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I was the featured vocalist on Andrew Rayels "Once In A Lifetime Love" that was put out by Armada records, and the song has gathered several million plays on Spotify!
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The new data centers in the US, the UK and Australia come as IBM looks to its cloud business to accelerate growth of its "strategic imperatives" revenue.
IBM announced Wednesday that it's opening four new IBM Cloud data centers across the globe, bringing its total data center footprint to 59 across 19 countries.
Two of the new data centers are in London, giving IBM a total of five data centers in the UK. An additional data center is slated to launch later this year. These new facilities are part of a plan IBM announced last year to triple its cloud data center capacity in the UK.
The third new data center now open is in San Jose, Calif., marking its third facility in the region and 23rd data center in the US. Lastly, IBM opened a new data center in Sydney, giving it four across Australia.
The expansion comes a day after IBM discussed its second quarter financial results, which delivered revenues below expectations. Investors are still waiting for IBM's investments in "strategic imperatives" such as AI and cloud computing to pay off, which should help make up for its weakening legacy businesses. SVP and CFO Martin Schroeter said Tuesday that he expects revenues from "strategic imperatives" to accelerate in the second half of 2017, in part because of new cloud contracts.
IBM's cloud revenue in Q2 came to $3.9 billion, up 17 percent year over year adjusting for currency. Over the last 12 months, its cloud revenue has totaled $15.1 billion. Schroeter cited some of IBM's recent wins in the cloud market, such as the 10-year agreement it signed with Lloyd's Banking Group, one of IBM's largest cloud agreement ever.
Amazon Web Services continues to dominate the public cloud infrastructure market, while other competitors like Google and Microsoft are making aggressive efforts to catch up. Schroeter said IBM's cloud business will benefit from its sophisticated capabilities, such as Watson-powered services.
"Enterprises will move to cognitive on the cloud with someone they trust, who has leading tools and industry expertise and a data model and business model consistent with their goals," he said.
"That is the IBM Cloud plus Watson."
Some analysts, however, have questioned the value Watson can generate for shareholders, given factors like the need for consulting to implement Watson and growing competition from cloud competitors.
IBM is also stressing its focus on security in the cloud, highlighting that it is one of the first global cloud companies to adopt the EU's Data Protection Code of Conduct for Cloud Service Providers. Its cloud facilities "are designed to be state of the art in terms of privacy and security," IBM boasts, promising to help European clients prepare for the implementation of the General Data Protection Regulations (GDPR).


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Trevor Noah speaks onstage during the 11th Annual Stand Up for Heroes Event presented by The New York Comedy Festival and The Bob Woodruff Foundation at The Theater at Madison Square Garden on November 7, 2017 in New York City.
JOHANNESBURG – "You don't overthrow a 93-year-old man, you just don't wake him up," quipped The Daily Show host Trevor Noah this week in reference to the current political situation in Zimbabwe.
Zimbabwe was thrown into political turmoil on Tuesday when the army seized control of the country and confined 93-year-old President Robert Mugabe into his house.
READ: Looks like a coup, smells like a coup, but is it a coup?
Mugabe, who has ruled the southern African country for 37 years, has reportedly agreed to step down next year.
But the army wants him to relinquish power immediately, with Emmerson Mnangagwa, the vice-president Mugabe fired last week, taking over as president.
South African born comedian Noah commented on the situation and compared it to the current political mood in the United States.


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Top 20 Highest Points On Islands Worldwide [41 PICS]
Tagged beaches, environment, islands, mountain peaks, mountains, nature, UNESCO World Heritage Site, volcanoes
When you think of islands, do you picture a tropical scene with beautiful warm beaches? Yet some islands are bitterly cold and covered in ice and snow. Some islands seem to have it all, from wonderful crashing waves to high peaks that are covered in snow. This is a list of high points for the top 20 islands in the world ordered by their highest point. As we did to illustrate the 20 U.S. National Parks by elevation, including the 20 highest points found in that park, here starting at #20 and counting down to #1 we show the immense diversity of islands. We've included an image from each island as well as a photo of the mountain or volcano peak that is the highest point of that island. [41 Photos]
Starting at #20 and counting down to #1, this is a list of the top 20 'highest' islands in the world, ordered by their highest points. Photo #1 by Paul (dex) Bica
Ranked at #20 on the list is Seram, Indonesia, where Mount Binaiya (Gunung Binaiya) stretches up to 9,931 feet, or 3,027 meters, making it one of the 100 most topographically prominent peaks on Earth. Jean-Claude Latombe's Mountaineering, Climbing, and Trekking Page has fascinating photos and the story of treking through Seram island and ascending Mt. Binaiya. Photo #2 by Jean-Claude Latombe / Standford
#20: Seram from the sea on top. But trekking deep in the jungle to climb Mount Binaiya, the photographer said, "Encounter with a python that had swallowed a deer. Extremely impressive. It takes about a month for a python to digest such a big animal." Photo #3 by Jean-Claude Latombe & #4 by Jean-Claude Latombe
Sunset with Mount Agung in Amed, Bali. Indonesia, the 19th highest point on islands worldwide. As we pointed out when showing off the 7 Sea Temples of Beautiful Bali, it is also known as the island paradise of 1,000 temples. Photo #5 by Mikaku
#19: Rice fields in Bali with Mount Agung in background. Mount Agung stands tall at 9,994 feet, or 3,031 meters, at the peak. Photo #6 by Global Images
#18: The Island of Maui ranks at #18 due to Haleakala. This is the scene at the big beach at Makena in Makena State Park, Maui. Photo #7 by Laszlo Ilyes
Staring down into the crater of Mount Haleakala from the summit during sunrise. In Haleakala National Park on the island of Maui in Hawaii, Haleakala is 10,023 feet (3,055 m) high, making it the 18th highest island point in the the world and the 2nd tallest point of USA islands. Haleakala also takes the prize as 18th for the list of U.S. National Parks by highest elevation. Photo #8 by The Q
Beach on Réunion, France, the island with the 17th highest point due to Piton des Neiges. Photo #9 by Samuel Hoarau
#17 highest point is Piton des Neiges peak on Réunion island. It reaches 10,072 feet, or 3,070 meters. Sunrise on the Piton des Neiges is rarely clear. The photographer added, "We were lucky because few clouds are shown." Photo #10 by Caroline et Louis VOLANT
#16 on the list of islands by highest points is Hispaniola in the Dominican Republic with Pico Duarte. Photo #11 by MichelleWalz
Pico Duarte is the highest peak in all the Caribbean islands; it reaches 10,164 feet, or 3,098 meters, coming in at 16th. Photo #12 by Best Home Decorators
#15 is Siple Island with Mount Siple, a potentially active Antarctic shield volcano, rising to 10,203 ft (3,110 meters) and dominating the northwest part of Siple Island. Photo #13 by Amar Andalkar
Siple Dome on Mount Siple, Antarctica, the 15th highest point on islands. Photo #14 by chinook
Sicily, Italy, with Mount Etna is ranked 14th for highest island point worldwide. Photo #15 by ISS / NASA
#14: At 10,912 ft, or 3,326 meters high, Mount Etna is one of the most active volcanoes in the world and is in an almost constant state of activity. It's located on the east coast of Sicily. Photo #16 by Josep Renalias
#13: Sulawesi island, Indonesia, because of the Mount Rantemario peak. Photo #17 by Clifton Beard
Summit of Mount Rantemario on the island of Sulawesi, reaches 11,411 ft (3,478 m) to rank 13th. Photo #18 by Me Irmawar
#12: Colors of a mystic Java sunset. The Semeru volcano on Java is 12th when it comes to highest points of islands worldwide. Photo #19 by Dennis Stauffer
Semeru volcano, or Mount Semeru, rises to 12,060 ft (3,676 m) on the island of Java, Indonesia. The photographer said of this Semeru background with stunning star trails, "Another freezing night in the Bromo Tengger Semeru National Park." Photo #20 by Dennis Stauffer
Ranked 11th: Clouds and Northern Lights aka Aurora Borealis over Greenland, which comes in at #11 in highest points of islands. Greenland is also famous for extraordinarily incredible icebergs. Photo #21 by Trine og Mads
#11: Gunnbjørn Fjeld, or Gunnbjørn, 12,119 ft (3,694 m) is the highest point on Greenland. It is also the highest mountain north of the Arctic circle. Wikipedia adds, "It is a nunatak, a rocky peak protruding through glacial ice. In the Icelandic Sagas, and thus by the Norsemen, the mountain was called Hvitserk, literally meaning 'whiteshirt'. It is named after Gunnbjorn Ulfsson, the Viking who was the first European to reach Greenland." The photographer noted, that this shot of Gunnbjørn was "taken during a joint British-Russian expedition in May 2011." Photo #22 by Maxim Bouev
#10 ranked island by highest point is Tenerife, Spain. This is one of the Canary Islands. Teide Peak stretches to 12,198 ft (3,718 m). The 73 sq mi (18,900 hectares) area surrounding the Teide volcano makes up Teide National Park, which was named a UNESCO World Heritage Site. This rare capture includes Teide, Luna, Venus and Jupiter. Photo #23 by Santiago Atienza
Mount Teide, the volcano on Tenerife is ranked 10th. The photographer wrote, "you can gaze at certain times like this in the early morning or in the evening still enjoy almost alone. The picture shows the highest mountain in Spain, the 3718 m high Pico del Teide. The imposing rock in the foreground is the Los Roques." Photo #24 by Christian Abend from Laufen an der Salzach, Bayern / Deutschland
#9 highest point is on the island of Lombok, Indonesia. This sunrise on Gili Trwangan has a view of Mount Rinjani in the background. Photo #25 by Fotoherby
A small eruption of #9 Mount Rinjani, with volcanic lightning. Lomvok's Mount Rinjani, or Gunung Rinjani, active volcano rises 12,224 ft (3,726 m) and has a crater lake (caldera) known as 'Child of the Sea.' The lake is about 660 ft, 200 meters, deep; the caldera also has a hot springs. Photo #26 by Oliver Spalt
#8 South Island, New Zealand, with Aoraki / Mt. Cook in the background and Lake Pukaki in the foreground. Mt. Cook National Park has more than 140 peaks standing over 6,600 ft (2,000 m). Photo #27 by Mike Bischoff
South Island ranked 8th due to 12,320 ft (3,755 m) peak of Aoraki / Mount Cook, seen here with lenticular clouds over the summit. Photo #28 by SW arts
#7: Endless fields of purple flowers in front of the 7th highest point on an island, Mount Fuji, which is still capped with snow. Photo #29 by Mt. Fuji
Honshu Island, Japan, comes in ranked 7th, since the towering 12,388 feet (3,776 meters) active stratovolcano Mount Fuji. Photo #30 by Fujisan
#6 is Ross Island, Antarctica, with Mt. Erebus seen here with a plume. Photo #31 by Johannes Zielcke
Ross Island is 6th due to the 12,448 ft (3,794 m) summit of Mount Erabus. Although considered Antarctica, Ross Island did not belong to a country or territory, so it was claimed by New Zealand. Photo #32 by T.J. and Jessie
#5 on the list of islands by highest point is Sumatra, Indonesia, with Mount Kerinci. The photographer said, "Sampan traversing the lake in Kerinci Seblat National Park." Photo #33 by Tim Mowrer
At 12,484 ft (3,805 m), Mount Kerinci is the highest volcano in Indonesia, the highest peak on the island of Sumatra, and the 5th highest point of any island globally. According to Wikipedia, "It is surrounded by the lush forest of Kerinci Seblat National Park, home to the endangered species of Sumatran Tiger and Sumatran Rhinoceros." Photo #34 by buitenzorger
#4 is Taiwan, Republic of China, with Yushan, also called Jade Mountain, looming up to 12,966 ft (3,952 m). Photo #35 by new7wonders
Sunrise over Yushan, 4th highest point among islands. In the winter, Yushan is often capped with thick snow, making the entire peak shine like stainless jade thus the name Jade Mountain. Photo #36 by new7wonders
The 3rd highest point on an island is Borneo, Malaysia, with Mount Kinabalu. Photo #37 by Kevin Wasilin
#3 Mount Kinabalu rises to 13,435 ft (4,095 m) and is protected as Kinabalu National Park, a World Heritage Site. The photographer called this, "Welcome in Mordor." Photo #38 by Bousure
#2: Hawaii has Mauna Kea. This little slice of paradise is Green Sand Beach, also called Papakolea Beach, Big Island, Hawaii. Photo #39 by Steve Dunleavy
Mauna Kea stretches to 13,803 ft (4,207 m), making it the 2nd highest point of islands worldwide. Hawaii seems to have it all, from gorgeous green or even black sandy beaches, to lava entering the Pacific at Hawaii Volcanoes National Park, to the high peak covered in snow as seen here at the Mauna Kea Observatories telescopes. Photo #40 by Joe Philipson
#1 on list of islands by highest point is New Guinea, Indonesia, with the Puncak Jaya summit at 4,884 m, 16,024 ft on Mount Carstensz. The peak is also called Carstensz Pyramid, locally referred to as peak Jaya; it is the highest peak in Australia and Oceania. The mountain was named Cartensz after Dutch sailor John Carstensz, who saw that the mountain on the equator was covered by ice. In 1623, no one would believe Carstensz's story about the the existence of the ice caps along the tropical equator. His reports were first met with laughter. These are reportedly the only tropical glaciers in Indonesia, and they are disappearing in what some claim is due to global warming. Photo #39 by Firman / Papua Cloud
To underscore the incredible shrinking Puncak Jaya Glacier, NASA said the top images were photographed by J.J. Dozy in 1936; by the United States Geological Survey during the Carstensz Glaciers Expeditions in 1972; and by a NASA astronaut on the International Space Station in 2005. The bottom Landsat 7 image shows the 4,884-meter-tall (16,024-foot) summit of Puncak Jaya, one of the Seven Summits, the nearby open pit of Freeport mine – one of the world's largest producers of copper and gold – and three glaciers. The Earth Observatory said, "While Puncak Jaya's peak is free of glaciers, there are several on its slopes, including the Carstensz Glacier, the Meren Glacier, and Northwall Firn (hanging glaciers)." In this image, "snow and ice appears blue while clouds are white, bare unvegetated land such as the rocky limestone summit ridges appear red or pinkish, the deep purple and grey is the open pit copper mine while the lush rainforest of the lowlands appears green." Photo #40 by NASA/JSC/USGS/ISS Crew Earth Observations experiment and the Image Science & Analysis Group & #41 by Earth Observatory / NASA
3 Responses to "Top 20 Highest Points On Islands Worldwide [41 PICS]"
TΑ 20 ΥΨΗΛΟΤΕΡΑ ΣΗΜΕΙΑ ΑΠΟ ΝΗΣΙΑ ΣΕ ΟΛΟ ΤΟ ΚΟΣΜΟ Νο 1 « A Green Observer says:
Really beautifully done photography, incredibly clear…thank you so much!

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text:
Trump Rejects Russian Hacking Report as Senators Launch Probe
President-elect says his kids will run business but won't be cutting deals
Then-Republican nominee Donald Trump addresses Fox News anchor Chris Wallace at an Oct. 19 debate in Las Vegas. Wallace interviewed the president-elect in a segment that aired Sunday. (Bill Clark/CQ Roll Call)
Posted December 11, 2016 at 11:18am
President-elect Donald Trump rejected a CIA finding that Russia intervened in the general election on his behalf, even as senior Senate Republicans and Democrats launch a joint review of Moscow's actions.
Trump dismissed reports of an intelligence agency assessment that found hackers linked to the Kremlin had interfered in the U.S. election with the goal of helping Trump win, calling the notion "ridiculous."
"I think it's just another excuse" being put forward by Hillary Clinton supporters, Trump said on "Fox News Sunday," adding, "I don't believe it."
Trump accused Democrats bitter over the general election outcome, of leaking the CIA assessment, suggesting such a move would be one part revenge and another pinning Clinton's loss on Moscow.
"They have no idea if it's Russia or China," Trump said of the CIA. "It could be somebody sitting in a bed someplace. They have no idea."
Four senior senators, however, believe it was Russia.
[(VIDEO) White House Watch: Generals, CEOs and Defense Spending Promises]
Incoming Senate Minority Leader Charles E. Schumer of New York and Senate Armed Services Ranking Democrat Jack Reed of Rhode Island joined Senate Armed Services Chairman John McCain, R-Ariz., and Lindsey Graham, R-S.C., a top panel member, in a rare joint statement saying the CIA report "should alarm every American."
"While protecting classified material, we have an obligation to inform the public about recent cyberattacks that have cut to the heart of our free society," the senators said Sunday. "Democrats and Republicans must work together, and across the jurisdictional lines of the Congress, to examine these recent incidents thoroughly and devise comprehensive solutions to deter and defend against further cyber-attacks.
"This cannot become a partisan issue. The stakes are too high for our country," they said, announcing a bipartisan review of Russia's actions. "We are committed to working in this bipartisan manner, and we will seek to unify our colleagues around the goal of investigating and stopping the grave threats that cyberattacks conducted by foreign governments pose to our national security."
By joining with the Democrats, McCain and Graham are setting up an early clash with Trump, who has been accused of being soft on Russia. The president-elect has said the United States and Russia should be allies, and that a better relationship would help each achieve its strategic gains.
President Obama has ordered his own review of Russia's alleged actions.
Separately, Trump refused to say whether he would nominate ExxonMobil executive Rex Tillerson for secretary of state, as has been reported. Trump mentioned Sen. Bob Corker and Mitt Romney as other possible nominees.
In a later tweet, Trump wrote "stay tuned!" on whether he will indeed nominate Tillerson, calling him a "world class player and dealmaker."
The real estate mogul also said he would have no role with his business once he is sworn in as president, saying the firm's executives and his adult children would run the company. He did not spell out any legal arrangement he would make to divest himself of decision-making or profits, though he has pledged to do so on Dec. 15.
Trump said his children won't be "making deals" with foreign governments or other entities that could trigger possible conflicts of interest. He did not elaborate.
Notably, however, Trump said his team is looking at existing laws to determine whether his children can advise him while also running his business.
Trump also continued to make what appeared contradictory statements about his guiding principles.
He declared himself a "big free-trader" and called U.S. companies moving production overseas to cut costs the result of "the dumb market." He vowed to make it difficult on companies that try to offshore jobs in lower-tax countries, by levying a 35 percent tax.
"They won't leave," he predicted of companies deciding between offshoring savings and such a massive tax.
But he also vowed to help businesses by slashing the time it would take to obtain regulatory approvals from federal agencies such as the EPA.
"When [companies] have to wait 10 and 15 years for an approval, and then you don't get the approval, it's not good," he said. "We're going to clean it up and speed it up."
On defense, Trump reiterated his call to ban government employees who make contract decisions on major programs like the Pentagon's F-35 fighter from ever working for the companies that benefit from those deals.
He called defense contracting overruns and the revolving door between the Defense Department and arms manufacturers a bigger problem than the the one between government service and lobbying, a major part of his campaign platform.
[Trump Openly Clashes With CIA Before Taking Office]
Many former officials and analysts predict Trump will immediately and aggressively begin dismantling most of outgoing President Barack Obama's policy legacy.
"I don't want to do that at all," said Trump, who in a series of one-on-one conversations as developed a professional relationship with the president he harshly criticized for years. "I just want to do what's right."
In one example, Trump declined to say he would withdraw the United States from the Paris climate pact that, if kept in place, would be a major part of Obama's legacy.
"I'm studying it," the incoming chief executive said. "I do say this: I don't want that agreement to put us at a competitive disadvantage. … I don't want that to give China or other countries signing agreements an advantage over us."


text:
\section{Another description of $\tau$-forms}
\label{tauinv}

In this section, we give a new algebraic characterization of the
strongly minimal sets living on a curve $C$, using the theory of $\tau$-differentials
developed in \cite{Ros}.   Of course, it is equivalent
to the earlier formulation, but has the advantage that it does not involve
considering equivalence classes, so it is easier to work with.
Below, we use the language of schemes, so a variety is an integral separated
scheme of finite type over an algebraically closed field.
We continue to assume that curves are smooth and projective.

Recall that there is a natural bijection between rank $n$ locally free
sheaves on a variety and rank $n$ vector bundles over it.
For example, the sheaf of differential forms on a variety $X$,
$\Om_X$, corresponds to the cotangent bundle $T^*X$,
and can be viewed as the sections of this bundle.  Given a curve $C$,
the locally free rank 2 sheaf of $\tau$-differentials on $C$, as defined in \cite{Ros},
and also denoted $\Omt_C$,
corresponds exactly to the set of rational $\tau$-forms, as defined above,
with a $\tau$-form $\omt$ corresponding to a section $s \in \Omt(U)$,
for open $U \sub C$.

Given a rank 2 bundle, one can consider all the rank 1 subbundles.
(To be precise, a rank 1 subbundle is a rank 1 bundle together
with a bundle embedding.)  Passing to sheaves, rank
1 subbundles correspond to invertible (i.e., rank 1) subsheaves.
Note that it is possible for one rank 1 sheaf to be properly embedded in
another (in which case the quotient will be a torsion sheaf), essentially
because this can also happen with free rank 1 modules.  Nevertheless,
in our situation, any invertible subsheaf of a rank 2 locally free sheaf
embeds in a unique maximal invertible subsheaf, as described below.

We have seen that \taufs\ correspond to sections of a rank 2 locally
free sheaf.  Since the $\seq$-equivalence class of such a form $\omt$
consists of those forms $g\omt, g \in K(C)$, such a class should correspond
to a (unique) maximal subsheaf containing $\omt$ as a section on an open set.
The following easy lemma makes this precise.

\bla
Let $C$ be a smooth curve, $\MH$ a rank 2 locally free sheaf on $C$,
and $s \in \MH(U)$ a section.  Then there is a unique maximal
invertible subsheaf $\MF \sub \MH$ such that $s \in \MF(U)$.
Further, the quotient $\MH / \MF$ is an invertible sheaf.
\ela

\prf
By possibly passing to a smaller open set, we may assume that $U = \Spec\ R$
is affine, $\MH(U) \cong \tilde{M}$, $M = \<m,n \>$ a free rank 2
$R$-module, and $s = am + bn\in M, a,b \in R$.  Since $R$ is a UFD
(e.g.\ \cite{Mat}, Thm.\ 20.1), $a$ and $b$ have a greatest common divisor
$c$.  Letting $a' = a/c$, $b'=b/c$, and $s'=a'm + b'n$, the module
$N = Rs' \sub M$ is the unique largest rank 1 free submodule of $M$
containing $s$.  Let $d,e \in R$ be such that $da'+eb'=1$, and let
$L = R(em-dn)$.  Then it is easy to check that $M  = N \op L$, and hence that
there is a short exact sequence $0 \lra N \lra M \lra L \lra 0$.

We want to define $\MF$ such that $\MF(U) = \tilde{N}$.  Since a sheaf is
determined by its stalk at each point $P$, it suffices to show how, for each
$P \in C - U$, $\MF(U)$ determines a unique maximal stalk $\MF_P$
that will make $\MF$ locally free.

For each such $P$, choose an affine scheme $\ W \sub U \cup \{P\}$
containing $P$, $W \cong \Spec\ S$, such that $\MH(W)$ is free on $W$.
Note that $ W':= W - P =
\Spec\ S_f$, for some $f \in S$, and $P = V(f)$.  Let $\MH(W') = \tilde{J}$
and choose basis elements $\<i,j\> \in J$ such that $\MF(W') = \tilde{I}$,
where $I := Ri \sub J$.  We then define $\MF_P$ to be the submodule of
$\tilde{J}_P$ generated by $\<i\>$.
Clearly, this determines an extension of $\MF$ to $U \cup P$ and thus,
continuing in this way to all of $C$.  (As an aside, note that one could
also define $\MF_P$ to be generated by $f^mi$, for any positive $m$.
Choosing $m > 1$ would yield a non-maximal invertible subsheaf.)
\qed

\msk

It is easy to see that two sections determine the same invertible subsheaf
if and only if, on some non-empty open set on which they are both defined,
each is a multiple of the other.  In particular, $\sim$ equivalence classes
of \taufs\ correspond to maximal invertible subsheaves, as desired.

\bdf
Let $C$ be a curve.  A {\em $\tau$-invertible sheaf} on $C$, denoted $\MFT$,
is a maximal invertible subsheaf of $\Omt_C$, i.e, an invertible sheaf $\MF$,
together with an embedding $f:\MF\ra\Omt_C$.
\edf

\brm
For any curve $C$, there is a {\em trivial} \tis, corresponding to
the natural embedding of $\O_C$ into $\Omt_C$.  Clearly, this \tis\
corresponds to the $\sim$-equivalence class of trivial \taufs.
Below, by \tis\ we will always mean non-trivial \tis.
\erm

We can reformulate Corollary~\ref{formz} as follows.

\bpro
Let $C$ be a curve.  There is a natural bijection between strongly
minimal sets living on $C$ and $\tau$-invertible sheaves on $C$.
\epro

As before, given a \tis\ $\MFT$ on $C$, let $\Xi(C,\MFT)$ denote the
corresponding strongly minimal\ set.

We now introduce the analog of a global \tauf.
Since we are dealing with equivalence classes of \taufs,
the correct notion is that of being a \tauf\ that is $\seq$-equivalent
to a global \tauf.  Passing to $\tau$-invertible sheaves, these are
the sheaves $\MFT$ that contain a global section.

\bdf
Let $C$ be a curve, $\MFT$ a \tis. Say that $\MFT$ is a {\em global \tis}
if $H^0(C,\MFT) \neq 0$, that is, if $\MFT$ has a global section.
\edf

Next, we define the appropriate notion of a $\tau$-invertible
sheaf being essential, which is more subtle than for 1-forms.
First, we establish the following results.

\bpro
\label{injpull}
Let $f: X \ra Y$ be a finite morphism of curves.
There is an exact sequence of sheaves on $X$,
$$
0 \lra f^*\Omt_Y \sr{\al}{\lra} \Omt_X \lra \Om_{X/Y} \lra 0
$$
\epro

\prf
To check that $\alpha$ is injective, it suffice to check this at the
stalk of the generic point.  Here, injectivity is obvious.
\qed

The following proposition follows immediately.

\bpro
\label{tauinvpull}
Let $f:X \ra Y$ be a finite morphism of curves, and $\MFT$ a (non-trivial)
\tis\ on $Y$.
Then there is a unique  \tis\ $\MGT$ on $Y$ such that $\al(f^*\MFT) \sub \MGT$.
Further, there is the following commutative diagram, with exact rows,
$$
\xymatrix{
0 \ar[r] & f^*\MFT \ar[r]\ar[d] & \MGT \ar[r]\ar[d] & \Om_{X/Y} \ar[r]\ar[d] & 0 \\
0 \ar[r] & f^*\Omt_Y \ar[r]^\al & \Omt_X \ar[r] & \Om_{X/Y} \ar[r] & 0,\\
}
$$
such that each vertical arrow is an injection, and $\MH$ is a torsion sheaf.
\epro

\bpr
The only point that needs to be checked is that the cokernel of the map
$f^*\MFT \lra \MGT$ is $\Om_{X/Y}$.  To verify this, it suffices to compute
the map locally around each ramification point of $X$.
\epr

\brm
This proposition says that, given a morphism of curves,
$f:X\ra Y$, the pullback of a \tis\ is not necessarily
a \tis, even though the pullback of a $\tau$-form is a $\tau$-form.
By analogy, the pullback of a 1-form on $Y$ is a 1-form on $X$
though, in general,  $f^*\Om_Y \neq \Om_X$.  But there is an
injective homomorphism $f^*\Om_Y \ra \Om_X$, just as there is
an injective homomorphism $f^*\MFT \ra \MGT$.

Note that given a morphism between curves (or more generally varieties),
$f: C_1 \lra C_2$, the pullback of the trivial \tis\ on $C_2$ is equal to
the trivial \tis\ on $C_1$, precisely because $f^*\O_{C_2} = \O_{C_1}$.
\erm

\bdf
In the notation of the previous proposition, we say that $\MGT$ is the
{\em weak pullback} of $\MFT$ (along $f$).
In case the natural map $f^*\MFT \ra \MGT$ is an isomorphism, we say
that $\MGT$ is the {\em strong pullback} of $\MFT$.
\edf

The following lemma follows immediately from the definitions.

\bla
Let $f: X \ra Y$ be a morphism of curves, $\omt$ a $\tau$-form on $Y$,
and $\MFT$ the \tis\ corresponding to $\omt / \sim$.  Then $\MGT$, the weak
pullback of $\MFT$, is the \tis\ corresponding to $f^*\omt$.
\ela

We can now reformulate Lemma~\ref{taupullback} in terms of \tiss.
\bla
Let $f: X \ra Y$ be a morphism of curves, $\MFT$ a \tis\ on $Y$,
and $\MGT$ the weak pullback of $\MFT$ on $X$.  Then
$f^{-1}\Xi(Y,\MFT)=\Xi(X,\MGT)$.
\ela

\bdf
Let $X$ be a curve.  A \tis\ $\MGT$ on $X$ is {\em essential}
if there does not exist a curve $Y$ and a morphism $f:X\ra Y$
such that $\MGT$ is the weak pullback of some \tis\ on $Y$.
\edf

\brm
Let $X$ be a curve, and $f:X \ra Y$ a regular map to another curve $Y$.
By Proposition~\ref{tauinvpull}, there is a natural embedding of $\O_X$-sheaves,
$\alpha: f^*\Omt_Y \lra \Omt_X$.  Notice that for any \tis\
$\MGT\sub \Omt_X$, there is maximal invertible subsheaf of
$\alpha(f^*\Omt_Y)$ that embeds in $\MGT$,
namely $\MGT \cap \alpha(f^*\Omt_Y)$.
Let us call this subsheaf $\MH$.  If $\MGT$ is essential, then
$\MH$ cannot be of the form $f^*\MFT$, for $\MFT$ a \tis\ on $Y$.

In particular, although every section of $f^*\Omt_Y$ is naturally a
function on $\tau X$, not every such function is the pullback along
$f$ of a $\tau$-form on $Y$.  Functions arising in this way
naturally correspond instead to sections of $f^{-1}\Omt_Y$, but this
is not an $\O_X$-sheaf.  Recall that by definition,
$f^*\Omt_Y = f^{-1}\Omt_Y \ot_{f^{-1}\O_Y}\O_X$.

As sections of a \tis\ $\MFT$ on $Y$ correspond to a $\sim$-equivalence class of functions
on $\tau Y$, sections of $f^{-1}\MFT$ correspond to the pullback of these
functions to $\tau X$ along the lifting map $f\prl$, as defined in Section~\ref{prolong}.
As $f^*\MFT = f^{-1}\MFT\ot_{f^{-1}\O_Y}\O_X$, sections of $f^*\MFT$ correspond to
products of functions from $f^{-1}\MFT$ with rational functions on $X$.
\erm

\brm
Observe that a $\tau$-form $\omt$ is $\sim$-essential, as defined in
Section~\ref{algcurves},
if and only if the \tis\ corresponding to $\omt / \sim$ is essential.
\erm

We can now restate Theorem~\ref{genprop}.

\bthm
Let $C$ be a smooth projective curve of genus $g \geq 1$.  Let $\MFT$ be
an essential \tis.  Then there is a finite set $A \sub \Xi(C,\MFT)$
such that $\Xi(C,\MFT) - A$ is strictly minimal. Further, for any
definable equivalence relation with finite classes on $\Xi(C,\MFT)$,
over any set of parameters, almost every class has one element.
Two such sets, $\Xi(C_1,\mathcal{F}^\tau_1)$ and
$\Xi(C_2,\mathcal{F}^\tau_2)$ are
non-orthogonal if and only if there is an isomorphism $f:C_1 \ra C_2$ such that
$\mathcal{F}^\tau_1$ is the weak pullback of $f^{*}\mathcal{F}^\tau_2$.
\ethm

We now complete the proof of Proposition~\ref{globdim}.  Given a (locally free)
sheaf $\MF$ on a projective variety $V$, let $\hh^0(V,\MF)= \dim_K H^0(V,\MF)$,
the dimension of the space of global sections of $\MF$.

Given a curve $C$, we will call $\hh^0(C,\Omt_C)$ the {\em $\tau$-rank} of $C$,
denoted $g^\tau_C$ or simply $g^\tau$.  The following proposition shows that $g^\tau$
can be easily determined by the Kodaira-Spencer rank of $C$, which we now explain.

By \cite{Ros}, we have the following short exact sequence
$$
\xi:\qquad \ 0 \lra \O_C \lra \Omt_C \lra \Om_{C/K} \lra 0.
$$
Recall that $\xi$ is called an extension of $\Om_{C/K}$ by $\O_C$,
and that there is a natural bijection between the set of such extensions and
the cohomology class $\Ext^1(\Om_{C/K}, \O_C)$, such that
the split extension corresponds to $0\in \Ext^1(\Om_{C/K}, \O_C)$
(e.g., \cite{CW}, Theorem 3.4.3).  In our case, this mapping is given as follows.
Given $f:C \lra \Spec\, K$, consider the sequence
$$
\eta: \qquad 0 \lra f^*\Om_K \lra \Om_C \lra \Om_{C/K} \lra 0
$$
(with $\Om_C$ projective).  Applying $\Hom(-,\O_C)$ yields the exact sequence
$$
\Hom(\Om_C,\O_C) \lra \Hom(f^*\Om_K, \O_C) \sr{\partial}{\lra}
\Ext^1(\Om_{C/K},\O_C) \lra 0
$$
Given $x\in \Ext^1(\Om_{C/K},\O_C)$, choose $\beta \in \Hom(f^*\Om_K, \O_C)$
such that $\pd(\beta)=x$.  We then let $\zeta$, the pushout of $\eta$ along $\beta$,
be the extension corresponding to $x$.
$$
\xymatrix{
\eta: & 0 \ar[r] & f^*\Om_K \ar[r]\ar[d]^\beta
& \Om_C \ar[r]\ar[d] & \Om_{C/K} \ar[r]\ar[d] & 0 \\
\zeta: & 0 \ar[r] & \O_C \ar[r] & \MG \ar[r] & \Om_{C/K} \ar[r] & 0
}
$$
Thus, by \cite{Ros}, the extension $\xi$ corresponds to the element
$\pd(\dltl) \in \Ext^1(\Om_{C/K},\O_C)$, where $\dltl$ is determined by the derivation
on $K$.

There are natural isomorphisms $\Ext^1(\Om_{C/K},\O_C) \cong
\Ext^1(\O_C, \Theta_{C/K}) \cong H^1(C,\Theta_{C/K})$,
with $\Theta_{C/K} = \Om_{C/K}^\vee$, the tangent sheaf
of $C$ (the dual of $\Om_{C/K}$) (e.g., \cite{RH}, pages 234-5).
Thus, there is a natural map, which we also denote $\pd$,
$\pd:\Hom(f^*\Om_K,C) \lra H^1(C,\Theta_{C/K})$,
which Buium~(\cite{Bui93}, p. 1395) calls the Kodaira-Spencer map of $C$.
The element $\pd(\dltl) \in \Ext^1(\Om_{C/K},\O_C)$, or $H^1(C,\Theta_{C/K})$,
is the Kodaira-Spencer class of $C$.

From the sequence $\eta$, corresponding to the Kodaira-Spencer class of $C$,
one obtains a long exact sequence of cohomology groups.
$$
0 \lra \HH^0(C,\O_C) \lra \HH^0(C,\Omt_C) \lra \HH^0(C,\Om_{C/K})
\stackrel{\partial}{\lra}
\HH^1(C,\O_C) \lra \ldots
$$
Let us call $\dim_K \partial (\HH^0(C,\Om_{C/K}))$ the {\em Kodaira-Spencer} rank of
$C$, denoted $KS_C$.  Clearly, if the Kodaira-Spencer class of $C$ is trivial,
that is, if $\eta$ splits, then $KS_C = 0$, though the converse does not hold in
general.
For curves of genus 1, though, $\eta$ splits if and only if $KS_C = 0$.
This follows, for example,
from Atiyah's classification of rank 2 bundles over elliptic curves \cite{At}.
This yields a complete understanding of global $\tau$-forms on elliptic curves.

The following proposition is equivalent to Proposition~\ref{globdim} above.

\bpro
\label{cohom}
Let $C$ be a smooth projective curve.
Then $1 \leq h^0(C,\Omt_C)\leq g + 1$.
Precisely, $h^0(C,\Omt_C) = g+1 - KS_C$.
In particular, if $C$ is defined over $k$, the field of constants, then
$h^0(C,\Omt_C) = g+1 $.
\epro

\prf
The first two statements follow from remarks above.  The third follows from
the fact that a curve defined over $k$ has trivial Kodaira-Spencer class.
(Since $H^1(C,\Theta_{C/K})$ classifies $TC$-torsors (e.g.,
see~\cite{JM}), this is equivalent to the fact that for such curves,
$\tau C \cong_C TC$.)
\qed

\ignore{
To prove the third,
recall that the class $H^1(C,\Theta_{C/K})$ classifies $TC$-torsors (e.g.,
see~\cite{JM}), that and Buium~(\cite{Bui93}, p. 1396) proves that the
{\em Kodaira-Spencer class}\
$\pd(\dltl)$ corresponds with the $TC$-torsor $\tau C$.  The torsor is
trivial, that is, isomorphic to $TC$, if and only if $\pd(\dltl) =0$, which
holds exactly when the original sequence $\xi$ splits.  Putting everything together,
one gets that $\xi$ splits if and only if $\tau C \cong_C TC$, which happens exactly
when $C$ is defined over $k$, as desired.  This completes the proof.
\qed
}

\subsection{Definability results}

Recall that the {\em degree} of an invertible sheaf on a variety is defined to
be the degree of the corresponding divisor.  We now observe that there is
a (uniform) bound on the degree of a $\tau$-invertible sheaf on a curve.

\bla
Let $V$ be a projective curve, $\MF$ a coherent sheaf on $V$.  Then for any
invertible subsheaf $\MG$ of $V$, $\deg \MG \leq \hh^0(\MF) + g(C) - 1$.
\ela

\bpr
By the Riemann-Roch theorem,
$\deg \MG = \hh^0(\MG) - \hh^0(\om \otimes \MG^{-1}) + g - 1 \leq
\hh^0(\MG) + g - 1 $.  Since $\hh^0(\MG) \leq \hh^0(\MF)$, this yields the
desired bound.
\epr

\bco
For any curve $C$, the degree of any \tis\ is $\leq 2g$.
\eco

The next result then follows by standard arguments.

\bco
For any curve $C$, the set of global $\tau$-forms on $C$ is definable,
as is the equivalence relation $\sim$ on these forms.  Further, this remains
true for families of curves (of fixed genus).  Finally, the dimension of
the space of global $\tau$-forms, as a $K$-vector space, is uniformly definable.
\eco

Observe that, by Proposition~\ref{cohom}, this last claim is equivalent to the known fact
that the Kodaira-Spencer rank of a curve is definable.

\ignore{
\newpage

We can now state one of our main results.
All curves are smooth projective curves over an uncountable algebraically
closed field $K$.  Given a curve $C$ and a sheaf $\MF$, recall that
$h^i(C,\MF) = \dim_K(H^i(C,\MF))$.  (I'll specify the notion of
generic curve later.)

\bthm
Let $C$ be a generic curve of genus $g \geq 2$.  Then every 'non-trivial'
global $\tau$-invertible sheaf is essential.
\ethm

The proof uses the following proposition, which was provided by L. Ein.

\bpro
Let $C$ be a generic curve of genus $g \geq 2$.  For any curve $C'$,
morphism $f:C \ra C'$, and invertible sheaf $\MF$ on $C'$,
if $h^0(C',\MF) = 0$, then $h^0(C,f^*\MF) = 0$.
\epro

\prf
The following fact is well-known.  (The proof involves a counting
argument.  The moduli space of curves of genus $g \geq 2$ is
$(3g-3)$-dimensional and, by the Hurwitz formula, if $f: C_1 \ra C_2$
is a morphism between curves, then $g(C_2) \leq g(C_1)$.)

\bla
Let $C$ be a generic curve of genus $g \geq 2$, $C$ any curve, and
$f:C\ra C'$ a morphism of degree $\geq 2$.  Then $C' \cong \P^1$.
\ela

We now assume that $C'\cong\P^1$, and let $\MF$ be an invertible sheaf
on $\P^1$, $\MF=\O_{\P^1}(n), n\in\Z$.  Then
$
h^0(C,f^*\MF)=h^0(\P^1,f_*f^*\MF)=h^0(C',\MF\otimes f_*\O_C).
$
The first equality is obvious, and the second follows immediately from
the  Projection Formula (\cite{RH}, p.\ 124).

We claim that $f_*\O_C$ is a locally free sheaf of the form
$\O_{\P^1}\oplus(\oplus_i\O_{\P^1}(a_i)),$
with $a_i < 0$, for all $i$.  We have a short exact sequence,
$$
0\lra \O_{\P^1}\stackrel{\alpha}\lra f_*\O_C\lra \ME \lra 0.
$$
Let $\beta: f_*\O_C \ra \O_{\P^1}$ be the map $\beta = \frac{1}{d}t$,
where $t$ is the standard trace map.  Then $\beta\circ\alpha$ is the
identity on $\O_{\P^1}$, so the above sequence splits,
i.e., $f_*\O_C=\O_{\P^1}\oplus\ME$.

Next, observe that $\ME$ is locally free.  To show this, it is sufficient
to show that $\ME_x$ is locally free, for all $x\in \PO$
(\cite{RH}, p.\ 124).  For the generic point $x_0\in \PO$,
we get an exact sequence of vector spaces over the field $K(y)$.
$$
0\lra \O_{\P^1,x_0}\lra f_*\O_{C,x_0}\lra \ME_{x_0} \lra 0
$$
as desired.  For non-generic $x\in\PO$, we get a corresponding short
exact sequence, with $f_*\O_{C,x}$ a finitely generated module over
$\O_{\PO,x}$, which is a discrete valuation ring, and hence a
principal ideal domain.  Clearly, $f_*\O_{C,x}$ is torsion free, and hence free
([L], p.\ 147).  Likewise, one can easily show that $\ME_x$ is torsion free,
and hence free.

Every locally free sheaf on $\P^1$ is of the form $\oplus_i\O_{P_1}(a_i)$,
$a_i\in\Z$ (\cite{RH}, p.\ 384).
Observe that $\ME$ has no global sections, so it must be of
the form $\oplus_i\O_{P_1}(a_i), a_i < 0$.  Indeed, one has the following exact
sequence.
$$
0\lra H^0(\PO,\O_\PO)\lra H^0(\PO,f_*\O_C)\lra H^0(\PO,\ME)
\lra H^1(\PO,\O_\PO)
$$
Since $h^0(\PO,f_*\O_C)=1$ and  $h^1(\PO,\O_\PO)=\textup{genus}(\PO)=0$,
we get $h^0(\PO,\ME)=0$, as desired.  This completes the proof of the claim.

Finally, let $\MF$ be an invertible sheaf on $\PO$ with $h^0(\PO,\MF)=0$,
so $\MF=\O_\PO(b), b < 0$.  By above, to show that $h^0(C,f^*\MF)=0$,
it suffices to show $h^0(C,\MF\otimes f_*\O_C)=0$.  But
$$
\MF\otimes f_*\O_C=\O_\PO(b)\otimes(\O_{\P^1}\oplus(\oplus_i\O_{P_1}(a_i)))=
\O_\PO(b) \oplus (\oplus_i\O_{P_1}(a_i+b))
$$
which obviously has no global sections,
completing the proof of the proposition.
\qed

We introduce a second, similar equivalence relation on \taufs\
that looks somewhat less natural but will be more useful for
later applications.

\bdf
Let $C$ be a curve, $\omt \in \Omt(C)$.  The {\em unary set} of $\omt$ is
$$
M_{\omt} = \{(a, u): a \in C-Z_{\omt}, u\in \tau C_a \text{ and }\omt_a(u) = 1\}.
$$
This is a rational section of the algebraic variety $\tau C$,
and thus birational to $C$.

Say that $\omt_1, \omt_2 \in \Omt (C)$ are {\em $\seq$-equivalent},
written $\omt_1\teq\omt_2$, if
$M_{\omt_1}$ and $M_{\omt_2}$ are almost equal.
\edf

\bla
\label{teqcl}
Let $C$ be a curve, $\omt_1, \omt_2 \in \Omt(C)$, both non-trivial.
Then $\omt_1$ and $\omt_2$ are $\seq$-equivalent if and only if there is a rational
function $f\in K(C)$ such that $\omt_1=f(\omt_2-1) + 1$.
\ela

\prf
If $\omt_1=f(\omt_2-1) + 1$, then it is clear $M_{\omt_1}$ and
$M_{\omt_2}$ are almost equal.  The other direction follows as
with $\seq$-equivalence.
\qed

\bla
\label{taunull}
Let $C$ be a curve, $\omt_1, \omt_2 \in \Omt(C)$, both non-trivial.
Suppose that $M_{\omt_1} \cap M_{\omt_2}$ is infinite.  Then
$\omt_1$ and $\omt_2$ are $\seq$-equivalent.  In other words,
if  $M_{\omt_1} \cap M_{\omt_2}$ is infinite, then
$M_{\omt_1}$ and $M_{\omt_2}$ are almost equal.
\ela

\prf
Like the proof of Lemma~\ref{sigmanull}.
\qed

To prove the second statement, we first determine $\ker(\dltl)$ and $\ker(\be)$.
Setting $R = K$ in the above diagram, one obtains
$$
\xymatrix{
0 \sr[r] &\Om_K \ar[r]^\alpha \ar[d]^{\tilde{\dl}} & \Om_K \ar[r]
\ar[d]^\beta & \Om_{K/K}\ar[r]\ar[d]^{=} & 0 \\
0 \ar[r] &K \ar[r]^{\iota} & \Omt_K \ar[r]^{\lm} & \Om_{K/K} \ar[r] & 0
}
$$
with $\Om_{K/K}=0$, $\dltl$ and $\beta$ surjective.
By definition, $\ker(\be)=\< \dl(a)db-\dl(b)da | a,b\in K\>\sub \Om_K$,
which we call $M$.  By the Snake Lemma, $\ker(\dltl)= M$ also,
so one gets a short exact sequence,
$0\lra M \lra \Om_K \lra K \lra 0$.

Now let $R$ be any $K$-algebra.  Tensoring the previous sequence with $R$,
one gets, $0\lra R\ot_K M \lra R \ot_K\Om_K \lra R \lra 0$.  By the definition
of $\Omt_R$, one also has $0 \lra R\ot_KM\lra\Om_R\lra\Omt_R\lra 0$.
By Lemma~\ref{tdfeq}, these sequences fit together to form the following
diagram.
$$
\xymatrix{
0 \ar[r] & R \ot_K M \ar[r] \ar[d]^= & R\ot_K \Om_K \ar[r]^\dltl \ar[d]^\al
& R \ar[r] \ar[d]^\iota & 0 \\
0\ar[r] & R \ot_K M \ar[r] & \Om_R \ar[r]^\be & \Omt_R \ar[r] & 0
}
$$
By the Snake Lemma again, $\Ker(\al) \cong \Ker (\iota)$, which completes the
proof.

There is also hidden my original construction of $\tau$-differentials.

\brm
There is a more algebraic way to construct the module of 1-forms
on a variety, via the notion of K\"{a}hler differentials.
Let $R = K[V]$ be a $K$-algebra, e.g.\ the ring of regular functions on an
affine  $K$-variety.
Then one defines the $R$-module $\Omega_{R/K}$ as a certain universal object.
For more information, see Eisenbud~(\cite{Eis}, Chapter 16)
or Hartshorne~(\cite{RH}, Chapter~II.8).
We introduce K\"{a}hler $\tau$-differentials, which should
provide a purely algebraic construction of \taufs.  I will develop
and investigate these ideas later.
\erm
Let $A$ be a ring, $B$ an $A$-algebra, and $M$ a $B$-module. Recall that a
derivation is an abelian group homomorphism $d:B\lra M$ satisfying the Leibniz
rule, $d(bc)=bdc+cdb$.  The derivation is $A$-linear if it is a homomorphism of
$A$-modules.
\bdf
Let $(A,\dl)$ be a differential ring and $B$ an $A$-algebra.
\ben
\item
A $B$-$\tau$-module is a pair $M^\tau=(M,\iota_M)$ such that $M$ is a
$B$-module and $\iota_M$ is an injective $B$-module map from $B$ itself
to $M$, $\iota_M : B\lra M$.  (Thus, a $B$-$\tau$-module is a
$B$-module together with a distinguished free rank 1 submodule of $M$.)
Given two $B$-$\tau$-modules $M_0^\tau$ and $M_1^\tau$, a
homomorphism $f$
from $M_0^\tau$ to $M_1^\tau$ is a $B$-module homomomorphism from $M_0$
to $M_1$ such that for all $b\in B$, $f\circ \iota_{M_0} (b)=\iota_{M_1}(b)$.
\item
A derivation $\tau$ from $B$ to a $B$-$\tau$-module $M^\tau$ is
$A$-quasilinear if for all $a\in A$, $\tau a = \iota (\dl (a))$.
\een
\edf
The following lemma explains the choice of the term `quasilinear'.
(The formula for $\tau(ab)$ bears a certain resemblance to Pillay \&
Ziegler's definition of a $\partial$-module in their paper,
beginning of Section 3.  They have an additive endomorphism $D:V\lra V$
such that for $c\in K$, $v \in V$, $D(cv)= cDv+\dl(c)v$.)
\bla  Let $(A,\dl)$ be a differential ring, $B$ an $A$-algebra, $M^\tau$ a
$B$-$\tau$-module, and $\tau :B\lra M^\tau$ an $A$-quasilinear derivation.
Then for $a\in A$, $b\in B$, $\tau(ab)=a\tau b + \dl (a) \iota b$.
In particular, if $\dl$ is the trivial derivation on $A$, then
$\tau$ is an $A$-linear derivation.
\ela
\prf
Let $a\in A, b \in B$.  Then $\tau(ab)=$
$$
a\tau(b) + b\tau(a)=
a\tau b + b\iota(\dl(a))=
a\tau b + \iota(b\dl(a))=
a\tau b + \dl(a)\iota(b).
$$
The second statement follows immediately.
\qed
\bdf
The module of K\"{a}hler $\tau$-differentials of $B$ over $A$,
written $\Omt_{A/B}$, is the $B$-$\tau$-module generated by
$\{\tau b : b \in B\} \cup \{\iota b : b \in B\}$, with the following relations
holding for all $a\in A$, $b,b' \in B$.
$$
\tau(bb')=b\tau b' + b'\tau b
$$
$$
\tau(a)=\iota(\dl(a))
$$
The map $\tau : B \lra\Omt_{A/B}$, sending $b$ to $\tau b$ is an
$A$-quasilinear derivation, called the universal
$A$-quasilinear derivation.
\edf
The next lemma follows immediately from the previous definition.
\bla
$\Omt_{A/B}$ has the expected universal property.  That is,
given a $B$-$\tau$-module $M^\tau$ and a $\tau$-derivation
$\sigma: B \lra M$, there is a unique homomorphism of $B$-$\tau$-modules
$\phi: \Omt_{A/B}\lra M$ such that $\sigma=\phi\circ\tau$.
Equivalently, there is a natural bijection,
$$
\text{Der}^\tau_B(B,M)\cong\text{Hom}^\tau(\Omt_{B/A},M).
$$
\ela
The significance of the following lemma will become clearer after
the introduction below of the $\Lm$-map from \taufs\ to 1-forms
in Lemma~\ref{lambda}.  The proof is immediate.
\bla
Let $(A,\dl)$ be a differential ring, $B$ an $A$-algebra.  There
is a canonical homomorphism $\widehat{\Lambda}$ from $\Omt_{B/A}$ to $\Om_{B/A}$,
that sends each $\tau b \in \Omt_{B/A}$ to $db \in \Om_{B/A}$ and has kernel
$\iota (B)$.
Thus, there is a $B$-module isomorphism, $\Omt_{B/A}\cong\Om_{B/A}\oplus B.$
\ela
}

redpajama_set_name: RedPajamaArXiv


Merrilee Campbell thinks boys are better in books. While she immerses herself in Pride and Prejudice and Romeo and Juliet, she transfers to Reignald R. Hero High, and all her fantasies come true. A great comparison between classic literature and real life, Bookish Boyfriends is a must read!
Millie Porter started an online vlog, or video-blog, to give advice. She loves doing it for the most part, but she doesn't appreciate the trolls who write negative comments on her videos. In keeping up with her advice vlogs, Millie learns that maybe she's the one that needs it the most.
Tim Tebow started out as a college football champion and went on to be drafted to play professionally for the Denver Broncos, New England Patriots and Philadelphia Eagles. Among all of this success, however, are trials and failures, neither of which define him. In this memoir for young people, Tebow proclaims his love for God and the faith that he has within himself to live through both the positives and negatives that life has to offer.
Zoey Miller loves spending her winters in Aspen with the Madison brothers, Parker and Luke. This year is no different. However, this time, she wants to win one of their hearts. Will she be able to do it?
Paige Westfield has just gotten a D on her calc test, and is desperate to raise her grade. It seems the only option is for her to contract Ben Franklin to be her tutor. And yes, he's related to THE Ben Franklin. Paige knows Ben is trouble, but she's willing to help him get his act together if he helps her. It's the perfect arrangement – or is it? Read Finding Perfect to find out!


text:
As at the report concerning Egypt, so shall they be sorely pained at the report of Tyre.
When the report cometh to Egypt, they shall be sorely pained at the report of Tyre.
When the report reaches Egypt, They will be in agony at the report about Tyre.
When the report comes to Egypt, they will be sorely pained over the report about Tyre.
When the news reaches Egypt, they will be in anguish over the news about Tyre.
When the Egyptians hear, they will be in anguish at the news about Tyre.
When the report reaches Egypt, they will be in anguish at the fate of Tzor.
When Egypt hears about Tyre, it will tremble.
—When the report came into Egypt, they were sorely pained at the news of Tyre.
When it shall be heard in Egypt, they will be sorry when they shall hear of Tyre:
When Egypt hears the news about Tyre, it will feel the pain of sorrow.
When the news comes to Egypt, they will agonize over the report from Tyre.
When the report comes to Egypt, they will be in anguish over the report about Tyre.
Egypt will hear the news about Tyre, and it will make Egypt ·hurt with sorrow [be in anguish].
When the fame cometh to the Egyptians, they shall be sorry, concerning the rumor of Tyre.
When the news reaches Egypt, the Egyptians will shudder over the news about Tyre.
Even the Egyptians will be shocked and dismayed when they learn that Tyre has been destroyed.
Egypt will hear the news about Tyre. And it will make Egypt hurt with sorrow.
When the news reaches Egypt, they will be in anguish at the report about Tyre.
When the report comes unto Egypt, they shall be sorely pained at the news from Tyre.
When the report comes to Egypt, they will be in anguish at the report about Tyre.
When Egypt hears the news, there will be great sorrow.
When Egypt gets the report on Tyre, what wailing! what wringing of hands!
When the report reaches Egypt, they shall be sorely pained at the report of Tyre.
When the report reaches Egypt they shall be in anguish at the report about Tyre.
When the report reaches Egypt, They will be in anguish over the report of Tyre.
When the report reaches Egypt, They will be in anguish at the report of Tyre.
Egypt will hear the news about Tyre, and it will make Egypt hurt with sorrow.
When the news reaches Egypt, they will be shaken by what has happened to Tyre.
The Egyptians will hear about what has happened to Tyre. They'll be very sad and troubled.
When word comes to Egypt, they will be in anguish at the report from Tyre.
When the report reaches Egypt, They also will be in agony at the report of Tyre.
When the news comes to Egypt, they will be in pain when they hear about Tyre.
When Egypt hears the news about Tyre, there will be great sorrow.
As at the report concerning Mitzrayim, so shall they be sorely pained at the report of Tzor.
When the Egyptians hear it, they will be stunned over the destruction of Tyre.
When the report reaches Egypt, they will be in anguish at the report about Tyre.
Egypt will be terribly upset, too, when they learn about Tyre's destruction.
When the report comes to Egypt, they will be in anguish at the report of Tyre.
When it shall be heard in Egypt, they shall make sorrow, when they hear of Tyre.
As [at] the report of Egypt they are pained, So [at] the report of Tyre.
Isaiah 23:4Isaiah 23:6
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text:
January 18, 2020 Crypto Bit News Staff
Recently, the cryptocurrency mining community has been shaken with rumors of Chinese miners leaving Iran — where crypto mining is authorized as an industrial activity — for Central Asia. The move has ostensibly been taken in an attempt to find a new safe haven amid the tensions between the United States and Iran, as well as rising oil and energy prices. Iran's attraction for crypto mining operations lies in subsidized electricity rates — as of July, 0.7 cent per kilowatt-hour (kWh), — which had purportedly even prompted miners from mining…
Rewards received by bitcoin miners form a major chunk of the salaries paid across major proof of work (PoW) blockchains, according to Yassine Elmandjra, a cryptocurrency analyst from ARK Invest. For example, bitcoin miners were paid over $15 million worth of the cryptocurrency as incentive to mine blocks and secure the network on Jan. 9. Meanwhile, the total rewards paid across bitcoin (BTC), ethereum (ETH) and other major PoW cryptocurrencies including zcash (ZEC), litecoin (LTC), ethereum classic (ETC), bitcoin cash (BCH) and bitcoin SV (BSV) were just a bit more…
Bitcoin News News
Former Venezuelan Gold Mining Company Wants to Centralize Bitcoin ATM Infrastructure
First Bitcoin Capital Corp. wants to unite the providers of bitcoin ATMs around the world. His plan: link thousands of ATMs run by dozens of often small-time operators with a comprehensive know-your-customer, transaction monitoring and regulatory compliance solution that streamlines every aspect of the back-end business. "We want to create a consortium of bitcoin ATM operators and put them on the global ATM network," Rubin said. He said the consortium will roll out this year. First Bitcoin's network hinges on its recently acquired 2015 patent. U.S. 9,135,787 B1, otherwise known…
January 9, 2020 Crypto Bit News Staff
The Securities and Exchange Commission (SEC) has filed suit against Donald G. Blakstad for scamming investors partly through an allegedly fraudulent crypto mining operation. In a filing Jan 8, the SEC alleged that Blackstad, 60, bilked over $3.5 million from investors of three separate firms: an oil and gas company; a vehicle parts holding company; and "Energy Sources International" (ESI), a purported cryptocurrency mining company whose Las Vegas datacenter has only one employee: Blakstad. The crypto mining operation specifically took in some $550,000 from five separate investors, all of whom…
Canadian tech company DMG Blockchain Solutions has installed 1,000 new Bitcoin (BTC) mining machines at its Christina Lake mining-as-a-service facility in British Columbia. Per a Jan. 6 press release, DMG has purchased the new miners from Chinese mining giant Bitmain, with the total power consumption of approximately 1.5 megawatts. Initially announced in December of last year, the newly installed mining equipment will serve a U.S.-based client, whose name was not revealed in the release. DMG operates as an industrial scale crypto mine hosting company, allowing clients to mine crypto through…
Japanese financial services giant SBI and internet provider GMO have reportedly agreed to process crypto transactions at what is set to become the world's largest mining site Source link
Micree Ketuan Zhan, the co-founder of Bitmain who was recently dismissed by Bitmain's current CEO Jihan Wu, has publicly opposed layoffs at the firm. As Bitmain — the world's largest cryptocurrency mining firm — is reportedly planning to cut its workforce by 50% before the next Bitcoin (BTC) halving, Zhan claimed that he is "firmly opposed to layoffs" in a Jan. 6 Weibo post addressed to Bitmain employees. Zhan argues that Bitmain cannot lose its leadership in a highly competitive market Apart from claiming that Bitmain does not need to…
Bitcoin Mining Power Hits Fresh All-Time High
Bitcoin's hash rate reached record highs this week, amid rising prices and anticipation of the miner reward halving later this year. Based on a seven-day average, the hash rate has risen sharply from approximately 93 exahashes per second (EH/s) on Dec. 30 to more than 106 EH/s on Sunday. The best day overall was Jan. 1 when the hashing power exceeded 119 EH/s, surpassing the previous record of 114 EH/s set back in October. Bitcoin's hash rate has increased considerably over 2019, rising from a weekly average of 40 EH/s…
Cryptocurrency mining giant Bitmain reportedly has plans to reduce its workforce by another fifty percent. On Dec. 2, Chinese news media outlet Wushuo Blockchain reported that Bitcoin (BTC) mining rig maker Bitmain has launched a "personnel optimization plan" that will layoff a significant portion of its employees before the next BTC halving, which is expected in May. The world's largest producer of cryptocurrency mining machines will reportedly hold an annual meeting on January 17, and it is expected that the layoffs will have been completed before that date. People familiar…
Hashing Ahead: A Look at Bitcoin Mining in 2020
December 30, 2019 Crypto Bit News Staff
The December 2019 report on the Bitcoin mining network from CoinShares Research, a division of the digital asset management firm, presented an industry in good health at the end of the year, with a hash rate that had almost doubled in the previous six months, a new generation of more powerful and efficient technology on the market and the continued use of sustainable, renewable energy. The report indicated that, at this year's average bitcoin price, fee ratio and block frequency, miners were on their way to making $5.4 billion in…


The Christmas tree outside the Capitol has to be my favorite in the city. It's actually real and unlike the one at the White House, doesn't have a weird shaped net over it. It's more natural and has more character. Every time I go to see this tree, it's absolutely freezing and windy. Thankfully, when I visited this year, my mom was visiting which meant we had a car aka no freezing sprint up the hill to the metro!


First, congratulations for being on the vineyard website! Nothing better than a nice wine paired with the perfect food……a marriage made in heaven. You will compliment each other beautifully…..nice job.
Then I have to tell you that you are always teaching me something I didn't know, like the PB2 thing……so cool, I will definitely give it a try. We swear by a smoothie everyday as well and I love, love peanut butter!
BTW Ghee is now a staple in our household, love it! Thanks for that one too.
Thank you for the kind words D! PB2 is a great product and I'm glad I could shed some light on it. Their website has some recipe's for inspiration on other ways to use it but be warned – it does not taste like regular peanut butter when you add water.
I cannot wait to try this. Haven't tried PB2 yet but been so interested. Where do you find yours?
I find it at my smaller local grocer, they tend to have neat things like PB2. I hope you can get your hands on some!
I swear every day is some food holiday. I apparently am so behind on my national food holidays!


text:
Q: 'Group by' calculations don't match expected results Here is my query:
select product_name, (sum(item_price - discount_amount) * (quantity)) AS product_total
from products inner join order_items
using(product_id)
group by product_name

My query's output:

Expected Output:

I got almost everything to display but Gibson Les Paul product_total is wrong and the last row doesn't show the Null with its product_total.

A: An inner join means that you will only see order_items that have a product_name.  Use an outer join to see order_items that do not have a product_name. Also, your sum should be the outer most function. Maybe something like:
select product_name, sum((item_price - discount_amount) * (quantity)) AS product_total
from order_items left join products
  on order_items.product_id=products.product_id
group by product_name

If needed, look up an explanation of inner and outer joins, like this question for example.
But, looking a little closer at your data, the type of join is probably not the issue for the null row. The bottom row is merely a total of all the other rows, so adding with rollup after the group by clause would give you that row.

redpajama_set_name: RedPajamaStackExchange


Turkmenistan agreed Nov. 27 to supply the Taliban, the Islamic movement controlling most of Afghanistan, with electricity and gas.
Turkmenistan is only the fourth country worldwide to recognize the Taliban and the second country in the region to do so after Pakistan. Other nations, most notably Russia, still support Afghanistan's opposition Northern Alliance, led by Ahmed Shah Massood and ousted Afghan president Burhanuddin Rabbani.
Turkmenistan's government hopes to use Afghanistan as an export route for its large deposits of natural gas. Until recently, Russia was the only viable route for this export, but Ashgabat wants economic independence from Moscow. If Turkmenistan opens friendly relations with the Taliban, which controls more than 90 percent of Afghanistan, then it has the first shot at exporting gas south through Afghanistan and Pakistan when peace is declared.
Turkmenistan took a risk in openly supporting the Taliban. Iran currently is the only feasible option for Turkmenistan's gas exports besides Russia, but Tehran has concerns over border security and economics. As long as Iran and other nations continued to shun the Taliban - and prolong the war in Afghanistan -Turkmenistan would not be able to use Afghanistan as an export route. Turkmenistan would be cut off from Russia and Iran and would have no way to export its gas.
Furthermore, Russia has worked to prolong the war in Afghanistan, threatening the deal Turkmenistan made with the Taliban. As long as Russia opposes the Taliban and Iran's security is threatened, Russia remains the only secure route for energy exports from the Caspian nations. If countries begin to support the Taliban move toward a peace process, however, then Iran would become a serious threat to Russian hegemony in the Caspian.
Iran, though, has signed on with Turkmenistan. On Nov. 28, Agence France-Presse reported Turkmenistan and Iran supported each other in the debate with Russia over the exploitation of the Caspian Sea. Then the BBC reported Dec. 9 Turkmenistan and Iran had opened a new gas pipeline between the two countries and were signing new export and transport deals. Iran, through this economic deal, also indicates it may be ready to work with the Taliban and support an Afghan peace process.
Turkmenistan's support of the Taliban indicates support is growing for a peace settlement in Afghanistan. More importantly, Turkmenistan has gained an important ally in Iran. Ashgabat is able to avoid isolation, and the alliance may provide an incentive for other nations to pledge their support to the Turkmen-Iranian alliance. This economic pact also suggests Iran is ready to start dealing with the Taliban.
Iran's tacit support of the Taliban may provide the incentive for other Central Asian nations to break from the Russian position and sign on with Iran and Turkmenistan. Kazakhstan and Uzbekistan, for example, also depend on energy transport routes through Russia. The Turkmenistan-Iran route provides an enticing alternative export corridor that would help wrench these nations from Russian regional hegemony. Furthermore, the countries will be poised to take advantage of an Afghan transport corridor once peace is declared.


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Summer Dresses, Yellow Hunters, Fairies and Jolly Phonics. Add this all together and you have yourself a fun child portrait session!
I have named this set the Pastel Twins as I just loved the pastel tones of pinks and blues that the girls arrived wearing which lead to the inspiration of a subtle pastel finish. As the cozied up and played together on our country walk in Horsham, they were full of giggles and laughter.
What great fun it was to be at Buddy's Minion Birthday party as he celebrated turning into a big boy of 3 years old! I first met Buddy on one of our Autumn Mini Shoots earlier on in the year and he was such a happy character, so full of energy and such a beautiful expressive personality - I knew it was going to be an exciting day.


I'd heard about destination weddings, but I never really knew destination birthday's were a thing. Until my oldest friend from school invited me to spend 4 days in Queensland to explore the magnificent Whitsunday islands with her. I was chuffed, I got to spend time with my buddy on her thirtieth day on earth and we got to experience the beauty that is tropical Queensland. Including a stunning scenic flight over the reef. It was honestly sensational.
Without boring you with what we did and how we did it day by day, I'll just give you the facts, when we went, how long for and who we booked with. I can't recommend this experience enough. All I can say is: Do. It.
How many days were we away? We took the Friday and Monday off work, therefore 4 days and 3 nights.
Can I scuba dive? Yes, for an additional $65. Snorkelling is free!
Would you book through this company again? To be honest, probably not. It felt more suitable for backpackers even though it was reasonably expensive. The food was cheap branded, there was never enough food (always limited) and the hygiene could have been improved in regards to washing dishes and the bathroom situation. In saying that, 23 year old me (7 years ago) would have absolutely loved it without question. So, I guess as they say: you get what you pay for. And hey! Maybe I'm just getting a little snobby as I've aged (equally true) I will say though, the host and captain were great, very accommodating and the booking staff were fine.
Was the plane ridiculously tiny, weren't you scared? Yes and not really. I mean at first you feel like it should probably just be a toy, but once you're in the air. The views are captivating!
But, really what more can I say? Let's let the pictures do the talking! Check out my gallery below. All taken with either an underwater camera or my iPhone6.