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Liveblog: Golden Globe Dust - Macleans.ca
Liveblog: Golden Globe Dust
Kate Winslet and Slumdog Millionaire are the night's big winners
The Golden Globes embrace film and TV, but the TV types definitely feel like second-class citizens at Hollywood's primo party. Last night, as a hyperventilating Kate Winslet accepted her second Golden Globe (she won best actress for Revolutionary Road and best supporting for The Reader), she looked like a woman labouring under the delusion she had just been named Queen of Heaven for Life. Then Rian Wilson from The Office came on with a girl from Gossip Girls and deflated the emotional bombast with a grin and a cheeky reminder: "Hello, we're TV actors."
But like the Globes, I'm more interested in film than TV. And the big winners on the movie side of the ledger last night, aside from Winslet, were Slumdog Millionaire and The Wrestler. Mickey Rourke's speech was the highlight, as he thanked Bruce Springsteen and Axel Rose, who gave him songs for free. Director Darren Aronofsky had trouble getting this movie about a washed-up wrestler financed because the studios had no faith in Rourke, a washed-up star. Rourke helped out by bringing rock stars on board. The irony is that he met Bruce Springsteen via Sean Penn, who was Rourke's chief rival for best actor with his performance in Milk. We'll see if Mickey takes Sean to the mat at the Oscars. Meanwhile, here's my running real-time account of the Globes, typed last night in a live blog…..
What begins is a half-assed live blog of the Golden Globes. Don't know how long I'll keep it up. But I'm sitting here at the TV with a laptop on my knees. Might as well type during the commercials. I'll try to limit myself to the movie awards, because I don't watch a lot of TV, except for the late night stuff. I'm going to tack new posts onto the bottom, rather than add fresh posts to the top, which may be a hassle if you're actually reading this live, instead of watching TV—why aren't you watching TV??—but it will read more coherently in the morning.
So far we've seen Kate Winslet bring an Oscar-scale speech to her victory for best supporting actress, which is strange on so many fronts. For one thing, she played the lead role in The Reader. I mean, if she wasn't the lead, who was? (I guess this is so she wouldn't have to compete with herself in the Best Actress category for Revolutionary Road) Also, she broke the Golden Globe rule: if you're any kind of star, you act blase about winning, because everyone knows the only award that counts is the Oscar, and the Globes are part dress-rehearsal, part promotional campaign. Two thumbs up for Kate's titanic passion.
Bruce Springsteen won best song for The Wrestler. Sting, looking strangely unblond and bearded, like he'd just come out of a marathon tantric session in the woods, handed him the statuette with a hug. Bruce offered the first decent quip of the night: "This is the only time I'm going to be in competition with Clint Eastwood. It felt pretty good too." Expect the Boss to win the Oscar
I said I wasn't going to write about the TV folks. But how could I ignore Tom Wilkinson's sputtering acceptance of Best Supporting Actor for the TV mini series, John Adams. He could barely bring himself to get the words out, as if to say, "I'm better than this." Hey, after all, he played a Nazi supporting Tom Cruise in Valkyrie.
Don Cheadle introduces a clip for Burn Before Reading, wraps his mouth around the phrase "dysfunctional glib gym employees," calls Brad Pitt stupid, and says, "The Coen brothers who, incidentally, have never hired me. . . are at it again. "
Everybody has fabulous tans. The breasts match the faces. These Golden Globes are truly golden.
Cool. Of my limited TV diet, the one show I have watched every episode of is In Treatment. Gabriel Byrne just won, but wasn't around to accept.
Ah, a live can't-believe-he-said-said-that moment. This is the reason we like the Globes more than the Oscars. Ricky Gervais just came onstage with a beer in his hand and started riffing dangerously. Started out by reprimanding the crowd for being so rowdy, telling the to "shush," and questioning the credibility of an awards show that failed to nominate him. Congratulated Kate Winslet, with an I-told-you-so comment: "Do a Holocaust movie, the awards come. The trouble is with Holocaust films, there's never any gag reel on the DVD." Even Spielberg is laughing, eventually, although he looks a little nervous about it. Then Gervais goes back to squeezing humour from sour grapes and slagging the Hollywood Foreign Press Association: "I can't believe I'm not nominated. What a waste of a campaign. This is the last time I have sex with 200 middle-aged journalists. It was horrible. A lot of them didn't even speak English. Europeans with wispy beards. The men were worse."
Sally Hawkins wins best actress for motion picture musical and comedy and can barely contain herself. She's so "I'll try to get through as muh as my nerves and knees will let me." Praises the "exceptionally phenomenal goddesses" nominated with her. She's beginning to break down. Emma Thompson looks like she might run up and embrace her like an anxious stage mum. No one is happier to to win a Golden Globe than an actor who has come out of nowhere. I mean, if Meryl Streep had won this award for Mamma Mia— and she was nominated—what on earth would she have been able to add to her legacy of acceptance speeches in accepting a minor award for a minor role. That would have required some serious acting.
Tom Hanks offers a brisk, exuberant acceptance for John Adams, its second award tonight. Our own Sarah Polley played a role in that mini-series, Sarah who was our honoured guest our own Toronto Film Critics Awards last Tuesday night. I see this as a clearcut victory for the TFCA. I noticed the Globes have bottles of Moet et Chandon champagne on every table. At the TFCA, we had bottles of Moet et Chandon on every table. I'm sure they stole the idea from us.
Heath Ledger, predictably and post-humously, wins best supporting actor for playing The Joker in The Dark Knight, as he did at the TFCA. He will win every nomination for this role, deservedly, all the way to the Oscar.
Colin Farrell presents the award for Best Foreign Language Film to Israel's Waltz With Bashir, which invents its own genre as an animated documentary about Israel's war with Lebanon. We wait for the director to say something about the current carnage in Gaza. He says, "I want to dedicate this prize as we promised to do to the eight beautiful production babies born [as it was being filmed] and . I hope one day when they grow up they will watch this film together and they see this war in the film and it will look like an ancient video game that ha nothing to do with their lives whatsoever." Close enough.
Laura Linney wins best actress for the mini-series John Adams, in what is beginning to look like a sweep.
Seth Rogen, looking radically slim and virtually handsome (honing himself for his superhero role in Green Hornet). Presenting with Elizabeth Banks, he says he regrets that he wasn't here in the' 80s: "Instead of being drunk with Mickey Rourke tonight, I'd be doing cocaine with Mickey Rourke tonight." Rogen and Banks present the screenwriting award to Slumdog Millionaire.
Alec Baldwin wins best actor for 30 Rock, thanks Tina Fey three times, thanks his family once, mentioning his daughter last, and does not call her "a rude, thoughtless little pig." What a guy.
Yet another win for John Adams. The star, Paul Giamatti, wins for best acdtor. This was a hell of a job, this little costume drama we put on. It seemed like there were thousands of people who saved my ass each and every day. Says Tom Wilkinson got him smoking again and thanks "the great nation of Hungary," where this story of American greatness was shot.
Tina Fey, televison's It Girl, demurs as 30 Rock wins the Globe for best comedy series. African American cast member Keith Powell accepts the award for her as she stands by smirking. "Tina Fey and I had an agreement," he says, "that if Barack Obama won I would speak for the show from now on. Welcome to post-racial America." I guess being on the cover of Vanity Fair as a star-spangled sex-bomb was recognition enough for theis thinking woman's comedienne, who has redefined the female face of TV comedy more dramatically than anyone since Mary Tyler Moore. (Hey, I said I wasn't going to type about TV, and now I'm unleashing vast, un-thought-out generalizations! Gotta love the Globes.)
Slumdog Millionaire wins for best score. A sign, I think, that it will win best picture. And, between you and me, I predict at this early stage that it will win the Oscar in the same category.
Maybe Tina Fey was just holding her fire. Now she wins best actress in a comedy or musical for 30 Rock and takes the podium. Looking seriously thin—after going on about how her campaign to lose weight in Vanity Fair—she's wearing a slinky black dress with glittery vertical cleavage. (Don't asked me what that means. I've never been able to describe women's clothes. What do they call it when a dress exposes the sides of women's breasts as opposed to the tops?) With her cat-that-ate-the-canary Mona Lisa smile of hers, she said, "I've always loved the Hollywood Foreign Press. I have all the Hollywood foreign press action figures." Then thanked someone else for writing the joke. After saying how lucky she's been this year, she adds, " If you ever start to feel to good about yourself they have this thing called the Intenet and you can find a lot of people who don't like you." Then she starts naming bloggers. Does anyone not like Tin Fey? Apparently there's an audience for every taste on the Web.
"When I think of Steven Spielberg, I think of history," says Martin Scorsese, introducing a tribute to the man who is, by implication, the other great living American director. Time to hit the fridge.
Spielberg talks. Thanking Scorsese, he says, "We have known each other for going on 39 years. I was going to talk about a bit about C.C. De Mille, but there's my inspiration right there [pointing at Scorsese]." Fittingly, in accepting the Cecil B. De Mille award, Spielberg says the first movie his dad saw him to see was Cecil B. De Mille's 'The Greatest Show on Earth" in 1952." He says the train wreck made an indelible impression on him: "C.B. De Mille got me into model railroads. Running my trains at full speed into whatever obstacle I could put in front of them." Once his dad warned him that he would wreck his train set if he kept staging crashes, he says, he got an 8 mm movie camera, shot one final wreck "and got the same feeling of gratification watching it over and over again."
The funny thing about Hollywood is that no matter how safe or commercial the filmmaker, he or she sees himself as a risk taker. And I guess Spielberg once was on the edge. Yet even now says he never stops feeling, "Am I going to get away with this? Whenever I tried to tell a risky story, whether it's about sharks or aliens, I always thought, "Am I going to get away with this?"
He didn't just get away with it. He ran away with it.
Danny Boyle wins best director for Slumdog Millionaire, which will go on, I predict, to win Best Picture. This Dickensian tale of a Mumbai underdog is the ultimate underdog movie. The ultimate Little Movie That Could. And its success began right here, at the Toronto International Film Festival, where it won distribution and took the Cadillac People's Choice Award after its premiere.
The first real upset of the evening. Colin Farrell wins best actor in a musical or comedy motion picture for In Bruges. He seems as surprised as anyone, saying, "They must have done the counting in Florida." On the other hand, this is not an Oscar category, and it was baffling to begin with : it might have been surprising to see any of the nominees win, with the exception of Javier Bardem for Vicky Cristina Barcelona. Farrell thanks his director, "for not listening to me when I asked you to cast someone else."
Finally another blast of bad taste.It's been a while since Ricky Gervais made people laugh and wince at the same time with a string of Holocaust gags. You gotta hand it the Brits. Now Borat's Sasha Baron Cohen unloads three one-liners about the recession affecting everyone, even celebrities I paraphrase: Victoria Beckham has had to go three days without eating. Charlie Sheen has had to have sex without paying for it. Even Madonna has been forced to give up one of her personal assistants—our thoughts go out to you Guy Ritchie." In the audience a kind of celebrity patriotism prevails, and there are more groans than laughs.
And now the Golden Globe for Best Actress in a a drama. Wow. This is an upset. Overrriding Anne Hathaway and Meryl Streep, Kate Winslet wins for Revolutionary Road. Her speech is a total emotional meltdown. This time she's really hyperventilating. "I'm so sorry, Ann, Meryl, Kristin! . . . Oh God, who's the other one. . . ? Angelina! Forgive me. Is this really happening? Thank you so much. Thank you so much. Please wrap up? You have no idea how I'm not wrapping up. . . ."
Kate, Kate. It's only the freakin' Golden Globes. You haven't won the Nobel Peace Prize. Save something for the Oscars, if this speech hasn't totally blown your opportunity of having another chance an accepting an award.
Perhaps the strangest part of the speech is where she rhapsodizes about her love for co-star Leonardo DiCaprio. About how she's loved him for so many years. About how she really, really loves him: "Leo, I love you with all my heart I really do." Uh, oh yeah, she's married to the director. She thanks her husband, Sam Mendes. "Thank you for directing this film and killing us every day . . . " Uh, yeah, and she says she loves him too. And her children, who were with them every day of the shoot and as much a part of it as anyone. Huh? This is a movie about adultery, and her children were as much a part of it as anyone else? I'm totally confused.
Apparently I'm not the only one. Right after her speech, Rian Wilson from The Office comes on to present an award and deadpans, "Hello, we're TV actors."
Mickey Rourke wins best actor for The Wrestler, a feat he's likely to repeat at the Oscars. He's takes his time with a nice, sensitive speech, but keep his perspective. "I'm not a really good public speaker," he says, "I was kinda hoping Robert Downey Jr. get up here and speak for me." He goes on to talk about how he was almost out of the business seven years ago, and how an agent took a chance on him. He thanks everyone he can think of, from Axel Rose to Bruce Springsteen, from his director to his dogs. About the director, Darren Aronofsky, he says, "He is one tough sonofabitch, and if you're not in shape, the man will bring you down, because he's smarter than the rest of us."
People used to thank God. Tonight no one thanked God. But Rourke may be the first actor who thanks God spelled backwards: "I'd like to thank all my dogs, the ones who are here and the ones who are not here any more."
Ah, as predicted here earlier this evening, Slumdog Millionaire takes best motion picture drama. Because Kate Winslet and Mickey Rourke took so long with their speeches, the filmmaker is cut off in mid-thank-you, just after expressing his gratitude to his girlfriend for dumping him just once. I, too, am signing off without further ceremony. I apologize for the errors and typos that undoubtedly litter this live blog. I'll clean it up in the morning. Good night and good luck.
Axel Rose
rian wilson
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Home > England > Cumbria > Historic Buildings > Rydal Mount
Rydal Mount
History, tourist information, and nearby accommodation
BY DAVID ROSS, EDITOR
Historic Connections
HERITAGE RATING: ?
Rydal Mount was the home of poet William Wordsworth from 1813 until his death in 1850, and of his wife Mary until her death in 1859. The house dates to the 16th century, and the interior features many of Wordsworth's personal effects, furniture, and mementoes. The house was purchased by Wordsworth's descendants in 1969 and is occasionally still used by the family.
The house is located at the end of a steep lane above the main A591 road out of Ambleside. There is very limited parking available at the house itself, though there is some on-street parking on the lane that leads to the house.
Part of the charm of Rydal Mount is the lovely garden in which it is set. Directly before the main entrance to the house is a mound believed to have been used as early as the 9th century as a look-out post. Stretching out to the north of the house are several informal terraces, including Dora's Terrace, which was only rediscovered quite recently.
Dora's Terrace was named for Wordsworth's daughter, as was Dora's Field. This was purchased by Wordsworth when he fell out with his landlady, Lady Diana le Fleming. Wordsworth anticipated having to leave Rydal, so he bought land next to the house, intending to build his own house there.
However, he patched up his differences with Lady le Fleming and scrapped his building plans. He then gave the property to his daughter, Dora. When she died, he and his wife Mary, planted Dora's Field with daffodils as a sign of remembrance. At the top of the garden is a rustic summerhouse built by Wordsworth. It is easy to imagine him looking out across the gentle slope of the garden to Rydal Water below.
The east side of the house
Though Rydal Mount has long been associated in public memory with Wordsworth, it is worth pointing out that the poet never actually owned the house; he rented it for all of his 37 years in residence here.
The 5 acres of gardens are a delight for keen gardeners. Much of the gardens are as they were planted by Wordsworth and his family. Wordsworth had strong ideas about what a garden should be; he thought that a garden should be composed of lawns, and trees planted so as to allow pleasant views of the natural surroundings.
The whole, he felt, should be informal, and as harmonious as possible with the natural surroundings. Among the many species planted about the garden are roses, cypress, magnolia, azalea, hydrangea, enkianthus, holly, clematis, and actinidia, among many others! The gardens are not overwhelming, but quiet, and peaceful, and there are very nice views out over the lake through carefully arranged gaps in the trees.
The house has 8 main rooms open to visitors. These include, on the ground floor, the Dining Room, Drawing Room, Library. The official Rydal Mount brochure recounts an amusing anecdote about the latter.
It seems that Wordsworth's servant was showing a visitor around the house. When they came to the Library, he said, "This is my Master's library, where he keeps his books; his study is out of doors." In the Library is a bookcase owned by Wordsworth. It was later purchased by Canon Rawnsley, co-founder of the National Trust, and is now back in the possession of the Wordsworth family.
The house front
The original house at Rydal was made up of just three of the current rooms; the Dining Room, and Dora and Dorothy's bedrooms. All the other rooms were added on later. Another original feature of the house is a mullioned window set in the east wall. This window was for many years hidden beneath whitewash. It was only during repairs to the house in 1979 that the hidden window was discovered. Why was it covered up for so many years?
The answer lies in government taxation. In the latter years of the 18th century, the government introduced a tax on windows as a way to finance the Napoleonic Wars. Taxes were assessed on a sliding scale so that houses with more windows had to pay more than houses with fewer windows.
The taxes were so high that many homeowners chose to cover over windows rather than pay the tax. We can safely assume that the hidden window at Rydal Mount was covered over to avoid taxes.
The house is full of objects owned and used by Wordsworth and his family. Among the furniture is the poet's favourite chair, a 'cutlass chair', with a curved arm built to allow men who wore a sword on either side to sit comfortably.
There are so many objects on display that it is difficult to pick out one or two for special mention. Upstairs, in Wordsworth's Study, is a sword belonging to the poet's younger brother, John, who was the captain of the 'Earl of Abergavenny'. The ship sank, and John lost his life. The sword was recovered from the wreckage.
In William and Mary's Bedroom is a framed letter from Wordsworth to Queen Victoria, in which he declines the post of Poet Laureate. The queen persisted, however, and Wordsworth was eventually persuaded to accept the post, but only on the condition that he was not required to write so much as a single line of verse for Her Majesty. Wordsworth is the only Poet Laureate ever to be afforded this privilege.
Rydal Mount does not get the number of visitors as nearby Dove Cottage, in Grasmere, but it well repays a visit. It is a lovely house, and the gardens are worth visiting on their own. Do take the time to walk to Rydal Falls (bring good waterproof footwear).
A winding garden walk
Wordsworth's summerhouse
Most photos are available for licensing, please contact Britain Express image library.
About Rydal Mount
Address: A591, Rydal, Ambleside, Lake District, Cumbria, England, LA22 9LU
Attraction Type: Historic Building
Location: On the north side of the A591 2 miles west of Ambleside. Limited parking along the lane.
Website: Rydal Mount
OS: NY363063
Photo Credit: David Ross and Britain Express
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We've 'tagged' this attraction information to help you find related historic attractions and learn more about major time periods mentioned.
Find other attractions tagged with:
16th century (Time Period) -
NEARBY HISTORIC ATTRACTIONS
Heritage Rated from 1- 5 (low-exceptional) on historic interest
Rydal Falls - 0.2 miles (Countryside)
Rydal Water - 0.6 miles (Countryside)
High Sweden Bridge - 1 miles (Countryside)
Dove Cottage & Wordsworth Museum - 1.3 miles (Historic Building)
Loughrigg Fell - 1.4 miles (Countryside)
Grasmere - 1.5 miles (Countryside)
Bridge House - 1.6 miles (Historic Building)
Grasmere, St Oswald's Church - 1.6 miles (Historic Church)
Nearest Accommodation to Rydal Mount:
Self Catering - B&Bs/Guesthouses - Hotels
Nearest Self Catering Cottages
Hall Bank Cottage, Rydal - 0.4 miles
Rydal, Cumbria
Stay from: £812.00 - 3645.00
White Moss, Wood Close, Grasmere - 1.2 miles
Ramblers Roost, Grasmere - 1.2 miles
More self catering near Rydal Mount >>
Nearest Hotels
Glenridding House Hotel - 6.8 miles
Ullswater, Cumbria
Stay from: £155.00
The Rusland Pool Hotel - 14.2 miles
Ulverston, Cumbria
Stay from: £49.00
The George Hotel - 16.3 miles
Orton, Cumbria
More Hotels near Rydal Mount >>
Nearest Bed and Breakfasts
Foxghyll Country House B&B - 1.7 miles
Ambleside, Cumbria
Waterwheel Guesthouse - 1.9 miles
Wanslea Guest House - 2.1 miles
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redpajama_set_name: RedPajamaCommonCrawl


text:
/// <reference path="../typings/tsd.d.ts" />
/// <reference path="../build/sp.list.repository.d.ts" />

namespace MyApp{
	export class TestCategoriesBaseItem extends SPListRepo.BaseListItem{
		contentType: string;
		_UIVersionString: string;
		edit: string;
		linkTitleNoMenu: string;
		linkTitle: string;
		docIcon: string;
		itemChildCount: SP.FieldLookupValue;
		folderChildCount: SP.FieldLookupValue;
		appAuthor: SP.FieldLookupValue;
		appEditor: SP.FieldLookupValue;
		constructor(item?: SP.ListItem){
			super(item);
			if(item){
				this.mapFromListItem(item);
			}
		}

		mapFromListItem(item: SP.ListItem): void{
			super.mapFromListItem(item);

			this.contentType = this.getFieldValue("ContentType");
			this._UIVersionString = this.getFieldValue("_UIVersionString");
			this.edit = this.getFieldValue("Edit");
			this.linkTitleNoMenu = this.getFieldValue("LinkTitleNoMenu");
			this.linkTitle = this.getFieldValue("LinkTitle");
			this.docIcon = this.getFieldValue("DocIcon");
			this.itemChildCount = this.getFieldValue("ItemChildCount");
			this.folderChildCount = this.getFieldValue("FolderChildCount");
			this.appAuthor = this.getFieldValue("AppAuthor");
			this.appEditor = this.getFieldValue("AppEditor");
		}

		mapToListItem(item: SP.ListItem): void{
			super.mapToListItem(item);

		}
	}
}

redpajama_set_name: RedPajamaGithub


text:
I.O.M.
Editorial/Features
We realise that not all visitors to this website are gun savvy, so here you will find some basic information on firearms and the firearms debate. If you are interested in further information, we strongly suggest you go to the links page and surf through some of the sites listed.
What is a firearm?
How does a firearm work?
Aren't firearms dangerous?
If firearms are dangerous, why do we need them?
Won't the handgun ban prevent another "Dunblane"?
In law, a firearm is defined as "a lethal barrelled weapon of any description from which any shot, bullet or missile can be discharged".
However, a better definition is that a firearm is an object with a barrel, through which a projectile or shot is discharged through the action of expanding gases from the burning of an explosive, or through the action of compressed air or gas which expands.
Firearms that are typically used by an individual can be divided into four categories. Each category is however broad:
1) "Handgun" – which includes a pistol, revolver, air pistol, muzzle-loading pistol, etc. This is a firearm without a shoulder stock which is held and fired from a person's hand or hands, and is typically smaller than other types of firearm. Also, most handguns have rifled barrels. Rifling is ridges cut into the barrel upon which the bullet bears while travelling down the barrel, causing it to spin in flight. Rifling usually increases the accuracy of the bullet for this reason.
2) "Rifle" – a rifle is a firearm with a stock and a rifled barrel which is fired using both hands and is shoulder-mounted.
3) "Shotgun" – a shotgun is a firearm that usually has a smoothbore barrel and is fired from the shoulder using both hands. Unlike rifles and handguns, shotguns are usually used with shot shells, which contain multiple projectiles instead of one. Shotguns have a much shorter range than most rifles or pistols.
4) "Machineguns" – this covers firearms such as assault rifles, submachineguns etc. A machinegun is a firearm capable of firing successive shots with a single pull on the trigger.
When the gun is fired, the bullet travels down the barrel, but in addition the force of the cartridge case going in the other direction enables the gun to eject the empty case, and strip another round off the top of the magazine and into the chamber of the firearm, thus enabling it to fire automatically. Some machineguns use the expanding gases to cycle the action instead, usually through a gas port in the barrel which allows the gases to bear against a gas piston attached to the bolt.
"Semi-automatic" firearms are not machineguns, though they function in a similar way, but do not fire continuously when the trigger is pulled. Instead they simply use the energy of the fired round to cycle the action automatically for the next shot, but are designed so that the operator of the firearm must deliberately pull the trigger for each shot to be fired.
Many assault rifles and submachineguns have a selector switch so that they can function both as machineguns and also semi-automatically.
The following illustrations and description show how a SIG-Sauer P225 9mm semi-automatic pistol work. There are dozens of different operating methods for firearms, but the basic principle of all is the same, the firearm must be loaded, the action cocked for firing, and the firing pin must strike the primer of the cartridge for it to fire:
1) A magazine is loaded with ammunition, and then inserted into the pistol until the magazine release button's stop is engaged (43).
2) The slide (4) is then pulled back, which cocks the hammer (29), and is then released, and is pushed forward by the recoil spring (3). A round of ammunition (P) is stripped from the magazine by the breech block (9) and is fed into the barrel's (1) chamber. The next round in the magazine is forced up by the magazine spring (52).
3) With the firearm now loaded, the trigger (20) can be pulled, which draws forward the trigger bar (22). This has the effect of pivoting the safety lever (28) forward which raises the firing pin safety block (13) allowing the firing pin (11) to move. At the same time, the trigger bar pulls forward the sear (24) from the notch (29a) on the hammer (29), releasing the hammer which is pushed forward by the main spring (34) and hammer strut (33), so that it strikes the rear of the firing pin (11) which hits the primer of the round of ammunition (P) causing it to fire.
4) After the round is fired, the force of the combustion while driving the bullet out of the barrel also forces the empty cartridge case backward, which is called "blowback". This thrusts the slide (4) to the rear. The safety lever (28) and firing pin safety block (13) are separated activating the firing pin safety block. A protrusion on the top of the trigger bar (22) resting in a recess in the slide is forced downward as the slide moves rearward, forcing the trigger bar downward and disengaging it from the safety lever (28). As the slide travels rearward, it takes the barrel with it, until the barrel is stopped by the locking insert (17) which tilts it down and away from the slide.
5) As the slide continues its rearward travel, the sear spring causes the sear (24) and safety lever (28) to return to their original positions. The empty case (H) is extracted by the extractor claw (10) mounted in the slide and upon striking the nose of the ejector on the slide catch lever is ejected out of the ejection port in the slide. As the slide (4) recoils up to its stop, it cocks the hammer (29) which is caught in its notch (29a) by the sear (24). The slide is propelled forward by the recoil spring (3), and the next round is fed into the chamber of the barrel (1) as the slide locks up with the the barrel again. The pistol is now ready to fire again.
6) After the last round is fired, the slide (4) travels rearward and is held in the rearward position by the slide release lever (18) which is forced upward by the action of the magazine spring (52) and magazine follower (53). The lever is forced into the slide release notch (4a) on the slide. The empty magazine is ejected by pressing the magazine release button, and a fresh loaded magazine can be inserted. The gun is then made ready to fire by pressing down on the slide release lever (18) which strips a round off the top of the magazine and into the barrel's chamber.
Most firearms, especially those designed for hunting and military use, are capable of causing fatal injuries, and are designed to do so. Safety rules for the handling of firearms should therefore be strictly adhered to:
1) Always keep your finger off the trigger until you are ready to shoot;
2) Never point a firearm at anything you do not wish to destroy;
3) Always be aware of whether your firearm is loaded or unloaded, and make sure by checking.
Firearms are needed for a wide variety of reasons. They are commonly used to destroy pests, from small animals such as rabbits and squirrels up to much larger animals such as deer. Firearms are usually considered the most humane method of dealing with animal pests which must be destroyed, as they are more discriminate and certain than other methods such as traps or poison.
Firearms are also used for self-defence, whether it be by police officers, soldiers or private individuals. A firearm is the most effective method of self-defence usually, because it is an "equaliser". A physically weak person would not ordinarily be able to defend themselves against a physically strong person, but a firearm enables the weak to defend themselves adequately.
Firearms also have other necessary uses, such as slaughtering instruments, or for veterinary use. Firearms also have a long-established use in a wide variety of sports, whether it be traditional bullseye target shooting, clay pigeon shooting, or deer stalking, or any of the many other shooting sports.
The ban on handguns will most certainly not prevent another mass murder like the one at Dunblane Primary School, for the exact same reason the ban on self-loading and pump-action rifles after the Hungerford killings did not stop Dunblane; that is because (a) there are a wide variety of other legally possessed firearms which are just as lethal and (b) the prohibition of legally possessed firearms does not address illegally possessed firearms.
There are in Great Britain at least 700,000 people licensed to own firearms of one type or another. In addition, there are estimated to be at least 850,000 illegally possessed firearms in Great Britain, though most estimates are higher.
A firearm of any description can be used to commit mass murder, especially in a situation like Dunblane where most of the victims were young children.
Mass shootings throughout the world have been committed with all manner of firearms. The worst mass shooting known for example, occurred in Norway in 2011 when a man opened fire with a semi-automatic hunting rifle and shot dead 69 people. However, at the other end of the firearm spectrum, a mentally disturbed man in New Zealand shot dead six and seriously injured four more with an illegally possessed single shot shotgun in late 1996. In France in 1990, a man shot dead 14 people with a double-barrel shotgun.
Prohibiting firearms of one type or another will do nothing to prevent further similar crimes, because nearly all firearms are capable of being used to commit these crimes.
As a practical matter, it is also not possible to prohibit all firearms, because they have certain essential agricultural uses in addition to their sporting use, and a complete prohibition would still not address illegally possessed firearms. (Of course illegal possession would probably grow if legal possession were completely outlawed).
In our opinion, the solution to the misuse of firearms is to have a sensible regulatory environment for legally possessed firearms which is not needlessly punitive, which would encourage benign owners of illegally possessed firearms to possess them legally, and to seek out and prosecute those who would use firearms for illegal purposes to the fullest extent of the law. Committing a violent offence with a firearm should be cause for the most severe criminal penalties, which is currently not the case in the British Isles.
It should be noted that according to Home Office statistics, 86-89% of firearm-related homicides in England & Wales are committed with illegally possessed firearms.
(Postscript: since this was originally written there has been another mass shooting in the UK in Cumbria that was committed using a .22 rifle and a double-barrel shotgun, so it is clear the handgun ban was pointless.)
"He who would surrender an essential liberty to purchase temporary safety deserves neither." – Benjamin Franklin (paraphrased into modern grammar)
The Tories' offensive Offensive Weapons Bill
The Tories are not our friends… and other tales of woe
Autopsy of a gun ban
The Law Commission reports
Brussels panics
P226 Sport II SL
Editorial and Features
Heckler & Koch USP9
Steyr M9
Colt M16A1
Colt AR-15A2 HBAR Sporter
Bushmaster M4 Carbine
Beretta 92FS
Colt M203
SIG-Sauer P225
Heckler and Koch P7
SIG Pro SP2009
S&W M&P 15-22
Field Sports and Pest Control
Gun and Ammunition Collecting
Send mail to editor@cybershooters.org with questions or comments about this web site.
Cybershooters Unless noted, all rights are reserved. Reproduction or redistribution without our express permission is prohibited.


It was a very productive day in the steam shop and, for once, the progress is very obvious.
Two new volunteers came to the shop for the first time, Ethan and Evelina. As it turned out, it was one of those days when there were a number of things going on in which they could be fully involved so they had busy days. Hope we did not put them off !.
· Phil and Brian did a thorough check of the smokebox, firebox and tubes to remove all tools and other items left over from tube fitting and other completed activities. A lot of what we are doing now seals up areas that we have been working in for months ........... so we certainly do not want foreign objects sealed up in there!.
· Ed worked on the cab lagging and located the connections needed to link in the pipe to the steam chest pressure gauge. This is important as it has to be connected and run thru the next section of sheet metal that has to be fitted.
The large casting of the fire hole door was lifted into the tender ready for refitting. Rick, who is our expert with the fireclay, was not around this week so hopefully it can be fitted next week.
Brian and Phil cleaned up the parts of the butterfly door ready for fitting once the casting is in place.
Much of the sheet metal for the backhead was sorted and moved up into the tender. This keeps it separate from the boiler lagging and convenient to where it must be fitted. In doing this we found the paint on a couple of sections was in bad shape. So Jane took charge of these, wire brushed and primed them.
· Jane and Evelina used the thread chaser on the hangers that support the air tanks under the walkways alongside the boiler. Another job to ensure that key parts are ready for immediate use when we come to final fitting.
· Eric continued work on the boiler lagging. Jim and Ethan ran a production line to split and taper the blocks to the required shape and Evelina worked with Eric to fit the blocks around the first two courses of the barrel.
This is a thoroughly dirty job as the lagging breaks up easily and generates a clinging white dust. However, they kept at it and, by end of day, had set the lagging blocks around both courses and applied steel banding to finally secure them into place.
In the evening, Jerry and I experimented with the lagging mud. This is another unpleasant material. It is a dusty large grained powder that must be mixed with water to produce a sticky sludge that has a reasonable tendency to stick to things (including you as well as the boiler!!). With this we were able to fill the gaps so, provided it stays in place as it dries, the first course is now ready for fitting the sheet metal.
· And then there was the major event!. As noted last week, we had set up the staging and brought in the super heater elements for the first row.
Everything was prepared in the morning to start fitting the super heater elements. All the ball ends of the elements and the sockets into which they fit were cleaned. The channels into which the bolt heads slide were checked and cleaned. All the nuts and bolts had been prepared some months ago.
In the afternoon, under Tom's supervision, the fitting started. Each element is secured by a special bolt whose square head locates into a channel in the super heater header. The bolt passes thru a yoke around the two pipes of the element. A specially shaped washer, designed to allow the yoke to bear evenly on both pipes is fitted and a special deep nut used to secure the yoke and so drive the balls of each pipe tightly and evenly into their sockets.
The ball joints at the end of each pipe must then be carefully checked for alignment with their sockets. The nut is then tightened, ensuring that the ball and socket remain accurately aligned. The tightening requires quite a bit of torque as the environment around this area is pretty brutal. In operation it is in the path of the fire gases!.
A production line developed. Cameron and Brian became the experts on the actual alignment and fitting. Sliding the element into place is tricky as the bolt must be in place in the yoke and its head must slide along the channel as the element is pushed into place. So you have the combination of sliding a 17 foot element along the flue and the bolt head along a tightly fitting channel.
Once in place, the balls are carefully aligned into their sockets and held there as the nut is tightened.
Here, Ethan, Jerry and Jim feed an element from the staging while Cameron and Brian locate it into its flue.
As it became clear that the process was going smoothly Ethan, Bob and I started fetching the elements for successive rows from the boxcar.
The longer down pipe are very obvious as Brian treats the bolt of one of the last elements with the anti-seize compound.
The change in the boxcar is remarkable. From being full of parts a few months ago, it is now substantially empty. So we took the opportunity to put some of the rods from #938 under cover for the winter.
By around 5:30 it was all done. 27 elements had been fitted. (Although built with 28, one was removed and blocked off before 1630 came to IRM so she now has 27). The view of the smokebox is now very different and 1630 now has a complete super heater again (subject to pressure testing in a couple of weeks).
· During the week Tom had worked on machining the throttle. That needs final lapping but is hopefully nearing completion.
I will now be out of circulation a great deal in the next couple of months, with a trip to the UK for a couple of weeks leaving today and for 4 weeks to Australia in mid October. Blog updates may be a little erratic for few weeks. I will need to rely on messages from the guys on what is happening to provide updates in the next couple of weeks. Then hopefully someone else can take over for a while. So, in the next couple of weeks, I will be awaiting the Sunday progress update like everyone else!.
Aren't the superheaters usually tested before they go back in?
Isn't there a concern that fired-up testing could show some leaks of tubes or flues behind the superheaters that would require their removal to correct?


This home has it all and will be ready to move in. Showings to start around April 26, 2019. Tile through out, All Stainless steel appliances included with gas top, gas stove and refrigerator, large capacity washer and dryer. Large back yard with privacy fence. Screened in porch with private back yard. Extended patio. Solar pleated window shades. Over sized master bedroom with large master shower and walk in closet. High hat lighting in family room, kitchen, and master bedroom. Pull down attic in garage with large storage area. Large screen tvs in all bedrooms and family room. Upgraded lighting fixtures. Fully sodded property with low maintenance Bermuda grass. CPI security system with cameras in family room and front door and Termite bond are transferrable to new home owner. All rooms are prewired for cable and internet. Home and tile flooring to be professionally Cleaned. Gutters cleaned and home power washed. More pictures to be uploaded. Home has been professionally measured.


text:
Q: Why are there some methods in java.base cannot be compiled by AOT compiler According to JEP-295 (http://openjdk.java.net/jeps/295)
There are some methods in java.base causing compilation failure, but why?
It will be appreciated that someone can help me so that I can realize limitation of Openjdk AOT compiler.

A: See the documentation you are linking to:

Current AOT limitations ...
May not compile java code which uses dynamically generated classes and bytecode (lambda expressions, invoke dynamic).
These limitations may be addressed in future releases.

Thus two potential explanations:

*

*the methods that do not compile fall into the known limitations

*otherwise, a bug in a "brand new, experimental" product

Most likely "option 2" is the better explanation - see again your link.
It contains a list of methods that fail compilation, together with the error that occurs:
cat java.base-list.txt

# jaotc: java.lang.StackOverflowError
exclude sun.util.resources.LocaleNames.getContents()[[Ljava/lang/Object;
exclude sun.util.resources.TimeZoneNames.getContents()[[Ljava/lang/Object;
exclude sun.util.resources.cldr.LocaleNames.getContents()[[Ljava/lang/Object;
exclude sun.util.resources..*.LocaleNames_.*.getContents\(\)\[\[Ljava/lang/Object;
exclude sun.util.resources..*.LocaleNames_.*_.*.getContents\(\)\[\[Ljava/lang/Object;
exclude sun.util.resources..*.TimeZoneNames_.*.getContents\(\)\[\[Ljava/lang/Object;
exclude sun.util.resources..*.TimeZoneNames_.*_.*.getContents\(\)\[\[Ljava/lang/Object;
# java.lang.Error: Trampoline must not be defined by the bootstrap classloader
exclude sun.reflect.misc.Trampoline.<clinit>()V
exclude sun.reflect.misc.Trampoline.invoke(Ljava/lang/reflect/Method;Ljava/lang/Object;[Ljava/lang/Object;)Ljava/lang/Object;
# JVM asserts
exclude com.sun.crypto.provider.AESWrapCipher.engineUnwrap([BLjava/lang/String;I)Ljava/security/Key;
exclude sun.security.ssl.*
exclude sun.net.RegisteredDomain.<clinit>()V
# Huge methods
exclude jdk.internal.module.SystemModules.descriptors()[Ljava/lang/module/ModuleDescriptor;

And when you check the sources, for example getContents() - no lambdas there.
So the methods that fail compilation are most likely caused by the fact that the new compiler is "not yet ready" to successfully compile arbitrary Java source code!

redpajama_set_name: RedPajamaStackExchange


Published 04/26/2019 11:40:55 am at 04/26/2019 11:40:55 am in Water Weir Fountain.
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AMX Mining is committed to building a sustainable lithium production Company, highly leveraged to the forecast growth in the lithium-ion battery sector.
AMX Mining is in the business of maximising shareholder return through the discovery and development of safe, efficient and environmentally responsible mining projects in secure tenure jurisdictions that offer a clear, short-term path forward to development. We will outperform our peers through ready access to existing infrastructure, low utility costs and recognised commodity exposure.
AMX Mining maintains a policy of openness and transparency in its plans for the exploration and development of its projects. The Company will progress its projects in careful consultation with all of its stakeholders.
It is AMX Mining's goal to establish new mining operations. These operations are envisaged to create employment for up to 200 operators and contractors.
Lithium is critical for the current global paradigm-shift to electrification of transport, stationary storage and renewable energy.
Lithium in the form of salt lake brine which can be easily processed using off-the-shelf, commercially proven technologies to produce lithium carbonate.
AMX Mining's flagship project is Beihuo Salt Lake Lithium Project, located in Chongqing, People`s Republic of China.
Copyright © AMX Mining All Rights Reserved.


Instylejackets.com offers you Justified TV Series Raylan Givens Trench Coat at a discounted rate. You can order this fabulous brand immediately on our online store without any delay. Additionally, we provide you with an opportunity to appear as a celebrity in your daily life.
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Justified TV Series Raylan Givens Trench Coat offers me with an utmost level of internal comfort that I just cannot ignore. This product offers me much differentiated look in my friends. I consider this brand as a source of comfort and style for myself.
A very cool looking dashing long woolen coat of the blackish color! A formal outfit and best for the winter cold season!
Best coat from the online store that I have seen! Selected this at the first sight when I saw it! The quality is great and the length is adequate!


text:
Las Cruces Indivisible group releases documentary
The Las Cruces chapter of the progressive Indivisible movement has documented its work in a free video.
Las Cruces Indivisible group releases documentary The Las Cruces chapter of the progressive Indivisible movement has documented its work in a free video. Check out this story on lcsun-news.com: https://www.lcsun-news.com/story/news/local/las-cruces/2018/07/26/las-cruces-indivisible-group-releases-documentary/843553002/
Sun-News report Published 1:00 p.m. MT July 26, 2018
Mary Martinez White, left, and Jennifer, who did not provide her last name, hold up signs protesting a proposed federal tax bill Monday Dec. 18, 2017 on the corner of Telshor Boulevard and Foothills Drive. The protest, put on by Indivisible Las Cruces, saw more than 15 protesters spread out on the four corners of the intersection, waving signs and cheering when cars would honk.(Photo: Josh Bachman/Sun-News)Buy Photo
LAS CRUCES - Local activist group Indivisible Las Cruces has documented its work in a 12-minute video titled "The Activists Next Door."
The documentary portrays the group's organization, starting in January 2017, in protest of Trump Administration policies on federal health care programs and environmental regulations. The group focused later on voter registration efforts and electoral campaigns.
The Indivisible movement is a politically progressive network of local groups initiated in 2016 inspired by a political pamphlet titled "Indivisible: A Practical Guide for Resisting the Trump Agenda."
The 12-minute video, a production of Indivisible Las Cruces and its volunteers, has been uploaded to YouTube where it may be viewed for free.
Read or Share this story: https://www.lcsun-news.com/story/news/local/las-cruces/2018/07/26/las-cruces-indivisible-group-releases-documentary/843553002/
University Avenue lane closures through Jan. 21
NM man in house arrest sues to use medical pot
2020 Las Cruces Space Festival poster announced
Vehicle crashes into apartment in Sunland Park
Turtle fountain falling apart at veterans home
Staybridge Suites installs EV charging station


text:
Q: Strapi: how to retrieve Content Type with empty related Content Type I have a Content Type called Question, it has a relation with Content Type called Answer (question has many answers). Here's how I retrieve it:
query Questions($locale:I18NLocaleCode) {
  questions(locale: $locale) {
    data {
      id
      attributes {
        text
        answers {
          data {
            id
            createdAt
            attributes {
              text
              createdAt
              user {
                ...user info
              }
            }
          }
        }
      }
    }
  }
}

So it returns all the questions with all the answers to each question.
However, now I want to pass some filter argument to the query to retrieve only question that have no answers (later I want to add more filters). I didn't find a way to do it with Strapi GraphQL API. Is it possible? And if yes - how? Or do I still have to retrieve all the questions and then filter them on the client side (very undesirable behavior)? Or perhaps write some custom route with custom handler and pass the filters in the request? But then again, not sure how to do it with Entity Service API or Query Engine API…


The Jonti-Craft Baltic Birch Large Wall-Mount Coat Rack provides handy storage for up to ten students. The unit mounts firmly on the wall at any height, so coats and backpacks are within reach on double hooks and in cubbies. The laminate finish keeps your sturdy birch plywood coat rack looking new. Rounded edges won't scratch skin. Available with or without cubby trays.


The Elektron Analog Four MKII is the new benchmark for analog tabletop synthesizers. Tweaked circuitry delivering deeper bass than ever, new powerful sound control possibilities, dedicated outputs for each voice. Not to mention the ultra durable buttons, the big OLED screen and the premium design. No stone has been left unturned. The cast aluminum enclosure and ultra-rugged backlit buttons (rated for 50 million presses) make this a road-ready performance tool. The vivid OLED screen offers better visual feedback than ever before. And precision hi-res encoders plus the addition of quick performance controls provide an amazing hands-on experience. If you've been waiting for the classic combination of a 4-voice analog synth and sequencer, wait no longer: the Elektron Analog Four MKII is better than ever.
Who doesn't love analog circuitry for its warm, punchy sound, but we also know that digital control of certain jobs provides unmatched precision — and the Elektron Analog Four MKII is a perfect example of engineering that takes full advantage of this. In the Analog Four MKII, both oscillators of each voice are analog, including their elaborate waveshaping. Their pitches, however, are controlled digitally — along with the noise generator.
The two filters of each voice (and the overdrive circuit between them) are fully analog — as are the amplifiers. The envelopes and LFOs — they're digital; a design that allows them to venture well into the audible range without nasty artifacts. The signal path from the oscillators to the audio outputs — that's totally analog, as well as the signal path from the external inputs to the audio outputs. The digital send effects run parallel to the main audio path; their wet signals are mixed with the dry analog signal before everything reaches the main output stage.
The Elektron Analog Four MKII's sequencer sports six tracks, the first four of which are synth tracks (with each track controlling a synth voice). Track 5 is your FX track, controlling movements of chorus, delay, and reverb parameters. Finally, you use the CV track to control the signals sent from the Analog Four MKII's two CV/Gate outputs. Naturally, the parameters of all track types can be parameter-locked. The Elektron sequencer lets you control your analog studio in the most efficient way possible.
You can sequence your analog gear with CV inputs using the Analog Four MKII's dedicated CV track. The four CV/Gate outputs pass control voltages for controlling any parameter or pitch (V/oct andHz/V), Trig and Gate pulses for trigging envelopes (V-trig and S-trig), and repetitive clock pulses for syncing analog sequencers, LFOs, drum machines, and more. You can use the Analog Four MKII's external inputs to filter external signals. You can even link the filters to form a parameter-lockable filter bank with up to eight filters (4 ladder and 4 multi-mode filters).


Although it's Wednesday here, it's sorta like Tuesday because we had Monday off and I have to work on Saturday. Ugh. I should consider myself lucky though, because a lot of ALT's have to go to school on Saturdays much more often than me. This will be my first time, even though my students and the other teachers are very often at school on Saturdays and holidays. From what I understand, it's sort of a Junior High Expo day, so the Jr. High kids can come and see what it's like to be a high schooler/decide if my high school is the one that they want to come to (kinda like a college visit, but at high school). They also have to take entrance examinations if they want to come here, so it's the first step in making a decision where to go. I don't really know if there will be Jr. High students in my class then or what, but I shall find out soon I am sure. Like Saturday morning right before class haha.
Really, I don't have much else to write. As I was trying to think of something to write, I realized that all I want to write is "I am getting my car on Monday" ... but you already know that if you talk to me on the phone or read my blog. So instead, I guess I will just say "Sayonara" and hope you are all at home watching the election results instead of reading my blog. I am checking on CNN as we speak and should go correct a few papers before class this afternoon. Might need to sneak off and take a nap - I love how acceptable it is to take small snoozers at work here in Japan!


Could Buildings Be Improved If They Were Designed Like A Web App?
David Galbraith has embarked on a fascinating journey, exploring the notion that the flow of people and their interactions inside buildings, is similar in design to the flow of data and user interaction of Web apps.
Could best practices in Web app design be applied to architectural design?
Dave is a buddy and he belongs to what is a very small group of people I know, who are both insightful and foresightful, about the tremendous changes that our digital technologies are creating around us.
The title is a play on the Case Study Houses of the 1940s. It's not a UX design but a UX inspired one.
Many architects tend to think of buildings as objects, the greatest ones, such as Frank Lloyd Wright, often thought about them as interconnected spaces but they focused on the spaces rather than the flow through them - this is analogous to looking at the stage set rather than the choreography.
I've often wondered what it would be like to design a building like you would choreograph a dance - so that the end design was a picture of a person moving rather than the environment and where if that was sophisticated enough the environment would be defined by the person's movement.
Web design is very linear, its all about flow and eliminating the niche, to get the bulk of people through a primary use case.
I'm intrigued by Dave's approach to architecture, it's one that is long overdue. Examples of great architecture seem to be always about image, loud statements about wealth, importance, or aesthetic panache. Architectural prizes reward a beauty that's nearly always skin deep, and rarely judge how well a design supports the actual work of a building. Apps are always judged on how well they work.
Dave's approach is sound and very sensible and it got me thinking: what would result if say you considered a home, or an office, as a container for a collection of life apps - a platform, an operating system (OS). Each app has a specific user interface but each having to share databases, processes, and the constraints of their hardware.
For example, a home or office building, is like an iPhone platform that runs apps such as Laundry, Cook, WatchTV, Exercise, Sleep, Work, etc. Each app has its own user interface designed for that specific activity but each shares the same restrictions defined by the building, the OS, such as the size of kitchen, building codes, etc.
Smaller homes would require different designs but each "Home OS" or "Office OS" would result in floor plans optimized to the available resources, and user loads, similar to the way apps are adapted for different sized displays, processor speeds, and network services.
Architectural designs that work well for one user, would surely work well for many, providing a scaling factor similar to the design of good software, which should attract entrepreneurial activity in licensing effective designs.
Testing out web apps is easy - testing out home or work spaces is not. Which means progress will be slow.
However, it should be possible to identify what works and what doesn't from what has already been built. Cheap sensors could capture masses of data about people flow in buildings and homes. Algorithms could quickly identify what works well and map it against its architecture. Best use cases could be quickly identified and improved in future designs.
It's using big datasets in a similar way that web app developers analyze Internet traffic flows and millions of user interactions, to identify what works best.
I hope other architects take notice of Dave's experiment. I hear that unemployment rates are the highest in their profession - which means there's a lot of them sitting around with time on their hands.


Suggested areas for parking are either the E-12 parking lot, near Husky Stadium (corner of Montlake Blvd. and NE Pacific St.) or Triangle Garage directly across from UW Medical Center (NE Pacific Street). Office of Admissions cannot validate parking.
Cross Montlake Blvd and NE Pacific Street, walk along the front of Medical Center and enter through main entrance on 3rd floor.
Enter the Health Sciences Building from the 2nd floor of the S-1 garage, then take the elevator from -1 to 3rd floor. Turn left out of the elevator, walk down the long hallway, then turn right and follow signs from "A-wing"
Turn right at the information desk on 3rd floor of Medical Center, walk down the long hallway, that leads you into the Health Science Building. Continue down this hallway and A-300 is located off the main lobby in the Health Sciences Building.
For more parking information and fees see Commuter Services for the Campus Parking Lots and Gatehouses.
Sound Transit has just opened the Link Light Rail route from Sea-Tac Airport to the University of Washington. Depending on the time of day, the trains leave every 6-10 minutes.
A shared shuttle bus offering door-to-door service. The approximate cost is $26.00 per passenger, and can be accessed outside of doors #6 and #26 on Sea-Tac's lower level.
All Seattle taxi cabs operate at the same rate set by King County. $2.60 meter drop, $2.70 per mile. Extra person charges may apply. Taxi service, one way, from Sea-Tac Airport - estimated cost $55.00 - $65.00.
Ask to be taken to 1959 NE Pacific Street or the Health Sciences/University of Washington Medical Center.
A complete list of rental car options and rates is available.
Further information regarding ground transportation from Sea-Tac Airport is also available.
These hotels/motels are in the University District and are either within walking distance of the Health Sciences Center or on a direct bus route. There are a number of other hotels in the area and this list is not comprehensive.
If you are interested in staying with a current student, the Medical Student Association has established an applicant hosting program for individuals interviewing in Seattle. If you would like to take advantage of this opportunity, please fill out this online form. If you have further questions about the program please email somhost@uw.edu. Please keep in mind a response may take up to a week. Accommodations will vary and are not guaranteed.


Preheat oven to 400F and lightly grease a baking sheet with oil.
In a large bowl, whisk together the mayonnaise, mustard and egg. Season to taste with salt and pepper, then fold in crab and parsley until just combined.
Gently form mixture into four 1-inch thick cakes (can use a small ramekin as a mold) and transfer to baking sheet.
Bake crab cakes in the middle of oven approximately 20 minutes, or until golden. Let crab cakes stand on baking sheet 5 minutes before.


text:
\section{Introduction} \setcounter{page}{1}
Let $X$ be a smooth projective variety.
The following definition was introduced by Sommese \cite{S1}.

\begin{Def}  A line bundle $L$ on $X$ is $k$-ample,
if for some $r>0, \, L^r $ is generated by sections and the fibers of the corresponding morphism
$$\phi:X \to {\mathbb P} H^0(X,L^r) $$
have  dimensions less or equal to $k. $

A vector bundle $E$ is said to be $k$-ample if ${\mathcal O}_{{\mathbb P} E}(1)$
is $k$-ample.
\end{Def}

Note that $0$-ample is the same as ample.

\begin{Not}\label{Notgbs}
Since we often need to use the phrase ``generated by sections'', we will abbreviate it by ``gbs''.
\end{Not}

Sommese proved in ( \cite{S1} p. 235, Corollary (1.10)) that direct sums and tensor products of $k$-ample vector bundles
are $k$-ample, if these vector bundles are generated by sections. The aim of the present article is to remove the latter
restriction.

\section{Generation by sections}
We start  by giving several interpretations of the statement:
         $$ {\mathcal O}_{{\mathbb P} E}(r)  \textrm {\ is gbs}.$$

1) \  Let $\pi:{\mathbb P} E\rightarrow X$ be the projection. The fibre of $\pi$ over a point $x \in X$ is given by the 1-codimensional
subspaces $V$ of $E_x$, where $E_x$ is the fibre of $E$ over $x$. Points of ${\mathbb P} E$ will be denoted by pairs $(x,V)$.
Let $V=v^\perp$ for $v \in E_x^*$. A section $s\in H^0({\mathbb P} E,{\mathcal O}_{{\mathbb P} E}(r))$ generates the fibre of ${\mathcal O}_{{\mathbb P} E}(r)$
over $(x,V)$ iff
$$\langle s(x,V), v^r\rangle\neq 0,$$
where $\langle\  , \ \rangle$ denotes the pairing of dual vector spaces.\\

2)\  Equivalently, let $\tilde s$ be the section in $H^0(X,S^rE)$ which corresponds to $s$ by the natural isomorphism
$$H^0({\mathbb P} E,{\mathcal O}_{{\mathbb P} E}(r))\simeq H^0(X,S^rE).$$
Then $s$ generates the fibre of ${\mathcal O}_{{\mathbb P} E}(r)$ over $(x,v^\perp)$ iff
$$\langle \tilde s(x), v^{\otimes r}\rangle\neq 0.$$

3)\ Given $x\in X$, the value $\tilde s(x)$ of each section $\tilde s \in H^0(X,S^rE)$ can be regarded as a homogeneous
polynomial map of degree $r$ from $E_x^*$ to ${\mathbb C}$. If the fibre of ${\mathcal O}_{{\mathbb P} E}(r)$ over $v^\perp$ is not generated by
any global section, then these polynomials have a common zero at $v$. If ${\mathcal O}_{{\mathbb P} E}(r)$ is gbs,
then 0 is the only common zero.\\

4)\ Since $v^{\otimes r}$ annihilates the kernel of the natural map $E^{\otimes r}\twoheadrightarrow S^rE$,
the fibre of ${\mathcal O}_{{\mathbb P} E}(r)$ over $(x,v^\perp)$ is generated by a section iff
there is a section $\tilde s \in H^0(X,E^{\otimes r})$ such that
$$\langle \tilde s(x), v^{\otimes r}\rangle\neq 0.$$

If ${\mathcal O}_{{\mathbb P} E}(r)$ is gbs, then ${\mathcal O}_{{\mathbb P} E}(kr)$ is gbs for every positive integer $k$. All these facts will be
used in the following without reference.

\begin{lem} \label{sumgbs} Let $X$ be a complex space not necessarily compact and $E,F$ vector bundles on $X$. Suppose
that ${\mathcal O}_{{\mathbb P} E}(r)$ and ${\mathcal O}_{{\mathbb P} F}(r)$ are gbs \rm (see Notation \ref{Notgbs}\rm ) for some $r>0$. Then
 ${\mathcal O}_{{\mathbb P}(E\oplus F)}(r)$ is gbs.
\end{lem}

Proof:
For each $x\in X$ and each non-zero $u\in (E_x^*\oplus F_x^*)$ we have to find a section
in $H^0(X,S^r(E\oplus F))$ which is not annihilated by $u^{\otimes r}$. Consider  $u=u_E+u_F$,
where $u_E,u_F$ are the components in $E_x^*, F_x^*$ resp. We may assume that $u_E\neq 0$. Since ${\mathcal O}_{{\mathbb P} E}(r)$
is gbs, there is a section in $H^0(X,S^rE)$ which is not annihilated by $u_E^{\otimes r}$. The image of this section
under the natural injection $H^0(X, S^rE)\hookrightarrow H^0(X, S^r(E\oplus F))$ is not annihilated by $u^{\otimes r}$.
$\hfill{\square}$

\begin{lem} \label{productgbs} Let $X$ be a complex space, not necessarily compact and $E,F$ vector bundles on
$X$. Suppose that
 ${\mathcal O}_{{\mathbb P} E}(r)$ and ${\mathcal O}_{{\mathbb P} F}(r)$ are gbs (see Notation \ref{Notgbs}) for some $r>0$. Let $e={\rm rk} E$. Then
 there exists a positive integer $n$ depending only on $r$ and $e$ such that ${\mathcal O}_{{\mathbb P} (E\otimes F)}(nr)$ is gbs.
\end{lem}

The proof of this lemma needs some preparations. We will use the terminology of decorated
oriented graphs (see Feynman diagrams in physics). Decoration means that to each vertex and to each arrow
another object is associated. The set of arrows of a graph $\gamma$ will be denoted by $A(\gamma)$.
Any arrow $a$ is said to have a head $h(a)$ and a tail $t(a)$. A vertex which is the head of exactly $n$ arrows
is said to have indegree $n$, a vertex which is the tail of exactly $n$ arrows is said to have outdegree $n$.
The bidegree of a vertex is written as (indegree, outdegree).

\begin{Def}
${}$

Let $V,W$ be vector spaces. Let $i=1,\ldots,n$, $\tilde v_i\in S^r V$, $\tilde w_i\in S^r W$, $u\in (V\otimes W)^*$. Then
$\Gamma_{V,W}(n,r,\tilde v_1,\ldots,\tilde v_n,\tilde w_1,\ldots,\tilde w_n,u)$ is the set of all decorated oriented graphs
with $2n$ fixed vertices $\alpha_i, \beta_i,$ such that for each $i$ the vertex $\alpha_i$ has bidegree $(0,r)$
and is decorated by $\tilde v(\alpha_i)=\tilde v_i$, the vertex $\beta_i$ has bidegree $(r,0)$ and is decorated by
$\tilde w(\beta_i)=\tilde w_i$,
and such that all $rn$ arrows are decorated by $u$. We will omit the indices $V,W$ of $\Gamma_{V,W}$, when no
ambiguity arises.
\end{Def}

\begin{Def}
 Let $\gamma\in \Gamma(n,1,v_1,\ldots,v_n,w_1,\ldots,w_n,u)$. The value $\vert\gamma\vert$ of $\gamma$ is defined by
$$\vert\gamma\vert = \prod_{a\in A(\gamma)} u\big(v(h(a))\otimes w(t(a))\big).$$
\end{Def}

\begin{Def}
 Let $\gamma\in \Gamma(n,r,\tilde v_1,\ldots,\tilde v_n,\tilde w_1,\ldots,\tilde w_n,u)$. Then the value $\vert\gamma\vert$ of
$\gamma$ is defined by the following two conditions:
Firstly, $\vert\gamma\vert$ depends linearly on all $\tilde v_i$ and $\tilde w_i$. Secondly, if
$\tilde v_i = v_i^{\otimes r}$, $\tilde w_i = w_i^{\otimes r}$ for $i=1,\ldots,n$, $v_i\in V$,  $w_i\in W$ let the expanded
graph
$$\gamma_{ex}\in \Gamma(rn,1,\underbrace{v_1,\ldots,v_1}_{r\ times},\ldots,\underbrace{v_n,\ldots,v_n}_{r\ times},
\underbrace{w_1,\ldots,w_1}_{r\ times},\ldots,\underbrace{w_n,\ldots,w_n}_{r\ times},u)$$
be a decorated oriented graph with vertices $\alpha_i^l, \beta_i^l,$ $i=1,\ldots,n$, $l=1,\ldots,r,$ such that
there is a bijection $\xi:A(\gamma_{ex})\rightarrow A(\gamma)$ with
$h(a)=\alpha_i$ if $h(\xi(a))=\alpha_i^l$, $t(a)=\beta_i$ if $t(\xi(a))=\beta_i^l$. Moreover, let
$\alpha_i^l$ be decorated by $v_i$, $\beta_i^l$ by $w_i$, for $i=1,\ldots,n$, $l=1,\ldots,r,$. Then
$$\vert\gamma\vert = \vert\gamma_{ex}\vert.$$
\end{Def}

For later use, note that $\gamma_{ex}$ has a symmetry group of order $(r!)^{2n}s_\gamma^{-1}$, where $s_\gamma$ is
the order of the group of vertex preserving symmetries of $\gamma$.\\

\begin{prop}\label{nu} There is a function $\nu:N \times N \rightarrow  N$ with the following property.
Consider finite dimensional vector spaces $V,W$ with $d={\rm dim} V$ and subspaces $A\subset S^rV$,
$B \subset S^rW$, such that the corresponding spaces of polynomial maps have 0 as only common zero.
Let $n=\nu(r,d)$. Then for any non-trivial $u \in (V\otimes W)^*$ there is a positive integer $m$ with $m\leq n$,
elements $\tilde v_i\in A$, $\tilde w_i\in B$ for $i=1,\ldots,m$ and a decorated directed graph
$\gamma\in \Gamma_{V,W}(m,r,\tilde v_1,\ldots,\tilde v_m,\tilde w_1,\ldots,\tilde w_m,u)$
such that  $\vert\gamma\vert\neq 0$.
\end{prop}

Proof:
We will construct a function  $\nu$ which is far from optimal, but  sufficient for our purpose.

We first show that it suffices to prove the proposition for the case that the map $\hat u\in Hom(V,W^*)$ corresponding to
$u$ is bijective. Otherwise, let $V'=V/ker(\hat u)$,  $W'=W/ker(\hat u^*)$, let $\pi_V,\pi_W$ be the corresponding projections
$\pi_V:S^rV\twoheadrightarrow S^rV'$, $\pi_W:S^rW\twoheadrightarrow S^rW'$ and let $u'$ be the element of $(V'\otimes W')^*$
induced by $u$. If $$\gamma\in \Gamma(m,r,V_1,\ldots,V_m,W_1,\ldots,W_m,u), \rm {and} $$
$$\gamma'\in \Gamma(m,r,\pi_V(V_1),\ldots,\pi_V(V_m),\pi_W(W_1),\ldots,\pi_W(W_m),u')$$
have the same underlying undecorated graph, then $\vert\gamma'\vert=\vert\gamma\vert$.

If $u$ is bijective, we can identify $W$ with $V^*$ and write
$$u(v\otimes w)=\langle v,w\rangle$$
for $v\in V$, $w\in W$. Let $SV$ be the symmetric algebra over $V$.
There are natural multiplication maps  $$m^r:S^rV\rightarrow End(SV),$$
and contraction maps  $$i^r:S^rV \rightarrow End(SV)$$
which restrict to
$$m^r_N:S^rV\rightarrow Hom(S^{N-r}V,S^NV),$$
$$i^r_N:S^rW \rightarrow Hom(S^NV,S^{N-r}V),$$
with $S^NV=0$ for negative $N.$ For $r=1$ the maps can be characterized by multilinearity and the properties
$m^1(v')v^{\otimes N}=\pi_S (v'\otimes v^{\otimes N})$, $i^1(w)v^{\otimes N}=N\langle w,v\rangle v^{\otimes (N-1)}$
for $v',v\in V$, $w\in W$, where $\pi_S$ is the projection of the tensor algebra of $V$ to the symmetric algebra $SV$.
For general $r$ they are characterized by multilinearity and the properties $m^r(v^{\otimes r})=m^1(v)^r$,
$i^r(w^{\otimes r})=i^1(w)^r$.

For $r=1$ let $P$ be a product with the factors $m^1(v_i)$, $i^1(w_i)$, taken in any order, where $v_i\in V$, $w_i\in W$,
$i=1,\ldots,m$. For any non-negative integer $N$, the product $P$ restricts to a map $P_N\in End(S^NV)$. For the trace of $P_N$
one finds
$$tr (P_N) = \sum_{\gamma\in \Gamma_m} c_\gamma\vert\gamma\vert,$$
where
$$\Gamma_m = \Gamma_{V,W}(m,1,v_1,\ldots,v_m,w_1,\ldots,w_m,\langle,\rangle)$$
and
$$c_\gamma={d+\rho+N-1\choose d+m-1},$$
where $\rho$ is the cardinality of the set of arrows $a$ of $\gamma$ such that the factor $m^1(v(h(a)))$ lies
to the right of the factor  $i^1(w(t(a)))$ in $P$. For $\rho>0$ this follows by induction on $\rho$
from
$i^1(w)m^1(v)=m^1(v)i^1(w)+\langle w,v\rangle$ and for $\rho=0$ from cyclic invariance of the trace and induction on $N$.
For $n=0$ and $N=0$ the statement is obvious.

For arbitrary $r$ the calculation can be reduced to the case $r=1$ as in the definition of $\vert\gamma\vert$.
Let  $P$ be a product with the factors $m^r(\tilde v_i)$, $i^r(\tilde w_i)$, taken in any order,
where $\tilde v_i\in S^rV$, $\tilde w_i\in S^rW$, $i=1,\ldots,m$. For any $N$, $P$ restricts to a map $P_N\in End(S^NV)$.
To calculate the trace of $P_N$ it is sufficient to consider the case $\tilde v_i = v_i^{\otimes r}$,
$\tilde w_i = w_i^{\otimes r}$ for $i=1,\ldots,n$, $v_i\in V$,  $w_i\in W$. One finds
$$tr( P_N) = \sum_{\gamma\in \Gamma^r_m} c_\gamma\vert\gamma\vert,$$
where
$$\Gamma^r_m = \Gamma_{V,W}(m,r,\tilde v_1,\ldots,\tilde v_m,\tilde w_1,\ldots,\tilde w_m,\langle,\rangle)$$
and
$$c_\gamma=(r!)^{2m}s_\gamma^{-1}{d+r\rho+N-1\choose d+rm-1}.$$
Here $\rho$ is the cardinality of the set of arrows $a$ of $\gamma$ such that the factor $m^r(\tilde v(h(a)))$ lies
to the right of the factor  $i^r(\tilde w(t(a)))$ in $P$, and  $s_\gamma$ is the order of the group of vertex preserving
symmetries of $\gamma$.

Let ${\mathcal A}^r_N$ be the subalgebra of ${\rm End}(S^NV)$ generated by the elements
$\{m^r_N(\tilde v) i^r_N(\tilde w)\vert \tilde v\in A, \tilde w\in B\} $. This algebra is spanned by products
$P_N=m^r_N(\tilde v_1) i^r_N(\tilde w_1)\cdots m^r_N(\tilde v_m) i^r_N(\tilde w_m)$, where $\tilde v_i\in A$,
$\tilde w_i\in B$ for $i=1,\ldots,m$.
We have seen that the trace of $P_N$ is given by a linear combination of numbers
$\vert\gamma\vert$, where $\gamma\in \Gamma(m,r,\tilde v_1,\ldots,\tilde v_m,\tilde w_1,\ldots,\tilde w_m,\langle,\rangle)$.
If those traces all vanish, then $ {\mathcal A}^N$  is a nilalgebra. In particular, there is a non-trivial subspace $V_0\subset S^NV$
such that
$$m^r_N(\tilde v) i^r_N(\tilde w)V_0=0$$
for all $\tilde v\in A$, $\tilde w\in B$.
Since the kernel of $m^r_N(\tilde v)$ vanishes for $\tilde v\neq 0$, this means that
$i^r_N(\tilde w)V_0=0$ for all $\tilde w\in B$.
Since the homogeneous polynomial maps in $B$ have 0 as only common zero, one can apply
a theorem of Macaulay (\cite{M} Sec. 6, or \cite{S2} p. 85 theorem 4.48). According to this theorem the ideal
in $SW$ generated by $B$ contains $S^NW$ if $N\geq rd$, which implies $i^N_N(\tilde w)V_0=0$
 for all $\tilde w\in S^NW$ and
yields a contradiction. Thus ${\mathcal A}^N$ is not nil for $N\geq rd$.

Let ${\mathcal W}^N_m$ be the subspace of ${\mathcal A}^N$ generated by products of length $\leq m$ of elements
of the form $m^r_N(\tilde v) i^r_N(\tilde w)$.
For each $m$ one has either\\  ${\rm dim} ({\mathcal W}^N_{m+1})>{\rm dim} ({\mathcal W}^N_m$) or ${\mathcal W}^N_m = {\mathcal A}^N$.\\  Let
$D(N)=({\rm dim} \ S^N V)^2$. Since $D(N)\geq {\rm dim} {\mathcal A}^N$, one has ${\mathcal W}^N_{D(N)} = {\mathcal A}^N$. In particular,
${\mathcal W}^{rd}_{D(rd)}$ contains elements of non-vanishing trace. Thus it suffices to take for $\nu(r,d)$ the least
common multiple of all integers less or equal to $D(rd)$.

 $\hfill{\square}$\\

{\bf Proof of Lemma}  \ref{productgbs}:\\

For each $x\in X$ and each non-zero $u\in (E_x\times F_x)^*$ we have to find a section
in $H^0(X,(E\otimes F)^{\otimes nr})$ which is not annihilated by $u^{\otimes nr}$.
Let
$$I_E: H^0(X,S^rE)^{\otimes n}\hookrightarrow  H^0(X,E^{\otimes rn}),$$
$$I_F: H^0(X,S^rF)^{\otimes n}\hookrightarrow  H^0(X,F^{\otimes rn}),$$
$$J: H^0(X,E^{\otimes rn})\otimes H^0(X,F^{\otimes rn})\hookrightarrow H^0(X,(E\otimes F)^{\otimes rn})$$
be the canonical injections.
The permutation group $S(rn)$ acts on $E^{\otimes rn}$ in the standard way, which yields a map
$$\Sigma:S(rn)\times H^0(X,E^{\otimes rn}) \rightarrow  H^0(X,E^{\otimes rn}).$$
Let
$$\Phi:S(rn)\times H^0(X,S^rE)^{\otimes n}\otimes H^0(X,S^rF)^{\otimes n}\rightarrow H^0(X,(E\otimes F)^{\otimes nr})$$
be defined by $\Phi=J\circ(\Sigma\otimes Id)\circ(Id\times I_E\otimes I_F)$.
Though $\Phi$ factors through the coset map $S(rn)\rightarrow S(rn)/S(r)^{\times n}$, it yields sufficiently
many sections for our purpose.
For $x\in X$, $\sigma\in S(rn)$, $\tilde s_i\in H^0(X,S^rE)$, $\tilde t_i\in H^0(X,S^rF)$, there is an element
$$\gamma\in \Gamma_{E_x,F_x}(n,r,\tilde s_1(x),\ldots,\tilde s_n(x),\tilde t_1(x),\ldots,\tilde t_n(x),u),$$
such that

\begin{equation} \label{phi}
\langle\Phi(\sigma,\tilde s_1\otimes \cdots \otimes\tilde s_n\otimes \tilde t_1\otimes \cdots \otimes\tilde t_n)_x,
 u^{\otimes nr}\rangle=\vert\gamma\vert.
\end{equation}

Conversely, for each $\gamma\in \Gamma(n,r,\tilde s_1(x),\ldots,\tilde s_N(x),\tilde t_1(x),\ldots,\tilde t_n(x),u)$
one can find a permutation $\sigma\in S(rn)$ satisfying  the equation\eqref{phi}.

By Proposition \ref{nu} there exist sections $\tilde s_i$, $\tilde t_i$ and
$$\gamma\in \Gamma(n,r,\tilde s_1(x),\ldots,\tilde s_N(x),\tilde t_1(x),\ldots,\tilde t_n(x),u)$$
such that $\vert\gamma\vert\neq 0$. The image of the corresponding section
$\Phi(\sigma,\tilde s_1\otimes \cdots \otimes\tilde s_n\otimes \tilde t_1\otimes \cdots \otimes\tilde t_n)$
in $H^0(X,S^{\otimes nr}(E\otimes F))$ yields a section in  $H^0({\mathbb P} (E\otimes F),{\mathcal O}_{{\mathbb P} (E\otimes F)}(nr))$
which generates the fibre over $(x,u^\perp)$.
 $\hfill{\square}$

\section {Direct sums and tensor products of $k$-ample vector bundles}

\begin{thm}\label{sum}

If $E,F$ are  $k$-ample, then $E\oplus F$ is  $k$-ample.
\end{thm}

Proof: We use criterion (1.7.3) of proposition (1.7) of Sommese \cite{S1}.
First note that ${\mathcal O}_{{\mathbb P} E }(r_1)$ gbs (see Notation \ref{Notgbs}) and ${\mathcal O}_{{\mathbb P} F}(r_2)$ gbs imply
that ${\mathcal O}_{{\mathbb P} E }(r)$ and ${\mathcal O}_{{\mathbb P} F }(r)$ are gbs whenever $r$ is a common multiple of $r_1,r_2$.
By Lemma  \ref{sumgbs} this implies that ${\mathcal O}_{{\mathbb P} (E\oplus F)}(r)$ is gbs.
Now assume that there is a holomorphic finite to one map $\phi:Z\rightarrow X$
of a compact analytic space $Z$ to $X$, such that ${\rm dim} Z = k+1$ and that there is a surjective map
$\phi^*(E\oplus F)\twoheadrightarrow Q$
with a trivial bundle $Q$. In particular there is a non-trivial section of
$\phi^*(E^*\oplus F^*)$. We may assume that the component $s:Z \rightarrow \phi^* E^*$
of this section is non-trivial.
Since ${\mathcal O}_{{\mathbb P} E}(r)$ is gbs, for any $z\in Z$ such that $s(z)\neq 0$ there is a section
$\sigma: X\rightarrow S^rE$ such that $\langle \phi^*\sigma(z), s(z)^{\otimes r}\rangle\neq 0$.
In particular, $\langle \phi^*\sigma, s^r\rangle$ yields a non-trivial section
of the trivial line bundle over $Z$. Such a section must be constant, which implies
that $s$ cannot vanish anywhere. Consequently it yields a trivial quotient bundle of $\phi^*E$,
contrary to the assumption that $E$ is $k$-ample. $\hfill{\square}$

\begin{thm}\label{product}

If $E,F$ are $k$-ample, then $E\otimes F$  is $k$-ample.
\end{thm}

Proof: We use criterion (1.7.4), proposition (1.7)  of Sommese \cite{S1}. We first show that $S^2E$ is $k$ ample.
By Lemma 1 the bundle $E^{\oplus e}$ is $k$-ample, where $e$ is the rank of $E$.
Since each irreducible summand of $S^n(S^2E)$ is isomorphic to an irreducible summand of $S^{2n}(E^{\oplus e}),$
we have $H^j(X,F\otimes S^n(S^2E))=0$ for $j>k$, for any coherent sheaf $F$ on $X$ and $n>>0$.
Since ${\mathcal O}_{{\mathbb P} (S^2E)}(1)$ is a restriction of ${\mathcal O}_{{\mathbb P} (E^{\otimes 2})}(1)$, by Lemma \ref{productgbs}
${\mathcal O}_{{\mathbb P} (S^2E)}(N)$ is gbs for some N. Thus $S^2E$ is $k$ ample.
Since $E\otimes F$ is a quotient of the $k$-ample bundle $S^2(E\oplus F)$, the theorem follows.
$\hfill{\square}$

\begin{rem}
 Theorem \ref{product} implies that there is no need for the gbs conditions in Lemma (1.11.4) and Proposition (1.13) of
\cite{S1}. One obtains the following.
\end{rem}

\begin{lem}
 Let $E$ be a vector bundle on a compact analytic space $X$, $Gr(s,E)$ be the bundle of $s$-codimensional subspaces
of the fibres of $E$ and $\xi(s,E)$ the tautological line bundle of $Gr(s,E)$. Then $\xi(s,E)$ is $k$-ample if $E$ is
$k$-ample.
\end{lem}

\begin{prop} For $i=1,\ldots,n$ let $E_i$ be a vector bundle of rank $r_i$ on a projective manifold $X$, If for all $i$
$E_i$ is $k_i$-ample. Then
$$H^p(X,\wedge^qT^*_X\otimes \wedge^{s_1}E_1\otimes \cdots \otimes \wedge^{s_n}E_n)=0$$
if $p+q>{\rm dim} X + \sum_i s_i(r_i-s_i) + {\rm min}_i\{k_i\}.$
\end{prop}

redpajama_set_name: RedPajamaArXiv


text:
package com.lunatic.mlx.kddcup99.ml

import com.lunatic.mlx.kddcup99.mllib.metadata.KddRecord
import KddRecord._
import com.lunatic.mlx.kddcup99.ml.pipelines.KddPipeline
import com.lunatic.mlx.kddcup99.mllib.Configuration
import com.lunatic.mlx.kddcup99.mllib.modgen.SparkRunnable
import org.apache.spark.SparkContext
import org.apache.spark.sql.SQLContext

/**
 *
 */
object Main extends SparkRunnable {

  def run(implicit sc: SparkContext, appConf: Configuration) = {

    val sqc = new SQLContext(sc)

    val labeledData = sc.textFile(appConf.inputTrainingData).map(_.split(",")).cache

    val splitRatio = 0.7

    val labelColumn = 41

//    val (trainData, testData) = SplitTrainingData.splitEvenlyByLabel(splitRatio, labelColumn, labeledData)
//
//    val kdds = trainData.map(toKddRecord)

    val kdds = labeledData.map(toKddRecord)

    val data = sqc.createDataFrame(kdds, schema)

    val pipeline = KddPipeline()

//    println("---------------------------------")
//    println("Pipeline Description")
//    println(pipeline.toString())
//    println(pipeline.explainParams())
//    println(pipeline.extractParamMap())
//    println("---------------------------------")
//    println("Pipeline Stages")
//    pipeline.getStages.foreach{ stage =>
//      println(s"---")
//      println(s"* $stage")
//      println(stage.explainParams())
//      stage.extractParamMap().toSeq.foreach{ param =>
//        println(f"  - ${param.param.name}%-20s - ${param.value}")
//      }
//    }

    sys.exit

    // Split the data into training and test sets (5% held out for testing)
    val Array(trainingData, testData) = data.randomSplit(Array(0.2, 0.8))

    val model = pipeline.fit(data)

    val predictions = model.transform(trainingData)

    println("---------------------------------")
    println("Predictions Schema")
    predictions.printSchema()

    println("---------------------------------")
    println("Predictions Sample")

    predictions.take(10).foreach(println)

    println("---------------------------------")
    println("Pipeline Model Description")
    println(model.toString())
    println(model.explainParams())
    println(model.extractParamMap())
    model.extractParamMap().toSeq.foreach{ param =>
      println(f"  - ${param.param.name}%-20s - ${param.value}")
    }
    model.stages.foreach{ stage =>
      println(s"---")
      println(s"* $stage")
      println(stage.explainParams())
      stage.extractParamMap().toSeq.foreach{ param =>
        println(f"  - ${param.param.name}%-20s - ${param.value}")
      }
    }

  }

}