text:
Israel Pulls Funding From UN Following World Heritage Site Vote
The United Nations announced a secret vote making Hebron in the West Bank a Palestinian heritage site, leaving many Israelis upset.
Article by Mitchel Summers
The city of Hebron in the West Bank is now a United Nations World Heritage site — and Israel pulled $1 million in funding from the U.N. in response.
UNESCO, the U.N.'s cultural committee, designated the city as a heritage site in danger. The vote passed 12-3 by secret ballot, with six abstentions.
Trouble is, UNESCO declared the city a Palestinian heritage site. Over 200,000 Palestinians live in the city, but so do hundreds of Jewish settlers who claim it's Israeli land.
Israel said UNESCO is ignoring its mission to foster cultural cooperation by overlooking the city's significance to Judaism.
The decision covers a site known both as the "Tomb of the Patriarchs" and the "Ibrahimi Mosque." It's a key holy site in both Judaism and Islam.
President Trump Is Confident That Israel And Palestine Will Find Peace
President Donald Trump met with Palestinian President Mahmoud Abbas and expressed desire for peace between Israel and Palestine.
Jews and Muslims currently share the holy site, but anyone who visits it has to pass through Israeli checkpoints.
It's the second-holiest site in Judaism and the fourth-holiest site in Islam. Both religions agree the patriarch Abraham, as well as several other holy figures, are buried there.
Israeli Prime Minister Benjamin Netanyahu called the move "another delusional UNESCO decision." He said Israel's pulled U.N. funding would go to establishing a Jewish heritage museum instead.
Latest in World
Brazil's Bolsonaro applies for 6-month US visitor visa
The former president applied for the visa amid an investigation into whether he had any role in inciting an uprising in Brazil's capital.
By Eraldo Peres / AP
Labor unions aimed to mobilize more than 1 million protesters to kill a bill that would raise the retirement age from 62 to 64.
By Christophe Ena / AP
Israel's new government sparks concern for the future of its democracy
Far-right changes to Israel's government have some experts and citizens concerned for the future rule of law in the country.
By Reuters / AP
Nichols family to address public for first time since video release
Attorney Ben Crump said the news conference will include other community members who were "victims" of alleged brutality from Memphis Police.
By Gerald Herbert / AP
The Allegheny Family Screening Tool is designed to assess a family's risk level when they are reported for child welfare concerns.
By Keith Srakocic / AP


text:
Home / Team / Solicitors
Laura O'Farrell - Solicitor
Laura is a solicitor in the Commercial & Business team at Hayes Solicitors. She practices in commercial litigation and dispute resolution and acts for a variety of companies, financial institutions, State bodies and individuals in contract law cases, enforcement and recovery actions and general commercial litigation matters. Laura also provides advice on a wide range of commercial and business law matters and regulatory requirements including data protection law and reviewing and drafting commercial contracts.
Laura joined the firm as a trainee solicitor in 2017 and qualified in 2019.
Debt Proceedings - Laura advises clients in relation to debt recovery proceedings before the District Court, Circuit Court and High Court.
Enforcement of security - Laura has assisted in a number of high profile enforcement matters before the High Court. She also has experience of enforcing security in Australia, having acted for a number of top Australian banks prior to joining the firm. Laura also worked for the National Asset Management Agency assisting with the completion of large loan portfolio sales.
Professional regulatory matters - Laura has assisted in defending Fitness to Practice complaints and defence of civil and criminal matters in the professional sphere. She has assisted in defending professional negligence actions and internal appeals to committees of professional bodies.
Defamation - Laura has assisted in defending numerous high value defamation cases.
Personal injury – Laura has been involved in a number of personal injury cases, both defending and prosecuting, in the areas of employers' liability and public liability.
Data protection law - Laura advises companies and State bodies on general data protection obligations as well as specific queries in respect of data access requests.
In-house experience — Laura completed a secondment in the Employment Litigation Team of a high profile Irish commercial bank.
Qualifications/Education
BA (Legal Science and Economics), National University of Ireland Galway
LL.B, National University of Ireland Galway
Professional qualification as a solicitor, Law Society of Ireland, 2019
Member of the Law Society of Ireland
View Short Bio
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lofarrell@hayes-solicitors.ie
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redpajama_set_name: RedPajamaCommonCrawl


It's unbelievably bold! Playing by the rules?? That's not for you!!! Being a good girl? What for?!
Be yourself - bright, spectacular, defiantly bold! Be in the middle! Play!
Bright raspberry, burning sandalwood, sensual musk and fresh lily of the valley - a game of contrasts! Your game! Declare yourself! Everything is possible!
Top notes: raspberry, red apple, violet.
Heart notes: freesia, lotus flower, lilly of the valley.
Bottom notes: sandalwood, jasmine, musk.
Alcohol Denat., Parfum (Fragrance), Aqua (Water), Benzyl Benzoate, Butylphenyl Methylpropional, Limonene, Benzyl Salicylate, Hexyl Cinnamal, Linalool, Citronellol, Geraniol, Alpha Isomethyl Ionone, Anise Alcohol, Citral, Benzyl Alcohol, Amyl Cinnamal, Isoeugenol.


text:
Ellen R. Kordik
Lecturer in Law
Jackie Walters
Professor Ellen Kordik received her bachelor of science in biology from IIT in Chicago, her master of science in microbiology at the University of Illinois, and her JD in 1986 from the College of Law, where she was an articles editor for the University of Illinois Law Review.
In 1986, she joined the law firm of Jenner & Block in Chicago and was a litigation partner there until joining our faculty in 1997. Professor Kordik has previously taught as an adjunct at many of the Chicago-area law schools. She teaches complex litigation, professional responsibility, statutory interpretation (Americans with Disabilities Act), business organizations, and products liability.
MS, JD University of Illinois
BS Illinois Institute of Technology
Law Teaching Practicum


text:
Acts	 Genesis Exodus Leviticus Numbers Deuteronomy Joshua Judges Ruth 1 Samuel 2 Samuel 1 Kings 2 Kings 1 Chronicles 2 Chronicles Ezra Nehemiah Esther Job Psalms Proverbs Ecclesiastes Song of Solomon Isaiah Jeremiah Lamentations Ezekiel Daniel Hosea Joel Amos Obadiah Jonah Micah Nahum Habakkuk Zephaniah Haggai Zechariah Malachi Matthew Mark Luke John Acts Romans 1 Corinthians 2 Corinthians Galatians Ephesians Philippians Colossians 1 Thessalonians 2 Thessalonians 1 Timothy 2 Timothy Titus Philemon Hebrews James 1 Peter 2 Peter 1 John 2 John 3 John Jude Revelation
"Now when they heard this, they were pricked in their heart, and said unto Peter and to the rest of the apostles, Men and brethren, what shall we do?"
Other Translations of Acts 2:37
"� Now when they heard this, they were pricked in their heart, and said vnto Peter, and to the rest of the Apostles, Men and brethren, What shall we doe?"
King James Version (1611) - View original scan of Acts chapter 2
"Now when they heard {this,} they were pierced to the heart, and said to Peter and the rest of the apostles, "Brethren, what shall we do?""
"Now when they heard `this,' they were pricked in their heart, and said unto Peter and the rest of the apostles, Brethren, what shall we do?"
"Now when these words came to their ears their hearts were troubled, and they said to Peter and the other Apostles, Brothers, what are we to do?"
"And having heard [it] they were pricked in heart, and said to Peter and the other apostles, What shall we do, brethren?"
"Now when they had heard these things, they had compunction in their heart, and said to Peter, and to the rest of the apostles: What shall we do, men and brethren? "
"Now when they heard this, they were pricked in their heart, and said to Peter and to the rest of the apostles, Men, brethren, what shall we do? "
"Stung to the heart by these words, they said to Peter and the rest of the Apostles, "Brethren, what are we to do?""
"Now when they heard this, they were cut to the heart, and said to Peter and the rest of the apostles, "Brothers, what shall we do?""
"Whanne thei herden these thingis, thei weren compunct in herte; and thei seiden to Petre and othere apostlis, Britheren, what schulen we do?"
"And having heard, they were pricked to the heart; they say also to Peter, and to the rest of the apostles, `What shall we do, men, brethren?'"


Tension over the global economic slump have fuelled May Day protests and riots across Europe with trouble breaking out in Germany, Greece, Austria, Turkey and France.
The government had declared May Day, traditionally marked by rallies by labour unions, a public holiday this year under pressure from unions.
In Greece, officers fired flash grenades to disperse rioters in Athens after attacks on banks and traffic cameras. Transport strikes disrupted bus, train and ferry services as well as flights by Greek carrier Olympic Airlines.
Twenty people were injured and five arrested after police clashed with demonstrators at a traditional labour day rally in Linz in northern Austria. The incident came after police intercepted some 50 hooded protesters among a procession of up to 700 people at a Communist Party rally.
In France, unions joined forces for the first time since the Second World War, but turnout was not as high as a protest day in March in which up to three million attended 300 rallies against President Nicolas Sarkozy's economic recovery plan.
Some politicians have warned of the threat of "revolution", following radical action such as "boss-napping" factory executives and an ongoing strike that has crippled dozens of French universities.
They are wary of the growing power of leftist radicals linked to the New Anti-Capitalist Party of Olivier Besancenot, a Trotskyist postman.
There were also marches in big cities in Spain, burdened with the highest unemployment rate in Europe. More than 10,000 people gathered in the centre of Madrid in a demonstration organised by the country's two largest trade unions.
In Italy, union leaders shifted rallies from major cities to the earthquake-stricken town of L'Aquila as a sign of solidarity with thousands who lost their jobs after last month's deadly quake.
In Russia, about 2,000 demonstrators gathered by a statue of Karl Marx in Moscow waving banners and red Soviet flags and calling for a return of communism.
In St Petersburg, birthplace of the 1917 Bolshevik Revolution, police arrested about 120 members far right militants armed with knives and knuckle dusters, police said.


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Fact-checking the 2nd night of the Democratic debate in Miami
By PolitiFact Staff June 27, 2019
This story will be updated with more fact-checks.
Democrats let loose their anti-Trump passions while attempting to showcase their diverse vantage points on gun control, race relations, immigration, healthcare, tax reform and more during the second night of debates at Miami's Adrienne Arsht Center.
Former Vice President Joe Biden and Sen. Bernie Sanders were the two most recognizable men positioned center-stage. But eight lesser-known candidates in the race to be the next president of the United States sparred with them.
Sen. Kamala Harris of California and South Bend, Ind., Mayor Pete Buttigieg delivered passionate remarks that especially sought to challenge Biden as the face of an old-guard lacking the progressive vision the party needs to move America forward.
But when passions flare, sometimes facts fizzle. Here's a look at some of the statements that got our attention as we monitored the second night of the 2020 Democratic debates.
Joe Biden: "I did not oppose busing in America. What I opposed is busing ordered by the Department of Education. That's what I opposed."
We found that Biden tried to oversimplify his complicated record on busing. We rated his statement Mostly False.
Biden won his Senate seat in 1972 on a platform of integration, but once in office, he faced white constituents who hated the idea of busing. A federal court order in 1974 to integrate Wilmington schools brought the issue home for Biden. In the Senate, Biden began supporting much of the anti-busing legislation.
University of New Hampshire historian Jason Sokol, who tracked Biden's response to local opposition, has told us the amendments that Biden supported "paint a somewhat muddled picture." Biden proposed some amendments that would have hamstrung desegregation while at other times he backed off.
Sokol told us after the debate that Biden's position on busing evolved but by the late 1970s he was decidedly against busing.
Biden's claim that he was for integration but against busing doesn't make a lot of sense, he told PolitiFact. "By that point in history, there were very few school districts voluntarily integrating by other means, which is why judges were ordering busing. He is using disingenuous logic," he said.
— Amy Sherman
Bernie Sanders: "Today the worker in the middle of our economy is making no more money than he or she made 45 years ago."
This is exaggerated.
For instance, according to data analyzed by the left-of-center Economic Policy Institute, the median hourly wage in 1973 was $16.96. Adjusted for inflation, the median wage rose to $18.80 by 2018.
While that's not a dramatic rise — especially over 45 years — it amounts to an 11% increase beyond inflation.
Another study, by the nonpartisan Congressional Budget Office, found that the cumulative growth in inflation-adjusted income for those in the middle 60% of the income distribution rose by 46% between 1979 and 2015. That was much slower than it was for the top 1% (which was a whopping 242%), but it was still an increase beyond inflation.
— Louis Jacobson
John Hickenlooper: "I share the sense of urgency. I'm a scientist, so I recognize that we're within 10 or 12 years of actually suffering irreversible damage (of climate change)."
While over 90% of publishing climate scientists say humans are causing global warming, the idea that scientists agree on a hard deadline lacks nuance. (Sanders also made reference to a 12-year deadline)
A 2018 United Nations Intergovernmental Panel on Climate Change report notes 2030 as a benchmark to deal with climate change "because signatories to the Paris agreement have pledged emission cuts by then," according to the Associated Press.
The panel's report predicts that, if warming continues at its current rate, global temperatures are likely to reach 1.5 degrees Celsius above pre-industrial levels between 2030 and 2052.
James Skea, a co-chairman of the report and sustainable energy professor at Imperial College London, told the Associated Press that the panel "did not say we have 12 years left to save the world."
And scientists have not agreed that humans have until 2030 to address climate change.
Kristie L. Ebi, director of the Center for Health and the Global Environment at the University of Washington in Seattle, told the Associated Press that, while adding greenhouse gases into the atmosphere will result in a continued rise in temperatures, "the earth does not reach a cliff at 2030 or 2052."
— Sophie Austin
Bernie Sanders: "President Trump, you're not standing up for working families when you try to throw 32 million people off the health care that they have."
This is one of Sanders' favorite lines, but it falls short of giving the full story of the Republican effort to repeal and replace Obamacare. We rated a similar claim Half True.
Scrapping the Affordable Care Act was a key campaign promise for President Donald Trump. In 2017, as the Republican-led Congress struggled to deliver, Trump tweeted, "Republicans should just REPEAL failing Obamacare now and work on a new health care plan that will start from a clean slate."
The Congressional Budget Office estimated that would lead to 32 million more people without insurance by 2026. But some portion of that 32 million would have chosen not to buy insurance due to the end of the individual mandate, which would happen under repeal. (It happened anyway, but that was part of the 2017 tax law.)
In the end, full repeal didn't happen. Instead, Trump was only able to zero out the fines for people who didn't have insurance. Insurance coverage has eroded. The latest survey shows about 1.3 million people have lost insurance since Trump took office.
— Jon Greenberg
Michael Bennet: "Bernie mentioned that the taxes that we would have to pay — because of those taxes, Vermont rejected Medicare for All."
This is true, though it could use some context.
Vermont's effort to pass a state-based single-payer health plan — which the state legislature approved in 2011 — officially fell flat in December 2014. Financing the plan ultimately required an 11.5% payroll tax on all employers, plus raising the income tax by as much as 9.5%. The governor at the time, Democrat Peter Shumlin, declared this politically untenable.
That said, some analysts suggest other political factors may have played a role, too — for instance, fallout after the state launched its Affordable Care Act health insurance website, which faced technical difficulties.
Nationally, when voters are told Medicare for All could result in higher taxes, support declines.
— Shefali Luthra, Kaiser Health News
Eric Swalwell: "This president, though, for immigrants, there is nothing he will not do to separate a family, cage a child, or erase their existence by weaponizing the census."
This frequent Democratic attack against Trump needs context, as people use different words to describe the enclosed facilities that hold immigrant children arriving at the southwest border and are waiting to be processed. The physical structures separate children based on age or gender.
Trump's critics using the harshest terms: "cages" or "dog kennels." The federal government calls them "barriers."
It's important to note that the Obama administration was also criticized for using "cages," when record numbers of unaccompanied minors from Central America were apprehended at the southwest border.
The Arizona Republic in 2014 showed a photo of child migrants sleeping on mats behind a fence in Nogales, Az. The story referred to the migrants as "children in cages."
But immigration experts have told us that family separations were relatively rare under Obama and other past administrations. They did not happen at nearly the scale that they are happening under the Trump administration.
It's unclear exactly how many children were separated from their parents during Trump's administration, which has acknowledged problems in its logistics and record-keeping. Under a court order, around 2,800 children have been reunited with their parents or otherwise discharged from federal custody.
The controversial family separations under Trump's watch happened as a result of a new policy introduced in April 2018 by Trump's then-Attorney General Jeff Sessions.
— Amy Sherman and Miriam Valverde
Andrew Yang: "We automated away 4 million manufacturing jobs in Michigan, Ohio and Pennsylvania and Wisconsin, and we're about to do the same thing to millions of retail jobs, call center jobs, fast food jobs, truck driving jobs and on and on through the economy."
That is mostly accurate on the number. But not all of the jobs were eliminated due to automation.
From 2000 to 2017, about 5.5 million American manufacturing jobs were lost, according to a 2018 report by the Bureau of Labor Statistics. The report lists as some of the reasons competition with China, the gap between the skills workers have and the skills employers need; and cross-regional migration.
Another study by Ball State University Center for Business and Economic Research attributed around 88 percent of job losses in manufacturing in recent years to productivity growth due to automation.
That number is also expected to rise. According to a 2017 report by the McKinsey Global Institute, between 39 to 73 million jobs may be eliminated in the United States by 2030, with fast food workers (as Yang said) and machinery operators projected to being hit the hardest.
— Samantha Putterman
Marianne Williamson: "So many Americans have unnecessary chronic illnesses — so many more compared to other countries."
There is evidence for this, at least for older Americans.
A November 2014 study by the Commonwealth Fund found that 68% of Americans 65 and older had two or more chronic conditions, and an additional 20% had one chronic condition.
No other country studied — the United Kingdom, New Zealand, Sweden, Norway, France, Switzerland, the Netherlands, Germany, Austria, or Canada — had a higher rate of older residents with at least two chronic conditions. The percentages ranged from 33 percent in the United Kingdom to 56 percent in Canada.
An earlier study published in the journal Health Affairs in 2007 found that "for many of the most costly chronic conditions, diagnosed disease prevalence and treatment rates were higher in the United States than in a sample of European countries in 2004."
Pete Buttigieg: (Addressing his response to a police shooting of a black man in his city) "Look, we've taken so many steps toward police accountability that, you know, the (Fraternal Order of Police) just denounced me for too much accountability."
This needs context. Buttigieg, mayor of South Bend, Ind., is under criticism over the handling of a fatal police shooting in his city, involving a white police officer and a black man. The Executive Board of the Fraternal Order of Police Lodge No. 36 was upset with Buttigieg for how he talked about police work in a campaign message about the shooting.
In a June 24 Facebook post, the board said Buttigieg was making "disparaging" remarks and focusing on one incident and not on other major shootings in the city.
Buttigieg had addressed the police involved shooting in a campaign email, saying an investigation was underway and that Eric Logan's death cast a "bright light" on a subject impacting Americans from all walks of life.
"All police work and all of American life takes place in the shadow of racism, which hurts everyone and everything it touches," Buttigieg's email said. "Historic racism, present-day racism, and generational racism — they all secrete a kind of poison into the bloodstream of this country. And we must join together to make things right, no matter how demanding that process may be."
The FOP executive board said Buttigieg's comments and actions were "driving a wedge" between law enforcement and South Bend. Buttigieg's focus on the incident was "solely for his political gain and not the health of the city," the board said.
"For Mayor and Presidential Candidate Pete Buttigieg to make disparaging remarks such as 'all police work and all of American life takes place in the shadow of racism' is divisive," the FOP executive board said in the Facebook post.
The FOP executive board said it stood in support of the officer involved, Sgt. Ryan O'Neill, and that until a thorough and complete investigation was done, it would not waver. Buttigieg said during the debate that he could not take the side of the victim or the police pending the investigation.
— Miriam Valverde
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The "ever-ready" -USA, China and Russia countries are not afraid for the WW3 as they know they may not win,but they know with they can bear suffering and enjoy giving a worst suffering to each other. USA and Russia are two countries that instigates a war with their invisible presence in the name of dictatorship oppression or hedgemony happening in other countries.
In fact, these countries' leaders are acting as dictatorship in place of peace process. The word "Democratic" or "Republic" doesn't make any difference to public, as dictators always have been oppressing people in some way. In a nation, people diversity is very natural, however, they need to understand that "If u a group needs a space, then, so does other groups".I want to know why MUSLIMS are targeted in media badly? When only 2% of crimes are conducted by them, where 98 % are others. U want to WW3 by instigating to attack US Soil by ARAB nations like the Pearl Harbour.
Rule No. 1 : Each human holds a right to choose a religion. Do not force others to choose or reprimand others' religion (Just follow what you like the best). Dialogue is better than war.
Rule No. 2 : Human wants to be like a Free bird flying in 3 dimensional atmosphere. So, let their inner dreams, desires and capacity to germinate.
Rule No.3: Human should value its virtue rather than anything. Everybody should understand it. It expounds the meaning of Helping others. Earning good Karma. Everything is weighed as good or bad. Financial interest makes life comfortable, doesn't give happiness totally.
Rule No. 4: If there are ever occur WW3, there are 190 nations worldwide. All other nation should unite and start a dialogue against it because there's no point of WW3. Only 3 people's decision can not bring millions of human loss for no apparent human development. What have we gained from WW1 and WW2? Nothing but pains and sorrows. If there again ever be war, then, the front-liners should be President, Vice-presidents, Government administrators, then, Military leaders and then, Public. Will they be ready? it is easy to place others in front-line of a war. Display your prowess of forgiveness, empathy and compassion. Not in raging a war.
5. Few people that can be easily counted on finger tips (politicians, finance leaders etc..), their ego & greed make WW3. 99% public, normal people do not like war. Now, these people should wake up.
I just want to say, "People Wake Up".
The ultimate truth is that "There is no real threat from each other. But, the fear hidden elusively inside in mind, will bring forth WW3".


text:
\section{Introduction}

In their fundamental work \cite{bk93}, Bloch and Kato used and developed
many techniques of what is now usually called $p$-adic Hodge theory,
initiated before in large parts by Fontaine. Bloch and Kato's focus was
the development of a general conjecture concerning special values of
$L$-functions, which culminated in their formulation of a version of
the Tamagawa number conjecture.

Working locally at a fixed prime $p$ and a fixed finite extensions $K
/ \mb{Q}_p$ with absolute Galois group $G_K$, we take a closer look at the
computations done in sections 3 and 4 in loc.cit., which depend in
certain situations on the property that the $p$-adic representation
$V$ of $G_K$ under consideration is ``in the Fontaine-Laffaille range''. This
is a condition on the filtration of the filtered $\varphi$-module
associated to $V$.

We introduce an auxillary integral ring $\A_{\trm{exp}}$, which, after
inverting $p$, computes the module $\D_{\mbox{\scriptsize cris}}(V)$ of a crystalline
representation, if one fixes a $G_K$-equivariant lattice $T \subset V$.  A
nice property of this integral version is that it contains already (up
to some $p$-powers) the inverses of the fundamental period $t$, so
that no awkward twisting to a positive representation is necessary.
Note that simply inverting $t$ in for example $\A_{\mbox{\scriptsize cris}}$ implies that
$p$ is then also already inverted, which leaves the integral world.

Using this ring, we show that one can construct a finite rank
Fontaine-Laffaille module $\D_{\trm{exp}}(T)$ out of $T$, which is used to connect the
$p$-adic valuation of the special value at $s = 0$ of the local
$L$-function $P(V, p^{-s})$ to a certain measure on this Fontaine-Laffaille module
(via Bloch-Kato's fundamental exact sequence), without any condition
on the filtration range of $V$:

\begin{thm}
  Asumme $K / \mb{Q}_p$ is unramified and let $V$ be a crystalline
  representation. Fix a $G_K$-equivariant lattice $T$ in $V$ and
  assume further that $P(V, 1) \not= 0$. Then:
  \begin{enumerate}
  \item $H^1(\D_{\trm{exp}}(T))[1/p] \cong H^1_e(K, V)$.
  \item $H^1(\D_{\trm{exp}}(T)) \cong H^1_e(K, T)$.
  \item $\exp_e : \D_{\mbox{\scriptsize dR}}(V) / \D_{\mbox{\scriptsize dR}}^0(V) \rightarrow H^1(K, V)$ coincides with the composite map
    \begin{align*}
      (\D_{\trm{exp}}(T) / \D_{\trm{exp}}^0(T)) [1/p] \overset{1 - \varphi}{\longrightarrow}
      (\D_{\trm{exp}}(T) / (1 - \varphi^0) \D_{\trm{exp}}(T)) [1/p] \\=
      H^1(\D_{\trm{exp}}(T))[1/p] \cong H^1(K, V),
    \end{align*}
    where the last canonical identification is explained in the proof.
  \end{enumerate}
\end{thm}
Here, $H^1_e(K, -)$ denotes the exponential part of $H^1(K, -)$, that
is, the image of the Bloch-Kato exponential map. As a corollary, we
obtain:
\begin{cor}
  Let $\mu$ be the Haar-measure on the finite-dimensional $K$-vector
  space $H^1(K, V)$ such that the image of the lattice
  \[
    \D_{\trm{exp}}(T) / \D_{\trm{exp}}^0(T) \subset \D_{\mbox{\scriptsize dR}}(V) / \D_{\mbox{\scriptsize dR}}^0(V) \overset{\sim}{\longrightarrow} H^1_e(K, V) = H^1_f(K, V)
  \]
  has measure $1$. Then
  \[
    \mu(H^1_e(K, T)) = | P(V, 1) |_p^{-1}.
  \]
\end{cor}

\section{Basic concepts from $p$-adic Hodge theory}

Fix a prime number $p$. Let $K / \mb{Q}_p$ be a finite extension of the
$p$-adic numbers, and denote by $G_K$ the absolute Galois group of
$K$. Let $K_0$ be the maximal unramified subextension of $K / \mb{Q}_p$.
Usually, $V$ will denote a $p$-adic representation, that is, a finite
dimensional $\mb{Q}_p$-vector space equipped with a continuous and linear
$G_K$-action. Similarly, $T$ will usually denote a $G_K$-stable
$\mb{Z}_p$-lattice in $V$. Such lattices always exist. One is interested in
the classification of such $V$ and $T$, and Fontaine's rings have
proven to be a powerful tool for this. We refer to \cite{fontouy} as a
basic reference.

Let $\mc{O}_{\mb{C}_p}$ be the ring of integers of the completion of an algebraic
closure $\mb{C}_p$ of $\mb{Q}_p$. Let $\widetilde{\E}^+ := \varprojlim_{x \mapsto x^p}
\mc{O}_{\mb{C}_p} / p$, which is a ring of charakteristic $p$, equipped with a
Frobenius $\varphi : x \mapsto x^p$, and a Galois action of $\sigma
\in G_K$ via $(x_n) \mapsto (\sigma(x_n))$. If $x = (x_k) \in
\widetilde{\E}^+$, let $x^{(0)} = \lim_{m \to \infty} \hat{x}_m^{p^m}$, where
$\hat{x}_m \in \mc{O}_{\mb{C}_p}$ is any lift of $x_m \in \mc{O}_{\mb{C}_p} / p$. This
defines a non-archimedean valuation $v : x \mapsto v_p(x^{(0)})$
on $\widetilde{\E}^+$.

Let $\widetilde{\A}^+ = W(\widetilde{\E}^+)$, the ring of Witt vectors of
$\widetilde{\E}^+$. This makes sense since $\widetilde{\E}^+$ is perfect, since it is a
perfection of the non-perfect ring $\mc{O}_{\mb{C}_p} / p$. $\widetilde{\A}^+$ is
a ring of charakteristic $0$ with Frobenius $\varphi :
\sum_{n \geq 0} [x_n] p^n \mapsto \sum_{n \geq 0} [\varphi(x_n)] p^n$,
and an action of $G_K$ that is defined analogously.

One has the important ring homomorphism
\[
\theta: \widetilde{\A}^+ \rightarrow \mc{O}_{\mb{C}_p},~~~ \sum [x_n] p^n \mapsto \sum x_n^{(0)}
p^n,
\]
which arises conceptually in Fontaine's theory of universal
thickenings.

We fix a system of $p^n$-th roots of unity $\varepsilon^{(n)} \in \mb{C}_p$
with $\varepsilon^{(0)} = 1, \varepsilon^{(1)} \not= 1$ and
$(\varepsilon^{(n+1)})^p = \varepsilon^{n}$. Then $\varepsilon =
(\overline{\varepsilon^{(n)}}) \in \widetilde{\E}^+$, where $\overline{x}$
means reduction $\mod p$. Let $\overline{\pi} := \varepsilon - 1$
(this notation is slightly unfortunate, but standard). One can show
that $v(\overline{\pi}) = \frac{p}{p-1}$.
Let $\pi := [\varepsilon] - 1 \in \widetilde{\A}^+$. Observe that $\pi \equiv
\overline{\pi} \mod p$, but $\pi \not= [\overline{\pi}]$.

Further let $\tilde{p} \in \widetilde{\E}^+$ with $\tilde{p}^{(0)} = -p$. Set
$\xi = [\tilde{p}] + p \in \widetilde{\A}^+$, i.e. $\theta(\xi) = -p + p =
0$. One can show that $\ker \theta = (\xi)$, since $\ker \theta
\subset (\xi, p)$, using the fact that $\widetilde{\A}^+$ is $p$-adically
complete and that $\mc{O}_{\mb{C}_p}$ does not have any $p$-torsion. More
generally, if $\xi' \in \widetilde{\A}^+$ such that $\theta(\xi') = 0$ and
$v(\overline{\xi'}) = 1$, then $\ker \theta = (\xi')$.

One defines $\B_{\mbox{\scriptsize dR}}^+ = \varprojlim_n \widetilde{\A}^+[1/p] / (\ker
\theta)$. $\theta$ extends to $\B_{\mbox{\scriptsize dR}}^+$, where $\ker \theta = (\xi)$
still holds. Let
\[
t = \log (1+\pi) = \sum_{n \geq 1}(-1)^{n+1}\frac{\pi^n}{n} \in
\B_{\mbox{\scriptsize dR}}^+,
\]
the fundamental period of $p$-adic Hodge theory. $t$ only depends on
the choice of a compatible system of $p^n$-th roots of unity.
Interestingly, one has $\ker \theta = (t)$. This shows that $\B_{\mbox{\scriptsize dR}}^+$
is a complete discrete valuation ring with maximal ideal $(t)$. $\B_{\mbox{\scriptsize dR}}
= \B_{\mbox{\scriptsize dR}}^+ [1/t]$ possesses the filtration $\textrm{Fil}^i \B_{\mbox{\scriptsize dR}} = t^i
\B_{\mbox{\scriptsize dR}}^+, i \in \mathds{Z}$. Since $\B_{\mbox{\scriptsize dR}}^+ / (t) \cong \mb{C}_p$, induced by
the map $\theta$, one has (algebraically, non-canonically) $\B_{\mbox{\scriptsize dR}} \cong
\mb{C}_p[[t]]$. But observe that the topology on $\B_{\mbox{\scriptsize dR}}^+$ is defined via
the inverse limit topology and the topology on $\widetilde{\A}^+$, which is
induced via the Witt-construction by the valuation topology on
$\widetilde{\E}^+$. With this topology, one still has a continuous action of
$G_K$, but the action of $\varphi$ does not extend to $\B_{\mbox{\scriptsize dR}}^+$.

This being the case one considers the ring $\A_{\mbox{\scriptsize cris}}$, which is defined
as the $p$-adic completion of the divided power envelope of $\widetilde{\A}^+$
with respect to the ideal $\ker \theta$, i.e.
\[
\widetilde{\A}^+[\frac{a^m}{m!}; a \in \ker \theta]^{\wedge} = \widetilde{\A}^+ \{
\frac{\xi^m}{m!} \} \subset \B_{\mbox{\scriptsize dR}}^+.
\]
If $x \in \A_{\mbox{\scriptsize cris}}$, then we may write (non-uniquely) $x = \sum_{n} a_n
\frac{\xi^n}{n!}$ with $a_n \in \widetilde{\A}^+$ and $a_n \to 0$ $p$-adically.
The map $\theta$ and the $\varphi$ and $G_K$-action extend to $\A_{\mbox{\scriptsize cris}}$.
Further, $t \in \A_{\mbox{\scriptsize cris}}$, since $\pi = b \xi$ ($\theta(\pi) = 0$) and
\[
  \frac{\pi^n}{n} = (n-1)! b^n \frac{\xi^n}{n!},~~ (n-1)! \to 0.
\]
Let $\B_{\mbox{\scriptsize cris}}^+ = \A_{\mbox{\scriptsize cris}}[1/p]$, $\B_{\mbox{\scriptsize cris}} = \B_{\mbox{\scriptsize cris}}^+[1/t] \subset
\B_{\mbox{\scriptsize dR}}$. $\B_{\mbox{\scriptsize cris}}^{G_K} = K_0$. Two facts about $\A_{\mbox{\scriptsize cris}}$ are:
$\varphi(\xi) \in p \A_{\mbox{\scriptsize cris}}$, and $t^{p-1} \in p \A_{\mbox{\scriptsize cris}}$, hence $\B_{\mbox{\scriptsize cris}}
= \A_{\mbox{\scriptsize cris}}[1/t]$. Two caveats about $\B_{\mbox{\scriptsize cris}}$ are: it has a funny
topology, since one can show that the topology induced on $\B_{\mbox{\scriptsize cris}}^+$
by $\B_{\mbox{\scriptsize cris}}$ (which comes equipped with the locally convex final
topology) is not the natural topology on $\B_{\mbox{\scriptsize cris}}^+$.  Furthermore,
$\B_{\mbox{\scriptsize cris}}^+ \subsetneq \textrm{Fil}^0 \B_{\mbox{\scriptsize cris}}$.

Let $V$ now be a $p$-adic representation. Let $\D_{\mbox{\scriptsize dR}}(V) = (\B_{\mbox{\scriptsize dR}}
\otimes_{\mb{Q}_p} V)^{G_K}$. This is a $K$-vector space, and the injectivity
of the canonical map
\[
  \alpha: \B_{\mbox{\scriptsize dR}} \otimes_K \D_{\mbox{\scriptsize dR}}(V) \rightarrow \B_{\mbox{\scriptsize dR}} \otimes_{\mb{Q}_p} V,~~~ b \otimes (\sum
  b_n \otimes v_n) \mapsto \sum b b_n \otimes v_n
\]
shows that $\dim_K \D_{\mbox{\scriptsize dR}}(V) \leq \dim_{\mb{Q}_p} V$. If equality holds, we
call $V$ \textbf{de Rham}. $\D_{\mbox{\scriptsize dR}}(V)$ comes equipped with a seperated
and exhaustive $K$-vector space filtration, given by $\textrm{Fil}^i \D_{\mbox{\scriptsize dR}}(V) =
(\textrm{Fil}^i \B_{\mbox{\scriptsize dR}} \otimes V)^{G_K}$.

Similarly, we let $\D_{\mbox{\scriptsize cris}}(V) := (\B_{\mbox{\scriptsize cris}} \otimes_{\mb{Q}_p} V)^{G_K}$. This
is a $K_0$-vector space, and the injectivity of the analogous
$\alpha$-map for $\B_{\mbox{\scriptsize cris}}$ shows that $\dim_{K_0} \D_{\mbox{\scriptsize cris}}(V) \leq
\dim_{\mb{Q}_p} V$. If equality holds, we call $V$
\textbf{crystalline}. $\D_{\mbox{\scriptsize cris}}(V)$ comes equipped with a $K_0$-linear
$\varphi$-action. Further, $K \otimes_{K_0} \D_{\mbox{\scriptsize cris}}(V)$ comes equipped
with a $K$-vector space filtration. If $V$ is crystalline then $V$ is
de Rham.

One fundamental theorem of Colmez and Fontaine states: the assignment
$V \mapsto \D_{\mbox{\scriptsize cris}}(V)$ induces an equivalence of categories between the
crystalline representations of $G_K$ and the category of $K$-filtered
admissible (i.e. $t_H(D) = t_N(D)$ and $t_H(D') \leq t_N(D')$ for all
subobjects $D' \subset D$, where $t_H$ resp. $t_N$ are the Hodge
number resp. the Newton number) $\varphi$-modules.  This equivalence heavily
uses the fact that the map $\alpha$ above in the $\B_{\mbox{\scriptsize cris}}$-case is
actually an isomorphism.

\section{The period ring $\A_{\trm{exp}}$}

\begin{defi}
  Let
  \[
    \widetilde{\A}^+ \left\{ \frac{\pi}{p} \right\} := \widetilde{\A}^+ \{X\} / (pX - \pi),
  \]
  where $A\{ X \}$ denotes the $p$-adic completion of $A[X]$ for any
  ring $A$, equipped with quotient topology.
\end{defi}

If $x \in \widetilde{\A}^+ \{ \pi / p \}$, we may write (non-uniquely) $x
= \sum_{n \geq 0} a_n (\pi / p)^n$ in $\widetilde{\A}^+[1/p]$. The natural
actions of $\varphi$ and $G_K$ extend to actions on $\widetilde{\A}^+ \{
\pi / p \}$.

\begin{lem}
  \label{lempit}
  In $\widetilde{\A}^+ \{ \pi / p \} \subset \B_{\mbox{\scriptsize dR}}^+$ one has the relation
  $t/p = \pi / p \cdot v$ mit $v \in \widetilde{\A}^+ \{ \pi / p \}^\times$, i.e.
  $\widetilde{\A}^+ \{ \pi / p \} = \widetilde{\A}^+ \{ t / p \}$. In particular, $t \in \widetilde{\A}^+ \{ \pi / p \}$.
\end{lem}
\begin{proof}
  First, we observe that
  \[
  \frac{t}{p} = \sum_{n \geq 1} (-1)^{n+1} \cdot \frac{\pi^n}{p \cdot n} = \frac{\pi}{p} \cdot
  \left( \sum_{n \geq 1} a_n \left(\frac{\pi}{p} \right)^n \right) = \pi / p \cdot v,
  \]
  with $a_n \to 0$ $p$-adically.

  Now, since $v \mod p \in \widetilde{\E}^+[X] / (\overline{\pi})$ is -1, hence a unit, we have that $v
  \in \widetilde{\A}^+ \{ \pi / p \}^\times$. Hence the claim.
\end{proof}

\begin{defi}
  Let $A$ be a subring of $\B_{\mbox{\scriptsize dR}}$, such that the Frobenius $\varphi$
  acts on $A$ (e.g. $\A_{\mbox{\scriptsize cris}}$). Set
  \[
    \textrm{Fil}^i_p A := \{ x \in \textrm{Fil}^i A|~ \varphi(x) \in p^i A \},
  \]
  where $(\textrm{Fil}^i A)_{i \in \mathds{Z}}$ is the filtration induced by $\B_{\mbox{\scriptsize dR}}$.
\end{defi}

\begin{defi}
  Let $\A_{\trm{exp}} := \widetilde{\A}^+\{\pi / p \} [p / t]$. The Frobenius $\varphi$
  on $\widetilde{\A}^+ \{ \pi / p \}$ extends to $\A_{\trm{exp}}$. We equip $\A_{\trm{exp}}$ with
  the filtration given by
  \[
  \textrm{Fil}^i \A_{\trm{exp}} := \bigcup_{i + k \geq 0} \left(\frac{p}{t}\right)^{k} \textrm{Fil}_p^{i +
    k} \widetilde{\A}^+ \left\{ \frac{\pi}{p} \right\}.
  \]
  Since $\textrm{Fil}^0_p \widetilde{\A}^+ \{ \pi / p \} = \widetilde{\A}^+ \{ \pi / p \}$, this
  filtration is seperated and exhaustive.
\end{defi}

\begin{prop}
  \label{exseq}
For every $k \geq 0$ we have the exact $G_K$-equivariant sequence
\[
0 \longrightarrow \left(\frac{t}{p}\right)^k \cdot \mb{Z}_p \longrightarrow \textrm{Fil}^k_p \widetilde{\A}^+\{ \pi / p \} \overset{1 - p^{-k} \varphi
}{\longrightarrow} \widetilde{\A}^+ \{ \pi / p \} \longrightarrow 0,
\]
which admits a continuous (not necessarely $G_K$-equivariant)
splitting $\widetilde{\A}^+ \{ \pi / p \} \rightarrow \textrm{Fil}^k_p \widetilde{\A}^+ \{ \pi / p \}$.
\end{prop}
\begin{proof}
  Obviously,
  \[
    \left(\frac{t}{p}\right)^k \cdot \mb{Z}_p \subset \ker (1 - p^{-k} \varphi).
  \]
  On the other hand, if $x \in \ker (1 - p^{-k} \varphi)$ then $x =
  \sum_n a_n (t / p)^n$ (see Lemma \ref{lempit}), with $\widetilde{\A}^+ \ni a_n \rightarrow 0$
  $p$-adically. For any $n \in \mathds{N}$ we have $(p^{-k} \varphi)^n(x)
  \equiv \varphi^n(a_n) (t/p)^k \mod p \widetilde{\A}^+ \left\{ \pi /
    p\right\}$, hence $x = y (t/p)^k$, with $y \in \widetilde{\A}^+$ and
  $\varphi(y) = y$, that is, $y \in \mb{Z}_p$ as is well-known.

  We now prove that $\textrm{Fil}^k_p \A_{\trm{exp}}$ is the $p$-adic closure of the module
  \[
    \widetilde{\A}^+[\xi^i \cdot (t / p)^j ; i + j \geq k ],
  \]
  which we denote by $N$. If $i + j \geq k$ one has
  \[
    \varphi \left( \xi^i \cdot \left( \frac{t}{p} \right)^j \right) = p^{i+j} \cdot
    \left(1 + \frac{\pi_0}{p}\right)^i \cdot \left( \frac{t}{p} \right)^j,
  \]
  where $\pi_0$ is the trace from $K((t))$ to $K((t^p))$ of
  $\pi$. Here we recall that $\pi = \exp t - 1 \in K((t))$.  Obviously
  $(t / p)^k \cdot \mb{Z}_p \subset N$, so we have to prove that for any $y
  \in \widetilde{\A}^+ \left\{ \pi / p \right\}$ there exists an $x \in
  \textrm{Fil}^k_p \widetilde{\A}^+ \left\{ \pi / p \right\}$ with $(1 - p^{-k}
  \varphi)(x) = y$. Since $N$ and $\widetilde{\A}^+ \left\{ \pi / p \right\}$
  are seperated and complete with respect to the $p$-adic topology, it
  suffices to show this result $\mod p$.
  If $y = \sum_{n > k} a_n (t / p)^n$  then $x = -y$ will do the job.

  Thus it remains to show that if $y \in \widetilde{\A}^+$ and $j \leq $k there
  exists an $x \in N$ such that
  \[
    (1 - p^{-k} \varphi)(x) - y \left( \frac{t}{p} \right)^j \in
    \left( p, \left( \frac{t}{p}\right)^i; i > n  \right) \unlhd
    \widetilde{\A}^+ \left\{ \pi / p \right\}.
  \]
  One checks that
  \[
    x = y' (q')^{k - j} \left( \frac{t}{p} \right)^{i},
  \]
  where $y' \in \widetilde{\A}^+$ is a solution of
  \[
    \varphi(y') - q'^{k - j} y' = b
  \]
  in $\widetilde{\A}^+$, satisfies this property (recall that $q' = \varphi^{-1}(q)$).
\end{proof}

\begin{cor}
  Dividing out $(t/p)^{-k}$ and taking the the direct limit over the
  sequence in Proposition \ref{exseq} we obtain an exact sequence
  \[
    0 \longrightarrow \mb{Z}_p \longrightarrow \textrm{Fil}^0 \A_{\trm{exp}} \overset{1 - \varphi}{\longrightarrow} \A_{\trm{exp}} \longrightarrow 0,
  \]
  where $\varphi$ is the extension of the $\varphi$ on $\widetilde{\A}^+$.
\end{cor}

\begin{prop}
  \label{invaexp}
  $(\A_{\trm{exp}})^{G_K} = \mathcal{O}_{K_0}$
\end{prop}
\begin{proof}
  Obviously, $\mathcal{O}_{K_0} \subset (\A_{\trm{exp}})^{G_K}$, since
  $(\widetilde{\A}^+)^{G_K} = \mathcal{O}_{K_0}$. We have the inclusions
  \[
  \varphi\left( \widetilde{\A}^+ \left\{ \frac{\pi}{p} \right\} \right) =
  \widetilde{\A}^+ \left\{ \frac{\pi^p}{p} \right\} \subset \A_{\mbox{\scriptsize cris}} \subset
  \widetilde{\A}^+ \left\{ \frac{\pi}{p} \right\},
  \]
  which leads to
  \[
    \varphi( \A_{\trm{exp}}[1/p] ) \subset \B_{\mbox{\scriptsize cris}} \subset \A_{\trm{exp}}[1/p].
  \]
  Since $(\varphi(\B_{\mbox{\scriptsize cris}}))^{G_K} = (\B_{\mbox{\scriptsize cris}})^{G_K} = K_0$, we have that
  $(\varphi( \A_{\trm{exp}}[1/p] ))^{G_K} = K_0$. Since $\varphi$ is injective
  on $\A_{\trm{exp}}$, and hence on $\A_{\trm{exp}}[1/p]$, we obtain $(\A_{\trm{exp}}[1/p])^{G_K} = K_0$.

  Note that by the above exact sequence, $1/p \not \in \A_{\trm{exp}}$:
  otherwise $1/p^n \in \A_{\trm{exp}}$ for all $n \geq 0$. But taking
  $G_{\mb{Q}_p}$-invariants gives an injection
  \[
  \A_{\trm{exp}} \hookrightarrow H^1(G_{\mb{Q}_p}, \mb{Z}_p) = \textrm{Hom}_{\textrm{cts}}(G_{\mb{Q}_p},
  \mb{Z}_p).
  \]
  The $\mb{Z}_p$-module on the right hand side is finitely generated, which
  would lead to a contradiction.

  Alternatively, one can use the exact sequence of \ref{exseq}, the filtration on $\A_{\trm{exp}}$ and a
  limit argument to proceed as in the proof of the statement
  \[
    H^0(K, \textrm{Fil}^i \B_{\mbox{\scriptsize dR}} / \textrm{Fil}^j \B_{\mbox{\scriptsize dR}}) = K,
  \]
  if $i \leq 0 < j$.
\end{proof}

\begin{defi}
  Let $T$ be a full $\mb{Z}_p$-lattice of $V$ that is invariant under the
  action of $G_K$ (such lattices always exist). We define the modules
  \[
  \D_{\trm{exp}}(T) := (\A_{\trm{exp}} \otimes_{\mb{Z}_p} T)^{G_K}
  \]
  and
  \[
    \D_{\trm{exp}}^0(T) := (\textrm{Fil}^0 \A_{\trm{exp}} \otimes_{\mb{Z}_p} T)^{G_K}.
  \]
\end{defi}

\begin{prop}
  \label{dexpfd}
  If $T$ is as before, $\D_{\trm{exp}}(T)$ is free $\mathcal{O}_{K_0}$-module of
  finite rank less or equal to $\textrm{rk}_{\mb{Z}_p} T$.
\end{prop}
\begin{proof}
  This is a variaton of the proof one usually encounters in Fontaine's
  theory of $B$-admissible rings. We outline the idea:

  Let $B$ be a topological integral domain, equipped with a continuous action of a topological
  group $G$.  Set $C = \textrm{Frac}(B)$ and $S = B^G$, which is again an integral domain, and fix
  a closed subring $R \subset S$. Assume $T$ is a finite free $R$-module with continuous
  $G$-action, so that $V = \textrm{Frac}(R) \otimes_R T$ is a finite-dimensional
  $\textrm{Frac}(R)$-vector space. Set $D_B(T) = (B \otimes_R T)^{G}$, $D_C(V) = (C
  \otimes_{\textrm{Frac(R)}} V)^G$ and assume $C^G = \textrm{Frac}(S)$.
  We want to prove the injectivity of the map
  \[
  B \otimes_{S} D_B(T) \longrightarrow B \otimes_R T .
  \]
  The inclusion $B \hookrightarrow C$ and the freeness of $T$ gives a diagram
  \[
    \xymatrix{
    B \otimes_{S} D_B(T) \ar[r] \ar@{^{(}->}[d] & B \otimes_R T \ar@{^{(}->}[dd] \\
    B \otimes_{S} D_C(V) \ar@{^{(}->}[d] \\
    C \otimes_{\textrm{Frac}(S)} D_C(V) \ar[r] & C \otimes_{\textrm{Frac}(R)} V
    },
  \]
  so that we are reduced to the case where all the rings are fields. Now one proceeds exactly
  as in \cite{fontouy}, Theorem 2.13.

  The above situation applies with $B = \A_{\trm{exp}}$, $R = \mb{Z}_p \subset \mathcal{O}_{K_0} = S$. The
  injectivity of the above map implies, by using the above notation and going to the quotient
  field $C$, that $D_B(T)$ is of $S$-rank smaller or equal than the $R$-rank of $T$. Since we
  these latter rings are discrete valuation rings, we are done.
\end{proof}

\begin{prop}
  \label{dexpdcris}
  If $V$ and $T$ are as above, we have
  \[
    \D_{\trm{exp}}(T)[1/p] = \D_{\trm{exp}}(V) = (\A_{\trm{exp}}[1/p] \otimes_{\mb{Q}_p} V)^{G_K} = \D_{\mbox{\scriptsize cris}}(V),
  \]
  and this identification is compatible with the $K$-filtrations and
  the action of the Frobenius $\varphi$.
\end{prop}
\begin{proof}
  The proof is similarly as in Proposition \ref{invaexp}. We have inclusions
  \[
  (\varphi(\A_{\trm{exp}}) \otimes T)^{G_K} \subset (\A_{\mbox{\scriptsize cris}}
  \otimes T)^{G_K} \subset (\A_{\trm{exp}} \otimes T)^{G_K},
  \]
  and since $\varphi$ is bijective on $\D_{\mbox{\scriptsize cris}}(V)$ and injective on $\A_{\trm{exp}}$, we conclude as
  before $(\A_{\trm{exp}}[1/t] \otimes T)^{G_K} = \D_{\mbox{\scriptsize cris}}(V)$.

  The compatibility with filtration and Frobenius can be checked by the construction.
\end{proof}

\section{Categories in integral $p$-adic Hodge theory}

Let $K / \mb{Q}_p$ be unramified for this section.

\begin{defi}
  A \textbf{Fontaine-Laffaille module} over $\mc{O}_K$ is a triple $(M, ( M^i )_{i \in \mathds{Z}},\\
  (\varphi^i)_{i \in \mathds{Z}} )$, which we also denote simply by $M$, consisting of
  \begin{itemize}
    \item an $\mc{O}_K$-module $M$,
    \item an exhausting and seperated decreasing filtration (of $\mc{O}_K$-modules) $M^i$ of $M$,
    \item a family of $\sigma$-semilinear maps $\varphi^i : M^i \rightarrow M$
      with the property $\varphi^i|_{M^{i+1}} = p \cdot \varphi^{i+1}$.
  \end{itemize}
  A morphism of Fontaine-Laffaille modules $f: M \rightarrow N$ is a
  $\mc{O}_K$-linear map $f$ such that $f(M^i) \subset N^i$ and $f \circ
  \varphi^i_M = \varphi^i_N \circ f$.  We denote by
  $\underline{\textrm{MF}}_{\mc{O}_K}$ the exact category of all
  Fontaine-Laffaille modules over $\mc{O}_K$.
\end{defi}

\begin{defi}
  A \textbf{filtered Dieudonn\'e module} over $\mc{O}_K$ is a Fontaine-Laffaille module $M$ such that
  \begin{itemize}
    \item $M$ is of finite type over $\mc{O}_K$,
    \item $M^i = M$ for $i \ll 0$ and $M^j = 0$ for $j \gg 0$,
    \item $M_\sigma = \sum_{i \in \mathds{Z}} \varphi^i(M^i)$.
  \end{itemize}
  We denote by $\underline{\textrm{MF}}^{\textrm{fd}}_{\mc{O}_K}$ (fortement divisible) the category of
  all filtered Dieudonn\'e modules.
\end{defi}

Here, $M_\sigma$ denotes the underlying module $M$, where $W$ acts via
$\sigma$.  Note that contrary to the usual convention we allow our
Dieudonn\'e modules to contain torsion.

\begin{ex}
  We have $\A_{\trm{exp}} = (\A_{\trm{exp}}, (\textrm{Fil}^i \A_{\trm{exp}})_{i \in \mathds{Z}},
  (\varphi^i)_{i \in \mathds{Z}}) \in \underline{\trm{MF}}$, where $\A_{\trm{exp}}$ and the
  filtration are given as before, and $\varphi^i = 1/p^i \cdot
  \varphi$, with $\varphi$ induced from the Frobenius on $\widetilde{\A}^+$.
\end{ex}

\begin{thm}
  The category $\underline{\textrm{MF}}^{\textrm{fd}}_{\mc{O}_K}$ is abelian.
\end{thm}
\begin{proof}
  This follows from the fact that $\underline{\textrm{MF}}^{\textrm{fd},
    \textrm{tor}}_{\mc{O}_K}$, the subcategory of all torsion $\mc{O}_K$-modules
  (\cite{fontlaf}, Proposition 1.8), is abelian, and completeness: let $f: M
  \rightarrow N$ be a map in $\underline{\trm{MF}}_{\mathcal{O_K}}$ with $M, N \in
  \underline{\textrm{MF}}^{\textrm{fd}}_{\mc{O}_K}$. This gives us, for any $n
  \in \mathds{N}$, a map $f_n : M / p^n \rightarrow N / p^n$ in
  $\underline{\textrm{MF}}^{\textrm{fd}, \textrm{tor}}_{\mc{O}_K}$, since $M$ and
  $N$ are of finite type, hence kernel and cokernel of $f_n$ exist.

  Since $M = \varprojlim M / p^n$, in a compatible way with the
  filtration and the Frobenii, we obtain, by going to the limit, the
  kernel and the cokernel of the map $f$. The normality of mono- and
  epimorphisms is an easy consequence again of the property that
  $\underline{\textrm{MF}}^{\textrm{fd}, \textrm{tor}}_{\mc{O}_K}$ is abelian.
\end{proof}

\begin{defi}
  A \textbf{filtered $\varphi$-module} over $K$ is a triple $(D, (D^i)_{i \in \mathds{Z}}, \varphi)$,
  consisting of
  \begin{itemize}
  \item a $K$-vector space $D$,
  \item an exhausting and seperated decreasing filtration (of $K$-vector spaces) $D^i$ of $D$,
  \item a $\sigma$-semilinear map $\varphi: D \rightarrow D$.
  \end{itemize}
  A morphism of filtered $\varphi$-modules is, similarly as before, a morphism of $K$-vector spaces
  compatible with the filtration and Frobenius $\varphi$. We denote by $\underline{\trm{MF}}_{K}$ the category of
  all $\varphi$-modules.

  A finite-dimensional filtered $\varphi$-module is called
  \textbf{admissible} if $t_N(D) = t_H(D)$ and $t_N(D') \leq t_H(D')$
  for all subobjects $D' \subset D$ in $\underline{\trm{MF}}_{K}$, where $t_N$
  resp. $t_H$ are the Newton- resp. Hodge number of $D$ (see \cite{fontouy}, 6.4.2.).
\end{defi}

\begin{ex}
  If $M \in \underline{\trm{MF}}_{\mc{O}_K}^{\textrm{fd}}$, one can naturally associate a
  finite-dimensional $\varphi$-module $D$ to $M$, namely $D := K
  \otimes_{\mc{O}_K} M$, with the filtration induced by $M^i$, and
  Frobenius $\varphi := 1/p^n \cdot \varphi^n$ for $n \ll 0$. We call
  $M$ \textbf{admissible} if $D$ is admissible.
\end{ex}

\begin{prop}
  \label{critfd}
  Let $D \in \underline{\trm{MF}}_{K}$ be admissible. Then an $\mc{O}_K$-lattice $M$,
  equipped with a filtration $M^i$ such that $M^i[1/p] = D^i$ can
  be considered as an object of $\underline{\trm{MF}}^{\textrm{fd}}_{\mc{O}_K}$ if and only if
  $\varphi(M^i) \subset p^i M$ (that is, one puts $\varphi^i := p^{-i}
  \varphi$).
\end{prop}
\begin{proof}
  The only thing we have to check is that the condition on $\varphi$
  holds, then $D$ is already in $\underline{\trm{MF}}^{\textrm{fd}}_{\mc{O}_K}$.  This can be
  infered by the proof of \cite{fontlaf}, Theorem 3.2, after reducing
  to the case where all weights are $\geq 0$ (see also \cite{laf},
  Proposition 7.8).
\end{proof}

\begin{prop}
  If $V$ is crystalline and $T$ as above, then $\D_{\trm{exp}}(T) \in \underline{\trm{MF}}_{\mc{O}_K}^{\textrm{fd}}$.
\end{prop}
\begin{proof}
  We know from proposition \ref{dexpfd} that $\D_{\trm{exp}}(T)$ is a free $\mc{O}_K$-module of finite
  rank and that $\D_{\trm{exp}}(T)[1/p] = \D_{\mbox{\scriptsize cris}}(V)$ (\ref{dexpdcris}) is admissible,
  since $V$ is crystalline. Since $\D_{\trm{exp}}^i(T)[1/p] = \textrm{Fil}^i \D_{\mbox{\scriptsize cris}}(V)$
  and $\varphi^i (\D_{\trm{exp}}^i(T)) \subset \D_{\trm{exp}}(T)$, the requirements of
  proposition \ref{critfd} are fulfilled, hence the claim.
\end{proof}

\section{Computation of $H^1_e(K, T)$}

Assume again that $K / \mb{Q}_p$ is unramified. We collect some facts
from sections 3 and 4 of \cite{bk93}.

\begin{prop}
  Let $M \in \underline{\trm{MF}}_{\mc{O}_K}^{\textrm{fd}}$ and put
  \[
    H^0(M) = \textrm{ker} ((1 - \varphi^0) : M^0 \rightarrow M),~~~
    H^1(M) = \textrm{coker} ((1 - \varphi^0) : M^0 \rightarrow M).
  \]
  and $H^i(M) = 0$ for $i \geq 2$. Then $(H^i)_{i \in \mathds{N}}$ is a
  cohomological $\delta$-functor.
\end{prop}
\begin{proof}
  This is abundantly clear by the snake lemma.
\end{proof}

Recall (\cite{bk93}, Proposition 1.17) the Bloch-Kato fundamental exact
sequences
\[
0 \longrightarrow \mb{Q}_p \longrightarrow \B_{\mbox{\scriptsize cris}}^{\varphi = 1} \oplus \B_{\mbox{\scriptsize dR}}^+ \overset{f}{\longrightarrow}
\B_{\mbox{\scriptsize dR}} \longrightarrow 0,
\]
where $f(x, y) = x - y$, and
\[
0 \longrightarrow \mb{Q}_p \longrightarrow \B_{\mbox{\scriptsize cris}} \oplus \B_{\mbox{\scriptsize dR}}^+ \overset{f}{\longrightarrow} \B_{\mbox{\scriptsize cris}} \oplus
\B_{\mbox{\scriptsize dR}} \longrightarrow 0,
\]
where $f(x,y) = ((1 - \varphi)(x), x - y)$. If $V$ is a $p$-adic
representation, we call the maps
\[
  \exp_e: \D_{\mbox{\scriptsize dR}}(V) \overset{\delta}{\rightarrow} H^1(K, V),~~~
  \exp_f: \D_{\mbox{\scriptsize cris}}(V) \oplus \D_{\mbox{\scriptsize dR}}(V) \overset{\delta}{\rightarrow}
  H^1(K, V)
\]
the \textbf{Bloch-Kato exponential maps}, which are induced by the
connecting homomorphism of Galois cohomology. We set
\[
  H^1_e(K, V) := \textrm{Im}(\exp_e),~~~ H^1_f(K, V) := \textrm{Im}(\exp_f).
\]
If $T \subset V$ is a $G_K$-equivariant $\mb{Z}_p$-lattice in $V$, denote
by $\iota: H^1(K, T) \rightarrow H^1(K, V)$ the canonical map. Set
\[
  H^1_f(K, T) := \textrm{Im}(\exp_f)
\]

Recall that the local $L$-function in the case $p = l$ is defined as
\[
  P(V, X) := \textrm{det}_{K}(1 - f \cdot X|~ \D_{\mbox{\scriptsize cris}}(V)),
\]
where $f$ denotes the $K$-linear map $\varphi^{[K : \mb{Q}_p]}$.

Assume now that $P(V, 1) \not= 0$. Then (cf.~the first lines in the
proof of Theorem 4.1, \cite{bk93}) $H^1_f(K, V) = H^1_e(K, V)$.

\begin{thm}
  Let $V$ be a crystalline representation, fix a $G_K$-equivariant
  lattice $T$ in $V$, and assume $P(V, 1) \not= 0$. Then:
  \begin{enumerate}
  \item $H^1(\D_{\trm{exp}}(T))[1/p] \cong H^1_e(K, V)$.
  \item $H^1(\D_{\trm{exp}}(T)) \cong H^1_e(K, T)$.
  \item $\exp_e : \D_{\mbox{\scriptsize dR}}(V) / \D_{\mbox{\scriptsize dR}}^0(V) \rightarrow H^1(K, V)$ coincides with the composite map
    \begin{align*}
      (\D_{\trm{exp}}(T) / \D_{\trm{exp}}^0(T)) [1/p] \overset{1 - \varphi}{\longrightarrow} (\D_{\trm{exp}}(T) / (1 - \varphi^0) \D_{\trm{exp}}(T)) [1/p]\\
      =  H^1(\D_{\trm{exp}}(T))[1/p] \cong H^1(K, V),
    \end{align*}
    where the last canonical identification is explained in the proof.
  \end{enumerate}
\end{thm}
\begin{proof}
  The proof is similar to \cite{bk93}, Lemma 4.5.
  We have the following commutative diagram:
  \[
\xymatrix{
  0 \ar[r] & \mb{Z}_p \ar[d] \ar[r] & \textrm{Fil}^0 \A_{\trm{exp}} \ar[r] \ar[d]^{x \mapsto (x,x)} & \A_{\trm{exp}} \ar[r]
  \ar[d]^{x \mapsto (x,0)} & 0 \\
  0 \ar[r] & \mb{Q}_p \ar[r] & \A_{\trm{exp}}[1/p] \oplus \B_{\mbox{\scriptsize dR}}^+ \ar[r] &
  \A_{\trm{exp}}[1/p] \oplus \B_{\mbox{\scriptsize dR}} \ar[r] & 0 }.
\]
Inverting $p$, tensoring with $V$ over $\mb{Q}_p$ and taking $G_K$-cohomology, we obtain a diagram
\[
\!\!\!\!\!\!\xymatrix@C=8pt{ 0 \ar[r] & H^0(\D_{\trm{exp}}(T))[1/p] \ar[r] \ar[d]^{=} & \D_{\trm{exp}}^0(T)[1/p] \ar[r] \ar[d] &
  \D_{\trm{exp}}(T)[1/p] \ar[d] \ar[r] & H^1(\D_{\trm{exp}}(T))[1/p] \ar[r] \ar[d] & 0 \\
  0 \ar[r] & H^0(K, V) \ar[r] & \D_{\mbox{\scriptsize cris}}(V) \oplus \D_{\mbox{\scriptsize dR}}^+(V) \ar[r] & \D_{\mbox{\scriptsize cris}}(V) \oplus \D_{\mbox{\scriptsize dR}}(V)
  \ar[r] & H^1(K, V) }
\]
with exact rows. Since $\D_{\trm{exp}}(T)[1/p] = \D_{\mbox{\scriptsize cris}}(V) = \D_{\mbox{\scriptsize dR}}(V)$ and
$\textrm{Im}(\exp_f) = \textrm{Im}(\exp_e)$, we have the claimed
identification $H^1(\D_{\trm{exp}}(T))[1/p] \cong H^1_f(K, V)$.

The exact sequence
\[
  0 \longrightarrow \D_{\trm{exp}}(T) \overset{p}{\longrightarrow} \D_{\trm{exp}}(T) \longrightarrow \D_{\trm{exp}}(T) / p \longrightarrow 0
\]
(which exists in $\underline{\trm{MF}}^{\textrm{fd}}_{\mc{O}_K}$, since $\D_{\trm{exp}}(T) \in
\underline{\trm{MF}}^{\textrm{fd}}_{\mc{O}_K}$) induces a sequence
\[
  H^0(\D_{\trm{exp}}(T) / p) \longrightarrow H^1(A \otimes T) \longrightarrow H^1(A \otimes T) \longrightarrow
  H^1(A \otimes T / p).
\]
Recall that $H^1_f(K, T) = \iota^{-1} (H^1_f(K,V))$, where $\iota :
H^1(K, T) \rightarrow H^1(K, V)$.  Since $H^1(\D_{\trm{exp}}(T)) \subset H^1(K, T)$ it
suffices to show that the cokernel of this inclusion does not have any
$p$-torsion. But this follows from the the commutative diagram
\[
\xymatrix{
  H^0(\D_{\trm{exp}}(T) / p) \ar[r] \ar[d]^\cong & H^1(A \otimes T) \ar[r]^-{p} \ar[d]^{\subset} &
  H^1(A \otimes T) \ar[r] \ar[d]^{\subset} & H^1(A \otimes T / p) \ar[d]^{\subset} \\
  H^0(T / p) \ar[r] & H^1(T) \ar[r]^p & H^1(T) \ar[r] & H^1(T / p)  }.
\]
\end{proof}

\begin{cor}
  Let $\mu$ be the Haar-measure on the finite-dimensional $K$-vector
  space $H^1(K, V)$ such that the image of the lattice
  \[
    \D_{\trm{exp}}(T) / \D_{\trm{exp}}^0(T) \subset \D_{\mbox{\scriptsize dR}}(V) / \D_{\mbox{\scriptsize dR}}^0(V) \overset{\sim}{\longrightarrow} H^1_e(K, V) = H^1_f(K, V)
  \]
  has measure $1$. Then
  \[
    \mu(H^1_f(K, T)) = | P(V, 1) |_p^{-1}.
  \]
\end{cor}
\begin{proof}
  This follows from the definition of $P(V, X)$ and the b), c) from the previous theorem.
\end{proof}

redpajama_set_name: RedPajamaArXiv


text:
Menu CIO
Being There - How to be a 24-hour 3-D marketer
John Sviokla (CIO) 15 May, 2001 15:37
Customers don't care what time it is when they do business-companies make them care. If you want to win them for life, be there whenever and wherever they want you to be.
Robert Woodruff, the man who took brown, bubbly sugar water-Coca-Cola-and turned it into a global cash machine, had a famous saying: "I want Coke to be within arm's length of desire." With that simple statement, Woodruff set out to make Coca-Cola the first ubiquitous product in the world.
Today, Coca-Cola is the best example of what I call a three-dimensional marketer. The three dimensions are who, where and when. Coca-Cola does a pretty good job of knowing who buys Coke, and making it available wherever and whenever customers want it-mostly through vending machines and an extraordinary distribution network.
Until the Internet came along, this "anywhere" strategy seemed relevant to just a narrow range of marketing companies like Coca-Cola. Yet with cheap, ubiquitous technology available to any company today, there's no reason why everyone can't try to be a 3-D leader with an anywhere, anytime strategy based on clear customer knowledge and targeting.
We're already seeing market leaders such as AOL and Fidelity follow Coca-Cola's lead. You can currently get AOL via its own ISP, over the Web, through the television, and on cell phones and PDAs; the company has plans to put AOL in physical locations such as schools, ATMs and Kinko's. Fidelity was one of the first companies to make trading and financial information available to customers on all forms of wireless devices. Recently, it announced a joint effort with GM to offer access to Fidelity accounts in all GM cars.
Why are AOL and Fidelity heading this way? If all people will have the potential to be connected at all times in all settings through the Internet and wireless phones, then you always want your product or service available to them. Imagine the power of taking a category of valuable products, such as financial transactions, and making it available all the time, everywhere. Not only will products and services be accessible at the consumer's moment of greatest interest, but the potential exists to stimulate further demand and alter patterns of usage.
Product or service ubiquity is a race where being first means everything. The company that reaches far beyond its nearest competitor in terms of anywhere and anytime access will gain virtually unshakeable market share. Pepsi has been chasing Coca-Cola for 80 years and has not yet made a significant dent in Coke's market share. Likewise, those financial services companies that try to compete with Fidelity will lose if Fidelity makes its way to everywhere status.
With ubiquitous availability comes an understanding of your customer that is simply not possible in a place-and-time constrained context. Letting people do business with you whenever and wherever they want gives you understanding of their natural patterns of buying activity-their true desires. But you can't do that without making your products available 24/7. Banks have gained a glimpse of the far-reaching power of 24/7 availability: Online customers are 50 percent less likely to switch banks, and those who pay bills electronically are even less likely to leave. Today there are companies that excel on one or two of the who-where-when marketing dimensions, but not all three.
Direct marketing companies such as Lands' End know who reads its catalogs and a little bit about where. These companies analyze the who so that they can create new offers and achieve more yield from their existing customers. They know where someone lives, but they do not know where they usually shop, where they buy goods.
Where can also mean big business: The great retailing companies have the where down to a science. Ever since the days when Sam Walton piloted his small propeller plane across the heartland of America to survey potential new sites for a Wal-Mart, companies like his and McDonald's have turned the issue of where into high art.
These companies' locations have made them staggeringly rich. Famed for its ability to crack the code of site selection, McDonald's sees 45 million customers a day-Wal-Mart more than 90 million a week. Yet these two companies know virtually nothing about their individual customers. There is no individual profile of each customer, only aggregate statistics to predict what type of customer will visit and where. Because McDonald's does not know who that individual is and how to contact him cheaply and efficiently, it cannot influence his buying decision to come to McDonald's on a particular day instead of going to Burger King.
Not surprisingly, when is the most misunderstood and least pursued dimension of marketing. The legacy of traditional shopping and working hours has led most companies to ignore its potential. Yet there is nothing more important to revenues and market share than understanding when someone is in the mood to buy. Some mainstream companies have made blunt efforts to discover the when of their customers' desires. For example, 7-Eleven in Japan restocks 20 percent of its stores five times a day to meet the morning, lunch, after school, dinner and after-dinner demands of its customers. This builds trust in the minds of its customers because 7-Eleven will have what the customers need, when they need it. Other nascent instruments include timing car company ad banners to appear when someone is searching in the automobile category on Yahoo.
When you put a 3-D strategy in place, make sure you put concrete measures in place. You need to show that this increased clarity in marketing can attract more customers to more profitable products more efficiently.
Technology is key to this who-where-when initiative, and there are three steps you can take to integrate a 3-D strategy with your technology strategy.
First, recognize that each new interface you create is a new challenge-and opportunity. You must have a technology architecture that can help create unity for your customer, not new and different interfaces each time. Imagine if Coke came in different size bottles or in different colors every time you bought it; there would be no brand as we know it today. That is what you need to keep in mind as you take your customer through different menu trees, voice mails, e-mails and Web interfaces.
Second, make sure you link your interface architecture to your customer value management approach. Each interface point is a potential control point for cross-selling, quality and service.
Third, show your senior management that your information systems control or influence what your management systems can do. Always drive to create management systems that are at least as fast and flexible as the customer's buying cycle. Cisco can close its books every night. Its business cycle is that fast. Can you?
John Sviokla, vice chairman of Diamond Cluster International, a Chicago-based consultancy, operates in 3-D all the time. Reach him anytime, anywhere at getreal@cio.com.
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John Dennen and Matt Christie are reunited in for the latest edition of the Opening Bell podcast.
The Boxing News team reflect on momentous weekend for British boxing, with Ricky Burns becoming a three-weight world champion and Tony Bellew triumphing at Goodison Park. They consider what should come next for both men and whether taking on David Haye really is a viable fight for Bellew.
As well as previewing the weekend's upcoming action, reviewing the latest Olympic qualification event, we discuss a big change for boxing as pros become eligible to enter Rio 2016.


In 2011, with the purchase of a parcel of land adjacent to our existing lot, construction began for another 87,000 square foot (82,000 sq ft footprint) addition at Londonderry Freezer Warehouse. The new building features state of the art refrigerated storage, conference facilities, corporate offices, and Quality Assurance inspection rooms. This addition officially opened in early 2012 and will ensure that our two companies are well positioned to serve our customers as they continue to grow and prosper.


If you never want to lose your messages again, either by switching to a new device or by ignorantly hitting the delete button, this article will be of great help to you so make sure to read it all to the last dot.
To be able to recover your messages when you loose them, you'll need to back 'em up, read on to see how to do that.
In other to easily back up and restore your messages, you need to download SMS backup & Restore app by Carbonite from Google play store and also the Add-on.
Before we proceed, note that the 'Add-on for SMS Backup & Restore' app doesn't appear on the list of apps in your phone as it is only used by the SMS Backup & Restore app on your phone.
3. You can choose the defualt file path or choose a custom folder if you want to save your backup to SD card or a particular folder in your phone.
4. Click on the check box next to the text message you want to backup, you can decide to only select the important ones or select all.
5. Click on 'local backup only' or 'local backup and Upload', the later will bring out options to upload to google drive, drop box or to email.
To be able to upload your backup, the Add-on for SMS Backup & Restore app will be needed so make sure you already have it installed.
6. Click on OK after selecting your preferred options.
7. You message(s) will automatically backup to wherever you selected.
You can now effortlessly restore your messages after backup, on a new device or same device whenever you need it, you just need to have the 'SMS backup & restore' app installed and then follow the below steps.
2. Click on the check box of the backup you want to restore.
4. Click 'OK' and you will get a message telling you to set the app as default messaging app.
Also note: If you don't want the SMS backup & Restore app to be your default messaging app, you can change this by going to 'Setting>> Apps>> Reset App preferences'.
You can now easily backup and restore your SMS messages without worrying about losing them, if you've backed-up your messages to Cloud, you can restore them back using the same process.
SMS messages can be restored to any Android device since data is stored in XML.


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For thousands of families in British Columbia, dementia is an uphill journey. It takes courage to make it up a mountain. The Climb for Alzheimer's is a fun, invigorating hike up the Grouse Grind® – but it's also an opportunity to help the Alzheimer Society of B.C. achieve its vision of a truly dementia-friendly society, where people living with dementia, their caregivers and their families are acknowledged, welcomed and included.
September 30, 2018 marks the seventh annual Climb. Teams fundraise and hike the Grouse Grind® together on event day. Don't have anyone to hike with? No problem – individual participants can still sign up and we'll match you with an established team where you can meet like-minded people who want to make a difference.
Who will you climb for? Register today or contact Special Events Officer Theresa Frazao at 604-742-4947 at tfrazao@alzheimerbc.org.


text:
The Wall Street Journal – Paul Elio's Quest to Build a Three-Wheel Car
Paul Elio's Quest to Build a Three-Wheel Car
Auto entrepreneur Paul Elio seeks government loan and other funding to start production next year
Paul Elio's car will carry two people, sell for $6,800 and get 84 miles per gallon—if it moves into production. PHOTO: ELIO MOTORS
June 2, 2015 11:13 p.m. ET
SOURCE: The Wall Street Journal
Paul Elio isn't a billionaire like Elon Musk.
In fact, at one point the budding auto entrepreneur was in such need of cash that he had to take a job roofing houses in Arizona just to pay the bills.
Now, five years later, the man behind Phoenix-based startup Elio Motors believes he is just months away from beginning production on a new three-wheel car at a former General Motors Co. assembly plant located just outside Shreveport, La.
The vehicle, which has two seats arranged front to back and one door, promises to deliver 84 miles to the gallon and cost about $6,800. It will be sold directly to consumers through company-owned stores instead of through dealers, which is similar to how Mr. Musk's Tesla Motors Inc. does business.
Whether Mr. Elio starts production in the first half of 2016 as planned, however, depends on whether he can raise an additional $230 million in capital, which he is hoping to do through a combination of outside investment and a loan from the federal government. He already had $70 million in funding, thanks in part to an angel investor and deposits from people seeking to reserve one of his vehicles. Once production starts, his goal is to churn out 250,000 cars annually.
"It's a lot of vehicles right?" Mr. Elio says. "But I believe we can sell to people in the new-car market, used-car market, those who drive clunkers and those who just want it, too.… The Elio is personal transportation, and people are going to want one even though they own other cars."
Skeptics abound
The idea of a three-wheeler roaming the nation's roads in any large quantity has already garnered a long list of skeptics, some of whom point out that traditional four-wheel small cars such as Daimler AG's Smartcar and the Fiat 500 have failed to stir strong interest.
"There has always been a dream by entrepreneurs to do a cheap car for the masses," says Matt DeLorenzo, managing editor for auto-tracking firm Kelley Blue Book. "But the complexity of cars makes it a pipe dream."
Although he has already had to push back the production date once, there is no denying the 51-year-old Mr. Elio has come a long way since his days as a roofer—and he has done it without venture capital or personal wealth.
"Our capital markets just aren't set up to fund a new car company," Mr. Elio says, adding that when he showed venture capitalists that he needed to spend as much as $300 million before producing any revenue, "their heads would pop off."
Saved by an angel
Mr. Elio's dream began in 2008 when he was working as an automotive engineer. Using revenue from his engineering firm, ESG Engineering, he founded Elio Motors and set to work on the design of the three-wheel car. But when ESG's revenue disappeared during the recession, he struggled for almost three years to find funding for his project.
"All my assets were gone. I had hundreds of résumés out for everything from high-end jobs down to Starbucks.… It took me working as a roofer before I was ultimately led to the support of Stuart Lichter."
Mr. Lichter is the billionaire real-estate developer and founder of Industrial Realty Group LLC. Mr. Elio crossed paths with Mr. Lichter in 2012 while looking around for a future production site. Mr. Lichter, who was involved in acquiring some of the assets spun off from GM during its bankruptcy, including the plant in Louisiana, talked to Mr. Elio about his project and was sold. He has invested $20 million in debt and equity so far.
"What is amazing to me is the number of nonrefundable reservations Paul has received, and the car isn't even in production," Mr. Lichter says.
The two-tier reservation system Mr. Elio set up in January 2013 allows potential Elio buyers to pay an amount ranging from $100 to $1,000 to secure a spot on a vehicle waiting list. They can make a refundable or nonrefundable deposit; those making nonrefundable deposits get a priority spot in line. Mr. Elio says the company has collected more than 41,000 reservations so far—most of them nonrefundable—locking in more than $280 million in product orders and $17 million in capital. It "is really a crowdfunding program," he says.
Betting on a loan
For his next cash infusion, Mr. Elio is turning to a provision in the 2012 federal JOBS Act, known as Rule 506(c), that allows a startup to offer accredited investors the chance to buy into the company with a minimum investment of $15,000. Accredited investors are those with $1 million in assets or income of at least $200,000 in each of the prior two years for an individual ($300,000 for a family).
"Before [the rule change] if people asked if you are raising money for your company you had to say no because you couldn't publicly disclose your private placements," Mr. Elio says. "Being able to publicly disclose the fundraise is a huge deal for us. It democratizes it."
He aims to raise $30 million through the strategy, he says.
Meanwhile, Mr. Elio also is seeking an Energy Department Advanced Technology Vehicles Manufacturing loan, which could amount to about $185 million. The program is designed to support the development of clean energy or more fuel-efficient technologies. Late last year, Elio's application moved into the second phase of the three-phase approval process, Mr. Elio says.
Kelley Blue Book's Mr. DeLorenzo applauds Mr. Elio's progress, saying "there is capital out there for people who have dreams." But in terms of revolutionizing travel, he isn't sold.
Mr. Bennett is a reporter for The Wall Street Journal in Detroit. Email him at [email protected]
https://www.eliomotors.com/the-wall-street-journal-paul-elios-quest-to-build-a-three-wheel-car/
Paul Elio's Quest to Build a Three-Wheel Car Auto entrepreneur Paul Elio seeks government loan […]


WhiskyIntelligence.com » Blog Archive » A New Malt Maniac – Whisky News - whisky industry press releases, newsletters, events, tasting notes, bottlings and comments.
The Malt Maniacs have inducted kidnapped a new member; Rich Howard of Minneapolis/St. Paul, Minnesota into the cult team.
Congrats to the Malt Maniacs on snagging your new member! Rich is a Malt Maniac in every sense, and an invaluable asset to maniacles everywhere.


The Birmingham Small Arms Company (BSA) factory was located in Armoury Road, Small Heath. This images shows women working in Machine Shop One. The factory produced rifles, Lewis guns, shells and vehicles. The BSA factory was founded in 1861.


text:
13-year-old Boy Dies Hit By Driver Who Runs Away
Video: 13-year-old Boy Dies Hit By Driver Who Runs Away
Video: Father Meets DRUNK DRIVER who KILLED his Son!!!!! You WON'T Believe What he Does Next 2023, January
A family in Florida has suffered a double tragedy. On Saturday morning, Tyler Pittard, 13, died after being struck by a van and the driver fled. The boy was walking with other people on a busy road. The teenager was accompanied by Donald Charles Keefer, 47, a family friend, and his aunt. They were returning from a fishing trip and were walking to a nearby gas station, local media WFTS, WFLA and Bay9News reported.
"He was with adults, he felt safe," Bruce Branch, the boy's great-uncle, told WFTS. However, the driver of a Chevy Silverado van hit the group that was walking, killing Tyler and Keefer. The boy's aunt survived and is hospitalized with serious injuries, according to reports. The driver fled, the Florida Highway Patrol reported.
Tyler Pittard
On Sunday a driver called the authorities saying that he suspected that he had been the person involved in the crash, however he assured that he thought he had run over a deer. His vehicle has a description compatible with that of the Chevy Silverado van and is now in the possession of the authorities investigating the case. The driver has not yet been arrested.
Another alarming fact is that Tyler died on the same stretch of US Highway 19 where months ago his mother, Mindy O'Neill, lost her life after being rammed by another car while riding a bicycle. According to a GoFundMe account that was created to help the family with the teenager's funeral expenses, Tyler lived with her grandmother Tammy after losing her mother. May they rest in peace.
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Q: converting the sum of numbers to logarithms If I have three number, $[0.2, 0.3, 0.4]$ then I can take the sum simply by adding the term: $0.2+0.3+0.4=0.9$. The proportion of the first element is then $0.2/0.9 = 2/9$.
Now I don't know what the three numbers are but I do know the logarithms (which are so small I cannot take the exponentiation of it). How can I get the proportion of the elements?
So given: $[-1.6094,   -1.2040,   -0.9163]$, how do I find out that the first component has a proportion of $2/9$?

A: If this is a practical problem then note you can use an arbitrary precision floating point library ( like MFPR , which is free ) to write a short application to do the exponentiation and addition and division to any precision you care to.

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Regular portions of Mewar and its rich legacy
to keep the 'Living Heritage' social and global!
www.eternalmewar.in
Eternal Mewar
HRH Group of Hotels
HRH InfoTech
Kartik Poornima
Rulers of Mewar
Maharana Bhagwat Singh
Ruling Period: 1955 – 1984
Acc. No.: 2011.T.0022_R
The Maharana was a title which came to be known as the Hindupats or the head of Hindu nation. Ever since the precinct of Rajputana history, the Maharanas of Mewar symbolised their pride with the rulers who fought for their people and died for the respect of their clan. But India was changing. In 1970 when the title hood was abolishing, Maharana Bhagwat Singh was on the throne.
Maharana Bhagwat Singh was just another seventeen year old boy from the Shivrati branch of Mewar, the descendants of Maharaj Arjun Singh, the fourth son of Maharana Sangram Singh II; when Maharana Bhopal Singh adopted him as his heir. Ascending the throne, Maharana Bhagwat Singh witnessed the Indian history shifting.
After ascending the throne, the state of Rajasthan was formed. The rulers of Rajasthan enjoyed the privy purses till the Indian government decided to abolish the institution of royalty. The royal titles and grants were eliminated and Maharana Bhagwat Singh was now Mr. Bhagwat Singh Mewar. But this did not deter him from his duty of serving his people. His main focus was social welfare of his kingdom. He was a visionary who realised the potential of selling palaces and converting forts into business endeavours. The remuneration was given to charitable trusts used for community welfare and education of his people. He converted Jag Niwas into Lake Palace Hotel, a charitable trust and sold Jag Mandir, Fateh Prakash and other estates on the shores of Lake Pichola to make sure his property was well kept.
Maharana Bhagwat Singh steered his people through transforming India and like his ancestors devoted his measures for the interests of his community. Through the time when the finances of the royal family were depleting, he did not let this come in his way and made policies that support the people of Mewar till date. He thus ended up being the last Maharana, a ruler who like his ancestors fulfilled his duty to respect and protect the legacy of his forefathers.
Maharana Bhagwat Singh, Maharanas of Mewar, Rajputana history, Rulers of mewar, rulers of Rajasthan
India's idol of courage that rose from Udaipur – Maharana Pratap
Haqiqat Bahida – an autobiography by Maharana Sajjan Singh of Udaipur
Shriji Arvind Singh Mewar's aid in Water Reservation, Conservation, and Management
479th Birth Anniversary of Maharana Pratap: A Brief Insight
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Newly minted University of Iowa graduate brings his passion for education to Iowa City schools
Aimee Breaux
Relief washed over Kedibona Ochs when, at 15, he first confided in a teacher that he was struggling with housing.
As a teenager, Ochs managed significant feats by himself.
He moved between Namibia and Iowa as a child twice, but still managed to keep his grades up. When his dad was unable to return to the U.S. after attending a funeral in Africa, he and his brother covered his newspaper delivery route from midnight to 7 a.m. before heading to school.
But at 15 he found himself homeless in Dubuque, bouncing between the homes of friends and family.
He marks telling a teacher of his struggles as a kind of turning point, when school staff began referring him to resources and encouraging him on campus.
"It showed me that people cared; people are there to help," he said. "I'm stubborn to the core. I need a lot of help in so many aspects of my life, but I'm the last one to ask for the help."
He jokes that from that point on, he became much more involved in school and became a "mascot" for campus.
As of this week, his passion for giving back to his school has become a professional endeavor. A newly minted University of Iowa graduate, Ochs has accepted a teaching job in the Iowa City Community School District.
Ochs gave the UI College of Education commencement speech Thursday. The speech could be described as an argument against the expression "those who can't, teach."
"I wanted top let the audience know, teaching is a profession," he explained. "Not every one is going to support it; they might think its an easy job, but I wanted to help them understand there's a lot that goes into this job."
Ochs will go on to teach freshman language arts at Liberty High school next school year. He'll replace a retiring teacher.
His hiring comes as the school district is expected to fill fewer teacher positions. After a wave of budget cuts across the district, administrators say they are hesitant to fill vacant positions.
The district usually hires between 80 and 100 teachers annually, though the district's human resources officer expects it will hire fewer teachers this year.
READ: Iowa City schools feel financial pressure to cut librarian positions
Staff at Liberty reached out to Ochs about the position on the recommendation of principals at West High, where Ochs worked as a student teacher.
Ochs clearly made an impression on students, teachers and administrators at West. The campus paper, West Side Story, featured Ochs in an article titled "Providing a bright spot." Walking the halls of West Friday, Ochs couldn't make it a dozen steps without a student stopping him for a high-five.
He's excited — and anxious — about having his own classes and starting off his career at a newer campus.
"You gotta be nervous for the things that are important," he said. "You might not realize it but you are shaping their beliefs, you're helping them figure out the world. Of course I'm nervous."
He said his philosophy about teaching in the field is "everybody's capable." Teaching, he says, is about helping all students find success.
"This is my way of giving back for everything the teachers did for me in high school," he said. "Now I want to give back to some student that may need me, of just need somebody."


Ikea is letting stray dogs find shelter at THIS store – and the pictures are so heartwarming!
There's nothing more we love than a feel good story. And when it involves our four-legged friends, our heartstrings get a second pull.
Both of these boxes were recently ticked when news emerged that an Italian branch of Swedish homeware giant Ikea had thrown open its doors to stray dogs in a search of a place to lay their heads.
The city of Catania – on Sicily's eastern coast – had experienced torrential rainfall which led to serious flooding over the past two weeks. Knowing that the area's homeless dogs would be overwhelmed by the deluge, an Ikea store allowed the presumably morose mutts some respite by enabling them to find shelter amongst its displays.
The good deed came to the attention of world when a video of a contented pair of doggies taking nap in a mock living room set up was shared on Facebook by Vittoria Taccia Gabrielli.
The Gazzetta del Sud newspaper confirmed that 'store staff let the strays wander around the aisles' and that the move had been met with approving nods from customers.
Since that date, pet lovers have taken to social media to share their own in-store snaps, and they've received scores of positive comments in response.
Is that a dog bed we can spy? If so, it seems that this little doggy was so pooped that he caught some much needed shut eye on the super soft rug rather than make that extra stretch to it. That hint of a sleepy smile seems to have really touched the hearts of commentators.
What do you think of this IKEA store's initiative?


Smalts Imported, Smaltz Imported or Osmalto Imported from Germany is the Highest grade of Sign Smalts and is a time old tradition of making a long lasting textured Sign background by embeding aggregate smalt made fine pigmented glass or colored sands. Smaltings were most popular from 1865-1935. Originally Buildings exterior walls were coated with Smaltz from the second floor on up to extend the life of the paint job out 20-25 years. Eventually Smalts worked it's way up to the roof, and that's were we see Smalting on a mass with modern roofing tiles.


UC Irvine's School of Education is continuing to receive national recognition for the quality and innovation of its programs and research. Since first ranked in 2007 at 88th among Graduate Schools of Education, the School has advanced each year, achieving this year's US News & World Report ranking of 14th among public universities.
Join us for bi-monthly talks by experts from across the country discussing the latest findings in education research.
Read about our faculty, research, projects, labs, students, alumni, and donors in our Fall 2018 magazine.
USDE EIR Expansion Grant will address reading and writing achievement gaps, grades 7-11, in six additional states.
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