
long_comment text after
# comment
comment foo bar
|
easyblock = "PythonPackage"
name = 'h5py'
version = '2.5.0'
homepage = 'http://www.h5py.org/'
description = """HDF5 for Python (h5py) is a general-purpose Python interface to the Hierarchical Data Format library,
version 5. HDF5 is a versatile, mature scientific software library designed for the fast, flexible storage of enormous
amounts of data."""
toolchain = {'name': 'intel', 'version': '2016a'}
toolchainopts = {'usempi': True}
source_urls = [PYPI_SOURCE]
sources = [SOURCE_TAR_GZ]
local_hdf5ver = '1.8.16'
versionsuffix = '-Python-%%(pyver)s-HDF5-%s' % local_hdf5ver
# to really use mpi enabled hdf5 we now seem to need a configure step
prebuildopts = ' python setup.py configure --mpi --hdf5=$EBROOTHDF5 && '
dependencies = [
('Python', '2.7.11'),
('HDF5', local_hdf5ver),
('pkgconfig', '1.1.0', '-Python-%(pyver)s'),
]
sanity_check_paths = {
'files': [],
'dirs': ['lib/python%(pyshortver)s/site-packages/'],
}
moduleclass = 'data'
|
Mr. Derrick will be working in London July 17, 1999 through August 2, 1999.
Mr. Derrick will be back in the office August 4, 1999.
July 18, 1999 through July 29, 1999
47 Park Street Hotel - Mayfair
47 Park Street
London, WIY 4EB
Telephone: 011-44-171-491-7282
Fax: 011-44-171-491-7281
After July 29, 1999, Mr. Derrick may be traveling to other offices in Europe
to be determined at a later time. You may want to be sure to get your things
signed while he is at the 47 Park Street Hotel. As always, please don't fax
things to him unless they absolutely cannot wait until his return or are of
an urgent nature. |
Description
The Callaway XR OS Women's graphite individual iron features a wide sole design making this club easy to hit and more forgiving. With the multi-piece construction you'll get more ball speed and better feel. Add this SW to complete your existing Callaway XR OS set.
Features include:
Wider sole design for more forgiveness and more distance from any impact location. |
Fri December 23, 2016 - 9:30 AM
Phoenix, AZ, 2016
401 E. Jefferson
Visit one of the most impressive and unique Major League Baseball facilities in the country -- Chase Field, home of the 2001 World Champion Arizona Diamondbacks and the 2011 MLB All-Star Game. Chase Field has several unique...See More
Visit one of the most impressive and unique Major League Baseball facilities in the country -- Chase Field, home of the 2001 World Champion Arizona Diamondbacks and the 2011 MLB All-Star Game. Chase Field has several unique features including dbTV, one of the largest high definition video scoreboards in MLB, a retractable roof to keep fans cool during the summer months, the RamTrucks.com Pool in right field, the Coors Light Strike Zone, and several restaurants including Friday's Front Row and The Draft Room. Make the experience memorable and take a guided tour of Chase Field. Even if you've been to the ballpark before, you're sure to learn and see something new! Tours are conducted year-round, for both individuals and groups. Tour space is limited -- it is strongly recommended that you purchase tickets in advance to guarantee your spot! See Less |
Q:
How do you profile your web site/web application?
This is to collect from the experience that the community has done information on the aspect of Profiling web Application.
Some years ago I worked at a very large project in C++/Java with the a CORBA ORB and we were using Rational Purify/CodeCoverage to instrument, detect memory leaks and discover bottlenecks on server code. From that time I did not have any experience on using tools like that on the .NET platform either working on pure c# or with a web application
Do you use tools?
Do you estimate traffic and do calculations on the expected bandwidth needed?
Do you profile differently server code and web page rendering?
What code coverage tool are you using?
I know this is a very big topic. Some information I have are from the book "Performance Analysis for Java WebSites", who is fo the Java platform and reference tools for that platform but ises an approach that is transferable and so the core ideas apply generally.
A:
As a free load testing solution I have used Pylot. I am sure there are better paid solutions if you have a budget.
If you can estimate traffic, this is the tool whose output you evaluate the scalabiltiy of your project against. Using the the asp.net output cache can improve your site performance under load significantly, so try this if your page views a second are less than you require.
For optimising your client side rendering speed use:
YSlow firefox plug-in
PageSpeed firefox plug-in by Google
Firebug firefox plug-in to check the number of HTTP requests are not excessive
and js/css resources are being cached etc.
If developing a asp.net web forms app you can enable page tracing by modifying your page directive so it contains
<%@ Page Trace="true">
This will help you find controls which take longer to render.
If you have an issue with server side code being slow I have found it is almost always the database causing the issue. You need to check for SQL which is slow to return a result; if you find any you need to look at applying new indexes to your tables. If your app is too chatty with the database you need to look at reducing the number of calls to the database. To find these problems you can use SQL Server Profiler; this comes bundled with SQL Server 2005/2008 Developer edition.
If you have the budget, you definitely want to check of out Redgate ANTS Performance Profiler for profiling your server side code.
|
BORIS Expression in Ovarian Cancer Precursor Cells Alters the CTCF Cistrome and Enhances Invasiveness through GALNT14.
High-grade serous carcinoma (HGSC) is the most aggressive and predominant form of epithelial ovarian cancer and the leading cause of gynecologic cancer-related death. We have previously shown that CTCFL (also known as BORIS, Brother of the Regulator of Imprinted Sites) is expressed in most ovarian cancers, and is associated with global and promoter-specific DNA hypomethylation, advanced tumor stage, and poor prognosis. To explore its role in HGSC, we expressed BORIS in human fallopian tube secretory epithelial cells (FTSEC), the presumptive cells of origin for HGSC. BORIS-expressing cells exhibited increased motility and invasion, and BORIS expression was associated with alterations in several cancer-associated gene expression networks, including fatty acid metabolism, TNF signaling, cell migration, and ECM-receptor interactions. Importantly, GALNT14, a glycosyltransferase gene implicated in cancer cell migration and invasion, was highly induced by BORIS, and GALNT14 knockdown significantly abrogated BORIS-induced cell motility and invasion. In addition, in silico analyses provided evidence for BORIS and GALNT14 coexpression in several cancers. Finally, ChIP-seq demonstrated that expression of BORIS was associated with de novo and enhanced binding of CTCF at hundreds of loci, many of which correlated with activation of transcription at target genes, including GALNT14. Taken together, our data indicate that BORIS may promote cell motility and invasion in HGSC via upregulation of GALNT14, and suggests BORIS as a potential therapeutic target in this malignancy. IMPLICATIONS: These studies provide evidence that aberrant expression of BORIS may play a role in the progression to HGSC by enhancing the migratory and invasive properties of FTSEC. |
What if you redesigned Darth Vader, but you only had a few sentences of description from the script to go on? Artist Carlos Villagra (and a few others) tried. Here's what they dreamed the Dark Lord could have looked like.
Here's the little bit of script (taken from A New Hope) that the artists had to go on:
INTERIOR: REBEL BLOCKADE RUNNER - MAIN HALLWAY. The awesome, seven-foot-tall Dark Lord of the Sith makes his way into the blinding light of the main passageway. This is Darth Vader, right hand of the Emperor. His face is obscured by his flowing black robes and grotesque breath mask, which stands out next to the fascist white armored suits of the Imperial stormtroopers. Everyone instinctively backs away from the imposing warrior and a deathly quiet sweeps through the Rebel troops. Several of the Rebel troops break and run in a frenzied panic.
And the first design, by Carlos Villagra.
By Jorge Lacera.
By James Groman.
By Hydro74.
[More designs at The Swedish Bed via The Daily What] |
I[NTRODUCTION]{.smallcaps} {#sec1-1}
==========================
Improved clinical outcomes for patients with acute emergent large vessel thromboembolic cerebrovascular accidents (CVA) at high-volume centers have been demonstrated. Prolonged transfer times to such institution are associated with less favorable outcomes. Direct access to smaller thrombectomy-capable low-volume centers is in most cases the only immediate viable revascularization option. The development of primary stroke centers and intravenous tissue plasminogen activator (IV tPA) administration protocols have established clear metrics for treatment with improved clinical outcomes' event at these smaller centers.\[[@ref1]\] The efficacy of endovascular reperfusion in emergent large vessel occlusion has been verified by five landmark studies. Rapid complete revascularization with minimizing risk is a prerequisite for favorable clinical outcomes.\[[@ref2][@ref3]\] IV tPA administration before endovascular intervention at thrombectomy-capable low-volume centers and how this affects procedural aspects and patient outcomes have not been investigated. This study was performed to determine if patients who qualify for endovascular intervention benefit from prior IV tPA administration at low-volume thrombectomy-proficient centers.
M[ETHODS]{.smallcaps} {#sec1-2}
=====================
This was an Institutional Review Board-approved study with waived individual consent. Retrospective chart review of all consecutive CVA patients treated with endovascular therapy at two select rural primary stroke centers between 2011 and 2015 was performed. Inclusion criteria were all anterior circulation large vessel occlusions within an 8 h time window from symptom onset with subsequent endovascular intervention.\[[@ref4]\] Patients' data regarding age, sex, and medical history, as well as thrombus location by catheter-based cerebral angiography and postprocedural reperfusion status were reviewed. Computed tomography (CT) head to procedure commencement time was determined from the time of CT acquisition recorded to the time documented in the chart as the time of procedure commencement. Total procedural times were obtained from the chart as noted by the time the procedure commenced to the time the procedure was considered complete, inclusive of anesthesia time and groin arteriotomy closure device placement. Revascularization was performed with a mechanical stent retriever, aspiration or a combination of local aspiration, and mechanical thrombectomy. Successful recanalization was defined as modified thrombolysis in cerebral infarction (TICI) score of 2b or 3. The initial NIHSS and the 90-day functional outcome were assessed using the modified Rankin scale (MRS) at a follow-up neurology visit. The primary outcome measure of the study was a comparison of MRS at 90 days in patients with IV tPA administration before thrombectomy versus those who did not qualify for IV tPA administration. A Modified Rankin Score of ≤2 at 90 days was considered a good functional outcome. All data points above were stratified into IV tPA versus no-IV tPA for comparison.
Statistical analysis {#sec2-1}
--------------------
The data were analyzed using the social science statistics web-based calculator. The Chi-square test and the Fisher\'s exact test were used to compare categorical variables between the non-IV tPA and the IV tPA groups, and the one-way ANOVA was used to compare distributions of continuous variables. *P* \< 0.05 was considered statistically significant.
R[ESULTS]{.smallcaps} {#sec1-3}
=====================
Demographics {#sec2-2}
------------
A total of 65 patients were initially assessed for inclusion in this study through retrospective chart review. After application of the set inclusion and exclusion criteria, 46 remained. Twenty-three patients (49%) received IV tPA before thrombectomy and 23 patients did not qualify for IV tPA (51%). There was no statistical difference in medical comorbidities (atrial fibrillation, carotid stenosis, hypertension, diabetes, or hyperlipidemia) between the two groups. Average age was 69.2 (±14.7) years in the preprocedural IV tPA group and 71.1 (±16.2) years in the no-IV tPA group.
Intervention {#sec2-3}
------------
Thromboembolic occlusion at the carotid terminus was identified in 26% (6/23) of patients in both groups. Middle cerebral artery (MCA) M1 occlusion was noted in 47.8% (11/23) of patients in no-IV tPA group and 43.4% (10/23) patients in the IV tPA group. MCA M2 and other small branch occlusions in the no-IV tPA group constituted 21% (5/23) and 30% (7/23) in the group that received IV tPA. TICI 2b/3 recanalization was achieved 86% (20/23 patients) in the no-IV tPA arm and 82% (19/23 patients) in the IV tPA arm (odds ratio \[OR\]: 0.03, confidence interval \[CI\] 95%: 0.062--0.16, *P* \< 0.0001). Average CT to procedure commencement time was 126.6 min (±43.01) in the no-IV tPA group and 142.2 (±42.87) min in the IV tPA group (*P* = 0.2). Procedural duration for the no-IV tPA group was 126.6 min (±43.01) and in the IV tPA group was 121.8 (±42.87) (*P* = 0.8). Mechanical thrombectomy alone was performed in 43% (10/23) of patients in the non-IV tPA group and in 56% (13/23) of patients in the IV tPA group. Aspiration with combined mechanical thrombus extraction was performed in 39% (9/23) of patients in the non-IV tPA group and in 22% (5/23) of patients in the IV tPA group. The remainder in each category was aspiration alone.
Clinical {#sec2-4}
--------
Average NIHSS was 17.6 in the IV tPA group and 16.6 in the no-IV tPA group (*P* = 0.7). MRS of 2 or less at 90 days was 10/23 patients (43.4%) in the non-IV tPA arm and 9/23 (39.1%) in the IV tPA group (OR: 0.84, CI: 0.26--2.70, *P* = 0.8). The mean MRS score for thrombectomy without tPA was 3.6 and with IV tPA was 3.3. Intracranial hemorrhage on CT was identified in 1/23 (4.3%) of patients in the no-IV tPA group and in 5/23 (21%) of patients in the IV tPA group (OR: 0.16, CI: 0.017--1.5, *P* = 0.07). Symptomatic intracranial hemorrhage with a fatal outcome was present in 1/23 (4.3%) of patients in the IV tPA group and none in the non-IV tPA group.
D[ISCUSSION]{.smallcaps} {#sec1-4}
========================
Patients undergoing endovascular thrombectomy for large vessel occlusion at select low-volume rural centers in our study showed benefit from this treatment regardless of IV tPA administration. The benefit of prethrombectomy IV tPA in improving clinical outcomes has been controversial.\[[@ref5][@ref6]\] A prospective study identified a dramatic improved functional outcome of 51.5% with prethrombectomy IV tPA administration versus 18.2% for endovascular intervention alone.\[[@ref7]\] A meta-analysis of five studies comparable to our study period identified 37.2% good functional outcomes with thrombectomy alone and 49.2% with combined therapy.\[[@ref8]\] Functional outcome in the thrombectomy alone group in our study was 43.4% and 39.1% with combined therapy, which was not statistically significant. In comparison to studies at a similar time period,\[[@ref2][@ref3][@ref9]\] reperfusion rates and clinical outcomes of our data are comparable to most comprehensive stroke centers with large volumes as well as the landmark studies proving the efficacy of endovascular treatment.
IV tPA theoretically facilitates thrombus dissolution and may address more distal or procedural new territory small vessel thromboembolism. This may translate to increased recanalization rates and decreased procedural times. In our study, the overall time from CT scan to procedure start in each group was not statistically significant. Administration of IV tPA requires clinical decision-making and drug preparation that delays noncontrast head CT to procedure commencement time. In our study, an average delay of 15.6 min was identified with IV tPA administration that was not statistically significant. At low-volume centers, prolonged symptom onset time to endovascular reperfusion has been identified, without published data identifying the root causes.\[[@ref10]\] Potential reasons for workflow inefficiency include delays in considering endovascular therapy, staffing with nondedicated endovascular neurointerventional call teams, the use and availability of general anesthesia, operator experience, and lack of dedicated neurocritical care. Given that the workflow times may be longer at low-volume centers, the addition of IV tPA bridging to mitigate this potential delay did not appear to be advantageous in our study. Our data identified increased symptomatic and nonsymptomatic intracranial hemorrhages in the combined therapy group, which was not statistically significant. The only intracranial hemorrhage-related mortality postthrombectomy was in the thrombectomy with IV tPA group. This is comparable to data from a published study of 1275 patients.\[[@ref8]\]
Comprehensive high-volume center data have shown a superiority in measured efficiency for times to recanalization and outcomes when compared to low-volume centers.\[[@ref10]\] Access to comprehensive centers may not always be a feasible option. Seventy percent of high-volume centers are more likely to be urban teaching hospitals.\[[@ref11]\] For multiple centers' concentrated large urban environments, it is a plausible premise to direct patient care to the highest volume center. In nonurban rural areas, often further than an hour travel time from a comprehensive center, a thrombectomy-capable primary care center is a more viable alternative. Theoretically, only 56% of the US population have access to endovascular-capable hospital by road in 1 h.\[[@ref12]\] For every 1 h delay in reperfusion, the odds of good clinical outcome decreases by 38%.\[[@ref13]\] A prolonged symptom onset to reperfusion time has been shown to be an independent limiting factor for favorable outcomes.\[[@ref14]\] Transfer delay is a major factor limiting the use of intra-arterial treatment in acute ischemic stroke and significantly limits favorable outcomes.\[[@ref9][@ref15]\] One in four patients becomes ineligible for endovascular thrombectomy during transfer with a trend toward poorer outcomes once transferred.\[[@ref16]\] The mortality rate is significantly lower in directly admitted patients as compared to transferred patients.\[[@ref17]\] Primary care centers capable of thrombectomy, with established and sustained treatment time proficiency protocols, will reduce the number of patients requiring transfer and reduce symptom onset to reperfusion times with resultant improved functional outcomes.
A larger proportion of ischemic stroke patients will be treated at low-volume centers. The incidence of large vessel occlusion has been determined to be 24 per 100,000 person-years and, the most recently estimated annual thrombectomy rate in 2015 of three procedures per 100,000 people indicates that there will be a significant increase in the volume of endovascular procedures.\[[@ref18]\] The mean number of thrombectomy cases per institution was 19.3 cases per year in an outcome report database. Nearly 89% of institutions reporting to this database were university hospitals.\[[@ref17]\] This falls well short of the 38 thrombectomy cases recommended for thrombectomy-capable centers and far short of the 50 cases suggested as a designate high-volume center.\[[@ref10][@ref19]\] Almost a decade ago, only 0.4%--2.6% of hospitals met the procedural volume recommended by various professional bodies for endovascular thrombectomy.\[[@ref11]\] In spite of less favorable outcomes with carotid stenting, a lower proportion of cases are currently treated at high-volume centers.\[[@ref20]\] In the state of New Jersey as an example, 60% of stroke admissions were at a primary stroke center as compared to 40% at a comprehensive center.\[[@ref21]\] As with carotid stenting, an increase in procedures for large vessel occlusions in thrombectomy-capable low-volume centers is probable with the anticipated increase in procedural volume.
The study is limited by the nonrandomized retrospective nature and the small number of patients in each cohort. The results of select rural centers that have clinical neuroscience services may not reflect outcomes at all rural-based hospitals. The findings of this study may be reflective of other rural institutions that have available resources including appropriate equipment as well as skilled technical and physician staff that will and can promote aspects of care for better functional outcomes. These centers should rather be considered as thrombectomy-proficient centers.
C[ONCLUSION]{.smallcaps} {#sec1-5}
========================
Optimizing care at thrombectomy-proficient low-volume institutions with high-volume center partnership and collaboration is a more tangible goal than promoting transfers out of primary stroke centers as the only viable option for favorable outcomes. Empowering these rural-based thrombectomy-proficient centers with preparedness tools, fostering a vigilant mindset, and frequent staff and physician training to focus on efforts directed at achieving faster recanalization times may allow for continued improvement in patient outcomes that will remain consistent with national data.\[[@ref22]\] Further studies are needed to identify the patient population that would benefit from endovascular intervention without tPA administration when compared to the currently favored model of combined therapy.
Financial support and sponsorship {#sec2-5}
---------------------------------
Nil.
Conflicts of interest {#sec2-6}
---------------------
There are no conflicts of interest.
|
Mapping a marsupial X chromosome using kangaroo-mouse somatic cell hybrids.
A series of marsupial-eutherian somatic cell hybrids was produced by fusion between lymphocytes from the red kangaroo (Macropus rufus) and HPRT-deficient mouse cells. The hybrids lost marsupial chromosomes and could therefore be used to map marsupial genes. Several of the hybrids contained a complete red kangaroo X chromosome and expressed the kangaroo form of the enzymes HPRT, G6PD, and PGKA. A number of HPRT-deficient revertant cell lines were derived from the hybrids. These possessed a variety of partially deleted X chromosomes. With these cell lines, it has been possible to establish the X-linkage of the genes for HPRT, G6PD, and PGKA in this marsupial and to localize these three genes to the terminal portion of the euchromatic arm of the red kangaroo X chromosome. |
Royal Bank of Scotland Stock Gains as it Swings into Profit
Royal Bank of Scotland (RBS) shares were up in the first minutes of trading after the U.K. lender reported it had swung into profit in the third quarter and expects to return to full-year profit in 2018.
RBS shares were up 2.17%, extending a three month gain of 14%, to change hands at 287.12 pence.
The majority state-owned bank reported an attributable profit of £392 million in the three months to the end of September, compared with a £469 million reported in the same time period last year and beating analyst forecast of £310 million. This brings the bank's total attributable profit for the year to £1.33 billion.
RBS, with is 71% owned by the U.K. government after a bailout during the financial crisis, said the results were buoyed by lower costs. The bank has cut £708 million in costs since the beginning of 2017.
The bank said that it was on track to deliver its 2017 financial targets. RBS's stated ambition for the year is to grow income, cut costs and use less capital across its core businesses and to make progress on resolving legacy issues. Adjusted income has increased by 5.6% in in the third quarter compared with the same quarter in 2016 and has increased by 7.5% for the year to date.
The bank has been distracted attempting to resolve legacy issues. It still has an impending settlement with the Department of Justice over misselling of mortgage-backed securities. The bank gave no update of the DoJ settlement but indicated that further provisions may be taken in 2017.
"We retain the 2017 full year financial guidance and medium term financial outlook we provided in the 2016 Annual Results document. In addition, and subject to any further provisions for the investigations of the US Department of Justice into the Group's historic RMBS related activities being substantially taken in 2017, our expectation remains that we will be profitable in 2018," the bank said. |
Sunday, November 10, 2013
The Arizona Capitol Times had another hard-hitting article on the prosecution of Tom Horne, the liberal Republican Attorney General who has no chance of winning re-election, according to powerful D.C. consultants who have taken polls. Prosecutor Sheila Polk is not the only one who has found wrongdoing by Horne, Maricopa County Attorney Bill Montgomery has also found reason to civilly prosecute Horne. This is why conservatives and Tea Party activists are supporting Mark Brnovich instead. Brnovich has been endorsed by Montgomery as well as Rep. Trent Franks.
Here are some excerpts from the article by Jeremy Duda -
The
crux of Polk’s case is a series of phone calls between Horne and Winn
that occurred while Winn was communicating via email with Brian Murray,
BLA’s political consultant. While Polk and Montgomery have presented
the calls and emails as evidence of coordination, Horne and Winn say
prosecutors can’t prove they were talking about the attorney general’s
race or BLA activities. They say the calls were about Winn assisting
Horne with a real estate deal.
As evidence of
coordination, Polk cited the timing of Winn’s phone calls with Horne and
her emails with BLA’s consultant. In some cases, the calls closely
followed Murray’s emails to Winn about the BLA ad, and
Winn’s responses to her consultant came just after her conversations
with Horne ended. Horne also forwarded Winn an email several days later
in which one of his campaign staffers discussed Horne’s campaign
strategy, and urged her to raise more money from a
national Republican group. And Polk’s case is bolstered by the fact
that Winn worked on Horne’s campaign, and didn’t leave to work on her IE
until a few weeks after the GOP primary....Several
elections attorneys emphasized that Polk will have to meet a relatively
low standard of proof. While prosecutors in criminal cases must
convince juries that a defendant is guilty beyond a reasonable
doubt, civil cases such as Horne and Winn’s are held to a lower
standard called “preponderance of evidence,” meaning that Polk only has
to show that a violation was more likely than not.--In
her order, Polk said the evidence against Horne and Winn is compelling.
She said the phone conversations between them while Winn and Murray
discussed the content of the ad via email shows that they coordinated
their efforts. Winn’s comments to Murray about how “we” wanted certain
changes to the ad show she was consulting with someone else, and Polk
insisted that person must be Horne.
“Winn almost always consulted with Horne prior to instructing Murray,” Polk wrote in her Oct. 17 order....“When
Horne sent strategic information to a supposedly independent campaign,
he intentionally and blatantly broke the barrier that was supposed to
exist between his campaign and BLA. The breach is so clear that
Horne must have recognized it to be improper,” Polk wrote....
Polk
ordered Horne and Winn to repay about $400,000 that the prosecutor
alleges was illegally contributed to BLA, and they could face a fine of
up to $1.2 million if they lose the case. |
---
abstract: |
Convex rank tests are partitions of the symmetric group which have desirable geometric properties. The statistical tests defined by such partitions involve counting all permutations in the equivalence classes. Each class consists of the linear extensions of a partially ordered set specified by data. Our methods refine existing rank tests of non-parametric statistics, such as the sign test and the runs test, and are useful for exploratory analysis of ordinal data. We establish a bijection between convex rank tests and probabilistic conditional independence structures known as semigraphoids. The subclass of submodular rank tests is derived from faces of the cone of submodular functions, or from Minkowski summands of the permutohedron. We enumerate all small instances of such rank tests. Of particular interest are graphical tests, which correspond to both graphical models and to graph associahedra.\
[**Keywords:**]{} braid arrangement, graphical model, permutohedron, polyhedral fan, rank test, semigraphoid, submodular function, symmetric group.
author:
- 'Jason Morton, Lior Pachter, Anne Shiu, Bernd Sturmfels, and Oliver Wienand'
title: Convex Rank Tests and Semigraphoids
---
Introduction
============
The non-parametric approach to statistics was introduced by [@Pitman1937SignificanceI] via the method of permutation testing. Subsequent development of these ideas revealed a close connection between non-parametric tests and [*rank tests*]{}, which are statistical tests suitable for ordinal data. Beginning in the 1950s, many rank tests were developed for specific applications, such as the comparison of populations or testing hypotheses for determining the location of a population. The geometry of these tests was explored in [@Cook]. More recently, the search for patterns in large datasets has spurred the development and exploration of new tests. For instance, the emergence of microarray data in molecular biology has led to tests for identifying significant patterns in gene expression time series; see e.g. [@Willbrand2005]. This application motivated us to develop a mathematical theory of rank tests. We propose that a [*rank test*]{} is a partition of $S_n$ induced by a map $\, \tau : S_n \rightarrow T\,$ from the symmetric group of all permutations of $[n]=\{1,\ldots,n\}$ onto a set $T$ of statistics. The statistic $\tau(\pi)$ is the [*signature*]{} of the permutation $\pi \in S_n$. Each rank test defines a partition of $S_n$ into classes, where $\pi$ and $\pi'$ are in the same class if and only if $\tau (\pi) = \tau(\pi')$. We identify $T = {\rm image}(\tau)$ with the set of all classes in this partition of $S_n$. Assuming the uniform distribution on $ S_n$, the probability of seeing a particular signature $t \in T$ is $\,1/n! \,$ times $| \tau^{-1}( t)|$. The computation of a $p$-value for a given permutation $\pi \in S_n$ leads to the problem of summing $$\label{Pvalue}
{\rm Pr}(\pi') \quad = \quad
\frac{1}{n !} \cdot |\, \tau^{-1} \bigl( \tau(\pi') \bigr)\, |$$ over permutations $\pi'$ with ${\rm Pr}(\pi') \leq {\rm Pr}(\pi)$, a computational task to be addressed in Section 6.
This paper is an expanded version of our note “Geometry of Rank Tests” which was presented in September 2006 in Prague at the conference [*Probabilistic Graphical Models (PGM 3)*]{}. The emphasis of our discussion is on the mathematics underlying rank tests, and, in particular, on the connection to statistical learning theory (semigraphoids). We refer to [@cyclohedron] for details on how to use our rank tests in practice, and how to interpret the p-values derived from (\[Pvalue\]).
The five subsequent sections are organized as follows. In Section 2 we explain how existing rank tests in non-parametric statistics can be understood from our geometric point of view, and how they are described in the language of algebraic combinatorics [@Stanley1997]. In Section 3 we define the class of [*convex rank tests*]{}. These tests are most natural from both the statistical and the combinatorial point of view. Convex rank tests can be defined as polyhedral fans that coarsen the hyperplane arrangement of $S_n$. Our main result (Theorem \[fantheorem\]) states that convex rank tests are in bijection with conditional independence structures known as [*semigraphoids*]{} [@Dawid; @Pearl; @Studeny2005Probabilistic].
Section 4 is devoted to convex rank tests that are induced by submodular functions. These [*submodular rank tests*]{} are in bijection with Minkowski summands of the $(n{-}1)$-dimensional permutohedron and with [*structural imset models*]{}. These tests are at a suitable level of generality for the biological applications [@cyclohedron; @Willbrand2005] that motivated us. The connection between polytopes and independence models is made concrete in the classification of small models in Remarks \[rmk1\]–\[rmk3\].
In Section 5 we study the subclass of [*graphical tests*]{}. In combinatorics, these correspond to graph associahedra, and in statistics to graphical models. The equivalence of these two structures is shown in Theorem \[maingraphical\]. The implementation of convex rank tests requires the efficient enumeration of linear extensions of partially ordered sets. Our algorithms and software are discussed in Section 6. A key ingredient is the efficient computation of distributive lattices.
Acknowledgments {#acknowledgments .unnumbered}
===============
Our research on rank tests originated in discussions with Olivier Pourquié and Mary-Lee Dequéant as part of the DARPA Program [*Fundamental Laws of Biology*]{}, that supported Jason Morton, Lior Pachter, and Bernd Sturmfels. Anne Shiu was supported by a Lucent Technologies Bell Labs Graduate Research Fellowship, and Oliver Wienand by the Wipprecht foundation. We thank Milan Studený and František Matúš for helpful comments.
Rank tests and posets
=====================
A permutation $\pi$ in $S_n$ is a total order on the set $[n] := \{1,\ldots,n\}$. This means that $\pi$ is a set of $\binom{n}{2}$ ordered pairs of elements in $[n]$. For example, $\pi = \{ (1,2), (2,3), (1,3) \}$ represents the total order $1>2>3$. If $\pi$ and $\pi'$ are permutations then $\,\pi \cap \pi'\,$ is a partial order.
In the applications we have in mind, the data are vectors $u \in {\mathbb{R}}^n$ with distinct coordinates. The permutation associated with $u$ is the total order $\,\pi = \{ \,(i,j)\in [n] \times [n] \,: \, u_i < u_j\,\}$. We shall employ two other ways of writing a permutation. The first is the [*rank vector*]{} $\,\rho = (\rho_1,\ldots,\rho_n)$, whose defining properties are $\{\rho_1,\ldots,\rho_n\} = [n]$ and $\rho_i < \rho_j$ if and only if $u_i < u_j$. That is, the coordinate of the rank vector with value $i$ is at the same position as the $i$th smallest coordinate of $u$. The second is the [*descent vector*]{} $\delta = (\delta_1 | \delta_2 | \ldots | \delta_n)$. The descent vector is defined by $u_{\delta_i} > u_{\delta_{i+1}}$ for $i=1,2,\ldots, n {-} 1$. Thus the $i$th coordinate of the descent vector is the position of the $i$th largest value of the data vector $u$. For example, if $\,u = (11,7,13)\,$ then its permutation is represented by $\, \pi = \{ (2,1),(1,3),(2,3)\}$, by $\,\rho = (2,1,3)$, or by $\,\delta = (3|1|2)$.
A permutation $\pi $ is a [*linear extension*]{} of a partial order $P$ on $[n]$ if $P \subseteq \pi$, i.e. $\pi$ is a total order that refines the partial order $P$. We write $\mathcal{L}(P) \subseteq S_n$ for the set of linear extensions of $P$. A partition $\tau$ of the symmetric group $S_n$ is a [*pre-convex rank test*]{} if the following axiom holds: $$\begin{array}{ccc}
{\rm (PC)} && \begin{array}{c} \text{If }\tau(\pi) = \tau(\pi') \text{ and } \pi'' \in \mathcal{L} (\pi \cap \pi') \text{ then } \tau(\pi) \! = \! \tau(\pi') \!= \!\tau(\pi''). \end{array}\end{array}$$ Note that $\, \pi'' \in \mathcal{L} (\pi \cap \pi') \,\,$ means $\pi \cap \pi' \subseteq \pi''$. The number of all rank tests $\tau$ on $[n]$ is the [*Bell number*]{} $B_{n!}$, which is the number of set partitions of a set of cardinality $n!$.
\[bell\] For $n=3$ there are $B_6 = 203$ rank tests, or partitions of the symmetric group $S_3$, which consists of six permutations. Of these $203$ rank tests, only $40$ satisfy the axiom (PC). One example is the pre-convex rank test in Figure 1. Here the symmetric group $S_3$ is partitioned into the four classes $\,\bigl\{ (1|2|3)\bigr\}$, $\,\bigl\{(2|1|3)\bigr\}$, $\,\bigl\{(2|3|1)\bigr\}$, and $\,\bigl\{(1|3|2), (3|1|2), (3|2|1)\bigr\}$.
Each class $C$ of a pre-convex rank test $\tau$ corresponds to a poset $P$ on the ground set $[n]$; namely, the partial order $P$ is the intersection of all total orders in that class: $P=\bigcap_{\pi \in C} \pi$. The axiom (PC) ensures that $C$ coincides with the set $\mathcal{L}(P)$ of all linear extensions of $P$. The inclusion $C \subseteq \mathcal{L}(P)$ is clear. The proof of the reverse inclusion $\mathcal{L}(P) \subseteq C$ is based on the fact that, from any permutation $\pi$ in $ \mathcal{L}(P)$, we can obtain any other $\pi'$ in $ \mathcal{L}(P)$ by a sequence of reversals $(a,b) \mapsto (b,a)$, where each intermediate $\hat{\pi}$ is also in $ \mathcal{L}(P)$. Consider any $\pi_0 \in \mathcal{L}(P)$ and suppose that $\pi_1 \in C$ differs by only one reversal $(a,b)\in \pi_0$, $(b,a) \in \pi_1$. Then $(b,a) \notin P$, so there is some $\pi_2 \in C$ such that $(a,b) \in \pi_2$; thus, $\pi_0\in \mathcal{L} (\pi_1 \cap \pi_2)$ by (PC). This shows $\pi_0 \in C$.
A pre-convex rank test therefore can be characterized by an unordered collection of posets $P_1,P_2,\ldots,P_k$ on $[n]$ that satisfies the property that the symmetric group $S_n$ is the disjoint union of the subsets $\mathcal{L}(P_1),\mathcal{L}(P_2), \ldots, \mathcal{L}(P_k)$. This structure was discovered independently and studied by Postnikov, Reiner and Williams [@PRW §3] who used the term [*complete fan of posets*]{} for what we shall call a convex rank test in Section 3. The posets $\,P_1, P_2, \ldots, P_k\,$ that represent the classes in a pre-convex rank test capture the shapes of data vectors. In graphical rank tests (Section \[sec:graphical\]), this shape can be interpreted as a smoothed topographic map of the data vector.
\[sign\_test\] The *sign test* is performed on data that are paired as two vectors $u=(u_1,u_2, \dots,u_m)$ and $ v = ( v_1, v_2, \dots, v_m)$. The null hypothesis is that the median of the differences $u_i - v_i$ is 0. The test statistic is the number of differences that are positive. This test is a rank test, because $u$ and $v$ can be transformed into the overall ranks of the $n=2m$ values, and the rank vector entries can then be compared. This test coarsens the convex rank test which is the MSS test of Section 4 with $\,\mathcal{K} \,= \,\{\{1,m+1\},\{2, m+2\}, \dots \}$.
A *runs test* can be used when there is a natural ordering on the data points, such as in a time series. The data are transformed into a sequence of ‘pluses’ and ‘minuses,’ and the null hypothesis is that the number of observed runs is no more than that expected by chance. Common types of runs tests include the sequential runs test (‘plus’ if consecutive data points increase, ‘minus’ if they decrease), and the runs test to check randomness of residuals, i.e. deviation from a curve fit to the data. A runs test is a coarsening of a convex rank test, known as [*up-down analysis*]{} [@Willbrand2005 §6.1.1], which is described in Example \[ex.updwn\] below.
\[fig:nonconvexpreconvex\] $$\begin{xy}<15mm,0mm>:
(0,0) ="origin" ;
({0.866025404}, 0.5) ="uprt" ;
({0.866025404}, -0.5) ="dwnrt" ;
(-{0.866025404}, 0.5) ="upl" ;
(-{0.866025404}, -0.5) ="dwnl" ;
(0, 1) ="up" ;
(0,-1) ="dwn" ;
"origin";"uprt" **@{--};
"origin";"dwnrt" **@{--};
"origin";"upl" **@{-};
"origin";"dwnl" **@{-};
"origin";"up" **@{-};
"origin";"dwn" **@{-};
(-.5,{0.866025404}) ="123" *+!DR{1|2|3} ;
(.5 ,{0.866025404}) ="132" *+!DL{1|3|2} ;
(1,0) ="312" *+!L{3|1|2} ;
(.5,-{0.866025404}) ="321" *+!UL{3|2|1} ;
(-.5,-{0.866025404}) ="231" *+!UR{2|3|1} ;
(-1,0) ="213" *+!R{2|1|3} ;
\end{xy}$$
These two examples suggest that many rank tests from classical non-parametric statistics have a natural refinement by a pre-convex rank test. However, not all tests have this property. Because many classical rank tests apply to loosely grouped data (e.g. data which are divided into two samples), the axiom (PC) is not always satisfied. In such cases, the pre-convex rank test is a first step, after which permutations are grouped together under additional symmetries, e.g., the permutations $\,\delta=\, (1|2|3|4|5)\,$ and $\,\delta'=\, (5|4|3|2|1)\,$ might be identified.
The adjective “pre-convex” refers to the following interpretation of the axiom (PC). Consider any two data vectors $u$ and $u'$ in ${\mathbb{R}}^n$, and a convex combination $u'' = \lambda u +
(1-\lambda) u'$, with $0 < \lambda < 1$. If $\,\pi, \pi', \pi'' \,$ are the permutations of $\, u,u',u'' \,$ then $\, \pi'' \in \mathcal{L}(\pi \cap \pi')$. Thus the equivalence classes in ${\mathbb{R}}^n$ specified by a pre-convex rank test are convex cones. In the next section, we shall remove the prefix from “pre-convex” if the faces of these cones fit together well.
Convex rank tests
=================
A [*fan*]{} in ${\mathbb{R}}^n$ is a finite collection $\mathcal{F}$ of polyhedral cones [@Ziegler1995] which satisfies the following properties:
- if $C \in \mathcal{F}$ and $C'$ is a face of $C$, then $C' \in \mathcal{F}$,
- if $C, C' \in \mathcal{F}$, then $C \cap C'$ is a face of $C$.
Two vectors $u$ and $v $ in ${\mathbb{R}}^n$ are [*permutation equivalent*]{} when $u_i < u_j$ if and only if $v_i < v_j$, and $u_i = u_j$ if and only if $v_i = v_j$ for all $i,j \in [n]$. Note that for two data vectors, each with distinct coordinates, they are permutation equivalent if and only if they have the same rank vector. The permutation equivalence classes (of which there are $13$ for $n=3$) induce a fan called the [*$S_n$-fan*]{}. The arrangement of hyperplanes $\{x_i =x_j \}$ that defines these classes is also known as the [*braid arrangement*]{}, and its regions as the [*Weyl chambers*]{} of the Lie algebra $\mathfrak{sl}(n)$. The maximal cones in the $S_n$-fan, which are the closures of the permutation equivalence classes, are indexed by permutations $\delta$ in $ S_n$. A [*coarsening*]{} of the $S_n$-fan is a fan $\mathcal{F}$ such that each permutation equivalence class of ${\mathbb{R}}^n$ is fully contained in a cone $C$ of $\mathcal{F}$. Such a fan $\mathcal{F}$ defines a partition of $S_n$ because each maximal cone of the $S_n$-fan is contained in some cone $C \in \mathcal{F}$.
A [*convex rank test*]{} is a partition of the symmetric group $S_n$ which is induced by a coarsening of the $S_n$-fan. We identify the fan with that rank test.
We say that two maximal cones, indexed by $\delta$ and $\delta'$, of the $S_n$-fan [*share a wall*]{} if there exists an index $k$ such that $\delta_k = \delta'_{k+1}$, $ \delta_{k+1} = \delta'_k$, and $\delta_i = \delta'_i$ for $\,i \not\in \{k,k+1\}$. This condition means that the corresponding permutations $\delta$ and $\delta'$ differ by an adjacent transposition. To such an unordered pair $\{\delta,\delta'\}$, we associate the following [*(elementary) conditional independence (CI) statement*]{}: $$\label{CIStatement}
\delta_k \perp \!\!\! \perp \delta_{k+1} \,|\, \{\delta_1 , \ldots, \delta_{k-1} \}.$$ The notation was coined by Dawid [@Dawid], where it is used to formally describe conditional independence among sets of random variables; we will see the connection shortly. For $k=1$ we use the standard convention to abbreviate $\, \delta_1 \perp \!\!\! \perp \delta_{2} \,|\, \{\, \} \,$ by $\, \delta_1 \perp \!\!\! \perp \delta_{2}$.
\[twenty-two\] For $n=3$ there are $40$ pre-convex rank tests (Example \[bell\]), but only $22$ of them are convex rank tests. The corresponding CI models are shown in Figure 5.6 on page 108 in [@Studeny2005Probabilistic].
The formula (\[CIStatement\]) defines a map from the set of walls of the $S_n$-fan onto the set $$\mathcal{T}_n \,\, := \,\, \bigl\{
\, i \perp \!\!\! \perp j \,|\, K \,: \, K \subseteq [n] \backslash \{i,j\} \bigr\}.$$ of all elementary CI statements. In this manner, each wall of the $S_n$-fan is labeled by a CI statement. The map from walls to CI statements is not injective; there are $(n-k-1)!(k-1)!$ walls which are labeled by .
The $S_n$-fan is the normal fan [@Ziegler1995] of the [*permutohedron*]{} ${\bf P}_n$, which is the $(n-1)$-dimensional convex hull of the vectors $(\rho_1,\ldots,\rho_n) \in {\mathbb{R}}^n$, where $\rho$ runs over all rank vectors of permutations in $S_n$. Each edge of ${\bf P}_n$ joins two permutations if they differ by an adjacent transposition. In other words, each edge corresponds to a wall and is thus labeled by a CI statement. A collection of parallel edges of ${\bf
P}_n$ that are perpendicular to a given hyperplane $\{x_i=x_j\}$ corresponds to the set of CI statements $i {\! \perp \!\!\! \perp \!}j |K$, where $K$ ranges over all subsets of $[n] \backslash \{i,j\}$.
The two-dimensional faces of ${\bf P}_n$ are squares and regular hexagons, and two edges of ${\bf P}_n$ have the same label in $\mathcal{T}_n$ if, but not only if, they are opposite edges of a square. Figure 2(c) depicts the subset of ${\bf P}_5$ in which the last two coordinates of $u \in {\mathbb{R}}^n$ are less than or equal to all other coordinates. It consists of two copies of the hexagon in 2(a), with the final two entries of the descent vector either $4|5$ (in the top hexagon) or $5|4$ (in the bottom hexagon). All vertical edges are labeled by the CI statement $4 {\! \perp \!\!\! \perp \!}5 | \{1,2,3\}$.
\[UpDown\] $$\begin{array}{ccc}
\begin{xy}<15mm,0cm>:
(-.9,1.3) *+!{++};
(-1.5,-.8) *+!{+-};
(1.5,.8) *+!{-+};
(.9,-1.4) *+!{--};
(-.5,{0.866025404}) ="123" *+!DR{1|2|3} *{\bullet};
(.5 ,{0.866025404}) ="132" *+!DL{1|3|2} *{\bullet};
(1,0) ="312" *+!L{3|1|2} *{\bullet};
(.5,-{0.866025404}) ="321" *+!UL{3|2|1} *{\bullet};
(-.5,-{0.866025404}) ="231" *+!UR{2|3|1} *{\bullet};
(-1,0) ="213" *+!R{2|1|3} *{\bullet};
"123";"132" **@{.};
"132";"312" **@{-};
"312";"321" **@{.};
"321";"231" **@{.};
"231";"213" **@{-};
"213";"123" **@{.};
({0.866025404}, 0.4) *+!{1 {\! \perp \!\!\! \perp \!}3 | \emptyset} ;
(-{0.866025404}, -0.5) *+!{1 {\! \perp \!\!\! \perp \!}3 | \{ 2\} } ;
\end{xy}
&
\begin{xy}
<15mm,0mm>:
(0,0) ="origin" ;
({0.866025404}, 0.5) ="uprt" *+!DL{1 {\! \perp \!\!\! \perp \!}3 | \emptyset} ;
({0.866025404}, -0.5) ="dwnrt" ;
(-{0.866025404}, 0.5) ="upl" ;
(-{0.866025404}, -0.5) ="dwnl" *+!UR{1 {\! \perp \!\!\! \perp \!}3 | \{ 2\} } ;
(0, 1) ="up" ;
(0,-1) ="dwn" ;
"origin";"uprt" **@{--};
"origin";"dwnrt" **@{-};
"origin";"upl" **@{-};
"origin";"dwnl" **@{--};
"origin";"up" **@{-};
"origin";"dwn" **@{-};
\end{xy}
&
\begin{xy}<19mm,0cm>:
(-.5,.43) ="123B" *{\bullet};
(.5 ,.43) ="132B" *{\bullet};
(1,0) ="312B" *{\bullet};
(.5,-.43) ="321B" *{\bullet};
(-.5,-.43) ="231B" *{\bullet};
(-1,0) ="213B" *{\bullet};
"123B";"132B" **@{.};
"132B";"312B" **@{.};
"312B";"321B" **@{-};
"321B";"231B" **@{-};
"231B";"213B" **@{-};
"213B";"123B" **@{.};
(-.5,.83) ="123T" *{\bullet};
(.5 ,.83) ="132T" *{\bullet};
(1,.4) ="312T" *{\bullet};
(.5,-.03) ="321T" *{\bullet};
(-.5,-.03) ="231T" *{\bullet};
(-1,.4) ="213T" *{\bullet};
"123T";"132T" **@{-};
"132T";"312T" **@{-};
"312T";"321T" **@{-};
"321T";"231T" **@{-};
"231T";"213T" **@{-};
"213T";"123T" **@{-};
"123T";"123B" **@{.};
"132T";"132B" **@{.};
"312T";"312B" **@{-};
"321T";"321B" **@{-};
"231T";"231B" **@{-};
"213T";"213B" **@{-};
(-.075,0.58) *+!{4 {\! \perp \!\!\! \perp \!}5 |\{1,2,3\}};
(0,1) *+!{*|\!*\!|\!*\!|4|5};
(0,-.65) *+!{*|\!*\!|\!*\!|5|4};
\end{xy}\\
\mbox{\bf (a)} & \mbox{\bf (b)} & \mbox{\bf (c)}
\end{array}$$
Any convex rank test $\mathcal{F}$ is characterized by the collection of walls $\{\delta,\delta'\}$ that are removed when passing from the $S_n$-fan to $\mathcal{F}$. So, from (\[CIStatement\]), any convex rank test $\mathcal{F}$ maps to a set $\mathcal{M}_\mathcal{F}\,$ of CI statements corresponding to missing walls, or a set $\mathbf{M}_\mathcal{F}$ of edges of the permutohedron. For example, if $\mathcal{F}$ is the fan obtained by removing the two dashed rays in Figure 2 (b) then the corresponding set of CI statements is $\,\mathcal{M}_\mathcal{F} \,= \,
\bigl\{ 1 {\! \perp \!\!\! \perp \!}3 | \emptyset , \,1 {\! \perp \!\!\! \perp \!}3 | \{ 2\} \bigr\}$.
Conditional independence statements [@Dawid] describe the dependence relationship among random variables. A [*semigraphoid*]{} is a set $\mathcal{M}$ of [*general*]{} conditional independence statements satisfying certain properties [@Pearl]. These general conditional independence statements, in contrast to the elementary CI statements already introduced, can take subsets of $[n]$ in their first two arguments. The conditions are, for $X,Y,Z$ pairwise disjoint subsets of $[n]$, $$\begin{aligned}
{\rm (SG 1)} & X \perp \!\!\! \perp Y \, |\, Z \in \mathcal{M} \implies Y \perp \!\!\! \perp X \, |\, Z \in \mathcal{M}\\
{\rm (SG 2)} & X \perp \!\!\! \perp Y \, |\, Z \in \mathcal{M} \mbox{ and } U \subset X \implies U \perp \!\!\! \perp Y \, |\, Z \in \mathcal{M}\\
{\rm (SG 3)} & X \perp \!\!\! \perp Y \, |\, Z \in \mathcal{M} \mbox{ and } U \subset X \implies X \perp \!\!\! \perp Y \, |\, (U \cup Z) \in \mathcal{M}\\
{\rm (SG 4)} & X \perp \!\!\! \perp Y \, |\, Z \in \mathcal{M} \mbox{ and } X \perp \!\!\! \perp W \,|\, (Y \cup Z) \implies X \perp \!\!\! \perp (W \cup Y) \, |\, Z \in \mathcal{M}.\end{aligned}$$ It was shown by Studený [@Studeny1990] that these are not a complete set of axioms for probabilistic conditional independence, although they are true of any probabilistic model. A semigraphoid is determined by its [*trace*]{} among statements of the form $ i \perp \!\!\! \perp j \, |\, K $ where $i$ and $j$ are singletons. Namely, $I {\! \perp \!\!\! \perp \!}J | K$ holds if and only if $i {\! \perp \!\!\! \perp \!}j |L$ for all $i \in I, j \in J$ and $L$ such that $K \subseteq L \subseteq (I \cup J \cup K) \setminus ij$; see [@Matus1992Equivalence]. Casting the semigraphoid axiom in terms of the trace, we say that a subset $\mathcal{M} $ of $\mathcal{T}_n$ is a [*semigraphoid*]{} if $\, i \perp \!\!\! \perp j \, |\, K \in \mathcal{M}\,$ implies $\, j \perp \!\!\! \perp i \, |\, K \in \mathcal{M}\,$ and the following axiom holds: $$\begin{aligned}
{\rm (SG)} && \qquad \quad i \perp \!\!\! \perp j \, |\, K \cup {\ell}\, \in \mathcal{M}
\quad \, \mbox{and} \,\quad
i \perp \!\!\! \perp \ell \, |\, K\, \in \mathcal{M} \\
&& \!\!\! \mbox{implies } \,\,\,\,\,
i \perp \!\!\! \perp j \,|\, K \in \mathcal{M}
\qquad \mbox{and } \quad i \perp \!\!\! \perp \ell \,| \, K \!\cup\! j \in \mathcal{M}.\end{aligned}$$ This axiom is stated in [@Matus2004; @Studeny2005Probabilistic]. Our first result is that semigraphoids and convex rank tests are the same combinatorial object:
\[fantheorem\] The map $\mathcal{F} \mapsto \mathcal{M}_\mathcal{F}$ is a bijection between convex rank tests and semigraphoids.
Before presenting the proof of this theorem, we shall discuss an example.
\[ex.updwn\] Let $\mathcal{F}$ denote the convex rank test called up-down analysis [@Willbrand2005]. In this test, each permutation $\pi \in S_n$ is mapped to the sign vector of its first differences, or, equivalently, its descent set. Thus this test is the natural map $\,\tau : S_n \rightarrow \{-,+\}^{n-1}$. The corresponding semigraphoid $ \mathcal{M}_\mathcal{F}$ consists of all CI statements $\, i \perp \!\!\! \perp j \,| \, K \,$ where $\, | i -j | \geq 2 $.
This convex rank test is visualized in Figure 2(a,b) for $n=3$. Permutations are in the same class (have the same sign pattern) if they are connected by a solid edge; there are four classes. In the $S_3$-fan, the two missing walls are labeled by conditional independence statements as defined in (\[CIStatement\]). For $n=4$ the up-down analysis test $\mathcal{F}$ is depicted in Figure 3. The double edges correspond to the twelve CI statements in $\mathcal{M}_\mathcal{F}$. There are eight classes; e.g., the class $\{3|4|1|2,3|1|4|2,1|3|4|2,1|3|2|4,3|1|2|4\}$ consists of the five permutations in $S_4$ which have the up-down pattern $(-,+,-)$.
\[UpDown4\] $$\begin{xy}<25mm,0cm>:
(1,0) ="3214" *+!U{3214} *{\bullet};
(1.8,0) ="2314" *+!U{2314} *{\bullet};
(.7,.18) ="3241" *+!R{3241} *{\bullet};
(1.5,.18) ="2341" *+!L{2341} *{\circ}; (.86,.5) ="3124" *+!R{3124} *{\bullet};
(2.45,.5) ="2134" *+!DR{2134} *{\bullet};
(.2,.8) ="3421" *+!R{3421} *{\bullet};
(1.8 ,.8) ="2431" *+!U{2431} *{\circ}; (1.5,1) ="1324" *+!R{1324} *{\bullet};
(2.29,1) ="1234" *+!DR{1234} *{\bullet};
(.35,1.14) ="3142" *+!L{3142} *{\bullet};
(2.75,1.14) ="2143" *+!U{2143} *{\bullet};
(.02,1.29) ="3412" *+!DR{3412} *{\bullet};
(2.44,1.29) ="2413" *+!DR{2413} *{\circ}; (.5,1.45) ="4321" *+!DR{4321} *{\circ}; (1.34,1.45) ="4231" *+!L{4231} *{\circ}; (1,1.6) ="1342" *+!DR{1342} *{\bullet};
(2.6,1.6) ="1243" *+!L{1243} *{\bullet};
(.35,1.92) ="4312" *+!DR{4312} *{\bullet};
(1.97,1.92) ="4213" *+!DR{4213} *{\circ}; (1.3,2.26) ="1432" *+!UR{1432} *{\bullet};
(2.11,2.26) ="1423" *+!L{1423} *{\bullet};
(.99,2.41) ="4132" *+!DR{4132} *{\bullet};
(1.81,2.41) ="4123" *+!DL{4123} *{\bullet};
"3214";"2314" **@{-}; "3241";"2341" **@{.}; "3241";"3214" **@{=}; "2341";"2314" **@{:}; "2134";"1234" **@{-}; "2143";"1243" **@{-}; "1234";"1243" **@{-}; "2134";"2143" **@{-}; "4132";"4123" **@{-}; "1432";"1423" **@{-}; "1432";"4132" **@{=}; "4123";"1423" **@{=}; "4312";"3412" **@{-}; "4321";"3421" **@{.}; "4312";"4321" **@{.}; "3412";"3421" **@{-}; "4213";"2413" **@{:}; "4231";"2431" **@{:}; "4213";"4231" **@{:}; "2413";"2431" **@{:}; "1342";"1324" **@{=}; "3142";"3124" **@{=}; "1342";"3142" **@{=}; "1324";"3124" **@{=}; "2314";"2134" **@{=}; "3124";"3214" **@{-}; "3421";"3241" **@{=}; "3412";"3142" **@{=}; "1324";"1234" **@{-}; "1432";"1342" **@{-}; "4312";"4132" **@{=}; "1423";"1243" **@{=}; "2341";"2431" **@{.}; "4321";"4231" **@{.}; "2413";"2143" **@{:}; "4123";"4213" **@{.}; \end{xy}$$
Our proof of Theorem \[fantheorem\] rests on translating the semigraphoid axiom (SG) into geometric statements about edges of the permutohedron. Recall that a semigraphoid $\mathcal{M}$ can be identified with the set $\mathbf{M}$ of edges of the permutohedron whose CI statement labels are those of $\mathcal{M}$.
\[obs:SqHexAxioms\] A set $\mathbf{M}$ of edges of the permutohedron ${\bf P}_n$ is a semigraphoid if and only if the set $\mathbf{M}$ satisfies the following two geometric axioms:\
[**Square axiom:**]{} Whenever an edge of a square is in $\mathbf{M}$, then the opposite edge is also in $\mathbf{M}$.\
$$\begin{xy}<5mm,0cm>:
(-.7,.7) ="TL" *{\bullet};
(.7,.7) ="TR" *{\bullet};
(-.7,-.7) ="BL" *{\bullet};
(.7,-.7) ="BR" *{\bullet};
"TL";"BL" **@{-};
"TR";"BR" **@{.};
"BR";"BL" **@{.};
"TL";"TR" **@{.};
\end{xy} \quad
\implies \quad
\begin{xy}<5mm,0cm>:
(-.7,.7) ="TL" *{\bullet};
(.7,.7) ="TR" *{\bullet};
(-.7,-.7) ="BL" *{\bullet};
(.7,-.7) ="BR" *{\bullet};
"TL";"BL" **@{-};
"TR";"BR" **@{-};
"BR";"BL" **@{.};
"TL";"TR" **@{.};
\end{xy}$$ [**Hexagon axiom:**]{} Whenever two adjacent edges of a hexagon are in $\mathbf{M}$, then the two opposite edges of that hexagon are also in $\mathbf{M}$. $$\begin{xy}<5mm,0cm>:
(-.5,{0.866025404}) ="123" *{\bullet};
(.5 ,{0.866025404}) ="132" *{\bullet};
(1,0) ="312" *{\bullet};
(.5,-{0.866025404}) ="321" *{\bullet};
(-.5,-{0.866025404}) ="231" *{\bullet};
(-1,0) ="213" *{\bullet};
"123";"132" **@{-};
"132";"312" **@{.};
"312";"321" **@{.};
"321";"231" **@{.};
"231";"213" **@{.};
"213";"123" **@{-};
\end{xy} \quad
\implies \quad
\begin{xy}<5mm,0cm>:
(-.5,{0.866025404}) ="123" *{\bullet};
(.5 ,{0.866025404}) ="132" *{\bullet};
(1,0) ="312" *{\bullet};
(.5,-{0.866025404}) ="321" *{\bullet};
(-.5,-{0.866025404}) ="231" *{\bullet};
(-1,0) ="213" *{\bullet};
"123";"132" **@{-};
"132";"312" **@{.};
"312";"321" **@{-};
"321";"231" **@{-};
"231";"213" **@{.};
"213";"123" **@{-};
\end{xy}$$
Let $\mathbf{M}$ be the subgraph of the edge graph of ${\bf P}_n$ defined by the statements in $\mathcal{M}$; that is, $\mathbf{M}$ consists of edges whose labels are in $\mathcal{M}$. Each class of the rank test defined by $\mathcal{M}$ consists of the permutations in some connected component of $\mathbf{M}$. We regard a path from $\delta$ to $\delta'$ on ${\bf P}_n$ as a word $\sigma^{(1)} \cdots
\sigma^{(l)}$ in the free associative algebra $\mathcal{A}$ generated by the adjacent transpositions of $[n]$. For example, the transposition $\sigma_{23} := (23)$ gives the path from $\delta$ to $\delta'=\sigma_{23} \delta = \delta_1 | \delta_3 | \delta_2 | \delta_4
| \dots | \delta_n$. The following relations in $\mathcal{A}$ define a presentation of the group algebra of $S_n$ as a quotient of $\mathcal{A}$:
$$\begin{array}{ccc}
{\rm (BS)} & \; \sigma_{i, i+1}\cdot \sigma_{i+k+1, i+k+2}\, -\, \sigma_{i+k+1, i+k+2} \cdot \sigma_{i, i+1}, & \\
{\rm (BH)} & \; \sigma_{i, i+1} \cdot \sigma_{i+1, i+2} \cdot \sigma_{i, i+1} - \sigma_{i+1, i+2}\cdot
\sigma_{i, i+1} \sigma_{i+1, i+2}, & \qquad \mbox{ and}
\\
{\rm (BN)} & \; \sigma_{i, i+1}^2 -1, &
\end{array}$$ where suitable $i$ and $k$ vary over $[n]$. The first two are the *braid relations*, and the third represents the idempotency of each transposition.
Now, we regard these relations as properties of a set of edges of ${\bf P}_n$, by identifying a word and a permutation $\delta$ with the set of edges that comprise the corresponding path in ${\bf P}_n$. For example, a set satisfying (BS) is one such that, starting from any $\delta$, the edges of the path $\sigma_{i, i+1} \sigma_{i+k+1, i+k+2}$ are in the set if and only if the edges of the path $ \sigma_{i+k+1, i+k+2} \sigma_{i,
i+1}$ are in the set. Note then, that (BS) is the square axiom, and (BH) is a weaker version of the hexagon axiom of semigraphoids. That is, implications in either direction hold in a semigraphoid. However, (BN) holds only directionally in a semigraphoid: if an edge lies in the semigraphoid, then its two vertices are in the same class; but the empty path at some vertex $\delta$ certainly does not imply the presence of all incident edges in the semigraphoid. Thus, for a semigraphoid, (BS) and (BH) hold, but (BN) must be replaced with the directional version
${\rm (BN')} \;\; \
\qquad \sigma_{i, i+1}^2 \rightarrow 1.$
We now consider a path $p$ from $\delta$ to $\delta'$ in a semigraphoid. Here is a crucial lemma for our proof:
\[lem.allshortestpaths\] Suppose that $\mathcal{M}$ is a semigraphoid. If $\delta$ and $\delta'$ lie in the same class of $\mathcal{M}$, then so do all shortest paths on ${\bf P}_n$ between them.
The lemma in turn depends on the following version of a classical result due to Jacques Tits. This result, which can be found in [@Brown1989 p. 49-51]), essentially states that the relations (BS),(BH),(BN) form a Gröbner basis for the two-sided ideals they generate in $\mathcal{A}$.
\[thm:Tits\] Let $p$ and $q$ be words representing paths on $\mathbf{P}_n$.
- A word $p$ is (BS),(BH),(BN)-reduced if and only if it is (BS),(BH),(BN’)-reduced.
- If $p$ and $q$ are reduced, then they represent the same element of the symmetric group $S_n$ if and only if $p$ can be transformed to $q$ by the the application of (BS) and (BH) only.
Theorem \[thm:Tits\] (1) says that if there is any path connecting $\delta$ and $\delta'$, then there is a shortest path connecting them. Thus if $\delta$ and $\delta'$ lie in the same class of $\mathcal{M}$, some shortest path $\delta \rightarrow \delta'$ also lies in that class. Now (2) says that if $p$ and $q$ are both shortest paths, then $q$ can be obtained from $p$ by application of only the square and hexagon axioms, (BS) and (BH). Thus if any shortest path $\delta \rightarrow \delta'$ lies in the class of $\mathcal{M}$ containing them both, so do all other shortest paths connecting them.
We need one lemma to deal with intersections of nonmaximal cones. Denote by ${\prec}$ the transitive relation “is a face of” and write $F_w(C)$ for the face of a cone $C$ at which $w$ is minimized.
\[lem:reduction\] If the intersection of two cones $C_1$ and $C_2$ is a face of both, then the intersection of any faces $D {\prec}C_1$ and $E {\prec}C_2$ is a face of both.
By transitivity of ${\prec}$ and the hypothesis it suffices to show $D \cap E {\prec}C_1 \cap C_2$. Since $D {\prec}C_1$, there exists a linear functional $w$ such that the face $F_w(C_1)$ equals $D$ and $C_1 \cap C_2 \subset C_1 \subset H_w^+$. Then $F_w(C_1 \cap C_2)=D \cap C_2$ so $D \cap C_2 {\prec}C_1 \cap C_2$. Similarly, $E \cap C_1 {\prec}C_1 \cap C_2$. Then since the intersection of any two faces of $C_1 \cap C_2$ is also a face, $D \cap E {\prec}C_1 \cap C_2$ as desired.
Both semigraphoids and convex rank tests can be regarded as sets of edges of ${\bf P}_n$. We first show that a semigraphoid satisfies (PC). Consider $\delta, \delta'$ in the same class $C$ of a semigraphoid, and let $\delta'' \in
\mathcal{L}(\delta \cap \delta')$. Further, let $p$ be a shortest path from $\delta$ to $\delta''$ (so, $p \delta = \delta''$), and let $q$ be a shortest path from $\delta''$ to $\delta'$. We claim that $qp$ is a shortest path from $\delta$ to $\delta'$, and thus $\delta'' \in C$ by Lemma \[lem.allshortestpaths\]. Suppose $qp$ is not a shortest path. Then, we can obtain a shorter path in the semigraphoid by some sequence of substitutions according to (BS), (BH), and (BN’). Only (BN’) decreases the length of a path, so the sequence must involve (BN’). Therefore, there is some $i$, $j$ in $[n]$, such that their positions relative to each other are reversed twice in $qp$. But $p$ and $q$ are shortest paths, hence one reversal occurs in each of $p$ and $q$. Then $\delta$ and $\delta'$ agree on whether $i>j$ or $j>i$, but the reverse holds in $\delta''$, contradicting $\delta'' \in \mathcal{L}(\delta \cap \delta')$. Thus every semigraphoid is a pre-convex rank test.
Now, we show that a semigraphoid corresponds to a fan. We first argue that we may reduce to the case of two maximal cones, each coming from a class in the semigraphoid, whose intersection is codimension one in both. By Lemma \[lem:reduction\], we can consider maximal cones only. Suppose two maximal cones $C_1$, $C_k$ have intersection $C_1 \cap C_k$ which is not codimension one. Then there exists a sequence of maximal cones $C_1, C_2, \dots, C_k$ such that $C_i \cap C_{i+1}$ is codimension one, $C_1 \cap C_k \subset C_i \cap C_{i+1}$ for all $i = 1, \dots k-1$, and in fact $C_1 \cap C_k = C_1 \cap C_2 \cap \cdots \cap C_k$. We have that $(C_i \cap C_{i+1}) \cap (C_{i+1} \cap C_{i+2})$ is a face of $C_{i+1}$ and $C_{i+2}$ by Lemma \[lem:reduction\], and also is a face of $C_i$. Thus $C_i \cap C_{i+1} \cap C_{i+2} {\prec}C_i, C_{i+1}, C_{i+2}$; continuing in this manner, we eventually get that $C_1 \cap C_2 \cap \cdots \cap C_k {\prec}C_1, C_k$ as required.
Consider the cone corresponding to a class $C$. We need only show that its codimension one intersection with another maximal cone is a shared face. Since $C$ is a cone of a coarsening of the $S_n$-fan, each facet of $C$ lies in a hyperplane $H =\{x_i=x_j\}$. Suppose a face of $C$ coincides with the hyperplane $H$ and that $i>j$ in $C$. A vertex $\delta$ borders $H$ if $i$ and $j$ are adjacent in $\delta$. We will show that if $\delta,\delta' \in C$ border $H$, then their reflections $\widehat{\delta} = \delta_1 | \dots |j|i| \dots
| \delta_n$ and $\widehat{\delta'}= \delta'_1 | \dots |j|i| \dots |\delta'_n$ both lie in some class $C'$. Consider a ‘great circle’ path between $\delta$ and $\delta'$ which stays closest to $H$: all vertices in the path have $i$ and $j$ separated by at most one position, and no two consecutive vertices have $i$ and $j$ nonadjacent. This is a shortest path, so it lies in $C$, by Lemma \[lem.allshortestpaths\]. Using the square and hexagon axioms (Observation \[obs:SqHexAxioms\]), we see that the reflection of the path across $H$ is a path in the semigraphoid that connects $\widehat{\delta}$ to $\widehat{\delta'}$ (Figure 3). This shows that the intersection of $C$ and $C'$ is a face of both. Thus a semigraphoid is a convex rank test.
Finally, if $\mathbf{M}$ is a set of edges of ${\bf P}_n$ representing a convex rank test, then it is easy to show that $\mathbf{M}$ satisfies the square and hexagon axioms.
\[fig:reflection\] $$\begin{xy}<15mm,0cm>:
(0,0); p+ ({0.866025404}, 0.5) *{\bullet}; p + (0,-1) *{\bullet} **@{.}; p + (-{0.866025404},-.5) *{\bullet} **@{-}; p + (-{0.866025404},+.5) *{\bullet} *+!UR{\widehat{\delta}} **@{-}; p + (0,1) *{\bullet} *+!DR{\delta} **@{.};
p+ ({0.866025404}, 0.5) *{\bullet} **@{-};
p+ ({0.866025404}, -0.5) *{\bullet} **@{-};
({0.866025404}, 0.5);
p + (1,0) **@{-};
({0.866025404}, -0.5);
p + (1,0) **@{-};
(2.73205081,0); p+ ({0.866025404}, 0.5) *+!DL{\delta'} *{\bullet}; p + (0,-1) *{\bullet} *+!UL{\widehat{\delta'}} **@{.}; p + (-{0.866025404},-.5) *{\bullet} **@{-}; p + (-{0.866025404},+.5) *{\bullet} **@{-}; p + (0,1) *{\bullet} **@{.};
p+ ({0.866025404}, 0.5) *{\bullet} **@{-};
p+ ({0.866025404}, -0.5) *{\bullet} **@{-};
(-1.5,0);
p+(5.8,0) *+!UL{x_i = x_j}**@{--};
\end{xy}$$
The submodular cone
===================
In this section we focus on a subclass of the convex rank tests. Let $2^{[n]}$ denote the collection of all subsets of $[n] = \{1,2,\ldots,n\}$. Any real-valued function $\, w : 2^{[n]} \rightarrow {\mathbb{R}}\, $ defines a convex polytope $Q_w$ of dimension $\leq n-1$ as follows: $$\begin{aligned}
Q_w \,\,\, := &
\bigl\{ \, x \in {\mathbb{R}}^n \,: \,
x_1 + x_2 + \cdots + x_n = w([n]) \\
& \text{\ \,and } \sum\nolimits_{i \in I} x_i \leq w(I)\,\,
\hbox{for all} \,\, \emptyset\neq I \subseteq [n] \,\bigr\}.\end{aligned}$$ A function $\, w : 2^{[n]} \rightarrow {\mathbb{R}}\, $ is called [*submodular*]{} if $\,w(I) + w(J)\, \geq\, w(I \cap J) + w(I \cup J)\,$ for $I,J \subseteq [n]$. The [*submodular cone*]{} is the cone ${\bf C}_n$ of all submodular functions $w :
2^{[n]} \rightarrow {\mathbb{R}}$. Working modulo its lineality space $\,{\bf C}_n \cap (-{\bf C}_n) $, we regard ${\bf C}_n$ as a pointed cone of dimension $2^n-n-1$.
Studying functions $w$ means that in considering the normal fan of a polytope $Q_w$, we want to retain information about non-binding inequalities that are just barely so, i.e. that hold with equality. For this reason we define the [*vector (normal) fan*]{} [@BGS]. The indicator function of each $I \in 2^{[n]}$ defines a vector $e_I$ in the $1$-skeleton of the $S_n$-fan, understood modulo $e_{[n]}$; for example, these vectors for $n=3$ are $e_{001}, e_{010}, e_{100}, e_{011}, \dots, e_{111}$. A [*vector fan*]{} ${\mathfrak{F}}$ is a collection of subsets of $\{e_I: I \in 2^{[n]}\}$ such that $U, V \in {\mathfrak{F}}$ implies $U \cap V \in {\mathfrak{F}}$. A vector fan defines a usual fan by taking the maximal cones of the fan to be the cones generated by the vector sets in the vector fan. We say that a vector fan is *complete* if its fan is. A vector fan ${\mathfrak{F}}$ *coarsens* another vector fan ${\mathfrak{G}}$ if for all $U \in {\mathfrak{G}}$, there exists $V \in {\mathfrak{F}}$ with $U \subset V$.
Given a function $w:2^{[n]} \rightarrow {\mathbb{R}}$, each $I\in 2^{[n]}$ defines an inequality $\sum_{i \in I} x_i \leq w_I$ appearing in the definition of $Q_w$; the vector normal fan tells us which of these inequalities holds with equality on some face of $Q_w$. We define the [*vector normal fan*]{} of a function $w:2^{[n]} \rightarrow {\mathbb{R}}$ as the set $\{ \{e_I: I \in 2^{[n]}, \sum_{i \in I}x_i =w_I$ for all $x \in F\}$ for each face $F \in Q_w \}$. The vector normal fan of $w$ defines a fan which is the normal fan of $Q_w$ and retains additional information.
\[prop:submodularnormal\] A function $\,w: 2^{[n]} \rightarrow {\mathbb{R}}\, $ is submodular if and only if the vector normal fan of $w$ is a coarsening of the vector $S_n$-fan.
Let $w_1 = w_2 =w_3 =1, w_{12}=w_{13}=w_{23}=w_{123}=3$. The polytope $Q_w$ is the point $(1,1,1)$ but the function $w$ is not submodular. The vector normal fan ${\mathfrak{F}}$ of $w$ is $\{\{e_{001},e_{010},e_{100}\}\}$ and the normal fan is all of ${\mathbb{R}}^3 / (1,1,1)$. ${\mathfrak{F}}$ does not coarsen the $S_n$-fan since, for example, $e_{110}$ is not contained in any set in ${\mathfrak{F}}$.
However, if we change $w$ slightly to define the same $Q_w$ but with the inequalities corresponding to $011,101$, and $110$ also holding with equality, e.g. $w_1 = w_2 =w_3 =1, w_{12}=w_{13}=w_{23}=2$, and $w_{123}=3$, the resulting vector normal fan of $w$ is a coarsening of the (vector) $S_n$-fan.
We show only the if direction of Proposition \[prop:submodularnormal\]. Suppose $w$ is not submodular. Then there exist $I,J \subset 2^{[n]}$ such that$$w_I + w_J < w_{I \cap J} + w_{I \cup J}$$ We also have that $$\begin{aligned}
\sum_{i \in I \cup J} x_i + \sum_{i \in I \cap J} x_i & = & \sum_{i \in I} x_i + \sum_{i \in J} x_i\\
&\leq & w_I + w_J < w_{I \cap J} + w_{I \cup J}\end{aligned}$$ So $\sum_{i \in I \cup J} x_i < w_{I \cup J} + (w_{I \cap J}- \sum_{i \in I \cap J} x_i)$ and similarly $\sum_{i \in I \cap J} x_i < w_{I \cap J} + (w_{I \cup J}- \sum_{i \in I \cup J} x_i)$, so that at most one of the inequalities corresponding to $I \cup J$ and $I \cap J$ can hold with equality at any point of $Q_w$. Then any set in the vector normal fan of $w$ either fails to contain $e_{I \cap J}$ or fails to contain $e_{I \cup J}$.
Proposition \[prop:submodularnormal\] can be paraphrased as follows: the function $w$ is submodular if and only if the optimal solution of $$\mbox{maximize $u \cdot x$ subject to $x \in Q_w$}$$ depends only on the permutation equivalence class of $u$. Thus, solving this linear programming problem constitutes a convex rank test. Any such test is called a [*submodular rank test*]{}.
A convex polytope is a [*(Minkowski) summand*]{} of another polytope if the normal fan of the latter refines the normal fan of the former. The polytope $Q_w$ that represents a submodular rank test is a summand of the permutohedron ${\bf P}_n$.
The following combinatorial objects are equivalent for any positive integer $n$:\
$1.$ submodular rank tests, $2.$ summands of the permutohedron $\mathbf{P}_n$, $3.$ structural conditional independence models [@Studeny2005Probabilistic], $4.$ faces of the submodular cone ${\bf C}_n$ in ${\mathbb{R}}^{2^n}$.
We have 1$\iff$2 from Proposition \[prop:submodularnormal\], and 1$\iff$3 follows from [@Studeny2005Probabilistic]. Further, 1$\iff$4 is a direct consequence of our definition of submodular rank tests.
All $22$ convex rank tests for $n=3$ are submodular. The submodular cone ${\bf C}_3$ is a $4$-dimensional cone whose base is a bipyramid. Its f-vector is $(1,5,9,6,1)$. The polytopes $Q_w$, as $w$ ranges over representatives of the faces of ${\bf C}_3$, are all the Minkowski summands of ${\bf P}_3$.
\[notsubmodular\] For $n \geq 4$, there exist convex rank tests that are not submodular rank tests. Equivalently, there are fans that coarsen the $S_n$-fan but are not the normal fan of any polytope.
This result is well-known. It is stated in Section 2.2.4 of [@Studeny2005Probabilistic] in the following form: “There exist semigraphoids that are not structural.”
An interesting example which also proves Proposition \[notsubmodular\] is the following semigraphoid: $$\mathcal{M} \quad = \quad
\bigl\{
2 \perp \!\!\! \perp 3 | \{1,4\},\,
1 \perp \!\!\! \perp 4 | \{2,3\}, \,
1 \perp \!\!\! \perp 2 | \emptyset,\,
3 \perp \!\!\! \perp 4 |\emptyset \,\bigr\}.$$ The corresponding fan consists of unimodular cones, or, equivalently, the posets $P_i$ representing this non-submodular convex rank test are all trees. This example answers a question posed in the first version of [@PRW]. A systematic method for showing that a semigraphoid is not submodular can be found in [@counterexamples]. Results in that paper include an example of a coarsest semigraphoid which is not submodular and a proof that the semigraphoid semigroup is not normal.
\[rmk1\] For $n=4$ there are $22108$ submodular rank tests, one for each face of the $11$-dimensional cone ${\bf C}_4$. The base of this submodular cone is a $10$-dimensional polytope with $f$-vector $
(1,37, 356, $ $ 1596, 3985, 5980, 5560, 3212, 1128, 228, 24,1)$. The $37 $ vertices of this polytope correspond to the maximal semigraphoids. These come in seven symmetry classes up to the $*$ involution (\[starinvolution\]) and the $S_4$-action. The types of maximal semigraphoids for $n=4$ are displayed in the following table:
\[rmk2\] For $n=5$ there are $117978$ coarsest submodular rank tests, in $1319$ $S_5$ symmetry classes. We confirmed this result of [@Studeny2000] with [POLYMAKE]{} [@Gawrilow2000].
We now define a class of submodular rank tests, which we call [*Minkowski sum of simplices (MSS) tests*]{}. Note that each subset $K$ of $[n]$ defines a submodular function $w_K$ by setting $w_K (I) = 1$ if $K \cap I $ is non-empty and $w_K(I) = 0$ if $K \cap I $ is empty. The corresponding polytope $Q_{w_K}$ is the simplex $\Delta_K = {\rm conv} \{ e_k : k \in K \}$.
Now consider an arbitrary subset $\,\mathcal{K} = \{K_1,K_2,\ldots,K_r \}\,$ of $2^{[n]}$. It defines the submodular function $\,w_{\mathcal{K}} = w_{K_1} + w_{K_2} + \cdots + w_{K_r}$. The corresponding polytope is the Minkowski sum $$\Delta_\mathcal{K} \quad = \quad \Delta_{K_1} + \Delta_{K_2} + \cdots + \Delta_{K_r}.$$ The associated MSS test $\tau_\mathcal{K}$ is defined as follows. Given $\rho \in S_n$, we compute the number of indices $j \in [r]$ such that $\,{\rm max}\{ \rho_k \,: \, k \in K_j \}\, = \,
\rho_i $, for each $i \in [n]$. The signature $\tau_\mathcal{K}(\rho)$ is the vector in ${\mathbb{N}}^n$ whose $i$th coordinate is that number. Few submodular rank tests are MSS tests:
\[rmk3\] For $n = 3$, there are $22$ submodular rank tests, but only $15$ of them are MSS tests. For $n=4$, there are $22108$ submodular rank tests, but only $1218$ of them are MSS tests.
In light of Theorem \[fantheorem\], it is natural to ask which semigraphoids correspond to an MSS test. Geometrically, we wish to know which edges of the permutohedron ${\bf P}_n$ are contracted when passing to the polytope $Q_{w_{\mathcal{K}}}$. To be precise, let $\mathcal{M}_\mathcal{K}$ denote the semigraphoid derived from $\mathcal{F}_{w_{\mathcal{K}}}$ using the bijection in Theorem \[fantheorem\]. We then have the following result:
\[CIsetfam\] The semigraphoid $\mathcal{M}_\mathcal{K}$ is the set of CI statements of the form $\, i \perp \!\!\! \perp j \, |\, K \,$ where all sets containing $\{i,j\}$ and contained in $\{i,j\} \cup [n] \backslash K \,$ are not in $\mathcal{K}$.
Consider two permutations $\delta$ and $\delta'$ which are adjacent on the permutohedron ${\bf P}_n$, and let $\,i \perp \!\!\! \perp j \, |\, K \,$ be the label of the edge that connects $\delta$ and $\delta'$. That CI statement is in $\mathcal{M}_\mathcal{K}$ if and only if $\delta$ and $\delta'$ are mapped to the same vertex in $\Delta_{\mathcal{K}}$ if and only if $\delta$ and $\delta'$ are mapped to the same vertex in each simplex $\Delta_{K_l}$ for $l=1,2,\ldots,r$. For each $l$, this means that the leftmost entry of the descent vector $\delta$ that lies in $K_l$ agrees with the leftmost entry of the other descent vector $\delta'$ that lies in $K_l$. This condition is equivalent to $$K_l \,\, \cap \,\, (\,K \,\cup \,\{i,j\} \,) \quad \not= \quad \{i,j\} \qquad \qquad
\hbox{for}\, \,\,\, l =1,2,\ldots,r .$$ Thus $\,i \perp \!\!\! \perp j \, |\, K \,$ is in the semigraphoid $\,\mathcal{M}_{\mathcal{K}}\,$ associated with the set family $\mathcal{K}$ if and only if $\mathcal{K}$ contains no set whose intersection with $\,K\,\cup \,\{i,j\} \,$ equals $\,\{i,j\}$. This is precisely our claim.
There is a natural involution $*$ on the set of all CI statements which is defined as follows: $$\label{starinvolution}
( i \perp \!\!\! \perp j \,|\, C)^* \quad := \quad
i \perp \!\!\! \perp j \,|\, [n]\backslash (C \cup \{i,j\}) .$$ If $\mathcal{M}$ is any semigraphoid, then the semigraphoid $\mathcal{M}^*$ is obtained by applying the involution $*$ to all the CI statements in the model $\mathcal{M}$. This involution is referred to as [*duality*]{} in [@Matus1992Ascending]. In the [*boolean lattice*]{}, whose elements are the subsets of $[n]$, the involution corresponds to switching the role of set intersection and set union.
The MSS test $\tau_{\mathcal{K}}$ was defined above in terms of weight functions $w$. What follows is a similar construction for the duals of MSS tests. Let $z_{\mathcal{K}}(J)=1$ for $J \in
\mathcal{K}$ and $z_{\mathcal{K}}(J) =0$ otherwise. Then the function $\,w^*: 2^{[n]} \rightarrow {\mathbb{R}}\, $ defined by $\,w_{\mathcal{K}}^* (I) := \sum_{J \subset I} z_{\mathcal{K}}(J)\,$ is supermodular. We set $$\begin{aligned}
Q_w^* & := \,
\bigl\{ \, x \in {\mathbb{R}}^n \,: \,
x_1 + x_2 + \cdots + x_n = w([n]) \\
& \text{\ \,and } \sum\nolimits_{i \in I} x_i \geq w(I)\,\,
\hbox{for all} \,\, \emptyset\neq I \subseteq [n] \,\bigr\}.\end{aligned}$$ Then the equality $\,Q_{w_{\mathcal{K}}}^* \,=\, \Delta_K\,$ holds for $\,\Delta_{\mathcal{K}} = \Delta_{\mathcal{K}_1} + \Delta_{K_2} + \cdots + \Delta_{K_r}$. This equality is precisely the statement in Proposition 6.3 of Postnikov’s paper [@Postnikov2005].
Graphical tests {#sec:graphical}
===============
We have seen that semigraphoids are equivalent to convex rank tests. We now explore the connection to graphical models. Let $G$ be a graph with vertex set $[n]$ and $\mathcal{K}(G)$ the collection of all subsets $K \subseteq
[n]$ such that the induced subgraph of $G|_K$ is connected. The [*undirected graphical model*]{} (or [*Markov random field*]{}) derived from the graph $G$ is the set $\mathcal{M}^G$ of CI statements: $$\label{noPath}
\mathcal{M}^G \,\,\, = \,\,\,
\bigl\{\, i \perp \!\!\! \perp j \,|\, C \,\, :\,\,
\mbox{the restriction of $\,G\,$ to} \,\,\, [n] \backslash C \,\,
\mbox{ contains no path from $i$ to $j$} \bigr\}.$$
\[Jasonslemma\] The set $\mathcal{M}^G$ of CI statements in the graphical model $G$ is equal to the semigraphoid $\,\mathcal{M}_{\mathcal{K}(G)}\,$ associated with the family $\mathcal{K}(G)$ of connected induced subgraphs of $G$.
The defining condition in (\[noPath\]) is equivalent to saying that the restriction of $G$ to any node set containing $\{i,j\}$ and contained in $\,\{i,j\} \cup ([n] \backslash C)\,$ is disconnected. With this observation, Theorem \[Jasonslemma\] follows directly from Proposition \[CIsetfam\].
The polytope $\Delta_G = \Delta_{\mathcal{K}(G)}$ associated with the graph $G$ is the [*graph associahedron*]{}. This is a well-studied object in combinatorics [@Postnikov2005; @Carr2004]. Carr and Devadoss [@Carr2004] showed that $\Delta_G$ is a simple polytope whose faces are in bijection with the tubings of the graph $G$. Tubings are defined as follows. Two subsets $A,B$ $\subset [n]$ are *compatible* for $G$ if one of the following conditions holds: $A\subset B$, $B\subset A$, or $A\cap B = \emptyset$, and there is no edge between any node in $A$ and $B$. A [*tubing*]{} of the graph $G$ is a subset ${\bf T}$ of $2^{[n]}$ such that any two elements of ${\bf T}$ are compatible. The set of all tubings on $G$ is a simplicial complex; it is dual to the face lattice of the simple polytope $\Delta_G$.
For any graph $G$ on $[n]$ we now have two convex rank tests. First, there is the [*graphical model rank test*]{} $\,\tau_{\mathcal{K}(G)}$, which is the MSS test of the set family $\mathcal{K}(G)$. Second, we have the [*graphical tubing rank test*]{} $\,\tau^*_{\mathcal{K}(G)}$, which is the convex rank test associated with the semigraphoid $\,(\mathcal{M}^G)^*\,$ dual to $\,\mathcal{M}^G$. Explicitly, that dual semigraphoid is given by $$\label{noPath2}
\! (\mathcal{M}^G)^* \,\, = \,\,
\bigl\{\, i \perp \!\!\! \perp j \,|\, C \,\, :\,
\mbox{the restriction of $\,G\,$ to} \,\,\, C \cup \{i,j\} \,
\mbox{ contains no path from $i$ to $j$} \bigr\}.$$
\[GM\] $$\begin{xy}<25mm,0cm>:
(1,0) ="3214" *+!U{3214} *{\bullet};
(1.8,0) ="2314" *+!U{2314} *{\bullet};
(.7,.18) ="3241" *+!R{3241} *{\bullet};
(1.5,.18) ="2341" *+!L{2341} *{\circ}; (.86,.5) ="3124" *+!R{3124} *{\bullet};
(2.45,.5) ="2134" *+!DR{2134} *{\bullet};
(.2,.8) ="3421" *+!R{3421} *{\bullet};
(1.8 ,.8) ="2431" *+!U{2431} *{\circ}; (1.5,1) ="1324" *+!R{1324} *{\bullet};
(2.29,1) ="1234" *+!DR{1234} *{\bullet};
(.35,1.14) ="3142" *+!L{3142} *{\bullet};
(2.75,1.14) ="2143" *+!U{2143} *{\bullet};
(.02,1.29) ="3412" *+!DR{3412} *{\bullet};
(2.44,1.29) ="2413" *+!DR{2413} *{\circ}; (.5,1.45) ="4321" *+!DR{4321} *{\circ}; (1.34,1.45) ="4231" *+!L{4231} *{\circ}; (1,1.6) ="1342" *+!DR{1342} *{\bullet};
(2.6,1.6) ="1243" *+!L{1243} *{\bullet};
(.35,1.92) ="4312" *+!DR{4312} *{\bullet};
(1.97,1.92) ="4213" *+!DR{4213} *{\circ}; (1.3,2.26) ="1432" *+!UR{1432} *{\bullet};
(2.11,2.26) ="1423" *+!L{1423} *{\bullet};
(.99,2.41) ="4132" *+!DR{4132} *{\bullet};
(1.81,2.41) ="4123" *+!DL{4123} *{\bullet};
"3214";"2314" **@{-}; "3241";"2341" **@{.}; "3241";"3214" **@{=}; "2341";"2314" **@{:}; "2134";"1234" **@{-}; "2143";"1243" **@{-}; "1234";"1243" **@{-}; "2134";"2143" **@{-}; "4132";"4123" **@{-}; "1432";"1423" **@{-}; "1432";"4132" **@{*}; "4123";"1423" **@{*}; "4312";"3412" **@{-}; "4321";"3421" **@{.}; "4312";"4321" **@{.}; "3412";"3421" **@{-}; "4213";"2413" **@{o}; "4231";"2431" **@{o}; "4213";"4231" **@{:}; "2413";"2431" **@{:}; "1342";"1324" **@{=}; "3142";"3124" **@{=}; "1342";"3142" **@{*}; "1324";"3124" **@{*}; "2314";"2134" **@{=}; "3124";"3214" **@{-}; "3421";"3241" **@{=}; "3412";"3142" **@{=}; "3412";"3142" **@{*}; "1324";"1234" **@{-}; "1432";"1342" **@{-}; "4312";"4132" **@{*}; "1423";"1243" **@{*}; "2341";"2431" **@{.}; "4321";"4231" **@{.}; "2413";"2143" **@{:}; "2413";"2143" **@{o}; "4123";"4213" **@{.}; \end{xy}$$
We summarize our discussion in the following theorem:
\[maingraphical\] The following four combinatorial objects are isomorphic for any graph $G$ on $[n]$: $\bullet$ the graphical model rank test $\tau_{\mathcal{K}(G)}$, $\bullet$ the graphical tubing rank test $\tau^*_{\mathcal{K}(G)}$, $\bullet$ the fan of the graph associahedron $\Delta_G$, $\bullet$ the simplicial complex of all tubings on $G$.
We note that when the graph $G$ is a path of length $n$, $\Delta_G$ is the [*associahedron*]{}, and when it is an $n$-cycle, $\Delta_G$ is the [*cyclohedron*]{}. The number of classes in either the MSS test $\tau_{\mathcal{K}(G)}$ or the tubing test $\tau^*_{\mathcal{K}(G)}$ is the [*$G$-Catalan number*]{} of [@Postnikov2005]. This number is the classical Catalan number $\frac{1}{n+1} {2n \choose n}$ for the associahedron test. It equals ${2n-2 \choose n-1}$ for the cyclohedron test [@cyclohedron].
Let $n=4$ and let $G $ be the $4$-chain $\, 1$—$2$—$3$—$4$. Then $$\begin{matrix}
\mathcal{M}^G \!\! & = & \bigl\{
1 \perp \!\!\! \perp 3 \,|\, 24, &
1 \perp \!\!\! \perp 4 \,|\, 23, &
2 \perp \!\!\! \perp 4 \,|\, 13, &
1 \perp \!\!\! \perp 3 \,|\, 2, &
1 \perp \!\!\! \perp 4 \,|\, 2, &
1 \perp \!\!\! \perp 4\,|\, 3, &
2 \perp \!\!\! \perp 4 \,|\, 3 \bigr\},\\
(\mathcal{M}^G)^* \!\! & = & \bigl\{
1 \perp \!\!\! \perp 3 \, ,&
1 \perp \!\!\! \perp 4 \, , &
2 \perp \!\!\! \perp 4 \,, &
1 \perp \!\!\! \perp 3 \,|\, 4, &
1 \perp \!\!\! \perp 4 \,|\, 3, &
1 \perp \!\!\! \perp 4\,|\, 2, &
2 \perp \!\!\! \perp 4 \,|\, 1 \bigr\}.
\end{matrix}$$ The corresponding tests $\tau_{\mathcal{K}(G)}$ and $\tau_{\mathcal{K}(G)}^*$ are depicted in Figure \[GM\]. Note that contracting either class of marked edges on the permutohedron in Figure \[GM\] leads to the $3$-dimensional associahedron $\Delta_G$. The associahedron $\Delta_G$ is the Minkowski sum of the simplices $\Delta_K$ where $K$ runs over $$\mathcal{K}(G) \quad = \quad \bigl\{ \{1\},
\{2\}, \{3\}, \{4\}, \{1,2\}, \{2,3\}, \{3,4\}, \{1,2,3\}, \{2,3,4\}, \{1,2,3,4\} \bigr\} .$$ The $3$-dimensional simple polytope $\Delta_4$ has $14$ vertices, one for each of the $14$ tubings of $G$.
In our application of graphical rank tests, we found it more natural to work with the tubing test $\tau^*_{\mathcal{K}(G)} $ instead of the MSS test $\tau_{\mathcal{K}(G)}$. We refer to our companion paper [@cyclohedron] which gives a detailed discussion of the cyclohedron test and its applications. By the cyclohedron test we mean the tubing test $\,\tau^*_{\mathcal{K}(G)}\, $ where the graph $G$ is a cycle of length $n$.
Applying the tubing test to a data vector $u \in {\mathbb{R}}^n$ can be viewed as an iterative procedure for drawing a topographic map on the graph $G$. Namely, we encircle the vertices of $G$ by sets $U_1, \dots, U_n$ in the order $\delta_1, \delta_2, \dots, \delta_{n-1}$, with the following provision: if $\delta_i$ is next to be encircled and shares an edge with some vertex $j$ which has already been encircled by some $U_j$, then $U_i$ must also contain the circle $U_j$. The result is a collection $U$ of $n-1$ encircled sets $U_1,U_2,\ldots,U_{n-1}$, and this unordered collection of sets is the signature of $v$. The height $h_i$ of the $i$-th node in the topographic map for $v$ is the number of sets $U_j$ which contain $i$. We can identify the signature $U$ with the [*height vector*]{} $h = (h_1,h_2,\ldots,h_n)$, since $U$ can be recovered uniquely from the vector $h$. The map $u \mapsto h(u)$ can be interpreted as a [*smoothing of the data*]{}. Figure 6 displays the topographic map when the data vector is $\,u=(2.1,0.3 ,1.8,,2.0,1.1,0.1)$. Here $G$ is the $6$-chain $\, 1$—$2$—$3$—$4$—$5$—$6$. and the descent vector of $u$ equals $\delta=(1|5|3|2|4|6)$.
On counting linear extensions
=============================
In this paper, we have introduced a hierarchy of rank tests, which range from pre-convex to graphical. Convex rank tests are applied to data vectors $u \in {\mathbb{R}}^n$, or permutations $\pi \in S_n$, and determine their cones in a fan $\mathcal{F}$ which coarsens the $S_n$-fan. The significance of a data vector in such a test is measured by a certain p-value, whose precise derivation is described in [@cyclohedron]. Computation of that p-value rests on our ability to compute the quantity $\,|\, \tau^{-1} \bigl(
\tau(\pi) \bigr)\,|$, which is the number of permutations in the maximal cone of $\mathcal{F}$ corresponding to $\pi$. Recall that the cones of a convex rank test are indexed by posets $P_1,P_2,\ldots,
P_k$ on $[n]$, and our computations amount to finding the cardinality of the set $\mathcal{L}(P_i)$ of linear extensions of $P_i$.
The problem of computing linear extensions of general posets is \#P-complete [@Brightwell1991], so our task is an intractable problem when $n$ grows large. However, for special classes of posets, and for moderate values of $n$, the situation is not so bad. For example, in the up-down analysis of Willbrand [*et al.*]{} (see Example \[ex.updwn\]), we need to count all permutations with a fixed descent set, a task for which an explicit determinantal formula appears in Stanley [@Stanley1997 page 69]. We refer to [@Brown2007] for a detailed study of the combinatorics of these [*up-down numbers*]{}.
Likewise, there is an efficient (and easy-to-implement) method for the computing quantities $\,|\, \tau^{-1} \bigl(\tau(\pi) \bigr)\,|\,$ for any graphical graphical tubing test $\,\tau^*_{\mathcal{K}(G)}$, as defined in Section 5. Indeed, here the fan $\mathcal{F}$ is unimodular, and hence the posets $P_i$ are all trees. The special trees arising from a graph $G$ in this manner are known as [*$G$-trees*]{} [@Postnikov2005; @Carr2004]. The $G$-tree of a permutation $\pi$ is a representation of the poset $P_i$ as a tree $\,T \,=\,\tau^*_{\mathcal{K}(G)}(\pi)\,$ with the minimum value as the root and maximal values as the leaves. Suppose the root of the tree $T$ has $k$ children, each of which is a root of a subtree $T^i$ for $i=1,\ldots,k$. Writing $|T^i |$ for the number of nodes in $T^i$, we have $$|\, \tau^{-1}(T ) \, |
\quad = \quad \binom{\sum_{i=1}^k |T^i |}{ |T^1|, \ldots, |
T^k|} \left( \prod_{i=1}^k |\tau^{-1}( T^{i})| \right).$$ This recursive formula translates into an efficient iterative algorithm. Our implementation of this algorithm, when $G$ is the $n$-cycle, is the workhorse behind our computations in [@cyclohedron]. For a graph $G$, let ${\operatorname{nbhd}}(i)$ be the set of vertices $j$ such that there is an edge $(i,j)$ in $G$.
\[alg:TGMpermCount\](Permutation Counting)
[*Input:*]{} A data point $u$ as a descent permutation $\delta$ and a graph $G$.\
[*Output:*]{} The number of permutations with the same signature as $\delta$, $|\, \tau^{-1}\tau(\pi(u)) \, |$.
> [**Initialize:**]{}\
> An indexed set of largest enclosing sets $LE_1=\dots=LE_n=\emptyset$, and counter $c=1$\
> [**for**]{} $\delta_i$ in $\delta$:\
> Initialize $\ell$ an empty list of enclosed tree lengths\
> $LE_{\delta_i}=\{\delta_i\}$\
> $j$ in ${\operatorname{nbhd}}(\delta_i)$:\
> $LE_j \neq \emptyset$ and $j \notin LE_{\delta_i}$:\
> $LE_{\delta_i} = LE_{\delta_i} {\sqcup}LE_j$\
> append $|LE_j|$ to $\ell$\
> $c = c \cdot {\sum_i(\ell_i) \choose \ell}$\
> $j$ in $LE_{\delta_i}$:\
> $LE_j = LE_{\delta_i}$\
> [**Return**]{} the permutation count $c$
In the remainder of this section we discuss our method for performing these computations for an arbitrary convex rank test. The test is specified (implicitly or explicitly) by a collection of posets $P_1,\ldots,P_k$ on $[n]$. From the given permutation, we identify the unique poset $P_i$ of which that permutation is a linear extension, and we construct the [*distributive lattice*]{} $L(P_i)$ whose elements are the order ideals of $P_i$. Recall that an [*order ideal*]{} of $P_i$ is a subset $O$ of $[n]$ such that if $l \in O$ and $(k,l) \in P_i$ then $k \in O$. The set of all order ideals is a distributive lattice with meet and join operations given by set intersection $O \cap O'$ and set union $O \cup O'$.
The distributive lattice $L(P_i)$ is a sublattice of the Boolean lattice $\,2^{[n]}$, whose nodes are the $2^n$ subsets of $[n] = \{1,2,\ldots,n\}$, and we represent $L(P_i)$ by its nodes and edges (cover relations) in $\,2^{[n]}$. We write each edge in $2^{[n]}$ as a pair $(K,l)$ where $K \subset [n]$ and $l \in [n] \backslash K$. The edge in the Boolean lattice $2^{[n]}$ represented by the pair $(K,l)$ is the cover relation $\,K\, \subset \, K \cup \{l\}$.
Permutations in $S_n$ are in natural bijection with maximal chains in the Boolean lattice $2^{[n]}$. For example, the descent permutation $\delta=(4|2|1|3)$ corresponds to the maximal chain $\,\bigl(\emptyset, \{4\}, \{2, 4\}, \{1,2,4\}, \{1,2,3,4\}\bigr)\,$ in the Boolean lattice $\,2^{[4]}$. If the poset $P_i$ is the linear order $\delta$ then $L(P_i)$ is the subgraph of $2^{[4]}$ consisting of the five nodes in the chain and the four edges $\,(\emptyset,4) ,\,(\{4\}, 2),\, (\{2,4\},1)\,$ and $\, (\{1,2,4\},3)\,$ which connect them. The maximal chains in $2^{[n]}$ that lie in the sublattice $L(P_i)$ are precisely the permutations that are linear extensions of $P_i$. Therefore our task is to construct $L(P_i)$ and then count its maximal chains.
\[remlinex\] The linear extensions of the poset $P_i$ are in bijection with the maximal chains in the distributive lattice $L(P_i)$. See [@Stanley1997 Section 3.5] for further information on this bijection.
In general, $L(P_i)$ is the graph whose nodes are those subsets of $[n]$ which are order ideals in $P_i$, and the edges are $(K,l)$ where both $K$ and $K \cup \{l\} $ are order ideals in $P_i$. Our strategy in computing the graph which represents $L(P_i)$ is as follows. We start with a given permutation $\delta$ which lies in the class indexed by $P_i$. That permutation determines a maximal chain in $2^{[n]}$ which must lie in $L(P_i)$. We then compute a certain closure of that subgraph in $2^{[n]}$ with respect to the semigraphoid $\mathcal{M}$ under consideration. This is precisely what is done in Algorithm 21 below. Knowledge of the distributive lattice $L(P_i)$ solves our problem since the number of maximal chains of $L(P_i)$ can be read easily from the representation of $L(P_i)$ in terms of nodes and edges.
\[DistLattice\](Building the Distributive Lattice)
[*Input:*]{} A data point as a descent permutation $\delta$ and a semigraphoid $\mathcal{M}$.\
[*Output:*]{} A distributive lattice $L(P_i)$ representing the class of $\delta$ in the convex rank test $\mathcal{M}$.
> [**Initialize:**]{}\
> A set of confirmed lattice nodes, $\,\mathbb{H} =
> \bigl\{ \emptyset, \{\delta_1\}, \{\delta_1, \delta_2\}, \dots, \{\delta_1, \dots, \delta_n\} \bigr\}$\
> A set of checked lattice edges, $\,E \,=\, \bigl\{ (\{\delta_1, \dots, \delta_{n-1}\}, \delta_n) \bigr\}$,\
> where each pair has the form (history, next position).\
> A stack of edges waiting to be checked:\
> $\,W \,= \,
> \bigl[(\emptyset,\delta_1), (\{\delta_1\}, \delta_2), (\{\delta_1, \delta_2\},\delta_3), \dots,(\{\delta_1, \dots, \delta_{n-2}\}, \delta_{n-1}) \bigr]
> $\
> \
> [**While**]{} $W \neq \emptyset$:\
> Pop $(H,i)$ from the stack $W$\
> Add $(H,i)$ to $E$\
> $j$ such that $(H {\cup}\{i\}, j) \in E$:\
> $i {\! \perp \!\!\! \perp \!}j | H \in \mathcal{M}$:\
> Add $(H,j)$ to $E$\
> $H {\cup}\{j\} \notin \mathbb{H}$:\
> Add $H {\cup}\{j\}$ to $\mathbb{H}$\
> Push $(H {\cup}\{j\}, i)$ onto $W$\
> [**Return**]{} the distributive lattice $\,L(P_i)\,= \, \bigl(\mathbb{H},E \bigr) $\
Our program for performing rank tests implements Algorithm \[DistLattice\]. It accepts a permutation $\delta$ and a rank test $\tau$, which may be specified either
- by a list of posets $P_1,\ldots,P_k$ (pre-convex),
- or by a semigraphoid $\mathcal{M}$ (convex rank test),
- or by a submodular function $w : 2^{[n]} \rightarrow {\mathbb{R}}$,
- or by a collection $\mathcal{K}$ of subsets of $[n]$ (MSS),
- or by a graph $G$ on $[n]$ (graphical test).
The output of our program has two parts. First, it gives the number $|\mathcal{L}(P_i)|$ of linear extensions, where the poset $P_i$ represents the equivalence class of $S_n$ specified by the data $\pi$. It also gives a representation of the distributive lattice $L(P_i)$, in a format that can be read by the [maple]{} package [posets]{} [@Stembridge2004]. Our software for Algorithms \[alg:TGMpermCount\] and \[DistLattice\] and, more generally, for applying convex rank tests $\tau $ to data vectors $u \in {\mathbb{R}}^n$ is available at $\, {\tt bio.math.berkeley.edu/ranktests/} $.
In closing let us give a concrete illustration of our current ability to count linear extensions. We computed the number of linear extensions of the Boolean poset $\,P = 2^{[5]}\,$ consisting of all subsets of $\{1,2,3,4,5\}$. Our program ran in less than one second on a laptop and found that $$|L(2^{[5]})| \quad = \quad
14,807,804,035,657,359,360.$$ This computation was inspired by work in population genetics by Daniel Weinreich [@Weinreich2005] who reports the analogous calculation for $P = 2^{[4]}$.
Conclusions {#conclusions .unnumbered}
===========
This work describes the connections among algebraic combinatorics, non-parametric statistics and graphical models (statistical learning theory). Specifically, we have proved the equivalence between semigraphoids and convex rank tests. This result provides the background for the counterexamples given in [@counterexamples] and the rank tests which were applied to biological data in [@cyclohedron].
[99]{} L. Billera and I. Gelfand and B. Sturmfels, [*Duality and minors of secondary polyhedra,*]{} Journal of Combinatorial Theory Ser. B, 57 (1993) pp. 258-268.
G. Brightwell and P. Winkler, [*Counting linear extensions,*]{} Order, 8 (1991) pp. 225-242.
F. Brown, T. Fink and K. Willbrand, [*On arithmetic and asymptotic properties of up-down numbers,*]{} Discrete Mathematics, 307 (2007) pp. 1722-1736.
K. Brown, . Springer, New York, 1989.
M. Carr and S. Devadoss, [*Coxeter complexes and graph associahedra,*]{} Topology and its Applications, 153 (2006), pp. 2155-2168.
W. Cook and L. Seiford, [*The geometry of rank-order tests,*]{} The American Statistician, 37 (1983) pp. 307-311.
A. Dawid, [*Conditional independence in statistical theory,*]{} Journal of the Royal Statistical Society B, 41 (1979) pp. 1-31.
E. Gawrilow and M. Joswig, [*Polymake: a framework for analyzing convex polytopes,*]{} in Polytopes – Combinatorics and Computation, eds. G. Kalai and G. M. Ziegler, Birkhäuser, 2000, pp. 43-74.
R. Hemmecke, J. Morton, A. Shiu, B. Sturmfels and O. Wienand, [*Three counterexamples on semigraphoids*]{}, To appear in Combinatorics, Probability, and Computing (2008).
L. Lovász, [*Submodular functions and convexity,*]{} in Mathematical Programming: The State of the Art, eds. A. Bachem, M. Groetschel, and B. Korte, Springer, 1983, pp. 235-257.
F. Matúš, [*Ascending and descending conditional independence relations,*]{} in Proceedings of the Eleventh Prague Conference on Inform. Theory, Stat. Dec. Functions and Random Proc., Academia, B, 1992, pp. 189-200.
F. Matúš, [*On equivalence of Markov properties over undirected graphs,*]{} Journal of Applied Probability, 29 (1992) pp. 745-749.
F. Matúš, [*Towards classification of semigraphoids,*]{} Discrete Mathematics, 277 (2004), pp. 115-145.
J. Morton, A. Shiu, L. Pachter and B. Sturmfels, [*The cyclohedron test for finding periodic genes in time course expression studies,*]{} Statistical Applications in Genetics and Molecular Biology 6 (2007) pp. 1-21.
J. Pearl, [*Probabilistic Reasoning in Intelligent Systems: Networks of Plausible Inference,*]{} Morgan Kaufman, San Mateo CA, 1988.
E. J. G. Pitman, [*Significance tests which may be applied to samples from any populations,*]{} Supplement to the Journal of the Royal Statistical Society, 4 (1937) pp. 119-130.
A. Postnikov, [*Permutohedra, associahedra, and beyond,*]{} Preprint (2005), available at [http://arxiv.org/abs/math/0507163]{}.
A. Postnikov, V. Reiner, and L. Williams, [*Faces of simple generalized permutohedra,*]{} Preprint (2006), available at [http://arxiv.org/abs/math.CO/0609184]{}.
R. P. Stanley, [*Enumerative Combinatorics*]{}, Volume I. Cambridge University, 1997.
J. Stembridge, [*Maple packages for symmetric functions, posets, root systems, and finite Coxeter groups.*]{} Available at [www.math.lsa.umich.edu/$\sim$jrs/maple.html]{}.
M. Studený, [*Conditional independence relations have no finite completeness characterization,*]{} Kybernetika, 25 (1990) pp. 72-79.
M. Studený, RR. Bouckaert, and T. Kocka, [*Extreme supermodular set functions over five variables*]{}, Institute of Information Theory and Automation, Research report n. 1977, Prague, 2000.
M. Studený, [*Probabilistic Conditional Independence Structures*]{}, Springer Series in Information Science and Statistics, Springer-Verlag, London, 2005.
J. Tits, [*Le problème des mots dans les groupes de Coxeter*]{}, Symposia Math., 1, (1968) pp. 175-185.
D. Weinreich, [*The rank ordering of genotypic fitness values predicts genetic constraint on natural selection on landscapes lacking sign epistasis,*]{} Genetics 171 (2005) pp. 1397-1405.
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|
Background
==========
Depression is frequent and chronic in older adults. According to research on community-dwelling older adults, the proportion of individuals reporting depressive symptoms is 2.8% to 35% \[[@B1]\]. The natural course of later-life depressive disorders is poor: a 6-year follow-up study showed that 76% of patients followed an unfavorable but fluctuating course or a severe chronic course of depression, and only 23% of patients experienced full remission \[[@B2]\].
Depression in older adults deteriorates the sufferers' quality of life (QOL) more than many other chronic diseases \[[@B3]\]. It gives a negative impact on patients' QOL in various ways, including wellbeing, perceived physical functioning, bodily pain, and general health perceptions \[[@B4]\]. The mortality rate of people with depression was found to be 1.8 times larger than that of non-depressed subjects due to suicide, unhealthy habits, and medical illnesses \[[@B5]\].
Depression is also costly. Depressed older adults use more outpatient resources than those without depression, including frequent appointments and increased laboratory and radiographic tests. They also have more non-specific medical complaints, and this is associated with increased total ambulatory care costs \[[@B6]\]. A study in the United States found that the additional medical cost per one depressed older adult was USD 686 for 1 year and USD 5,271 for 4 years \[[@B7]\].
As the world population continues to age, there is an urgent need therefore for medicine and social policy to find ways to reduce and prevent depression in older adults in the community.
However, to the best of the authors' knowledge, no simple, effective interventions currently exist for the prevention of depression in the elderly population \[[@B8]\]. The existing prevention studies have limitations in study design or rely on time-consuming psychotherapy, which is unrealistic for a community prevention program. They need weekly sessions with a duration of 45 to 120 minutes for 6 to 10 weeks \[[@B9]-[@B12]\], and trained workers or specialists \[[@B9],[@B11]-[@B14]\]. The subjects of most of the studies were not general people in community but those with specific disease or physical symptoms such as diabetes \[[@B10]\], macular degeneration \[[@B11]\], hip fracture \[[@B15]\], chronic pain \[[@B12]\], and most of the studies recruited subjects in clinical settings \[[@B10],[@B11],[@B13],[@B15]\]. Some studies lacks sample size calculation \[[@B9],[@B10],[@B12]\] and were quasi-randomized controlled trials \[[@B10],[@B14]\].
A postcard intervention was first carried out in the United States in 1976 for suicide prevention among discharged major depression patients. Researchers sent 24 letters over 5 years and reported that this significantly decreased suicide rates for the first 2 years and tended to lower suicide rates up to 13 years in total \[[@B16],[@B17]\]. Three more postcard intervention trials were conducted in Israel and Australia in 2005, 2010, and 2011, that focused on the prevention of drug overdose or self-harm. The results showed significant decrease in the number of drug overdose episodes, and the rates of suicide ideation and suicide attempts \[[@B18]-[@B21]\]. The prevention of depression in patients with a recent stroke by postcard is also planned \[[@B22]\].
The advantage of the postcard intervention is its low personal and financial cost: it only requires paper, pencil, and postage. Therapists are not required to visit the participants and vice versa. If the postcards do not contain medical and related information, a wide range of people such as elementary school students can take part in the intervention program.
This paper describes the study protocol for a pragmatic, randomized controlled trial designed to examine the effectiveness of the postcard intervention for improvement of depression in community-dwelling individuals aged 65 years or older. This study will focus in particular on those who have increased depressive symptoms and insufficient social support at baseline, because it is expected that the intervention is more effective among such individuals.
Objectives
----------
For community-dwelling older adults (aged 65+ years) reporting symptoms of depression and limited social support, this study aims to: (1) examine the effectiveness of a postcard intervention for the improvement of depressive symptoms; (2) evaluate the effectiveness of a postcard intervention in global geriatric health indicators such as quality of life (QOL) and the activities of daily living (ADL); and (3) assess the acceptability of the postcard intervention.
Methods and design
==================
Ethical approval
----------------
The Institutional Review Board (IRB) of the Graduate School of Medicine, Kyoto University has reviewed and approved the study protocol and informed consent documents (E1658, 12 February 2013).
Study setting
-------------
The study will be conducted in the community of a rural town, located in the center of Shikoku, one of the four main islands in Japan. Its main industries are agriculture and forestry. It has a population of 4,407, of whom 1,711 (38.8%) are aged 65 years or older.
Our study team has been conducting a longitudinal observational study in this community since 2004, in which we administer comprehensive geriatric assessments and report results and make referral to physicians as necessary. This observational study has been approved by the IRB of the Graduate School of Medicine, Kyoto University (E-18), and written informed consent has been obtained from all the participants.
Study design
------------
### Design overview
We will conduct a pragmatic, non-blinded, parallel comparison, randomized controlled trial using Zelen's design in this community. Figure [1](#F1){ref-type="fig"} depicts the participants' flow. Participants will be selected based on the questionnaire surveys including a self-rated depression scale. Participants will then be randomized to the intervention or no-intervention groups at a 1:1 ratio using computer-generated random numbers. Randomization will be stratified by gender and self-rated depression scale score. To conceal group assignments, random number generation and group allocation will be conducted at the same time by an independent epidemiologist not involved in the participant recruitment or intervention or assessments. Informed consent forms will be mailed to those assigned to the intervention group. Postcards will be sent to the consenting participants once a month for eight consecutive months. They will be enclosed with a self-addressed stamped envelope to facilitate non-obligatory replies by mail. As this is a pragmatic study, any treatment outside the trial will be permitted. Self-reported outcomes will be measured at baseline and post intervention; the questionnaire about participants' impression of effectiveness of the intervention and their recollection of the number of postcards received will be measured after treatment.
{#F1}
### Zelen's design
The study uses a randomized controlled trial with the single consent version (Zelen's design) \[[@B23],[@B24]\]. This is a variation of the standard randomized controlled design in which participants are randomized to intervention or control before consent is sought. Consent is obtained from the intervention group only after the randomization. The most important advantage of this method is that participants know the intervention they will receive at the time of consent. In a conventional randomization, participants who agree to join the study may retract their consent or continue participation with reluctance after finding out their assigned intervention, whereas the Zelen's method requires a decision only on the allocated intervention.
The main ethical concern is that consent is obtained only from the intervention group. To overcome this point, the revised Zelen method has been proposed \[[@B25]\]. This method is a combination of an observational study and a randomized controlled trial. Eligible participants first consent to an observational study, and then they are randomly assigned to intervention and control groups; those in the intervention group are asked to consent to participate in the study. Those in the control arm are not informed of this, but will be followed in the observational study if they agreed. Our protocol will follow this method.
Participants
------------
### Inclusion criteria
Participants will meet the following criteria: (1) being 65 years of age or older; (2) exhibiting symptoms of depression with a score of ≥4 on the self-rated 15-item Geriatric Depression Scale (GDS-15); and (3) reporting that they eat meals alone in the questionnaire.
The study will include individuals with sub-threshold depression. Indicated prevention aimed at sub-threshold depression is said to be most efficient in terms of costs and benefit \[[@B26]\]. As there are no agreed-upon definitions of sub-threshold depression on GDS-15, the study will include those with scores ≥4 points on the GDS-15, which is 1 point below the established cutoff to detect major depression \[[@B27]\].
Considering the nature of the intervention, the study will target those who are at risk of social isolation. We hypothesize that eating alone, rather than living alone, better represents the risk of isolation. In fact, eating alone was more strongly associated with depression than living alone among community-dwelling older adults \[[@B28]\].
### Exclusion criteria
Participants will be excluded if they cannot understand and sign the informed consent form. Those who currently reside in a hospital or institution will be excluded.
Sample size
-----------
To detect an effect size of 0.5 with *P* = 0.05 at 80% power, 63 participants are required per group. Assuming a non-consent and dropout rate of 30%, a total of 180 subjects are needed. Based on the results of our previous observational study performed in the same town in 2012, this sample size is believed to be feasible.
Intervention
------------
Letters written on A4 paper with some colorful illustrations will be sent in a sealed envelope once a month for 8 months. The letter will be composed of two parts: the first part will be a handwritten reply to messages returned from the participants if there are replies or comments, which aims to increase social connectedness and to enhance their self-respect; the second part will be seasonal greetings or news of the month from Kyoto, Japan, where the study authors are located, printed by computer. Kyoto is one of the most famous cultural centers in Japan and hosts various historical events that we believe will be of interest to participants living far from Kyoto.
Although a self-addressed stamped reply card will be enclosed with the letter, replying is not mandatory; this will be indicated on the reply card.
Outcomes
--------
### Primary outcomes
Primary outcomes will be the change in GDS-15 score as the measure of effectiveness.
### Secondary outcomes
Secondary outcomes will be self-rated QOL as evaluated by visual analogue scales, self-rated basic ADL, and self-rated advanced ADL.
### Other outcomes
The subjective sense of effectiveness of the intervention, recollection of the number of intervention mailings received, and the number of mailed replies will be evaluated to measure acceptability of the postcard intervention.
Outcome measures
----------------
### GDS-15
The GDS-15 is a validated depression scale comprised of 15 items. This scale was developed to exclude the effects of non-specific somatic symptoms such as anorexia and insomnia, which are frequently observed among elderly populations \[[@B29],[@B30]\]. Each item can have two answers: yes or no. The highest possible score is 15, indicating the most severe depressive state. Using a cutoff point of 5, the GDS-15 has a sensitivity of 92% and a specificity of 81% to detect major depression as ascertained by a structured clinical interview \[[@B27]\].
### QOL
Subjective QOL will be assessed using a 100-mm visual analogue scale (lowest QOL on the left end of the scale, and highest on the right) for the following five items: subjective sense of health; relationship with family; relationship with friends; financial state; and subjective happiness \[[@B31],[@B32]\].
### Basic ADL (BADL)
Each participant will rate his or her independence with respect to seven items corresponding to basic activities of daily living (BADL). Specifically, these items are as follows: walking, ascending and descending stairs, feeding, dressing, going to the toilet, bathing, and grooming. Each BADL item is evaluated based on four levels of competence: 3, completely independent; 2, requiring some assistance; 1, requiring much assistance; 0, completely dependent. The scores for the seven BADL items will be summed for a total score of 0 to 21 \[[@B33],[@B34]\].
### Advanced ADL (AADL)
For higher-level functional capacity, the Tokyo Metropolitan Institute of Gerontology Index of Competence rating scale will be used to measure competence \[[@B35],[@B36]\]. This scale consists of 13 items encompassing three sublevels of competence: (1) instrumental ADL (five items: the ability to use public transport, buy daily necessities, prepare a meal, pay bills, and handle banking matters); (2) intellectual activities (four items: the ability to complete forms, read newspapers, read books or magazines, and show interest in television programs or news articles on health-related matters); and (3) social roles (four items: the ability to visit friends, give advice to relatives and friends in confidence, visit someone at the hospital, and initiate conversation with younger people). Because each item is rated as 'yes' or 'no', instrumental ADL has a score range of 0 to 5, intellectual ADL 0 to 4, and social role ADL 0 to 4.
### Sociodemographic and other information
Data about age, sex, eating alone, and living alone will be obtained through a self-reported questionnaire. Participants' subjective sense of the effectiveness of the intervention will be evaluated on a four-point scale ranging from 0 (not effective) to 4 (very effective).
Management of adverse events
----------------------------
We expect that no adverse events will result from the postcards. However, if an emergent situation such as a high risk of suicide is suspected based on the reply card, a certified psychiatrist will evaluate the participant and refer him/her to the hospital if needed.
Statistical analysis
--------------------
The time × group interaction for baseline and post intervention will be analyzed using a generalized linear mixed model, which enable us to analyze data even when they have missing values. Sensitivity analysis will be conducted by way of ANCOVA with data imputed by a multiple imputation method and with completer's data, using baseline data alone or with ADL score as a covariate and post-intervention data as dependent variables. The homogeneity of variance will be analyzed with Leven's test. Statistical analysis will be performed using SPSS ver. 20.0 (IBM Inc., Armonk, NY, USA).
Discussion
==========
The study protocol describes the design of a pragmatic randomized controlled trial to verify the efficacy of postcard intervention to prevent and improve depression among community-dwelling older adults in Japan. This is the first application of the postcard intervention for depression of community-dwelling older adults. The advantage of the postcard intervention is its low human and financial cost, which cannot be matched by other existing approaches such as psychotherapy. This intervention can be carried out by anyone who can write a letter. Its application will be broad.
There are three advantages to the study. First, the study is set in the community, whereas previous postcard interventions were conducted in clinical settings \[[@B16]-[@B19],[@B21]\]. Considering the potential of postcard interventions, however, application to a broader field is desirable; the application in community settings targeting local residents is but one of them. Second, the study focuses on people who eat alone, not those who live alone. The supposed effect of postcard intervention is to make a connection with people, thereby reducing feelings of isolation. Even if a person is surrounded by many other people, he or she will be lonely unless people pay attention to him or her. In this sense, eating alone rather than living alone can reflect true isolation \[[@B28]\]. Third, the study will evaluate ADLs. Previous prevention studies did not consider their effect on ADLs. However, as the participants in this study are older adults whose ADLs have bidirectional interactions with depressive mood, the influence of ADLs should be considered and the effects of the intervention on AADL in particular should be evaluated.
If the efficacy of the postcard intervention for depression in community-dwelling older adults is verified, it will be a milestone in community intervention.
Trial status
============
Participant recruitment will begin in June 2013.
Abbreviations
=============
GDS-15: 15-Item geriatric depression scale; AADL: Advanced activities of daily living; ADL: Activities of daily living; BADL: Basic activities of daily living; IRB: Institutional review board; QOL: Quality of life; VAS: Visual analogue scale.
Competing interests
===================
TAF has received honoraria for speaking at CME meetings sponsored by Asahi Kasei, Eli Lilly, GlaxoSmithKline, Mochida, MSD, Otsuka, Pfizer, Shionogi, and Tanabe-Mitsubishi. He is a diplomate of the Academy of Cognitive Therapy. He has received royalties from Igaku-Shoin, Seiwa-Shoten, and Nihon Bunka Kagaku-sha. He is on the advisory board for Sekisui Chemicals and Takeda Science Foundation. The Japanese Ministry of Education, Science, and Technology, the Japanese Ministry of Health, Labor and Welfare, and the Japan Foundation for Neuroscience and Mental Health have funded his research projects. All the other authors report no competing interests.
Authors' contributions
======================
HI and KM made substantial contribution to the conception and design of the study, were involved in drafting the manuscript, and will be responsible for the administration and direction of the study as well as the analysis and interpretation of data. TAF made a substantial contribution to the conception of the study, was involved in drafting the manuscript, and will be responsible for the analysis and interpretation of data. TF, KO, and EF assisted with the negotiation with the local government, and will be responsible for the preparation of the study materials as well as the analysis and interpretation of data. YI, YK, WC, and MM will be responsible for data collection. RS contributed to the conceptualization of the study design and will be responsible for data analysis and interpretation. All authors read and approved the final manuscript.
Acknowledgements
================
This study is supported by a grant from the Japan Society of the Promotion of Science: Scientific Research A (23241079).
|
Welcome
This advice-column-style blog for SLPs was authored by Pam Marshalla from 2006 to 2015, the archives of which can be explored here. Use the extensive keywords list found in the right-hand column (on mobile: at the bottom of the page) to browse specific topics, or use the search feature to locate specific words or phrases throughout the entire blog.
Small, Gentle Jaw Control
By Pam Marshalla
Q: My client lowers the jaw too much when he speaks, and he has a frontal lisp. He tends to clench the jaw when I tell him to hold the jaw up. I may have taught him this when using a bite stick to position the jaw. Not sure what to do now.
As you have discovered, making a strong crushing bite on a firm object is not what he needs. The term “jaw stability” does not mean “jaw rigidity.”
Our clients need to learn how to hold the jaw high position that is still flexible. The jaw moves in a very small subtle up-down pattern during speech. It does not clench.
I teach this with a firm flexible straw instead of a bite stick. Place the straw between the molars on one side along the length of the dental line. Have the client bite gently into the straw, and then lower the jaw again all the while holding on to the straw with the molars.
This will teach him the very small movements the jaw needs to make––a little higher for /s/ and a little lower for the vowels.
The jaw moves in big up-down oval-shaped patterns during chewing, but it moves in very small up-down patterns during speech. A chewing activity could help him experience the gross motor pattern, and the straw then can help him feel it in the small, restricted range of jaw movements needed for speech. — Use a mirror throughout so he can see what’s going on.
Many therapists are trying to teach “jaw strength” but strength of the jaw muscles have nothing to do with speech. The motor pattern is what needs to be taught––the soft, gentle, very small up-down movements the jaw makes during speech.
Straws come in basically four sizes– coffee stirrer straws, soda straws, McDonald’s Straws, and Milkshake Straws. Paper ones are no good. Use plastic… and they come in various stiffnesses. I usually keep all kinds around to experiment with what I need. |
Wisconsin native, conservative critic of everything.
"Once abolish the God, and the government becomes the God." ---G K Chesterton
"The only objective of Liberty is Life" --G K Chesterton
"Fallacies do not cease to be fallacies because they become fashions" --G K Chesterton
"A man can never have too much red wine, too many books, or too much ammunition." -- Rudyard Kipling
Monday, January 22, 2007
Solzhenitsyn on Law
People in the West have acquired considerable skill in using, interpreting and manipulating law, even though laws tend to be too complicated for an average person to understand without the help of an expert. Any conflict is solved according to the letter of the law and this is considered to be the supreme solution. If one is right from a legal point of view, nothing more is required, nobody may mention that one could still not be entirely right, and urge self-restraint, a willingness to renounce such legal rights, sacrifice and selfless risk: it would sound simply absurd. One almost never sees voluntary self-restraint. Everybody operates at the extreme limit of those legal frames. An oil company is legally blameless when it purchases an invention of a new type of energy in order to prevent its use. A food product manufacturer is legally blameless when he poisons his produce to make it last longer: after all, people are free not to buy it.I have spent all my life under a communist regime and I will tell you that a society without any objective legal scale is a terrible one indeed. But a society with no other scale but the legal one is not quite worthy of man either. A society which is based on the letter of the law and never reaches any higher is taking very scarce advantage of the high level of human possibilities.So what IS 'the meaning of the word is....'? |
Voices of Impact – Dr. Judy Salerno / Health of Women [HOW] Study
Research, Leadership
By: Judy Salerno, President & CEO
Susan G. Komen
Hello and thanks to all of you for such a warm welcome as I assume my new role as President and CEO of Susan G. Komen. I’ll be posting on our blog frequently, and I hope you’ll come back many times for regular updates.
Although I’ve been at Komen for just a few weeks, I’ve been familiar with Komen’s work for many years in my roles at the Institute of Medicine, the Department of Veterans Affairs and the National Institutes of Health. I thought I knew a lot about this remarkable organization before I joined it. But I’m learning quickly that Komen has a depth and breadth that many people aren’t even aware of.
There’s news here on several fronts, starting with the innovation of several major breast cancer organizations working together on a single research project. There are well over 1,000 breast cancer organizations in the U.S. Many were founded locally to help a local hospital or to honor a family member or friend with breast cancer. The larger organizations, like Komen, YSC, Dr. Love’s foundation and others, were established with specific goals. Some fund just research; others are active only in public policy; others reach out to specific groups of breast cancer survivors. Komen works in all of these areas: research, public policy and community health, and we do this in 30 countries worldwide.
Whatever the specific focus of breast cancer organizations, our primary goals are the same: to end breast cancer and save lives. Ultimately, we seek not just treatments, but scientifically proven prevention strategies. There are many opportunities to advance our mutual goals by working together, and this new collaboration of the Dr. Susan Love Research Foundation, Komen and the Young Survival Coalition is a great start. We hope other breast cancer organizations will join us as well.
We have two priorities in this partnership. We want all people to participate in the Health of Women [HOW] breast cancer study, developed by Dr. Love and Dr. Leslie Bernstein in 2012 as a means of identifying the causes of, and prevention strategies for, breast cancer. The HOW project is precisely the kind of large-scale, long-term study we need to identify patterns common to breast cancer patients – medical history, family history, geographic location, reproductive history, ethnicity or lifestyle, for example. Dr. Love is “democratizing” the research by inviting everyone – male, female, old, young, breast cancer survivor or not – to participate in the HOW study. Even if you are not a breast cancer patient, comparing information about you can help us identify areas which make a person at greater risk for developing breast cancer.
Our second objective is aimed at improving breast cancer treatment by identifying issues that breast cancer survivors are facing today. These include side effects of current treatments such as lymphedema, depression, mobility issues, “chemo brain,” loss of sexual appetite, etc. Almost every breast cancer survivor knows of these side effects, but many don’t talk about them.
I’ve seen this phenomenon myself as a practicing physician. Many patients are made to feel that they should be grateful to be alive, and so they’re reluctant to report other effects of treatments that are saving their lives. For the medical community, there are few places where such side effects are widely documented. This means that healthcare providers themselves may not know how widespread some side effects really are, or which of their patients are likely to experience them. |
Graphene is an allotrope of carbon comprising a sheet of carbon atoms generally including a single atomic layer thickness. Graphene possesses exceptional electronic and material properties, including an ultra-high electron mobility. However, graphene alone is not generally considered to be suitable as a switching medium because it lacks a bandgap. A bandgap can be established structurally in graphene, however such a structural bandgap can degrade or destroy the band structure, which hinders the ultra-high electron mobility otherwise provided by graphene. Accordingly, graphene devices having structurally-established bandgaps generally deliver a poor ON-state current. |
Article content
Serious collisions closed down sections of highways in eastern Ontario on Monday, as freezing rain made driving treacherous in some parts of the province.
Both Highway 401 westbound and Highway 7 were closed in sections at times, leaving frustrated commuters stuck in back-to-back traffic with delayed or cancelled plans between Ottawa and Toronto.
We apologize, but this video has failed to load.
tap here to see other videos from our team. Try refreshing your browser, or Highways reopen after icy day in Eastern Ontario Back to video
Highway 401 westbound at Brighton was closed for hours due to numerous collisions, including a tractor trailer that dumped diesel fuel, but reopened shortly before 7 p.m.
Highway 7, which had been closed since mid-afternoon between Highway 37 and Highway 41, also reopened to traffic shortly after 9 pm. |
The transport of octamethylcyclotetrasiloxane (D4) and polydimethylsiloxane (PDMS) in lightly cross-linked silicone rubber.
The transport of octamethylcyclotetrasiloxane (D4), one of the major constituents of silicone fluids and rubbers, and low viscosity polydimethylsiloxane oil into a silica filled cross-linked silicone elastomeric rubber was measured as a function of temperature, cross-link density of the rubber, and concentration of the D4 in methanol solution. A small amount of material, approximately 3 wt%, is extracted from the rubber with hexane. The extraction process has a large effect upon D4 solubility in the rubber, increasing from approximately 160 to 180 wt% after extraction. The heats of solution for both penetrants into the rubber are essentially zero and the activation energies for diffusion are small, approximately 8 and 15 kJ molt(-1) for D4 and PDMS, respectively. The diffusion process is Fickian and the diffusion coefficient of D4 into silicone/silica rubbers is essentially independent of concentration over the concentration investigated, i.e. from 1 to 100 vol% D4 in methanol. The permeability, i.e. the product of the diffusion coefficient and the solubility, decreases rapidly for D4 concentrations less than 50 vol% (0.1 mol fraction). This suggests that the permeation of D4 out of any encapsulation device, such as a silicone breast implant, is linearly dependent upon the concentration of D4 in the prosthesis. Swelling is isotropic and was measured by dimensional changes in rectangular samples and correlates well with the volume of D4 sorbed. |
Yoona, who starred in KBS 2TV’s Monday-Tuesday drama, “Love Rain”, (script by Oh Sooyeon, directed by Yoon Sukho), talked with Star News on May 30th through SM Entertainment. She said, “During ‘Love Rain’, I reflected on myself and considered it to be valuable time.”
Yoona continued, “It’s been a while since I was in a drama. ‘Love Rain’ is a drama that has allowed me to experience a lot of different emotions which I couldn’t feel before in my previous works. But above all, it was good to have time to reflect on myself once more.”
Yoona also mentioned, “I was even happier that I got to work with good actors and staff. I will try my best to grow and improve for everyone.”
She also said to the viewers, “Thank you so much to those who showed their support for ‘Love Rain’.” She told them, “Look forward to Yoona’s different image,” adding that, “She will become a more mature actress.”
The last episode of “Love Rain” aired on the May 29th, which brought a conclusion to the love story between Jang Geunsuk and Yoona’s characters, Seo Joon and Jung Hana.
Source: Star News
Translated by: jyhwang@soshified
Edited by: nicholys@soshified, minigiglo@soshified, SeraphKY@soshified, Best Served Soup@soshified, MoonSoshi9@soshified
Have a news item that you think Soshified should know about? Leave us a tip or e-mail us at tip@soshified.com.
Follow us on Twitter: http://twitter.com/soshified for the latest on Girls’ Generation. |
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• Explicit risk controls: We adhere to clearly defined risk controls that are linked to market volatility and manager performance. These dynamic risk measures allow us to limit portfolio risk when needed, and increase risk when warranted.
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The strategy invests across a range of sectors, including Treasurys and agencies, investment grade credit, high yield and bank loans, emerging market debt, and securitized debt.
Allocations to each sector can be customized to meet your specific investment objectives, with the ultimate goal of providing a portfolio that offers multiple sources of alpha. |
Jamie Condliffe
Bike theft in San Francisco is high and rising, so the city's decided to take a technological approach to catching the culprits: tagging expensive bikes with GPS trackers so it can find out who's stealing them—and then charge them with felony.
The New York Times reports that a small team at the San Francisco Police Department are trading traditional techniques for transponders and iPhones, tracking crooks in real time and posting the results to the @SFPDBikeTheft Twitter account. The officers use expensive bicycles—to ensure that they're taken and that the thieves can be charged with a felony—which are locked up then inevitably nabbed, then traced. The Times describes one recent success:
Recently, for example, a thief took a $1,500 bicycle from outside a train stop and pedaled off into the sunset. But 30 minutes later, Officer Friedman and his team, having tracked the bike, converged on the rider at a park.
Similar schemes have been put to use on college campuses and in cities like Vancouver and Sacramento. But with San Francisco bike theft rising 70 percent from 2006 to 2012, with 4,035 bicycles a year now being stolen, clearly it's time for the city to throw whatever it can at the problem.If your city isn't helping you out with GPS bait bikes, though, you better know how to get your bicycle back once it's stolen. Good luck! [New York Times] |
I agree with Irvine, I don't think Lecavalier is going anywhere now that Yzerman is on board. Lecavalier has a highly unmovable contract, TB will never get fair value for him. Best to keep him and build your team around him I think.
Guest5526
Posted - 06/05/2010 : 09:55:38 But really, who cares! he has done nothing for the past couple years!
Guest5526
Posted - 06/05/2010 : 09:54:07
quote:Originally posted by Guest7281
quote:Originally posted by Guest9262
quote:Originally posted by Axey
quote:Originally posted by Guest9262
Lecavalier is going to Montréal but this wont make Montréal a better team.
But in the long term this will really help Tampa Bay.
How won't it make the Habs a better team? riddle me that one
Well the trade is not made yet but the Habs would have to trade an aweful lot to get him. Draft picks, good young players and a couple prospects. And plus they would have to depart some players to make room for his salary.
And Lecavalier even though he brings a lot to a team and is one of the best forwards in the NHL, he can't do it all by himself. They need depth and Montréal has that right now but if they give what Tampa Bay is asking they will lose a lot of depth.
So this is why. But like I said the trade isn't made yet so lets just wait and see IF it happens and what Montréal will give.
Actually Montreal has over 30 million in cap space so they wont have to move any salary should they pick him, especially considering Tampa would get a couple of roster players in return freeing up even more cap room.
Actualy the Habs dont have 30 mill in cap space, where did you get that number from? They have 16 mill to play with, it's an actual fact! But they have to sign both young goalies, that should take up at least 7 to 8 mill. , and I know alot of ppl on here are claiming the Habnots will be trading Price, but that's not so clear cut, cuz his value has gone down with 2 average seasons!If they did covet Vinny, I'm just speculating Gomez hefty contract would have to go, as they need to fill roster spots, thru FA, trades, or minors!
Guest4266
Posted - 06/05/2010 : 08:29:22 I think that he should go to the Toronto Maple Leafs.
With new owners, new general manger & a new (soon-to-be named) head coach arriving in Tampa this year
Will Vinny be traded? If so, to who?-
I'm of the frame of mind that Yzerman keeps Lecavlier. In fact, I believe if anybody (as a GM/Coach), can get Lecavlier back to his old form, it is Stevie.
Perhaps with all of these changes, and owners who seem to actually care about more than just the money coming in/out, about creating a real team, maybe Vincent can rejuvinate his on-ice success.
So, I believe Vinny stays. Also, I think with Yzerman at the reigns, Vinny can come back to form.
Irvine/prez.
Deaner
Posted - 06/04/2010 : 14:10:14 Vinny aint going to the habs...I can see Yzerman gettin rid of his ass though.
Rambo2305
Posted - 07/06/2009 : 07:39:48 Going back to the draft....was anyone else annoyed with the fact that the Habs fans cheered like hell when Bettman was getting ready to annouce a trade, and gave a nice "booo" when they realized it wasn't for Vinny?
Get over yourselves, Lecavalier will never wear the Habs jersey, correction, maybe when his contract with TB ends.
"Most people spend time and energy going around problems, rather than trying to solve them" - Henry Ford
Posted - 07/05/2009 : 13:16:37 Vinny will stay with the lightning,they wont give away a good player.Every team wants him though but hes worth alot
PhillyFan12Philly Rules!!!
brentrock2
Posted - 07/03/2009 : 06:24:40 I'm sure now that Vinny will stay with tampa bay because if Montreal got him they would be very close over the salary cap.Montreal won't get him because they already have Gomez, Gionta and Cammaleri.
Guest8571
Posted - 06/30/2009 : 17:38:11
quote:Originally posted by Porkchop73
Doubt Vinny goes to Montreal after the Habs traded for Gomez. Too much money would be tied up in two players.
I agree totally. They cannot tie that much money into two older players and expect to succeed next year and in the future. Especially with half their team being FAs.
Porkchop73
Posted - 06/30/2009 : 17:03:23 Doubt Vinny goes to Montreal after the Habs traded for Gomez. Too much money would be tied up in two players.
Guest7281
Posted - 06/29/2009 : 13:02:51
quote:Originally posted by Guest9262
quote:Originally posted by Axey
quote:Originally posted by Guest9262
Lecavalier is going to Montréal but this wont make Montréal a better team.
But in the long term this will really help Tampa Bay.
How won't it make the Habs a better team? riddle me that one
Well the trade is not made yet but the Habs would have to trade an aweful lot to get him. Draft picks, good young players and a couple prospects. And plus they would have to depart some players to make room for his salary.
And Lecavalier even though he brings a lot to a team and is one of the best forwards in the NHL, he can't do it all by himself. They need depth and Montréal has that right now but if they give what Tampa Bay is asking they will lose a lot of depth.
So this is why. But like I said the trade isn't made yet so lets just wait and see IF it happens and what Montréal will give.
Actually Montreal has over 30 million in cap space so they wont have to move any salary should they pick him, especially considering Tampa would get a couple of roster players in return freeing up even more cap room.
Guest9262
Posted - 06/28/2009 : 12:46:44
quote:Originally posted by Axey
quote:Originally posted by Guest9262
Lecavalier is going to Montréal but this wont make Montréal a better team.
But in the long term this will really help Tampa Bay.
How won't it make the Habs a better team? riddle me that one
Well the trade is not made yet but the Habs would have to trade an aweful lot to get him. Draft picks, good young players and a couple prospects. And plus they would have to depart some players to make room for his salary.
And Lecavalier even though he brings a lot to a team and is one of the best forwards in the NHL, he can't do it all by himself. They need depth and Montréal has that right now but if they give what Tampa Bay is asking they will lose a lot of depth.
So this is why. But like I said the trade isn't made yet so lets just wait and see IF it happens and what Montréal will give.
Axey
Posted - 06/28/2009 : 10:40:50
quote:Originally posted by Guest9262
Lecavalier is going to Montréal but this wont make Montréal a better team.
But in the long term this will really help Tampa Bay.
How won't it make the Habs a better team? riddle me that one
JOSHUACANADA
Posted - 06/26/2009 : 11:37:57
quote:Originally posted by brentrock2
I think Vincent will go to Montreal because that is really the only team really wanting him. There is a 75% chance he will go to Montreal than any other team.(Hopefully Montreal gets him)
brentrock2
Actually, I would bet if we had the 30 GM on this site picking there fantasy teams with a salary Cap, 2/3 would make room for Vinny. He is just not available to other teams. Montreal is the only team with a snowballs chance in hell. Tell me Burke wouldn't trade Kaberle, Kubina and a 1st round for Lecavalier. You just can't find guys with 10 years in the league, 100 point seasons, 40-50 goal multiple seasons and a bonified leader in the room. He is everything they billed Sundin to be and he's Canadian with a stanley cup ring.
Guest9262
Posted - 06/26/2009 : 11:06:17 Lecavalier is going to Montréal but this wont make Montréal a better team.
But in the long term this will really help Tampa Bay.
Pasty7
Posted - 06/26/2009 : 08:35:01
quote:Originally posted by Rambo2305
Wow, this one should be locked up...
Vinny's already said he isn't going anywhere, management said he wasn't going anywhere..
Montreal, get over yourself, might as well ask which of the 30 teams wants Lecavalier???
Face it, your team choked like they were...can't finish that nvm... lol
"Most people spend time and energy going around problems, rather than trying to solve them" - Henry Ford
actually bettman had to meet with the co owners of tampa to tell them to come to an areement one of the two wanted a personal cap in the 35 to 40 million range (meaning he would have to trade vinny) and the other wanted to go full out,,, so this is not unfounded and not obserd i don't know if he is coming to montreal espcially since gainy called lawton's actions when they were intially shopping vinny this season disgracefull... remeber lawton denied it then to said there was no way,,, then it comes out there was a deal in place
Pasty
Rambo2305
Posted - 06/26/2009 : 05:49:39 Wow, this one should be locked up...
Vinny's already said he isn't going anywhere, management said he wasn't going anywhere..
Montreal, get over yourself, might as well ask which of the 30 teams wants Lecavalier???
Face it, your team choked like they were...can't finish that nvm... lol
"Most people spend time and energy going around problems, rather than trying to solve them" - Henry Ford
brentrock2
Posted - 06/26/2009 : 04:13:29 I think Vincent will go to Montreal because that is really the only team really wanting him. There is a 75% chance he will go to Montreal than any other team.(Hopefully Montreal gets him) |
# coding=utf-8
# Copyright 2020 The Google Research Authors.
#
# Licensed under the Apache License, Version 2.0 (the "License");
# you may not use this file except in compliance with the License.
# You may obtain a copy of the License at
#
# http://www.apache.org/licenses/LICENSE-2.0
#
# Unless required by applicable law or agreed to in writing, software
# distributed under the License is distributed on an "AS IS" BASIS,
# WITHOUT WARRANTIES OR CONDITIONS OF ANY KIND, either express or implied.
# See the License for the specific language governing permissions and
# limitations under the License.
r"""Load notMNIST and covert it to numpy arrays.
Download notMNIST dataset from
http://yaroslavvb.com/upload/notMNIST/notMNIST_small.tar.gz and unzip the file
"""
from __future__ import absolute_import
from __future__ import division
from __future__ import print_function
import os
from absl import app
from absl import flags
import numpy as np
from PIL import Image
import tensorflow.compat.v1 as tf
flags.DEFINE_string('out_dir', '/tmp/image_data', 'Directory to save datasets.')
flags.DEFINE_string('raw_data_dir', '/tmp/notMNIST_small',
'Directory to raw data')
FLAGS = flags.FLAGS
def load_non_mnist(raw_data_dir):
"""load not_mnist raw data and save to numpy arrays."""
max_count = 0
for (root, _, files) in os.walk(raw_data_dir):
for f in files:
if f.endswith('.png'):
max_count += 1
print('Found %s files' % (max_count,))
images_np = np.zeros((max_count, 28, 28), dtype=np.int32)
labels_np = np.zeros((max_count,), dtype=np.int32)
count = 0
for (root, _, files) in os.walk(raw_data_dir):
for f in files:
if f.endswith('.png'):
try:
img = Image.open(os.path.join(root, f))
images_np[count, :, :] = np.asarray(img)
surround_folder = os.path.split(root)[-1]
assert len(surround_folder) == 1
labels_np[count] = ord(surround_folder) - ord('A')
count += 1
except OSError:
pass
print('All non-mnist loaded.')
return images_np, labels_np
def main(unused_argv):
images_np, labels_np = load_non_mnist(FLAGS.raw_data_dir)
with tf.compat.v1.gfile.Open(
os.path.join(FLAGS.out_dir, 'notmnist.npy'), 'wb') as f:
np.save(f, np.expand_dims(images_np, axis=3))
np.save(f, labels_np)
print('Saved np arrays to %s' % FLAGS.out_dir)
if __name__ == '__main__':
app.run(main)
|
Background {#Sec1}
==========
The dynamic post-translational modification of nucleosomal histones plays a critical role in transcriptional regulation. Hyperacetylation of nucleosomal core histones results in transcriptional activation, while their hypoacetylation leads to transcriptional repression. Modifications of nucleosomal histone acetylation and deacetylation affect the chromatin structure and related gene expression, and thus regulate various cellular processes, including DNA synthesis, cell division and differentiation, apoptosis, and others \[[@CR1], [@CR2]\]. The level of histone acetylation is determined by histone acetyltransferase (HAT) and histone deacetylase (HDAC) activities \[[@CR3], [@CR4]\]. Impaired HDAC activity could interfere with the balance between HATs and HDACs and thus alter the transcriptional status of many genes, in particular those related to disease. Therefore, HDACs have become promising therapeutic targets for the treatment of cancer, diabetes, and other human diseases \[[@CR5], [@CR6]\]. HDACs are classified into four classes (Classes I--IV) depending on their sequence identity and domain organization. Classes I (HDACs 1--3 and 8), II (HDACs 4--7, 9, and 10), and IV (HDAC 11) are referred to as classical HDACs and are generally simultaneously targeted by most HDAC inhibitors \[[@CR7]\]. Class III HDACs include Sirt1--7; they are nicotinamide (NAD)-dependent and are insensitive to HDAC inhibitors \[[@CR8]\]. To date, a number of HDAC inhibitors have been reported and they can be divided into several structural categories: hydroxamic acids, aliphatic acids, benzamides, cyclic peptides and others \[[@CR9]--[@CR11]\]. HDAC inhibitors have emerged as a new class of therapeutic agents and have generated much interest among pharmacologists, and cancer and diabetes researchers \[[@CR5], [@CR12], [@CR13]\]. Three HDAC inhibitors, vorinostat (SAHA) \[[@CR14]\], cyclic peptide FK228 (romidepsin) \[[@CR15], [@CR16]\] and belinostat \[[@CR17]\], have been approved by the U.S. Food and Drug Administration (FDA) for the treatment of cutaneous and peripheral T cell lymphoma. However, most HDAC inhibitors, including the clinically approved agents, non-selectively inhibit the deacetylase activity of class I and II HDACs, and many suffer from metabolic instability. These characteristics have been associated with reduced potency and toxic side effects *in vivo* \[[@CR18]\]. Significant efforts are ongoing to address these and other deficiencies of HDAC inhibitors to improve their HDAC inhibitory potential for the treatment of cancer and other diseases \[[@CR19]--[@CR21]\]. In addition, substantial efforts have been made to develop new HDAC inhibitors with potential therapeutic applications \[[@CR22]\]. In the present study, we present a hierarchical virtual screening protocol with SYBYL-*X*2.0 \[[@CR23]\] and Gold 5.2 \[[@CR24]\] software suites for the identification of compounds as potential HDAC inhibitors. It provides a stable and reliable solution for virtual screening of HDAC inhibitors based on commercial software's of drug discovery. A pharmacophore model was built and used for virtual screening to identify candidate compounds from the enamine dataset in the ZINC database \[[@CR25]\]. Then, the remaining compounds were docked into the active site of HDAC8. Finally, 22 compounds were identified as the final hit compounds. Enzyme inhibition assays with the HDAC inhibitor drug screening kit showed that three of the 22 compounds had HDAC inhibitory properties. Among these three compounds, ZINC12555961 was confirmed to have significant inhibitory activity against HDACs. Further *in vitro* cell experiments demonstrated that ZINC12555961 can selectively inhibit proliferation and promote apoptosis of cancer cells.
Methods {#Sec2}
=======
Pharmacophore modeling {#Sec3}
----------------------
The GALAHAD module in SYBYL-X 2.0 was adopted for ligand-based pharmacophore modeling. Seven hydroxamic acid inhibitors (marked with \* in Table [1](#Tab1){ref-type="table"}) with structural diversity were selected as representative compounds. All parameters were set to their default values (such as aligning molecules with pharmacophore features, no molecular template used, etc.) with the exception of 150 generations and a population size of 100. In the virtual screening process performed with the UNITY module in SYBYL-*X*2.0, at least five out of seven features in the pharmacophore model had to be matched.Table 1Seven compounds for generating pharmacophore modelsNameStructureRef.BindingDB_50114811{#d30e490}\[[@CR45], [@CR48]\]BindingDB_50114835{#d30e509}\[[@CR31], [@CR45], [@CR48]\]BindingDB_50123975{#d30e531}\[[@CR44]\]BindingDB_50214436{#d30e547}\[[@CR31]\]NK308{#d30e563}\[[@CR46]\]SAHA{#d30e579}\[[@CR31], [@CR46], [@CR47]\]TSA{#d30e601}\[[@CR31], [@CR47]\]
Molecular docking {#Sec4}
-----------------
GOLD 5.2 was adopted for molecular docking screening. HDAC8 (PDB id: 1 T69) was selected as the docking target. All the water molecules in HDAC8 were removed and hydrogen atoms were added. The binding site of HDAC8 was defined as those residues within 10 Å from the ligand in the X-ray structures. The parameters of the genetic algorithm (GA) were used in default values (such as the population size of 100, the selection pressure of 1.1, etc.) except that ligands were subjected to 30 GA runs, the number of operations was set to 300,000, and the early termination option was turned off. The three top scoring conformations of every ligand were retained at the end of the calculation. Two of the fitness functions implemented in GOLD 5.2, ChemPLP and ChemScore were used in our experiments.
HDAC inhibitory activity assay {#Sec5}
------------------------------
The HDAC inhibitor drug screening kit (k340-100, BioVision, CA, USA) was used to measure HDAC inhibitory activities of the candidate compounds according to the manufacturer's instructions. The candidate compounds, assay buffer, and HDAC fluorometric substrate, which comprises an acetylated lysine side chain, were added to HeLa nuclear extracts in a 96-well plate and incubated at 37 °C for 30 min. The reaction was stopped by adding lysine developer, and the mixture was incubated for another 30 min at 37 °C. An additional positive control included incubation with double-distilled water, and the inhibitor control consisted of incubation with Trichostatin A (TSA) at 20 μM. HDAC activities were quantified by a fluorescence plate reader (POLARstar OPTIMA, BMG, BRD) with excitation at 370 nm and emission at 450 nm.
Cell lines {#Sec6}
----------
Four cell lines, namely HepG2 (human hepatocellular carcinoma cell line), L02 (human normal liver cell line), MDA-MB-231 (human breast cancer cell line), and MCF-10A (human normal breast cell line), were obtained from the Cell Bank of the Chinese Academy of Sciences (Shanghai, China). Cells were cultured in an appropriate medium supplemented with 10 % fetal bovine serum (TBD Science, Tianjin, China), 100U/ml penicillin and 100 mg/mL streptomycin (Ameresco, US) at 37 °C and 5 % CO2.
MTT assay {#Sec7}
---------
The MTT (3-(4, 5-dimethylthia-zol-2-yl)-2, 5-diphenyl tetrazolium bromide) assay was used to examine the effects of the candidate compounds on cell viability. The candidate compounds were dissolved in DMSO (dimethyl sulfoxide) as 10 mM/L stock solutions. Cells were plated in 96-well plates (1 × 10^4^ cells/well) in 100 μL of growth medium and allowed to grow for 24 h. The cells were then treated with 0, 1, 10, 50 and 100 μM of each candidate compound in the presence of 3 % serum. After 44 h of treatment, 20 μL of MTT \[5 mg/mL in phosphate-buffered saline (PBS); Sigma Chemical Co.\] were added to each well for an additional 4 h of incubation. The blue MTT formazan precipitate was dissolved in 100 μL of DMSO. The optical density of samples was measured at 570 nm using a micro ELISA reader (Bio-Rad, Hercules, CA). Cell viability was expressed as a percentage relative to the untreated control cells.
DAPI staining assay {#Sec8}
-------------------
A DAPI staining assay was performed to reveal the presence of condensing nuclei and apoptotic bodies in compound-treated cells. HepG2 and MDA-MB-231 cells were treated with the candidate compounds (60 and 90 μM) for 48 h, and then harvested, fixed with 4 % paraformaldehyde for 30 min, washed with PBS, and stained with DAPI at a final concentration of 0.5 μg/mL for 15 min at room temperature. The cells were then analyzed using a fluorescence microscope. Three independent experiments were performed, and at least four different fields with a minimum of 100 cells/field were scored.
Apoptosis assay {#Sec9}
---------------
Annexin V-FITC/PI (propidium iodide) assay was performed to evaluate apoptosis of cancer cells induced by the hit compound ZINC12555961. HepG2 and MDA-MB-231 cells were seeded on 6-well plates at a density of 1 × 10^6^ cells/well, and incubated with 90 μM of ZINC12555961 for 48 h. Then, the cells were harvested by trypsinization, washed in ice-cold PBS, and resuspended in 190 μL binding buffer containing 5 μL Annexin V and 10 μL PI (Beyotime, China). The cells were incubated in the dark for 10 min and then analyzed by flow cytometry (BD FACSCanto™).
Cell cycle analysis {#Sec10}
-------------------
DNA staining with PI (Beyotime, China) was used to determine the cell cycle distribution of compound-treated cells. The number of cells at specific phases of the cell cycle was analyzed and sorted using flow cytometry. HepG2 and MDA-MB-231 cells were seeded at a density of 1 × 10^6^ cells/well. After treatment, the cells were collected, washed with PBS, fixed with 50 % alcohol and stained with PI at a final concentration of 1 mg/mL for 30 min. The percentages of cells in different phases of the cell cycle were measured with a flow cytometer (BD FACSCantoTM) and analyzed with the Modfit software (Verity Software House, Topsham, USA).
Results {#Sec11}
=======
Pharmacophore-based virtual screening {#Sec12}
-------------------------------------
The Enamine dataset in the ZINC database, containing collection of 1.8 million structurally diverse compounds, were used as the screening compound set \[[@CR25]\]. As listed in Table [1](#Tab1){ref-type="table"}, seven compounds were used to generate pharmacophore models. 50 hydroxamic acid-based HDAC inhibitors collected from the literature \[[@CR26]--[@CR31]\] and Enamine_p0.18, a subset of the enamine dataset \[[@CR25]\], which contains 22,565 compounds, were combined as the test dataset. Moreover, the maximal unbiased benchmarking data sets for HDACs (MUBD-HDACs) that cover all classical HDACs, composed of 631 HDAC inhibitors and 24,609 unbiased decoys \[[@CR32]\], were further used to validate our pharmacophore models. The HDAC inhibitor data sets in MUBD-HDACs have been validated extensively as chemically diverse, while the decoy sets are shown to be property-matching with ligands but have no HDAC inhibitory activities. MUBD-HDACs is freely available at <http://www.xswlab.org/>. All compounds were minimized under the Tripos Standard (TS) force field with Gasteiger-Hückel atomic partial charges. Minimizations were done using the Powell method and terminated at an energy gradient value of 0.01 kcal/mol. All models derived from at least six ligands of the training set (N_NITS ≥ 6) are listed in Table [2](#Tab2){ref-type="table"}, except those models with high energies. Constructing a good pharmacophore model requires balancing among various criteria such as maximizing steric consensus (STERICS), maximizing the pharmacophore consensus (HBOND), and minimizing strain energy (ENERGY) \[[@CR33]\]. SPECIFICITY is a logarithmic indicator of the expected discrimination for each model. It is based on the number of features contained, their distribution across any partial match constraints, and the degree to which the features are separated in space. A good pharmacophore model usually has a higher SPECIFICITY value, a higher steric score and a lower energy value. MODEL_001, MODEL_002, MODEL_005, MODEL_006, MODEL_010, MODEL_021, MODEL_030, and MODEL_041 were selected to further validate their screening abilities for the test dataset and the decoy dataset. The enrichment factor (EF) was calculated using equation ([1](#Equ1){ref-type=""}):Table 2Pharmacophore models generated by GALAHADNameSPECIFICITYN_NITSENERGYSTERICSHBONDMODEL_0012.762659.041807.4255.3MODEL_0022.321744.531724.8240.0MODEL_0053.842642.641585.6241.8MODEL_0065.225655.971822.2239.9MODEL_0102.208748.491714.7243.7MODEL_0214.220642.251551.0224.6MODEL_0302.821644.051610.4231.5MODEL_0351.094640.951630.8235.7MODEL_0371.651644.931714.6233.4MODEL_0412.545639.681521.0202.5$$\documentclass[12pt]{minimal}
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\begin{document}$$ \mathrm{E}\mathrm{F} = \left(Ha\times D\right)\ /\ \left(Ht\times A\right) $$\end{document}$$
Where *D* indicates the total number of compounds in the test datasets; *A* means the total number of known inhibitors in the test datasets; *Ht* is the hit number of compounds retrieved from the test datasets; and *Ha* represents the number of known inhibitors in the hit compounds.
As listed in Table [3](#Tab3){ref-type="table"}, the calculation results indicated that MODEL_006 had the best EF values. Moreover, MODEL_006 had the highest SPECIFICITY value, a moderate steric score, and an acceptable energy value. Therefore, it was selected as the final pharmacophore model. As shown in Fig. [1](#Fig1){ref-type="fig"}, MODEL_006 included seven pharmacophore features as follows: three hydrophobes (HY5, HY6 and HY7), two hydrogen bond (HB) acceptors (AA3 and AA4), and two HB donors (DA1 and DA2). Note that the pharmacophore AA_4 and DA_2 were overlapped each other. The hydrophobic moieties of the pharmacophore reflect the need for a hydrophobic region such as the linker domain or the cap group domain \[[@CR33]\]. The HB acceptor and donor moieties of the pharmacophore reflect the need for the ZBG domain \[[@CR27]\]. As a result, MODEL_006 was used as a 3D query to screen the Enamine database using the UNITY search module in SYBYL-X 2.0. In the seven features of MODEL_006, the maximum omitted features were set to two. The final 11,905 hits were retrieved.Table 3The EF values of the pharmacophore models for the test and decoy datebasesNameTest datasetDecoy datasetHtHaEFHtHaEFMODEL_0012354076.9872387433981.82089MODEL_0022693660.5308695863651.52305MODEL_0051452474.8634586524031.86315MODEL_0061984398.2267787525432.48172MODEL_0102274079.7004479624122.06983MODEL_0211542882.2363689654742.11489MODEL_0301782563.5252890453871.71144MODEL_0412183878.8412896544551.88523Fig. 1Pharmacophore MODEL_006 and its molecular alignment derived from the representative compounds. **a** Molecular alignment of 7 representative compounds. **b** Pharmacophore model (length unit: angstrom): three hydrophobes (HY5, HY6 and HY7), two hydrogen bond (HB) acceptors (AA3 and AA4), and two HB donors (DA1 and DA2). Cyan spheres represent hydrophobes; green spheres indicate HB acceptors; and magenta spheres indicate HB donors. Note that the pharmacophore AA_4 and DA_2 were overlapped each other
Molecular docking-based virtual screening {#Sec13}
-----------------------------------------
Initial validation of the docking protocol was performed by re-docking the ligand extracted from the HDAC 8 crystal structure (PDB id: 1 T64) to HDAC 8 itself. The top conformation of the ligand produced by GOLD 5.2 was very close to the crystal structure-bound conformation of the ligand. The root-mean-square deviation between the docked pose and its bound pose in the crystal structure was 0.75 Å. This indicated that GOLD 5.2 is able to reproduce the correct binding pose of an HDAC inhibitor ligand to its receptor. Next, the decoy dataset MUBD-HDACs \[[@CR32]\] were further used to validate our docking protocol. 1 T64, 1 T67 and 1 T69 were used as the receptor protein and the four docking functions in GOLD 5.2 (ChemPLP, Goldscore, Chemscore and ASP) were used as scoring functions. The EF values and functional thresholds of ChemPLP, Goldscore, Chemscore and ASP for the top 1, 5, 10 and 20 % of ranked compounds in the decoy database were listed in Table [4](#Tab4){ref-type="table"}. From the Table [4](#Tab4){ref-type="table"}, we can see that the EF values of ChemPLP and Chemscore are obviously higher than that of Goldscore and ASP. The previous literature also pointed out that ChemPLP demonstrated the best results for both pose prediction and virtual screening \[[@CR34]\]. So we chose ChemPLP and Chemscore together to further evaluate their screening results. As listed in Table [5](#Tab5){ref-type="table"}, the combination of the two scoring functions is better in screening results than the two separate functions. Finally, ChemPLP was chosen as the main function and Chemscore was chosen as as the secondary scoring function for practical screening. After validation of the docking protocol, all 11,905 hits retrieved by pharmacophore model-based screening were docked into the active site of three crystal structures of HDAC8 (1 T64, 1 T67 and 1 T69). According to the actual situation, the about top 1 % of ranked compounds in the hit database were decided to remain for experimental validation. The score 72 for ChemPLP and 23 for ChemScore were selected as the final score thresholds. As a results, 154 compounds were selected as the hit compounds. Finally, the Selector module in SYBYL-X 2.0 was adopted for the clustering analysis by creating and comparing diverse subsets of the 154 hit compounds. 22 of the 154 hit compounds with diverse structures were selected as the final hits (listed in Table [6](#Tab6){ref-type="table"}). We further examined the binding patterns of the 22 final hits (Additional file [1](#MOESM1){ref-type="media"}: Figure S1). As listed in Table [7](#Tab7){ref-type="table"}, the binding patterns of the 22 hit compounds can be divided into five types. For the first class of compounds, their ZBG domain may form covalent metal chelate complexes and hydrogen bond interactions with HDAC residues. For the second class of compounds, their ZBG domain and the linker domain may form covalent metal chelate complexes and hydrogen bond interactions with HDAC residues. For the third class of compounds, its ZBG domain and cap group domain may form covalent metal chelate complexes and hydrogen bond interactions with HDAC residues GLY151 and TYR306. The fourth class of compounds included ZINC03260906 and ZINC09715944. The ZBG domain of ZINC03260906 may form covalent metal chelate complexes and its linker domain and the cap group domain may form hydrogen bond interactions with HDAC residues HIS180 and PHE208. The ZBG domain of ZINC09715944 may form covalent metal chelate complexes and its linker domain and the cap group domain may form hydrogen bond interactions with HDAC residues LYS33, HIS143, and SER150. The fifth class of compounds included ZINC02627831 and ZINC12581173. The linker domain of ZINC02627831 may form hydrogen bond interactions with HDAC residue ASP101. The ZBG domain of ZINC12581173 may form hydrogen bond interactions with HDAC residue LYS33.Table 4The EF values and functional thresholds for the top 1, 5, 10 and 20 % of the decoy database in individual docking function testDocking functionsTop (%)Functional thresholdsEF1T641T671T69ChemPLP172.94.674.214.87568.34.474.154.681062.84.314.094.522056.94.144.024.44Goldscore160.11.871.371.14555.41.711.231.041050.11.681.190.982046.21.651.141.03Chemscore123.74.083.944.31519.13.983.814.181015.13.943.784.152012.73.873.724.06ASP135.31.451.211.78531.31.311.111.561028.41.291.031.382024.11.230.951.15Table 5The EF values and functional thresholds for the top 1, 5, 10 and 20 % of the decoy database in combined docking function testTop (%)Docking functionsEFChemPLPChemscore1T641T671T69171.822.84.844.544.98567.417.64.374.214.431062.114.23.533.843.812057.512.33.243.313.42Table 6Structure of the 22 final compoundscmpd.ᅟcmpd.ᅟZINC01895726{#d30e1719}ZINC12555961{#d30e1728}ZINC02627831{#d30e1738}ZINC12581173{#d30e1747}ZINC02639234{#d30e1757}ZINC23140995{#d30e1766}ZINC03260906{#d30e1776}ZINC23141716{#d30e1785}ZINC03307410{#d30e1795}ZINC23141899{#d30e1804}ZINC06178852{#d30e1814}ZINC23143331{#d30e1823}ZINC06415107{#d30e1833}ZINC23886004{#d30e1842}ZINC06497704{#d30e1852}ZINC58161863{#d30e1861}ZINC09350495{#d30e1871}ZINC60063267{#d30e1880}ZINC09715944{#d30e1890}ZINC67907864{#d30e1899}ZINC11325463{#d30e1909}ZINC84111476{#d30e1918}Table 7Different binding patterns of 22 hit compoundsBinding patternsHit namesInteraction typesPharmacophore interaction regions1ZINC01895726, ZINC02639234, ZINC06178852, ZINC23140995, ZINC23141716, ZINC23143331, ZINC23886004, ZINC58161863, ZINC60063267, ZINC67907864, ZINC84111476metal chelate bonds and hydrogen bondsZBG domain2ZINC03307410, ZINC06415107, ZINC06497704, ZINC09350495, ZINC12555961, ZINC23141899metal chelate bonds and hydrogen bondsZBG domain and linker domain3ZINC11325463metal chelate bonds and hydrogen bondsZBG domain and cap group domain4ZINC03260906, ZINC09715944metal chelate bonds and hydrogen bondsZBG domain, linker domain and cap group domain5ZINC02627831, ZINC12581173hydrogen bondsZBG domain or linker domain
Inhibitory enzymatic activity evaluation {#Sec14}
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Based on the in silico results, additional *in vitro* studies were performed to evaluate the activity of the final 22 hit compounds. A fluorometric HDAC activity assay was firstly performed to examine the inhibitory activity of the 22 hit compounds against HDACs in HeLa nuclear extracts (Biovision K340-100). The experimental results are depicted in Fig. [2](#Fig2){ref-type="fig"}, which shows that three compounds, namely ZINC12555961, ZINC02639234, and ZINC09715944, inhibited the enzymatic activity of HDACs. Their relative enzymatic activities were 52 % (*P* = 0.008), 76 % (*P* = 0.006) and 82 % (*P* = 0.011), respectively, whereas that of the control inhibitor TSA was 12 % (*P* = 0.0003). The other 19 hit compounds did not show significant inhibitory activity against HDACs.Fig. 2Inhibitory activity of the 22 hit compounds against HDACs. The enzymatic activities of the hit compounds are expressed as percentages of the positive control. Black bars represent the positive control, white bars represent the inhibitor control, and gray bars indicate the hit compounds being treated. Results are expressed as the mean ± SD (n ≥ 3). \* mean *P* \< 0.05 and \*\* mean *P* \< 0.01
The three active compounds did not belong to either of the four main classes of HDAC inhibitors, namely hydroxamic acids, aliphatic acids, benzamides, and cyclic peptides \[[@CR9]--[@CR11]\]. ZINC12555961 has four main functional groups: fluorophenyl, methoxyphenyl, acrylamide cyanide and nitrophenyl; ZINC02639234 has three main functional groups: benzothiazole, triazole and dihydroxy phenyl; and ZINC09715944 has three main functional groups: benzene and pyridazine ketone, pyrrole and methoxypheny. The docking poses of the three active compounds to the active site of HDAC8 (1 T69) are shown in Fig. [3](#Fig3){ref-type="fig"}. The two hydroxyls of the dihydroxy phenyl group of ZINC02639234 formed covalent chelate complexes with zinc ions and hydrogen bond interactions with HDAC residues ASP178 and ASP267. In addition, the benzothiazole group of ZINC02639234 had hydrophobic contact with HDAC residue PHE208 (Fig. [3a](#Fig3){ref-type="fig"}). For ZINC09715944, its methoxyphenyl group formed covalent chelate complexes with zinc ions, the carbonyl of its linker domain formed hydrogen bond interactions with HDAC residue LYS33, and the carbonyl of its benzene and pyridazine ketone group was not only able to form hydrogen bond interactions with HDAC residues HIS143 and SER150, but was also able to form hydrophobic contact with HDAC residue PHE208 (Fig. [3b](#Fig3){ref-type="fig"}). For ZINC12555961, the hydroxyl of its nitrophenyl group was not only able to form covalent chelate complexes with zinc ions, but was also able to form hydrogen bond interactions with HDAC residue ASP267; the nitro of its nitrophenyl group could form hydrogen bond interactions with HDAC residues GLY304, GLN263, GLY140 and HIS142; the carbonyl group of its acrylamide cyanide was able to form hydrogen bond interactions with HDAC residue HIS180, and its methoxyphenyl group was in hydrophobic contact with HDAC residue PHE208 (Fig. [3c](#Fig3){ref-type="fig"}). The three compounds were selected for further cytotoxicity assays.Fig. 3Molecular docking results. Docked orientations of **a** ZINC02639234, **b** ZINC09715944, and **c** ZINC12555961. Active site residues are shown by lines and the metal ion (Zn^2+^) is shown by a grey sphere. The hydrogen bond network with protein residues and the metal ion is represented by a yellow dotted line
Anti-proliferative activity and apoptosis-inducing mechanism {#Sec15}
------------------------------------------------------------
HDAC inhibitors selectively induce cell growth arrest and apoptosis in a wide variety of cancer cells. To test the cytotoxicity of the three hit compounds with HDAC inhibitory activities, MTT, DAPI staining, and Annexin V-FITC assays were designed and performed according to the procedures described in the Materials and methods section. In the MTT assay, HepG2, L02, MDA-MB-231, and MCF-10A cells were cultured in 3 % serum-supplemented medium and treated with four different concentrations (1, 10, 50, and 100 μM) of the three hit compounds and SAHA. The viabilities of the four cell lines after 48 h of treatment were measured using the MTT assay. The IC50 values for ZINC12555961, ZINC02639234, ZINC09715944 and SAHA against the four cell lines were calculated and are listed in Table [8](#Tab8){ref-type="table"}, which shows that ZINC02639234 and ZINC09715944 exhibited stronger toxicity towards normal cells than against cancer cells, whereas ZINC12555961 showed stronger toxicity towards cancer cells than normal cells. Increasing ZINC12555961 concentrations (from 0 to 100 μM) led to a steady decrease in the viability of MDA-MB-231cells (IC50 = 57 μM), but had a less toxic effect on MCF-10A cells (IC50 = 142 μM). In the HepG2 and L02 cell lines, ZINC12555961 exhibited nearly the same toxicity, with IC50 values of 87 and 85 μM, respectively. Furthermore, in our experiments, ZINC12555961 was more potent than SAHA in inhibiting the viability of cancer cells (Fig. [4](#Fig4){ref-type="fig"}). The IC50 values of SAHA in HepG2, L02, MDA-MB-231, and MCF-10A cells were 166 μM, 85 μM, 178 μM, and 59 μM, respectively. The results indicated that ZINC12555961 has inhibitory activity against all four cells and displays promising and selective inhibitory activity against cancer cell viability, ZINC09715944 has inhibitory activity against HepG2, L02, and MCF-10A cells but not against MDA-MB-231 cell and does not display selective inhibitory activity against cancer cell viability, and zinc02639234 has no inhibitory activity against all four cells.Table 8Comparison of the IC50 values of SAHA, ZINC12555961, ZINC02639234 and ZINC09715944 against the HepG2, L02, MDA-MB-231 and MCF-10A cell linesChemicalsIC50(μM)HepG2L02MDA-MB-231MCF-10ASAHA166 ± 985 ± 2178 ± 1359 ± 10ZINC1255596187 ± 1085 ± 1557 ± 7142 ± 17ZINC02639234\>200\>200\>200\>200ZINC09715944157 ± 1265 ± 3\>20047 ± 7Data are expressed as the mean ± SD from at least three independent experimentsFig. 4Comparison of the cytotoxicity of ZINC12555961 and SAHA against cancer cells and normal cells. **a** MDA-MB-231 and MCF-10A cells were treated with 0, 10, 50 and 100 μM SAHA. **b** HepG2 and L02 cells were treated with 0, 10, 50 and 100 μM SAHA. **c** MDA-MB-231 and MCF-10A cells were treated with 0, 10, 50 and 100 μM ZINC12555961. **d** HepG2 and L02 cells were treated with 0, 10, 50 and 100 μM ZINC12555961. Significance was determined by the Student's *t*-test. The values represent as the mean ± S.D. \* means *P* \< 0.05
As ZINC12555961 could remarkably suppress the viability of HepG2 and MDA-MB-231 cells, a DAPI staining assay was performed to determine whether its inhibitory effect on cell viability is associated with the induction of cell apoptosis. Fluorescent microscopic images of DAPI stained nuclei of HepG2 and MDA-MB-231 cells treated with ZINC12555961 or DMSO for 48 h are shown in Fig. [5](#Fig5){ref-type="fig"}. The concentrations of zinc12555961 used in HepG2 and MD-231 cells were 90 μM and 60 μM, respectively. Apoptotic nuclei in both cell lines were split into several nuclear apoptotic bodies, and apoptotic cells are shown in deep white by DAPI staining, as indicated by the red arrows (Fig. [5b and d](#Fig5){ref-type="fig"}), whereas cells treated with DMSO exhibited round intact nuclei. These results indicate that ZINC12555961 may play an important role in inducing cancer cell apoptosis.Fig. 5Nuclear morphological changes and apoptotic HepG2 and MDA-MB-231 cells treated with ZINC12555961 (90 μM) and DMSO for 48 h. Arrows indicate apoptotic nuclei. **a** HepG2 cells by DAPI staining treated with DMSO. **b** HepG2 cells by DAPI staining treated with ZINC12555961. **c** MDA-MB-231 cells by DAPI staining treated with DMSO. **d** MDA-MB-231 cells by DAPI staining treated with ZINC12555961
Flow cytometric analysis with Annexin V-FITC conjugated to PI was performed to further examine the effect of ZINC12555961 on cancer cell apoptosis. HepG2 and MDA-MB-231 cells were treated with DMSO (90 μM) or ZINC12555961 (90 μM) for 48 h. Apoptotic cells were stained and monitored by flow cytometry (Fig. [6](#Fig6){ref-type="fig"}). In both HepG2 and MDA-MB-231 cell lines, ZINC12555961 significantly induced the late apoptotic stage. The apoptosis rates of HepG2 cells treated with DMSO and ZINC12555961 were 5.1 and 47.9 %, respectively, and those of MDA-MB-231 cells treated with DMSO and ZINC12555961 were 6.4 and 25.2 %, respectively. ZINC12555961 also affected the early apoptotic stage in both cell lines. The early apoptosis rates of the control group vs. the experimental group in HepG2 and MDA-MB-231 cells were 3.5 % vs. 4.5 % and 3.9 % vs. 8.7 %, respectively. These results indicate that the rates of apoptosis induced by ZINC12555961 were significantly higher than those of the control groups. ZINC12555961 may play an important role in inducing cancer cell apoptosis.Fig. 6Quantitative analysis of the effects of ZINC12555961 on the apoptosis of human cancer cell lines. HepG2 cells **a** and MDA-MB-231 cells **b** were treated with DMSO or 90 μM ZINC12555961 for 48 h. Cells were harvested by trypsinization and centrifugation, stained with Annexin V-FITC and PI, and analyzed by flow cytometry. Representative results are shown
As shown above, the effects of ZINC12555961 on apoptosis, as measured by flow cytometry, do not explain the decrease of cell viability measured by MTT assay. This suggests that ZINC12555961 may affect cell viability through other mechanisms. Therefore, we tested the effect of ZINC12555961 on cell cycle distribution by flow cytometric analysis. HepG2 and MDA-MB-231 cells were treated with ZINC12555961 for 32 and 48 h and analyzed by flow cytometry (Fig. [7](#Fig7){ref-type="fig"}). In the HepG2 cell line, the percentage of cells in the G~1~ phase decreased from 70.3 to 66.4 % in response to ZINC12555961, and this decrease was accompanied by an increase in the proportion of cells in the G~2~ phase from 3.5 to 8.3 % (Fig. [7a](#Fig7){ref-type="fig"}). Similarly, in MDA-MB-231 cells, the percentage of cells in the G~1~ phase decreased from 61.5 to 51.6 % in response to ZINC12555961, and this decrease was accompanied by an increase in the proportion of cells in the G~2~ phase from 12.3 to 24.9 % (Fig. [7b](#Fig7){ref-type="fig"}). This indicates that the inhibitory effect of ZINC12555961 on the proliferation of HepG2 and MDA-MB-231 cells may be associated with G~2~/M phase cell cycle arrest.Fig. 7Effects of ZINC12555961 on cell cycle progression in cancer cells. HepG2 cells **a** and MDA-MB-231 cells **b** were treated with DMSO or ZINC12555961 for 32 and 48 h. At the end of treatment, cells were trypsinized, incubated with RNase, stained with PI, and analyzed by flow cytometry. Representative results are shown
Discussions {#Sec16}
===========
HDAC enzymes have emerged as exciting and promising novel targets for the treatment of cancer, diabetes and other human diseases. HDAC inhibitors, as a new class of potential therapeutic agents, have attracted a great deal of interest both for research and clinical applications. Computer aided drug design (CADD) and virtual screening have been applied in the development of new HDAC inhibitors. Many HDAC inhibitors were designed and synthesized based on CADD approaches \[[@CR35]--[@CR41]\]. Certain potent HDAC inhibitors with novel structures were identified by virtual screening approaches \[[@CR31], [@CR42], [@CR43]\]. Vadivelan et al. developed a pharmacophore model based on common chemical features of HDAC inhibitors \[[@CR44]\]. Melagrakia et al. developed a linear five-parameter quantitative structure-activity relationship (QSAR) model of HDAC inhibitors \[[@CR45]\]. Xiang et al. developed a pharmacophore model and three QSAR models for a series of benzimidazole and imidazole inhibitors of HDAC2 \[[@CR46]\]. Zhao et al. used a two-step modeling approach to study the selectivity and activity of HDAC inhibitors \[[@CR47]\]. Thangapandian et al. used pharmacophore modeling and molecular docking approaches for the identification of potential HDAC8 inhibitors \[[@CR48]\]. More recently, Thangapandian et al. used a combined pharmacophore modeling, molecular docking and molecular dynamics (MD) simulation approach for the identification of potential HDAC8 inhibitors \[[@CR49]\]. Nair et al. used a combined pharmacophore modeling, flexible docking, and three-dimensional QSAR (3D--QSAR) approach for the identification of benzimidazole and imidazole derivatives \[[@CR50]\]. Although these studies did not experimentally validate the activities of their candidate compounds, their use of virtual screening approaches for HDAC inhibitors provides support for further computational and experimental research. Park et al. identified novel classes of HDAC inhibitors with new zinc-chelating groups using docking simulations, and experimentally validated the activities of their candidate compounds \[[@CR42]\]. Tang et al. identified three hit compounds using a combinatorial QSAR screening model based on support vector machine and k-Nearest Neighbors algorithms, and experimentally confirmed the inhibitory activities of the compounds against HDAC1 \[[@CR31]\]. Zhang et al. identified a potent HDAC inhibitor with a novel scaffold using ZBG (zinc-binding group)-based virtual screening, and experimentally confirmed the inhibitory activities of the compounds against HDAC8 \[[@CR43]\]. In the present study, we developed a hierarchical virtual screening protocol for the identification of potential HDAC inhibitor compounds. The multistage virtual screening workflow was used to screen and identify 22 final hit compounds, and the HDAC inhibitory activities of three of the 22 compounds, namely ZINC12555961, ZINC02639234 and ZINC09715944, were experimentally validated by *in vitro* enzyme inhibition assays. The results confirmed the efficacy and validity of our screening method. The three active compounds showed a novel structure that does not belong to the previously reported four classes of HDAC inhibitors. All three active hits showed different scaffolds, thereby providing wide opportunities for future HDAC inhibitor design. The novelty of the 22 final hit compounds was assessed using SciFinder scholar (<https://scifinder.cas.org/>). The SciFinder results confirmed that these compounds were not previously tested for HDAC inhibitory activity.
We further examined the cytotoxicity of the three hit compounds with HDAC inhibitory activities against the human normal liver cell line, L02, and the liver cancer cell line, HepG2, as well as the human breast cancer cell line, MDA-MB-231, and the human breast epithelial cell line, MCF-10A. The MTT assay results demonstrated that the active compound ZINC12555961 could selectively suppress the viability of human cancer cell lines (HepG2 and MDA-MB-231 cells). Staining with DAPI and Annexin V-FITC/PI flow cytometry assays revealed that the effect of ZINC12555961 on cancer cell death may be mediated by the induction of apoptosis and G~2~/M phase cell cycle arrest. These results indicate that ZINC12555961 is a promising HDAC inhibitor and has anti-tumor potential. Future studies will be aimed at elucidating the molecular mechanisms underlying ZINC12555961-induced selective cancer cell apoptosis and evaluating the isoform-selective HDAC inhibitory effects of ZINC12555961. ZINC12555961-focused virtual screening will also be further developed in the future.
Conclusions {#Sec17}
===========
In conclusion, the study identified three new HDAC inhibitors. The new-found HDAC inhibitors are worthy to further investigations.
Abbreviations {#Sec18}
=============
AA, Hydrogen bond acceptors; CADD, Computer aided drug design; DMSO, Dimethyl sulfoxide; EF, Enrichment factor; ELISA, Enzyme-linked immunosorbnent assay; FITC, Fluorescin isothiocyanate; GA, Genetic algorithm; HB, Hydrogen bond donors; HB, Hydrogen bond; HDAC, Histone deacetylases; HepG2, Human hepatocellular carcinoma cell line; HY, Hydrophobes; L02, Human normal liver cell line; MCF-10A, Human normal breast cell line; MDA-MB-231, Human breast cancer cell line; MTT, 3-(4, 5-dimethylthia-zol-2-yl)-2, 5-diphenyl tetrazolium bromide; PBS, Phosphate-buffered saline; PDB, Protein data bank; PI, Propidium iodide; SAHA, Suberoylanilide hydroxamic acid; TSA, Trichostatin A; ZBG, Zinc-binding group
Additional file {#Sec19}
===============
Additional file 1: Figure S1.Comparisons of 22 ligand's binding poses against 1T69. (DOCX 8808 kb)
Funding {#FPar1}
=======
This work was supported by the National Natural Science Foundation of China (no. 61172183), the Natural Science Foundation of Jilin Province, China (no. 20130101148JC), the Changchun Science and Technology Bureau, China (no. 12ZX55), Science and Technology Development Program of Jilin Province (No. 20150309003YY), Key Science and Technology Project of Jilin Province (20150224038YY), the 2014 Industrial Technology Research and Development Special Project of Jilin Province, China (no. 2014Y100), and the 2015 Department of Education 12th Five-Year Science and Technology Research Planning Projects of Jilin Province, China (no. 2014B053).
Availability of data and materials {#FPar2}
==================================
Data used in this study are given as tables and additional files. Details of materials used are available in the reference list.
Authors' contributions {#FPar3}
======================
YXH and YXL conceived and designed the research. JZ, QHS, LHZ, CF and TTL performed the research including data collection, experiments and analysis. YLB, LGS and LBZ suggested extension and modifications to the research. YXH supervised the whole research and revised the manuscript critically. All authors have read and approved the final manuscript.
Competing interests {#FPar4}
===================
The authors declare that they have no competing interests.
Consent for publication {#FPar5}
=======================
Not applicable.
Ethics approval and consent to participate {#FPar6}
==========================================
Not applicable.
|
Review by Lewis B
Brink
Developer Splash Damage
Publisher Bethesda
Released May 2011
Available for PC (version reviewed) Xbox 360 and PS3
Time Played “Finished” campaign mode and all challenges up to ***. Rank 15 (20+ hours).
Verdict: 5/5 Gold Star
The finest team-shooter since Team Fortress 2. Need I say more?
The Ugly
Brink hasn’t had the best of starts. Sound disappearing, vanishing text, lag and frame rate issues so bad many simply cannot play are just some of its many faults. In spite of all this, I absolutely love it. I have been fortunate in not experiencing the vast majority of problems people are reporting. The primary one I did experience, though, was probably the worst. Trying to achieve a stable 30 frames per second in Brink was quite the challenge, and despite what I consider a good ‘rig,’ at times it was exhausting to achieve. Fresh driver installations, driver rollbacks, overclocking my CPU, my GPU, my RAM. I’ve done it all. My PC is certainly better for it (60% more oomph!) and thankfully, after the latest driver release from ATI, Brink now runs at over 60fps consistently.
It’s frustrating for many. Not that I’ve doubted my computer’s capabilities (Battlefield: Bad Company 2 and Crysis 2 both run at 60fps on maximum detail) but because there are those with computers worse than mine (and better) who have abandoned the game entirely due to the inability to achieve playable frame rates. In many respects I can completely understand why; I spent more time trying to achieve consistent frame rates than playing the actual game. Something, somewhere, went wrong in that scenario.
Whilst I wouldn’t want to point fingers, Brink clearly wasn’t ready to be launched when it did. I’d estimate it’s three patches away from where it should be, and this is unacceptable. To make matters worse, Splash Damage and Bethesda actually brought the release forward. I’ll repeat that. They brought the release date forward!
Thankfully the companies are communicating well with the Brink community through the Bethesda Blog , and are hard at work attempting to rectify any outstanding issues as quickly as possible, whilst working closely with Nvidia and ATI to continually improve frame rates.
But if you are fortunate enough like me to persevere, you will be rewarded with a stunning team experience.
The Good
In Brink, two factions, Resistance and Security, battle in a once-utopian city called The Ark, a floating city above the waters of a flooded Earth. When you begin the game you are presented with a choice. To defend the ark or to leave it, each determining which faction your character fights for. After deciding, you’ll be greeted with an ingame cutscene specific to your faction (which I might add are visually fantastic and voice acted brilliantly) but also a video tutorial which, if you choose to watch, will earn you 1,000 experience points. It is surprising that Splash Damage chose to use a video as a tutorial, but the game does inform you that this is aimed towards console users (lord knows why). Whether you decide to watch it or not is up to you, as it’s really not as painful as reviewers have made out, but afterwards comes the fun part; customising your appearance.
As Resistance your appearance is rough and ready, with torn clothes and makeshift outfits. As Security you’re quite the opposite: clean, well armoured and well equipped. Both display the sheer effort Splash Damage have gone to in not only creating stunning character models, but polished and solid outfits, hair styles, faces and colour schemes. As someone who has played many MMOGs over the years and toyed with countless character creators, Brink has to be one of my all time favourites, even without the ability to manually change my character’s physical proportions.
After deciding your appearance it’s time to get in game. Brink is split into three elements: Campaign (single player, but with A.I. bots); Freeplay (online, with human and bot players); and Challenges (various scenarios to teach and test you on mechanics of the game). I really can’t stress enough how you should complete all the challenge modes first. Not only will they teach you the basics of Brink first hand, as well as the classes, but they are also a great deal of fun and difficult, allowing you to unlock weapon attachments and earn a good quantity of experience. Having weapon scopes and different magazine types makes a huge difference when fighting against human players, so make it your first port of call and get adjusting those guns.
The single player follows either Resistance or Security, played from either perspective, across 8 maps. Each map is narrated whilst it loads, followed by an in game cutscene. It’s a strange choice by Splash Damage, as the cutscenes are acted and animated beautifully, yet the single player campaign is little more than the multiplayer maps strung together. This doesn’t mean the single player isn’t enjoyable, as the maps are fantastic (we’ll get to those later) and the cutscenes highly enjoyable, but it does make you wonder why they didn’t choose to develop a full campaign. Although, that would be like suggesting people purchased Battlefield: Bad Company 2 for the single player. With the quality of the script, acting and animations, it is a little sad to see Splash Damage not develop this side of the game more. The Ark has so many possibilities that a sequel has to be made.
As far as guns are concerned, there are a huge array, all varying in statistical goodness such as range, damage, equip speed and rate of fire. These can all be customised through unlocked weapon attachments. What is surprising is that I’ve found all the weapons to be usable and balanced. Although at first it might seem that pistols have no place when you can carry a second submachine gun, they are the only secondary weapon light body types can use. To balance this out, a pistol user also carries a knife, making for a deadly melee attack in replace of other body types’ extra firepower. It’s a neat touch. Brink has drawn criticism from some for the fact it only has bullet-based weapons (rifles, pistols, automatic pistols and shotguns) yet there are around 17 to choose from, from the very start, dependent on body type and with 24 in total.
I didn’t hear people complaining that Call of Duty had only bullet-based weapons or Team Fortress 2 only having 3 weapons per class when it first launched. It’s entirely irrelevant. Splash Damage have made great efforts in the number of guns available and they all look and sound exceptional.
In terms of classes, there are four to choose from: Soldier, Engineer, Spy and Medic, which all fill a specific role to not only overcome your opponents, but to complete maps and objectives. I’ll pre-warn now – a balanced team is a must. More so than any other team-oriented shooter I’ve ever played, but on the flip side (lord knows why people choose to see this as a negative element of the game) it encourages teamwork, a spread of class types and gives focus to your role at various points throughout a map. Thankfully if you are ever short of a class you or teammates can switch instantly at control stations, though I’ve yet to encounter a need to change.
As an Engineer in many team shooters I play, I’ve really been enjoying Brink’s version. The ability to directly support my team by boosting their weapon damage, as well as my own, combined with turrets, mines and Kevlar vest hand-outs gives a real sense of purpose, that I am actually making a difference to my team. The fact that experience is instantly awarded based on supporting others is also great touch. Nothing annoyed me more in Team Fortress 2 where you’d receive no points for your dispenser helping others, yet in many cases they were pivotal to a team’s success. That I’m also actively involved in objectives, whether primary or secondary, and that I am as component with weapons as a Soldier, makes for an even but different kind of playing field, depending on the class you choose.
Objectives, like in Quake Wars, earn you experience but are also pivotal to the success of your team. A map will always have a primary objective which can be selected at any time through the objective wheel (you hold down the middle mouse button; which is another fabulous thing that DICE dropped from Battlefield: Bad Company 2). When you use the objective wheel, your character automatically faces in the right direction you want to be heading in. It keeps play straight forward and doesn’t leave you needing to scour a mini map or the horizon to see what it is you’re actually aiming for, although you can turn this option off. What I also appreciate (and contrary again to many reviews) is that secondary objectives are purposeful and entirely class dependent; they are all part of the bigger picture. Escorting a computer-controlled character, defending an area, repairing or constructing submachine guns or capturing command points. Without doing them, you can still win. By doing them, it makes life that much easier for you and your team.
Which brings me nicely onto three elements of Brink I adore. SMART, experience points and map design.
The SMART bit.
I really love SMART (Smooth Movement Across Random Terrain). Games feel boring without it. Battlefield: Bad Company 2 is officially dull as a result of not having SMART and my dream is for games to have destructible scenery and SMART combined, at which point I’d be in heaven. But for the time being, I’m genuinely finding it difficult to revert back to the standard control system other first person shooters offer. SMART is incredibly simple and has arguably been lifted from Mirror’s Edge, but where DICE abandoned the best creative idea they’ve ever had, Splash Damage have ran with it.
In a nutshell, players use SMART while sprinting. When you approach an obstacle, be it a wall, crate or barrier, SMART will assess the type of obstacle and the direction the player is looking and attempt to navigate the obstacle based on this information. For example, a player can either climb over or slide under an obstacle depending on whether they approach looking at the top or bottom of it. It’s fast, slick and incredibly satisfying.
One scenario saw me jump down from a railing, vault over a crate, slide under some piping and kill two players while skidding out the other side on my backside. BOO-YA! is exactly the noise I made whilst waving my arms in the air. It’s just so satisfying and becomes second nature so incredibly quickly that if I don’t see it in other games, I’ll be very upset indeed.
The Experience
Unlike other team based shooters which reward players based on points per kill, Brink rewards players with experience points based on how many times they hit an opponent, or completing objectives/supporting teammates. Each bullet that contacts an enemy results in scrolling ‘XP’ text on your screen. If you unload an entire clip into someone, you might earn 40xp. If you hit them with only two bullets, you might earn 4xp. What this results in is an proportional share of experience, rather than an absolute figure. Those who do the most work are rewarded and those who sneak in at the last minute earn just the scraps. Not only does this allow you to earn experience points rapidly, but highlights on the score board those undertaking the most objectives, supporting team mates and dealing the most damage to opponents. The leader board as a result is very different to say, Call of Duty, which is based entirely on kill counts alone.
For anyone who accidentally team kills or damages team mates, you’ll lose significant quantities of experience, making for careful play at times but also making the Operative’s disguises actually useful. Unlike in Team Fortress 2 where everyone ‘spy-checks’ with no repercussions, it’s simply not worth being constantly paranoid and shooting teammates in Brink.
With a little bit of experience under your belt, you’ll also unlock two additional body types: Light and Heavy (Medium is default), which offer differences in terms of speed, agility and health. I settle for the Medium on my Engineer (and Operative) as it’s a fine balance between the three, although I have been tempted to try a Heavy combination for the extra health and back-line support. Light I found a little too flimsy for my play style as they die easily, but in the right hands they can be formidable, the slippery little buggers.
There are abilities which you can invest in based on your rank and experience. Each class has a reasonably large quantity of these that improve their equipment and play style, but they can only be unlocked through experience and by levelling up your character. For example, a heavy sentry gun as the Engineer is not immediately available until you’ve earned your stripes, but in the meantime you can invest in a light or medium versions or ‘universal abilities’, such as ‘Sprinting Reload’ and ‘Battle Hardened’ which provide subtle, worthwhile permanent improvements to your character. My particular favourite, ‘Combat Intuition,’ sees a directional warning appear on your screen if you are being targeted by an opposing player (but not fired upon), giving you a split second to react and hopefully defend yourself.
With experience also comes visual rewards. Costumes, hairstyles and face paint doesn’t cost physical experience, as they are unlocked based on the more experience you earn, but there are hundreds of extra pieces for both Resistance and Security, all of which are modelled to perfection.
What Splash Damage now need to do is open up the floor to the community and allow them to create their own, or follow the lead of Valve and receive community submissions to place in game. They could even charge for them if they really wanted to emulate Valve; they could also do with adding female characters whilst they are at it.
The Maps
I have a very good memory for routes, but I must admit I am still learning Brink’s maps. They are expansive horizontally (as opposed to vertically) but focused at the same time. Multiple routes and scenery to clamber over does make maps feel busy and at times confusing, whilst there are still certain locations on maps I can’t figure out how to reach. I should point out however, that this shouldn’t be looked on as a form of criticism, as it demonstrates the effort that Splash Damage has made to add an additional layer of depth to the game.
Many are reasonably straight forward and there are some bottlenecks, but these bottlenecks have never stopped me enjoying the game. If my team lost it was through our own failures as a team, not that of the map design. What I have found annoying though is that I don’t see people complaining about Team Fortress 2’s map design, which purposefully contains constant bottlenecks on 99% of maps. Why single Brink out?
Two maps in particular stand out for me (although all are brilliant). The first is Breakout, which sees Resistance attacking a Security fortification to rescue a pilot . As Resistance you have to assault the complex and gain access, before downloading the data key, freeing the pilot and working your way back through the level with him in tow, to exit the way you came in. It’s brilliantly designed, from the multiple routes to the different layers and angles of attack. You genuinely feel penned in and on edge. The entire dynamics of fighting also change considerably on the way out as you are forced to approach the map from an entirely different angle.
My second favourite is Attack On CCity, where Resistance must defend a gate system against Security. Not only does the map look absolutely phenomenal, but the shanty town image, the nooks and crannies and the multiple routes everywhere are so enjoyable I’ve genuinely never encountered a better map in a team-based shooter. It’s an absolute triumph and is so exciting to play through as you desperately defend or attack.
The Conclusion
I won’t name and shame other review sites, but I cannot understand where the criticism of Brink stems from. Brink hasn’t had a perfect launch, but to receive the treatment it has is absolutely mystifying. Brink is the finest team based shooter to arrive in years and having sunk hundreds of hours into Battlefield: Bad Company 2, Team Fortress 2 and Call of Duty (amongst countless others) it easily surpasses them all on almost every level. Its learning curve is steep, and team work is entirely pivotal to your success, but at what stage did we approach team-based shooters as solo affairs?
I want to be encouraged to work together, I want to achieve something as a team and Brink does just that and offers me plenty of choices along the way. From the incredible map design, to smaller changes such as rewarding experience points based on the damage you did as opposed to the killing blow, an incredible character customiser and some of the punchiest, satisfying weapons I’ve ever used in a first person shooter, Splash Damage deserve your money.
Both Team Fortress 2 and Battlefield: Bad Company 2 feel stale against Brink (though neither are bad games!). SMART is a triumph, and the fact that playing games without it feels so alien is a testament to the product Splash Damage have produced. Whilst Brink may always remain niche, I’m so glad it has come along and I cannot wait for its first free expansion pack due in June.
Email the author of this post at lewisb@tap-repeatedly.com
Website: Brink Official Site
Minimum System Requirements (PC):
OS: Windows XP SP3, Vista or Windows 7
CPU: Intel Core 2 Duo @ 2.4 Ghz / AMD Athlon 64 X2 5200+
RAM: 2 GB
HDD: 8 GB free disk space
Graphics: 512 MB Graphics Memory
Sound Card: DirectX 9 Compatible
DirectX: Version 9.0c
Reviewer’s System: |
package middleware
import (
"database/sql"
"errors"
"github.com/DATA-DOG/go-sqlmock"
"github.com/HFO4/cloudreve/models"
"github.com/HFO4/cloudreve/pkg/auth"
"github.com/HFO4/cloudreve/pkg/cache"
"github.com/HFO4/cloudreve/pkg/serializer"
"github.com/HFO4/cloudreve/pkg/util"
"github.com/gin-gonic/gin"
"github.com/jinzhu/gorm"
"github.com/qiniu/api.v7/v7/auth/qbox"
"github.com/stretchr/testify/assert"
"io/ioutil"
"net/http"
"net/http/httptest"
"strings"
"testing"
)
var mock sqlmock.Sqlmock
// TestMain 初始化数据库Mock
func TestMain(m *testing.M) {
var db *sql.DB
var err error
db, mock, err = sqlmock.New()
if err != nil {
panic("An error was not expected when opening a stub database connection")
}
model.DB, _ = gorm.Open("mysql", db)
defer db.Close()
m.Run()
}
func TestCurrentUser(t *testing.T) {
asserts := assert.New(t)
rec := httptest.NewRecorder()
c, _ := gin.CreateTestContext(rec)
c.Request, _ = http.NewRequest("GET", "/test", nil)
//session为空
sessionFunc := Session("233")
sessionFunc(c)
CurrentUser()(c)
user, _ := c.Get("user")
asserts.Nil(user)
//session正确
c, _ = gin.CreateTestContext(rec)
c.Request, _ = http.NewRequest("GET", "/test", nil)
sessionFunc(c)
util.SetSession(c, map[string]interface{}{"user_id": 1})
rows := sqlmock.NewRows([]string{"id", "deleted_at", "email", "options"}).
AddRow(1, nil, "admin@cloudreve.org", "{}")
mock.ExpectQuery("^SELECT (.+)").WillReturnRows(rows)
CurrentUser()(c)
user, _ = c.Get("user")
asserts.NotNil(user)
asserts.NoError(mock.ExpectationsWereMet())
}
func TestAuthRequired(t *testing.T) {
asserts := assert.New(t)
rec := httptest.NewRecorder()
c, _ := gin.CreateTestContext(rec)
c.Request, _ = http.NewRequest("GET", "/test", nil)
AuthRequiredFunc := AuthRequired()
// 未登录
AuthRequiredFunc(c)
asserts.NotNil(c)
// 类型错误
c.Set("user", 123)
AuthRequiredFunc(c)
asserts.NotNil(c)
// 正常
c.Set("user", &model.User{})
AuthRequiredFunc(c)
asserts.NotNil(c)
}
func TestSignRequired(t *testing.T) {
asserts := assert.New(t)
auth.General = auth.HMACAuth{SecretKey: []byte(util.RandStringRunes(256))}
rec := httptest.NewRecorder()
c, _ := gin.CreateTestContext(rec)
c.Request, _ = http.NewRequest("GET", "/test", nil)
SignRequiredFunc := SignRequired()
// 鉴权失败
SignRequiredFunc(c)
asserts.NotNil(c)
c.Request, _ = http.NewRequest("PUT", "/test", nil)
SignRequiredFunc(c)
asserts.NotNil(c)
}
func TestWebDAVAuth(t *testing.T) {
asserts := assert.New(t)
rec := httptest.NewRecorder()
AuthFunc := WebDAVAuth()
// options请求跳过验证
{
c, _ := gin.CreateTestContext(rec)
c.Request, _ = http.NewRequest("OPTIONS", "/test", nil)
AuthFunc(c)
}
// 请求HTTP Basic Auth
{
c, _ := gin.CreateTestContext(rec)
c.Request, _ = http.NewRequest("POST", "/test", nil)
AuthFunc(c)
asserts.NotEmpty(c.Writer.Header()["WWW-Authenticate"])
}
// 用户名不存在
{
c, _ := gin.CreateTestContext(rec)
c.Request, _ = http.NewRequest("POST", "/test", nil)
c.Request.Header = map[string][]string{
"Authorization": {"Basic d2hvQGNsb3VkcmV2ZS5vcmc6YWRtaW4="},
}
mock.ExpectQuery("SELECT(.+)users(.+)").
WillReturnRows(
sqlmock.NewRows([]string{"id", "password", "email"}),
)
AuthFunc(c)
asserts.NoError(mock.ExpectationsWereMet())
asserts.Equal(c.Writer.Status(), http.StatusUnauthorized)
}
// 密码错误
{
c, _ := gin.CreateTestContext(rec)
c.Request, _ = http.NewRequest("POST", "/test", nil)
c.Request.Header = map[string][]string{
"Authorization": {"Basic d2hvQGNsb3VkcmV2ZS5vcmc6YWRtaW4="},
}
mock.ExpectQuery("SELECT(.+)users(.+)").
WillReturnRows(
sqlmock.NewRows([]string{"id", "password", "email", "options"}).AddRow(1, "123", "who@cloudreve.org", "{}"),
)
// 查找密码
mock.ExpectQuery("SELECT(.+)webdav(.+)").WillReturnRows(sqlmock.NewRows([]string{"id"}))
AuthFunc(c)
asserts.NoError(mock.ExpectationsWereMet())
asserts.Equal(c.Writer.Status(), http.StatusUnauthorized)
}
//未启用 WebDAV
{
c, _ := gin.CreateTestContext(rec)
c.Request, _ = http.NewRequest("POST", "/test", nil)
c.Request.Header = map[string][]string{
"Authorization": {"Basic d2hvQGNsb3VkcmV2ZS5vcmc6YWRtaW4="},
}
mock.ExpectQuery("SELECT(.+)users(.+)").
WillReturnRows(
sqlmock.NewRows(
[]string{"id", "password", "email", "group_id", "options"}).
AddRow(1,
"rfBd67ti3SMtYvSg:ce6dc7bca4f17f2660e18e7608686673eae0fdf3",
"who@cloudreve.org",
1,
"{}",
),
)
mock.ExpectQuery("SELECT(.+)groups(.+)").WillReturnRows(sqlmock.NewRows([]string{"id", "web_dav_enabled"}).AddRow(1, false))
// 查找密码
mock.ExpectQuery("SELECT(.+)webdav(.+)").WillReturnRows(sqlmock.NewRows([]string{"id"}).AddRow(1))
AuthFunc(c)
asserts.NoError(mock.ExpectationsWereMet())
asserts.Equal(c.Writer.Status(), http.StatusForbidden)
}
//正常
{
c, _ := gin.CreateTestContext(rec)
c.Request, _ = http.NewRequest("POST", "/test", nil)
c.Request.Header = map[string][]string{
"Authorization": {"Basic d2hvQGNsb3VkcmV2ZS5vcmc6YWRtaW4="},
}
mock.ExpectQuery("SELECT(.+)users(.+)").
WillReturnRows(
sqlmock.NewRows(
[]string{"id", "password", "email", "group_id", "options"}).
AddRow(1,
"rfBd67ti3SMtYvSg:ce6dc7bca4f17f2660e18e7608686673eae0fdf3",
"who@cloudreve.org",
1,
"{}",
),
)
mock.ExpectQuery("SELECT(.+)groups(.+)").WillReturnRows(sqlmock.NewRows([]string{"id", "web_dav_enabled"}).AddRow(1, true))
// 查找密码
mock.ExpectQuery("SELECT(.+)webdav(.+)").WillReturnRows(sqlmock.NewRows([]string{"id"}).AddRow(1))
AuthFunc(c)
asserts.NoError(mock.ExpectationsWereMet())
asserts.Equal(c.Writer.Status(), 200)
_, ok := c.Get("user")
asserts.True(ok)
}
}
func TestRemoteCallbackAuth(t *testing.T) {
asserts := assert.New(t)
rec := httptest.NewRecorder()
AuthFunc := RemoteCallbackAuth()
// 成功
{
cache.Set(
"callback_testCallBackRemote",
serializer.UploadSession{
UID: 1,
PolicyID: 513,
VirtualPath: "/",
},
0,
)
cache.Deletes([]string{"1"}, "policy_")
mock.ExpectQuery("SELECT(.+)users(.+)").
WillReturnRows(sqlmock.NewRows([]string{"id", "group_id"}).AddRow(1, 1))
mock.ExpectQuery("SELECT(.+)groups(.+)").
WillReturnRows(sqlmock.NewRows([]string{"id", "policies"}).AddRow(1, "[513]"))
mock.ExpectQuery("SELECT(.+)policies(.+)").
WillReturnRows(sqlmock.NewRows([]string{"id", "secret_key"}).AddRow(2, "123"))
c, _ := gin.CreateTestContext(rec)
c.Params = []gin.Param{
{"key", "testCallBackRemote"},
}
c.Request, _ = http.NewRequest("POST", "/api/v3/callback/remote/testCallBackRemote", nil)
authInstance := auth.HMACAuth{SecretKey: []byte("123")}
auth.SignRequest(authInstance, c.Request, 0)
AuthFunc(c)
asserts.NoError(mock.ExpectationsWereMet())
asserts.False(c.IsAborted())
}
// Callback Key 不存在
{
c, _ := gin.CreateTestContext(rec)
c.Params = []gin.Param{
{"key", "testCallBackRemote"},
}
c.Request, _ = http.NewRequest("POST", "/api/v3/callback/remote/testCallBackRemote", nil)
authInstance := auth.HMACAuth{SecretKey: []byte("123")}
auth.SignRequest(authInstance, c.Request, 0)
AuthFunc(c)
asserts.True(c.IsAborted())
}
// 用户不存在
{
cache.Set(
"callback_testCallBackRemote",
serializer.UploadSession{
UID: 1,
PolicyID: 550,
VirtualPath: "/",
},
0,
)
cache.Deletes([]string{"1"}, "policy_")
mock.ExpectQuery("SELECT(.+)users(.+)").
WillReturnRows(sqlmock.NewRows([]string{"id", "group_id"}))
c, _ := gin.CreateTestContext(rec)
c.Params = []gin.Param{
{"key", "testCallBackRemote"},
}
c.Request, _ = http.NewRequest("POST", "/api/v3/callback/remote/testCallBackRemote", nil)
authInstance := auth.HMACAuth{SecretKey: []byte("123")}
auth.SignRequest(authInstance, c.Request, 0)
AuthFunc(c)
asserts.NoError(mock.ExpectationsWereMet())
asserts.True(c.IsAborted())
}
// 签名错误
{
cache.Set(
"callback_testCallBackRemote",
serializer.UploadSession{
UID: 1,
PolicyID: 514,
VirtualPath: "/",
},
0,
)
cache.Deletes([]string{"1"}, "policy_")
mock.ExpectQuery("SELECT(.+)users(.+)").
WillReturnRows(sqlmock.NewRows([]string{"id", "group_id"}).AddRow(1, 1))
mock.ExpectQuery("SELECT(.+)groups(.+)").
WillReturnRows(sqlmock.NewRows([]string{"id", "policies"}).AddRow(1, "[514]"))
mock.ExpectQuery("SELECT(.+)policies(.+)").
WillReturnRows(sqlmock.NewRows([]string{"id", "secret_key"}).AddRow(2, "123"))
c, _ := gin.CreateTestContext(rec)
c.Params = []gin.Param{
{"key", "testCallBackRemote"},
}
c.Request, _ = http.NewRequest("POST", "/api/v3/callback/remote/testCallBackRemote", nil)
AuthFunc(c)
asserts.NoError(mock.ExpectationsWereMet())
asserts.True(c.IsAborted())
}
// Callback Key 为空
{
c, _ := gin.CreateTestContext(rec)
c.Request, _ = http.NewRequest("POST", "/api/v3/callback/remote", nil)
AuthFunc(c)
asserts.True(c.IsAborted())
}
}
func TestQiniuCallbackAuth(t *testing.T) {
asserts := assert.New(t)
rec := httptest.NewRecorder()
AuthFunc := QiniuCallbackAuth()
// Callback Key 相关验证失败
{
c, _ := gin.CreateTestContext(rec)
c.Params = []gin.Param{
{"key", "testQiniuBackRemote"},
}
c.Request, _ = http.NewRequest("POST", "/api/v3/callback/remote/testQiniuBackRemote", nil)
AuthFunc(c)
asserts.True(c.IsAborted())
}
// 成功
{
cache.Set(
"callback_testCallBackQiniu",
serializer.UploadSession{
UID: 1,
PolicyID: 515,
VirtualPath: "/",
},
0,
)
cache.Deletes([]string{"1"}, "policy_")
mock.ExpectQuery("SELECT(.+)users(.+)").
WillReturnRows(sqlmock.NewRows([]string{"id", "group_id"}).AddRow(1, 1))
mock.ExpectQuery("SELECT(.+)groups(.+)").
WillReturnRows(sqlmock.NewRows([]string{"id", "policies"}).AddRow(1, "[515]"))
mock.ExpectQuery("SELECT(.+)policies(.+)").
WillReturnRows(sqlmock.NewRows([]string{"id", "access_key", "secret_key"}).AddRow(2, "123", "123"))
c, _ := gin.CreateTestContext(rec)
c.Params = []gin.Param{
{"key", "testCallBackQiniu"},
}
c.Request, _ = http.NewRequest("POST", "/api/v3/callback/qiniu/testCallBackQiniu", nil)
mac := qbox.NewMac("123", "123")
token, err := mac.SignRequest(c.Request)
asserts.NoError(err)
c.Request.Header["Authorization"] = []string{"QBox " + token}
AuthFunc(c)
asserts.NoError(mock.ExpectationsWereMet())
asserts.False(c.IsAborted())
}
// 验证失败
{
cache.Set(
"callback_testCallBackQiniu",
serializer.UploadSession{
UID: 1,
PolicyID: 516,
VirtualPath: "/",
},
0,
)
cache.Deletes([]string{"1"}, "policy_")
mock.ExpectQuery("SELECT(.+)users(.+)").
WillReturnRows(sqlmock.NewRows([]string{"id", "group_id"}).AddRow(1, 1))
mock.ExpectQuery("SELECT(.+)groups(.+)").
WillReturnRows(sqlmock.NewRows([]string{"id", "policies"}).AddRow(1, "[516]"))
mock.ExpectQuery("SELECT(.+)policies(.+)").
WillReturnRows(sqlmock.NewRows([]string{"id", "access_key", "secret_key"}).AddRow(2, "123", "123"))
c, _ := gin.CreateTestContext(rec)
c.Params = []gin.Param{
{"key", "testCallBackQiniu"},
}
c.Request, _ = http.NewRequest("POST", "/api/v3/callback/qiniu/testCallBackQiniu", nil)
mac := qbox.NewMac("123", "123")
token, err := mac.SignRequest(c.Request)
asserts.NoError(err)
c.Request.Header["Authorization"] = []string{"QBox " + token + " "}
AuthFunc(c)
asserts.NoError(mock.ExpectationsWereMet())
asserts.True(c.IsAborted())
}
}
func TestOSSCallbackAuth(t *testing.T) {
asserts := assert.New(t)
rec := httptest.NewRecorder()
AuthFunc := OSSCallbackAuth()
// Callback Key 相关验证失败
{
c, _ := gin.CreateTestContext(rec)
c.Params = []gin.Param{
{"key", "testOSSBackRemote"},
}
c.Request, _ = http.NewRequest("POST", "/api/v3/callback/oss/testQiniuBackRemote", nil)
AuthFunc(c)
asserts.True(c.IsAborted())
}
// 签名验证失败
{
cache.Set(
"callback_testCallBackOSS",
serializer.UploadSession{
UID: 1,
PolicyID: 517,
VirtualPath: "/",
},
0,
)
cache.Deletes([]string{"1"}, "policy_")
mock.ExpectQuery("SELECT(.+)users(.+)").
WillReturnRows(sqlmock.NewRows([]string{"id", "group_id"}).AddRow(1, 1))
mock.ExpectQuery("SELECT(.+)groups(.+)").
WillReturnRows(sqlmock.NewRows([]string{"id", "policies"}).AddRow(1, "[517]"))
mock.ExpectQuery("SELECT(.+)policies(.+)").
WillReturnRows(sqlmock.NewRows([]string{"id", "access_key", "secret_key"}).AddRow(2, "123", "123"))
c, _ := gin.CreateTestContext(rec)
c.Params = []gin.Param{
{"key", "testCallBackOSS"},
}
c.Request, _ = http.NewRequest("POST", "/api/v3/callback/oss/testCallBackOSS", nil)
mac := qbox.NewMac("123", "123")
token, err := mac.SignRequest(c.Request)
asserts.NoError(err)
c.Request.Header["Authorization"] = []string{"QBox " + token}
AuthFunc(c)
asserts.NoError(mock.ExpectationsWereMet())
asserts.True(c.IsAborted())
}
// 成功
{
cache.Set(
"callback_TnXx5E5VyfJUyM1UdkdDu1rtnJ34EbmH",
serializer.UploadSession{
UID: 1,
PolicyID: 518,
VirtualPath: "/",
},
0,
)
cache.Deletes([]string{"1"}, "policy_")
mock.ExpectQuery("SELECT(.+)users(.+)").
WillReturnRows(sqlmock.NewRows([]string{"id", "group_id"}).AddRow(1, 1))
mock.ExpectQuery("SELECT(.+)groups(.+)").
WillReturnRows(sqlmock.NewRows([]string{"id", "policies"}).AddRow(1, "[518]"))
mock.ExpectQuery("SELECT(.+)policies(.+)").
WillReturnRows(sqlmock.NewRows([]string{"id", "access_key", "secret_key"}).AddRow(2, "123", "123"))
c, _ := gin.CreateTestContext(rec)
c.Params = []gin.Param{
{"key", "TnXx5E5VyfJUyM1UdkdDu1rtnJ34EbmH"},
}
c.Request, _ = http.NewRequest("POST", "/api/v3/callback/oss/TnXx5E5VyfJUyM1UdkdDu1rtnJ34EbmH", ioutil.NopCloser(strings.NewReader(`{"name":"2f7b2ccf30e9270ea920f1ab8a4037a546a2f0d5.jpg","source_name":"1/1_hFRtDLgM_2f7b2ccf30e9270ea920f1ab8a4037a546a2f0d5.jpg","size":114020,"pic_info":"810,539"}`)))
c.Request.Header["Authorization"] = []string{"e5LwzwTkP9AFAItT4YzvdJOHd0Y0wqTMWhsV/h5SG90JYGAmMd+8LQyj96R+9qUfJWjMt6suuUh7LaOryR87Dw=="}
c.Request.Header["X-Oss-Pub-Key-Url"] = []string{"aHR0cHM6Ly9nb3NzcHVibGljLmFsaWNkbi5jb20vY2FsbGJhY2tfcHViX2tleV92MS5wZW0="}
AuthFunc(c)
asserts.NoError(mock.ExpectationsWereMet())
asserts.False(c.IsAborted())
}
}
type fakeRead string
func (r fakeRead) Read(p []byte) (int, error) {
return 0, errors.New("error")
}
func TestUpyunCallbackAuth(t *testing.T) {
asserts := assert.New(t)
rec := httptest.NewRecorder()
AuthFunc := UpyunCallbackAuth()
// Callback Key 相关验证失败
{
c, _ := gin.CreateTestContext(rec)
c.Params = []gin.Param{
{"key", "testUpyunBackRemote"},
}
c.Request, _ = http.NewRequest("POST", "/api/v3/callback/upyun/testUpyunBackRemote", nil)
AuthFunc(c)
asserts.True(c.IsAborted())
}
// 无法获取请求正文
{
cache.Set(
"callback_testCallBackUpyun",
serializer.UploadSession{
UID: 1,
PolicyID: 509,
VirtualPath: "/",
},
0,
)
cache.Deletes([]string{"1"}, "policy_")
mock.ExpectQuery("SELECT(.+)users(.+)").
WillReturnRows(sqlmock.NewRows([]string{"id", "group_id"}).AddRow(1, 1))
mock.ExpectQuery("SELECT(.+)groups(.+)").
WillReturnRows(sqlmock.NewRows([]string{"id", "policies"}).AddRow(1, "[519]"))
mock.ExpectQuery("SELECT(.+)policies(.+)").
WillReturnRows(sqlmock.NewRows([]string{"id", "access_key", "secret_key"}).AddRow(2, "123", "123"))
c, _ := gin.CreateTestContext(rec)
c.Params = []gin.Param{
{"key", "testCallBackUpyun"},
}
c.Request, _ = http.NewRequest("POST", "/api/v3/callback/upyun/testCallBackUpyun", ioutil.NopCloser(fakeRead("")))
AuthFunc(c)
asserts.NoError(mock.ExpectationsWereMet())
asserts.True(c.IsAborted())
}
// 正文MD5不一致
{
cache.Set(
"callback_testCallBackUpyun",
serializer.UploadSession{
UID: 1,
PolicyID: 510,
VirtualPath: "/",
},
0,
)
cache.Deletes([]string{"1"}, "policy_")
mock.ExpectQuery("SELECT(.+)users(.+)").
WillReturnRows(sqlmock.NewRows([]string{"id", "group_id"}).AddRow(1, 1))
mock.ExpectQuery("SELECT(.+)groups(.+)").
WillReturnRows(sqlmock.NewRows([]string{"id", "policies"}).AddRow(1, "[520]"))
mock.ExpectQuery("SELECT(.+)policies(.+)").
WillReturnRows(sqlmock.NewRows([]string{"id", "access_key", "secret_key"}).AddRow(2, "123", "123"))
c, _ := gin.CreateTestContext(rec)
c.Params = []gin.Param{
{"key", "testCallBackUpyun"},
}
c.Request, _ = http.NewRequest("POST", "/api/v3/callback/upyun/testCallBackUpyun", ioutil.NopCloser(strings.NewReader("1")))
c.Request.Header["Content-Md5"] = []string{"123"}
AuthFunc(c)
asserts.NoError(mock.ExpectationsWereMet())
asserts.True(c.IsAborted())
}
// 签名不一致
{
cache.Set(
"callback_testCallBackUpyun",
serializer.UploadSession{
UID: 1,
PolicyID: 511,
VirtualPath: "/",
},
0,
)
cache.Deletes([]string{"1"}, "policy_")
mock.ExpectQuery("SELECT(.+)users(.+)").
WillReturnRows(sqlmock.NewRows([]string{"id", "group_id"}).AddRow(1, 1))
mock.ExpectQuery("SELECT(.+)groups(.+)").
WillReturnRows(sqlmock.NewRows([]string{"id", "policies"}).AddRow(1, "[521]"))
mock.ExpectQuery("SELECT(.+)policies(.+)").
WillReturnRows(sqlmock.NewRows([]string{"id", "access_key", "secret_key"}).AddRow(2, "123", "123"))
c, _ := gin.CreateTestContext(rec)
c.Params = []gin.Param{
{"key", "testCallBackUpyun"},
}
c.Request, _ = http.NewRequest("POST", "/api/v3/callback/upyun/testCallBackUpyun", ioutil.NopCloser(strings.NewReader("1")))
c.Request.Header["Content-Md5"] = []string{"c4ca4238a0b923820dcc509a6f75849b"}
AuthFunc(c)
asserts.NoError(mock.ExpectationsWereMet())
asserts.True(c.IsAborted())
}
// 成功
{
cache.Set(
"callback_testCallBackUpyun",
serializer.UploadSession{
UID: 1,
PolicyID: 512,
VirtualPath: "/",
},
0,
)
cache.Deletes([]string{"1"}, "policy_")
mock.ExpectQuery("SELECT(.+)users(.+)").
WillReturnRows(sqlmock.NewRows([]string{"id", "group_id"}).AddRow(1, 1))
mock.ExpectQuery("SELECT(.+)groups(.+)").
WillReturnRows(sqlmock.NewRows([]string{"id", "policies"}).AddRow(1, "[522]"))
mock.ExpectQuery("SELECT(.+)policies(.+)").
WillReturnRows(sqlmock.NewRows([]string{"id", "access_key", "secret_key"}).AddRow(2, "123", "123"))
c, _ := gin.CreateTestContext(rec)
c.Params = []gin.Param{
{"key", "testCallBackUpyun"},
}
c.Request, _ = http.NewRequest("POST", "/api/v3/callback/upyun/testCallBackUpyun", ioutil.NopCloser(strings.NewReader("1")))
c.Request.Header["Content-Md5"] = []string{"c4ca4238a0b923820dcc509a6f75849b"}
c.Request.Header["Authorization"] = []string{"UPYUN 123:GWueK9x493BKFFk5gmfdO2Mn6EM="}
AuthFunc(c)
asserts.NoError(mock.ExpectationsWereMet())
asserts.False(c.IsAborted())
}
}
func TestOneDriveCallbackAuth(t *testing.T) {
asserts := assert.New(t)
rec := httptest.NewRecorder()
AuthFunc := OneDriveCallbackAuth()
// Callback Key 相关验证失败
{
c, _ := gin.CreateTestContext(rec)
c.Params = []gin.Param{
{"key", "testUpyunBackRemote"},
}
c.Request, _ = http.NewRequest("POST", "/api/v3/callback/upyun/testUpyunBackRemote", nil)
AuthFunc(c)
asserts.True(c.IsAborted())
}
// 成功
{
cache.Set(
"callback_testCallBackUpyun",
serializer.UploadSession{
UID: 1,
PolicyID: 512,
VirtualPath: "/",
},
0,
)
cache.Deletes([]string{"1"}, "policy_")
mock.ExpectQuery("SELECT(.+)users(.+)").
WillReturnRows(sqlmock.NewRows([]string{"id", "group_id"}).AddRow(1, 1))
mock.ExpectQuery("SELECT(.+)groups(.+)").
WillReturnRows(sqlmock.NewRows([]string{"id", "policies"}).AddRow(1, "[657]"))
mock.ExpectQuery("SELECT(.+)policies(.+)").
WillReturnRows(sqlmock.NewRows([]string{"id", "access_key", "secret_key"}).AddRow(2, "123", "123"))
c, _ := gin.CreateTestContext(rec)
c.Params = []gin.Param{
{"key", "testCallBackUpyun"},
}
c.Request, _ = http.NewRequest("POST", "/api/v3/callback/upyun/testCallBackUpyun", ioutil.NopCloser(strings.NewReader("1")))
AuthFunc(c)
asserts.NoError(mock.ExpectationsWereMet())
asserts.False(c.IsAborted())
}
}
func TestCOSCallbackAuth(t *testing.T) {
asserts := assert.New(t)
rec := httptest.NewRecorder()
AuthFunc := COSCallbackAuth()
// Callback Key 相关验证失败
{
c, _ := gin.CreateTestContext(rec)
c.Params = []gin.Param{
{"key", "testUpyunBackRemote"},
}
c.Request, _ = http.NewRequest("POST", "/api/v3/callback/upyun/testUpyunBackRemote", nil)
AuthFunc(c)
asserts.True(c.IsAborted())
}
// 成功
{
cache.Set(
"callback_testCallBackUpyun",
serializer.UploadSession{
UID: 1,
PolicyID: 512,
VirtualPath: "/",
},
0,
)
cache.Deletes([]string{"1"}, "policy_")
mock.ExpectQuery("SELECT(.+)users(.+)").
WillReturnRows(sqlmock.NewRows([]string{"id", "group_id"}).AddRow(1, 1))
mock.ExpectQuery("SELECT(.+)groups(.+)").
WillReturnRows(sqlmock.NewRows([]string{"id", "policies"}).AddRow(1, "[702]"))
mock.ExpectQuery("SELECT(.+)policies(.+)").
WillReturnRows(sqlmock.NewRows([]string{"id", "access_key", "secret_key"}).AddRow(2, "123", "123"))
c, _ := gin.CreateTestContext(rec)
c.Params = []gin.Param{
{"key", "testCallBackUpyun"},
}
c.Request, _ = http.NewRequest("POST", "/api/v3/callback/upyun/testCallBackUpyun", ioutil.NopCloser(strings.NewReader("1")))
AuthFunc(c)
asserts.NoError(mock.ExpectationsWereMet())
asserts.False(c.IsAborted())
}
}
func TestIsAdmin(t *testing.T) {
asserts := assert.New(t)
rec := httptest.NewRecorder()
testFunc := IsAdmin()
// 非管理员
{
c, _ := gin.CreateTestContext(rec)
c.Set("user", &model.User{})
testFunc(c)
asserts.True(c.IsAborted())
}
// 是管理员
{
c, _ := gin.CreateTestContext(rec)
user := &model.User{}
user.Group.ID = 1
c.Set("user", user)
testFunc(c)
asserts.False(c.IsAborted())
}
// 初始用户，非管理组
{
c, _ := gin.CreateTestContext(rec)
user := &model.User{}
user.Group.ID = 2
user.ID = 1
c.Set("user", user)
testFunc(c)
asserts.False(c.IsAborted())
}
}
|
Why Men Always Get the “S” Jobs
I’ve had enough. For far too long, me and my fellow menfolk have been forced into cruel, subservient tasks: The s— jobs. My latest epiphany reveals why it occurs, and one brave man’s attempt to right this messy wrong.
Be forewarned: This epiphany is laden with gender stereotypes. If you are a woman, and you happen to also get the s— jobs, you have my sympathies. This epiphany is not directed at you.
But in my neck of the woods, the man not only gets the shaft, he gets the s—. Here are a few examples:
• When the toilet backs up, we have to plunge the s—
• When the dog craps on the lawn, we have to pick up the s—
• When the garbage has to go out, we have to put out the s—
• When its redecorating time (groan), we have to move all the heavy s—
Now I know my wife could do these things. She’s an exercise nut, and more than capable of plunging, pick up, putting out and moving s—. But I always get the job. In fact, it’s always been assumed that men get the job.
Forget the “Glass Ceiling.” This is the “S— Basement.” And after extensive research, I’ve uncovered why it exists.
Downfall via Diapers
In previous generations, men were the lone bread earners in the household. After busting their humps all day, aproned wives rewarded them a martini and a warm meal. And these guys didn’t change diapers. At all.
But mom did have a little pull, even back then. To compensate for not handling their kid’s s—, dads had to handle everyone else’s s—. They accepted the s— jobs, and all was well.
Fast forward to today’s dual-income household, where both moms and dads bust humps all day. In these household, men change diapers. They don’t dare do otherwise.
Yet we don’t see a re-alignment of the s— jobs. Why?
I explained the inequities of the system to my wife. She listened patiently, nodding as I described the insurmountable horrors of plunging, scooping, and picking up s—.
“I used to dream of you,” I said. “And now I dream of poop.”
I proposed that we redistribute the workload. We would share s— jobs together, just as we had shared diaper changing. Together we would scrape dog excrement from our shoe soles. Together we would move that freakin’ heavy hide-a-bed for the thousandth time. Together we would contract e coli from fecal matter run amuck.
That toilet is scary-looking. I sympathize. Not only do I do all the cooking, cleaning, and laundry, but I do the S— work too. But only because I don’t have a husband to do it for me. Oops! Isn’t it written somewhere in the wedding vows? |
Q:
How to write Regex to replace some words and add "" if missing in JavaScript. Unexpected behavior
I have to find some key strings and surrond them with quote if they are not:
`[aa,bb,cc, "aa","bb","cc"]`.replace(/[^"](aa|bb)[^"]/g, `"$1"`)
expected:
"["aa","bb",cc, "aa","bb","cc"]"
but I got this:
""aa""bb",cc, "aa","bb","cc"]"
What was happend with '[' and comma ','?
A:
You need to capture [^"] too and use back reference while replacing, as you're not capturing them and replacing only the matched value with the captured group so you end up loosing the value matched by [^"]
let final = `[aa,bb,cc "aa","bb","cc"]`.replace(/([^"])(aa|bb)(?!")/g, `$1"$2"`)
console.log(final)
|
A community worker checks body temperature for a resident at a street near the Yellow Crane Pavilion in Wuhan, central China's Hubei Province, on Friday. Photo: Xinhua |
Hazard and mode of action of disinfection by-products (DBPs) in water for human consumption: Evidences and research priorities.
Disinfection of water system is an essential strategy to protect human health from pathogens and prevent their regrowth during water distribution, but the reaction of disinfectant agents with organic matter can lead to the formation of disinfection by-products (DBPs). Given their widespread occurrence, potential human health impacts and (eco)toxicity associated with exposure to DBPs are of particular interest due to their potential carcinogenicity and vary non-carcinogenic effects, such as endocrine disruption. Understanding the public health implications of this emerging issue is crucial for societies and decision-makers, supporting more effective water safety plans. Here, we review the recent literature on the effects of DBPs presented in drinking water and treated swimming pools water, focusing particularly in unregulated compounds and the putative underlying mode of action, linking the available data with adverse health outcomes. Overall, the majority of studies highlight the limited knowledge in the understanding of the underlying mode of action of DBPs. Yet, available evidences indicate that different signaling pathways seem to be involved in the adverse outcomes associated with distinct DBPs classes. The main knowledge gaps in this field are also identified, and future research priorities discussed. |
Abstract
Bacteria contain DNA polymerase I (PolI), a single polypeptide chain consisting of ∼930 residues, possessing DNA-dependent DNA polymerase, 3′-5′ proofreading and 5′-3′ exonuclease (also known as flap endonuclease) activities. PolI is particularly important in the processing of Okazaki fragments generated during lagging strand replication and must ultimately produce a double-stranded substrate with a nick suitable for DNA ligase to seal. PolI's activities must be highly coordinated both temporally and spatially otherwise uncontrolled 5′-nuclease activity could attack a nick and produce extended gaps leading to potentially lethal double-strand breaks. To investigate the mechanism of how PolI efficiently produces these nicks, we present theoretical studies on the dynamics of two possible scenarios or models. In one the flap DNA substrate can transit from the polymerase active site to the 5′-nuclease active site, with the relative position of the two active sites being kept fixed; while the other is that the 5′-nuclease domain can transit from the inactive mode, with the 5′-nuclease active site distant from the cleavage site on the DNA substrate, to the active mode, where the active site and substrate cleavage site are juxtaposed. The theoretical results based on the former scenario are inconsistent with the available experimental data that indicated that the majority of 5′-nucleolytic processing events are carried out by the same PolI molecule that has just extended the upstream primer terminus. By contrast, the theoretical results on the latter model, which is constructed based on available structural studies, are consistent with the experimental data. We thus conclude that the latter model rather than the former one is reasonable to describe the cooperation of the PolI's polymerase and 5′-3′ exonuclease activities. Moreover, predicted results for the latter model are presented.
There are no comments yet on this publication. Be the first to share your thoughts. |
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50+ unique gifts for under $50: Gift ideas for absolutely everyone
What's This?
By Mashable Deals2018-12-03 16:43:46 UTC
Some people possess an innate talent for thoughtful gift-giving. No matter the occasion or recipient, they somehow nail their gift selections year after year, cementing their status as favorite aunt/coworker/second-cousin-twice-removed.
On the other end of the spectrum, we all know that one misguided friend or family member who gifts the same scented lotion for every special occasion. Their go-to presents are boring at best, cringe-worthy at worst, and almost always end up relegated to the re-gift pile.
For those who aren’t naturally gifted at the art of giving, we've done the heavy lifting for you. Below are more than 50 options guaranteed to help you rise through the ranks of gift-giver hierarchy — all for under $50.
This fun-sized waffle maker is ideal for anyone who takes breakfast seriously — or really, for anyone who likes to get creative in the kitchen. A recipe book that outlines how to make everything from fluffy Belgian waffles to grilled sandwiches to flavorful fajitas is included with purchase, and a non-stick surface makes cooking easy and cleanup painless.
From AMAZON
$14.99
See Details
Forget about braving mind-boggling brunch lines for a savory breakfast. This ingenious device lets you whip up pro-level brekkie sandwiches — complete with eggs, cheese, veggies, or whatever inventive ingredients you care to include — in five minutes flat.
From AMAZON
$35.99
See Details
Regardless of which side of the cilantro debate you’re on, we can all agree that stale herbs are the pits. The Herb Savor can help you prolong the life of an array of kitchen herbs by up to three weeks, and it’s specially designed to fit in a fridge door for easy storage.
From AMAZON
$24.95
See Details
Slice, dice, and chop your way to Iron Chef status with this kitchen gadget that packs big punch in a bite-sized package. Whip up pico de gallo, mince garlic with the press of a button, or chop onions tears-free with this handy machine.
From AMAZON
$49.87
See Details
Take your fiestas up a notch with this adorable taco holder that’s shaped like a Tyrannosaurus (also available in other dino-species). Perfect for kids or fun-loving adults, this unique utensil is the kitchen accessory you never knew you were missing in your life.
From AMAZON
$12.99
See Details
Gift this NutriChef Electric Crepe Maker & Griddle to your friends with creative culinary inclinations. From crepes and blintzes to home-style pancakes, this gadget will be a welcome addition to any brunch-lover’s kitchen collection.
From AMAZON
$29.99
See Details
Nostalgia Electronics’ retro-style hot dog toaster will revolutionize your indoor BBQ game. The machine makes two hot dogs at a time while simultaneously toasting the buns — and it’s sure to be a hit with the whole family.
From AMAZON
$14.99
See Details
There are casual coffee drinkers, and then there are those who can’t see colors in the morning before they’ve had their cup of joe. For the latter, this self-stirring mug will be a much-appreciated gift they’ll get use out of daily.
From AMAZON
$9.99
See Details
From the brand behind one of the hottest blenders on the market, this Ninja coffee bean grinder is a pro-grade accessory that’s perfect for any coffee connoisseur. It’s simple to use and grinds up to 60 grams of coffee at once.
Go ahead and hit that snooze button — this cold brew iced coffee maker lets you create cold brew at home, so you can skip the morning shuffle through that outrageous line at the corner café.
From AMAZON
$29.99
See Details
Who needs the spa when you can get the same treatment at home? This dual-speed handheld spin brush has three interchangeable heads and works with the cleanser of your choice. It can even be used in the shower. Gift it to the hard-to-buy-for beauty guru on your list and rest assured it'll be her favorite present all season.
From VANITY PLANET
$39.99
See Details
Need a time out? There’s something truly mesmerizing about this futuristic-looking hourglass. Its wooden stand includes a strong magnet that molds magnetic sand into intriguing patterns at the base. A great gift idea for that friend who is known office-wide for their quirky desk tchotchkes.
From AMAZON
$9.98
See Details
It's hard to keep track of little things like keys (and phones and wallets...) which is why the Tile Mate makes a great gift. Loop the Tile Mate through any keychain or attach it to anything you don't want to lose. Then, when something's nearby but out of sight, you can make the Tile ring to find it. Can’t find your phone? The Tile can make it ring — even if it’s on silent.
From AMAZON
$49.99
/ Four
See Details
Constantly-working people need a mouse that works just as hard as they do — we’re thinking a mouse that only needs to be charged once a month could do the trick. The tiny ergonomic shape optimizes comfort even for long periods of computer time and can be stored in a pocket for travel. The nearly 5,000 5-star reviews on Amazon mention that clicks are quiet and satisfying, likely adding some fun to a boring workday.
From AMAZON
$15.99
See Details
If the sound of your smartphone’s default alarm prompts a Pavlovian reaction of undiluted rage, it’s time to reevaluate how you wake up in the morning. This alarm clock gradually increases or decreases brightness to mimic the rising or setting of the sun, and includes six adjustable, natural sounds for a much more Zen experience.
From AMAZON
$25.99
See Details
These cool-looking lamps are more than just soothing to look at. It is believed that Himalayan salt lamps have a purifying effect on the air and emit calming vibes via ionization. Each of these lamps is uniquely crafted by hand and comes with a hand-carved wooden base. (Comes with two lamps!)
From AMAZON
$24.99
See Details
For your friend or family member who has been trying to “get into” meditation for the past three years, this miniature Zen garden is a solid first step. The 9-inch by 9-inch rosewood frame makes a handsome addition to any desk, coffee table, or bookshelf.
From AMAZON
$21.99
See Details
This aromatherapy oil diffuser is great for anyone who can use a calming presence in their life. The diffuser features seven color settings and a lovely wood-grain look that will subtly complement any home’s aesthetic.
From AMAZON
$19.99
See Details
Get mom up to speed on the bath bomb trend with this set by ArtNaturals. The kit includes 12 essential oil bath bombs with scents like coconut, lavender, aloe vera, seagrass, and green tea — and it’s perfect for a DIY spa day.
From AMAZON
$16.95
See Details
For those interested in dabbling in the world of smart home accessories, a great place to start is with these easy-to-install smart outlets. Use these Wi-Fi outlets for tasks like brewing morning coffee from bed, putting nightlights on a timer, and monitoring household data.
From AMAZON
$19.98
See Details
Looking for a gift for that friend who can’t even keep succulents alive? Try opting for something a little more mechanical. These robotic dogs respond to commands to sit, walk, beg, sing, and even jump — no puppy potty pads required.
From AMAZON
$29.82
See Details
This kit lets kids virtually connect with the world around them by turning everyday objects into connected keyboards. There are thousands of possibilities for the future engineer, inventor, or programmer to fiddle with — and it’s also a fun twist on family game night.
From AMAZON
$49.95
See Details
This beautiful keyboard lends an island aesthetic to any home office. Both the keyboard and accompanying mouse wirelessly connect to your computer or gaming system, so you can keep your work station clear from cords and clutter.
From AMAZON
$47.99
See Details
This speaker integrates with home voice-assistant technologies to bring you a fusion of smart-home seamlessness and high-tech tunes. It makes the perfect gift for your buddies on a first-name basis with Siri, Alexa, Cortana, and the whole AI gang.
From AMAZON
$34.99
See Details
This waterproof portable speaker can withstand sun, sand, and surf. It’s sure to be appreciated by any frequent traveler, beach-junkie, camping buddy, or that friend who treats the shower like their personal concert hall.
From AMAZON
$34.99
See Details
Stuck on what to get your cousin who’s going through a serious hipster phase? Made by heritage brand Victrola, the portable record player comes in an attractive suitcase-style case and includes a turntable, speakers, and built-in Bluetooth technology.
From AMAZON
$44.18
See Details
A smart home isn't complete without an Echo — and the newest Dot makes it easier than ever to add the convenience of Alexa to any room. The third generation Dot is the cutest yet, complete with better sound and microphones than earlier ones. They can control smart devices like lights or blinds, become a household intercom, order pizza or an Uber, and more. You're basically gifting an easier (lazier) life, and who wouldn't want that?
From AMAZON
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Mashable
is a global, multi-platform media and entertainment company. Powered by its own proprietary technology, Mashable is the go-to source for tech, digital culture and entertainment content for its dedicated and influential audience around the globe. |
Relationship between N-methyl-D-aspartate receptor NR1 splice variants and NR2 subunits.
The N-methyl-D-aspartate (NMDA) subtype of ionotropic glutamate receptor is assembled from one NR1 subunit, expressed in eight splice variants, and four NR2 subunits (NR2A-D). The combination of subunits and splice variants determines the pharmacological and physiological properties of the receptor. In the present study we investigated the relationship between NR1 splice variants and NR2 subunits in rat brain using a series of antibodies selective for the four NR1 cassettes, which vary in the NR1 splice variants, and for NR2A and NR2B. Sodium deoxycholate at pH 9.0 solubilized about 35% of the receptor, which was intact based on co-immunoprecipitation of NR1 and NR2 subunits and chemical cross-linking of the solubilized receptor. The cross-linked product contained three high molecular weight components, Mr = 603,000, 700,000, and 750,000, which were immunolabeled with antibodies to NR1 and to NR2 subunits. Immunoprecipitation analyses using antibodies selective for NR2A and NR2B showed no preferential assembly between NR2 subunits and NR1 splice variants. There was little co-immunoprecipitation of NR2A and NR2B, suggesting that most NMDA receptor complexes contain only one of these subunits. However, receptor complexes can contain at least two different NR1 splice variants. In developing conditions for the solubilization of intact NMDA receptor complexes, we observed a differential solubilization of NR1 and NR2 subunits. NR2 was nearly insoluble in Triton X-100 in both microsomal and synaptic membrane fractions, while NR1 was readily soluble in the microsomal fraction but insoluble in the synaptic membrane fraction. These results suggest that the NR1 subunit is modified when it is incorporated into the synaptic membrane, possibly by strengthening its interaction with NR2 or another synaptic protein. |
California is working with federal officials to expand the testing of possible coronavirus patients, Gov. Gavin Newsom announced Thursday.
“We are currently in deep partnership with the [Centers for Disease Control and Prevention] on one overriding protocol that drives our principal focus right now, and that’s testing,” Newsom said, calling expanded testing “our top priority, not just in the state of California but, I imagine, across the United States.”
Such an expansion includes both broadening the criteria that a person must meet to be tested for COVID-19, as well as getting more coronavirus test kits sent to California, he said. The state has 200 kits for both diagnostic and surveillance purposes, but federal officials say more will arrive in the coming days, he said.
In his first news conference to address the coronavirus issue, the governor sought to calm concerns and reassure Californians that the state will be prepared if the virus spreads. Despite several California communities having declared local emergencies over the outbreak, Newsom said there are currently no plans to declare a statewide emergency.
“We’re meeting this moment,” Newsom said. “We have been in constant contact with federal agencies. We have history and expertise in this space. We are not overreacting, but nor are we underreacting to the understandable anxiety that many people have as it relates to this novel virus.”
The announcement comes after a Solano County woman became the country’s first coronavirus case in a patient who neither recently traveled out of the country nor was in contact with someone who did, raising the possibility that the virus is already spreading within the community. The woman was not tested for several days until after she was hospitalized because she did not fit CDC criteria, which include both symptoms of the virus and either a recent history of travel to China or close contact with another coronavirus patient.
The case demonstrates the need for more rapid testing of all coronavirus patients, Sen. Dianne Feinstein (D-Calif.) wrote Thursday in a letter to Vice President Mike Pence, who has been appointed the Trump administration’s coronavirus czar.
“I understand the newly diagnosed patient was not tested immediately for coronavirus, despite the request by her attending physicians at the University of California-Davis Medical Center,” she wrote. “It is unclear whether the delay was caused by overly restrictive CDC testing criteria or whether the CDC was unable to process the diagnostic test faster.”
A total of 33 people have been diagnosed with coronavirus in California, and five have since left the state, Newsom said. Of the confirmed cases, 24 were either evacuated from the Diamond Princess cruise ship or returned on repatriation flights from Wuhan, China — the epicenter of the coronavirus outbreak.
Newsom said more than 8,400 people are being monitored in 49 local jurisdictions.
The state is “doing those protocols and monitoring as it relates to more traditional commercial flights that came in from points of concern and potential points of contact, particularly in Asia,” he said.
COVID-19 has infected more than 80,000 people in about three dozen countries since it was discovered in late December. More than 2,700 have died, most of them in mainland China.
There are currently 60 confirmed cases in the United States, and officials continue to stress that the risk to the general public remains low. Still, they are warning people to be prepared to see more U.S. cases.
“Ultimately, we expect we will see coronavirus spread in this country,” CDC Director Nancy Messonnier said Tuesday. “It’s not so much a question of if, but a question of when.”
The warnings already are scaring away visitors to California. On Thursday, Microsoft and Epic Games announced they were pulling out of next month’s Game Developers Conference in San Francisco.
Facebook also announced it has canceled the in-person portion of its annual F8 software developer conference, which was scheduled to take place May 5-6 in San Jose.
UC Davis officials said the Solano County coronavirus patient arrived at its medical center from another hospital Feb. 19 but was not tested until Sunday. The hospital said that precautions had been put in place because of healthcare workers’ concerns about the patient’s condition, but that it has asked a “small number” of employees to stay home and monitor their temperature.
The woman was “in her community” for a number of days before accessing care, California Health and Human Services Agency Secretary Dr. Mark Ghaly said Thursday.
Investigators are now working to find and contact any individuals who may have come in contact with the woman. The CDC has sent 10 staffers to help trace her contacts, Newsom said.
Sacramento County health officials said that dozens of people might be quarantined at home based on contact with the woman. UC Davis spokesman Steve Telliano disputed that the number could be higher than 70, but refused to say if more or fewer people were quarantined.
Sacramento County Director of Health Services Peter Beileson said the patient had been transferred to the hospital with virus containment protocols, but additional measures, such as isolation in a negative pressure room, which prevents air from escaping, were now in place. Beileson said he did not expect additional cases of the virus based on those in contact with the patient in Sacramento, where she is receiving treatment, but did expect more cases in Solano County, where she is from.
Solano County is also home to Travis Air Force Base, where several hundred people were quarantined after returning on repatriation flights from Wuhan and the Diamond Princess cruise ship in Japan. Multiple positive coronavirus cases have been reported among those groups. But Dr. Sonia Angell, director of the California Department of Public Health, said there is no evidence of a connection between the woman and the returning travelers.
Newsom said the state expected that the virus would transmit within the community.
“What happened in Solano County did not surprise anybody,” Newsom said. “I think the only thing ... that surprised folks was it didn’t happen sooner.”
On Thursday, Solano County announced that it had issued a proclamation of local emergency due to the virus.
“We are taking this situation seriously and are taking steps necessary to protect the health and safety of Solano County residents,” Dr. Bela Matyas, Solano County health officer, said in a statement. “It is important to recognize that we have moved from containment to mitigation. We are investigating potential exposures and ensuring that proper evaluation and care are provided if they become sick.”
The county has activated its Department Operations Center to bolster its efforts in identifying, screening and contacting people who were exposed to the virus. The declaration allows the county to better collaborate with local, state and federal agencies to implement containment efforts, officials said.
Some other local governments have also declared states of emergency amid coronavirus fears, including Santa Clara County on Feb. 10, San Diego County last week, San Francisco on Tuesday and Orange County on Wednesday.
Such a declaration creates a mechanism for officials to ask for mutual aid from state and federal partners in the event that resources are exhausted, Orange County health officer Nichole Quick said Wednesday.
Los Angeles County officials also are discussing whether to declare a public health emergency.
There is no defined criteria, such as a certain number of coronavirus cases, for what merits an emergency declaration, said L.A. city emergency management department General Manager Aram Sahakian in a meeting with City Council members on Wednesday.
“If the core group preparing for this incident feels and sees the need for additional resources and additional multiagency coordination, they can pull the trigger on it,” he said. “It doesn’t mean it’s out of control. It doesn’t mean we’re going to panic.”
The last time L.A. County declared a public health emergency was in 2017 during a hepatitis A outbreak.
The hepatitis A outbreak required extraordinary measures because the disease was difficult to contain and required widespread vaccination efforts, said L.A. County health department epidemiologist Dr. Prabhu Gounder in a meeting with City Council members on Wednesday. California officials declared a state of emergency at that time as well.
“At the time, there was a shortage of hepatitis A vaccine. We would not have been able to get this additional vaccine for the response without the state declaration,” Gounder said.
Times staff writers Anita Chabria and Soumya Karlamangla contributed to this report. |
---
abstract: |
In this work we develop a new criterion for the existence of topological horseshoes for surface homeomorphisms in the isotopy class of the identity. Based on our previous work on forcing theory, this new criterion is purely topological and can be expressed in terms of equivariant Brouwer foliations and transverse trajectories. We then apply this new tool in the study of the dynamics of homeomorphisms of surfaces with zero genus and null topological entropy and we obtain several applications. For homeomorphisms of the open annulus ${\mathbb{A}}$ with zero topological entropy, we show that rotation numbers exists for all points with nonempty omega limits, and that if ${\mathbb{A}}$ is a generalized region of instability then it admits a single rotation vector. We also offer a new proof of a recent result of Passegi, Potrie and Sambarino, showing that zero entropy dissipative homeomorphisms of the annulus having as an atractor a circloid have a single rotation number.
Our work also studies homeomorphisms of the sphere without horseshoes. For these maps we present a structure theorem in terms of fixed point free invariant sub-annuli, as well as a very restricted description of all possible dynamical behavior in the transitive subsets. This description ensures, for instance, that transitive sets can contain at most $2$ distinct periodic orbits and that, in many cases, the restriction of the homeomorphism to the transitive set must be an extension of an odometer. In particular, we show that any nontrivial and stable transitive subset of a dissipative diffeomorphism of the plane is always infinitely renormalizable in the sense of Bonatti-Gambaudo-Lion-Tresser.
address:
- 'Sorbonne Université, Institut de Mathématiques de Jussieu-Paris Rive Gauche, CNRS, Univ Paris Diderot, Campus Pierre et Marie Curie, case 247, 4 place Jussieu, 75252 Paris cedex 5, France'
- 'University of São Paulo, Instituto de Matemática e Estatística, Rua do Matão 1010, 05508-090, São Paulo, Brasil & Friedrich-Schiller-Universit''’at, Institute of Mathematics, Ernst-Abbe-Platz 2, 07743, Jena, Germany'
author:
- Patrice Le Calvez
- Fabio Tal
title: Topological horseshoes for surface homeomorphisms
---
[^1]
Introduction
============
In the whole text, we will define the entropy of a homeomorphism $f$ of a Hausdorff locally compact topological space $X$ as being the topological entropy of the extension of $f$ to the Alexandrov compactification of $X$, that fixes the point at infinity. We will say that $f$ is [*topologically chaotic*]{} if its entropy is positive and if the number of periodic points of period $n$ for some iterate of $f$ grows exponentially in $n$. An important class of topologically chaotic homeomorphisms is the class of homeomorphisms that possesses a [*topological horseshoe*]{}. Let us give a precise definition of what we have in mind: a compact subset $Y$ of $X$ is a topological horseshoe if it is invariant by a power $f^r$ of $f$ and if $f^r\vert_Y$ admits a finite extension $g:Z\to Z$ on a Hausdorff compact space such that:
- $g$ is an extension of the Bernouilli shift $\sigma :\{1,\dots, q\}^{{\mathbb{Z}}}\to \{1,\dots, q\}^{{\mathbb{Z}}}$, where $q{\geqslant}2$;
- The preimage of every $s$-periodic sequence of $\{1,\dots,q\}^{{\mathbb{Z}}}$ by the factor map $u:Z\to \{1,\dots, q\}^{{\mathbb{Z}}}$ contains a $s$-periodic point of $S$.
By an extension we mean a homeomorphism semi-conjugated to a given homeomorphism, by a finite extension we mean an extension such that the fibers of the factor map are all finite with an uniform bound in their cardinality. Note that, if $h(f)$ denotes the topological entropy of $f$, then $$rh(f)=h(f^r){\geqslant}h(f^r\vert_Y)=h(S){\geqslant}h(\sigma)=\log q,$$and that $f^{rn}$ has at least $q^n/m$ fixed points for every $n{\geqslant}1$, if the cardinality of the fibers of the factor map $v:Z\to Y$ are uniformly bounded by $m$.[^2]
Since the groundbreaking work of Smale [@smale], the presence of horseshoes have been a paradigmatic feature of dynamical systems, and its prevalence as a phenomena is well known. Their role in the particular case of surface dynamics is even larger, as demonstrated by the results of Katok in [@Katok], showing by Pesin’s Theory that any sufficiently smooth diffeomorphism of a surface with strictly positive topological entropy has a topological horseshoe. Nonetheless, this direct relationship between existence of horseshoes and topological entropy is not valid in the context of homeomorphisms (see [@Rees], [@BeguinCrovisierLeRoux1]).
This article is a natural continuation of [@LeCalvezTal], where the authors gave a new orbit forcing theory for surface homeomorphisms related to the notion of [*maximal isotopies*]{}, [*transverse foliations*]{} and [*transverse trajectories*]{}. A classical example of forcing result is Sharkovski’s theorem: there exists an explicit total order $\preceq$ on the set of natural integers such that every continuous transformation $f$ on $[0,1]$ that contains a periodic orbit of period $m$ contains a periodic orbit of period $n$ if $n\preceq m$. Recall that if $f$ admits a periodic orbit whose period is not a power of $2$, one can construct a Markov partition and codes orbits with a sub-shift of finite type: more precisely there exists a topological horseshoe. The simplest situation from which the existence of a topological horseshoe can be deduced is the existence of a point $x$ such that $f^3(x){\leqslant}x<f(x)<f^2(x)$. The fundamental theorem of this paper, which will be announced at the end of this introduction, is a simple criterion of existence of a topological horseshoe stated in terms of transverse trajectories. We already gave a partial result in this direction in [@LeCalvezTal] but it required stronger hypothesis on the transverse trajectories. Moreover the conclusion was weaker: we proved that the map was topologically chaotic but did not get a topological horseshoe.
The main result can be used as an interesting tool in the study of surface homemorphisms, with relations to horseshoes or to entropy. It has the advantage of being a purely topological technique, and as so can be applied with no restrictions to homeomorphisms, but the theorem has also new and relevant consequences even in the presence of additional smoothness. We apply our main result do deduce many applications to the study of homeomorphisms of a surface of genus zero isotopic to the identity. This will be done usually by studying the complementary problem of describing the behaviour of homeomorphisms with no topological horseshoes, and consequently of $\mathcal{C}^{1+\epsilon}$ diffeomorphisms with zero topological entropy. This subject has been previously investigated for instance by Handel in [@Handel], who studied among other things the existence of rotation numbers for maps of the annulus, and by Franks and Handel in [@FranksHandel], who presented a structure theorem for $\mathcal{C}^{\infty}$ diffeomorphisms of the sphere preserving the volume measure, a result we extended to the $\mathcal{C}^{0}$ context in [@LeCalvezTal]. Another related topic that has received considerable attention is the study of dissipative diffeomorphisms of the plane with no entropy, in particular in the Henon family, with an emphasis in describing the dynamics of diffeomorphisms that can be approximated by maps with strictly positive entropy (see, for instance, [@CLM; @LM; @HMT]). Recent progress, which includes a description of dynamical behavior of transitive sets for the subset of [*strongly dissipative diffeomorphisms*]{}, was obtained by Crovisier, Kocsard, Koropecki and Pujals (see [@CKKP]), which enabled a breakthrough in the understanding of the boundary of entropy zero for a region of the parameter plane for the Henon family (see [@CPT]).
Applications
------------
Our first application, one that is a main tool in obtaining the other results, is about rotation numbers for annular homeomorphisms with no topological horseshoes. We will use the definition introduced by Le Roux [@LeRoux] and developed by Conejeros [@Conejeros]. Write ${\mathbb{T}}^1={\mathbb{R}}/{\mathbb{Z}}$, denote by $\check\pi: {\mathbb{R}}^2\to {\mathbb{A}}$ the universal covering projection of the open annulus ${\mathbb{A}}={\mathbb{T}}^1\times{\mathbb{R}}$ and by $\pi_1:{\mathbb{R}}^2 \to {\mathbb{R}}$ the projection in the first coordinate. Write $T:(x,y)\mapsto (x+1,y)$ for the fundamental covering automorphism. Let $f$ be a homeomorphism of ${\mathbb{A}}$ isotopic to identity and $\check f$ a lift to ${\mathbb{R}}^2$. Say that a point $z$ is an [*escaping*]{} point of $f$, if the sequence $(f^n(z))_{n{\geqslant}0}$ converges to an end of ${\mathbb{A}}$ or equivalently if its $\omega$-limit set $\omega(z)$ is empty. We will denote $\mathrm{ne}^+(f)$ the complement of the set of escaping points. Denote by $\mathrm{ne}^-(f)=\mathrm{ne}^{+}(f^{-1})$ the set of points $z$ such that the $\alpha$-limit set $\alpha(z)$ is non empty. We will define $ \mathrm{ne}(f)=\mathrm{ne}^+(f)\cup \mathrm{ne}^-(f^{-})$ and denote $\Omega(f)$ the set of non-wandering points. We say that $z\in\mathrm{ne}^+(f)$ has a [*rotation number $\mathrm{rot}_{\check f}(z)$*]{} if, for every compact set $K\subset{\mathbb{A}}$ and every increasing sequence of integers $(n_k)_{k{\geqslant}0}$ such that $f^{n_k}(z)\in K$, we have $$\lim_{k\to+\infty}\frac{1}{n_k}\left( \pi_1(\check f^{n_k}(\check z))- \pi_1(\check z)\right)=\mathrm{rot}_{\check f}(z),$$ if $\check z$ is a lift of $z$. [^3]
It is straightforward that if $h$ is a homeomorphism of ${\mathbb{A}}$ then $z$ belongs to $\mathrm{ne}^+(f)$ is and only if $h(z)$ belongs to $\mathrm{ne}^+(h\circ f\circ h^{-1})$. Moreover, if $\check h$ is a lift of $h$ to ${\mathbb{R}}^2$, then $\mathrm{rot}_{\check h\circ\check f\circ\check h^{-1}}(h(z))$ exists if and only if $\mathrm{rot}_{\check f}(z)$ exists and we have $$\mathrm{rot}_{\check h\circ\check f\circ\check h^{-1}}(h(z)) = \begin{cases}
\mathrm{rot}_{\check f}(z) &\mathrm{if }\,\, h\mathrm{\,\, induces \,\, the \,\, identity \,\, on\,\, }H_1({\mathbb{A}},{\mathbb{Z}}),\\
-\mathrm{rot}_{\check f}(z) &\mathrm{otherwise.}\\
\end{cases}$$ Consequently, one can naturally define rotation numbers for a homeomorphism of an abstract annulus $A$ and a given lift to the universal covering space, as soon as a generator of $H_1(A,{\mathbb{Z}})$ is chosen. More precisely, let $f$ be homeomorphism of $A$ isotopic to the identity and $\check f$ a lift of $f\vert_A$ to the universal covering space $\check A$ of $A$. Let $\kappa$ be a generator of $H_1(A,{\mathbb{Z}})$ and $h:A\to {\mathbb{A}}$ a homeomorphism such that $h_*(\kappa)=\kappa_*$, where $\kappa_*$ is the generator of $H_1({\mathbb{A}},{\mathbb{Z}})$ induced by the loop $\Gamma_* :t\mapsto (t,0)$. If $\check h:\check A\to {\mathbb{R}}^2$ is a lift of $h$, then $\check h\circ \check f\circ\textcolor{black}{\check{h}}^{-1}$ is a lift of $h\circ f\vert_A\circ h^{-1}$, independent of the choice of $\check h$. If $z$ belongs to $\mathrm{ne}^+(f\vert_A)$, then $h(z)$ belongs to $\mathrm{ne}^+(h\circ f\vert_A\circ h^{-1})$. The existence of does not depend on the choice of $h$. In case such a rotation number is well defined, we will denote it $\mathrm{rot}_{\check f, \kappa}(z)$ because it does not depend on $h$. Note that $$\mathrm{rot}_{\check f, -\kappa}(z)=-\mathrm{rot}_{\check f, \kappa}(z).$$ We will also define $$\mathrm{rot}_{f, \kappa}(z)=\mathrm{rot}_{\check f, \kappa}(z)+{\mathbb{Z}}\in {\mathbb{R}}/{\mathbb{Z}}.$$
\[thmain:rotation-number\] Let $f$ be a homeomorphism of ${\mathbb{A}}$ isotopic to the identity and $\check f$ a lift of $f$ to ${\mathbb{R}}^2$. We suppose that $f$ has no topological horseshoe. Then
1. each point $z\in\mathrm{ne}^+(f)$ has a rotation number $\mathrm{rot}_{\check f}(z)$;
2. for every points $z,z'\in\mathrm{ne}^+(f)$ such that $z'\in\omega(z)$, we have $\mathrm{rot}_{\check f}(z')=\mathrm{rot}_{\check f}(z)$;
3. if $z\in \mathrm{ne}^+(f)\cap \mathrm{ne}^-(f)$ is non-wandering, then $\mathrm{rot}_{\check f^{-1}}(z)=-\mathrm{rot}_{\check f}(z)$;
4. the map $\mathrm{rot}_{\check f^{\pm}} : \Omega(f) \cap \mathrm{ne}(f)\to{\mathbb{R}}$ is continuous, where $$\mathrm{rot}_{\check f^{\pm}}(z) = \begin{cases}\mathrm{rot}_{\check f}(z) &\mathrm{if}\enskip z\in \Omega(f) \cap \mathrm{ne}^+(f),\\
-\mathrm{rot}_{\check f^{-1}}(z) &\mathrm{if}\enskip z\in \Omega(f) \cap \mathrm{ne}^-(f).
\end{cases}$$
In [@LeCalvezTal], we proved the existence of the rotation number $\mathrm{rot}_{\check f}(z)$ for every bi-recurrent point $z$, assuming that $f$ is not topologically chaotic (which is a stronger hypothesis). Similar results were previously proved by Handel [@Handel].
The map $\check f :(x,y)\mapsto (x+y, \,y\vert y\vert)$ lifts a homeomorphism $f$ of ${\mathbb{A}}$ isotopic to the identity such that $$\mathrm{ne}^+(f)={\mathbb{T}}^1\times[-1,1],\enskip \mathrm{ne}^-(f)={\mathbb{A}}, \enskip \Omega(f)={\mathbb{T}}^1\times\{-1,0,1\}.$$ Note that $$\mathrm{rot}_{\check f}(z) = \begin{cases}
-1 &\mathrm{if}\enskip z\in {\mathbb{T}}^1\times\{-1\},\\
0 &\mathrm{if}\enskip z\in {\mathbb{T}}^1\times(-1,1),\\
1 &\mathrm{if}\enskip z\in {\mathbb{T}}^1\times\{-1\},\\
\end{cases}$$ and $$\mathrm{rot}_{\check f^{-1}}(z) = \begin{cases}
1 &\mathrm{if}\enskip z\in {\mathbb{T}}^1\times(-\infty,0),\\
0 &\mathrm{if}\enskip z\in {\mathbb{T}}^1\times\{0\},\\
-1 &\mathrm{if}\enskip z\in {\mathbb{T}}^1\times(0,+\infty).\\
\end{cases}$$ The maps $ \mathrm{rot}_{\check f}$ and $\mathrm{rot}_{\check f^{-1}}$ are non continuous on their domains of definition and the equality $ \mathrm{rot}_{\check f}(z)=-\mathrm{rot}_{\check f^{-1}}(z)$ is true only if $z\in\Omega(f)$.
The next result is a strong improvement of the assertion (3) of Theorem \[thmain:rotation-number\], it will be expressed in terms of [*Birkhoff cycles*]{} and [*Birkhoff recurrence classes*]{}. Let $X$ be a metric space, $f$ a homeomorphism of $X$, and $z_1$, $z_2$ two points of $X$. We say that there exists a [*Birkhoff connection*]{} from $z_1$ to $z_2$ if for every neighborhood $W_1$ of $z_1$ and every neighborhood $W_2$ of $z_2$, there exists $n{\geqslant}1$ such that $W_1\cap f^{-n}(W_2)\not=\emptyset$. A [*Birkhoff cycle*]{} is a finite sequence $(z _i)_{i\in{\mathbb{Z}}/p{\mathbb{Z}}}$ such that for every $i\in{\mathbb{Z}}/p{\mathbb{Z}}$, there exists a Birkhoff connection from $z_i$ to $z_{i+1}$. A point $z$ is said to be [*Birkhoff recurrent*]{} for $f$ if there exists a Birkhoff cycle containing $z$. Note that a point $z$ is non-wandering if and only if it defines a Birkhoff cycle , so every non wandering point is Birkhoff recurrent. One can define an equivalence relation in the set of Birkhoff recurrent points. Say that $z_1$ is [*Birkhoff equivalent*]{} to $z_2$ if there exists a Birkhoff cycle containing both points. The equivalence classes will be called Birkhoff recurrence classes. Note that every $\omega$-limit set or $\alpha$-limit set is contained in a single Birkhoff recurrence class. In particular, every transitive set, which means every set $\Lambda$ such that there exists $z\in\Lambda$ satisfying $\omega(z)=\Lambda$, is contained in a single Birkhoff recurrence class. On the other hand, it is easily seen that any Birkhoff recurrent class is contained in a chain recurrent class, but the converse does not need to hold, for instance when $f$ is the identity.
If $f$ is a homeomorphism of ${\mathbb{A}}$ isotopic to the identity, we denote $f_{\mathrm{sphere}}$ the continuous extension of $f$ to the sphere obtained by adding the two ends $N$ and $S$ of ${\mathbb{A}}$.
\[thmain:birkhoffcycles\] Let $f$ be a homeomorphism of ${\mathbb{A}}$ isotopic to the identity and $\check f$ a lift of $f$ to ${\mathbb{R}}^2$. We suppose that $f$ has no topological horseshoe. If $\mathcal B$ is a Birkhoff recurrence class of $f_{\mathrm{sphere}}$, there exists $\rho\in{\mathbb{R}}$ such that, $$\begin{cases} \mathrm{rot}_{\check f}(z) = \rho&\mathrm{if}\enskip z\in {\mathcal B}\cap\mathrm{ne}^+(f),\\
\mathrm{rot}_{\check f^{-1}}(z) = -\rho&\mathrm{if}\enskip z\in {\mathcal B}\cap\mathrm{ne}^-(f).
\end{cases}$$
We immediately deduce:
\[corollarymain:birkhoffclasses\] Let $f$ be a homeomorphism of ${\mathbb{A}}$ isotopic to the identity and $\check f$ a lift of $f$ to ${\mathbb{R}}^2$. We suppose that $f$ has no topological horseshoe. Then for every Birkhoff cycle $(z _i)_{i\in{\mathbb{Z}}/p{\mathbb{Z}}}$ of $f$ in $\mathrm{ne}(f)$, there exists $\rho\in{\mathbb{R}}$ such that, for every $i\in{\mathbb{Z}}/p{\mathbb{Z}}$: $$\begin{cases} \mathrm{rot}_{\check f}(z_i) = \rho&\mathrm{if}\enskip z_i\in \mathrm{ne}^+(f),\\
\mathrm{rot}_{\check f^{-1}}(z_i) = -\rho&\mathrm{if}\enskip z_i\in \mathrm{ne}^-(f).
\end{cases}$$
If $f$ is a homeomorphism of ${\mathbb{A}}$ isotopic to the identity and $\check f$ a lift of $f$ to ${\mathbb{R}}^2$, we will say that $\rho\in{\mathbb{R}}$ is a rotation number of $\check f$ if either there exists $z\in \mathrm{ne}^+(f)$ such that $\mathrm{rot}_{\check f}(z) =\rho$ or if there exists $z\in \mathrm{ne}^-(f)$ such that $\mathrm{rot}_{\check f^{-1}}(z) =-\rho$. We can prove another result.
\[propositionmain\_regionofinstability\] Let $f$ be a homeomorphism of ${\mathbb{A}}$ isotopic to the identity, $\check f$ a lift of $f$ to ${\mathbb{R}}^2$. We suppose that:
1. $\check f$ has at least two rotations numbers;
2. $N$ and $S$ belong to the same Birkhoff recurrence class of $f_{\mathrm{sphere}}$ .
Then $f$ has a topological horseshoe.
It was already known by Birkhoff [@Birkhoff] that the hypothesis of the theorem are satisfied for an area preserving positive twist map of the annulus, restricted to a Birkhoff region of instability. The positiveness of the entropy was already stated in this case (see Boyland-Hall [@BoylandHall] and Boyland [@Boyland] for a topological proof, see Angenent [@Angenent] for a more combinatorial proof). The previous corollary extends the result for situations including a generalized region of instability: the area preserving hypothesis can be replaced by the existence of a cycle of Birkhoff connections containing the two ends, the twist condition by the existence of two different rotation numbers.
The third application is a new proof of a recent result of Passegi, Potrie and Sambarino [@PassegiPotrieSambarino]. Say that a homeomorphism $f$ of ${\mathbb{A}}$ is [*dissipative*]{} if for every non empty open set $U$ of finite area, the area of $f(U)$ is smaller than the area of $U$. Say that an invariant compact set $X$ is [*locally stable*]{} if $X$ admits a fundamental system of forward invariant neighborhoods. Say that a compact set $X\subset{\mathbb{A}}$ is a [*circloid*]{} if
- it is the intersection of a nested sequence of sets homeomorphic to ${\mathbb{T}}^1\times[0,1]$ and homotopic to ${\mathbb{T}}^1\times\{0\}$;
- it is minimal for the inclusion among sets satisfying this property.
Its complement has two connected components that are non relatively compact. A classical example is a [*cofrontier*]{}: $X$ is a compact set whose complement has exactly two connected components, they are not relatively compact and $X$ is their common boundary.
\[thmain:circloids\] Let $f$ be a homeomorphism of ${\mathbb{A}}$ isotopic to the identity and $\check f$ a lift of $f$ to ${\mathbb{R}}^2$. We suppose that $f$ has no topological horseshoe. Let $X$ be an invariant circloid. If $f$ is dissipative or if $X$ is locally stable, then the function $\rho_{\widetilde f}$, which is well defined on $X$, is constant on this set. More precisely, the sequence of maps $$\check z\mapsto \frac{ \pi_1(\check f^{n}(\check z))- \pi_1(\check z)}{n}$$ converges uniformly to this constant on $\check\pi^{-1}(X)$.
Nevertheless, is not clear that they are necessary to get the conclusion. So, it is natural to ask whether the conclusion holds without supposing $f$ dissipative or $X$ locally stable.
We can conclude this description of results related to rotation number with the following result which has its own interest:
\[prop:zero-rotation\]Let $f$ be a homeomorphism isotopic to the identity of ${\mathbb{A}}$ and $\check f$ a lift of $f$ to ${\mathbb{R}}^2$. We suppose that $\check f$ is fixed point free and that there exists $z_*\in\Omega(f)\cap \mathrm{ne}^+(f)$ such that $\mathrm{rot}_{\check f}(z_*)$ is well defined and equal to $0$.
- Then, at least one of the following situation occurs:
1. there exists $q$ arbitrarily large such that $\check f^q\circ T^{-1}$ has a fixed point;
2. there exists $q$ arbitrarily large such that $\check f^q\circ T$ has a fixed point.
- Moreover, if $z_*$ is positively recurrent then $f$ has a topological horseshoe.
An important section of this article is devoted to a structural study of the homeomorphisms of the $2$-sphere ${\mathbb{S}}^2$ with no horseshoe. Let us state the fundamental result.
\[thmain:global-structure\] Let $f:{\mathbb{S}}^2\to{\mathbb{S}}^2$ be an orientation preserving homeomorphism that has no topological horseshoe. Then, the set $$\Omega'(f)=\{z\in\Omega(f)\,\vert\, \alpha(z)\cup\omega(z)\not\subset\mathrm{fix}(f)\}.$$ can be covered by a family $(A_{\alpha})_{\alpha\in\mathcal A}$ of invariant sets such that:
1. $A_{\alpha}$ is a fixed point free topological open annulus and $f\vert_{A_{\alpha}}$ is isotopic to the identity;
2. if $\kappa$ is a generator of $H_1(A_{\alpha},{\mathbb{Z}})$, there exists a lift of $f\vert_{A_{\alpha}}$ to the universal covering space of $A_{\alpha}$ whose rotation numbers are all included in $[0,1]$;
3. $A_{\alpha}$ is maximal for the two previous properties.
Such a result was already proved by Franks and Handel in the case of smooth area preserving diffeomorphisms. In that case, the first condition implies the second one. Moreover, the family $(A_{\alpha})_{\alpha\in\mathcal A}$ is explicitly defined in terms of [*disk recurrent points*]{} (see [@FranksHandel]) and the annuli are pairwise disjoint. This result is the main building block in a structure theorem for area preserving diffeomorphisms given by the two authors (one should apply the previous result to all iterates of $f$). We extended Franks-Handel results in the case of homeomorphisms with no wandering points in [@LeCalvezTal]. In the present article, there is an attempt to give a similar structure theorem with no assumption about wandering points. It appears that some of the results remain (usually with proofs technically tougher) and some not (for example, the annuli appearing in the previous theorem are not necessarily disjoint). There is probably some progress to do to obtain a complete description of homeomorphisms with no topological horseshoes. Nevertheless we were able to state some results on the structure of Birkhoff recurrence classes and of transitive sets.
A property of Birkhoff recurrence classes, related to rotation numbers, has already been stated in Theorem \[thmain:birkhoffcycles\]. Let us continue to describe the behavior of Birkhoff recurrence classes. Say a set $X\subset \mathrm{fix}(f)$ is [*unlinked*]{} if the restriction of $f$ to ${\mathbb{S}}^2\setminus X$ is still homotopic to the identity. Let us also say a homeomorphism $f$ of ${\mathbb{S}}^2$ is an [*irrational pseudo-rotation*]{} if:
- $f$ has exactly two periodic points, $z_0$ and $z_1$, which are both fixed;
- $\mathrm{ne}(f\vert _{{\mathbb{S}}^2\setminus\{z_0, z_1\}})$ is not empty;
- if $\check f$ is a lift of $f\vert _{{\mathbb{S}}^2\setminus\{z_0, z_1\}}$ to the universal covering space of ${\mathbb{S}}^2\setminus\{z_0, z_1\}$, its unique rotation number is an irrational number $\rho$.
\[propmain:birkoffclasses3fixedpoints\] Suppose that $f$ is an orientation preserving homeomorphism of ${\mathbb{S}}^2$ with no topological horseshoe. If $\mathcal{B}$ is a Birkhoff recurrence class containing two fixed points $z_0$ and $z_1$, then
- either there exists $q{\geqslant}1$ such that every periodic point of $f$ distinct from $z_0$ and $z_1$ has a period equal to $q$, all recurrent points in $\mathcal B$ are periodic points and $\mathrm{fix}(f^q)\cap\mathcal{B}$ is an unlinked set of $f^q$;
- or $f$ is an irrational pseudo-rotation.
\[crmain:periodsofbirkhoffclasses\] Let $f$ be an orientation preserving homeomorphism of ${\mathbb{S}}^2$ with no topological horseshoe. Let $\mathcal{B}$ be a Birkhoff recurrence class containing periodic points of different periods, then:
- there exists integers $q_1$ and $q_2$, with $q_1$ dividing $q_2$, such that every periodic point in $\mathcal{B}$ has a period either equal to $q_1$ or to $q_2$
- if $q_1{\geqslant}2$, there exists a unique periodic orbit of period $q_1$ in $\mathcal B$;
- if $q_1=1$, there exist at most two fixed points in $\mathcal B$.
As an illustration, note that if the restriction of $f$ to a transitive set is topologically chaotic, then $f$ has a topological horseshoe.
Let us now enumerate the possible dynamics of transitive sets for homeomorphisms with no topological horseshoes. The following corollary shows that, if there is no topological horseshoe, transitive sets almost always must contain at most a single periodic orbit.
\[crmain:transitivewithperiodicorbit\] Let $f$ be an orientation preserving homeomorphism of ${\mathbb{S}}^2$ with no topological horseshoe. Assume that there exists a transitive set containing two distinct periodic orbits. Then $f$ is an irrational pseudo-rotation.
Before providing a more precise description of the dynamics of transitive sets for homeomorphisms with no topological horseshoes, let us say, following [@BGLT], that $f:{\mathbb{S}}^2\to{\mathbb{S}}^2$ is [*topologically infinitely renormalizable*]{} over a closed, nonempty invariant set $\Lambda$, if there exists an increasing sequence $(q_n)_{n{\geqslant}0}$ of positive integers and a decreasing sequence $(D_n)_{n{\geqslant}0}$ of open disks such that:
- $q_n$ divides $q_{n+1}$;
- $D_n$ is $f^{q_n}$ periodic;
- the disks $f^k(D_n)$, $0{\leqslant}k<q_n$ are pairwise disjoint;
- $\Lambda\subset \bigcup_{0{\leqslant}k<q_n}f^{k}(D_n)$.
We note that, if $f$ is topologically infinitely renormalizable over $\Lambda$, then the restriction of $f$ to $\Lambda$ is semi-conjugated to an odometer transformation and one can show, by standard Brouwer Theory arguments, that $f$ has periodic points of period $q_n$ for each $n\in {\mathbb{N}}$.
\[prmain:transitivesetsgeneralcase\] Let $f:{\mathbb{S}}^2\to{\mathbb{S}}^2$ be an orientation preserving homeomorphism with no horseshoe and $\Lambda$ a transitive set. Then:
1. either $\Lambda$ is a periodic orbit;
2. or $f$ is topologically infinitely renormalizable over $\Lambda$;
3. or $\Lambda$ is of irrational type.
Let us precise the last case. It means that $\Lambda$ is the disjoint union of a finite number of closed sets $\{\Lambda_1, ..., \Lambda_q\}$ that are cyclically permuted by $f$, and that there exists an irrational number $\rho\in{\mathbb{R}}$ such that if $z_0$, $z_1$ are different fixed points of $f^{qn}$, $n{\geqslant}1$, there exists a lift $\check g$ of $f^{qn}\vert_{{\mathbb{S}}^2\setminus\{z_0,z_1\}}$ and a generator $\kappa$ of $H_1({\mathbb{S}}^2\setminus\{z_0,z_1\},{\mathbb{Z}})$ such that:
- either the unique rotation number of $\Lambda$ for $\check g$ is
- or the unique rotation number of $\Lambda$ for $\check g$ is
Among classical examples of transitive sets of irrational type one can mention: periodic closed curves with no periodic orbits, Aubry-Mather sets, the sphere itself for a transitive irrational pseudo-rotation (Anosov-Katok examples for instance), and Perez-Marco hedgehogs.
Finally, we can improve the previous proposition when considering $\mathcal{C}^{1}$ dissipative diffeomorphisms, a class that has been extensively studied in the context of the Hénon family and the transition to positive entropy.
\[prmain:dissipativediffeomorfisms\] Let $f:{\mathbb{R}}^2\to{\mathbb{R}}^2$ be an orientation preserving diffeomorphism with no horseshoe, such that $\vert det Df(x)\vert<1$ for all $x\in{\mathbb{R}}^2$. Let $\Lambda$ be a compact transitive set that is locally stable. Then either $\Lambda$ is a periodic orbit, or $f$ is topologically infinitely renormalizable over $\Lambda$.
Forcing results
---------------
Our two fundamental results deal with mathematical objects (maximal isotopies, transverse foliations, transverse trajectories, $\mathcal F$-transverse intersection) that will be reminded in the next section. We will detail the theorems in the same section. The first result is a realization theorem improving results stated in [@LeCalvezTal]. The second theorem is the key point in the proofs of all statements above.
\[th: realization\] Let $M$ be an oriented surface, $f$ a homeomorphism of $M$ isotopic to the identity, $I$ a maximal isotopy of $f$ and $\mathcal F$ a foliation transverse to $I$. Suppose that $\gamma:[a,b]\to \mathrm{dom}(I)$ is an admissible path of order $r$ with a $\mathcal{F}$-transverse self intersection at $\gamma(s)=\gamma(t)$, where $s<t$. Let $\widetilde \gamma$ be a lift of $\gamma$ to the universal covering space $\widetilde{\mathrm{dom}}(I)$ of $\mathrm{dom}(I)$ and $T$ the covering automorphism such that $\widetilde\gamma$ and $T(\widetilde\gamma)$ have a $\widetilde{\mathcal F}$-transverse intersection at $\widetilde\gamma(t)=T(\widetilde\gamma)(s)$. Let $\widetilde f$ be the lift of $f\vert_{\mathrm{dom}(I)}$ to $\widetilde{\mathrm{dom}}(I)$, time-one map of the identity isotopy that lifts $I$, and $\widehat f$ the homeomorphism of the annular covering space $ \widehat{\mathrm{dom}}(I)=\widetilde {\mathrm{dom}}(I)/T$ lifted by $\widetilde f$. Then we have the following:
1. For every rational number $p/q\in(0,1]$, written in an irreducible way, there exists a point $\widetilde z\in \widetilde{\mathrm{dom}}(I)$ such that $\widetilde f^{qr}(\widetilde z)=T^p(\widetilde z)$ and such that $\widetilde I_{\widetilde{\mathcal F}}^{{\mathbb{Z}}}(\widetilde z)$ is equivalent to $\prod_{k\in{\mathbb{Z}}} T^k(\widetilde\gamma_{[s,t]})$.
2. For every irrational number $\rho\in[0,1/r]$, there exists a compact set $\widehat Z_{\rho}\subset\widehat{\mathrm{dom}}(I)$ invariant by $\widehat f$, such that every point $\widehat z\in \widehat Z_{\rho}$ has a rotation number $\mathrm{rot}_{\widetilde f}(\widehat z)$ equal to $\rho$. Moreover if $\widetilde z\in\widetilde{\mathrm{dom}}(I)$ is a lift of $\widehat z$, then $\widetilde I_{\widetilde{\mathcal F}}^{{\mathbb{Z}}}(\widetilde z)$ is equivalent to $\prod_{k\in{\mathbb{Z}}} T^k(\widetilde\gamma_{[s,t]})$.
\[th: horseshoe\]Let $M$ be an oriented surface, $f$ a homeomorphism of $M$ isotopic to the identity, $I$ a maximal isotopy of $f$ and $\mathcal F$ a foliation transverse to $I$. If there exists a point $z$ in the domain of $I$ and an integer $q{\geqslant}1$ such that the transverse trajectory $I_{\mathcal F}^{q}(z)$ has a $\mathcal F$-transverse self-intersection, then $f$ has a topological horseshoe. Moreover, the entropy of $f$ is at least equal to $\log 4/3q$.
In [@LeCalvezTal] we stated a weaker version of this result. Supposing that $z$ was a recurrent point of $f$ and $f^{-1}$, we proved that $f$ was topological chaotic, without displaying any horseshoe. The fact that the recurrence is no more required is what permit us to get the applications announced in this introduction.
Notations and organization of the paper
---------------------------------------
We will use the following notations:
- If $Y$ is a subset of a topological space $X$, we will write $\overline Y$, $\mathrm{Int}(Y)$, $\mathrm{Fr}(Y)$ for its closure, its interior and its frontier, respectively.
- We will denote $\mathrm{fix}(f)$ the set of fixed points of a homeomorphism defined on a surface $M$.
- A set $X\subset M$ will be called $f$-free if $f(X)\cap X=\emptyset$.
- A line of $M$ is a proper topological embedding $\lambda:{\mathbb{R}}\to M$ (or the image of this embedding). If the complement of $\lambda$ has two connected components, we will denote $R(\lambda)$ the component lying on the right of $\lambda$ and $L(\lambda)$ the component lying on its left.
- If $\mathcal F$ is a topological foliation defined on $M$, we will denote $\phi_z$ the leaf passing through a given point $z\in M$.
The paper is organized as follows: In Section 2 we recall the basic features of Brouwer equivariant theory and of the forcing theory, as well as some basic results on rotations numbers and Birkhoff recurrence classes. In Section 3 we construct the topological horseshoes and prove Theorems \[th: realization\] and \[th: horseshoe\]. Section 4 describes properties of transverse trajectories with no transverse intersection in the sphere, Section 5 is dedicated to proving the applications concerning rotation numbers, including Theorems \[thmain:rotation-number\], \[thmain:birkhoffcycles\] and \[thmain:circloids\], as well as Propositions \[propositionmain\_regionofinstability\] and \[prop:zero-rotation\]. In Section 6 we derive our structure theorem for homeomorphisms of the sphere with no topological horseshoes and prove Theorem \[thmain:global-structure\]. Lastly, Section 7 deals with the final applications for Birkhoff recurrence classes and transitive sets, proving Propositions \[propmain:birkoffclasses3fixedpoints\], \[prmain:transitivesetsgeneralcase\] and \[prmain:dissipativediffeomorfisms\] and related results.
We would like to thank A. Koropecki for several discussions and suggestions regarding this work.
Preliminaries
=============
Maximal isotopies, transverse foliations and transverse trajectories
--------------------------------------------------------------------
### Maximal isotopies
Let $M$ be an oriented surface (not necessarily closed, not necessarily of finite type) and $f$ a homeomorphism of $M$ that is isotopic to the identity. We denote by $\mathcal I$ the space of [*identity isotopies of $f$*]{} which means the set of continuous paths defined on $[0,1]$ that join the identity to $f$ in the space of homeomorphisms of $M$, furnished with the $C^0$ topology (defined by the uniform convergence of maps and their inverse on compact sets). If $I=(f_t)_{t\in[0,1]}$ is such an isotopy, we define the [*trajectory*]{} $I(z)$ of a point $z\in M$ to be the path $t\mapsto f_t(z)$. For every integer $n{\geqslant}1$, we can define by concatenation $$I^n(z)=\prod_{0{\leqslant}k<n} I(f^k(z)).$$ Furthermore we can define $$I^{{\mathbb{N}}}(z)=\prod_{k{\geqslant}0} I(f^k(z)),
\enskip I^{-{\mathbb{N}}}(z)=\prod_{k<0} I(f^k(z)),
\enskip I^{{\mathbb{Z}}}(z)=\prod_{k\in{\mathbb{Z}}} I(f^k(z)),$$ the last path being called the [*whole trajectory*]{} of $z$.
The [*fixed point set of $I$*]{} is the set $\mathrm{fix}(I)=\bigcap_{t\in[0,1]} \mathrm{fix}(f_t)$ and the [*domain of $I$*]{} its complement, that we will denote $\mathrm{dom}(I)$. We have a preorder on $\mathcal I$ defined as follows: say that $I\preceq I'$ if
- $\mathrm{fix}(I)\subset\mathrm{fix}(I')$;
- $I'$ is homotopic to $I$ relative to $\mathrm{fix}(I)$.
We have the recent following result, due to Béguin-Crovisier-Le Roux [@BeguinCrovisierLeRoux2]:
For every $I\in\mathcal I$, there exists $I'\in\mathcal I$ such that $I\preceq I'$ and such that $I'$ is maximal for the preorder.
A weaker version of the theorem was previously proved by Jaulent [@Jaulent] and stated in terms of singular isotopies. While sufficient for the applications we are looking for, it is less easy to handle. Note that we used Jaulent’s formalism in [@LeCalvezTal].
An isotopy $I$ is maximal if and only if, for every $z\in\mathrm{fix}(f)\setminus\mathrm{fix}(I)$, the closed curve $I(z)$ is not contractible in $\mathrm{dom}(I)$ (see [@Jaulent]). Equivalently, if we lift the isotopy $I\vert_{\mathrm{dom}(I)}$ to an identity isotopy $\widetilde I=(\widetilde f_t)_{t\in[0,1]}$ on the universal covering space $\widetilde{\mathrm{dom}}(I)$ of $\mathrm{dom}(I)$, the maximality of $I$ means that $\widetilde f_1$ is fixed point free. Note that every connected component of $\widetilde{\mathrm{dom}}(I)$ must be a topological plane.
### Transverse foliation
We keep the same notations as above. We have the following result (see [@LeCalvez1]):
\[th.transversefoliation\]If $I\in\mathcal I$ is maximal, there exists a topological oriented singular foliation $\mathcal F$ on $M$ such that
- the singular set $\mathrm{sing}(\mathcal F)$ coincides with $\mathrm{fix}(I)$;
- for every $z\in \mathrm{dom}(I)$, the trajectory $I(z)$ is homotopic in $\mathrm{dom}(I)$, relative to the ends, to a path $\gamma$ [positively transverse [^4]]{} to $\mathcal F$, which means locally crossing each leaf from the right to the left.
We will say that $\mathcal F$ is [*transverse to $I$*]{}. It can be lifted to a non singular foliation $\widetilde{\mathcal F}$ on $\widetilde{\mathrm{dom}}(I)$ which is transverse to $\widetilde I$. This last property is equivalent to saying that every leaf $\widetilde\phi$ of $\widetilde{\mathcal F}$, restricted to the connected component of $\widetilde{\mathrm{dom}}(I)$ that contains it, is a [*Brouwer line*]{} of $\widetilde f_1$: its image is contained in the connected component $L(\widetilde\phi)$ of the complement of $\widetilde\phi$ lying on the left of $\widetilde\phi$ and its inverse image in the connected component $R(\widetilde\phi)$ lying on the right (see [@LeCalvez1]). The path $\gamma$ is not uniquely defined. When lifted to $\widetilde{\mathrm{dom}}(I)$, one gets a path $\widetilde\gamma$ from a lift $\widetilde z$ of $z$ to $\widetilde f_1(\widetilde z)$:
- that meets once every leaf that is on the left of the leaf $\phi_{\widetilde z}$ containing $\widetilde z$ and on the right of the leaf $\phi_{\widetilde f(\widetilde z)}$ containing $\widetilde f(\widetilde z)$,
- that does not meet any other leaf.
Every other path $\widetilde\gamma'$ satisfying these properties projects onto another possible choice $\gamma'$. We will say that two positively transverse paths $\gamma$ and $\gamma'$ are [*equivalent*]{} if they can be lifted to $\widetilde{\mathrm{dom}}(I)$ into paths that meet exactly the same leaves. We will write $\gamma=I_{\mathcal F}(z)$ and call this path the [*transverse trajectory of $z$*]{} (it is defined up to equivalence). For every integer $n{\geqslant}1$, we can define by concatenation $$I_{\mathcal F}^n(z)=\prod_{0{\leqslant}k<n} I_{\mathcal F}(f^k(z)).$$ Furthermore we can define $$I_{\mathcal F}^{{\mathbb{N}}}(z)=\prod_{k{\geqslant}0} I_{\mathcal F}(f^k(z)),
\enskip I_{\mathcal F}^{-{\mathbb{N}}}(z)=\prod_{k<0} I_{\mathcal F}(f^k(z)),
\enskip I_{\mathcal F}^{{\mathbb{Z}}}(z)=\prod_{k\in{\mathbb{Z}}} I_{\mathcal F}(f^k(z)),$$ the last path being called the [*whole transverse trajectory*]{} of $z$.
The following proposition, that will be useful in the article, is easy to prove (see [@LeCalvezTal]):
\[prop:stability\] We have the following:
- for every $z\in\mathrm{dom}(I)$ and every $n{\geqslant}1$, there exists a neighborhood $U$ of $z$ such that $I_{\mathcal F}^n(z)$ is a subpath (up to equivalence) of $I_{\mathcal F}^{n+2}(f^{-1}(z'))$, if $z'\in U$;
- for every $z\in\mathrm{dom}(I)$, every $z'\in\omega(z)$, every $m{\geqslant}0$ and every $n{\geqslant}1$, $I_{\mathcal F}^n(z')$ is a subpath (up to equivalence) of $I_{\mathcal F}^{{\mathbb{N}}}(f^m(z))$.
We will say that a path $\gamma:[a,b]\to {\mathrm{dom}}(I)$, positively transverse to $\mathcal F$, is [*admissible of order $n$*]{} if it is equivalent to a path $I_{\mathcal{F}}^n
(z)$, $z\in\mathrm{dom}(I)$. It means that if $\widetilde \gamma:[a,b]\to \widetilde {\mathrm{dom}}(I)$ is a lift of $\gamma$, there exists a point $\widetilde z\in\widetilde{\mathrm{dom}}(I)$ such that $\widetilde z\in\phi_{\widetilde \gamma (a)}$ and $\widetilde f^n(\widetilde z)\in\phi_{\widetilde \gamma (b)}$, or equivalently, that $\widetilde f^n(\phi_{\widetilde \gamma (a)})\cap \phi_{\widetilde \gamma(b)}\not=\emptyset$.
$\mathcal F$-transverse intersection
------------------------------------
Let us begin with a definition. Given three disjoint lines $\lambda_0,\lambda_1, \lambda_2:{\mathbb{R}}\to \widetilde{\mathrm{dom}}(I)$ contained in the same connected component of $\widetilde{\mathrm{dom}}(I)$, we will say that that $\lambda_2$ is [*above $\lambda_1$ relative to $\lambda_0$*]{} (and that $\lambda_1$ is [*below $\lambda_2$ relative to $\lambda_0$*]{}) if none of the lines separates the two others and if, for every pair of disjoint paths $\gamma_1, \gamma_2$ joining $z_1=\lambda_0(t_1)$ to $z_1'\in \lambda_1$ and $z_2=\lambda_0(t_2)$ to $ z_2'\in\lambda_2$ respectively, such that the paths do not meet the lines but at their endpoints, one has that $t_1< t_2$.
![Order of lines relative to $\lambda_0$.[]{data-label="figure_order_lines"}](Figure_order_lines.pdf){height="48mm"}
Let $\gamma_1:J_1\to \mathrm{dom}(I)$ and $\gamma_2:J_2\to \mathrm{dom}(I)$ be two transverse paths defined on intervals $J_1$ and $J_2$ respectively. Suppose that there exist $t_1\in J_1$ and $t_2\in J_2$ such that $\gamma_1(t_1)=\gamma_2(t_2)$. We will say that $\gamma_1$ and $\gamma_2$ have a [*$\mathcal F$-transverse intersection*]{} at $\gamma_1(t_1)=\gamma_2(t_2)$ if there exist $a_1, b_1\in J_1$ satisfying $a_1<t_1<b_1$ and $a_2, b_2\in J_2$ satisfying $a_2<t_2<b_2$ such that if $\widetilde\gamma_1,\widetilde\gamma_2$ are lifts of $\gamma_1, \gamma_2$ to $\widetilde{\mathrm{dom}}(I)$ respectively, verifying $\widetilde\gamma_1(t_1)=\widetilde\gamma_2(t_2)$, then
- $\phi_{\widetilde\gamma_1(a_1)}\subset L(\phi_{\widetilde\gamma_2(a_2)}),\enskip \phi_{\widetilde\gamma_2(a_2)}\subset L(\phi_{\widetilde\gamma_1(a_1)})$;
- $\phi_{\widetilde\gamma_1(b_1)}\subset R(\phi_{\widetilde\gamma_2(b_2)}),\enskip \phi_{\widetilde\gamma_2(b_2)}\subset R(\phi_{\widetilde\gamma_1(b_1)})$
- $\phi_{\widetilde\gamma_2(b_2)}$ is below $\phi_{\widetilde\gamma_1(b_1)}$ relative to $\phi_{\widetilde\gamma_1(t_1)}$ if $\phi_{\widetilde\gamma_2(a_2)}$ is above $\phi_{\widetilde\gamma_1(a_1)}$ relative to $\phi_{\widetilde\gamma_1(t_1)}$ and above $\phi_{\widetilde\gamma_1(b_1)}$ if $\phi_{\widetilde\gamma_2(a_2)}$ is below $\phi_{\widetilde\gamma_1(a_1)}$.
![$\mathcal{F}$-transverse intersection. []{data-label="figure1"}](figure_transverse_intersection.pdf){height="48mm"}
Roughly speaking, it means that there is a “crossing” in the space of leaves of $\widetilde{\mathcal F}$ (which is a one-dimensional non Hausdorff manifold): a path joining $\phi_{\widetilde\gamma_1(a_1)}$ to $\phi_{\widetilde\gamma_1(b_1)}$ and a path joining $\phi_{\widetilde\gamma_2(a_2)}$ to $\phi_{\widetilde\gamma_2(b_2)}$ must intersect. If $\gamma_1=\gamma_2$ one speaks of a [*$\mathcal F$-transverse self-intersection*]{}. This means that if $\widetilde \gamma_1$ is a lift of $\gamma_1$, there exist a covering automorphism $T$ such that $\widetilde\gamma_1$ and $T\widetilde\gamma_1$ have a $\widetilde{\mathcal F}$-transverse intersection at $\widetilde\gamma_1(t_1)=T\widetilde\gamma_1(t_2)$.
The next proposition is the key result of [@LeCalvezTal]. It permits to construct new admissible paths from a pair of admissible paths.
\[pr: fundamental\] Suppose that $\gamma_1: [a_1,b_1]\to M$ and $\gamma_2: [a_2,b_2]\to M$ are transverse paths that intersect $\mathcal{F}$-transversally at $\gamma_1(t_1)=\gamma_2(t_2)$. If $\gamma_1$ is admissible of order $n_1$ and $\gamma_2$ is admissible of order $n_2$, then $\gamma_1\vert_{[a_1,t_1]}\gamma_2\vert_{[t_2,b_2]}$ and $\gamma_2\vert_{[a_2,t_2]}\gamma_1\vert_{[t_1,b_1]}$ are admissible of order $n_1+n_2$. Furthermore, either one of these paths is admissible of order $\min(n_1, n_2)$ or both paths are admissible of order $\max(n_1, n_2)$.
So, the meaning of Theorem \[th: horseshoe\] is that the existence of an admissible path of order $q$ with a self intersection implies the existence of a horseshoe. Moreover it gives a lower bound $\log 4/3q$ of the entropy.
Rotation numbers
----------------
As explained in the introduction, rotation numbers can be naturally defined in an abstract annulus $A$ for a homeomorphism $f$ isotopic to the identity and a given lift to the universal covering space, as soon as a generator of $H_1(A,{\mathbb{Z}})$ is chosen. We will state two useful propositions. In this article a [*loop*]{} on the sphere is a continuous map $\Gamma:{\mathbb{T}}^1\to{\mathbb{S}}^2$. One can define a [*dual function*]{} $\delta$ defined up to an additive constant on ${\mathbb{S}}^2\setminus\Gamma$ as follows: for every $z$ and $z'$ in ${\mathbb{R}}^2\setminus\Gamma$, the difference $\delta(z')-\delta(z)$ is the algebraic intersection number $\Gamma\wedge \gamma'$ where $\gamma'$ is any path from $z$ to $z'$.
\[prop: rotation numbers\] Let $f$ be a homeomorphism of the sphere ${\mathbb{S}}^2$ and $(z_i)_{i\in {\mathbb{Z}}/3{\mathbb{Z}}}$ three distinct fixed points. For every $i\in {\mathbb{Z}}/3{\mathbb{Z}}$, set $A_i={\mathbb{S}}^2\setminus\{z_{i+1},z_{i+2}\}$ and fix as a generator of $H_1(A_i,{\mathbb{Z}})$ the oriented boundary of a small closed disk containing $z_{i+1}$ in its interior. Denote $f_i$ the restriction of $f$ to $A_i$ and write $\check f_i$ for the lift of $f_i$ to the universal covering space of $A_i$ that fixes the preimages of $z_i$. Let $z\in {\mathbb{S}}^2$ be a point such that the sequence $(f^n(z))_{n{\geqslant}0}$ does not converge to $z_i$, for every $i\in{\mathbb{Z}}/3{\mathbb{Z}}$. If the three rotations numbers $\mathrm {rot}_{\check f_i}(z)$, $i\in{\mathbb{Z}}/3{\mathbb{Z}}$, are well defined then $\sum_{i\in{\mathbb{Z}}/3{\mathbb{Z}}}\mathrm {rot}_{\check f_i}(z) =0$.
Fix an open topological disk $D$ that contains a point $z'\in\omega(z)$ and no point $z_i$. Then fix a path $\gamma$ joining $z'$ to $z$ that does not contain any $z_i$. It is a classical fact that there exists an identity isotopy $I$ that fixes each $z_i$. Moreover $I\vert_{A_i}$ is lifted to an identity isotopy of $\check f_i$. Let $(n_k)_{k{\geqslant}0}$ be an increasing sequence such that $f^{n_k}(z)\in D$, for every $k{\geqslant}0$. Consider the loop $\Gamma_k$ naturally defined by $I^{n_k}(z)\gamma_k\gamma$, where $\gamma_k$ is a path included in $D$ that joins $f^{n_k}(z)$ to $z'$. Fix a dual function $\delta_k$ of $\Gamma_k$. It is defined up to an additive constant, and its values on each $z_i$ is independent of the choice of $\gamma_k$. By definition of $\mathrm {rot}_{\check f_i}(z)$, one has $$\mathrm {rot}_{\check f_i}(z) =\lim_{k\to+\infty} {\delta_k(z_{i+1})- \delta_k(z_{i+2})\over n_k}.$$ The conclusion is immediate.
The proof of the next result can be found in [@LeRoux].
\[prop: rotation numbers powers\] Let $f$ be a homeomorphism of the annulus ${\mathbb{A}}$ and $\check f$ a lift to the universal covering space. Fix two integer numbers $p\in{\mathbb{Z}}$ and $q{\geqslant}1$. Then, one has $\mathrm{ne}(f)= \mathrm{ne}(f^q)$. Moreover $\mathrm{rot}_{\check f^q\circ T^{p}}(z)$ is defined if and only if it is the case for $\mathrm{rot}_{\check f}(z)$ and one has $\mathrm{rot}_{\check f^q\circ T^{p}}(z)=q\mathrm{rot}_{\check f}(z)+p$.
Birkhoff cycles and Birkhoff recurrence classes
-----------------------------------------------
Let $X$ be a Hausdorff topological space and $f:X\to X$ a homeomorphism. We recall from the introduction that, given points $z_1$ and $z_2$ in $X$, we say that there exists a Birkhoff connection from $z_1$ to $z_2$ if, for every neighborhood $W_1$ of $z_1$ in $X$ and every neighborhood $W_2$ of $z_2$, there exists $n>0$ such that $W_1\cap f^{-n}(W_2)$ is not empty. We will write $z_1\underset{f}\rightarrow z_2$ if there is a Birkhoff connection from $z_1$ to $z_2$ and will write $z\underset{f}\preceq z'$ if there exists a family $(z_i)_{0{\leqslant}i{\leqslant}p}$ such that $$z=z_0\underset{f}\rightarrow z_1\underset{f}\rightarrow \dots \underset{f}\rightarrow z_{p-1}\underset{f}\rightarrow z_p=z'.$$
This same concept appears for instance in [@Arnaud] or in [@crovisier_birkhoff], where it is said that $z_2$ is a weak iterate of $z_1$. We have defined in the introduction the concepts of Birkhoff cycles, Birkhoff recurrent points and Birkhoff recurrent classes for $f$. Note that in [@Arnaud] a Birkhoff cycle was named an $(\alpha,\omega)$ closed chain, and Birkhoff recurrent points were named $(\alpha, \omega)$-recurrent. Note that the set of Birkhoff recurrent points of $f$ coincides with the set of Birkhoff recurrent points of $f^{-1}$ and that the Birkhoff recurrent classes are the same for $f$ and $f^{-1}$. Let us state some other simple additional properties that will be used frequently in the remainder of the article.
\[prop: birkhoff connexions\] Let $X$ be a Hausdorff topological space and $f$ a homeomorphism of $X$.
1. If there exists a Birkhoff connection from $z_1$ to $z_2$, then there exists a Birkhoff connection from $z_1$ to $f(z_2)$ and, if $z_2$ is not $f(z_1)$, then there exists a Birkhoff connection from $f(z_1)$ to $z_2$. Furthermore, if $z_2\notin \{ f^n(z_1), n{\geqslant}1\}$ , then there exists a Birkhoff connection from any point $z\in\omega(z_1)$ to $z_2$.
2. Let $\mathcal{B}$ be a nonempty Birkhoff recurrence class for $f$. Then $\mathcal{B}$ is invariant and, for all $z\in\mathcal{B}$ the closure of the orbit of $z$ is included in $\mathcal{B}$.
3. If $f$ has a Birkhoff connection from $z_1$ to $z_2$, then given $q{\geqslant}1$, there exists some $0{\leqslant}j < q$ such that $f^q$ has a Birkhoff connection from $z_1$ to $f^j(z_2)$.
4. The set of Birkhoff recurrent point of $f$ is equal to the set of Birkhoff recurrent point of $f^q$ for every $q{\geqslant}2$. Moreover, if $z_1$ and $z_2$ are in the same Birkhoff recurrence class of $f$, then given $q{\geqslant}2$, there exists some $0{\leqslant}j < q$ such that $z_1$ and $f^j(z_2)$ are in the same Birkhoff recurrence class of $f^q$. More precisely, if $(z_i)_{i\in Z/r{\mathbb{Z}}}$ is a Birkhoff circle of $f$, there exist $p{\geqslant}1$ and a Birkhoff circle $(z'_j)_{j\in {\mathbb{Z}}/pr{\mathbb{Z}}}$ of $f^q$ such that $z_j$ is in the $f$-orbit of $z_i$ if $j\equiv i \pmod r$.
Suppose that $f$ has a Birkhoff connection from $z_1$ to $z_2$. If $W_1$ is a neighborhood of $z_1$ and $W_2$ is a neighborhood of $f(z_2)$, then $f^{-1}(W_2)$ is a neighborhood of $z_2$ and so, there exists $n>0$ such that $W_1\cap f^{-n}(f^{-1}(W_2))=W_1\cap f^{-n-1}(W_2)$ is not empty. So, there is a Birkhoff connection from $z_1$ to $z_2$. Suppose now that $f(z_1)\not=z_2$. If $W_1$ is a neighborhood of $f(z_1)$, if $W_2$ is a neighborhood of $z_2$ and if $W_1\cap W_2=\emptyset$, there exists $n>0$ such that $f^{-1}(W_1)\cap f^{-n}(W_2)\not=\emptyset$ because $f^{-1}(W_1)$ is a neighborhood of $z_1$, and so $W_1\cap f^{-n+1}(W_2)\not=\emptyset$. Note that $n\not=1$, which implies that there is a Birkhoff connection from $f(z_1)$ to $z_2$. Finally, suppose that $z_2\notin \{ f^n(z_1), n{\geqslant}1\}$ and that $z$ belongs to $\omega(z_1)$. Let $W_1$ be a neighborhood of $z$ and $W_2$ a neighborhood of $z_2$. There exists $n_1{\geqslant}0$ such that $f^{n_1}(z)\in W_1$. As explained above, there is a Birkhoff connection from $f^{n_1}(z_1)$ to $z_2$, and so, there exists $n>0$ such that $W_1\cap f^{-n}(W_2) \not=\emptyset$. The assertion (1) is proven.
Let us prove (2). Of course, the set of Birkhoff recurrent points is invariant by $f$. Moreover (2) is obviously true if $z$ is a periodic point. Suppose now that $z$ is a Birkhoff recurrent point that is not periodic and let us prove that $z$ and $f(z)$ are in the same Birkhoff recurrence class. There exists a sequence $(z_i)_{0{\leqslant}i{\leqslant}p}$ such that $$z=z_0\underset{f}\rightarrow z_1\underset{f}\rightarrow \dots \underset{f}\rightarrow z_{p-1}\underset{f}\rightarrow z_p=z.$$ As $z$ is not periodic, there exists $i_0<p$ such that $z_{i+1}\not=f(z_i)$. So, using (1), one deduces that $$z\underset{f}\rightarrow f(z)\underset{f}\rightarrow f(z_1)\underset{f}\rightarrow \dots \underset{f}\rightarrow f(z_{i_0})\underset{f}\rightarrow z_{i_0+1}\underset{f}\rightarrow \dots \underset{f}\rightarrow z_{p-1}\underset{f}\rightarrow z_p=z,$$ which implies that $z$ and $f(z)$ are in the same Birkhoff recurrence class. Let us prove now that $\omega(z)$ is included in the same class by fixing $z'\in\omega(z)$. Let $i_0$ be the smallest integer such that $z_{i_0+1}$ is not in the strict positive orbit of $z_{i_0}$. Using (1), one deduces that $$z=z_0\underset{f}\rightarrow z_1\underset{f}\rightarrow\dots \underset{f}\rightarrow z_{i_0} \underset{f}\rightarrow z'\underset{f}\rightarrow z_{i_0+1}\underset{f}\rightarrow \dots \underset{f}\rightarrow z_{p-1}\underset{f}\rightarrow z_{p}=z,$$ which implies that $z$ and $z'$ are in the same Birkhoff recurrence class. We prove in a similar way that $\alpha(z)$ is included in the Birkhoff recurrence class of $z$.
Let us prove (3) by contraposition. Suppose that for every $0{\leqslant}j<q$, there exists a neighborhood $W_1^j$ of $z_1$ and a neighborhood $W_2^j$ of $f^j(z_2)$ such that $W_1^j\cap f^{-qs}(W_2^{j})=\emptyset$ for every $s>0$. The set $W_1=\bigcap_{0{\leqslant}j<q} W_1^j$ is a neighborhood of $z_1$ and the set $W_2=\bigcap_{0{\leqslant}j<q} f^{-j}(W_2^j)$ a neighborhood of $z_2$. Note that $$W_1\cap f^{-qs}(W_2)\subset W^0_1\cap f^{-qs}(W^0_2)=\emptyset,$$ for every $s>0$ and that $$W_1\cap f^{-qs-r}(W_2)\subset W_1^{q-r}\cap f^{-q(s+1)} (W_2^{q-r})=\emptyset,$$ for every $s{\geqslant}0$ and every $0<r<q$ . Consequently, $W_1\cap f^{-n}(W_2)=\emptyset$ for every $n>0$.
Let us prove (4). The fact that a Birkhoff recurrent point of $f^q$ is a Birkhoff recurrent point of $f$ is obvious. Let us prove the converse. Suppose that $z$ is a Birkhoff recurrent point of $f$ and fix $q{\geqslant}2$. If $z$ is a periodic point of $f$, of course it is a Birkhoff recurrent point of $f^q$. Suppose now that $z$ is not periodic and consider a sequence $(z_i)_{0{\leqslant}i{\leqslant}p}$ such that $$z=z_0\underset{f}\rightarrow z_1\underset{f}\rightarrow \dots \underset{f}\rightarrow z_{p-1}\underset{f}\rightarrow z_p=z.$$ Applying (3), one deduces that there exists a sequence $(s_i)_{1{\leqslant}i{\leqslant}p}$, such that $0{\leqslant}s_{i+1}-s_i<q$ if $i<p$ and $$z=z_0\underset{f^q}\rightarrow f^{s_1}(z_1)\underset{f^q}\rightarrow \dots \underset{f^q}\rightarrow f^{s_{p-1}}(z_{p-1})\underset{f^q}\rightarrow f^{s_{p}}(z_p)=f^{s_{p}}(z),$$which implies that $z\underset{f^q}\preceq f^{s_{p}}(z)$, where $0{\leqslant}s_q{\leqslant}p(q-1)$. Let $i_0<p$ be an integer such that $z_{i_0+1}$ is not in the strict positive orbit of $z_{i_0}$. Using (1), one knows that $$z=z_0\underset{f}\rightarrow z_1\underset{f}\rightarrow\dots \underset{f}\rightarrow z_{i_0} \underset{f}\rightarrow f^{-pq}(z_{i_0+1})\underset{f}\rightarrow \dots \underset{f}\rightarrow f^{-pq}(z_{p-1})\underset{f}\rightarrow f^{-pq}(z_{p})=f^{-pq}(z).$$ Applying (3) again as above, one deduces that there exists $0{\leqslant}s'_q{\leqslant}p(q-1)$ such that $z\underset{f^q}\preceq f^{s'_{p}-pq}(z)$. Note know that $$\label{eq:birkhoffcycleiterate}
z\underset{f^q}\preceq f^{s_{p}}(z)\underset{f^q}\preceq f^{2s_{p}}(z)\underset{f^q}\preceq \dots \underset{f^q}\preceq f^{(pq-s'_q)s_{p}}(z)\underset{f^q}\preceq f^{(pq-s'_q)(s_{p}-1)}(z)\underset{f^q}\preceq\dots \underset{f^q}\preceq z,\tag{*}$$ which implies that $z$ is a Birkhoff recurrent point of $f^q$.
Suppose now that $z$ and $z'$ are two Birkhoff recurrent points of $f$ in the same Birkhoff recurrence class. We want to prove that $z$ and $f^j(z')$ are in the same Birkhoff recurrence class of $f^q$ for some $0{\leqslant}j < q$. Using the fact that the Birkhoff recurrence class of $f^q$ are invariant by $f^q$, it is sufficient to prove that $z$ and $f^j(z')$ are in the same Birkhoff recurrence class of $f^q$ for some $j\in{\mathbb{Z}}$. Of course, one can suppose that $z$ and $z'$ are not on the same orbit of $f$. There exists a sequence $(z_i)_{0{\leqslant}i{\leqslant}p}$ such that $$z=z_0\underset{f}\rightarrow z_1\underset{f}\rightarrow \dots \underset{f}\rightarrow z_{p-1}\underset{f}\rightarrow z_p=z$$ and such that $z'$ is equal to a certain $z_i$. We have seen that there exists a sequence $(s_i)_{1{\leqslant}i{\leqslant}p}$, such that $0{\leqslant}s_{i+1}-s_i<q$ if $i<p$ and $$z=z_0\underset{f^q}\rightarrow f^{s_1}(z_1)\underset{f^q}\rightarrow \dots \underset{f^q}\rightarrow f^{s_{p-1}}(z_{p-1})\underset{f^q}\rightarrow f^{s_{p}}(z_p)=f^{s_{p}}(z).$$ Moreover we have seen that $z$ and $f^{s_{p}}(z)$ are in the same Birkhoff recurrence class of $f^q$ (the argument above works even if $z$ is periodic). So $z$ is in the same Birkhoff recurrence class of $f^q$ that a point in the orbit of $z'$. Finally, one observes that from (\[eq:birkhoffcycleiterate\]) one obtains a Birkhoff cycle for $f^q$ with the properties stated in the last assertion of (4).
We can note that, in general, the class itself does not need to be closed. For instance, let $X$ be the space obtained by identifying, in $[0,1]\times\{0,1\}$, the points $(x,0)$ and $(x,1)$ if $x\in F=\{0\}\cup\{1/n, n{\geqslant}1\}$. Let $T_1:[0,1]\to [0,1]$ be a homeomorphism such that $T_1(x)=x$ if $x\in F$ and $T_1(x)>x$ otherwise. Let $T:X\to X$ be such that $T(x,0)=(T_1(x),0)$ and $T(x,1)= (T_1^{-1}(x), 1)$. Note that, for every $n{\geqslant}1$, $(1/n,0)=(1/n,1)$ and $(1/(n+1),0)=(1/(n+1),1)$ belong to the same Birkhoff class. Note also that the Birkhoff class of $(0,0)$ is reduced to this point.
The set of Birkhoff recurrent points is not necessarily closed. For instance, let $(h_n)_{n{\geqslant}1}$ be a sequence of orientation preserving homeomorphisms of ${\mathbb{T}}^1={\mathbb{R}}/{\mathbb{Z}}$ such that $h_n$ has exactly $n$ fixed points and has a lift $\check h_n:{\mathbb{R}}\to{\mathbb{R}}$ having fixed points and satisfying $\check h_n(x){\geqslant}x$ for all $x\in{\mathbb{R}}$. Assume furthermore that $h_n$ converges uniformly to a homeomorphism $h_0$ that is not the identity, but has infinitely many fixed points. Consider the homeomorphism $T$ of $X={\mathbb{T}}^1\times F$ defined as a $T(x, 1/n)= (h_n(x), 1/n)$ in case $n>0$ and $T(x,0)=(h_0(x), 0)$. Every point $(x, y)$ with $y\not=0$, is a Birkhoff recurrent point, but $(x,0)$ is a Birkhoff recurrent point if and only if $x$ is fixed by $h_0$.
We will conclude this section by studying what happens, when looking at the iterates.
\[prop: birkhoff classes iterate\] Let $X$ be a Hausdorff topological space and $f$ a homeomorphism of $X$. We suppose that $\mathcal B$ is a Birkhoff recurrence class of $f$ and fix $q{\geqslant}2$. Then, there exist $q'\in\{1,\dots, q\}$ dividing $q$ and a family of pairwise disjoint sets $({\mathcal B}'_i)_{i\in{\mathbb{Z}}/q'{\mathbb{Z}}}$ such that:
- ${\mathcal B}'_i$ is a Birkhoff recurrence class of $f^q$ for every $i\in{\mathbb{Z}}/q'{\mathbb{Z}}$;
- $\mathcal B=\bigsqcup_{i\in{\mathbb{Z}}/q'{\mathbb{Z}}} {\mathcal B}'_i$;
- $f({\mathcal B}'_i)={\mathcal B}'_{i+1}$, for every $i\in{\mathbb{Z}}/q'{\mathbb{Z}}$.
Moreover, if the closure of the $f$-orbit of $z_*\in{\mathcal B}$ is compact, each set $\overline{ O_f(z_*)}\cap \mathcal B'_i$ is compact and is the closure of the $f^{q'}$-orbit of a point of the orbit of $z_*$. In addition, there exists a family of pairwise disjoint open sets $(U_i)_{i\in{\mathbb{Z}}/q'{\mathbb{Z}}}$ such that:
- $\overline{ O_f(z_*)}\cap \mathcal B'_i\subset U_i$ for every $i\in{\mathbb{Z}}/q'{\mathbb{Z}}$;
- $f(U_i)=U_{i+1}$, for every $i\in{\mathbb{Z}}/q'{\mathbb{Z}}$.
We have seen that every Birkhoff recurrent point of $f$ is a Birkhoff recurrent point of $f^q$ and that every Birkhoff recurrence class of $f$ is invariant by $f$. Fix $z\in \mathcal B$ and denote ${\mathcal B}'_k$, the Birkhoff recurrence class of $f^k(z)$ for $f^q$. Of course $f( {\mathcal B}'_k)= {\mathcal B}'_{k+1}$ for every $k\in{\mathbb{Z}}$. Moreover one has $ {\mathcal B}'_q= {\mathcal B}'_{0}$. We deduce that there exists a divisor $q'$ of $q$ such that the sequence $({\mathcal B}'_k)_{k\in{\mathbb{Z}}}$ is $q'$-periodic. We can index it with $i\in{\mathbb{Z}}/q'{\mathbb{Z}}$ to get our family $({\mathcal B}'_i)_{i\in{\mathbb{Z}}/q'{\mathbb{Z}}}$. The three first items are satisfied. Note also that we cannot have $z\underset{f^q}\preceq f^r(z')$ if $z$ and $z'$ belong to ${\mathcal B}'_{0}$ and $r\not\in q'{\mathbb{Z}}$. Indeed, we would have $$z\underset{f^q}\preceq f^r(z')\underset{f^q}\preceq f^{r}(z)\underset{f^q}\preceq f^{2r}(z')\underset{f^q}\preceq f^{2r}(z)\underset{f^q}\preceq\dots \underset{f^q}\preceq f^{q'r}(z')\underset{f^q}\preceq z,$$ contradicting the fact that $z$ and $f^r(z')$ are not in the same Birkhoff recurrence class of $f^q$. In particular there is no Birkhoff connection of $f^q$ from $z$ to $f^r(z')$ and no Birkhoff connection of $f^q$ from $z$ to $f^r(z')$. One deduces that there exist an open neighborhood $V^r_{z,z'}$ of $z$ and an open neighborhood $W^r_{z,z'}$ of $z'$ such that $f^{qk}(V^r_{z,z'})\cap f^r(W^r_{z,z'})=\emptyset$, for every $k\in{\mathbb{Z}}$. Setting $$V_{z,z'}=\bigcap_{0<r<q, q'\nmid r}V^r_{z,z'}\,,\enskip W_{z,z'}=\bigcap_{0<r<q, q'\nmid r}W^r_{z,z'},$$ one gets an open neighborhood $V_{z,z'}$ of $z$ and an open neighborhood $W_{z,z'}$ of $z'$ such that $$k\not\in q'{\mathbb{Z}}\rightarrow f^{k}(V_{z,z'})\cap W_{z,z'}=\emptyset.$$ Suppose that the closure of the $f$-orbit of $z_*\in \mathcal B$ is compact. One can suppose that $z_*$ belongs to ${\mathcal B}'_0$. Write $O_f(z_*)=\bigcup_{k\in{\mathbb{Z}}} O_{f^q} (f^k(z_*))$. The family $(O_{f^q} (f^k(z_*))_{k\in{\mathbb{Z}}}$ is $q$-periodic and $O_{f^q} (f^k(z_*))\subset {\mathcal B}'_i$, if $k+q'{\mathbb{Z}}=i$. As explained in Proposition \[prop: birkhoff connexions\], the closure of $O_{f^q} (f^k(z_*))$ is included in the Birkhoff recurrence class of $f^k(z_*)$, which means in ${\mathcal B}'_i$, if $k+q'{\mathbb{Z}}=i$. Moreover $\overline {O_{f^q}(f^k(z_*))}$ is compact, being a closed subset of $\overline{O_f(z_*)}$. So, one gets the following decomposition in disjoint compact sets: $$\overline{O_f(z_*)}=\bigsqcup_{i\in {\mathbb{Z}}/q'{\mathbb{Z}}} \overline{O_f(z_*)}\cap {\mathcal B}'_i=\bigsqcup_{i\in {\mathbb{Z}}/q'{\mathbb{Z}}}\left( \bigcup_{k\in{\mathbb{Z}},\, k+q'{\mathbb{Z}}=i} \overline {O_{f^q}(f^k(z_*))}\right).$$ By compactness of $ \overline{O_f(z_*)}\cap {\mathcal B}'_0$, for every point $z\in \overline{O_f(z_*)}\cap {\mathcal B}'_0$ one can find a finite subset $I'_z$ of $\overline{O_f(z_*)}\cap {\mathcal B}'_0$ such that $ \overline{O_f(z_*)}\cap {\mathcal B}'_0\subset\bigcup_{z'\in I'_z} W_{z,z'}$. The set $V_z=\bigcap _{z'\in I'_z} V_{z,z'}$ is an open neighborhood of $z$, the set $W_z=\bigcup _{z'\in I'_z} W_{z,z'}$ an open neighborhood of $\overline{O_f(z_*)}\cap {\mathcal B}'_0$ and one knows that $$k\not\in q'{\mathbb{Z}}\rightarrow f^{k}(V_{z})\cap W_{z}=\emptyset.$$ Using again the compactness of $ \overline{O_f(z_*)}\cap {\mathcal B}'_0$, one can find a finite subset $I$ of $\overline{O_f(z_*)}\cap {\mathcal B}'_0$ such that $ \overline{O_f(z_*)}\cap {\mathcal B}'_i\subset\bigcup_{z\in I} V_{z}$. The sets $V=\bigcup _{z\in I} V_{z}$ and $W=\bigcap _{z\in I} W_{z}$ are open neighborhoods of $\overline{O_f(z_*)}\cap {\mathcal B}'_0$ and one knows that $$k\not\in q'{\mathbb{Z}}\rightarrow f^{k}(V)\cap W=\emptyset.$$ For every $i\in{\mathbb{Z}}/q'{\mathbb{Z}}$, define $$U_i=\bigcup_{k\in{\mathbb{Z}}, \, k+q'{\mathbb{Z}}=i} f^k(V\cap W).$$The family $(U_i)_{i\in{\mathbb{Z}}/q'{\mathbb{Z}}}$ satisfies the two last items.
Suppose that the orbit of $z_*\in \mathcal B$ is compact. Set $U=\bigcup_{i\in{\mathbb{Z}}/q{\mathbb{Z}}} U_i$. The map $f$ acts naturally as a permutation on the set of connected components of $U$ that meets $\overline {O_f(z_*)}$ and this action is transitive. Moreover this set is finite because $\overline {O_f(z_*)}$ is compact. So, this permutation has a unique cycle whose length is a multiple of $q'$ because the $U_i$ are pairwise disjoint and $f(U_i)=U_{i+1}$. In other terms, there exists an integer $r{\geqslant}1$ and a family $(U'_j)_{j\in {\mathbb{Z}}/ rq'{\mathbb{Z}}}$ of distinct connected components of $U$ that cover $\overline {O_f(z_*)}$ and satisfy $f(U'_j)=U'_{j+1}$.
There are two kinds of Birkhoff recurrent points: the points such that there exists an integer $M$ that bounds, for every $q$, the integer $q'$ appearing in the proof of Proposition \[prop: birkhoff classes iterate\] applied to the Birkhoff recurrence class of $z$; the points where such an integer $M$ does not exist. For such a point, $f$ is “abstractly” topologically infinitely renormalizable. This dichotomy will appear in the final section of our article.
Construction of horseshoes
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Preliminary results
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We suppose in this section that
- $M$ is an oriented surface;
- $f$ is a homeomorphism of $M$ isotopic to the identity;
- $I=(f_t)_{t\in[0,1]}$ is a maximal isotopy of $f$;
- $\mathcal F$ is a foliation transverse to $I$;
- $\gamma=[a,b]\to \mathrm{dom}(I)$ is an admissible transverse path or order $1$;
- $\gamma$ has a $\mathcal F$-transverse self intersection at $\gamma(s)=\gamma(t)$, where $a<s<t<b$.
We want to show that $f$ has a topological horseshoe, as defined in the introduction. We start by describing the dynamics that can be deduced from our assumptions, stating some useful lemmas. We will mainly work in the universal covering space $\widetilde{\mathrm{dom}}(I)$ of the domain of $I$ and occasionally in other covering spaces and will construct there topological horseshoes with a precise geometric description. We denote $\widetilde I=(\widetilde f_t)_{t\in[0,1]}$ the identity isotopy of $\widetilde{\mathrm{dom}}(I)$ that lifts $I\vert_{ \mathrm{dom}(I)}$ and set $\widetilde f=\widetilde f_1$. We denote $\widetilde{\mathcal F}$ the non singular foliation of $\widetilde{\mathrm{dom}}(I)$ that lifts $\mathcal F\vert_{ \mathrm{dom}(I)}$. Fix a lift $\widetilde\gamma$ of $\gamma$. By assumptions, there exists a non trivial covering transformation $T$ such that $T(\widetilde\gamma)$ and $\widetilde\gamma$ have a $\widetilde{\mathcal F}$-transverse intersection at $T(\widetilde\gamma(s))=\widetilde\gamma(t)$, where $s<t$. Since $\widetilde \gamma$ and $T(\widetilde \gamma)$ have a $\widetilde{\mathcal{F}}$-transverse intersection, the sets $R(\widetilde\phi_{\widetilde\gamma(a)}), R(T(\widetilde\phi_{\widetilde\gamma(a)})), L(\widetilde\phi_{\widetilde\gamma(b)})$ and $L(T(\widetilde\phi_{\widetilde\gamma(b)}))$ are all disjoint. Consider the annular covering space $\widehat{\mathrm{dom}}(I)=\widetilde{\mathrm{dom}}(I)/T$. Denote by $\pi: \widetilde{\mathrm{dom}}(I)\to \widehat{\mathrm{dom}}(I)$ the covering projection, by $\widehat I$ the induced identity isotopy, by $\widehat f$ the induced lift of $f$, by $\widehat{\mathcal{F}}$ the induced foliation and by $\widehat \gamma$ the projection of $\widetilde \gamma$. The fact that $R(\widetilde\phi_{\widetilde\gamma(a)})$ and $R(T(\widetilde\phi_{\widetilde\gamma(a)}))$ are disjoint implies that $\widehat\phi_{\widehat\gamma(a)}$, the projection of $ \widetilde\phi_{\widetilde\gamma(a)}$, is homoclinic to an end of $\widehat{\mathrm{dom}}(I)$ and that the sets $\overline{R(T^k(\widetilde\phi_a))}$, $k\in{\mathbb{Z}}$, are pairwise disjoint. The fact that $L(\widetilde\phi_{\widetilde\gamma(b)})$ and $L(T(\widetilde\phi_{\widetilde\gamma(b)}))$ are disjoint implies that $\widehat\phi_{\widehat\gamma(b)}$, the projection of $ \widetilde\phi_{\widetilde\gamma(b)}$, is homoclinic to an end of $\widehat{\mathrm{dom}}(I)$ and that the sets $\overline{L(T^k(\widetilde\phi_{\widetilde\gamma(b)}))}$, $k\in{\mathbb{Z}}$ are pairwise disjoint. The fact that $T(\widetilde\gamma)$ and $\widetilde\gamma$ have a $\widetilde{\mathcal F}$-transverse intersection implies that the two previous ends are equal. Otherwise, one could find a path $\widetilde\delta$ joining $\widetilde\phi_{\widetilde\gamma(a)}$ to $\widetilde\phi_{\widetilde\gamma(b)}$ and projecting onto a simple path of $\widehat{\mathrm{dom}}(I)$ and consequently such that $\widetilde\delta\cap T(\widetilde\delta)=\emptyset$. We will denote $S$ this common end and $N$ the other one. Note that $\overline{R(T^k(\widetilde\phi_a))}\cap\overline{ L(T^{k'}(\widetilde\phi_b))}=\emptyset$ for every integers $k$ and $k'$. The complement of $\overline{R(\widehat\phi_{\widehat\gamma(a)})}\cup\overline{L(\widehat\phi_{\widehat\gamma(b)})}$ is an essential open set of $\widehat{\mathrm{dom}}(I) $. If $\Gamma^*$ is a simple essential loop in this complement that is lifted to a line $\gamma^*:{\mathbb{R}}\to\widetilde{\mathrm{dom}}(I)$ satisfying $\gamma^*(t+1)=T(\gamma^*(t))$ for every $t\in{\mathbb{R}}$, there exists a unique $k_0{\geqslant}1$ such that $T^{-k_0}(\widetilde\phi_{\widetilde{\gamma}(b)})$ is above $\widetilde\phi_{\widetilde{\gamma}(a)}$ but below $T(\widetilde\phi_{\widetilde{\gamma}(a)})$ relative to $\widetilde\gamma^*$. The integer $k_0$ can be defined by the fact that $\gamma$ and $T^k(\gamma)$ have a $\widetilde{\mathcal{F}}$-transverse intersection if $1{\leqslant}k{\leqslant}k_0$ and no transverse intersection if $k>k_0$. We will suppose by now and until giving the proofs of theorems K and L that $k_0=1$. For convenience, we will denote $\widetilde\phi_a=\widetilde\phi_{\widetilde\gamma(a)}$ and $\widetilde\phi_b=T^{-1}(\widetilde\phi_{\widetilde\gamma(b)})$. We will often use the following result: if $\sigma$ is a path connecting $\widetilde \phi_a$ to $T(\widetilde \phi_a)$ and contained in $\bigcup_{k\in{\mathbb{Z}}} T^{k}(R(\widetilde \phi_b))$ and if $j\not=0$, then $\widetilde \phi_b$ and $T^{j}(\widetilde \phi_b)$ lie in different connected components of the complement of $\widetilde \phi_a\cup\sigma\cup T(\widetilde \phi_a)$.
![Lemma \[lm:pqintersections\] []{data-label="figureintersectionlemma"}](figure_intersection_lemma.pdf){height="48mm"}
\[lm:pqintersections\] \[lemma:accessibility\]For every $q{\geqslant}2$ and every $p\in\{0,\dots, q\}$, one has $\widetilde f^q(\widetilde\phi_a)\cap T^p(\widetilde\phi_b)\not=\emptyset$.
Let us explain first why $$\widetilde f^q(\widetilde\phi_a)\cap T^p(\widetilde\phi_b)\not=\emptyset\enskip \mathrm{and} \enskip q'>q \Rightarrow\enskip\widetilde f^{q'}(\widetilde\phi_a)\cap T^p(\widetilde\phi_b)\not=\emptyset.$$ Recall that, for every leaf $\widetilde\phi$ of $\widetilde{\mathcal F}$, one has
$$\overline{R(\widetilde\phi)}\subset \widetilde f(R(\widetilde \phi)), \enskip \widetilde f(\overline{L(\widetilde\phi)})\subset L(\widetilde \phi),$$ which implies that $$\widetilde{f}^{q}(\overline{R(\widetilde\phi)})\subset \widetilde f^{q'}(R(\widetilde \phi)), \enskip \widetilde f^{q'}(\overline{L(\widetilde\phi)})\subset \widetilde{f}^{q}(L(\widetilde \phi)),$$if $q'> q{\geqslant}0$. We deduce that $$\widetilde f^q(\widetilde\phi_a)\cap T^p(\widetilde \phi_b)\not=\emptyset \Rightarrow\widetilde f^q\left(\overline{R(\widetilde\phi_a)}\right)\cap \overline{L(T^p(\widetilde \phi_b)})\not=\emptyset \Rightarrow\widetilde f^{q'}\left(\overline{R(\widetilde\phi_a)}\right)\cap \overline{L(T^p(\widetilde \phi_b)})\not=\emptyset.$$ It means that there exists a point on the right of $\widetilde\phi_a$ that is sent by $\widetilde f^{q'}$ on the left of $T^p(\widetilde\phi_b)$. But this implies that there is a point on $\widetilde\phi_a$ that is sent by $\widetilde f^{q'}$ on $T^p(\widetilde\phi_b)$. The proof is easy and can be found in [@LeCalvezTal]. The reason is the following: in case $\widetilde f^{q'}(\widetilde\phi_a)\cap T^p(\widetilde\phi_b)=\emptyset$, one should have $$\widetilde f^{q'}(R(\widetilde\phi_a))\cup L(T^p(\widetilde \phi_b))=\widetilde{\mathrm{dom}}(I),$$ but this is prohibited by the the fact that $$T(R(\widetilde\phi_a))\cap\left(f^{q'}(R(\widetilde\phi_a))\cup L(T^p(\widetilde \phi_b))\right)=\emptyset.$$
The path $\widetilde\gamma$ is an admissible path of $\widetilde f$ of order $1$, which means that $\widetilde f(\widetilde\phi_a)\cap T(\widetilde \phi_b)\not=\emptyset$. So the lemma is proved if $p=1$. Applying Proposition \[pr: fundamental\], one deduces that boths paths $$\widetilde\gamma\vert_{[a,t]} \,T(\widetilde\gamma)\vert_{[s,b]}, \enskip T(\widetilde\gamma)\vert_{[a,s]}\, \widetilde\gamma\vert_{[t,b]}$$ are admissible paths of $\widetilde f$ of order $2$, which means that $$\widetilde f^2(\widetilde\phi_a)\cap T^2(\widetilde\phi_b)\not=\emptyset, \enskip \widetilde f^2(T(\widetilde\phi_a))\cap T(\widetilde\phi_b)\not=\emptyset.$$ Consequently the lemma is proven if $p\in\{0,2\}$.
Let us suppose now that $p{\geqslant}3$. It is also proved in [@LeCalvezTal] that $$\widetilde\gamma_{[a,s]}\prod_{0{\leqslant}i<p} T^i(\widetilde\gamma_{[s,t]}) T^{p-1} (\widetilde\gamma_{[t,b]})$$ is admissible of order $p$, which means that $\widetilde f^p(\widetilde\phi_a)\cap T^p(\widetilde\phi_b)\not=\emptyset$. Consequently the lemma is proved for every $p$.
Let us state another important lemma:
\[lemma:following\]Every simple path $\delta:[c,d]\to \widetilde{\mathrm{dom}}(I)$ that joins $ T^{-p_0}(\widetilde\phi_a)$ to $ T^{p_1}(\widetilde\phi_a)$, where $p_0$, $p_1$ are positive, and which is $T$-free, meets $R(\widetilde\phi_a)$.
Recall first, by standard Brouwer theory arguments, that $\delta$ is $T^k$-free for every $k{\geqslant}1$, because $T$ is fixed point free. There exists $c'\in[c,d)$ and $p'_0>0$, uniquely defined, such that $\delta(c')$ belongs to $ T^{-p'_0}(\widetilde\phi_a)$ and such that $\delta_{(c',d]}$ is disjoint from every leaf $ T^{-p}(\widetilde\phi_a)$, $p>0$. Then there exists $d'\in(c',d]$, and $p'_1>0$ uniquely defined, such that $\delta(d')$ belongs to $T^{p'_1}(\widetilde\phi_a)$ and such that $\delta_{[c',d')}$ is disjoint from every leaf $ T^{p}(\widetilde\phi_a)$, $p>0$. Note that $\delta_{(c',d')}$ is on the left of every leaf $T^{p}(\widetilde\phi_a)$, $p\not=0$. One can extend $\delta_{[c',d']}$ as a line $\lambda:{\mathbb{R}}\to \widetilde{\mathrm{dom}}(I)$ such that $\lambda\vert_{(-\infty, c')}$ is included in $R(T^{-p'_0}(\widetilde\phi_a))$ and $\lambda\vert_{(d', +\infty)}$ is included in $R(T^{p'_1}(\widetilde\phi_a))$. We will prove by contradiction that $\delta_{[c',d']}$ meets $R(\widetilde\phi_a)$. If it is not the case, then $$R(\widetilde\phi_a)\subset R(\lambda), \enskip R(T^{p'_0+p'_1}(\widetilde\phi_a))\subset L(\lambda).$$ The point $T^{p'_0}(\delta(c'))$ being on $\widetilde\phi_a$, belongs to the closure of $R(\widetilde\phi_a)$. It does not belong to $\delta_{[c',d']}$ (because $\delta$ is $T^{p'_0}$-free), neither to $R(T^{-p'_0}(\widetilde\phi_a))$, nor to $R(T^{p'_1}(\widetilde\phi_a))$. So it does not belong to $\lambda$. Consequently it belongs to $R(\lambda)$. Similarly, $T^{p'_0}(\delta(d'))$ does not belong to $\lambda$ and belongs to the closure of $R(T^{p'_0+p'_1}(\widetilde\phi_a))$, so its belongs to $L(\lambda)$. The path $T^{p'_0}(\delta_{[c',d']})$, being disjoint from $\delta_{[c',d']}$, from $R(T^{-p'_0}(\widetilde\phi_a))$ and from $R(T^{p_1}(\widetilde\phi_a))$, is disjoint from $\lambda$. We have found a contradiction.
We can prove similarly that every $T$-free simple path that joins $ T^{-p_0}(\widetilde\phi_b)$ to $ T^{p_1}(\widetilde\phi_b)$, where $p_0$, $p_1$ are positive, meets $L(\widetilde\phi_b)$.
For every path $\delta: [c,d]\to \widetilde{\mathrm{dom}}(I)$, we define its [*interior*]{} as being the path $\dot\delta=\delta\vert_ {(c,d)}$. The following sets $$L_a=\bigcap_{k\in{\mathbb{Z}}} L(T^k(\widetilde\phi_a)), \enskip R_b=\bigcap_{k\in{\mathbb{Z}}} R(T^k(\widetilde\phi_b))$$ are open sets homeomorphic to the plane. For every $q{\geqslant}1$, one can look at the relatively compact connected components of $\widetilde f^{-q}(\widetilde\phi_b)\cap L_a$. By Lemma \[lemma:following\], every such a component is the interior of path joining $T^{p}(\widetilde\phi_a)$ to $T^{p'}(\widetilde\phi_a)$, where $\vert p-p'\vert{\leqslant}1$. For every $p{\geqslant}0$ and $q{\geqslant}1$, we will define the set ${\mathcal X}_{p,q}$ of paths joining $\widetilde\phi_a$ to $T(\widetilde\phi_a)$ whose interior is a connected component of $T^p\circ f^{-q}(\widetilde\phi_b)\cap L_a$. We denote ${\mathcal X}=\bigcup_{p{\geqslant}0, q{\geqslant}1} {\mathcal X}_{p,q}$. Note that two different paths in ${\mathcal X}$ have disjoint interiors, and are disjoint if they do not belong to the same set ${\mathcal X}_{p,q}$. Fix a real parametrization $t\mapsto \widetilde\phi_a(t)$ of $\widetilde\phi_a$. If $\delta_0\in\mathcal X$ joins $\widetilde\phi_a(v_0)$ to $T(\widetilde\phi_a)(v'_0)$, and if $\delta_1\in\mathcal X$ joins $\widetilde\phi_a(v_1)$ to $T(\widetilde\phi_a)(v'_1)$, then $(v_1-v_0)(v'_1-v'_0){\leqslant}0$. Moreover, if $\delta_0\not=\delta_1$ at least one of the number $v_1-v_0$, $v'_1-v'_0$ is not zero. So one gets a natural total order on ${\mathcal X}$ by setting $$\delta_0{\leqslant}\delta_1\enskip\mathrm{if}\enskip v_1{\leqslant}v_0\enskip\mathrm{and} \enskip v'_0{\leqslant}v'_1.$$ One can extend this order to every set $\mathcal Y$ of paths containing $\mathcal X$, provide the paths of $\mathcal Y$ join $\widetilde\phi_a$ to $T(\widetilde\phi_a)$, have their interior included in $L_a$, and have pairwise disjoint interiors.
\[lemma: paths\] We have the following:
1. The set $\mathcal X_{p,q}$ is not empty if $q{\geqslant}2$ and $1{\leqslant}p{\leqslant}q$.
2. If $\delta_0$ and $\delta_1$ are two paths in $\mathcal X$, there are finitely many paths in $\mathcal X_{p,q}$ between $\delta_0$ and $\delta_1$.
3. Suppose that $q{\geqslant}3$ and $1{\leqslant}p_0<p_2< p_1{\leqslant}q$. For every $\delta_0\in \mathcal X_{p_0,q}$ and every $\delta_1\in \mathcal X_{p_1,q}$, there exist at least two paths in $\mathcal X_{p_2,q}$ between $\delta_0$ and $\delta_1$.
To get $(1)$, apply Lemma \[lemma:accessibility\] that asserts that $\widetilde f^{-q}\circ T^p(\widetilde\phi_b)$ meets $\widetilde \phi_a$ and $T(\widetilde \phi_a)$ if $q{\geqslant}2$ and $1{\leqslant}p{\leqslant}2$, and then apply Lemma \[lemma:following\].
Let us prove $(2)$. Suppose that $\delta_0$ and $\delta_1$ are two paths in $\mathcal X$, and write $(\delta_i)_{i\in I}$ the family of paths in $\mathcal X_{p,q}$ that are between $\delta_0$ and $\delta_1$. It consists of subpaths of the line $\widetilde f^{-q}\circ T^p(\widetilde\phi_b)$ with pairwise disjoint interiors, whose union is relatively compact. If this family is infinite, it contains a sequence that converges to a point in the Hausdorff topology. Such a point should belong both to $\widetilde \phi_a$ and $T(\widetilde \phi_a)$. We have a contradiction.
Let us prove $(3)$. Suppose that $q{\geqslant}3$ and $1{\leqslant}p_0<p_2< p_1{\leqslant}q$ and fix $\delta_0\in \mathcal X_{p_0,q}$ and $\delta_1\in \mathcal X_{p_1,q}$. By $(2)$, one can suppose that there is no path in $\mathcal X_{p_0,q}\cup \mathcal X_{p_1,q}$ between $\delta_0$ and $\delta_1$. Suppose for example that $\delta_0\prec\delta_1$. The path $\delta_0$ joins $\widetilde\phi_a(v_0)$ to $T(\widetilde\phi_a)(v'_0)$, and $\delta_1$ joins $\widetilde\phi_a(v_1)$ to $T(\widetilde\phi_a)(v'_1)$. Denote $\beta=\widetilde\phi_a\vert_{[t_1,t_0]}$ and $\beta'=T(\widetilde\phi_a)\vert_{[t'_0,t'_1]}$. Consider the simple closed curve $C=\beta\delta_0\beta'\delta_1^{-1}$ and the relatively compact connected component $\Delta$ of its complement.
Let us prove that $C$ is $T$-free. Note first that $T(C)\cap C\subset \beta'\cap T(\beta)$. So, one must prove that $\beta'\cap T(\beta)=\emptyset$. Each path $\delta_1^{-1}\beta\delta_0$ and $T(\delta_0)T(\beta')T(\delta_1^{-1})$ join $T(\widetilde\phi_a)$ to itself. Moreover, they are disjoint and their interior are included in $L(T(\widetilde\phi_a))$. So, either one of the paths $\beta'$ or $T(\beta)$ is included in the other one, or they are disjoint. But in the first case, $\overline\Delta$ has to be forward or backward invariant by $T$, which contradicts the fact that $T$ is fixed point free. One deduces that $\beta'\cap T(\beta)=\emptyset$ and consequently that $C$ is $T$-free.
So, $\overline\Delta$ is also $T$-free and $\Delta$ is included in $L_a$. Applying Lemma \[lemma:following\] to $ \widetilde f^q(\beta'')$, one deduces that every path $\beta''\subset\overline\Delta$ that joins $\delta_0$ to $\delta_1$ meets $\widetilde f^{-q}\circ T^{p_2}(L(\widetilde\phi_b)))$. So, there exists a simple path $\sigma\subset\widetilde f^{-q}\circ T^{p_2}(L(\widetilde\phi_b)))$ that joins $\widetilde\phi_a$ to $T(\widetilde\phi_a)$ and whose interior is contained in $\Delta$. Consequently, there exists at least a path $\delta_2$ of $\mathcal X_{p_2,q}$ between $\delta_0$ and $\sigma$ and another path of $\delta_2'$ of $\mathcal X_{p_2,q}$ between $\sigma$ and $\delta_1$, see Figure \[figureinbetweenpaths\].
![Lemma \[lemma: paths\] []{data-label="figureinbetweenpaths"}](figure_middle_path.pdf){height="48mm"}
Weak version of the Realization Theorem (Theorem \[th: realization\])
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Let us begin with the following preliminary version of Theorem \[th: realization\].
\[prop: preliminary-realization\] If $q{\geqslant}2$ and $1{\leqslant}p{\leqslant}q$, then $\widetilde f^q\circ T^{-p}$ has a fixed point.
We set $\widetilde g=\widetilde f^q\circ T^{-p}$. The hypothesis implies that either $1{\leqslant}p+1{\leqslant}q$ or $1{\leqslant}p-1{\leqslant}q$. We will study the first case. The sets $\mathcal X_{p,q}$ and $\mathcal X_{p+1,q}$ are non empty by assertion $(1)$ of Lemma \[lemma: paths\]. Choose $\delta_0\in \mathcal X_{p,q}$ and $\delta_1\in \mathcal X_{p+1,q}$. Suppose that $\delta_0$ joins $\widetilde\phi_a(t_0)$ to $T(\widetilde\phi_a(t'_0))$ and that $\delta_1$ joins $\widetilde\phi_a(t_1)$ to $T(\widetilde\phi_a(t'_1))$. We remark that, as $\widetilde g(\delta_0)$ is contained in $\widetilde \phi_b$ and as $\widetilde g(\delta_1)$ is contained in $T(\widetilde \phi_b)$, then $\widetilde g(\delta_0)$ and $\widetilde g(\delta_1)$ lie in different connected components of the complement of $\widetilde \phi_a\cup \delta_0\cup T(\widetilde \phi_a)$ as well as in different connected components of the complement of $\widetilde \phi_a\cup \delta_1\cup T(\widetilde \phi_a)$.
Suppose first that $\delta_0\prec\delta_1$. Denote $\beta=\widetilde\phi_a\vert_{[t_1,t_0]}$ and $\beta'=T(\widetilde\phi_a)\vert_{[t'_0,t'_1]}$ and consider the simple closed curve $C=\beta\delta_0\beta'\delta_1^{-1}$. The index $i(\widetilde g, C)$ is well defined, we will prove that it is equal to $+1$. It will imply that $\widetilde g$ has a fixed point inside the topological disk bounded by $C$. Recall that to define $i(\widetilde g, C)$, one needs to choose a homeomorphism $h$ between $\widetilde{\mathrm{dom}}(I)$ and the Euclidean plane and consider the winding number of the vector field $\xi: z\mapsto h\circ \widetilde g\circ h^{-1}(z)-z$ on $h(C)$, this integer being independent of the choice of $h$. To make the computation, one can use Homma’s theorem [@Homma] that asserts that $h$ can be chosen such that $$h(\widetilde\phi _a)=\{0\}\times{\mathbb{R}}, \enskip h(T(\widetilde\phi _a))=\{1\}\times{\mathbb{R}}, \enskip h(\delta_0)=[0,1]\times\{0\}, \enskip h(\delta_1)=[0,1]\times\{1\}.$$ Note now that $\xi(z)$ is pointing:
- on the right when $z$ belongs to $h(\beta)=\{0\}\times[0,1]$,
- on the left when $z$ belongs to $h(\beta')=\{1\}\times[0,1]$,
- below when $z$ belongs to $h(\delta_0)=\{0\}\times[0,1]$,
- above when $z$ belongs to $h(\delta_1)=\{1\}\times[0,1]$.
This implies that the winding number is equal to $-1$.
In case where $\delta_1\prec\delta_0$, define $$\beta=\widetilde\phi_a\vert_{[t_0,t_1]}, \enskip \beta'=T(\widetilde\phi_a)\vert_{[t'_1,t'_0]}, \enskip C=\beta\delta_1\beta'\delta_0^{-1}$$ and choose $h$ such that $$h(\widetilde\phi _a)=\{0\}\times{\mathbb{R}}, \enskip h(T(\widetilde\phi _a))=\{1\}\times{\mathbb{R}}, \enskip h(\delta_0)=[0,1]\times\{1\}, \enskip h(\delta_1)=[0,1]\times\{0\}.$$ Note now that $\xi(z)$ is pointing:
- on the right when $z$ belongs to $h(\beta)=\{0\}\times[0,1]$,
- on the left when $z$ belongs to $h(\beta')=\{1\}\times[0,1]$,
- below when $z$ belongs to $h(\delta_0)=\{1\}\times[0,1]$,
- above when $z$ belongs to $h(\delta_1)=\{0\}\times[0,1]$.
This implies that the winding number is equal to $+1$.
In case $1{\leqslant}p-1{\leqslant}q$, we choose $\delta_0\in \mathcal X_{p,q}$ and $\delta_1\in \mathcal X_{p-1,q}$. The arguments are similar.
![Proposition \[prop: preliminary-realization\]. Relative position of $\widetilde g(\beta), \widetilde g(\delta_0), \widetilde g(\beta')$ and $\widetilde g(\delta_1)$. []{data-label="figure_index_1"}](figure_index_2.pdf){height="48mm"}
\[cor: preliminary-realization\] For every $p/q\in(0,1]$, such that $p$ and $q$ are relatively prime, there exists a point $\widetilde z$ such that $\widetilde f^q(\widetilde z)=T^p(\widetilde z)$.
In case $p/q<1$ we just apply Proposition \[prop: preliminary-realization\]. The case $p/q=1$ is particular, as not being an immediate consequence of Proposition \[prop: preliminary-realization\]. Nevertheless, we know that $(\widetilde f\circ T^{-1})^2=\widetilde f^2\circ T^{-2}$ has a fixed point. Brouwer’s theorem asserts that the existence of a periodic point implies the existence of a fixed point, for an orientation preserving plane homeomorphism. So $\widetilde f\circ T^{-1}$ has a fixed point.
Existence of horseshoe in the universal covering space
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If $\delta_0$ and $\delta_1$ are two disjoint paths in $\mathcal X$, we denote $\overline{\Delta}_{\delta_0,\delta_1}$ the closed “rectangle” bounded by $\delta_0$, $\delta_1$, $\beta$ and $\beta'$, where $\beta$ is the subpath of $\widetilde \phi_a$ joining the ends of $\delta_0$, $\delta_1$ that lie on $\widetilde \phi_a$ and $\beta'$ is the subpath of $\widetilde T(\widetilde \phi_a)$ joining the two other ends. We will call [*horizontal sides*]{} the paths $\delta_0$, $\delta_1$, and vertical sides the paths $\beta$, $\beta'$. We will prove the following result.
\[prop: horseshoe-universal\] If $q{\geqslant}3$ and $1< p<q$, there exists a compact set $\widetilde Z_{p,q}\subset\widetilde{\mathrm{dom}}(I)$, invariant by $\widetilde f^q\circ T^{-p}$ such that
- the restriction of $\widetilde f^q\circ T^{-p}$ to $\widetilde Z_{p,q}$ is an extension of the Bernouilli shift $\sigma :\{1,2\}^{{\mathbb{Z}}}\to \{1,2\}^{{\mathbb{Z}}}$;
- every $r$-periodic sequence of $\{1,2\}^{{\mathbb{Z}}}$ has a preimage by the factor map $\widetilde H_{p,q}:\widetilde Z_{p,q}\to \{1,2\}^{{\mathbb{Z}}}$ that is a $r$-periodic point of $\widetilde f^q\circ T^{-p}$.
We suppose that $q{\geqslant}3$ and fix $1<p<q$. Using Lemma \[lemma: paths\] we can find four paths $\delta_1\in\mathcal X_{p-1,q}$, $\delta_2\in\mathcal X_{p,q}$, $\delta_3\in\mathcal X_{p,q}$ and $\delta_4\in\mathcal X_{p+1,q}$ all pairwise disjoint such that $\delta_1\prec\delta_2\prec\delta_3\prec\delta_4$ or $\delta_4\prec\delta_3\prec\delta_2\prec\delta_1$.
We set $$\overline\Delta_1=\overline\Delta_{\delta_1,\delta_2}, \enskip\overline\Delta_2=\overline\Delta_{\delta_3,\delta_4},\enskip \overline\Delta=\overline\Delta_{\delta_1,\delta_4}$$ and $\widetilde g=\widetilde f^q\circ T^{-p}$. Note again that $\widetilde g(\delta_1)\subset T^{-1}(\widetilde \phi_b)$, that both $\widetilde g(\delta_2)$ and $\widetilde g(\delta_3)$ are contained in $\widetilde \phi_b$ and that $\widetilde g(\delta_4)\subset T(\widetilde \phi_b)$.
Let us explain first why $\overline\Delta$ is a topological horseshoe in the sense of Kennedy-Yorke (see [@KennedyYorke]). Every continuum $K\subset \overline\Delta$ that meets $\delta_{1}$ and $\delta_{4}$ contains a continuum $K_1\subset \overline\Delta_1$ that meets $\delta_{1}$ and $\delta_{2}$ and a continuum $K_2\subset \overline\Delta_2$ that meets $\delta_{3}$ and $\delta_4$. The set $\widetilde g(K_1)$ is a continuum that meets $\widetilde\phi_b$ and $T^{-1}(\widetilde\phi_b)$ and that does not meet neither $R(\widetilde\phi_a)$ nor $R(T(\widetilde\phi_a))$. This implies that it contains a continuum included in $\overline\Delta$ that meets $\delta_{1}$ and $\delta_{4}$. Similarly, $\widetilde g(K_2)$ is a continuum that meets $\widetilde\phi_b$ and $T(\widetilde\phi_b)$ and that does not meet neither $R(\widetilde\phi_a)$ nor $R(T(\widetilde\phi_a))$. So, it contains a continuum included in $\overline\Delta$ that meets $\delta_{1}$ and $\delta_{4}$. Summarizing, every continuum $K\subset \overline\Delta$ that meets $\delta_{1}$ and $\delta_{4}$ contains two disjoint sub-continua whose images by $\widetilde g$ are included in $\overline\Delta$ and meet $\delta_{1}$ and $\delta_{4})$. This is the definition of a topological horseshoe in the sense of Kennedy-Yorke. According to [@KennedyYorke], there exists a compact set $\widetilde Z$ invariant by $\widetilde g$ such that $\widetilde g\vert_{\widetilde Z}$ is an extension of the two sided Bernouilli shift on $\{1,2\}^{{\mathbb{Z}}}$, see Figure \[figure\_horseshoe\_covering\].
![Horseshoe in the universal covering space[]{data-label="figure_horseshoe_covering"}](figure_horseshoe_covering.pdf){height="48mm"}
To get our proposition it remains to prove that to every periodic sequence corresponds at least one periodic point in $\widetilde Z$. We will use the [*Conley index theory*]{}, or at least a simplified version of it. If $\widetilde z$ belongs to $\partial(\overline\Delta_1\cup \overline\Delta_2)$, then either its image by $\widetilde g$ or its inverse image is outside $\overline\Delta_1\cup \overline\Delta_2$. The first situation occurs when $z$ belongs to an horizontal side, the second one when $z$ belongs to a vertical side. This means that the set $\widetilde Z_{p,q}=\bigcap_{k\in{\mathbb{Z}}} \widetilde g^{-k}(\overline\Delta_1\cup \overline\Delta_2)$ is a compact set included in $\Delta_1\cup \Delta_2$. According to the discrete Conley index theory, $\widetilde Z_{p,q}$ is a [*locally maximal invariant set*]{} and $\overline\Delta_1\cup \overline\Delta_2$ an [*isolating block*]{} of $\widetilde Z_{p,q}$. Note that there exists a map $\widetilde H_{p,q}: \widetilde Z_{p,q}\to \{1,2\}^{{\mathbb{Z}}}$ that associates to a point $\widetilde z\in\widetilde Z_{p,q}$ the sequence $(\varepsilon_k)_{k\in{\mathbb{Z}}}$ such that $\widetilde g^{i}(\widetilde z)\in \Delta_{\varepsilon_k}$. The map $\widetilde H_{p,q}$ is continuous and satisfies $\widetilde H_{p,q}\circ\widetilde g=\sigma\circ \widetilde H_{p,q}$. We will prove that every periodic sequence of period $r$ is the image by $\widetilde H_{p,q}$ of a periodic point of $\widetilde g$ of period $r$. This will imply that $\widetilde H_{p,q}$ is onto, because its image is compact, and we will get a complete proof of Proposition \[prop: horseshoe-universal\]. Different mathematical objects can be defined in Conley index theory. According to Franks and Richeson [@FranksRicheson], one can associate to every locally maximal invariant set a pointed compact space $(X,*)$ and a continuous transformation $F:X\to X$, fixing $*$ and locally constant in a neighborhood of $*$. The shift equivalence class of such a map is uniquely defined (the definition is given in [@FranksRicheson]): it is the Conley index of $X$. In our situation it can be easily described. Identify $\delta_1\cup\delta_2\cup\delta_3\cup\delta_4$ to a point $*$, denote $\overline\Delta_1\cup \overline\Delta_2/\sim$ the quotient space and consider the map $$F: \overline\Delta_1\cup \overline\Delta_2/\sim \to \overline\Delta_1\cup \overline\Delta_2/\sim$$ such that
- $F(*)=*$;
- $F(z)=*$, if $z\in \overline\Delta_1\cup \overline\Delta_2$ and $\widetilde g(z)\not\in \overline\Delta_1\cup \overline\Delta_2$;
- $F(z)$ is the class of $\widetilde g(z)$, if $z\in \overline\Delta_1\cup \overline\Delta_2$ and $\widetilde g(z)\in \overline\Delta_1\cup \overline\Delta_2$.
The map $F$ is well defined, continuous, fixes $*$ and is locally constant in a neighborhood of $*$. In fact it is representative of the Conley index (see [@HernandezLeCalvezRuizdelPortal]). One can look at the action $F_*$ of $F$ in homology, getting the [*homological Conley index*]{}, which is a shift equivalence of finite ranks endomorphisms. In our situation it is easy to compute. The space $H_0(\overline\Delta_0\cup \overline\Delta_2/\sim,{\mathbb{Z}})$ is one-dimensional and the induced endomorphim $F_{*,0}$ is the identity. The space $H_1(\overline\Delta_0\cup \overline\Delta_2/\sim,{\mathbb{Z}})$ is two-dimensional and generated by $e_1$ and $e_2$, where $e_1$ is the homology class of a path in $\overline\Delta_1$ joining $\delta_{1}$ to $\delta_2$ and $e_2$ the homology class of a path in $\overline\Delta_2$ joining $\delta_{3}$ to $\delta_4$. The matrix $M=(m_{i,j})$ of the induced endomorphism $F_{*,1}$ in this basis can be easily computed: if $\delta_0\prec\delta_1\prec\delta_2\prec\delta_3$, then $$M=\left(\begin{matrix} 1& -1\\ 1& -1\end{matrix}\right),$$ if $\delta_3\prec\delta_2\prec\delta_1\prec\delta_0$, then $$M=\left(\begin{matrix} -1& 1\\ -1& 1\end{matrix}\right).$$ Let us remind some facts on the Lefschetz index (see [@Brown], [@Dold], [@Nussbaum]). The Lefschetz index of a continuous map $G:N\to N$ at an isolated fixed point is well known when $N$ is manifold. The notion has been extended by Dold to the case where $N$ is an absolute neighborhood retract (ANR). For example the Lefschetz index $i(F,*)$ is equal to $1$ because $*$ is an attracting fixed point of $F$. Moreover, if $Y\subset N$ is a compact invariant set and if there is a neighborhood $V$ of $Y$ whose fixed points are all contained in $Y$, one may define the index $i(G,Y)\in {\mathbb
Z}$. In the case where there are finitely many fixed points in $Y$, it is equal to the sum of all Lefschetz indices of fixed points that are in $Y$. A particular case is the case where $N$ is compact and $Y=N$. The number $i(G,Y)$ is called the [*Lefschetz number*]{} of $G$ and denoted by $\Lambda(G)$. It is related to the action on the singular homology groups by the Lefschetz-Dold formula $$\Lambda(G)=\sum_{i=0}^n (-1)^{i}{\rm tr}(G_{*,i}),$$ where $G_{*,i}$ is the morphism induced by $G$ on $H_i(N,{\mathbb
R})$ and $n$ the largest integer such that $H_i(N,{\mathbb
R})\not=0$. The Leschetz number of $F$ is equal to $$\mathrm{tr}(F_{*,0})-\mathrm{tr}(F_{*,1})= 1-\mathrm{tr}(M).$$ Every fixed point of $F$ different from $*$ being in $\widetilde Z_{p,q}$, one deduces that $$L(F)=i(F,*)+i(F,\widetilde Z_{p,q})= 1+i(\widetilde g, \widetilde Z_{p,q})$$ which implies that $$i(\widetilde g, \widetilde Z_{p,q})=-\mathrm{tr}(M).$$ More can be said:
- The sequences $(i(\widetilde g^n, \widetilde Z_{p,q}))_{n{\geqslant}1}$ and $(-\mathrm{tr}(M^n))_{n{\geqslant}0}$ coincide.
- If ${\bf e}\in\{1,2\}^{{\mathbb{Z}}}$ is the constant sequence equal to $i\in\{1,2\}$, then $\widetilde H_{p,q}^{-1}(\{{\bf e}\})$ is a locally maximal invariant set and the rectangle $\overline\Delta_{i}$ an isolating block of $H^{-1}(\{{\bf e}\})$. The $(1,1)$ matrix $(m_{i,i})$ represents the homological Conley index of rank $1$, and consequently, $i(\widetilde g, H^{-1}(\{{\bf e}\}))=-m_{i,i}$. Note that this is the integer found by computation in the proof of Proposition \[prop: preliminary-realization\].
- If ${\bf e}=(\varepsilon_k)_{k\in{\mathbb{Z}}}\in\{1,2\}^{{\mathbb{Z}}}$ is a periodic sequence of periodic $s$, $\widetilde H_{p,q}^{-1}(\{{\bf e}\})$ is itself a locally maximal invariant set of $\widetilde g^s$ and its homological Conley index of rank $1$ is the $(1,1)$ matrix $(\prod_{0{\leqslant}k<s} m_{\varepsilon_k,\varepsilon_{k+1}})$.
Of course, $\prod_{0{\leqslant}k<s} m_{\varepsilon_k,\varepsilon_{k+1}}\not=0$ and so $\mathrm{fix}(g^s)\cap \widetilde H_{p,q}^{-1}(\{{\bf e}\})\not=\emptyset$. But this exactly what we want to prove.
Existence of horseshoe in an annular covering space
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We will construct other “geometric horseshoes” in this subsection, not in the universal covering space of the domain of a maximal isotopy but in an annular covering space. The first interest of these constructions is that they will permit us to get the constant $\log 4/3q$ in the statement of Theorem \[th: horseshoe\]. The second interest is that the horsehoes that we will construct are [*rotational horseshoes*]{}, as defined for instance in [@PassegiPotrieSambarino], and we will deduce some additional dynamical properties related to rotation numbers.
We consider in this subsection, the annular covering $\mathring{\mathrm{dom}}(I)=\widetilde{\mathrm{dom}}(I)/T^2$. Denote $\mathring I$ the induced identity, $\mathring f$ the induced lift of $f$, and $\mathring{\mathcal{F}}$ the induced foliation.
\[prop: horseshoe-annular\] If $q{\geqslant}2$, there exists a compact set $\mathring Z_{q}\subset\mathring{\mathrm{dom}}(I)$, invariant by $\mathring f^q$ such that
- the restriction of $\mathring f^q$ to $\mathring Z_{q}$ is an extension of the Bernouilli shift $\sigma :\{1,\dots, 2q-2\}^{{\mathbb{Z}}}\to \{1,\dots, 2q-2\}^{{\mathbb{Z}}}$;
- every $s$-periodic sequence of $\{1,\dots, 2q-2\}^{{\mathbb{Z}}}$ has a preimage by the factor map $\mathring H_{q}:\mathring Z_{q}\to \{1,\dots, 2q-2\}^{{\mathbb{Z}}}$ that is a $s$-periodic point of $\mathring f^q$.
We suppose that $q{\geqslant}2$. By Lemma \[lemma: paths\], we can find a decreasing or increasing sequence of pairwise disjoint paths in $\mathcal X$, denoted $(\delta_i)_{1{\leqslant}i{\leqslant}2q-2}$, such that, for every $l\in\{1,\dots, q-1\}$, one has $$\delta_{2l-1}\in \mathcal X_{l,q},\enskip \delta_{2l}\in \mathcal X_{l+1,q}.$$ The rectangles of the family $\left(T^k\left(\overline\Delta_{\delta_{2l-1,2l}}\right)\right)_{1{\leqslant}l<q, k\in{\mathbb{Z}}}$ are pairwise disjoint and define by projection in $\mathring{\mathrm{dom}}(I)$ a family $(\mathring\Delta_i)_{1{\leqslant}i{\leqslant}2q-2}$ of rectangles such that
- $\mathring\Delta_i$ is the projection of $\overline\Delta_{\delta_{2i-1,2i}}$ if $1{\leqslant}i{\leqslant}q-1$,
- $\mathring\Delta_i$ is the projection of $T(\overline\Delta_{\delta_{2(i-q+1)-1,2(i-q+1}})$ if $q{\leqslant}i{\leqslant}2q-2$.
See Figure \[figure\_horseshoe\_annular2\].
![Horseshoe in the annular covering $\mathring{\mathrm{dom}}(I)$ - Case $q$=2.[]{data-label="figure_horseshoe_annular2"}](figure_horseshoe_annular2.pdf){height="48mm"}
Here again $\bigcup_{1{\leqslant}i{\leqslant}2q-2} \mathring\Delta_i$ is an isolating block and $\mathring Z_q=\bigcap_{k\in{\mathbb{Z}}} \mathring f^{-kq}(\bigcup_{1{\leqslant}i{\leqslant}2q-2} \mathring\Delta_i)$ is a compact set included in the interior of $\bigcup_{1{\leqslant}i{\leqslant}2q-2} \mathring\Delta_i$. Denote $\mathring H_q: \mathring Z_q\to \{1,\dots 2q-2\}^{{\mathbb{Z}}}$ the map that associates to a point $\mathring z\in\mathring Z_q$ the sequence $(\varepsilon_k)_{k\in{\mathbb{Z}}}$ such that $\mathring f^{qk}(\mathring z)\in \Delta_{\varepsilon_k}$. It is continuous and satisfies $\mathring H_q\circ\mathring f^q=\sigma\circ \mathring H_q$. We want to prove that every periodic sequence of period $s$ is the image by $\mathring H_q$ of a periodic point of $\mathring f^q$ of period $s$. This will imply that $\mathring H_q$ is onto, because its image is compact and we will get a complete proof of Proposition \[prop: horseshoe-annular\]. We identify the union of the horizontal sides to a point $*$ and we have a continuous map $F$, defined on the quotient space $(\bigcup_{1{\leqslant}i{\leqslant}2q-2} \mathring\Delta_i)/\sim$ fixing $*$ and locally constant in a neighborhood of $*$. The space $H_1(\bigcup_{1{\leqslant}i{\leqslant}2q-2} \mathring\Delta_i/\sim,{\mathbb{R}})$ is $2q-2$-dimensional and generated by the $e_i$, $1{\leqslant}i{\leqslant}2q-1$, where:
- $e_i$ is the homology class of a path in $\mathring\Delta_i$, which is the projection of path in $\overline\Delta_i$ joining $\delta_{2i-1}$ to $ \delta_{2i}$, if $1{\leqslant}i{\leqslant}q-1$;
- $e_i$ is the homology class of a path in $\mathring\Delta_i$, which is the projection of path in $T(\overline\Delta_{i-q+1})$ joining $T(\delta_{2(i-q+1)-1})$ to $ T(\delta_{2(i-q+1)})$, if $q{\leqslant}i{\leqslant}2q-2$.
In case where $(\delta_i)_{1{\leqslant}i{\leqslant}2q-2}$ is decreasing, the matrix $M=(m_{i,j})$ of $F_{*,1}$ in the basis $(e_i)_{1{\leqslant}i{\leqslant}2q-2}$ is
$$M=\left(\begin{matrix} 1& -1&\dots&(-1)^{q-1}&1& -1&\dots&(-1)^{q-1}\\
1& -1&\dots&(-1)^{q-1}&1& -1&\dots&(-1)^{q-1}\\
\dots& \dots&\dots& \dots& \dots& \dots&\dots& \dots\\
-1& 1&\dots&(-1)^{q}&-1& 1&\dots&(-1)^{q}\\
-1& 1&\dots&(-1)^{q}&-1& 1&\dots&(-1)^{q}\\
\dots& \dots&\dots& \dots& \dots& \dots&\dots& \dots\end{matrix}\right),$$
In the case where $(\delta_i)_{1{\leqslant}i{\leqslant}2q-2}$ is increasing,
$$M=\left(\begin{matrix} -1& 1&\dots&(-1)^{q}&-1& 1&\dots&(-1)^{q}\\
-1& 1&\dots&(-1)^{q}&-1& 1&\dots&(-1)^{q}\\
\dots& \dots&\dots& \dots& \dots& \dots&\dots& \dots\\
1& -1&\dots&(-1)^{q-1}&1& -1&\dots&(-1)^{q-1}\\
1& -1&\dots&(-1)^{q-1}&1& -1&\dots&(-1)^{q-1}\\
\dots& \dots&\dots& \dots& \dots& \dots&\dots& \dots
\end{matrix}\right).$$ Here again, if ${\bf e}=(\varepsilon_k)_{k\in{\mathbb{Z}}}\in\{1,\dots, 2q-2\}^{{\mathbb{Z}}}$ is a periodic sequence of period $s$, $\mathring H_q^{-1}(\{{\bf e}\})$ is a locally maximal invariant set of $\mathring f^{qs}$ and its homological Conley index of rank $1$ is the $(1,1)$ matrix $(\prod_{0{\leqslant}i<s} m_{\varepsilon_k,\varepsilon_{k+1}})$. It is not equal to zero and so $\mathrm{fix}(\mathring f^{qs})\cap \mathring H_q^{-1}(\{{\bf e}\})\not=\emptyset$.
The full topological horseshoe $\mathring Z_q$ constructed in Proposition \[prop: horseshoe-annular\] on the double annular covering $\mathring{\mathrm{dom}}(I)$ is relevant to obtain the bound on the topological entropy, but we are also interested in applications concerning rotation numbers. This is easier to do on the annular covering space $\widehat{\mathrm{dom}}(I)=\widetilde{\mathrm{dom}}(I)/T$ for which $\mathring{\mathrm{dom}}(I)$ is a $2$-fold covering. One can consider the projection $P:\mathring{\mathrm{dom}}(I)\to\widehat{\mathrm{dom}}(I)$, and the set $\widehat Z_q=P(\mathring Z_q)$, which is invariant by $\widehat f^q$, where $\widehat f$ is the natural lift of $f\vert_{\mathrm {dom}(I)}$ to $\widehat{\mathrm{dom}}(I)$. Furthermore, if we define $\widehat H_q:\widehat Z^q\to \{1, \dots, q-1\}^{{\mathbb{Z}}}$, so that the $l$-th letter of $\widehat H_q(\widehat z)$ is $i$ if the $l$-th letter of $\mathring H_q(\mathring z)$ is either $i$ or $i+q-1$, where $\mathring{z}\in P^{-1}(\widehat{z})$, then the restriction of $\widehat f^q$ to $\widehat Z_q$ is semi-conjugated, by $\widehat H_q$, to the Bernouilli shift on $\{1, \dots, q-1\}^{{\mathbb{Z}}}$.
We will identify a compact invariant subset $\widehat Z_q'$ of $\widehat Z_q$, such that the restriction of $\widehat{f}^q$ to this set is still semi-conjugated by $\widehat{H_q}$ to the Bernouilli shift, and that is a [*rotational horseshoe*]{}, in the sense that the existence of a rotation number for a point $\widehat z\in\widehat Z_q'$ (and its value) can be determined by the knowledge of the sequence $\widehat H_q(\widehat z)$.
Keeping the notations of Proposition \[prop: horseshoe-annular\], let $\overline \Delta_0\subset \widetilde{\mathrm{dom}}(I)$ be the set bounded by $\delta_1, \delta_{2q-2},\widetilde \phi_a$ and $T(\widetilde \phi_a)$, and note that, for $1{\leqslant}i{\leqslant}q-1$, $\widetilde f^{q}(\overline \Delta_{\delta_{2i-1},\delta_{2i}})$ intersects both $T^{i}(\Delta_0)$ and $T^{i+1}(\Delta_0)$. We set $\overline \Delta_i'= \overline \Delta_{\delta_{2i-1},\delta_{2i}}\cap \widetilde f^{-q}(T^{i}(\Delta_0))$, which projects, in $\widehat{\mathrm{dom}}(I)$ to a subset $\widehat \Delta_i'$ of $P(\mathring \Delta_i)$, see figure \[rotationalHorseshoe\]. The set $\widehat Z_q'$ is defined as $\bigcap_{k\in{\mathbb{Z}}}\widehat f^{-kq}\left(\bigcup_{1{\leqslant}i{\leqslant}q-1}\widehat \Delta_i'\right)$. Fix $\widehat z\in \widehat Z_q'$ and write $(\varepsilon_k)_{k\in{\mathbb{Z}}}=\widehat H_q(\widehat z)$. Fix $k{\geqslant}1$ and set $\alpha_n=\sum_{0{\leqslant}k<n} \varepsilon_k$. If $\widetilde z$ is the lift of $\widehat z$ that belongs to $\overline\Delta_{\delta_{2\varepsilon_0-1,2\varepsilon_0}}$, then $\widehat f^{nq}(\widehat z)$ belongs to $T^{\alpha_n}(\overline\Delta_{\delta_{2\varepsilon_n-1,2\varepsilon_n-1}})$. In particular $\widehat z$ has a rotation number $\rho$ (for the lift $\widetilde f$) if and only if the sequence $(\alpha_n/n)_{n{\geqslant}1}$ converges to $q\rho$. We deduce the following result:
![Rotational horseshoe for $q=3$. The sets $\overline{\Delta_1}'$ and $\overline{\Delta_2}'$ appear respectively as a light yellow shaded region and a light green shaded region.[]{data-label="rotationalHorseshoe"}](figure_rotationalhorseshoe3.pdf){height="78mm"}
\[prop: irrational compact set\] For any $0<\rho{\leqslant}1$, there exists a nonempty minimal set $\widehat N_{\rho}$ such that, for each $\widehat z\in \widehat N_{\rho}$, one has that $\mathrm{rot}_{\widetilde f}(\widehat z)=\rho.$
If $\rho$ is rational, the result is an immediate consequence of Proposition \[prop: preliminary-realization\] and $\widehat N_{\rho}$ can be chosen as a periodic orbit. We suppose by now that $\rho$ is irrational. It suffices to show that there exists an integer $q{\geqslant}3$ and a nonempty closed set $X_{\rho}$ of $\{1, \dots, q-1\}^{{\mathbb{Z}}}$, invariant by the Bernouilli shift and such that for every $(\varepsilon_k)_{k\in{\mathbb{Z}}}\in X_{\rho}$, one has $$\lim_{n\to+\infty} {1\over n} \sum_{k=0}^n \varepsilon _k=q\rho.$$ Indeed, every minimal subset of the compact set $\bigcup_{0{\leqslant}l<q}\widehat f^l(\widehat H_q^{-1}(X_{\rho}))$ will satisfy the prescribed properties.
The existence of $X_{\rho}$ is a well known fact concerning Sturmian sequences. Set $$\rho'= {q\over q-2}\,\rho-{1\over q-2},$$choosing $q$ large enough to insure that $\rho'\in(0,1)$. Consider the partition of $[0,1[=[0, 1-\rho'[\cup [1-\rho', 1[$ and define a sequence ${\bf e}'=(\varepsilon'_k)_{k\in{\mathbb{Z}}}$ in the following way $$\varepsilon'_k= \begin{cases}0 & \mathrm {if}\enskip k\rho'-[k\rho']\in [0,1-\rho'[,\\1 & \mathrm {if}\enskip k\rho'-[k\rho']\in [1-\rho',1[.\\
\end{cases}$$ Every sequence ${\bf e}=(\varepsilon_k)_{k\in{\mathbb{Z}}}$ in the closure $X'$ of the orbit of ${\bf e}'$ by the Bernouilli shift defined on $\{0,1\}^{{\mathbb{Z}}}$ satisfies $$\lim_{n\to+\infty} {1\over n} \sum_{k=0}^n \varepsilon _k=\rho'.$$ So the set $$X_{\rho}=\{(1+(q-2)\varepsilon_k)_{k\in{\mathbb{Z}}}\,\vert \, (\varepsilon_k)_{k\in{\mathbb{Z}}}\in X'\}$$ satisfies the prescribed properties.
Proofs of Theorem \[th: realization\] and Theorem \[th: horseshoe\]
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Let us begin by proving the realization theorem (Theorem \[th: realization\]). Let us remind the statement:
Let $M$ be an oriented surface, $f$ a homeomorphism of $M$ isotopic to the identity, $I$ a maximal identity isotopy of $f$ and $\mathcal F$ a foliation transverse to $I$. Suppose that $\gamma:[a,b]\to \mathrm{dom}(I)$ is an admissible path of order $r$ with a $\mathcal{F}$-transverse self intersection at $\gamma(s)=\gamma(t)$, where $s<t$. Let $\widetilde \gamma$ be a lift of $\gamma$ to the universal covering space $\widetilde{\mathrm{dom}}(I)$ of $\mathrm{dom}(I)$ and $T$ the covering automorphism such that $\widetilde\gamma$ and $T(\widetilde\gamma)$ have a $\widetilde{\mathcal F}$-transverse intersection at $\widetilde\gamma(t)=T(\widetilde\gamma)(s)$. Let $\widetilde f$ be the lift of $f\vert_{\mathrm{dom}(I)}$ to $\widetilde{\mathrm{dom}}(I)$, time-one map of the identity isotopy that lifts $I$, and $\widehat f$ the homeomorphism of the annular covering space $ \widehat{\mathrm{dom}}(I)=\widetilde {\mathrm{dom}}(I)/T$ lifted by $\widetilde f$. Then we have the following:
1. For every rational number $p/q\in(0,1]$, written in an irreducible way, there exists a point $\widetilde z\in \widetilde{\mathrm{dom}}(I)$ such that $\widetilde f^{qr}(\widetilde z)=T^p(\widetilde z)$ and such that $\widetilde I_{\widetilde{\mathcal F}}^{{\mathbb{Z}}}(\widetilde z)$ is equivalent to $\prod_{k\in{\mathbb{Z}}} T^k(\widetilde\gamma_{[s,t]})$.
2. For every irrational number $\rho\in[0,1/r]$, there exists a compact set $\widehat Z_{\rho}\subset\widehat{\mathrm{dom}}(I)$ invariant by $\widehat f$, such that every point $\widehat z\in \widehat Z_{\rho}$ has a rotation number $\mathrm{rot}_{\widetilde f}(\widehat z)$ equal to $\rho$. Moreover if $\widetilde z\in\widetilde{\mathrm{dom}}(I)$ is a lift of $\widehat z$, then $\widetilde I_{\widetilde{\mathcal F}}^{{\mathbb{Z}}}(\widetilde z)$ is equivalent to $\prod_{k\in{\mathbb{Z}}} T^k(\widetilde\gamma_{[s,t]})$.
We keep the notations introduced at the beginning of the section (but replacing $f$ with $f^r$). In particular we have a foliation $\widehat{\mathcal F}$ defined on the sphere $\widehat{\mathrm{dom}}(I)\cup\{S,N\}$ with two singular points $S$ and $N$. We will frequently use classical properties of foliations on a sphere in what follows. The path $\widetilde\gamma\vert_{[s,t]}$ projects onto a closed path of $\widehat{\mathrm{dom}}(I)$ that defines naturally an essential loop $\widehat{\Gamma}$. This loop is simple because it is transverse to $\widehat{\mathcal F}$. There is no loss of generality by supposing that that $S$ is on the right of $\widehat\Gamma$ and $N$ on its left. Write $U_{\widehat\Gamma}$ for the union of leaves that meet $\widehat\Gamma$, it is an open annulus. We have the following results:
1. Every leaf in $U_{\widehat\Gamma}$ has a $\omega$-limit set equal to $S$.
2. Every simple loop transverse to $\widehat{\mathcal F}$ that is not equivalent to $\widehat \Gamma$ is essential, disjoint from $U_{\widehat\Gamma}$ and separated from this set by a closed leaf.
3. Either every leaf of $U_{\widehat\Gamma}$ is adherent to $N$, or there exists a closed leaf $\widehat\phi$ in the frontier of $U_{\widehat \Gamma}$ and this leaf is the common $\alpha$-limit set of every leaf of $U_{\widehat \Gamma}$.
The first property can be deduced from the fact that $\widehat \gamma$ joins $\widehat\phi_{\widehat\gamma(a)}$, which is homoclinic to $S$, to $\widehat\phi_{\widehat\gamma(b)}$ which is also homoclinic to $S$.
Recall that there exists $k_0{\geqslant}1$ such that $\widetilde\gamma$ and $T^k(\widetilde\gamma)$ have a $\widetilde{\mathcal{F}}$-transverse intersection if $1{\leqslant}k{\leqslant}k_0$ and no transverse intersection if $k>k_0$. As explained above, one gets a simple loop $\widehat\Gamma_{k_0}$ of $\widehat{\mathrm{dom}}(I)_{k_0}=\widetilde{\mathrm{dom}}(I)/T^{k_0}$ transverse to the foliation $\widehat{\mathcal F}_{k_0}$ naturally defined on $\widehat{\mathrm{dom}}(I)_{k_0}$. Of course $\widehat{\mathrm{dom}}(I)_{k_0}$ is the $k_0$-fold finite covering of $\widehat{\mathrm{dom}}(I)$ and $\widehat{\mathcal F}_{k_0}$ the lifted foliation. Applying $(1)$ both to $\widehat\Gamma$ and $\widehat\Gamma_{k_0}$ and $(2)$ to $\widehat\Gamma_{k_0}$, one deduces that $\widehat\Gamma_{k_0}$ is equivalent to the lift of $\widehat\Gamma$ to $\widehat{\mathrm{dom}}(I)_{k_0}$. This means that if $\widetilde\gamma$ and $T^k(\widetilde\gamma)$ have a $\widetilde{\mathcal{F}}$-transverse intersection at $\widetilde \gamma(t') =T^{k_0}(\widetilde \gamma(s'))$, then $s'<t'$ and $\prod_{k\in{\mathbb{Z}}} T^k(\widetilde\gamma\vert_{[s,t]})$ and $\prod_{k\in{\mathbb{Z}}} T^{kk_0}(\widetilde\gamma\vert_{[s',t']})$ are equivalent.
One can apply what has been done in this section to $\widetilde f^r$ and $T^{k_0}$. Applying Corollary \[cor: preliminary-realization\], one can find for every $p/q\in(0,1]$ written in an irreducible way, a point $\widetilde z_{k_0}\in \widetilde{\mathrm{dom}}(I)$ such that $\widetilde f^{qk_0r}(\widetilde z_{k_0})=T^{pk_0}(\widetilde z_{k_0})$. Indeed $p/qq_0\in(0,1]$. This implies that the $k_0$-th iterate of $\widetilde f^{qr}\circ T^{-p}$ has a fixed point. By Brouwer theory, one deduces that $f^{qr}\circ T^{-p}$ itself has fixed point, which means a point $\widetilde z\in \widetilde{\mathrm{dom}}(I)$ such that $\widetilde f^{qr}(\widetilde z)=T^p(\widetilde z)$. The point projects onto a periodic point $\widehat z$ whose transverse trajectory defines an essential transverse loop.
Suppose first that every leaf of $U_{\widehat\Gamma}$ is adherent to $N$. In this case, there is no essential transverse loop but the powers of $\widehat\Gamma$, so $\widetilde I_{\widetilde{\mathcal F}}^{{\mathbb{Z}}}(\widetilde z)$ is equivalent to $\prod_{k\in{\mathbb{Z}}} T^k(\widetilde\gamma_{[s,t]})$.
Suppose now that there is closed leaf $\widehat\phi$ in the frontier of $U_{\widehat \Gamma}$. One of the sets $\widehat f(\overline{U_{\widehat\Gamma}})$ or $\widehat f^{-1}(\overline{U_{\widehat\Gamma}})$ is included in $U_{\widehat \Gamma}$. In that case, $\widehat A=\bigcup_{k\in{\mathbb{Z}}} \widehat f^{r}(U_{\widehat\Gamma})$ is an essential invariant annulus and its preimage $\widetilde A=\widehat\pi^{-1}(\widehat A)$ is a connected and simply connected domain. But $\widetilde A$ contains the closed curve $C$ that appears in the proof of Proposition \[prop: preliminary-realization\] and so the point $\widetilde z_{k_0}$ can be chosen to belong to $\widetilde A$. It is the same for $\widetilde z$ because one can apply Brouwer theory to the restriction of $\widetilde f^{qr}\circ T^{-p}$ to $\widetilde A$. But this implies that the orbit of $\widetilde z$ projects onto a periodic orbit included in $U_{\widehat\Gamma}$. Applying $(2)$, one deduces that $\widetilde I_{\widetilde{\mathcal F}}^{{\mathbb{Z}}}(\widetilde z)$ is equivalent to $\prod_{k\in{\mathbb{Z}}} T^k(\widetilde\gamma_{[s,t]})$.
The item (2) can be proven similarly using Proposition \[prop: irrational compact set\]. $\Box$
Let us conclude by proving Theorem \[th: horseshoe\] and remind its statement.
\[th: horseshoe2\]Let $M$ be an oriented surface, $f$ a homeomorphism of $M$ isotopic to the identity, $I$ a maximal identity isotopy of $f$ and $\mathcal F$ a foliation transverse to $I$. If there exists a point $z$ in the domain of $I$ and an integer $r{\geqslant}1$ such that the transverse trajectory $I_{\mathcal F}^{r}(z)$ has a $\mathcal F$-transverse self-intersection, then $f$ has a topological horseshoe. Moreover, the entropy of $f$ is at least equal to $\log 4/3r$.
Let us apply Proposition \[prop: horseshoe-universal\]: there exist a covering transformation $T$ of the universal covering space $\widetilde{\mathrm{dom}}(I)$ and for every $q{\geqslant}3$ and $2{\leqslant}p<q$ a compact set $\widetilde Z_{p,q}\subset\widetilde{\mathrm{dom}}(I)$, invariant by $\widetilde f^{qr}\circ T^{-p}$ such that
- the restriction of $\widetilde f^{qr}\circ T^{-p}$ to $\widetilde Z_{p,q}$ is an extension of the Bernouilli shift $\sigma :\{1,2\}^{{\mathbb{Z}}}\to \{1,2\}^{{\mathbb{Z}}}$;
- every $s$-periodic sequence of $\{0,1\}^{{\mathbb{Z}}}$ has a preimage by the factor map $\widetilde H_{p,q}:\widetilde Z_{p,q}\to \{1,2\}^{{\mathbb{Z}}}$ that is a $s$-periodic point of $\widetilde f^{qr}\circ T^{-p}$.
The covering projection $\widetilde\pi: \widetilde{\mathrm{dom}}(I)\to \mathrm{dom}(I)$ induces a semi-conjugacy between $\widetilde f^{qr}\circ T^{-p}\vert _{\widetilde Z_{p,q}}$ and $f^{qr}\vert _{\widetilde\pi(\widetilde Z_{p,q})}$. The set $\widetilde Z_{p,q}$ being compact, projects onto a compact subset $Z_{p,q}$ of $\mathrm{dom}(I)$ and every point in $Z_{p,q}$ has finitely many lifts in $\widetilde Z_{p,q}$, with an uniform upper bound. The set $Z_{p,q}$ is a topological horseshoe, as defined in the introduction.
One could have applied Proposition \[prop: horseshoe-annular\]: for every $q{\geqslant}2$, there exists a topological horseshoe $Z_{q}$, image by the covering projection $\mathring \pi: \mathring{\mathrm{dom}}(I)\to \mathrm{dom}(I)$ of the set $\mathring Z_{q}$ defined by Proposition \[prop: horseshoe-annular\]. One has $$h(f){\geqslant}{1\over qr} h(f^{qr}\vert_{Z_q})={1\over qr} h(\mathring f^{qr}\vert_{\mathring Z_q})={1\over qr}\log (2q-2).$$ Noting that the function $$q\mapsto {1\over q}\log (2q-2)$$ reaches its maximum for $q=3$, one concludes that $$h(f){\geqslant}{\log 4\over 3r}.$$
Paths on the sphere with no $\mathcal F$ transverse self-intersection and applications
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Definitions
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We suppose in this section that $\mathcal F$ is a singular oriented foliation on the sphere ${\mathbb{S}}^2$. The [*natural lift*]{} of a loop $\Gamma:{\mathbb{T}}^1\to{\mathbb{S}}^2$ is the path $\gamma:{\mathbb{R}}\to{\mathbb{S}}^2$ such that $\gamma(t)=\Gamma(t+{\mathbb{Z}})$. One can define a [*dual function*]{} $\delta$ defined up to an additive constant on ${\mathbb{S}}^2\setminus\Gamma$ as follows: for every $z$ and $z'$ in ${\mathbb{R}}^2\setminus\Gamma$, the difference $\delta(z')-\delta(z)$ is the algebraic intersection number $\Gamma\wedge \gamma'$ where $\gamma'$ is any path from $z$ to $z'$. Say that $\Gamma$ is [*transverse*]{} to $\mathcal F$ if it is the case for $\gamma$. In that case $\delta$ decreases along each leaf with a jump at every intersection point. One proves easily that $\delta$ is bounded and that the space of leaves that meet $\Gamma$, furnished with the quotient topology is a (possibly non Hausdorff) one dimensional manifold (see [@LeCalvezTal]). In particular, there exists $n_0$ such that $\Gamma$ meets each leaf at most $n_0$ times. If $\Gamma$ is a simple loop (which means that it is injective), $\delta$ takes exactly two values and $\Gamma$ meets each leaf at most once. The converse is obviously true. In that case we will write $U_{\Gamma}$ for the union of leaves that meet $\Gamma$. It is an open annulus. We can define the dual function of a multi-loop $\Gamma=\sum_{1{\leqslant}i{\leqslant}p}\Gamma_i$ and have similar results in case each $\Gamma_i$ is transverse to $\mathcal F$.
Let $J_1$ and $J_2$ be two real intervals. We will say that $J_1$ is [*on the left*]{} of $J_2$ (equivalently $J_2$ is [*on the right*]{} of $J_1$) if the lower end of $J_2$ is not smaller than the lower end of $J_1$ and the upper end of $J_2$ is not smaller that the upper end of $J_1$. In particular $J_1$ is on the left and on the right of itself.
Let $\gamma:J\to\mathrm {dom} ({\mathcal F})$ be a transverse path and $\Gamma_0$ a transverse loop.
- The path $\gamma$ [*is drawn on*]{} $\Gamma_0$ if it is equivalent to a sub-path of the natural lift $\gamma_0$ of $\Gamma_0$;
- The path $\gamma$ [*draws*]{} $\Gamma_0$ if there exist $a<b$ in $J$ and $t\in{\mathbb{R}}$ such that $\gamma\vert_{[a,b]}$ is equivalent to $\gamma_0\vert_{[t,t+1]}$.
- The path $\gamma$ [*draws infinitely*]{} $\Gamma_0$ if there exist $c\in J$ and $t\in{\mathbb{R}}$ such that $\gamma\vert_{(\inf J,c]}$ is equivalent to $\gamma_0\vert_{(-\infty, t]}$, or if $\gamma\vert_{[c,\sup J)}$ is equivalent to $\gamma_0\vert_{[t,+\infty)}$.
- The path $\gamma$ [*exactly draws*]{} $\Gamma_0$ if it draws $\Gamma_0$ and is drawn on $\Gamma_0$.
- The path $\gamma$ [*exactly draws infinitely* ]{} $\Gamma_0$ if it draws infinitely $\Gamma_0$ and is drawn on $\Gamma_0$.
In case $\Gamma_0$ is a transverse simple loop, $\gamma$ draws $\Gamma_0$ if there exist $a<b$ in $J$ such that $\gamma\vert_{[a,b]}$ is included in $U_{\Gamma_0}$ and meets every leaf at least once (or equivalently meets a leaf twice). One proves easily that a path that draws a transverse loop draws a transverse simple loop (see [@LeCalvezTal]).
Let $\gamma:J\to\mathrm {dom} ({\mathcal F})$ be a transverse path and $\Gamma_0$ a transverse simple loop. We set $$J_{\Gamma_0}=\{t\in J\,\vert\, \gamma(t)\in U_{\Gamma_0}\}.$$
- A connected component $J_0$ of $J_{\Gamma_0}$ is a [*drawing component*]{} if $\gamma\vert_{J_0}$ draws $\Gamma_0$.
- A connected component $J_1$ of $J_{\Gamma_0}$ is a [*crossing component*]{} and $\gamma\vert_{J_0}$ [*crosses*]{} $\Gamma_0$ if both ends $a$, $b$ of $J_1$ are in $J$ and $\gamma(a)$ and $\gamma(b)$ belong to different components of ${\mathbb{S}}^2\setminus U_{\Gamma}$.
- The path $\gamma$ [*crosses*]{} $\Gamma_0$ is there is at least one crossing component.
Intersection of paths and loops
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\[prop:drawing-crossing\] Suppose that $\gamma:J\to\mathrm {dom} ({\mathcal F})$ is a transverse path with no $\mathcal F$-transverse self-intersection. Then:
1. if $\gamma$ draws a transverse simple loop $\Gamma_0$, there exists a unique drawing component of $J_{\Gamma_0}$;
2. if $\gamma$ draws and crosses a transverse simple loop $\Gamma_0$, there exists a unique crossing component of $J_{\Gamma_0}$ and it is the drawing component of $J_{\Gamma_0}$;
3. if $\gamma$ draws a transverse simple loop $\Gamma_0$ and does not cross it, then the drawing component coincides with $J$ in a neighborhood of at least one end of $J$;
4. if $\gamma$ draws two non-equivalent transverse simple loops $\Gamma_0$ and $\Gamma_1$, then the drawing component of $J_{\Gamma_0}$ is on the right of the drawing component of $J_{\Gamma_1}$ or on its left.
Let us suppose that $\gamma$ draws and crosses $\Gamma_0$. Denote $\gamma_0$ the natural lift of $\Gamma_0$. Consider a drawing component $J_0$ and a crossing component $J_1$. Let us prove that $J_0=J_1$ (See Figure \[figure\_crossingcomponent\] for a depiction of construction below). If not, one can suppose for instance that $J_1=(a_1,b_1)$ is on the right of $J_0$. This implies that the upper end $b_0$ of $J_0$ is finite, belongs to $J$, and that $\gamma(b_0)\not\in U_{\Gamma_0}$. Suppose first that $\gamma(b_0)$ and $\gamma(a_1)$ belong to the same connected component of ${\mathbb{S}}^2\setminus U_{\Gamma_0}$. In that case, the leaves $ \phi_{\gamma(b_0)}$ and $\phi_{\gamma(a_1)}$ are on the frontier of $U_{\Gamma_0}$, and this annulus is locally on the right of the first leaf and on the left of the second one. This implies that for every $c_0\in J_0$ and $c_1\in J_1$, the paths $\gamma\vert_{[c_0,b_0)}$ and $\gamma\vert_{(a_1, c_1]}$ meet every leaf of $U_{\Gamma_0}$ finitely many times (there is no spiraling phenomena). Moreover, one can choose $c_0$ such that every leaf is met exactly once by $\gamma\vert_{[c_0,b_0)}$. Consequently, there exist $t_0$, $t_1$, $t'_1$, with $t_0{\leqslant}t_1<t_0+1$ such that $\gamma\vert_{[c_0,b_0)}$ is equivalent to $\gamma_0\vert_{[t_0,t_0+1)}$ and $\gamma\vert_{(a_1,b_1)}$ is equivalent to $\gamma_0\vert_{(t_1,t'_1)}$. It implies that there exist $d_0\in (c_0,b_0)$ and $d_1\in (a_1,b_1)$ such that $\gamma(d_0)=\gamma(d_1)$. Note now that $\gamma_{[c_0,b_0]}$ and $\gamma_{[a_1, b_1]}$ have a $\mathcal F$-transverse intersection at $\gamma(d_0)=\gamma(d_1)$. Indeed, if we lift these two paths to the universal covering space into paths $\widetilde\gamma_0=[c_0,b_0]\to\widetilde{\mathrm {dom}}({\mathcal F})$ and $\widetilde\gamma_1:[a_1,b_1]\to\widetilde{\mathrm {dom}}({\mathcal F})$, such that $\widetilde\gamma_0(d_0)=\widetilde\gamma_1(d_1)$, these paths are transverse to the lifted foliation $\widetilde{\mathcal F}$. Moreover, one of the leaves $\phi_{\widetilde\gamma_0(c_0)}$, $\phi_{\widetilde\gamma_1(a_1)}$ is above the other one relative to $\phi_{\widetilde\gamma_0(d_0)}$, one of the leaves $\phi_{\widetilde\gamma_0(b_0)}$, $\phi_{\widetilde\gamma_1(b_1)}$ is above the other one relative to $\phi_{\widetilde\gamma_0(d_0)}$, and if $\phi_{\widetilde\gamma_0(c_0)}$ is below (respectively above) $\phi_{\widetilde\gamma_1(a_1)}$, then $\phi_{\widetilde\gamma_0(b_0)}$ is below (respectively above) $\phi_{\widetilde\gamma_1(b_1)}$. This implies that $\gamma$ has a $\mathcal F$-transverse self-intersection, which contradicts the hypothesis.
In the case where $\gamma(b_0)$ and $\gamma(a_1)$ do not belong to the same connected component of ${\mathbb{S}}^2\setminus U_{\Gamma}$, there exists another crossing component $J_2=(a_2,b_2)$ of $J_{\Gamma_0}$ which is on the right of $J_0$ and on the left of $J_1$ and such that $\gamma(b_0)$ and $\gamma(a_2)$ belong to the same connected component of ${\mathbb{S}}^2\setminus U_{\Gamma}$. We still have a contradiction. One deduces that $(1)$ is satisfied in case $\gamma$ crosses $\Gamma$ and that $(2)$ is satisfied.
![Proposition \[prop:drawing-crossing\]. Depiction of why item (2) holds. Leafs of the foliation are drawn in red, transversal paths in black, and the annulus $U_{\Gamma_0}$ is the shaded region.[]{data-label="figure_crossingcomponent"}](figure_crossingcomponent.pdf){height="48mm"}
To get $(1)$, it remains to study the case when $\gamma$ does not cross $\Gamma_0$. Suppose that $J_0$ and $J_1$ are two drawing components of $J_{\Gamma_0}$ and that $J_1$ is on the right of $J_0$. The upper end $b_0$ of $J_0$ and the lower end $a_1$ of $J_1$ are finite, belong to $J$, and none of the points $\gamma(b_0)$, $\gamma(a_1)$ belong to $U_{\Gamma_0}$. The fact that $\gamma$ does not cross $\Gamma_0$ implies that $\gamma(b_0)$ and $\gamma(a_1)$ belong to the same connected component of ${\mathbb{S}}^2\setminus U_{\Gamma_0}$. Here again, for every $c_0\in J_0$ and $c_1\in J_1$, the paths $\gamma\vert_{[c_0,b_0)}$ and $\gamma\vert_{(a_1, c_1]}$ meet every leaf of $U_{\Gamma_0}$ finitely many times and one can choose $c_0$ and $c_1$ such that every leaf is met exactly once by $\gamma_{[c_0,b_0)}$ and $\gamma\vert_{(a_1,c_1]}$. Consequently, there exist $t_0$, $t_1$, with $t_0{\leqslant}t_1<t_0+1$ such that $\gamma\vert_{[c_0,b_0)}$ is equivalent to $\gamma_0\vert_{[t_0,t_0+1)}$ and $\gamma\vert_{(a_1,c_1]}$ is equivalent to $\gamma_0\vert_{(t_1,t_1+1]}$. It implies that there exist $d_0\in (c_0,b_0)$ and $d_1\in (a_1,c_1)$ such that $\gamma(d_0)=\gamma(d_1)$. Like in the first situation, we observe that $\gamma\vert_{[c_0,b_0]}$ and $\gamma\vert_{[a_1, c_1]}$ have a $\mathcal F$-transverse intersection at $\gamma(d_0)=\gamma(d_1)$. This contradicts the hypothesis.
Let us prove $(3)$. Suppose that $\gamma$ draws a transverse simple loop $\Gamma_0$ and does not cross it. If the drawing component $J_0$ of $J_{\Gamma_0}$ does not coincide with $J$ in a neighborhood of at least one end of $J$, then the two ends $a_0$ and $b_0$ of $J_0$ are finite and $\gamma(a_0)$ and $\gamma(b_0)$ do not belong to $U_{\Gamma}$. More precisely $\gamma(a_0)$ and $\gamma(b_0)$ belong to the same connected component of ${\mathbb{S}}^2\setminus U_{\Gamma}$ because $\gamma$ does not cross $\Gamma_0$. Here again, for every $c_0\in J_0$ the paths $\gamma\vert_{(a_0,c_0]}$ and $\gamma\vert_{[c_0, b_0)}$ meet every leaf of $U_{\Gamma_0}$ finitely many times and one can choose $c_0$ and $c'_0$ in $J_0$ such that every leaf is met exactly once by $\gamma\vert_{(a_0,c_0]}$ and $\gamma\vert_{[c'_0, b_0)}$. Consequently, there exist $t_0$, $t'_0$, with $t'_0{\leqslant}t_0< t'_{0}+1$ such that $\gamma\vert_{(a_0,c_0]}$ is equivalent to $\gamma_0\vert_{(t_0,t_0+1]}$ and $\gamma\vert_{[c'_0,b_0)}$ is equivalent to $\gamma_0\vert_{[t'_0,t'_0+1)}$. It implies that there exist $d_0\in (a_0,c_0)$ and $d'_0\in (c'_0,b_0)$ such that $\gamma(d_0)=\gamma(d'_0)$. Note now that $\gamma\vert_{[a_0,c_0]}$ and $\gamma\vert_{[c'_0,b_0]}$ have a $\mathcal F$-transverse intersection at $\gamma(d_0)=\gamma(d'_0)$. This contradicts the hypothesis.
It remains to prove (4). Suppose that $\gamma$ draws two non equivalent transverse simple loops $\Gamma_0$, $\Gamma_1$. Denote $J_0$, $J_1$ the drawing components of $J_{\Gamma_0}$, $J_{\Gamma_1}$ respectively. The loops are not equivalent, so there is a leaf that is met by $\gamma\vert _{J_{\Gamma_0}}$ and not by $\gamma\vert _{J_{\Gamma_1}}$ and a leaf that is met by $\gamma\vert _{J_{\Gamma_1}}$ and not by $\gamma\vert _{J_{\Gamma_0}}$. Consequently, none of the inclusions $J_0\subset J_1$ or $J_1\subset J_0$ occurs and the conclusions follows.
\[cor:decomposition\] Suppose that $\gamma:[a,b]\to\mathrm {dom} ({\mathcal F})$ is a transverse path with no $\mathcal F$-transverse self-intersection such that $\phi_{\gamma(a)}=\phi_{\gamma(b)}$. Then, there exists a subdivision $(c_i)_{0{\leqslant}i{\leqslant}r}$ of $[a,b]$ such that:
1. $\phi_{\gamma(c_i)}=\phi_{\gamma(a)}$ for every $i\in\{0,\dots,r\}$;
2. $\phi_{\gamma(t)}\not=\phi_{\gamma(t')}$ if $c_i{\leqslant}t<t'<c_{i+1}$ and $i\in\{0,\dots,r-1\}$.
Suppose that $\gamma$ draws a transverse simple loop $\Gamma_0$ such that $\phi_{\gamma(a)}\not\subset U_{\Gamma_0}$. It must cross $\Gamma_0$ by assertion $(3)$ of Proposition \[prop:drawing-crossing\]. But it must have at least two crossing components because $\phi_{\gamma(a)}=\phi_{\gamma(b)}$ and this contradicts assertion $(2)$ of Proposition \[prop:drawing-crossing\]. Consequently, if $a'<b'$ in $[a,b]$ satisfy $
\phi_{\gamma(a')}=\phi_{\gamma(b')}$, then there exists $c'\in[a',b']$ such that $
\phi_{\gamma(c')}=\phi_{\gamma(a)}$. Indeed, there exist $a''<b''$ in $[a',b']$ such that $
\phi_{\gamma(a'')}=\phi_{\gamma(b'')}$ and such that $\phi_{\gamma(t)}\not=\phi_{\gamma(t')}$ if $a''{\leqslant}t<t'<b''$ (see [@LeCalvezTal].) The path $\gamma\vert_{[a'',b'']} $ defines naturally a simple loop $\Gamma_0$ and $\gamma$ draws $\Gamma_0$. So, there exists $c''\in[a'',b'']$ such that $
\phi_{\gamma(c'')}=\phi_{\gamma(a)}$.
The path $\gamma$ defines naturally a transverse loop met by $\phi_{\gamma(a)}$. As reminded in the beginning of the section, this loop meets every leaf finitely many times. Consequently, there exists a subdivision $(c_i)_{0{\leqslant}i{\leqslant}r}$ of $[a,b]$ such that:
- $\phi_{\gamma(c_i)}=\phi_{\gamma(a)}$ for every $i\in\{0,\dots,r\}$;
- $\phi_{\gamma(t)}\not=\phi_{\gamma(a)}$ if $c_i{\leqslant}t<c_{i+1}$ and $i\in\{0,\dots,r-1\}$.
The argument above tells us that $\phi_{\gamma(t)}\not=\phi_{\gamma(t')}$ if $c_i{\leqslant}t<t'<c_{i+1}$.
Note that the conclusion of Corollary \[cor:decomposition\] does not say that the paths $\gamma\vert_{[c_i,c_{i+1}]}$ must be equivalent.
Fix a transverse path $\gamma:J\to\mathrm {dom} ({\mathcal F})$ with no $\mathcal F$-transverse self-intersection. Suppose that $\gamma$ draws a transverse simple loop $\Gamma_0$ but is not drawn on it. Denote $J_0$ the drawing component of $J_{\Gamma_0}$. If the lower end $a_0$ of $J_0$ is finite and belongs to $J$, then $\gamma(a_0)$ does not belong to $U_{\Gamma_0}$. We denote $K^{\alpha}_{\Gamma_0}$ the connected component of ${\mathbb{S}}^2\setminus U_{\Gamma_0}$ that contains $\gamma(a_0)$ and $K^{\omega}_{\Gamma_0}$ the other one. Similarly, if the upper end $b_0$ of $J_0$ is finite and belongs to $J$, we denote $K^{\omega}_{\Gamma_0}$ the connected component of ${\mathbb{S}}^2\setminus U_{\Gamma_0}$ that contains $\gamma(b_0)$ and $K^{\alpha}_{\Gamma_0}$ the other one. Note that if the two ends are finite, the notations agree because $J_0$ must be a crossing component by assertion $(3)$ of Proposition \[prop:drawing-crossing\]. Note that in all cases, $$\begin{cases}\gamma(t)\in U_{\Gamma_0}\cup K^{\alpha}_{\Gamma_0},& \mathrm{if}\enskip t<a_0,\\ \gamma(t)\in U_{\Gamma_0}\cup K^{\omega}_{\Gamma_0},& \mathrm{if}\enskip t>b_0,\\ \end{cases}$$ otherwise $\gamma$ would have at least two crossing components.
\[prop:order-on-loops\]Suppose that $\gamma:[a,b]\to\mathrm {dom} ({\mathcal F})$ is a transverse path with no $\mathcal F$-transverse self-intersection that draws two transverse simple loops $\Gamma_0$ and $\Gamma_1$ and denote $J_0$, $J_1$ the drawing components of $J_{\Gamma_0}$, $J_{\Gamma_1}$ respectively. If $J_1$ is on the right of $J_0$, then $K^{\alpha}_{\Gamma_0}\subset K^{\alpha}_{\Gamma_1}$ and $K^{\omega}_{\Gamma_1}\subset K^{\omega}_{\Gamma_0}$.
We can suppose that $\Gamma_0$ and $\Gamma_1$ are not equivalent, otherwise the result is trivial as: $$J_0=J_1, \enskip K^{\alpha}_{\Gamma_0}=K^{\alpha}_{\Gamma_1}, \enskip K^{\omega}_{\Gamma_0}=K^{\omega}_{\Gamma_1}.$$ Note that, by item (4) of Proposition \[prop:drawing-crossing\], either $J_0$ is on the right of $J_1$ or the converse is true, and we are assuming the latter. By the comments made just before the statement of Proposition \[prop:order-on-loops\], we have: $$U_{\Gamma_1}\cap K^{\alpha}_{\Gamma_0}= U_{\Gamma_0}\cap K^{\omega}_{\Gamma_1}=\emptyset.$$ The set $K^{\alpha}_{\Gamma_0}$ is included in ${\mathbb{S}}^2\setminus U_{\Gamma_1}$. Being connected, it is contained in $K^{\alpha}_{\Gamma_1}$ or in $K^{\omega}_{\Gamma_1}$. Suppose that it is contained in $K^{\omega}_{\Gamma_1}$. It is a closed and open subset of $K^{\omega}_{\Gamma_1}$ because $U_{\Gamma_0}\cup K^{\alpha}_{\Gamma_0}$ is a neighborhood of $K^{\alpha}_{\Gamma_0}$ and $U_{\Gamma_0}\cap K^{\omega}_{\Gamma_1}=\emptyset$. So, $K^{\alpha}_{\Gamma_0}$ and $K^{\omega}_{\Gamma_1}$ are equal. Since the loop $\Gamma_1$ is simple, it separates the two ends of $U_{\Gamma_1}$ and intersects every leaf in $U_{\Gamma_1}$ exactly once. Therefore, every leaf $\phi\in U_{\Gamma_1}$ is adjacent to both connected components of the complement of $U_{\Gamma_1}$, and we deduce that every leaf $\phi\in U_{\Gamma_1}$ is adjacent to $K^{\alpha}_{\Gamma_0}$ but disjoint from this set. So $\phi$ meets $U_{\Gamma_0}$, because $U_{\Gamma_0}\cup K^{\alpha}_{\Gamma_0}$ is a neighborhood of $K^{\alpha}_{\Gamma_0}$, and this implies that $\phi$ is contained in $U_{\Gamma_0}$. One deduces that $U_{\Gamma_1}$ is included in $U_{\Gamma_0}$, which is impossible because $\Gamma_0$ and $\Gamma_1$ are not equivalent. In conclusion, one gets the inclusion $K^{\alpha}_{\Gamma_0}\subset K^{\alpha}_{\Gamma_1}$. The proof of the other inclusion is similar.
Applications
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\[prop:non-wandering\] Let $f$ be an orientation preserving homeomorphism of ${\mathbb{S}}^2$ with no topological horseshoe. Let $I$ be a maximal isotopy of $f$ and $\mathcal F$ a foliation transverse to $I$. If $z\in\mathrm{dom}(I)$ is a non-wandering point of $f$, then:
- either $I^{{\mathbb{Z}}}_{\mathcal F} (z)$ never meets a leaf twice,
- or $I^{{\mathbb{Z}}}_{\mathcal F} (z)$ draws a unique transverse simple loop $\Gamma_0$ (up to equivalence) and exactly draws it.
Suppose that $I^{{\mathbb{Z}}}_{\mathcal F} (z)$ meets a leaf twice. In that case, it draws a transverse loop $\Gamma_0$, that can be supposed simple by Corollary \[cor:decomposition\]. If $I^{{\mathbb{Z}}}_{\mathcal F} (z)$ is not drawn on $\Gamma_0$, one can find integers $m<n$ such that $\gamma=I_{\mathcal F}^{{\mathbb{Z}}} (z)\vert_{[m,n]}=I_{\mathcal F}^{n-m}(f^m(z))$ draws $\Gamma_0$ and is not drawn on $\Gamma_0$: there exists $t\in[m,n]$ such that $\phi_{\gamma(t)}\not\subset U_{\Gamma_0}$. By Proposition \[prop:stability\], there exists a neighborhood $W$ of $z$ such that for every $z'\in W$, $\gamma$ is a sub-path of $I_{\mathcal F}^{{\mathbb{Z}}} (z)\vert_{[m-1,n+1]}$. Consequently, $I_{\mathcal F}^{{\mathbb{Z}}} (z')\vert_{[m-1,n+1]}$ draws $\Gamma_0$ and is not drawn on $\Gamma_0$. The point $z$ being non-wandering, one can find $z'\in W$ and $l>n-m+2$ such that $f^{l}(z')\in W$. So $I_{\mathcal F}^{{\mathbb{Z}}} (z')\vert_{[m-1+l,n+1+l]}$ draws $\Gamma_0$ and is not drawn on $\Gamma_0$. More precisely, there exists an interval $J_0\subset [m-1,n+1]$ and an interval $J_1\subset [m-1+l,n+1+l]$ such that $I_{\mathcal F}^{{\mathbb{Z}}} (z')\vert_{J_0}$ and $I_{\mathcal F}^{{\mathbb{Z}}} (z')\vert_{J_1}$ are equivalent to $\gamma$. There exists at least two drawing components, separated by the point $t_0\in J_0$ corresponding to $t$, if $t$ is at the right of the drawing component of $\gamma$, separated by the point $t_1\in J_1$ corresponding to $t$, if $t$ is at the left of the drawing component of $\gamma$. By assertion $(1)$ of Proposition \[prop:drawing-crossing\], we deduce that $I^{{\mathbb{Z}}}_{\mathcal F} (z)$ has a $\mathcal F$-transverse self intersection. This contradicts Theorem \[th: horseshoe\].
The first case only appears in case $\alpha(z)$ and $\omega(z)$ are included in $\mathrm{fix}(I)$. We will define $$\mathrm{ne}(I)=\{z\in \mathrm{dom}(I) \, \vert \, \alpha(z)\cup\omega(z)\not\subset\mathrm{fix}(I)\}.$$ Note that if $z\in \mathrm{ne}(I)$, there exists at least one leaf that is met infinitely many times by $I^{{\mathbb{Z}}}_{\mathcal F} (z)$. By Corollary \[cor:decomposition\], $I^{{\mathbb{Z}}}_{\mathcal F} (z)$ draws at least one transverse simple loop. Moreover, this loop is unique and drawn infinitely if $z\in\Omega(f)$.
We can generalize the previous result.
\[prop:Birkhoff-cycles\]Let $f$ be an orientation preserving homeomorphism of ${\mathbb{S}}^2$ with no topological horseshoe. Let $I$ be a maximal isotopy of $f$ and $\mathcal F$ a foliation transverse to $I$. Let $(z_i)_{i\in{\mathbb{Z}}/r{\mathbb{Z}}}$ be a Birkhoff cycle in $\mathrm{ne}(I)$. Then there exists a transverse simple loop $\Gamma$ such that, for every $i\in {\mathbb{Z}}/r{\mathbb{Z}}$, $\Gamma$ is the unique simple loop (up to equivalence) drawed by $I^{{\mathbb{Z}}}_{\mathcal F} (z_i)$.
In the case where there exists a periodic orbit that contains all the $z_i$, we can apply Proposition \[prop:non-wandering\]. Otherwise, taking a subsequence if necessary, one can suppose that for every $i\in{\mathbb{Z}}/r{\mathbb{Z}}$, the point $z_{i+1}$ is not the image of $z_i$ by a positive iterate of $f$. Likewise, we may assume that $r{\geqslant}2$, otherwise the unique point $z_0$ must be non-wandering and the result follows again by Proposition \[prop:non-wandering\]. Note that, since each $z_i\in \mathrm{ne}(I)$, then each $I^{{\mathbb{Z}}}_{\mathcal{F}}(z_i)$ draws at least a simple transverse loop. It suffices to show that, if $0{\leqslant}i<j{\leqslant}r-1$ and $I^{{\mathbb{Z}}}_{\mathcal{F}}(z_i)$ draws the loop $\Gamma_i$ while $I^{{\mathbb{Z}}}_{\mathcal{F}}(z_j)$ draws the loop $\Gamma_j$, then $\Gamma_i$ and $\Gamma_j$ are equivalent, because, as $r{\geqslant}2$, this also implies that $I^{{\mathbb{Z}}}_{\mathcal{F}}(z_i)$ draws a unique simple loop up to equivalence.
Let us then choose for every $i$ a simple transverse loop $\Gamma_i$ that is drawn by $I^{{\mathbb{Z}}}_{\mathcal{F}}(z_i)$. The fact that $(z_i)_{i\in{\mathbb{Z}}/r{\mathbb{Z}}}$ is a Birkhoff cycle implies that, if we choose for every $i$ a neighborhood $W_i$ of $z_i$ sufficiently small, there exists $z'_i\in W_i$ such that $f^{n_i}(z'_i)\in W_{i+1}$, where $n_i$ can be supposed arbitrarily large. In particular, $I^{{\mathbb{Z}}}_{\mathcal F} (z'_i)$ draws $\Gamma_i$ and $\Gamma_{i+1}$ and moreover, the drawing component of $\Gamma_{i+1}$ is on the right of the drawing component of $\Gamma_{i}$. Suppose for a contradiction that there exists $i'$ such that $\Gamma_{i'}\not=\Gamma_{i'+1}$. As explained in Proposition \[prop:order-on-loops\], for each $i$ such that $\Gamma_{i}\not=\Gamma_{i+1}$, exactly one of the two connected components of the complement of $U_{\Gamma_i}$ meets $\Gamma_{i+1}$, which we denote it $K^{\omega}_{i\to i+1}$, and exactly one of the two connected components of the complement of $U_{\Gamma_{i+1}}$ meets $\Gamma_{i}$, which we denote it $K^{\alpha}_{i\to i+1}$. We can extend the above sets in a unique way to obtain two global families $(K^{\alpha}_{i\to i+1})_{i\in{\mathbb{Z}}/r{\mathbb{Z}}}$ and $(K^{\omega}_{i\to i+1})_{i\in{\mathbb{Z}}/r{\mathbb{Z}}}$ by imposing that, whenever $$\Gamma_i=\Gamma_{i+1}\Rightarrow K^{\alpha}_{i\to i+1}=K^{\alpha}_{i-1\to i} \enskip\mathrm{and} \enskip K^{\omega}_{i\to i+1}=K^{\omega}_{i+1\to i+2}.$$
There exists $i$ such that $K^{\alpha}_{i-1\to i}=K^{\omega}_{i\to i+1}$.
If $K^{\alpha}_{i-1\to i}\not=K^{\omega}_{i\to i+1}$ for every $i\in{\mathbb{Z}}/rZ$, then, the sequence $(K^{\alpha}_{i\to i+1})_{i\in{\mathbb{Z}}/r{\mathbb{Z}}}$ is non decreasing while the sequence $(K^{\omega}_{i\to i+1})_{i\in{\mathbb{Z}}/r{\mathbb{Z}}}$ is not increasing by Proposition \[prop:order-on-loops\]. So the two sequences are constant, as is the sequence $(U_{\Gamma_i})_{i\in Z/r{\mathbb{Z}}}$ because $U_{\Gamma_i}={\mathbb{S}}^2\setminus \left(K^{\omega}_{i\to i+1}\cup K^{\alpha}_{i\to i+1}\right)$. This contradicts our assumption that there exists $i$ such that $\Gamma_{i}\not=\Gamma_{i+1}$.
We deduce that there exists a transverse simple loop $\Gamma$, a connected component $K$ of ${\mathbb{S}}^2\setminus U_{\Gamma}$ and $i_0{\leqslant}i{\leqslant}i_1$ such that
- $\Gamma_{i}=\Gamma$ for every $i\in\{i_0,\dots,i_1\}$;
- $\Gamma_{i_0-1}\not=\Gamma$;
- $\Gamma_{i_1+1}\not=\Gamma$;
- the point $z'_{i_0-1}$ draws $\Gamma_{i_0-1}$ and $\Gamma$ and the drawing component of the first loop is on the left of the drawing component of the second one;
- the point $z'_{i_1}$ draws $\Gamma$ and $\Gamma_{i_1+1}$ and the drawing component of the first loop is on the left of the drawing component of the second one;
- $K^{\alpha}_{i_0-1\to i_0}=K^{\omega}_{i_1\to i_1+1}=K$.
For every $i\in{\mathbb{Z}}/p{\mathbb{Z}}$ we denote:
- $(a_i,b_i)$ the drawing component of $I^{{\mathbb{Z}}}_{\mathcal F} (z_i)$ with $\Gamma_i$,
- $(a'_i,b'_i)$ the drawing component of $I^{{\mathbb{Z}}}_{\mathcal F} (z'_i)$ with $\Gamma_i$,
- $(a''_i,b''_i)$ the drawing component of $I^{{\mathbb{Z}}}_{\mathcal F} (z'_{i-1})$ with $\Gamma_i$.
We know that $$-\infty<a''_{i_0}, \enskip b'_{i_1}<+\infty, \enskip I^{{\mathbb{Z}}}_{\mathcal F} (z'_{i_0-1})(a''_{i_0})\in K, \enskip I^{{\mathbb{Z}}}_{\mathcal F} (z'_{i_1})(b'_{i_1})\in K.$$
Suppose that $i_0<i_1$. Then $b_i=+\infty$ if $i_0{\leqslant}i <i_1$ and $a_i=-\infty$ if $i_0<i{\leqslant}i_1$.
Consider $i\in\{i_0,\dots, i_1-1\}$ and suppose that $b_i<+\infty$. Fix $c_i$ in $(a_i, b_i)$ and $c_{i+1}$, $c'_{i+1}$ in $(a_{i+1}, b_{i+1})$ satisfying $c_{i+1}<c'_{i+1}$, such that $I^{{\mathbb{Z}}}_{\mathcal F} (z_i)\vert_{[c_i, b_i]}$ and $I^{{\mathbb{Z}}}_{\mathcal F} (z_{i+1})\vert_{[c_{i+1}, c'_{i+1}]}$ draw $\Gamma$. If the neighborhoods $W_{i}$ and $W_{i+1}$ are chosen sufficiently small and $n_i$ sufficiently large, there exist $c'_i< d'_i<e'_i<f'_i$ such that $I^{{\mathbb{Z}}}_{\mathcal F} (z'_i)\vert_{[c'_i, d'_i]}$ is equivalent to $I^{{\mathbb{Z}}}_{\mathcal F} (z_i)\vert_{[c_i, b_i]}$ and $I^{{\mathbb{Z}}}_{\mathcal F} (z'_i)\vert_{[e'_i, f'_i]}$ is equivalent to $I^{{\mathbb{Z}}}_{\mathcal F} (z_{i+1})\vert_{[c_{i+1}, c'_{i+1}]}$. Therefore both $[c'_i, d'_i]$ and $[e'_i, f'_i]$ intersect a drawing component of $I^{{\mathbb{Z}}}_{\mathcal F} (z'_i)$ with $\Gamma$. But since $I^{{\mathbb{Z}}}_{\mathcal F} (z_i)(b_i)$ does not belong to $U_{\Gamma}$, $I^{{\mathbb{Z}}}_{\mathcal F} (z'_i)(d'_i)$ also does not belong to $U_{\Gamma}$, and therefore there must be at least two drawing component of $I^{{\mathbb{Z}}}_{\mathcal F} (z'_i)$ with $\Gamma$, a contradiction with the assertion $(1)$ of Proposition \[prop:drawing-crossing\]. One proves in the same way the second statement.
There exists $i\in\{i_0,\dots, i_1\}$ such that:
- $-\infty<a''_i$;
- $ I^{{\mathbb{Z}}}_{\mathcal F} (z'_{i-1})(a''_{i})\in K$;
- $I^{{\mathbb{Z}}}_{\mathcal F} (z'_{i-1})\vert_{(a''_i, b''_i)}$ meets every leaf of $U_{\Gamma}$ at least twice.
Suppose first that $i_0=i_1$. We know that $a''_{i_0}$ is finite and that $I^{{\mathbb{Z}}}_{\mathcal F} (z'_{i_0-1})(a''_{i_0})\in K$. We will prove that $I^{{\mathbb{Z}}}_{\mathcal F} (z_{i_0})\vert_{(a_{i_0},b_{i_0})}$ meets every leaf of $U_{\Gamma}$ at least twice. It will also be the case for $I^{{\mathbb{Z}}}_{\mathcal F} (z'_{i_0-1})\vert_{(a''_{i_0}, b''_{i_0})}$ if the neighborhood $W_{i_0}$ is chosen sufficiently small and $n_{i_0-1}$ sufficiently large. So, the lemma will be proved. In fact we will prove that $I^{{\mathbb{Z}}}_{\mathcal F} (z_{i_0})$ is drawn on $\Gamma$. The point $z_{i_0}$ belonging to $\mathrm{ne}(f)$, $I^{{\mathbb{Z}}}_{\mathcal F} (z_{i_0})$ will draw infinitely $\Gamma$.
Suppose that $a_{i_0}$ is finite. In that case, $a'_{i_0}$ is also finite and $I^{{\mathbb{Z}}}_{\mathcal F} (z'_{i_0-1})(a''_{i_0})$, $I^{{\mathbb{Z}}}_{\mathcal F} (z_{i_0})(a_{i_0})$ and $I^{{\mathbb{Z}}}_{\mathcal F} (z'_{i_0})(a' _{i_0})$ are on the same leaf. One deduces that $I^{{\mathbb{Z}}}_{\mathcal F} (z'_{i_0})(a' _{i_0})\in K$. By hypothesis, one knows that $I^{{\mathbb{Z}}}_{\mathcal F} (z'_{i_0})(b' _{i_0})\in K$. This contradicts the assertion $(3)$ of Proposition \[prop:drawing-crossing\]. Suppose that $b_{i_0}$ is finite. In that case, $b''_{i_0}$ is also finite and $I^{{\mathbb{Z}}}_{\mathcal F} (z'_{i_0-1})(b''_{i_0})$, $I^{{\mathbb{Z}}}_{\mathcal F} (z_{i_0})(b_{i_0})$ and $I^{{\mathbb{Z}}}_{\mathcal F} (z'_{i_0})(b' _{i_0})$ are on the same leaf. One deduces that $I^{{\mathbb{Z}}}_{\mathcal F} (z'_{i_0})(b'' _{i_0})\in K$. By hypothesis, one knows that $I^{{\mathbb{Z}}}_{\mathcal F} (z'_{i_0})(a'' _{i_0})\in K$. We get the same contradiction. In conclusion, $I^{{\mathbb{Z}}}_{\mathcal F} (z_{i_0})$ is drawn on $\Gamma$.
Suppose now that $i_0<i_1$. We have to study the case where $I^{{\mathbb{Z}}}_{\mathcal F} (z_{i_0})\vert_{(a_{i_0},b_{i_0})}$ does not meet every leaf of $U_{\Gamma}$ at least twice. We have seen in the previous lemma that $b_{i_0}$ is infinite. So $a_{i_0}$ is finite. In that case, $a'_{i_0}$ is also finite and $I^{{\mathbb{Z}}}_{\mathcal F} (z'_{i_0-1})(a''_{i_0})$, $I^{{\mathbb{Z}}}_{\mathcal F} (z_{i_0})(a_{i_0})$ and $I^{{\mathbb{Z}}}_{\mathcal F} (z'_{i_0})(a' _{i_0})$ are on the same leaf and so $I^{{\mathbb{Z}}}_{\mathcal F} (z'_{i_0})(a' _{i_0})\in K$. Replacing $i_{0}$ by $i_0+1$, all what was done above is still valid. If $I^{{\mathbb{Z}}}_{\mathcal F} (z_{i_0+1})$ is drawn on $\Gamma$, the lemma will be valid with $i_0+1$. If not, we go to the following index. The process will stop until reaching $i_1$.
By a similar reasoning, there exists $j\in\{i_0,\dots, i_1\}$ such that
- $b'_j<+\infty$,
- $ I^{{\mathbb{Z}}}_{\mathcal F} (z'_{j})(b'_{j})\in K$,
- $I^{{\mathbb{Z}}}_{\mathcal F} (z'_{j})\vert_{(a'_j, b'_j)}$ meets every leaf of $U_{\Gamma}$ at least twice.
We set $\gamma'_i= I^{{\mathbb{Z}}}_{\mathcal F} (z'_{i})$ and $\gamma'_j= I^{{\mathbb{Z}}}_{\mathcal F} (z'_{j})$ and denote $\gamma$ the natural lift of $\Gamma$. Here again, for every $c''_i\in (a''_i, b''_i)$ and $c'_j\in (a'_j, b'_j)$, the paths $gamma'_i\vert_{(a''_i, c''_i]}$ and $\gamma'_j\vert_{[c'_j, b'_j)}$ meet every leaf of $U_{\Gamma}$ finitely many times and one can choose $c''_i$ and $c'_j$ such that every leaf is met exactly twice by $\gamma'_i\vert_{(a''_i, c''_i]}$ and by $\gamma'_j\vert_{[c'_j, b'_j)}$. Consequently, there exist $t''_i$ and $t'_j$, with $t''_i-1<t'_j{\leqslant}t''_i$, such that $\gamma'_i\vert_{(a''_i, c''_i]}$ is equivalent to $\gamma\vert_{(t''_i,t''_i+2]}$ and $\gamma'_j\vert_{[c'_j, b'_j)}$ is equivalent to $\gamma\vert_{[t'_j,t'_{j+2})}$. Let consider $d''_i\in (a''_i, c''_i)$ and $d'_j\in (c'_j, b'_j)$ uniquely defined by the fact that $\gamma'_i\vert_{(a''_i, c''_i]}$ is equivalent to $\gamma\vert_{(t''_i,t''_i+1]}$ and $\gamma'_j\vert_{[d'_j, b'_j)}$ is equivalent to $\gamma\vert_{[t''_i+1,t'_j+2)}$. One can suppose, by eventually changing $\gamma'_i$ in its equivalence class, that $\gamma'_i(d_i)=\gamma'_j(d_j)$. Note that the paths $\gamma'_i\vert_{[a''_i, c''_i]}$ and $\gamma'_j\vert_{[d'_j, b'_j]}$ have a $\mathcal F$-transverse intersection at $\gamma'_i(d_i)=\gamma'_j(d_j)$. By Proposition \[pr: fundamental\], the path $\gamma'_i\vert_{[a''_i, d''_i]}\gamma'_j\vert_{[d'_j, b'_j]}$ is admissible, and contains a subpath that is equivalent to $\gamma\vert_{(t''_i, t'_j+2)}$. In particular, since $t'_j>t_i-1$, this path draws $\Gamma$, does not cross it and its ends are not in $U_{\Gamma}$. By Proposition \[prop:drawing-crossing\], it has a $\mathcal F$-transverse self intersection. By Theorem \[th: horseshoe\], this implies that $f$ has a horseshoe. We have got a contradiction.
Paths on the annulus with no $\mathcal F$ transverse self-intersection
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Definitions, positive and negative loops
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A singular oriented foliation $\mathcal F$ on the annulus ${\mathbb{A}}={\mathbb{T}}^1\times{\mathbb{R}}$ can be extended to a singular foliation on ${\mathbb{S}}^2$ by adding two singular points, the upper end $N$ and the lower end $S$. So one can apply the results of the previous section. Let us study now this particular class of foliations. Write $[\Gamma]\in H_1({\mathbb{A}},{\mathbb{Z}})$ the homology class of a loop $\Gamma$ of ${\mathbb{A}}$. Consider the loop $\Gamma_*:x\mapsto(x,0)$, which means the circle ${\mathbb{T}}^1\times\{0\}$ oriented with an increasing angular coordinate, so that the upper end is on its left side and the lower end on its right side. One gets a generator of $H_1(M,{\mathbb{Z}})$ by considering $[\Gamma_*]$. Say that a loop $\Gamma$ is essential if $[\Gamma]\not=0$ and unessential otherwise. Say that $\Gamma$ is a [*positive*]{} loop if there exists $p>0$ such that $[\Gamma]=p[\Gamma_*]$ and define in the same way [*negative, non positive and non negative*]{} loops. Note that if $\Gamma$ is a simple loop, there exists $p\in\{-1,0,1\}$ such that $[\Gamma]=p[\Gamma_*]$. Let us remind that if a multi-loop $\Gamma=\sum_{1{\leqslant}i{\leqslant}p}\Gamma_i$ in ${\mathbb{A}}$ is homologous to zero in ${\mathbb{A}}$, one can choose the dual function (defined on ${\mathbb{S}}^2)$ to vanish in a neighborhood of $N$ and $S$. If $\check {\mathbb{A}}={\mathbb{R}}\times{\mathbb{R}}$ is the universal covering space of ${\mathbb{A}}$, we will denote $T:(x,y)\mapsto(x+1, y)$ the covering automorphism naturally associated with the loop $\Gamma_*$.
Suppose now that $\gamma:[a,b]\to\mathrm{dom}(I)$ is a transverse path such that $\phi_{\gamma(a)}=\phi_{\gamma(b)}=\phi$. One can find a transverse path $\gamma':[a,b]\to\mathrm{dom}(I)$ equivalent to $\gamma$ such that $\gamma'(a)=\gamma'(b)$. The leaf $\phi_{\gamma(a)}$ cannot be a closed leaf and so, the equivalence class of the transverse loop naturally defined by $\gamma'$ depends only on $\gamma$: we will called such a transverse loop, the [*natural transverse loop*]{} defined by $\gamma$.
\[prop:positive-negative\] Suppose that $\gamma:J\to\mathrm {dom} ({\mathcal F})$ is a transverse path with no $\mathcal F$-transverse self-intersection that draws a positive transverse simple loop $\Gamma_0$ and a negative transverse simple loop $\Gamma_1$. Then $U_{\Gamma_0}\cap U_{\Gamma_1}=\emptyset$.
Of course $\Gamma_0$ and $\Gamma_1$ are not equivalent. Denote $J_0$, $J_1$ the drawing components of $J_{\Gamma_0}$, $J_{\Gamma_1}$ respectively. There is no loss of generality by supposing that $J_1$ is on the right of $J_0$. The path $\gamma\vert_{J_1}$ does not cross $\Gamma_0$. Otherwise $J_1$ would contain a crossing component of $J_{\Gamma_0}$, but the only eventual crossing component of $J_{\Gamma_0}$ is $J_0$, the drawing component, by assertion $(2)$ of Proposition \[prop:drawing-crossing\]. So, there exists a transverse path equivalent to $\gamma\vert_{J_1}$ that is disjoint from $\Gamma_0$ and consequently a transverse loop $\Gamma'_1$ equivalent to $\Gamma_1$ that is disjoint from $\Gamma_0$. The multi loop $\Gamma_{0}+\Gamma'_{1}$ is homologous to zero and its dual function can be chosen to take its values in $\{0,1\}$ because the loops are disjoints. This implies that a leaf cannot meet both loops.
\[cor:positive-negative1\] Suppose that $\gamma:J\to\mathrm {dom} ({\mathcal F})$ is a transverse path with no $\mathcal F$-transverse self-intersection such that $\phi_{\gamma(t_0)}=\phi_{\gamma(t_1)}=\phi_{\gamma(t_2)}$ where $t_0<t_1<t_2$. If the transverse loop naturally defined by $\gamma\vert_{[t_0,t_1]}$ is positive, then the transverse loop naturally defined by $\gamma\vert_{[t_1,t_2]}$ is non negative. Similarly if the first loop is negative, the second one is non positive.
Suppose that $\gamma\vert_{[t_0,t_1]}$ is positive and $\gamma\vert_{[t_1,t_2]}$ negative. Using Corollary \[cor:decomposition\], one can find $t_0{\leqslant}t'_0<t'_1{\leqslant}t_1$ and $t_1{\leqslant}t''_1<t''_2{\leqslant}{t_2}$ such that
- $\phi_{\gamma(t'_0)}=\phi_{\gamma(t'_1)}=\phi_{\gamma(t''_1)}=\phi_{\gamma(t''_2)}=\phi_{\gamma(t_0)}$;
- the transverse loop naturally defined by $\gamma\vert_{[t'_0,t'_1]}$ is a positive transverse simple loop;
- the transverse loop naturally defined by $\gamma\vert_{[t''_1,t''_2]}$ is a negative transverse simple loop;
- these two last loops are drawn by $\gamma$.
It remains to note that $\phi_{\gamma(t_0)}$ meets both loops, which contradicts Proposition \[prop:positive-negative\].
\[cor:positive-negative2\] Suppose that $\gamma:J\to\mathrm {dom} ({\mathcal F})$ is a transverse path with no $\mathcal F$-transverse self-intersection such that $\phi_{\gamma(t_0)}=\phi_{\gamma(t_1)}$ and $\phi_{\gamma(t_2)}=\phi_{\gamma(t_3)}$, where $t_0<t_2<t_3<t_1$. If the transverse loop naturally defined by $\gamma\vert_{[t_0,t_1]}$ is positive, then the transverse loop naturally defined by $\gamma\vert_{[t_2,t_3]}$ is non negative. Similarly if the first loop is negative, the second one is non positive.
Suppose that $\gamma\vert_{[t_0,t_1]}$ is positive and $\gamma\vert_{[t_2,t_3]}$ non negative. Using Corollary \[cor:decomposition\], one can find $t_0{\leqslant}t'_0<t'_1{\leqslant}t_1$ and $t_2{\leqslant}t'_2<t'_3{\leqslant}{t_3}$ such that
- $\phi_{\gamma(t'_0)}=\phi_{\gamma(t'_1)}=\phi_{\gamma(t_0)}$;
- $\phi_{\gamma(t'_2)}=\phi_{\gamma(t'_3)}=\phi_{\gamma(t_2)}$;
- the transverse loop naturally defined by $\gamma\vert_{[t'_0,t'_1]}$ is a positive transverse simple loop;
- the transverse loop naturally defined by $\gamma\vert_{[t'_2,t'_3]}$ is a negative transverse simple loop;
- these two last loops are drawn by $\gamma$.
As explained in the proof of Corollary \[cor:decomposition\], the leaf $\phi_{\gamma(t_0)}$ meets both loops, which contradicts Proposition \[prop:positive-negative\].
Let us give an application.
\[cor:free-disks\] Let $f$ be a homeomorphism of ${\mathbb{A}}$ isotopic to the identity and $\check f$ a lift of $f$ to $\check {{\mathbb{A}}}$. Let $D$ be a topological open disk of ${\mathbb{A}}$ and $\check D$ a lift of $D$. We suppose that $\check D$ is a free disk of $\check f$ and that there exists a point $\check z\in \check D$ and positive integers $n_0,n_1,m_0,m_1$ such that $$\check z\in \check D, \,\check f^{n_0}(\check z)\in T^{m_0}(\check D),\, \check f^{n_0+n_1}(\check z)\in T^{m_0-m_1}(\check D).$$Then $f$ has a topological horseshoe.
Choose an identity isotopy $I^*$ of $f$ that is lifted on $\check {\mathbb{A}}$ to an identity isotopy of $\check f$. Then choose a maximal isotopy $I$ such that $I^*\preceq I$. Set $z=\pi(\check z)$. As explained in [@LeCalvezTal], one can find a singular foliation ${\mathcal F}$ transverse to $I$ such $z$, $f^{n_0}(z)$ and $f^{n_0+n_1}(z)$ are on the same leaf of the restricted foliation ${\mathcal F}\vert_{D}$. Setting $\gamma=I^{n_0+n_1}_{{\mathcal F}}(z)$, one gets
- $\phi_{\gamma(0)}=\phi_{\gamma(n_0)}=\phi_{\gamma(n_0+n_1)}$;
- the transverse loop naturally defined by $\gamma\vert_{[0,n_0]}$ is positive;
- the transverse loop naturally defined by $\gamma\vert_{[n_0,n_0+n_1]}$ is negative.
By Corollary \[cor:positive-negative1\], one deduces that $I^{n_0+n_1}_{{\mathcal F}}(z)$ has a $\mathcal F$-transverse self-intersection, and by Theorem \[th: horseshoe2\], that $f$ has a topological horseshoe.
Rotation numbers
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Let us remind the theorem we want to prove, Theorem \[thmain:rotation-number\] of the introduction.
\[th:rotation-number\] Let $f$ be a homeomorphism of ${\mathbb{A}}$ isotopic to the identity and $\check f$ a lift of $f$ to ${\mathbb{R}}^2$. We suppose that $f$ has no topological horseshoe. Then
1. each point $z\in\mathrm{ne}^+(f)$ has a [well defined]{} rotation number $\mathrm{rot}_{\check f}(z)$;
2. for every points $z,z'\in\mathrm{ne}^+(f)$ such that $z'\in\omega(z)$ we have $\mathrm{rot}_{\check f}(z')=\mathrm{rot}_{\check f}(z)$;
3. if $z\in \mathrm{ne}^+(f)\cap \mathrm{ne}^-(f^{-})$ is non-wandering, then $\mathrm{rot}_{\check f^{-1}}(z)=-\mathrm{rot}_{\check f}(z)$;
4. the map $\mathrm{rot}_{\check f^{\pm}} : \Omega(f) \cap \mathrm{ne}(f)\to{\mathbb{R}}$ is continuous, where $$\mathrm{rot}_{\check f^{\pm}}(z) = \begin{cases}\mathrm{rot}_{\check f}(z) &\mathrm{if}\enskip z\in \Omega(f) \cap \mathrm{ne}^+(f),\\
-\mathrm{rot}_{\check f^{-1}}(z) &\mathrm{if}\enskip z\in \Omega(f) \cap \mathrm{ne}^-(f).
\end{cases}$$
We will begin by proving $(1)$ by contradiction. Suppose that the $\omega$-limit set of $z$ is not empty and that $z$ has no rotation number. There are two possibilities.
[Case 1:]{}there exists a compact set $K\subset{\mathbb{A}}$, two numbers $\rho_{-}<\rho_+$ in ${\mathbb{R}}\cup\{-\infty,\infty\}$ and two increasing sequences of integers $(n_k)_{k{\geqslant}0}$ and $(m_k)_{k{\geqslant}0}$ such that $f^{n_k}(z)\in K$, $f^{m_k}(z)\in K$ and such that for any lift $\check z$ of $z$, one has $$\lim_{k\to+\infty}\frac{1}{n_k}\left( \pi_1(\check f^{n_k}(\check z)- \pi_1(\check z)\right)=\rho_{-}$$ and $$\lim_{k\to+\infty}\frac{1}{m_k}\left( \pi_1(\check f^{m_k}(\check z)- \pi_1(\check z)\right)=\rho_{+}.$$
[Case 2:]{}there exists a compact set $K\subset{\mathbb{A}}$, a number $\rho\in \{-\infty,\infty\}$ and an increasing sequence of integers $(n_k)_{k{\geqslant}0}$ such that $f^{n_k}(z)\in K$ and such that for any lift $\check z$ of $z$, one has $$\lim_{k\to\infty}\frac{1}{n_k}\left( \pi_1(\check f^{n_k}(\check z)- \pi_1(\check z)\right)=\rho.$$
Let us begin by the first case. Taking subsequences if necessary, one can suppose that there exist $z_-$ and $z_+$ in $K$ such that $\lim_{k\to+\infty} f^{n_k}(z)=z_-$ and $\lim_{k\to+\infty} f^{m_k}(z)=z_+$. Let us choose two integers $p$ and $q>0$ relatively prime such that $\rho_-<p/q<\rho_+$ and such that both $z_-$ and $z_+$ are not fixed points of $f^q$ with rotation number $p/q$. Write $n_k=qn'_k+r_k$ and $m_k=qm'_k+s_k$, where $r_k$ and $s_k$ belong to $\{0,\dots, q-1\}$. Taking subsequences if necessary, one can suppose that there exists $r$ and $s$ in $\{0,\dots, q-1\}$ such that $r_k=r$ and $s_k=s$ for every $k\in{\mathbb{Z}}$. Setting $$z'_-=f^{-r}(z_-),\,\,z'_+=f^{-s}(z_+),\,\,f'=f^q,\,\check f'=\check f^q\circ T^{-p},$$ one knows that $$\lim_{k\to+\infty} f'^{n'_k}(z)=z'_-,\,\,\lim_{k\to+\infty} f'^{m'_k}(z)=z'_+.$$Moreover $$\lim_{k\to+\infty}\frac{1}{n'_k}\left( \pi_1(\check f'^{n'_k}(\check z)- \pi_1(\check z)\right)=q\rho_{-}-p<0$$ and $$\lim_{k\to+\infty}\frac{1}{m'_k}\left( \pi_1(\check f'^{m'_k}(\check z)- \pi_1(\check z)\right)=q\rho_{+}-p>0.$$
Choose an identity isotopy $I'^*$ of $f'$ that is lifted to an identity isotopy of $\check f'$, then a maximal isotopy $I'$ such that $I'^*\preceq I'$, and finally a singular foliation ${\mathcal F}'$ transverse to $I'$. The singular points of $I'$ are periodic points of $f$ of period $q$ and rotation number $p/q$. So $z'_-$ and $z'_+$ are non singular points. The trajectory $\gamma'=I'^{{\mathbb{N}}}_{{\mathcal F}'}(z)$ meets the leaves $\phi_{z'_-}$ and $\phi_{z'_+}$ infinitely many times. Denote $(t_{k}^-)_{k{\geqslant}0}$ and $(t_{k}^+)_{k{\geqslant}0}$ the (increasing) sequences of meeting times. One can find $t_{k_0}^+< t_{k_2}^-<t_{k_3}^-<t_{k_1}^+$ such that $\gamma\vert_{[t_{k_0}^+,t_{k_1}^+]}$ is positive and $\gamma\vert_{[t_{k_2}^-,t_{k_3}^-]}$ negative. By Corollary \[cor:positive-negative2\], one deduces that $\gamma'$ has a ${\mathcal F}'$-transverse self-intersection, and by Theorem \[th: horseshoe2\], that $f$ has a topological horseshoe.
Let us study now the second case. We will suppose that $\rho=-\infty$, the case where $\rho=+\infty$ being analogous. Taking a subsequence if necessary, one can suppose that there exists $z'\in K$ such that $\lim_{k\to+\infty} f^{n_k}(z)=z'$.
Suppose first that $z'$ is a fixed point and denote $p\in{\mathbb{Z}}$ its rotation number. Choose an identity isotopy $I'^*$ of $f$ that is lifted to an identity isotopy of $\check f'=\check f\circ T^{-p+1}$, then a maximal isotopy $I'$ such that $I'^*\preceq I'$, and finally a singular foliation ${\mathcal F}'$ transverse to $I'$. One knows that $\rho_{\check f'} (z')=1$. This means that the transverse loop $\Gamma$ associated with $z'$ is a positive simple loop. The fact that $z'\in\omega(z)$ implies that for every $k{\geqslant}0$, the path $I'^{{\mathbb{N}}}_{{\mathcal F}'}(f^k(z))$ draws $\Gamma$. One deduces from Proposition \[prop:drawing-crossing\] that $I'^{{\mathbb{N}}}_{{\mathcal F}'}(f^k(z))$ draws infinitely $\Gamma$, if $k$ is sufficiently large. This contradicts the fact that $\rho=-\infty$.
Suppose now that $z'$ is not a fixed point and consider a free disk $D$ containing $z'$. The orbit $(f^n(z))_{n{\geqslant}0}$ meets the disk $D$ infinitely many times. Denote $(n_{k})_{k{\geqslant}0}$ the (increasing) sequences of meeting times. Fix a lift $\check z$ of $z$. For every $k{\geqslant}0$, there exists $p_k\in{\mathbb{Z}}$ such that $\check f^{n_{k}}(\check z)\in T^{p_k}(\check D)$. Moreover, $\lim_{k\to+\infty} p_k/n_k=-\infty$. Choose an integer $r$ such that $(p_1-p_0)+r(n_1-n_0)>0$. Consider the point $\check z''=\check f^{n_0}(\check z)$, the disk $\check D''=T^{p_0}(\check D)$ and the lift $\check f''=\check f\circ T^{r}$. We have $$\check z''\in \check D'', \,\check f''^{n_1-n_0}(\check z'')\in T^{p_1-p_0+r(n_1-n_0)}(\check D''),\, f''^{n_k-n_0}(\check z'')\in T^{p_k-p_0+r(n_k-n_0)}(\check D'').$$ So $p_k-p_0+r(n_k-n_0)<0$, if $k$ is large enough. By Corollary \[cor:free-disks\], one deduces that $f$ has a horseshoe.
We will also prove $(2)$ by contradiction. Suppose that $z$ and $z'$ belong to $\mathrm{ne}^+(f)$ and that $z'\in\omega(z)$ and suppose for instance that $\mathrm{rot}_{\check f}(z)<\mathrm{rot}_{\check f}(z')$. Choose two integers $p$ and $q>0$ relatively prime such that $\mathrm{rot}_{\check f}(z)<p/q<\mathrm{rot}_{\check f}(z')$. There exists $r\in\{0,\dots, q-1\}$ such that $f^r(z')$ is in the $\omega$-limit set of $z$ for $f^q$. Choose a maximal isotopy $I'$ of $f^q$ that is lifted to an identity isotopy of $\check f'=\check f^q\circ T^{-p}$ and a singular foliation ${\mathcal F}'$ transverse to $I'$. The point $f^r(z')$ being non-wandering and having a positive rotation number for $\check f'$, one deduces, by Proposition \[prop:non-wandering\] that there exists a positive simple transverse loop $\Gamma$ such that $I'^{{\mathbb{Z}}}_{{\mathcal F}'}(f^r(z'))$ is drawn on $\Gamma$ and that $I'^{{\mathbb{N}}}_{{\mathcal F}'}(f^r(z')))$ draws infinitely $\Gamma$. The fact that $z'\in\omega_{f'}(z)$ implies that for every $k{\geqslant}0$, $I'^{{\mathbb{N}}}_{{\mathcal F}'}(f'{}^k(z))$ draws $\Gamma$. By Proposition \[prop:drawing-crossing\], there exists $k{\geqslant}0$ such that $I'^{{\mathbb{N}}}_{{\mathcal F}'}(f'{}^k(z))$ is drawn on $\Gamma$. This contradicts the fact that $\mathrm{rot}_{\check f'}(z)<0$.
The proof of $(3)$ is very close. Let $z$ be in $\Omega(f)\cap \mathrm{ne}^+(f)\cap \mathrm{ne}^-(f)$. Suppose for instance that $-\mathrm{rot}_{\check f^{-1}}(z)<\mathrm{rot}_{\check f}(z)$ and choose two integers $p$ and $q>0$ relatively prime such that $-\mathrm{rot}_{\check f^{-1}}(z)<p/q<\mathrm{rot}_{\check f}(z)$. Choose a maximal isotopy $I'$ of $f^q$ that is lifted to an identity isotopy of $\check f'=\check f^q\circ T^{-p}$ and a singular foliation ${\mathcal F}'$ transverse to $I'$. The point $z$ being non-wandering for $f$ is also non-wandering for $f^q$. One deduces, by Proposition \[prop:non-wandering\] that there exists a simple transverse loop $\Gamma$ such that $I'^{{\mathbb{Z}}}_{{\mathcal F}'}(z')$ is equivalent to the natural lift of $\Gamma$. This contradicts the fact that$-\mathrm{rot}_{\check f'^{-1}}(z)<0<\mathrm{rot}_{\check f'}(z)$.
It remains to prove $(4)$. Fix $z\in\Omega(f)\cap \mathrm{ne}(f)$. Suppose that $p$ and $q>0$ are relatively prime and that $p/q<\mathrm{rot}_{f^{\pm}}(z)$. Choose a maximal isotopy $I'$ of $f^q$ that is lifted to an identity isotopy of $\check f^q\circ T^{-p}$ and a singular foliation ${\mathcal F}'$ transverse to $I'$. As explained in Proposition \[prop:non-wandering\], $I'{}^{{\mathbb{Z}}}_{{\mathcal F}'} (z)$ draws a unique transverse simple loop $\Gamma$. There exists a neighborhood $W$ of $z$ such that $I'{}^{{\mathbb{Z}}}_{{\mathcal F}'} (z')$ draws $\Gamma$ for every $z'\in W$. In particular, $I'{}^{{\mathbb{Z}}}_{{\mathcal F}'} (z')$ is drawn on $\Gamma$ if $z\in\Omega(f)\cap\mathrm{ne}(f)\cap W$ , which implies that $\mathrm{\rho}(z'){\geqslant}p/q$. We have a similar result supposing $p/q>\mathrm{rot}(z)$. So the function $\mathrm{rot}_{f^{\pm}}$ is continuous on $\Omega(f)\cap \mathrm{ne}(f)$.
Birkhoff recurrent points and Birkhoff recurrent classes
--------------------------------------------------------
Let us remind the theorem we want to prove, Theorem \[thmain:birkhoffcycles\] of the introduction.
\[th:birkhoffcycles\] Let $f$ be a homeomorphism of ${\mathbb{A}}$ isotopic to the identity and $\check f$ a lift of $f$ to ${\mathbb{R}}^2$. We suppose that $f$ has no topological horseshoe. If $\mathcal B$ is a Birkhoff recurrence class of $f_{\mathrm{sphere}}$, there exists $\rho\in{\mathbb{R}}$ such that, $$\begin{cases} \mathrm{rot}_{\check f}(z) = \rho&\mathrm{if}\enskip z\in {\mathcal B}\cap\mathrm{ne}^+(f),\\
\mathrm{rot}_{\check f^{-1}}(z) = -\rho&\mathrm{if}\enskip z\in {\mathcal B}\cap\mathrm{ne}^-(f).
\end{cases}$$
Of course one can suppose that ${\mathcal B}\cap\mathrm{ne}(f)$ is not empty. Otherwise there is nothing to prove, every $\rho\in{\mathbb{R}}$ is suitable. We will divide the proof into three cases, the case where $\mathcal B$ contains neither $N$ nor $S$, the case where $\mathcal B$ contains both $N$ and $S$ and the remaining case, the most difficult one, where $\mathcal B$ contains exactly one end. Let us begin with the following lemma, that is nothing but Corollary \[corollarymain:birkhoffclasses\] of the introduction.
\[lemma:Birkhoff-cycles\] Let $f$ be a homeomorphism of ${\mathbb{A}}$ isotopic to the identity with no topological horseshoe and $\check f$ a lift of $f$ to the universal covering space. We suppose that $(z_i)_{i\in{\mathbb{Z}}/r{\mathbb{Z}}}$ is a Birkhoff cycle in $\mathrm{ne}(f)$ and $(\rho_i)_{i\in{\mathbb{Z}}/r{\mathbb{Z}}}$ a family of real numbers such that:
- either $z_i\in\mathrm{ne}^+(f)$ and $\rho_i=\mathrm{rot}_{\check f}(z_i)$,
- or $z_i\in\mathrm{ne}^-(f)$ and $\rho_i=-\mathrm{rot}_{\check f^{-1}}(z_i)$.
Then all $\rho_i$ are equal.
We will argue by contradiction. Suppose that the $\rho_i$ are not equal. One can find $p\in{\mathbb{Z}}$ and $q>0$ relatively prime such that
- $\rho_i\not=p/q$, for every $i\in {\mathbb{Z}}$;
- there exist $i_0,i_1$ in ${\mathbb{Z}}$ satisfying $\rho_{i_0}<p/q<\rho_{i_1}$.
Choose a maximal isotopy $I'$ of $f^q$ that is lifted to an identity isotopy of $\check f^q\circ T^{-p}$ and a singular foliation ${\mathcal F}'$ transverse to $I'$. The orbit of each $z_i$ is included in the domain of $I'$ because the singular points of $I'$ are periodic points of period $q$ and rotation number $p/q$. Using assertion (4) of Proposition \[prop: birkhoff connexions\], one can construct a Birkhoff cycle $(z'_j)_{j\in{\mathbb{Z}}/mr{\mathbb{Z}}}$ of $f^q$ such that $z'_j$ belongs to the orbit of $z_i$ if $j \equiv i \pmod r$. By Proposition \[prop: rotation numbers powers\], one knows that
- $z'_j\in\mathrm{ne}^+(f^q)$ if $z_i\in\mathrm{ne}^+(f)$ and $\mathrm{rot}_{\check f^q\circ T^{-p}}(z'_j)= q\rho_i-p$;
- $z'_j\in\mathrm{ne}^-(f^q)$ if $z_i\in\mathrm{ne}^-(f)$ and $-\mathrm{rot}_{\check f^{-q}\circ T^p}(z'_j)= q\rho_i-p$.
Consequently, $I'^{{\mathbb{Z}}}_{{\mathcal F}'}(z'_j)$ draws a negative transverse simple loop if $j \equiv i_0 \pmod r$ and $I'^{{\mathbb{Z}}}_{{\mathcal F}'}(z'_j)$ draws a positive transverse simple loop if $j \equiv i_1 \pmod r$. This contradicts Proposition \[prop:Birkhoff-cycles\].$\Box$
We will also use the following result
\[lemma:periodic-orbits\] Let $f$ be a homeomorphism of ${\mathbb{A}}$ isotopic to the identity and $\check f$ a lift of $f$ to ${\mathbb{R}}^2$. Suppose that there exist $z_0$ and $z_1$ in $ \mathrm{ne}(f)$ and $\rho_0<\rho_1$ such that such for every $i\in\{0,1\}$:
- either $z_i\in\mathrm{ne}^+(f)$ and $\rho_i=\mathrm{rot}_{\check f}(z_i)$,
- or $z_i\in\mathrm{ne}^-(f)$ and $\rho_i=-\mathrm{rot}_{\check f^{-1}}(z_i)$.
Suppose moreover that $z_0$ and $z_1$ are in the same Birkhoff recurrence class of $f_{\mathrm{sphere}}$, or that $N$ and $S$ are in the same Birkhoff recurrence class of $f_{\mathrm{sphere}}$. Then, for every rational number $p/q\in(\rho_0,\rho_1)$, written in an irreducible way, there exists a periodic point $z$ of period $q$ and rotation number $\mathrm{rot}_{\check f}(z)=p/q$.
Fix a rational number $p/q\in(\rho_0,\rho_1)$, written in an irreducible way, then choose a maximal isotopy $I'$ of $f^q$ that is lifted to an identity isotopy of $\check f^q\circ T^{-p}$ and a foliation ${\mathcal F}'$ transverse to $I'$. It is sufficient to prove that if ${\mathcal F}'$ is non singular, then $z_0$ and $z_1$ are not in the same Birkhoff recurrence class of $f_{\mathrm{sphere}}$, and $N$ and $S$ are not in the same Birkhoff recurrence class of $f_{\mathrm{sphere}}$. By Proposition \[prop: rotation numbers powers\], if $z'_i$ belongs to the orbit of $z_i$, then
- $z'_i\in\mathrm{ne}^+(f^q)$ if $z_i\in\mathrm{ne}^+(f)$ and $\mathrm{rot}_{\check f^q\circ T^{-p}}(z'_i)= q\rho_i-p$;
- $z'_i\in\mathrm{ne}^-(f^q)$ if $z_i\in\mathrm{ne}^-(f)$ and $-\mathrm{rot}_{\check f^{-q}\circ T^p}(z'_i)= q\rho_i-p$.
Consequently, $I'^{{\mathbb{Z}}}_{{\mathcal F}'}(z'_0)$ draws a negative transverse simple loop $\Gamma_0$ and $I'^{{\mathbb{Z}}}_{{\mathcal F}'}(z'_1)$ draws a positive transverse simple loop $\Gamma_1$. One can find positive integers $n_0$ and $n_1$ such that the multi-loop $n_0\Gamma_0+n_1\Gamma_1$ is homological to zero. Write $\delta$ for its dual function that vanishes in a neighborhood of the ends of ${\mathbb{A}}$. Either the minimal value $m_-$ of $\delta$ is negative or the maximal value $m_+$ is positive. We will look at the first case, the second one being analogous. The fact that $\delta$ decreases along the leaves of ${\mathcal F}'$ implies that the closure of a connected component $U$ of ${\mathbb{A}}\setminus (\Gamma_0\cup\Gamma_1)$ where $\delta$ is equal to $m_-$ is a topological manifold whose boundary is transverse to ${\mathcal F}'$, the leaves going inside $U$. By connectedness of $\Gamma_0$ and $\Gamma_1$, we know that $\overline U$ is a topological disk or an essential annulus. The first situation is impossible because we should have a singular point of ${\mathcal F}'$ inside $U$. So $\overline U$ is an annulus and by Poincaré-Bendixson Theorem, it contains an essential closed leaf. This leaf separates $z'_0$ and $z'_1$ and is disjoint from its image by $f^q$. So $z'_0$ and $z'_1$ are not in the same Birkhoff recurrence class of $f_{\mathrm{sphere}}^q$. By assertion (4) of Proposition \[prop: birkhoff connexions\], one deduces that $z_0$ and $z_1$ are not in the same Birkhoff recurrence class of $f_{\mathrm{sphere}}$. Note also that $N$ and $S$ are not in the same Birkhoff recurrence class of $f_{\mathrm{sphere}}$.
Let us go now to the proof of Theorem \[th:birkhoffcycles\]. We suppose that there exist $z^*_0$ and $z^*_1$ in ${\mathcal B}\cap \mathrm{ne}(f)$ and $\rho_0<\rho_{1}$ such that such for every $i\in\{0,1\}$:
- either $z^*_i\in\mathrm{ne}^+(f)$ and $\rho_i=\mathrm{rot}_{\check f}(z^*_i)$,
- or $z^*_i\in\mathrm{ne}^-(f)$ and $\rho_i=-\mathrm{rot}_{\check f^{-1}}(z^*_i)$.
We want to find a contradiction.
[*First case: the Birkhoff class $\mathcal B$ does not contain $N$ or $S$.*]{}
There exists a Birkhoff cycle $(z_i)_{i\in{\mathbb{Z}}/r{\mathbb{Z}}}$ of $f_{\mathrm{sphere}}$ that contains $z^*_0$ and $z^*_1$ and does not contain $N$ and $S$. So, it is a Birkhoff cycle of $f$. Using item (1) of Proposition \[prop: birkhoff connexions\], one knows that for every $i\in{\mathbb{Z}}/r{\mathbb{Z}}$, the closure of $O_{f_{\mathrm{sphere}}}(z_i)$ is contained in $\mathcal B$ and so does not contain neither $N$, nor $S$. In particular $z\in ne^+(f)\cap ne^-(f)$. We can apply Lemma \[lemma:Birkhoff-cycles\] and obtain a contradiction.
In that case, we will get the contradiction from the following result that is nothing but Proposition \[propositionmain\_regionofinstability\] and does not use the fact that $z^*_0$ and $z^*_1$ are in the same Birkhoff recurrence class than $N$ and $S$.
\[prop:PB\] Let $f$ be a homeomorphism of ${\mathbb{A}}$ isotopic to the identity, $\check f$ a lift of $f$ to the universal covering space and $f_{\mathrm{sphere}}$ be the continuous extension of $f$ to ${\mathbb{S}}^2={\mathbb{A}}\cup\{N,S\}$. We suppose that $\rho_0<\rho_1$ are two real numbers and that:
- there exists $z_0^*\in\mathrm{ne}^+(f)$ such that $\rho_0=\mathrm{rot}_{\check f}(z_0^*)$ or there exists $z_0^*\in\mathrm{ne}^{-}(f)$ such that $\rho=-\mathrm{rot}_{\check f^{-1}}(z_0^*)$;
- there exists $z_1^*\in\mathrm{ne}^+(f)$ such that $\rho_1=\mathrm{rot}_{\check f}(z_1^*)$ or there exists $z_1^*\in\mathrm{ne}^-(f)$ such that $\rho_1=-\mathrm{rot}_{\check f^{-1}}(z_1^*)$;
- $N$ and $S$ are in the same Birkhoff recurrence class of $f_{\mathrm{sphere}}$.
Then, $f$ has a topological horseshoe.
We will argue by contradiction, supposing that $f$ has no topological horseshoe. In that case, there exists a Birkhoff cycle $(z_i)_{i\in{\mathbb{Z}}/r{\mathbb{Z}}}$ of $f_{\mathrm{sphere}} $ that contains $N$ and $S$. Let $p/q<p'/q'$ be rational numbers in $(\rho_0,\rho_1)$ such that
- there is no $i\in{\mathbb{Z}}/r{\mathbb{Z}}$ satisfying $z_i\in\mathrm{ne}^+(f)$ and $\mathrm{rot}_{\check f}(z_i)\in\{p/q,p'/q'\}$;
- there is no $i\in{\mathbb{Z}}/r{\mathbb{Z}}$ satisfying $z_i\in\mathrm{ne}^-(f)$ and $\mathrm{rot}_{\check f^{-1}}(z_i)\in\{-p/q,-p'/q'\}$.
Since $N$ and $S$ belong to the same Birkhoff recurrence class of $f_{\mathrm{sphere}}$, we can apply Lemma \[lemma:periodic-orbits\] and find a periodic [point]{} $z\in {\mathbb{A}}$ of period $q$ and rotation number $\mathrm{rot}_{\check f}(z)=p/q$, a periodic point $z'\in{\mathbb{A}}$ of period $q'$ and rotation number $\mathrm{rot}_{\check f}({z'})=p'/q'$, such that for every $i\in{\mathbb{Z}}/r{\mathbb{Z}}$ and every $k\in{\mathbb{Z}}$, one has $$\omega_{f_{\mathrm{sphere}}^{qq'}}(f^{k}(z_i))\cap\{z,z'\}=\alpha_{f_{\mathrm{sphere}}^{qq'}}(f^{k}(z_i))\cap\{z,z'\}=\emptyset.$$ By Proposition \[prop: birkhoff connexions\], one can construct a Birkhoff cycle $(z'_j)_{j\in{\mathbb{Z}}/mr{\mathbb{Z}}}$ of $f_{\mathrm{sphere}}^{qq'}$ such that $z'_j$ belongs to the orbit of $z_i$ if $j \equiv i \pmod r$, and we know that $z'_j$ belong to $\mathrm{ne}(f_{\mathrm{sphere}}^{qq'}\vert_{{\mathbb{S}}^2\setminus\{z, z'\}})$. Applying Lemma \[lemma:Birkhoff-cycles\], we deduce that if $\check g$ is a lift of $f_{\mathrm{sphere}}^{qq'}\vert_{{\mathbb{S}}^2\setminus\{z, z'\}}$ to the universal covering space, if $\kappa\in H_1({\mathbb{S}}^2\setminus\{z, z'\},{\mathbb{Z}})$ is the generator given by the oriented boundary of a small disk containing $z$ in its interior, then $\mathrm{rot}_{\check g,\kappa}(N)=\mathrm{rot}_{\check g,\kappa}(S)$. The contradiction will come from the following equality $$\mathrm{rot}_{\check g,\kappa}(S)-\mathrm{rot}_{\check g,\kappa}(N)= \mathrm{rot}_{\check f^{qq'}}(z)-\mathrm{rot}_{\check f^{qq'}}(z')= q'p-qp'\not=0.$$ There are different explanations of the previous equality (see Franks \[F\] for example). Let us mention the following one. By conjugacy, one can always suppose that ${\mathbb{S}}^2$ is the Riemann sphere. So one can define the cross ratio $[z_1,z_2,z_3,z_4]$ of four distinct points. Let us choose an identity isotopy $(g_t)_{t\in[0,1]}$ of $g_{\mathrm{sphere}}$. The closed path $$t\mapsto[g_t(z), g_t(z'), g_t(S),g_t(N)]$$ defines a loop $\Gamma$ on ${\mathbb{S}}^2\setminus \{0,1,\infty\}$ whose homotopy class is independent of the chosen isotopy, two isotopies being homotopic. The integer $r=\delta_{\Gamma}(0)-\delta_{\Gamma}(\infty)$, where $\delta_{\Gamma}$ is a dual function of $\Gamma$, also does not depend on the isotopy. The integer $\mathrm{rot}_{\check f^{qq'}}(z)-\mathrm{rot}_{\check f^{qq'}}(z')$ is independent of the lift $\check f$. In fact it is equal to $r$. Indeed, one can suppose that $(z',S,N)=(1,0,\infty)$ and that $(g_t)_{t\in[0,1]}$ fixes $z'$, $S$ and $N$. So it defines a lift $\check g^*$ of $f^{qq'}$ to the universal covering space of ${\mathbb{A}}$ such that $\mathrm{rot}_{\check g^*}(z')=0$. But one has $[g_t(z), g_t(z'), g_t(S),g_t(N)]=g_t(z)$. So $\mathrm{rot}_{\check g^*}(z)$ is nothing but $r$. As we know that $[g_t(z), g_t(z'), g_t(S),g_t(N)]=[g_t(S), g_t(N), g_t(z),g_t(z')]$, we deduce for the same reasons that $\mathrm{rot}_{\check g,\kappa}(S)-\mathrm{rot}_{\check g,\kappa}(N)=r$.
Let us state now a corollary that will useful while studying the last case and that says that $z_0^*$ and $z_1^*$ cannot be periodic.
\[cor:periodic\] Let $f$ be a homeomorphism of ${\mathbb{A}}$ isotopic to the identity, $\check f$ a lift of $f$ to the universal covering space and $f_{\mathrm{sphere}}$ be the continuous extension of $f$ to ${\mathbb{S}}^2={\mathbb{A}}\cup\{N,S\}$. We suppose that there exist two periodic points $z_0^*$ and $z_1^*$ of $f$ such that
- $\mathrm{rot}_{\check f}(z_0^*)<\mathrm{rot}_{\check f}(z_1^*)$;
- $z_0^*$ and $z_1^*$ are in the same Birkhoff recurrence class of $f_{\mathrm{sphere}}$.
Then, $f$ has a topological horseshoe.
Taking a power of $f$ instead of $f$, one can suppose that $z_0^*$ and $z_1^*$ are fixed points. As explained in the proof of the Proposition \[prop:PB\], if $\check g$ is a lift of $f_{\mathrm{sphere}}\vert_{{\mathbb{S}}^2\setminus\{z_0^*, z_1^*\}}$ to the universal covering space, if $\kappa\in H_1({\mathbb{S}}^2\setminus\{z_0^*, z_1^*\},{\mathbb{Z}})$ is the generator given by the oriented boundary of a small disk containing $z_0^*$ in its interior, then $$\mathrm{rot}_{\check g,\kappa}(S)-\mathrm{rot}_{\check g,\kappa}(N)= \mathrm{rot}_{\check f}(z_0^*)-\mathrm{rot}_{\check f}(z_1^*)\not=0.$$ So, one can apply Proposition \[prop:PB\] to deduce that $f$ has a topological horseshoe.
Applying Lemma \[lemma:periodic-orbits\] and Corollary \[cor:periodic\], one can find a rational number $p/q\in(\rho_0,\rho_1)$ such that if $z\in {\mathbb{A}}$ is a periodic point of period $q$ and rotation number $\mathrm{rot}_{\check f}(z)=p/q$, then:
- for every $k\in{\mathbb{Z}}$, one has $\omega_{f_{\mathrm{sphere}}^q}(f^k(z_0^*))\not=\{z\}$ and $\alpha_{f_{\mathrm{sphere}}^q}(f^k(z_0^*))\not =\{z\}$;
- for every $k\in{\mathbb{Z}}$, one has $\omega_{f_{\mathrm{sphere}}^q}(f^k(z_1^*))\not=\{z\}$ and $\alpha_{f_{\mathrm{sphere}}^q}(f^k(z_1^*))\not =\{z\}$;
- $z$ does not belong to $\mathcal B$.
By Proposition \[prop: birkhoff connexions\], $z^*_0$ and $z_1^*$ belong to the same Birkhoff recurrence class of $f^q_{\mathrm{sphere}}$ [that $S$ belongs to]{}. Choose a maximal isotopy $I'$ of $f^q$ that is lifted to an identity isotopy of $\check f^q\circ T^{-p}$ and a singular foliation ${\mathcal F}'$ transverse to $I'$. The singular points of ${\mathcal F}'$ are periodic point of period $q$ and rotation number $\mathrm{rot}_{\check f}(z)=p/q$, plus the two ends $N$ and $S$ if we consider the foliation as defined on the sphere. If $z$ and $z'$ are two singular points, the isotopy $I'\vert_{{\mathbb{S}}^2\setminus\{z,z'\}}$ can be lifted to an identity isotopy of a lift of $f^q\vert_{{\mathbb{S}}^2\setminus\{z,z'\}}$. We write $\check g_{z,z'}$ for this lift. We will consider a small disk containing $z$ in its interior to define rotation number. Of course $\mathrm{rot}_{\check g_{z,z'}}(z'')=0$ is $z''$ is a third singular point. The first case studied above tell us that if neither $z$ nor $z'$ is equal to $S$, then $z^*_0$, $z^*_1$ and $S$ belong to $\mathrm{ne}(f^q\vert_{{\mathbb{S}}^2\setminus\{z,z'\}})$ and that $$\mathrm{rot}_{\check g_{z,z'}}(z^*_0)=\mathrm{rot}_{\check g_{z,z'}}(z^*_1)=\mathrm{rot}_{\check g_{z,z'}}(S)=0.$$ By Proposition \[prop: rotation numbers\], one deduces that if $z'$ is a singular point distinct from $S$, one has $$\mathrm{rot}_{\check g_{S,z'}}(z^*_0)=\mathrm{rot}_{\check g_{S,N}}(z^*_0)-\mathrm{rot}_{\check g_{z',N}}(z^*_0)=q\rho_0-p.$$ Similarly, one gets $$\mathrm{rot}_{\check g_{S,z'}}(z^*_1)=q\rho_1-p,$$ and consequently $$\mathrm{rot}_{\check g_{S,z'}}(z^*_0)<\mathrm{rot}_{\check g_{S,z'}}(z^*_1).$$
By assumption, $\mathrm{dom}(I')\cap (\alpha_{f_{\mathrm{sphere}}^q}(z^*_0)\cup \omega_{f_{\mathrm{sphere}}^q}(z^*_0))$ is non empty. Choose a point $z_0$ in this set and denote $\phi_0$ the leaf of ${\mathcal F}'$ that contains this point. Similarly, choose a point $z_1$ in $\mathrm{dom}(I')\cap (\alpha_{f_{\mathrm{sphere}}^q}(z^*_1)\cup \omega_{f_{\mathrm{sphere}}^q}(z^*_1))$ and denote $\phi_1$ the leaf that contains this point. Note that both $z_0$ and $z_1$ belong to the same Birkhoff recurrence class of $f_{\mathrm{sphere}}^q$ as $S$, $z_0^*$ and $z_1^*$. The fact that $\mathrm{rot}_{\check g_{S,z'}}(z^*_0)<0<\mathrm{rot}_{\check g_{S,z'}}(z^*_1)$ implies that for every singular point $z'\not=S$ there exists a simple transverse loop $\Gamma_{0}$ crossing $\phi_0$ and such that $z'\in L(\Gamma_0)$ and $S\in R(\Gamma_0)$ and a simple transverse loop $\Gamma_{1}$ crossing $\phi_1$ and such that $z'\in R(\Gamma_1)$ and $S\in L(\Gamma_1)$. Denote $\widehat\omega(\phi_0)$ the complement of the connected component of ${\mathbb{S}}^2\setminus\omega(\phi_0)$ that contains $\phi_0$. It is a saturated cellular closed set that contains $\omega(\phi_0)$ and that contains at least one singular point. Define in similar way $\widehat\alpha(\phi_0)$, $\widehat\omega(\phi_1)$ and $\widehat\alpha(\phi_1)$. The fact that for every singular point there exists a simple transverse loop $\Gamma_{0}$ crossing $\phi_0$ and such that $z'\in L(\Gamma_0)$ and $S\in R(\Gamma_0)$ implies that $S$ is the unique singular point in $\widehat\omega(\phi_0)$. Indeed $\Gamma_0$ is contained in ${\mathbb{S}}^2\setminus\omega(\phi_0)$, as is $L(\Gamma_0)$. Similarly $S$ is the unique singular point contained in $\widehat\alpha(\phi_1)$. We deduce that $\phi_1$ does not meet $\Gamma_0$, otherwise $\Gamma_0$ is contained in ${\mathbb{S}}^2\setminus\alpha(\phi_1)$, as is $L(\Gamma_0)$. Similarly $\phi_0$ does not meet $\Gamma_1$.
Suppose that $S\not\in\omega(\phi_0)$. In that case there is no singular point in $\omega(\phi_0)$, which implies that $\omega(\phi_0)$ is a closed leaf $\phi'_0$. But $\phi_0'$ separates $z_0$ and $S$, and this is impossible since both belong to the same Birkhoff recurrence class. Therefore $S\in{\omega}(\phi_0)$ and one can likewise show that $S\in\alpha(\phi_1)$. Let us prove now that $\alpha(\phi_0)$ does not contain any singular point. Indeed if $z'\in\alpha(\phi_0)\cap \mathrm{Fix}(I')$, there exists a transverse simple loop $\Gamma_1$ that separates $S$ and $z'$ and that does not meet $\phi_0$. So $\alpha(\phi_0)$ is a closed leaf $\phi''_0$. Similarly, $\omega(\phi_1)$ is a closed leaf $\phi''_1$. Note that there is no closed leaf that separates $\phi_0$ and $\phi_1$ because both $z_0$ and $z_1$ are in the same Birkhoff recurrence class of $f^q$. So the leaves $\phi''_0$ and $\phi''_1$ are equal. But this is impossible because this leaf cannot be both attracting and repelling from the side that contains $S$.
The following Corollary will be later used to prove a broader result in Corollary \[cr:onlytwoperiods\]
\[cr:divideperiods\] Let $f$ be an orientation preserving homeomorphism of ${\mathbb{S}}^2$. We suppose that there exists a Birkhoff recurrence class $\mathcal{B}$ which contains two periodic points, $z_0, z_1$, with respective smallest periods $q_0<q_1$, and that $q_{0}$ does not divide $q_{1}$. Then $f$ has a topological horseshoe.
We suppose for a contradiction that $f$ has no topological horseshoe. By Proposition \[prop: birkhoff connexions\], one can find a point in the orbit of $z_0$ and a point in the orbit of $z_1$ that belong to the same Birkhoff recurrence class of $f^{q_1}$ and there is no loss of generality by supposing that this is the case for $z_0$ and $z_1$. Note that for every $k\in{\mathbb{Z}}$, the points $f^k(z_0)$ and $f^k(z_1)$ belong to the same Birkhoff recurrence class of $f^{q_1}$. Suppose that $q_0$ does not divide $q_1$ and denote $s$ the remainder of the Euclidean division of $q_{1}$ by $q_{0}$. It implies that $f^{q_1}(z_{0})=f^s(z_{0})\not=z_0$. Since $q_{1}$ is not a multiple of $q_{0}$, it is larger than two and $f^{q_{1}}$ must have at least three distinct fixed points. Choose a maximal isotopy $I'$ of $f^{q_{1}}$ whose singular set contains contains $z_1$, $f^{s}(z_1)$ and at least a third fixed point of $f^{q_{1}}$, then consider a singular foliation ${\mathcal F}'$ transverse to $I'$. The point $z_{0}$ belongs to the domain of $I'$ and $I'_{{\mathcal F}'}{}^{{\mathbb{Z}}}(z_{0})$ is equivalent to the natural lift of a simple transverse loop $\Gamma$. Otherwise, one knows by Proposition \[prop:non-wandering\] that $f$ has a horseshoe (see also [@LeCalvezTal])\]. Moreover $I'_{{\mathcal F}'}{}^{{\mathbb{Z}}}(z_{0})$ coincides with $I'_{{\mathcal F}'}{}^{{\mathbb{Z}}}(f^s(z_{0}))$ because $f^s(z_{0})=f^{q_{1}}(z_{0})$. One can find two fixed points $z$ and $z'$ of $I'$ separated by $\Gamma$, and since $I'$ has at least three fixed points, one can assume that $\{z, z'\}\not=\{z_{1},f^s(z_{1})\}$. The isotopy $I'$ can be lifted to an identity isotopy of a certain lift $\check g$ of $f^{q_{1}}\vert_{{\mathbb{S}}^2\setminus\{z, z'\}}$ and $\mathrm{rot}_{\check g}(z_0)=\mathrm{rot}_{\check g}(f^s(z_0))\not=0.$ Moreover, $\mathrm{rot}_{\check g}(z'')=0,$ for every fixed point $z''$ of $I'$ that is different from $z$ and $z'$. So, if $z_{1}\not\in\{z, z'\}$, we have $\mathrm{rot}_{\check g}(z_0)\not=\mathrm{rot}_{\check g}(z_1)$ and if $f^s(z_{1})\not\in\{z, z'\}$, we have $\mathrm{rot}_{\check g}(f^s(z_0))\not=\mathrm{rot}_{\check g}(f^s(z_1))$. By Theorem \[th:birkhoffcycles\], this contradicts the fact that $f$ has no topological horseshoe.
Circloids
---------
Let us remind the statement of Theorem \[thmain:circloids\] of the introduction.
\[th:circloids\] Let $f$ be a homeomorphism of ${\mathbb{A}}$ isotopic to the identity and $\check f$ a lift of $f$ to ${\mathbb{R}}^2$. We suppose that $f$ has no topological horseshoe. Let $X$ be an invariant circloid. If $f$ is dissipative or if $X$ is locally stable, then the function $\rho_{\widetilde f}$, which is well defined on $X$, is constant on this set. Moreover, the sequence of maps $$\check z\mapsto \frac{ \pi_1(\check f^{n}(\check z))- \pi_1(\check z)}{n}$$ converges uniformly to this constant on $\pi^{-1}(X)$.
We suppose that $f$ and $X$ satisfy the hypothesis of the theorem. In particular $f$ is dissipative or $X$ is locally stable. We denote $U_S$ and $U_N$ the non relatively compact connected components of ${\mathbb{A}}\setminus X$, neighborhoods of $S$ and $N$ respectively. By Theorem \[th:rotation-number\], the function $\rho_{\widetilde f}$ is well defined on $X$, and it is continuous when restricted to $\Omega(f)\cap X$. To prove that it is constant, it is sufficient to prove that the sequence of maps $$\check z\mapsto \frac{ \pi_1(\check f^{n}(\check z))- \pi_1(\check z)}{n}$$ converges uniformly to a constant on $\pi^{-1}(X)$. Assume for a contradiction that it is not the case. Koropecki proved in [@Koropecki] that there exist real numbers $\rho<\rho'$ such that, for every $p/q\in (\rho,\rho')$ written in an irreducible way, there exists a periodic point of period $q$ and rotation number $p/q$, extending the method of Barge-Gillette used in the special case of cofrontiers (see [@BargeGillette]). Let us consider in $(\rho,\rho')$ a decreasing sequence $(p_n/q_n)_{n{\geqslant}0}$ converging to $\rho$ and an increasing sequence $(p'_n/q'_n)_{n{\geqslant}0}$ converging to $\rho'$. For every $n$, choose a periodic point $z_n$ of period $q_n$ and rotation number $p_n/q_n$ and denote $O_n$ its orbit. Similarly, choose a periodic point $z'_n$ of period $q'_n$ and rotation number $p'_n/q'_n$ and denote $O'_n$ its orbit. Taking subsequences if necessary, one can suppose that the sequences $(z_n)_{n{\geqslant}0}$ and $ (z'_n)_{n{\geqslant}0}$ converge to points $z$ and $z'$ in $X$, respectively. Since the rotation number is continuous on the nonwandering set of $f$ restricted to $X$, we get that $\rho_{\check f}(z)=\rho$ and $\rho_{\check f}(z)=\rho'$. We will prove now that there is a Birkhoff connection from $z$ to $z'$. Of course, one could prove in the same way that there is a Birkhoff connection from $z'$ to $z$. So, by Theorem \[th:birkhoffcycles\], it implies that $f$ has a topological horseshoe, in contradiction with the hypothesis.
Say an open set $U\subset{\mathbb{A}}$ is [*essential*]{} if it contains an essential loop. Then
\[lemma:essential\] If $W$ is a forward (respectively backward) invariant open set that contains $z$, then the connected component of $z$ in $W$ is essential and also forward (respectively backward) invariant.
Let $W$ be a forward invariant open set that contains $z$. The connected component $V$ of $W$ that contains $z$ contains the periodic points $z_n$ if $n$ is large, so it is forward invariant by a power $f^q$ of $f$ and contains the $f^q$-orbit of $z_n$. Let $\check V$ be a connected component of $\check \pi^{-1}(V)$. If $n$ is large, $\check V$ contains a point $\check z_n$ such that $\check f^{qq_n}(\check z_n)=T^{qp_n}(\check z_n)$ and so $\check f^{qq_n}(\check V)\subset T^{qp_n}(\check V)$. Let us fix $n_0<n_1$ large enough. One deduces that $\check f^{qq_{n_0}q_{n_1}}(\check V)$ is included both in $T^{qp_{n_0}q_{n_1}}(\check V)$ and in $T^{qq_{n_0}p_{n_1}}(\check V)$, which implies that $T^{qp_{n_0}q_{n_1}}(\check V)=T^{qq_{n_0}p_{n_1}}(\check V)$. As $qp_{n_0}q_{n_1}\not=qq_{n_0}p_{n_1}$, it implies that $V$ is essential. Furthermore, since $V$ is a connected component of $W$ and $W$ is forward invariant, then either $f(V)\subset V$, or $f(V)$ is disjoint from $V$. But any open essential set of ${\mathbb{A}}$ whose image is disjoint from itself must be wandering, and since $V$ has periodic points, this implies that $V$ is forward invariant. We have a similar proof in case $W$ is backward invariant.
Let $U$ be a neighborhood of $z$ and $U'$ be a neighborhood of $z'$ such that $U\cap U'=\emptyset$. Set $V$ to be the connected component of $\bigcup_{n{\geqslant}0} f^n(U)$ that contains $z$ and $V'$ to be the connected component of $\bigcup_{n{\geqslant}0} f^{-n}(U')$ that contains $z'$. We will prove that $V\cap V'\not=\emptyset$. By Lemma \[lemma:essential\], $V$ is a connected essential and forward invariant open set that contains $z$. Similarly, $V'$ is an essential connected backward invariant open set that contains $z'$.
We will suppose from now that $V\cap V'\not=\emptyset$ and will find a contradiction. The connected component $K'$ of ${\mathbb{A}}\setminus V$ that contains $V'$ is a neighborhood of one end of the annulus, $N$ for instance, and is backward invariant. Its complement $W$ is an open neighborhood of $S$ that contains $V$ and is disjoint from $V'$. More precisely, it is a forward invariant open annulus. Similarly, the connected component $K$ of ${\mathbb{A}}\setminus V'$ that contains $V$ is a neighborhood of $S$ and is forward invariant. Moreover it contains $W$ because $V'\subset K'$. Its complement $W'$ is a backward invariant annulus, neighborhood of $N$ that contains $V'$. Moreover, one has $W\cap W'=\emptyset$ because $W\subset K$.
![Theorem \[th:circloids\], the case where $f$ is dissipative.[]{data-label="figure_circloid"}](circloid.pdf){height="48mm"}
The fact that $X$ is a circloid implies that $\overline U_S$ intersects $V'$. So $V'\cap U_S$ is a non empty open set, relatively compact and backward invariant (see Figure \[figure\_circloid\]). Consequently, $f$ is not dissipative. The open set $W'\cap U_S$ is not empty because $V'\subset W'$. It implies that $W'\cup U_S$ is connected and contains a topological line $\lambda$ that joins $N$ to $S$. If $U$ is a sufficiently small neighborhood of $X$, then $U\cap\lambda\subset W'$. So, $W\cap U$ does not meet $\lambda$ because $W\cap U\cap\lambda\subset W\cap W'=\emptyset$, and is not essential. If $X$ is locally stable, one can choose such a neighborhood $U$ to be forward invariant. This contradicts Lemma \[lemma:essential\] because $W\cap U$ contains $z$ and is not essential.
Annular homeomorphisms with fixed point free lifts
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Let us finish this section with the proof of Proposition \[prop:zero-rotation\] that we remind:
\[prop:recurrentzero-rotation\] Let $f$ be a homeomorphism isotopic to the identity of ${\mathbb{A}}$ and $\check f$ a lift of $f$ to the universal covering space. We suppose that $\check f$ is fixed point free and that there exists $z_*\in\Omega(f)\cap \mathrm{ne}^+(f)$ such that $\mathrm{rot}_{\check f}(z_*)$ is well defined and equal to $0$.
- Then, at least one of the following situation occurs:
1. there exists $q$ arbitrarily large such that $\check f^q\circ T^{-1}$ has a fixed point;
2. there exists $q$ arbitrarily large such that $\check f^q\circ T$ has a fixed point.
- [Moreover, if $z_*$ is recurrent, then $f$ has a horseshoe.]{}
We begin with the proof of $i)$. Let $I$ be a maximal isotopy of $f$ that is lifted to an identity isotopy $\check I$ of $\check f$. Choose a transverse foliation $\mathcal F$ of $I$ and write $\check {\mathcal F}$ for the lifted foliation. Note that $\mathcal F$ and $\check {\mathcal F}$ are not singular because $\check f$ has no fixed point. By hypothesis $\omega_f(z_*)$ is non empty and $I_{\mathcal F}^{{\mathbb{N}}}(z_*)$ meets infinitely often every leaf that contains a point of $\omega_f(z_*)$. Fix such a leaf $\phi$ and a lift $\check z_*$ of $z_*$. There exists two integers $p_0$ and $q_0{\geqslant}2$ and a lift $\check \phi$ of $\phi$ such that $\check I_{\check{\mathcal F}}^{q_0-1}(\check f(\check z_*))$ contains as a sub-path a transverse path that goes from $\check \phi$ to $T^{p_0}(\check \phi)$. The foliation $\check {\mathcal F}$ being non singular, every transverse path meets a leaf at most once and so $p_0\not=0$. For the same reason, $\check z_*$ is not periodic and consequently $z_*$ itself is not periodic (because $\mathrm{rot}_{\check f}(z_*)=0$). We will prove the first assertion in case $p_0>0$ (and would obtain the second one in case $p_0<0$).
One can find an open disk $D$ containing $z_*$ such that:
- $f^k(D)\cap D=\emptyset$ for every $k\in\{1,\dots, q_0\}$;
- if $\check D$ is the lift of $D$ that contains $\check z_*$, then for every $\check z\in\check D$, the path $\check I_{\check{\mathcal F}}^{q_0-1}(\check f(\check z_*))$ is a sub-path of $\check I_{\check{\mathcal F}}^{q_0+1}( \check z)$ (up to equivalence) .
The second item tells us that there exists a transverse path from every point of $\check D$ to $\check \phi$ and a transverse path from $T^{p_0}(\check \phi)$ to every point of $\check f^q(D)$ if $q>q_0$.
The point $z_*$ being non-wandering, there exists $q>q_0$ arbitrarily large such that $f^{q}(D)\cap D\not=\emptyset$. We will prove that $\check f^q\circ T^{-1}$ has a fixed point. There exists $p\in{\mathbb{Z}}$ such that $\check f^q(\check D)\cap T^p(\check D)\not=\emptyset$. Let us prove that $p>p_0$. There exists a transverse path from $\check \phi$ to $T^{p_0}(\check \phi)$ and so a transverse path from $\check \phi$ to $T^{np_0}(\check \phi)$ for every $n{\geqslant}1$. There exists a transverse path from every point of $\check D$ to $\check \phi$ and a transverse path from $T^{p_0}(\check \phi)$ to every point of $\check f^q(D)$. So, if $\check f^q(\check D)\cap T^p(\check D)\not=\emptyset$, there exists a transverse path from $T^{p_0}(\check\phi)$ to $T^{p}(\check\phi)$, which implies that there is a transverse path from $\check\phi$ to $T^{p-p_0}(\check\phi)$ and for every $m{\geqslant}1$ a transverse path from $\check\phi$ to $T^{m(p-p_0)}(\check\phi)$. Consequently, there exists a transverse path from $\check\phi$ to $T^{np_0+m(p-p_0)}(\check\phi)$. If $p{\leqslant}p_0$, one can choose suitably $n$ and $m$, such that $np_0+m(p-p_0)=0$, but this is impossible, as otherwise there would be a transverse path beginning and ending at $\check \phi$, which contradicts the fact that $\check{\mathcal{F}}$ is nonsingular. In particular, $\check f^q(\check D)\cap T^p(\check D)=\emptyset$ if $p{\leqslant}1$.
Let $I'$ be a maximal isotopy of $f'=f^q$ that is lifted to an identity isotopy $\check I'$ of $\check f'=\check f^q\circ T^{-1}$. By hypothesis, $z_*\in\mathrm{ne}^+(f')\cap\Omega(f')$ and $\mathrm{rot}_{\check f'}(z_*)=-1$. By definition of $q$, the set $D\cap f'{}^{-1}(D)$ is not empty. Fix a point $z$ in this set. As explained in [@LeCalvezTal], one can find a foliation ${\mathcal F}'$ to $I'$ such that $z_*$, $z$ and $f'(z)$ belong to the same leaf $\phi'_*$ of $\mathcal F'\vert_{D}$. The reason being that $\check D$ is a free disk of $\check f'$. We want to prove that ${\mathcal F}'$ has singularities. Write $\check z$ for the lift of $z$ that belong to $\check D$ and $\check \phi'_*$ for the lift of $\phi'_*$ that contains $\check z_*$ and $\check z$. Note that $\check f'(\check z')\in T^{p-1}(\check \phi_*)$, because $\check f'(\check z')\in T^{p-1}(\check D)$ by hypothesis. So there exists a transverse path from $\check \phi'_*$ to $T^{p-1}(\check \phi'_*)$ and consequently a positive loop $\Gamma_+$ transverse to ${\mathcal F}'$ that meets $\phi'_*$. The fact that $z_*\in\mathrm{ne}^+(f')\cap\Omega(f')$ and $\mathrm{rot}_{\check f'}(z_*)<0$ implies that $I'{}_{\mathcal F'}^{{\mathbb{Z}}}(z_*)$ is drawn on a negative transverse simple loop $\Gamma_-$. There exist $n_-{\geqslant}1$ such that the multi-loop $\Gamma_++n_-\Gamma_-$ is homological to zero. As explained in the proof of Proposition \[prop:PB\], we can suppose that its dual function $\delta$ that vanishes in a neighborhood of the ends of ${\mathbb{A}}$ has a negative minimal value $m_-$. We have seen in the proof of Proposition \[prop:PB\] that a connected component $U$ of ${\mathbb{A}}\setminus (\Gamma_-\cup\Gamma_+)$ where $\delta$ is equal to $m_-$ contains a singular point of ${\mathcal F}'$ if it is a topological disk. If it is not disk, it is an open annulus separating $\Gamma_-$ and $\Gamma_+$. This is impossible because the leaves intersecting the boundary of $U$ are going inside, which forbids $\check \phi'_*$ to meet both $\Gamma_-$ and $\Gamma_+$.
To get the second assertion of the proposition, note that the point $z$ can be chosen equal to $z_*$ in case $z_*$ is recurrent. This implies that $I'{}_{\mathcal F'}^{{\mathbb{Z}}}(z_*)$ draws $\Gamma_+$ and $\Gamma_-$. The point $z_*$ being non-wandering, we have seen before that it implies that $f$ has a topological horseshoe. Note that we can also give a proof by using Corollary \[cor:free-disks\] applied to $\check D$ and $\check f'$.
Note that the first item is not true if we suppose that $z\in\mathrm{ne}^+(f)\setminus\Omega(f)$. Consider on ${\mathbb{S}}^2={\mathbb{A}}\cup\{N,S\}$ a flow satisfying:
- the end $N$ and $S$ are the only fixed points,
- there exist an orbit $O$ homoclinic to $S$,
- the disk bounded by $O\cup \{S\}$ that does not contain $N$ is foliated by orbits homoclinic to $S$,
- the $\alpha$-limit set of a point disjoint from this disk is equal to $N$ and its $\omega$- limit set to $O\cup \{S\}$ if the orbit is not reduced to $N$,
Now, choose $f$ to be the time one map of this flow, reduced to ${\mathbb{A}}$, and $\check f$ the time one map of the lifted flow.
Homeomorphisms of the sphere with no topological horseshoe
===========================================================
The goal of this section is to prove Theorem \[thmain:global-structure\] of the introduction, the fundamental result that permits to describe the structure of homeomorphisms of the sphere with no topological horseshoe. We will introduce the notion of positive an squeezed annuli in Sub-section \[subsection: annuli\], which will be fundamental in our study, then will state and prove a “local version” of the theorem in Sub-sections \[subsection: prooflocal\] and [\[subsection:auxilliary\]]{}, which means a version related to a maximal isotopy. The proof of Theorem \[thmain:global-structure\] will be given in Sub-section \[subsection: proofglobal\]
Positive and squeezed annuli {#subsection: annuli}
----------------------------
Let $f$ be a homeomorphism of ${\mathbb{A}}$ isotopic to the identity and $\check f$ a lift of $f$ to the universal covering space. A [*$\check f$-positive disk*]{} is a topological open disk $D\subset{\mathbb{A}}$ such that:
- every lift $\check D$ of $D$ is $\check f$-free;
- there exist $p>0$ and $q>0$ satisfying $f^q(\check D)\cap T^p(\check D)\not=\emptyset$.
Replacing the condition $p>0$ by $p<0$ in the definition one defines similarly a [*$\check f$-negative disk*]{}.
Of course a disk $D\subset{\mathbb{A}}$ that contains a $\check f$-positive disk is a $\check f$-positive disk, if it is lifted to $\check f$-free disks. Franks’ lemma (see [@Franks]) asserts that if $\check f$ is fixed point free, then a disk $D\subset{\mathbb{A}}$ that is lifted to $\check f$-free disks cannot be both $\check f$-positive and $\check f$-negative. Moreover, every lift of $D$ is wandering. This implies, again assuming $\check f$ is fixed point free, that every point $z\in\Omega(f)$ admits a fundamental system of neighborhoods by $\check f$-positive disks or a fundamental system of neighborhoods by $\check f$-negative disks. In the first situation, every disk containing $z$ and lifted to $\check f$-free disks is positive, and we will say that $z$ is [*$\check f$-positive*]{}. We will use the following easy result:
\[prop: positive\] Let $f$ be a homeomorphism of ${\mathbb{A}}$ isotopic to the identity and $\check f$ a lift of $f$ to the universal covering space. We suppose that $f$ has no horseshoe and that $\check f$ is fixed point free. Let $I$ be a maximal isotopy of $f$ that is lifted to an identity isotopy of $\check f$ and $\mathcal F$ a transverse foliation of $I$. Then if $z\in\Omega(f)$ is $\check f$-positive, there exists a positive transverse loop that meets the leaf $\phi_z$.
Write $\check{\mathcal F}$ for the lifted foliation to ${\mathbb{R}}^2$. Suppose that $z\in\Omega(f)$ is $\check f$-positive. If $D$ is a topological open disk containing $z$ sufficiently small, if $\check z\in{\mathbb{R}}^2$ is a lift of $z$ and $\check D$ the lift of $D$ containing $z$, then $I^{2}_{\check {\mathcal F}} ( \check f^{-1}(\check z'))$ meets $\phi_{\check z}$ for every $\check z'\in\check D$. By hypothesis, one can find $\check z'\in \check D$, $n{\geqslant}2$ and $p>0$ such $\check f^p(\check z')\in T^p(\check D)$. Consequently $I^{n+2}_{\check {\mathcal F}} ( \check f^{-1}(\check z'))$ contains as a sub-path a transverse path joining $\phi_{\check z}$ to $T^p(\phi_{\check z})$.
Let $f$ be a homeomorphism of ${\mathbb{A}}$ isotopic to the identity such that $\Omega(f)\not=\emptyset$. We will say that a lift $\check f$ of $f$ to the universal covering space, is a [*positive lift*]{} of $f$ if:
- $\check f$ is fixed point free,
- every point $z\in\Omega(f) $ is positive.
We can define similar objects replacing “positive” by “negative”. Note that if $\check f$ is a positive lift of $f$, then the rotation number of a non-wandering point is non negative when it is defined.
Let $f$ be a homeomorphism of ${\mathbb{A}}$ isotopic to the identity. We will say that $f$ is [*squeezed*]{} if there exists a lift $\check f$ such that
- $\check f$ is positive,
- $\check f\circ T^{-1}$ is negative.
Note that a squeezed homeomorphism is fixed point free.
Let $f$ be an orientation preserving homeomorphism of ${\mathbb{S}}^2$. A [*squeezed annulus* ]{} of $f$ is an invariant open annulus $A\subset {\mathbb{S}}^2$ such that $f\vert _A$ is conjugate to a squeezed homeomorphism of ${\mathbb{A}}$.
Note that if $A$ is a fixed point free invariant annulus, then the two connected components of ${\mathbb{S}}^2\setminus A$ are fixed. Otherwise there are permuted by $f$, which implies that $f$ is fixed point free, in contradiction with the fact that it preserves the orientation. Consequently, $f\vert_A$ is isotopic to the identity. Note that if $A\subset A' $ are squeezed annuli, then $A$ is essential in $A'$ (which means that it contains a loop non homotopic to zero in $A$). Otherwise, if we add to $A$ the compact connected component $K$ of $A'\setminus A$, we obtain an invariant disk included in $A'$ that contains a non-wandering point (because $K$ is compact and invariant) and such a disk must contain a fixed point. So, if $(A_j)_{j\in J}$ is a totally ordered family of squeezed annuli, then $\bigcup_{j\in J} A_j$ is a squeezed annulus. Consequently, by Zorn’s lemma, every squeezed annulus is included in a maximal squeezed annulus. It is a classical fact that if $f$ has no wandering point, the squeezed annuli are the fixed point free invariant annuli. Let us explain why. As explained above, if $A$ is a fixed point free invariant annulus, then $f\vert_A$ is isotopic to the identity. So to prove that $A$ is squeezed, it is sufficient to prove that a homeomorphism $f$ of ${\mathbb{A}}$ isotopic to the identity that has no wandering point and is fixed point free, is squeezed. Fix a lift $\check f$ and a free disk $D$ and consider the first integer $q>1$ such that $f^q(D)\cap D\not=\emptyset$. If $\check D$ is a lift of $D$, there exists $p\in{\mathbb{Z}}$ such that $\check f^q(\check D)\cap T^p(\check D)\not=\emptyset$. Replacing $\check f$ by another lift if necessary, one can suppose that $0{\leqslant}p<q$. As explained above, $p$ is different from $0$ and $D$ is a $\check f$-positive disk and a $\check f\circ T^{-1}$-negative disk. The set of positive points of $\widetilde f$ is open, being the union of $\widetilde f$-positive disks. Similarly, the set of negative points of $\widetilde f$ is open. So, by connectedness of ${\mathbb{A}}$, one deduces that every point of ${\mathbb{A}}$ is a positive point of $\widetilde f$ and for the same reason is a negative point of $\widetilde f\circ T^{-1}$.
Let us define now positive invariant annuli of an orientation preserving homeomorphism $f$ of ${\mathbb{S}}^2$. They will be defined relative to a maximal isotopy $I$ of $f$. It has been explained in the introduction (just before the statement of Theorem \[thmain:rotation-number\]) that if $A\subset {\mathbb{S}}^2$ is a topological open annulus invariant by $f$, such that $f$ fixes the two connected components of ${\mathbb{S}}^2\setminus A$, one can define rotation numbers of $f\vert_A$ as soon a generator $\kappa$ of $H_1(A,{\mathbb{Z}})$ is chosen. Remind that $\kappa_*$ is the generator of $H_1({\mathbb{A}},{\mathbb{Z}})$ induced by the loop $\Gamma_* :t\mapsto (t,0)$. If $U\subset V$ are two open sets of ${\mathbb{S}}^2$ we will denote $\iota_*: H_1(U,{\mathbb{Z}})\to H_1(V,{\mathbb{Z}})$ the morphism induced by the inclusion map $\iota: U\to V$. Let $I$ be a maximal isotopy of $f$. Remind that $\widetilde \pi :\widetilde{\mathrm{dom}}(I)\to\mathrm{dom}(I)$ is the universal covering projection and that $I$ can be lifted to $\widetilde{\mathrm{dom}}(I)$ into an identity isotopy of a certain lift $\widetilde f$ of $f\vert_{\mathrm{dom}(I)}$. Let $A\subset \mathrm{dom}(I)$ be a topological open annulus invariant by $f$ such that each connected component of ${\mathbb{S}}^2\setminus A$ contains at least a fixed point of $I$. This implies that the two connected components of ${\mathbb{S}}^2\setminus A$ are fixed. Let $\widetilde A$ be a connected component of $\widetilde\pi^{-1}(A)$. Consider a generator $T$ of the stabilizer of $\widetilde A$ in the group of covering automorphisms. There exists a covering automorphism $S$ such that $\widetilde f(\widetilde A)=S(\widetilde A)$ and one must have $S\circ T\circ S^{-1}=T$ because $\widetilde f$ commutes with the covering automorphisms. Of course, one can suppose that $\mathrm{dom}(I)$ is connected, which implies that the group of covering automorphisms is a free group. Consequently $S$ is a power of $T$ and $\widetilde A$ is invariant by $\widetilde f$. Moreover $\widetilde A$ is the universal covering space of $A$ and $\widetilde f\vert_{\widetilde A}$ is a lift of $f\vert_A$.
Let $f$ be an orientation preserving homeomorphism of ${\mathbb{S}}^2$ and $I$ a maximal isotopy of $f$. A [*positive annulus* ]{} of $I$ is an open annulus $A\subset {\mathbb{S}}^2$ invariant by $f$ verifying:
- each connected component of ${\mathbb{S}}^2\setminus A$ contains at least a fixed point of $I$;
- there exists a homeomorphism $h:A\to {\mathbb{A}}$ such that $\widetilde f\vert_{\widetilde A}$ is a positive lift of $h\circ f\vert_A\circ h^{-1}$, where $\widetilde h:\widetilde A\to {\mathbb{R}}^2$ is a lift of $h$ between the two universal covering spaces (we keep the notations introduced above to define $\widetilde f$ and $\widetilde A$).
In that case the [*positive class*]{} of $ H_1(A,{\mathbb{Z}})$ is the generator $\kappa$ such that $h_*(\kappa)=\kappa_*$, its inverse being the [*negative class*]{}.
If $A\subset A' $ are two positive annuli, then $A$ is essential in $A'$. So, if $(A_j)_{j\in J}$ is a totally ordered family of positive annuli, then $\bigcup_{i\in I} A_i$ is a positive annulus, which implies that every positive annulus of $I$ is included in a maximal positive annulus. Note also that if $A\subset A' $ are two positive annuli, the positive class of $H_1(A,{\mathbb{Z}})$ is sent onto the positive class of $H_1(A',{\mathbb{Z}})$ by the morphism $\iota_*: H_1(A,{\mathbb{Z}})\to H_1(A',{\mathbb{Z}})$.
Local version of Theorem \[thmain:global-structure\] {#subsection: prooflocal}
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If $f$ is an orientation preserving homeomorphism of ${\mathbb{S}}^2$ we set $$\Omega'(f)=\{z\in\Omega(f)\,\vert\, \alpha(z)\cup\omega(z)\not\subset\mathrm{fix}(f)\}.$$Moreover, if $I$ is a maximal isotopy of $f$, we set $$\Omega'(I)=\{z\in\Omega(f)\,\vert\, \alpha(z)\cup\omega(z)\not\subset\mathrm{fix}(I)\}.$$ In this long sub-section, we will prove the following result:
\[th:local-structure\] Let $f:{\mathbb{S}}^2\to{\mathbb{S}}^2$ be an orientation preserving homeomorphism that has no topological horseshoe, and $I$ a maximal isotopy of $f$. Then $\Omega'(I)$ is covered by positive annuli.
We suppose in \[subsection: prooflocal\] that $f$ is an orientation preserving homeomorphism of ${\mathbb{S}}^2$ with no topological horseshoe and that $I$ is a maximal isotopy of $f$. We want to cover $\Omega'(I)$ by positive annuli. To construct these annuli, we need to use a foliation $\mathcal F$ transverse to $I$. Recall that the whole transverse trajectory of a point $z\in\Omega'(I)$ draws a unique transverse simple loop and exactly draws it infinitely. In particular, it is contained in the open annulus $U_{\Gamma}$, union of leaves that meet $U$. We want to construct a positive annulus that contains $z$. The simplest case where it is possible to do so is the case where $U_{\Gamma}$ itself is invariant. This is a very special case because it happens if and only if $U_{\Gamma}$ is a connected component of $\mathrm{dom}(I)$. Another case where it is not difficult to construct such an annulus is the case where $U_{\Gamma}$ is bordered by two closed leaves $\phi$ and $\phi'$ . The reader should convince [himself]{} that the set of points whose whole transverse trajectory meets $U_{\Gamma}$, which is nothing but the set $$U_{\Gamma}\cup \left(\bigsqcup_{k\in{\mathbb{Z}}} R(f^{k+1}(\phi))\setminus L(f^k(\phi))\right)\cup\left(\bigsqcup_{k\in{\mathbb{Z}}} R(f^{k+1}(\phi'))\setminus L(f^k(\phi'))\right)$$ is a positive annulus containing $z$. The general case is much more complicated. Roughly speaking[, one]{} will consider[, for a given connected component of the complement of $U_{\Gamma}$, either]{} the set of points that “enter in the annulus $U_{\Gamma}$” [from that connected component or the set that “leave it” to that component, but not both, and similarly the points that enter or leave from the other connected component of the complement of $U_{\Gamma}$]{}. We will need many of the results previously stated in this article, like the result about general regions of instability (Proposition \[prop:PB\])for instance, to succeed in constructing our annulus.
For such a loop $\Gamma$, define the following sets
- $r_{\Gamma}$ is the connected component of ${\mathbb{S}}^2\setminus
U_{\Gamma}$ that lies on the right of $\Gamma$;
- $l_{\Gamma}$ is the connected component ${\mathbb{S}}^2\setminus
U_{\Gamma}$ that lies on the left of $\Gamma$;
- $\Omega'(I)_{\Gamma}$ is the set of points of $z\in\Omega'(I)$ such that $I_{\mathcal F}^{{\mathbb{Z}}}(z)$ draws $\Gamma$;
- $V_{\Gamma}$ is the set of points $z$ such that $I_{\mathcal F}^{{\mathbb{Z}}}(z)$ draws $\Gamma$;
- $W_{\Gamma}$ is the set of points $z$ such that $I_{\mathcal F}^{{\mathbb{Z}}}(z)$ meets $U_{\Gamma}$;
- $W_{\Gamma}^{\rightarrow r}$ is the set of points $z$ such that $I_{\mathcal F}^{{\mathbb{Z}}}(z)$ meets a leaf $\phi\subset r_{\Gamma}\cap \overline U_{\Gamma}$ where $U_{\Gamma}$ is locally on the right of $\phi$;
- $W_{\Gamma}^{r\rightarrow }$ is the set of points $z$ such that $I_{\mathcal F}^{{\mathbb{Z}}}(z)$ meets a leaf $\phi\subset r_{\Gamma}\cap \overline U_{\Gamma}$ where $U_{\Gamma}$ is locally on the left of $\phi$;
- $W_{\Gamma}^{\rightarrow l}$ is the set of points $z$ such that $I_{\mathcal F}^{{\mathbb{Z}}}(z)$ meets a leaf $\phi\subset l_{\Gamma}\cap \overline U_{\Gamma}$ where $U_{\Gamma}$ is locally on the right of $\phi$;
- $W_{\Gamma}^{l\rightarrow }$ is the set of points $z$ such that $I_{\mathcal F}^{{\mathbb{Z}}}(z)$ meets a leaf $\phi\subset l_{\Gamma}\cap \overline U_{\Gamma}$ where $U_{\Gamma}$ is locally on the left of $\phi$.
Equivalently, $W_{\Gamma}^{\rightarrow r}$ is the set of points $z$ such that there exist $s<t$ verifying $$I_{\mathcal F}^{{\mathbb{Z}}}(z)(s)\in U_{\Gamma}, \enskip I_{\mathcal F}^{{\mathbb{Z}}}(z)(t)\in r_{\Gamma}$$ and $W_{\Gamma}^{r\rightarrow }$ is the set of points $z$ such that there exist $s<t$ verifying $$I_{\mathcal F}^{{\mathbb{Z}}}(z)(s)\in r_{\Gamma},\enskip I_{\mathcal F}^{{\mathbb{Z}}}(z)(t)\in U_{\Gamma}.$$ Note that the six last sets are open and invariant, and that:
- the whole transverse trajectory of $z\in W_{\Gamma}\setminus (W_{\Gamma}^{r\rightarrow}\cup W_{\Gamma}^{\rightarrow l})$ meets $U_{\Gamma}$ in a unique real interval $(a,b)$, moreover $I_{\mathcal F}^{{\mathbb{Z}}}(z)(a)\in l_{\Gamma}$ if $a>-\infty$ and $I_{\mathcal F}^{{\mathbb{Z}}}(z)(b)\in r_{\Gamma}$ if $b<+\infty$;
- the whole transverse trajectory of $z\in W_{\Gamma}\setminus (W_{\Gamma}^{l\rightarrow}\cup W_{\Gamma}^{\rightarrow r})$ meets $U_{\Gamma}$ in a unique real interval $(a,b)$, moreover $I_{\mathcal F}^{{\mathbb{Z}}}(z)(a)\in r_{\Gamma}$ if $a>-\infty$ and $I_{\mathcal F}^{{\mathbb{Z}}}(z)(b)\in l_{\Gamma}$ if $b<+\infty$;
- the whole transverse trajectory of $z\in W_{\Gamma}\setminus (W_{\Gamma}^{\rightarrow r}\cup W_{\Gamma}^{\rightarrow l})$ meets $U_{\Gamma}$ in a unique real interval $(a,+\infty)$ where $a{\geqslant}-\infty$;
- the whole transverse trajectory of $z\in W_{\Gamma}\setminus (W_{\Gamma}^{r\rightarrow}\cup W_{\Gamma}^{l\rightarrow })$ meets $U_{\Gamma}$ in a unique real interval $(-\infty,b)$ where $b{\leqslant}+\infty$.
We will define the four following invariant open sets:
$$W_{\Gamma}^{l\rightarrow r}=W_{\Gamma}\setminus (\overline{W_{\Gamma}^{r\rightarrow}}\cup \overline{W_{\Gamma}^{\rightarrow l}})$$ $$W_{\Gamma}^{r\rightarrow l}=W_{\Gamma}\setminus (\overline{W_{\Gamma}^{l\rightarrow}}\cup \overline{W_{\Gamma}^{\rightarrow r}}),$$ $$W_{\Gamma}^{r,l\rightarrow }=W_{\Gamma}\setminus (\overline{W_{\Gamma}^{\rightarrow r}}\cup \overline{W_{\Gamma}^{\rightarrow l}}),$$ $$W_{\Gamma}^{\rightarrow r,l}=W_{\Gamma}\setminus (\overline{W_{\Gamma}^{r\rightarrow}}\cup \overline{W_{\Gamma}^{l\rightarrow }}).$$
We will prove the following result, which immediately implies Theorem \[th:local-structure\]:
\[prop:local-structure\] There exists a positive annulus $A_{\Gamma}$ that contains $\Omega'(I)_{\Gamma}$ and whose positive generator is the homology class of a simple loop freely homotopic to $\Gamma$ in $\mathrm{dom}(I)$. More precisely, at least one of the set $W_{\Gamma}^{l\rightarrow r}$, $W_{\Gamma}^{r\rightarrow l}$, $W_{\Gamma}^{r,l\rightarrow }$ or $W_{\Gamma}^{\rightarrow r,l }$ has a connected component that contains $\Omega'(I)_{\Gamma}$, and if we add to this component the connected components of its complement that have no singular points, we obtain an open annulus $A_{\Gamma}$ verifying the properties above.
Before proving \[prop:local-structure\], we will study more carefully the sets related to $\Gamma$ that we have introduced. Replacing $\mathrm{dom}(I)$ by a connected component, we can always suppose that $\mathrm{dom}(I)$ is connected. Denote by $\widetilde\pi: \widetilde{\mathrm{dom}}(I)\to\mathrm{dom}(I)$ the universal covering projection, by $\widetilde I$ the lifted identity isotopy, by $\widetilde f$ the induced lift of $f$, by $\widetilde{\mathcal{F}}$ the lifted foliation.
Let $\gamma$ be the natural lift of $\Gamma$. For every lift $\widetilde\gamma$ of $\gamma$ in $\widetilde{\mathrm{dom}}(I)$, define the following objects:
- $T_{\widetilde \gamma}$ is the covering automorphism such that $\widetilde\gamma(t+1)=T_{\widetilde \gamma}(\widetilde\gamma(t))$ for every $t\in{\mathbb{R}}$;
- $U_{\widetilde\gamma}$ is the union of leaves that meet $\widetilde\gamma$;
- $r_{\widetilde\gamma}$ is the union of leaves that are not in $U_{\widetilde\gamma}$ and are on the right of $\widetilde \gamma$;
- $l_{\widetilde\gamma}$ is the union of leaves that are not in $U_{\widetilde\gamma}$ and are on the left of $\widetilde \gamma$;
- $\Omega'(I)_{\widetilde\gamma}$ is the set of points $\widetilde z$ that lift a point of $\Omega'(I)$ and such that $\widetilde I_{\widetilde{\mathcal F}}^{{\mathbb{Z}}}(\widetilde z)$ is included in $U_{\widetilde\gamma}$;
- $V_{\widetilde\gamma}$ is the set of points $\widetilde z$ such that $\widetilde I_{\widetilde{\mathcal F}}^{{\mathbb{Z}}}(\widetilde z)$ contains a sub-path of $\widetilde\gamma$ of length ${\geqslant}1$;
- $W_{\widetilde\gamma}$ is the set of points $\widetilde z$ such that $\widetilde I_{\widetilde{\mathcal F}}^{{\mathbb{Z}}}(\widetilde z)$ meets $U_{\widetilde\gamma}$;
- $W_{\widetilde\gamma}^{\rightarrow r}$ is the set of points $\widetilde z\in W_{\widetilde\gamma}$ such that $\widetilde I_{\widetilde{\mathcal F}}^{{\mathbb{Z}}}(\widetilde z)$ meets a leaf $\widetilde\phi\subset r_{\widetilde\gamma}\cap \overline U_{\widetilde\gamma}$ where $U_{\widetilde\gamma}$ is on the right of $\widetilde \phi$;
- $W_{\widetilde\gamma}^{r\rightarrow }$, $W_{\widetilde\gamma}^{\rightarrow l}$ and $W_{\widetilde\gamma}^{l\rightarrow}$ are defined in a similar way;
- $W_{\widetilde\gamma}^{l\rightarrow r}$ is the set of points $\widetilde z\in W_{\widetilde\gamma}$ that lift a point of $W_{\Gamma}^{l\rightarrow r}$;
- $W_{\widetilde\gamma}^{r\rightarrow l}$, $W_{\widetilde\gamma}^{r,l\rightarrow }$ and $W_{\widetilde\gamma}^{\rightarrow r,l }$ are defined in a similar way.
Note that, since $U_{\Gamma}$ is an annulus, $U_{\widetilde \gamma}$ is a connected component of $\widetilde \pi^{-1}(U_{\Gamma})$ and therefore if $T$ is a covering transformation, then $T(U_{\widetilde \gamma})\cap U_{\widetilde \gamma}$ is not empty if and only $T$ is a power of $T_{\widetilde\gamma}$. Furthermore, as told above, if $z$ belongs to $W_{\Gamma}^{l\rightarrow r}$, it meets $U_{\Gamma}$ in a unique real interval. This implies that, given a lift $\widetilde z$ of a point $z$ in $ W_{\Gamma}^{l\rightarrow r}$ that belongs to $W_{\widetilde\gamma}$, the whole transverse trajectory $\widetilde I_{\widetilde{\mathcal F}}^{{\mathbb{Z}}}(\widetilde z)$ can only meet a single connected component of $\widetilde\pi^{-1}(U_{\Gamma})$. Therefore, the only posssible lifts of $z$ that are in $W_{\widetilde\gamma}$ are the $T_{\widetilde\gamma}^k(\widetilde z)$, $k\in{\mathbb{Z}}$. Note that for every covering transformation $T$, we have $T(W_{\widetilde\gamma}^{l\rightarrow r})=W_{T(\widetilde\gamma)}^{l\rightarrow r}$ and that $W_{\widetilde\gamma}^{l\rightarrow r}\cap T(W_{\widetilde\gamma}^{l\rightarrow r})=\emptyset$ if $T$ is not a power of $T_{\widetilde \gamma}$. We have similar properties for the three other sets $W_{\widetilde\gamma}^{r\rightarrow l}$, $W_{\widetilde\gamma}^{r,l\rightarrow }$ and $W_{\widetilde\gamma}^{\rightarrow r,l }$.
Suppose that $V_{\Gamma}\cap U_{\Gamma}\not=\emptyset$ and fix $z\in V_{\Gamma}\cap U_{\Gamma}$. Then the image of the morphism $i_*:\pi_1(z,V_{\Gamma})\to \pi_1(z,\mathrm{dom}(I))$ induced by the inclusion $i:V_{\Gamma}\to \mathrm{dom}(I)$ is included in the image of the morphism $j_*:\pi_1(z,U_{\Gamma})\to \pi_1(z,\mathrm{dom}(I))$ induced by the inclusion $j:U_{\Gamma}\to \mathrm{dom}(I)$.
Note that $V_{\widetilde \gamma}$ is invariant by $T_{\widetilde\gamma}$ and projects onto $V_{\Gamma}$ and conversely that $\widetilde \pi^{-1}(V_{\widehat\Gamma})$ is the union of the $V_{\widetilde \gamma}$, indexed by the set of lifts. For every point $ z\in V_{\Gamma}$, the path $I_{{\mathcal F}}^{{\mathbb{Z}}}(z)$ has no ${\mathcal F}$-transverse self-intersection and draws $\Gamma$. By Proposition \[prop:drawing-crossing\], there is a unique drawing component and so, for every lift $\widetilde z$ of $z\in V_{\Gamma}$, there exists a unique lift $\widetilde \gamma$ of $\gamma$ (up to a composition by a power of $T_{\widetilde\gamma}$) such that $\widetilde z$ belongs to $V_{\widetilde \gamma}$. In particular $T(V_{\widetilde \gamma})\cap V_{\widetilde \gamma}=\emptyset$ if $T$ is not a power of $T_{\widetilde\gamma}$. This implies that $V_{\widetilde \gamma}$ is open and closed in $\widetilde \pi^{-1}(V_{\Gamma})$. Consequently, the stabilizer in the group of covering automorphisms of a connected component of $\widetilde \pi^{-1}(V_{\Gamma})$ is a sub-group of the stabilizer of a lift $\widetilde\gamma$.
To prove Proposition \[prop:local-structure\] we need to work on the annular covering $\widehat{\mathrm{dom}}(I)=\widetilde{\mathrm{dom}}(I)/T_{\widetilde \gamma}$. Denote by $\pi: \widetilde{\mathrm{dom}}(I)\to \widehat{\mathrm{dom}}(I)$ and $\widehat\pi: \widehat{\mathrm{dom}}(I)\to\mathrm{dom}(I)$ the covering projections, by $\widehat I$ the induced identity isotopy, by $\widehat f$ the induced lift of $f$, by $\widehat{\mathcal{F}}$ the induced foliation. The line $\widetilde\gamma$ projects onto the natural lift of a transverse simple loop $\widehat\Gamma$. Write $$U_{\widehat\Gamma}, \enskip r_{\widehat\Gamma}, \enskip l_{\widehat\Gamma}, \enskip \Omega'(I)_{\widehat\Gamma}, \enskip V_{\widehat\Gamma},\enskip W_{\widehat\Gamma},\enskip W_{\widehat\Gamma}^{\rightarrow r}, \enskip W_{\widehat\Gamma}^{r\rightarrow }, \enskip W_{\widehat\Gamma}^{\rightarrow l}, \enskip W_{\widehat\Gamma}^{l\rightarrow}, \enskip W_{\widehat\Gamma}^{l\rightarrow r},\enskip W_{\widehat\Gamma}^{l\rightarrow r}, \enskip W_{\widehat\Gamma}^{r,l\rightarrow }, \enskip W_{\widehat\Gamma}^{\rightarrow r,l }$$ for the respective projections of $$U_{\widetilde\gamma}, \enskip r_{\widetilde\gamma}, \enskip l_{\widetilde\gamma}, \enskip \Omega'(I)_{\widetilde\gamma},\enskip W_{\widetilde\gamma}, \enskip W_{\widetilde\gamma},\enskip W_{\widetilde\gamma}^{\rightarrow r},\enskip W_{\widetilde\gamma}^{r\rightarrow },\enskip W_{\widetilde\gamma}^{\rightarrow l},\enskip W_{\widetilde\gamma}^{l\rightarrow},\enskip W_{\widetilde\gamma}^{l\rightarrow r},\enskip W_{\widetilde\gamma}^{l\rightarrow r},\enskip W_{\widetilde\gamma}^{r,l\rightarrow },\enskip W_{\widetilde\gamma}^{\rightarrow r,l }.$$
We have the following:
- The set $U_{\widehat\Gamma}$ is the union of leaves that meet $\widehat\Gamma$ and is the unique annular connected component of $\widehat\pi^{-1}(U_{\Gamma})$.
- The two connected components of $\widehat{\mathrm{dom}}(I)_{\mathrm {sph}}\setminus U_{\widehat \Gamma}$ are $r_{\widehat\Gamma}\sqcup\{R\}$ and $l_{\widehat\Gamma}\sqcup\{L\}$, where $\widehat{\mathrm{dom}}(I)_{\mathrm {sph}}$ is the sphere obtained by adding the end $R$ of $\widehat{\mathrm{dom}}(I)$ on the right of $\widehat\Gamma$ and the end $L$ on its left.
We will often refer to the the following classification, for the right side of $U_{\Gamma}$:
1. the set $\mathrm{Fr}(U_{\widehat \Gamma})\cap r_{\widehat\Gamma}$ is empty;
2. the set $\mathrm{Fr}(U_{\widehat \Gamma})\cap r_{\widehat\Gamma}$ is reduced to a closed leaf $\widehat\phi$ such that $U_{\widehat\Gamma}\subset R(\widehat\phi)$;
3. the set $\mathrm{Fr}(U_{\widehat \Gamma})\cap r_{\widehat\Gamma}$ is reduced to a closed leaf $\widehat\phi$ such that $U_{\widehat\Gamma}\subset L(\widehat\phi)$;
4. the set $\mathrm{Fr}(U_{\widehat \Gamma})\cap r_{\widehat\Gamma}$ is a non empty union of leaves homoclinic to $R$.
In situation (1), one has $$\widehat W^{r\to}=\widehat W^{\to r}=\emptyset.$$ In situation (2), one has $$\widehat W^{r\to}=\emptyset,\enskip \widehat W^{\to r} =\bigsqcup_{k\in{\mathbb{Z}}} R(\widehat f^{k+1}(\widehat \phi))\setminus L(\widehat f^k(\widehat\phi)).$$ In situation (3), one has $$\widehat W^{\to r}=\emptyset,\enskip \widehat W^{r\to } =\bigsqcup_{k\in{\mathbb{Z}}} R(\widehat f^{k+1}(\widehat \phi))\setminus L(\widehat f^k(\widehat\phi)).$$ In situation (4), one has $$W_{\widehat\Gamma}^{\rightarrow r}\not=\emptyset, \enskip W_{\widehat\Gamma}^{r\rightarrow }\not=\emptyset.$$
Note that in all situations, every whole transverse trajectory meets $\mathrm{Fr}(U_{\widehat \Gamma})\cap r_{\widehat\Gamma}$ at most twice. Of course one has a similar classification for $\mathrm{Fr}(U_{\widehat \Gamma})\cap l_{\widehat\Gamma}$. Note also that every transverse trajectory that meets $U_{\widehat \Gamma}$ has a unique interval of intersection with $U_{\widehat \Gamma}$.
Now define the sets $X^{\rightarrow R}_{\widehat\Gamma}$, $X^{R\rightarrow }_{\widehat\Gamma}$, $X^{\rightarrow L}_{\widehat\Gamma}$, $X^{L\rightarrow }_{\widehat\Gamma}$ as follows: $$\widehat z\in X^{\rightarrow R}_{\widehat\Gamma}\Leftrightarrow \lim_{n\to+\infty} \widehat f^{n}(\widehat z)=R,$$ $$\widehat z\in X^{R\rightarrow }_{\widehat\Gamma}\Leftrightarrow \lim_{n\to+\infty} \widehat f^{-n}(\widehat z)=R,$$ $$\widehat z\in X^{\rightarrow L}_{\widehat\Gamma}\Leftrightarrow \lim_{n\to+\infty} \widehat f^{n}(\widehat z)=L,$$ $$\widehat z\in X^{L\rightarrow }_{\widehat\Gamma}\Leftrightarrow \lim_{n\to+\infty} \widehat f^{-n}(\widehat z)=L.$$ So we have $$\widehat{\mathrm{dom}}(I)\setminus \left(X^{\rightarrow R}_{\widehat\Gamma}\cup X^{\rightarrow L}_{\widehat\Gamma}\right)=\mathrm{ne}^{+}(\widehat f), \enskip \widehat{\mathrm{dom}}(I)\setminus \left(X^{R\rightarrow }_{\widehat\Gamma}\cup X^{L\rightarrow }_{\widehat\Gamma}\right)=\mathrm{ne}^{-}(\widehat f).$$In situation (4), one has $$W_{\widehat\Gamma}^{\rightarrow r}\subset X^{\rightarrow R}_{\widehat\Gamma}, \enskip W_{\widehat\Gamma}^{r\rightarrow }\subset X^{R\rightarrow }_{\widehat\Gamma},\enskip r_{\widehat \Gamma}\subset X^{\rightarrow R}_{\widehat\Gamma}\cup X^{R\rightarrow }_{\widehat\Gamma},\enskip r_{\widehat \Gamma}\setminus W_{\widehat\Gamma}\subset X^{\rightarrow R}_{\widehat\Gamma}\cap X^{R\rightarrow }_{\widehat\Gamma}.$$ Indeed, if $z\in W_{\widehat\Gamma}^{\rightarrow r}$, then the whole transverse trajectory of $z$ meets a leaf $\widehat\phi\subset \mathrm{Fr}(U_{\widehat \Gamma})\cap r_{\widehat\Gamma}$, uniquely defined, homoclinic to $R$ and such that $L(\widehat \phi)$ is simply connected. There exists $n$ such that $\widehat I^{{\mathbb{N}}}_{\widehat{\mathcal F}}(\widehat{f}^n(z))\subset L(\phi)$. In particular $\widehat I^{{\mathbb{N}}}_{\widehat{\mathcal F}}(\widehat{f}^n(z))$ never meets a leaf more than once because $L(\widehat\phi)$ is simply connected and $\widehat{\mathcal F}$ non singular. So $\omega(z)=\emptyset$. But this means that $\lim_{n\to+\infty} \widehat f^{n}(\widehat z)=R$.
In the rest of the section, when $\widehat\phi$ is a closed leaf of $\widehat{\mathcal F}$ we will write $$A_{\widehat\phi}= \bigsqcup_{k\in{\mathbb{Z}}} R(\widehat f^{k+1}(\widehat \phi))\setminus L(\widehat f^k(\widehat\phi)),$$ noting that we get an invariant open annulus. Similarly, if $\phi$ is a closed leaf of $\mathcal F$ we will write $$A_{\phi}= \bigsqcup_{k\in{\mathbb{Z}}} R(f^{k+1}(\phi))\setminus L(f^k(\phi)).$$ Let us state three preliminary results:
The set $\Omega'(I)_{\widehat \Gamma}$ is included in $\Omega'(\widehat f)$.
First note that, as $\Omega'(I)_{\widetilde\gamma}\subset U_{\widetilde\gamma}$, we have that $\Omega'(I)_{\widehat\Gamma}\subset U_{\widehat\Gamma}$. Let $\widehat z$ be in $\Omega'(I)_{\widehat \Gamma}$. It projects by $\widehat \pi$ onto a point $z \in \Omega'(I)_{\Gamma}$. One can suppose that $z\in\Omega(I)$ and that $\omega(z)\not\in\mathrm{fix}(I)$ for instance. This implies that there exists a point $z'\in\ U_{\Gamma}$ and a subsequence $(f^{n_k}(z))_{k{\geqslant}0}$ of $(f^n(z))_{n{\geqslant}0}$ that converges to $z'$. The trajectory $I^{n_k}(z)$ is homotopic to $I^{n_k}_{\mathcal F}(z)$ and this last path is contained in $U_{\Gamma}$. Lifting our paths to $\widehat{\mathrm{dom}}(I)$ we deduce that $(\widehat f^{n_k}(\widehat z))_{k{\geqslant}0}$ converges to $\widehat z'$, the lift of $z'$ that belongs to $U_{\widehat \Gamma}$. So $\widehat z$ belongs to $\mathrm{ne}^{+}(\widehat f)$. Moreover, if $W$ is a neighborhood of $z$ sufficiently small, there exists $m{\geqslant}1$ such that $I_{\mathcal F}^{m}(z')$ is included in $U_{\Gamma}$ and draws $\Gamma$, for every $z'\in W$. We deduce that if $z'\in W\cap f^{-n}(W)$, both $I_{\mathcal F}^m(z')$ and $I_{\mathcal F}^m(f^n(z'))$ draw $\Gamma$, and since $I_{\mathcal F}^{{\mathbb{Z}}}(z')$ has a single drawing component for $\Gamma$, then $I_{\mathcal F}^n(z')$ is included in $U_{\Gamma}$. Lifting our paths to to $\widehat{\mathrm{dom}}(I)$ and using the fact that $I^{n}(z')$ is homotopic to $I^{n}_{\mathcal F}(z')$, we deduce that $z\in\Omega(\widehat f)$. Consequently, $\Omega'(I)_{\widehat \Gamma}$ is included in $\Omega'(\widehat f)$.
\[lemma:existence\] If $\Omega'(I)_{\widehat \Gamma}\not=\emptyset$, there is a point $\widehat z\in \Omega'(\widehat f)$ such that $\mathrm{rot}_{\widetilde f, [\widehat\Gamma]}(\widehat z)>0$, where $[\widehat\Gamma]\in H_1(\widehat{\mathrm{dom}}(I),{\mathbb{Z}})$ is the homology class of $\widehat\Gamma$.
Observe first that the lift $\widetilde f$ of $\widehat f$ is fixed point free and that $\widehat f$ has no topological horseshoe because it is the case for $f$. So, every point in $\mathrm{ne}(\widehat f)$ has a rotation number by Theorem \[th:rotation-number\]. Choose a point $\widehat z\in\Omega'(I)_{\widehat \Gamma}$. Its rotation number is well defined because $\widehat z$ belongs to $\mathrm{ne}(\widehat f)$. The rotation number cannot be negative because $\widehat I_{\widehat{\mathcal F}}^{{\mathbb{Z}}}(\widehat z)$ is drawn on $\widehat\Gamma$. If it is positive, the conclusion of the lemma holds. If if it zero, one can apply Proposition \[prop:recurrentzero-rotation\] to ensure the existence of a periodic point whose rotation number is non zero, noting that the proof of Proposition \[prop:recurrentzero-rotation\] tells us that this rotation number is positive.
\[lemma:no-Birkhoff-cycle\]If $\Omega'(I)_{\widehat \Gamma}\not=\emptyset$ and $U_{\widehat\Gamma}\not= \widehat{\mathrm{dom}}(I)$, then there is no Birkhoff connection from $R$ to $L$ or no Birkhoff connection from $L$ to $R$.
By Lemma \[lemma:existence\], there is a point $\widehat z\in \Omega'(\widehat f)$ such that $\mathrm{rot}_{\widetilde f, [\widehat\Gamma]}(\widehat z)>0$. The assumption $U_{\widehat\Gamma}\not= \widehat{\mathrm{dom}}(I)$ means that the sets $r_{\widehat\Gamma}$ and $l_{\widehat\Gamma}$ are not both empty. Let us suppose for instance that $r_{\widehat\Gamma}\not=\emptyset$. This means that we are not in situation (1). In case we are in situation (2) or (3), there exists an essential closed leaf $\widehat \phi$ and this leaf is disjoint from its image by $\widehat f$. In that case the conclusion is immediate. It remains to study the case where situation (4) occurs. In that case, there exists a leaf $\widehat \phi$ homoclinic to $R$. We deduce that $\widehat\phi$ is disjoint from all its images by the iterates of $\widehat f$. By a result of Béguin-Crovisier-Le Roux [@LeRoux], one can blow up the end $R$ by adding a circle $\widehat \Sigma_R$ to $\widehat {\mathrm {dom}}(I)$ in a neighborhood of $R$ to get a semi-open annulus $\widehat {\mathrm {dom}}(I)_{\mathrm {ann}}$ such that $\widehat f$ extends to a homeomorphism $\widehat f_{\mathrm {ann}}$ having a unique fixed point $\widehat z_*$ on $\widehat \Sigma_R$. Write $\widetilde{\mathrm {dom}}(I)_{\mathrm {ann}}=\widetilde{\mathrm {dom}}(I)\sqcup\widetilde\Sigma_S$ for the universal covering space of $\widehat {\mathrm {dom}}(I)_{\mathrm {ann}}$, where $\widetilde\Sigma_R$ is the universal covering space of $\widehat\Sigma_R$ and keep the notation $T_{\widetilde \gamma}$ for the natural covering automorphism. There is a unique lift $\widetilde f_{\mathrm {ann}}$ of $\widehat f_{\mathrm {ann}}$ that extends $\widetilde f$ and that fixes the preimages of $\widehat z_*$. One can construct an open annulus $\widehat {\mathrm {dom}}(I)_{\mathrm {double}}$ by pasting two copies of $\widehat {\mathrm {dom}}(I)_{\mathrm {ann}}$ on $\widehat \Sigma_{R}$ and naturally define a homeomorphism $\widehat f_{\mathrm {double}}$ on $\widehat {\mathrm {dom}}(I)_{\mathrm {double}}$. Its universal covering space $\widetilde {\mathrm {dom}}(I)_{\mathrm {double}}$ can be constructed by pasting two copies of $\widetilde {\mathrm {dom}}(I)_{\mathrm {ann}}$ on $\widetilde \Sigma_{R}$ and one gets a natural lift $\widetilde f_{\mathrm {double}}$. We keep the notation $T_{\widetilde\gamma}$ for the natural covering automorphism. Suppose that there exists a Birkhoff connection from $R$ to $L$ and a Birkhoff connection from $L$ to $R$. In that case $\widehat f$ satisfies the intersection property. It is straightforward to prove that $\widehat f_{\mathrm {double}}$ itself has the intersection property. Consequently, for every integers $p$ and $q>0$ relatively prime, such that $p/q\in(0,\mathrm{rot}_{\widetilde f} (\widehat z))$, there exists a periodic point of $\widehat f_{\mathrm {double}}$ of period $q$ and rotation number $p/q$ and consequently there exists a periodic point of $\widehat f$ of period $q$ and rotation number $p/q$. Applying Proposition \[prop:PB\], one deduces that $\widehat f$ has a horseshoe. We have got a contradiction.
We will suppose from now on that $U_{\widehat\Gamma}\not= \widehat{\mathrm{dom}}(I)$, or equivalently that the boundary of $U_{\widehat\Gamma}$ is not empty and contains at least one leaf of $\widehat{\mathcal F}$. In the other case, $U_{\widehat\Gamma}$ coincides with $U_{\Gamma}$, it is a connected component of $ {\mathrm{dom}}(I)$. Note that it is a positive annulus.
Let us state now the most important lemma.
\[lemma:homoclinic\] At least one of the sets $\overline{W_{\widehat\Gamma}^{r\rightarrow }}$, $\overline {W_{\widehat\Gamma}^{\rightarrow r}}$ is contained in $X^{R\rightarrow }_{\widehat\Gamma}\cap X^{\rightarrow R}_{\widehat\Gamma}$, and one of the sets $\overline{W_{\widehat\Gamma}^{l\rightarrow }}$, $\overline {W_{\widehat\Gamma}^{\rightarrow l}}$ is contained in $X^{L\rightarrow }_{\widehat\Gamma}\cap X^{\rightarrow L}_{\widehat\Gamma}$.
We can suppose that both sets $W_{\widehat\Gamma}^{r\rightarrow }$ and $W_{\widehat\Gamma}^{\rightarrow r}$ are non empty, otherwise the statement is obvious. Consequently we are in situation (4): the set $U_{\widehat\Gamma}$ is adherent to $R$ and each leaf on the right of $U_{\widehat\Gamma}$ is a leaf homoclinic to $R$. Moreover, one has $W_{\widehat\Gamma}^{r\rightarrow }\subset X^{R\rightarrow }_{\widehat\Gamma}$ and $W_{\widehat\Gamma}^{\rightarrow r}\subset X^{\rightarrow R}_{\widehat\Gamma}$. We will consider two cases, depending whether there exists a whole transverse trajectory that meets four times a leaf of $U_{\widehat \Gamma}$ or not.
By Lemma \[lemma:no-Birkhoff-cycle\], we know that at least one of the sets $X^{R\rightarrow }_{\widehat\Gamma}$ or $X^{\rightarrow R}_{\widehat\Gamma}$ is not adherent to $L$. Let us suppose for example that this is the case for $X^{R\rightarrow }_{\widehat\Gamma}$. The set $W_{\widehat\Gamma}^{r\rightarrow }$ shares the same property because $W_{\widehat\Gamma}^{r\rightarrow }\subset X^{R\rightarrow }_{\widehat\Gamma}$. Of course its closure $\overline{W_{\widehat\Gamma}^{r\rightarrow }}$ is not adherent to $L$. Let us suppose that $\overline{W_{\widehat\Gamma}^{r\rightarrow }}\not\subset (X^{R\rightarrow }_{\widehat\Gamma}\cap X^{\rightarrow R}_{\widehat\Gamma})$ and choose $\widehat z\in \overline{W_{\widehat\Gamma}^{r\rightarrow }}\setminus (X^{R\rightarrow }_{\widehat\Gamma}\cap X^{\rightarrow R}_{\widehat\Gamma})$. The fact that the orbit of $\widehat z$ is non adherent to $L$ and that $\widehat z\not\in X^{R\rightarrow }_{\widehat\Gamma}\cap X^{\rightarrow R}_{\widehat \Gamma}$ implies that $\widehat z\in \mathrm{ne}(\widehat f)$. Moreover, $\omega(\widehat z)$ and $\alpha(\widehat z)$ are not included in $\widehat U_{\Gamma}$ if not empty, otherwise $\widehat I_{\widehat F}(\widehat z)$ would draw $\widehat\Gamma$ infinitely, in contradiction with the hypothesis. The limit sets are included in $l_{\widehat\Gamma}$, more precisely, they are contained in $l_{\widehat\Gamma}\setminus \mathrm{Fr}(U_{\widehat\Gamma})$ and so the whole transverse trajectory of $\widehat z$ meets $l_{\widehat\Gamma}$. Of course $\widehat z\not\in W_{\widehat\Gamma}^{l \rightarrow }$ because $\widehat z\in \overline{W_{\widehat\Gamma}^{r\rightarrow }}$ and because $W_{\widehat\Gamma}^{r\rightarrow }$ and $ W_{\widehat\Gamma}^{l\rightarrow }$ are disjoint open sets. So $\widehat z$ belongs to $W_{\widehat\Gamma}^{\rightarrow l}$ and $\omega(\widehat z)\not=\emptyset$. As explained previously, this implies that situation (3) occurs on the left side: there exists a closed essential leaf $\widehat\phi\subset \mathrm{Fr}(U_{\widehat \Gamma})\cap l_{\widehat \Gamma}$ such that $U_{\widehat \Gamma}\subset R(\widehat\phi)$. But this implies that $X^{\rightarrow R}_{\widehat\Gamma}$ is not adherent to $L$. Let us explain now why $\overline{W_{\widehat\Gamma}^{\rightarrow r}}\subset X^{R\rightarrow }_{\widehat\Gamma}\cap X^{\rightarrow R}_{\widehat\Gamma}$. Indeed, replacing $\widehat z$ by $\widehat z'\in \overline{W_{\widehat\Gamma}^{\rightarrow r}}\setminus (X^{R\rightarrow }_{\widehat\Gamma}\cap X^{\rightarrow R}_{\widehat\Gamma})$ in the previous argument will lead to a contradiction because $W_{\widehat\Gamma}^{l\rightarrow }=\emptyset$ and $\overline{W_{\widehat\Gamma}^{\rightarrow r}}\cap W_{\widehat\Gamma}^{\rightarrow l}=\emptyset$.
Let suppose that $\widehat\phi\subset U_{\widehat \Gamma}$ is met four times by a whole transverse trajectory. Fix a lift $\widetilde\phi\subset U_{\widetilde \gamma}$. We know that $$\bigcup_{n{\geqslant}1} \widetilde f^{-n}(L(T^3(\widetilde \phi))\cap R(\widetilde \phi)\not=\emptyset.$$
If $\overline{W_{\widehat\Gamma}^{r\rightarrow }}\not\subset (X^{R\rightarrow }_{\widehat\Gamma}\cap X^{\rightarrow R}_{\widehat\Gamma})$, there exists an admissible path $\widetilde\gamma_0: [a_0,b_0]\to \widetilde{\mathrm{dom}}(I)$ such that $$\widetilde\gamma_0(a_0)\in r(\widetilde\gamma), \enskip
\widetilde\gamma_0(a_0)\in L(\widetilde\phi)\cap R(T(\widetilde\phi)), \enskip
\widetilde\gamma_{(a_0,b_0]}\subset U_{\widetilde\gamma}, \enskip \widetilde\gamma(b_0)\in T^3(\widetilde\phi).$$
One can construct a line $\widetilde\lambda\subset\bigcup_{n{\geqslant}1} \widetilde f^{-n}(L(T^3(\widetilde \phi))\cup R(\phi)$ that coincides with $\widetilde \gamma$ outside a compact set, then one can find a neighborhood $V$ of $L$ in $\widehat {\mathrm{dom}}(I)$ whose preimage in $\widetilde {\mathrm{dom}}(I)$ belongs to $L(\widetilde\lambda)$. Suppose that $\overline{W_{\widehat\Gamma}^{r\rightarrow }}\not\subset (X^{R\rightarrow }_{\widehat\Gamma}\cap X^{\rightarrow R}_{\widehat\Gamma})$, and choose $\widehat z\in \overline{W_{\widehat\Gamma}^{r\rightarrow }}\setminus (X^{R\rightarrow }_{\widehat\Gamma}\cap X^{\rightarrow R}_{\widehat\Gamma})$. As explained in the proof given in the first case, there are three possibilities
- the orbit of $\widehat z$ is adherent to $L$;
- there exists a closed essential leaf $\widehat\phi'\subset \mathrm{Fr}(U_{\widehat \Gamma})\cap l_{\widehat \Gamma}$ such that $U_{\widehat \Gamma}\subset R(\widehat\phi')$;
- $\widehat I_{\widehat F}(\widehat z)$ draws $\widehat\Gamma$ infinitely.
In the last case, one can find $\widehat z'\in W_{\widehat\Gamma}^{r\rightarrow }$ such that $\widehat I_{\widehat F}(\widehat z')$ meets $\widehat\phi$ at least three times. Denote $\widehat\phi''$ the unique leaf on the boundary of $U_{\widehat\Gamma}$ met by $\widehat I_{\widehat F}(\widehat z')$ such that $U_{\widehat \Gamma}$ is on its left and let $\widetilde\phi''$ be the unique lift of $\widehat\phi''$ between $\widetilde\phi$ and $T(\widetilde\phi)$. The lift of $\widehat I_{\widehat F}(\widehat z')$ that intersects $\widetilde\phi''$ contains, as a sub-path, a path satisfying the conclusion of the sub-lemma.
In the first case the orbit of $\widehat z$ meets $V$ and so, one can find $\widehat z'\in W_{\widehat\Gamma}^{r\rightarrow }$ whose orbit meets $V$. In the second case, one can find $\widehat z'\in W_{\widehat\Gamma}^{r\rightarrow }$ whose orbit meets $L(\widehat \phi')$ and one can observe that the lift $\widetilde \phi'$ of $\widehat \phi'$ is included in $L(\widetilde\lambda)$. In both cases, denote $\widehat\phi''$ the unique leaf on the boundary of $U_{\widehat\Gamma}$ met by $\widehat I_{\widehat F}(\widehat z')$ such that $U_{\widehat \Gamma}$ is on its left and let $\widetilde\phi''$ be the unique lift of $\widehat\phi''$ lying between $\widetilde\phi$ and $T(\widetilde\phi)$. The set $\bigcup_{n{\geqslant}0} \widetilde f^n(R(\widetilde\phi'')) $ meets $\widetilde \lambda'$ but does not meet $R(\widetilde\phi)$, so it meets $\bigcup_{n{\geqslant}1} \widetilde f^{-n}(L(T^3(\widetilde \phi)))$. It implies that the conclusion of the sub-lemma is true in both cases.
Similarly, if $\overline{W_{\widehat\Gamma}^{\rightarrow r}}\not\subset (X^{R\rightarrow }_{\widehat\Gamma}\cap X^{\rightarrow R}_{\widehat\Gamma})$, there exists an admissible path $\widetilde\gamma_1: [a_1,b_1]\to \widetilde{\mathrm{dom}}(I)$ such that $$\widetilde\gamma_1(b_1)\in r(\widetilde\gamma), \enskip
\widetilde\gamma_1(b_1)\in L(T^2(\widetilde\phi))\cap R(T^3(\widetilde\phi)), \enskip
\widetilde\gamma_{[a_1,b_1)}\subset U_{\widetilde\gamma}, \enskip \widetilde\gamma(a_1)\in \widetilde\phi.$$ The paths $\widetilde\gamma_0$ and $\widetilde\gamma_1$ project onto transverse loops that meet every leaf of $U_{\widehat\gamma}$ at least twice. Such a situation has been studied in the proof of Proposition \[prop:Birkhoff-cycles\]. We can construct an admissible path with a $\widehat{\mathcal F}$-transverse self-intersection, in contradiction with the fact that $f$ has no horseshoe.
We are ready now to construct the positive annulus we are looking for. Let us begin with a simple result.
\[lemma:fundamental-group\]There exists at most one fixed essential connected component of $W_{\widehat\Gamma}^{l\rightarrow r}$. We have a similar result for $W_{\widehat\Gamma}^{r\rightarrow l}$, $W_{\widehat\Gamma}^{\rightarrow r, l}$ and $W_{\widehat\Gamma}^{r,l \rightarrow }$.
Let us prove by contradiction that there exists at most one essential connected component of $W_{\widehat\Gamma}^{l\rightarrow r}$ that is fixed. Indeed, if $V_1$ and $V_2$ are two such components, the complement of $V_1\cup V_2$ in $\widehat{\mathrm{dom}}(I)$ has a compact connected component $K$ and this component is fixed. This implies that for every $\widehat z\in K$, the sequences $(\widehat f^n(\widehat z))_{n{\geqslant}0}$ and $(\widehat f^{-n}(\widehat z))_{n{\geqslant}0}$ do not converge to an end of $\widehat{\mathrm{dom}}(I)$.
Let us prove that $K\cap r_{\widehat\Gamma}=\emptyset$. This is clear in situation (1). It is also true in situation (4) because $K\cap X^{R\rightarrow }_{\widehat\Gamma}= K\cap X^{\rightarrow R }_{\widehat\Gamma} =\emptyset$ and $r_{\widehat\Gamma}\subset X^{R\rightarrow }_{\widehat\Gamma}\cup X^{\rightarrow R }_{\widehat\Gamma}$. Suppose now that we are in situation (2) and write $\widehat\phi$ for the closed leaf included in $\mathrm{Fr}(U_{\widehat\Gamma})\cap r_{\widehat\Gamma}$. Choose an essential loop $\Gamma_1$ in $V_1$ and an essential loop in $\Gamma_2$ in $V_2$. Using the fact that $\Gamma_1$ and $\Gamma_2$ are compact and included in $W_{\widehat\Gamma}$, one knows that there exists $n{\geqslant}0$ such that $\widehat f^{-n}(\Gamma_1)$ and $\widehat f^{-n}(\Gamma_2)$ are both contained in $L(\widehat\phi)$. This implies that $K$ is contained in $L(\widehat\phi)$. In case situation (3) occurs, we conclude with a similar argument.
A similar proof tells us that $K\cap l_{\widehat\Gamma}=\emptyset$ and consequently that $K$ is included in $U_{\widehat \Gamma}$. This implies that $K$ is contained in $W_{\widehat\Gamma}$ and disjoint from $W_{\widehat\Gamma}^{r\rightarrow }$ and $W_{\widehat\Gamma}^{\rightarrow l}$. But $V_1\cup V_2\cup K$ is a neighborhood of $K$ that is disjoint from $W_{\widehat\Gamma}^{r\rightarrow }$ and $W_{\widehat\Gamma}^{\rightarrow l}$, and so $K$ is disjoint from $\overline{W_{\widehat\Gamma}^{r\rightarrow }}$ and $\overline{W_{\widehat\Gamma}^{\rightarrow l}}$. Consequently, $V_1\cup V_2\cup K$ is included in $W_{\widehat\Gamma}^{l\rightarrow r}$. We have got our final contradiction.
The proof for the other sets is similar.
\[prop: existence\] If not empty, the set $\Omega'(I)_{\widehat \Gamma}$ is contained in an essential connected component of $W_{\widehat\Gamma}^{l\rightarrow r}$, $W_{\widehat\Gamma}^{r\rightarrow l}$, $W_{\widehat\Gamma}^{r,l\rightarrow }$ or $W_{\widehat\Gamma}^{\rightarrow r,l }$. Moreover this component will contain $A_{\widehat\phi}$, if $\widehat\phi$ is a closed leaf in $\mathrm{Fr}(U_{\widehat \Gamma})$.
[*Proof.*]{} We will suppose that $\Omega'(I)_{\widehat \Gamma}\not=\emptyset$ and that $$\overline{W_{\widehat\Gamma}^{r\rightarrow }}\subset X^{R\rightarrow }_{\widehat\Gamma}\cap X^{\rightarrow R}_{\widehat\Gamma}, \enskip \overline {W_{\widehat\Gamma}^{\rightarrow l}}\subset X^{L\rightarrow }_{\widehat\Gamma}\cap X^{\rightarrow L}_{\widehat\Gamma}.$$ We will prove that
- there exists a fixed essential component $W$ of $W_{\widehat\Gamma}^{l\rightarrow r}$ (unique by Lemma \[lemma:fundamental-group\]);
- $W$ contains $\Omega'(I)_{\widehat \Gamma}$;
- $W$ contains $A_{\widehat\phi}$, if $\widehat\phi$ is a closed leaf in $\mathrm{Fr}(U_{\widehat \Gamma})$.
If $\widehat\phi$ is a closed leaf in $\mathrm{Fr}(U_{\widehat \Gamma})\cap r_{\widehat \Gamma}$, the assumption $\overline{W_{\widehat\Gamma}^{r\rightarrow }}\subset X^{R\rightarrow }_{\widehat\Gamma}\cap X^{\rightarrow R}_{\widehat\Gamma}$ tells us that we are in situation (2) on the right side and we know that $A_{\widehat\phi}\subset W_{\widehat\Gamma}^{l\rightarrow r}$. So there exists a fixed essential component of $W_{\widehat\Gamma}^{l\rightarrow r}$ and this component contains $A_{\widehat\phi}$. Similarly, if $\widehat\phi$ is a closed leaf in $\mathrm{Fr}(U_{\widehat \Gamma})\cap l_{\widehat \Gamma}$, we are in situation (3) on the left side and there exists a fixed essential component of $W_{\widehat\Gamma}^{l\rightarrow r}$ that contains $A_{\widehat\phi}$.
Note now that every essential connected component of $W_{\widehat\Gamma}^{l\rightarrow r}$ that meets $\Omega'(I)_{\widehat \Gamma}$ is fixed. Indeed, remember that $\Omega'(I)_{\widehat \Gamma}\subset \Omega(\widehat f)$. Moreover the set $W_{\widehat\Gamma}^{l\rightarrow r}$ being invariant by $\widehat f$, every connected component of $W_{\widehat\Gamma}^{l\rightarrow r}$ that meets $\Omega(\widehat f)$ is periodic. Consequently it is fixed if supposed essential. We know that $$\Omega'(I)_{\widehat \Gamma} \subset W_{\widehat\Gamma}\cap \mathrm{ne}(\widehat f)\subset W_{\widehat\Gamma}\setminus (X^{R\rightarrow }_{\widehat\Gamma}\cap X^{L\rightarrow }_{\widehat\Gamma})\subset W_{\widehat\Gamma}^{l\rightarrow r}.$$ Choose $\widehat z\in \Omega'(I)_{\widehat \Gamma}$ and denote $\widehat W$ the connected component of $ W_{\widehat\Gamma}^{l\rightarrow r}$ that contains $\widehat z$. We will prove that $\widehat W$ is essential. Choosing another point of $\Omega'(I)_{\widehat \Gamma}$ would give us the same set $\widehat W$ by Lemma \[lemma:fundamental-group\]. So, $\widehat W$ contains $\Omega'(I)_{\widehat \Gamma}$. We will argue by contradiction and suppose that $\widehat W$ is not essential. The point $\widehat z$ being a non-wandering point of $\widehat f$, we know that $\widehat W$ is periodic. We denote $q$ its period. One can fill $\widehat W$, adding the compact connected components of its complement, to get a disk $\widehat W'$. The fact that $\widehat z$ is a non-wandering point of $\widehat f$ implies that $\widehat z$ is a non-wandering point of $\widehat f^{q}_{\widehat W'}$. So, $\widehat f^q\vert _{\widehat W'}$ has a fixed point $\widehat z'$ by Brouwer’s theory. There exists an integer $p\in{\mathbb{Z}}$ such that every lift $\widetilde W'$ of $\widehat W'$ satisfies $\widetilde f^{q}(\widetilde W')=T_{\widetilde \gamma}^p(\widetilde W')$. Moreover, the lift of $\widetilde z'$ that is in $\widetilde W'$ verifies $\widetilde f^q(\widetilde z')=T^p(\widetilde z')$. In other terms, there exists a periodic point $\widehat z'\in \widehat W'$ of period $q$ and rotation number $p/q$. Note that $p$ is positive. Indeed it is not negative because $\widehat z\in \Omega'(I)_{\widehat \Gamma}$ and does not vanish vanish because $\widehat z$ cannot be lifted to a non wandering point of $\widetilde f$, this map being fixed point free.
A point of the frontier of $W_{\widehat\Gamma}^{l\rightarrow r}$ in $\widehat{\mathrm{dom}}(I)$ belongs to $\overline{W_{\widehat\Gamma}^{r\rightarrow }}$, to $\overline{W_{\widehat\Gamma}^{\rightarrow l}}$, or does not belong to $W_{\widehat \Gamma}$. In the last case it belongs to $\bigcap_{k\in{\mathbb{Z}}} \widehat f^k(r_{\widehat\Gamma})$ or to $\bigcap_{k\in{\mathbb{Z}}} \widehat f^k(l_{\widehat\Gamma})$. By assumption, $\overline{W_{\widehat\Gamma}^{r\rightarrow }}$ is included in $X^{R\rightarrow }_{\widehat\Gamma}\cap X^{\rightarrow R}_{\widehat\Gamma}$ and $\overline{W_{\widehat\Gamma}^{\rightarrow l}}$ is included in $X^{L\rightarrow }_{\widehat\Gamma}\cap X^{\rightarrow L}_{\widehat\Gamma}$. So the frontier of $\widehat W'$ in $\widehat{\mathrm{dom}}(I)_{\mathrm {sph}}$ is contained in the frontier of $W_{\widehat\Gamma}^{l\rightarrow r}$, which means in the union of the two disjoint closed sets $\{R\}\cup\overline{W_{\widehat\Gamma}^{r\rightarrow }}\cup\left(\bigcap_{k\in{\mathbb{Z}}} \widehat f^k(r_{\widehat\Gamma})\right)$ and $\{L\}\cup\overline{W_{\widehat\Gamma}^{\rightarrow l}}\cup\left(\bigcap_{k\in{\mathbb{Z}}} \widehat f^k(l_{\widehat\Gamma})\right)$. Moreover, one knows that this frontier is connected because $\widehat W'$ is a topological disk. So it is contained in one of the two sets and there is no loss of generality by supposing that it is contained in $\{R\}\cup\overline{W_{\widehat\Gamma}^{r\rightarrow }}\cup\left(\bigcap_{k\in{\mathbb{Z}}} \widehat f^k(r_{\widehat\Gamma})\right)$. Of course the frontier is not reduced to $R$, because $\widehat W'$ does not contain $L$ and so $r_{\widehat\Gamma}$ is not empty, we are not in situation (1). The set $\mathrm{Fr}(U_{\widehat \Gamma})\cap r_{\widehat\Gamma}$ cannot be reduced to a closed leaf $\widehat\phi$. Indeed, as explained above, we would be in situation (2) and $W$ should be disjoint from $A_{\widehat\phi}$ because it is not essential. Its closure should be disjoint from $r_{\widehat\Gamma}$ because $\widehat W$ meets $U_{\widehat \Gamma}$ (recall that $\widehat z$ belongs to $U_{\widehat \Gamma}$) and the same occurs for $\widehat W'$. We conclude that we are in situation (4), the set $\mathrm{Fr}(U_{\widehat \Gamma})\cap r_{\widehat\Gamma}$ is a non empty union of leaves homoclinic to $R$. Recall that in this situation, we have $\bigcap_{k\in{\mathbb{Z}}} \widehat f^k(r_{\widehat\Gamma})\subset X^{R\rightarrow }_{\widehat\Gamma}\cap X^{\rightarrow R}_{\widehat\Gamma}$ and so the frontier of $\widehat W'$ in $\widehat{\mathrm{dom}}(I)$ is contained in $X^{R\rightarrow }_{\widehat\Gamma}\cap X^{\rightarrow R}_{\widehat\Gamma}$.
As in the proof of Lemma \[lemma:no-Birkhoff-cycle\] we blow up the end $R$ by adding a circle $\widehat \Sigma_R$ to $\widehat {\mathrm {dom}}(I)$, we construct $\widehat f_{\mathrm {ann}}$ on the semi-open annulus $\widehat {\mathrm {dom}}(I)_{\mathrm {ann}}$ and $\widehat f_{\mathrm {ann}}$ on the annulus $\widehat {\mathrm {dom}}(I)_{\mathrm{double}}$. We remind that $\widehat f_{\mathrm {double}}$ has a unique fixed point $\widehat z_*$ on $\widehat \Sigma_R$ and denote $\widetilde f_{\mathrm {double}}$ the lift that fixes the preimages of $\widehat z_*$. Consider a maximal isotopy $\widehat I'$ of $\widehat f^q_{\mathrm {double}}$ that is lifted to a maximal isotopy $\widetilde I'$ of $T_{\widetilde\gamma}^{-p}\circ\widetilde f^q_{\mathrm {double}}$ such that $\widehat z'\in \mathrm{sing}(\widehat I')$ and then a foliation $\widehat{\mathcal{F}'}$ transverse to $\widehat I'$. The point $\widehat z_*$ belongs to the domain of $\widehat I'$. Being a fixed point, its whole trajectory is equivalent to the natural lift of a simple transverse loop $\widehat \Gamma'$, because $\widehat f_{\mathrm double}$ has no horseshoe. The $\alpha$-limit set and the $\omega$-limit set of a point of $\Sigma_{R}$ being reduced to $\widehat z^*$ and $\widehat f_{\mathrm double}$ having no horseshoe, we know that the whole transverse trajectory of any point of $\Sigma_{R}$ is equivalent to the natural lift of $\widehat \Gamma'$. The same occurs for a point on $\mathrm{Fr}(\widehat W')$ because its $\alpha$-limit set and its $\omega$-limit set are contained in $\Sigma_{R}$. Consequently, the set $V_{\widehat\Gamma'}$ of points whose whole transverse trajectory draws $\widehat\Gamma'$ is an open set that contains $\Sigma_R$ and $\mathrm{Fr}(\widehat W')$. As $V_{\widehat \Gamma'}$ is contained in the open annulus $U_{\widehat \gamma'}$ of leaves met by $\Gamma'$ and as $\widehat W'$ is a topological disk with $\mathrm{Fr}(\widehat W') \subset U_{\widehat \gamma'}$, it follows that $\widehat W'$ itself must be contained in this annulus, a contradiction since $\widehat z'$ is a singular point. The proof is finished in this case.
Suppose now that $$\overline{W_{\widehat\Gamma}^{r\rightarrow }}\subset X^{R\rightarrow }_{\widehat\Gamma}\cap X^{\rightarrow R}_{\widehat\Gamma}, \enskip \overline {W_{\widehat\Gamma}^{l\rightarrow }}\subset X^{L\rightarrow }_{\widehat\Gamma}\cap X^{\rightarrow L}_{\widehat\Gamma}.$$ We will prove that
- there exists a fixed essential component $W$ of $W_{\widehat\Gamma}^{\rightarrow r,l}$ (unique by Lemma \[lemma:fundamental-group\]);
- $W$ contains $\Omega'(I)_{\widehat \Gamma}$;
- $W$ contains $A_{\widehat\phi}$, if $\widehat\phi$ is a closed leaf in $\mathrm{Fr}(U_{\widehat \Gamma})$.
If $\widehat\phi$ is a closed leaf in $\mathrm{Fr}(U_{\widehat \Gamma})\cap r_{\widehat \Gamma}$, the assumption $\overline{W_{\widehat\Gamma}^{r\rightarrow }}\subset X^{R\rightarrow }_{\widehat\Gamma}\cap X^{\rightarrow R}_{\widehat\Gamma}$ tells us that we are in situation (2) on the right side and the assumption $\overline {W_{\widehat\Gamma}^{l\rightarrow }}\subset X^{L\rightarrow }_{\widehat\Gamma}\cap X^{\rightarrow L}_{\widehat\Gamma}$ tells us that $A_{\widehat\phi}$ does not meet $\overline {W_{\widehat\Gamma}^{l\rightarrow }}$ and so that $A_{\widehat\phi}\subset W_{\widehat\Gamma}^{\rightarrow r,l}$. So there exists a fixed essential component of $W_{\widehat\Gamma}^{\rightarrow r, l}$ and this component contains $A_{\widehat\phi}$. Similarly, if $\widehat\phi$ is a closed leaf in $\mathrm{Fr}(U_{\widehat \Gamma})\cap l_{\widehat \Gamma}$, we are in situation (2) on the left side and there exists a fixed essential component of $W_{\widehat\Gamma}^{\rightarrow r,l}$ that contains $A_{\widehat\phi}$. The rest of the proof is similar. The remaining cases can be proven in the same way. We can conclude by Lemma \[lemma:homoclinic\].
Note that if $U_{\widehat\Gamma}$ is bordered by two closed leaves $\widehat\phi$ and $\widehat\phi'$, then the annulus $A_{\widehat\phi}\cup U_{\widehat\Gamma}\cup A_{\widehat\phi'}$ coincides with one the sets $W_{\widehat\Gamma}^{l\rightarrow r}$, $W_{\widehat\Gamma}^{r\rightarrow l}$, $W_{\widehat\Gamma}^{r,l\rightarrow }$ or $W_{\widehat\Gamma}^{\rightarrow r,l }$ and is equal to $\widehat W$. In case $\mathrm{Fr}(U_{\widehat \Gamma})$ contains a closed leaf $\widehat\phi$, then $A_{\widehat\phi}$ does not separate $\widehat W\setminus A_{\widehat\phi} $.
[*Proof of Proposition \[prop:local-structure\]*]{}Here again we suppose that $\overline{W_{\widehat\Gamma}^{r\rightarrow }}$ is contained in $X^{R\rightarrow }_{\widehat\Gamma}\cap X^{\rightarrow R}_{\widehat\Gamma}$ and $\overline {W_{\widehat\Gamma}^{\rightarrow l}}$ contained in $X^{L\rightarrow }_{\widehat\Gamma}\cap X^{\rightarrow L}_{\widehat\Gamma}$. Write $\widehat W$ for the essential component of $W_{\widehat\Gamma}^{l\rightarrow r }$ that contains $\Omega'(I)_{\widehat \Gamma}$. One gets an invariant annulus $\widehat A$ by filling $\widehat W$, which means by adding the compact connected components of its complement. The choice of the class $[\widehat \Gamma]$ as a generator of $H_1(\widetilde{\mathrm{dom}}(I),{\mathbb{Z}})$ permits us to give a partition of $\Omega(\widehat f)$ in $\widetilde f$-positive or $\widetilde f$-negative points. Let us prove that $\widehat A$ is a positive annulus of $\widehat I$ and $\kappa\in H_{1}(\widehat A,{\mathbb{Z}})$ its positive generator, where $\iota_*(\kappa)=[\widehat \Gamma]$. Suppose that $\widehat z\in\Omega(\widehat f)$ is $\widetilde f$-negative. As stated in Proposition \[prop: positive\], the transverse loop associated with $\widehat z$ is negative. This loop cannot meet $U_{\widehat \Gamma}$, so it is contained in $r_{\widehat\Gamma}$ or in $l_{\widehat \Gamma}$. Suppose for instance that it is contained in $r_{\widehat\Gamma}$. This is possible only if $\mathrm{Fr}(U_{\widehat \Gamma})\cap r_{\widehat\Gamma}$ is reduced to a closed leaf $\widehat\phi$, more precisely when situation (2) occurs. As explained in Proposition \[prop: existence\], the annulus $A_{\widehat\phi}$ is included in $\widehat W$. So all compact components you need to add to $\widehat W$ to get $\widehat A$ are included in $R(\widehat\phi)$ and so $z$ does not belong to $\widehat A$. The fact that every point of $\widehat A\cap \Omega(\widehat f)$ is $\widetilde f$-positive, implies that $\widehat A$ is positive and $\kappa\in H_{1}(\widehat A,{\mathbb{Z}})$ its positive generator, where $\iota_*(\kappa)=[\widehat \Gamma]$.
Write $\widetilde W=\pi^{-1}(\widehat W)$ and $\widetilde A=\pi^{-1}(\widehat A)$. As a connected component of $\widetilde\pi^{-1}({W_{\Gamma}^{l\rightarrow r }})$ one knows that the stabilizer of $\widetilde W$ in the group of covering automorphisms is generated by $T_{\widetilde \gamma}$ and, as remarked after the definition of $W_{\widetilde\gamma}^{l\rightarrow r}$, that $T(\widetilde W)\cap \widetilde W=\emptyset$ if $T$ is not power of $T_{\widetilde\gamma}$. One deduces similar results for $\widetilde A$. This implies that $\widetilde A$ projects onto a positive annulus $A$ of $I$ that contains $\Omega'(I)_{\Gamma}$ and whose positive generator is given by the homology class of a simple loop freely homotopic to $\Gamma$ in $\mathrm{dom}(I)$. In fact $\widetilde W$ projects onto a connected component of $W_{\widehat\Gamma}^{l\rightarrow r }$ and $A$ is obtained from this component by adding the connected components of the complement that have no singular points. $\Box$
Additional results {#subsection:auxilliary}
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We would like to state a result relied only on the isotopy $I$ and not on the transverse foliation $\mathcal F$. In this subsection, $f$ is an orientation preserving homeomorphism of ${\mathbb{S}}^2$ and $I$ a maximal isotopy of $f$. We keep the usual notations such as $\mathrm{dom}(I)$, $\widetilde{\mathrm{dom}}(I)$, $\widetilde \pi$, $\widetilde f$, .... Let us begin with an useful lemma.
\[lemma:homotopic annuli\] If $A$ and $A'$ are open annuli in $\mathrm{dom}(I)$ invariant by $f$ such that:
- the two connected components of ${\mathbb{S}}^2\setminus A$ contain a fixed point of $I$;
- the two connected components of ${\mathbb{S}}^2\setminus A'$ contain a fixed point of $I$;
- $A\cap A'\cap \Omega(f)\not=\emptyset$.
Then $A$ and $A'$ are homotopic in $\mathrm{dom}(I)$ (meaning that an essential simple loop $\Gamma$ of $A$ and an essential loop $\Gamma'$ of $A'$ are freely homotopic in $\mathrm{dom}(I)$, up to a change of orientation).
Let us fix $z\in A\cap A'\cap\Omega(f)$ and a lift $\widetilde z\in \widetilde{\mathrm{dom}}(I)$. Write $\widetilde A$ and $\widetilde A'$ for the connected components of $\widetilde\pi^{-1}(A)$ and $\widetilde\pi^{-1}(A')$ respectively, that contains $\widetilde z$. Fix a generator $T$ of the stabilizer of $\widetilde A$ in the group of covering automorphisms and a generator $T'$ of the stabilizer of $\widetilde A'$. We have seen at the beginning of this section that $\widetilde A$ and $\widetilde A'$ are invariant by $\widetilde f$. Let $D\subset A\cap A'$ be a topological $I$-free disk containing $z$. One can find an integer $n{\geqslant}1$ and a point $z'\in D\cap f^{-n}(D)$. Write $\widetilde D$ for the lift of $D$ that contains $\widetilde z$, and $\widetilde z'$ the lift of $z'$ that lies in $\widetilde D$. There exists a non trivial covering transformation $T''$ such that $\widetilde f^n(\widetilde z)\in T''(\widetilde D)$ and so $T''(\widetilde D)\subset \widetilde A\cap\widetilde A'$. This implies that $T''$ is a non trivial power of $T$ and a non trivial power of $T'$. In other terms, the stabilizers of $\widetilde A$ and $\widetilde A'$ are equal. But this means that $A$ and $A'$ are homotopic.
Let us explain now how to associate simple loops to points of $\Omega'(I)$ independently of a transverse foliation. Note first that for every point $z\in \Omega'(I)$, there exists $I$-free disks $D$ such that $f^k(z)\in D$ for infinitely many $k\in{\mathbb{Z}}$. Indeed, consider a point $z'\in\alpha(z)\cup\omega(z)$ that is not fixed by $I$. Any topological open disk containing $z'$ is $I$-free if sufficiently small, and meets the orbit of $z$ infinitely often.
\[prop: free disks\] For every $z\in\Omega'(I)$, there exists a simple loop $\Gamma \subset\mathrm{dom}(I)$, uniquely defined up to homotopy, such that:
1. there exists a positive annulus $A$ containing $z$ such that an essential simple loop of $A$ defining the positive class of $A$ is homotopic to $\Gamma$;
2. every invariant annulus $A'$ containing $z$ and such that the two connected components of its complement contain a fixed point of $I$ is homotopic to $A$ in $\mathrm{dom}(I)$;
3. for every $I$-free disk $D$ that contains two points $f^n(z)$ and $f^m(z)$, $n<m$, and every path $\delta\subset D$ joining $f^m(z)$ to $f^n(z)$, the loop $I^{m-n}(f^n(z))\delta$ is homotopic to a positive power of $\Gamma$.
We have seen in Theorem \[th:local-structure\] that there exists a positive annulus $A$ that contains $z$. More precisely, if $\mathcal F$ is a transverse foliation of $I$, one can choose for $\Gamma$ the transverse simple loop associated with $z$ (or any simple loop homotopic in $\mathrm{dom}(I)$). The assertion (1) is proved and the assertion (2) is an immediate consequence of Lemma \[lemma:homotopic annuli\]. It remains to prove (3). Suppose that $D$ is a $I$-free disk that contains two points $f^n(z)$ and $f^m(z)$, $n<m$. It has been proved in [@LeCalvezTal] that there exists a transverse foliation ${\mathcal F}'$ of $I$ such that $f^n(z)$ and $f^m(z)$ belong to the same leaf of the restricted foliation ${\mathcal F}'\vert_D$. Let $\phi'$ be the leaf of ${\mathcal F}'$ that contains $f^n(z)$ and $f^m(z)$ and $\Gamma'$ the transverse loop associated with $z$. The homotopy class of $I^{m-n}(f^n(z))\delta$ does not depend on the choice of the path $\delta\subset D$ joining $f^m(z)$ to $f^n(z)$. Moreover, choosing for $\delta$ the segment of leaf of $\phi'\cap D$ that joins $f^m(z)$ to $f^n(z)$, and noting that $I^{m-n}(f^n(z))\delta$ is homotopic to $I^{m-n}_{{\mathcal F}'}(f^n(z))\delta$, one deduces that $I^{m-n}(f^n(z))\delta$ is homotopic to a positive power of $\Gamma'$. As seen in the proof of Theorem \[th:local-structure\], there exists a positive annulus $A'$ that contains $z$, such that $\Gamma'$ is homotopic in $\mathrm{dom}(I)$ to an essential simple loop of $A'$ that defines the positive class of $A'$. By $(2)$, one deduces that $\Gamma'$ is homotopic to $\Gamma$ or to its inverse. If we want to prove that $\Gamma'$ is homotopic to $\Gamma$, we have to look carefully at the proof of Lemma \[lemma:homotopic annuli\]. If the disk $D$ is chosen very small, then the path $I^{n+2}_{\widetilde{\mathcal F}} (\widetilde f^{-1}(\widetilde z'))$ contains as a sub-path a transverse path joining the leaf $\widetilde\phi$ of $\widetilde{\mathcal F}$ that contains $\widetilde z$ to $T''(\widetilde\phi)$. Similarly, the path $I^{n+2}_{\widetilde{\mathcal F}'} (\widetilde f^{-1}(\widetilde z'))$ contains as a sub-path a transverse path joining the leaf $\widetilde\phi'$ of $\widetilde{\mathcal F}'$ that contains $\widetilde z$ to $T''(\widetilde\phi')$. We deduce that $T''$ is a positive power of $T$ and a positive power of $T'$. So $T$ and $T'$ are equal, which means that $\Gamma$ and $\Gamma'$ are homotopic in $\mathrm{dom}(I)$.
A simple loop $\Gamma\subset\mathrm{dom}(I)$ that satisfies the properties stated in Proposition \[prop: free disks\] is said to be [*associated with $z\in\Omega'(I)$*]{}.
\[th:familyofannulilocal\] One can cover $\Omega'(I)$ by invariant annuli $A\subset\mathrm{dom}(I)$ satisfying the following: a point $z\in\Omega'(I)$ belongs to $A$ if and only one can choose as a loop associated with $z\in\Omega'(I)$ an essential simple loop of $A$.
Let $\mathcal F$ be a foliation transverse to $I$. For every $z\in\Omega'(I)$, write $\Gamma_z$ for the transverse simple loop associated to $z$ and set $z\sim z'$ if $\Gamma_z$ and $\Gamma_{z'}$ are homotopic, up to the sign. If $\widehat{\mathrm{dom}}(I)$ is an annular covering space associated to $\Gamma_z$, as defined in the previous sub-section, then $z'\sim z$ if and only if $\Gamma_{z'}$ can be lifted as a simple loop of $\widehat{\mathrm{dom}}(I)$. In case, $z\sim z'\Leftrightarrow \Gamma_z=\Gamma_{z'}$, there is nothing to prove, one can choose for $A$ the annulus given by Proposition \[prop:local-structure\]. We will suppose from now that this does not happen. This implies that for every $z'\sim z$, the annulus $U(\Gamma_{z'})$ has a closed leaf on its boundary. Moreover, there exist at most two different loops $\Gamma_{z'}$, $z'\sim z$, that are not bordered by two closed leaves. Write $\mathcal C$ for the set of closed annulus bordered by two closed leaves homotopic to $\Gamma_z$, up to the sign, and eventually reduced to a single loop. For every $C\in{\mathcal C}$, define $$A_C=A_{\phi}\cup C\cup A_{\phi'},$$ where $C$ is bordered by $\phi$ and $\phi'$. One gets an invariant open annulus homotopic to $U(\Gamma_{z})$. Now define $$C^*=\bigcup_{C\in\mathcal C} C, \, A^*=\bigcup_{C\in\mathcal C} A_C.$$ The first set is an annulus, that can be open, semi open and bordered with a closed leaf, compact and bordered by two closed leaves or degenerate and reduced to unique closed leaf. The second set is an an invariant open annulus.
In case $C^*$ is open, it contains all sets $U_{\Gamma_{z'}}$, $z'\sim z$, and consequently all points $z'\sim z$. We can choose $A= A^*$.
In case $C^*$ is semi open and contains all sets $U_{\Gamma_{z'}}$, $z'\sim z$, we can choose $A=A^*$.
In case $C^*$ is semi open and does not contain all sets $U_{\Gamma_{z'}}$, $z'\sim z$, there exists a unique [$U_{\Gamma_{z_*'}}$, with $z_*\sim z$, such that $U_{\Gamma_{z_*'}}$]{} is not contained in $C^*$ and the closed leaf $\phi$ that borders $C^*$ also borders $U_{\Gamma_{z_*}}$. Using Proposition \[prop: existence\] and the remark that follows, we know that the positive annulus positive $A_{\Gamma_{z_*}}$ constructed in proposition Proposition \[prop:local-structure\] contains $A_{\widehat\phi}$, more precisely $A_{\Gamma_{z_*}} \cap A^*= A_{\widehat\phi}$. So we have two homotopic annuli whose intersection is a annulus homotopic to both of them, so the union is an annulus and we can choose $A=A_{\Gamma_{z_*}} \cup A^*$.
In case $C^*$ is compact, we similarly construct $A$ as the union of $A^*$ and two annuli of type $A_{\Gamma}$.
Proof of Theorem \[thmain:global-structure\] {#subsection: proofglobal}
--------------------------------------------
Using the fact that every squeezed annulus is contained in a maximal squeezed annulus, Theorem \[thmain:global-structure\] can be written as below:
\[th:global-structure2\] Let $f:{\mathbb{S}}^2\to{\mathbb{S}}^2$ be an orientation preserving homeomorphism that has no topological horseshoe. Then $\Omega'(f)$ is covered by squeezed annuli.
Let $z$ be a point of $\Omega'(f)$. Suppose first that $z$ belongs to an invariant open annulus $A_0$ such that:
- the two connected components of ${\mathbb{S}}^2\setminus A_0$ are fixed by $f$;
- if $\kappa_0$ is a generator of $H_1(A,{\mathbb{Z}})$, then $\mathrm{rot}_{ f\vert_{A_0},\kappa_0}(z)\not=0+{\mathbb{Z}}$.
We write $f_0=f\vert_{A_0}$ and denote $\check f_0 $ the lift of $f_0$ to the universal covering space $\check A_0$, such that $\mathrm{rot}_{\check f_0,\kappa_0}(z)\in(0,1)$. Let $I_0$ be a maximal isotopy of $f_0$ that is lifted to an identity isotopy of $\check f_0$. By Theorem \[th:local-structure\], there exists a positive annulus $A_1$ of $I_0$ that contains $z$. The fact that $\mathrm{rot}_{\check f_0,\kappa_0}(z)>0$ implies that $A_1$ is essential in $A_0$ and that the positive generator $\kappa_1$ of $H_1(A_1,{\mathbb{Z}})$ is sent onto $\kappa_0$ by the morphism $i_*: H_1(A_1,{\mathbb{Z}})\to H_1(A_0,{\mathbb{Z}})$. Let $\check f_1$ be a lift of $f_1=f\vert_{A_1}$ to the universal covering space of $A_1$ such that $\mathrm{rot}_{\check f_1,\kappa_1} (z)=\mathrm{rot}_{\check f_0, \kappa_0}(z)-1$ and $I_1$ be a maximal isotopy of $f_1$ lifted to an identity isotopy of $\check f_1$. By Theorem \[th:local-structure\], there exists a positive annulus $A_2$ of $I_1$ containing $z$. The fact that $\mathrm{rot}_{\check f_1,\kappa_1} (z)<0$ implies that $A_2$ is essential in $A_1$ and that the positive generator $\kappa_2$ of $H_1(A_2,{\mathbb{Z}})$ is sent onto $-\kappa_1$ by the morphism $i_*: H_1(A_2,{\mathbb{Z}})\to H_1(A_1,{\mathbb{Z}})$. The annulus $A_2$ is a squeezed annulus containing $z$.
Suppose now that such an annulus $A_0$ does not exist. We will slightly change the proof. By assumption, $\alpha(z)\cup\omega(z)$ is not included in $\mathrm{fix}(f)$. Choose a point $z'$ of $\alpha(z)\cup\omega(z)$ that is not fixed. Any topological open disk containing $z'$ is free if sufficiently small, and meets the orbit of $z$ infinitely often. So, there exists a free open disk $D$ containing $z$ and $f^q(z)$, where $q{\geqslant}2$. One can find a simple path $\delta$ in $D$ that joins $f^q(z)$ to $z$. There exists an identity isotopy $I_*=(h_t)_{t\in[0,1]}$ supported on $D$, such that $h=h_1$ sends $f^{q}(z)$ on $z$. Moreover, one can suppose that the trajectory $t\mapsto h_t(z)$ is the subpath of $\delta$ that joins $f^q(z)$ to $z$. The point $z$ is a periodic point of $h\circ f$, of period $q$. Moreover $f$ and $h\circ f$ have the same fixed points because $D$ is free. Fix $z_*\in\mathrm{fix}(f)$. The main result of [@LeCalvez3] tells us that $f$ has a fixed point $z_0\not=z_*$ such that $\mathrm{rot}_{h\circ f\vert_{A_0}, \kappa_0}(z)\not =0+{\mathbb{Z}}$, where $A_{0}={\mathbb{S}}^2\setminus\{z_*,z_0\}$ and $\kappa_0$ is a generator of $H_1(A_0,{\mathbb{Z}})$. Write $f_0=f\vert_{A_0}$, $h_0=h\vert _{A_0}$, denote $\check h_0$ the lift of $h_0$ to the universal covering space $\check A_0$ of $A_0$ naturally defined by $I_*$ and $\check f_0$ the lift of $f_0$ to $\check A_0$ such that $\mathrm{rot}_{\check f_0, \kappa_0}(z)=0$, which exists since by assumption $A_0$ does not satisfy the properties of the previous case. Replacing $\kappa_0$ with $-\kappa_0$ if necessary, one can suppose that $\mathrm{rot}_{\check h_0\circ\check f_0, \kappa_0}(z)>0$. Let $I_0$ be a maximal isotopy of $f_0$ lifted to an identity isotopy of $\check f_0$. As explained in [@LeCalvezTal], there exists a foliation $\mathcal F_0$ transverse to $I_0$ such that $\delta$ is included in a leaf of $\mathcal F_0$. The proof of Theorem \[th:local-structure\] tells us the following:
- $I_0{}_{{\mathcal F}_0}^{{\mathbb{Z}}}(z)$ strictly draws infinitely a transverse simple loop $\Gamma_1$;
- $z$ belongs to a positive annulus $A_1$;
- the image of the positive class $\kappa_1$ of $A_1$ is sent onto the class of $\Gamma_1$ by the morphism $i_*: H_1(A_1,{\mathbb{Z}})\to H_1(\mathrm{dom}(I),{\mathbb{Z}})$.
The homology class in $A_0$ of the loop naturally defined by $ I_0{}_{\mathcal F_0}^q(z)\delta$ is a positive multiple of $\kappa_0$ because $\mathrm{rot}_{\check h_0\circ\check f_0, \kappa_0}(z)>0$ but it is a positive multiple of the class of $\Gamma_1$. So $A_1$ is essential in $A_0$ and the positive generator $\kappa_1$ of $H_1(A_1,{\mathbb{Z}})$ is sent onto $\kappa_0$ by the morphism $i_*: H_1(A_1,{\mathbb{Z}})\to H_1(A_0,{\mathbb{Z}})$. Set $f_1=f\vert_{A_1}$. The rotation number of $z$ for the lift of $f_1$ to the universal covering space of $A_1$ naturally defined by $I_0$ (as explained at the beginning of this section) is equal to zero because $\mathrm{rot}_{\check f_0, \kappa_0}(z)=0$. Let $\check f_1$ be the lift of $f_1$ such that $\mathrm{rot}_{\check f_1,\kappa_1} (z)=-1$ and $I_1$ be a maximal isotopy of $f_1$ that is lifted to an identity isotopy of $\check f_1$. By Theorem \[th:local-structure\], there exists a positive annulus $A_2$ of $I_1$ that contains $z$. To ensure that $A_2$ is a squeezed annulus, one must prove that the positive generator $\kappa_2$ of $H_1(A_2,{\mathbb{Z}})$ is sent onto $-\kappa_1$ by the morphism $i_*: H_1(A_2,{\mathbb{Z}})\to H_1(A_1,{\mathbb{Z}})$. It cannot be sent on $\kappa_1$ because $\mathrm{rot}_{\check f_1,\kappa_1} (z)<0$ so it remains to prove that $A_2$ is essential in $A_1$. If it is not essential, the compact connected component $K'$ of $A_1\setminus A_2$ contains a fixed point $z''$ of $I_1$ and we have $\mathrm{rot}_{\check f_1,\kappa_1} (z'')=0$. Let $\check{f}_0'$ be the lift of $f_0$ such that $\mathrm{rot}_{\check{f}_0',\kappa_0} (z)=-1$, which implies that $\mathrm{rot}_{\check{f}_0',\kappa_0} (z'')=0$. The set $A_2\cup K'$ being an invariant open disk in $A_1\subset A_0$ containing $z$ and $z''$, one deduces that the pre-images of $A_2\cup K'$ by the projection from $\check{A}_0$ to $A_0$ are all invariant by $\check{f}_0'$, and as $z$ is nonwandering, any point $\check{z}$ projecting onto $z$ is also nonwandering for $\check{f_0}'$. Since $\mathrm{rot}_{\check{f}_0',\kappa_0} (z)=-1$, one can find a disk $\check D$ that is free for $\check{f}_0'$ and that contains $\check z$, and positive integers $n_0,m_0$ such that $(\check{f}_0')^{n_0}(\check z)\in T^{-m_0}(\check D)$. Since $\check z$ is nonwandering, there exists some point $\check z_1$ sufficiently close to $\check z$ and a positive integer $n_1$ such that both $\check z_1, (\check{f}_0')^{n_0+n_1}(\check z_1)$ belong to $\check D$ and such that $(\check{f}_0')^{n_0}(\check z_1)\in T^{-m_0}(\check D)$. We have got a contradiction with Corollary \[cor:free-disks\].
The following result can be easily deduced from the proof above.
\[prop:linking-recurrent\] Let $f:{\mathbb{S}}^2\to{\mathbb{S}}^2$ be an orientation preserving homeomorphism that has no topological horseshoe. Then for every recurrent point $z\not\in\mathrm{fix}(f)$ and every fixed point $z_*$, there exists a fixed point $z_0\not=z_*$ such that $\mathrm{rot}_{f\vert_{{\mathbb{S}}^2\setminus\{z_*,z_0\}},\kappa}(z)\not=0+{\mathbb{Z}}$, if $\kappa$ is a generator of $H_1({\mathbb{S}}^2\setminus\{z_*,z_0\},{\mathbb{Z}})$.
Indeed, if the conclusion fails, the second part of the proof of Theorem \[th:global-structure2\] tells us that there exists a fixed point $z_0\not=z_*$, a maximal isotopy $I_0$ of $f\vert_{{\mathbb{S}}^2\setminus\{z_*,z_0\}}$ and a positive annulus $A_1$ of $I_0$ such that $\mathrm{rot}_{\check f_1,\kappa_1} (z)=0$, where $\check f_1$ is the lift of $f_1\vert_{A_1}$ to the universal covering space of $A_1$ defined by $I_0$ (as explained at the beginning of the section) and $\kappa_1$ the positive generator of $H_1(A_1,{\mathbb{Z}})$. We have seen in Proposition \[prop:zero-rotation\] that it implies that $f$ has a topological horseshoe.
Birkhoff Recurrence classes and recurrent points of homeomorphisms of ${\mathbb{S}}^2$ with no topological horseshoe
====================================================================================================================
In this section we examine the possible dynamical behaviour of Birkhoff recurrence classes and transitive sets for homeomorphisms and diffeomorphisms of ${\mathbb{S}}^2$ with no topological horseshoe.
Birkhoff recurrence classes
---------------------------
We start by restating and proving Proposition \[propmain:birkoffclasses3fixedpoints\].
\[prop\_birkoffclasses3fixedpoints\] Suppose that $f$ is an orientation preserving homeomorphism of ${\mathbb{S}}^2$ with no topological horseshoe. If $\mathcal{B}$ is a Birkhoff recurrence class containing two fixed points $z_0$ and $z_1$, then
- either there exists $q{\geqslant}1$ such that every periodic point of $f$ distinct from $z_0$ and $z_1$ has a period equal to $q$, all recurrent points in $\mathcal B$ are periodic points and $\mathrm{fix}(f^q)\cap\mathcal{B}$ is an unlinked set of $f^q$;
- or $f$ is an irrational pseudo-rotation.
For a given generator of $H_1({\mathbb{S}}^2\setminus\{z_0,z_1\},{\mathbb{Z}})$ and a given lift of $f\vert_{{\mathbb{S}}^2\setminus\{z_0,z_1\}}$, there exists a unique rotation number by Proposition \[prop:PB\]. If this number is irrational, $f$ is an irrational pseudo-rotation. Suppose that it is rational and can be written $p/q$ in an irreducible way, then all periodic points have period $q$ and rotation number $p/q$ (which does not mean that such periodic points exist). If there is no recurrent point but $z_0$ and $z_1$, item (1) is true with any integer $q$. Suppose now that such a recurrent point $z$ exists. By Proposition \[prop:recurrentzero-rotation\], one knows that the annulus ${\mathbb{S}}^2\setminus\{z_0,z_1\}$ contain fixed points of $f^q$. Let $z_2$ be such a point. The class $\mathcal B$, containing fixed points, is a Birkhoff recurrence class of $f^q$. Let us prove by contradiction, that $z$ is a fixed point of $f^q$. By Proposition \[prop:linking-recurrent\], there exists a fixed point $z_3$ such that $\mathrm{rot}_{f\vert_{{\mathbb{S}}^2\setminus\{z_2,z_3\}},\kappa}(z)\not=0+{\mathbb{Z}}$, if $\kappa$ is a generator of $H_1({\mathbb{S}}^2\setminus\{z_2,z_3\},{\mathbb{Z}})$. Of course either $z_3\not=z_0$ or $z_3\not=z_1$.The assumption $z_3\not=z_0$ contradicts Theorem \[th:rotation-number\] because we would have $$\mathrm{rot}_{f^q\vert_{{\mathbb{S}}^2\setminus\{z_2,z_3\}},\kappa}(z)\not=0+{\mathbb{Z}}, \enskip\mathrm{rot}_{^qf\vert_{{\mathbb{S}}^2\setminus\{z_2,z_3\}},\kappa}(z_0)=0+{\mathbb{Z}}.$$ The assumption $z_3\not=z_1$ contradicts Theorem \[th:rotation-number\], because we would have $$\mathrm{rot}_{f^q\vert_{{\mathbb{S}}^2\setminus\{z_2,z_3\}},\kappa}(z)\not=0+{\mathbb{Z}}, \enskip\mathrm{rot}_{f^q\vert_{{\mathbb{S}}^2\setminus\{z_2,z_3\}},\kappa}(z_1)=0+{\mathbb{Z}}.$$
To prove the second assertion of the proposition, consider a maximal isotopy $I$ of $f^q$ that fixes $z_0$, $z_1$ and $z_2$. If $z'_0$ and $z'_1$ are two different fixed points of $I$ and $z'_2$ is a third fixed point of $f^q$, denote $\mathrm{rot}_{I, z'_0,z'_1}(z'_2)$ the rotation number of $z'_2$ defined for the lift of $f^q\vert_{{\mathbb{S}}^2\setminus\{z'_0,z'_1\}}$ to the universal covering space that fixes the lifts of the remaining points of $\mathrm{fix}(I)$, a generator of $H_1({\mathbb{S}}^2\setminus\{z'_0,z'_1\})$ being given by the oriented boundary of a small disk containing $z'_0$ in its interior. In particular, one has $\mathrm{rot}_{I, z'_0,z'_1}(z'_2)=0$ if $z'_2\in \mathrm{fix}(I)$. By Proposition \[prop:PB\], one knows that $\mathrm{rot}_{I, z_0,z_1}(z'_2)=0$ for every fixed point $z'_2\in {\mathbb{S}}^2\setminus\{z_0,z_1\}$ of $f^q$ because $\mathrm{rot}_{I, z_0,z_1}(z_2)=0$. By Theorem \[th:rotation-number\], one knows that $\mathrm{rot}_{I, z'_0,z'_1}(z'_2)=0$ for every fixed point $z'_2\in \mathcal B\setminus\{z'_0,z'_1\}$ of $f^q$ if exactly one of the points $z'_0$, $z'_1$ is equal to $z_0$ or $z_1$. Indeed if, for instance, one of the points $z'_0$, $z'_1$ is equal to $z_0$, then $\mathrm{rot}_{I, z'_0,z'_1}(z_1)=0$. One deduces that $\mathrm{rot}_{I, z'_0,z'_1}(z'_2)=0$ for every fixed point $z'_2\in \mathcal B\setminus\{z'_0,z'_1\}$ of $f^q$ if none of the points $z'_0$, $z'_1$ is equal to $z_0$ or $z_1$. Indeed, by Proposition \[prop: rotation numbers\], if $z'_2\not=z_0$, then $$\mathrm{rot}_{I, z'_0,z'_1}(z'_2)= \mathrm{rot}_{I, z'_0,z_0}(z'_2)-\mathrm{rot}_{I, z'_1,z_0}(z'_2)=0$$ and a similar argument, introducing $z_1$, can be done if $z'_2\not=z_1$.
Assume now that $\mathcal{B}\cap\mathrm{fix}(f^q)$ is not unlinked. Then, one can find a fixed point $z'_2\in{\mathcal B}$ of $f^q$ which belongs to the domain of $I$. Let $\mathcal F$ be a foliation transverse to $I$. The closed curve $I_{\mathcal F}(z'_2)$ defines naturally a loop $\Gamma$. Let $\delta_{\Gamma}$ be a dual function. As seen before, one can find a point $z'_0\in\mathrm{fix}(I)$ where the minimal value $k_-$ of $\delta_{\Gamma}$ is reached and a point $z'_1\in\mathrm{fix}(I)$ where its maximal value $k_+$ is reached. Note now that $$\mathrm{rot}_{I, z'_0,z'_1}(z'_2)=k_+-k_->0.$$We have found a contradiction.
As a consequence, we obtain Corollary \[crmain:periodsofbirkhoffclasses\] from the introduction, which we restate here:
\[cr:onlytwoperiods\] Let $f$ be an orientation preserving homeomorphism of ${\mathbb{S}}^2$ with no topological horseshoe. Let $\mathcal{B}$ be a Birkhoff recurrence class containing periodic points of different periods, then:
- there exists integers $q_1$ and $q_2$, with $q_1$ dividing $q_2$, such that every periodic point in $\mathcal{B}$ has a period either equal to $q_1$ or to $q_2$
- if $q_1{\geqslant}2$, there exists a unique periodic orbit of period $q_1$ in $\mathcal B$;
- if $q_1=1$, there exist at most two fixed points in $\mathcal B$.
By Corollary \[cr:divideperiods\], we already know that, if $\mathcal{B}$ has periodic points of periods $q_1<q_2$, then $q_1$ divides $q_2$. Assume, for a contradiction, that $\mathcal{B}
$ contains a periodic point $z_1$ of period $q_1$, a periodic point $z_2$ of period $q_2$ and a periodic point $z_3$ of period $q_3$, where $q_1<q_2< q_3$. The homeomorphism $f^{q_2}$ has at least three fixed points ($z_1$, $z_2$ and $f(z_2)$). Moreover, by the item (4) of Proposition \[prop: birkhoff connexions\], there exists a Birkhoff recurrence class of $f^{q_2}$ that contains a point $z'_1$ in the orbit of $z_1$, a point $z'_2$ in the orbit of $z_2$ and a point $z'_3$ in the orbit of $z_3$. Since $z'_1$ and $z'_2$ are fixed by $f^{q_2}$, and $z'_3$ is periodic but not fixed, $f^{q_2}$ must have a topological horseshoe by Proposition \[prop\_birkoffclasses3fixedpoints\], as does $f$.
Suppose that $q_1{\geqslant}2$ and that $\mathcal B$ contains at least two periodic orbits of period $q_1$. One can find a Birkhoff recurrence class of $f^{q_1}$ that contains a point in each of these two orbits plus a point in a periodic orbit of period $q_2$. This last point being not fixed by $f^{q_1}$, this contradicts Proposition \[prop\_birkoffclasses3fixedpoints\]. The last item is an immediate consequence of Proposition \[prop\_birkoffclasses3fixedpoints\].
Recurrent points and transitive sets.
-------------------------------------
Let $f:{\mathbb{S}}^2\to{\mathbb{S}}^2$ be an orientation preserving homeomorphism with no horseshoe. If $z_0$ and $z_1$ are periodic points of $f$ and $z$ is recurrent and not periodic, we define $\mathrm{rot}_{f,z_0,z_1}(z)\in{\mathbb{T}}^1$ in the following way: we denote $q$ the smallest common period of $z_0$ and $z_1$, we choose the class $\kappa$ of the boundary of a small topological disk containing $z_0$ as a generator of $H_1({\mathbb{S}}^2\setminus \{z_0,z_1\},{\mathbb{Z}})$, and we set $$\mathrm{rot}_{f,z_0,z_1}(z)=\mathrm{rot}_{f^{q}\vert_{{\mathbb{S}}^2\setminus \{z_0,z_1\}},\kappa}(z),$$ which is well defined by Theorem \[th:rotation-number\]. We will say that a recurrent and non periodic point $z$ has a [*rational type*]{} if $\mathrm{rot}_{f,z_0,z_1}(z)\in{\mathbb{Q}}/{\mathbb{Z}}$ for every choice of $z_0$ and $z_1$ and has an [*irrational type*]{} otherwise.
### Recurrent points of rational type
Let us show that, if $z$ is recurrent and of rational type, then $f$ is topologically infinitely renormalizable over $\Lambda=\omega(z)$.
\[prop:rational\]Let $f:{\mathbb{S}}^2\to{\mathbb{S}}^2$ be an orientation preserving homeomorphism with no topological horseshoe, $z$ a recurrent point of $f$ of rational type, and $\Lambda=\omega(z)$. There exists an increasing sequence $(q_n)_{n{\geqslant}0}$ of positive integers and a decreasing sequence $(D_n)_{n{\geqslant}0}$ of open disks containing $z$ such that:
- $q_n$ divides $q_{n+1}$;
- $D_n$ is $f^{q_n}$ periodic;
- the disks $f^k(D_n)$, $0{\leqslant}k<q_n$ are pairwise disjoint;
- $\Lambda\subset \bigcup_{k=0}^{q_n-1}f^{k}(D_n)$.
Furthermore, $f$ has periodic points of arbitrarily large period.
We will construct the sequence by induction. The key point is the fact that $f$ is “abstractly” renormalizable over $\overline{O_f(z)}$, as explained in the remark following Proposition \[prop: birkhoff classes iterate\]. We fix $z_*$ in $\mathrm{fix}(f)$ and set $$q_0=1, \enskip D_0={\mathbb{S}}^2\setminus\{z_*\}.$$ Let us construct $q_1$ and $D_1$. By Proposition \[prop:linking-recurrent\], there exists $z_0\in \mathrm{fix}(f)\cap D_0$ such that $\mathrm{rot}_{f,z_*,z_0}(z)\not=0+{\mathbb{Z}}$. By hypothesis, this number is rational and can be written $p/q+{\mathbb{Z}}$ where $p$ and $q>0$ are relatively prime. Let us state the key lemma, that will use Theorem \[th:global-structure2\]
\[lemma:differentclasses\]The Birkhoff recurrence classes of $f^i(z)$, $0{\leqslant}i<q$, for $f^q$, are all distinct.
Set $A_0={\mathbb{S}}^2\setminus\{z_*,z_0\}$ and write $\kappa_0$ for the generator of $H_1(A_0,{\mathbb{Z}})$ defined by the boundary of a small topological disk containing $z_0$ in its interior. By Theorem \[th:global-structure2\], there exists a squeezed annulus $A$ of $f^q$ that contains $z$. Using Proposition \[prop:recurrentzero-rotation\], one deduces that $\mathrm{rot}_{f^q\vert A,\kappa}(z)\not=0+{\mathbb{Z}}$, if $\kappa$ is a generator of $H_1(A,{\mathbb{Z}})$. We know that $\mathrm{rot}_{f^q\vert A_0,\kappa_0}(z)=0+{\mathbb{Z}}$ and so $A$ is not essential in $A_0$. The compact connected component $K$ of $A_0\setminus A$ contains at least a fixed point of $f^q$, because $A\cup K$ is a $f^q$-invariant open disk that contains a $f^q$-invariant compact set. Moreover, there exists $\rho\in{\mathbb{R}}/{\mathbb{Z}}$, equal to $\mathrm{rot}_{f^q\vert A,\kappa}(z)$ up to the sign, such that $$\begin{cases}
\mathrm{rot}_{f^q,z',z^*} (z)=\rho&\mathrm{if }\,\, z'\in\mathrm{fix}(f^q)\cap K,\\
\mathrm{rot}_{f^q,z',z^*} (z)=0+{\mathbb{Z}}&\mathrm{if }\,\, z'\in\mathrm{fix}(f^q)\setminus K,\\
\end{cases}$$ the last equality being due to the fact that $A\cup K$ is a $f^q$-invariant disk included in the annulus ${\mathbb{S}}^2\setminus\{z',z^*\}$ if $z'\in\mathrm{fix}(f^q)\setminus K$. Let us consider $0<i<q$. We immediately deduce that $$\begin{cases}
\mathrm{rot}_{f^q,f^i(z'),z^*} (f^i(z))=\rho&\mathrm{if }\,\, z'\in\mathrm{fix}(f^q)\cap K,\\
\mathrm{rot}_{f^q,f^i(z'),z^*} (f^i(z))=0+{\mathbb{Z}}&\mathrm{if }\,\, z'\in\mathrm{fix}(f^q)\setminus K.\\
\end{cases}$$ To prove the lemma, it is sufficient to prove that the Birkhoff recurrence classes of $z$ and $f^i(z)$ for $f^q$ are distinct if $1{\leqslant}i<q$. We will argue by contradiction and suppose that there exists $i\in\{1,\dots, q-1\}$ such that the Birkhoff recurrence classes of $z$ and $f^i(z)$ for $f^q$ are equal. Let $\mathring A_0$ be the $q$-tuple cover of $A_0$, let $\mathring \pi: \mathring A_0\to A_0$ be the covering projection and $\mathring T$ the generator of the group of covering automorphisms naturally defined by $\kappa_0$. Denote $\mathring z^* $ the end of $\mathring A_0$ corresponding to $z^*$ and $\mathring z_0 $ the end of $\mathring A_0$ corresponding to $z_0$. Fix a lift $\mathring{f}$ of $f\vert_{A_0}$ to $\mathring A_0$ and set $\mathring g=\mathring f^q\circ \mathring T^{-p}$. A simple finiteness argument permits us to say that if there exists a Birkhoff connection for $f^q$ from a point $z'\in A_0$ to a point $z''\in A_0$, there exists a Birkhoff connection for $\mathring g$ from a given lift $\mathring z'\in \mathring{\pi}^{-1}(\{z'\})$ to a certain lift $\mathring z''\in \mathring{\pi}^{-1}(\{z''\})$. Fix $\mathring z\in\mathring{\pi}^{-1}(\{z\})$. One deduces that there exists $0 {\leqslant}j {\leqslant}q-1$ such that $\mathring z\underset{\mathring f^q}\preceq \mathring T^j(\mathring f^i(\mathring z))$. The maps $\mathring f$ and $\mathring T$ commuting, one gets $$\mathring z\underset{\mathring g}\preceq \mathring T^j(\mathring f^i(\mathring z))\underset{\mathring g}\preceq \mathring T^{2j}(\mathring f^{2i}(\mathring z))\underset{\mathring g}\preceq \dots \underset{\mathring g}\preceq \mathring T^{(q-1)j}(\mathring f^{(q-1)i}(\mathring z))\underset{\mathring g}\preceq \mathring T^{qj}(\mathring f^{qi}(\mathring z))= \mathring f^{qi}(\mathring z)=\mathring g^i\circ \mathring T^{pi}(\mathring z).$$ Similarly, there exists $0 {\leqslant}j' {\leqslant}q-1$ such that $ \mathring T^{j'}(\mathring f^i(\mathring z))\underset{\mathring g}\preceq\mathring z$ and consequently $$\mathring g^i\circ \mathring T^{pi}(\mathring z)=\mathring f^{qi}(\mathring z)=\mathring T^{qj'}(\mathring f^{qi}(\mathring z))\underset{\mathring g}\preceq \mathring T^{(q-1)j'}(\mathring f^{(q-1)i}(\mathring z))\underset{\mathring g}\preceq\dots \underset{\mathring g}\preceq \mathring T^{2j'}(\mathring f^{2i}(\mathring z))\underset{\mathring g}\preceq \mathring T^{j'}(\mathring f^i(\mathring z))\underset{\mathring g}\preceq \mathring z.$$
Consequently $\mathring z$ is a Birkhoff recurrent point of $\mathring g$, and so is every point $\mathring f^{m}\circ \mathring T^n(\mathring z)$, where $m$ and $n$ are integers. Moreover the Birkhoff recurrence class of $\mathring z$ is equal to the Birkhoff recurrence class of $\mathring g^i\circ \mathring T^{pi}(\mathring z)$ and so is equal to the Birkhoff recurrence class of $\mathring T^{pi}(\mathring z)$. For every $l\in\{0,\dots,q\}$ denote $\mathring A^l$ the connected component of $\mathring\pi^{-1}(A)$ that contains $\mathring T^l(\mathring z)$ and $\mathring K^l$ the compact connected component of $\mathring A_0\setminus \mathring A^l$ (which is itself a connected component of $\mathring\pi^{-1}(K)$). We write $\mathring g_{\mathrm{sphere}}$ the extension of $g$ to $\mathring A_0\cup\{\mathring z^*,\mathring z_0\}$. Every annulus $\mathring A^l$ is a squeezed annulus of $\mathring g_{\mathrm{sphere}}$. Moreover every set $\mathring K^l$ contains a least one fixed point of $\mathring g$ and like in the annulus $A_0$, one has $$\begin{cases}
\mathrm{rot}_{\mathring g,\mathring z',\mathring z^*} (\mathring T^l(\mathring z))=\rho&\mathrm{if }\,\, \mathring z'\in\mathrm{fix}(\mathring g)\cap \mathring K^l,\\
\mathrm{rot}_{\mathring g,\mathring z',\mathring z^*} (\mathring T^l(\mathring z))=0+{\mathbb{Z}}&\mathrm{if }\,\, \mathring z'\in\mathrm{fix}(\mathring g)\setminus \mathring K^l.\\
\end{cases}$$ Note now that $pi$ is not a multiple of $q$ because $p$ and $q$ are relatively prime, and so the Birkhoff recurrence class of $\mathring z$ for $\mathring g$ contains a translate $\mathring T^l(\mathring z)$, where $l\in\{1,\dots ,q-1\} $. The equalities above contradict Theorem \[th:birkhoffcycles\]
Now one can apply Proposition \[prop: birkhoff classes iterate\] and the remarks that follow this proposition: there exists an integer $r{\geqslant}1$ and a covering of $\Lambda$ by a family $(W^j)_{j\in {\mathbb{Z}}/rq{\mathbb{Z}}}$ of pairwise disjoint open sets of ${\mathbb{S}}^2$ that satisfies $f(W^j)=W^{j+1}$ for every $j\in{\mathbb{Z}}/rq{\mathbb{Z}}$. These open sets are included in $A_0$ and are not essential, because $rq{\geqslant}2$ and $f(W^j)=W^{j+1}$ for every $j\in{\mathbb{Z}}/rq{\mathbb{Z}}$. For the same reason, if $j'\not=j$, the compact connected components of $A_0\setminus W^i$ do not contain any $W^{j'}$, $j'\not=j$. Adding the compact connected components of $A_0\setminus W^j$ to $W^j$, one gets a disk $D^j$. We have got a covering of $\Lambda$ by a family $(D^j)_{j\in {\mathbb{Z}}/rq{\mathbb{Z}}}$ of pairwise disjoint open disks of ${\mathbb{S}}^2$ that satisfies $f(D^j)=D^{j+1}$ for every $j\in{\mathbb{Z}}/rq{\mathbb{Z}}$. One can suppose that $z$ belongs to $D^0$. We set $D_1=D^0$ and $q_1=rq$. To construct $D_2$ and $q_2$ we do exactly the same replacing $D_0$ by $D_1$ and $f$ by $f^{q_1}\vert_{D_1}$ and continue the process.
Note that, since $D_n$ is homeomorphic to the plane and invariant by $f^{q_n}$, and since $f^{q_n}$ has a recurrent point in $D_n$, then by standard Brouwer Theory arguments $f^{q_n}$ must have a fixed point in $D_n$. Since the disks $f^k(D_n)$, $0{\leqslant}k<q_n$, are pairwise disjoint, this implies that $f$ has a periodic point of prime period $q_n$.
As a direct consequence, we have:
Let $f:{\mathbb{S}}^2\to{\mathbb{S}}^2$ be an orientation preserving homeomorphism with no horseshoe, $z$ a recurrent point of $f$ of rational type, and $\Lambda=\omega(z)$. Then the restriction of $f$ to $\Lambda$ is semiconjugated to an odometer.
### Recurrent points of irrational type
Let us look now at some particular recurrent points of irrational type.
\[prop:irrational\] Let $f:{\mathbb{S}}^2\to{\mathbb{S}}^2$ be an orientation preserving homeomorphism with no horseshoe and $z$ a recurrent point of $f$ of irrational type. We suppose that there exist two fixed points $z_0$ and $z_1$ such that $\mathrm{rot}_{f,z_0,z_1}(z)=\alpha+{\mathbb{Z}}$, where $\alpha\not\in{\mathbb{Q}}$. We denote $\check f$ the lift of $f\vert_{{\mathbb{S}}^2\setminus\{z_0,z_1\}}$ such that $\mathrm{rot}_{\check f}(z)=\alpha$ and such that $\kappa$ is the generator of $H_1({\mathbb{S}}^2\setminus\{z_0,z_1\},{\mathbb{Z}})$ defined by the boundary of a small topological disk containing $z_0$ in its interior. Let $(p_n/q_n)_{n{\geqslant}0}$ and $(p'_n/q'_n)_{n{\geqslant}0}$ be two sequences of rational numbers converging to $\alpha$, the first one increasing, the second one decreasing. There exists a decreasing sequence $(A_n)_{n{\geqslant}0}$ of invariant open annuli containing $z$ such that
- $A_{n}$ is essential in ${\mathbb{S}}^2\setminus\{z_0,z_1\}$,
- $A_n$ is a positive annulus of $\check f^{q_n}\circ T^{-p_n}$, with a positive generator sent onto $\kappa$ by the morphism $\iota_*: H_1(A_n,{\mathbb{Z}})\to H_1({\mathbb{S}}^2\setminus\{z_0,z_1\},{\mathbb{Z}})$;
- $A_n$ is negative annulus of $\check f^{q'_n}\circ T^{-p'_n}$, with a negative generator sent onto $\kappa$ by the morphism $\iota_*: H_1(A_n,{\mathbb{Z}})\to H_1({\mathbb{S}}^2\setminus\{z_0,z_1\},{\mathbb{Z}})$.
Let $I_0$ be a maximal isotopy of $f^{q'_0}\vert_{{\mathbb{S}}^2\setminus\{z_0,z_1\}}$ that is lifted to an identity isotopy of $\check f^{q'_0}\circ T^{-p'_0}$. By Theorem \[th:local-structure\], there exists a positive annulus $A'_0$ containing $z_0$. It must be essential and its positive class is sent onto $-\kappa$ by the morphism $\iota_*: H_1(A'_0,{\mathbb{Z}})\to H_1({\mathbb{S}}^2\setminus\{z_0,z_1\},{\mathbb{Z}})$. Let $I'_0$ be a maximal isotopy of $f^{q_0}\vert_{A'_0}$ that is lifted to an identity isotopy of the restriction of $\check f^{q_0}\circ T^{-p_0}$ to the lift of $A'_0$. By Theorem \[th:local-structure\], there exists a positive annulus $A_0$ containing $z_0$. It must be essential and its positive class is sent onto $\kappa$ by the morphism $\iota_*: H_1(A_0,{\mathbb{Z}})\to H_1({\mathbb{S}}^2\setminus\{z_0,z_1\},{\mathbb{Z}})$. We continue this process and construct alternately two sequences $(A'_n)_{n{\geqslant}0}$ and $(A_n)_{n{\geqslant}0}$. The sequence $(A_n)_{n{\geqslant}0}$ satisfies the conclusion of the proposition.
Let $X=\bigcap_{n{\geqslant}0}A_n$. Note that the complement of $A_n$ has exactly two connected components, $F_n$ that contains $z_0$ and $F_n'$ that contains $z_1$. Let us define $F=\bigcup_{n{\geqslant}0}F_n$ and $F'=\bigcup_{n{\geqslant}0}F'_n$, which are connected, invariant and disjoint. This implies that the complement of $X$ is either connected (if $F\cup F'$ is connected) or has two connected components (if $F\cup F'$ is not connected). Note also that if $z'_0$ and $z'_1$ are periodic points that belong to $F$, then $\mathrm{rot}_{f,z'_0,z'_1}(z)=0+{\mathbb{Z}}$, because there exists $n$ such that $z$ belongs to an invariant open disk $A_n\cup F_n'$ of ${\mathbb{S}}^2\setminus\{z'_0,z'_1\}$. In case $z'_0$ belongs to $F$ while $z'_1$ belongs to $F'$, then $\mathrm{rot}_{f,z'_0,z'_1}(z)=q\mathrm{rot}_{f,z_0,z_1}(z)$, where $q$ is the smallest common period of $z'_0$ and $z'_1$. Finally, one knows that every point of $z'\in X$ has a rotation number equal to $\alpha$ when this rotation number is defined. This is not necessary the case for every point in $\overline X$. Indeed let us suppose that $f$ is a homeomorphism of ${\mathbb{S}}^2$ that contains an invariant closed annulus $A$ satisfying the following:
- every point of the boundary of $A$ is fixed;
- there exists an essential invariant loop $\Gamma$ in $A$ of irrational rotation number;
- the dynamics is north-south in the open annuli bounded by $\Gamma$ and the boundary circles.
If $z_0$ and $z_1$ are chosen on different boundary components of $A$ and $z$ chosen on $\Gamma$, all sets $A_n$ coincide with the interior of $A$, for $n$ large enough, and consequently, there exist infinitely many fixed points in $\overline X$.
In the previous example, there exists a compact invariant annular set of irrational rotation number. In the following situation, this will not be the case. Starting with a Denjoy counterexample, one can construct a homeomorphism $f$ of ${\mathbb{S}}^2$ that contains an invariant closed annulus $A$ satisfying the following:
- every point of the boundary is fixed;
- there exists an invariant Cantor set $K$ of irrational rotation number;
- the orbit of a point $z$ in the interior of $A$ that do not belong to $K$ has either its $\alpha$ limit or its $\omega$ limit contained in one of the boundary circles.
There is still some more information one can derive about the Birkhoff recurrence class of a recurrent point of irrational type:
\[co: B.classes\] Let $f:{\mathbb{S}}^2\to{\mathbb{S}}^2$ be an orientation preserving homeomorphism with no topological horseshoe and $z$ a recurrent point of $f$ of irrational type. We suppose that there exist two fixed points $z_0$ and $z_1$ such that $\mathrm{rot}_{f,z_0,z_1}(z)=\alpha+{\mathbb{Z}}$, where $\alpha\not\in{\mathbb{Q}}$. Let $\mathcal{B}$ be the Birkhoff recurrence class of $z$. Then:
- $\mathcal{B}$ does not contain any periodic point that is neither $z_0$ nor $z_1$;
- if $\mathcal{B}$ contains $z_0$, then for any pair of periodic points $z'_0, z'_1$ not containing $z_0$, $\mathrm{rot}_{f,z'_0,z'_1}(z)=0+{\mathbb{Z}}$;
- if $\mathcal{B}$ contains both $z_0$ and $z_1$, then $f$ is a irrational pseudo-rotation.
The first item is a direct consequence of Theorem \[th:birkhoffcycles\] , since $\mathrm{rot}_{f,z_0,z_1}(z')\in{\mathbb{Q}}/{\mathbb{Z}}$ if $z'$ is periodic. The second item is also a consequence of Theorem \[th:birkhoffcycles\], since $\mathrm{rot}_{f,z'_0,z'_1}(z_0)=0+{\mathbb{Z}}$. The third item is a direct consequence of Proposition \[prop:PB\], where $z_0$ and $z_1$ play the role of $N$ and $S$.
Let us explain now what happens for a general recurrent point of irrational type. Let $f:{\mathbb{S}}^2\to{\mathbb{S}}^2$ be an orientation preserving homeomorphism with no horseshoe and $z$ a recurrent point of irrational type. If there exist two fixed points $z_0$ and $z_1$ such that $\mathrm{rot}_{f,z_0,z_1}(z)\not\in{\mathbb{Q}}/{\mathbb{Z}}$ we can apply Proposition \[prop:irrational\]. Suppose now that $\mathrm{rot}_{f,z_0,z_1}(z)\in {\mathbb{Q}}/{\mathbb{Z}}$ for every fixed points $z_0$ and $z_1$. Then starting from any fixed point $z_*$, one can construct a disk $D_1$ containing $z$ and an integer $q_1$ such that $f^{q_1}(D_1)=D_1$ and $f^k(D_1)\cap f^{k'}(D_1)=\emptyset$ if $0{\leqslant}k<k'<q_1$, like in the proof of Proposition \[prop:rational\]. Moreover we get that $\omega_{f^{q_1}}(z)$ is also contained in $D_1$.
One deduces that $\mathrm{rot}_{f,z'_0,z'_1}(z)\in {\mathbb{Q}}/{\mathbb{Z}}$ for every periodic points $z'_0$ and $z'_1$ that do not belong to $D_1$. In particular $\mathrm{rot}_{f,z'_0,z'_1}(z)\in {\mathbb{Q}}/{\mathbb{Z}}$ if the periods of $z'_0$ and $z'_1$ are both smaller than $q_1$. Let us consider the map $f^{q_1}\vert_{D_1}$ or more precisely its extension $f_1$ to the Alexandroff compactification of $D_1$ writing $z_{1,*}$ the point at infinity. Suppose that $\mathrm{rot}_{f_1,z_{1,*},z_1}(z)\in {\mathbb{Q}}/{\mathbb{Z}}$ for every fixed point $z_1\in D$ of $f_1$. Then we can continue the process and construct a disk $D_2$ containing $z$ and an integer $q_2>q_1$, multiple of $q_1$, such that $f^{q_2}(D_2)=D_2$ and $f^k(D_1)\cap f^{k'}(D_1)=\emptyset$ if $0{\leqslant}k<k'<q_2$. We know that $\mathrm{rot}_{f,z'_0,z'_1}(z)\in {\mathbb{Q}}/{\mathbb{Z}}$ if the periods of $z'_0$ and $z'_1$ are smaller than $q_2$. This implies that the process of Proposition \[prop:rational\] must stop at finite time. Consequently, one can apply Proposition \[prop:irrational\] to a power $f^q$ of $f$ and construct a decreasing sequence $(A_n)_{n{\geqslant}0}$ of annuli fixed by $f^q$ and such that $f^k(A_n)\cap A_n=\emptyset$ if $0<k<q$. Then one can consider $X(z)=\bigsqcup_{0{\leqslant}k<q} f^k\left(\bigcap_{n{\geqslant}0} A_n\right)$.
### Transitive Sets
We say that a transitive set $\Lambda$ is [*of irrational type*]{} if $\Lambda=\omega(z)$ for some recurrent point $z$ of irrational type. In particular $\Lambda$ is contained in the Birkhoff recurrence class of $z$. Let us now restate Proposition \[prmain:transitivesetsgeneralcase\] from the introduction.
\[pr:transitivesetsgeneralcase\] Let $f:{\mathbb{S}}^2\to{\mathbb{S}}^2$ be an orientation preserving homeomorphism with no horseshoe and $\Lambda$ a closed and invariant transitive set. Then:
1. either $\Lambda$ is a periodic orbit;
2. or $f$ is topologically infinitely renormalizable over $\Lambda$;
3. or $\Lambda$ is of irrational type.
Furthermore, if the second possibility holds, then $f$ has periodic points of arbitrarily large prime periods.
If the recurrent points in $\Lambda$ all are periodic, we are in case 1. If $\Lambda$ has a recurrent point of rational type then, by Proposition \[prop:rational\] we are in case 2. If $\Lambda$ has a recurrent point of irrational type then by Proposition \[prop:irrational\] we are in case 3.
We also obtain directly a proof of Corollary \[crmain:transitivewithperiodicorbit\] of the introduction as, if the second possibility in Proposition \[pr:transitivesetsgeneralcase\] holds, then $\Lambda$ does not contain any periodic point, and if $\Lambda$ is of irrational type and contains two distinct periodic orbits, then applying Corollary \[co: B.classes\] one may deduce that $f$ is a irrational pseudo-rotation.
Let us conclude this section with an application that concerns dissipative $\mathcal{C}^1$ diffeomorphisms of the plane ${\mathbb{R}}^2$. Let us recall Proposition \[prmain:dissipativediffeomorfisms\]
\[pr:dissipativediffeomorfisms\] Let $f:{\mathbb{R}}^2\to{\mathbb{R}}^2$ be an orientation preserving diffeomorphism with no topological horseshoe, such that $0<\mathrm{det} Df(z)<1$ for all $z\in{\mathbb{R}}^2$. Let $\Lambda$ be a compact, invariant, and transitive subset that is locally stable. Then either $\Lambda$ is a periodic orbit, or $f$ is topologically infinitely renormalizable over $\Lambda$.
Let $f_{\mathrm{sphere}}$ be the natural extention of $f$ to the Alexandrov compactification of ${\mathbb{R}}^2$ by adding a point $\infty$ at infinity. By Proposition \[pr:transitivesetsgeneralcase\], it suffices to show that $\Lambda$ cannot have a recurrent point of irrational type.
Assume, for a contradiction, this is false and let $z$ be a recurrent point of irrational type such that $\Lambda=\omega(z)$. There exist $q{\geqslant}1$ and $z_0, z_1$ in $\mathrm{fix}(f^q_{\mathrm{sphere}})$ such that $\mathrm{rot}_{f^q_{\mathrm{sphere}}, z_0, z_1}(z)\not \in{\mathbb{Q}}/{\mathbb{Z}}$. By Theorem \[th:global-structure2\], there exists a squeezed annulus $A$ of $f^q_{\mathrm{sphere}}$ that contains $z$. Moreover $z_0$ and $z_1$ are not in the same component of the complement of $A$. Denote $z^*$ the point $z_0$ or $z_1$ that is not in the same component as $\infty$. As explained before, one has $\mathrm{rot}_{f^q_{\mathrm{sphere}}, z^*,\infty}=\mathrm{rot}_{f^q\vert_{{\mathbb{R}}^2\setminus\{z^*\}}}(z)\not \in{\mathbb{Q}}/{\mathbb{Z}}$. Write $\Lambda'=\omega_{f^q_{\mathrm{sphere}}}(z)$. Since $\Lambda'$ is a compact subset of ${\mathbb{R}}^2$ and $\Lambda'\subset \Lambda$, one knows that $\infty$ does not belong to $\Lambda'$. We claim that $z^*$ also does not belong to $\Lambda'$. To see this, we argue by contradiction and suppose that $z^*\in\Lambda'$. Note first that $z^*$ cannot be neither a sink nor a source because $\Lambda'$ is a transitive set of $f^q$. We deduce that the eigenvalues of $Df^q(z^*)$ are distinct real numbers, otherwise $z$ would be a sink because $0<\mathrm{det} Df^q(z^*)<1$. Therefore one may blowup the point $z^{*}$ to a circle $\Sigma$ and extend $f^q\vert_{{\mathbb{R}}^2\setminus\{z^*\}}$ to a homeomorphism $g_{\mathrm{ann}}$ of ${\mathbb{R}}^2_{\mathrm{ann}}=({\mathbb{R}}^2\setminus\{z^*\})\sqcup\Sigma$ admitting on $\Sigma$ either a fixed point (if the eigenvalues are both positive), or a periodic point with period two (if the eigenvalues are both negative). We paste two copies of ${\mathbb{R}}^2_{\mathrm{ann}}$ on $\Sigma$ to get an open annulus ${\mathbb{R}}^2_{\mathrm{double}}$ and construct by reflection a homeomorphim $g_{\mathrm{double}}$ on ${\mathbb{R}}^2_{\mathrm{double}}$. The set $\omega_{g_{\mathrm{double}}}(z)$ contains the point $z$, with $\mathrm{rot}_{g_{\mathrm{double}}}(z)\not \in{\mathbb{Q}}/{\mathbb{Z}}$, and a periodic point $z'\in\Sigma$, with $\mathrm{rot}_{g_{\mathrm{double}}}(z')=0+{\mathbb{Z}}$ or $\mathrm{rot}_{g_{\mathrm{double}}}(z')=1/2+{\mathbb{Z}}$. By Theorem \[th:rotation-number\], we deduce that $g_{\mathrm{double}}$ has a topological horseshoe, which is also the case for $f$.
Since $\Lambda$ is locally stable for $f$, so is $\Lambda'$ for $f^q$ and one can find a forward invariant neighborhood $U$ of $\Lambda'$ such that $z^{*}$ does not belong to $\overline U$. The connected component of $U$ that contains $z$, denoted $U'$, is forward invariant by a power of $f^q$, because $z$ is recurrent. It is essential in the annulus ${\mathbb{R}}^2\setminus\{z^*\}$, otherwise $\mathrm{rot}_{f^q\vert_{{\mathbb{R}}^2\setminus\{z^*\}}}(z)$ would be rational. This implies that $U'$ is forward invariant by $g$ and that the connected component of ${\mathbb{R}}^2\setminus\overline{U'}$ that contains $z^*$ is backward invariant and relatively compact. This contradicts the fact that $f^q$ decreases the area.
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[^1]: F. A. Tal was partially supported by the Alexander von Humboldt foundation as well as by CAPES, FAPESP and CNPq-Brasil.
[^2]: The reader will notice that the topological horseshoes that will be constructed in this article possess the following additional stability property: if $U$ is neighborhood of $Y$ and $\mathcal U$ a neighborhood of $f$ for the compact open topology, then every map $f'\in\mathcal U$ admits a topological horseshoe $Y'\subset U$ satisfying the same properties as $Y$.
[^3]: In the article, when writing $\mathrm{rot}_{\check f}(z)=\rho$, we implicitly suppose that $z$ belongs to $\mathrm{ne}^+(f)$ and has a rotation number equal to $\rho$.
[^4]: in the whole text, “transverse” will mean “positively transverse”
|
Pages
Wednesday, July 11, 2012
Hi Again!!
Yikes! It's been a LONG time since I've posted!! Dave's on vacation, and he's home and catching up on household projects and trying to get his office/studio organized, plus he's trying to get familiar with his new gear....so, in light of him doing some work on the house, my home has been torn apart!
We decided to lay a new floor in our kitchen. We had green and white marble tile, laid like a checkerboard, and let me tell you, that floor was the most dangerous floor I've ever walked on!! Super impractical for a busy kitchen! If I spilled something on it, I couldn't see it, and invariably someone would walk in and walk right into the spill, slipping and falling! We lived with that green and white floor for 9 years before we moved to the country, and when we came back, we saw that in front of the sink, in a 4x4 foot section, the tiles had been completely shattered! It was just time to lay a new floor. So, Hannah and Russell tore up the floor last month and we lived with a bare, dusty floor for a month or so. Dave put in 16 HOURS NON-STOP, without sleep, working on my floor so that I could have my kitchen back. It's always interesting and character building to try to live and make your home in only two rooms, without a stove! My husband is a true man and an awesome provider. During the time he worked on the floor, we couldn't go in the kitchen at all, especially when he was grouting the tiles. After you grout tiles you have to let them sit for a specific number of hours, usually at least 12 hours. Zoyks! That was tough for all of us! But we made it and my floor is beautiful. It's a gorgeous, natural looking, light terra cotta tile look with light clay-colored grout. I'll post pictures of the before-and-after, as soon as we find them!!
The next Big Event around our house is.....Hannah's birthday! She's going to be old enough that she can start learning how to drive. She really already knows how, from driving tractors since she was small! She is going to be a great driver. I'm excited to see what this next year holds for her.
We absolutely LOVE it when Daddy has vacations! We love having him home all day, every day! However, he may feel differently, as he ends up needing a vacation from his vacation! He'll go back to work to relax!
And, as though the Lord were spilling out bowls of blessing on us, the weather has been absolutely fabulous and fine. Clear skies, gentle breezes, temps in the low 80's....ahhhhhh, summer has come to us at last!
1 comment:
Hi LeanneI would love to see your floor! I know what it's like to live in a half done house ( our renovations have been going on well it seems forever). Lovely to catch up on your news ~ enjoy your summer weather!!BlessingsRenata:)
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My Supplications
Welcome to my blog! I've been married to my Musical Mailman for 21 years and we're still going strong. We have 9 kids here on earth, and in my mansion in heaven, 6 more babies are waiting for me! We're a homeschool family and we've been homeschooling for almost 15 years. I love to sing and have been singing since I was 6 years old. I hope you feel free to grab your favorite cup of coffee and a cozy blanket and browse my blog!! |
Bifacial exemptions are back (briefly)
Invenergy has been awarded a temporary restraining order on the basis that the removal of the bifacial exemption was done without allowance for notice or comment prior to pulling the exclusion. With the decision going before the courts and the Section 201 mid-term review underway, tensions are at an all-time high.
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In the latest development in the dramatic saga of the Section 201 tariffs, the U.S. Court of International Trade has granted Solar Energy Industries Association (SEIA) and developer Invenergy a temporary restraining order (TRO) – which is already effective – against the withdrawal of the 201 bifacial exemption.
For anybody who has not been following along, bifacial modules became exempt from the 25% tariff in June, under the logic that the supply of such modules to the United States was “highly limited.” The exemption, however, proved to be a brief one and was overturned last month, becoming effective as of October 28.
This TRO stems from a complaint filed by Invenergy shortly after the reinstitution of the tariff. The complaint alleged that the U.S. Trade Representative’s action on the exemption was “unlawfully entered” because it didn’t allow for notice or comment prior to pulling the exclusion. Other critics of the move have been harsher, with John Williamson, the founder and CEO of engineering consulting and software company KiloNewton describing the removal as “a nonsense reaction to a nonsense exemption to a nonsense policy.”
The TRO is effective for the 14 days following its issuance, meaning it ends on November 21st, unless the court rules on the matter earlier.
The exemption is a tricky topic within the industry, as it is opposed by two of the biggest module manufacturers Hanwha Q Cells and First Solar. This is because Hanwha has a 1.7 GW solar factory in Georgia, whereas First Solar makes thin film and has always been free from the tariffs, which apply only to crystalline silicon solar. As such, the giants argued that the exemption was a direct undermining of trade protections.
There was hope that the exemption (are you getting tired of that phrase yet?) would expedite the growth of the bifacial market, as it would be economically advantageous for developers to use bifacial modules in projects, regardless of whether or not they were planning on seeing the benefits of the technology.
Even if bifacial modules do not get that expected kickstart in financial feasibility, the technology’s day in the sun (pun absolutely intended) still may not be that far down the line. Wood Mackenzie Senior Research Analyst Xiaojing Sun says that bifacial modules will be cost-competitive again in 2021, even with the Section 201 tariffs. She describes this as a “temporary setback” for the technology, noting that “long-term growth is still expected.”
It should also be noted that the tariffs are set to step down, decreasing 5% annually before disappearing entirely after 2022. That is unless the Trump Administration makes any major changes during the tariffs’ mid-term review. As Standard Solar Chief Development Officer Tony Clifford wrote for pv magazine, the outcome will likely happen in one of three ways:
The most likely possibility: the tariffs stay in place as is.
The worst-case scenario: the implementation of the original tariff percentage, 50%, which would be catastrophic for the solar industry and would cripple a fair bit of companies.
The best-case scenario: the president declares victory in the tariff battle and removes them entirely. Also unlikely, but wouldn’t it be fun?
With the Section 201 mid-term review underway and Invenergy’s ongoing legal fight against the tariffs, expect much more drama to come before a solutions rears itself.
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Tim Sylvia
Tim Sylvia is an associate editor at pv magazine USA. A recent graduate of Hood College, Tim has been with pv magazine since May 2018.
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Purpose, Character of Use, and Authorization for the NIEHS Kids' Pages
The NIEHS Kids' Pages is a government (non-profit) educational website authorized by a 1997 Presidential Memorandum. It is designed to:
teach children about the connections between their health and the environment;
encourage children to pursue careers in health, science, and the environment;
explain the mission of the National Institute of Environmental Health Sciences; and
assist children in learning to read and master challenging mathematics and science.
The NIEHS Kids' Pages include a wide variety of fun activities designed to attract children to this website where they can learn about the impact of the environment on human health, the NIEHS mission, and possible careers in health, medicine, science, mathematics, and the environment.
So why do we include a Sing-Along section?How is that "educationally" relevant?
In support of the White House Memorandum guidelines that were issued, the Sing-Along pages are included because sing-along activities are particularly useful for motivating young children to learn to read and to improve their reading skills. Since starting the site we have received numerous letters from educators and other visitors supporting the value of sing-along materials as an educational tool; many others also indicated that the music site is particularly helpful for children and adults with special needs.
We greatly appreciate the input and ideas we have received from parents, teachers, and caretakers about how our site can help with educational initiatives. More information and tools relating to the educational purpose of music can be found at
The Educational Benefits of Music
, including links to Songs for Teaching. In addition, here are a few excerpts on this subject that were taken from our visitors' email messages:
"Most of my kids are poor readers BECAUSE reading was something 'you must do because I say so' and not because it's fun. Their environment never taught them that reading is fun. That's my job. And thanks to your site, I'm able to do my job better."
"I have been NYS certified in both Special Education and Educating the Emotionally Disturbed for over twenty-five years. I have worked in reformatories for teenage boys and girls (where I am currently employed), public schools, and a NYS prison. Music is a universal language—cross cultural, multi-ethnic, etc. Most of our troubled youth (all income brackets, all ethnic groups, racial groups, etc.) are often difficult to reach because teachers/social workers/caring adults do not have a common language with these youngsters. Music, and music lyrics, are an excellent first step in establishing communications. By using the lyrics on your site, I have been able to capture my students' interests. From there I can lead them to vocabulary improvement, poetic ideas, dealing with feelings—in short, coping with reality, even if it is an ugly reality."
All of the materials presented on this website were collected from other online resources and are used solely for non-commercial educational purposes. No commercial gains will be generated by the contents of this website and materials herein are considered to be available to NIEHS for use solely in a non-profit manner as an educational tool for children. No harmful effect on the market is anticipated and, if anything, it is our understanding that organizations or entities represented by these materials might inadvertently be helped by their inclusion on this site. If you have any concerns with our use of any of these materials, please so that we can rectify the situation. (Also see
Music Assistance
, and
Assistance, Privacy Policies, Accessibility Issues, Copyrights, and Other Disclaimers
. )
White House Press Release of April 18, 1997 (presented below), authorizes the development of this website.
Authorization
White House Press ReleaseMEMORANDUM FOR THE HEADS OF EXECUTIVE DEPARTMENTS AND AGENCIES
My number one priority for the next 4 years is to make sure that all Americans have the best education in the world.
One of the goals of my Call to Action for American Education is to bring the power of the Information Age into all of our schools. This will require connecting every classroom and library to the Internet by the year 2000; making sure that every child has access to modern, multimedia computers; giving teachers the training they need to be as comfortable with the computer as they are with the chalkboard; and increasing the availability of high-quality educational content. When America meets the challenge of making every child technologically literate, children in rural towns, the suburbs, and inner city schools will have the same access to the same universe of knowledge.
I believe that Federal agencies can make a significant contribution to expanding this universe of knowledge. Some agencies have already launched a number of exciting projects in this area. The White House has a special "White House for Kids" home page with information on the history of the White House. NASA's K-12 initiative allows students to interact with astronauts and to share in the excitement of scientific pursuits such as the exploration of Mars and Jupiter and with experiments conducted on the Space Shuttle. The AskERIC service (Education Resources Information Center), supported by the Department of Education, has a virtual library of more than 900 lesson plans for K-12 teachers, and provides answers to questions from educators within 48 hours— using a nationwide network of experts and databases of the latest research. Students participating in the Vice President's GLOBE project (Global Learning and Observation for a Better Environment) collect actual atmospheric, aquatic, and biological data and use the Internet to share, analyze, and discuss the data with scientists and students all over the world. With support from the National Science Foundation, the Department of Energy, and the Department of Defense's CAETI program (Computer-Aided Education and Training Initiative), the Lawrence Berkeley Laboratory has developed a program that allows high school students to request and download their own observations of the universe from professional telescopes.
We can and should do more, however. Over the next 3 months, you should determine what resources you can make available that would enrich the Internet as a tool for teaching and learning, and produce and make available a new or expanded version of your service within 6 months.
You should use the following guidelines to support this initiative:
Consider a broad range of educational resources, including multimedia publications, archives of primary documents, networked scientific instruments such as telescopes and supercomputers, and employees willing to serve as tele-mentors or answer student and teacher questions.
Expand access not only to the information and other resources generated internally, but by the broader community of people and institutions that your agency works with and supports. For example, science agencies should pursue partnerships with professional societies, universities, and researchers to expand K-12 access to scientific resources.
Update and improve your services in response to comments from teachers and students, and encourage educators to submit curricula and lesson plans that they have developed using agency material.
Focus on the identification and development of high-quality educational resources that promote high standards of teaching and learning in core subjects. Of particular importance are resources that will help students read well and independently by 4th grade, and master challenging mathematics, including algebra and geometry, by 8th grade.
Make sure the material you develop is accessible to people with disabilities. Earlier this month, I announced my support for the Web Accessibility Initiative, a public-private partnership that will make it easier for people with disabilities to use the World Wide Web.
I am also directing the Department of Education to develop a "Parents Guide to the Internet," that will explain the educational benefits of this exciting resource, as well as steps that parents can take to minimize the risks associated with the Internet, such as access to material that is inappropriate for children.
The Department of Education will also be responsible for chairing an interagency working group to coordinate this initiative to ensure that the agency-created material is of high quality, is easily accessible, and promotes awareness of Internet-based educational resources among teachers, parents, and students. |
from csc.conceptnet.models import *
from csc.util import foreach
target_frame = Frame.objects.get(language=en, relation__name='UsedFor', text='{1} is for {2}')
def queryset():
frame = Frame.objects.get(text='{1} are {2}', language=en, relation__name='IsA')
got = RawAssertion.objects.filter(language=en, frame=frame)
return got
def fix(s):
if s.surface2.text.startswith('for '):
print s
newsurf = SurfaceForm.get(s.surface2.text[4:], 'en', auto_create=True)
print "=>",
print s.correct_assertion(target_frame, s.surface1, newsurf)
foreach(queryset(), fix)
|
Saturday, November 11, 2006
A 57-year-old man was dead on a hospital toilet for one and a half days before being discovered by staff, it has emerged.
The man had been admitted to Södersjukhuset's casualty department in Stockholm on 26th October suffering from pains in the chest and abdomen. Doctors took samples from him, and decided to keep him in overnight.
But when a nurse came to his bed at 7:30pm to give him his test results, his bed was empty and the patient had disappeared. Staff searched for the man in the immediate vicinity of the ward, but did not find him.
Nurses assumed that the man had left the hospital, so they discharged him. It was only nearly two days later that the man was found in a toilet in the adjacent x-ray department. He appeared to have gone there on his own, locked himself in the toilet and died.
"He had been there for one and a half days," said hospital spokeswoman Ulrica Franzén to The Local.
Franzén said she could not reveal whether doctors had considered the man's condition to be life-threatening, but she did say that he had not been confused.
"If someone is in a confused state then we keep an eye on them, but that was not judged to be the case here."
She said that a more thorough search of the hospital for the man had not been judged necessary.
"We are a very big hospital, and it's not unusual for people to just leave. We didn't look for him because he was not confused."
The hospital has reported the incident to the National Board for Social Welfare and to the police. |
Q:
Carregar Combobox a partir de um List<>
Estou tentando carregar itens em um ComboBox a partir de um List<>, deu certo porém, posteriormente eu iria precisar pegar o código referente ao item selecionado, mas não esta dando certo.
O código para carregar o ComboBox é este:
private void frmCadProduto_Load(object sender, EventArgs e)
{
Model.CadProdutoBD cadProdutoDB = new Model.CadProdutoBD();
List<Control.CadCategoriaProduto> produto = new List<Control.CadCategoriaProduto>();
produto = cadProdutoDB.carregaCategoriaProduto();
foreach (Control.CadCategoriaProduto p in produto)
{
cmbCategoria.Items.Add(p.Categoria);
}
}
Vi que tem como fazer o seguinte:
cmbCategoria.DisplayMember = p.Categoria;
cmbCategoria.ValueMember = p.Categoria_id;
Porém não esta trazendo os itens desta forma.
A:
Use a propriedade DataSource do Combobox.
// ...
List<Control.CadCategoriaProduto> produto = new List<Control.CadCategoriaProduto>();
// ...
cmbCategoria.DataSource = produto;
cmbCategoria.DisplayMember = "Categoria";
cmbCategoria.ValueMember = "Categoria_id";
|
Q:
Fourier Transform of Constant Function
One of the requirements for the existence of Fourier transform of $f(x)$ is that:
$\int_{-\infty}^{\infty} |f(x)| dx $ exists.
However, the table says that the Fourier transform of constant functions (\emph{i.e.}, $f(x)=1$) do exist and it is $\delta(k)$ although $\int_{-\infty}^{\infty} 1 dx = \infty$ .
Could anyone can help me to understand this? Thanks in advance.
A:
The Fourier transform defined by an ordinary (Riemann or) Lebesgue integral only exists when $f \in L^1$.
It is however possible to extend the definition to tempered distributions (for example, every locally integrable function that "doesn't grow too fast" can be identified with a tempered distribution). The Fourier transform of such a thing is not in general a function though, as witnessed by your example.
A:
I think it might be good to think the dirac as the limit of a some know function. For example when we take a zero mean Gaussian density $f(x;0, \sigma)$ with variance $\sigma^2$, we know that when $\lim_{\sigma\rightarrow 0}f(x;0, \sigma)=\delta(x)$. This is one type of the definition of the dirac delta function.
When we take the fourier transform of this function $$\int_{-\infty}^{\infty}f(x)e^{itx}\mbox{d}x=e^{i\mu t}e^{-\frac{1}{2}(\sigma t)^2}$$ and when $\sigma\rightarrow\infty$ we have $$|e^{i\mu t}e^{-\frac{1}{2}(\sigma t)^2}|\rightarrow 1$$
Another way would be to take a rectangular function on $[-t,t]$ at the frequency domain $t$ and let $t\rightarrow\infty$. At the time domain $x$ one would get a sinc function with a single main lobe. The number of zero crossings will increase, the power of side lobes will decrease and eventually go to zero, what remains will be a dirac delta function.
Mathematically, if absolute summability is not satisfied we also get something which is strangely defined; something which has an infinite power at frequency $t=0$, from Perseval's energy preservation rule.
|
Q:
How to add Attribute Support to out-of-box Sales Order Entity?
Out-of-box Acumatica Sales Order (SO301000) does not have attribute support. How to extend attribute support to Sales Order entity in Acumatica?
A:
At the very core, your entity main DAC must have GUID column (NoteID) to reference CSAnswers table and must have field that identify the class of the Entity.
We will make use of Order Type to define list of attributes to gather particular order type-specific information.
Create a Graph Extension for SOOrderTypeMaint Graph and declare data view to define list of attributes for a particular order type. We will be using out-of-box CSAttributeGroupList<TEntityClass, TEntity>
public class SOOrderTypeMaintPXExt : PXGraphExtension<SOOrderTypeMaint>
{
[PXViewName(PX.Objects.CR.Messages.Attributes)]
public CSAttributeGroupList<SOOrderType, SOOrder> Mapping;
}
Create a Graph Extension for SOOrderEntry Graph and declare data view for attributes specific to current order type.
public class SOOrderEntryPXExt : PXGraphExtension<SOOrderEntry>
{
public CRAttributeList<SOOrder> Answers;
}
Create DAC Extension for SOOrder DAC and declare user defined field decorated with CRAttributesField attribute and specify the ClassID field – in our case it is OrderType.
public class SOOrderPXExt : PXCacheExtension<SOOrder>
{
#region UsrAttributes
public abstract class usrAttributes : IBqlField { }
[CRAttributesField(typeof(SOOrder.orderType))]
public virtual string[] UsrAttributes { get; set; }
#endregion
}
Modify Order Types page (SO201000) as below using Customization Engine
<px:PXTabItem Text = "Attributes" >
<Template>
<px:PXGrid runat = "server" BorderWidth="0px" Height="150px" SkinID="Details" Width="100%" ID="AttributesGrid"
MatrixMode="True" DataSourceID="ds">
<AutoSize Enabled = "True" Container="Window" MinHeight="150" />
<Levels>
<px:PXGridLevel DataMember = "Mapping" >
<RowTemplate>
<px:PXSelector runat = "server" DataField="AttributeID" FilterByAllFields="True" AllowEdit="True"
CommitChanges="True" ID="edAttributeID" /></RowTemplate>
<Columns>
<px:PXGridColumn DataField = "AttributeID" Width="81px" AutoCallBack="True" LinkCommand="ShowDetails" />
<px:PXGridColumn DataField = "Description" Width="351px" AllowNull="False" />
<px:PXGridColumn DataField = "SortOrder" TextAlign="Right" Width="81px" />
<px:PXGridColumn DataField = "Required" Type="CheckBox" TextAlign="Center" AllowNull="False" />
<px:PXGridColumn DataField = "CSAttribute__IsInternal" Type="CheckBox" TextAlign="Center" AllowNull="True" />
<px:PXGridColumn DataField = "ControlType" Type="DropDownList" Width="81px" AllowNull="False" />
<px:PXGridColumn DataField = "DefaultValue" RenderEditorText="False" Width="100px" AllowNull="True" />
</Columns>
</px:PXGridLevel>
</Levels>
</px:PXGrid>
</Template>
</px:PXTabItem>
Modify Sales Orders page (SO301000) as below using Customization Engine
<px:PXTabItem Text="Attributes">
<Template>
<px:PXGrid runat="server" ID="PXGridAnswers" Height="200px" SkinID="Inquire"
Width="100%" MatrixMode="True" DataSourceID="ds">
<AutoSize Enabled="True" MinHeight="200" />
<ActionBar>
<Actions>
<Search Enabled="False" />
</Actions>
</ActionBar>
<Mode AllowAddNew="False" AllowDelete="False" AllowColMoving="False" />
<Levels>
<px:PXGridLevel DataMember="Answers">
<Columns>
<px:PXGridColumn TextAlign="Left" DataField="AttributeID" TextField="AttributeID_description"
Width="250px" AllowShowHide="False" />
<px:PXGridColumn Type="CheckBox" TextAlign="Center" DataField="isRequired" Width="80px" />
<px:PXGridColumn DataField="Value" Width="300px" AllowSort="False" AllowShowHide="False" />
</Columns>
</px:PXGridLevel>
</Levels>
</px:PXGrid>
</Template>
</px:PXTabItem>
Download deployment package
|
But back on topic, I chased in Cat5 round our place as I was plissed off with some wifi devices working in the lounge but others wouldn't. Hardwiring is more secure and if you haven't started redecorating your new gaff yet.........?
We use the Billion here in Oz - had two dud units supplied first off, but ours hasn't been rebooted in more than two years and we never have a peep out of it. Looks to be ultimately configurable too (though aside from setting up WPA2 etc, we've not really touched ours) |
Tag Archives: voting
We’re all supposed to vote a week from Tuesday. And that’s great, just really terrific. But what a tease, an election with nothing at stake. I can’t wait until these phony elections are over, so we can start doing some serious politicking two years from now. Hell yeah, I’m talking presidential elections. They’re like the Olympics, or the World Cup, only they last for like a year and a half, and instead of focusing all of our energies toward faraway countries in a spirit of mostly benign sports competition, we get to wage personal warfare against friends, family, and strangers alike.
I’m serious, I’m talking all out war. It starts innocently enough. Sure, at this point in time we only have an idea about some of the men and women thinking about how they potentially might want to start considering setting up an exploratory committee to test the waters regarding the viability of a book tour to measure a theoretical dropping of the hat into the presidential contest. But in the coming months, once this midterm nonsense is out of the way, we’re going to start hearing from all sorts of people who think they have what it takes.
They’re going to start scheduling debates on both sides of the political spectrum, and sure, you’ll see the big names, I’m sure Hillary Clinton and Chris Christie will be standing out front. But there’s also going to be like all of these governors and senators and other random jokers that you’ve never heard of before.
And they’re all going to dig deep and start piling on the front-runners. It’s going to be a classic race to the bottom, with everyone trying to out-America one another, only talking about the most contentious of popular talking points: abortion, guns, taxes, Christmas.
By the time they narrow down the playing field, everyone paying attention is going to be foaming at the mouth, convinced that this is the year that the fate of our nation will be irreparably sealed. And even though every election comes down to these so-called “independent voters,” everybody already has their minds made up. Right now, right this second, even though nobody is officially in the race, I promise you that everyone knows exactly who they’re going to vote for come 2016.
Sure, everything looks calm now, but get ready, because this time next year, you’ll go on the Internet, you’ll log onto Facebook, and everyone you know is going to be putting up recycled headlines and overblown mischaracterizations about the other side. People you haven’t spoken to in years will be popping up on your news feed giving the world their expert two cents on why everything that you believe in shows that you’re an idiot.
I’m sure I’ll be doing it too. Right now I’m acting like I’m above all of this stuff, and sure, maybe I’ll pay some lip service to being respectful and keeping my opinions to myself, but there are always at least a few points during every campaign cycle where everybody gets caught up, a particular controversy or a quote taken totally out of context, and I’ll dive in, guns blazing, family lines forgotten, friendship irreparably destroyed.
And then the election will be over and nothing is going to change at all. Because look at what we’re dealing with today, Ebola, celebrity plastic surgery, none of this stuff has anything to do with politics. But whatever, like I said, it’s easy to talk like I’m above the fray when there’s nothing else going on. Midterm elections are boring. Nobody ever goes out to vote, and you wind up with only the most cranky senior citizens dictating who goes to Congress. I’m done ranting. If you need me, I’ll be outside, washing my car, polishing my bumper to get it ready for all of those inflammatory 2016 bumper stickers, hopefully I’ll get to really piss off some complete stranger behind me paying five bucks a gallon at the gas station.
Don’t tell me what to do. You’re not the boss of me. There’s only one person that’s the boss of me. There’s only one person that I’ll listen to. And that one person is nobody. I won’t listen to anybody. If you ever tell me what to do I’ll just do the exact opposite. Unless of course you’re thinking that can fool me into doing whatever you want by telling me to do the opposite. In this case I’ll recognize your true intent behind the clever semantic trick, and I’ll do what you’re telling me to do, but only because I’ll know that you’re really wanting me to do the opposite and so, yeah, just do me a favor and don’t tell me to do or not to do anything, because I’ll never listen. Like I’ll listen, I’ll hear you, but I’ll willfully do whatever it is that you don’t want me to do, regardless of how you phrase it. What I’m getting at here is, you’re not in charge of me.
I’m not being defiant. Well, I am being defiant. Or maybe you’re the one being defiant. And don’t even try to be nice to me, because it’s still bossing me around, and I’m not somebody that you can just come up to and say, “Hey Rob, it’s so nice to see you. Have a seat and let me get you a drink.” All right? Because, one, don’t tell me to sit down. If I want to sit down, I’ll sit down without you having to tell me to have a seat. And two, let you get me a snack? How about let me let you watch me get myself a snack. Because who says you’re in charge? What are you the mayor?
I went to vote for President last month, and when I got the ballot, it told me to “fill in the box completely.” Stupid piece of paper. If I’m not going to listen to a person, if I’m going to ignore that joker who tried to get me to stand in line, to sign my name at the bottom of that form, to stop asking other people who they were going to vote for, to please stand behind the curtain, to give the other people their privacy while casting their votes, what makes you think I’m going to listen to a dumb piece of paper? I don’t get bossed around by people, I’m definitely not getting bossed around by a piece of paper. You know who wrote that piece of paper? Some clown, trying to tell everybody what to do. You know where that piece of paper came from? A tree. Who do you think I am, standing on line all day so I can get told what to do by a tree? By a dead tree?
Come on, I bought a can of Coke and I was just about to take a big sip when I noticed on top of the can, on top of the logo it said, “Enjoy” Coca-Cola. Get the fuck out of here. Why don’t you enjoy Coca-Cola? I’m the one who paid a dollar for that can. And now I’m getting forced into reading some sort of a simplified instruction manual? Enjoy? I made a grimace, a really strong face and I choked that Coke down, purposefully making myself laugh halfway through that big gulp, laughing so hard that the Coke, all of those bubbles, they got caught up in my nose and started spraying everywhere, and it hurt, it was all up my sinuses, and there was Coke all over my hands, and it dried and just got really sticky. Nosiree, I most certainly did not enjoy that can of Coke. And I went online afterwards to write a strongly worded email to Coke, to tell them to just sell me a can of soda without all of the fascism, the bossing around, but I got distracted by a feature on the web site that showed pictures of Coke cans throughout the ages, and this one can from a long time ago, it didn’t even say “Enjoy,” it just said, “Drink.” Fuck that shit.
Do you know how much of a fit I used to throw in kindergarten when my teacher put on the hokey pokey? Put this in, put that in, do this, do that. I can’t bring anything on an airplane, because I’m not about to let some flight attendant tell me to put it away during takeoff. When I’m driving I’ll stop at every green light and go at every red one. What else … do you know how many trains I’ve missed, walking up to the car when that conductor goes, “All aboard!” Chill out dude, and don’t tell me what to do. So I’ll just turn around. You get all aboard. Bossy control freak jerk.
I just donated a pint of blood and smoked a whole pack of cigarettes right after. Because try and guess what that nurse told me to avoid for about an hour or so after the donation. Just guess. Yep. Smoking. Just, seriously, don’t boss me around. Just stop telling me what to do. Just leave me alone and don’t talk to me and don’t tell me to do or not do anything.
I just rocked the vote. It’s always such a surreal experience. You put so much weight on one little action, and it only happens once in a while, so that when you finally do it, it just comes and goes so fast, leaving you feeling almost a little hollow afterward. That’s kind of dramatic. But wouldn’t you agree that there’s a ton of build up for a two minute procedure that, when you step back from it, never really feels as grand as you thought it would be?
I imagined myself heroically. First of all, I got up much earlier than normal. I guess it really depends on what your definition of early is, but for a guy who normally wakes up closer to ten everyday, I thought seven-thirty was pretty impressive. Yeah, I had set the alarm to seven, and another one for seven-fifteen, but whatever, seven-thirty was, wow, I’m still pinching myself to make sure this isn’t one of those snooze-button induced dreams where you think you’re getting up, going into the shower, making breakfast, going to vote, and then all of the sudden you roll over and it’s nine and you’re still in bed. Does that happen to anybody else? I’m just going to go ahead and assume that, of course it does.
So I get to the polling place and there are lines of people waiting to direct everybody else. I’m pretty sure the poll workers outnumbered the voters by a margin of two to one. And it goes without saying, because everybody always talks about this, but all of the volunteers are senior citizens, and you can just tell how pumped they are to be running the show.
“This is it,” I could imagine themselves getting all motivated for the big day, “Once every four years, we’re back in charge! No computers! No cell phones! Just mountains of paperwork and lots of people to be corralled into long lines! Let’s go out there and show these whippersnappers how to get some work done!”
The system is, whatever, it seems stupid to complain about a process that wasn’t really that bad or that long, one that only happens once in a while anyway. But still, just like the last time I voted, I walked through the door and there were more than a few different lines of people. They asked me for my address and the first letter of my last name. And then they told me to wait in line B. There was only one person in line B, which was great for me. But I looked around, and line D had people snaking outside of the polling place. It didn’t make any sense. But I wasn’t complaining, because I got out very quick.
This was the first time that I voted that there weren’t any giant voting machines. It used to be so cool, you’d switch in all of the levers for the candidates you wanted to vote for, and then you had to pull a giant mechanical arm, so the machine could tally all of your votes. You could feel the whole thing rocking from the inside, this big metal booth, bigger than a soda machine.
This time it was just a scantron. You filled out the bubbles, and then slid it into a little voting scanner machine. I missed that visceral sensation of having voted, having made voting this physical exercise. With the old machines, you really needed to pull, making it feel like you had actually accomplished something. After I slid in my paper, I actually said, “That’s it? I’m done?” and had to be pointed towards the exit, “Yeah, that’s it. Now move it, line D is getting restless.”
A couple of things. Not that it’s anybody’s business, but not like it’s a state secret either, I voted down party lines, not because I’m a loyal partisan, but just because I didn’t really feel like it would be worth it to vote for any third party candidates. Best case scenario, I’d be helping the opposition. Another minor point, there were two judge elections where the Democrats were running unopposed. I filled in the bubble for the first one, but then I immediately wished that I hadn’t. If this guy is going to win anyway, why give him my vote? It’s just a huge joke, really. Why even have an election for that position? I could imagine this guy running for reelection years from now claiming, “The people love me. Look how many votes I got!” when it was really just a matter of default luck.
But voting is a good thing. I always like to vote. I always like feeling like I’m marginally a part of the political process, of America. I wish it were more based on the popular vote, because I hate to think that my presidential vote “doesn’t count.” Because yeah, it was much more satisfying to vote for Senator, to vote for those incompetent clowns that run the show in Albany.
Random: For the 2004 election I was still a college student. I had this one class called American Pluralism, all about America and stuff. Anyway, right before the election, the professor held an informal vote, everybody wrote down who they were voting for and passed it to the prof, who tallied it up, wasting a solid ten minutes of class time. He announced the results, “OK, so we have X for Bush, X for Kerry, and … one vote for MacGyver.” And he said it totally straight-faced, as if he had no idea that he had just been punked. The whole class burst out into laughter and, I have to tell you, it was such a satisfying laugh, like I felt like my insides were being massaged and worked out.
Before I wrap it up here, I do have one suggestion. There should be a voting machine, but it’s like a wall, and there will be spots in the wall painted red or blue or whatever color your party is. And you have to punch through the wall (it’s going to be plywood, nothing too strong,) to retrieve your ballot for that party. And that way, you could really feel like you voted, even more than the old voting machines, like your hand will be really sore for the rest of the day and maybe even a little cut up. “A small price to pay,” you could tell everyone. And that could just be maybe an optional method, you know, only if you wanted to. So, just for the future, that’s something that the board of elections should consider. |
True molecular solutions of natural cellulose in the binary ionic liquid-containing solvent mixtures.
Evidence is presented for the first time of true molecular dissolution of cellulose in binary mixtures of common polar organic solvents with ionic liquid. Cryogenic transmission electron microscopy, small-angle neutron-, X-ray- and static light scattering were used to investigate the structure of cellulose solutions in mixture of dimethyl formamide and 1-ethyl-3-methylimidazolium acetate. Structural information on the dissolved chains (average molecular weight ∼ 5 × 10(4)g/mol; gyration radius ∼ 36 nm, persistence length ∼ 4.5 nm), indicate the absence of significant aggregation of the dissolved chains and the calculated value of the second virial coefficient ∼ 2.45 × 10(-2)mol ml/g(2) indicates that this solvent system is a good solvent for cellulose. More facile dissolution of cellulose could be achieved in solvent mixtures that exhibit the highest electrical conductivity. Highly concentrated cellulose solution in pure ionic liquid (27 wt.%) prepared according to novel method, utilizing the rapid evaporation of a volatile co-solvent in binary solvent mixtures at superheated conditions, shows insignificant cellulose molecular aggregation. |
Role of DNA methylation and methyl-DNA binding proteins in the repression of 5-lipoxygenase promoter activity.
Human 5-lipoxygenase (5-LO) is the key enzyme in the formation of inflammatory leukotrienes. 5-LO gene expression is mainly restricted to B cells and cells of myeloid origin. It is known that basal 5-lipoxygenase promoter activity is regulated by DNA methylation. In this study we investigated the impact of the DNA methylation status of the 5-LO promoter on its activity and the role of methyl DNA binding proteins (MBDs) in transcriptional silencing of the 5-LO promoter. Using ChIP assays, we found that the methyl-DNA binding proteins MBD1, MBD2 and MeCP2 bind to the methylated 5-LO core promoter in U937 cells. Knock down of each of the MBDs upregulates 5-LO mRNA expression in U937 cells indicating that these proteins are involved in silencing of the 5-LO gene. In reporter gene assays with in vitro methylated 5-LO promoter constructs, the extent of 5-LO promoter methylation inversely correlated with its activity. Furthermore, we found that MBD1 overexpression repressed 5-LO promoter activity when the CpG sites at the Sp1 binding site close to the transcriptional start site (GC4) were methylated. Gel shift data indicate that recruitment of Sp1 to this binding site is prevented by methylation. |
Hulu is said to be working on a live-streaming service that it would use to provide broadcast and cable TV channels like ESPN and the Fox network.
The Wall Street Journal reports that the plan is for the streaming company to position it as a “competitor to traditional pay-TV providers.” Hulu is conveniently part-owned by Walt Disney and 21st Century Fox, and the Journal claims that it’s on the brink of singing deals to provide access to live content from the pair as early as next year. Indeed, the newspaper reckons that Disney’s ABC, ESPN, Disney Channel, Fox News and FX would all be among the channels available via the service.
Interestingly, NBCUniversal also owns a part of Hulu, but it hasn’t agreed to signing over access to any content yet. But that might not matter too much: The newspaper’s sources tell it that Hulu “isn’t looking to offer all the hundreds of channels found in the traditional cable bundle.”
It’s thought that Hulu would provide an online DVR system as part of the service, as well as providing on-demand access to old episodes of some shows. This all blurs the boundaries between streaming and cable provision, and you could accordingly expect the service to cost more than current offerings from the likes of Netflix—an “executive close to Hulu” told the Journal that around $40 per month “was in the ballpark.”
It’s not clear of course if the as-yet unnamed services will definitely make it into the wild, let alone by the start of next year. We’ll have to wait and see.
[WSJ] |
Q:
How to know if an instance has been initialized?
How can we do:
instance = Object.allocate ## Doesn't call initialize(). Object is just a sample class.
...
# Somewhere instance would get itself initialized. I do magic.
...
instance.initialized? ## There's no method like that.
?
Of course I ask this not wanting to use flags anywhere. I want to know a general solution for default behaviours like it would work even for a simple instance of Object.
A:
So I actually ended up with this solution. I'm still open for suggestions.
class ::Object
def object(name, klass = Object, &block)
instance = klass.allocate
if instance.method(:initialize).owner == BasicObject
instance.instance_exec(&block) if block_given?
else
raise "No block given." unless block_given?
instance.instance_exec do
def initialize(*args)
@__initialized = true
super
end
end
instance.instance_exec(&block)
instance.instance_exec do
raise "Instance of #{klass} needs to be initialized." unless @__initialized
remove_instance_variable :@__initialized
end
end
(self.is_a?(Module) ? self : self.class).const_set(name, instance)
end
end
I could use it like
object :Hey, SomeClassWithInitialize do
puts "Perhaps sometimes I'd be doing some things first before initializing."
initialize
puts "And other things after initializing."
def hey!
puts "Hey!"
end
end
Hey.hey!
Update: Turns out that there really is no function in Ruby that declares a flag that initialize is called. All calls from rb_obj_call_init to rb_obj_dummy does nothing significant to this.
̶I̶ ̶m̶a̶y̶ ̶j̶u̶s̶t̶ ̶c̶o̶n̶s̶i̶d̶e̶r̶ ̶u̶s̶i̶n̶g̶ ̶s̶i̶m̶p̶l̶e̶ ̶s̶i̶n̶g̶l̶e̶t̶o̶n̶s̶ ̶f̶o̶r̶ ̶i̶n̶s̶t̶a̶n̶c̶e̶s̶ ̶t̶h̶a̶t̶ ̶d̶o̶e̶s̶n̶'̶t̶ ̶i̶n̶h̶e̶r̶i̶t̶ ̶a̶n̶y̶t̶h̶i̶n̶g̶,̶ ̶a̶n̶d̶ ̶̶S̶i̶n̶g̶l̶e̶t̶o̶n̶#̶i̶n̶s̶t̶a̶n̶c̶e̶̶ ̶f̶o̶r̶ ̶a̶n̶y̶t̶h̶i̶n̶g̶ ̶t̶h̶a̶t̶ ̶d̶o̶e̶s̶ ̶o̶v̶e̶r̶ ̶t̶h̶i̶s̶ ̶m̶e̶t̶h̶o̶d̶ ̶b̶u̶t̶ ̶i̶t̶ ̶w̶o̶u̶l̶d̶ ̶o̶n̶l̶y̶ ̶b̶e̶ ̶c̶o̶n̶c̶l̶u̶d̶e̶d̶ ̶w̶h̶e̶n̶ ̶I̶ ̶a̶c̶t̶u̶a̶l̶l̶y̶ ̶u̶s̶e̶ ̶o̶n̶e̶.̶ Classes declared with Singleton does not inherit. I forgot.
Using an alias and an extra method was actually not needed.
|
Cytotoxic T cells to an epitope in the islet autoantigen IA-2 are not disease-specific.
Cytotoxic CD8 T lymphocytes (CTL) are effectors of pancreatic islet beta-cell destruction in type 1 diabetes but, with the exception of a single report, CTL to islet antigen peptides have not been identified. We used autologous blood monocyte-derived dendritic cells to elicit HLA-A2-restricted CTL to a peptide, MVWESGCTV (aa 797-805), that is contiguous with a dominant CD4 T-cell epitope in the islet antigen tyrosine phosphatase IA-2. IA-2 peptide-specific CTL activity measured as 51Cr release from autologous lymphoblasts was detected in 2/6 islet antibody-positive relatives at high risk for type 1 diabetes but also in 2/6 closely HLA-matched controls. All subjects had CTL activity to an HLA-A2-restricted Epstein-Barr virus peptide. CTL to the IA-2 self-peptide were therefore not disease-specific, consistent with other evidence that autoreactive T cells are present in healthy individuals. |
Interesting Facts about the Golem
Golem is a large Pokemon and the final evolution of Geodude. It looks unlike its previous evolutions, as it has a large rock formation that makes up its entire body except for its limbs and head. Each hand has three fingers, all of which possess claws, and each foot has four toes, which again all have the same kind of claws. It has the face of a turtle, with two fangs protruding from its lower jaw. They are capable of retracting their arms, legs and head so that they can roll at high speeds inside their hollow shell. The Alolan variant of this Pokemon has deposits of iron sand on its face that give the impression of hair, eyebrows and a beard. The Alola Golem also has two large, black rocks that protrude from its back. There is a small rock between them that it can fire at foes. But if it cannot for any reason, they can also fire any Geodude that is around.
Did you know?
?
This Pokemon is sometimes called The Megaton Pokemon.
They shed their exterior shell every year so that they can grow bigger, much like snakes do. The shed shell hardens and becomes part of the Earth almost immediately.
Because of how hard its outer body is, Golem can withstand dynamite blasts without so much as a scratch.
Golem weigh over 660 pounds!
They are capable of blowing themselves up, using this explosive reaction to vault from place to place.
Additional Facts about Golem : Like other evolutions of this Pokemon, Golem live high up in the mountains. In the event of an earthquake, they will roll down these mountains to reside in the foothills where it is safer. Because this can cause damage to the homes of people living there, these communities have dug depressions into the ground that help to guide the Pokemon’s course when it rolls down from the mountains. They have also been seen living inside volcanic craters. |
---
author:
- John McDonald
title: 'RH Sneutrino Condensate CDM and the Baryon-to-Dark Matter Ratio'
---
Introduction {#intro}
============
A striking feature of the observed Universe is the similar mass density in baryons and cold dark matter. (The ’Baryon-to-Dark Matter’ (BDM) ratio.) From the WMAP three-year results for the $\Lambda$CDM model, $\Omega_{DM}/\Omega_{B} = 5.65 \pm 0.58$ [@wmap]. However, in most models the physics of baryogenesis and dark matter production are unrelated. So why is the mass density in baryons within an order of magnitude of that of dark matter? There are three possibilities: (i) A remarkable coincidence. (ii) Some anthropic selection mechanism, usually assumed but undefined (e.g. in the case of thermal relic neutralino dark matter). (iii) The mechanisms for the origin of the baryon asymmetry and dark matter are related.
The latter possibility seems the simplest interpretation of the BDM ratio. Indeed, we may be ignoring a [*big clue*]{} to the nature of the correct BSM particle theory. It is highly non-trivial for a particle physics theory to have within its structure (without contrivance) a mechanism that can account for the BDM ratio. Therefore if the BDM ratio is due to such a mechanism it would provide us with a powerful principle by which to select the best canadiate particle physics models.
BDM models broadly divide into two classes: The dark matter particle and baryon number are related by a conserved charge, $Q_{B} + Q_{cdm} = 0 \Rightarrow n_{cdm} \sim n_{B}$. The CDM particle mass satisfies $ m_{cdm} = m_{n}n_{B}/n_{cdm}$ and so $m_{cdm} \sim 1 GeV$ is necessary. However, this does fit well with SUSY if the LSP mass is $O(m_{W})$ or larger. In this case the dark matter and baryon densities are related by [*similar*]{} physical mechanisms for their origin. This implies a less rigid relation between $n_{B}$ and $n_{CDM}$, which may allow us to understand why it is the mass rather than number densities that are observed to be similar.
SUSY BDM Models {#sec:2}
===============
It is perhaps significant that many of the BDM models studied in the past are based on SUSY and the Affleck-Dine (AD) baryogenesis mechanism [@susybdm]. Moreover, the SUSY BDM models tend to be far less baroque than the non-SUSY schemes [@nsbdm]. So SUSY already appears favoured by the ’BDM principle’. However, the previously considered schemes have all been of the charge conservation type, which are not compatible with SUSY breaking schemes which have $O(m_{W})$ soft SUSY-breaking masses. So is there a compelling SUSY dynamical BDM model? In [@dcdm] we suggested that there is:
****
d=4 $\left(H_{u}L\right)^2$ Affleck-Dine leptogenesis
+
RH Sneutrino Condensate CDM with d=4 superpotential
The key ingredient is that the potentials of the AD scalar and the RH sneutrino condensate are both lifted by non-renormalizable terms of the same mass dimension, in this case $d = 4$. In this case the number densities in baryons and dark matter are related, but not by the tight condition of charge conservation [@dcdm]. This allows the dynamics to account for $n_{B} \sim 100 n_{cdm}$ and so why the mass densities are related.
The RH sneutrinos are introduced via the superpotential $$W = W_{MSSM} + W_{\nu} \;\;,\;\;
W_{\nu} = \lambda_{\nu} N H_{u}L + \frac{M_{N}}{2} N^2 ~.$$ If $M_{N} < M_{W}$ then the RH sneutrino can be the LSP[^1]. We will consider the simplest case where $M_{N} = 0$ and the neutrinos are pure Dirac, with $\lambda_{\nu} < 10^{-13}$. The small coupling ensures that the RH sneutrinos are out of thermal equilibirium and so a condensate formed in the early Universe will exist today. The dynamics of the scalar fields follows the now-standard flat direction scenario (Figure \[fig2\]). With a Planck-suppressed non-renormalizable term of the form ($M = M_{Pl}/\sqrt{8\pi}$) $$W = \frac{\lambda_{N} N^4}{4! M} ~,$$ the potential is (neglecting unimportant A-terms) $$V(N) \approx \left( m_{N}^{2} - c_{N}H^{2} \right) |N|^{2} + \frac{\lambda_{N}^{2} |N|^{2\left(n-1\right)}}{M^{2\left(n-3\right)}} ~,$$ where $m_{N} \approx 100 GeV$ is from SUSY breaking. This non-renormalizable term is of the lowest possible dimension and so perhaps the most natural one to consider. The RH sneutrino will begin to oscillate once $H^{2} = H^2_{osc} \approx m_{N}^{2}/c_{N}$, with initial amplitude $$|N|_{osc} \approx |N|_{min} [H \approx H_{osc \;N}] = \left(\frac{12}{\lambda_{N}^{2}}\right)^{1/4} \left(m_{N}M\right)^{1/2}
~.$$ The resulting energy density today is then $$\rho_{N\;o}
= \frac{\sqrt{12} \pi^2}{45} \frac{c_{N} T_{\gamma}^{3}
T_{R} m_{N} }{\lambda_{N} M} ~,$$ and the observed dark matter density is obtained if the reheating temperature satisfies $$T_{R} \approx 2.6 \times 10^{7} \; \frac{\lambda_{N}}{c_{N}}
\left(\frac{h}{0.7}\right)^{2}
\left(\frac{\Omega_{N}}{0.23}\right)
\left(\frac{100 GeV}{m_{N}}\right) GeV ~.$$ This is consistent with the most recent thermal gravitino upper bound [@grav], $T_{R} < 10^{6-8} GeV$, when $\lambda_{N}/c_{N} \; ^{<}_{\sim} \; 0.1-1$.
The Affleck-Dine mechanism is arguably the simplest and most natural way to generate the matter-antimatter asymmetry in the context of the MSSM. The lowest dimension (B-L)-violating operator capable of doing this is $\left(H_{u}L\right)^{2}$. The corresponding $d=4$ flat direction potential is $$V(\Phi) = \left( m_{\Phi}^{2} - c_{\Phi}H^{2}\right)
\left|\Phi\right|^{2}$$ $$\;\;\;\;\;\;\;\; + \left( A_{\Phi}\frac{ \lambda_{\Phi}}{4!M} \Phi^{4}
+ h.c. \right) + \frac{ \left|\lambda_{\Phi}
\right|^{2}}{3!^{2} M^{2}} \left|\Phi\right|^{6} ~.$$ At $cH^2 \sim m_{\Phi}^2$, the CP- and L-violating A-term kicks the real and imaginary oscillations out of phase, leaving the $\Phi$ field in a ellipitical orbit in the complex $\Phi$ plane. This corresponds to a L-asymmetry in the condensate, which is conserved once the $\Phi$ field and A-term diminishes via expansion. Once the scalars in the condensate decay perturbatively, spahleron processes will convert the L-asymmetry into an equivalent B-asymmetry, $n_{B} = -(8/23) n_{L\;initial}$. The resulting B-asymmetry is then [@dcdm] $$n_{B} \approx \frac{f_{A}}{4} \frac{\rho_{\Phi\;o}}{m_{\Phi}}\;\;;\;\; f_{A} = \frac{16}{23} \sin\left(2 \theta\right)
\sin\left(\delta\right) ~,$$ where $\theta$ and $\delta$ are CP-violating angles. The key point is that $$\rho_{\Phi \; o} = m_{\Phi}^{2} \phi(t_{o})^{2}/2$$ is the density the $d=4$ $\Phi$ condensate would have in the Universe at present if the $\Phi$ condensate did not decay. [*Therefore there is a direct connection between the B asymmetry generated along a $d=4$ flat direction and the dark matter density due to condensate lifted by a $d=4$ non-renormalizable term.*]{} This connection is purely due to the similar dynamics of the scalar field-based mechanisms for the origins of the baryon asymmetry and of dark matter. As a result of its dynamical nature, the relationship between the baryon and dark matter number densities is looser than the strict relation found in charge conservation-based BDM models. The model has pluses and minuses. On the plus side, for reasonable values of the parameters $\lambda_{i},c_{i}$ we can account for the observed BDM ratio; a minus is that the parameter dependence of the dynamics makes precise prediction impossible without a theory of $\lambda_{i},c_{i}$. However, these should eventually be calculable within a complete theory of Planck-scale physics.
The resulting BDM ratio is given by $$\frac{\Omega_{B}}{\Omega_{DM}} \approx
\frac{f_{A}}{4} \frac{m_{n}}{m_{\Phi}}
\frac{\rho_{\Phi \; o}}{\rho_{N \; o}}
= \frac{f_{A}}{400}
\left(\frac{100 GeV}{m_{N}}\right)
\left[ \frac{c_{\Phi}}{c_{N}} \frac{\lambda_{N}}{\lambda_{\Phi} } \right] ~.$$ The square-bracketed terms reflect the dynamical nature of the BDM ratio in this model. If $c_{\Phi}/c_{N}$ and $\lambda_{N}/\lambda_{\Phi}$ are both $\sim 1$ then we would get the same kind of relation as in the charge-conservation type models, $n_{B}/n_{DM} \sim 1$. However, thanks to the freedom offered by the dynamics, the observed BDM ratio can be understood if there is a small hierarchy between $\lambda_{\Phi}$ and $\lambda_{N}$ e.g. $$\lambda_{\Phi} \sim 0.01 \lambda_{N}\;,\;\; f_{A} \sim 0.5\;,\;\;
c_{\Phi} \sim c_{N} \;,\;\; m_{N} \sim 100 GeV$$ $$\Rightarrow \frac{\Omega_{B}}{\Omega_{DM}}\approx \frac{1}{6} ~.$$ It is worth emphasizing what has been gained here: $\bullet$ The ’big mystery’ of why completely unrelated mechanisms of baryogenesis and dark matter generation produce mass densities within an order of magnitude has been reduced to a simple hierarchy of couplings. Such a hierarchy of non-renormalizable Yukawas is entirely plausible, given the range of values observed for the Standard Model Yukawa couplings. $\bullet$ No new physics has been added to the MSSM beyond the RH neutrinos required to account for neutrino masses. The Planck-suppressed non-renormalizable terms are of the lowest order and therefore what we would add to the model on general terms anyway.
It is quite remarkable that the MSSM has within its structure such a simple and economical mechanism for generating and relating the dark matter and baryon densities. As such, the MSSM with neutrino masses appears to be strongly favoured by the ’BDM principle’. Note also the power of the BDM ratio as a selection rule: it not only identifies the particle physics model, but also the mechanisms of baryogenesis and dark matter generation and the identity of the dark matter particle itself.
Observable Features of the Model
================================
The model has a number of potentially observable features: These arise either because the $N$ and $\Phi$ are effectively massless during inflation (as in D-term inflation models) or because the phase field is effectively massless (as in F-term inflation models with suppressed $H$ coreections to the A-terms)[@dcdm]. Therefore the observability of isocurvature perturbations depends on the nature of SUSY inflation. The CDM isocurvature (CDI) contribution to the CMB power spectrum is $$C_{l} = (1 - \alpha) C_{l}^{ad} + \alpha C_{l}^{iso}$$ $$\alpha_{CDI} \approx
\frac{H_{I}^{2}}{\pi^{2} P_{\cal{R}} N_{I}^{2}}$$ where $N_{I}$, $H_{I}$ are the values duing inflation. Using this, and comparing with the present upper bound [@bean], $\alpha < 0.26$, we obtain an upper bound on $H_{I}$ $$H_{I} \;^{<}_{\sim} \; \left(\frac{48}{5}\right)^{1/2} \frac{\pi^{2}
P_{{\cal R}} M \alpha_{lim}}{\lambda_{N}}$$ $$\equiv 4.4 \times 10^{11}
\left(\frac{0.1}{\lambda_{N}}\right)
\left(\frac{\alpha_{lim}}{0.26}\right)
\left(\frac{N_{I}}{N_{*}}\right)^{2} GeV ~.$$ Here $N_{*}$ is the upper limit on $N_{I}$ where the non-renormalizable potential gives $N$ an effective mass of order $H_{I}^2$. (Figure \[fig1\].) The upper bound on $H_{I}$ is close to typical values in SUSY inflation models e.g. D-term hybrid inflation has $H_{I} \approx 1.1 \times 10^{13} g GeV$.
In addition to CDM isocurvature perturbations of the RH sneutrino density, there can also be baryon isocurvature (BI) perturbations due to quantum fluctuations of $\Phi$. The ratio of baryon to CDM isocurvature perturbation is $$\frac{\alpha_{BI}}{\alpha_{CDI}}
= \frac{f_{\theta}^{2} N_{I}^{2}}{4 \phi_{I}^{2}}
\left(\frac{\Omega_{B}}{\Omega_{CDM}} \right)^{2}$$ $$\approx
8 \times 10^{-3} f_{\theta}^{2} \left( \frac{N_{I}}{\phi_{I}} \right)^{2} \;\;;\;\;\;\; f_{\theta} = \frac{2}{\tan 2\theta} \sim 1 ~.$$ The CDM and baryon isocurvature perturbations cannot be distinguished by CMB observations, but could eventually be observed via the 21cm background [@21cm]. The ratio then becomes a direct indicator of the nature of SUSY inflation. F-term inflation implies $\Phi_{I}$ and $N_{I}$ are at the minimum of their potentials (Figure \[fig1\]), so the baryon isocurvature contribution is negligible in this case [@dcdm] $$\frac{\alpha_{BI}}{\alpha_{CDI}}
\approx
8 \times 10^{-3} f_{\theta}^{2} \left( \frac{c_{N}}{c_{\phi}} \right)^{1/2} \frac{\lambda_{\phi}}{\lambda_{N}} \sim 10^{-3}- 10^{-4} ~.$$ On the other hand, in D-term inflation $N_{I}$ and $\Phi_{I}$ are undetermined, so a large or even dominant baryon isocurvature perturbation is possible in this case. The RH sneutrino can be regarded as a third member of the family of very weakly interacting SUSY dark matter candidates, joining the gravitino [@gravitino] and axino [@axino]. In common with these better-known candidates, MSSM collider pheomenology will be quite different from conventional (thermal-relic) neutralino LSPs, including the possibility of charged NLSPs. To distinguish RH sneutrino LSPs from gravitinos and axinos, the best possibility would be if the stau was the NLSP (MSSM-LSP) and if its decay mode to LSPs could be observed by trapping and observing its lifetime and final states. Final states for stau decay to gravitinos and axinos generally have $\tau^{-}$ leptons, whereas decay to RH sneutrinos typically will have a charged Higgs $h_{u}^{-}$. More detailed study of NLSP decays is called for, to ensure that the lifetime of the stau is short enough to avoid disrupting light element abundances. The NLSP phenomenology of RH sneutrino condensate CDM can also be distinguised from RH sneutrino dark matter from thermal relic NLSP decay [@trsn], since the latter is restricted to the parameter region where $\Omega_{NLSP} > \Omega_{cdm}$, whereas there is no restriction on $\Omega_{NLSP}$ in the case of RH sneutrino condensate dark matter, allowing the MSSM-LSP to exist in the region of MSSM parameter space where $\Omega_{NLSP} \ll \Omega_{cdm}$. Observation of a CDM isocurvature perturbation would also distinguish between condensate and thermal relic RH sneutrinos.
Summary
=======
RH sneutrino condensate CDM combined with Affleck-Dine baryogenesis can plausibly account for the observed similarity of the baryon and dark matter mass densities in the Universe. Seen as a selection principle, the requirement that a particle physics model can [*without contrivance*]{} account for the BDM ratio favours the MSSM with neutrino masses and RH sneutrino condensate CDM. It is quite remarkable that the MSSM with neutrino masses has the ability to account for the BDM ratio as a natural consequence its structure.
CDM and baryon isocurvature perturbations are possible, the ratio of which gives information on the nature of SUSY inflation. Long-lived NLSP phenomenology is also expected, which can be distinguished from gravitino and axino LSP phenomenology via trapped stau final states, and from thermal relic RH sneutrino LSP phenomenology once the parameters of the MSSM are known.
There are a number of issues which remain to be addressed. One is the possibility that $M_{N} \neq 0$. In this case it is possible that the heavier generation condensate RH sneutrinos, which will have a lifetime longer than the age of the Universe so long as $\lambda_{\nu}$ is still small, would decay into the LSP RH sneutrino plus $e^{+}e^{-}$. This could produce a potentially observable diffuse $\gamma$-ray background. In addition, the phenomenology and cosmology of MSSM-LSPs in this model should be studied in detail.
Acknowledgement {#acknowledgement .unnumbered}
===============
This work was supported (in part) by the European Union through the Marie Curie Research and Training Network “UniverseNet” (MRTN-CT-2006-035863) and by STFC (PPARC) Grant PP/D000394/1.
[999]{}
D.N.Spergel et al, [*astro-ph/0603449*]{}.
S.D.Thomas, ; K.Enqvist and J.McDonald, ; D.Hooper, J.March-Russell and S.M.West, ; S.Abel and V.Page, [*JHEP*]{} [**0605**]{} (2006) 024; L.Roszkowski and O.Seto, [*hep-ph/0608013*]{}.
S.M.Barr, R.S.Chivukula and E.Fahri, ; S.M.Barr, ; D.B.Kaplan, ; R.Kitano and I.Low, , [*hep-ph/0503112*]{}; N.Cosme, L.Lopez Honorez and M.H.G.Tytgat, ; G.R.Farrar and G.Zaharijas, .
J. McDonald, JCAP [**0701**]{}, 001 (2007) \[arXiv:hep-ph/0609126\].
T.Asaka, K.Ishiwata and T.Moroi, .
K.Kohri, T.Moroi and A.Yotsuyanagi, .
R.Bean, J.Dunkley and E.Pierpaoli, .
R. Barkana and A. Loeb, Mon. Not. Roy. Astron. Soc. Lett. [**363**]{}, L36 (2005) \[arXiv:astro-ph/0502083\].
W. Buchmuller, K. Hamaguchi, M. Ratz and T. Yanagida, Phys. Lett. B [**588**]{}, 90 (2004) \[arXiv:hep-ph/0402179\].
A. Brandenburg, L. Covi, K. Hamaguchi, L. Roszkowski and F. D. Steffen, Phys. Lett. B [**617**]{}, 99 (2005) \[arXiv:hep-ph/0501287\].
[^1]: RH sneutrino dark matter with $M_{N} = 0$ was first considered in a thermal relic model in [@trsn].
|
# Copyright (c) Open Enclave SDK contributors.
# Licensed under the MIT License.
option(ENABLE_REFMAN "Enable Doxygen reference manual generation" ON)
if (ENABLE_REFMAN)
find_package(Doxygen)
if (NOT DOXYGEN_FOUND)
message(
FATAL_ERROR
"Doxygen not found! Skip generation with '-DENABLE_REFMAN=OFF'")
endif ()
set(DOXYGEN_PROJECT_NAME "Open Enclave")
set(DOXYGEN_USE_MDFILE_AS_MAINPAGE "MainPage.md")
set(DOXYGEN_OUTPUT_DIRECTORY "${OE_DOCDIR}/openenclave")
set(DOXYGEN_RECURSIVE NO) # CMake enables this by default.
# While Doxygen would automatically strip the working directory, we
# have to specify it because we have another directory to strip.
set(DOXYGEN_STRIP_FROM_PATH
"${PROJECT_SOURCE_DIR}/include/openenclave/;${PROJECT_SOURCE_DIR}/docs/refman"
)
# Allow @experimental to be used in headers to mark an API as experimental and add it to a list of
# experimental APIs.
# Also allow @advanced to be used in headers to mark an API as advanced and add it to a list of
# advanced APIs.
set(DOXYGEN_ALIASES
"experimental = \\xrefitem experimental \\\"This feature is marked as experimental\\\" \\\"Experimental List\\\""
"experimental{1} = \\xrefitem experimental \\\"Experimental\\\" \\\"Experimental List\\\" \\1"
"advanced = \\xrefitem advanced \\\"This feature is marked as advanced and is not intended for use by apps\\\" \\\"Advanced List\\\""
)
# NOTE: These were set to their non-default values in the existing
# configuration file, so that configuration has been copied here.
# However, they may or may not be desired.
set(DOXYGEN_JAVADOC_AUTOBRIEF YES)
set(DOXYGEN_QT_AUTOBRIEF YES)
set(DOXYGEN_SEPARATE_MEMBER_PAGES YES)
set(DOXYGEN_OPTIMIZE_OUTPUT_FOR_C YES)
set(DOXYGEN_GENERATE_LATEX NO)
set(DOXYGEN_MACRO_EXPANSION YES)
set(DOXYGEN_EXPAND_ONLY_PREDEF YES)
set(DOXYGEN_PREDEFINED "=OE_ALIGNED(x)=;OE_EXTERNC_BEGIN=")
doxygen_add_docs(
doxygen
# This list of files is explicit because we disable recursion.
enclave.h
host.h
attestation/attester.h
attestation/custom_claims.h
attestation/verifier.h
attestation/sgx/evidence.h
bits/evidence.h
bits/properties.h
bits/report.h
bits/result.h
bits/types.h
bits/exception.h
bits/module.h
../../docs/refman/MainPage.md
WORKING_DIRECTORY ${PROJECT_SOURCE_DIR}/include/openenclave/
COMMENT "Generating refman HTML documentation")
endif ()
|
Q:
Python - split, regex and condition
I have a target artist and would like to fetch its correspondent id, like so:
import re
target = 'Portishead'
videos = ['Portishead - Roads (Vg1jyL3cr60)', 'Portishead - Roads - (WQYsGWh_vpE)', 'Need For Speed (Linkin Park - Roads Untraveled) Music Video (7Lkq7bf6kU8)', 'Lawson - Roads (I-SOaSU0ieA)', 'Vargas & Lagola - Roads (Audio) (Kd3s20GmPVE)']
for item in videos:
artist = item.split('-')[0]
# here I get whats inside parenthesis, not always an id
video_id = re.findall('\(([^)]+)', item)
# and here the id, which is always the last split item
id_ = (video_id[-1])
if artist == target:
print id_
but my if condition is not working for the target artist. I print no results.
what is the best way of achieving this using a for loop or otherwise, considering that the real list is very large?
I want to fetch above "Vg1jyL3cr60"
EDIT: @Alexandre Cécile. I post here the entire function that calls youtube API, if you are interested in perfecting the function that narrows down the search for artist videos, once you pass track title and artist name. You will need a key for that, though.
from google.oauth2 import service_account
def youtube_id(track_name, target_artist):
GET_CREDENTIALS = os.environ.get('GOOGLE_APPLICATION_CREDENTIALS')
PASS_CREDENTIALS =
service_account.Credentials.from_service_account_file(GET_CREDENTIALS)
YOUTUBE_API_SERVICE_NAME = "youtube"
YOUTUBE_API_VERSION = "v3"
DEVELOPER_KEY = "mykey"
youtube = build(YOUTUBE_API_SERVICE_NAME, YOUTUBE_API_VERSION, credentials=PASS_CREDENTIALS,
developerKey=None)
# Call the search.list method to retrieve results matching the specified
# query term.
search_response = youtube.search().list(
q=track_name,
part="id,snippet",
#maxResults=track_name.max_results
).execute()
videos = []
videos_ids = []
channels = []
playlists = []
# Add each result to the appropriate list, and then display the lists of
# matching videos, channels, and playlists.
for search_result in search_response.get("items", []):
if search_result["id"]["kind"] == "youtube#video":
videos.append("%s (%s)" % (search_result["snippet"]["title"],
search_result["id"]["videoId"]))
videos_ids.append("%s" % (search_result["id"]["videoId"]))
elif search_result["id"]["kind"] == "youtube#channel":
channels.append("%s (%s)" % (search_result["snippet"]["title"],
search_result["id"]["channelId"]))
elif search_result["id"]["kind"] == "youtube#playlist":
playlists.append("%s (%s)" % (search_result["snippet"]["title"],
search_result["id"]["playlistId"]))
print ("Videos:\n", "\n".join(videos), "\n")
print ("Channels:\n", "\n".join(channels), "\n")
print ("Playlists:\n", "\n".join(playlists), "\n")
ids=[]
for video in videos:
artist = re.split(r'\s*-\s*', video)[0]
id = re.search(r'.*\(([^)]+)', video)[1]
if id and artist == target_artist:
videos_ids.append(id)
print ('VIDEOS IDS', videos_ids)
return videos_ids[-1]
A:
The issue you're having is mainly due to a space being present at the end of your match (since - splits on - and leaves the space behind it). Tcode below should work for you. It uses re.split to split on \s*-\s* (any number of spaces, followed by -, followed by any number of spaces).
I also cleaned up some of the other parts of your code. I added .* to the start of your second regex to capture only the last instance (and changed [0] to [1] to get the captured contents instead of the entire match).
The last portion checks to see if id exists and if artist == target before printing.
See code in use here
import re
target = 'Portishead'
videos = [
'Portishead - Roads (Vg1jyL3cr60)',
'Portishead - Roads - (WQYsGWh_vpE)',
'Need For Speed (Linkin Park - Roads Untraveled) Music Video (7Lkq7bf6kU8)',
'Lawson - Roads (I-SOaSU0ieA)',
'Vargas & Lagola - Roads (Audio) (Kd3s20GmPVE)'
]
for video in videos:
artist = re.split(r'\s*-\s*', video)[0]
id = re.search(r'.*\(([^)]+)', video)[1]
if id and artist == target:
print(id)
Results in:
Vg1jyL3cr60
WQYsGWh_vpE
Explanation of regex patterns:
\s*-\s* this pattern matches - and any whitespace around them
\s* matches any whitespace character any number of times
- matches this character literally
\s* matches any whitespace character any number of times
.*\(([^)]+) this pattern matches the last instance of the left parenthesis in a string
.* match any character any number of times (this is how we can ensure we match the last parenthesis as it's greedy and will match as many characters as possible)
\( match ( literally
([^)]+) captures the following
[^)]+ matches one or more of any character except )
|
Q:
Improper advertising identifier [IDFA] usage (Limit Ad Tracking)
I'm submitting an ios game update to the app store.
xcode shows the following error:
"Improper advertising identifier [IDFA] usage. Your app contains the Advertising Identifier [IDFA] API but your app is not respecting the Limit Ad Tracking setting in iOS."
I haven't changed anything regearding my ad settings and there were no problems with the last version of my app.
I have been trying to find a workaround for the last 20 hours and I am starting to become insane.
What do I have to change and how? Please help me!
A:
I had a similar error in an update to an app. I wasn't using any ads but do have Facebook integration (which needs the AdSupport framework). I believe, after searching the net, that Facebook uses the advertising ID for its own analysis purposes so, even though I'm not including ads in my app, the validation and upload processes through xCode were failing with the error "Your app contains the Advertising Identifier [IDFA] API..."
I searched and found that I needed to download the Facebook SDK source code, update the FBUtility.m to remove the references to the advertisingID but, in fact, I simply needed to:
1) download the source code for the latest SDK, which I did from here: https://github.com/facebook/facebook-ios-sdk (I downloaded the zip file from github to my documents folder)
2) build the framework - open the terminal. Use cd documents at the command prompt, then use this command: sudo scripts/build_framework.sh, which will run the build_framework.sh script that is in the scripts subfolder within the downloaded Facebook SDK folder
3) Remove the old FacebookSDK.framework from your Xcode project and add the new one (in my case, I navigated to documents/facebook-ios-sdk/build & choose the FacebookSDK.framework folder
4) Archive the project and it should (it was in my case) be good to upload
Hope that helps someone along the way - I've been at this for days!!
A:
I had the same problem and was able to solve it by removing AdMob ads and leaving only iAds.
I know that this isn't a perfect solution but afterwards you should at least be able to upload your app.
|
package away3d.materials.passes;
import away3d.cameras.Camera3D;
import away3d.core.base.IRenderable;
import away3d.core.managers.Stage3DProxy;
import away3d.core.math.Matrix3DUtils;
import away3d.textures.Texture2DBase;
import openfl.display3D.Context3D;
import openfl.display3D.Context3DProgramType;
import openfl.display3D.Context3DTextureFormat;
import openfl.geom.Matrix3D;
import openfl.geom.Vector3D;
import openfl.Vector;
/**
* DistanceMapPass is a pass that writes distance values to a depth map as a 32-bit value exploded over the 4 texture channels.
* This is used to render omnidirectional shadow maps.
*/
class DistanceMapPass extends MaterialPassBase
{
public var alphaThreshold(get, set):Float;
public var alphaMask(get, set):Texture2DBase;
private var _fragmentData:Vector<Float>;
private var _vertexData:Vector<Float>;
private var _alphaThreshold:Float;
private var _alphaMask:Texture2DBase;
/**
* Creates a new DistanceMapPass object.
*/
public function new()
{
super();
_fragmentData = Vector.ofArray([ 1.0, 255.0, 65025.0, 16581375.0,
1.0/255.0, 1.0/255.0, 1.0/255.0, 0.0,
0.0, 0.0, 0.0, 0.0]);
_vertexData = new Vector<Float>(4, true);
_vertexData[3] = 1;
_numUsedVertexConstants = 9;
}
/**
* The minimum alpha value for which pixels should be drawn. This is used for transparency that is either
* invisible or entirely opaque, often used with textures for foliage, etc.
* Recommended values are 0 to disable alpha, or 0.5 to create smooth edges. Default value is 0 (disabled).
*/
private function get_alphaThreshold():Float
{
return _alphaThreshold;
}
private function set_alphaThreshold(value:Float):Float
{
if (value < 0)
value = 0;
else if (value > 1)
value = 1;
if (value == _alphaThreshold)
return value;
if (value == 0 || _alphaThreshold == 0)
invalidateShaderProgram();
_alphaThreshold = value;
_fragmentData[8] = _alphaThreshold;
return value;
}
/**
* A texture providing alpha data to be able to prevent semi-transparent pixels to write to the alpha mask.
* Usually the diffuse texture when alphaThreshold is used.
*/
private function get_alphaMask():Texture2DBase
{
return _alphaMask;
}
private function set_alphaMask(value:Texture2DBase):Texture2DBase
{
_alphaMask = value;
return value;
}
/**
* @inheritDoc
*/
@:allow(away3d) override private function getVertexCode():String
{
var code:String;
code = "m44 op, vt0, vc0 \n" +
"m44 vt1, vt0, vc5 \n" +
"sub v0, vt1, vc9 \n";
if (_alphaThreshold > 0) {
code += "mov v1, va1\n";
_numUsedTextures = 1;
_numUsedStreams = 2;
} else {
_numUsedTextures = 0;
_numUsedStreams = 1;
}
return code;
}
/**
* @inheritDoc
*/
@:allow(away3d) override private function getFragmentCode(animationCode:String):String
{
// TODO: not used
var code:String;
var wrap:String = _repeat? "wrap" : "clamp";
var filter:String;
if (_smooth)
filter = _mipmap? "linear,miplinear" : "linear";
else
filter = _mipmap? "nearest,mipnearest" : "nearest";
// squared distance to view
code = "dp3 ft2.z, v0.xyz, v0.xyz \n" +
"mul ft0, fc0, ft2.z \n" +
"frc ft0, ft0 \n" +
"mul ft1, ft0.yzww, fc1 \n";
if (_alphaThreshold > 0) {
var format:String;
switch (_alphaMask.format) {
case Context3DTextureFormat.COMPRESSED:
format = "dxt1,";
case Context3DTextureFormat.COMPRESSED_ALPHA:
format = "dxt5,";
default:
format = "";
}
code += "tex ft3, v1, fs0 <2d," + filter + "," + format + wrap + ">\n" +
"sub ft3.w, ft3.w, fc2.x\n" +
"kil ft3.w\n";
}
code += "sub oc, ft0, ft1 \n";
return code;
}
/**
* @inheritDoc
*/
@:allow(away3d) override private function render(renderable:IRenderable, stage3DProxy:Stage3DProxy, camera:Camera3D, viewProjection:Matrix3D):Void
{
var context:Context3D = stage3DProxy._context3D;
var pos:Vector3D = camera.scenePosition;
_vertexData[0] = pos.x;
_vertexData[1] = pos.y;
_vertexData[2] = pos.z;
_vertexData[3] = 1;
var sceneTransform:Matrix3D = renderable.getRenderSceneTransform(camera);
context.setProgramConstantsFromMatrix(Context3DProgramType.VERTEX, 5, sceneTransform, true);
context.setProgramConstantsFromVector(Context3DProgramType.VERTEX, 9, _vertexData, 1);
if (_alphaThreshold > 0)
renderable.activateUVBuffer(1, stage3DProxy);
var matrix:Matrix3D = Matrix3DUtils.CALCULATION_MATRIX;
matrix.copyFrom(sceneTransform);
matrix.append(viewProjection);
context.setProgramConstantsFromMatrix(Context3DProgramType.VERTEX, 0, matrix, true);
renderable.activateVertexBuffer(0, stage3DProxy);
stage3DProxy.drawTriangles(renderable.getIndexBuffer(stage3DProxy), 0, renderable.numTriangles);
}
/**
* @inheritDoc
*/
@:allow(away3d) override private function activate(stage3DProxy:Stage3DProxy, camera:Camera3D):Void
{
var context:Context3D = stage3DProxy._context3D;
super.activate(stage3DProxy, camera);
var f:Float = camera.lens.far;
f = 1/(2*f*f);
// sqrt(f*f+f*f) is largest possible distance for any frustum, so we need to divide by it. Rarely a tight fit, but with 32 bits precision, it's enough.
_fragmentData[0] = 1*f;
_fragmentData[1] = 255.0*f;
_fragmentData[2] = 65025.0*f;
_fragmentData[3] = 16581375.0*f;
if (_alphaThreshold > 0) {
context.setTextureAt(0, _alphaMask.getTextureForStage3D(stage3DProxy));
context.setProgramConstantsFromVector(Context3DProgramType.FRAGMENT, 0, _fragmentData, 3);
} else
context.setProgramConstantsFromVector(Context3DProgramType.FRAGMENT, 0, _fragmentData, 2);
}
} |
Q:
Secure ASP.NET MVC controller method from random posts?
I have a controller method for an ajax request that I'm curious as to how I can stop random folks from submitting data to that url. I want to make sure it's coming from that specific form.
[HttpPost]
public JsonResult DemographicsSave(FormCollection collection)
{
return Json("Patient Saved", JsonRequestBehavior.AllowGet);
}
A:
This is a NuGet package created by my friend for just this very purpose and shores up some security "holes" by default. It should save you some time in implementation!
http://nuget.org/packages/ValidationAndAuthenticationByDefault.MVC3
Alternatively you can add to your controller :
[HttpPost]
[ValidateAntiForgeryToken]
Followed by the following in your view:
@Html.AntiForgeryToken()
|
Q:
Unix: Using date in alias doesn't output the current date/time
In my bash_profile, I'm referencing an external alias.sh file, which has the following alias:
alias date="echo `date "+%Y-%m-%d at %H:%M":%S`"
If I issue date in a new terminal session, it constantly outputs the specific date at the time when the alias.sh file was sourced (i.e. when I started the new session) …
How do I make an alias that actually outputs the current date, when executing the aliased command?
A:
Uhh, why not just?
alias date='date "+%Y-%m-%d at %H:%M":%S'
No need to echo it. When you use the backwards tick ( ` ), whatever's in it gets evaluated when alias.sh is sourced.
|
The Tigers felt they needed a second baseman. Talented laden Omar Infante had not lived up to expectations and Jason Smith was not a starting caliber infielder. At the same time, the team dealt with troubled reliever Ugueth Urbina. The pitcher proved a distraction that hurt the team more than he helped. Luckily, the Tigers found a willing trade partner with the Philadelphia Phillies. Detroit shipped Urbina and infielder Ramon Martinez to Philadelphia for Placido Polanco. The deal became one of the great steals in Tiger history. Polanco made All Star teams, won Gold Gloves, set fielding records, and helped Detroit to the pennant while Urbina landed in prison for attempted murder.
The Tigers pinned their second base hopes on Omar Infante. Despite his talent, the youngster did not seem to deal well with the pressure. Infante performed respectably in 2004 with a .264 average, 16 home runs, 55 RBI, and .766 OPS. His production dropped dramatically in 2005 and Detroit gave up. In 121 games, Infante batted .222 with 9 home runs and 43 RBI. The team began to search for a replacement. Eventually, Infante shifted to a utility role before moving to Atlanta.
As the team began to scour rosters for a second baseman, their closer displayed disturbing behavior. Ugueth Urbina signed a two-year deal with Detroit in March 2004. He saved 21 games in 2004 and pitched well in 2005. However, rumors swirled about a midair incident in which Urbina attempted to open the airplane door resulting in some sort of altercation. Shortly after the reports surfaced, Detroit traded Urbina and infielder Ramon Martinez to Philadelphia for Placido Polanco. Polanco blocked Chase Utley's elevation in Philadelphia, so the Phillies willingly parted with him in return for a solid reliever and reserve infielder.
Polanco solidified Detroit's infield and proved a godsend. The second baseman played five seasons with the Tigers. Over that period, Polanco hit .338, .295, .341, .307, and .285. Along the way, Polanco accumulated two Gold Gloves, a Silver Slugger, and appeared in the 2007 All Star Game. Additionally, he won the 2006 ALCS MVP with a .529 average and 1.167 OPS. On top of this, Polanco set the Major League record for 149 errorless games at second, most consecutive chances without an error, and became the first second baseman in history to play an entire season without committing an error.
While Polanco became Detroit's breakout star, Urbina unknowingly played out his career with the 2005 Phillies. The pitcher appeared in 56 games for Philadelphia, compiled a 4-3 record with 1 save and 1.146 WHIP. He made his final big league appearance on October 2, 2005. Off season legal issues soon intervened with the pitcher's career.
Urbina expected to continue with his career in 2006, but a bizarre incident intervened. On November 7, 2005, Venezuelan police arrested the pitcher for attempted murder. Apparently, Urbina charged five farm workers with a machete and tried to dump gasoline upon them. A court convicted the former pitcher of attempted murder and sentenced him to nearly 15 years in prison. Urbina was released on Christmas Eve 2012 after serving nearly six years. Philadelphia's other acquisition, Ramon Martinez, played only 33 games for the Phllies and then moved onto the Dodgers and Mets. As a result, the Phillies managed to clear space for Utley, but got next to nothing in return for a record setting second baseman.
The Detroit Tigers stole Placido Polanco from the Phillies. In fairness, few would have foreseen Polanco's success with Detroit. Even fewer would have predicted Ugueth Urbina's meltdown and incarceration. The Tigers acquired an All Star Gold Glove second baseman that helped win a pennant. The Phillies got 56 games from a relief pitcher and 33 games from a backup infielder.
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Don Keko earned his M.A. in history from Central Michigan University and a teaching certificate from the University of Michigan. He has taught history for the past decade. The lifelong Tiger baseball fan is working on his first book, which is on popular music and blogs on popular culture and history at http://cicero390.blogspot.com/.
He lives near Flint, Michigan. He can be reached by e-mail at dck1971@aol.com, or follow him on Twitter at http://twitter.com/dck1971. |
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1. Field of the Invention
The present invention relates to a disk drive and more particularly to a magnetic disk drive having a head stack assembly with a plastic inner sleeve.
2. Description of the Prior Art and Related Information
A typical hard disk drive includes a head disk assembly ("HDA") and a printed circuit board assembly ("PCBA"). The HDA includes at least one magnetic disk ("disk"), a spindle motor for rotating the disk, and a head stack assembly ("HSA") that includes a head with at least one transducer for reading and writing data. The HSA is controllably positioned by a servo system in order to read or write information from or to particular tracks on the disk. The typical HSA has three primary portions: (1) an actuator assembly that moves in response to the servo control system; (2) a head gimbal assembly ("HGA") that extends from the actuator assembly and biases the head towards the disk; and (3) a flex cable assembly that provides an electrical interconnect with minimal constraint on movement.
The industry presently prefers a "rotary" or "swing-type" actuator assembly which conventionally comprises a body portion that rotates on a pivot bearing cartridge between limited positions, a coil portion that extends from one side of the body portion to interact with one or more permanent magnets to form a voice coil motor, and an actuator arm that extends from an opposite side of the body portion to support the HGA.
A typical HGA includes a load beam, a gimbal attached to an end of the load beam, and a head attached to the gimbal. The load beam has a spring function which provides a "gram load" biasing force and a hinge function which permits the head to follow the surface contour of the spinning disk. The load beam has an actuator end that connects to the actuator am and a gimbal end that connects to the gimbal which carries the head and transmits the gram load biasing force to the head to "load" the head against the disk. A rapidly spinning disk develops a laminar air flow above its surface that lifts the head away from the disk in opposition to the gram load biasing force. The head is said to be "flying" over the disk when in this state.
Early head stack assemblies typically included an all metal "E-block" which included an all metal body portion and metal actuator arms that were formed from a single block of metal, e.g. aluminum. A metal pivot bearing cartridge was installed in the bore of the HSA via a side-pull screw. Such E-blocks provided consistent grounding of the actuator arms and unitary movement of the actuator arms such that relative radial movement between the arms, due to, for example temperature changes, was insignificant.
Later head stack assemblies, often called "hybrid" HSAs, were typically formed by overmolding a plastic body portion over several individual metal actuator arms. Hybrid HSAs were an improvement over the E-block HSAs from an assembly point of view because the plastic body portion allowed metal pivot bearing cartridges to be press-fit into the bore of the plastic body portion. Such press-fitting reduced the cost of assembling the components since screws and the associated threaded openings in the pivot bearing cartridge and the body portion were eliminated. Also, particulates due to inserting the screw into the associated threaded openings were not generated and press-fitting the pivot bearing cartridge into the bore was quicker which reduced assembly time of the components.
In hybrid HSAs, a vertical ground pin was needed to connect the individual actuator arms to one another within the plastic body portion to provide grounding to the actuator arms. Hybrid HSAs, however, were subject to inconsistent grounding because of inadequate contact between the ground pin and the actuator arms after the ground pin was inserted in respective holes in the arms. Moreover, thermal relaxation of the plastic body portion could cause the metal actuator arms to move out of radial alignment relative to one another such as during drive operation when significant temperature changes occurred within the disk drive. |
/*
* BSD-style license; for more info see http://pmd.sourceforge.net/license.html
*/
package net.sourceforge.pmd.lang.plsql.ast;
import net.sourceforge.pmd.annotation.InternalApi;
import net.sourceforge.pmd.lang.dfa.DFAGraphMethod;
public class ASTTypeMethod extends AbstractPLSQLNode implements ExecutableCode, DFAGraphMethod {
@Deprecated
@InternalApi
public ASTTypeMethod(int id) {
super(id);
}
@Deprecated
@InternalApi
public ASTTypeMethod(PLSQLParser p, int id) {
super(p, id);
}
@Override
public Object jjtAccept(PLSQLParserVisitor visitor, Object data) {
return visitor.visit(this, data);
}
/**
* Gets the name of the method.
*
* @return a String representing the name of the method
*/
@Override
public String getMethodName() {
ASTMethodDeclarator md = getFirstChildOfType(ASTMethodDeclarator.class);
if (md != null) {
return md.getImage();
}
return null;
}
@Override
public String getName() {
return getMethodName();
}
}
|
INTRODUCTION
============
Population of adults with cerebral palsy
----------------------------------------
Cerebral palsy (CP) is the most common motor disability in children with an estimated prevalence of 2.6 to 2.9 per 1,000 births in the United States between 2011 to 2013 \[[@b1-arm-2019-43-3-241]\]. According to a study conducted in South Korea, the prevalence of CP was 3.2 per 1,000 children in 2008 \[[@b2-arm-2019-43-3-241]\]. Improvements in the level of medical care have resulted in a steady decline in the incidence and severity of CP in prematurely born children since 1990 \[[@b3-arm-2019-43-3-241]\]. However, the population of adults with CP is increasing along with the increased survival rate of infants born with the disability \[[@b4-arm-2019-43-3-241]\]. A survey of individuals with CP born after 1990 revealed that 60% of the children at Gross Motor Functional Classification System (GMFCS) level V and 96% of the children at GMFCS level I or II survived until 19 years of age \[[@b5-arm-2019-43-3-241]\]. According to data based on the surveillance rates in Sweden up to 50 years of age \[[@b6-arm-2019-43-3-241]\], the risk of death in all age groups was consistently high in adults with CP. In a long-term study of patients with CP between 1959 and 2002, it was reported that 180 of 1,880 (9.6%) individuals died during the follow-up period. Individuals with mild-to-moderate motor impairment had a cumulative survival rate of more than 95% at age 50, compared with 74.5% among those with severe motor impairment. It was estimated that approximately 400,000 adults with cerebral palsy lived in the United States in 1995, which is gradually increasing due to medical advances and prolonged life expectancy, with an estimated number of more than 1,000,000 adults with CP in the US currently \[[@b7-arm-2019-43-3-241]\].
TRANSITION TO ADULTHOOD
-----------------------
CP is superimposed on the dynamic process of development and aging despite being defined as static motor impairment \[[@b8-arm-2019-43-3-241]\]. Adults with CP need health services for the continued monitoring and management of their condition. Furthermore, the development of additional health problems in adulthood increases the need for ongoing access to health services.
Health challenges diagnosed in adults with CP differ from those in childhood and adolescence in patients with CP. Individuals with CP manifest an increased risk of death due to cancer including breast cancer and cancers of digestive and genitourinary organs \[[@b9-arm-2019-43-3-241]\]. In addition to ongoing needs traced to childhood, regular reviews of cardiovascular health, cervical cytology and mammography in women, access to dental care and screening is important in adults diagnosed with CP. Adults with CP have been additional diagnosed with chronic pain and fatigue, osteoarthritis and osteoporosis, and an overall reduction in mobility \[[@b10-arm-2019-43-3-241]\]. According to Young et al. \[[@b11-arm-2019-43-3-241]\], adults with complex physically disabling conditions acquired in childhood including CP exhibit ongoing health issues and require frequent medical care. Their hospitalization rate was 9-fold higher than that of adults without disability. The pediatric and adult healthcare systems are structurally distinct in most countries, and the transition to adult healthcare occurs at 18 or 19 years old \[[@b12-arm-2019-43-3-241]\]. This transition is determined by age alone and not based on the patient's ability to direct their own care. Therefore, at a point when most adults are not ready to inherit further responsibility for their own medical care, they are switched to a more independent adult-oriented healthcare system.
It is also reported that young people diagnosed with CP do not receive sufficient information prior to transitioning underscoring the need to foster knowledge and improve their skills during the transition period \[[@b13-arm-2019-43-3-241]\]. To overcome these apparent challenges, we need to fully understand the healthcare needs of adults with CP in order to guide the development of adult-focused health services \[[@b14-arm-2019-43-3-241]\]. The transition from childhood to adulthood in CP has been debated for the past 20 years with aging now emerging as a new and independent challenge. Also, the current knowledge of physiatrists among elderly adults with CP is an important issue.
GENERAL HEALTH ISSUES
---------------------
### Cardiometabolic and pulmonary morbidity
In a population-representative sample in United States, adults with CP were found to carry a higher rate of chronic conditions \[[@b15-arm-2019-43-3-241]\] such as diabetes, asthma, hypertension, other heart conditions, stroke, and emphysema than adults without CP. In middle-aged adults with CP between the ages of 40 and 60 years \[[@b16-arm-2019-43-3-241]\] and young adults with CP \[[@b17-arm-2019-43-3-241]\], the proportion of individuals with multiple morbidity (≥2 among 12 chronic conditions) was high. Furthermore, obesity and a high GMFCS level were linked to increased risk of multimorbidity.
Risk factors for cardiovascular disease exist among young adults diagnosed with CP \[[@b18-arm-2019-43-3-241]-[@b20-arm-2019-43-3-241]\]. Age has been identified as the strongest independent predictor of vascular health \[[@b19-arm-2019-43-3-241],[@b20-arm-2019-43-3-241]\], with a higher prevalence of non-ambulatory than ambulatory CP \[[@b21-arm-2019-43-3-241]\]. Also, fatigue \[[@b22-arm-2019-43-3-241],[@b23-arm-2019-43-3-241]\] and bowel symptoms \[[@b24-arm-2019-43-3-241]\] were prevalent in adults with CP. According to Ryan et al. \[[@b25-arm-2019-43-3-241]\] in Ireland, the prevalence of metabolic syndrome was high in 55 young adults with CP with an average age of 37.5, with a prevalence of 20.5% among ambulatory adults and 28.6% in non-ambulatory adults. In a study of young adults with ambulatory CP, moderate physical activity was reported to be associated with a lower cardiometabolic risk suggesting that fitness and physical activity were important measures to reduce non-communicable disease \[[@b25-arm-2019-43-3-241]-[@b28-arm-2019-43-3-241]\].
### Unmet healthcare needs
Rehabilitation facilities were insufficient to manage aging in adults diagnosed with CP in spite of reduced functional abilities \[[@b29-arm-2019-43-3-241]\]. Surveys conducted in Sweden during the year 2001 suggested that 60% of the cohorts investigated were engaged in some kind of physical training \[[@b30-arm-2019-43-3-241]\]. As reported by Balandin and Morgan \[[@b31-arm-2019-43-3-241]\] in 1997, 20% of adults with CP have difficulties accessing medical services and 41% lack access to facilities. Furthermore, the survey also highlighted difficulties in communication and the need for external assistance. According to a recent report in South Korea \[[@b32-arm-2019-43-3-241]\], the medical check-up rate including private health screening, workplace health checks, medical check-ups provided by national health insurance services, and free medical check-ups in adults with CP were lower than in the total population with disabilities, with financial burden cited as the biggest factor preventing hospital attendance. According to the study, 53.2% of the individuals underwent a medical check-up in the past 2 years, while 44.2% did not \[[@b32-arm-2019-43-3-241]\]. Only one-third (37.0%) of individuals with CP received rehabilitation therapy in South Korea \[[@b32-arm-2019-43-3-241]\]. Individuals with CP reported the need for the following medical treatments: pain treatment (42.9%), additional physical therapy (35.7%), examination by a physiatrist (27.3%), orthosis prescriptions (14.3%), occupational therapy (11%), and surgery (3.9%) \[[@b32-arm-2019-43-3-241]\]. In addition, due to the financial burden and lack of knowledge of patients diagnosed with CP, the demand for rehabilitation services was not met in one-third of all adults with CP.
DETERIORATION IN PHYSICAL ACTIVITY
----------------------------------
In adults with CP, physical activity was reported to decrease with increasing GMFCS levels \[[@b33-arm-2019-43-3-241]\]. There is a lack of evidence supporting the efficacy of the intervention to sustain and increase habitual physical activities in children and youth with CP \[[@b34-arm-2019-43-3-241]-[@b36-arm-2019-43-3-241]\]. The effect of deterioration in physical activity or exercise intervention on long-term health in patients with CP has yet to be elucidated. In line with the previous review, the Cochrane study group revealed that there is low-to-very low-quality evidence supporting the benefit of physical activity in children in terms of improved gross motor function and gait speed \[[@b35-arm-2019-43-3-241]\]. Although few studies have examined the effects of physical activity in adulthood, increased physical fitness in young adults with spastic CP was effective in improving fatigue, mental health, and social participation without a significant change in gross motor function \[[@b23-arm-2019-43-3-241]\]. Although physical activity is expected to have a significant long-term impact on adults compared with pediatric populations, studies have yet to investigate the effects of physical activity in adults. A comprehensive review and further analysis of the effects of physical activity on adults with CP are needed.
MUSCULOSKELETAL ISSUES
----------------------
Osteoporosis and arthritis
--------------------------
Adults with CP may be at an increased risk of hypovitaminosis D-induced osteopenia and impaired bone mass \[[@b37-arm-2019-43-3-241]\]. Adults with CP are more frequently deficient in vitamin D due to the use of anticonvulsants or fewer outdoor activities. Recent studies have shown an insufficient or deficient vitamin D levels in more than 50% of the adults with CP based on abdominal obesity, which is an independent predictor of lowered vitamin D levels \[[@b38-arm-2019-43-3-241]\]. Limited weight-bearing, inappropriate nutrition, medications (especially for the treatment of epilepsy), and other factors represent potential risk factors for the early onset of osteoporosis \[[@b10-arm-2019-43-3-241]\].
According to a study regarding the age-related trajectory in adults with CP, the odds ratios of osteoporosis, osteoarthritis, and rheumatoid arthritis in adults aged 30 years and older were higher than in those aged 18--30 years \[[@b39-arm-2019-43-3-241]\]. Multiple musculoskeletal morbidities among adults aged 31 to 40 years were 1.9-fold higher compared with adults aged 18--30 years, 4.3-fold in adults aged 41 to 50 years, and 6.1-fold higher than those above 50 years. Low bone mineral density was also observed in ambulatory adults with CP \[[@b40-arm-2019-43-3-241],[@b41-arm-2019-43-3-241]\], along with frequent fragility fractures \[[@b42-arm-2019-43-3-241]\]. Because adults with CP are more vulnerable to osteoporosis, more frequent check-ups and active interventions are required starting at a younger age. Evaluation of body composition using dual-energy X-ray absorptiometry (DXA) or lean tissue mass is clinically important for optimal outcomes following feeding and exercise interventions \[[@b42-arm-2019-43-3-241]\]. In addition, interventions to increase weight-bearing or muscle mass are preferred \[[@b42-arm-2019-43-3-241]\]. It is also particularly important to reduce the risk of falls in ambulatory individuals with CP \[[@b42-arm-2019-43-3-241]\]. Since a greater risk has been reported in adults with CP, the development of a protocol for the diagnosis and treatment of osteoporosis in individuals with CP is needed.
Sarcopenia
----------
Botulinum toxin injections, orthopedic surgery, and neurosurgical interventions are occasionally performed in children and adolescents with CP due to muscle imbalance during their developmental period. Increasing muscle mass can be a challenge due to the lack of muscle mass when children become adults.
Adults with CP carry smaller and less dense psoas major \[[@b41-arm-2019-43-3-241]\], suggesting greater muscle fat infiltration, poor muscle quality, and less contractile tissue in the muscles \[[@b37-arm-2019-43-3-241]\]. Even ambulatory young adults with CP presented with a calf area 45% smaller than in a typically developing population \[[@b43-arm-2019-43-3-241]\].
Because individuals with CP show a low basal metabolic rate, obesity may occur even with a calorie-controlled diet \[[@b41-arm-2019-43-3-241]\]. Since premature aging associated with sarcopenia in CP may cause acute functional deficits and disabilities \[[@b44-arm-2019-43-3-241]\] and sarcopenia is frequently detected in adults with CP, dietary modification, nutritional supplements and exercise therapy are needed. The development of a detailed exercise protocol for adults with CP is required because increased muscle mass improves strength, functionality, endurance, and general metabolic health in individuals with CP \[[@b37-arm-2019-43-3-241]\]. Protein intervention stimulates skeletal muscle protein synthesis and inhibits protein breakdown resulting in positive protein balance and a net gain in muscle mass \[[@b37-arm-2019-43-3-241]\]. Frequent exercise with protein ingestion accelerates muscle synthesis, accretion of muscle protein, and facilitates muscle hypertrophy \[[@b37-arm-2019-43-3-241]\]. Regular exercise is important to improve the muscle changes associated with aging in individuals with CP \[[@b37-arm-2019-43-3-241]\]. Supplementation with at least 800 IU of vitamin D is effective in improving muscle strength and preventing falls and fractures, and therefore, vitamin D supplementation is also important in this population \[[@b37-arm-2019-43-3-241]\]. A well-designed research protocol highlighting the intervention in sarcopenia treatment is also needed.
Pain
----
According to Murphy et al. \[[@b7-arm-2019-43-3-241]\], adults with CP aged below 50 years frequently reported cervical pain, back pain, muscle pain, joint pain, hand paresthesia, and overuse syndrome. A study demonstrating a 10-year long-term deterioration of perceived health and functioning found that pain and fatigue were the most common health challenges faced by adults with CP \[[@b45-arm-2019-43-3-241]\]. Another study revealed chronic pain (lasting more than 3 months) in 75% of the adults diagnosed with CP \[[@b22-arm-2019-43-3-241]\]. Multiple studies revealed that the back, hip, and the lower limbs were the most common pain locations in adults with CP \[[@b10-arm-2019-43-3-241],[@b11-arm-2019-43-3-241],[@b22-arm-2019-43-3-241],[@b32-arm-2019-43-3-241],[@b45-arm-2019-43-3-241]-[@b50-arm-2019-43-3-241]\]. Adults with CP also manifested pain associated with contractures, orthopedic deformities, fractures, pressure from sitting on bony prominences, as well as spasticity \[[@b44-arm-2019-43-3-241]\]. Despite widespread recognition of pain in adults with CP, the impact of pain-related quality of life (QOL) has not been studied adequately.
However, a large number of adults with CP cannot access medical services in South Korea \[[@b32-arm-2019-43-3-241]\] and as a result, are not adequately treated for pain. Adults with CP need access to medical facilities for appropriate evaluation and treatment because chronic pain may result in decreased gait function \[[@b46-arm-2019-43-3-241]\] and reduced QOL.
Neurological challenges associated with myelopathy
--------------------------------------------------
Cervical myelopathy is often induced by early degeneration of the cervical spine due to abnormal movements in adults with CP especially in adults with dyskinetic CP \[[@b51-arm-2019-43-3-241]-[@b54-arm-2019-43-3-241]\]. Surgical intervention has also been used to reduce exacerbation of weakness and paresthesia \[[@b55-arm-2019-43-3-241],[@b56-arm-2019-43-3-241]\]. The surgical outcome was not as favorable as expected most likely due to continued and persistent neck movements after surgery.
Therefore, in order to improve the surgical outcome and to prevent the recurrence of myelopathy, dyskinetic movements should be reduced following perioperative botulinum toxin injections \[[@b52-arm-2019-43-3-241]\], used to reduce cervical dystonia-related pain and disability in adults with dyskinetic CP \[[@b57-arm-2019-43-3-241]\].
NUTRITIONAL CHALLENGES AND DYSPHAGIA
====================================
Malnutrition has been reported in adults with severe functional disability \[[@b58-arm-2019-43-3-241]\]. Due to the geographic dependence and altered economic status, obesity accounts for a large proportion of the nutritional challenges reported recently \[[@b17-arm-2019-43-3-241],[@b38-arm-2019-43-3-241]\]. However, further studies are needed because of the paucity of evidence supporting the effect of interventions targeting obesity and malnutrition in adults with CP.
Children with CP are reported to manifest a higher prevalence of dysphagia. Adults with CP, despite maintaining the same dietary patterns as in childhood, may experience gradual deterioration in their swallowing and mealtime capabilities.
A qualitative study investigating swallowing difficulty in adults with CP \[[@b59-arm-2019-43-3-241]\] reported gradual changes in swallowing capabilities starting as early as 30 years of age. Even in adults with dyskinetic CP exposed to unrestricted diets and not previously evaluated for swallowing function, a videofluoroscopic swallowing study revealed frequent aspiration without cough reflex \[[@b60-arm-2019-43-3-241]\] probably due to abnormal sensorimotor integration or chronic aspiration-induced desensitization of the laryngeal airway. Considering the impact of aspiration risk on general health, the swallowing function is a challenging issue in all CP populations.
However, quantitative studies investigating the prevalence and the effect of therapeutic intervention on swallowing function in this population have yet to be conducted despite the high frequency of swallowing problems directly affecting the QOL.
FUNCTIONAL LIMITATION
=====================
According to a study conducted in the Netherlands, 70% of young adults with CP between the ages of 18 and 22 reported challenges with activities of daily living \[[@b61-arm-2019-43-3-241]\] including difficulties in self-care, productivity, and leisure activities, especially involving recreation, leisure, meal preparation, and housework.
Adolescents with CP gradually exhibit a progressive decline in strength and functional reserve through adult life \[[@b37-arm-2019-43-3-241]\]. Prior to the age of 35 years, the ability to walk decreases in adults with CP despite acquired ambulation during adolescence. Deterioration in GMFCS levels is most evident in the late 20s and early 30s, and dependence and perceived difficulties in activity influence adults with CP \[[@b45-arm-2019-43-3-241],[@b46-arm-2019-43-3-241],[@b62-arm-2019-43-3-241]\]. It has been reported that the ambulatory function deteriorates in adulthood \[[@b7-arm-2019-43-3-241],[@b30-arm-2019-43-3-241],[@b63-arm-2019-43-3-241]\], which is likely due to new medical age-related challenges in patients with CP based on the fact that the GMFCS level remained almost stable in individuals with CP until the age of 21 years \[[@b64-arm-2019-43-3-241]\]. If GMFCS I-III declined to IV-V with age, it is likely that adults with CP represent a burden for their family members and caregivers, and increasingly need assistive device use. Therefore, functional limitation should be addressed in health policy regulations for adults with CP.
Limitations in functional activity were found to be a major restricting factor for social participation in young adults with CP. Although intellectual disability rather than GMFCS level in children is known to have a significant impact on social participation \[[@b65-arm-2019-43-3-241],[@b66-arm-2019-43-3-241]\], there is a lack of evidence to support this finding in adults with CP. Further, work participation is restricted in adults with CP who do not suffer from an intellectual disability \[[@b67-arm-2019-43-3-241]\], and further research is required to encourage increased participation in society and in the workplace.
The International Classification of Functioning, Disability and Health (ICF) guidelines were developed to accurately assess the function of individuals with disabilities. The ICF Children and Youth version (ICF-CY) core set was used to measure functional limitation in children and adolescents with CP below 18 years of age \[[@b68-arm-2019-43-3-241]\]. The ICF core set in adults with CP has yet to be developed and the need to use ICF to accurately assess adults with CP has been highlighted \[[@b10-arm-2019-43-3-241]\]. Recently, there has been a move to develop the ICF core sets for adults with CP.
HEALTH-RELATED QUALITY OF LIFE AND SOCIAL PARTICIPATION
=======================================================
Adults with spastic bilateral CP are reported to have difficulty engaging socially and have a low health-related quality of life (HRQOL) \[[@b69-arm-2019-43-3-241]\]. According to a 8-year follow-up study conducted in Canada on adults with CP, HRQOL deterioration was most evident in their late 20s and 30s \[[@b70-arm-2019-43-3-241]\]. However, a longitudinal multicenter study conducted in the Netherlands reported lower HRQOL in adults with CP than in populations without disability and found that HRQOL and social participation were fairly stable for many years \[[@b71-arm-2019-43-3-241]\].
To date, a 36-item Short Form Health Survey (SF-36) has been used to evaluate HRQOL in adults with CP \[[@b26-arm-2019-43-3-241],[@b45-arm-2019-43-3-241],[@b63-arm-2019-43-3-241],[@b69-arm-2019-43-3-241],[@b71-arm-2019-43-3-241]\] although they have not been validated in this population. The survey consists of physical functioning, physical role, bodily pain, general health perception, vitality, social functioning, emotional role and mental health. Each area is assigned a maximum score of 100 points. Among these items, physical functioning and physical role are reported to be low in adults with CP in the Netherlands \[[@b69-arm-2019-43-3-241]\]. Although the SF-36 is applicable only to people without cognitive impairment and cannot be used to evaluate individuals with intellectual disabilities, it can be suggested as a useful assessment tool for adults with CP. Unmet needs for medical and rehabilitation treatment in South Korea represent a possible barrier to improved QOL and social participation among individuals with CP, and therefore, appropriate financial and technical resources are required to address the unmet needs of these populations in the healthcare system \[[@b32-arm-2019-43-3-241]\].
CONCLUSION
==========
In addition to early detection and habilitation or rehabilitation of children with CP, transition to adulthood has been highlighted as an important issue in the past 10 to 20 years. Aging in this population is an emerging issue. Physiatrists require adequate knowledge to prepare for the aging population of adults with CP and further studies are needed to investigate the impact of physical activity, nutrition, sarcopenia, myeloradiculopathy, and swallowing function in these individuals.
No potential conflict of interest relevant to this article was reported.
Conceptualization: Bang MS. Writing -- original draft: Yi YG, Bang MS. Writing -- review and editing: Yi YG, Jung SH, Bang MS. Approval of final manuscript: all authors.
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Q:
Need to import .dmp file into Oracle Express, export it as a flat CSV or TXT file for importation into SAS
Does anyone know how to write a specific command line that imports a .dmp file into an Oracle Express database, and then exports the data as a CSV file?
Any help would be greatly appreciated. Thank you!
A:
Import using impdp:
impdp scott/tiger@db10g tables=EMP,DEPT directory=TEST_DIR dumpfile=EMP_DEPT.dmp logfile=impdpEMP_DEPT.log
Then spool to csv:
set heading off
spool myfile.csv
select col1|','||col2 from my_tables;
set colsep ','
select * from my_table;
spool off;
Follow below links for help.
import: https://oracle-base.com/articles/10g/oracle-data-pump-10g
spool to csv: http://www.dba-oracle.com/t_export_to_csv_file.htm
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The present invention relates in general to ink or printing fluid input systems for use in a rotary printing press and, in particular, to an ink or printing fluid input system in which different ink colors are applied to a single metering roller.
In normal printing practice a single rotary plate cylinder is intended to print only a single ink color. For certain printing operations it is often desired to apply ink in two or more different colors to a single metering roller at different areas along the axial length of the metering roller. For that purpose, two or more ink or printing fluid input devices must be associated with the metering roller and must be separated from each other so that ink of different colors can be applied to the metering roller in the desired areas and so that smearing or mixing of the different colored inks will be avoided.
It is well known in the prior art to use different types of physical barriers between the ink input devices to insure that the different colored inks are kept separated. For example, it is known to use an ink separator plate that extends at right angles to the axis of the metering roller. Such an ink separator plate has a concavely curved end face having the same radius of curvature as that of the metering roller, and is brought into sealing contact with the surface of the metering roller. Also known is the use of sponge type materials in which water is circulated, these sponge type materials contacting the surface of the metering roller at a position for dividing one colored ink from another colored ink.
A drawback of the prior art is that these prior art seals of necessity form mechanical contact with the metering roller. These seals are difficult to set and require constant attention due to wear. Ink containment on anilox rolls used in flexographic and keyless printing press applications are also typically constructed using mechanical type seals which have these same drawbacks. |
THE UPS OF DOWNS
Directed by Ali Palmer-Smith 10 min / documentary / 2002 / UK
Despite being born with Down syndrome, Danny lives life to the fullest. This short film shows how independent and informed a young man with Danny’s condition can be. The film gives a glimpse into a life to be proud of. |
[1][ISMAP]-[2][Home]
### GUIDE ### [3][Background] [4][Synopsis] [5][Credits] [6][Episode
List] [7][Previous] [8][Next]
_Contents:_ [9]Overview - [10]Backplot - [11]Questions - [12]Analysis
- [13]Notes - [14]JMS
_________________________________________________________________
Overview
Babylon 4 returns as abruptly as it vanished, but its reappearance
may bode ill for the future. Delenn receives a momentous offer.
[15]Kent Broadhurst as Major Krantz. [16]Tim Choate as Zathras.
[17]Denise Gentile as Lise Hampton.
Sub-genre: Intrigue/mystery
[18]P5 Rating: [19]8.57
Production number: 118
Original air date: August 10, 1994
Written by J. Michael Straczynski
Directed by Jim Johnston
Watch For:
* A man shouts at Garibaldi and Sinclair; what he says might provide
clues about the nature of the opponents in another scene.
* Look closely at what's inside a transparent case given to Delenn.
It's an object that's been shown in a previous episode.
_________________________________________________________________
Backplot
* Babylon 4 was stolen by people from the future, apparently at
Sinclair's behest during that time period, to act as a base of
operations in a tremendous war being fought between the forces of
light and darkness.
* Sinclair will be a great leader, possibly _the_ leader, of the
forces of light in that war.
* At some point, long before he participates in Babylon 4's
disappearance, Sinclair will flee a place (most likely Babylon 5,
cf. [20]"Signs and Portents") that is about to be overrun by some
evil creatures. Garibaldi will stay behind to fight, but will
force Sinclair to leave. A [21]transcript of the scene in question
is available.
* The Grey Council stopped the war because of a prophecy. Valen (a
revered figure, see [22]jms speaks) said that humans, or some
among them, had a destiny with which the Minbari could not be
allowed to interfere.
* Delenn is on Babylon 5 to study humanity, to determine whether the
prophecy is correct.
Unanswered Questions
* Will Babylon 4 appear again? If so, when?
* Was it really Sinclair in the suit? If not, who or what was it?
* What was Sinclair trying to prevent from happening?
* Who was waiting for Delenn and Sinclair?
* Will Delenn keep her position on the council?
* What is the purpose of the triluminary?
* What was happening in Sinclair's flashforward? Who or what was
attacking the station? Is it related to the destruction of the
station as foretold in [23]"Signs and Portents?"
* What was the crazed man referring to when he shouted about
"monsters" and said, "I see you... you think I can't?" (see
[24]Analysis)
* Is the Grey Council's cruiser the same place Sinclair was taken
during the Battle of the Line? (cf. [25]"And the Sky Full of
Stars")
* Why is Delenn so convinced she must remain on Babylon 5, even at
the risk of her standing in the Council?
* What "change" does Delenn believe is coming?
Analysis
* None of the races have demonstrated an ability to travel through
time. Yet within Sinclair's lifetime, such technology will either
be developed, discovered, or introduced by people from the distant
past or future. Does it exist already? If so, who has it? Zathras'
people may be the ones to provide the technology.
* The voice that speaks to Sinclair sounds like Delenn's, but her
face is intentionally not shown. Presumably there is a reason for
that; Delenn may be due to change in some way that will alter her
appearance.
* After Babylon 4 completes its time jump, a voice (presumably a
computer) announces that the atmosphere was breathable. Why wasn't
it breathable before? Zathras clearly had no trouble breathing in
the past, so is something about the future Sinclair different that
prevents him from breathing a normal atmosphere?
* In the past, when we've seen Grey Council members, they have had
silver triangles on their foreheads ([26]"And the Sky Full of
Stars" and [27]"Signs and Portents" during Morden's visit to
Delenn.) Yet no such triangles were visible this time. What do the
triangles mean, and what causes them to appear? (see [28]jms
speaks)
* The attacking force in Sinclair's vision of the future seemed to
be invisible. Witness the fact that Garibaldi's men were firing in
seemingly random directions, as if they didn't know where the
enemy was. It also seems unlikely that they'd use a flamethrower
if they could aim at their opponents. When the unknown force
finally cut through the wall, it was forced inward, but nothing
could be seen forcing it. This also explains what the crazed man
on B4 was talking about; he'd seen visions of a battle against
invisible foes too.
The only instance of invisibility seen in the series up until this
episode was in [29]"The War Prayer," and it was developed by the
Earth Alliance military, suggesting perhaps that the attackers
might be humans.
* Garibaldi flashed back to an event two years earlier. That may
suggest that Sinclair's flashforward (if indeed that's what it
was) was to two years in the future, which would put the scene
somewhere in the year 2260, season three of the series.
Notes
* Babylon 4 is larger than Babylon 5.
jms speaks
* The one I'm most looking forward to writing just now, though, is
"Babylon Squared," in which we finally show what happened to
Babylon 4, and in the process ask more questions than we answer
(though at least we DO answer the questions we asked about the
fate of that station in general...you'll know what happened to it,
just not yet what it means). The end of this episode will cause
more speculation and consternation and astonishment than anything
you've seen on TV in a long, long, very long time.
* What a weird day...filming "Babylon Squared," and one minute I'm
standing in the anteroom/hallway of a Minbari cruiser that leads
into the Great Hall and the chambers of the Grey Council...a few
minutes later I'm standing in a section of Babylon 4, and the
whole atmosphere of the crew is *very* different, the whole
sensibility is strange... very strange.
"Babylon Squared" has a *very* different look to it, and a very
eerie and foreboding feel about it, which I like a lot. Jim
Johnston, who directed "Soul Hunter" and several others is doing
it. Very moody.
* Yesterday, I got the final air-check versions of "Babylon Squared"
and "Chrysalis" to QC before delivering them to PTEN. Watched both
of them three times in the same day. They're just stunning.
Probably the two best episodes of the entire season.
* Yes, you will see the Major Conflict that leads to the situation
with Babylon 4. We're building toward a massive conflagration
here.
* Yes, you will definitely, at some point, see the flip side of the
B2 episode.
* No, actually, B2 was structured for maximum jarring effect, thus
the sudden cuts back and forth, the sickly green light in
B4...makes the person watching feel unexplainably anxious, which
was a subliminal but definite intent. So no, nothing much was cut.
And yes, eventually we will see the flip-side of the B4 story.
* In B-squared, we saw the present events in the vanishment of B4;
in a future episode, we'll actually see our characters make the
decision to go back in time and yank B4 forward, what went wrong,
and so on.
* _The One's suit was very similar to the suits in "2001."_
Re: the suit...that wasn't an intentional 2001 nod...we went to
Modern Props to get a space suit for Babylon Squared, and the only
one they had on hand that would work for us was one left-over from
2010, which I asked the folks in costume to change as much as
possible...though it was pretty much what it was regardless. So
that one wasn't intentional.
* Nope; Zathras is one of his race, which aren't offspring of any
other two groups.
* When Zathras shows up in time, it'll definitely be recognizeable
as Zathras.
* B5 is smaller than B4 because they sunk most of their budget into
B4; on B5 they had to get outside funding, and scrimped.
* B1-B4 were located in roughly the same sector, with B4 using some
of the materials from 1-3 leftover. B5 was constructed about 3
hours (traveling time in real-space) from the location of B4.
* No commander had yet been assigned to Babylon 4. One Major Krantz
had been assigned to oversee the final stages of construction, and
was on board -- along with about 1300 others in the construction
crew -- when the station vanished. The station had only been on-
line 24 hours, and the discussions of a commanding officer had
just begun when it disappeared.
* Major Krantz wasn't so much in charge of B4 as he was (as noted in
dialogue) assigned to oversee the final stages of construction.
His job was to get the station finished, then turn it over to
someone else to run.
* Why no triangles on the Council's heads?
While the triangle is one element of the Grey Council symbology,
it is not present and visible at all times and under all
circumstances; it has a particular purpose or meaning.
* The triangle only manifests itself for specific reasons, at
specific times, neither of which were appropriate to that moment.
And yes, the Triluminary is much cooler...and does something quite
interesting.
* Valen was the one who brought Minbari civilization together, he is
their Christ-figure. And yes, the heavyset Grey Council member is
the same one as in "Sky."
* _Does the Grey Council live permanently on that ship?_
They stay on the cruiser almost entirely during their tenure in
the council, only leaving for personal family crisis/situations
and the like.
* Garibaldi's closing lines in Sinclair's flashforward are
reminiscent of "Aliens."
When you're shooting a show, invariably you get to the stage and
find that you have, for instance, three lines, one per character
in the room...and you're trying to get them out the door, and it
moves better if you give one line to one character and the other
two to the other character. That sometimes happens. But rarely. In
the Garibaldi's yell case, it was written as a quick shot, he
yells and we're out. The director wanted to extend the shot a bit,
visually. I wasn't in the studio at the time, so Jerry improvised
a series of yells.
This sort of thing is *extremely* rare on the show; the actors and
directors know they *cannot* change dialogue on the set without
approval from me or Larry. On any given script, no more than about
3-6 lines get modified for staging purposes once we get to the
set. And always with approval required. This is an absolute, hard
and fast rule. The only reason the Garibaldi thing happened is
that they figured it was just a yell, so nothing could get messed
up story-wise (which is the primary reason this is so strict;
change one word in a line and it could screw up plot points three
episodes down the road) by having him yell a few specific lines.
If I'd been there for that scene, I would've written him something
a little less reminiscent of "Aliens."
* The script called for Garibaldi to take up the Big Massive Gun and
fire, with a primal YELL that went on forever. Any dialogue at
that point which replaced the yell came from the actor. The
"you're already dead" was only relevant to the scene, not T2.
* Re: Garibaldi in the flash-forward scene...no, it wasn't any kind
of "homage" to Aliens. (And for the most part, I try and stay
clear of any kind of homage unless it's primarily a throwaway; I
want my story to be MY story, not a bunch of homages.)
The single most moving kind of story for me is the "last man on
the bridge"...the last defender who has to hold the line while
others get away, knowing he will probably not survive it. This has
great power for me, and for many others, which is why it shows up
again and again in films, literature, TV and other venues. The
Garibaldi scene has NOTHING to do with Aliens, and everything to
do with that figure.
Re: *why* it is that humans are special...has nothing to do with
sacrifice, or dedication (well, that's not quite true, it has
something to do with it), but that's not the totality of it.
There's one more element you don't know about yet, that won't be
revealed until season two, episode one, "Points of Departure."
Once you see that episode, you'll fully understand that there is
one very particular thing about humans that is very special
indeed.
* I kinda *have* to play fair with the story; if you hear Delenn's
voice, then you can be sure it's Delenn.
In one form or another.
* "So who IS the One? Some of the evidence points to Sinclair, but
other bits seem to indicate Delenn. Yet neither seems to fit all
the facts above."
Exactly.
What you have here in your message are two pieces of the puzzle.
You're confounded by the fact that somehow they don't quite seem
to fit into one another. That's because there's one last piece
missing in this part of the picture, which fits in between them.
The intent is to put this piece into clear view in year three,
probably between episodes 8 and 11 approximately. At that point,
the question of the One will be fully answered.
* Re: Sinclair as the One...funny how all this time very few folks
have really commented much on how it was that Zathras could look
right into Sinclair's face and say, "NOT the One."
* _Garibaldi's eyes glow for one frame in the flashforward scene._
(sigh) Our rotoscope EFX guy was waiting for a bunch of PPG EFX to
finish rendering in that battle scene, and was bored, and like
many such EFX types, filled in the eyes of Garibaldi with weird
stuff while waiting around. When the other scene finished
rendering, he got out, believing that he had not saved that one
frame. Unknowingly, he had.
Nobody caught it until after broadcast.
We talked.
* "It has been divulged that Sinclair is coexisting in a parallel
dimension Babylon 4."
Actually, this has *not* been divulged...what it is is a
speculation based on an offhand comment by Michael at a
convention. I jump in here only because, well, that ain't it. B4
is not in an alternate dimension, neither is there an alternate
Sinclair. Just a course correction to the discussion.
* With only one exception, you won't see time travel anywhere in the
five-year run of the B5 story.
* _Which do you do first?_
Fasten, button.
Levi's Jeans forever!
* _Why wasn't Franklin delivering the autopsy report?_
Garibaldi is head of security, and Franklin would likely give him
the report, which Garibaldi then relays. In such things there is a
chain of command. And as you say, it seemed pointless to bring in
the actor just for one half-page scene.
* _Credits for two Gray Council members?_
Mark Henrickson was the...rounder of the two Minbari. The one with
the staff wanted to go uncredited.
No real reason, he just felt it would be better for the character
to remain mysterious; and since it really wasn't a big part, it
wouldn't make a real difference one way or another in his credits
and resume. (I know that sounds weird, but as near as I can
determine, that's the reason. He did a great job, and we're
looking forward to having him again.)
* Does the triluminary have anything to do with the sculpture in
Delenn's quarters?
Yes, the Triluminary does have a function in the device she's been
making.
[35][Next]
[36]Last update: October 27, 1997
References
1. file://localhost/cgi-bin/imagemap/titlebar
2. LYNXIMGMAP:file://localhost/lurk/maps/maps.html#titlebar
3. file://localhost/home/woodstock/hyperion/docs/lurk/background/020.shtml
4. file://localhost/home/woodstock/hyperion/docs/lurk/synops/020.html
5. file://localhost/home/woodstock/hyperion/docs/lurk/credits/020.html
6. file://localhost/home/woodstock/hyperion/docs/lurk/episodes.php
7. file://localhost/home/woodstock/hyperion/docs/lurk/guide/019.html
8. file://localhost/home/woodstock/hyperion/docs/lurk/guide/021.html
9. file://localhost/home/woodstock/hyperion/docs/lurk/guide/020.html#OV
10. file://localhost/home/woodstock/hyperion/docs/lurk/guide/020.html#BP
11. file://localhost/home/woodstock/hyperion/docs/lurk/guide/020.html#UQ
12. file://localhost/home/woodstock/hyperion/docs/lurk/guide/020.html#AN
13. file://localhost/home/woodstock/hyperion/docs/lurk/guide/020.html#NO
14. file://localhost/home/woodstock/hyperion/docs/lurk/guide/020.html#JS
15. http://us.imdb.com/M/person-exact?+Broadhurst,+Kent
16. http://us.imdb.com/M/person-exact?+Choate,+Tim
17. http://us.imdb.com/M/person-exact?+Gentile,+Denise
18. file://localhost/lurk/p5/intro.html
19. file://localhost/lurk/p5/020
20. file://localhost/home/woodstock/hyperion/docs/lurk/guide/013.html
21. file://localhost/home/woodstock/hyperion/docs/lurk/guide/020.flash.html
22. file://localhost/home/woodstock/hyperion/docs/lurk/guide/020.html#JS:valen
23. file://localhost/home/woodstock/hyperion/docs/lurk/guide/013.html
24. file://localhost/home/woodstock/hyperion/docs/lurk/guide/020.html#AN:inv
25. file://localhost/home/woodstock/hyperion/docs/lurk/guide/008.html
26. file://localhost/home/woodstock/hyperion/docs/lurk/guide/008.html
27. file://localhost/home/woodstock/hyperion/docs/lurk/guide/013.html
28. file://localhost/home/woodstock/hyperion/docs/lurk/guide/020.html#JS:triangle
29. file://localhost/home/woodstock/hyperion/docs/lurk/guide/007.html
30. file://localhost/lurk/lurker.html
31. file://localhost/home/woodstock/hyperion/docs/lurk/guide/020.html#TOP
32. file://localhost/cgi-bin/uncgi/lgmail
33. file://localhost/home/woodstock/hyperion/docs/lurk/episodes.php
34. file://localhost/home/woodstock/hyperion/docs/lurk/guide/019.html
35. file://localhost/home/woodstock/hyperion/docs/lurk/guide/021.html
36. file://localhost/lurk/lastmod.html
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/*********************************************************************/
/* */
/* Optimized BLAS libraries */
/* By Kazushige Goto <kgoto@tacc.utexas.edu> */
/* */
/* Copyright (c) The University of Texas, 2009. All rights reserved. */
/* UNIVERSITY EXPRESSLY DISCLAIMS ANY AND ALL WARRANTIES CONCERNING */
/* THIS SOFTWARE AND DOCUMENTATION, INCLUDING ANY WARRANTIES OF */
/* MERCHANTABILITY, FITNESS FOR ANY PARTICULAR PURPOSE, */
/* NON-INFRINGEMENT AND WARRANTIES OF PERFORMANCE, AND ANY WARRANTY */
/* THAT MIGHT OTHERWISE ARISE FROM COURSE OF DEALING OR USAGE OF */
/* TRADE. NO WARRANTY IS EITHER EXPRESS OR IMPLIED WITH RESPECT TO */
/* THE USE OF THE SOFTWARE OR DOCUMENTATION. */
/* Under no circumstances shall University be liable for incidental, */
/* special, indirect, direct or consequential damages or loss of */
/* profits, interruption of business, or related expenses which may */
/* arise from use of Software or Documentation, including but not */
/* limited to those resulting from defects in Software and/or */
/* Documentation, or loss or inaccuracy of data of any kind. */
/*********************************************************************/
#define ASSEMBLER
#include "common.h"
#define M r3
#define N r4
#define A r5
#define LDA r6
#define B r7
#define AO1 r8
#define AO2 r9
#define J r10
#define B1 r11
#define B2 r28
#define M4 r29
#define INC r30
#define INC2 r31
#define c01 f0
#define c02 f1
#define c03 f2
#define c04 f3
#define c05 f4
#define c06 f5
#define c07 f6
#define c08 f7
PROLOGUE
PROFCODE
stwu r31, -4(SP)
stwu r30, -4(SP)
stwu r29, -4(SP)
stwu r28, -4(SP)
slwi LDA, LDA, ZBASE_SHIFT
slwi M4, M, 1 + ZBASE_SHIFT
li r9, -2
and B2, N, r9
mullw B2, B2, M
slwi B2, B2, ZBASE_SHIFT
add B2, B2, B
cmpwi cr0, M, 0
ble- LL(99)
cmpwi cr0, N, 0
ble- LL(99)
subi B2, B2, 2 * SIZE
subi M4, M4, 6 * SIZE
li INC, 1 * SIZE
li INC2, 2 * SIZE
andi. r0, A, 2 * SIZE - 1
bne LL(100)
subi A, A, 2 * SIZE
srawi. J, M, 1
ble LL(20)
.align 4
LL(10):
mr AO1, A
add AO2, A, LDA
add A, AO2, LDA
sub B1, B, M4
addi B, B, 8 * SIZE
srawi. r0, N, 1
mtspr CTR, r0
ble LL(15)
.align 4
LL(12):
LFPDUX c01, AO1, INC2
LFPDUX c02, AO1, INC2
LFPDUX c03, AO2, INC2
LFPDUX c04, AO2, INC2
STFPDUX c01, B1, M4
STFPDUX c02, B1, INC2
STFPDUX c03, B1, INC2
STFPDUX c04, B1, INC2
bdnz LL(12)
.align 4
LL(15):
andi. r0, N, 1
ble LL(19)
LFPDUX c01, AO1, INC2
LFPDUX c02, AO2, INC2
STFPDUX c01, B2, INC2
STFPDUX c02, B2, INC2
.align 4
LL(19):
addic. J, J, -1
bgt LL(10)
.align 4
LL(20):
andi. J, M, 1
addi M4, M4, 4 * SIZE
ble LL(99)
mr AO1, A
sub B1, B, M4
srawi. r0, N, 1
mtspr CTR, r0
ble LL(23)
.align 4
LL(22):
LFPDUX c01, AO1, INC2
LFPDUX c02, AO1, INC2
STFPDUX c01, B1, M4
STFPDUX c02, B1, INC2
bdnz LL(22)
.align 4
LL(23):
andi. r0, N, 1
ble LL(99)
LFPDUX c01, AO1, INC2
STFPDUX c01, B2, INC2
.align 4
LL(99):
addi SP, SP, -4
lwzu r28, 4(SP)
lwzu r29, 4(SP)
lwzu r30, 4(SP)
lwzu r31, 4(SP)
addi SP, SP, 4
blr
.align 4
LL(100):
subi A, A, SIZE
srawi. J, M, 1
ble LL(120)
.align 4
LL(110):
mr AO1, A
add AO2, A, LDA
add A, AO2, LDA
sub B1, B, M4
addi B, B, 8 * SIZE
srawi. r0, N, 1
mtspr CTR, r0
ble LL(115)
.align 4
LL(112):
LFDUX c01, AO1, INC
LFDUX c02, AO1, INC
LFDUX c03, AO1, INC
LFDUX c04, AO1, INC
LFDUX c05, AO2, INC
fsmfp c01, c02
LFDUX c06, AO2, INC
fsmfp c03, c04
LFDUX c07, AO2, INC
fsmfp c05, c06
LFDUX c08, AO2, INC
fsmfp c07, c08
STFPDUX c01, B1, M4
STFPDUX c03, B1, INC2
STFPDUX c05, B1, INC2
STFPDUX c07, B1, INC2
bdnz LL(112)
.align 4
LL(115):
andi. r0, N, 1
ble LL(119)
LFDUX c01, AO1, INC
LFDUX c02, AO1, INC
LFDUX c03, AO2, INC
LFDUX c04, AO2, INC
fsmfp c01, c02
fsmfp c03, c04
STFPDUX c01, B2, INC2
STFPDUX c03, B2, INC2
.align 4
LL(119):
addic. J, J, -1
bgt LL(110)
.align 4
LL(120):
andi. J, M, 1
addi M4, M4, 4 * SIZE
ble LL(999)
mr AO1, A
sub B1, B, M4
srawi. r0, N, 1
mtspr CTR, r0
ble LL(123)
.align 4
LL(122):
LFDUX c01, AO1, INC
LFDUX c02, AO1, INC
LFDUX c03, AO1, INC
LFDUX c04, AO1, INC
fsmfp c01, c02
fsmfp c03, c04
STFPDUX c01, B1, M4
STFPDUX c03, B1, INC2
bdnz LL(122)
.align 4
LL(123):
andi. r0, N, 1
ble LL(999)
LFDUX c01, AO1, INC
LFDUX c02, AO1, INC
fsmfp c01, c02
STFPDUX c01, B2, INC2
.align 4
LL(999):
addi SP, SP, -4
lwzu r28, 4(SP)
lwzu r29, 4(SP)
lwzu r30, 4(SP)
lwzu r31, 4(SP)
addi SP, SP, 4
blr
EPILOGUE
|
The Screamer is a monster in Silent Hill: Downpour and has the ability to stun Murphy Pendleton with a piercing scream. Her attacks vary from concentrated jabs to wild swinging with her fists. She can also alert or call other Screamers to aid her in battle, making her a tough enemy if surrounded. They are the weakest enemies in the game.
Contents
Appearance
The Screamer is humanoid in appearance and resembles a woman wearing a tattered dress and high-heeled boots. They have dark, gray-toned skin with deformed, stretched-tight flesh on their face, inhumanly wide mouths, thin, wispy hair, and empty, bleeding eyes. Screamers' weapons are their hands, which are permanently affixed in a fist thanks to the several nails driven from the top of their hands through their fingers, which they jab and slash with. Their garb is ripped in many places, exposing tattered flesh underneath. A Screamer's dress is also hugged close to the figure, as too is the monster's hair, which suggests her being soaked by rain.
Character
Screamers are among the most common enemies in the game. The player will encounter more powerful enemies, but the Screamer's unique abilities make her a dangerous opponent. At close range, the Screamer usually just attacks with quick strikes. From farther away, however, this creature tends to make use of her incapacitating scream. When the player sees a Screamer inhale, they should run in to interrupt the attack, run out of range, or take cover behind a suitable object. If Murphy allows a Screamer to slip behind him, the creature might jump on his back. When this happens, the player must follow the on-screen prompts to shake the Screamer loose. The Screamer's incapacitating attacks are particularly troublesome when other creatures are in the area. When the player faces multiple opponents, they should consider dealing with any Screamers before they deal with more imposing enemies.
Symbolism
This article or section contains speculation.
The Screamer may represent:
The alarm sirens at Murphy's prison and how they would ring if he tried to escape his cell, and, by extension, police sirens as well.
A murder victim. Her body is covered in decaying flesh, and her screams may reflect the screams that a victim lets out during their struggle.
Murphy's wife, Carol Pendleton. When idle, the Screamer sometimes holds her arms close, rocking back and forth with head slightly tilted, as if cradling a child. This could represent Murphy's concern about how Carol dealt with Charlie Pendleton's death and Murphy's fear of Carol, thinking that she blames and hates him for their son's death. Thus, he sees the woman he loved now as a woman that is twisted, distorted, and hostile.
Part of Sanchez's psyche, a possible rapist or sexual predator due to the way he treats Anne Marie Cunningham during the bus transfer representing his possible misdeeds. This seems to be supported by his use of derogatory language towards the Screamer ("Bitch!"). The Screamer is not overtly sexual (she has a very flat chest beneath a heavy coat) and this may demonstrate his own predatory nature of possibly forcing sexuality upon them. Sanchez is shown repeatedly beating her face, which is heavily scarred with slashes, and her skin is stretched very tightly as if she has had plastic surgery; this could suggest that Sanchez has previously beaten his victim(s) so badly that they required facial reconstructive surgery. Her flesh-like coat is torn most around where her breasts would be, possibly hinting at a link to sexual violence and sadism. As Murphy displays no signs of sexual aggression, this further supports that she is Sanchez's monster.
Anne Cunningham's response to her father's murder. It is given in the game that she is distrustful of Murphy and suspects him to be Frank's murderer, and the various cutscenes between the two are characterized by volatility. The Screamer's intense shriek, followed by her vicious attack on Murphy, could be representative of Anne taking out her anger upon the protagonist.
Trivia
Killing or incapacitating 10 Screamers in the game unlocks the Achievement/Trophy "Silence is Golden".
During the point in the game when demonic police patrol vehicles chase Murphy through the streets of Silent Hill, large groups of Screamers are called in by the car's siren when the car catches up to Murphy. The summoning of these creatures by the siren is further evidence that the Screamer may represent the alarms and sirens from police vehicles.
When a Screamer is killed, her dying scream is cut short by a static sound, like a police communication radio turning off. |
GEN-27 exhibits anti-inflammatory effects by suppressing the activation of NLRP3 inflammasome and NF-κB pathway.
Prolonged inflammation and deregulated cytokine production are associated with diversified inflammatory diseases. Genistein (GEN), the active and predominant isoflavonoid in dietary soybean, possesses anti-inflammatory activity. Our study aimed to assess the anti-inflammatory effects of GEN-27, a derivative of GEN, as well as explore the potential molecular mechanisms using lipopolysaccharide (LPS)-induced RAW264.7 cells. In our study, we demonstrated that GEN-27 administration (1, 5, or 10 μM) dose-dependently inhibited nitrite and nitric oxide (NO) levels in LPS-stimulated RAW264.7 cells. Also, GEN-27 suppressed the release of LPS-induced pro-inflammatory cytokines including tumor necrosis factor-α (TNF-α), interleukin-1β (IL-1β), IL-6, and IL-18. Moreover, GEN-27 attenuated LPS-induced inducible NO synthase (iNOS), and cyclooxygenase-2 (COX-2) expressions at messenger RNA and protein levels, and reversed the promoter activity of iNOS in RAW264.7 cells. Mechanistically, GEN-27 abated LPS-induced reactive oxygen species production, as well as mitigated LPS-induced increase of caspase 1 activity and the protein levels of NOD-like receptor 3 (NLRP3), anti-apoptosis-associated speck-like protein-containing a CRAD (ASC), and caspase 1 in RAW264.7 cells in a dose-dependent manner. Similarly, GEN-27 dose-dependently weakened adenosine triphosphate-induced NLRP3 and IL-1β in RAW264.7 cells. In addition, GEN-27 treatment significantly suppressed LPS-induced phosphorylation of nuclear factor-κB (NF-κB) p65 and alleviated LPS-induced increase of transcriptional activity of NF-κB in RAW264.7 cells. In summary, these results revealed that GEN-27 exhibited anti-inflammatory effects by suppressing the activation of NLRP3 inflammasome and NF-κB pathway, suggesting that GEN-27 may be served as a promising therapeutic agent for the prevention and therapy of inflammatory-associated diseases. |
The 82-year-old driver involved in a crash that killed a bicyclist west of Omaha in March has entered a guilty plea.On Wednesday, Ruth Jeffers pleaded guilty to misdemeanor motor vehicle homicide in connection with the March 13 crash.-- Video: Driver pleads guilty in fatal bicycle crashInvestigators said Jeffers was driving the Ford Taurus that struck and killed former Millard West teacher Jim Johnston near 260th and West Center Road that day.After entering the plea, Judge Jeffrey L. Marcuzzo found Jeffers guilty.Jeffers is scheduled to be sentenced in August.Misdemeanor motor vehicle homicide carries a penalty of up to one year in prison, a $1,000 fine or both. There is no minimum sentence for this charge.Johnston’s family was at the court hearing Wednesday.“March wasn't a very good month for us,” said Jed Johnston, his brother. “We are trying to hold it together and move forward, but these things bring little reminders back up and it does hurt.”The ghost bike memorial still stands on West Center Road, a visible reminder for the Johnston family that healing is a long way off.“There is no closure. Jim was out on a wonderful day doing what he loved, riding where he was supposed to ride, being safe, and his life was ended -- that hurts and it's going to take a long, long time for all of us to heal,” Johnston said.
The 82-year-old driver involved in a crash that killed a bicyclist west of Omaha in March has entered a guilty plea.
On Wednesday, Ruth Jeffers pleaded guilty to misdemeanor motor vehicle homicide in connection with the March 13 crash.
-- Video: Driver pleads guilty in fatal bicycle crash
Investigators said Jeffers was driving the Ford Taurus that struck and killed former Millard West teacher Jim Johnston near 260th and West Center Road that day.
After entering the plea, Judge Jeffrey L. Marcuzzo found Jeffers guilty.
Jeffers is scheduled to be sentenced in August.
Misdemeanor motor vehicle homicide carries a penalty of up to one year in prison, a $1,000 fine or both. There is no minimum sentence for this charge.
Johnston’s family was at the court hearing Wednesday.
“March wasn't a very good month for us,” said Jed Johnston, his brother. “We are trying to hold it together and move forward, but these things bring little reminders back up and it does hurt.”
The ghost bike memorial still stands on West Center Road, a visible reminder for the Johnston family that healing is a long way off.
“There is no closure. Jim was out on a wonderful day doing what he loved, riding where he was supposed to ride, being safe, and his life was ended -- that hurts and it's going to take a long, long time for all of us to heal,” Johnston said. |
Brownies and onion dip
Pages
Sunday, February 8, 2015
This weekend was stake conference (church meetings where multiple congregations come together). Last night was the adult session. As I pulled into the parking lot, I saw multiple couples walking in holding hands. I was instantly very aware of my single status. It's unusual for me to struggle with this at all. I'm very happy with my life. Last night, though, I felt like an outsider walking into that chapel. I saw the neighborhood grandpa and his wife, and they made room for me to sit. I mentioned the single-status-reminder to him, and bless his heart, he reached out and held my hand for a second. We had a good laugh about it. After the meeting, two couples decided to go out for pie and invited me. I declined, because I needed to do more homework, but it was nice of them to make me feel included.
Lesson learned: I'm only alone if I choose to be. I have a great support system.
Today I had a headache and I had to force myself to go to stake conference, because it's always worth it. The stake president said how great it was to see everyone there with their families, and my immediate response was positive. I realized that yes, I was there as a family! Either as a family of one or with my ward family.
I heard so many things that touched my heart this weekend, both spoken words and thoughts that came as a result of being in the right place at the right time. It would have been so easy to turn around and leave last night, or to stay in bed today, but I was blessed as a result of making the right choice.
2. Spotify. I like Pandora (with Chrome blocking the ads) but I'm really starting to dig Spotify. Currently enjoying Premium access for 3 months for only 99 cents. Even better!
3. Nostril waxing. Sounds nuts, but I think many men and some women would benefit from it. My friend told me about it and I tried it a couple months ago. You barely notice any pain - definitely the easiest area I've ever had waxed. And if you're in the Salt Lake area and have any waxing needs, I love Carly at Wax Me Too.
Sunday, January 4, 2015
I had a change in my church calling (assignment) effective today. I've been serving with the ward young women for almost five years, and now I'm going to be serving as the secretary for the stake young women presidency instead. I found out last Sunday and it's been an odd week, trying to ease out of my responsibilities but not knowing my new ones. Today was an emotional day for me. I shared my testimony with the girls and then went to the stake offices to get set apart. Then we had two meetings in a row. I've immediately been put to work, for which I'm grateful.
I've never before spent this long in the same calling. This is not the hardest time I've had getting released, probably because I'll still be around the girls to some degree, but it is an adjustment. I'm hopeful that I'll have more Tuesday nights available for dancing (yay!) but it will certainly be work in different ways.
I'm grateful for this new opportunity to serve and grow. I'm grateful for the gospel and the organization of the Church. I am incredibly blessed and I hope I remember to recognize my gratitude for all that I have.
Saturday, December 27, 2014
It was a fantastic Christmas season. Festival of Trees, a live nativity (with a camel!), California with family, sewing, baking, Messiah sing-along, looking at lights, ice skating, dates, helping with missionary care packages, a party with the young women at church, etc.
I've already put photos on Facebook or Instagram, and I get so focused on enjoying family that I slack at taking pictures while I'm with them. I did take some baking pictures and I'll share recipe links in a later post. Here's a blurry picture with my sister and mom.
For the last year or so, I've been asking my grandma questions in order to write her life story. On this visit, I did some of the interviewing as a video. It increasingly hits me that I don't know how much longer I'll have, and I want to remember her voice.
Saturday, November 29, 2014
I'm stressing over the end of the semester, but I'm starting to plan Christmas baking. This is my favorite source of recipes for Christmas cookies. It's the most adorable book (as all Gooseberry Patch books are) with cute ideas and cutouts for cookie exchanges.
I always completely skip over this chapter, because cut-out cookies are too much hassle for me.
Tuesday, November 25, 2014
It's November and I love posting one thing each day for which I am thankful, but I have friends on Facebook who hate it. I'm going to update this blog post every few days instead.
I'm thankful for...
1: The beauty of God's earth. There are so many amazing places to explore within a reasonable traveling distance.
2: Getting home safely even though we drove through rain and snow storms and lost traction a few times. Time to replace my tires!
3: People who let me into their lives.
4: How fluffy my carpet feels after vacuuming with the vacuum I just bought.
5: Knowing exactly where to turn for peace when dating brings disappointments.
6: Scoring 86 on my last exam! Way higher than the first two!
7: The peace and inspiration I find in the temple.
8: Beautiful fall weather.
9: Laughter. I laughed my way to tears several times today.
10: Rain on my birthday. I love it.
11: My sweet sister doing yard work for me. It's ready for winter!
12: My parents' 32nd anniversary. I'm grateful they stuck it out on this go-round.
13: Sweet friends who make me sweet birthday treats.
14: New tires and the $ to purchase them, so I can drive safely this winter.
15: The empowering feeling that comes from repairing things in my home.
16: Heat. Baby, it's cold outside. (There's ice inside my windows.)
17: The flexible schedule of my new job position. I love the reduced stress.
18: The talents and accomplishments of the lovely young (and not-so-young) women at church and their willingness to share.
19: The semester being almost over! Only one more new concept to learn and then it's all review and assignments!
20: The repair to my water heater being simple and inexpensive (and my banging pipes being fixed as a side benefit).
21: The opportunity to appreciate good music with a fellow music-lover.
22: Men who act as gentlemen on dates.
23: Sisters who remain dear to my heart across the miles and through divorce.
24: Living in a city where I feel safe. (Praying for those in rioting areas.)
25: The beautiful view of lights across the valley from my dining room.
26: The healing power of the Atonement.
27: Peace.
28: The Christmas season.
29: Leaves to rake - just the right amount to keep it enjoyable!
30: The opportunity to learn by preparing a talk for church.
About Me
Yes, I like brownies with onion dip. I pretty much love dip of any kind - it makes me happy. I also love the sound of my windchimes blowing, the feel of the fairly constant breeze, and walking through crunchy leaves. I'm on a bread hiatus right now, but if you're wondering why I was baking so much, read this. |
//
// Copyright 2011-2012 Ettus Research LLC
// Copyright 2018 Ettus Research, a National Instruments Company
//
// SPDX-License-Identifier: GPL-3.0-or-later
//
#include "convert_common.hpp"
#include <uhd/utils/byteswap.hpp>
#include <emmintrin.h>
using namespace uhd::convert;
DECLARE_CONVERTER(sc16_item32_le, 1, fc32, 1, PRIORITY_SIMD)
{
const item32_t* input = reinterpret_cast<const item32_t*>(inputs[0]);
fc32_t* output = reinterpret_cast<fc32_t*>(outputs[0]);
const __m128 scalar = _mm_set_ps1(float(scale_factor) / (1 << 16));
const __m128i zeroi = _mm_setzero_si128();
// this macro converts values faster by using SSE intrinsics to convert 4 values at a time
#define convert_item32_1_to_fc32_1_nswap_guts(_al_) \
for (; i + 3 < nsamps; i += 4) { \
/* load from input */ \
__m128i tmpi = _mm_loadu_si128(reinterpret_cast<const __m128i*>(input + i)); \
\
/* unpack + swap 16-bit pairs */ \
tmpi = _mm_shufflelo_epi16(tmpi, _MM_SHUFFLE(2, 3, 0, 1)); \
tmpi = _mm_shufflehi_epi16(tmpi, _MM_SHUFFLE(2, 3, 0, 1)); \
__m128i tmpilo = _mm_unpacklo_epi16(zeroi, tmpi); /* value in upper 16 bits */ \
__m128i tmpihi = _mm_unpackhi_epi16(zeroi, tmpi); \
\
/* convert and scale */ \
__m128 tmplo = _mm_mul_ps(_mm_cvtepi32_ps(tmpilo), scalar); \
__m128 tmphi = _mm_mul_ps(_mm_cvtepi32_ps(tmpihi), scalar); \
\
/* store to output */ \
_mm_store##_al_##ps(reinterpret_cast<float*>(output + i + 0), tmplo); \
_mm_store##_al_##ps(reinterpret_cast<float*>(output + i + 2), tmphi); \
}
size_t i = 0;
// need to dispatch according to alignment for fastest conversion
switch (size_t(output) & 0xf) {
case 0x0:
// the data is 16-byte aligned, so do the fast processing of the bulk of the
// samples
convert_item32_1_to_fc32_1_nswap_guts(_) break;
case 0x8:
// the first sample is 8-byte aligned - process it to align the remainder of
// the samples to 16-bytes
item32_sc16_to_xx<uhd::htowx>(input, output, 1, scale_factor);
i++;
// do faster processing of the bulk of the samples now that we are 16-byte
// aligned
convert_item32_1_to_fc32_1_nswap_guts(_) break;
default:
// we are not 8 or 16-byte aligned, so do fast processing with the unaligned
// load and store
convert_item32_1_to_fc32_1_nswap_guts(u_)
}
// convert any remaining samples
item32_sc16_to_xx<uhd::htowx>(input + i, output + i, nsamps - i, scale_factor);
}
DECLARE_CONVERTER(sc16_item32_be, 1, fc32, 1, PRIORITY_SIMD)
{
const item32_t* input = reinterpret_cast<const item32_t*>(inputs[0]);
fc32_t* output = reinterpret_cast<fc32_t*>(outputs[0]);
const __m128 scalar = _mm_set_ps1(float(scale_factor) / (1 << 16));
const __m128i zeroi = _mm_setzero_si128();
// this macro converts values faster by using SSE intrinsics to convert 4 values at a time
#define convert_item32_1_to_fc32_1_bswap_guts(_al_) \
for (; i + 3 < nsamps; i += 4) { \
/* load from input */ \
__m128i tmpi = _mm_loadu_si128(reinterpret_cast<const __m128i*>(input + i)); \
\
/* byteswap + unpack -> byteswap 16 bit words */ \
tmpi = _mm_or_si128(_mm_srli_epi16(tmpi, 8), _mm_slli_epi16(tmpi, 8)); \
__m128i tmpilo = _mm_unpacklo_epi16(zeroi, tmpi); /* value in upper 16 bits */ \
__m128i tmpihi = _mm_unpackhi_epi16(zeroi, tmpi); \
\
/* convert and scale */ \
__m128 tmplo = _mm_mul_ps(_mm_cvtepi32_ps(tmpilo), scalar); \
__m128 tmphi = _mm_mul_ps(_mm_cvtepi32_ps(tmpihi), scalar); \
\
/* store to output */ \
_mm_store##_al_##ps(reinterpret_cast<float*>(output + i + 0), tmplo); \
_mm_store##_al_##ps(reinterpret_cast<float*>(output + i + 2), tmphi); \
}
size_t i = 0;
// need to dispatch according to alignment for fastest conversion
switch (size_t(output) & 0xf) {
case 0x0:
// the data is 16-byte aligned, so do the fast processing of the bulk of the
// samples
convert_item32_1_to_fc32_1_bswap_guts(_) break;
case 0x8:
// the first sample is 8-byte aligned - process it to align the remainder of
// the samples to 16-bytes
item32_sc16_to_xx<uhd::htonx>(input, output, 1, scale_factor);
i++;
// do faster processing of the bulk of the samples now that we are 16-byte
// aligned
convert_item32_1_to_fc32_1_bswap_guts(_) break;
default:
// we are not 8 or 16-byte aligned, so do fast processing with the unaligned
// load and store
convert_item32_1_to_fc32_1_bswap_guts(u_)
}
// convert any remaining samples
item32_sc16_to_xx<uhd::htonx>(input + i, output + i, nsamps - i, scale_factor);
}
DECLARE_CONVERTER(sc16_chdr, 1, fc32, 1, PRIORITY_SIMD)
{
const sc16_t* input = reinterpret_cast<const sc16_t*>(inputs[0]);
fc32_t* output = reinterpret_cast<fc32_t*>(outputs[0]);
const __m128 scalar = _mm_set_ps1(float(scale_factor) / (1 << 16));
const __m128i zeroi = _mm_setzero_si128();
// this macro converts values faster by using SSE intrinsics to convert 4 values at a time
#define convert_item32_1_to_fc32_1_guts(_al_) \
for (; i + 3 < nsamps; i += 4) { \
/* load from input */ \
__m128i tmpi = _mm_loadu_si128(reinterpret_cast<const __m128i*>(input + i)); \
\
/* unpack + swap 16-bit pairs */ \
__m128i tmpilo = _mm_unpacklo_epi16(zeroi, tmpi); /* value in upper 16 bits */ \
__m128i tmpihi = _mm_unpackhi_epi16(zeroi, tmpi); \
\
/* convert and scale */ \
__m128 tmplo = _mm_mul_ps(_mm_cvtepi32_ps(tmpilo), scalar); \
__m128 tmphi = _mm_mul_ps(_mm_cvtepi32_ps(tmpihi), scalar); \
\
/* store to output */ \
_mm_store##_al_##ps(reinterpret_cast<float*>(output + i + 0), tmplo); \
_mm_store##_al_##ps(reinterpret_cast<float*>(output + i + 2), tmphi); \
}
size_t i = 0;
// need to dispatch according to alignment for fastest conversion
switch (size_t(output) & 0xf) {
case 0x0:
// the data is 16-byte aligned, so do the fast processing of the bulk of the
// samples
convert_item32_1_to_fc32_1_guts(_) break;
case 0x8:
// the first sample is 8-byte aligned - process it to align the remainder of
// the samples to 16-bytes
chdr_sc16_to_xx(input, output, 1, scale_factor);
i++;
// do faster processing of the bulk of the samples now that we are 16-byte
// aligned
convert_item32_1_to_fc32_1_guts(_) break;
default:
// we are not 8 or 16-byte aligned, so do fast processing with the unaligned
// load and store
convert_item32_1_to_fc32_1_guts(u_)
}
// convert any remaining samples
chdr_sc16_to_xx(input + i, output + i, nsamps - i, scale_factor);
}
|
<!DOCTYPE html>
<html>
<head>
<meta charset=utf-8 />
<title>Overthrow Demo</title>
<meta name="viewport" content="width=device-width,initial-scale=1">
<link href="demo.css" rel="stylesheet">
<style>
.overthrow-enabled .overthrow {
-webkit-overflow-scrolling: touch;
}
.overthrow-enabled .sidescroll ul {
width: 300%;
}
@media (min-width: 60em){
.overthrow-enabled .sidescroll ul {
width: 100%;
}
}
</style>
<script src="../../../node_modules/toss/toss.js"></script>
<script src="../../../src/overthrow-detect.js"></script>
<script src="../../../src/overthrow-polyfill.js"></script>
<script src="../../../src/overthrow-toss.js"></script>
<script src="../../../src/overthrow-init.js"></script>
<script src="../../../extensions/overthrow-sidescroller.js"></script>
</head>
<body>
<p>Basic side-scroller w/arrows</p>
<div class="sidescroll-nextprev">
<div class="overthrow sidescroll">
<ul>
<li><img src="img/angkor.jpg"></li>
<li><img src="img/monkey.jpg"></li>
<li><img src="img/monks.jpg"></li>
</ul>
</div>
</div>
<p>Basic side-scroller w/arrows</p>
<div class="sidescroll-nextprev">
<div class="overthrow sidescroll sidescroll-3">
<ul>
<li><img src="img/angkor.jpg"></li>
<li><img src="img/monkey.jpg"></li>
<li><img src="img/monks.jpg"></li>
</ul>
</div>
</div>
<script>
overthrow.sidescroller( document.querySelectorAll( ".overthrow-enabled .sidescroll-nextprev" ) );
</script>
</body>
</html>
|
What They're Saying
By the Staff of Cougfan.com
11/11/2002
From CNN to the Lewiston Trib, words of praise for Ol' Wazzu -- now ranked No. 3 in the land --- aren't hard to come by.
After Washington State suffered an ugly loss at Ohio State in September, who could've guessed that in November, the Cougars may be a Buckeye loss away from the Fiesta Bowl? WSU is among a handful of team peaking at just the right time -- Stewart Mandel, CNNSI.com
This defense isn't as good as the '94 team, the offense probably not as good as the last Rose Bowl team. But put them together, covering up for each other as they did in beating back Oregon, 32-21, this might prove to be the best WSU team in history. -- Blaine Newnham, Seattle Times
So here I sit, stuffing it in the face of traditional columnist thinking, writing about the Cougs again.Why not? They have climbed to third in the rankings. They have the inside track to the Rose Bowl. They have won 19 of their past 22 games. The following statement is amazingly possible: The Cougs -- yes, the Cougs -- have a reasonable shot at playing for the national championship. And someone wants me to stop writing about them now? As Lucy used to say: I don't think so, Ricky. -- Jim Moore, Seattle P-I
The Cougars seem to thrive on hardship, whether it comes from outside or inside their control, and they've now won four times after trailing in the second half. Even their least tenable dream took a step toward plausibility: Oklahoma's loss to Texas A&M on Saturday improves WSU's long-shot chances of a national title. -- Dale Grummert, Lewiston Morning Tribune
The true believers were well beyond this one. They had already locked in their airfares and booked their rooms in Pasadena and maybe in Tempe…. No, this one was for everybody else -- those who merely want to believe, and even those who don't. This was for those either resigned or dead sure that the Cougs would concoct some calamity on themselves in this stroll to destiny because, well, it just can't be as easy as it's looked and because that's the way it always happens. -- John Blanchette, Spokesman-Review
THE DRIVE will live long in the memories of the fans and coaches and players, a determined drive of 96 yards in the fourth quarter, a drive from behind to take the lead, a drive by a team playing for a berth in the Rose Bowl and maybe something even larger. The amazing play will endure for years in the winner's highlight films -- the defensive back tipping the ball, the wide receiver following it, catching it, scoring the touchdown that puts the game away. For the past few years, that was Oregon, making that kind of drive, and that kind of play, in the game's decisive moments. Saturday afternoon in Martin Stadium, it was Washington State. The Ducks weren't bad, but the Cougars were just better. It happens. -- Ron Bellamy, Eugene Register-Guard
This was WSU's payback to Gesser. This was the day the Cougars thanked him for all those quick-release throws, all those shows of leadership through three years of being a co-captain, all those early mornings in the summer when he rustled guys out of bed to go work out. This was a day on which the WSU defense was simply extraordinary, and a running back named Jermaine Green showed why the Cougars were so agog about him last spring. All of them had Jason Gesser's back. Good thing, because for three quarters Gesser played like Chris Simms plays against Oklahoma. To be charitable, he was awful. -- Bud Withers, Seattle Times.
"He's one the toughest guys I've ever met in my life -- maybe the toughest." Legendary tough-guy Jason Gesser talking about receiver Mike Bush's ability to play through an infected molar that was so tender it hurt to put on his helmet. |
(a) Field of the Invention
The present invention relates to a transmission line, and more particularly to a transmission line with better characteristic impedance (Z0) and high flexibility.
(b) Description of the Related Art
With reference to FIG. 1 for transm+itting a LVDS signal, the signal communication between a liquid crystal display (LCD) and a host system involves a huge volume and a very high frequency, and thus the high frequency signal transmission established between an LCD interface 6 and a system motherboard interface 7 adopts a low voltage differential signal (LVDS) transceiver 9 with a super high speed (1.4 Gb/s), a low power consumption and a low electromagnetic interference (EMI) as the signal transmission interface of the LCD interface 6, and a signal transmission line 9 is provided for connecting the signal transmission interface of the system motherboard interface 7 (which is a connector socket 71 of the system motherboard interface 7) for a conventional LVDS signal transmission.
According to a LVDS interface standard defined by ANSI-YUA-EIA-644-I995, a signal transmission line 9 for a LVDS signal transmission must be a signal transmission line 9 with a characteristic impedance (Z0) equal to 100Ω±5% before the impedance (Z) of a circuit between the LCD interface 6 and the system motherboard interface 7 can be matched, and the LVDS signal transmission must satisfy this condition to achieve the effects of reducing the electromagnetic inference and noises, correctly executing the signal transmission between the LCD interface (or LVDS interface) 6 and the system motherboard interface 7 and preventing errors. If the aforementioned condition is not satisfied, signal reflections, noises, data loses, deformations or distortions may occur in signal transmissions between the LCD interface 6 and the system motherboard interface 7.
With reference to FIG. 2A for a schematic view of a conventional signal transmission line, a thicker insulating layer 92 and a metal layer 93 are attached sequentially on a surface of a flexible flat cable 91. With reference to FIG. 2B, the thickness of a plastic film layer 911 of a flexible flat cable 91 is increased to improve the insulating thickness of the flexible flat cable 91 for producing a compliant characteristic impedance (Z0). Regardless of increasing the thickness of the plastic film layer 911 or adding the insulating layer 92, a specific thickness of a poly (ethylene terephthalate (PET) material is required for complying with the requirements of the characteristic impedance (Z0), and its hardness will be relatively higher, and thus the flexible flat cable 91 complies with the required characteristic impedance (Z0) but the flexibility becomes lower, and the operation of the whole signal transmission line 9 becomes less flexible. |
[**On the density and temperature of neutron-rich systems at the energy of vanishing flow in heavy-ion collisions**]{}\
[^1]\
[*Department of Physics, Panjab University, Chandigarh -160 014, India.\
*]{}
We study nuclear dynamics at the the energy of vanishing flow for neutron-rich systems. In particular, we shall study the collision rate, density and temperature reached in a heavy-ion reaction with neutron-rich systems. We shall also study the mass dependence of these quantities. Our results indicate nearly mass independent nature for the density reached whereas a significant mass dependence exists for the temperature of neutron-rich systems.
Introduction
============
The collective transverse in-plane flow has been used extensively over the past three decades to study the properties of hot and dense nuclear matter., i.e., the nuclear matter equation of state (EOS) and in-medium nucleon-nucleon cross section [@sch]. It has been reported to be highly sensitive to the above mentioned properties and also to the entrance channel parameters like incident energy, colliding geometry and system size [@ogli89; @andro03; @luka05; @zhang06; @luka08]. The energy dependence of flow led to its disappearance at a particular incident energy called energy of vanishing flow (EVF) or balance energy (E$_{bal}$) [@krof89]. A large number of theoretical studies have been carried out in the past studying the sensitivity of E$_{bal}$ to the system size and colliding geometry [@mag00; @sood1; @mag100].
Role of isospin degree of freedom in collective transverse in-plane flow and its disappearance has also been a matter of great interest for the past decade [@pak97; @li]. Isospin degree of freedom plays its role in determining the nuclear equation of state of asymmetric nuclear matter. The availability of radioactive ion beams (RIBs) [@rib1; @rib2] around the world helps in carrying out the studies on the matter lying far away from the stability line. A number of studies have been carried out in the recent past to see the role of isospin degree of freedom in collective flow and its disappearance [@pak97; @li; @gaum1]. In Ref. [@gaum2] author and others studied the isospin effects in E$_{bal}$ at all the colliding geometries. A very few studies have been carried out to study other related phenomena at E$_{bal}$ of the neutron-rich systems. In Ref. [@sood2] Sood and Puri have presented a complete study of the nuclear dynamics at E$_{bal}$ for stable systems. The study includes participant-spectator matter, density and temperature reached in a heavy-ion reaction at E$_{bal}$. Motivated by this, author and others presented a study of participant-spectator matter of neutron-rich systems at E$_{bal}$ in Ref. [@gaum3]. The study revealed a similar behaviour of participant-spectator matter for neutron-rich systems as for the stable systems. Another important quantity which reflects the dynamics in a heavy-ion collision is the density and temperature reached in a reaction. In the present paper, we study the density and temperature reached in heavy-ion reactions of neutron-rich matter at E$_{bal}$. We also aim to see the role of isospin degree of freedom in the density and temperature reached in the reactions of neutron-rich systems and to see if the behaviour of these properties at balance energy differs from that for systems lie close to the stability line.
The present study is carried out within the framework of isospin-dependent quantum molecular dynamics (IQMD) model [@hart98]. Section 2 describes the model in brief. Section 3 explains the results and gives our discussion, and Sec. 4 summarizes the results.
The model
=========
The IQMD model [@hart98] which is the extension of quantum molecular dynamics (QMD) [@aichqmd] model treats different charge states of nucleons, deltas, and pions explicitly, as inherited from the Vlasov-Uehling-Uhlenbeck (VUU) model. The IQMD model has been used successfully for the analysis of a large number of observables from low to relativistic energies. Puri and coworkers have demonstrated that QMD, IQMD carries essential physics needed to demonstrate the various phenomena such as collective flow, multifragmentation and particle production [@dhawan; @batko]. The isospin degree of freedom enters into the calculations via symmetry potential, cross sections, and Coulomb interaction.
In this model, baryons are represented by Gaussian-shaped density distributions
$$f_{i}(\vec{r},\vec{p},t) =
\frac{1}{\pi^{2}\hbar^{2}}\exp(-[\vec{r}-\vec{r_{i}}(t)]^{2}\frac{1}{2L})
\times \exp(-[\vec{p}- \vec{p_{i}}(t)]^{2}\frac{2L}{\hbar^{2}})$$
Nucleons are initialized in a sphere with radius R = 1.12 A$^{1/3}$ fm, in accordance with liquid-drop model. Each nucleon occupies a volume of *h$^{3}$*, so that phase space is uniformly filled. The initial momenta are randomly chosen between 0 and Fermi momentum ($\vec{p}$$_{F}$). The nucleons of the target and projectile interact by two- and three-body forces, potential, and interactions. In addition to the use of explicit charge states of all baryons and mesons, a symmetry potential between protons and neutrons corresponding to the Bethe-Weizsäcker mass formula has been included. The hadrons propagate using Hamilton equations of motion:
$$\begin{aligned}
\frac{d\vec{{r_{i}}}}{dt} = \frac{d\langle H
\rangle}{d\vec{p_{i}}};& & \frac{d\vec{p_{i}}}{dt} = -
\frac{d\langle H \rangle}{d\vec{r_{i}}}\end{aligned}$$
with
$$\begin{aligned}
\langle H\rangle& =&\langle T\rangle+\langle V \rangle
\nonumber\\
& =& \sum_{i}\frac{p^{2}_{i}}{2m_{i}} + \sum_{i}\sum_{j>i}\int
f_{i}(\vec{r},\vec{p},t)V^{\textrm{ij}}(\vec{r}~',\vec{r})
\nonumber\\
& & \times f_{j}(\vec{r}~',\vec{p}~',t) d\vec{r}~ d\vec{r}~'~
d\vec{p}~ d\vec{p}~'.\end{aligned}$$
The baryon potential V$^{\textrm{ij}}$ in the above relation, reads as
$$\begin{aligned}
\nonumber V^{\textrm{ij}}(\vec{r}~'-\vec{r})& =&V^{\textrm{ij}}_{\textrm{Skyrme}} + V^{\textrm{ij}}_{\textrm{Yukawa}} +
V^{\textrm{ij}}_{\textrm{Coul}} + V^{\textrm{ij}}_{\textrm{sym}}
\nonumber\\
& =& [t_{1}\delta(\vec{r}~'-\vec{r})+t_{2}\delta(\vec{r}~'-\vec{r})\rho^{\gamma-1}(\frac{\vec{r}~'+\vec{r}}{2})]
\nonumber\\
& & +t_{3}\frac{\exp(|(\vec{r}~'-\vec{r})|/\mu)}{(|(\vec{r}~'-\vec{r})|/\mu)}+
\frac{Z_{i}Z_{j}e^{2}}{|(\vec{r}~'-\vec{r})|}
\nonumber \\
& & +t_{4}\frac{1}{\varrho_{0}}T_{\textrm{3i}}T_{\textrm{3j}}\delta(\vec{r_{i}}~'-\vec{r_{j}}).
\end{aligned}$$
Here *Z$_{i}$* and *Z$_{j}$* denote the charges of *ith* and *jth* baryon, and *T$_{3i}$* and *T$_{3j}$* are their respective *T$_{3}$* components (i.e., $1/2$ for protons and $-1/2$ for neutrons). The parameters *$\mu$* and *t$_{1}$,...,t$_{4}$* are adjusted to the real part of the nucleonic optical potential. For the density dependence of the nucleon optical potential, standard type parametrization is employed. We use a soft equation of state along with the standard isospin- and energy-dependent cross section reduced by 20$\%$, i.e. $\sigma$ = 0.8 $\sigma_{nn}^{free}$. In a recent study, Gautam *et al*. [@gaum1] has confronted the theoretical calculations of IQMD with the data of $^{58}Ni+^{58}Ni$ and $^{58}Fe+^{58}Fe$ [@pak97]. The results with the soft EOS (along with the momentum-dependent interactions) and above choice of cross section are in good agreement with the data at all colliding geometries. The details about the elastic and inelastic cross sections for proton-proton and proton-neutron collisions can be found in [@hart98; @cug]. The cross sections for neutron-neutron collisions are assumed to be equal to the proton-proton cross sections. Two particles collide if their minimum distance *d* fulfills $$d \leq d_{0} = \sqrt{\frac{\sigma_{tot}}{\pi}}, \sigma_{tot} =
\sigma(\sqrt{s}, type),$$ where ’type’ denotes the ingoing collision partners (N-N....). Explicit Pauli blocking is also included; i.e. Pauli blocking of the neutrons and protons is treated separately. We assume that each nucleon occupies a sphere in coordinate and momentum space. This trick yields the same Pauli blocking ratio as an exact calculation of the overlap of the Gaussians will yield. We calculate the fractions P$_{1}$ and P$_{2}$ of final phase space for each of the two scattering partners that are already occupied by other nucleons with the same isospin as that of scattered ones. The collision is blocked with the probability $$P_{block} = 1-[1 - min(P_{1},1)][1 - min(P_{2},1)],$$ and, correspondingly is allowed with the probability 1 - P$_{block}$. For a nucleus in its ground state, we obtain an averaged blocking probability $\langle P_{block}\rangle$ = 0.96. Whenever an attempted collision is blocked, the scattering partners maintain the original momenta prior to scattering.
Results and discussion
======================
We simulate the reactions of Ca+Ca, Ni+Ni, Zr+Zr, Sn+Sn, and Xe+Xe series having N/Z = 1.0, 1.6 and 2.0. In particular, we simulate the reactions of $^{40}$Ca+$^{40}$Ca (105), $^{52}$Ca+$^{52}$Ca (85), $^{60}$Ca+$^{60}$Ca (73); $^{58}$Ni+$^{58}$Ni (98), $^{72}$Ni+$^{72}$Ni (82), $^{84}$Ni+$^{84}$Ni (72); $^{81}$Zr+$^{81}$Zr (86), $^{104}$Zr+$^{104}$Zr (74), $^{120}$Zr+$^{120}$Zr (67); $^{100}$Sn+$^{100}$Sn (82), $^{129}$Sn+$^{129}$Sn (72), $^{150}$Sn+$^{150}$Sn (64) and $^{110}$Xe+$^{110}$Xe (76), $^{140}$Xe+$^{140}$Xe (68) and $^{162}$Xe+$^{162}$Xe (61) at an impact parameter of b/b$_{\textrm{max}}$ = 0.2-0.4 and at the incident energies equal to balance energy. The values in the brackets represent the balance energies for the systems. The reactions are followed till the transverse in-plane flow saturates. It is worth mentioning here that the saturation time varies with the mass of the system. It has been shown in Ref. [@sood4] that the transverse in-plane flow in lighter colliding nuclei saturates earlier compared to heavy colliding nuclei. Saturation time is about 100 (150 fm/c) in lighter (heavy) colliding nuclei in the present energy domain. We use the quantity “*directed transverse momentum $\langle p_{x}^{dir}\rangle$*” to define the nuclear transverse in-plane flow, which is defined as [@hart98; @aichqmd; @leh]
$$\langle{p_{x}^{dir}}\rangle = \frac{1} {A}\sum_{i=1}^{A}{sign\{
{y(i)}\} p_{x}(i)},$$
where $y(i)$ and $p_{x}$(i) are, respectively, the rapidity (calculated in the center of mass system) and the momentum of the $i^{th}$ particle. The rapidity is defined as
$$Y(i)= \frac{1}{2}\ln\frac{{\vec{E}}(i)+{\vec{p}}_{z}(i)}
{{\vec{E}}(i)-{\vec{p}}_{z}(i)},$$
where $\vec{E}(i)$ and $\vec{p_{z}}(i)$ are, respectively, the energy and longitudinal momentum of the $i^{th}$ particle. In this definition, all the rapidity bins are taken into account.
-1cm ![(Color online) The time evolution of spectator matter (left panels) and participant matter (right panels) for systems having N/Z = 1.0, 1.6 and 2.0. Lines are explained in the text.[]{data-label="fig1"}](fig1.eps "fig:"){width="10cm"}
![ (Color online) The N/Z dependence of participant and spectator matter. Symbols are explained in the text.[]{data-label="fig2"}](fig2.eps "fig:"){width="10cm"} -0cm
0.5cm ![ The system size dependence of participant and spectator matter for different N/Z ratios. Various symbols are explained in the text.[]{data-label="fig3"}](fig3.eps "fig:"){width="8cm"} 0.5cm
In fig. 1(a), we display the time evolution of average density ($\rho^{avg}/\rho_{0}$) whereas fig. 1(b) displays the time evolution of maximum density ($\rho^{max}/\rho_{0}$) for the systems having N/Z = 1.0, i.e, we display the reactions of $^{40}$Ca+$^{40}$Ca, $^{58}$Ni+$^{58}$Ni, $^{81}$Zr+$^{81}$Zr, $^{100}$Sn+$^{100}$Sn, and $^{110}$Xe+$^{110}$Xe at energy equal to balance energy. Lines represent the different systems. Solid, dashed, dotted, dash-dotted, and short dotted lines represent the reactions of Ca+Ca, Ni+Ni, Zr+Zr, Sn+Sn, and Xe+Xe, respectively. From figure, we find that maximal value of $\rho^{avg}/\rho_{0}$ is higher for lighter systems as compared to the heavier ones. Moreover, the density profile is more extended in heavier systems indicating that the reaction finishes later in heavier systems. This is because of the fact that the heavier reaction occurs at low incident energy. Also the $\rho^{avg}/\rho_{0}$ and $\rho^{max}/\rho_{0}$ are nearly same for heavier systems but differ for lighter systems as in Ref. Further, the maximum and average densities are comparable for medium and heavy mass systems indicating that the dense matter is formed widely and uniformly in the central zone of the reaction. On the other hand, the substantial difference in two densities for the lighter colliding nuclei has been explained in Ref. [@sood2] and indicates the non-homogeneous nature of dense matter. It is worth mentioning that collective flow saturates at higher densities whereas multifragmentation occurs at sub-density zone. Other phenomena such as fusion-fission are very low density phenomena [@puril].
![ The N/Z dependence of participant and spectator matter. Symbols are explained in the text.[]{data-label="fig2"}](fig4.eps "fig:"){width="12cm"} -0cm
The quantity which reflects the density achieved in a reaction is the collision rate. In fig. 2, we display the time evolution of the collision rate for various systems having N/Z = 1.0, 1.6 and 2.0. Solid, dashed and dotted lines corresponds to systems having N/Z = 1.0, 1.6 and 2.0, respectively. From figure, we see that collision rate first increases with time, reaches maximum at around 20-40 fm/c (which is the high dense phase of the reaction) and then finally decreases and becomes constant at around 80 fm/c. We also find that the maximum value of the collision rate also increases with the system mass. Moreover, the effect of N/Z ratio on the collision rate is very less.
In fig. 3 we display the system size dependence of maximal value of the maximum ($\rho^{max}$) and average density ($\rho^{avg}$) for the systems having N/Z = 1.0, 1.6 and 2.0. We see that the maximal value of $\rho^{max}$ and $\rho^{avg}$ follows a power law behaviour proportional to A$^{\tau}$. The power law factor is 0.01$\pm$ 0.04 (-0.03$\pm$ 0.04), -0.04$\pm$ 0.01 (-0.08$\pm$ 0.02), and -0.07$\pm$ 0.008 (-0.10$\pm$ 0.02) for $\rho^{avg}$ ( $\rho^{max}$) for systems with N/Z = 1.0, 1.6 and 2.0, respectively. It shows that the dependence of maximal value of $\rho^{avg}$ and $\rho^{max}$ is very weak on the system size for all the N/Z ratios. This was also predicted in Ref.cite[sood2]{} for stable systems.
-1cm ![ The system size dependence of anisotropy ratio for various N/Z ratios.[]{data-label="fig5"}](fig5.eps "fig:"){width="8cm"}
-1cm ![The system size dependence of anisotropy ratio for various N/Z ratios.[]{data-label="fig5"}](fig6.eps "fig:"){width="12cm"}
In fig. 4 we display the N/Z dependence of maximal value of $\rho^{avg}$ and $\rho^{max}$. Solid (open) symbols display the results for $\rho^{avg}$ ( $\rho^{max}$). From figure we see that both $\rho^{avg}$ and $\rho^{avg}$ decreases slightly with N/Z of the system for all the system masses. A slight exception to this is there for the lighter mass of Ca+Ca.
The another associated quantity linked with the dense matter is the temperature. In principle, a true temperature can be defined only for a thermalized and equilibrated matter. Since in heavy-ion collisions the matter is non-equilibrated, one can not talk of “temperature”. One can, however, look in terms of the local environment only. In our present case, we follow the description of the temperature given in Refs. [@khoa921; @khoa922]. In the present case, extraction of the temperature [*T*]{} is based on the local density approximation, i.e., one deduces the temperature in a volume element surrounding the position of each particle at a given time step [@khoa921; @khoa922]. Here, we postulate that each local volume element of nuclear matter in coordinate space and time has some “temperature” defined by the diffused edge of the deformed Fermi distribution consisting of two colliding Fermi spheres, which is typical for a nonequilibrium momentum distribution in heavy-ion collisions. In this formalism (dubbed the hot Thomas-Fermi approach [@khoa921]), one determines extensive quantities like the density and kinetic energy as well as entropy with the help of momentum distributions at a given temperature. Using this formalism, we also extracted the average and maximum temperature within a central sphere of 2 fm radius as described in the case of density.\
In fig. 5 we display the maximal value of $<T^{avg}>$ and $<T^{max}>$ as a function of the composite mass of the system. From figure, we see that $<T^{avg}>$ and $<T^{max}>$ follows a power law behaviour with system mass for all the N/Z ratios. The power law factor is -0.16$\pm$ 0.06 (-0.08$\pm$ 0.05), -0.15$\pm$ 0.07 (-0.16$\pm$ 0.06), and -0.19$\pm$ 0.09 (-0.15$\pm$ 0.07) for for $<T^{avg}>$ ($<T^{max}>$) for systems having N/Z = 1.0, 1.6 and 2.0, respectively. Similar power law behaviour was also predicted in Ref. [@sood2] for stable systems (N/Z $\simeq$ 1). This system size dependence of temperature is sharp in contrast with the density. This is because of the fact that the temperature depends on the kinetic energy of the system.
In fig. 6 we display the N/Z dependence of maximal value of $<T^{avg}>$ and $<T^{max}>$ for various system masses. Solid (open) symbols represent the maximal value of $<T^{avg}>$ ($<T^{max}>$). From figure, we see that for all the system masses, $<T^{avg}>$ and $<T^{max}>$ decreases with N/Z of the system.
Summary
=======
We studied the collision rate, density and temperature reached in reactions of neutron-rich systems at energy of vanishing flow. Our results pointed the similar behvaiour for neutron-rich systems as for the stable systems. We also investigated the mass dependence of these quantities. We found a very weak mass dependence of the density although the temperature follows a significant mass dependence.
This work has been supported by a grant from Centre of Scientific and Industrial Research (CSIR), Govt. of India. Author is thankful to Profs. J. Aichelin and R. K. Puri for enlightening discussions on the present work.
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[^1]: Email: sakshigautm@gmail.com
|
1. Field of the Invention
The present invention relates to an improved pulse width modulation technique. The invention is particularly advantageous when applied in a method or system for the efficient generation of a switched power supply. The invention finds particular advantageous application in mobile communication handsets, although the invention is not limited to such an application.
2. Description of the Related Art
It is known in the art that efficient power supply generation requires some form of switching means rather than a linear regulator providing a constant voltage. Switched mode power supplies are well-known in the art. Switched mode power supplies have, however, been typically used for static or slowly varying DC voltages. In such applications, the main consideration for the transient response is to respond to rapid changes in the load current rather than rapid changes in output voltage.
With recent advances in radio communications, and more particularly in mobile telecommunications, efficient operation of transmitters is required for applications in which there are rapidly varying output amplitudes. It is known in the art to address the need for such efficient operation using two techniques. A first technique is known as envelope tracking (ET), and a second technique is known as envelope elimination and restoration (EER). These techniques have required the development of a power supply requiring the efficient provision of a wide output range where the output voltage is expected to show rapid and large signal variations, in addition to variations in the load current.
A typical switch mode implementation is to include a highly non-linear switching block inside a feedback loop, such that the output is effectively linearised by the feedback for signals at a frequency much less than the switching frequency. There are three commonly used methodologies for switch mode implementations: i) Pulse Width Modulation (PWM); ii) Hysteretic (also known as Bang Bang); and iii) Delta Sigma.
A typical PWM technique comprises providing a ramp that is initiated at the start of each clock cycle. At the start of each clock cycle the switch output is set to be “1”. When the ramp crosses the signal, the signal output switches to “0”. This means that over a single cycle, the average of the output signal is equal to the control inputs, resulting in quasi-linear operation that provides inherently accurate tracking. Conventional PWM algorithms begin to fail when there is a significant change of signal level within one ramp cycle. When the signal changes rapidly, the quasi-linearity is lost and the function becomes a unit level quantiser with delay. This increased delay can result in large signal limit cycle oscillations occurring with wide bandwidth loops.
One problem of prior art techniques is to provide good switching accuracy at large bandwidths. One solution is to increase the switching rate. This means that switcher states are updated at a faster rate and hence are more able to follow fast changing signals.
However when the switched mode supply is implemented using CMOS technology, the switching elements have to be made large enough to pass the required current. Energy is consumed from the supply to switch the large bank of transistors, and this energy is proportional to switcher update rate. A more complicated controller is therefore justified if it increases the number of small geometry devices while simultaneously reducing the number of switching rates at the large transistor banks.
With reference to FIG. 1a, there is illustrated an example output achieved for an exemplary input in a known PWM arrangement. FIG. 1a illustrates the input and output signals over two cycles, cycle 1 between time t0 and t1, and cycle 2 between time t1 and t2. The time instances t0, t1, t2 represent time instances at which there are successive rising edges of a clock signal. FIG. 1a illustrates a simple example in which the output voltage level can switch between one of two levels, V0 and V1. V0 may be a ground level, and V1 may be a positive voltage level. The input signal waveform is identified by reference number 1012. The output signal is identified by reference numeral 1002.
As illustrated in FIG. 1a, in the first cycle there is illustrated a rising ramp 1004, which rises from voltage V0 at time t0 to voltage V1 at time t1. Similarly in cycle 2 there is a rising ramp 1006 which raises from voltage V0 at time t1 to the voltage V1 at time t2. The output starts at voltage V1, and also starts each subsequent cycle at voltage V1.
With further reference to FIG. 1a, when the rising ramp signal 1004 crosses the falling input signal 1012, as denoted by reference numeral 1008, the output 1002 is transitioned to the voltage level V0. The output signal 1002 remains at voltage V0 until the end of the cycle at time t1, at which time the output is forced to voltage V1 for the start of the second cycle. In the second cycle when the rising ramp at 1006 crosses the falling input signal 1012, as indicated by reference numeral 1010, a transition to voltage V0 again occurs. The operation continues in this way.
As can be seen from FIG. 1a, the output signal 1002 is a poor representation of the input signal 1012. A problem in achieving an accurate representation of the original signal is that the PWM process can very accurately replicate the time instant of the falling edge of the waveform, but cannot replicate the rising edge of the waveform at all. The only way in which a rising edge can be carried forward is by setting the voltage to a fixed level, V1, at the beginning of each cycle.
With reference to FIG. 1b, it can be seen that when applying the conventional PWM technique to a multi-level buck converter, the accuracy for fast transitions degrades further.
With reference to FIG. 1b, there is again shown the input signal and output signal over a two cycle period from time t0 to time t2. In this multi-level example, the output signal can switch between one of five voltage levels: V0, V1, V2, V3, V4. The input signal waveform is denoted by reference numeral 2012, and the output signal is denoted by reference numeral 2002.
In this example, at the beginning of the first cycle the output voltage is at voltage level V1.
As shown in FIG. 1b, there is a ramp in each cycle for each of the voltage bands (five voltages providing four voltage bands). Thus in each cycle, for each band, the ramp starts at the lower voltage of the band, and terminates at the end of the cycle at the higher voltage of the band.
Immediately at the start of the cycle, a rising ramp 2011, being the ramp associated with the first band (between voltage levels V0 and V1) of the first cycle crosses the falling input signal 2012, and thus the output signal immediately transitions to voltage V0.
If the input signal is rising faster than the ramp, then quasi-linear PWM operation is not possible. This limit is reached when a slew rate of one full transition in (n−1) clock cycles, where n is the number of voltage levels. Therefore one full cycle of signal cannot be processed in less than pi*(n−1) clock cycles. For four voltage levels, the maximum bandwidth is one tenth of the clock rate.
At time t1 the voltage transitions to the highest voltage level V4 as this is the voltage level which most closely approximates the input signal waveform at that time. Thereafter the falling input signal 2012 crosses the rising ramp between voltage levels V3 and V4 of the second cycle at a given time, as denoted by reference numeral 2010, and at that time the output voltage is transitioned to the lower voltage level V3.
In a typical prior art PWM technique only a single transition is permitted in a given cycle, and thus once the transition from V4 to V3 takes place, the output signal remains at level V3 for the remainder of the cycle.
As can be seen from FIG. 1b, as there is no quasi-linear operation a substantial error is incurred.
The above described PWM technique incorporates the conventional method of obtaining PWM in which the ramp rises across the clock cycle, and the output signal is at an upper voltage level at the start of each cycle. This reproduces only falling edges.
The inverse to the conventional method for obtaining a PWM signal is that in which the ramp falls across the clock cycle, and the output waveform at the start of a cycle is set to the lower voltage level. This reproduces only rising edges.
It is an aim of the present invention to provide an improved switch mode voltage supply in which an accurate tracking of an input signal is provided. |
Boehner Can’t Get Majority of GOP to Back Deal
American Majority Action, the conservative group that’s leading the charge against House Speaker John Boehner continuing in his current position, said Monday evening that Boehner doesn’t have at least 50 percent of his House GOP members’ support for the fiscal cliff deal that Senate leaders and the White House have reportedly cut.
The deal, between Senate Minority Leader Mitch McConnell and Vice President Joe Biden, reportedly raises spending and hikes taxes on all Americans making more than $450,000. According to AMA, it raises spending by delaying sequester budget cuts.
In an email to reporters late Monday, AMA spokesman Ron Meyer said he’s “heard directly from senior GOP conservative members in the House that Speaker Boehner does not have a majority of support from the GOP caucus–not even close.” |
Bopindolol and atenolol in patients with stable angina pectoris. Double-blind randomized comparative trial.
32 patients with stable angina and a positive symptom-limited exercise test (SLET) were investigated in a double-blind randomized trial in order to assess the therapeutic efficacy of bopindolol or atenolol on the incidence of angina pectoris and on angina pectoris threshold heart rate (ATHR). After a washout and placebo period of 2 weeks, each of the two drugs were applied in dosages of 0.5 mg bopindolol and 50 mg atenolol. The dosage was increased every month up to 2 mg bopindolol and 200 mg atenolol. At the end of every period, the patients were retested by SLET. After 3 months of active treatment, we noticed that the incidence of anginal attacks was lower in the bopindolol group (2.45 vs. 3.29). The resting heart rate was also lower in the bopindolol group (55.89 vs. 63.38). No statistical significance was found between the peak work rate, ATHR, exercise duration and S-T depression. The rate-pressure product was lower in the bopindolol group. We concluded from this that bopindolol and atenolol are active in decreasing the incidence of angina, the former being more effective. Exercise performance and cardiocirculatory parameters did not differ between the two groups. |
Moslehi made the announcement Sunday at an event marking the first anniversary of the death of 35-year-old physicist Daryoush Rezaie.
In addition to the suspected "terrorists" who shot and killed Rezaie outside his Tehran home on July 23, 2011, "two groups in charge of training terrorists were arrested inside and outside Iran," Moslehi said, according to the semiofficial Fars news agency. Moslehi did not provide details.
Rezaie taught at Mohaqeq Ardebili Technical College, in the northwestern city of Ardebil, according to the semiofficial Mehr news agency.
His death followed bombing attacks on two other Iranian physicists in November 2010, Iranian media said. Professor Majid Shahriari was killed, but professor Fereydoun Abbasi and his wife escaped with only minor wounds. Shahriari's wife and driver were injured. Both professors were on the faculty of Tehran's Shahid Beheshti University. Assailants on motorcycles attached bombs to cars carrying the professors at separate locations.
In January 2010, Iranian elementary-particle physicist Massoud Ali-Mohammadi was also killed in a bomb attack, according to Mehr. Ali-Mohammadi, 50, was riding his motorcycle near his Tehran home when a remote-controlled bomb planted on the bike detonated, according to Iranian media.
Majid Jamali Fashi was convicted in that killing and also found guilty of spying. Prosecutors accused him of working for the Mossad, Israel's national intelligence agency, and said Israel paid him $120,000 to carry out the hit. He was hanged in May.
Iran has blamed spy agencies in Israel, the United States and the United Kingdom for the attacks. |
package com.refactor.repository;
import com.refactor.repository.model.Book;
import java.util.List;
import io.reactivex.Single;
public interface BooksRepository {
Single<List<Book>> getBooks();
}
|
Review: Vitality
Daily Greens Vitality is a cold-pressed blend of kale, jalapeño, pineapple, cucumber, cilantro, lime and Himalayan pink salt. It definitely skews towards the “not so sweet” end of the spectrum of green juices. At this stage of the cold-pressed juice category, seems like a smart place to be — at least for green juice. In terms of flavor, the first things that we taste are the kale, celery, cilantro, and a little bit of heat from the jalapeno. While you can taste the pineapple, we honestly wish that there was just a tiny bit more of it as this would help balance out the taste (the product has only 18 g of sugar per bottle, so there's probably room for a little more). Lastly, there's the Himalayan sea salt, which gives the product its slightly salty finish and helps cut the vegetable flavor just a little bit. All in all, it's a nice tasting drink, and we really do appreciate the layers of flavor that the company has managed to create. On the packaging front, the product uses a 16 oz. round plastic bottle. It's the same one that is quickly becoming synonymous with high pressure processed juices. As far as the branding goes, the name Daily Greens is a good description of what the product is all about. However, it feels a little bit lacking in the personality department, despite the use of the word Vitality along the bottom. We only point this out in that we fear that a product like this could quickly become commoditized against more strongly branded competition. Overall, we're definitely fans what the company put inside the bottle, but we feel as though the opportunity for this product might be a little bit bigger if the branding were just a little more appealing. |
Dancing With The Stars Updates GOOD NEWS / BAD NEWS Derek Hough: Derek may be getting ready to do a stint on Broadway, so we don’t know whether he’ll be a pro on the show during spring season. Julianne Hough: According to what I’ve heard, Julianne will NOT return as a judge with her DWTS family each week this season. She’s getting married pretty soon, and she has many other projects in the works. Len Goodman will return from England to reclaim his seat at the judges’ table. It’s possible that she may return for a cameo now and then. Meanwhile, look for her to cover the March issue of “Redbook.” I think she was on February’s “Cosmopolitan.” “Grease”: It’s being said that the “Grease Live” soundtrack is soaring on iTunes and the DVD will be available March 8 at Amazon and Best Buy. You can preorder the DVD. Noah Galloway: The talented and inspiring Army veteran, who reached third place in the finals of season 20 with Sharna Burgess, has been visiting and lifting the spirits of hospitalized Army vets. What a great guy! “Dancing Pros Live”: ATTENTION, Niagara Falls residents and visitors! At the Fallsview Casino Resort, February 18-25, you can see stars Edyta Sliwinska, Cheryl Burke, Gleb Savchenko, and Elena Samodanova, along with dancers from all over the world. The show comes highly recommended. “Shall We Dance on Ice” will show again on Saturday, February 27 on ABC, starring pros Chelsie Hightower, Dmitry Chaplin, Anna Trebunskaya, and Damian Whitewood, accompanied by former winners Rumer Willis and Meryl Davis with former contestant Charlie White. Maksim (Maks) and Valentin (Val) Chmerkovskiy: “Our Way!” This summer the brothers will present the story they’ve always wanted to tell about their lives, highlighting their journey and rise through the world of dance. Although no other names are available yet, the show will include many other dancers, likely many from their DWTS family, and be staged by some of the top choreographers. Without a doubt it will wow us with the creativity, elegance, passion, and seduction for which the brothers are well known. And, don’t forget the unparalleled skill and professionalism that we always see. Watch for “Our Way!” to be scheduled at a theater near you and buy tickets early! Which Celebs Will Dance in Spring Season? The rumors are swirling! One says that Jodie Sweetin definitely will dance this season while another says it’s a maybe. Another says that Scott Disick is in...
NEWS: http://www.people.com/article/dancing-stars-val-chmerkovskiy-sued-posting-meme-mocking-girl “Val Chmerkovskiy Sued for Allegedly Posting Meme Mocking Girl with Down Syndrome Posted: 01 Feb 2016 09:46 PM PST People is reporting that a family is suing Dancing With The Stars pro Val Chmerkovskiy and CBS. Below is more detail from the People article… A family is suing Val Chmerkovskiy for allegedly sharing a meme that mocks their daughter, who has Down syndrome. According to the lawsuit obtained by PEOPLE, Chmerkovskiy shared a meme of the girl, which includes a photo of her drinking a soda along with the text, “Letting your kid become obese should be considered child abuse.” The lawsuit claims the photo was taken in June 2008 while the girl was at a baseball game without the family’s knowledge. The photographer allegedly posted the photo soon after on his Flickr account with the text, “Everything that’s wrong with America.” The photo eventually became a meme in 2014, and the lawsuit claims that the Dancing with the Stars pro, 29, helped it go viral when he allegedly shared it on his Facebook page in January. The lawsuit says the meme also appeared on cbsnews.com, and is naming CBS as a defendant, as well as the original photographer. “Although we have not yet been served a copy of the complaint, we have removed the photo in question from a gallery that was published in 2011,” a rep for CBS told PEOPLE in a statement. To read more, see People.” VICKIE’S REACTION: Val merely attempted to promote good health and did not know that the girl has Down Syndrome. Of his own accord three days later he removed the comment. He’s very much into good health for everyone, especially for children so that they’ll have a fair chance at a bright future. The girl’s family is suing the photographer, CBS, and Val for something like $6 million each, according to what I heard, and something was said about the possibility of settling out of court for at least a few hundred thousand dollars. Seriously? How ridiculous is all that? If anyone could really be somewhat at fault here, it might be the photographer—but not to such an exaggerated degree. This is just another frivolous lawsuit. I suggest that the girl’s parents and their attorney(s) should step back and try to take an objective view of how their actions are representing them; the view from out here is that they are ridiculous, unreasonable, and disgustingly greedy—off the charts. I also recommend that they should remember and invoke the...
Dancing With The Stars Live! “Dance All Night Tour” This posting covers news from mainly December and January. My Christmas gift from my family was two tickets and VIP passes to the Live! “Dance All Night” tour performance in Cincinnati for last Sunday, January 24. OMG! I was totally surprised. How wonderful! How exciting! How lucky I am! On the tour: Pros: Val, Peta, Keo, Sharna, Lindsay, Emma. Troupe: Brittany Cherry, Jenna Johnson, Alan Bersten, Paul Barris, Julz Tocker Special Guest Star: Alek Skarlatos Dances Star and are Choreographed by: The Pros, Alan, Jenna, and Brittany Cherry VIP ROOM, BEFORE THE SHOW: A repeating video ran of the dancers rehearsing, teaching certain dance moves to viewers, informing us of other tidbits, etc. On the video Emma says that with partners, whether young or much older, it’s all about retaining what she’s teaching and being willing to put yourself out there. To me, Emma and Sharna are always especially engaging. Jenna Johnson: Our first photo opp that day. VIPs’ photos, whether alone or with Jenna, were taken in front of a green wall and then superimposed onto a DWTS set. “I’m Jenna,” she said, offering her hand for a shake. I said, “We know who YOU are, sweetie. I’m Vickie and this is my husband, Bill.” Very sweet and cute. Live! dance captain. Comes across as being confident and strong. This is her second tour. Alan Bersten: Our second photo opp. Walked around the room in loose pants, tennis shoes, a baseball cap turned backwards, and a sleeveless shirt that revealed bulging biceps. Comfortable in any setting and genuinely sweet, like “the boy next door.” Spent two hours in the VIP room talking with fans, taking countless photos with fans (adults and children) and with Jenna in regular and dance-style poses. Was interested in my book, “Dancing in the Stars: Carroll Webster, International Dance Luminary of Vaudeville & Hollywood,” which I was autographing for the dancers, and said he wants to read it. He’s had two seasons in the pro troupe, and this is his second time on tour. For our third photo opp in the VIP area, we were told that photos would be available with four of the pros and that we could choose to have a photo with one or any number of them. My hubby and I agreed that we should be photo’d with all of them. When our turn came I left two copies of my book on the designated table just outside...
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About Me
Professionally, after decades of employment for others, Vickie Weaver founded Writestyle in 1996 to provide writing, editing, proofreading, training, and more to clients worldwide. She has written and/or edited for “Coexistence Magazine” (national), “Ohio Magazine,” and various newspapers. In addition, she has edited or contributed to the writing of numerous books. Personally, she has won awards and publication for some of her poems and is compiling a book of her poetry. She enjoys music and dance, and strives for beauty and harmony in life. |
This invention relates generally to devices for detecting sound pressure waves in water and, in particular, to large scale passive sonar arrays capable of detecting weak sound waves.
To detect weak coherent sound signals in an ocean environment, a sound detecting apparatus must overcome many obstacles a few of which are ambient noise, attenuation of sound as a function of frequency and array gain.
Ambient noise in the ocean comes from many sources including biological noise such as whales and other underwater animals, miscellaneous ship noises such as gear and propeller noises, wind generated wave noise, and storm generated (rain) noise.
This ambient noise level tends to mask the noise from a particular sound source.
For long range (over 50 miles) sonar detection of a man-made sound signal in an ocean environment, almost 100% of the sound propagating in the vertical direction is ambient noise while 90-100% of the sound propagating in the horizontal direction is man-made sound signal. In other words, the main long range signal path of man-made sound in an ocean environment is in a generally horizontal plane.
Long range sound propagation in the ocean depends upon many factors such as sound frequency, ocean depth, salinity water temperature, surface waves and nature of underwater terrain.
In deep water, in temperate latitudes where there is a temperature gradient from warm surface water to deeper colder water, sound velocity will gradually decrease with depth up to a particular depth below which sound will gradually increase in velocity due to steady increase in water pressure. In the region where sound velocity reaches a minimum, underwater sound signals will be refracted back and forth across this region and can travel great distances. Depending upon the temperature gradient of the water, latitude and season, this minimum sound velocity region, defining generally horizontal plane or "sound channel", can vary in depth from 0 to 3,000 feet.
Where the sound source is located on the surface of the water or below the surface but above the sound channel, this phenomenon can produce what are known as "shadow zones" in the region above the sound channel measuring 20 to 50 miles in width where the sound intensity of the source will be very weak or non-existent. Between these shadow zones are "convergence zones" where the sound signals tend to concentrate and are more readily detected.
To avoid these shadow zones, sonar detectors should be placed at or near the convergence zone or submerged to the sound channel depth. Few prior art sonar devices used apparatus for controlling the depth of a towed detector. Most prior art sonar devices were either placed or towed on the surface or in upper several 100 feet of the ocean or placed on the ocean floor.
In relatively shallow water, that is, water shallower than the depth of the sound channel, the sound pressure wave can propagate over relatively long distances by multiple reflections from the water surface and ocean bottom. In this situation the ocean acts in the manner of a waveguide permitting long wavelength sound waves to be attenuated less than shorter wave length sound waves.
In the process of being reflected, the sound wave is further distorted and modified depending upon the size and period of surface waves and the topography of the ocean bottom.
Propagation of sound in the ocean is also very frequency dependent. For frequencies below approximately 100 Hz, sound energy is attenuated, for the most part, by losses due to reflection from the ocean surface and bottom. This would be represented by an absorption coefficient of about 0.001 to 0.003 dB per kiloyard.
For frequencies ranging from 100 Hz to about 10 KHz, sound energy is attenuated primarily by B(OH).sub.3 in the water. This would be represented by an absorption coefficient of about 0.001 to 1.0 dB per kiloyard. Above 10 KHz sound energy is principally attenuated by MgSO.sub.4. This would be represented by an absorption coefficient of about 0.001 to 1.0 dB per kiloyard at 10 KHz to 200 dB per kiloyard at about 500 KHz.
Thus, the transmission losses for an unknown sound source will be lowest for frequencies of about 1,000 Hz and below. This would be represented by an absorption coefficient of about 0.10 dB per kiloyard down to about 0.001 dB per kiloyard or less.
Because most man-made sounds in the ocean fall below 2,000 Hz, the frequency range of most underwater sound detectors or hydrophones fall between 600 Hz and 2,000 Hz with the most used range being 600 Hz to 1,200 Hz.
Since ambient noise and sound propagation in the ocean are an uncontrollable parameters, the sonar devices of the prior art have all attempted to increase array gain by various techniques including increasing the length or size of a linear array of hydrophones, design circuits that analyze the incoming signal by frequency domain techniques using Fourier analysis, or use electronic circuits to discriminate the direction of the incoming signal by various beam, aperture or receiving lobe forming techniques.
Nearly all of the prior art signal processing circuits used analog circuits and methods to increase the signal-to-noise ratio based on the RMS value of the hydrophone output signal.
Although increasing the number of hydrophones will increase hydrophone array signal-to-noise ratio, there is, however, a practical limit as to how many hydrophones can be used both as to cost and ability of the electronic circuits to control beam direction and signal processing.
Since control of beam or aperture lobe direction for receiving an incoming signal is important for increasing the signal-to-noise ratio, most prior art sonar arrays have used this technique.
The primary technique for discriminating signal direction is to measure the time delay of the pressure wave between hydrophones. Most of these circuits measure such time delay in real time, as opposed to initially storing the time delay data for later processing.
For sound pressure waves arriving perpendicular, 90 degrees, to the array, there will be no time delay detected between hydrophones for a linear array.
For sound pressure waves arriving at an angle between 0 and 90 degrees or between 90 and 180 degrees to the array, there will be a measurable time delay between adjacent hydrophones, the maximum delay being at 0 and 180 degrees.
One prior art apparatus utilized a towed linear array of equally spaced hydrophones several thousand feet long in order to increase array gain according to the theory that array gain is equal to 10 times the log of the number of hydrophones used.
This array was generally towed behind a ship. In some cases the linear array was provided with a means for controlling its depth but not necessarily its horizontal attitude.
The problem with a linear array is its flexibility and inability to always deploy into a straight line, particularly when the towing vessel makes a turning maneuver.
If a time delay method of determining signal direction is used, the ability to discriminate direction is substantially reduced while the array is in the curved condition.
Since the ability to discriminate direction greatly increases the signal-to-noise ratio of the array, such a maneuver substantially reduces the effectiveness of such a linear configuration.
In addition, for such a linear array, there is an ambiguity as to which side of the array the signal is coming from. The pressure wave front arriving from a sound source 45 degrees to the right in front of the array will produce a received signal identical to a sound source 45 degrees to the left in front of the array.
The only way for such a linear array to discriminate as to which side the signal is coming from is to use individual hydrophones sensitive to sound pressure waves only in one hemisphere.
Furthermore, many of the prior art sonar array systems are adapted to simultaneously receive signals propagating along the vertical as well as the horizontal direction. Because of the very high noise level in the vertical direction produced during a storm by the rain striking the surface of the water, sonar systems adapted to detect sound waves without distinguishing between vertical and horizontal direction are virtually useless during even a light rain storm at sea.
Because a vertically disposed linear array of hydrophones cannot be towed horizontally and maintain its linearity, such an array configuration would not be practical as a mobile sonar listening platform. |
Category: Solar Lights
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Q:
Why doesn't this windowed expression result in a divide by zero error?
I came across this answer on Programming Puzzles & Code Golf. In it, the author uses the expression (though the answer has since been edited to use a different solution):
row_number()over(order by 1/0)
I would have expected the 1/0 to result in a divide by zero exception, but it doesn't.
When I asked the author on PPCG, they replied "because 1/0 is not being calculated. where exists(select 1/0) will have the same effect". This leaves me a bit nonplussed, because where exists(select 1) is valid syntax, but row_number()over(order by 1) is not. So why is the expression in the order by not calculated? The result of the expression is an integer, and an integer is not allowed in the order by. How does SQL Server handle the order by in that case? I assume the effect is the same as row_number()over(order by (SELECT NULL)), but if we give an expression, I'd expect that expression to be evaluated.
Coincidentally, if one uses something like:
SELECT ROW_NUMBER() OVER ( ORDER BY A.x )
FROM (
SELECT *
, 1 / 0 x
FROM master..spt_values
) A
Again, no divide by zero error is reported (when the x column is not selected, naturally). So why is this allowed when an integer is not?
A:
You're not allowed integer literals in this ORDER BY but you are allowed expressions returning integers - otherwise you wouldn't be able to use e.g. CASE expressions.
So that's why you're not allowed OVER (ORDER BY 1) but are apparently allowed OVER (ORDER BY 1/0). I don't think it's a deliberate feature to allow you to write this. You're certainly not allowed a divide by zero if your expression is actually dependent on any columns from the rows - this generates an error:
select name,object_id,ROW_NUMBER() OVER (ORDER BY 1/(object_id-3))
from sys.objects
So, I'd put it down to "the optimizer is smart enough to realise that this expression doesn't depend on any row values, therefore it's the same value for all rows so we do not need to compute it". In a sane language, this would generate a warning.
A:
Let's try a few more examples...
ROW_NUMBER() OVER (ORDER BY 2/1)
Windowed functions and NEXT VALUE FOR functions do not support integer
indices as ORDER BY clause expressions.
The problem with 2/1 is that it gets constant folded to 2 early in the optimisation process so is treated the same as ROW_NUMBER() OVER (ORDER BY 2) which is not permitted.
ROW_NUMBER() OVER (ORDER BY LOG(1))
Windowed functions and NEXT VALUE FOR functions do not support
constants as ORDER BY clause expressions.
Again constant folding kicks in - this time the result is not an integer index but SQL Server doesn't allow constants anyway.
ROW_NUMBER() OVER (ORDER BY LOG(-1))
This succeeds on recent versions of SQL Server - on old versions such as SQL Server 2008 you will see An invalid floating point operation occurred.. This specific case is mentioned in the context of CASE here. The compile time constant folding broke the semantics of CASE and this was fixed in more recent versions.
The suppression of constant folding in these error cases (LOG(-1) and 1/0) is enough for it to bypass the checks that give the error messages above. However SQL Server still recognizes that the expression is in fact a constant and can be optimized out later (so you don't get a sort operation to order the rows by the result of this expression).
During the simplification phase ROW_NUMBER ORDER BY non_folded_const_expression is simplified to just ROW_NUMBER and non_folded_const_expression removed from the tree as no longer referenced. Thus it causes no problems at runtime as it does not even exist in the final execution plan.
|
Secure APIs Deployed to Kubernetes with OAuth2 Using JWT Tokens Leased by Vault - matyix
https://banzaicloud.com/blog/oauth2-vault/
======
benmmurphy
don't people use JWT because they don't want to have some external dependency
like a database. i'm not sure what using JWT + storing authentication state in
a database gives you. couldn't it just be easier to store the authentication
state in the database and give the user a 128 bit random token pointing to the
state. if you are worried about someone compromising the DB you can also key
the lookup by cryptographic hash instead of the storing the plaintext token.
~~~
matyix
We use JWT because the user data is already stored upront (w/out needing to
check a database) and it’s a well defined storage mechanism. With Vault we
check only the token ID not the full token, so it’s not a heavyweight select.
On the other hand we have 3d party (used internally) systems using the same
mechanism (e.g. Drone).
~~~
givehimagun
But then you have to handle expiration and revocation synchronously. Is the
extra network + cpu to unmarshall a token that much of a savings over 1 redis
call on the server for the user details?
------
matyix
The Helm chart with role binding and service account support to deploy Vault
is open sourced as well.
------
zie
We do something very similar, tho we don't use JWT, just a token, and store
the userID in the KV store.
How do you handle cleaning up the vault KV store? We have a separate process
that runs around cleaning up old tokens, but am open to a better way.
~~~
ah-
Vault does the expiration/cleaning up for you.
~~~
zie
You are mistaken: "Even will a ttl set, the secrets engine never removes data
on its own. The ttl key is merely advisory." \-
[https://www.vaultproject.io/docs/secrets/kv/index.html](https://www.vaultproject.io/docs/secrets/kv/index.html)
------
justonepost
Interesting, but you could just rely on short expiry with refresh as well.
------
tty7
Nice! I'm looking at doing something similar
|
About
Translations
The Responsive Heart
The Responsive Heart
December 1st, 1985 @ 8:15 AM
Matthew 9:35-38
And Jesus went about all the cities and villages, teaching in their synagogues, and preaching the gospel of the kingdom, and healing every sickness and every disease among the people. But when he saw the multitudes, he was moved with compassion on them, because they fainted, and were scattered abroad, as sheep having no shepherd. Then saith he unto his disciples, The harvest truly is plenteous, but the labourers are few; Pray ye therefore the Lord of the harvest, that he will send forth labourers into his harvest.
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Play Audio
The title of the sermon this morning is The Responsive Heart, and the reading of the background text is in the ninth chapter of the Book of Matthew, beginning in verse 35 and reading to the end of the chapter. Matthew 9:35-38:
And Jesus went about all the cities and villages, teaching in their synagogues, and preaching the gospel of the kingdom, and healing every disease and every sickness among the people.
But when He saw the multitudes, He was moved with compassion on them, because they fainted, and were scattered abroad, as sheep having no shepherd.
Then saith He unto His disciples, The harvest truly is plenteous, but the laborers are few;
Pray ye therefore the Lord of the harvest, that He will send forth laborers into His harvest.
[Matthew 9:35-38]
Jesus speaks of the great need, then of the few who respond, and out of that came the subject: The Responsive Heart, those who do respond.
On the first Sunday in January—this coming 1986, it will be dated January 5—the pastor always delivers a message on The State of the Church. This Sunday is the first Sunday in December, and, responding to the need of the church, I have prepared a special address entitled The Responsive Heart. We shall review first some of these days that are past. And out of all of the things that have happened in my purview in these last several years, the greatest and most miraculous of all was our liberation appeal.
A few years ago, our church owed $10,500,000. We had bought the Spurgeon Harris Building, we had built the Ross Avenue parking building, and there were some other things attendant to the expansion of the facilities of our great congregation. And instead of pegging that interest, we allowed it to float with the prime, paying to the bank one point above prime. Prime reached upward to 22.5%, and that meant that we were paying 23.5% interest on a $10,500,000 loan. It was devastating and confiscatory in the life of our church.
Upon a morning, the executive committee of the fellowship of deacons, a committee comprised of the chairman of the fellowship and all of the previous vice chairmen, they called me to eat breakfast with them at the Dallas Country Club. And at that meeting, they announced to me that we had to sell one of our buildings, and the building the men had chosen was the Spurgeon Harris Building. That building, as you know, houses our lower parking building, it houses our Meridian Adult division, it houses our Singles division, it houses our First Baptist Academy secondary school. I never faced an agony of prayer like I did in those days, and it was then that I announced that we would have a liberation appeal here in our church. And we raised in that liberation appeal $2,200,000.00, which staved off the foreclosure on our property.
In those days, the Criswell Foundation bought the block immediately in front of our church: the YMCA, the street, the Baptist Building, and from the church the night school of SMU. And on that property, Lincoln Properties built that beautiful, impressive, gorgeous forty-five story office building immediately in front of our church, and in the arrangement, in the deal we made with the Lincoln Properties, they paid our debt. And the deal was made in an astonishing way: we kept our two parking buildings, the Spurgeon Harris Building and the Ross Avenue building. We not only kept the buildings, but we receive all of the revenue from those buildings. All of it was paid, the debt was paid, on account of the necessity of Lincoln Properties using, having to have our two parking buildings. It was a miracle. I still cannot believe that such a thing could come to pass: that we paid the debt, ten and a half million dollars; that we own the buildings, and that we receive the revenue from them.
Now I speak of the days that are present. The rentals from those two parking buildings are at present about $1,200,000 a year; and it goes up each year. That money, under no conditions, ever—I think—ought to be used for the support of the ministry. Our people ought to support the ministry of the church. And if the day ever comes when we refuse to support the ministry of the church, I think the church ought to be closed down. I think we ought to right “Ichabod” over it; we ought to admit to the Lord that we are unworthy to be called after the name of Christ, and let’s let somebody else bear His name and do His work. Under no conditions ever are the incomes that are brought into this church to be used for the paying of the pastor’s salary, for the staff, and for the outreach ministries of our dear people.
That $1,200,000 a year that comes to the church ought to be used for these physical facilities—not to buy light bulbs, or to sweep it out, or to clean, but for repair and for upkeep, that our five blocks of property here might be an example of excellence. When you come into the church, it doesn’t look like a dirty, neglected congregation; but when you come into the church, it is up-kept, it is beautifully kept, it is marvelously kept, it is excellently kept, and you sense that when you walk into any one of the five buildings.
There is always a great outlay to keep up any kind of a physical property, and how much more so when they cover five city blocks. For example, just recently the boiler went out in this Truett Building, and we are now constructing a new boiler. It has to be paid for. For example, recently the big compressor went out in the Chapel Building; we have to buy a new compressor. If we use the Burt Building, we must raise $500,000 for new elevators. We must also install window sills in that Burt Building. And recently we had to put a new roof on the Burt Building. Recently we had to put a new roof on the Veal Building. There is no such thing as having anything without an upkeep assignment. That’s true with your house where you live: if you don’t keep it up, it will finally be destroyed by the weather, the wind and the rain. It’s true with your car: you have to keep it up if you drive it. It’s true with your human anatomical body: you have to take care of it. That $1,200,000.00 a year income that we have ought to be used for the facilities, the upkeep, the excellence of our properties.
Then it came to pass that the Salvation Army offered us their building for sale—we own that entire block except that corner—and they asked $4,600,000.00 for that building. We did not own all the land underneath the Spurgeon Harris Building; part of it was on a lease. And in order to buy that lease, we had to spend $1,900,000.00 to own the land underneath that Spurgeon Harris Building. That comprised a total of $6,500,000 that we used to buy the Salvation Army building and the land underneath the Spurgeon Harris Building. And the choice was made that we would pay that $6,500,000 out of the income that we receive from our rental properties, besides some money left over from Lincoln Properties when they bought the block in front of us. Out of the rentals, the decision was made to pay for the Salvation Army building and to pay for the land underneath the Spurgeon building. We did not make any appeal at all. We did not bring to the people any burden of prayer. We asked no sacrifice in giving. We just took an easy vote, and the vote was that we pay for the Salvation Army building and the land under the Spurgeon Harris Building out of the rentals that we receive from our properties.
The repercussion of such an easy decision as that is found in the life of our congregation. You cannot take $1,200,000 out of a sustaining program and not feel it. You have to have a fund for facility repair, for the operation and upkeep of so great an assignment. And when we took that $1,200,000 out of our programming here in the church, we created a deficit. I don’t care who you are, if you would take $1,200,000 out of your programming and your work, you would feel it. That’s what we did, and we created in this year a current deficit, and the men said that by the end of the year, we would owe one million dollars. We would be in debt $1,000,000. That’s why I made that appeal the first Sunday in October, on the sixth day of October of this year. And in that appeal, our people brought to the house of the Lord, in cash, $1,850,000. One million and fifty thousand dollars of that was designated; $800,000 dollars of it was for our debt. But our finance committee says that despite the large offering we brought the first Sunday in October, that we are coming to the end of the year with a deficit of from $250,000.00 to $400,000.00.
Now I speak of days future. In the twenty-fourth chapter, in the last chapter of 2 Samuel, there is told the dramatic story of the Lord’s judgment upon David [2 Samuel 24:1]. The Bible doesn’t say why: was it he failed to trust in the Lord? Was it pride? However, David numbered the people, his army, took a census of the nation against the will of God [2 Samuel 24:1-9]. And the Lord confronted him, and the Lord said to him, “You have a choice of three things: either seven years of famine, or to flee before your enemies for three months, or a pestilence of three days, and you choose” [2 Samuel 24:11-13]. And David, in an agony, said, “Lord, may we not fall into the hands of men; but may we fall into the hands of the Lord” [2 Samuel 24:14]. And he chose three days pestilence. And from Dan to Beersheba, there died seventy thousand men [2 Samuel 24:15].
And David saw the destroying angel of the Lord standing above Jerusalem, with his sword drawn, and he cried unto God, “O Lord, O God!” [2 Samuel2 4:16-17]. And the Lord sent Gad the prophet to him, and said, “David, you go to Mount Moriah,” that’s where Abraham offered up Isaac [Genesis 22:1-10], “you go to mount Moriah, and there the threshing floor on top of the mount is owned by Araunah the Jebusite. And you build an altar there, and you placate the judgment of Almighty God” [2 Samuel 24:18-19]. And David came to Araunah the Jebusite, the stranger, and said to him, “God hath commanded me thus to intercede for our people, to build here an altar unto the Lord” [2 Samuel:24:20-21]. And Araunah said, “Take this mount, take this threshing floor, I give it to you. I give it to you [2 Samuel 24:22-23]. I give you these oxen for sacrifice, and I give you these implements of threshing, this wood, for wood to burn. And may the king be pleased to take it; I give it to you.” And ennobled David turned to Araunah and said, “Nay, nay, not so: neither will I offer unto the Lord my God of that which doth cost me nothing. I will buy it of thee of a price” [2 Samuel 24:24]. And David bought the mount and threshing floor, and there he built his altar [2 Samuel 24:25]; and there he told Solomon to build the temple of God [2 Chronicles 3:1].
When I look at the Salvation Army building, I think we ought to write over it, “This is an offering to God of that which doth cost us nothing!” When you look at the Spurgeon Harris Building, I think we ought to write over the front of it, “This is an offering to God of that which doth cost us nothing!” It is a rebuke from heaven itself that our people would take and assume a property, and offer nothing for it. I think it is an insult to heaven, and unworthy of our name, and I think it is a judgment of God upon us. Our people are giving more than we’ve ever given, but for us to plan a program to buy a great building, to buy the land downtown for a sustaining building, and we pay nothing for it, and the deficit we incur, and the handicap under which we labor is a judgment of God.
That brings me to the responsive heart; I am praying that this month of December our people will bring a special gift to the church. We should have done it years ago, but this month of December, we’re going to bring a special gift to the church. We’re going to ask God that we finish the year without a deficit; a special offering to the Lord. And I am praying that in 1986 there will be a set day for a building fund appeal. We owe several millions of dollars. We set a day in 1986 for a building fund appeal. For us to use the Army Building, the old Salvation Army Building, for our teenagers, we must renovate it; we must make it usable for them, and that will be a cost of something like $800,000. For us to use the Burt Building—and I think we ought to use it—it’s a small lot, and if we tear it down, you just add a few feet to that plaza. The building is built like a fort: built there to stand forever. It’s built for an office building. We can use it for offices. I asked our First Baptist Academy leadership: they would love to have it. Our Junior children are in it. They ought to be taken out of it, and for us to take our Junior children out of the Burt Building, we have to build another place for them. These things are bound up in that building fund appeal we ought to make in 1986, and if we do it, I think God will bless us.
The Christian faith is nothing, it is dust and ashes in our hands, if we take out of it tears, and blood, and sacrifice, and trial, and cost. It’s a country club and not a church. It’s a convenience and not a necessity. It’s an “at ease in Zion” [Amos 6:1], and not a devotion to the Lord God. I often think of our children: why is it that God made it that when a child is born, the child is born in travail? I think it’s because the cost of the birth of a child does something to the heart of the parents. Our freedom was bought in blood and in sacrifice. The worth and the depth of meaning of the Christian faith is found in the sacrifice and the tears and the blood that we pour into it.
We’re not to give to God that which is worthless, the leftovers and the unwanted; to tip Him. When the Lord sat over against the treasury in the temple, He watched the people as they gave. And there came by a poor widow who cast in two mites, a fourth of a cent. And the Lord condemned her because of her poor judgment? The Lord said, “Look, she has cast in everything that she has,” and the Lord commended her [Mark 12:41-44]. When Mary of Bethany broke over our Savior the alabaster box of ointment, Judas voiced the opinion of all of the apostles: “What a waste.” God said, “Not so. Not so” [Matthew 26:6-13; Mark 14:3-9; John 12:1-7].
A visiting pastor was invited by the pastor of a church to go see a man in the congregation. They called him “Old Skinflint”; they called him “Old Tight-Fisted”; they called him “Old Scrooge.” The pastor said to his friend, “We keep our giving records secret. We don’t publish them to the church. I just want you to see this man. I thought we’d have a prayer in his home.” And the reason: this man, this godly man, took all of that he possessed and gave it to God, all of it, every month, except what was necessary for him to eat and to wear in the hovel in which he lived. Everything he gave to the Lord God.
I counted dollars
While God counted crosses.
I counted gains
While He counted losses.
I counted my wealth
By the things gained in store,
But He valued me
By the scars that I bore.
I counted the honors
And sought for ease;
He wept while He counted
The hours on my knees.
And I never knew,
Until one day by a grave,
How vain are these things
We spend a life to save.
[author and work unknown]
And could I add my own conclusion:
And I never knew,
Until one day as I wept,
How empty are the things
I have selfishly kept!
The responsive heart: “Pastor, by God’s grace and with His help, I’m going to change, we’re going to change, the church is going to change; and we’re going to give our lives and everything we have to the Lord.” And then God will bless us with all of the necessities that we have need of in our lives.
So this month of December, coming to the end of the year triumphantly, we’ve been honest with God, we’ve paid our debts, and in 1986 we’re going to have a great day, and whatever we owe, we’re going to pay it. The Lord grant it to us as a church and as a people.
And we’re going to sing us a hymn now of invitation and appeal. “This is God’s day for me, pastor, and I’m giving my heart and life to the Lord Jesus [Romans 10:9-10], and I’m coming forward.” “This is my family, pastor, and we’re call coming today.” “Want to take the Lord as my Savior, and here I stand.” “Want to put my life in the fellowship of this wonderful church, and here I come.” “God’s called me, spoken to my heart in a deep way; and I’m answering with my life.” As the Spirit shall press the appeal to your soul, make the answer: “Yes, Lord, here I am, and here I stand.” On the first note of the first stanza, in the balcony round, down one of these stairways; in the press of people on this lower floor, down one of these aisles: “This is God’s day for me, pastor, and here I am.” Welcome, in the name of the Lord and of His Holy Spirit, welcome, while we stand and while we sing. |
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