On March 7, 2017, Viet Nam News published an article titled "Mekong water diplomacy vital". The article quoted several statements by Dr Lê Anh Tuấn, Deputy Director of the Research Institute for Climate Change at Cần Thơ University.
The Mekong River Commission (MRC) Secretariat takes this opportunity to respond to some of the statements in the article and highlight where water diplomacy is influencing outcomes.
The above statement was made in response to the drought in 2016, which was considered a meteorological drought year.
Construction of hydropower plants on the river does not necessarily result in the scarcity of fresh water in the lower Mekong Delta, particularly during the dry season. Conversely, hydropower plants can contribute to increased flow during the dry season as they discharge water for energy production, contributing to increased flow in the river channel. However, the flow may reduce during the wet season as the dams store surplus water.
Two information sources indicate increases in dry-season flows due to operation of dams.
In the upper Mekong basin, known as the Lancang in China, the 2016 drought event resulted in 16 per cent less flows compared to the long term average. As such, this would ordinarily translate to lower flows in the lower Mekong basin in 2016. However, because of the water stored in dams in China, emergency water releases from these dams increased dry season flows to mitigate the effects of the drought.
For instance, despite the drought conditions during the months of March – April 2016, flows in the Mekong were higher than average flows over the period 1960 – 2015. This is directly because of emergency water supplements from cascade dams from China.
A total of 12.65 billion cubic meters of water was discharged from Jinghong hydropower reservoir over the period March to May 2016. These releases amounted to between 40 – 89 per cent of flows along various sections of the Mekong river. The emergency water supplement increased water levels or discharge along the Mekong mainstream to an overall extent of 0.18-1.53m or 602-1,010m3/s.
If these emergency releases would not have happened, flows would have been 47per cent lower at Jinghong, 44 per cent lower at Chiang Saen, 38 per cent lower at Nong Khai and 22% lower at Stung Treng. This additional flow has also alleviated salinity intrusion in the Mekong Delta.
A more detailed description of the emergency operations is provided in the Joint Observation and Evaluation Technical Report produced between the Mekong River Commission and the Chinese Ministry of Water Resources. The report is available at: http://www.mrcmekong.org/assets/Publications/Final-Report-of-JOE.pdf.
At present, storage dams on the mainstream are only in the Lancang portion of the Basin. None of the Laos mainstream dams are fully constructed so are not impeding flows, and are not true storage dams (they are classified as run-of-river dams).
Since 1995, with dam development including the more recent construction of the Lancang dams, there has been an upward trend in dry-season flows. This is shown in the figure below which shows flows monitored in the northern part of the Basin at Chiang Saen gauge in Laos near the Chinese / Myanmar border, and also at Kratie gauge in Southern Cambodia in the Southern part of the Basin.
Flows at Chiang Saen show a moderate upward trend in dry season flows. This is attributable to increased dry season releases directly from Lancang dams. We see an even greater upward trend in dry-season flows at Kratie which is presumed to be primarily a result of dam releases from tributary dams in the Lower Mekong.
Conversely, in the figure below, we see a clear downward trend in wet season flows at Chiang Saen, and less pronounced at Kratie, which is likely to be a result of dam operators storing water in wet seasons and possibly an increase in extractions of wet season flows for productive purposes.
Hydropower dams may have the potential to trap sediment in the reservoir and thereby may cause some adverse effects. However, through the MRC PNPCA (Procedure for Notification, Prior Consultation, and Agreement) process, a thorough review of dam design and functionality is considered.
Through this process, we have seen, for example, developers of the Xayaburi dam modify the design to improve movement of silt and facilitate the migration of fish.
Viet Nam is a signatory to the 1995 Mekong River Agreement and has ministerial level representation on the Council and head of department level representation on the Joint Committee. Further representation is also provided through government officials within various committees and expert groups that work with the MRC Secretariat.
The MRC website provides further information on the governance and organisational structure of the Mekong River Commission showing where representation occurs.
Over the last five years, Mekong partner countries have participated in the MRC Climate Change and Adaptation initiative (CCAI) which was a direct response to the member states' call for a collaborative regional adaptation initiative to climate change and its challenges.
Climate change adaptation is one of the priority areas of work in the Basin Development Strategy 2016-2020, and the Strategic Plan 2016-2020.
The MRC is currently developing a Mekong Adaptation Strategy and Action Plan (MASAP), expected to be finalised later this year. The MASAP is a transboundary climate change adaptation strategy that will offer strategic directions in climate change adaptation work in the Lower Mekong Basin countries and support relevant national strategies and processes.
In the preparation of the MASAP, a series of basin-wide assessments of climate change impacts on water and water resources and related sectors in the Lower Mekong Basin has been conducted. The climate change assessments contain seven themes and sectors. They assess impacts of climate change on hydrology, flood behaviour, drought behaviour, hydropower production, ecosystem services and biodiversity, food security, and socio-economic and vulnerability.
With these assessments, the MRC CCAI defines a wide range of potential future changes projected to occur over the next 20 to 50 years according to the GCM-based regional climate change scenarios.


Topic: [Obsolete] [Q&A] Which cities/ferries/roads are included?
Re: Q&A Which cities/ferries/roads are included?
Sorry if this has been asked already but now Going East! has been released, will you be adding any ferry connections for Poland? I believe there are freight connections to Copenhagen and Helsinki in real life.
Like ScuL has said, extra cities is a lot of work (if we're going for realism) and if you have too many ferries (which is only two days' work), you end up with Euro Ferry Simulator!
I posted a crappy MS Paint extension to the original ETS2 map a while back on the official SCS forum for ETS2. The region we're concerned with is a bit small and has too many ferry connections but you get the idea.


text:
package discovery

import (
	"bytes"
	"strings"

	"golang.org/x/crypto/openpgp"
)

// Verify the data using the provided openpgp detached signature and the
// embedded hashicorp public key.
func verifySig(data, sig []byte, armor string) (*openpgp.Entity, error) {
	el, err := openpgp.ReadArmoredKeyRing(strings.NewReader(armor))
	if err != nil {
		return nil, err
	}

	return openpgp.CheckDetachedSignature(el, bytes.NewReader(data), bytes.NewReader(sig))
}

redpajama_set_name: RedPajamaGithub


Booking the perfect Keaau, Hawaii videographer is probably the most important step for your Keaau, Hawaii wedding or special event. Our local directory of Keaau, Hawaii Videographers lists some of the best videographers in Hawaii. Talented videographers will record and preserve your event for all time. Hence, its crucial to hire a local Keaau, Hawaii videographer.
Much of the work you'll see from your Keaau, HI Videographer won't be available until after the event has completed. This means you'll need to carefully research your Keaau, HI Videographer selections. Consider the following four suggestions before hiring your Keaau, HI videographer.
Our website has a number of methods to instantly check the availability of your selected videographers near Keaau, HI. Each thumbs up icon denotes an instant availability since these members use Gigbuilder availability checkers powered by WeDJ.com There is no need to spend hours needlessly contacting numerous companies that may not be available.
We make it easy for Keaau, HI Videographers to contact you. Simply click the contact button next to any Keaau Videographer listing and provide your contact information in the pop-up window for your selected videographer. Alternatively, click the save button and contact all Videographers at once using the quick quote system at the top of the page.
Most Keaau, HI Videographers will require a deposit/retainer fee and signed contract before they will work an event. Make sure to fully read the terms and conditions before signing.


There are differing stigmas of teenage pregnancy, which can affect teenage mothers. Teenage mothers face the risk of not being accepted by their peers and community. Acquiring a position in society, that is often the subject of negative criticism and judgment, can profoundly impact teenage mothers (Luttrell, 2014). This research paper will analyze teenage pregnancy stigmas and how they can impact young mothers. This paper will include a literature review, analyzing adolescent sexuality and behavior, teenage pregnancy and motherhood and perceptions of teenage pregnancy. Lastly, recommendations for future practice will be discussed.
The Australian teenage fertility rates, have been on the decline for many years. In 2014, there was approximately 9,204 teenagers giving birth in Australia (ABS, 2014). In 2006, in South Australia alone, 982 women aged 15 to 19 had abortions (ABS, 2014). These kinds of statistics, mean there is still a large amount of teenage mothers that require support and assistance. This paper will look at making recommendations in practice and service development. The paper includes: examining statistical information and reviewing literature and research.


The trucking industry in the USA is growing bigger and bigger each year. As opposed to two or three decades ago, dry bulk trailers and put to more and more use in the present era. You can see this with more and more trucking companies, as they are starting to implement dry bulk transportation in their trucking business. The demand for dry bulk transportation has never been higher and conservative estimates state that it has increased at a 5% minimum when compared to last year. There are many things to know about this specialized service when it comes to freight transportation. Hiring a company that knows what they are doing is a must.
The trucking industry's demand for dry bulk transportation has increased exponentially in the past few years. This is due to an increase in the need for dry goods due to a variety of factors. Increasing populations and the demand for dry goods such as concrete and flour are a necessary to keep a society spinning. Dry bulk transportation is used most notably for finely grated or granulated substances, as well as a variety of other mixable and pourable goods. This can include food products, chemical products, and a wide array of building materials. Dry bulk tanker trailers are generally vacuum sealed to protect the contents of the tanker. This protects the product from contaminants and adverse weather systems.
For dry bulk, manufacturers' commonly use aluminum or steel. Generally, dry bulk trailers vary between 560-3120 cubic feet. Dry bulk trailers can handle cement, ash, sand, lime, sugar, grains, starch, flour and, industrial alcohol (among others).
At TPS, we understand and possess the processing technology to ensure safe cement from quarry to transport to a processing plant. We even possess the capability to carry the process through the final shipment of ready-to-mix concrete for operation plants. Transporting raw materials and finished products – such as cement – can be a large percentage of your total operational cost. Choosing a company that can efficiently and safely move large bulk cargo can save thousands upon thousands of dollars when it comes to your bottom line. From belt conveyors to trailers to hoppers and chutes, there are many moving parts when it comes to bulk material handling and processing. We deliver the results and experience needed to get the job done.


They are great beards and braids. You can do any character using the putty. I would like to try it. What about try a complete chain-mail body of a war lord? Thanks for sharing.
Thanks, I have been trying with mixed results with the chain mail. I am going to try something new, most likely next week with the chain mail. Hope it works.


text:
Why Anti-Abortion Groups Should Not Have Charitable Tax Status
Currently, there are about 170 anti-abortion groups in Canada that have charitable tax status—about 70 groups that appear to be largely political in nature, and 100 anti-abortion "counselling" centres:
25 Pro-Life Societies
35 Right-to-Life Societies
10 Other anti-abortion organizations (approximately)
46 Birthrights (anti-abortion counselling)
34 Crisis Pregnancy Centres (anti-abortion counselling)
15 Other anti-abortion counselling centres (approximately)
We believe that anti-abortion groups inherently do not qualify for charitable tax status, for the following reasons.
1. Anti-abortion groups espouse a specific cause and seek to sway
the public to their point of view.
A Revenue Canada court decision said: "I believe that jurisprudence generally supports the proposition that activities primarily designed to sway public opinion on social issues are not charitable activities." The decision also said: "An organization such as HLIC which espouses a specific cause and seeks to sway the public to its way of thinking would not qualify as charitable under the category of advancing education."[1]
HLIC is an anti-abortion group that had their tax status revoked in 1998. It was found that its publications were "strongly worded to promote HLIC's views on the abortion issue and other controversial social issues." The decision also said, "There is no case law... that would support a finding that promoting an organization's position on such issues as abortion...is charitable. In fact, the courts have found that purposes that are related to promoting one side of a controversial issue or cause are not charitable at law." It was found that HLIC was "devoting substantial resources on political activities which are not incidental and ancillary to charitable objects." These political activities were deemed to include mailing "shock value" postcards to Members of Parliament in Ottawa, organizing a March for Life on Parliament Hill, and promoting its views through tendentious publications, brochures, and advertisements.
All anti-abortion groups are dedicated primarily to the cause of protesting abortion. They seek to persuade the public to the point of view that abortion is wrong and should be illegal.
2. Promoting the anti-choice or "pro-life" view is a political
purpose in itself. In contrast, the pro-choice position is not
political—it represents mainstream society and exemplifies
professional health care standards.
Revenue Canada decided that a medical abortion clinic, Everywoman's Health Centre, is not a trust for "the political purpose of promoting the pro-choice view."[2] The clinic was given charitable tax status because it provided necessary medical services to the community, often to poor and vulnerable women who might not otherwise be able to access such services.
However, contrary to this statement by Revenue Canada, we argue that the "pro-choice view" is not a one-sided political view, nor is it the opposite of the "anti-choice view." The pro-choice position is the broad, middle-ground view shared by a large majority of Canadians, whether or not they personally agree with abortion. In contrast, anti-choice groups wish to use the law to force women to bring unwanted pregnancies to term (the opposite of that is forced abortion). The pro-choice view opposes this extremist, discriminatory position, and says that women should have information on and equal access to all pregnancy options, in a safe, non-judgmental atmosphere. In fact, pro-choice health care is the professional standard in this country for all health care—patients must be respected as responsible decision-makers, be given unbiased and accurate information on all options, and not be morally judged for the choices they make. Anything less is unethical and unprofessional.
But such questionable practices are exactly what anti-abortion groups do, uniformly. They promote their "pro-life" view in a narrow, one-sided manner. They believe that childbirth is the only choice for pregnant women and that abortion is evil. They do not respect or trust women's own medical decisions about their bodies, and they promote sexist attitudes about women, insisting that motherhood is the only proper role for women. They are almost exclusively preoccupied with demonizing abortion and abortion providers, working to remove the right to legal abortion, and persuading women not to have abortions by whatever means. Much of the information they present about abortion is false, inflammatory, scare-mongering, and/or biased.
In contrast, the only charitable group that might be seen as being "pro-choice" is Childbirth by Choice Trust in Toronto, but this educational charity is non-political. It provides non-judgmental, objective information on all available pregnancy and family planning options, including abortion. This approach, in fact, exemplifies the mainstream pro-choice position and standards of professional health care; it does not promote a politically-biased view.
3. Anti-choice groups distort the issue of abortion when
presenting "the other side."
Revenue Canada Charities Division Director Ron Davis said (Ottawa Citizen, April 22, 1995): "As long as they present both sides of the story, [then they are acceptable]. Providing pregnancy counselling is a charitable purpose. I suppose that an organization that tends to favour one side over another can still be considered as charitable as long as it's not really distorting the issue."
Anti-choice groups, especially anti-abortion counselling agencies, engage in scare-mongering and guilt-inducing techniques to dissuade women seeking abortions. Their literature and brochures are uniformly anti-abortion, and information presented about abortion is usually false and/or distorted. They also promote abstinence as the only birth control option, with little or no information on other methods, except to say that condoms are unreliable. They scare-monger on sexually-transmitted diseases, exaggerating statistics and not putting risks into context.
4. Anti-abortion groups do not qualify as educational because their so-called educational activities are mostly unstructured, and consist largely of tendentious propaganda, opinion, misinformation, and appeals to emotion.
Most anti-abortion groups claim education and/or research as their main charitable activities. But the "education" generally consists of distributed pamphlets and newsletters. Some groups also hold conferences, workshops, etc., but these are primarily aimed at their membership, not the general public. This is generally the extent of their "education." It is not structured education and it does not advance people's knowledge.
Revenue Canada has said that for a group to be eligible for charitable tax status, the information it provides must be "presented in an unbiased manner so as to allow the reader to make up his/her own mind on the position being advocated."[3] But the literature of all anti-abortion groups tends to be one-sided, emotional, and rife with unsubstantiated opinions and misinformation. In fact, much of their "education" is devoted to condemning abortion (or euthanasia). For examples, refer to almost any anti-choice brochure.
Some anti-abortion propaganda is so extreme (e.g., calling doctors "baby butchers" and "killers"), that it serves to incite hatred and possibly violence against abortion providers. Three Canadian doctors have been shot since 1994 and one has been stabbed—an American anti-abortion extremist has been charged with one shooting, is the main suspect in the other shootings, and has been convicted of the brutal murder of a New York doctor. Abortion providers in Canada must take comprehensive measures to protect the safety of themselves, as well as their patients, staff, and families. No anti-abortion group should have charitable tax status if they use inflammatory language that demonizes or incites hatred against providers (or that discriminates against women).
5. Anti-abortion groups are against abortion (believing it to be murder), but there are no laws against abortion in Canada. Therefore, the de facto purpose of all anti-abortion groups is to make abortion illegal again.
The ultimate, implicit goal of all anti-abortion groups is to restrict abortion by law, or at the very least, change current government policies to make abortion less accessible. Campaign Life Coalition, which calls itself the "political arm" of Canada's anti-abortion movement, has made that very clear. "We are working to restore the right to life, from conception to natural death, at all levels of government - federal, provincial, and municipal. We support any legislative measure to end anti-life practices, either directly or indirectly..." (http://www.campaignlifecoalition.com) By extension, all anti-abortion groups support and work towards this goal, even if they claim to be using "educational" means to achieve it.
6. The purpose of anti-choice groups has become largely detrimental to society since 1988, because access to legal abortion has been deemed by the courts to be a constitutional right guaranteeing women's equality and liberty.
A charitable tax status may be revoked by the Minister ... "if with increasing knowledge it appears that a purpose once thought beneficial is truly detrimental to the community."[4]
In the 1988 Morgentaler decision that struck down Canada's abortion law, the Supreme Court justices ruled that access to abortion is guaranteed to women based on Section 7's guarantee of "security of the person" (Charter of Rights and Freedoms). Justices also cited other provisions of the Charter, including "freedom of conscience" (section 2a), "liberty" (section 7), and "equality" (section 15). One of the justices said in a concurring opinion: "The right to liberty...guarantees a degree of personal autonomy over important decisions intimately affecting his or her private life. ... The decision whether or not to terminate a pregnancy is essentially a moral decision and in a free and democratic society, the conscience of the individual must be paramount to that of the state." Access to legal abortion is therefore a constitutional right for women and must not be abridged.
Abortion is also a medically necessary service endorsed by the Canadian Medical Association and covered under the Canada Health Act. No law exists to prohibit a woman's access to this service. The maintenance of legal abortion services is supported by both federal and provincial governments, and every major political party, including the new Conservative party.
However, anti-choice groups do not accept the legal right to abortion, or the right to non-judgmental information about abortion. They seek to remove or restrict these rights—and the only way to do this is to change laws or public policy—thereby discriminating against women and removing an important constitutional right. What harms women harms their families and all of society. The work of anti-abortion groups therefore has no public benefit and is detrimental to society.
7. International human rights documents recognize that women have a basic right to reproductive healthcare, and related information and education.
The United Nations Convention on the Elimination of All Forms of Discrimination against Women (CEDAW), ratified by Canada in 1981, says that signatory states will take measures to eliminate discrimination by ensuring that women have the same rights as men to "decide freely and responsibly on the number and spacing of their children and to have access to the information, education and means to enable them to exercise these rights."
The United Nations Beijing Declaration and Platform for Action (1995) states: "Women who have unwanted pregnancies should have ready access to reliable information and compassionate counselling. ... In circumstances where abortion is not against the law, such abortion should be safe."
Women's basic human rights include the right to unbiased and accurate information about reproductive healthcare services, and the right to access such services, including contraception and abortion (where legal). Pro-choice groups provide such information and services, but anti-abortion organizations and counselling agencies do not. Therefore, the latter are in direct violation of international human rights codes.
8. Anti-abortion groups claim to be advocates for the "unborn", which is not an allowable charitable purpose. They also work to persuade women not to have abortions.
According to Revenue Canada guidelines, advocacy must not be on behalf of "narrow, sectional interests," and it must not try to change people's behaviour or their opinions. Most of society disputes whether the "unborn" should have any moral or legal status (or to what degree). In fact, the "unborn" have no legal status in Canada—various court decisions have declared they are not legal persons. Further, all anti-abortion groups (especially counselling agencies) strive to change women's minds about having an abortion, and to persuade them of the "immorality" of abortion. This is not a charitable purpose.
9. The views espoused by so-called "educational" anti-abortion groups and by anti-abortion counselling centres are religious in nature, not educational.
Anti-abortion groups are entirely religiously-based and motivated, because the anti-choice viewpoint is fundamentally a religious doctrine.[5] Even though many groups proselytize openly (often to unsuspecting and vulnerable clients), they obtained their charitable status on claims of being "educational" or engaged in "research" or "family/crisis counselling."
Canadian anti-abortion counselling centres are Christian ministries, as they themselves admit (Christian Info News, Feb 7, 1988). American anti-abortion counseling centres are also self-admitted Christian ministries, and Canadian groups are based on exactly the same model. In fact, Birthright is a Canadian group that spread to the U.S.[6]
10. Because of their charitable status, anti-abortion groups enjoy an unfair and unethical tax advantage and higher donation rates compared to pro-choice groups.
Research shows that people donate three times as much money when they can claim a charitable tax credit than when they can't (Ottawa Citizen, April 22, 1995).
Pro-choice groups such as ourselves, the Pro-Choice Action Network, would not even consider applying for charitable status because of our political work, even though we also carry out a large amount of educational work and provide services to women. Pro-choice groups are being penalized financially because of their honesty in not applying for charitable tax status, while 170 anti-choice groups in Canada are enjoying the fruits of this status even though they primarily just disseminate propaganda disguised as education, and often engage in too much political activity.
Most of these 170 anti-abortion groups have never been audited. In contrast, the charitable educational group, Childbirth by Choice Trust in Toronto, was audited in the mid-1990's by Revenue Canada after a probably complaint by an anti-abortion group. It retained its status after making its literature more neutral. (Planned Parenthood and some family planning agencies also have charitable tax status, but their primary focus is family planning and birth control, as well as information and referrals for all pregnancy options).
[1] Reasons for Judgment, A-288-94. March 18, 1998, Human Life International in Canada (HLIC) and Minister of National Revenue.
[2] Reasons for Judgment, A-129-90. Oct 24, 1991. Federal Court of Appeal. Everywoman's Health Centre Society and Minister of National Revenue.
[3] Federal Court of Appeal, A-94-96. Alliance for Life v. Minister of National Revenue, 1999, para. 11.
[4] National Anti-Vivisection Society v. Inland Revenue Commissioners, 1948, as cited in Federal Court of Appeal. A-129-90. Reasons for Judgment. Oct 24, 1991. Between Everywoman's Health Centre Society and Minister of National Revenue.
[5] Arthur, Joyce. Abortion and Religion, Pro-Choice Press, Autumn 2001, www.prochoiceactionnetwork-canada.org/prochoicepress/01autumn.html
[6] Cooper, Cynthia L. Funds to Pregnancy Centers Blur Church-State Line. WomensNews, Oct, 2001.

redpajama_set_name: RedPajamaCommonCrawl


This entry was posted on 26 Agosto, 2012 at 12:01 am and is filed under Empreendedorismo, Gestão Carreira, Videos Formação.	You can follow any responses to this entry through the RSS 2.0 feed. You can leave a response, or trackback from your own site.


text:
Q: Authenticating in Email HTML Form I'm building a web service for internal use and one feature I would like to implement is having users be able to fill out HTML forms and submit from email so that they don't necessarily have to go to the web page to use its functionality. The problem is that the web page requires authentication via nginx so a simple redirect won't work. Accessing the email account requires authenticating into Okta which is also the same creds that will get you into nginx, so is it possible to somehow pass along that information to the form?
The web page is running on flask, just in case that's important.

A: In short the answer is no. In most modern day mailing services such as gmail the mail will be counted as spam because you are trying to gather "sensitive" information(even if its something as simple as a name)
Trying to gather information through email is a security risk and you should do so only through your secure server

redpajama_set_name: RedPajamaStackExchange


Tom Eltang, one of the most famous and representative Danish pipemaker, was born in 1958 and from an early age, he is attracted by the world of the slow smoke and, in particular, by tobacco pipes.
He is intrigued by this wonderful instrument for smoking when he was only eleven, and he earns experience in small start-up facilities and even in pipe factories.
However he is not interested in industrial production, which does not give him satisfaction and does not allow him to express his creativity.
Thus, at the age of sixteen he is presented to Anne Julie, a famous painter and pipemaker known for her inventiveness and originality of his creations, who is impressed by a pipe made by the young Eltang.
They start three years of apprenticeship in Anne workshop in Copenhagen, which followed by three more in the laboratory Pipe Dan, where Tom will repair pipes and will have the opportunity to put to use the lessons learned in previous experiences also realizing some of his pipe.
In these years that Eltang learns deeply all the details of the operation of a pipe, after which, in previous years, had developed a taste for the forms for processing and for the colors, especially the contrast that' he would make him famous in later years.
Tom goes to work at Stanwell, where he remains until 1980, becoming know personally in slow smoking world. Indeed the well-known brand often sends him at fairs and tobacco shops in Germany, to demonstrate the art of making a pipe.
In 1980 Tom starts his own company, while continuing to work with the Danish brand which still sells some of his drawings.
The first ten years on his company are difficult, so much so that Tom joins the production of pipe other activities, less imaginative, but more profitable. Since 1990, however, he has made exclusively his pipe.
Today he is considered one of the most important pipemakers in the world, famous for the well-known colour of the smooth pipes called "golden contrast", learned during his apprenticeship with Anne Julie, but perfected at the highest levels.
In any case, the production of Tom is not limited to the smooth pipes, but also extends to rusticate and sandblasted pipes, in which Tom Eltang is a master.


text:
Q: Python: get first elements from tuple of strings I have the following query: is it possible to get tuple of first symbols for every string in tuple by slice?
I've tried the following code:
t = ("aaa", "aab", "abc", "aba", "bcc")
print(t[:][0])

and I got "aaa". What I want is ['a', 'a', 'a', 'a', 'b'].
Sure it can be done by generator:
letters = (x[0] for x in t)

But is it possible to do by slices?
Thank you.

A: The following code does not explicitly use slicing, but it is very efficient:
letters = next(zip(*t))
letters
#('a', 'a', 'a', 'a', 'b')

This method is 25% faster than the list comprehension with selection ([0]) and 48% faster than the list comprehension with slicing ([:1]).

A: The classic solution is to use a list comprehension. An alternative, functional, method is to use operator.itemgetter.
But you can get closer to vectorised functionality via 3rd party NumPy. Here you should see large performance improvements. The slicing is trivial; the cost is largely from conversion to a NumPy array.
The NumPy solution assumes your strings are of the same length and you are comfortable with having your output as an array rather than a list:
res = np.array(t).view('<U1')[::len(t[0])]

Some performance benchmarking:
import numpy as np
from operator import itemgetter

t = ("aaa", "aab", "abc", "aba", "bcc")
t = t*500000
t_arr = np.array(t)

def comp(tup):
    return [x[0] for x in tup]

def func(tup):
    return list(map(itemgetter(0), tup))

def nump(tup):
    return np.array(tup).view('<U1')[::len(tup[0])]

def nump2(arr):
    return arr.view('<U1')[::len(arr[0])]

def dyz(tup):
    letters, *_ = zip(*tup)
    return letters

def dyz2(tup):
    return next(zip(*tup))

%timeit comp(t)       # 276 ms per loop
%timeit func(t)       # 338 ms per loop
%timeit nump(t)       # 174 ms per loop
%timeit nump2(t_arr)  # 2.86 µs per loop
%timeit dyz(t)        # 351 ms per loop
%timeit dyz2(t)       # 245 ms per loop

A: No it's not possible because a Slice can't return more than one String.
You need to loop on your Tuple : [x[0] for i in t].


text:
Obama deficit plan includes $1.3B cuts to imaging
By Kate Madden Yee, AuntMinnie.com staff writer
September 21, 2011 -- President Barack Obama's plan to tame the U.S. budget deficit includes $320 billion in health savings over the next 10 years -- $1.3 billion of which would come from cuts to medical imaging, according to the proposal released September 19 by the White House.
"The plan [builds] on the Affordable Care Act to strengthen Medicare and Medicaid by reducing wasteful spending and erroneous payments, and supporting reforms that boost the quality of care," the White House said in a statement.
The cuts would come from updates to Medicare payments to "more appropriately account" for utilization of advanced imaging, as well as mandatory prior authorization. The president's proposal is part of ongoing negotiations on deficit reduction mandated by the Budget Control Act of 2011, which are to be handled by a special 12-member congressional "super committee." The committee must find an additional $1.5 trillion in federal savings to allow for another debt ceiling increase by December 23, or automatic cuts in both discretionary and mandatory programs will be triggered.
Medicare spending for advanced diagnostic imaging services paid for under the Medicare Physician Fee Schedule has grown dramatically in recent years due to an increase in the number and intensity of those services, the proposal stated. According to the Medicare Payment Advisory Commission (MedPAC), this volume growth may signify these services are mispriced.
What's the bottom line? Beginning in 2013, the president's proposal would implement a "payment adjustment" for advanced imaging equipment to account for higher levels of utilization, which will save $400 million over 10 years.
There's more: Also in 2013, the president's proposal would mandate prior authorization for the most expensive imaging services, to "ensure that these services are used as intended and protect the Medicare program and its beneficiaries from unwarranted use." Total savings would be $900 million over 10 years.
The proposal does not fix the sustainable growth rate (SGR) formula, which the U.S. Centers for Medicare and Medicaid Services (CMS) uses to set the conversion factor for the Medicare Physician Fee Schedule; the projected reduction to the fee schedule for 2012 is 29.5%, unless Congress manages to intervene.
In addition, the proposal would find $500 million in savings through penalties against those Medicare providers that fail to show meaningful use of electronic health records. The penalties would be credited to a special account beginning in 2020, and would be applied toward deficit reduction in 2021.
Copyright © 2011 AuntMinnie.com
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I love the winter. That's because something in the winter is different to other times.
In cold winters, people's kindness is like the warmth of family. I feel happier than normal.
The peak of the mountain we reached by ropeway was the coldest place so far.
The trees were covered in lots of snow and ice, and didn't look like trees at all.
The snow changes the appearance of everything.
And the snow itself will change into something special.
Wearing traditional foot and leg protectors made from rope and wood, I walked on top of the deep snow.
Wearing this handmade footwear, it was easy to walk on top of the snow.
Though I did sometimes fall over.
Since I like snow, I slowly picked myself up, and continued walking through the pure-white world.
The man who was my guide stayed with me the whole time.
The udon noodles we ate at the rest-stop at the top of the mountain had a nostalgic flavor. It was a warm, happy taste.
"Are you PIMTHA?" a voice said, in Thai.
I looked around to see a Thai father and his cute daughter.
They knew who I was, and we took a photo together.
It was a nice surprise in a place that was minus 17 degrees.
I've tried to challenge myself by going to driving school three times in Thailand, but I've always fail.
That someone like me could enjoy a snowmobile…!
The snowmobile coach was like a father helping someone learning how to ride a bicycle for the first time.
"Calm down and relax," he said, and my body, which had been tense, suddenly lightened.
When I rode the snowmobile I felt free.
The snow was very soft, and always caught me.
I could go anywhere I wanted to.
What kinds of snow have you seen before?
The snow in Tohoku has now become something special to me.
It's very cold here, and the snow is very deep.
It took a bit of courage to come here.
But I haven't been lonely here.
The people who live with this snow welcome you like family.
And, along with the beautiful snow of Tohoku, you will surely have an unforgettable trip.


text:
Pioneers – game based on "The Settlers of Catan"
Pioneers is a GNOME computer implementation of the Settlers of Catan board game. It can be played over the internet.
The game includes a server, a client, a computer player, an editor and a meta-server (for locating running servers).
The GTK server has a user interface in which you can customise the game parameters. Once you are happy with the game parameters, press the Start Server button, and the server will start listening for client connections.
When you start the client program, it displays a connect dialog. You must define the hostname and port where the server can be located. You can enter these manually, or press the Meta Server button. The meta server maintains a list of all registered Pioneers servers that are currently running a game. This allows you to join a game anywhere on the Internet.
Pioneers is most fun with three or four players, but two players is still ok, or you can add some computer players if you like.
Pioneers was formerly known as Gnocatan.
Play with up to 8 players.
Artificial Intelligence.
Default boards are included, and SeaFarers.
Website: sourceforge.net/projects/pio
Support: Manual
Developer: Dave Cole, Andy Heroff, Roman Hodek, Dan Egnor, and many others
License: GNU GPL v2
Pioneers is written in C. Learn C with our recommended free books and free tutorials.
Return to Games Home Page | Return to Puzzle Games Home Page


text:
John W. Lister, AIA, Principal
John W. Lister, Firm Principal
John W. Lister
NCARB, LEED AP, GGP
Born and raised in southern California, John knew from the time he was in middle school and took his first drafting class, that he wanted to be an architect. After completing his education at University of Southern California, John started his career with the Los Angeles firms of Skidmore Owings and Merrill, notably working on the Universal Studios Amphitheater and Getty Oil World headquarters. Before founding JL Architects, John was the lead project architect on estate homes in Mexico and Maryland and the youngest architect ever named associate at Robinson-Willis, PC in Washington, DC. Since founding JL Architects in 1988, the firm's focus has evolved to include retail centers, restaurants, entertainment, office buildings, manufacturing, industrial, and multi-family housing.
With licenses in 38 states, John's diverse experience gives him a unique and fresh perspective. In the digital era, he has expertise in optimizing consumer experiences in-store, at home, and at work. His team is active in sustainability and green building certification, consulting for design professionals and building owners with a guaranteed method for success. John is a member of The American Institute of Architects and ICSC. He serves on the Green Building Initiative's national Board of Directors leading the Government Engagement Committee. The current Chair of the West Chester Area Council of Governments Building Code Variance Board, John also held the position of Chair of East Bradford Planning Commission. He holds certifications from NCARB, is a Green Globes Professional, and a LEED AP. Having earned multiple degrees from the University of Southern California, John is a rabid Trojans fan and has adopted a loyalty to Philadelphia sports as well.
John is also always quick to share that he has been blessed with 4 children, each of whom he will say he loves and admires the most. He feels particularly fortunate to live in a home of his own design and work in an office building of his own design. Acting as owner/architect/developer of the JLA office building provided the opportunity to learn about development on another level. That experience, along with the rest of life's journey helps to bring balance of understanding to each client, and each project.
ContactLicensure
© JL Architects All Rights Reserved.


text:
\section{Introduction}

\subsection{Motivation}
Consider a smooth projective curve $X$ of genus $g$ over an algebraically closed field $k$. The {\em canonical representation} of $G:=\operatorname{Aut}(X)$ is the $g$-dimensional $k$-representation of $G$ induced by the natural action of $G$ on the space of global regular 1-forms on $X$. In characteristic zero, the canonical representation has been determined in the work of Chevalley and Weil, responding to a query by Hecke (see \cite{ChWe}, \cite{Weil}), and these methods extend to positive characteristic if the characteristic of $k$ does not divide $|G|$. If the characteristic divides $|G|$, Kani and Nakajima have separately resolved the case where $X \rightarrow X/G$ is tamely ramified (see \cite{Kani}, \cite{Nak1}, \cite{Nak2}), while Valentini and Madan resolved the case where $G$ is cyclic of a $p$-power degree (see \cite{ValMad}) (a result that Nakajima later generalized to any invertible $G$-sheaf, see \cite{Nak3}). Most recently, K\"ock has generalized these results to the situation where the cover is weakly ramified, i.e. when all the higher ramification groups vanish at all points (see \cite{Ko}).

In this paper, we focus on the natural question of when the canonical representation is irreducible. If it is and $k$ has characteristic zero, then $g^2 \le |G|$. Together with the Hurwitz bound $|G| \le 84(g-1)$, this implies that $g \le 82$. This situation is studied in \S19 of \cite{Bre}, which includes a list of all possible genera and automorphism groups for which the canonical representation is irreducible. The Klein quartic $X(7)$ is a particularly important example of this phenomenon in characteristic zero (see \cite{Elk}).

The situation is quite different in positive characteristic. The Hurwitz bound continues to hold for characteristic $p>0$, provided that $2 \le g<p$, with the sole exception of the curve $y^2=x^p-x$ (see \cite{Roq}). However, in the general case, the bounds on irreducible representations are weaker, and thus we get no upper bound independent of $|G|$ on the genera of curves with irreducible canonical representation. This presents the possibility of having curves with arbitrarily large genus and an irreducible canonical representation. This paper shows that this holds for the projective smooth curve corresponding to $y^2=x^p-x$, which has genus $(p-1)/2$ (this curve is also studied over finite fields in \cite{GJK}). The question remains open whether for fixed $p$ there exist curves over $k$ with irreducible canonical representation of arbitrary large dimension.

\subsection{Summary of Results}

Let $k$ be an algebraically closed field of characteristic $p > 2$, and let $X$ be the unique smooth projective curve over k with affine equation $y^2=x^p-x$. Set $G:=\operatorname{Aut}_k(X)$, which is a degree $2$ central extension of $\operatorname{PGL}_2({\bf F}_p)$ (discussed in \cite{Roq} \S5). Explicitly, let $\widetilde{G} \subseteq GL_2({\bf F}_p) \times {\bf F}_{p^2}^\times$ be the subgroup of elements $(\sigma, u_\sigma)$ with $\det \sigma = u_\sigma^2$. Then ${\bf F}_p^\times$ acts diagonally on $\widetilde{G}$ via \[\lambda (\sigma, u_{\sigma})= \left( \left(\begin{smallmatrix}
\lambda & 0\\
0 & \lambda \\
\end{smallmatrix}\right) \sigma, \lambda^{(p+1)/2} u_\sigma \right),\] and $G= {\bf F}_p^\times \backslash \widetilde{G}$. We will consistently represent elements of $G$ by chosen lifts in $\widetilde{G}$. Then if $\sigma = \left(\begin{smallmatrix}
a & b\\
c & d\\
\end{smallmatrix}\right) \in GL_2({\bf F}_p)$, and $u_\sigma$ is a choice of square root of $\det \sigma$, we can write the action of $G$ on the affine coordinates of $X$ as
\begin{equation*}
(x,y) \mapsto \left(\frac{ax+b}{cx+d}, u_{\sigma}\frac{y}{(cx+d)^{(p+1)/2}}\right).
\end{equation*}

In this paper, we show that $H^0(X, \Omega^1_{X/k})$ is irreducible as a $k[G]$-module. We do this by considering a small index subgroup $H$ of $G$ that is a quotient of $\operatorname{SL}_2({\bf F}_p)$. Noting that $H^0(X, \Omega^1_{X/k})$ has dimension $g=(p-1)/2$ as a $k$-vector space, we get that the restriction of the canonical representation to this subgroup is the $(g-1)^{st}$ symmetric power of the standard representation of $\operatorname{SL}_2({\bf F}_p)$ on $k^2$, which is well-known to be irreducible. It is not difficult to then describe $H^0(X, \Omega^1_{X/k})$ as a $k[G]$-submodule of the induced representation of $\operatorname{Sym}^{g-1}(k^2)$.

Note that since $G$ is not a cyclic $p$-group, this is beyond the results of Valentini and Madan mentioned earlier. Furthermore, the cover $X \rightarrow X/G$ is wildly (but not weakly) ramified: the point at infinity and $\{ (x, 0) \in X : x \in {\bf F}_p \}$ are an orbit of $X$ under $G$, and since this orbit has $p+1$ elements, each is ramified of degree $2p(p-1)$. The previous work of Kani and Nakajima applies only where $X \rightarrow X/G$ is tamely ramified, and thus not to this case.

In the second half of this paper, we consider the de Rham hypercohomology as a $k[G]$-module. In characteristic $p$, Maschke's Theorem does not apply if $p$ divides $|G|$ and it is possible to find representations that are reducible but not semisimple. If a nonzero representation is not the direct sum of two proper subrepresentations, call it {\em indecomposable}. The de Rham hypercohomology of any curve $C$ is clearly reducible, since $H^0(C, \Omega_{C/k}^1)$ embeds as a $k[\operatorname{Aut}(C)]$-submodule, but we show that in the case of the curve $X$, the de Rham cohomology is indecomposable as a $k[G]$-module.

It has been previously shown that the crystalline cohomology of $X$ tensored with ${\bf Q}_p$ is irreducible (see \cite{Katz}, Proposition 7.2). Knowing that the crystalline cohomology is irreducible as a $k[G]$-module, it is not necessarily true that $H^1_{dR} (X/k)$ should be indecomposable as a $k[G]$-module, since its $G$-invariant subrepresentations do not lift to the crystalline cohomology. Thus we do not necessarily expect $H^1_{dR} (X/k)$ to be indecomposable as a $k[G]$-module, especially since it is an unusual (and distinctly positive characteristic) phenomena to have a reducible indecomposable representation.

\section{The action of $G$ on $H^0(X, \Omega^1_{X/k})$}\label{sec:HD}

As $H^0(X, \Omega^1_{X/k})$ is functorial in $X$, it has a $k$-linear action of $G$. To determine the structure of of $H^0(X, \Omega^1_{X/k})$ as a $k[G]$-module, we will use the following classical result about its structure as a $k$-vector space. Throughout the paper, we use the notation that $P_t= (t,0)$ for $t=0,1,\dots,p-1,$ and $P_{\infty}$ is the ``point at infinity" of $X$.

\begin{lemma}\label{lem:HDbasis}
Let
\begin{equation}\label{HDbasis}
\tilde{\tau}_i = \frac{x^idx}{y}.
\end{equation}
The set $\{ \tilde{\tau}_i \mid i= 0, 1, \dots, g-1\}$ is then a $k$-basis for $H^0(X, \Omega^1_{X/k})$.
\end{lemma}

\begin{proof} This follows from the calculations below, which we will use later in the paper:
\begin{align*}
\div(x) &= 2P_0 + (-2)P_{\infty}\\
\div(y) &= P_0 + P_1 + ... + P_{p-1} + (-p)P_{\infty}\\
\div(dx) &= P_0 + P_1 + ... + P_{p-1}+ (-3)P_{\infty}.
\end{align*}
So $\div(\tilde{\tau}_i)=2iP_0 + (p-2i-3)P_\infty$, and $\tilde{\tau}_i$ is regular for $i=0, \ldots, g-1$. They are linearly independent since they have different pole orders at $P_\infty$, and thus a $k$-basis.
\end{proof}

To study $H^0(X, \Omega^1_{X/k})$ as a $G$-representation, we begin by considering a specific subgroup of small index. Consider the morphism
$\theta: \operatorname{SL}_2({\bf F}_p) \rightarrow G$ that sends $\sigma \in \operatorname{SL}_2({\bf F}_p)$ to the equivalence class of $(\sigma, 1)$, and let $H =\operatorname{im}\theta$.

One can check that
\begin{equation*}
\operatorname{ker}\theta =
\begin{cases}
I & \text{ if } p \equiv 1 \text{ mod } 4 \\
\pm I & \text{ if } p \equiv 3 \text{ mod } 4.
\end{cases}
\end{equation*}
As $|G|=2(p-1)p(p+1)$ and $|\operatorname{SL}_2({\bf F}_p)|=(p-1)p(p+1)$, we conclude $|G:H|=2$ if $p\equiv 1$ mod 4, and that $|G:H|=4$ if $p\equiv 3$ mod 4.

Let $\operatorname{Sym}^{g-1}(k^2)$ be the $(g-1)^{st}$ symmetric power of the standard representation of $\operatorname{SL}_2({\bf F}_p)$ on $k^2$. We identify $\operatorname{Sym}^{g-1}(k^2)$ with the $k$-subspace of $k[u,v]$ consisting of homogeneous polynomials with degree $g-1$. This space has basis $\{ v^{g-1}, v^{g-2}u, \dots , u^{g-1}\}$ and $\operatorname{SL}_2({\bf F}_p)$-action given by \[
\left( \begin{array}{cc}
a & b\\
c & d
\end{array}
\right)
\left( \begin{array}{c}
u \\
v
\end{array}
\right) =
\left( \begin{array}{c}
au+bv \\
cu+dv
\end{array}
\right).
\]
Note that if $p \equiv 3$ mod $4$, then $g-1$ is even, and the action of $\operatorname{SL}_2({\bf F}_p)$ on $\operatorname{Sym}^{g-1}(k^2)$ is trivial on $\pm I$. Because of this, we can define an $H$-action on $\operatorname{Sym}^{g-1}(k^2)$ where if $h \in H$ such that $\theta(\sigma)=h$, then $h$ acts as $\sigma$ does. This is well-defined since the kernel of $\theta$ acts trivially. Let $V$ be this $H$-representation.

\begin{prop}\label{irrHom}
The map \begin{equation*}
\varphi: \operatorname{res}_H(H^0(X, \Omega^1_{X/k})) \rightarrow V \qquad \text{ given by }\qquad \frac{x^idx}{y}\mapsto u^iv^{g-i-1}
\end{equation*} is a $k[H]$-module isomorphism.
\end{prop}

\begin{proof}
Clearly $\varphi$ is an isomorphism of vector spaces over $k$, so we need only check that $\varphi$ is $H$-equivariant. Let $h=(\gamma,1) \in H$ for $\gamma= \left(\begin{smallmatrix}
a & b\\
c & d\\
\end{smallmatrix}\right) \in \operatorname{SL}_2({\bf F}_p)$. We calculate
\begin{align*}
h \circ \varphi\left(x^iy^{-1}dx\right) &= h(u^iv^{g-i-1})\\
&=(au+bv)^i(cu+dv)^{g-i-1}
\end{align*} and
\begin{align*}
\varphi \circ h(x^iy^{-1}dx) &= \varphi \left(\left(\frac{ax+b}{cx+d}\right)^i\left(\frac{y}{(cx+d)^{g+1}}\right)^{-1}d\left(\frac{ax+b}{cx+d}\right)\right)\\
&= \varphi((ax+b)^i(cx+d)^{g-i-1}y^{-1}dx)\\
&= (au+bv)^i(cu+dv)^{g-i-1},
\end{align*}
since $\displaystyle d\left(\frac{ax+b}{cx+d}\right) = (cx+d)^{-2} dx$.
\end{proof}

\begin{cor}\label{irreducible}
As a $G$-representation, $H^0(X, \Omega^1_{X/k})$ is irreducible. Moreover, $H^1(X, \mathcal{O}_X)$ is naturally the contragredient of $H^0(X, \Omega^1_{X/k})$, hence irreducible as well.
\end{cor}
\begin{proof} It is well-known that $\operatorname{Sym}^{g-1}(k^2)$ is irreducible as a representation of $\operatorname{SL}_2({\bf F}_p)$ (see the discussion following Corollary 3 of Chapter 3 in \cite{Alp} 14-16). It is clear from the way we defined $V$ that it is thus irreducible as an $H$-representation, and this implies it is irreducible as a $G$-representation. The Serre duality pairing $H^0(X, \Omega^1_{X/k}) \times H^1(X, \mathcal{O}_X) \rightarrow k$ is $G$-equivariant, so $H^1(X, \mathcal{O}_X)$ is canonically the contragredient of $H^0(X, \Omega^1_{X/k})$, and thus irreducible. \end{proof}

We can also explicitely describe the structure of $H^0(X, \Omega^1_{X/k})$ as a subrepresentation of the induced representation of $V$.

\begin{prop}
If $p \equiv 1$ mod 4, then $\operatorname{Ind}_H^G(V) \cong V^2$ and $H^0(X, \Omega^1_{X/k})$ can be embedded as the subrepresentation \[\{ (v,-v): v \in V\}.\] If $p \equiv 3$ mod 4, then $\operatorname{Ind}_H^G(V) \cong V^4$, then $H^0(X, \Omega^1_{X/k})$ can be embedded as the subrepresentation spanned by \[ \{(\varphi(\tilde{\tau}_j), (-1)^ji \varphi(\tilde{\tau}_j), -\varphi(\tilde{\tau}_j), (-1)^{j+1}i\varphi(\tilde{\tau}_j)) \in V^4 \mid j=0, \dots, g-1\},\]  where $i \in k$ such that $i^2=-1$, and using the same notation as in Lemma~\ref{HDbasis} and Proposition~\ref{irrHom} for $\tilde{\tau}_j$ and $\varphi$.
\end{prop}

\begin{proof}Recall that the induced representation $\operatorname{Ind}_{H}^G(V)$ is $k[G] \otimes_{k[H]} V$. Frobenius reciprocity tells us that
\[ \operatorname{Hom}_G(H^0(X, \Omega^1_{X/k}), \operatorname{Ind}_H^G(V)) = \operatorname{Hom}_H(\operatorname{res}_H(H^0(X, \Omega^1_{X/k})), V).\]
However, since Corollary~\ref{irreducible} tells us that $\operatorname{res}_H(H^0(X, \Omega^1_{X/k}))$ and $V$ are isomorphic irreducible $H$-representations, this means that the right hand side is one dimensional. Thus there is a nonzero $k[G]$-linear map from $H^0(X, \Omega^1_{X/k})$ into $\operatorname{Ind}_H^G(V)$ and it is unique up to scaling. (For more on irreducible and induced representations, see sections 1 and 8 respectively of \cite{Alp}.)

In the case where $p\equiv 1$ mod 4, $|G:H|=2$ so as a vector space the induced representation is $V \oplus V$. If $\alpha \in G$ corresponds to the equivalence class of $(\left( \begin{smallmatrix} 1 & 0 \\ 0 & 1\end{smallmatrix}\right), -1) \in \tilde{G}$, then $G = H \sqcup H \alpha$. Note in particular that if $\omega \in H^0(X, \Omega^1_{X/k})$, the $\alpha \omega = -\omega$ and if $(v_1, v_2) \in \operatorname{Ind}_H^G(V)$, then $\alpha(v_1, v_2) = (v_2, v_1)$. Note that we can easily calculate the $G$-action on the induced representation from that.

In particular the map
\begin{align*}
T:H^0(X, \Omega^1_{X/k}) &\rightarrow \operatorname{Ind}_H^G(V)\\
\omega &\mapsto (\varphi(\omega), -\varphi(\omega))
\end{align*}
is $G$-equivariant. Thus $H^0(X, \Omega^1_{X/k})$ embeds into $\operatorname{Ind}_H^G(V)$ as the $k[G]$-submodule $\{(v, -v)\in V \oplus V \}$.

If $p \equiv 3$ mod 4, then $|G:H|=4$ so as a vector space the induced representation is $V \oplus V \oplus V \oplus V$. If $\beta \in G$ corresponds to the equivalence class of $((\begin{smallmatrix} -1 & 0 \\ \phantom{-}0 & 1\end{smallmatrix}), i)$ where $i \in k$ such that $i^2=-1$, then $G = H \sqcup H \beta \sqcup H \beta^2 \sqcup H \beta^3$. Note in particular that if $\tilde{\tau}_j$ is a basis element of $H^0(X, \Omega^1_{X/k})$ as in Lemma~\ref{HDbasis}, then $\beta\tilde{\tau}_j = (-1)^ji \tilde{\tau}_j$. If $(v_1, v_2, v_3, v_4) \in \operatorname{Ind}_H^G(V)$, then $\beta(v_1, v_2, v_3, v_4) = (\beta v_4, \beta v_1, \beta v_2, \beta v_3)$; note we can easily calculate the $G$-action on the induced representation from this. In particular, the injective map
\begin{align*}
T: H^0(X, \Omega^1_{X/k}) &\rightarrow \operatorname{Ind}_H^G(V)\\
\tilde{\tau}_j &\mapsto (\varphi(\tilde{\tau}_j), (-1)^ji \varphi(\tilde{\tau}_j), -\varphi(\tilde{\tau}_j), (-1)^{j+1}i\varphi(\tilde{\tau}_j))
\end{align*}
is $G$-equivariant.\end{proof}

\section{The action of $G$ on $H^1_{dR} (X/k)$}

For any smooth proper curve $Y$ over a field $L$, $H^0(Y, \Omega_{Y/L})$ embeds as strict subrepresentation of $H_{dR}^1(Y/L)$. If $L$ is of characteristic zero, it follows from Maschke's Theorem that $H^0(Y, \Omega_{Y/L})$ is a $L[\operatorname{Aut}(Y)]$-module direct summand of $H_{dR}^1(Y/L)$. However, Maschke's Theorem does not apply in positive characteristic if the characteristic of $L$ divides $|\operatorname{Aut}(Y)|$ (as in our case), so it need not be the case that reducible implies decomposable. In this section we will show that, in fact, $H^1_{dR} (X/k)$ is indecomposable as a $k[G]$-module.

For these computations we will use the \v{C}ech cohomology with the open affine cover $\mathcal{U}=\{U_1, U_2\}$, where $U_1=X-P_0$ and $U_2=X-P_{\infty}$ (for background on \v{C}ech hypercohomology, see EGA $0_{\text{III}}$ \S12 \cite{EGA} and SGA Exp V \cite{SGA}). The method used here is largely based on that in \cite{HaaJan}.  By definition, $\check{H}^1_{dR}(\mathcal{U})$ is the quotient of the $k$-vector space
\begin{equation*}
\{ (\omega_1, \omega_2, f_{12})\mid \omega_i \in \Omega_{X/k}(U_i), f_{12} \in \mathcal{O}_X(U_1 \cap U_2), df_{12}=\omega_1-\omega_2 \}
\end{equation*} by the $k$-subspace
\begin{equation*}
\{(df_1, df_2, f_1-f_2)\mid f_i \in \mathcal{O}_X(U_i) \}.
\end{equation*}

\begin{theorem}\label{nosplit}
The canonical exact sequence of $k[G]$-modules
\begin{equation} \label{eq:exactseq}
0 \rightarrow H^0(X, \Omega^1_{X/k}) \rightarrow H^1_{dR} (X/k) \rightarrow H^1(X, \mathcal{O}_X) \rightarrow 0
\end{equation}
does not split.\end{theorem}

To prove this, we will first find a $k$-basis for $\check{H}^1_{dR}(\mathcal{U})$ and study the action of $G$ on this basis.

\begin{lemma}\label{lem:basisdR}
For $i=0,1,\dots, g-1$, let $\tau_i$ be the class of \begin{equation}(x^iy^{-1}dx, x^iy^{-1}dx, 0),
\end{equation} in $\check{H}^1_{dR}(\mathcal{U})$. For $i=1,2,\dots, g$, let $\eta_i$ be the class of \begin{equation}((1-2i)x^{1-g-i}d(yx^{-g-1}), -2ix^{2g-i}dy, yx^{-i})
\end{equation} in $\check{H}^1_{dR}(\mathcal{U})$. Together, these form a $k$-basis for $\check{H}^1_{dR}(\mathcal{U})$.
\end{lemma}
\begin{proof} We need to first determine that these are well-defined elements in $\check{H}^1_{dR}(\mathcal{U})$.
It is clear the $\tau_i$ are well defined in $H^1_{dR} (X/k)$ since $x^iy^{-1}dx$ is regular for $i=0, 1, \dots, g-1$ (note: the $\tau_i$ are the images of the $\tilde{\tau}_i~\in~\check{H}^1(\mathcal{U}, \Omega^1_{X/k})$ under the canonical map). To show that $\eta_i$ is well defined, we first calculate that $d(yx^{-g-1})= x^{g-1}dy$. So
\begin{align*}
(1-2i)x^{1-g-i}d(yx^{-g-1}) - (-2ix^{2g-i}dy) &= (1-2i)x^{-i}dy+2ix^{2g-i}dy\\
&= d(yx^{-i}).
\end{align*}
The requirement that $\omega_1-\omega_2 = df_{12}$ for an element $(\omega_1, \omega_2, f_{12}) \in~\check{H}^1_{dR}(\mathcal{U})$ follows from this, but we still need to ensure that the $\omega_1$, $\omega_2$ and $f_{12}$ are regular on the appropriate open sets.  To do this, it is enough to calculate their divisors, which are:
\begin{align*}
\div((1-2i)x^{1-g-i}d(yx^{-g-1})) &= (-2i)P_0 + (2g+2i -2)P_{\infty}\\
\div(-2ix^{2g-i}dy) &= 2(2g-i)P_0 + (2i-2g-2)P_{\infty}\\
\div(yx^{-i}) &= (1-2i)P_0 + \sum_{j=1}^{p-1}P_j + (2i-p)P_{\infty}.
\end{align*}
Notice that the first equation refers to a 1-form regular on $U_1$, the second a 1-form regular on $U_2$, and the third a function regular on $U_1 \cap U_2$; we conclude that the $\eta_i$s are well-defined in $\check{H}^1_{dR}(\mathcal{U})$.

Since by Lemma \ref{lem:HDbasis}, the $\tilde{\tau}_i$s form a basis for $\check{H}^0(\mathcal{U}, \Omega^1_{X/k})$, and the $\tau_i$s are their images in $\check{H}^1_{dR}(\mathcal{U})$, it suffices to show that the image of $\{ \eta_i | i=1, \dots, g\}$ in $\check{H}^1(\mathcal{U}, \mathcal{O}_X)$ is a basis.

Recall that $H^1(X, \mathcal{O}_X)$ is the dual of $H^0(X, \Omega^1_{X/k})$, and that under the canonical Serre-duality pairing there is a map $H^0(X, \Omega^1_{X/k}) \times H^1(X, \mathcal{O}_X) \rightarrow k$ given by the residue map (see \cite{Tate} or Appendix B of \cite{Con} for a complete discussion). We can use this to show that $\displaystyle -2yx^{-i}$ is the dual element to $y^{-1}x^{i-1}dx$.

For $i=0, \dots, g-1$ and $j= 1, \dots, g$ the pairing gives
\begin{equation}
\langle y^{-1}x^idx, yx^{-j} \rangle = \operatorname{res}(x^{i-j}dx).
\end{equation}
This can be calculated by summing the residues at points $P$ where $P \in U_1$, or equivalently (by applying the Residue Theorem), the negative of the residue at $P_0$. Note in particular this means for the inner product to be non-zero, $x^{i-j}dx$ must have a pole somewhere on $U_1$ and a pole at $P_0$. By the calculations in Lemma~\ref{lem:HDbasis},
\begin{equation*}
\div(x^{i-j}dx) = (2i-2j+1)P_0 + P_0 + P_1 + \dots + P_{p-1} +(2j-2i- 3)P_\infty.
\end{equation*}
This has a pole at $P_0$ if $j \ge i+1$ and a pole at $P_\infty$ if $j \le i+1$. It follows that $\langle y^{-1}x^idx, yx^{-j} \rangle=0$ if $i \not = j-1$. If $i=j-1$, then $\langle y^{-1}x^idx, yx^{-j} \rangle=\operatorname{res}_{P_\infty}(x^{-1}dx)$. To calculate this residue, note that $\displaystyle t=\frac{y}{x^{(p+1)/2}}$ is a uniformizer at $P_\infty$ (using the proof of Lemma~\ref{lem:HDbasis}). Since \[ t^2 = \frac{y^2}{x^{p+1}} = \frac{x^p-x}{x^{p+1}} = \frac{1}{x} - \frac{1}{x^p}\] we find
\begin{equation*}
\frac{1}{x} = t^2 + \left(\frac{1}{x}\right)^p,
\end{equation*}
from which it follows (by repeatedly substituting) that $\displaystyle x^{-1} = \sum_{i=0}^\infty t^{2p^i}$. It is straightforward to show from this that $\operatorname{res}_{P_\infty}(x^{-1}dx)=-2$.

Thus $\{-2yx^{-i}\mid i=1, \dots, g\}$ is the dual basis of the $\{ \tilde{\tau}_{i-1}\mid i =1, \dots, g\}$, and in particular a basis for $H^1(X, \mathcal{O}_X)$.
\end{proof}

{\em Proof of Theorem~$\ref{nosplit}$.} Now that we have a basis for $H^1_{dR} (X/k)$ as a split vector space, we need to consider how our group $G$ acts on these basis vectors.  Specifically, we are going to consider $\sigma \in G$, the equivalence class of  $\left((\begin{smallmatrix}
1 & 1\\
0 & 1\\
\end{smallmatrix}), 1\right)$, and its action on the span of $\{ \eta_i \mid i=1, \dots, g \}$, the subspace (non-canonically) isomorphic to $H^1(X, \mathcal{O}_X)$. We will calculate the image of each $\eta_i$ under $\sigma$, and use this to show that there is no splitting of the sequence as $k[G]$-modules.

Here we run into the problem that our cover $\mathcal{U}$ is not preserved under the group action.  While $\sigma U_2=U_2$, we get that $\sigma U_1= X - P_{p-1}$.  To account for this, we will refine our cover.  Let $U_3=X-P_1$.  Note that then $U_3=\sigma^{-1}U_1$.  Define the covers $\mathcal{U}'=\{U_2, U_3\}$ and $\mathcal{U}''=\mathcal{U} \cup \mathcal{U}'$.

We can then calculate the action of $\sigma$ on $\check{H}^1_{dR}(\mathcal{U})$ using the following commutative diagram:

\centerline{
\begin{xy}
(0,3)*+{}="A";
(0,17)*+{}="B";
(10,0)*+{H^1_{dR} (X/k) \cong \check{H}^1_{dR}(\mathcal{U})}="LD";
(10,20)*+{H^1_{dR} (X/k) \cong \check{H}^1_{dR}(\mathcal{U})}="LU";
(60,0)*+{\check{H}^1_{dR}(\mathcal{U}')}="RD";
(60,20)*+{\check{H}^1_{dR}(\mathcal{U}'')}="RU";
{\ar@{->}^{\sigma} "A";"B"};
{\ar@{->}_{\qquad\cong}^{\qquad\rho} "RU";"LU"};
{\ar@{->}^{\rho'}_{\cong} "RU";"RD"};
{\ar@{->}^{\qquad\sigma} "LD";"RD"};
\end{xy}}
\noindent where $\rho, \rho'$ are the restriction maps, which give isomorphisms of the spaces. As we did with $\check{H}^1_{dR}(\mathcal{U})$, we have a similar definition for $\check{H}^1_{dR}(\mathcal{U''})$, namely as the quotient of
\begin{multline*} \{(\omega_1, \omega_2, \omega_3, f_{12}, f_{13}, f_{23}) \mid \omega_j \in \Omega_{X/k}(U_j), f_{jk} \in \mathcal{O}_X(U_j \cap U_k)\\ df_{jk}=\omega_j-\omega_k, f_{23}-f_{13}+f_{12}=0\},
\end{multline*}
by the subspace spanned by \[ \{(df_1, df_2, df_3, f_1-f_2, f_1-f_3, f_2-f_3) \mid f_j \in \mathcal{O}_X(U_j)\}.\] Then $\rho$ and $\rho'$ are respectively the projections to $(\omega_1, \omega_2, f_{12})$ and $(\omega_2, \omega_3, f_{23})$.

\begin{lemma}\label{lem:nubasisHdR}
For $i = 1, \dots, g$, let
\begin{align*}
\omega_{1i} &= (1-2i)x^{-g-i+1}d(yx^{-g-1})\\
\omega_{2i} &= -2ix^{p-i-1}dy\\
\omega_{3i} &= \sum_{m=1}^{p-i} \binom{p-i}{m} (1+2m)(x-1)^{m-3g} d(y(x-1)^{-g-1})\\
f_{12i} &= yx^{-i}\\
f_{23i} &= \sum_{m=1}^{p-i} \binom{p-i}{m}y(x-1)^{m-p}=\left(\frac{x^m-1}{x^p-1}\right)y\\
f_{13i} &= yx^{-i} +\left(\frac{x^m-1}{x^p-1}\right)y.
\end{align*}
Let $\nu_i =(\omega_{1i}, \omega_{2i}, \omega_{3i}, f_{12i}, f_{13i}, f_{23i})$. If $i=1, \dots, g$, then $\nu_i$ is a well-defined hyper 1-cocycle for the covering $\mathcal{U}''$. Moreover, under the canonical refinement map $\check{H}^1_{dR}(\mathcal{U}'') \stackrel{\sim}{\to} \check{H}^1_{dR}(\mathcal{U})$, the image of each $\nu_i$ is $\eta_i$.
\end{lemma}

\begin{proof} Since it is clear that the projection of $\nu_i$ to $\check{H}^1_{dR}(\mathcal{U})$ is $\eta_i$, it suffices to show that each $\nu_i$ is well-defined.

Since we know from Lemma \ref{lem:basisdR} that the $\eta_i$ are well-defined, we can conclude that $\omega_{1i} \in \Omega_{X/k}(U_1), \omega_{2i} \in \Omega_{X/k}(U_2)$, $f_{12i} \in \mathcal{O}_X(U_1 \cap U_2)$, and $df_{12i}= \omega_{1i}-\omega_{2i}$ for $i=1, \dots, g$. It then remains to verify the following conditions:
\begin{enumerate}
\item $\omega_{3i} \in \Omega_{X/k}(U_2)$
\item $f_{23i} \in \mathcal{O}_X(U_2~\cap~U_3)$
\item $f_{13i} \in \mathcal{O}_X(U_1~\cap~U_3)$
\item $df_{23i} = \omega_{2i} - \omega_{3i}$
\item $df_{13i} = \omega_{1i}-\omega_{3i}$
\item $f_{23i} - f_{13i} + f_{12i} = 0$
\end{enumerate}
Note that 6 is clear from the definition, and that with the other conditions it will imply 5. Next we will show that conditions 1, 2 and 4 on $f_{23i}$ and $\omega_{3i}$ follow from previous calculations. We know from Lemma~\ref{lem:basisdR}, that $(1-2i)x^{1-g-i}d(yx^{-g-1}) \in \Omega_{X/k}(U_1)$, $-2ix^{2g-i}dy \in \Omega_{X/k}(U_2)$, and that their difference is $d(yx^{-i}) \in \Omega_{X/k}(U_1 \cap U_2)$. Changing coordinates by replacing $x$ with $x-1$, we can conclude that 1-forms that previously were regular on $U_1$ will now be regular on $U_3$, while those regular on $U_2$ remain so, and that the equality still holds. Thus we get that
\begin{equation}\label{eq:calcbasis1}\displaystyle
(1-2i)(x-1)^{1-g-i}d(y(x-1)^{-g-1})+2i(x-1)^{2g-i}dy =d(y(x-1)^{-i})
\end{equation}
where \begin{align*}
(1-2i)(x-1)^{1-g-i}d(y(x-1)^{-g-1}) &\in \Omega_{X/k}(U_3)\\
-2i(x-1)^{2g-i}dy &\in \Omega_{X/k}(U_2)\\
y(x-1)^{-i} &\in \mathcal{O}_X(U_2~\cap~U_3).
\end{align*} If we substitute $\displaystyle i=p-m$ into (\ref{eq:calcbasis1}) (recalling that $p=2g+1$), it becomes
\begin{equation*}
(1+2m)(x-1)^{m-3g}d(y(x-1)^{-g-1}) - 2m (x-1)^{m-1}dy = d(y(x-1)^{m-p}).
\end{equation*}
Now take, multiply it by $\displaystyle \binom{p-i}{m}$, take the sum of these from $m=1$ to $m=p-i$. Referencing back to the definitions of $\omega_{3i}$ and $f_{23i}$, we find
\begin{equation}\label{eq:interm}
-\omega_{3i} + \sum_{m=1}^{p-1} \binom{p-1}{m} 2m (x-1)^{m-1}dy = df_{23i}.
\end{equation}
It follows from this that conditions 1 and 2 hold, that is $\omega_{3i} \in \Omega_{X/k}(U_2)$ and $f_{23i} \in \mathcal{O}_{X/k}(U_2 \cap U_3)$. Now note that
\begin{align*}
\omega_{2i} &= -2i x^{p-i-1}dy\\
&= -2i \sum_{\ell=0}^{p-i-1} \binom{p-i-1}{\ell} (x-1)^{\ell}dy\\
&= \sum_{m=1}^{p-i} \binom{p-i}{m} 2m (x-1)^{m-1}dy.
\end{align*}
So we can substitute this into (\ref{eq:interm}), and it follows that $\omega_{2i}-\omega_{3i}=df_{23}$, which is condition 4.

The only thing that remains is to check that $f_{13i} \in \mathcal{O}_X(U_1 \cap U_3)$. Since $f_{23i}$ and $f_{12i}$ are regular on $U_1 \cap U_2 \cap U_3$, we need only check that $f_{13i}$ is regular at $P_\infty$. We can do this by calculating the image of $f_{13i}$ in $\operatorname{Frac}(\mathcal{O}_{X,P_\infty})$. Recall from the proof of Lemma~\ref{lem:basisdR}, that $\displaystyle t = \frac{y}{x^{(p+1)/2}}$ is a uniformizer at $P_\infty$. Using this and the definition of $f_{13i}$, we calculate that $f_{13i}=O(t)$. Since this means $f_{13i}$ has no pole at $P_\infty$, we can conclude $f_{13i}$ is regular on $U_1 \cap U_3$. So $f_{13i} \in \mathcal{O}_X(U_1 \cap U_3)$.
\end{proof}

Now we conclude the proof of Theorem \ref{nosplit}. By Lemma \ref{lem:nubasisHdR}, each $\nu_i$ is the inverse image of $\eta_i$ under the canonical restriction map $\check{H}_{dR}^1(\mathcal{U}'') \stackrel{\sim}{\to} \check{H}_{dR}^1(\mathcal{U})$. The projection of $\nu_i$ onto $\check{H}_{dR}^1(\mathcal{U}')$ gives us $(\omega_{2i}, \omega_{3i}, f_{23i})$, and to understand $\sigma \eta_i$, we need only calculate $\sigma(\omega_{2i}, \omega_{3i}, f_{23i})$.

So we need to consider the image of $(\omega_{3i}, \omega_{2i}, f_{23i})$ under $\sigma$:
\begin{align*}
\sigma \omega_{3i} &= -\frac{i}{p-i}\sum_{j=1}^{p-i}\binom{p-i}{j}(1+2j)x^{j-3g}d(yx^{-g-1})\\
\sigma \omega_{2i} &= -2i(x+1)^{p-i-1}dy = -\frac{i}{p-i}\sum_{j=1}^{p-i}\binom{p-i}{j}2jx^{j-1}dy\\
\sigma f_{23i} &= -\frac{i}{p-i}\sum_{j=1}^{p-i}\binom{p-i}{j}yx^{j-p}.
\end{align*}
We can restate this as
\begin{equation}\label{eq:sigmadoes}
\sigma(\omega_{3i}, \omega_{2i}, f_{23i}) = -\frac{i}{p-i}\sum_{j=1}^{p-i}\binom{p-i}{j} \tilde{\eta}_{p-j}
\end{equation}
where
\begin{equation*}
\tilde{\eta}_\ell = ((1-2\ell)x^{1-g-i}d(yx^{-g-1}), -2\ell x^{2g-\ell}dy, yx^{-\ell}).
\end{equation*}
We would like to be able to write this in terms of the basis from Lemma~\ref{lem:basisdR}. If $g+1 \le j \le p-1$, then each $\tilde{\eta}_{p-j} = \eta_{p-j}$, which is such a basis element. However, for $1 \le j \le g$, this is not the case and so we need to rewrite $\tilde{\eta}_{p-j}$ in terms of the basis. We find that for $1 \le j \le g$,
\begin{equation*}
\tilde{\eta}_{p-j}-(d(yx^{j-p}), 0, yx^{j-p})=(2j x^{j-1}dy, 2j x^{j-1}dy, 0) = -2j\tau_{j-1}.
\end{equation*}
Since the above is in the same equivalence class as $\tilde{\eta}_{p-j}$ within $\check{H}_{dR}^1(\mathcal{U})$, for $1 \le j \le g$, $\tilde{\eta}_{p-j}$ is in the image of $H^0(X, \Omega^1_{X/k})$.  This shows that the space spanned by the $\eta_i$ is not stable under the action of $\sigma$.

Suppose there is a $k[G]$-linear map $\alpha: H^1(X, \mathcal{O}_X) \rightarrow H^1_{dR} (X/k)$ that splits the exact sequence in Equation (\ref{eq:exactseq}). Let $\{ \tilde{\tau_\ell}^* \in H^1(X, \mathcal{O}_X) \mid \ell=0, \dots, g-1 \}$ be a dual basis of $\{\tilde{\tau_\ell} \in H^0(X, \Omega^1_{X/k}) \mid \ell=0, \dots, g-1 \}$ under the Serre duality pairing. Then define $\tau_\ell^* = \alpha(\tilde{\tau_\ell}^*) \in H^1_{dR} (X/k)$. We would like to know how to write $\tau_\ell^*$ in terms of the $\tau_i$ and $\eta_i$, the basis given in Lemma~\ref{lem:basisdR}.

Fix $t \in {\bf F}_p^\times$. and let $T$ indicate the element of $G$ such that \[(x,y) \mapsto (t^2x, ty).\]
It is easy to calculate that $T\tau_i = t^{2i+1}\tau_i$ and $T\eta_{j} = t^{1-2j}\eta_j$. Since the pairing is $G$-equivariant, as is the map $\alpha$, it has to be the case that $T\tau_\ell^* = t^{-2\ell-1}\tau_\ell^*$. This implies that $\tau_\ell^*$ is in the $t^{-2\ell-1}$-eigenspace of $T$, which is spanned by $\tau_{g-\ell-1}$ and $\eta_{\ell+1}$. So there exist unique nontrivial $a_\ell, b_\ell \in k$ such that $\tau_\ell^* = a_\ell \tau_{g-\ell-1} + b_\ell \eta_{\ell+1}$.

Since $\alpha$ is $G$-equivariant, $\sigma \tau_\ell^* = \sigma \alpha(\tilde{\tau_\ell}^*) =\alpha(\sigma \tilde{\tau_{\ell}}^*)$, so it must be that
\begin{equation}\label{asbs} a_\ell (\sigma \tau_{g-\ell-1}) + b_\ell (\sigma \eta_{\ell+1}) = \sigma \tau_\ell^* \in \operatorname{im} \alpha.\end{equation} However, we know from the proof of Theorem~\ref{irrHom} and Equation~\ref{eq:sigmadoes} that
\[ \sigma \tau_{g-\ell-1} = \sum_{i=0}^{g-\ell-1} \binom{g-\ell-1}{i} \tau_i\]
and
\[\sigma \eta_{\ell+1} = -\frac{\ell+1}{p-\ell-1}\left(-2 \sum_{i=1}^{g}\binom{p-\ell-1}{i}i \tau_{i-1}+\sum_{i=g+1}^{p-\ell-1} \binom{p-\ell-1}{i} \eta_{p-i}\right).\]

It is simple to show, by plugging this into Equation (\ref{asbs}), that no nontrivial $a_\ell, b_\ell$ satisfy this, giving a contradiction. Thus the exact sequence in Equation (\ref{eq:exactseq}) does not split under the action of $G$. \qed

\begin{cor}
$H^1_{dR} (X/k)$ is a non-projective indecomposable $k[G]$-module.
\end{cor}
\begin{proof}
Corollary 7 on page 33 of \cite{Alp} states that the order of a Sylow $p$-group of $G$ divides the dimension of any projective $k[G]$-module. A Sylow $p$-group of $G$ has order $p$ and the dimension of $H^1_{dR} (X/k)$ is $2g=p-1$. So $H^1_{dR} (X/k)$ is non-projective.

Suppose there exist nontrivial $k[G]$-submodules $M$ and $N$ such that \[H^1_{dR} (X/k) = M \oplus N. \]
Consider the sequence from Theorem~\ref{nosplit}:
\begin{equation}\label{eq:unsplittable}
0 \rightarrow H^0(X, \Omega^1_{X/k}) \xrightarrow{\psi} H^1_{dR} (X/k) \xrightarrow{\psi'} H^1(X, \mathcal{O}_X) \rightarrow 0 \end{equation}
where $\psi, \psi'$ are the canonical maps. Suppose that $\operatorname{im} \psi \cap M=0$. Then $\operatorname{ker} \psi' \cap M = 0$, so $\psi'|M$ gives an isomorphism between $M$ and some submodule of $H^1(X, \mathcal{O}_X)$. However, since $H^1(X, \mathcal{O}_X)$ is irreducible by Corollary~\ref{irreducible}, this means that $M$ is isomorphic to $H^1(X, \mathcal{O}_X)$ as a $k[G]$-module. However, this gives a splitting of the exact sequence in (\ref{eq:unsplittable}), which contradictions Theorem~\ref{nosplit}.

Now suppose instead that $\operatorname{im} \psi \cap M$ is nonzero. Since $\operatorname{im} \psi \cong H^0(X, \Omega^1_{X/k})$ as a $k[G]$-module, and $H^0(X, \Omega^1_{X/k})$ is irreducible by Corollary~\ref{irreducible}, we can conclude that $\operatorname{im} \psi \subseteq M$. So $\operatorname{ker} \psi' \subseteq M$, and thus $\operatorname{ker} \psi' \cap N$ is zero. Thus, $N$ is isomorphic to a $k[G]$-submodule of $H^1(X, \mathcal{O}_X)$ through $\psi'$. Since $N$ is nontrivial and $H^1(X, \mathcal{O}_X)$ is irreducible by Corollary~\ref{irreducible}, this means that $\psi'$ induces an isomorphism between $N$ and $H^1(X, \mathcal{O}_X)$. However, this gives a splitting of the exact sequence in (\ref{eq:unsplittable}), which contradicts Theorem \ref{nosplit}.
\end{proof}

\bibliographystyle{plain}

redpajama_set_name: RedPajamaArXiv


text:
The 5 Most User-Friendly Music Festival Apps
By Julia Cook | June 3, 2014 | 9:30am
Photo via Flicker User Shawn Ahmed Design Lists
You're probably not going to decide which festival to attend based on the app you use to keep track of what bands to see where, but whether or not it crashes every five seconds definitely affects your experience. Choose wisely, unless you feel like a sharpie-filled arm or crumpled piece of paper is a better place for your festival agenda.
5. SXSWfm, by SXSW LLC
There's a reason the actual SXSW app doesn't make this list. It does every possible thing wrong: moves slowly, crashes often and doesn't sync your schedule between phone and computer (and sometimes fails to save it at all). The redeemer for the developers at SXSW is the radio app, which was a pioneer for other festival stations, and runs year round. The station features genre-specific shows, including Americana, Hip Hop and Indie. Chronolog, a program with Austin Chronicle's Chase Hoffberger and Kevin Curtin, showcases the local music scene, so you can hit up SXSW and ACL acting like a local, even when you're not. Their most popular programs are archived, and the app features an alarm clock so you can wake up in the know.
4. Newport Folk Festival, by Cantina
This refreshingly simple app has the essentials: map, news and lineup. You can't even buy your tickets on it, and that's what makes it great. Who downloads a fest app when they don't have tickets? Exactly. The schedule has the standard features (put a star next to the band you want to see), but with a literal twist: tilt your phone to switch between list and timeline format. It's a seemingly obvious touch that gives the app universal appeal.
3. Firefly, by Aloompa LLC
This one is the most extensive fest assistant that's made by the creators of the Governor's Ball, Electric Daisy and Hangout apps. These guys are known for their cute photo booth function, which lets you add souvenir festival filters to your photos and upload them directly to your social media outlets. This one has—you guessed it—fireflies to waft gracefully around you and your friends. Complete with festival map, Slacker radio station and an outline of the grounds' attractions, this app covers almost all the bases. Unfortunately, it lacks a live stream function, which the Aloompa guys included on their Hangout version. Another drawback is that the schedule doesn't include Thursday shows—yet. However, it does take you directly from your band of choice to their place on the map, which is good for those of us who aren't as proficient at remembering which part of the forest has which down-home-hipster name.
2. Coachella, by Aloompa and Goldenvoice
The Coachella app comes with photo booth courtesy of Aloompa (see above), and blows the rest out of the water in terms of setup. When you download the app, you're asked to specify your social media and notification preferences (emergencies, ticket info, etc). The app then uses this information to activate your wristband straight from your phone. Like the Hangout app, it includes the weather on the front page (why don't they all do this?), and you can view the lineup in grid, timeline, or list format, by day or stage. It improves on Firefly and SXSW with its Fun & Essentials page, which tells you where to find things like charging stations, dodgeball, and of course, the trademark ferris wheel.
1. Summerfest, by Crowd Torch
Arguably the world's first modern music festival (the inaugural Summerfest took place a month before Woodstock), Summerfest also boasts the most comprehensive festival app. Compiling the usual lineup, map, radio, photo booth, and social media, the Summerfest app also offers an admission promotions page, which allows visitors to scroll through and see if buying a ticket (also possible using the app) is really necessary that day. You can even donate to Goodwill, buy a motorcycle, or find a buddy in the military to be admitted for free on various days during the festival. The schedule is also extremely user-friendly, allowing users to flag their favorite artists, write notes on them, and share these with their social media followers. It beats scribbling on the back of your Shaky Knees booklet, that's for sure.
hangout fest
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Cybercom and our Innovation Zone are involved in creating the smart factories of the future. In concrete terms, the Smart Factories project involves students building two physical factories in Gothenburg and Skövde. They are "mini-factories", but the technology being loaded into them is anything but "mini". For instance, the factories are fully automated.
Over the long run, Smart Factories are meant to inspire more digitalised, automated and sustainable industry. The project is being run by Göteborgs Tekniska College and was made possible by Cybercom and other corporate partners.
Smart Factories is being run through student projects, specifically intended to inspire young people to choose technical studies. In total, there will be about 21,000 hours of student jobs in fields including virtual commissioning, sustainable production and programming of automated processes.
Cybercom is contributing to the Smart Factories project in several ways, including data collection in the Cloud, AR visualisation and the factory assembly guide, the Smart Factories mobile app and the Machinebook service.