Dictionary defines warriors as "a person who fights in battle and is known for having courage and skill." In Roots, reimagined Kizzy Kinte tells her dad "Reading is my way of being a warrior, my way of feeling free inside." Teaching enslaved people to read and write was illegal in most Southern states, so reading was an act of resistance, an act of rebellion, a warrior act that could get you sold, or worse. Reading material described as "subversive," such as David Walker's "Appeal," could get you killed.
While enslaved people deeply desired knowledge, and risked their lives to learn to read, an ugly saying emerged about African Americans and literacy. "If you want to hide something from a black people, put it in a book." I think the saying has less to do with Black folk and reading, and more to do with White people's wishful thinking. Still, the achievement gap suggests that we need more African American people to become literacy warriors.
The education system is replete with racial disparity and unfairness. African Americans, 18 percent of the pre-kindergarten population, garner 48 percent of the suspensions, and are three times as likely as White students to get expelled. Black girls are the most likely to experience expulsions. African American students are also three times as likely to be held back as other students.
African American students have less access to advanced courses, like Advanced Placement (AP) and International Baccalaureate (IB). Too often their schools simply don't offer the courses that may make a difference in college admissions. While one in five White students takes calculus in high school, just one in fifteen African American students take calculus. And, African American students are three times as likely to be taught by first year teachers as whites are.
There are all kinds of reasons for these educational disparities, some of which are a function of income, residential location, wealth, and parents' education (and all of these are correlated). There are multiple organizations that advocate for more educational equity and for closing the achievement gap. Billions of resources have gone into addressing the challenges that come with closing the achievement gap, but educational inequality is persistent. We need educational warriors, people who are committed to fighting the educational inequality with courage and skill.
It is going to take courage to call out the sacred cows that many dare not criticize. Who are the teachers who are inefficient? What can we do about them? Do programs like Teach for America, "highly regarded" programs that send inexperienced teachers to classrooms serve or disserve young people? It's going to take skill, too, to call people out in a way that motivates them toward reform, not resistance.
One of the statistics that bothered me most about the condition of education is the fact that young Black people start out behind when they get to kindergarten. Home literacy is an issue. Ninety-one percent of White children who are not enrolled in preschool are read to at least three times a week by family members, compared to 78 percent of African American students. Some say this gap in home literacy is among the factors in producing the word gap, which some say may be as large as 30 million words (some researchers dispute this compilation and its meaning). There is also a racial gap in the availability of books in homes, and African American youngsters may have less access to libraries.
Where are the literacy warriors who would read to the children who are experiencing gaps? Where are the ones who will ensure that libraries in some communities are adequately stocked? Are there warriors who will give children books to take home, warriors who will fight to expand the offerings of our nation's public schools?
Schools are letting out for the summer in the coming weeks, which means that millions of young people who are not engaged in summer educational activities will actually lose learning by the time they return to school in the fall. Are there literacy warriors who will stand in the gap and ensure that every child has some meaningful educational program during the summer? Or will we stand by and watch the achievement gap stay the same or even grow wider?
We need literacy warriors. These should not only be teachers in classrooms and parents at home, but preachers, politicians, and all others concerned with the achievement gap. We need folks who will fight for literacy with persistence, and courage, and skill. We need warriors as fierce as Kunta and Kizzy Kinte, as committed as Mary McLeod Bethune, as passionate as the teachers, black and white, who were paid little or nothing to go south to teach the newly emancipated. Where are the literacy warriors?
Julianne Malveaux is an author and economist. Her latest book "Are We Better Off? Race, Obama and Public Policy." Is available via juliannemalveaux.com or Amazon.com.


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Scotty's Contracting & Stone, LLC is a member of the largest 100% employee owned ESOP in the United States. Houchens Industries based in Bowling Green, KY has over 40 diversified companies that together employ over 18,000 employees. As an ESOP, Scotty's employees are provided a retirement plan through Houchens ESOP. Each year employees are awarded shares by the ESOP to each qualifying employees' account. An ESOP provides employees with an opportunity to contribute to the success of the retirement plan, since the value of the ESOP shares is determined by the total appraised value of the total companies.
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A fifth straight day of sun and the fields were dry enough for cultivation, one of my absolute favorite sports. The implement of choice, a Lilliston Rolling Cultivator, hitches to the tractor.
This tool is a very thorough and precise weeder that allows soil to be moved laterally so a heavy cut will hill the potatoes, a delicate cut will trickle soil around squash seedlings. The Lilliston saves dozens of hours of hand work.
Thirty-five years ago we purchased one of these new at a price that, to my thrifty Yankee father-in-law, seemed too steep.
Upon its delivery, we set it up and I fine-tuned it as Gary made practice passes through the strawberry field. "Not bad," he said after a second row was done.
"That looks damned good," he allowed a few rows later.
I watched him finish out the field, figuring I'd have my chance tomorrow.
Came the dawn, out to the tractor shed at 5:30 and no cultivator to be seen.


The chart below shows how much it costs to get one million British Thermal Units (MBTU) of heat value from new appliances using the types of fuels listed. Fuel prices shown are current as of April 1998.
The EFFICIENCY column shows the listed overall heat extraction efficiency of the appliance used. For this chart, we used the manufacturer's listed efficiency ratings for the Avalon 945/990 woodstove, the Waterford Erin pellet stove, the Hearthstone Sterling natural gas and LP stoves, and the Efel Harmony II oil stove.
The ANNUAL COST column shows the average fuel expense required to heat a typical 1500 sq. ft. house for one year (64.5 MBTU).
Because BTU content varies among species of fuel wood, the chart lists separate figures for the four most commonly available local fuel woods.
Note that the above chart compares NEW appliances. Efficiency ratings have improved dramatically in almost all appliance catagories in recent years: for example, if you're currently heating with a typical 10-year-old woodstove, you could save up to 30% on your fuel bill by replacing it with a new, high efficiency model.


Amazing custom built home situated on 1.45 acres!!! This home boasts a den and keeping room both with fireplaces, spacious master suite with HUGE walk-in closet, media area and bonus room upstairs!!! Hardwood floors in all living areas! Backyard is a dream come true to relax in!!


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Chuck Liddell
Going another round with MMA author Kelly Crigger
By Mark Chalifoux
It's always good to catch up with my guy, MMA author/writer Kelly Crigger, to debate the issues of the day. Who's the true face of MMA and is a new star on the rise? Should Fedor Emelianenko sign with the UFC? Will UFC 100 cripple other shows? We cover it all with this week's discussion.
Issue 1 – George St. Pierre has eclipsed Chuck Liddell as the face of MMA.
Kelly: True. Chuck Liddell helped get MMA where it is today, but Georges St. Pierre has taken the torch and run with it. His image is the face of a new MMA generation that graces the covers of a wide variety of MMA and mainstream magazines, from Black Belt to Men's Fitness. He's a squeaky-clean, non-trash talking national hero in Canada with crossover appeal in the US. He's a chick magnet who brings women into MMA and could probably get away with describing his crepe suzette recipe on Martha Stewart's show without his shirt (as long as Martha keeps hers on). Above all else his fights are damn exciting (GSP vs. Jon Fitch is fight of the year in my opinion). He's the next stage in the evolution of the athlete (sorry to inform you NFL) who can strike, wrestle, grapple, and stay cool under pressure. After his fight with BJ Penn, the crowd booed Thiago Alves the second his mug graced the jumbotron not because they disliked Alves, but because they've completely embraced GSP as the sport's Barack Obama. He even (allegedly) dated Mandy Moore for a brief time, which drives his stock even higher than the ailing S&P 500. The bottom line is GSP is the rock star of MMA who has eclipsed Chuck Liddell as the most recognizable and admired fighter we have. If the UFC is smart, they'll use his popularity and keep his face on the covers of as many magazines as possible, even if that means a photo shoot for Teen Beat.
Mark: False. The face of MMA is still Chuck Liddell. Until Georges St. Pierre is on Entourage or gracing the honorable pages of entertainment blogs for his Grammys appearance, he won't be the face of the sport. I agree that he's the best example of the evolution of MMA. He's way ahead of the game and he's one of the most terrifying champs in the UFC (I still think Anderson Silva is the UFC's best fighter). But face of the sport? Come on, not even close. GSP's hurt by his thick French-Canadian accent more than anything. Hell, I've got French-Canadian roots and even I had trouble understanding him the last time we spoke over the phone. He can't make that mainstream crossover without being able to dominate in a SportsCenter appearance just as well as in the Octagon. He may be change, but he's not change I can believe in (my hand was forced to use some sort of lame campaign slogan connection ever since you invoked the Commander-in-Chief). If anyone is going to eclipse Chuck, it will probably be Rampage Jackson.
Issue 2 – Strikeforce's new deal with Showtime will finally make them an enemy in Dana White's eyes.
Kelly: True. Dana White has long stated his respect for Strikeforce CEO Scott Coker because he genuinely cares about MMA, he's always known his place in relation to the UFC and he's always stayed true to his roots as a regional MMA promoter. But Strikeforce's purchase of the defunct EliteXC franchise coupled with their announcement that they have struck a deal with Showtime has raised their profile above the acceptable levels for the UFC, which paints a big red target on their backs. White has a history of crushing his competition like a Whack-A-Mole champion when they get out of line and become a threat to his brand. Until now, Strikeforce has never raised its heads high enough to feel the business end of Dana White's hammer and has even been praised by the UFC's president for its business model. But Strikeforce's recent purchases of several fighter's contracts (Jake Shields, Nick Diaz, Scott Smith, Benji Radach and Josh Thomson, among others) may have pushed them over the edge from friendly coexistence to MMA war. Strikeforce is clearly the second most powerful MMA organization now, especially with Affliction's future in doubt. I think what is especially painful for White to swallow is Strikeforce's Showtime contract (if that holds true). For years White negotiated with HBO for a broadcast deal only to come up empty-handed in the end. Suddenly Strikeforce comes along with enough capital to buy a slew of contracts and negotiates a deal that the UFC couldn't pull off. That's threatening to a man who's used to decimating his enemies and currently has few to slay.
Mark: False. An enemy? An organization that has more than one of a division's top five fighters would be an enemy. That would be an organization that could feasibly compete with the UFC. Strikeforce is still good for the sport because it's run by guys that know MMA and will help grow the sport, while the UFC will pick off any truly great stars Strikeforce develops. I really like Strikeforce, I'm jazzed about the deal, but they aren't a true enemy until Affliction falls apart and Strikeforce can pick up those fighters as well (who can wait for Kimbo Slice vs. Fedor? Over/under on that fight's length - three seconds). Strikeforce will be able to eclipse the WEC for the rights as the second best MMA promotion, but that's not enough to put them in adversary mode for White. There's no denying that Strikeforce will try to position itself as a competitor to the UFC, but enemy? The "t-shirt guys" are enemies. Strikeforce is merely a very good minor league system, and a potential threat down the road.
Issue 3 - What should the UFC do with Karo Parisyan?
Kelly: Karo is no longer the top ten welterweight he was when the UFC offered him a shot at then-champion Matt Hughes in 2006. Since then he's been injured, inconsistent, and allegedly faked an injury to pull out of his UFC 88 fight when he actually suffered from a panic attack. Now he's come up positive for three - not just one - but three banned painkillers after his UFC 94 fight against Dong Hyun Kim, a fight he barely won. Did that give him an unfair advantage in the fight? Absolutely, so if the NSAC is ever going to overturn a fight, it should be that one. Are those reasons to cut him from the UFC rolls? I think so, but the UFC has a history of keeping guys around when they're down on their luck. Chris Lytle has lost 17 fights, but he's still around because he gives it all he has in the octagon. Patrick Cote lost four in a row, but the UFC kept him around because he's also an exciting fighter. So losing fights is apparently not a criteria to get cut. But stepping outside Dana White's code of conduct is, as Jesse Taylor and John Koppenhaver (I refuse to call him War Machine) found out when they were cut for mouthing off. Jon Fitch got cut from the UFC for a day for refusing to sign a release for his likeness for a video game. In the TUF house ... well, forget trying to nail down what the "banned from the UFC" criteria is there. Karo's situation is a little different. It's not like he came up hot for one banned substance like Sean Sherk did (who was eventually reinstated after his suspension) or is an overly exciting fighter (all of his wins in the last two years were by decision). Parisyan has been caught with a host of banned substances in him and barely wins his fights. If these aren't grounds to cut a guy from the organization, then it's clear that the only standards for being released from the UFC are the ones in White's head.
Mark: I don't necessarily disagree with many of White's cuts. While cutting guys like Fabricio Werdum may be unfair, I can see how it could make sense from a business perspective. Parisyan is in a tough spot. If he was an exciting fighter, he'd have more of a leg to stand on. If it's me, I lose him. The UFC can certainly afford to lose another welterweight, especially since Parisyan isn't exactly a factor in the title picture. There's a healthy roster of young welterweights close to having break-out years, so I only see him going down in the welterweight picture. He's a talented fighter and he's an excellent gatekeeper at welterweight, that's for sure. He has a role in the UFC, but is it one that can withstand using banned substances? Probably not. Guys like Parisyan are the ones that are made examples of, while people at the top (like Alves) could get away with something like this.
Switch it up. Mark goes first.
Issue 4 - The UFC will have more than two title fights at UFC 100.
Mark: False. I think they will put together a terrific show for UFC 100 but the rumors about all the titles being up for grabs are just hopeful wishes. That would cripple the shows before and after UFC 100 just for one blow-out show. UFC 100 would be record-setting but it wouldn't do enough buys to justify the losses they would take on the shows in the months before and after UFC 100. Is anyone going to pay 50 bucks for a card headlined by Dan Henderson vs. Michael Bisping? What about Mike Swick vs. Jon Fitch? Lyoto Machida vs. Forrest Griffin? That's like making someone pay 50 bucks to watch Rampage fight Keith Jardine (whoops). Those are good fights, but not enough to drive PPV sales, so I think UFC 100 will be a little scaled back from what most people think. I wouldn't be surprised to see something similar to UFC 92. Dana White has already said GSP will fight Alves, and I would guess Henderson vs. Bisping would be a part of that card as well and that also might be a good time for Rashad Evans vs. Rampage. Put together a few interesting fights on the undercard and you can have a very, very good show. It doesn't need to be lights out from the get-go, though. It's just not worth it.
Kelly: It might not be worth it, but we're talking about the brainchild of an egomaniac. Sacrificing a couple of shows to put on a historical event is right up Dana White's alley. He's already proven that he can keep a consistently-entertaining business running when others fall by the wayside, so a huge event that no one thinks he can pull off is just the challenge he drools over. Whether or not he can pull it off is the question. We know the welterweights are probably going to fight and lightweight champion BJ Penn and top contender Kenny Florian will have plenty of time to prepare if White wants them to. Middleweight champ Anderson Silva could be ready for a new victim if he blows through Thales Leites like I expect him to, but he'll only have three months to prepare. The light heavyweights (Evans and Rampage) could easily be ready, though Rampage will be at a disadvantage by fighting Keith Jardine first. The heavyweights simply won't happen. There's no way the new unified heavyweight belt holder (Frank Mir or Brock Lesnar) will be ready after their UFC 98 fight for another one in less than 60 days. Besides the scheduling, there's another issue. Even if White demands five title fights, I can't envision all ten fighters agreeing to it. How many of these guys are willing to be demoted to non-main event status or even be the first fight of the televised card? I think in the end five title fights just won't be feasible, but they'll stack the undercard to make it the best UFC ever.
Issue 5 - Fedor Emelianenko should sign with the UFC .
Mark: True. I understand his desire to still compete in Sambo was a potential problem for the UFC and that Fedor's management has been difficult to deal with, but this is something Fedor really needs to start thinking about. Affliction is on its final legs and then Fedor will be forced to sign with Strikeforce. That's not a terrible fate, but Strikeforce won't be throwing money around like Affliction so he'll have to take a serious pay cut to go that route. On the other hand, he could go to the UFC and dominate the heavyweight division. More importantly, he could have the full power of the UFC marketing machine behind him. As much as Dana White has hated on Fedor in the past, he'd be the first one to jock him once Fedor signed with the UFC (and rightfully so, White is a shrewd businessman). With the UFC starting to expand more globally and signing merchandising deals and putting together UFC Primetime specials, Fedor would be walking into a gold mine. More importantly, it would introduce him to thousands of new MMA fans, cementing his future financially and his place in the sport forever.
Kelly: I won't argue with you here. The only reason a few guys in Nowhere, Idaho think Fedor's not the best in the world is because he hasn't fought in the UFC. But one look at his list of victims and you realize that it's void of only a couple guys of consequence -- Couture, Mir and Lesnar. The other UFC heavyweights aren't ready for him, but if he came over to the UFC they would spend millions convincing us that they were. The question is will Dana swallow his PRIDE and deal with Vadim Finkelstein? White got over his grudge against Tito and brought him back to the organization and got through his differences with Couture, so it's possible. But there's definite bad blood here to get past first.
Issue 6 - What should Strikeforce do with Kimbo Slice?
Mark: Now that Strikeforce has acquired the contracts of EliteXC stars, like Kimbo Slice, the question everyone has had is: what do you do with them? Scott Coker suggested he needs to have 10-12 fights before he headlines a show but it's not like Slice is a young guy. He's not going to get that much better as he ages. He was a huge draw but was exposed as a huge fraud. What EliteXC did with Slice is a huge black eye for the sport. Strikeforce could make some quick cash promoting him even if it's not in a main event, but the promotion is better off leaving him behind. Strikeforce has the star power and talent to put together great shows, they don't need a freak show. Let Kimbo go back to Japan ... or better yet, let him go back to YouTube.
Kelly: I, for one, would be very interested to see what Mr. Slice has been up to since his flash KO at the hands of Seth Petruzelli. Has Bas Rutten been developing him as a fighter or has he been getting fat while he gets more gold teeth molds made? Kinda like Brock Lesnar I'd like to see what he's really made of. You're right that he's not getting any younger, but there are plenty of talented, young heavyweights that Strikeforce can throw at him to test him out in short span. If he's got what it takes in MMA, then he'll be headlining a show in no time. If Strikeforce doesn't pick him up there are plenty of smaller organizations that will buy his contract to bring them some notoriety. He's good for that at least.
What do you think? Let us know.
Kelly Crigger is a freelance MMA writer and author of the book "Title Shot: Into the Shark Tank of Mixed Martial Arts". Contact him through his website at IntoTheSharkTank.com

redpajama_set_name: RedPajamaCommonCrawl


text:
Stock Control
Sales and Forecasting
Finance Config
Live Integrations
Caternet
Cultivating a Robust & Loyal Team - Zupa
Publication Date: 09 Dec 2021
Working life has changed for many, beyond all recognition over the last year or so. The Covid pandemic has led to many working remotely, and also to a rising talent war as more employees than ever are looking for new employment opportunities. The start of a New Year has always been a popular time to re-evaluate career options, this year promises to be even more prominent on that front. So how can business owners and managers improve staff loyalty and retain their people on a long-term basis?
We spoke to our CEO, Jerry Brand, who has been running successful businesses for more than 35 years to get his views on cultivating a robust and loyal team.
How do you encourage staff to stay at your company?
As an SME leader it is vital during challenging periods and beyond, to nurture loyalty by keeping in touch with your people on a regular basis and listening to their feedback. That means being present and visible in meetings wherever possible and working to build a progressive culture of positivity where people can grow and thrive. We have various teams at Zupa, and our team leaders are always working closely with their people, whether that is via one-to-one conversations or as part of a bigger group. It is crucial to retain a regular dialogue to better understand the individual members of your workforce. This close contact is even more important now that we have moved towards a hybrid remote business model. Having said that, as a business we are very conscious of the importance of retaining face to face interactions, so we are currently organising a physical meeting space so that we can all meet up together both socially and for in-person business interactions.
What issues are you currently having to deal with?
We have learned through leaning into deeper conversations with our teams, that even though some individuals are openly struggling with the move towards remote working and are quite vocal about it, there are others who will avoid saying anything; perhaps because they are embarrassed or are worried about making a fuss. This kind of thing can potentially lead to those employees leaving the business, in search of a more physical office environment. Even though there will be occasions where you can't avoid staff moving on, it is important to have had those conversations, firstly to see if anything can be done to improve the situation at the time, and secondly to learn from the experience for the future. It is important that we, as business leaders, continue to talk more to our people and to dig deeper into potential worries before they escalate.
In terms of the business landscape for 2022, do you see lots of movement coming up in terms of staff leaving?
As with any industry, people will come and go, that is the norm. Also, there will always be seasonal pressures that entice people to change focus or those who sometimes have a more specific focus on their career and/or where they work and are getting their working experience from. Right now though, things on that front seem to be pretty standard in my view, but I think for those businesses that are retaining good people, that will be down to their internal comms.
What is it making them think about looking for new roles in general i.e. what might trigger this?
The issues that are enticing people to leave their roles, tend to be lack of development opportunities, not being recognised for their contributions, lack of flexibility to improve work-life balance, environmental ethics of the company, battles with salary and remuneration, boredom and simply itchy feet – all of which are key triggers.
Procurement Trends for 2022 & Beyond - Zupa
Keeping Brakes On Top Of Europe's Foodservice Solutions
A fresh, hot cup of success for GRIND
general: hello@zupa.com
support: help@zupa.com


SAN ANTONIO - Registration is now open for local families to invite a Fort Sam Houston soldier into their home for Thanksgiving.
For many young soldiers, Thanksgiving will be their first major holiday away from their family, and for some it can be a lonely time.
Last year, 218 families and 750 soldiers participated in Mission Thanksgiving, which started more than 20 years ago.
"San Antonio is a special city with special people," said 32nd Medical Brigade Commander Col. Charles H. O'Neal. "Their willingness to open their doors to our soldiers, who are all just beginning their military careers, makes it clear why San Antonio is nicknamed 'Military City, USA.' Area families never disappoint, some having participated in this program for over 15 years."
Families interested in hosting a soldier can register online, send an email or call 210-669-7067.
Registration runs through Nov. 1.


Mesón Villa Salada restaurant has spacious terraces surrounded by a beautiful garden with capacity for 500 people. All year round it offers all kinds of celebrations and events such as weddings, birthdays, family dinners, romantic dinners, group dinners, business dinners , end of year dinners, bachelor parties, Christmas meals and New Year's Eve celebration.
Our specialty is grilled meats, paellas, fish and good seafood. We work with the best products on the market.
Coffee bar area: You can choose all kinds of coffees, national and reserve liquors, gin and tonic service, cocktails of great variety and exquisite homemade Sangria Winery: we have a great variety of wines and cavas.
Our cuisine stands out for its typical Mediterranean food, we have special menus, (celiac, vegetarian and vegan, we have a large lounge, large terrace and its outdoor gardens.
You will enjoy a perfect banquet with family and friends for this great day.
THIS IS THE MOST IMPORTANT DAY OF YOUR LIVES.
MENUS carefully studied for this great day, customize your minutes, decoration, floristry, cards, animation for adults and children.
The company has its own lounge, with capacity for 150 people, decorated with touches of flamenco airs, we have large terraces, very spacious place that can accommodate large numbers of customers.
An intimate and cozy atmosphere perfect for an occasion of this importance.
The landscaped area that surrounds it, conditioned according to the will of the couple, not only works as a ceremony space, but also as a reception point for the welcome cocktail or as an aperitif space.


The William M. Plater Civic Engagement Medallion has been established to recognize students who have demonstrated exemplary commitment to their communities during their years as an IUPUI student.
These are one-time service events that focus on a different social issue each month, ranging from hunger, to youth, to water access.
These are one-time service events held at various times throughout the year in which you can volunteer alongside community partners in the Indianapolis community.
Joining a service-based student organization is a great way to meet new friends and to make a difference in our community. Join us for our community service and civic engagement events held throughout the year.
The many Sam H. Jones scholarships offer financial awards and a variety of components to help you grow as a student, community member, and leader.
If you have ideas for service-related programs and events, use grants from IUPUI and other organizations to make your ideas happen.
Presenting a poster is a great opportunity to obtain interesting and valuable feedback for projects, research, or practices from a knowledgeable crowd. You can make connections with other students, faculty, staff, and community leaders.
SLA scholars gain valuable knowledge and hands-on experience in the application of their discipline in unscripted, real-world settings.
Sign up for the weekly e-newsletter that gives you up-to-date information on volunteer opportunities on campus and connects you with various opportunities to serve the Indianapolis community.
Using your federal work-study award at an approved community work-study agency is a great way to gain real-life experience at a nonprofit or government agency.


Read our guidelines for information on social media use for faculty, staff and students at the Albert Einstein College of Medicine.
Looking to set up a social media profile for an Einstein department or organization?
Tweets about "from:@EinsteinMed OR from:@EpgntxEinstein OR from:@EinsteinWell OR from:@WaspSystem"


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Increase sex drive male
WEDNESDAY, Feb. 17, 2016 (HealthDay News) - Testosterone treatment may restore some sexual desire and function in older men whose natural hormone levels have declined, suggest.
But the treatments did not do much to boost vigor or real function in males 65 or older, as much have actually thought, stated co-researcher Dr. Thomas Gill.
"guys who will be experiencing reduced desire or reasonable intercourse and generally are thinking about increasing in those areas, testosterone therapy will be reasonable to take into account, " Gill stated.
But, "your physician probably wouldn't be urged to prescribe testosterone exclusively for issues with real function or entirely for low-energy, predicated on these outcomes, " he included. Gill is a professor of geriatrics and epidemiology at Yale University and director of the Yale Program on the aging process in New Haven, Conn.
The application of testosterone replacement treatment has nearly doubled recently, from 1.3 million patients last year to 2.3 million in 2013, in line with the U.S. Food and Drug Administration.
The outcome of recent trials "really make it possible to put the effects of testosterone in some quantifiable framework, " said Dr. Eric Orwoll, associate dean for medical sciences at Oregon Health & Science University in Portland.
"I would hope this will bring an even more logical approach to your choice, " said Orwoll, who penned an editorial associated the test outcomes. "you cannot decorate testosterone supplementation as a panacea. The consequences were often not present or moderate. This isn't planning to provide countless impetus for folks to just take testosterone."
This new results, posted inside Feb. 18 issue of the New The united kingdomt Journal of Medicine, come out of the Testosterone tests - a federally funded pair of seven clinical studies being performed at 12 internet sites in the usa.
A total of 790 men 65 or older were signed up for the Testosterone tests, and had been assigned to utilize either-or a placebo solution for example year. Every one of the guys had to have as a result of, together with to have a minumum of one health condition caused by their testosterone deficiency.
The three primary clinical tests evaluated the potential good great things about testosterone therapy on sexual function, actual function and vitality. Four various other tests studied the results of testosterone on heart wellness, , mental capabilities and ; those results are going to be posted later on.
Men whom utilized testosterone solution for starters 12 months experienced modest but measurable improvements in sexual intercourse, sexual interest and erectile function, versus people who received a placebo gel, Gill stated.
Testosterone seems to be most readily useful as a treatment for paid down intercourse or need, since there at this time are not any alternative remedies for people issues, Gill added.
But testosterone couldn't outperform medications currently available on the market for the treatment of, such as for instance or, and wouldn't be a proper treatment for erectile problems alone, he said.
Results had been blended when it stumbled on the huge benefits for vitality and real purpose, the analysis authors said.
Testosterone would not notably enhance men's vigor, however some males reported a somewhat much better state of mind much less serious, the studies discovered.
In addition, the failed to notably improve the length of men especially assigned into actual function portion of the tests. However when researchers lumped together the males in all three primary trials, they discovered some benefit - 20.5 percent of testosterone people enhanced their walking length weighed against 12.6 percent of males just who got a placebo.
Based on these outcomes, medical practioners might think about testosterone treatment for males having problems with sexual function, using the knowing that the hormones may potentially boost their vitality and real work as really, Gill proposed.
"you could ponder over it something such as 'value added, ' " he stated.
Alternatively, guys would do better sticking with existing treatments for actual purpose and vigor instead of resorting to testosterone, Gill said.
Lasting security additionally stays a problem with testosterone remedies, Gill and Orwoll said.
Although these studies revealed no significant health risks over per year, health professionals are involved that prolonged usage of testosterone could boost a guy's risk of or heart disease.
In 2015, the FDA issued a warning that testosterone treatments could raise a man's threat of otherwise .
"The trial had been reassuring there weren't major risks over a-year, however some of the results need to be examined over a longer period, " Orwoll explained.
There is also some concern that some males getting testosterone therapy do not actually need it. At this time, approximately 70 % of males who obtain testosterone prescriptions through retail pharmacies are between 40 and 64 years of age, based on the Food And Drug Administration.
"Our outcomes, that have been moderate, really just pertain to men that 65 years or older, " Gill stated. "Younger men are not as likely than older men to have low levels of testosterone entirely due to their age."
Copyright © 2016 HealthDay. All liberties set aside.
SOURCES: Thomas Gill, M.D., teacher, geriatrics and epidemiology, Yale University, and manager, Yale Program on the aging process, brand new Haven, Conn.; Eric Orwoll, M.D., professor and connect dean, medical sciences, Oregon Health & Science University, Portland, Ore.; Feb. 18, 2016,
Source: www.medicinenet.com
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text:
\section{Introduction}

In  \cite {ah} the authors posed the  problem of studying, and possibly classifying, situations like this:
\begin{itemize}
\item [(*)] $C$ is a smooth, projective, complex curve of genus $g$,  $Z$ is an irreducible $r$--dimensional subvariety of a Brill--Noether locus $W^ s_d(C)\subsetneq J^ d(C)$, and $Z$ is stable under translations by the elements of an abelian subvariety $A\subsetneq J(C)$ of dimension $a>0$ (if so, we will say that $Z$ is $A$--\emph{stable}).
\end{itemize}

Actually in \cite {ah} the variety $Z$ \emph{is} the translate of a positive dimensional  proper abelian subvariety of $J(C)$, while the above slightly more general formulation  was given in \cite {df}.

 The motivation for studying  (*) resides, among other things,  in a theorem of Faltings (see \cite {f}) to the effect that if $X$ is an abelian variety defined over a number field $\mathbb K$, and $Z\subsetneq X$ is a subvariety  not containing any translate of a positive dimensional abelian subvariety of $X$, then the number of rational points of $Z$ over $\mathbb K$ is finite. The idea in  \cite {ah} was to apply Faltings' theorem to the $d$--fold symmetric product $C(d)$ of a curve $C$ defined over a number field $\mathbb K$.
If $C$ has no positive dimensional linear series of degree $d$, then $C(d)$ is isomorphic to its Abel--Jacobi image $W_d(C)$ in $J^ d(C)$. Thus $C(d)$ has finitely many rational points over $\mathbb K$ if $W_d(C)$ does not contain any translate of a positive dimensional abelian subvariety of $J(C)$. The suggestion in \cite {ah} is that, if, by contrast, $W_d(C)$ contains the translate of a positive dimensional abelian subvariety of $J(C)$, then $C$ should be \emph{quite special}, e.g., it should admit a map to a curve of lower positive genus (curves of this kind clearly are in situation (*)). This idea was tested in \cite {ah}, where a number of partial results  were proven for low values of $d$.

The problem was taken up in \cite {df}, see also \cite{df-errata},  where, among other things, it is proven that if (*) holds, then $r+a+2s\leqslant d$, and, if in addition $d+r\leqslant g-1$, then $r+a+2s= d$ if and only if:
\begin{itemize}
\item [(a)]  there is a degree 2 morphism $\varphi\colon C\to C'$, with $C'$ a smooth curve of genus $a$, such that
$A=\varphi^*(J(C'))$ and $Z=W_{d-2a-2s}(C)+\varphi^*(J^{a+s}(C'))$.
\end{itemize}

In \cite {df} there is also the following example with $(d,s)=(g-1,0)$:
\begin{itemize}
\item [(b)]  there is an (\'etale) degree 2 morphism $\varphi\colon C\to C'$, with $C'$ a smooth curve of genus $g'=r+1$, $A$ is  the  Prym variety of $\varphi$ and $Z\subset W_{g-1}(C)$ is the connected component of $\varphi_*^{-1}(K_{C'})$ consisting of  divisor classes $D$ with $h^0(\mathcal O_C(D))$ odd, where $\varphi_*\colon J^{g-1}(C)\to J^{g-1}(C')$ is the \emph{norm map}.  One has $Z\cong A$, hence $r=a$.\end{itemize}
One more family of examples is the following (see Corollary \ref {for:ten} below):
\begin{itemize}
\item [(c)]  $C$ is hyperelliptic, there is a degree 2 morphism $\varphi\colon C\to C'$ with $C'$ a smooth curve of genus $a$ such that $g>2a+1$,  $A=\varphi^*(J(C'))$, $0<s<g-1$ and $Z=\varphi^*(J^{a}(C'))+W_{d-2s-2a}(C)+ W^ s_{2s}(C)$ (notice  that $W^s_{2s}(C)$ is a point).
\end{itemize}

The aforementioned result in \cite {df} goes exactly in the direction indicated in \cite {ah}. The unfortunate feature of it is the hypothesis $d+r\leqslant g-1$ which turns out to be \emph{quite strong}. To understand how strong it is, consider the case $(d,s)=(g-1,0)$, which is indeed the crucial one (see \cite [Proposition 3.3] {df} and \S \ref {ssec:GBN} below) and in which Debarre--Fahaloui's theorem is void.

The aim of the present paper is to give the full classification of the cases in which (*) happens and $d=r+a+2s$. What we prove (see Theorem \ref{thm:main} and Corollary \ref {for:ten}) is that if (*) holds then, with no further assumption, either (a) or (b) or (c) occurs.

The idea of the proof is not so different, in principle, from the one proposed in \cite {df} in  the restricted situation considered there. Indeed, one uses the $A$--stability of $Z$ and its maximal dimension  to produce linear series on $C$ which are not birational, in fact  composed with a degree 2 irrational involution. The main tool in  \cite {df}, inspired by \cite {ah}, is a Castelnuovo's type of analysis for  the growth of the dimension of certain linear series.

Our approach also consists  in producing a non birational linear series on $C$, but it is in a sense more direct. We consider  (*) with $(d,s)=(g-1,0)$ and $a+r=g-1$, i.e., the basic case (the others follow from this), in which $Z$ is contained in $W_{g-1}(C)$, which is a translate of the theta divisor $\Theta\subset J(C)$. This
 immediately produces, restricting to $Z$  the Gauss map of $\Theta$, a base point free sublinear series $L$ of dimension $r$ of the canonical series of $C$. It turns out that $Z$ is birational to an irreducible component of the variety $C(g-1,L)\subset C(g-1)$ consisting of all  effective divisors of $C$ of degree $g-1$ contained in some divisor of $L$. The $A$--stability of $Z$ implies that $C(g-1,L)$ has some other component besides the one birational to $Z$, and this forces $L$ to be non--birational. Once one knows this, a (rather subtle) analysis of the map determined by $L$ and of its image leads to the conclusion.

As for the contents of the paper, \S \ref {sec:symm} is devoted to a few general facts about the varieties $C(k,L)\subset C(k)$ of effective  divisors of degree $k$ contained in a divisor of a linear series $L$ on a curve $C$. These varieties, as we said,  play a crucial role in our analysis. Section \ref {sec:absub} is devoted to the proof of our main result.

One final word about our own interest in this problem, which is quite different from the motivation of \cite {ah, df}. It is in fact  related to the study of irregular surfaces $S$ of general type, where situation (*) presents itself in a rather natural way. For example, let $C\subset S$ be a smooth,  irreducible curve, and assume that $C$ corresponds to the general point of an  irreducible component $\mathcal C$ of the Hilbert scheme of curves on $S$ which dominates $\Pic^ 0(S)$.  Recall that there is also only one irreducible component $\mathcal K$ of the Hilbert scheme of curves algebraically equivalent to canonical curves on $S$ which dominates $\Pic^ 0(S)$ (this is called the \emph{main paracanonical system}). The curves of $\mathcal C$ cut out on $C$ divisors which are residual, with respect to $\vert K_C\vert$, of divisors cut out by curves in  $\mathcal K$. Consider now the one of the two systems  $\mathcal C$ and $\mathcal K$ whose curves cut on $C$ divisors of minimal degree $d=\min\{C^ 2, K_S\cdot C\}$, and denote by $s$ the dimension of the general fibre of this system over $\Pic^ 0(S)$. Thus the image of
the  natural  map $\Pic^ 0(S)\to  J^d(C)$ defined by $\eta\mapsto \mathcal O_C(C+\eta)$  (or by $\eta\mapsto \mathcal O_C(K_S+\eta)$) is a $q$--dimensional abelian variety  contained in $W^s_d(C)$, which is what happens in (*).
 Thus  understanding  (*) would provide us with the understanding of (most) curves on irregular surfaces.

The results in this paper, even if restricted to the very special case of (*)  in which $Z$ has maximal dimension, turn out to be useful in surface theory. For example, if $S$ is a minimal irregular  surface of general type, then  $K^ 2_S\geqslant 2p_g$ (see \cite {debarre}). Using the results in this paper we are able to classify irregular minimal surfaces for which $K^ 2_S= 2p_g$.
This classification  is given   in \cite{cmp}.

\section{Linear series on curves and related subvarieties of symmetric products}\label{sec:symm}

\subsection{Generalities} Let $C$ be  a smooth, projective, irreducible   curve of genus $g$. For an integer $k\geqslant 0$, we denote by $C(k)$ the $k$-th symmetric product of $C$ (by convention, $C(0)$ is a point).

 Let $L$ be a base point free $g^r_d$ on $C$. We denote the corresponding morphism by $\phi_L\colon C\to \bar C \subset \mathbb{P}^ r$ and by  $C'$  the normalization of  $\bar C$. We let $f\colon C\to C'$ be the induced morphism and  $L'$ the linear series on $C'$ such that $L=f^ *(L')$. We set $\deg(f)=\nu\ge 1$, so that $d=\delta \nu$, with $\delta=\deg(\bar C)$. We say that $L$ is \emph{birational} if  $\phi_L\colon C\to \bar C$ is birational.

 Let $k\leqslant d$ be a positive integer. We consider the incidence correspondence
 $$\mathcal C{(k,L)} =\{(D,H)\in C(k)\times L \ | \ D\leqslant H\}$$
 with projections $p_i$ (with $1\leqslant i\leqslant 2$) to the first and second factor. Set
 $C(k,L)=p_1(\mathcal C(k,L))$, which has a natural scheme structure (see \cite {c83} or \cite [p. 341]  {acgh}).
  Note the isomorphism
 \[\mathfrak s\colon  (D,H)\in \mathcal C{(k,L)}\to (H-D,H)\in \mathcal C{(d-k,L)}.\]

   \begin{lem}\label{lem:irreducible}
    \begin{inparaenum}
   \item If $k\leqslant r$, then $C(k,L)=C(k)$.\\
   \item If $k\geqslant r$, then $C(k,L)$ has pure dimension $r$. If $V$ is an irreducible component of
   $C(k,L)$, there is a unique component $\mathcal V$ of $\mathcal C(k,L)$ dominating $V$ via $p_1$ and $p_1$ induces a birational morphism
   of $\mathcal V$ onto $V$. Finally $\mathcal V$ dominates $L$ via $p_2$.\\
   \item If $\min\{k,d-k\}\geqslant r$ the isomorphism $\mathfrak s$ induces a map $\mathfrak r\colon  C(k,L)\dasharrow C(d-k,L)$, which is componentwise birational.\\
      \item If $L$ is birational, then $C(k,L)$ is irreducible.
   \end{inparaenum}
 \end{lem}
 \begin{proof}
Part (i) is clear. The dimensionality assertion in (ii) follows from \cite[\S~1]{c83} or  \cite [Lemma (3.2), p. 342]  {acgh}. The rest of (ii) follows  from these facts: the fibres of $p_1$ are isomorphic to linear subseries of $L$ and $p_2$ is finite, so no component of $\mathcal C(k,L)$ has dimension larger than $r$. Part (iii) follows from (ii). Part (iv) follows from the Uniform Position Theorem (see \cite [p. 112] {acgh}). \end{proof}

 Next we look at the case  $L$ non--birational and $k\geqslant r$.
Consider the induced  finite morphism $f_*\colon C(k,L)\to C'(k)$.
For each partition $\underline m=(m_1, \dots m_s)$ of $k$ with  $\delta \geqslant s\geqslant r$ and $1\leqslant  m_1\leqslant m_2\leqslant \dots \leqslant m_{s}\leqslant \nu$, we denote by $C( \underline m, L)$ the closure in $C(k, L)$ of the inverse image via $f_*$  of the set of  divisors of the form $m_1y_1+\dots +m_sy_s$, with $y_1+\dots+ y_s$ a reduced divisor
in $C'(s,L')$.  To denote a partition $\underline m$ as above we may use the \emph{exponential notation} $\underline m=(1^{\mu_1},\ldots, \nu^ {\mu_\nu})$,
meaning that $i$ is repeated $\mu_i$ times, with $1\leqslant i\leqslant \nu$, and we  may omit $i^ {\mu_i}$ if $\mu_i=0$.  Note that $\sum_{i=1}^\nu  \mu_i=s\leqslant \delta$.
Set $\mu_0:=\delta-s$.  Then $\underline m^ c=(1^{\mu_{\nu-1}},\ldots, (\nu-1)^ {\mu_1}, \nu ^ {\mu_0})$ is a partition of $d-k$ which we call the \emph{complementary partition} of $\underline m$.

\begin{lem}\label{lem:reducible} In the above set up each irreducible component of  $C(\underline m, L)$ has dimension $r$, hence it is an irreducible component of  $C(k,L)$ and all irreducible components of $C(k,L)$ are of this type.  \end{lem}
 \begin{proof} Let $V$ be an irreducible component of $C(k,L)$: by  Lemma \ref{lem:irreducible}  there is a unique   irreducible component $\mathcal V$ of $\mathcal C(k,L)$ dominating it and $p_1|_{\mathcal V}\colon \mathcal V\to V$ is birational.
  Let $D\in V$ be a general point and let $(D,H)$ be the unique point of $\mathcal V$ mapping to $D$ via $p_1$, so that $H$ is a general divisor in $L$  (cf. Lemma \ref {lem:irreducible}, (ii)). Hence
 $H$ consists of $\delta$ distinct fibres $F_1,\ldots, F_\delta$ of $f$, each being a reduced divisor of degree $\nu$ on  $C$. Then $D$ consists of $m_i$ points in $F_i$, for $1\leqslant i\leqslant s\leqslant \delta$, where we may assume $1\leqslant  m_1\leqslant m_2\leqslant \dots \leqslant m_{s}\leqslant \nu$. Moreover, since $\dim(V)=\dim(\mathcal V)=r$ and $p_2$ is finite, one has $s\geqslant r$.  Hence $D\in C(\underline m,L)$, with $\underline m=(m_1, \dots m_s)$, i.e.  $V\subseteq C(\underline m,L)\subseteq C(k,L)$ hence $V$ is a component of $C(\underline m,L)$.

The above considerations and Lemma  \ref {lem:irreducible}, (ii), applied to $C'(s,L')$, imply that the image of $C(\underline m,L)$ in $C'(k)$ has dimension $r$, so  each component of $C(\underline m,L)$ has  dimension $r$. \end{proof}

\subsection{Abel--Jacobi images}

 We assume from now on  that $C$ has  genus $g>0$.
For an integer $k$, we denote  by $J^k(C)\subset \Pic(C)$ the set of linear equivalence classes of divisors of degree $k$ on $C$.  So $J(C):=J^ 0(C)$ is the Jacobian of $C$, which is a principally polarised abelian variety whose theta divisor class we denote by $\Theta_C$, or simply by $\Theta$.

The abelian variety $J(C)$  acts via translation on $J^ k(C)$ for all $k$.
If $X\subseteq J^ k(C)$ and $Y\subseteq J(C)$, we say that $X$ is \emph{$Y$--stable}, if for all $x\in X$ and for all $y\in Y$, one has $x+y\in X$.

For all integers $k$, fixing the class of a divisor of degree $k$  determines  an  isomorphism $J^ k(C)\cong J(C)$. Given a subvariety $V$ of  $J^ k(C)$, one says that it \emph{generates} $J^ k(C)$ if the image of $V$ via  one of the above isomorphisms generates  $J(C)$ as an abelian variety. This  definition     does not depend on the choice of the isomorphism  $J^ k(C)\cong J(C)$.

For every $k\geqslant 1$, we denote by $j_k \colon C(k)  \to J^ k(C)\cong J(C)$ (or simply by $j$) the Abel--Jacobi map. We denote  by $W^ s_k(C)$ the subscheme of $J^ k(C)$ corresponding to classes of divisors $D$ such that $h^ 0(\mathcal O_C(D))\geqslant s+1$ (these are the so--called {\em Brill--Noether loci}). One sets $W_k(C):=W^ 0_k(C)={\rm Im}(j_k)$ and $W_{g-1}(C)$ maps to  a theta divisor of $J(C)$, so we may abuse notation and write $W_{g-1}(C)=\Theta_C$.

We denote by $\Gamma_C(k, L)$ [resp. $\Gamma_C(\underline m, L)$]
the image in $W_k(C)$ of $C(k,L)$ [resp. $C(\underline m,L)$] (we may drop the subscript $C$ if there is no danger of confusion). The expected dimension of $\Gamma(k,L)$ is $\min\{r,g\}$
(by dimension of a scheme we mean the maximum of the dimensions of its components).

We set $\rho_C(k,L):= \dim(\Gamma_C(k,L))$ (simply denoted by $\rho(k,L)$ or by $\rho$ if no confusion arises). By Lemma \ref {lem:irreducible}, (i), one has:
\begin{equation}\label{eq:aiuto}
{\rm if} \,\,\, k\leqslant r \,\, \, {\rm then}\,\,\, \Gamma(k,L)=W_k(C),\,\,\, {\rm hence}\,\,\, \rho=\min\{k,g\}
\end{equation}
So we will  consider next the case $k>r$, in which  $\rho\leqslant \dim(C(k,L))=r$, by Lemma  \ref {lem:irreducible}, (ii).
Then the class $c (k,L)$ of $C(k, L)$ in the Chow ring of $C(k)$ is computed in \cite[ Lemma VIII.3.2] {acgh}. If  $x$ is the class of $C(k-1)\subset C(k)$ and $\theta:=j^ *(\Theta)$, one has
\begin{equation}\label{eq:class}
c(k,L)=\sum_{s=0}^{k-r} {{d-g-r}\choose{s}}\frac {x^s\theta^{k-r-s}}{(k-r-s)!} \ .
\end{equation}

\begin{lem}\label{class} Assume $k>r$  and $d-g-r \geqslant 0$. Then:\\
\begin{inparaenum}[(i)]
\item if $k-g\leqslant \min\{k-r, d-g-r\}$ one has $\rho=r\leqslant g$;\\
\item if $k-g\geqslant \min\{k-r, d-g-r\}=k-r$, one has $\rho=g\leqslant r$;\\
\item if $k-g\geqslant \min\{k-r, d-g-r\}=d-g-r$, one has $\rho=d-k\leqslant \min\{r,g\}$.
\end{inparaenum}
\end{lem}

\begin{proof} Note that $x^ s$ is the class of $C(k-s)\subset C(k)$, for $1\leqslant s\leqslant k$. Applying  the projection formula (cf. \cite{fulton}, Example 8.1.7)  to \eqref {eq:class}, we find the class $\gamma(k,L)$ of $\Gamma(k,L)$
\begin{equation}\label{eq:class2}
\gamma(k,L)=\sum_{s=0}^{k-r} {{d-g-r}\choose{s}}\frac {w_{k-s}\Theta^{k-r-s}}{(k-r-s)!} \
\end{equation}
where $w_i$ is the class of $W_i(C)$ for any $i\geqslant 0$. By Poincar\'e's formula (cf. \cite[\S 11.2]{birk}) one has
\[
w_{k-s}\Theta^{k-r-s}=
\begin{cases}
\Theta^{k-r-s}, & \text{if $k-g\geqslant s\geqslant 0$,} \\
\frac {\Theta^{g-r}}{(g-k+s)!}, & \text{if $\max\{0,k-g\}\leqslant s\leqslant  \min\{k-r, d-g-r\}$},
\end{cases}
\]
whence the assertion follows. \end{proof}

\begin{lem}\label{lem:class2}
\begin{inparaenum}[(i)]
\item If $\rho=g$ then $g\leqslant \min\{k,r\}$;\\
\item  if $r\geqslant k\geqslant g$, then $\rho=g$;\\
\item  if $k> r\geqslant g$ and $d\geqslant k+g$ then $\rho=g$;\\
\item  $\rho=0$ if and only if $k=d$.
\end{inparaenum}
\end{lem}

\begin{proof} Parts (i) and (ii) follow from Lemma \ref {lem:irreducible}.

(iii) In \eqref {eq:class2} one has the summand corresponding to the index $s=k-r>0$, which is $\Theta^0$ with the positive coefficient ${{d-g-r}\choose{k-r}}$, and no other summand in  \eqref {eq:class2} cancels it.

(iv) If $k=d$ then $C(k,L)=L$ and clearly $\rho=0$. Conversely, if $\rho=0$ then in \eqref {eq:class2} the term $\Theta^ g$ has to appear with non--zero coefficient and no other term $\Theta^ i$  with $0\leqslant i<g$  appears with non--zero coefficient.  By looking at the proof of Lemma \ref {class}, we see that  the summand $\Theta^ g$ appears in \eqref  {eq:class2} only if $0\leqslant s=k-r-g$. Then $d\geqslant k\geqslant r+g$. So we may apply Lemma \ref {class}, and conclude that $\rho=0$ occurs only in case (iii), if $k=d$. \end{proof}

\begin{lem}\label{lem:genera}
Let $A\subseteq J(C)$ be an  abelian subvariety of dimension $a$ and let $p\colon J^k(C)\to J':=J(C)/A$ be the map  obtained by composing an isomorphism $J^k(C)\cong J(C)$ with the quotient map $J(C)\to J'$. Then
$$\dim (p(\Gamma(k,L)))=\min\{g-a, \rho\}. $$

\end{lem}
\begin{proof}
If $\rho=g$ the statement is obvious, hence we assume $\rho<g$.

Consider first the case $k>r$.  Assume  by
 contradiction that $\dim (p(\Gamma(k,L))<\min\{\rho, g-a\}$.
Let $\xi$ be the class of the pull back to $J(C)$ of an ample line bundle of $J'$. We have $\bar \gamma(k,L)\xi^{\rho}=0$, where $\bar\gamma(k,L)$ is the $\rho$--dimensional part of $\gamma(k,L)$.
By \eqref {eq:class2} one has
$\bar \gamma(k,L)= \alpha \Theta^ {g-\rho}$,
where $\alpha\in \mathbb Q$ is positive because $\Gamma(k, L)$ is an effective non--zero cycle of dimension $\rho$.
Hence $\bar \gamma(k,L)\xi^{\rho}=\alpha  \Theta^ {g-\rho}\xi^{\rho}>0$, because $\Theta$ is ample. Thus we have a contradiction.

If $k\leqslant r$, then $\Gamma(k,L)=W_k(C)$, $\rho=k$ and  $\gamma(k,L)$  is again a rational multiple of $\Theta^ {g-k}$ (by Poincar\'e's formula), so the proof proceeds as above. \end{proof}

\begin{cor} \label {cor:gen} If $A\subseteq J(C)$ is an  abelian subvariety of dimension $a>0$ and $\Gamma(k, L)$ is $A$--stable, then the restriction of $p$ to $\Gamma(k,L)$ is surjective onto $J'=J(C)/A$, hence $\Gamma(k, L)=J^ k(C)=W_k(C)$, i.e., $\rho=g$.
\end{cor}

\begin{cor} \label {cor:gen2} If $d-1 \geqslant k\geqslant 1$, then $\Gamma(k,L)$ generates $J^ k(C)$.
\end{cor}
\begin{proof}  If $k\leqslant r$ then $\Gamma(k,L)=W_k(C)$ and the assertion is clear. Assume $k>r$.
By Lemma \ref {lem:genera}, $\Gamma(k,L)$ generates $J^ k(C)$ as soon as $\rho>0$, which is the case by Lemma \ref {lem:class2}, (iv). \end{proof}

\subsection{A useful lemma}

Let  $L$ be  a base point free $g^ 1_d$, let  $\phi_L\colon  C\to \mathbb{P}^ 1$ be the corresponding map and denote by $G_L$ the Galois group of $\phi_L$.
\begin{lem} \label{lem:tetra}
If  $L$ is a base point free  $g^1_4$, then one of the following occurs:\\
\begin{inparaenum}
\item[(a)]  $C(2,L)$ is irreducible.\\
\item[(b)]  $C(2,L)$ has two components.  This  occurs if and only if  $G_L\cong \mathbb{Z}_2, \,\, \mathbb{Z}_4$.\\
\item[(c)]  $C(2,L)$ has 3 components. This occurs if and only if  $G_L\cong \mathbb Z_2^2$.
\end{inparaenum}
\end{lem}
\begin{proof}
We argue as in the proof  of \cite[Lemma 12.7.1]{birk}. Let $\Delta\subset \mathbb P^1$ be the set of critical values of $\phi_L\colon  C\to \mathbb P^ 1$,  let $\rho\colon \pi_1(\mathbb P^1\setminus \Delta)\to \mathfrak S_4$ be the monodromy representation and let $\Sigma:={\rm Im}(\rho)$. The group $\Sigma$ acts transitively on $\mathbb I_4:=\{1,2,3,4\}$ (identified with the general divisor $x_1+\ldots+x_4\in L$)
and its order $s$  is divisible by 4. The irreducible components of  $C(2,L)$ are in 1--to--1 correspondence with the $\Sigma$--orbits of the order two subsets of $\mathbb I_4$.
If $\Sigma$ contains an element of order 3, then $s$ is divisible by 12.  Hence $\mathfrak A_4\subseteq \Sigma$, thus the action is transitive, $C(2,L)$ is irreducible and  (a)  holds.

Assume  $\Sigma$ is a $2$--group. If $s=8$ then $\Sigma$ is a $2$--Sylow subgroup of $\mathfrak S_4$, hence $\Sigma$ is  the dihedral group $D_4$. Then  $C$ is obtained from a $\Sigma$--cover $C_0\to \mathbb P^1$ by moding out
 by a reflection $\sigma\in \Sigma$. In the $\Sigma$-action  on $\mathbb I_4$ we may assume that an  element of order 4 acts by sending
 $i$ to ${i+1}$  modulo 4, for $1\leqslant i\leqslant 4$.  So the order 2 element in the center of  $\Sigma$ induces the involution $\iota$ of $C$ that maps $i$ to ${i+2}$ modulo 4, for $1\leqslant i\leqslant 4$, and $\iota$ generates $G_L\cong \mathbb{Z}_2$.  There are
two orbits for the $\Sigma$--action on the set of order two subsets of $\mathbb I_4$: one of order 2 given by $\{\{1,3\}, \{2,4\}\}$, the other of order 4 given by $\{ \{i,i+1\}, \,\, \text {for}\,\, 1\leqslant i\leqslant 4\}$ (here $i$ is taken modulo $4$). These orbits  respectively correspond to  two components
$E_1, E_2$ of $C(2,L)$ and we are in case (b).

Assume  $s=4$. Then $\phi_L$ is Galois with  $G_L=\Sigma$. If $G_L\cong \mathbb Z_4$, then the $\Sigma$--orbits on the set of order two subsets of $\mathbb I_4$ are as in the previous case.
If $G_L\cong \mathbb Z_2^2$, then
one has $G_L=\{\Id, (12)(34), (13)(24),(14)(32)\}$. There are then three orbits, corresponding to three components $E_1,E_2,E_3$.\end{proof}

One can be more precise about the components $E_i$ of $C(2,L)$ in Lemma \ref {lem:tetra}, whose geometric genera we denote by $g_i$ (with $1\leqslant i\leqslant 2+\epsilon$, and $\epsilon =0$ in case (b), $\epsilon =1$ in case (c)).

\begin{lem}\label {lem:tetra2} Same setting and notation as in Lemma \ref {lem:tetra} and its proof.
Then:\\
\begin{inparaenum}[(i)]
\item each component of $C(2,L)$ maps birationally to its image in $J^ 2(C)$ unless $C$ is hyperelliptic and $L$ is  composed with the hyperelliptic involution $\mathcal L$: in this case one of the components of $\Gamma(2,L)$ is $\mathcal L=C(2,\mathcal L)$, which is contracted to a point in $J^ 2(C)$;\\
\item in case  (b) one has  $E_1\cong C/\iota$ (where $\iota$ is the non--trivial involution in $G_L$), which is hyperelliptic, and the abelian subvariety of $J^2(C)\cong J(C)$ generated by $j(E_1)$ is  the $\iota$--invariant part of $J(C)$.  Moreover  $2g_2\geqslant g$;\\
\item In case (c) one has $E_i:=C/\iota_i$ where $\iota_i$ are the three nonzero elements of $G_L$, for $1\leqslant i \leqslant 3$.
\end{inparaenum}
\end{lem}

\begin{proof} We prove the only non--trivial assertion, i.e., $2g_2\geqslant g$  in part (ii).

First assume $G_L=\mathbb Z_4=\langle \rho\rangle$.
Consider in $C(2)$ the curves $E_1=\{P+\rho^2(P)\ |\ P\in C\}$ and  $E_2=\{P+\rho (P)\ |\ P\in C\}$.   One has  $C(2,L)=E_1\cup E_2$ and $E_1\cong C/\rho^2$.
The curve $E_2$ is the image in $C(2)$ of the graph of $\rho$, so  $g_2=g$.

Suppose now $G_L=\mathbb Z_2$ and $\Sigma=D_4$.  Recall that $C$ is obtained from a $D_4$--Galois cover $f\colon  C_0\to \mathbb P^1$ by moding out
 by a reflection $\sigma\in D_4$ (see the proof of Lemma \ref {lem:tetra}). Denote by $g_0$ the genus of $C_0$. Let $\rho\in D_4$ be an element  of order 4, so that $D_4=\langle \sigma,\rho\rangle$. Let $n$ be the number of points of $C_0$ fixed by $\sigma$, $n'$ the number of points fixed by $\sigma \rho$,  $m$ the number of points fixed by $\rho$ and $m+\varepsilon$ the number of points fixed by $\rho^2$.  The Hurwitz formula, applied to $C_0\to \mathbb P^ 1$ and to $C_0\to C=C_0/\sigma$, gives
\begin{equation}\label{eq:gC}
g_0=\frac 32m+n+n'+\frac \varepsilon 2-7, \,\,\,\, {\rm and} \,\,\,\,  g= \frac {n'} 2+\frac n 4+\frac \varepsilon 4-\frac 34 m-3.
\end{equation}

\begin{claim}\label{lem:rs}
\begin{inparaenum}
\item $m$, $n$, $n'$ and $\varepsilon$  are even.\\
\item $n+m\equiv n'+m\equiv n+n'\equiv 0 \mod 4$, and at most one among $m,n,n'$ can be 0.
\end{inparaenum}
\end{claim}
\begin{proof} [Proof of the Claim]  (i) The numbers $n, n'$ and $m+\varepsilon$ are even because $\sigma, \sigma \rho$ and $\sigma \rho^ 2$ are involutions. If $P\in C_0$ is fixed by $\rho$, then $\sigma(P)$ is also fixed by $\rho$. Since  the stabilizer of any point is cyclic, then $\sigma(P)\neq P$. This implies that  $m$ is even.

(ii) Consider the $\mathbb Z_2^ 2$--cover  $D:=C_0/\rho^2 \to\mathbb P^1$. The  cardinalities  of the images in $\mathbb P^1$ of the fixed  loci of the three involutions are  $n/2$, $n'/2$ and $m/2$.
Indeed, denote by $\gamma_1$ [resp. by $\gamma_2$]  the image  of $\sigma$ [resp.  of $\rho$]  in $D_4/\rho^2\cong \mathbb Z_2^2$. Let $Q\in \mathbb P^1$ be a branch point whose preimage in $D$  is fixed   by   $\gamma_1$. Then the preimage of $Q$ in $C_0$  consists of 4 points,  two of which fixed by $\sigma$ and   two  by $\sigma\rho^2$, so the number of such points $Q$ is $(2n)/4=n/2$.  Similarly, the image  in $\mathbb P^1$ of the set of points of $D$  fixed by $\gamma_2$ has cardinality $m/2$ and the image of the set of points fixed by $\gamma_1\gamma_2$ has cardinality $n'/2$.
Hurwitz formula for $D_1:=D/\gamma_1\to \mathbb P^1$ gives
 \[ 2g(D_1)-2=\frac m2+\frac {n'}2-4,
 \]
hence $m+n'>0$ is divisible by 4. Similarly  $n+n'$ and $m+n$ are positive and  divisible by 4.
\end{proof}

We compute now the ramification of $f\colon  C_0\to \mathbb P^ 1$ and of  $L$. Write the $D_4$--orbit of $P\in C_0$ general as
\begin{equation}\label{eq:fibra}
\begin{array}{llll}
P& \rho (P)&\rho^2(P)&\rho^3(P)\\
\sigma (P)& \sigma \rho (P)&\sigma \rho^2(P)&\sigma \rho^3(P).
\end{array}
\end{equation}
Denote by $Q_1,\dots, Q_4$ the images in $C$ of  the points in the first row (or, what is the same, in the second row) of \eqref {eq:fibra}. The singular fibers of $f$ occur when $P$ has non trivial stabilizer, i.e., when:\\
\begin{inparaenum} [$\bullet$]
\item  $P\in C_0$ is  fixed by $\rho$. The fiber of $f$ is  $4(P+\sigma(P))$ and the corresponding divisor of
$L$  is  $4Q_1$. There are $m/2$ divisors of $L$ of this type;\\
\item  $P$ is fixed by $\rho^2$ but not by $\rho$. The fiber of $f$ is  $2(P+\sigma(P)+\rho(P)+\sigma\rho(P))$.
Then $Q_1=Q_3$,  $Q_2=Q_4$, and the corresponding divisor of $L$ is $2(Q_1+Q_2)$. There are $\varepsilon/4$ such divisors;\\
\item  $P$ is fixed by $\sigma$.   The fibre of $f$ is $2(P+\rho(P)+\rho^ 2(P)+\rho^ 3(P))$, and
$Q_2=Q_4$, while $Q_1,  Q_2$ and $Q_3$  are distinct, so the corresponding divisor of $L$ is $Q_1+2Q_2+Q_3$ and there are $n/2$ such divisors; \\
\item $P$ is fixed by $\sigma\rho^2$. This is the same as the previous case;\\
\item $P$ is fixed by $\sigma \rho$.  The fibre of $f$ is again $2(P+\rho(P)+\rho^ 2(P)+\rho^ 3(P))$, and
\ $Q_1=Q_2$, $Q_3=Q_4$, so the corresponding divisor of $L$ is $2Q_1+2Q_3$ and there are $n'/2$ such fibers; \\
\item $P$ is fixed by $\sigma\rho^3$. This  is the same as the previous case.
\end{inparaenum}

Denote by $\iota$ the involution of $C$ induced by $\rho^2$. Then  $C(2,L)$ is the union of $E_1=\{P+\iota (P)\ |\ P\in C\}=C/\iota$ and of the irreducible curve $E_2$. Keeping the above notation, if $Q_1+\cdots+ Q_4$ is the general divisor of $L$, then $E_1$ is described by the divisors $Q_1+Q_3, Q_2+Q_4$ and $E_3$ by  $Q_1+Q_2, Q_1+Q_4, Q_2+Q_3, Q_3+Q_4$.

To compute $g_2$,
define a map $\phi\colon C_0\to E_2$  by sending $P\in C_0$ to the image via $C_0\to C$ of the divisor
$P+\rho(P)$, i.e.,  $Q_1+Q_2\in E_2$.

\begin{claim} \label{lem:gE_2}
\begin{inparaenum}
\item $\deg(\phi)=2$ and  $E_2$ is birational to $C_0/\sigma\rho$;\\
\item  $g_2=\frac 34 m+\frac n 2+\frac {n'}4+\frac \varepsilon 4-3$.
\end{inparaenum}
\end{claim}
\begin{proof} [Proof of the Claim]
(i)  Let $Q+Q'\in E_2$ be a general point and let  $P, \sigma (P)$ [resp. $P', \sigma (P')$] be the preimages of $Q$ [resp. of $Q'$] on $C_0$. Since $Q+Q'$ is of the form
$Q_1+Q_2, Q_1+Q_4, Q_2+Q_3$, or  $Q_3+Q_4$,
we may assume that $P'=\rho (P)$, so that $\psi^ {-1}(Q+Q')=\{P, \sigma\rho(P)\}$.

Part (ii) follows by applying Hurwitz formula.
\end{proof}

Finally, suppose by contradiction  that $2g_2\leqslant g-1$.
Then, by (ii) of Claim \ref{lem:gE_2} and by \eqref{eq:gC}, we would have $3(m+n)+\varepsilon\le 8$, hence $m+n\le 2$, contradicting (ii) of Claim \ref{lem:rs}.\end{proof}

\section{Abelian subvarieties of Brill--Noether loci}\label{sec:absub}

As in  \cite{ah,df}, we consider $Z\subseteq W^s_d(C)\subsetneq J^ d(C)$ an irreducible  $A$--stable variety of dimension $r$,  with $A\subsetneq J(C)$ an abelian subvariety of dimension $a>0$. Note that $r\geqslant a$, with equality if and only if $Z\cong A$. Moreover, since $W^s_d(C)\subsetneq J^ d(C)$,  the general linear series $L\in Z$ is special,  thus $s>d-g$ and $d\geqslant 2s$ by Clifford's theorem. From \cite[Proposition 3.3]{df},  we have
\begin{equation}\label{eq:DF}
r+a+2s\leqslant d.
\end{equation}

 In this section, we  classify the cases in which equality holds in \eqref {eq:DF}, thus improving the partial results in \cite {df} on this subject.  Note that if equality holds in  \eqref {eq:DF}, then  $Z\not \subseteq  W^{s+1}_d(C)$ and  $A$ is a maximal abelian subvariety of $J(C)$ such that $Z$ is $A$--stable.

\subsection{The Theta divisor case}  As in \cite {df}, we first consider the case $(d,s)=(g-1,0)$,
i.e.,  $Z\subset W_{g-1}(C)=\Theta$ is an irreducible $A$--stable variety of dimension $r=g-1-a$.
Then $Z\not\subseteq  W^ 1_{g-1}(C)={\rm Sing}(\Theta)$, i.e., $Z$ has a non--empty intersection with
$\Theta_{\rm sm}:=\Theta - {\rm Sing}(\Theta)$.

\begin{thm}\label{thm:main}
Let $C$ be a curve of genus $g$. Let $A\subsetneq J(C)$ be an abelian variety of dimension $a>0$ and $Z\subset \Theta$ an irreducible,  $A$--stable variety of dimension $r=g-1-a$.
Then there is a degree 2 morphism $\varphi\colon C\to C'$, with $C'$ smooth of genus $g'$, such that one of the following occurs:\\
\begin{inparaenum}
\item[(a)] $g'=a$, $A=\varphi^*(J(C'))$ and $Z=W_{g-1-2a}(C)+\varphi^*(J^a(C'))$;\\
\item[(b)] $g'=r+1$, $\varphi$ is \'etale, $A$ is  the  Prym variety of $\varphi$ and $Z\subset W_{g-1}(C)$ is the connected component of $\varphi_*^{-1}(K_{C'})$ consisting of  divisor classes $D$ with $h^0(\mathcal O_C(D))$ odd, where $\varphi_*\colon J^{g-1}(C)\to J^{g-1}(C')$ is the {norm map}.
\end{inparaenum}
In particular, $Z\cong A$ is an abelian variety if and only if either we are in case (a) and $g=2a+1$, or  in case (b).
\end{thm}

\begin{rem} Cases (a) and (b) of Theorem \ref{thm:main} are not mutually exclusive. Indeed,  if the curve $C$ in case (b) is hyperelliptic, then the Abel-Prym map $C\to A$  induces a $2$-to-$1$ map $\psi\colon C\to D$, where $D$ is  a smooth curve  embedded into    $A\cong J(D)$ by the Abel-Jacobi map. One can  check that $A=\psi^*(J^r(D))$, namely  this is also an instance  of   case (a) of Theorem \ref{thm:main}.
\end{rem}

The proof of Theorem \ref{thm:main}  requires various preliminary lemmas.  First, recall that the tangent space to $J(C)$ at $0$ can be identified with $H^ 1(\mathcal O_C)\cong H^0(K_C)^*$.
Denote by $T\subseteq H^0(K_C)^*$ the tangent space to $A$ at $0$ and  by $L$ the linear series $\mathbb P(T^{\perp})\subseteq |K_C|$.  One has $\dim(L)=g-1-a=r$.

Since $Z\cap \Theta_{\rm sm}\not= \emptyset$,   the Gauss map of $\Theta$ restricts to a  rational map $\gamma \colon Z \dasharrow  \mathbb P:=\vert K_C\vert$.

\begin{lem}\label{lem:im} One has $\gamma(Z)=L$.
\end{lem}

 \begin{proof}  Since $Z$ is $A$--stable, one has $\gamma(Z)\subseteq L$.  A point of $\Theta_{\rm sm}$ can be identified with a divisor $D$ of degree $g-1$ such that $h^ 0(\mathcal O_C(D))=h^ 0(\mathcal O_C(K_C-D))=1$.
 The Gauss map sends $D\in \Theta_{\rm sm}$ to the unique divisor of $\vert K_C\vert$ containing $D$.  Then $\gamma^ {-1}(\gamma(D))$ is finite. Hence $\dim(\gamma(Z))=\dim(Z)=r=\dim (L)$. The assertion follows. \end{proof}

 As in  \cite {df}, formula \eqref {eq:DF} for $(d,s)=(g-1,0)$ follows from the argument in the proof of Lemma \ref {lem:im}, and  the general case follows from this (see \S \ref {ssec:GBN} below).

Note
the birational involution $\sigma\colon  \Theta\dasharrow \Theta$,  defined on $\Theta_{\rm sm}$, sending a divisor $D\in \Theta_{\rm sm}$ to the unique effective divisor $D'\in \vert K_C-D\vert$. Then $\sigma$ restricts on  $Z$ to a birational map (still denoted by $\sigma$)  onto its image $Z'$, which is also $A$-stable.

\begin{lem}\label{lem:free}
 The linear series $L$ is base point free.
\end{lem}
\begin{proof}
Suppose  $P\in C$ is a base point of $L$.
If every $D\in Z$ contains $P$, then the map $D\to D-P$ defines an injection  $Z\hookrightarrow W_{g-2}(C)$, contradicting   \eqref {eq:DF}.
If  the general $D\in Z$ does not contain $P$, then $\sigma(D)$ contains $P$ for  $D\in Z$  general, and we can apply the previous  argument to  $Z'$.
\end{proof}

\begin{lem}\label{lem:nobir} One has:\\
\begin{inparaenum}
\item $Z$ is a component of $\Gamma(g-1, L)$.\\
\item $Z\subsetneq \Gamma(g-1, L)$.\\
\item  The linear series $L$ is not birational.
\end{inparaenum}
\end{lem}
\begin{proof} By Lemma \ref {lem:im}, we have $Z\subseteq\Gamma(g-1, L)$ and the dimension of $\Gamma(g-1, L)$ is equal to $r$ (by Lemma \ref {class}).  So (i) holds. If $Z= \Gamma(g-1, L)$, then $\Gamma(g-1, L)$ is $A$-stable. By
 Corollary \ref  {cor:gen}, we  have $r=g$, a contradiction. This proves (ii). Then (iii) holds by Lemma \ref{lem:irreducible}, (iv).
 \end{proof}

Recall the notation $\phi_L\colon C\to \bar C\subset \mathbb P^r$,  $C'$ for the normalization of  $\bar C$,  $f\colon C\to C'$ the induced morphism and  $\nu=\deg(f)>1$. For any integer $h$ one has the morphism $f^ *\colon J^ h(C')\to J^ {h\nu }(C)$. Let  $g'$ be the genus of $C'$ and $L'$ the (birational) linear series on $C'$ of dimension $r$ such that $L=f^*(L')$, whose degree is $\delta=\frac {2g-2}\nu$.

 \begin{lem} \label{lem:key}
 Assume $\nu \leqslant 3$ and let  $\underline m=(1^ {\mu_1}, \dots, \nu^ {\mu_\nu})$ be the partition of $g-1$ such that $Z\subseteq \Gamma(\underline m, L)$. Set $\mu_1=\mu$, $\mu_2=\mu'$. Then  $\nu=2$ and there are the following possibilities:\\
 \begin{inparaenum}
 \item $\mu=g-1$ (hence $\mu'=0$);\\
 \item $\mu=g-1-2a$, $\mu'=a$, $A=f^*(J(C'))$ (hence $g'=a$) and $Z=W_{\mu}(C)+f^*(J^a(C'))$.
 \end{inparaenum}
\end{lem}

 \begin{proof}
 We first consider the case $\nu=2$.

 If $\mu=0$, then $g-1=2\mu'$  and  $Z=f^*(\Gamma_{C'}(\mu',L'))$, since $\Gamma_{C'}(\mu',L')$ is irreducible by Lemma \ref{lem:irreducible}.    The general point of $Z$ corresponds to a linearly isolated divisor, hence $\mu'\leqslant g'$.
 Since $f^ *\colon J^ {\mu'}(C')\to J^ {g-1}(C)$ is finite, one has $r=\rho_{C'}(\mu',L')\leqslant \mu'\leqslant g'$. Since $Z$ is $A$--stable, we have an isogeny $A\to \bar A\subseteq J(C')$ (so that $a\leqslant g'$) and $\Gamma_{C'}(\mu',L')$ is $\bar A$--stable.
 Hence by  Corollary \ref {cor:gen} one has $\Gamma_{C'}(\mu',L')=J^ {\mu'}(C')=W_{\mu'}(C')$ and $g'\leqslant \min\{\mu', r\}$ (see Lemma \ref {lem:class2}, (i)). Then  $Z$ is $f^*(J(C'))$--stable,  and, since $a$ is the maximal dimension of an abelian subvariety of $J(C)$ for which $Z$ is stable
  and  $a=\dim (\bar A)\leqslant g'$, it follows   $a=g'$. In conclusion $\mu'=a=r=g'$, and we are in case (ii).

 Assume now  $\mu>0$.
 The general  point  of $Z$ (which is a component $\Gamma(\underline m,L)$)  is smooth for $\Theta$, hence it  corresponds to a linearly isolated, effective divisor $D$ of degree $g-1$, which is reduced (see Lemma \ref {lem:irreducible}, (ii)) and  can be written in a unique way  as $D=M+f^*(N)$, where $M$ and $N$ are effective divisors, with $\deg(M)=\mu$, $\deg (N)=\mu'$ and $M':=f_*(M)$   reduced.
  So there is a rational map $h\colon Z \dasharrow J^{\mu}(C)$ defined by $D=M+f^*(N)\mapsto M$.

  Assume $\mu\le r$. By Lemma \ref{lem:irreducible} the image of $h$ is $W_{\mu}(C)$. Identify $A$ with its general translate inside $Z$. Then we have a morphism  $h|_A\colon  A\to J(C)$ whose image we denote by $\bar A$. Then  $W_{\mu}(C)$ is $\bar A$--stable.
 Since  $W_{\mu}(C)$ is birational to $C(\mu)$, which is of general type because  $\mu<g$, then $\bar A=\{0\}$. It follows that  each component of the general fibre of $h$ is $A$--stable, in particular $r-\mu\geqslant a\geqslant 1$.

Take $M\in W_{\mu}(C)$ general and set $L'':=L'(-M')$. Since $L'$ is birational and $M'\in W_\mu(C')$ is general, with $\mu< r=\dim (L')$, then $L''$ has dimension $r-\mu\geqslant 1$ and it is base point free. It is also birational as soon as $r-\mu\geqslant 2$.

Assume first $r-\mu\geqslant 2$. Then $C'(\mu',L'')$ is irreducible by Lemma \ref {lem:irreducible}, (iv), and  there is a  birational morphism  $C'(\mu',L'')\to h^ {-1}(M)\subset J^{g-1}(C)$  factoring through the map $C'(\mu',L'')\to J^ {\mu'}(C')\hookrightarrow J^{g-1}(C)$, where the last inclusion is  translation by $M$. Namely, up to a translation, $h^{-1}(M)=\Gamma(\mu',L'')$. In particular, $\dim (\Gamma(\mu', L''))=\dim (C(\mu', L''))=r-\mu$, hence $\mu'\ge r-\mu$.

Remember that $h^{-1}(M)=\Gamma(\mu', L'')$ is $A$--stable for $M\in W_{\mu}(C)$ general. By Corollary \ref {cor:gen} one has $\Gamma_{C'}(\mu',L'')=W_{\mu'}(C')=J^ {\mu'}(C')$, so $g'\leqslant \min\{\mu', r-\mu\}=r-\mu$. On the other hand, since $h^ 0(\mathcal O_C(D))=1$, one has also  $h^ 0(\mathcal O_C(f^ *(N)))=1$, hence $\mu'\leqslant g'$ and we conclude
that  $\mu'=g'=r-\mu$.
 The same argument as above yields $a=g'$ and we are again in case (ii).

If $r-\mu=1$, then $a=1$, $r=g-2$, $\mu=g-3$ and $\mu'=1$. On the other hand  $L$ is cut out on the canonical image of $C$ by the hyperplanes through the point $p_A$ which is the projectivized tangent space to $A$ at the origin. Then $\phi_L: C\to  \bar C$ is the projection from $p_A$, $\bar C$ it is a normal elliptic curve, and we are again  in case (ii).

Assume now $\mu>r$ and keep the above notation. In this case the map $h\colon  Z\dasharrow \bar Z:=h(Z)$ is generically finite, $\bar A$ is isogenous to $A$ and $\bar Z$ is  $\bar A$--stable. By  \cite[Lemma 3.1] {df}, we have $g-1\geqslant \mu\geqslant \dim (\bar Z)+\dim (\bar A)=r+a=g-1$, thus $\mu=g-1$, so we are in case (i).

Finally consider the case $\nu=3$.   Write the general $D\in \Gamma(\underline m, L)$ as $D=M_1+M_2+f^*(N)$,   where $M_1$ is reduced of degree $\mu$, $f_*(M_2)=2M'_2$ with $M'_2$ reduced of degree $\mu'$ and, as above, $\mu_3=\deg(N)\leqslant g'$. Set $\tau=\mu+\mu'$
 and consider the rational map $h\colon Z \dasharrow W_{\tau}(C)\subseteq J^{\tau}(C)$ defined by $D\mapsto M_1+f^*(M'_2)-M_2$.

If $\tau\leqslant r$,  arguing as above (and keeping a similar notation) one sees that $h(Z)=W_{\tau}(C)$, the general fibre of $h$ is $A$--stable, hence $r-\tau\geqslant a$. However $a=1$ and $\phi_L$ non--birational, forces, as we have seen, $\nu=2$, which is not the case here. Hence we have $a\geqslant 2$. We consider now $L''=L'(-f_*(M_1)-M'_2)$, which has dimension $r-\tau\geqslant 2$ and is base point free and birational, so  $C'(\mu_3,L'')$ is irreducible.
The general fiber $h^ {-1}(h(D))$ of $h$ is isomorphic to $\Gamma_{C'}(\mu_3, L'')$ and is $A$--stable. In particular $\dim(\Gamma_{C'}(\mu_3, L''))=r-\tau$, hence $\mu_3\ge r-\tau$. By Corollary \ref {cor:gen}, we have $\Gamma_{C'}(\mu_3, L'')\cong W_{\mu_3}(C')\cong J^ {\mu_3}(C')$, so $g'\leqslant \min\{\mu_3,r-\tau\}=r-\tau$, hence $g'\le r-\tau \leqslant \mu_3\leqslant g'$, thus $\mu_3=r-\tau=g'$. In addition,
as above, we have $a=g'$, so that $A=f^*(J(C'))$.
Then $g-1-a=r= \tau+\mu_3=\mu+\mu'+\mu_3$. On the other hand $g-1=\mu+2\mu'+3\mu_3$. This yields $\mu'+2\mu_3= a= \mu_3$, hence $\mu'=\mu_3=0$, which is not possible.

If $\tau>r$ then $h\colon  Z\dasharrow \bar Z:=h(Z)$ is generically finite, $\bar A$ is isogenous to $A$ and $\bar Z$ is  $\bar A$--stable. By  \cite[Lemma 3.1] {df}, we have $g-1\geqslant \tau\geqslant \dim (\bar Z)+\dim (\bar A)=r+a=g-1$, thus $\tau=g-1$, contradicting $\tau\leqslant  \deg (L')=\frac 2 3(g-1)$. \end{proof}

  \begin{lem}\label{lem:degL}  If $\nu\geqslant 4$ then $\nu=4$ and
 either\\
 \begin{inparaenum} [(i)]
 \item there is a degree 2 map $\psi\colon C\to E_1$ with $E_1$ a genus $r$ hyperelliptic curve such that $Z=A=\psi^*(J^ r(E_1))$; or\\
 \item there is a faithful $\mathbb Z_2^2$--action on $C$ with rational quotient;  denoting by  $f_i\colon C\to E_i$ (for $1\leqslant i\leqslant 3$) the quotient map for the three non--trivial involutions of $\mathbb Z_2^2$, with $E_i$ of genus $g_i$, and $g_1\geqslant g_2\geqslant g_3$, then $g_1=r+1$, $g_2+g_3=r$ and  $Z=A=f_2^*(J^ {g_2}(E_2))\times f_3^*(J^ {g_3}(E_3))$ is the Prym variety associated to $f_1$.
 \end{inparaenum}

 \end{lem}
\begin{proof}
Since $L$ is base point free by Lemma \ref{lem:free} and it is not birational by Lemma \ref{lem:nobir}, we have $\delta=\frac {2g-2}\nu\geqslant r\geqslant  \frac{g-1}{2}$, hence   $\nu\leqslant 4$.

Assume $\nu=4$. Then  $\bar C$ is a   curve of degree $\frac {g-1}{2}$ spanning a projective space of dimension $r\geqslant \frac{g-1}{2}$. Hence $r=a=\frac{g-1}{2}$, $Z=A$,  and $\phi_L=f$ is the composition of a $g^1_4$ (that we denote by $\mathcal L$) with the degree $r$ Veronese embedding $\mathbb P^1\to \mathbb P^r$.

Let $\underline m=(1^ {\mu_1},\ldots, 4^ {\mu_4})$ be the partition of $2r$ such that $Z\subseteq \Gamma(\underline m,L)$.

\begin{claim}
One has $\underline m=(2^ r)$.
\end{claim}

\begin {proof} [Proof of the Claim] Assume by contradiction this is not the case, so that one among $\mu_1,\mu_3,\mu_4$ is non--zero.
We have  $r=\dim(\Gamma(\underline m,L))=\mu_1+\cdots+\mu_4$,
because $\bar C$ is a rational normal curve of degree $r$ in $\mathbb P^r$,   and $2r=g-1=\mu_1+2\mu_2+3\mu_3+4\mu_4$, hence
$\mu_1=\mu_3+2\mu_4$ and $\mu_1>0$.
So we may write the general divisor $D\in Z$ as $D=M+N$, where $M':=f_*(M)$ is reduced of degree $\mu:=\mu_1$.
Then we proceed as  in the proof of Lemma \ref{lem:key}.

Consider the rational map $h\colon  Z=A\dasharrow J^{\mu}(C)$ defined by $D=M+N\mapsto M$.  It extends to a morphism and $\bar A:=h(A)$ is an abelian variety contained in $W_{\mu}(C)$.
Since $\mu\leqslant r=\frac{g-1}{2}$, one has $\bar A=W_{\mu}(C)$, which is  impossible.
\end{proof}

Let $E:=C(2,L)$. Assume first $C$ is not hyperelliptic. Then there is a birational (dominant) morphism $E(r)\to \Gamma((2^ r), L)$ (see Lemma \ref {lem:tetra2}, (i)).
If $E$ is irreducible, then  by Corollary \ref {cor:gen2} its image generates $J^2(C)$, hence also  $\Gamma((2^ r), L)=Z$ generates $J^ {2r}(C)$ and we obtain a contradiction.

So $E$ is reducible and we apply Lemma \ref{lem:tetra} and  \ref{lem:tetra2}.  Suppose we are in case (b) of Lemma \ref{lem:tetra} and (ii) of Lemma \ref{lem:tetra2}. Then  $A=Z=\Gamma((2^ r), L)$ is birational to $E_1(r_1)\times E_2(r_2)$, for  non--negative integers $r_1,r_2$. However, if $r_i>0$, then $r_i=g_i$, for $i\in \{1,2\}$.
Since $2g_2\geqslant g$, one has $r_2=0$ because $2r=g-1<g$. Hence
$A$ is birational to $E_1(g_1)$ and we are in case (i).

Consider now case (c) of Lemma \ref{lem:tetra} and (iii) of Lemma \ref{lem:tetra2}.   We claim that  $J(C)$ is isogenous to $J(E_1)\times J(E_2)\times J(E_3)$.  Indeed, consider the representation of $\mathbb Z_2^2$ on $H^0(K_C)$. Since  $C/\mathbb Z_2^2$ is rational, we have $H^0(K_C)=V_{\chi_1}\oplus V_{\chi_2}\oplus V_{\chi_3}$, where for $i=1,2,3$ the non trivial character $\chi_i$  of $\mathbb Z_2^2$ is  orthogonal to the involution $\iota_i$ such that $C/\iota_i\cong E_i$, and $\mathbb Z_2^2$ acts on $V_{\chi_i}$ as  multiplication by $\chi_i$. Thus $V_{\chi_i}$ is the tangent space to $f_i^*(J(E_i))$.

Recall  that there are three non--negative integers $r_1,r_2,r_3$ such that $A$ is birational to $E_1(r_1)\times E_2(r_2)\times E_3(r_3)$. If $r_i>0$, then $r_i=g_i$ (for $1\leqslant i\leqslant 3$). Since $g_1+g_2+g_3=g=2r+1$, at least one of the integers $r_i$ is zero. If two of them are zero,  we are again in case (i). So assume  $r_1=0$ and $r_2,r_3$  non--zero. Then $r=g_2+g_3$ and $g_1=r+1$. Moreover  the involution $\iota_1$ acts on $A$ as multiplication by $-1$, hence $A$ is the Prym variety of $f_1\colon C\to E_1$, and we are in case (ii).

The case $C$ hyperelliptic can be treated similarly. In case (b) of Lemma \ref{lem:tetra} and (ii) of Lemma \ref{lem:tetra2}, $E_1$ is rational (hence it is contracted to a point  by $j$), and
$A=Z=\Gamma((2^ r), L)$ is birational to $E_2(r)$. Then $r=g_2$ and we reach a contradiction since $2g_2\geqslant g$.  In case
(c) of Lemma \ref{lem:tetra} and (iii) of Lemma \ref{lem:tetra2} one has $g_3=0$ and the above argument applies with no change.\end{proof}

\begin{proof}[Proof of Theorem \ref{thm:main}]
By Lemma \ref{lem:degL}, we may assume $\nu=2$.
Let  $\underline m$  be the partition of $2r$ such that $Z\subseteq \Gamma(\underline m, L)$. By Lemma \ref{lem:key}, it is enough to consider
 the case $\nu=2$ and   $\underline m=(1^ {g-1})$, i.e., case (i) of that lemma. Then $Z$ is contained in the kernel $P$ of $f_*\colon J^{g-1}(C)\to J^{g-1}(C')$ (namely $P$ is the generalized Prym variety associated with $f$).
  The space $H^0(K_C)$ decomposes under the involution $\iota$ associated with $f$ as  $H^0(K_{C'})\oplus V$, where $V$ is the space of antiinvariant $1$-forms. Hence $V^*$ is the tangent space to $P$,   the tangent space $T$ to $A$ is also contained in $V^*$ and the linear series $L$ is equal to $\mathbb P(T^{\perp})\supseteq \mathbb P(H^0(K_{C'}))$. On the other hand, by construction $\iota$ acts trivially on $L$, hence $T=V^*$ and thus $A=P$.  This implies that $Z=A=P$, $g=2r+1$ and $f$ is unramified with $g'=r+1$.  \end{proof}

  \subsection{General Brill--Noether loci}\label{ssec:GBN} The proof of the general formula \eqref {eq:DF} in \cite{df} uses an argument which is  useful to  briefly recall.
 Let $Z\subseteq W^ s_d(C)\subsetneq J^ d(C)$ be $A$--stable and $L\in Z$ general, so $L$ is a special  $g_d^s$, which we may assume to be complete, so $d\geqslant 2s$. Let $F_L$ be the fixed divisor of $L$ and, if $s>0$,
 let $L'$ be the base point free residual linear series. Set $d'=\deg(L')$. Since $L'$ is also special, we have $d'\geqslant 2s$.  Then, by Lemma \ref {lem:tetra}, (iii), one has the birational map $\mathfrak r\colon  C(d'-s,L')\dasharrow C(s,L')\cong C(s)$.

Consider the  morphism $j\colon  C(d'-s,L')\times C(g-1-d+s)\to J^{g-1}(C)$ such that $j(D,D')$ is the class of $D+F_L+D'$ (if $s=0$, we define $j\colon C(g-1-d)\to J^{g-1}(C)$ by $D'\mapsto F_L+D'$).
  If $(D,D')$ is general in $C(d'-s,L')\times C(g-1-d+s)$, the divisor $D+F_L+D'$ is linearly isolated, hence $j$  is generically finite onto its image $Z_L$, which  therefore has dimension $g-1-d+2s$. Consider the closure $Z'$ of  the union of the  $Z_L$'s, with $L\in Z$ general, which is $A$--stable. One has $Z'\subseteq \Theta$ and the above discussion yields $r':=\dim(Z')=r+g-1-d+2s$. Therefore $r'+a\leqslant g-1$ if and only if  \eqref {eq:DF} holds, with equality if and only if  equality holds in \eqref {eq:DF}.

  \begin{cor}\label{for:ten} Let $C$ be a curve of genus $g$. Let $A\subsetneq J(C)$ be an abelian variety of dimension $a>0$ and $Z\subseteq W^ s_d(C)
  \subsetneq J^ d(C)$ an irreducible,  $A$--stable variety of dimension $r=d-2s-a$.
 Assume $(d,s)\neq (g-1,0)$. Then
  there is a degree 2 morphism $\varphi\colon C\to C'$, with $C'$ a smooth curve of genus $a$ with $g>2a+1$, such that
$A=\varphi^*(J(C'))$ and either\\
\begin{inparaenum}[(i)]
\item $Z=W_{d-2a-2s}(C)+\varphi^*(J^{a+s}(C'))$, or\\
\item $C$ is hyperelliptic and $Z=\varphi^*(J^{a}(C'))+W_{d-2s-2a}(C)+ W^ s_{2s}(C)$.
\end{inparaenum}
\end{cor}

\begin{proof} We keep the same notation as above.

We apply Theorem \ref {thm:main}  to $Z'$. Since
$d=r+a+2s\geqslant 2(a+s)$ and $(d,s)\neq (g-1,0)$, then $2a<g-1$, hence case (b) of Theorem \ref {thm:main} does not occur for $Z'$.
So we have a degree 2 morphism $\varphi\colon C\to C'$, with $C'$ a smooth curve of genus $a$ with $g>2a+1$, such that
$A=\varphi^*(J(C'))$ and $Z'=W_{g-1-2a}(C)+\varphi^*(J^a(C'))$. The case $s=0$ follows right away and we are in case (i). So we assume $s>0$ from now on.

If $L'$ is composed with the involution $\iota$ determined by $\varphi$, we are again in case (i). So assume $L'$ is not composed with $\iota$. The general $D\in C(d'-s,L')$ contains no fibre of $\varphi$ and the same happens for the general $D'\in C(g-1-d+s)$. Since $D+F_L+D'$ corresponds to the general point of $Z$, and contains exactly $a$ general fibres of $\varphi$, then $F_L$ has to contain these $a$ fibres, whose union we denote by $F$. Moreover, the description of $Z$ implies that $D+D'+(F_L-F)$ is a general divisor of degree $g-1-2a$, in particular $D$ is a general divisor of degree $d'-s$. But $D$ is also general in $C(d'-s,L')$, and this implies $d'-s\leqslant s$. On the other hand $d'\geqslant 2s$, hence $d'=2s$. Then, either $L'$ is the canonical series of $C$ or $C$ is hyperelliptic and $L'$ is the $s$--multiple of the $g^ 1_2$. However the former case does not occur since by construction $d'=\deg (L')\le g-1$,
hence  we are in case (ii).   \end{proof}

redpajama_set_name: RedPajamaArXiv


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Three new confirmed cases of Zika virus in Missouri
Posted 3:44 pm, August 12, 2016, by Kevin S. Held
ST. LOUIS (KPLR) – The Missouri Department of Health and Senior Services confirmed three new cases of the Zika virus on Friday: two men and a woman.
Two of the individuals had traveled to Nicaragua while the other recently visited Puerto Rico. The woman who tested positive was not pregnant, health officials said.
With these latest confirmations, this brings the total number of travel-related cases of Zika virus in Missouri to 20.
According to the Missouri Department of Health, there have been no reported cases of the Zika virus being contracted from a mosquito bite in the state.
The Zika virus can be spread through a mosquito bite, unprotected sexual contact, and blood transfusion. A pregnant woman can pass the virus to her fetus during pregnancy. Nearly 80 percent of people who catch the virus will show no symptoms, which includes: fever, rash, joint soreness, or redness of the eyes.
International health officials suspect a connection between pregnant woman catching the virus and a birth defect called microcephaly in newborns. Babies with microcephaly will have smaller head sizes and their brains may not develop properly.
The CDC recommends pregnant women avoid traveling to Zika-affected areas, such as Mexico, the Caribbean, Central American, and South America.
At present, there is no vaccine for Zika virus.
Childcare worker charged with breaking child's leg at 'licensed exempt' facility
Wentzville School District confirms multiple cases of whooping cough
Three flu-related deaths reported in Missouri
Missouri sheriffs accuse Dept. of Corrections of catch and release; state says it's following the law
Old Monroe police chief indicted for exploitation of the elderly


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Navigate to your Gravity folder (typically C:\Program Files (x86)\Gravity\Ragnarok Online 2 or some variation) and find the file named Version.dat.
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This is by far one of the things that I believe God enjoys teaching me the most. There is some kind of phrase or story that goes something like this: If you pray for it, God is going to give you opportunities to practice it. In my case, I pray often for patience and God hears me every time. I know this because I have lots of opportunities (and challenges) to practice growing in patience.
Some lessons in patience are bigger or harder than others. Ben is on a unique deployment schedule. He deploys for 2 months at a time, but sometimes is only home for only 2 – 3 months before he has to deploy again. And while he is home, he has to go on any number of trips that can last from 5 days to unknown – literally. Each time he is gone, especially when its a trip, I can't know for certain when he will be home. I have to be patient and have trust. Some trips this is easier than others.
My lessons in patience aren't always this large. Just the other day, I found some beautiful meyer lemons at one of our local grocery stores. I was so excited. *Back story* We have a eureka lemon tree in our backyard and I got my first harvest a few months ago. I had so many lemons I spent quite a bit of time looking at different things to do with them. I found something I had never heard of before – preserved lemons. It is a staple in Moroccan and North African cooking (neither of which is familiar to me, but I love to try to make new things). Just salt and the lemons in a mason jar. Even though my lemons weren't the kind typically used, the recipe said my eurekas would be an ok replacement for the typical meyer lemons (who knew there were so many varieties of lemons!) For 3 weeks! After patiently waiting for these beautiful lemons, they were done. I had a recipe and couldn't wait to start. Got my lemons, got my roasted garlic, and my fettuccine (I went Italian, that's my wheelhouse). I figured I'd get braver and branch into the Moroccan after I knew these lemons were good. On my trip back to the refrigerator with the remaining lemons, time stopped. In the next moment, the jar, and all of those precious lemons, were on the floor. The glass had shattered.
I was devastated. My dinner was delicious. But my lemons were gone. All that waiting, and for what? Ben was deployed at the time so he didn't even get to taste the fruits of my waiting. I haven't attempted to make them again, it takes so long for them to be ready, and if they can be gone that quickly, is there really a point?
Then I saw those meyer lemons (the kind that the recipe would be amazing preserved). There was no question, I immediately grabbed them. Because even though I was still sore about my clumsiness, I knew I had to do it again. I can still taste that meal and am willing to wait the 3 weeks to have it again. So, presently we are a few days into our wait. Ben is not home, but he will be by the time they are done. I'm so glad that God gave me another opportunity to wait patiently.
It's incredible how patient we really can be, when we know what is waiting for us at the other end. I know that those lemons are going to be amazing, because I've tasted them already before. Being patient this time is much easier already than last time. What is cool about this experience of patience, I think, is that I'm choosing to see it as a lesson from God. It makes my waiting and patience more than just an act of willpower, it is a way I am training my soul and my heart to be patient in the bigger things, especially in times when I am waiting for God's timing to reveal His plan for my life and our family.
And if you're interested in the dinner I made after you practice your own patience skills, check out: http://www.thekitchn.com/recipe-fettucci-5560. I grilled some chicken and tossed that in as well. Enjoy!


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