PAWTUCKET, R.I. – The Little East Conference (LEC) Commissioner Cora H. Brumley announced that the league's live video streaming service will now be known as the LEC Network, which was developed in partnership with Skye Design Studios. The web address for the LEC Network will continue to be www.LittleEast.tv, and games will continue to be available for free to all viewers.
"The Little East Conference has a proud tradition of technological innovation for the benefit of the league's student-athletes. It was time to give our webcasting service a proper name and dynamic visual brand identity," said Commissioner Cora H. Brumley. "Skye Design Studios has a well-earned reputation for distinctive graphic design and all our expectations were exceeded."
The LEC Network provides a centralized home for the hundreds of live game brodcasts produced each year by the conference's nine member institutions – Castleton University, Eastern Connecticut State University, Keene State College, UMass Boston, UMass Dartmouth, Plymouth State University, Rhode Island College, University of Southern Maine and Western Connecticut State University.
The LEC Network will continue to be powered by Stretch Internet in a partnership that saw the LEC launch the first centralized live video streaming service of any NCAA Division I, II or III conference in New England in 2014. In addition, the LEC was the first conference in the region to debut on OTT apps such as Amazon Fire, Android TV, Apple TV and Roku.
At SDS we believe that each client is unique, and as such, each job requires a fresh perspective to ensure a final product that captures the true essence of the brand. This perspective is established through a full immersion into the client's culture, our overall branding experience, and a commitment to constant design innovation. With a background in sport identity both nationally and on a global scale for over ten years, SDS represents an incredible depth of design experience. We have developed memorable brands for a variety of clients, ranging from NCAA UEFA Champions League identity work to sport identities for Major League Baseball, NBA, NHL, and MLS.


On the 26th August the South African hip hop community will come together at Zone 6 Venue in Diepkloof, Soweto to honour and celebrate the late Linda ' PRO ' Mkhize in a tribute concert including the best of South Africa's hip hop artists and DJs.
The lineup will include AKA, HHP, Kwesta, Cassper Nyovest, K.O., Gigi LaMayne, Reason, Skwatta Kamp, Da Les, Stogie-T, Amunishn, L-Tido, Rouge, JR, Nadia Nakai, Jub Jub, Kid-X, Ma-E, Blaklez, Red Button, Yanga, Zubz, Ntukza, P Dot O, Nveigh, Solo and the BETR gang, Nothende, Max-Hoba and F-Eezy.
DJ lineup includes Citi Lyts, DJ Fresh, Vigilante, Major League DJs, DJ Dimplez, DJ Capital, DJ Switch, P Kuttah, C-live, DJ 1D, DJ Milkshake, Ms Cosmo, P-Kuttah, DJ Mr X, DJ ZanD , DJ Rizz and DJ Zai Maya.
The event will be hosted by MCs Moozlie, Lee Kasumba, Lootlove and Sandile GQ.
There is no loftier legacy to collectively pay homage to than that of larger than life, lyrical magician, Number 1 Soweto Boy Linda "PRO" Mkhize. The loss of this prolific storyteller has left an entire community of artists broken. Pro's contribution to the hip hop culture was so refreshingly new and unprecedented that the void will never be filled, but rather will be lauded with memories of Pro on and off stage.
The South African hip hop community, political leaders, friends, family and fans united at his memorial service held at Bassline, Johannesburg on the 16th August.
Many artists paid tribute to PRO including the likes of AKA, Anatii, Maggz, Khuli Chana, Slikour, Kwesta, Gigi Lamayne, DJ Citi Lyts, IVY League among many.
From his debut Heads & Tales all the way to his last album Continua, he never had a single moment of compromise, his honesty and creativity were consistently spell-binding. He was not just an outstanding artist, he was a loving human being with the biggest heart.
Tickets are available from Computicket from R100 for general access and R250 for VIP. All proceeds from the concert will go into an education fund for PRO's daughter, Nonkanyezi.


text:
Le tournoi de Marseille est un tournoi de tennis professionnel masculin. L'édition 2012, classée en catégorie ATP 250 Series, s'est disputée du 20 au .

Faits marquants
 Juan Martín del Potro remporte le tournoi en simple en battant Michaël Llodra en finale. Il empoche ainsi son  ATP.
 En double, Nicolas Mahut et Édouard Roger-Vasselin s'imposent face à Dustin Brown et Jo-Wilfried Tsonga.

Résultats en simple
 Têtes de série

 Qualifiés

 Wild cards (invités)

Tableau final

Résultats en double

Têtes de série

Tableau final

Notes et références

Voir aussi

Article connexe
 Saison 2012 de l'ATP

Liens externes
  Site officiel du tournoi

Open 13
ATP 2012
Tennis en France en 2012
Tennis en février 2012
Sport en France en février 2012

redpajama_set_name: RedPajamaWikipedia


Welcome to our website. This website is owned and operated by Outback Aromas (ABN: 34466756824), successors and assignees, (we or us). It is available at: http://www.outbackaromas.com.au (Site) and may be available through other addresses or channels.


CHECK OUT THIS FUNNY VIVINT SMARTHOME VIDEO!
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text:
Q: SQS all message goes to dead letter queue I've configured an SQS queue and an additional dead letter queue using terraform.
resource "aws_sqs_queue" "sqs_deadletter" {
  name = "worker-dead-letter"
}

resource "aws_sqs_queue" "sqs" {
  name = "worker"
/* TODO: If I enable this all messages goes to the dead letter queue
  redrive_policy = jsonencode({
    deadLetterTargetArn = aws_sqs_queue.sqs_deadletter.arn
    maxReceiveCount = 4
  })
*/
}

resource "aws_lambda_event_source_mapping" "sqs" {
  event_source_arn = aws_sqs_queue.sqs.arn
  function_name = aws_lambda_function.worker.arn
  enabled = true
  batch_size = var.batch_size
}

I use the below handler to process my messages.
@Introspected
class LegacyToModernRequestHandler : MicronautRequestHandler<SQSEvent, Unit>() {
    private val logger = KotlinLogging.logger {}

    override fun execute(input: SQSEvent) {
        input.records.forEach {
            handle(it)
        }
    }

    private fun handle(message: SQSMessage) {
        val key = message.body
        logger.info { "LegacyToModernRequestHandler($key)" }
    }
}

But all my messages goes to the DLQ. How can I indicate successful handling so that doesn't happen?

A: If you are not using AUTO_ACKNOWLEDGEMENT mode, you will have to explicitly acknowledge the message so that it is processed successfully. Otherwise it will go to DLQ. Can you show how have you configured your SQS queue?

redpajama_set_name: RedPajamaStackExchange


Despite generous domestic supplies, the imports of industrial wood increased.
In 2015, exports of roundwood from Slovenia reached 2.7 million m³, which is almost 13% higher than a year before. A comparative analysis of wood exports in 2014–2015 showed that the exports of round logs for sawmills and peeled veneer demonstrated the biggest growth.
According to the Statistics Service, in 2015, the domestic timber reserves are large enough to reduce the roundwood imports by about 2%. A large area of forestlands affected by black ice and bark beetles was cut down.
According to the data provided by the Foreign Trade Office in 2015, only 448,000 m³ of roundwood were imported to Slovenia. According to the same data, the exports of saw log and peeled veneer increased by 32%, while the exports of pulpwood and other industrial roundwood materials and stove woods decreased by 2% and 10%, respectively.
Dynamics of imports showed different results. As compared to the year 2014, the imports of stove woods decreased by 10%, and the imports of saw log and peeled veneer reduced by 17%. Last year, the imports of pulpwood, particleboards and other industrial roundwood materials rose by 10% in spite of the national timber reserves.


text:
Linówko (deutsch Klein Lienichen) ist ein Ort in der Gmina Ińsko (deutsch Nörenberg) in der polnischen Woiwodschaft Westpommern. Er liegt circa fünf Kilometer westlich von Ińsko zwischen der westlichen Spitze des Jezioro Ińsko (Enzigsee, Lienicher Lanke) und dem Lienischen See.

Geschichte

Aus vorgeschichtlicher Zeit stammten die Großsteingräber bei Klein Lienichen, zehn Grabanlagen der jungsteinzeitlichen Trichterbecherkultur. Sie wurden im 19. Jahrhundert zerstört.

Ehemals gehörte Klein-Lienichen zum Landkreis Dramburg der Neumark, dann zum Landkreis Saatzig im Regierungsbezirk Stettin der Provinz Pommern. Der Ort umfasste das Dorf und ein Lehns-Rittergut und bildete mit dem Vorwerk mit den Wohnplätzen Friedrichsthal und Karlsthal (auch Carlshof genannt) sowie mit dem nördlich des Dorfes im Klein Lienichener Wald gelegenen Gut Ziegenberg und mit dem Rittergut Langenhagen (Długie) A. B. einen ca. 6.000 Morgen umfassenden Güterkomplex, der seit länger als 400 Jahre in ununterbrochener Erbfolge in dem Besitz der Familie von Mellenthin war. Vormals ein Lehen derer von Wedel, wurde es der Familie von Mellenthin von den Kurfürsten zu Brandenburg direkt zum Lehen gegeben. Der Regierungsrat Friedrich von Mellenthin ließ 1852 ein neugotisches Herrenhaus umgeben von seinen Wirtschaftsgebäuden, über See und Wald gelegen, errichten. Er vererbte diesen Besitz an seinen Sohn Carl von Mellenthin.

Die letzte Eigentümerin war die Witwe Alexandrine Johanna Erdmuthe von Mellenthin, geborene von Tiedemann, die 1913 Maximilian von Mellenthin († 1922) geheiratet hatte. 1939 zählten zu dem 950 ha großen Rittergut 50 Pferde, 200 Rinder und 300 Schweine. 1942 wurde das Gut Ziegenberg an den Staatsforst verkauft.

Besonderheiten des Gemeindegebietes waren ein großer Findling in der Nähe des Lienicher Sees sowie die evangelische Dorfkirche, ein Fachwerkbau mit niedrigem Westturm. An der Altarwand war ein mittelalterlicher Altarschrein mit Schnitzfiguren der Anna, Maria und eines Märtyrers aufgestellt. Die Kanzel war eine schlichte Schnitzarbeit aus der Barockzeit, die eichenfarbig überstrichen war.
Klein Lienichen gehörte bis 1945 zum evangelischen Kirchspiel Steinhöfel (Kamienny Most) im Kirchenkreis Freienwalde (Chociwel) in der Kirchenprovinz Pommern der Kirche der Altpreußischen Union und ist heute in die Trinitatiskirchengemeinde in Stettin in der Diözese Pommern-Großpolen der Evangelisch-Augsburgischen Kirche in Polen eingepfarrt.

Literatur
 Paul Schulz (Hrsg.): Der Kreis Saatzig und die kreisfreie Stadt Stargard – Ein pommersches Heimatbuch. Rautenberg, Leer 1984, ISBN 3-7921-0307-9.
 Johannes Hinz: Pommern. Wegweiser durch ein unvergessenes Land. Flechsig-Buchvertrieb, Würzburg 2002, ISBN 3-88189-439-X, S. 174.

Weblinks
 Klein Lienichen bei Meyers Gazetteer (mit historischer Landkarte)
 Klein Lienichen beim Heimatkreis Saatzig

Ort der Woiwodschaft Westpommern
Ort in Pommern
Gmina Ińsko


text:
Ann Romney Happy with "Elegant, Consistent, Fabulous" Olympic Performance by Rafalca and Jan Ebeling
9 years ago StraightArrow Comments Off on Ann Romney Happy with "Elegant, Consistent, Fabulous" Olympic Performance by Rafalca and Jan Ebeling
At the Olympic dressage competition in London, Ann Romney hugs Jan Ebeling after his ride on the mare Rafalca of which the wife of US presidential candidate Mitt Romney is a part-owner. © 2012 Ken Braddick/dressage-news.com
LONDON, Aug. 2–Ann Romney, wife of U.S. presidential candidate Mitt Romney, was happy with the "elegant, consistent, fabulous" performance by the mare, Rafalca, of which she is a part-owner and came to London to watch compete for the United States team in the Olympic dressage competition
The partnership of Jan, Rafalca and the three women owners, Ann Romney, Beth Meyer and Amy Ebeling, is celebrated around the world as bringing more attention to the sport than in the entire 100 years it has been in the Olympics.
The ownership trio sat in the stands among the 23,000 spectators who ignored occasional heavy rain to watch the opening day of the dressage competition that will be the finale of equestrian sports at these Games with the dressage musical freestyle on Aug. 9.
Obviously moved by the first Olympic appearance for Rafalca and rider Jan Ebeling who helped Ann overcome crippling affects of multiple schlerosis with therapeutic riding lessons, the competition in London's Greenwich Park, the oldest royal park at 600 years, was the culmination of a dream.
Jan, 53, a German migrant who has lived in the U.S.more tha half his lietime and is based in Moorpark, California, with his wife, Amy, and soccer-playing son, Ben, said he was "really happy" with the ride that scored 70.423 per cent that put them midway in the first day's standings of the two days of Grand Prix.
The score did, however, get the three-member team off to a good start with Tina Konyot of Palm City, Florida, and Calecto V and Steffen Peters of San Diego, California, and Ravel scheduled to compete Friday.
Barring the unknown, the U.S. is on track to be among the seven nations whose teams move on to the Grand Prix Special next Tuesday that will decide the medals. Great Britain and Germany are locked in a struggle for gold and silver with Denmark, the Netherlands, Sweden and the U.S. contending for bronze.
Jan Ebeling riding Rafalca for the United States at the Olympic Games in London. © 2012 Ken Braddick/dressage-news.com
Because of media attention, the U.S. dressage team has maintained a low profile since moving to England early in July to prepare for the Olympics.
"There certainly was a lot of media attention going on," Jan said. "But in the end I think it's good for the sport and I don't really get distracted. I have a pretty good way of focusing."
This was Jan's first experience as an Olympian though he represented the United States at the Pan American Games in 2003 where the team won gold.
Rafalca is a 15-year-old Oldenbrg mare (Argentinius x Raton x Rubinstein) that Jan has competed in three World Cup Finals, in Las Vegas in 2009, Leipzig, Germany in 2011 and 's-Hertogenbosch, the Netherlands, last April.
Jan and Amy Ebeling, Ann Romney, Ben, and Bether Meyer. © 2012 Ken Braddick/dressage-news.com
Valentina Truppa & Eremo del Castegno Take Nice CDI3* Grand Prix Freestyle
NICE, France, Apr. 6, 2015--Valentina Truppa rode Fixdesign Eremo del Castegno to victory in the CDI3* Grand Prix Freestyle Monday in the first competition in six months for Italy's... Read More
British Dressage Seeks Views on Possible Mandatory Safety Helmet Rule
StraightArrow 11 years ago
British Dressage is seeking views from its members on whether to follow the United States and Canada to mandate the use of safety helmets beginning with the 2012 competition... Read More
Oded Shimoni Competing, Coaching & More Coaching
LEXINGTON, Kentucky, Apr. 21--Oded Shimoni dismounted from Granada after competing in the Kentucky Cup CDI3* World Equestrian Games Test Event in the Kentucky Horse Park's new main stadium, hurried... Read More
Cornerstone Event/Markel Insurance Launch New West Coast Young Talent Search
BURBANK, California, Feb. 5--Cornerstone Event Management, organizer of several top California international dressage competitions, on Tuesday announced a partnership with Markel Insurance to launch a Young Talent Search for... Read More

redpajama_set_name: RedPajamaCommonCrawl


text:
Q: Clearing textinput in QML I have textInput box:
TextInput {
    x: 5
    y: 2
    maximumLength: 16
    width: maximumLength * 20
    height: 17
    focus: false
    validator: RegExpValidator { regExp: /\d+/ }
    KeyNavigation.down: amount
}

And a clear button.
When I click on clear button it should clear the text input box. How to do it?

A: You have to add the id property to your TextInput element.
TextInput {
    id: mytextbox
    x: 5
    y: 2
    ...
}

And on the event click of your MouseArea for your clear button you can do:
    onClicked: {
        mytextbox.text = "0";
    }

A: Try this one with your MouseArea:
onClicked: {
  textEdit1.text = " "
  mouseArea4.visible = false
}


text:
Banking sector of Kazakhstan displayed an impressive growth all over the Commonwealth in 2006
Kommersant-Bank
Rates of growth and levels of developments in the meantime vary from country to country. Kazakh banks did the best and reached the level typical of the most advanced countries of the world (in the context of the national economy, of course). Russian banks remain at the average CIS level.
Level of national banking systems is evaluated by the ratio between their total assets and the GDP. Where availability of banking services is concerned, Kazakhstan is an unquestionable leader in the Commonwealth (101.7%). All other countries are far behind - Ukraine with 63.5%, Russia with 52.8%, and Moldova with 51.7%.
The Uzbek banking system is the outsider. The ratio between assets and capitals went down there from 37% to 34.7%, and the rate of growth was one of the lowest throughout the Commonwealth. This banking system is one of the least transparent in the post-Soviet zone. Aggregate portfolios of Uzbek banks' loans went down from $3.3 million to $3.2 million in 2005. This figure was not even published in 2006 (a clear indication of continued deterioration).
Analysis of the parameters of the national banking systems in the last two years explains the Kazakh success. The ratio between assets and GDP more than doubled (48.5% to 101.7%). The situation with the ratio between loans and the GDP (32.7% to 68.7%) and capitals and the GDP (6.3% to 13.4%) is similar.
As a matter of fact, this performance is quite outstanding even by East European standards. The average ratio between assets and the GDP there amounts to 60-100% with the exception of the leaders like the Czech Republic and Croatia with the coefficient approaching 110%. The Kazakh banking system formally matches its analogs in the advanced countries like the United States (70%) and Germany (137%) though the room for improvement is still considerable bearing in mind that in France this parameter amounts to 239% and in Great Britain to 340%. The record figures shown by the Kazakh banking system are ascribed to the growth of its total assets related to the institutional reforms under way.
The Kazakh Central Bank launched radical reforms and reduced the number of commercial banks in the republic to 33 by January 1, 2007 (against 1,189 in Russia). This dramatic reduction allowed for a more exacting and versatile supervision approaching international standards. Kazakh banks provide reports in accordance with internationally accepted requirements. That and the mandatory audit made the national banking system more transparent, increased domestic and foreign investors' confidence in it, and generally improved the overall performance. The growth was so rapid, in fact, that local regulators decided to curtail it somewhat and specified a connection between the loans made by the banks and size of their own capitals.
Parameters of sufficiency of capitals is another importance nuance. The norms established in Kazakhstan (5-6% for the ratio between first level capitals and total assets and 10-12% for the ratio between capitals and assets evaluated by degree of risk) do not impede the growth of loans and, consequently, total assets.
It should be added as well that the Kazakh banking system is highly concentrated. Ten largest banks control 94% of total assets. It means that the whole system enjoys the highest possible pre-investment credit ratings. The said ten largest banks therefore set the general trend. In Russia, ten largest banks control only 54% of all assets. Reduction of the number of financial establishments usually boosts investment attractiveness of the remaining ones in the eyes of foreign players in the market.
Aleksei Buzdalin of INTERFAX-CEA
Kommersant-Bank, April 5, 2007, p. 29
© Translate by Ferghana.Ru


Film director and producer Vidhi Kasliwal's next is a documentary on Jain monk Acharya Vidyasagar. Titled Vidyoday, the film, which will be produced under her banner Landmarc Films, will trace the life of one of the most revered Digambar Jain Munis (philosopher monks).
"The film also enlightens us upon the various facets of Jainism as a philosophy, the frugal yet fulfilling lives of Digambara Jain Monks, their main teachings and principles, such as respecting life of all species and 'ahimsa' (non-violence)," said an official statement from the makers.
She added that shooting the documentary wasn't easy. "It was a difficult project to helm, but I was fortunate to have a wonderful team and we got utmost co-operation from all of Acharya Shri's pupils and followers all over the country. It took extensive research, intensive fieldwork, strenuous shooting, intricate editing and post-production work by close to 80 people over 1000 days canning 200 hours of footage to complete this 108-minute documentary," she said.
Kasliwal also stated the importance of the project. "And I speak for my entire team when I say this – working on this documentary has been one of the highlights of our careers and it has certainly left each one of us elevated and enriched," she said.
Kasliwal's recent productions include acclaimed Marathi films like Ringan (2017), Gacchi (2017) and Pipsi (2018).
Watch the teaser of the documentary by clicking HERE.
Thanks for sharing such a useful information. I appreciate your post.


It's time to pack! As of this point, the forecast for Paris looks cool-to-mild, with a chance of rain on a few days. While I've put this wardrobe together with my upcoming trip to Paris in mind, it could easily work for any other city during the fall.
When selecting the clothes, I prioritize pieces that will pack and travel well, and will be easy to layer and coordinate. While this doesn't make for the most exciting collection, it will be one that feels smart, appropriate, and comfortable (both physically and emotionally).
My neutrals are black, navy and grey, and I've included a couple of accent pieces with color and/or print. I challenged myself to pack a little less black and more navy this time. I've stuck to a long-over-lean silhouette, and my usual ratio 2-to-1 ratio of tops to bottoms. I don't count shoes, accessories, outerwear or "underpinnings" as part of the 12 items. You can certainly pack more or less or include dresses or tweak the ratio to suit your preferences. This is just the formula that I've found works best for me.
I'll also allow myself to add a "wild card" piece to be chosen at the last minute, based on the current forecast and my mood.
Some of the pieces above are no longer available; in those instances I've linked to similar items.
I always start with the shoes when planning my travel wardrobes. I've decided to allow myself 3 pairs this time.
Slip-on sneakers. These are just so darn comfortable. Here's a similar style that's waterproof and budget-friendly.
Something funky. (Similar style) I purchased these Robert Clergerie brogues in New York last year, and logged several miles in them the first day. They've passed the Travel Comfort test with flying colors. They're not for rainy days, but give me a fun break from black footwear.
As mentioned above, I've kept to slim-leg or skinny styles. This silhouette works best for me and is easiest to style. But again, go with whatever silhouette works best for you.
Skinny jeans. Dark wash denim, no distressing.
Navy sateen slim leg jeans. A little dressier than the denim.
Slim-leg ponte knit pants. Comfortable but polished. Can be hand washed if necessary. These are my in-flight pants.
Floral top. The "throw on and go" option.
Black zip-front blouse (similar). I've had this one from Theory for a few years, and it's still one of my favorite pieces.
Cashmere cardigan (similar here and here).
I'm still deciding. My scarf wardrobe has dwindled a bit and could use a refresh. I'll be shopping in Paris for some solid, colored scarves.
The bag is the Pearl from Lo & Sons. And based on my experience in the spring with cooler-than-expected temperatures, I'll probably toss in my packable down parka (similar).
How far ahead do you plan and pack your travel wardrobes?
Room With A View: Paris!
I'm looking forward to your search for scarves in Paris. Although my budget doesn't often let me buy top-end, I'm an easy touch when Hermes scarves and shawls are available. I wear them every chance I get.
P.S. Those Robert Clergerie shoes are way cool!
Instead of our packable parka or lined rain coat, why not bring a packable down vest you can layer under the two coats you are bringing? That way you can take it off if it warms up during the day.
I bought a couple of scarves from Monoprix when I was there in September. They are pashima size. I bought a blue and a nice muted yellow (good for fall).
I agree with the vest idea. I usually go to Paris in September or October and the weather can range greatly. I had stressed over packing a large enough range of tops and coats to handle this, then swapped out some of the warm sweaters and heavier coat for a vest for layering. Lots more flexibility and less stuff.
I thought that was a good idea too. I've nixed the heavier raincoat idea and added a lightweight insulating vest to the mix.
I have just returned from a four week trip to the US and was thinking about and planning my wardrobe for months but failed miserably!
We knew it would be warm to hot for the first 9 days (from mid September) in California and Las Vegas but were completely thrown by this weather continuing when we flew to New England where we expected autumn leaves and mild to cool weather. The final week, last week, in New York was hot and unbearably humid.
Hi Susan! Just returned from Austria/France trip-it was cooler in Austria as expected but layers & EF charcoal knee length anorak worked great. Paris was warmer than usual so tunic length black knit jacket worked well. Did take 2 short knit dresses in charcoal & burgundy, wore w leggings & ankle boots or flats- saw lot of same in Vienna & Paris so I fit right in! Also the "new" green is everywhere in the good shops & that was my wild card piece, a long cardi so I was able to get some cool accessories. I took similar sneakers & saw often in Paris so ok with skinny jeans when foot fatigue sets in-have fun in Paris! Wonderful beyond words as always!!!!
Jan, I'm curious as to what kind of knit dresses you took?
I was interested in the "in-flight" pants comment. Are the EF ponte pants sturdy enough to avoid damage from the seats on a long flight? The in-flight outfit is a dilemma for me, because I want something polished, comfortable, with fabric that won't pill or get worn from my restless shifting. One flight to Asia wrecked one of my favorite pairs of pants!
Sue, I've found both these and the EF stretch crepe pants to work well for long flights. I've never had any problems with them snagging or pilling.
That looks like a good wardrobe. Down here in the south the weather is still balmy–windows open even at night–but Paris is considerably cooler.
I love your brogues. Interesting shoes are very in.
I love these travel wardrobe posts! Have you considered doing one for travel to a resort? I know that it should seem simple, but I've never seen anything that's "age appropriate." I can just bring a bikinis, cut off denim and flip flops. Please?
How were you able to "dwindle" your scarf collection?
And what makes a great scarf in your opinion? especially for travel?
Hi Mary, some just plain wore out, or were in colors that no longer suit me. What makes a great scarf for travel depends on the season, but I usually like to have one cashmere stole or large scarf for the plane, and then will choose the rest based on weather. I find oblong rather than squares are easiest for travel.
I am going to London and then Antwerp in early November and am really stressing about what to wear. Some people have said that jeans and sneakers (even nice ones) are not appropriate. But we are going to be doing a ton of walking and I know my feet will be killing me if I don't have the right kind of footwear.
And Jan, sorry you ran into our horrible humidity. I live in Mass. – it just now today is finally cooler out!
Wear what's comfortable – you'll see everything in London from sneakers to Laboutins.
Bonnie, there was a time (and still may be) when many Americans went all over the world wearing white athletic sneakers. This was really not what anyone else in the world was doing, hence the warning you received. I believe the more modern "city" sneakers are more acceptable now. My last trip to Europe was two years ago and I was surprised by how many people everywhere (on the way, airports, etc.) were wearing white-soled sneaker- type shoes.
New England was thankfully as beautiful as ever Bobbie!
@ Bonnie, I have never been to Antwerp but have visited London more than once as well as other European cities. I have never felt out of place wearing jeans or sneakers for sight-seeing. We saw so many Adidas sneakers in France this summer! Except for a very nice dinner or occasion, jeans and sneaks should be fine.
Thank you for sharing your travel wardrobe. And have a wonderful time in the most beautiful city I have seen.
The 12 piece travel wardrobe has been my travel saviour. The first time I did it, I approached packing like a military exercise – it took a month of choosing and honing – and I still took things I didn't wear. Now that I've used the rule a few times, I can pretty much leave the planning until the week before I leave.
Neutrals really are the key – it's like a travel uniform.
My only piece of advice is, like you, to pack tried and true footwear. We North Americans seem to obsess over whether we'll fit in footwear-wise – why wreck a trip you've paid thousands of dollars for to meet some stranger's perceived expectations? I've never noticed if anyone looks at my feet or cares – I'm too busy sightseeing/shopping/dining!
Haha Jane2, you've hit the nail on the head regarding footwear and perception! I have lived in New Zealand for over 40 years but still carry some silly British notions in my head about appearance.
I hope you're planning to see the Christian Dior exhibit. I was there yesterday, and it was out of this world. Truly, truly fabulous. I bought my ticket in advance so I could skip the long line. Have a great trip!
Yes! Josephine and I have tickets. Can't wait!
Enjoy your blog. I am currently in Venice…. I am with a carry on and find that my skirt/dresses are taking me everywhere. I see more skirts/dresses than I see pants. For me I am far more comfy in my skirts than pants. I as well see the Italian women in a skirt over pants. I even fly in my dress (with leggings and a wonderful jacket/leggings and scarf— I come away from the airport and am ready to visit the city, as my Hotel room is rarely ready). Works for me. Have a great time!
Mentally, my head is always in Paris 🙂 When physically going there I start planning six weeks prior and pack three days before my flight. When I was there in early September the weather was quiet cool with two rainy days. The Dior exhibit was the perfect antidote to the rain.
Your wardrobe looks lovely and appropriate. Jeans and sneakers were everywhere!
We go to France nearly every year because we have family there. Not every trip includes Paris, but we have been many times. I find that an issue I have with packing is over-thinking it. I will sometimes take an item because it works in the "theory" in my head, but it doesn't get enough wear in the real world to justify having packed it. I've been working on that one now for decades, and I am getting better! I take only one bag for a month-long trip, my trusty Lipault carry-on, doing lots of laundry along the way. One horror story I can share is the time my one bag did not arrive in Paris when I did and showed up four days later. My answer was Monoprix, and I got one scarf there I have now been wearing for years.
Can you fit all this, including one of the coats, in a carry-on?
I can fit all of this and one of the coats into a carry-on, yes. I use packing cubes for compression which helps save space. I'll wear some of these items on the flight, and usually the bulkiest of the shoes.
Your packing-for-travel posts are so helpful! How many days are you packing for here, and what size is your carry-on? I'm leaving for 6 days in Vancouver, BC, the day after tomorrow, and think I've fit everything into a 21"x 14" 4-wheel roller bag. But I have't fit in everything that you have.
Hi Martha, I'm going for 8 days, but this is the same wardrobe I'd pack for a two week trip, and probably just do some laundry at a halfway point. The carry-on I normally use is 20″ x 14″ x 10″. However, because I plan to do some shopping on this trip and will also be bringing home some skincare and other liquids, I'm taking a slightly larger hard-sided bag that I'll check.
I love your brogues. They will add a playful touch to your outfits. I'm happy that brogues/oxfords made a comeback. So comfy!
Can you recommend a packable rain jacket/coat for travelling?
I love your packing posts. It is always the most stressful part of the trip for me. Lay out…put away…add…take away… I'm exhausted before I board the plane.
Helpful guide and very similar to what I'm packing for New England! I'm just bringing one jacket and I'll use that for colder temps and sweater(s) for fall chill. I picked up a very soft merino wool sweater at Costco, of all places, while browsing as my daughter shopped (I tag along as my daughter's guest). I ordered EF denim leggings when they were on sale and we'll see how they fit when they arrive, and if they make the cut. Otherwise, no denim for this trip. I prefer cotton tops and layering for warmth over that. I cannot remember the last time I wore a wool sweater top on its own, probably when I lived in NH! Speaking of sweaters, so much no on the Gia sweater with holes. Even if I sound too prim and sensible, "holes in wool/cashmere" is all wrong, even for fashion, and it even looks strained, sagging open on the pressure of the shoulders. Ugh. Also, because we take so much care to not get any in our beautiful sweaters in the first place. I personally feel that holes in clothes, as in all the denim, is a conspiratorial joke on the masses by the industry, "let's see if we can get them to wear clothes that look trashed." And they did! Important to remember to not lose one's sense of style and individuality which, to me, is a fashion statement.
Just retuned from Paris this evening. The color that was everywhere was mustard gold/yellow. Handbags, scarves, sweaters and shoes. Looks great with navy or black.
Ooh, thanks for this! Mustard/gold is one of my favorite colors, and I wear it a lot for work (school colors). I'm glad to know I'll also be au courant while I'm wearing it!
I totally agree with the over-thinking it comment. Once, when my luggage remained state-side and I landed in Paris I had only what I was wearing and my carry-on with one change of clothes. Black pants, black cashmere crew neck and a Hermes scarf. (and undies) My checked bag took 3 days to arrive. I learned then my carry-on plus my great looking water-proof coat, great short boots and great bag were really all I needed and I looked as good as I did after my bag arrived. Tired, well worn (though comfy) jeans and tops are never a part of my city wardrobe. Simple, chic, monochromatic, pulled-together items are key.
By the way, have you taken the tram (T3) around the edge of Paris yet? It is cutting edge, and can provide surprising access to different parts of the city each a short walk towards the city centre. It is also a great boon for disabled people.
Having traveled in the autumn/spring to find unexpectedly cool weather I alway pack a pair of silk long underwear.- camisole and pants. They have saved me from freezing several times. Can also be pair of backup PJ's. Just a thought.
I always pack a silk camisole and a silk long sleeve, even when going to warmer climates. Silk breathes, and ya never know when you might need an extra layer. I suffer from Reynaud Syndrome, so even on warmer days if I sit in an overly air conditioned room too long my fingers and nose can go numb. I've also learned to pack a very light weight fleece "just in cases", because for me, July can turn into November in fifteen minutes or less.
Always better to pack too much and be over prepared than to pack too little and be left wishing you'd pack something you didn't!
My husband and I just returned from traveling for one month visiting Berlin, Paris and Dublin. We both carried an Osprey Farpoint 40 Travel backpack 21X13X9. We travel independently using public transportation in cities, trains between countries and airbnb. We are 59 years old. My husband always used the back pack, but I always used the rolling small carry thinking I was too "old" for a backpack and the rolling carry on was "easier". After several trips with the carry on, I decided this year to try the backpack. Too say the backpack is easier is an understatement! I know many women have back issues and cannot physically handle a back pack, but if your back is in good condition, I can heartedly recommend this alternative! The rolling carry on is not convenient for those of us who use public transportation. The buses, metros and cobblestones make life so difficult with a rolling carry on! I pack light when we travel to Europe in the fall with black, gray and burgundy neutrals: 1 black knee length dress, 1 dark denim skinny jean, 1 EF knit straight pant, leggings, tights, silk underwear, smart wool socks, 1 pr thick cushioned ankle socks (double as slippers in the airbnb apartments), compression socks, 1 pair ankle boots (Try English made Hotter boots for out of this world comfort and European chic!), 1 merino wool cardigan, 1 burgundy cashmere V-neck sweater, 2 tank tops, 1 short sleeve t's, 2 long sleeve t's, 1 yoga pant for bedtime, 1 solid burgundy, very large dense wool scarf (I call this scarf my "secret weapon" against any surprise cold weather in the fall) & one patterned burgundy cashmere & silk scarf. C'est tout! I take one dark navy slightly above knee belted trench coat with detachable hood and layer underneath as needed with the merino wool & cashmere sweaters. I have learned to layer for unexpected cold weather or just wear less underneath the trench coat for warm temps. I use a "travelon" black messenger bag from ebags that has the same safety features as the well known pacsafe at half the cost. I always bring a small packable tote to carry our groceries back to our apartment and a large packable gym bag to carry our souvenirs on the plane & we check our backpacks on our return flight. It is true as another reader pointed out, I just don't need that much! If anything, I tend to wind up wearing the same building block pieces & using a colored sweater or scarf to change the look. What I have learned over the years is that no one is looking at me! It is me having the privilege to meet and talk with many wonderful people, gaze at the awe inspiring architecture, learn about the centuries of rich European history, share the European culture, and eat sumptuous food! So long as I am comfortable, have somewhat a European "continental" style, and I'm wearing clean clothing which is appropriate for the weather, it doesn't matter to me that I am wearing the same outfit! I know this will not appeal to everyone….but going with a few neutral, comfortable, well made pieces of clothing has taken the anxiety, worry and spending too much money for "travel clothing" out of my travel!
Helas! I found beautiful EF slip on loafers in metallic at TJ Maxx today…the 7 was too short and the 7 1/2 were way too wide! Sad, as the price was amazing and they were of the softest possible leather in the uppers. I've decided what I need to explore the hinterlands in Provence (the Camargue, etc.) is a good, not too "trendy" white denim jacket. It will go will all my casual blacks and navies…Any suggestions for something that looks fairly classic? I ordered the Everlane jacket, but the fit doesn't quite work with my longer torso, but shorter arms.


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Behavioral Targeted Emails, or sending targeted emails to your contacts based on their actions and behaviors, is one of the most effective strategies an inbound marketer can adopt.
A lack of technical know-how and strategy makes behavioral emailing a difficult task. To make it worse, there is not enough quality educational material present to get you up to speed.
This post is exactly written as a countermeasure to help you reap the benefits of Behavioural Targeting.
What is Behavioral Email Targeting?
Behavioral email targeting is the practice of sending automated, targeted emails to the contacts in your database based on their interactions with your brand across multiple channels: on-site, email, your mobile app and beyond.
Behavioral Email Targeting is all about reacting the user with appropriate communication based on their interaction and engagement with your brand.
If a user is dealt with the proper messages and on the correct channel for interaction, the user is more likely to engage with your brand.
According to the DMA, Over 75% of email revenue is generated by triggered campaigns, rather than one-size-fits-all campaigns.
Behaviourally Targeted emails are based on user action and triggers, as discussed above. It is one of the most sorted out techniques to tackle instances like abandoned cart, pushing product updates news to users, etc.
For example, when you abandon any product on eCommerce websites like Amazon, Etsy, etc., you will start to notice that all their messaging is related to the product that you abandoned. Most of these brands will remind you via emails, push-notifications, etc. that you haven't checked out yet.
Behavior-based targeting solves a major problem for these brands by asking customers to return and complete a specific action.
According to the DMA report, in 2014, emails triggered by behavior were responsible for 30% of revenue – up from 17% in 2013.
Brands that are leveraging behavior based targeting are weaving their brand around the customer (or, according to the customer).
You know how Amazon suggests items that might interest you, based on the items you've browsed? That's how the entire Internet could work soon. With systems tracking your cursor and keeping tabs on your browsing history, the Internet can get to know you better — and be smarter than ever. In fact, Amazon's algorithm is a paragon of excellence for marketers.
We have already seen firms like Forbes, Chowhound and Snopes doing the same for their websites. Their newsletters and on-site recommendation is completely based on user activity and behavior.
Also building a user-experience that's ideal according to the user, gets them to relate with a brand on a personal level. It builds trust between you and the brand.
Marketers are targeting users by leveraging all their interests and activities. It helps marketers avoid the tunnel vision marketing technique. By tunnel vision, we mean, promoting a single product or a small group of products, because the user was interested in a product from that niche.
The problem traditional marketers face with such a technique is that such messaging would only irritate customers.
As humans, we have multiple interests and like a variety of things. A product that I was looking at today, a pair of Nike shoes, probably won't pique my interest a month later. Maybe I'll be onto a completely different product, like a smartphone.
As a result, Behaviour targeted algorithms were developed to understand a customer in a better manner. It takes in account all sort of details that it can get its hands on, from demographics to email interaction, web activity, messaging reaction and more. In the end, you know your customers better, and the customer also has a quality experience with your brand.
Behavior-based email targeting or targeting, in general, is immensely useful for marketers.The more users consume information on the Internet, the more we learn about them. That being said, don't become a creepy marketer. Data still is not identifiable personally, but one can still harness the strength of AI and machine learning to target users based on their behavior.
At the end, your brand focus would shift towards being more customer-centric—the ultimate goal that we are trying to achieve.


text:
Malaysia's royal council meets after king's surprise resignation
Reuters Mon, 07 Jan 2019 Mon, 07 Jan 2019
Malaysian King Muhammad V attends a welcome ceremony at the Parliament House in Kuala Lumpur, Malaysia February 26, 2017. (AFP FIle Photo)
Sultan Muhammad V (R), preparing to deliver his address during the opening ceremony of the parliament in Kuala Lumpur as Malaysian Prime Minister Mahathir Mohamad (L) looks on. Malaysia's King Sultan Muhammad V has abdicated, a statement from the National Palace said on January 6, 2019. (AFP FIle Photo)
KUALA LUMPUR - Members of Malaysia's royal families met on Monday to decide when a new king will be elected from among them after King Muhammad V unexpectedly resigned, the national
news agency Bernama reported. The king resigned on Sunday after just two years on the throne, the first time a monarch has stepped down before completing their five-year tenure.
No reason was given for the resignation. Malaysia is a constitutional monarchy and the king assumes a largely ceremonial role, including acting as custodian of Islam in the Muslim-majority country. The king's
assent is needed for the appointment of the prime minister and various senior officials.
Malaysia's nine royal households take turns to provide a king, who is chosen through a vote in a Council of Rulers, made up of the nine households, most of which are led by a sultan.
Heads of six of the nine households met on Monday at the national palace to decide a date for the election of a new king, Bernama reported. It was not clear why the other three did not attend.
A vote must be held within four weeks of the throne becoming vacant.
King Muhammad, 49, had resumed duties last week after spending two months on medical leave.
Images purporting to show him getting married in Russia appeared on social media in December and identified his new wife as Oksana Voevodina, who won a Miss Moscow contest in 2015.
The palace did not respond to earlier requests for comment on the photos or reports of a marriage.
Media has reported tension between the palace and the government of Prime Minister Mahathir Mohamad since the latter returned to power last year.
Mahathir led the opposition to a historic election win in May, defeating a coalition that had governed Malaysia for 60 years.
In June, the government and palace faced a near two-week impasse over a plan to appoint someone how was not from the majority ethnic Malay community as attorney-general.
The king eventually approved the appointment, though the incident had stoked ethnic tensions.
Mahathir, known for challenging royalty during his earlier 22-year tenure as prime minister, said in a blog post last week that everyone "from the Rulers to the Prime Minister and Ministers, to the civil servants
and ordinary citizens" are subject to the law. He did not elaborate.
Mahathir said on Monday the government hoped the council would elect a new king as soon as possible as the government needed to keep the king apprised on "certain matters", Bernama reported.


text:
\section{Introduction}

In the last decades a lot of work has been done
on so-called Hilbert-Speiser fields; in particular, criteria were developed which
insure that a given field is not Hilbert-Speiser. The framework is Galois module structure of extensions of number fields.
More precisely the object of our study is the property of an extension of
number fields to have a \textit{normal integral basis}: an extension $L/K$ of number fields admits a NIB if $\ensuremath{\mathscr{O}}_L$ is a rank $1$ free
 $\ensuremath{\mathscr{O}}_K[G]$-module, where we denote by $\ensuremath{\mathscr{O}}_L$ and $\ensuremath{\mathscr{O}}_K$ the corresponding rings of integers. The Hilbert-Speiser theorem states that for abelian number fields tameness is equivalent to the existence of a normal integral basis over $\Q$, while
 in general we only have that an extension with NIB has to be tame.
By definition
a number field $K$ is Hilbert-Speiser if every tame abelian extension $L/K$ has NIB,
 and $K$ is $C_l$-Hilbert-Speiser if every such extension with Galois group
isomorphic to $C_l$, the cyclic group of prime order $l$, has NIB.

 The first contribution to the topic of Hilbert-Speiser fields
came from the important result contained in \cite{GRRS99}:
$\Q$ is the only Hilbert-Speiser field,
i.e. for every number field $K\supsetneq \Q$ there exists a (cyclic of prime order) tame abelian extension that does not have NIB. Subsequent research went towards
the finer problem of finding criteria for $C_l$-Hilbert-Speiser fields. For instance, in \cite{Car03}, \cite{Ich04} and \cite{Yos09}
we find some conditions for abelian $C_2$ and $C_3$-Hilbert-Speiser
fields. On the other hand, we know from \cite{Her05} and \cite{GJ09}
that for arbitrary $l$, the prime $l$ cannot be
too highly ramified in $C_l$-Hilbert-Speiser fields.

In this work, we intend to consider a weakened version of NIB:
we will say that a tame abelian extension $L/K$ of number fields has
a \textit{weak normal integral basis} (WNIB for short)
if $\mathcal{M}\otimes_{\ensuremath{\mathscr{O}}_K[G]}\ensuremath{\mathscr{O}}_L$ is free of rank $1$ over $\mathcal{M}$, where $\mathcal{M}$ is the maximal order of $K[G]$.
WNIB's have been studied for instance in \cite{Gre90}, \cite{Gre97} and \cite{GJ12}.
We shall ask the same
questions as above, substituting ``WNIB'' for ``NIB'' everywhere, and a number
field $K$ ``Leopoldt field'' (instead of ``Hilbert-Speiser field'')
if every tame abelian extension $L/K$ has WNIB.
We are mainly going to establish necessary conditions for number fields
to be $C_l$-Leopoldt, where as before $l$ is a prime number. This
also gives criteria and (sometimes conditional) finiteness results
for $C_l$-Hilbert-Speiser fields; for instance we will see that this permits to
correct an oversight in the article \cite{Ich16}, whose techniques,
even though they were originally
conceived to deal with Hilbert-Speiser fields, turn out to be supple enough
to be applied to Leopoldt fields as well.
Ichimura informs us that the corrected proof of Proposition 2 of \cite{Ich16},
to appear in a corrigendum, will proceed by a slightly different
argument Ichimura already knew at that time, but which finally was not
used in the paper  \cite{Ich16}.

First of all we introduce some notation:

\begin{itemize}
 \item $\text{Cl}(\Lambda)$ denotes the locally free class group $\Lambda$, where
 $\Lambda$ is an order over a Dedekind domain (see for instance \cite[$\S$39A or $\S$49]{CR87})
 \item $h_K$ is the class number of the number field $K$
 \item $h_K^-$ is the relative class number of the CM-field $K$
 \item $h_l^-:=h_{\Q(\zeta_l)}^-$
 \item $h(d):=h_{\Q(\sqrt{d})}$
 \item $N_{L/K}$ stands for the norm map, defined either on the fractional ideals or directly between the ideal class groups of the number fields; namely the mutiplicative extension of the map $\mathcal{P}\mapsto\wp^{f(\mathcal{P}|\wp)}$ where $\mathcal{P}|\wp$ is an extension of prime ideals
 \item $H_K$ is the Hilbert class field of $K$
\end{itemize}

In the remainder of this introduction,
let us recall the concepts and formulate the questions we will
be dealing with in this paper, and review a few standard results
which are important in the sequel.

\begin{definition}
 Let $L/K$ be a tame $G$-Galois abelian extension of number fields, and let $\mathcal{M}$ be the maximal order of $K[G]$. The extension has a weak normal integral basis
 (WNIB)
 if $\mathcal{M}\otimes_{\ensuremath{\mathscr{O}}_K[G]}\ensuremath{\mathscr{O}}_L$ is free of rank $1$ over $\mathcal{M}$.
\end{definition}

Without any essential changes to the classical proofs on normal integral bases, we have the following
\begin{proposition}\label{disjointwnib}
 Let $L/K$ be an abelian tame extension of number field and $F$ an intermediate field. If $L/K$ has a WNIB, so does $F/K$.

 Let $L_1/K$ and $L_2/K$ be arithmetically disjoint extensions of number fields with $L_1/K$ tame abelian. If $L_1/K$ has WNIB, so does $L/L_2$, where $L=L_1L_2$.
\end{proposition}

We recall that there is the following surjection, with kernel denoted by $D(\ensuremath{\mathscr{O}}_K[G])$:
\begin{equation*}
 0\longrightarrow D(\ensuremath{\mathscr{O}}_K[G])\longrightarrow \textup{\text{Cl}}(\ensuremath{\mathscr{O}}_K[G])\longrightarrow\textup{\text{Cl}}(\mathcal{M})\longrightarrow 0.
\end{equation*}

\begin{definition}
 Let $K$ be a number field. We say that $K$ is Leopoldt if every tame abelian extension $L/K$ has WNIB, and
 $G$-Leopoldt fields if every tame abelian extension $L/K$ with Galois group isomorphic to $G$ has WNIB.
\end{definition}

\begin{remark}
  We note that other authors (see for example \cite{Car07} and \cite{BCGJ11}) employ the term
  ``Leopoldt field'' in a different (and stricter) sense, so that we briefly
  considered calling our definition ``weakly Leopoldt'';
  but we have decided against it for simplicity.
\end{remark}

As we stated in the abstract, $\Q$ is the only Hilbert-Speiser field. One may ask if this is still true in the weaker sense:
\begin{conjecture}
 $\Q$ is the only Leopoldt field.
\end{conjecture}
However, there is no chance to adapt the proof of \cite{GRRS99}. A fundamental ingredient is the study of objects related to
Swan modules, which live in the kernel group, and
this approach cannot work in our context as we are interested in a quotient by the kernel group (that is the class group of the maximal order).
Instead we study $C_l$-Leopoldt fields directly.

In two short sections we will deal with the simple cases $l=2$ and $l=3$,
for the sake of completeness.  In Section \ref{real}
we develop necessary conditions for totally real fields that intersect $\Q(\zeta_l)$ trivially
to be $C_l$-Leopoldt. In Section \ref{hilbert-speiser} we
prove that the list of real abelian $C_l$-Hilbert-Speiser fields for certain values of $l$ given in \cite{Ich16} is actually complete.
The final Section \ref{intersection} presents some results concerning
fields that may intersect $\Q(\zeta_l)$ nontrivially.

In the rest of the introduction we review some basic tools.
We can firstly use McCulloh's important theorem \cite{McC83} to speak about realizable classes. Let us recall it.
First of all, the augmentation map $\varepsilon:\ensuremath{\mathscr{O}}_K[G]\rightarrow \ensuremath{\mathscr{O}}_K$
defines a map
\begin{equation*}\begin{split}
 \varepsilon_*:\text{Cl}(\ensuremath{\mathscr{O}}_K[G])&\longrightarrow \text{Cl}(\ensuremath{\mathscr{O}}_K)\\
 [M]&\longmapsto[\ensuremath{\mathscr{O}}_K\otimes_{\ensuremath{\mathscr{O}}_K[G]}M],\end{split}
\end{equation*}
where by $[M]$ we mean the class of $M$ in the locally free class group.
We define $\text{Cl}^0(\ensuremath{\mathscr{O}}_K[G])$ to be the kernel of the above homomorphism. When $G$ is elementary abelian, we want to give the group $\text{Cl}(\ensuremath{\mathscr{O}}_K[G])$ a structure of
$\Z[\Delta]$-module, where $\Delta$ is the multiplicative group of the finite field whose additive structure is isomorphic to $G$:
if $G\cong \F{l^n}^+$, then $\Delta:=\F{l^n}^*$. This
is simply done noting that each element of $\Delta$ defines a ring automorphism of $\ensuremath{\mathscr{O}}_K[G]$ and hence a functorial endomorphism of $\text{Cl}(\ensuremath{\mathscr{O}}_K[G])$.
Now let
 \begin{equation*}
  \theta=\frac{1}{l}\sum_{g\in \Delta}t_{\F{l^n}/\F{l}}(g)g^{-1}\in\Q[\Delta]
 \end{equation*}
 be the Stickelberger element, where $t_{\F{l^n}/\F{l}}$ is the composition of the trace $\text{Tr}_{\F{l^n}/\F{l}}$ with the integer representative of the elements of
 $\F{l}\cong\Z/l\Z$ between $0$ and $l-1$, and
 \begin{equation*}
  \mathcal{J}=\Z[\Delta]\theta\cap\Z[\Delta]\subseteq\Z[\Delta]
 \end{equation*}
 the Stickelberger ideal.

 \begin{theorem}[McCulloh \cite{McC83}]\label{mcculloh}
 Let $K$ be a number field and $G$ an elementary abelian group. Then
 \begin{equation*}
  R(\ensuremath{\mathscr{O}}_K[G])=\textup{\text{Cl}}^0(\ensuremath{\mathscr{O}}_K[G])^{\mathcal{J}},
 \end{equation*}
 where we denote by $R(\ensuremath{\mathscr{O}}_K[G])$ the ``realizable classes'' in $\textup{\text{Cl}}(\ensuremath{\mathscr{O}}_K[G])$, i.e. given by the tame Galois extensions of $K$ with Galois group $G$.
\end{theorem}
Using a cancellation property for projective modules
(which can easily be established over commutative rings), we obtain:
\begin{corollary}
 A number field $K$ is $G$-Hilbert-Speiser with $G$ elementary abelian if and only if the Stickelberger ideal $\mathcal{J}$ annihilates $\textup{\text{Cl}}^0(\ensuremath{\mathscr{O}}_K[G])$.
\end{corollary}
This is of course the main ingredient for the research on Hilbert-Speiser fields that has been done.

Now let us return to our to our main focus of studying Leopoldt fields.
As before we can take $G$ to be any elementary abelian group and obtain
\begin{equation*}
 R(\mathcal{M}):=\left\{ \left[\mathcal{M}\otimes_{\ensuremath{\mathscr{O}}_K[G]}\ensuremath{\mathscr{O}}_L \right]:L/K \text{ } G \text{-tame}\right\}=\text{im}\left(\text{Cl}^0(\ensuremath{\mathscr{O}}_K[G])^{\mathcal{J}}\right)=
 \text{im}\left(\text{Cl}^0(\ensuremath{\mathscr{O}}_K[G])\right)^{\mathcal{J}}
\end{equation*}
(the action of $\Delta$ commutes with the map $\textup{\text{Cl}}(\ensuremath{\mathscr{O}}_K[G])\rightarrow\textup{\text{Cl}}(\mathcal{M})$).
The extension of scalars to the maximal order does not change the augmentation,
so from the surjectivity
we obtain that
\begin{equation*}
 R(\mathcal{M})=\text{Cl}^0(\mathcal{M})^{\mathcal{J}}.
\end{equation*}

We recall that we know what the maximal order $\mathcal{M}$ is. If $K\times K(\zeta_l)\times\cdots\times K(\zeta_l)$ is the Wedderburn decomposition of
$K[G]$, then $\mathcal{M}$ is simply $\ensuremath{\mathscr{O}}_K\times \ensuremath{\mathscr{O}}_{K(\zeta_l)}\times\cdots\times \ensuremath{\mathscr{O}}_{K(\zeta_l)}$ and thus the part annihilated by
augmentation is $\ensuremath{\mathscr{O}}_{K(\zeta_l)}\times\cdots\times \ensuremath{\mathscr{O}}_{K(\zeta_l)}$.
Therefore we get the following
\begin{corollary}\label{leocond}
  Let $K$ be a number field and $G$ an $l$-elementary abelian group. Then $K$ is $G$-Leopoldt if and only if the Stickelberger ideal $\mathcal{J}\subseteq\Z[\Delta]$
  annihilates $\textup{\text{Cl}}(K(\zeta_l))\times\cdots\times\textup{\text{Cl}}(K(\zeta_l))$.
\end{corollary}

\begin{remark}\label{leoremark}
 Assume $G=C_l$ (so that $\Delta\cong (\Z/l\Z)^*$) and let $K$ be a number field. When $K$ is disjoint from $\Q(\zeta_l)$, there is a simple consequence of the above
 corollary: $K$ is $C_l$-Leopoldt if and only if $\mathcal{J}$ annihilates $\text{Cl}(K(\zeta_l))$, where the module structure of $\text{Cl}(K(\zeta_l))$ over $\Z[\Delta]$
 is the classical one from Galois theory.

 More generally, let $H\cong\text{Gal}(K(\zeta_l)/K)$ be a proper subgroup of $\Delta=\{\sigma_1,...,\sigma_{l-1}\}$. We can identify
 $H$ with the subset of the $\sigma_j$'s such that there exists an automorphism of $\text{Gal}(K(\zeta_l)/K)$ which sends $\zeta_l\mapsto \zeta_l^j$.
 Looking at the structure of $K[G]\cong K[z]/(z^l-1)$, the set of the $(l-1)/d$ components of $\text{Cl}^0(\mathcal{M})$, which are equal to $\text{Cl}(K(\zeta_l))$, where $d=|H|$,
 is permuted via a free action of $\Delta/H$, while the module structure of $\text{Cl}(K(\zeta_l))$ over $\Z[H]$ given by restriction is the classical one,
 namely with the identification $H\cong \text{Gal}(K(\zeta_l)/K)$.

 Hence if $K$ is $C_l$-Leopoldt then $\pi_H(\mathcal{J})$ annihilates $\text{Cl}(K(\zeta_l))$, where $\pi_H:\Z[\Delta]\rightarrow \Z[H]$ is the map
 that only conserves the coefficients of the elements of $H$.
 One can easily verify that $\pi_H(\mathcal{J})$ is an ideal of $\Z[H]$ (although $\pi_H$ is not a ring homomorphism).
 The work by Ichimura and Sumida-Takahashi on the image of the Stickelberger ideal through this map will turn out to be very useful
 for our purposes.
\end{remark}

\section{$C_2$-Leopoldt fields}
In this situation for every field $K$ the component of the maximal order in $K[C_2]$ annihilated by the augmentation (let us call it zero-component) is $\ensuremath{\mathscr{O}}_K$. We need to know the Stickelberger ideal of $\Z$, but
in this case $\theta=\frac{1}{2}$
and so $1\in\mathcal{J}$. Therefore Corollary \ref{leocond} tells us that the class group has to be trivial and we have the following

\begin{proposition}
 A number field $K$ is $C_2$-Leopoldt if and only if $h_K=1$.
\end{proposition}

\section{$C_3$-Leopoldt fields}
Here we have to study the Stickelberger ideal of $\Z[C_2]$. But this is the unit ideal as before, because $1=(2j-1)\theta$.

If $\sqrt{-3}\in K$, then the zero-component of the maximal order in $K[C_3]$ is $\ensuremath{\mathscr{O}}_K\times \ensuremath{\mathscr{O}}_K$, so as before we have:

\begin{proposition}
 A number field $K$ with $\sqrt{-3}\in K$ is $C_3$-Leopoldt if and only if $h_K=1$.
\end{proposition}

In the other case, the zero-component of the maximal order is necessarily $\ensuremath{\mathscr{O}}_{K(\sqrt{-3})}$, so we obtain the following general proposition:

\begin{proposition}
A number field $K$ is $C_3$-Leopoldt if and only if $h_{K(\sqrt{-3})}=1$.
\end{proposition}

When $K$ is an abelian extension $K(\sqrt{-3})$ is an imaginary abelian extension of $\Q$, and from Yamamura \cite{Yam94} we know the imaginary abelian number fields
with class number $1$.

\begin{corollary}
 The abelian Leopoldt fields of the type $C_3$ are finite in number.
\end{corollary}

\begin{remark}
 Taking $K$ general, looking at the Hilbert class field of $K$ we conclude that $h_K$ can be $1$ or $2$. We cannot say that
 $h_K=1$, like the $C_2$ case: for example we know from \cite{Yam94} that $\Q(\sqrt{3},\sqrt{-7},\sqrt{-3})$ has class number $1$, while $\Q(\sqrt{3},\sqrt{-7})$ does not
 (and so the class number is $2$). However the conclusion holds whenever $K$ is not ramified in $3$ or it is not totally imaginary, because there is total ramification (in
 primes over $3$ or
 the real archimedean ones) and so $h_K=1$ by \cite[Theorem 10.1]{Was97}.
\end{remark}

There is one more situation in which we have a finiteness result on $C_3$-Leopoldt fields. Using \cite{Odl75}, which states that there are
finitely many normal CM-fields with relative
class number $1$, we have:
\begin{corollary}
 There is only a finite number of normal and totally real fields which are $C_3$-Leopoldt.
 The same holds for normal CM-fields which are $C_3$-Leopoldt.
\end{corollary}

If we assume the generalized Riemann hypothesis, by \cite{AD03} we find a finite number of totally real or CM-fields as well. That article will be very useful
again at a later point in the present paper.

\section{Totally real $C_l$-Leopoldt fields}\label{real}

We define $\Delta_l:=\text{Gal}(\Q(\zeta_l)/\Q)\cong (\Z/l\Z)^*$, with the elements $\{\sigma_1,\sigma_2,...,\sigma_{l-1}\}$, and
$\theta_l$, $\mathcal{J}_l$ respectively
the Stickelberger element and ideal in $\Q[\Delta_l]$.

We can state a general result
that holds when a number field $K$ is disjoint from $\Q(\zeta_l)$, using Iwasawa's class number formula \cite{Iwa62}
in a similar way to what happens in \cite{Gre97}.

\begin{proposition}
 If $K$ is a $C_l$-Leopoldt number field disjoint from $\Q(\zeta_l)$, denote
 by $J\in\textup{\text{Gal}} (K(\zeta_l)/K)$ the automorphism corresponding to the complex conjugation
 $\sigma_{l-1}\in\textup{\text{Gal}}(\Q(\zeta_l)/\Q)$. Then
 the exponent of
 $\textup{\text{Cl}}(K(\zeta_l))^{1-J}$ divides $h^-_l$.
\end{proposition}

\begin{proof}
 In this case we can see the class group $\text{Cl}(K(\zeta_l))$ as a $\Z[\Delta]$-module with the identification $\text{Gal}(\Q(\zeta_l)/\Q)\cong \text{Gal}(K(\zeta_l)/K)$.
 Moreover we can consider $\text{Cl}(K(\zeta_l))^{1-J}$
as a $\Z[\Delta]^-$-module, where of course
 \begin{equation*}
  \Z[\Delta]^-:=\text{Ker}\left((1+\sigma_{-1}):\Z[\Delta]\longrightarrow\Z[\Delta]\right)=\text{Im}\left((1-\sigma_{-1}):\Z[\Delta]\longrightarrow\Z[\Delta]\right).
 \end{equation*}
 Iwasawa's class number formula tells us that
 \begin{equation*}
 [\Z[\Delta]^-:\mathcal{J}_l^-]=h^-_{\Q(\zeta_{l^n})},
 \end{equation*}
 where $\mathcal{J}_l^-=\mathcal{J}_l\cap \Z[\Delta]^-$.

By Corollary \ref{leocond} we are done.
\end{proof}

We are more interested in the case when $K$ is totally real. In fact in this situation the automorphism $J$ continues to be a complex conjugation,
as it fixes $K\Q(\zeta_l)^+$, where
$\Q(\zeta_l)^+=\Q(\zeta_l+\zeta_l^{-1})$ is the maximal real subfield of $\Q(\zeta_l)$. Note that $K\Q(\zeta_l)^+$ is the maximal real subfield of $K(\zeta_l)$.
We are more interested in taking the minus part of the odd part:
\begin{equation*}
 \text{Cl}(K(\zeta_l))^-:=\text{Ker}\left((1+J):\text{Cl}(K(\zeta_l))_{odd}\longrightarrow\text{Cl}(K(\zeta_l))_{odd}\right);
\end{equation*}
of course $\text{Cl}(K(\zeta_l))^-$ continues to have a structure of $\Z[\Delta]^-$-module. Together with
the well-known fact that the order of $\text{Cl}(K(\zeta_l))^-$ is the odd part of $h^-_{K(\zeta_l)}$, this permits us to conclude the following

\begin{corollary}\label{minusresweak}
 If $K$ is a totally real $C_l$-Leopoldt number field disjoint from $\Q(\zeta_l)$, the exponent of
 $\textup{\text{Cl}}(K(\zeta_l))^-$ (we are taking the minus part of the odd part) divides $h^-_l$.
 In particular the odd primes dividing the relative class number
 $h^-_{K(\zeta_l)}$ divide $h^-_l$.
\end{corollary}

In the case $l\equiv 3\pmod 4$ we have that $K(\zeta_l)/K(\sqrt{-l})$ is an odd Galois extension of CM-fields,
and by \cite[Theorem 5]{LOO97} we deduce that $h_{K(\sqrt{-l})}^-\mid h_{K(\zeta_l)}^-$.
More precisely (even though forgetting about the prime $2$), we can easily state and prove the following, which was surely already known:

\begin{proposition}\label{oddCM}
 Let $L/K$ be an odd extension of CM-fields. Then the map $N^-_{L/K}:\textup{\text{Cl}}(L)^-\rightarrow \textup{\text{Cl}}(K)^-$
 induced by the norm is surjective. In particular the odd part
 of $h_K^-$ divides the odd part of $h_L^-$.
\end{proposition}
\begin{proof}
First of all we describe the norm map between the minus parts: the norm map between ideal class groups
commutes with the conjugation $J$, as for every pair of primes $\mathcal{P}|\wp$ of $L/K$ we have
\begin{equation*}
 N_{L/K}(\mathcal{P})^{J}=\wp^{f(\mathcal{P}|\wp)J}=(\wp^J)^{f(\mathcal{P}|\wp)}=(\wp^J)^{f(\mathcal{P}^J|\wp^J)}=N_{L/K}(\mathcal{P}^J);
\end{equation*}
moreover, $J$ commutes with every immersion of $L$ over $K$ for the same reason for which the conjugation is independent from the immersion in $\C$ of the CM-fields.
So if $[I]^{1+J}=0$, i.e. $I^{1+J}=(x)$, then
\begin{equation*}
 N_{L/K}([I])^{1+J}=N_{L/K}([I]^{1+J})=[N_{L/K}(x)]=0,
\end{equation*}
hence the norm map is also well defined between the minus parts of the odd parts.

 Now, considering also $N_{L/K}:\text{Cl}(L)\rightarrow \text{Cl}(K)$, we can verify that $\text{coker}(N_{L/K})^-=\text{coker}(N^-_{L/K})$.
 By the well-known commutative the diagram
given by Artin maps, namely
\[\begin{tikzcd}
\textup{\text{Cl}}(E) \arrow{r}{\psi_{H_E/E}} \arrow[swap]{d}{N_{E/F}} & \textup{\text{Gal}}(H_E/E) \arrow{d}{\textup{\text{ restriction}}} \\
\textup{\text{Cl}}(F) \arrow{r}{\psi_{H_F/F}} & \textup{\text{Gal}}(H_F/F)
\end{tikzcd}
\]
for any extension $E/F$ of number fields (see for instance the appendix in \cite{Was97} on class field theory),
 we deduce that $\text{coker}(N_{L/K})\cong \text{Gal}(L\cap H_K/K)$ and it is already of odd order; moreover,
 $J$, which is the complex conjugation, acts in the same way on the two groups. So $J$ acts trivially, i.e. there cannot be any automorphism of $L\cap H_K$
 in the kernel of $1+J$, whence we get $\text{coker}(N^-_{L/K})=0$.

\end{proof}

Hence we get:

\begin{corollary}\label{minusres}
 If $K$ is a totally real $C_l$-Leopoldt number field disjoint from $\Q(\zeta_l)$ and $l\equiv 3\pmod 4$,
 the exponent of
 $\textup{\text{Cl}}(K(\sqrt{-l}))^-$ divides $h^-_l$.
 In particular the odd primes dividing the relative class number
 $h^-_{K(\sqrt{-l})}$ divide $h^-_l$.
\end{corollary}

This is already sufficient to compute some cases of non-$C_l$-Leopoldt fields, but we can obtain some general information thanks to an idea
by Ichimura in \cite{Ich16}. In particular, our aim is to study totally real normal fields disjoint from $\Q(\zeta_l)$ which are
$C_l$-Leopoldt, when $l=7,11,19,43,67,163$. The approach is similar to Ichimura's, with two major differences:

\begin{itemize}
 \item we want to study Leopoldt fields and not only Hilbert-Speiser fields (indeed the approach followed by Ichimura is flexible enough to apply also
         to this more general task);
 \item we correct a mistake in the article: in the proof we need the image of the Stickelberger ideal in the group ring given by the quadratic subfield of $\Q(\zeta_l)$
 instead of its own Stickelberger ideal.
\end{itemize}

Let $\delta_l:=\text{Gal}(\Q(\sqrt{-l})/\Q)\cong \Z/2\Z$ with elements $\{1,j\}$.
If $l$ were not ramified in $K$, the extension $K(\zeta_l)/K(\sqrt{-l})$ would satisfy the hypotheses of
\cite[Theorem 10.1]{Was97}, and so the norm map between
class groups would be surjective. This is what Ichimura considered (indeed we already know that totally real $C_l$-Hilbert-Speiser fields
are not ramified in $l$,
by \cite{GJ09}), but actually for our purposes we can just assume non-arithmetic disjointness: by the commutative diagram
given by Artin maps, namely
\[\begin{tikzcd}
\textup{\text{Cl}}(E) \arrow{r}{\psi_{H_E/E}} \arrow[swap]{d}{N_{E/F}} & \textup{\text{Gal}}(H_E/E) \arrow{d}{\textup{\text{ restriction}}} \\
\textup{\text{Cl}}(F) \arrow{r}{\psi_{H_F/F}} & \textup{\text{Gal}}(H_F/F)
\end{tikzcd}
\]
for any extension $E/F$ of number fields, we
know that the order of the cokernel of $N:=N_{K(\zeta_l)/K(\sqrt{-l})}$ divides $(l-1)/2$, and for our aims this will be sufficient.

 If we denote by $r:\Z[\Delta_l]\rightarrow \Z[\delta_l]$
 the linear extension of the map $\Delta_l \twoheadrightarrow \delta_l$ given by
         restriction,
it is straightforward that, for $[I]\in\text{Cl}(K(\zeta_l))$ and $\sigma\in\Z[\Delta_l]$, $N([I]^\sigma)=N([I])^{r(\sigma)}$.
This, together with Corollary \ref{leocond}, permits us to conclude the following
\begin{lemma}
  If $K$ is $C_l$-Leopoldt with $l\equiv 3\pmod 4$ then $r(\mathcal{J}_l)$ annihilates a subgroup of $\textup{\text{Cl}}(K(\sqrt{-l}))$ with index dividing $(l-1)/2$.
\end{lemma}

Now we want to find useful elements in $r(\mathcal{J}_l)$, in particular rational integers; in fact, if $n\in r(\mathcal{J}_l)\cap \Z$, then by the
above lemma $n(l-1)/2$ annihilates $\text{Cl}(K(\sqrt{-l}))$ and this will be very useful.

\begin{lemma}
 If $l\equiv 3\pmod 4$ is a prime number, then
 \begin{equation*}
  \frac{l-1}{2}h(-l)=\left(\frac{1}{l}\sum_{\left(\frac{i}{l}\right)=-1}i\right)^2-\left(\frac{1}{l}\sum_{\left(\frac{i}{l}\right)=1}i\right)^2\in r(\mathcal{J}_l)
 \end{equation*}
  where in the sums we are taking the representatives modulo $l$ between $0$ and $l-1$.
\end{lemma}
\begin{proof}
 We know from \cite[Lemma 6.9]{Was97} that $\mathcal{J}=\mathcal{J}'\theta$ (we omit subscripts),
 where $\mathcal{J}'$ is the ideal in $\Z[\Delta]$ generated by the elements of the type $c-\sigma_c$ with $(c,l)=1$.
 We note that
 \begin{equation*}
  r(\theta)=\frac{1}{l}\sum_{\left(\frac{i}{l}\right)=1}i+\left(\frac{1}{l}\sum_{\left(\frac{i}{l}\right)=-1}i\right)j\in\Z[\delta].
 \end{equation*}

 Moreover, $l=l+1-\sigma_{l+1}$ and $r(4-\sigma_4)=3$. So there exist $a,b\in\Z$ such that $r(a(4-\sigma_4)+b(l+1-\sigma_{l+1}))=1$ (we assume $l\neq 3$); therefore
 $r(\theta)\in r(\mathcal{J})$. Multiplying $r(\theta)$ with its conjugate, we obtain the second equality. The first one comes from the class number formula, for example
 explained in \cite[Theorem 4.17]{Was97}.
\end{proof}

Putting the two lemmas together, we finally obtain:

\begin{corollary}\label{res}
 If $l\equiv 3\pmod 4$ is a prime number and $K$ is a $C_l$-Leopoldt field disjoint from $\Q(\zeta_l)$, then
 $\textup{\text{Cl}}(K(\sqrt{-l}))$ has exponent divisible by $\frac{(l-1)^2}{4}h(-l)$. In particular
 $h_{K(\sqrt{-l})}$ is only divisible by primes dividing
 $\frac{l-1}{2}h(-l)$.

 Note that when $h(-l)=1$ we have that $\frac{l-1}{2}h(-l)$ is odd.
\end{corollary}

For a moment we forget about the condition $l=7,11,19,43,67,163$.
The last corollary, modulo the generalized Riemann hypothesis, ensures us that for every $l\equiv 3\pmod 4$ there are
only finitely many number fields $K$ disjoint from $\Q(\zeta_l)$
such that $K(\sqrt{-l})$ is CM (i.e. $K$ is totally real or CM) and $K$ is $C_l$-Leopoldt: in fact Amoroso and Dvornicich in
\cite{AD03} proved that conditionally the exponent of the ideal class group of a CM-field tends to infinity with its discriminant.

We put it on record:
\begin{corollary}
 Assuming the generalized Riemann hypothesis, if $l\equiv 3\pmod 4$ there is only a finite number of totally real and CM fields disjoint from $\Q(\zeta_l)$
 that are $C_l$-Leopoldt.
\end{corollary}

If we find conditions for the class number to be $1$ (more precisely, we are going to study the relative class number), we get unconditional
finiteness properties for $C_l$-Leopoldt fields. This is why we assume $l=7,11,19,43,67,163$, as we will see.

\begin{remark}
 We briefly explain why we used the quadratic subfield of $K(\zeta_l)/K$. In Ichimura's article, the author claimed that in our cases $r(\mathcal{J})=\Z[\delta]$ looking
 at Sinnott's result in \cite{Sin80}; however this is not true (for example because the elements that come from the Stickelberger ideal, which in cyclotomic
 fields fits also Sinnott's definition,
 have too big augmentation) and the problem is that Sinnott's result concerns the Stickelberger ideal of the quadratic field,
 and not the the image from the cyclotomic one. So we are unable to prove that $1\in r(\mathcal{J})$,
 but we do find a certain natural number, and it will be very important to know that it is odd. This
 would not have been possible if we only considered the cyclotomic extension, because by Lemma 2.1 of Sinnott's article
 the elements of the Stickelberger ideal live in the set
 \begin{equation*}
  A:=\left\{\sigma\in\Z[\Delta]:(1+j)\sigma\in(\sigma_1+...+\sigma_{l+1})\Z\right\}.
 \end{equation*}
 In particular there are no non-zero elements in $\Z$ that are also in the Stickelberger ideal.
\end{remark}

Now we can put together our last result with Corollary \ref{minusres},
so that the case of $h(-l)=1$ will reduce our task to check a finite number of fields.

First of all, looking at any class number table (for example from \cite{Was97}), we note the following

\begin{lemma}
 If $h(-l)=1$, then $\left(h^-_l,\frac{l-1}{2}\right)=1$.
\end{lemma}

Hence we have finally obtained that:

\begin{theorem}\label{true}
 Let $K$ be a totally real $C_l$-Leopoldt field disjoint from $\Q(\zeta_l)$, where $h(-l)=1$. Then $h_{K(\sqrt{-l})}^-=1$. In particular,
 normal totally real $C_l$-Leopoldt fields unramified in $l$ form a finite set because of \textup{\cite{Odl75}}
 ($K(\sqrt{-l})$ is CM).
\end{theorem}

Of course we can find from the finite list of $C_l$-Leopoldt fields also the totally real
abelian $C_l$-Hilbert-Speiser fields, and this is a way to adjust Ichimura's result: it is still true that there are finitely many
totally real abelian $C_l$-Hilbert-Speiser fields.

\begin{remark}
 As we have the divisibility $h(-l)\mid h_l^-$, this method does not work if $h(-l)>1$.
\end{remark}

\section{Complete list of real abelian $C_l$-Hilbert-Speiser fields for $l=7,11,19,43,67,163$}\label{hilbert-speiser}
Here we prove that the list of totally real abelian $C_l$-Hilbert-Speiser fields for $l=7,11,19,43,67,163$ given in \cite{Ich16} is actually complete.
We have to study the discrepancy between Theorem \ref{true} and \cite[Corollary]{Ich16}: since the latter simply claims that
$h_{K(\sqrt{-l})}=1$, we have to check the fields such that $h_{K(\sqrt{-l})}^-=1$ and $h_K=h_{K(\sqrt{-l})^+}\neq1$; moreover, as we are interested in
Hilbert-Speiser fields, we already know that $K$ must be unramified in $l$.

Unfortunately there is no article with a complete table of the imaginary abelian fields with relative class number $1$. The article \cite{CK00} gives the non-cyclic ones,
while the list of cyclic imaginary fields is slightly more difficult to reconstruct: we have to look at the tables in \cite{PK97}, \cite{PK98} and \cite{CK98}.

Here is our analysis:
\begin{itemize}
 \item Imaginary cyclic sextic number fields: the list of sextic fields with relative class number $1$ can be consulted in table $3$ of \cite{PK97}.
 They have the following curious property: every such field $L$ with class number different from $1$ (i.e. $h_{L^+}> 1$) has the same conductor
 of $L^+$. Thus, if $K$ is $C_l$-Hilbert-Speiser and $L=K(\sqrt{-l})$ belongs to this list, $K$ is surely ramified in $l$ since $L$ is. So we do not
 gain anything new.
 \item Imaginary cyclic number fields of $2$-power degree: \cite{PK98} shows that these fields have degree up to $16$ and gives a complete list
 of the non-quadratic ones (since we can assume $K\supsetneq\Q$, we are not interested in them). From the tables we see that
 if a cyclic imaginary field of degree $4$, $8$ or $16$ has relative class number $1$, then it has class number $1$.
 \item Other imaginary cyclic number fields: in \cite{CK98} the authors showed that they have degree less or equal than $20$, and they are listed in Table I.
 There is only one such field with class number different from $1$, but both the conductors of $L$ and $K=L^+$ are equal to $91$. Like before, we
 obtain that this is not a new case of a quadratic extension of a Hilbert-Speiser field.
 \item Non-cyclic imaginary abelian number fields: the complete list is given in \cite[Table I]{CK00}. Here there are a lot
 of fields with class number greater than $1$, and unfortunately the authors do not give conductors of the fields, but even in this case
 it is not difficult to conclude that every totally real $C_l$-Hilbert-Speiser $K$ with $K(\sqrt{-l})$ in this table and $h_K>1$ is ramified in $l$.
 In fact almost all the fields $L$ in the table which contain $\sqrt{-l}$ for $l=7,11,19,43,67,163$ are of the form $F(\sqrt{-l})$, with
 $F$ imaginary and unramified in $l$ (this means that its characters, that can be read from the table, are not all even and have conductors coprime to $l$);
 so their maximal real subfield must be ramified in $l$ by an easy argument. All the exceptions concern $l=7$: $\langle \chi_3,\chi_7^3,\chi_7^4\psi_9\rangle$,
 $\langle \chi_7^3,\chi_3\chi_5^2,\chi_7^2\rangle$ and $\langle \chi_7^3,\chi_7^3\chi_{13}^6,\chi_7^4\chi_{13}^4\rangle$, for which
 $\langle\chi_7^4\psi_9\rangle$, $\langle\chi_7^2\rangle$ and $\langle\chi_7^4\chi_{13}^4\rangle$ respectively are real and
 hence contained in $K$ and ramified at $7$.
\end{itemize}

\begin{corollary}
 The only real abelian $C_7$-Hilbert-Speiser fields are $\Q(\sqrt{5})$ and $\Q(\sqrt{13})$. The only real abelian $C_{11}$-Hilbert-Speiser field
 is $\Q(\cos(2\pi/7))$. There is no real abelian $C_l$-Hilbert-Speiser field if $l=19,43,67,163$.
\end{corollary}

\section{$C_l$-Leopoldt fields that intersect $\Q(\zeta_l)$}\label{intersection}
Until now we only studied the property of being $C_l$-Leopoldt of fields that are disjoint from $\Q(\zeta_l)$. Using the observations
in Remark \ref{leoremark}, we can now concentrate on number fields which intersect $\Q(\zeta_l)$.

Suppose initially that $\zeta_l\in K$. Then the group $\Delta=\Delta_l=\text{Gal}(\Q(\zeta_l)/\Q)$ acts freely on the set of the $l-1$ components of
\begin{equation*}
 \text{Cl}^0(\mathcal{M})\cong\text{Cl}(K)\times\cdots\times\text{Cl}(K).
\end{equation*}
Since $l\theta\in\mathcal{J}$ and annihilates $\text{Cl}^0(\mathcal{M})$, for
every $[I]\in \text{Cl}(K)$ we know from Corollary \ref{leocond} that $l\theta$ annihilates
\begin{equation*}
 ([I],0,...,0)\in\text{Cl}(K)\times\cdots\times\text{Cl}(K).
\end{equation*}
But the first component of $([I],0,...,0)^{l\theta}$ is $[I]$ itself, because the coefficient of $1$ in $l\theta\in\Z[\Delta]$ is $1$. Hence we have obtained:
\begin{proposition}
 Let $K$ be a $C_l$-Leopoldt field that contains $\zeta_l$. Then $\textup{\text{Cl}}(K)=\{1\}$, i.e. $h_K=1$. Of course the converse holds.
\end{proposition}

A simple consequence of the above result is that there are no CM $C_l$-Leopoldt fields which are odd extensions of $\Q(\zeta_l)$ if $l\geq 23$, by
\cite[Theorem 5]{LOO97} (in odd extensions $E/F$ of CM-fields there is the divisibility $h_F^-\mid h_E^-$). Moreover, assuming the generalized Riemann hypothesis,
we conclude that there are only finitely many CM $C_l$-Leopoldt fields that contain $\zeta_l$, from \cite{AD03}.

In particular, $\Q(\zeta_l)$ is
$C_l$-Leopoldt if and only if $2\leq l\leq 19$. We also observe that the last proposition implies \cite[Proposition 3.3]{Her05}, apart from
a finite number of $l$'s.

Now we consider the situation in which $K\cap \Q(\zeta_l)=\Q(\zeta_l)^+=\Q(\zeta_l+\zeta_l^{-1})$.
Here the decomposition is
\begin{equation*}
 \text{Cl}^0(\mathcal{M})\cong\text{Cl}(K(\zeta_l))\times\cdots\times\text{Cl}(K(\zeta_l)),
\end{equation*}
where
we have $(l-1)/2$ copies of the ideal class group of $K(\zeta_l)$. From what we have already said in Remark \ref{leoremark}, each
component has a structure of $\Z[H]$-module, where
\begin{equation*}
 H=\text{Gal}(K(\zeta_l)/K)\cong \text{Gal}(\Q(\zeta_l)/\Q(\zeta_l)^+)=\{1,\sigma_{-1}\}.
\end{equation*}
Since $\pi_H(l\theta)=1+(l-1)\sigma_{-1}$ and $\pi_H((2-\sigma_2))\theta=\sigma_{-1}$, then $\pi_H(l\theta-(l-1)(2-\sigma_2)\theta)=1$ and annihilates
$\text{Cl}(K(\zeta_l))$ if $K$ is $C_l$-Leopoldt. Therefore we have an analogous conclusion:

\begin{proposition}
 Let $K$ be a number field such that $K\cap \Q(\zeta_l)=\Q(\zeta_l)^+$. Then $K$ is $C_l$-Leopoldt if and only if $h_{K(\zeta_l)}=1$.
\end{proposition}

\begin{remark}
 The above results are exactly the generalizations of our criteria about $C_2$ and $C_3$-Leopoldt fields.
 Moreover, modulo the generalized Riemann hypothesis and finitely many cases this result implies \cite[Theorem 1.1]{GJ09},
 when $[K(\zeta_l):K]=2$.
\end{remark}

At this point it is
tempting to look at the case $|H|=3$; and indeed, the conclusion continues to hold,
as explained in the following result:

\begin{proposition}
 Let $l\equiv 1\pmod 3$ be a prime and $K$ a number field with the property that $K\cap \Q(\zeta_l)$ is the subfield of $\Q(\zeta_l)$ with degree $(l-1)/3$
 over $\Q$. Then $K$ is $C_l$-Leopoldt if and only if $h_{K(\zeta_l)}=1$.
\end{proposition}
\begin{proof}
 In a similar way to the preceding cases, we want to prove that $\pi_H(\mathcal{J})=\Z[H]$. Let $a,b$ be the non-trivial representatives
 of the third roots of $1$ modulo $l$, i.e. such that $H=\{\sigma_1,\sigma_a,\sigma_b\}$. For every $2\leq c\leq l-1$ we have
 \begin{equation*}
  \pi_H((c-\sigma_c)\theta)=\left\lfloor\frac{ca}{l}\right\rfloor\sigma_a^{-1}+\left\lfloor\frac{cb}{l}\right\rfloor\sigma_b^{-1}=
  \left\lfloor\frac{cb}{l}\right\rfloor\sigma_a+\left\lfloor\frac{ca}{l}\right\rfloor\sigma_b.
 \end{equation*}

 If there exists $2\leq c\leq l-1$ such that $\left\lfloor\frac{ca}{l}\right\rfloor\neq\left\lfloor\frac{cb}{l}\right\rfloor$, taking the minimum $c_m$
 of such values, we notice that $\pi_H((c_m-\sigma_{c_m})\theta-(c_m-1-\sigma_{c_m-1})\theta)$ is $\sigma_a$ or $\sigma_b$,
 since every increase inside the integral part is $a/l<1$ and $b/l<1$ when $c$ rises by $1$, whence we obtain
 that $1\in\pi_H(\mathcal{J})$ as $\pi_H(\mathcal{J})$ is an ideal.

 Therefore we may assume that for every $2\leq c\leq l-1$ we have $\left\lfloor\frac{ca}{l}\right\rfloor=\left\lfloor\frac{cb}{l}\right\rfloor$. Taking
 $c=l-1$ this implies $a-1=b-1$, that is $a=b$, which is not possible.

\end{proof}

Our considerations about the projection of the Stickelberger ideal are also consequences of \cite[Theorem 2(III)]{IS06},
in a work by Ichimura and Sumida-Takahashi concerning $p$-normal integral bases. According to Lemma 1
of the same article, when $|H]$ is even, then $\pi_H(\mathcal{J})$ is contained in the ideal of $\Z[H]$ generated by $1+\rho+\rho^2+\cdots+\rho^{|H|/2-1}$,
where $\rho$ is any generator of $H$, and so cannot be the whole ring if $|H|\geq 4$ (e.g. look at the augmentation).

If $|H|$ is odd, then \cite[Theorem 2(I)]{IS06} and \cite[Theorem 2(II)]{IS06} imply
that the order of the quotient $\Z[H]/\pi_H(\mathcal{J})$ divides
\begin{equation*}
 \left[(1+\rho+\rho^2+\cdots+\rho^{(l-1)/2-1})\Z[\Delta]:\pi_\Delta(\mathcal{J})\right]=h_l^-,
\end{equation*}
where $\rho$ generates $\Delta$, so:
\begin{corollary}\label{corollarygain}
 Let $K$ be a $C_l$-Leopoldt field such that the degree $[\Q(\zeta_l):K\cap \Q(\zeta_l)]$ is odd.
 Then the exponent of $\textup{\text{Cl}}(K(\zeta_l))$ divides $h_l^-$. In particular if $3\leq l\leq 19$ then $K$ is $C_l$-Leopoldt if and only if $h_{K(\zeta_l)}=1$.
\end{corollary}

Remaining in the situation of $|H|$ being odd, we can say something more precise. Now we start from above, and assume $l\equiv 3\pmod 4$ and $|H|=(l-1)/2$,
i.e. $K\cap\Q(\zeta_l)=\Q(\sqrt{-l})$. In this case Ichimura's class number formula in \cite[Theorem]{Ich06a} tells us that
$[\Z[H]:\pi_H(\mathcal{J})]=h_l^-/h(-l)$. According to \cite[Theorem 2(II)]{IS06},
thanks to which we know that the index of $\pi_{H_1}(\mathcal{J})$ in $\Z[H_1]$
divides that concerning $H_2$ whenever $H_1<H_2$ are of odd order, this permits us to conclude
the following
\begin{corollary}
 Let $K$ be a $C_l$-Leopoldt field such that the degree $[\Q(\zeta_l):K\cap \Q(\zeta_l)]$ is odd, with $l\equiv 3\pmod 4$.
Then the exponent of $\textup{\text{Cl}}(K(\zeta_l))$ divides $h_l^-/h(-l)$.
\end{corollary}
In particular, if $l=7,11,19,23$ then $K$ is $C_l$-Leopoldt if and only if $h_{K(\zeta_l)}=1$.
Moreover in a lot of other cases we gain some information on divisibility respect to Corollary \ref{corollarygain},
i.e. we are able to find primes that have the same exponent
in $h(-l)$ and $h_l^-$; for example, the first cases for $l$ are $47$, $59$, $71$, $79$, $103$, $107$, $127$, $151$, $167$, $179$, $191$, $223$, $239$.

When $|H|$ is even there is no hope to obtain in this way an equivalence between being $C_l$-Leopoldt and divisibility properties of $h_{K(\zeta_l)}$:
the product
\begin{equation*}
 (a_0+a_1\rho+\cdots+a_{|H|-1}\rho^{|H|-1})\cdot (1+\rho+\rho^2+\cdots+\rho^{|H|/2-1})
\end{equation*}
cannot be a non-zero rational integer for any $a_0+a_1\rho+\cdots+a_{|H|-1}\rho^{|H|-1}\in\Z[H]$ by an easy polynomial divisibility argument.
Hence we do not find rational integers that annihilate $\text{Cl}(K(\zeta_l))$. However it is possible to get a conclusion in a similar fashion to Corollary
\ref{minusresweak}: since $(1-\rho)(1+\rho+\rho^2+\cdots+\rho^{|H|/2-1})=1-\rho^{|H|/2}$ we have that the index
$n_H:=[(1+\rho+\rho^2+\cdots+\rho^{|H|/2-1})\Z[H]:\pi_{H}(\mathcal{J})]$ is such that $n_H\Z[H]^-\subseteq \pi_{H}(\mathcal{J})$, where the minus is referred to
``conjugation``, that is the automorphism $J:=\rho^{|H|/2}\in H$ which corresponds to the conjugation $\sigma_{-1}\in\Delta$. Moreover by
Ichimura and Sumida-Takahashi's work $n_H|h_l^-$. Hence:
\begin{proposition}
 Let $K$ be a $C_l$-Leopoldt field such that $H=\text{Gal}(K(\zeta_l)/K)$ has even order.
 Then $n_H$ annihilates $\textup{\text{Cl}}(K(\zeta_l))^{1-J}$. In particular $h_l^-$ does.
\end{proposition}

As in Corollary \ref{minusresweak} we get the following
\begin{corollary}
 With the preceding hypotheses, let us assume that $K$ is totally real. Then the odd primes that divide $h_{K(\zeta_l)}^-$ also divide $h_l^-$.
\end{corollary}

Actually, without this generality we can get better results if $|H|<(l-1)/2$ and without assuming $l\equiv3\pmod4$. In fact if
$23\leq l\leq 499$, from \cite[Proposition 3]{IS06} we can read off the few cases in which $|H|<(l-1)/2$ and $\pi_H(\mathcal{J})\subsetneq \Z[H]$:
for instance if $|H|$ is odd we only have
$l=277$ and $|H|=69$, $l=349$ and $|H|=87$, $l=331$ and $|H|=33$. The authors expected in general that this situation is very rare, i.e.
we can conclude as well that in most cases a $C_l$-Leopoldt field with $|H|<(l-1)/2$ odd is such that $h_{K(\zeta_l)}=1$.
\cite[Theorem 2]{Ich06b} gives more evidence about this expectation: if, for $H$ of odd order, there is a prime $q|[\Z[H]:\pi_H(\mathcal{J})]$, then
$h_l^-$ is divisible by $q^{[\Delta:H]/2}$; on the contrary, it is quite common for a cyclotomic relative class number to have rather small exponents.
This is also true in the even case substituting $\Z[H]$ with $(1+\rho+\rho^2+\cdots+\rho^{|H|/2-1})\Z[H]$, but
as we have seen we get a much weaker annihilation/divisibility conclusion.

To sum up: in the context of
$C_l$-Leopoldt fields $K$ that intersect nontrivially with $\Q(\zeta_l)$, if we just use
ad hoc arguments and previous results of Ichimura and Sumida-Takahashi it appears that we are
able to obtain restrictions on the class number of $\Q(\zeta_l)$  only in the
case when $[\Q(\zeta_l):K\cap \Q(\zeta_l)]$ is equal to $2$ or an odd number, and thus finiteness conditions
if we assume the generalized Riemann hypothesis, by \cite{AD03} (without fixing $l$ if $[\Q(\zeta_l):K\cap \Q(\zeta_l)]$ is $1$, $2$ or $3$).
Returning to Hilbert-Speiser fields, from \cite{Her05} we already knew that we
do not have such odd cases of $C_l$-Hilbert-Speiser fields for $l\geq 5$, unless the intersection
is $\Q(\sqrt{-l})$ when $l\equiv 3\pmod4$. When instead $[\Q(\zeta_l):K\cap \Q(\zeta_l)]$ is even, i.e. $K\cap \Q(\zeta_l)$ is real,
then it is likely that if $K$ is $C_l$-Hilbert-Speiser then the intersection is trivial, by \cite{Ich07}.

\section{Acknowledgements}
 The majority of the results were
obtained in the first author's M.Sc. thesis, written under the joint
supervision of
Ilaria Del Corso at Pisa and the second author.

redpajama_set_name: RedPajamaArXiv


text:
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In 2000, Perry and Salim built their first hotel, a 79-room Holiday Inn Express & Suites off of the Dallas Tollway and Parker Road in Plano, TX. In 2006, the Molubhoys designed a 1940's Hollywood-themed Holiday Inn Express, located at Twin Creeks in Allen, TX. Well-exceeding industry benchmarks, and unique in nature, Intercontinental Hotels Group (IHG) was proud to announce the addition of this one-of-a-kind hotel to the franchise. Since the initial flagship hotels, the Molubhoy brothers have added several contemporary-themed hotels that have put Atlantic on the hotelier map, including; a Holiday Inn Express and Towneplace Suites in DeSoto, a Staybridge Suites in Irving, and a premier benchmark Holiday Inn Express & Suites near the Galleria Mall in Dallas. Combining one brother's sense of style and the others sense of operation, Atlantic Hotels, Inc. became internationally renowned.
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text:
Keltie Ferris, Joanne Greenbaum, Arlene Shechet & Jessica Stockholder
Informal Get Together
KELTIE FERRIS
HER*CU*LES
110 by 80 in. 279.4 by 203.2 cm.
)))(((sounds)))(((
Graphite on canvas in the artist's frame
Canvas: 72 by 60 in. 182.9 by 152.4 cm.
Framed: 75 by 63 by 3 in. 190.5 by 160 by 7.6 cm.
JOANNE GREENBAUM
Oil, acrylic and ink wash on canvas
72 by 70 in. 182.9 by 177.8 cm.
Cinopia
Oil and flashe on canvas
ARLENE SHECHET
Tell All
Glazed ceramic and painted steel
62 1/2 by 21 by 16 in. 158.8 by 53.3 by 40.6 cm.
Glazed ceramic, resin, paint and wood
13 by 13 by 10 in. 33 by 33 by 25.4 cm.
Informal Get Together [JS 623]
Aluminum tube, yellow webbing, pink plastic and fabric parts, baskets, 5 Ikea lamps, small rug, power bar and acrylic paint
96 by 85 by 24 in. 243.8 by 215.9 by 61 cm. Height variable
Installation view of Informal Get Together at Mitchell-Innes & Nash New York, 2022
Mitchell-Innes & Nash is pleased to present Informal Get Together, an exhibition bringing together painting and sculpture by Keltie Ferris, Joanne Greenbaum, Arlene Shechet and Jessica Stockholder on view from July 14 through August 19, 2022. Taking its title from the 2014 sculpture by Jessica Stockholder, Informal Get Together highlights Shechet and Stockholder's painterly approach to sculpture alongside Greenbaum and Ferris's three-dimensional and structural approach to painting.
Keltie Ferris has developed his exuberant and complex approach to abstract painting using a variety of methods including spray gun, dry pigment and hand-painted fields. On view in the present exhibition are two works spanning the most recent decade of Ferris's career. HER*CU*LES, 2016 demonstrates his range of mark making, scale shifts and close ups. Ferris couples syncopated brushstrokes with airbrushed oil paint to create surfaces that combine the topography of textiles with the banter of gestural mark making—a signature yet ever-evolving component of his practice. A more recent painting, )))(((sounds)))(((, is a tightly worked, grisalle composition which brims with contradiction – meditative and anxious, visually expanding and contracting.
Over the past thirty years, Joanne Greenbaum has become known for a distinct abstract visual language defined by an unconventional layering of forms in a complex pictorial space. Often using drawing as the springboard for her paintings, Greenbaum formulates a vital relationship between the two approaches which ultimately reflects her own unique vocabulary of line and volume. On view in Informal Get Together are two largescale canvases from 2003 in which underlying geometric structures, recalling architectural details such as staircases or scaffolding, balance the intuitive with the systematic. A third painting from 2011 is indicative of the artist's continued interest in the tension between figure, line and ground found within her earlier work.
Arlene Shechet teases out longstanding tensions between object and base, interior and exterior, figure and space, combining formal concerns with a playful sense of corporeality. Shechet's painted ceramic sculptures are suggestive of contorted torsos or deep-sea-creatures covered in drippy and crackled glaze. In Show Off, a small wall-bound sculpture, five tentacle-like arms curl downwards from the center of the sculpture, appearing to perilously dangle off it's wall support. A variety of surface finishes, from glossy and textured to monochrome and matte, force the viewer to consider the sculpture in the round and disrupt assumptions about the solidity of a sculptural object.
Jessica Stockholder's works have played a crucial role in expanding the dialogue between sculpture and painting. She often incorporates the architecture in which the work has been conceived while merging seemingly disparate, everyday objects that create coherent whole compositions. The quotidian goods she often uses make use of the aesthetic and formal qualities of these often-overlooked items while orchestrating an intersection of pictorial and physical space. She probes how meaning derives from physicality, and engages the sensuality and pleasure evoked by color and formal order in an effort to call attention to the edges of understanding.


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