From my earliest model building days, I've always been drawn to the beautiful box art on model kits. A quick search turned up the above photo of the box art from a kit I built as a kid. This illustration captures all the excitement of modern naval technology. The Enterprise is a commanding presence, carving a frothy white wake in an ocean of gorgeous blue. An A4 Skyhawk streaks overhead, reminding us that this is no ordinary ship, but a mobile airfield. So powerful is the illustration, I instantly recognized it out of hundreds of photos in the search results, even though the model is long gone.
Not all box art is created equal, and there's obviously no relationship between the quality of the illustration and the quality of the kit itself. Some excellent kits come in poorly illustrated boxes, and vice-versa. Japanese kit manufacturers usually get it right on both counts, with well engineered kits in finely illustrated boxes. I find Tamiya's box art to be exceptionally good.
Not all types of kits get the same artistic treatment. For a long time, science fiction kits were the poor cousins of the box art world. AMT and MPC, which brought us subjects from Star Trek, Star Wars, Space: 1999 and other iconic sci-fi shows and films, always had cheap looking box art. MPC's Eagle was a mess on all counts: one of the most inaccurately mastered kits ever produced, in a really ugly box!
Things have improved for sci-fi modelers, now that we have premium science fiction kits from companies like Fine Molds and Fujimi. The Star Wars line from Fine Molds features top notch box art. I especially like the box art on the Y-Wing kit. And Fujimi's Spinner from Blade Runner also comes in a nicely illustrated box.
Some box art is so good, you could cut it out and frame it if not for the type that covers the illustration. And some of the top artists in the field, like Roy Cross, Jack Leynnwood, Kihachiro Ueda and Shigeo Koike, have had their box art republished in prints and books. Koike's website at http://shigeokoike.com/en features some of the masterful work he's done for Hasegawa and Fuji Heavy Industries. If you've never heard of any of these artists, it's because they generally aren't permitted to sign their names to their box art illustrations. The kit producer will argue that they want to promote the kit, not the artist, although I see no reason why they couldn't do both.
If we think of the model building experience as a journey, then the journey begins when walking into a hobby store and seeing a new kit, or spotting it online. An attractive package will naturally generate more sales, and marketing departments know full well the importance of capturing the prospective buyer's imagination the first time he sets eyes on a new kit. It can make the difference between winning and losing the sale.
The worst thing a kit manufacturer can do is have no box art at all. Anyone who's ever bought a resin garage kit is familiar with the curiously anticlimactic experience of receiving their kit in the mail, unwrapping it, and being confronted with a plain cardboard box devoid of any graphics at all. Or a splotchy black and white photocopy stapled to the box, which is worse.
Of course it would be asking a lot to expect a garage kit producer to be a gifted professional artist as well as a good mold maker. Garage outfits are usually one man operations, and few people are that multi-talented.
What's harder to excuse is mass producers of kits who make bad decisions when it comes to box art. Remember when Airfix offered kits in plastic bags? The only visible artwork was a tiny piece of folded cardboard stapled to the top of the bag! You had to really want the kit badly to stomach such cheap packaging. Even worse, you'd be left wondering if the unprotected parts would still fit together when you got the kit home. Without the protection of a cardboard box, parts could come off sprues and get damaged or lost. But then quality has never been a priority for Airfix.
In the age of online shopping, one reason people still enjoy walking into a hobby store is because it's like entering an art gallery full of illustrations of your favourite planes/cars/boats/spaceships. The poster-size box art on large kits beats squinting at tiny website photos hands down. Shrink-wrapped kits are even better, adding a glossy coating which makes colours pop (and reassures you that no pieces are missing).
With few exceptions, kit producers prefer to use paintings of the "real thing" rather than photographs of the completed model for their box art. Paintings are better at generating interest for a potential buyer because they exploit his interest in the actual subject, which is the reason he's buying the kit. And a painter can place the subject in a dynamic scene, whereas a photo of the completed kit will always look static and dull in comparison.
It's the same subject in the same scale, but the illustrated package is far more appealing. And being human, we assume that what looks good on the outside will transfer to the inside. So most people will prefer the Revell kit, other things being equal.
If you have a closet full of old kit boxes and you're about to move, you may be contemplating throwing them in the recycling. But instead, consider cutting out the front illustrations and keeping them. You can discard the rest of the box, and still have all that beautiful artwork when you arrive at your new place.
The introduction of a new model kit based on a subject from a movie is an eagerly anticipated event for many hobbyists. I remember when MPC released its Millennium Falcon kit back in 1977, coinciding with the release of the first Star Wars movie. MPC's marketing department was on the ball and capitalized on the movie's success by being quick on the draw with the release of the kit.
Not all kit releases are timed as perfectly as MPC's Falcon release. The same subject was released by FineMolds nearly three decades later, in 2005. The FineMolds Falcon, although late to the game, was a dramatic improvement over the MPC kit. It featured the usual high level of accuracy and detail associated with the FineMolds brand, in correct 1:72 scale. The MPC kit did not even specify a scale, and Star Wars aficionados have enjoyed debating the scale of the kit since it was released.
When I first heard about the FineMolds Falcon release, my first thought was, "Wow, it sure took them long enough." But FineMolds actually did something very clever.
The price differential between the two kits is considerable, even with a few decades' worth of inflation factored in. The result is that they are not in direct competition with one another. The MPC kit was clearly aimed at kids and teens with little or no money. The FineMolds kit, on the other hand, was targeted at advanced modellers with generous budgets. The funny thing is, even though the markets for the two kits are distinct, the person buying them could quite often be one and the same. If you were a teen when Star Wars was released, and bought the MPC kit back then, there's a good chance that you'd be willing to upgrade to the FineMolds version several decades later. My guess is that this is exactly what FineMolds planned.
The FineMolds Falcon has been so successful that Revell is now re-releasing it under their Master Series label. It will still be manufactured in Japan by FineMolds but will have revised packaging and English instructions.
It's less clear what Bandai was thinking with their recent release of 1:72 models of the classic X-Wing and Y-Wing. Chances are, anyone who wanted a 1:72 scale kit of either of these subjects would have picked up the FineMolds version by now. Price, accuracy and quality for both brands are similar. There just isn't that much difference between them. Bandai will also be releasing a 1:144 Millennium Falcon, which has been done before by FineMolds as well. How Bandai will be able to turn a profit on these "me too" products is a mystery. A much better move would have been to produce a 1:48 Y-Wing to complement the FineMolds 1:48 X-Wing. This would have filled an empty niche in the market.
When Episode 7 of Star Wars comes out this December, a newly designed X-Wing will grace the screen. It looks good—smaller and sleeker—and can be glimpsed in the teaser trailer for the film. Fortunately, Bandai has made one good decision, which is to launch a kit of the new X-Wing just before Episode 7's release in December. Unless the movie is a complete disaster, this kit will be a guaranteed cash cow. The new X-Wing will be followed by a kit of the new TIE Fighter, which looks like a lot like the classic TIE but with an inverted colour scheme. This kit should be popular as well.


text:
Q: Resize a picture to fit a JLabel I'm trying to make a picture fit a JLabel.  I wish to reduce the picture dimensions to something more appropriate for my Swing JPanel.
I tried with setPreferredSize but it doesn't work.
I'm wondering if there is a simple way to do it? Should I scale the image for this purpose?

A: Outline
Here are the steps to follow.

*

*Read the picture as a BufferedImage.

*Resize the BufferedImage to another BufferedImage that's the size of the JLabel.

*Create an ImageIcon from the resized BufferedImage.

You do not have to set the preferred size of the JLabel.  Once you've scaled the image to the size you want, the JLabel will take the size of the ImageIcon.
Read the picture as a BufferedImage
BufferedImage img = null;
try {
    img = ImageIO.read(new File("strawberry.jpg"));
} catch (IOException e) {
    e.printStackTrace();
}

Resize the BufferedImage
Image dimg = img.getScaledInstance(label.getWidth(), label.getHeight(),
        Image.SCALE_SMOOTH);

Make sure that the label width and height are the same proportions as the original image width and height.  In other words, if the picture is 600 x 900 pixels, scale to 100 X 150.  Otherwise, your picture will be distorted.
Create an ImageIcon
ImageIcon imageIcon = new ImageIcon(dimg);

A: public static void main(String s[])
  {

    BufferedImage image = null;
    try
    {
        image = ImageIO.read(new File("your image path"));

    } catch (Exception e)
    {
        e.printStackTrace();
    }

    ImageIcon imageIcon = new ImageIcon(fitimage(image, label.getWidth(), label.getHeight()));
    jLabel1.setIcon(imageIcon);
}

private Image fitimage(Image img , int w , int h)
{
    BufferedImage resizedimage = new BufferedImage(w,h,BufferedImage.TYPE_INT_RGB);
    Graphics2D g2 = resizedimage.createGraphics();
    g2.setRenderingHint(RenderingHints.KEY_INTERPOLATION, RenderingHints.VALUE_INTERPOLATION_BILINEAR);
    g2.drawImage(img, 0, 0,w,h,null);
    g2.dispose();
    return resizedimage;
}

A: The best and easy way for image resize using Java Swing is:
jLabel.setIcon(new ImageIcon(new javax.swing.ImageIcon(getClass().getResource("/res/image.png")).getImage().getScaledInstance(200, 50, Image.SCALE_SMOOTH)));

For better display, identify the actual height & width of image and resize based on width/height percentage

A: You can try it:
ImageIcon imageIcon = new ImageIcon(new ImageIcon("icon.png").getImage().getScaledInstance(20, 20, Image.SCALE_DEFAULT));
label.setIcon(imageIcon);

Or in one line:
label.setIcon(new ImageIcon(new ImageIcon("icon.png").getImage().getScaledInstance(20, 20, Image.SCALE_DEFAULT)));

The execution time is much more faster than File and ImageIO.
I recommend you to compare the two solutions in a loop.

A: i have done the following and it worked perfectly
try {
        JFileChooser jfc = new JFileChooser();
        jfc.showOpenDialog(null);
        File f = jfc.getSelectedFile();
        Image bi = ImageIO.read(f);
        image1.setText("");
        image1.setIcon(new ImageIcon(bi.getScaledInstance(int width, int width, int width)));

    } catch (Exception e) {
    }

A: Or u can do it this way. The function u put the below 6 lines will throw an IOException. And will take your JLabel as a parameter.
BufferedImage bi=new BufferedImage(label.width(),label.height(),BufferedImage.TYPE_INT_RGB);

Graphics2D g=bi.createGraphics();

Image img=ImageIO.read(new File("path of your image"));

g.drawImage(img, 0, 0, label.width(), label.height(), null);

g.dispose();

return bi;

A: public void selectImageAndResize(){
    int returnVal = jFileChooser.showOpenDialog(this); //open jfilechooser
    if (returnVal == jFileChooser.APPROVE_OPTION) {    //select image
        File file = jFileChooser.getSelectedFile();    //get the image
        BufferedImage bi;
        try {
            //
            //transforms selected file to buffer
            //
            bi=ImageIO.read(file);
            ImageIcon iconimage = new ImageIcon(bi);

            //
            //get image dimensions
            //
            BufferedImage bi2 = new BufferedImage(iconimage.getIconWidth(), iconimage.getIconHeight(), BufferedImage.TYPE_INT_ARGB);
            Graphics g = bi.createGraphics();
            iconimage.paintIcon(null, g, 0,0);
            g.dispose();

            //
            //resize image according to jlabel
            //
            BufferedImage resizedimage=resize(bi,jLabel2.getWidth(), jLabel2.getHeight());
            ImageIcon resizedicon=new ImageIcon(resizedimage);
            jLabel2.setIcon(resizedicon);
        } catch (Exception ex) {
            System.out.println("problem accessing file"+file.getAbsolutePath());
        }
    }
    else {
        System.out.println("File access cancelled by user.");
    }
}

A: Assign your image to a string.
Eg image
Now set icon to a fixed size label.
image.setIcon(new javax.swing.ImageIcon(image.getScaledInstance(50,50,WIDTH)));

redpajama_set_name: RedPajamaStackExchange


Safetrac announces another successful year at the LearnX Impact Awards Asia Pacific 2015, winning Platinum and Silver awards.
Safetrac has claimed the Platinum award for Best Learning Compliance Program for the fifth successive year. This unprecedented record validates Safetrac as one of the leading compliance training companies in Asia/Pacific. Throughout the past five years Safetrac has adapted to the changing market by incorporating new technology, design and learning methods to continuously improve its compliance training courses.
In 2014, Safetrac launched a range of new courses to improve the learning outcomes of training which incorporates interactive tutorials, narration, embedded videos and animated scenarios. This continued development has led to Safetrac's ongoing success and consolidated its position as the benchmark in compliance training programs.
Additionally Safetrac has won a Silver award for Best eLearning Design, Audio. Safetrac implemented audio into a customised training program to add another dimension to the training and enhance the learning experience for both visual and audio learners. Audio was incorporated into the training via multiple interactions, animations and embedded videos.
"It is a fantastic achievement for Safetrac to be recognised for two awards at this year's LearnX Impact Awards. We are especially proud to once again be acknowledged for providing the best compliance training program in Asia/Pacific and for gaining that acknowledgement for an incredible five years in a row. Additionally, to be recognised for best audio for a custom course shows the diversity that Safetrac can bring to a bespoke compliance training program".
The LearnX Impact Awards is an annual event run by the LearnX Foundation, a not-for-profit organisation promoting innovative workforce learning and supporting technologies. The event attracts leading learning and training industry organisations.


Hi dog owners! My girlfriend has a ~80 black lab and Id like to get a cover for the back seat of my 2013 wrx sedan and I was wondering what other members are using? I was thinking something like this https://www.amazon.ca/seat-covers-cars-YoGi-Prime/dp/B01DZXYSJC/ref=sr_1_3?ie=UTF8&qid=1546907356&sr=8-3&keywords=backseat+dog+cover but I was thinking it would be nice if it covered the door panels a bit more. Also our lack of independent head rest would prevent me from using the hammock style function.
I have a similar one, make sure they cover the side of the seat, other wise they will still get filthy. Also depending on the dog (mine) they will high center themselves near the console making for an interesting emergence stop.
I have a similar one to that as well. They work great, and can be quickly undone from the front seats so the back seat is usable. just don't forget to order the door panels as well !!!!


text:
Republican Establishment Mobilized Black Mississippi Voters With Race-Baiting
If there were ever any doubt the Republican establishment will do and say anything to stay in power, look no further than Mississippi.
In a sign of desperation, the establishment outrageously used race card tactics to defeat Senator Thad Cochran's challenger Chris McDaniel in the recent Mississippi Republican primary.
While race card antics is the norm for Al Sharpton and Jesse Jackson, it's shocking that the Republican establishment would stoop to an all-time low by mimicking the actions of left-wing activists.
Robocalls sent out by Cochran supporters implied the Tea Party and Chris McDaniel were racists. The recording told voters to "say no to the Tea Party and their disrespectful treatment of the first African-American President."
Flyers were also sent out claiming Tea Party activists would try to stop black Americans from voting in the primary.
The New York Times reported on black Cochran supporters:
Roger Smith, a black Democrat who said he was being paid to organize for Mr. Cochran, said, "I don't know too much about McDaniel other than what McDaniel's saying: that he's Tea Party, he's against Obama, he don't like black people."
"You're going to get one of the white guys in there," he said. "You got to make a choice."
Kino Sintee, 17, and three black friends waved "Thad" signs on a street corner in a black Hattiesburg neighborhood. They said the preacher from Mount Olive Baptist Church asked them to help out.
"They're talking about taking everything away from us," he said. "People still need stuff."
Tragically, the Republican Establishment tactics worked. Cochran beat McDaniel because of black turnout. Instead of allowing grassroots conservative Mississippians to decide on their next senator, Democrats determined the Republican nominee.
Using the race card reinforces the negative image of the Tea Party and it will do lasting damage to the reputation of Republicans within the black community.
But then again, the Republican Establishment is only concerned about power today and not the future of the party or the country.

redpajama_set_name: RedPajamaCommonCrawl


text: == Definition of Preventive conservation == '''Preventive conservation '''can be defined as meaning all indirect actions aimed at increasing the life expectancy of objects and collections <bib id="Vantaa:2000"/><ref>http://www.restauratoren.de/fileadmin/red/pdf/Dokument_Vantaa_web.pdf</ref>; as opposed to the direct treatment of individual objects ('''remedial conservation'''). Preventive conservation requires the assessment of objects, deterioration agents, and the environmental context (surveys and investigations); as well as long-term judicious management and forward planning. The term covers all cultural heritage, be it movable or immovable, and aims to keep an object or group of objects in a preferred state where minimum damage and/or deterioration occurs, as well as addressing the assessment and management of potential risks to objects and collections and the formulation of emergency response strategies. It includes direct actions undertaken to avoid natural and man made deterioration agents, in addition to improving the environmental conditions for the storage and/or display of objects. Indeed, notions of preventive conservation influence contemporary museum design for new buildings, and modification where possible of historic buildings [PADFIED T.:2005]<ref>How to keep for a while what you want to keep for ever, http://www.padfield.org/tim/cfys/phdk/phdk_tp.php (seen 01.09.2009)</ref>. It should be noted, however, that the technical literature on preventive conservation mainly addresses the management requirements of museum and archive collections. Further examples of general preventive conservation literature and some more relevant for the issue of salt related deterioration are given below. '''Preventive conservation and salt deterioration''' Within the context of salts-related deterioration of cultural property, objects are often exposed to direct weathering (e.g. historic buildings, outdoor monuments and sculpture). Objects affected by salt deterioration within historic structures include architectural surfaces and wall paintings, flooring, sculpture and monuments. The potential for controlling the environmental conditions within historic structures is subject to significant ethical, financial, and technical restrictions. Objects in museum environments are also often affected by salts which develop from gaseous emissions from storage materials and showcases, however, in such cases environmental control can be achieved comparatively more easily.


2 A common set of standards to improve consultation, participation and dialogue with partners such as regional, local, urban and other public authorities, trade unions, employers, non-governmental organisations and bodies responsible for promoting social inclusion, gender equality and non-discrimination during the planning, implementation, monitoring and evaluation of projects financed by the European Structural and Investment Funds (ESIF) has been adopted today by the European Commission. La Commissione europea ha adottato oggi una serie comune di norme destinate a migliorare la consultazione, la partecipazione e il dialogo con i partner (autorità regionali, locali, cittadine e altre autorità pubbliche, sindacati, datori di lavoro, organizzazioni non governative e organismi di promozione dell'inclusione sociale, della parità di genere e della non discriminazione) nelle fasi di pianificazione, attuazione, sorveglianza e valutazione dei progetti finanziati dai Fondi strutturali e d'investimento europei (Fondi ESI).
3 These Funds comprise the European Regional Development Fund (ERDF), the European Social Fund (ESF), the Cohesion Fund (CF), the European Agricultural Fund for Rural Development (EAFRD) and the European Maritime and Fisheries Fund (EMFF). Tali Fondi comprendono il Fondo europeo di sviluppo regionale (FESR), il Fondo sociale europeo (FSE), il Fondo di coesione (FC), il Fondo europeo agricolo per lo sviluppo rurale (FEASR) e il Fondo europeo per gli affari marittimi e la pesca (FEAMP).
4 This European Code of Conduct on the Partnership Principle requires Member States to strengthen cooperation between their authorities responsible for spending EU structural and investment funds and project partners so as to facilitate the sharing of information, experience, results and good practices in the 2014-20 programming period, and so help to ensure that this money is spent effectively. In base a questo codice europeo di condotta sul principio del partenariato gli Stati membri sono tenuti a rafforzare la cooperazione tra le rispettive autorità responsabili per la spesa dei Fondi strutturali e d'investimento dell'UE e i partner dei progetti al fine di agevolare lo scambio di informazioni, esperienze, risultati e buone pratiche nel periodo di programmazione 2014-2020 e contribuire così a garantire che il denaro sia speso in modo efficace.
5 "We want to ensure that Member States work in constructive partnerships with representative stakeholders in the planning and implementation of programmes using European Structural and Investment Funds so as to maximise the impact of this money", said Commissioner Andor. "Vogliamo garantire che gli Stati membri collaborino in partenariati costruttivi con le parti interessate rappresentative per pianificare e attuare i programmi finanziati dai Fondi strutturali e d'investimento europei in modo da massimizzare l'impatto di questo denaro", ha detto il Commissario Andor.
6 The Code of Conduct, which takes the form of a legally-binding Commission Regulation, sets out objectives and criteria to ensure that Member States implement the partnership principle. Il codice di condotta, che assume la forma di un regolamento della Commissione giuridicamente vincolante, fissa gli obiettivi e i criteri per garantire che gli Stati membri applichino il principio del partenariato.
12 - create platforms for mutual learning and exchange of good practice and innovative approaches. - creare piattaforme per l'apprendimento reciproco e lo scambio di buone pratiche e di approcci innovativi.
13 The Regulation establishes the principles Member States must apply but leaves ample flexibility to Member States to organise the precise practical details for involving relevant partners in the different stages of the programming. Il regolamento stabilisce i principi che gli Stati membri devono applicare, ma lascia loro un ampio margine di flessibilità per organizzare le precise modalità pratiche della partecipazione dei partner alle varie fasi della programmazione.
15 Partnership, one of the key principles of the management of European Union funds, implies close cooperation between public authorities at national, regional and local levels in the Member States and with the private sector and other interested parties. Il partenariato, uno dei principi chiave della gestione dei fondi dell'Unione europea, implica una stretta cooperazione tra le autorità pubbliche a livello nazionale, regionale e locale negli Stati membri, il settore privato e altre parti interessate.
16 Until now, despite being an integral part of Cohesion Policy, feedback from stakeholders indicates that its implementation has varied greatly between different Member States, depending largely on whether the institutional and political culture in a Member State was already conducive to consultation, participation and dialogue with relevant stakeholders. Fino ad ora, benché tale principio sia parte integrante della politica di coesione, le osservazioni pervenute dalle parti interessate indicano che la sua attuazione è avvenuta in modo assai diverso da uno Stato membro all'altro, soprattutto a seconda che la cultura istituzionale e politica di uno Stato membro fosse o meno già favorevole alla consultazione, alla partecipazione e al dialogo con le parti interessate.
17 The new rules, in the form of a legally binding and directly applicable Commission Regulation (a so-called 'Delegated Act'), therefore strengthen the partnership requirement in Article 5 of the Common Provisions Regulation for the European Regional Development Fund (ERDF), the European Social Fund (ESF), the Cohesion Fund (CF), the European Agricultural Fund for Rural Development (EAFRD) and the European Maritime and Fisheries Fund (EMFF) for the period 2014-2020. Le nuove norme, sotto forma di un regolamento della Commissione giuridicamente vincolante e direttamente applicabile (un cosiddetto "atto delegato"), rafforzano quindi il requisito di partenariato di cui all'articolo 5 del regolamento recante disposizioni comuni sul Fondo europeo di sviluppo regionale (FESR), sul Fondo sociale europeo (FSE), sul Fondo di coesione (FC), sul Fondo europeo agricolo per lo sviluppo rurale (FEASR) e sul Fondo europeo per gli affari marittimi e la pesca (FEAMP) per il periodo 2014-2020.
These Funds comprise the European Regional Development Fund (ERDF), the European Social Fund (ESF), the Cohesion Fund (CF), the European Agricultural Fund for Rural Development (EAFRD) and the European Maritime and Fisheries Fund (EMFF).
This European Code of Conduct on the Partnership Principle requires Member States to strengthen cooperation between their authorities responsible for spending EU structural and investment funds and project partners so as to facilitate the sharing of information, experience, results and good practices in the 2014-20 programming period, and so help to ensure that this money is spent effectively.
"We want to ensure that Member States work in constructive partnerships with representative stakeholders in the planning and implementation of programmes using European Structural and Investment Funds so as to maximise the impact of this money", said Commissioner Andor.
The Code of Conduct, which takes the form of a legally-binding Commission Regulation, sets out objectives and criteria to ensure that Member States implement the partnership principle.
- ensure transparency in the selection of partners representing regional, local and other public authorities, social and economic partners and bodies representing the civil society, to be appointed as full members in the monitoring committees of the programmes.
- create platforms for mutual learning and exchange of good practice and innovative approaches.
The Regulation establishes the principles Member States must apply but leaves ample flexibility to Member States to organise the precise practical details for involving relevant partners in the different stages of the programming.
Partnership, one of the key principles of the management of European Union funds, implies close cooperation between public authorities at national, regional and local levels in the Member States and with the private sector and other interested parties.
Until now, despite being an integral part of Cohesion Policy, feedback from stakeholders indicates that its implementation has varied greatly between different Member States, depending largely on whether the institutional and political culture in a Member State was already conducive to consultation, participation and dialogue with relevant stakeholders.
The new rules, in the form of a legally binding and directly applicable Commission Regulation (a so-called 'Delegated Act'), therefore strengthen the partnership requirement in Article 5 of the Common Provisions Regulation for the European Regional Development Fund (ERDF), the European Social Fund (ESF), the Cohesion Fund (CF), the European Agricultural Fund for Rural Development (EAFRD) and the European Maritime and Fisheries Fund (EMFF) for the period 2014-2020.


text:
Cakrawalanews Collection of news and information from various reliable sources
Techno News
Home/Software/These five websites are ideal for repairing blurry photographs and graphics.
Credit Image: pixabay
These five websites are ideal for repairing blurry photographs and graphics.
administrator 25 January 2023 Software
One of the worst things that can happen is to take a great shot with ideal lighting or frame and have the photo come out blurry or out of focus. If you have enough time, you may attempt to retake the shot, but if not, developments in artificial intelligence and photo editing technologies can help you recover it.
Technology is growing at a breakneck pace, and what was once an unusable tangle of fuzzy pixels can now be fixed using a range of internet programs and services that enable you to remove blur from your photographs, and here are five of the finest you should try.
PicWish
Image Sharpener by VanceAI
PineTools
Image Sharpener by Media.io
Among the several image repair tools provided by PicWish, the "Unblur and Sharpen" tool should be considered. You will simply need to submit your fuzzy photograph, and the online service will automatically eliminate the blur as best it can. This is a lot faster and simpler method than learning how to blur a picture on your own.
The finest thing is that PicWish gets excellent outcomes in this area. PicWish reduces picture blur significantly, and you can clearly see this owing to a before and after slider. You may move this slider across the screen to quickly compare the before and after versions of your picture, enabling you to better understand the changes that have been done.
Finally, by registering and checking in, you may obtain a copy of your blur-corrected photograph.
This program, as the name implies, employs Artificial Intelligence to apply its numerous repairs and try to clear up fuzzy photographs with a single click. It includes tools for reducing motion blur, restoring blurred photos, and general sharpening procedures to handle the many of ways an image might be out of focus.
VanceAI compares before and after photographs side by side, and in most instances, the after images are considerably clearer and sharper than they would be otherwise. VanceAI is free to use for the first two photographs, but after that, credits must be purchased to continue correcting images.
PineTools is a little different from many other image-enhancing software. It provides you additional decision-making ability and enables you to choose the size of the focus masks as well as the overall effect intensity.
The effect itself is not as striking as some of the other alternatives, but if you find the right parameters for your picture, you may achieve a better outcome. If the other alternatives on this list don't work, PineTools is a more manual and less mechanical technique to restore your fuzzy photographs that you may want to try. It has no file size restriction and does not need registration.
Fotor Fotor is an image editing suite with a variety of functions, including picture blur removal. Simply following the link to Fotor's blur correction tool and submitting your picture will result in the online service automatically removing blur from your shot.
The tool also includes a before and after slider, which is really important for determining precisely what has been accomplished. The results are fairly amazing, with photographs that are highly crisp without being too manipulated.
The issue is that the only method to download your Fotor photograph is to subscribe to the service or start a free trial, and even paying the subscription contains a watermark.
Finally, the Media.io sharpening tool is always a fantastic alternative for correcting a fuzzy picture as long as it does not exceed 5 MB in size, which is the tool's limit.
That being said, the method is quite similar to previous ones, with the picture to be rectified being uploaded and a before and after button to click to see the changes between before and after, and you can quickly save your image with just one click.
Outlook, Skype, Teams, and Xbox Live are all unavailable due to a severe outage at Microsoft.
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These five websites are ideal for repairing blurry photographs and graphics. 25 January 2023
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Quinn Industrial Holdings is considering options including a potential stock market flotation or private equity investment, the Belfast Telegraph understands.
Formerly the heart of Sean Quinn's business empire, Quinn Industrial Holdings' activities include the manufacturing of plastic packaging and building products.
It has experienced a surge in revenues and profitability despite the impact of Brexit uncertainty and currency fluctuations.
An initial public offering (IPO) of the company would be a significant landmark for the business, which was acquired in 2014 for €98m (£84m) by a group of local businessmen - John McCartin, John Bosco O'Hagan and Ernie Fisher - backed by US hedge funds Silver Point Capital, Brigade Capital and Contrarian Capital.
The US firms own more than 80% of the company.
Shareholders also include Quinn Industrial Holdings' management team, among them chief executive Liam McCaffrey. The company employs 830 people.
While an IPO is an option for the firm, market sources have cast doubt over the move given stock market volatility.
The company declined to comment on any possible IPO or investment process.
However, it said that it "continues to assess a range of financing options to drive continuing growth".
Quinn Industrial Holdings, which has been subject to sabotage and intimidation over the past number of years, has invested €45m (£38.7m) in the business since 2014.
An additional €10m (£8.6m) in capital expenditure will be invested in it this year, compared to €21.7m (£18.7m) last year.
The investment has been fuelled almost exclusively from cashflow, the company's chief financial officer, Dara O'Reilly, said yesterday.
The business's turnover last year jumped 15% to €240m (£206.8m), while earnings before interest, tax, depreciation and amortisation (EBITDA) rose 10% to €26.4m (£22.7m).
Its products include plastic packaging for food, as well as cement and concrete product manufacturing. Revenue has risen 50% since the business was bought, while EBITDA has more than quadrupled.
The growth in revenue last year was driven by volumes on the island of Ireland, where the firm generates just under half its sales. The remainder are from Great Britain.
Speaking to the Belfast Telegraph, Mr McCaffrey said the group could consider the construction of a manufacturing plant in Britain and added that future consolidation within the concrete manufacturing sector in Ireland could also present acquisition opportunities.
The firm is also planning to build its first ready-mix concrete facility in the Dublin region, at a site near the Naas roundabout, with an investment of between €2m (£1.7m) and €3m (£2.6m). It will capitalise on opportunities on the east coast.
The group will eventually target a 10% to 15% share of the Dublin ready-mix concrete market.


» Flow requirement : in LPM/GPM,if possible.
» Application: Boiler tube , Heat Exchanger, if any.
» Existing pump detail , if any.
» Brief about existing cleaning method & details of tools used.
» Flow rate of the pump : LPM/GPM.
(If flow rate is not available then largest and smallest size/volume of equipment).
» Maximum pressure required which you want as factor of safety or future requirement.
» Would you like to stop the pump immediately after obtaining the pressure?


text:
Q: MySQL - Select Concat an entire row How can I select and concat every field in a row?
I want to do something similar to this:
SELECT concat(SELECT GROUP_CONCAT(COLUMN_NAME)
FROM INFORMATION_SCHEMA.COLUMNS
WHERE table_name = 'tbl_name')
as single FROM tbl_name

..but obviously the above doesn't work. Any suggestions?

A: You will have the build the query dynamically, i.e. list all columns in concat(...) explicitly. You can do that on the client side by running two queries or in a stored procedure using a prepared statement.

A: You can do it in 3 steps:
1- Build field list
SELECT group_concat( column_name ) as field_list
FROM INFORMATION_SCHEMA.COLUMNS
WHERE table_name = 'TABLE_NAME'
GROUP BY table_name
LIMIT 0 , 1

2- Copy the value of field_list field
3- Extract data:
SELECT CONCAT( "Paste value copied at previous step" )
FROM TABLE_NAME

If you create a stored procedure, than you can combine the above steps


text:
package main

import (
	"fmt"
	"net"
	"log"
)

func AppendIfMissingReq(dataList []Request, i Request) []Request {
	for _, ele := range dataList {
		if ele.SessionId == i.SessionId {
			return dataList
		}
	}
	return append(dataList, i)
}

func AppendIfMissingString(dataList []string, i string) []string {
	for _, ele := range dataList {
		if ele == i {
			return dataList
		}
	}
	return append(dataList, i)
}

func CheckExistingString(dataList []string, i string) bool {
	for _, ele := range dataList {
		if ele == i {
			return true
		}
	}
	return false
}

func CreateHost(_ip, _port string) string {
	testIp := net.ParseIP(_ip)
	if testIp.To4() == nil {
		return _ip
	} else {
		return fmt.Sprintf("%s:%s", _ip, _port)
	}
}

func GetSelectedResourceForRequest(records []SelectionResult, sessionId string, pickedResources []string) (resourceForRequest SelectedResource, isExisting bool) {
	for _, record := range records {
		if record.Request == sessionId {
			log.Println("AvailableResources: ", record.Resources)

			record.Resources.Priority = DiffArray(pickedResources, record.Resources.Priority)
			record.Resources.Threshold = DiffArray(pickedResources, record.Resources.Threshold)
			resourceForRequest = record.Resources
			isExisting = true
			return
		}
	}
	isExisting = false
	return
}

func DiffArray(a, b []string) []string {
	m := make(map[string]bool)
	for _, s := range a {
		m[s] = true
	}
	result := make([]string, 0)
	for _, s := range b {
		if !m[s] {
			result = append(result, s)
		}
	}
	return result
}

redpajama_set_name: RedPajamaGithub


text:
\section{#1}}
\renewcommand{\thesection.\arabic{equation}}{\arabic{section}.\arabic{equation}}
\def\thesection.\arabic{equation}{\thesection.\arabic{equation}}
\newcommand{\ell}{\ell}
\newcommand{\nonumber}{\nonumber}
\newcommand\para{\paragraph{}}
\newcommand{\ft}[2]{{\textstyle\frac{#1}{#2}}}
\newcommand{\eqn}[1]{(\ref{#1})}
\newcommand\mf{\mathcal{F}}
\newcommand\ma{\mathcal{A}}
\newcommand\Q{\mathcal{Q}}
\newcommand\R{\mathbb{R}}
\newcommand\Z{\mathbb{Z}}
\newcommand\F{\mathbb{F}}
\newcommand\ml{\mathcal{L}}
\newcommand\Pf{\mathrm{Pf}}
\newcommand\D{\mathcal{D}}
\newcommand\cp{\mathbb{CP}}
\newcommand\C{\mathbb{C}}
\newcommand\T{\mathbb{T}}
\newcommand\diff{\mathrm{d}}
\newcommand{\partial}{\partial}
\newcommand{\mathbb{E}}{\mathbb{E}}
\newcommand{\mathrm{d}}{\mathrm{d}}
\newcommand{\mathrm{e}}{\mathrm{e}}
\newcommand{\mathrm{i}}{\mathrm{i}}
\newcommand{\overline\nabla}{\overline\nabla}
\newcommand{\mathrm{e}}{\mathrm{e}}
\newcommand{\mathrm{vol}}{\mathrm{vol}}
\newcommand{\mathrm{Vol}}{\mathrm{Vol}}
\newcommand{\mathrm{Im}\, }{\mathrm{Im}\, }
\newcommand{\mathrm{Re}\, }{\mathrm{Re}\, }
\newcommand\g{\gamma}
\newcommand\e{\epsilon}
\newcommand\G{\Gamma}
\newcommand{\zeta}{\zeta}
\newcommand{{\cal P}_-}{{\cal P}_-}
\newcommand{\mg}[1]{\textbf{mg: {#1}}}

\newcommand\pamo{{\cal P}_\mathrm{A}(p)}
\newcommand\pb{{\cal P}_\mathrm{B}(p)}
\newcommand\qamo{{\cal Q}_\mathrm{A}(q)}
\newcommand\qb{{\cal Q}_\mathrm{B}(q)}
\newcommand\p{\mathcal{P}(p)}
\newcommand\q{\mathcal{Q}(q)}
\newlength{\sswidth}
\newcommand{\sla}[1]{
   \settowidth{\sswidth}{$#1$}
   \mbox{$\not{\hspace*{-0.3\sswidth}#1}$}}

\newcommand{\mathrm{bulk}}{\mathrm{bulk}}
\newcommand{\mathrm{grav}}{\mathrm{grav}}
\newcommand{\ga}{F}
\newcommand{\mathrm{bdry}}{\mathrm{bdry}}
\newcommand{\mathrm{ct}}{\mathrm{ct}}
\newcommand{\mathcal{F}}{\mathcal{F}}
\newcommand{\s}{s}

\numberwithin{equation}{section}

\begin{document}

\begin{titlepage}

\begin{center}

\today

\vskip 2.3 cm

{\Large \bf The gravity dual of supersymmetric gauge theories
\vskip 5mm
on a biaxially squashed three-sphere}

\vskip 2 cm

{Dario Martelli$^1$ and James Sparks$^2$\\}

\vskip 1cm

$^1$\textit{Department of Mathematics, King's College, London, \\
The Strand, London WC2R 2LS,  United Kingdom\\}

\vskip 0.8cm

$^2$\textit{Mathematical Institute, University of Oxford,\\
24-29 St Giles', Oxford OX1 3LB, United Kingdom\\}

\end{center}

\vskip 2 cm

\begin{abstract}
\noindent We present the gravity dual to a class of three-dimensional $\mathcal{N}=2$ supersymmetric gauge theories on
a biaxially squashed three-sphere, with a non-trivial background gauge field. This is described by a 1/2 BPS Euclidean solution of four-dimensional $\mathcal{N}=2$ gauged
supergravity, consisting of a Taub-NUT-AdS metric with a non-trivial instanton for the graviphoton field.
The holographic free energy of this solution agrees precisely with the large $N$ limit of the free energy obtained from the localized partition function
of a class of Chern-Simons quiver gauge theories. We also discuss a different supersymmetric
solution, whose boundary is a biaxially squashed Lens space $S^3/\Z_2$ with a topologically non-trivial background gauge field.
This metric is of Eguchi-Hanson-AdS type, although it is not Einstein, and has a single unit of gauge field flux through the $S^2$ cycle.
 \end{abstract}

\end{titlepage}

\pagestyle{plain}
\setcounter{page}{1}
\newcounter{bean}
\baselineskip18pt

\section{Introduction}

Supersymmetric gauge theories on compact curved backgrounds are interesting for various reasons. For example, supersymmetry may be combined with
localization techniques, allowing one to perform a variety of exact computations in strongly coupled field theories. The authors of
\cite{Hama:2011ea} presented a construction of ${\cal N}=2$ supersymmetric gauge theories in three dimensions in the background of a $U(1)\times U(1)$-invariant
squashed three-sphere and R-symmetry gauge field.
The gravity dual of this construction was recently given in  \cite{Martelli:2011fu}. It consists of a 1/4 BPS Euclidean solution of four-dimensional $\mathcal{N}=2$ gauged
supergravity, which in turn may be uplifted to a supersymmetric solution of eleven-dimensional supergravity. In particular, the bulk metric in
\cite{Martelli:2011fu} is simply AdS$_4$,  and the graviphoton field is an instanton with (anti)-self-dual field strength. The asymptotic metric and gauge field then reduce to the
background considered in \cite{Hama:2011ea}.

The purpose of this letter is to present the gravity dual to a different field theory construction, obtained recently in
\cite{Imamura:2011wg}. In  this reference the authors have constructed three-dimensional ${\cal N}=2$ supersymmetric gauge theories
in the background of the $SU(2)\times U(1)$-invariant squashed three-sphere (which we refer to as  \emph{biaxially} squashed)
and a non-trivial background $U(1)$  gauge field, and have computed the corresponding partition functions using localization.
Differently from a
similar construction discussed briefly in \cite{Hama:2011ea}, this partition function depends non-trivially on the squashing parameter.
As we will see, the gravity dual to this set-up will have some distinct features with respect to the solution in \cite{Martelli:2011fu}. In particular, the metric
is not simply AdS$_4$, although it will again be an Einstein metric, and there is  a self-dual graviphoton.

The plan of the rest of this paper is as follows. In section \ref{revsect} we review the construction
of \cite{Imamura:2011wg}. In section \ref{nuts} we discuss the gravity dual. In section \ref{bolty} we describe a different
supersymmetric solution, consisting of a non-Einstein metric and a non-instantonic graviphoton field.
Section \ref{discussione} concludes.

\section{Supersymmetric gauge theories on the  biaxially squashed $S^3$}

\label{revsect}

In the construction of  \cite{Imamura:2011wg}
the metric on the three-sphere is, up to an irrelevant  overall factor, given by
\begin{eqnarray}
\diff s^2_3 & = & \sigma_1^2 + \sigma_2^2 + \frac{1}{v^2}\sigma_3^2~,
\label{fametric}
\end{eqnarray}
where $\sigma_i$ are the standard $SU(2)$ left-invariant one-forms on $S^3$, defined as $\mathrm{i} \sigma_i \tau_i = -2 \mathtt{g}^{-1}\diff \mathtt{g}$, where
$\tau_i$ denote the Pauli matrices and  $\mathtt{g}\in SU(2)$. The background $U(1)$ gauge field reads
\begin{eqnarray}
A^{(3)} & = & \frac{\sqrt{v^2-1}}{2v^2}\sigma_3 ~,
\label{newgau}
\end{eqnarray}
and the spinors in the supersymmetry transformations obey the equation (setting the radius $r=2$ in \cite{Imamura:2011wg})
\begin{eqnarray}
\nabla_\alpha^{(3)}\chi- \frac{\mathrm{i}}{4v}\gamma_\alpha \chi - A^{(3)}_\beta \gamma_\alpha{}^\beta \chi & = & 0~,
\label{newks}
\end{eqnarray}
where $\nabla_\alpha^{(3)}$,   $\alpha=1,2,3$, is the spinor covariant derivative constructed from the metric (\ref{fametric}),  and
$\gamma_\alpha$ generate $\mathrm{Cliff}(3,0)$. There are \emph{two} linearly independent solutions to
(\ref{newks}), transforming as a doublet under $SU(2)$, whose explicit form is given in \cite{Imamura:2011wg}. This will be important for identifying the gravity dual.

 In  \cite{Imamura:2011wg} the authors
constructed Chern-Simons, Yang-Mills, and matter Lagrangians for  the ${\cal N}=2$
vector multiplets $V=(\mathscr{A}_\alpha,\sigma,\lambda,D)$ and chiral multiplets $\Phi = (\phi,\psi,F)$,
in the background of the metric  (\ref{fametric}) and R-symmetry  gauge field (\ref{newgau}).
These  are
invariant under a set of supersymmetry transformations, provided the spinorial parameters obey the equation (\ref{newks}).
The supersymmetric completion of the Chern-Simons Lagrangian contains new terms, in addition to those appearing in flat space,
proportional to $\sigma^2$ and $\sigma A^{(3)}\wedge \diff \mathscr{A}$ (\emph{cf}. eq. (32)
of \cite{Imamura:2011wg}). The Yang-Mills and matter Lagrangians are total supersymmetry variations (\emph{cf}. eq. (31) of \cite{Imamura:2011wg})
and therefore can be used to compute the partition function using localization. In particular, the partition function
localizes on supersymmetric configurations obeying
\begin{eqnarray}
\mathscr{A}_\alpha \ = \  D \ = \ 0 ~, \qquad \sigma  \ = \ u \ =  \ \mathrm{constant}~,
\label{newloc}
\end{eqnarray}
with the matter fields all being zero.
Notice that although $D=0$, the Chern-Simons Lagrangian
is non-zero because of the new  term proportional to $\sigma^2$, and therefore it contributes classically to the localized partition function, as in previous constructions.
The Yang-Mills and matter terms contribute one-loop determinants from the Gaussian
integration about the the classical solutions (\ref{newloc}). The final partition function may be expressed again in terms of double sine functions $s_b(z)$, and
for a $U(N)$ gauge theory at Chern-Simons level $k\in\Z$ reads
\begin{eqnarray}\label{partition}
Z & = & \int\prod_{\mathrm{Cartan}} \diff u\,  \exp\left(\frac{\mathrm{i} \pi k}{v^2}\mathrm{Tr}\, u^2\right)\frac{\prod_{\mathrm{Roots}\, \alpha} s_b\left(\frac{\alpha(u)-\mathrm{i}}{v}\right)}{
\prod_{\mathrm{Chirals}, \, \mathrm{rep}\, \mathcal{R}_a} \prod_{\rho\in \mathcal{R}_a} s_b\left(\frac{\rho(u)-\mathrm{i}(1-\Delta_a)}{v}\right)}~,
\end{eqnarray}
where $b=(1+\mathrm{i}\sqrt{v^2-1})/v$.
The exponential term is the classical contribution from the Chern-Simons Lagrangian, evaluated on (\ref{newloc}); the numerator is the one-loop
vector multiplet determinant and
involves a product over the roots $\alpha$ of the gauge group $G$; while the denominator is the one-loop
matter determinant and involves a product over chiral fields of R-charge $\Delta_a$ in representations $\mathcal{R}_a$, with $\rho$ running
over weights in the weight-space decomposition of $\mathcal{R}_a$.
Following \cite{Martelli:2011qj}, one can easily extract  the large $N$ behaviour of this partition function for a
class of non-chiral ${\cal N}=2$ quiver Chern-Simons-matter theories. The calculation was done in \cite{Imamura:2011wg}, and the result is that the leading contribution to the free energy (defined as $\mathcal{F}=-\log Z$) is given by
\begin{eqnarray}
\mathcal{F}_v & = & \frac{1}{v^2}\, \mathcal{F}_{v=1}~,
\label{newfree}
\end{eqnarray}
and thus depends very simply on the squashing parameter $v$. In the next section we will present the supergravity dual to this construction, in particular
showing that the holographic free energy precisely agrees with the field theory result (\ref{newfree}).

\section{The gravity dual}\label{nuts}

As anticipated in \cite{Martelli:2011fu}, we will show that the gravity dual to the set-up described in the previous section is
a supersymmetric solution of $d=4$, $\mathcal{N}=2$ gauged supergravity.
In \emph{Lorentzian} signature, the bosonic part of the action is given by
\begin{eqnarray}\label{4dSUGRA}
S_{\mathrm{Lorentzian}} &=& \frac{1}{16\pi G_4}\int \diff^4x\sqrt{-\det g_{\mu\nu}}\left[R + 6g^2 - (F^L)^2\right]~.
\end{eqnarray}
Here $R$ denotes the Ricci scalar of the four-dimensional metric $g_{\mu\nu}$, and the cosmological constant
is given by $\Lambda=-3g^2$. The graviphoton is an Abelian gauge field $A^L$ with field strength
$F^L=\diff A^L$; here the superscript $L$ emphasizes that this is a Lorentzian signature object.
A solution to the equations of motion derived from (\ref{4dSUGRA})
 is supersymmetric if there is a non-trivial spinor $\epsilon$
satisfying the Killing spinor equation
\begin{eqnarray}\label{LKSE}
\left[ \nabla_\mu + \tfrac{1}{2} g \Gamma_\mu - \mathrm{i} g A^L_\mu + \tfrac{\mathrm{i}}{4} F^L_{\nu\rho} \Gamma^{\nu\rho} \Gamma_\mu \right] \epsilon &=& 0~.
\end{eqnarray}
Here $\Gamma_\mu$, $\mu=0,1,2,3$, generate the Clifford algebra $\mathrm{Cliff}(1,3)$, so
$\{\Gamma_\mu,\Gamma_\nu\}=2g_{\mu\nu}$.

Since the background of \cite{Imamura:2011wg} preserves half of the maximal supersymmetry in three dimensions,
we should seek a \emph{1/2 BPS Euclidean} solution of $d=4$, $\mathcal{N}=2$ gauged supergravity,
whose metric has as conformal boundary the biaxially squashed metric on $S^3$ (\ref{fametric}),
and whose background $U(1)$ gauge field restricted to this asymptotic boundary
reduces to (\ref{newgau}). This very strongly suggests that the appropriate solution
is a Euclideanized version of the 1/2 BPS Reissner-Nordstr\"om-Taub-NUT-AdS solution discussed
 in \cite{AlonsoAlberca:2000cs}.

We will first present this Euclidean solution, and then discuss the Wick rotation
that leads to it. The metric reads
\begin{eqnarray}
\diff s^2_4 &= &  \frac{r^2-\s^2}{\Omega(r)}\diff r^2 +
 (r^2-\s^2)(\sigma_1^2+\sigma_2^2) +\frac{4\s^2\Omega(r)}{r^2-\s^2}\sigma_3^2 ~,
\label{TNAdS}
\end{eqnarray}
where
\begin{eqnarray}
\Omega (r) & = &  (\s - r)^2 \left[1 + g^2 (r - \s) (r+3 \s )\right]~,
\label{finalome}
\end{eqnarray}
and $\s$ is the NUT parameter.\footnote{This is denoted $N$ in  \cite{AlonsoAlberca:2000cs}.}
The  $SU(2)$ left-invariant one-forms  $\sigma_i$ may be written in terms of angular variables
as
\begin{eqnarray}
\sigma_1 + \mathrm{i} \sigma_2 \, =\, \mathrm{e}^{-\mathrm{i} \psi}  ( \diff \theta + \mathrm{i} \sin \theta \diff \varphi) ~,\qquad \sigma_3 \, =\, \diff \psi + \cos \theta \diff \varphi~.
\end{eqnarray}
The graviphoton field is
\begin{eqnarray}
A & = &  s\frac{r-s}{r+s}\sqrt{1-4g^2s^2}  \, \sigma_3~.
\label{gaugefield}
\end{eqnarray}
In the orthonormal frame
\begin{eqnarray}\label{frame}
e_1 &=& \sqrt{r^2-s^2}\, \sigma_1~, \qquad \ \ \  \, e_2 \ = \  \sqrt{r^2-s^2}\, \sigma_2~,\nonumber\\
e_3  & = & 2s \sqrt{\frac{\Omega(r)}{r^2-s^2}}\sigma_3~, \qquad e_4 \ = \ \sqrt{\frac{r^2-s^2}{\Omega(r)}}\diff r~,
\end{eqnarray}
the curvature may be written as
\begin{eqnarray}
F &=& \diff A \ = \ - \frac{\s\sqrt{1-4g^2\s^2}}{(r+\s)^2}\left(e_{12}+e_{34}\right)~.
\label{instantfield}
\end{eqnarray}
Thus the gauge field is an \emph{instanton}, as in the solution discussed in \cite{Martelli:2011fu}.
In particular, with our choice of orientation the curvature is self-dual, and the on-shell gauge field action is finite.
Since the stress-energy tensor of an instanton vanishes, the metric (\ref{TNAdS})  is accordingly  an Einstein metric.
However, differently from the solution in  \cite{Martelli:2011fu},
one can check that this metric is \emph{not} locally AdS$_4$.
It is in fact a Euclidean version of the well-known Taub-NUT-AdS metric, with a special value of the mass parameter. This metric is
locally asymptotically AdS$_4$, and therefore it can be interpreted holographically  \cite{Chamblin:1998pz}.
Notice that for $|s|\leq 1/(2g)$ the gauge field (\ref{gaugefield}) is \emph{real}, while for $|s|> 1/(2g)$ it is \emph{purely imaginary};
the intermediate case with  $|s|=1/(2g)$ has vanishing gauge field instanton and the metric
reduces to Euclidean AdS$_4$.

For large $r$  the metric becomes
\begin{eqnarray}\label{approxmetric}
\diff s^2_4 & \approx & \frac{\diff r^2}{g^2 r^2}+ r^2\left[\sigma_1^2+\sigma_2^2+{4g^2\s^2}\sigma_3^2\right]~,
\end{eqnarray}
while to leading order the gauge field reduces to
\begin{eqnarray}\label{A3}
A & \approx  & A^{(3)} \ \equiv \   s\sqrt{1-4g^2s^2}  \, \sigma_3~.
\end{eqnarray}
We see that the conformal boundary may be identified \emph{precisely} with the metric (\ref{fametric}), and the background gauge field with (\ref{newgau}),
by setting $s = \tfrac{1}{2gv}$. Recall here  that in order to uplift to eleven-dimensional supergravity one should also set $g=1$ \cite{Martelli:2011fu}.
Notice that when $|v|=1$ the boundary metric reduces to the round metric on $S^3$, and the background gauge field vanishes.
Correspondingly, in the bulk the instanton field vanishes, and the metric becomes AdS$_4$.

\subsection*{Wick rotation and regularity}

Let us discuss briefly  how this solution was obtained. The reader not interested in these details may safely jump to the discussion of the Killing spinors and the holographic free energy.

As we are interested in a 1/2 BPS solution, we
may begin by appropriately Wick rotating  the solution (2.1), (2.4) of  \cite{AlonsoAlberca:2000cs}. We take their parameter
$\aleph=+1$, so as to obtain a biaxially squashed $S^3$ as constant $r$ surface. The Wick rotation may then be taken to be
$t\rightarrow \mathrm{i}\tau$, $N\rightarrow \mathrm{i} s$, $Q\rightarrow \mathrm{i} Q$, together with a change in sign of the metric. This leads to the following
metric and gauge field
\begin{eqnarray}
\diff s^2_4 &=&  \frac{r^2-\s^2}{\Omega(r)}\diff r^2 + (r^2-\s^2)(\diff\theta^2+\sin^2\theta\diff\varphi^2) + \frac{\Omega(r)}{r^2-\s^2}\left(\diff \tau+ 2 \s \cos\theta\diff\varphi\right)^2~,\nonumber
\\\label{intermedi}
A^L &=& \frac{\s P - Qr}{r^2-\s^2}\diff\tau + \frac{P(r^2+\s^2)-2\s Qr}{r^2-\s^2}\cos\theta\diff\varphi~,
\end{eqnarray}
where
\begin{eqnarray}
\Omega(r) &=& g^2 (r^2-\s^2)^2 + (1-4g^2\s^2)(r^2+\s^2)-2Mr + (P^2-Q^2)~.
\label{genom}
\end{eqnarray}
This depends on the parameters $s,g,M,P,Q$. Notice we have kept a Lorentzian superscript on $A^L$ in (\ref{intermedi}) -- the reason for this will become
clear momentarily.

For the 1/2 BPS solution of interest,
the Euclideanized BPS equations  of \cite{AlonsoAlberca:2000cs} imply that
\begin{eqnarray}
M^2 &=& (1-4g^2\s^2)\left[\s^2(1-4g^2\s^2)+P^2-Q^2\right]~,\nonumber\\
\s^2 P(1-4g^2\s^2) &=& sMQ - P(P^2-Q^2)~,\label{BPS}
\end{eqnarray}
and the corresponding 1/2 BPS solution then depends on only \emph{two} parameters. We take these to be
$s$ and $Q$, with
\begin{eqnarray}
P & =& \mathrm{i}\s\sqrt{1-4g^2\s^2}~,\qquad M \  =\  -\mathrm{i} Q\sqrt{1-4g^2\s^2}~,
\label{jimllfixit}
\end{eqnarray}
then solving (\ref{BPS}). The factors of $\mathrm{i}$ in (\ref{jimllfixit}) may look problematic, but there are (at least) two different
ways of obtaining real solutions. We require $s$ and $M$ to be real in order that the metric in (\ref{intermedi}) is real.
If $|s| \leq 1/(2g)$ then $P$ and $Q$ will be purely imaginary, and we may write $P=\mathrm{i} p$, $Q=-\mathrm{i} q$
to obtain the \emph{real} gauge field
\begin{eqnarray}\label{Wiki}
A &\equiv & -\mathrm{i} A^L \ = \ \frac{\s p + qr}{r^2-\s^2}\diff\tau + \frac{p(r^2+\s^2)+2\s qr}{r^2-\s^2}\cos\theta\diff\varphi~.
\end{eqnarray}

Redefining  $\tau = 2\s \psi$, in terms of standard Euler angles $(\theta,\varphi,\psi)$  notice that the metric (\ref{intermedi}) takes the form
presented in  (\ref{TNAdS}), albeit with a more general form of the function $\Omega (r)$, given by
(\ref{genom}) and (\ref{jimllfixit}). That (\ref{TNAdS}) has only one free parameter $s$, and not the two we have above,
follows from imposing \emph{regularity} of the Euclidean metric.
At any fixed $r>s$ that is not a root of $\Omega(r)$, we obtain a smooth biaxially squashed $S^3$ metric.
In order to obtain a complete metric, the space must ``close off'' at the largest root $r_0$ of $\Omega(r)$, so that $\Omega(r_0)=0$.
More precisely, if $r_0>s$ this should be a single root, while if $r_0=s$ the metric will be regular only if $r_0=s$ is a double root of $\Omega(r)$.
We shall return to the former case in section \ref{bolty}, here focussing on the case $r_0=s$. The condition $\Omega(r_0=s)=0$ immediately fixes
\begin{eqnarray}
q &=& -\s \sqrt{1-4g^2\s^2}~,
\end{eqnarray}
so that now (see also \cite{Emparan:1999pm})
\begin{eqnarray}
p &=& \s\sqrt{1-4g^2\s^2} \ = \ -q~, \qquad  M \ = \ \s(1-4g^2\s^2)~.
\end{eqnarray}
It is then in fact automatic that $r=\s$ is a double root of $\Omega$.

In conclusion, we end up with the metric (\ref{TNAdS}), with $\Omega (r)$ given in (\ref{finalome}),
and gauge field  (\ref{gaugefield}). The gauge field  is manifestly non-singular and one can check that the
metric indeed  smoothly closes off at $r=\s$, giving the topology $M_4=\R^4$.

\subsection*{Killing spinors}

In this subsection we briefly discuss the supersymmetry of the Euclidean solution
 (\ref{TNAdS}), (\ref{gaugefield}), in particular reproducing the three-dimensional
 spinor equation (\ref{newks}) asymptotically.

  In Lorentzian signature the Killing spinor equation
 is (\ref{LKSE}). However, in Wick rotating we have introduced a factor of $\mathrm{i}$ into the gauge field in (\ref{Wiki}), so that $A^L=\mathrm{i} A$.
 Thus the appropriate Killing spinor equation to solve in this case is
\begin{eqnarray}\label{funnyKSE}
\left[ \nabla_\mu + \tfrac{1}{2} g \Gamma_\mu +  g A_\mu - \tfrac{1}{4} F_{\nu\rho} \Gamma^{\nu\rho} \Gamma_\mu \right] \epsilon &=& 0~.
\end{eqnarray}
This possibility of Wick rotating the gauge field (or not) was also discussed in \cite{Dunajski:2010uv}.
In particular, the authors of \cite{Dunajski:2010uv} pointed out that any Euclidean solution
with a real gauge field that solves (\ref{funnyKSE}) will automatically be 1/2 BPS.
The reason is simple: if $\epsilon$ solves (\ref{funnyKSE}), then so does its
conjugate $\epsilon^c$. We shall see this explicitly below.

We introduce the following representation for the generators of Cliff$(4,0)$
\begin{eqnarray}
\hat{\Gamma}_4 & =&  \begin{pmatrix} 0 & \mathrm{i} \mathbb{I}_2 \\ -\mathrm{i} \mathbb{I}_2 & 0 \end{pmatrix}~, ~~~~~~~~
\hat{\Gamma}_\alpha  \ =  \ \begin{pmatrix} 0 & \tau_\alpha \\ \tau_\alpha & 0 \end{pmatrix} ~,
\end{eqnarray}
where $\alpha \in 1,2,3$, $\tau_\alpha$ are the Pauli matrices, and hats denote tangent space quantities.
Decomposing the Dirac spinor $\epsilon$ into positive and negative chirality parts as
\begin{eqnarray}
\epsilon &=& \left(\begin{array}{c} \epsilon_+\\ \epsilon_-\end{array}\right)~,
\end{eqnarray}
where $\epsilon_\pm$ are two-component spinors, it is then straightforward, but tedious, to verify that in the orthonormal frame (\ref{frame})
\begin{eqnarray}\label{spinors}
\epsilon_+ &=& \left(\begin{array}{c} \lambda(r)\chi_+\\ \lambda^*(r)\chi_-\end{array}\right)~, \qquad \epsilon_- \ = \ \mathrm{i} \sqrt{\frac{r-\s}{r+\s}}\left(\begin{array}{c}
\lambda^*(r)\chi_+ \\ \lambda(r)\chi_-\end{array}\right)~,
\end{eqnarray}
is the general solution to the $\mu=r$ component of (\ref{funnyKSE}), where $\chi_\pm$ are independent of $r$
and we have defined
\begin{eqnarray}
\lambda(r) & \equiv & \left(g(r+\s)-\mathrm{i} \sqrt{1-4g^2\s^2}\right)^{1/2}~.
\end{eqnarray}
If we now define the charge conjugate spinor $\epsilon^c \equiv B\epsilon^*$, where
$B$ is the charge conjugation matrix defined in \cite{Martelli:2011fu},
then it is straightforward to see that taking the conjugate $\epsilon\rightarrow \epsilon^c$ simply
maps $\chi_+\rightarrow -\chi_-^*$, $\chi_-\rightarrow\chi_+^*$.

Let us analyze the large $r$ asymptotics of the Killing spinor equation (\ref{funnyKSE}), and its solutions (\ref{spinors}). We begin by expanding
\begin{eqnarray}
\epsilon_+ &=& \sqrt{g}r^{1/2}\left[\mathbb{I}_2+\left(\frac{\s}{2}\mathbb{I}_2-\frac{\mathrm{i}}{2g}\sqrt{1-4g^2\s^2}
\tau_3\right)r^{-1}+\mathcal{O}(r^{-2})\right]\chi~,\nonumber\\ \label{spinorexpand}
\epsilon_- &=& \mathrm{i}\sqrt{g}r^{1/2}\left[\mathbb{I}_2-\left(\frac{\s}{2}\mathbb{I}_2-\frac{\mathrm{i}}{2g}\sqrt{1-4g^2\s^2}
\tau_3\right)r^{-1}+\mathcal{O}(r^{-2})\right]\chi~,
\end{eqnarray}
where we have defined the $r$-independent two-component spinor
\begin{eqnarray}\label{chispinor}
\chi & \equiv & \left(\begin{array}{c}\chi_+\\ \chi_-\end{array}\right)~.
\end{eqnarray}
We then write the asymptotic expansion of the metric as
\begin{eqnarray}
\mathrm{d} s^2_4 &=& \frac{\diff r^2}{g^2 r^2}\left[1+\mathcal{O}(r^{-2})\right]+ \frac{r^2}{g^2}\left[\diff s^2_3+\mathcal{O}(r^{-2})\right]~,\\\label{threemetric}
\diff s^2_3 &\equiv & g^2\left[\sigma_1^2+\sigma_2^2+{4g^2\s^2}\sigma_3^2\right]~.
\end{eqnarray}
It is then straightforward to extract the coefficient of $r^{1/2}$ in the Killing spinor equation (\ref{funnyKSE}).
One finds that the positive and negative chirality projections lead to the \emph{same} equation
for $\chi$, namely
\begin{eqnarray}\label{3dkse}
\nabla_\alpha^{(3)}\chi+ gA^{(3)}_\alpha \chi - \frac{\mathrm{i} \s}{2}\gamma_\alpha \chi - \frac{1}{2g}\sqrt{1-4g^2\s^2}\gamma_\alpha\tau_3 \chi & = & 0~,
\end{eqnarray}
where $\nabla^{(3)}$ denotes the spin connection for the three-metric (\ref{threemetric}),
and $A^{(3)}$ is defined in (\ref{A3}). Using the explicit form for $A^{(3)}$ in (\ref{A3}), the identity $\gamma_\alpha\gamma_\beta=\gamma_{\alpha\beta}
+g^{(3)}_{\alpha\beta}$,  and
recalling that $s=1/(2gv)$, $g=1$, we precisely obtain the spinor equation (\ref{newks}). Finally, one can verify
that the $d=4$ spinors (\ref{spinors}), with $\chi$ satisfying (\ref{3dkse}), do indeed solve (\ref{funnyKSE}).

\subsection*{The holographic free energy}

The holographic free energy of the Taub-NUT-AdS solution was discussed in \cite{Emparan:1999pm},
but of course in this latter reference there was no instanton field, which is crucial for supersymmetry.
The calculation proceeds essentially as in section 2.5 of \cite{Martelli:2011fu}, except for the following caveat. The integrability condition for
the Killing spinor equation (\ref{funnyKSE}) gives the equations of motion following from the action
\begin{eqnarray}
S_{\mathrm{Euclidean}} &=& -\frac{1}{16\pi G_4}\int \diff^4x\sqrt{\det g_{\mu\nu}}\left(R + 6g^2 + F^2\right)~,
\end{eqnarray}
which has \emph{opposite} (relative) sign for the gauge field term compared with (\ref{4dSUGRA}) (see also \cite{Dunajski:2010uv}). This is clear from the fact that our equation
(\ref{funnyKSE}) was obtained from the Lorentzian form of the equation by  sending $A\to \mathrm{i} A$. It is therefore natural to expect that in the computation of the holographic free energy
we have to evaluate
the action $S_{\mathrm{Euclidean}}$ on shell.

Setting $g=1$ and cutting off the space at $r=R$, the bulk gravity contribution is given by
\begin{eqnarray}
I^{\mathrm{grav}}_{\mathrm{bulk}} &=& \frac{3}{8\pi G_4}\int \diff^4  x\sqrt{\det g_{\mu\nu}}\
 = \   \frac{4 \pi \s R^3}{G_4}  - \frac{12\pi \s^3  R}{G_4}   +  \frac{8 \pi \s^4 }{G_4} ~.
\end{eqnarray}
Denoting by $R[\gamma]$  the scalar curvature of the boundary metric, and by $K$ the trace of its second fundamental form,
the combined gravitational boundary terms
\begin{eqnarray}
I_\mathrm{ct}^\mathrm{grav} +I_\mathrm{bdry}^\mathrm{grav} & = &  \frac{1}{8\pi G_4}\int \mathrm{d}^3x \sqrt{\det \gamma_{\alpha\beta}}  \left(  2 + \frac{1}{2} R [\gamma]  -  K  \right)
\label{gravbdry}
\end{eqnarray}
have the following asymptotic expansion
\begin{eqnarray}
I_\mathrm{ct}^\mathrm{grav} +I_\mathrm{bdry}^\mathrm{grav} & = & -\frac{4 \pi \s R^3}{G_4}  + \frac{12\pi \s^3  R}{G_4}   + \frac{4\pi\s^2(1-4\s^2)}{G_4} + {\cal O}(1/R)~,
\end{eqnarray}
where in particular notice there is a non-zero \emph{finite} contribution.
The instanton action is\footnote{Notice that when $1-4s^2 \geq 0$ this term becomes negative. The calculation is however valid for any value of $s>0$.}
\begin{eqnarray}
I^F_{\mathrm{bulk}} &=& -\frac{1}{16\pi G_4}\int \mathrm{d}^4 x \sqrt{\det g_{\mu\nu}} F_{\mu\nu}F^{\mu\nu} \ =
  \ - \frac{2\pi \s^2(1-4\s^2)}{G_4}~.
\end{eqnarray}
Therefore the total on-shell action $S_{\mathrm{Euclidean}}$, obtained after removing the cut-off ($R \to \infty$), is given by
\begin{eqnarray}
I & = & I^{\mathrm{grav}}_{\mathrm{bulk}}  + I_\mathrm{ct}^\mathrm{grav} +I_\mathrm{bdry}^\mathrm{grav} + I^F_{\mathrm{bulk}} \ =\ \frac{2\pi \s^2}{G_4}~.
\end{eqnarray}
Since the round sphere result\footnote{To recover the result for $S^2\times S^1$ boundary, one should first change coordinates back to the form
in (\ref{intermedi}), and then set $\s=0$ there. In these coordinates, with $\tau \in [0, 2\pi]$
the gravitational contribution to the free energy is half that of the round sphere.}  is $\s=1/2$, we thus see that
\begin{eqnarray}
I_{s} & =&   \frac{2\pi \s^2}{G_4} \ = \ (2\s)^2 I_{s=1/2}~,
\end{eqnarray}
which  since $v=1/(2s)$ precisely agrees with the field theory result (\ref{newfree}).

\section{A supersymmetric Eguchi-Hanson-AdS solution}
\label{bolty}

In the previous section the Taub-NUT-AdS solution  existed for both $1-4g^2\s^2\geq 0$ and
$1-4g^2\s^2\leq 0$,
with the sign determining whether the gauge field is real or purely imaginary, in a
fixed choice of Wick rotation. However, in this section we consider a different
solution which exists only when $1-4g^2\s^2\leq 0$, or equivalently $|s|\geq 1/(2g)$.
In this case the Euclidean supersymmetry equation takes the same form as the Lorentzian
equation (\ref{LKSE}), namely
\begin{eqnarray}
\left[ \nabla_\mu + \tfrac{1}{2} g \Gamma_\mu - \mathrm{i} g A_\mu + \tfrac{\mathrm{i}}{4} F_{\nu\rho} \Gamma^{\nu\rho} \Gamma_\mu \right] \epsilon &=& 0~.
\label{KSE2}
\end{eqnarray}
We will show that  there is a one-parameter family of  regular solutions in this class, of topology $M_4=T^*S^2$, for which there are Killing spinors solving (\ref{KSE2}).

When $|s|\geq 1/(2g)$ we may rewrite (\ref{jimllfixit}) as
\begin{eqnarray}\label{realBPS}
P &=& -\s\sqrt{4g^2\s^2-1}~, \qquad M \ =\ Q\sqrt{4g^2\s^2-1}~,
\end{eqnarray}
which are now \emph{real}. Again setting  $\tau=2\s\psi$, the metric takes the form given in
(\ref{TNAdS})
where now
\begin{eqnarray}\label{unlikely}
\Omega(r) &=& g^2(r^2-\s^2)^2-\left[r\sqrt{4g^2s^2-1}+Q\right]^2~.
\end{eqnarray}
It will be useful to note that the four roots of $\Omega(r)$ in (\ref{unlikely}) are
\begin{eqnarray}\label{roots}
\left\{\begin{array}{c} r_4 \\ r_3\end{array}\right\} &=& \frac{1}{2g}\left[\sqrt{4g^2\s^2-1}\pm \sqrt{8g^2\s^2+4gQ-1}\right]~,\nonumber\\
\left\{\begin{array}{c} r_2 \\ r_1\end{array}\right\} &=& \frac{1}{2g}\left[-\sqrt{4g^2\s^2-1}\pm \sqrt{8g^2\s^2-4gQ-1}\right]~.
\end{eqnarray}
Notice that these are all complex if $|s|< 1/(2g)$.
The gauge field is given by (after a suitable gauge transformation)
\begin{eqnarray}\label{noninstanton}
A &=&  - \frac{s}{r^2-s^2}\left[2Qr+(r^2+s^2)\sqrt{4g^2\s^2-1}\right]\, \sigma_3~.
\end{eqnarray}
As $r\rightarrow \infty$ this tends to
\begin{eqnarray}\label{globalpart}
A &\approx & A^{(3)} \ \equiv \  -s\sqrt{4g^2s^2-1}\sigma_3~,
\end{eqnarray}
which is (up to analytic continuation) what we had in the previous example (\ref{A3}).

\subsection*{Killing spinors}

Taking the same Clifford algebra and spinor conventions as the previous section, and again using the orthonormal frame
(\ref{frame}), one can verify that the integrability condition for the Killing spinor equation (\ref{KSE2})
leads to the algebraic relation
\begin{eqnarray}
\epsilon_-&=& \mathrm{i} \sqrt{\frac{r-s}{r+s}}\left(\begin{array}{cc}\sqrt{\frac{(r-r_3)(r-r_4)}{(r-r_1)(r-r_2)}} & 0
\\ 0 & \sqrt{\frac{(r-r_1)(r-r_2)}{(r-r_3)(r-r_4)}}\end{array}\right)\epsilon_+~.
\end{eqnarray}
Here recall that $\epsilon_\pm$ are two-component spinors, and $r_i$, $i=1,2,3,4$, are the four roots of $\Omega$ in (\ref{roots}).
Substituting into the $\mu=r$ component of (\ref{KSE2}) then leads to decoupled first order ODEs, which may be solved to give
\begin{eqnarray}\label{spinors2}
\epsilon_+ &=& \left(\begin{array}{c}\sqrt{\frac{(r-r_1)(r-r_2)}{(r-s)}} \chi_+\\ \sqrt{\frac{(r-r_3)(r-r_4)}{(r-s)}} \chi_-\end{array}\right)~, \qquad \epsilon_- \ = \ \mathrm{i} \left(\begin{array}{c}
\sqrt{\frac{(r-r_3)(r-r_4)}{(r+s)}} \chi_+ \\ \sqrt{\frac{(r-r_1)(r-r_2)}{(r+s)}} \chi_-\end{array}\right)~,
\end{eqnarray}
where $\chi_\pm$ are independent of $r$. The large $r$ expansion of these is given by
\begin{eqnarray}
\epsilon_+ &=& r^{1/2}\left[\mathbb{I}_2+\left(\frac{\s}{2}\mathbb{I}_2+\frac{1}{2g}\sqrt{4g^2\s^2-1}
\tau_3\right)r^{-1}+\mathcal{O}(r^{-2})\right]\chi~,\\
\epsilon_- &=& \mathrm{i} r^{1/2}\left[\mathbb{I}_2-\left(\frac{\s}{2}\mathbb{I}_2+\frac{1}{2g}\sqrt{4g^2\s^2-1}
\tau_3\right)r^{-1}+\mathcal{O}(r^{-2})\right]\chi~,
\end{eqnarray}
where the two-component spinor $\chi$ is again given by (\ref{chispinor}). Notice this is the same as (\ref{spinorexpand}), up
to analytic continuation. Again using the metric expansion and three-metric in (\ref{threemetric}), we may
extract the coefficient of $r^{1/2}$ in (\ref{KSE2}). A very similar computation to that in the previous section
then leads to the three-dimensional Killing spinor equation
\begin{eqnarray}\label{3dkse2}
\nabla_\alpha^{(3)}\chi- \frac{\mathrm{i} \s}{2}\gamma_\alpha \chi + \mathrm{i} g A^{(3)}_\beta \gamma_{\alpha}^{\ \beta}\chi & = & 0~.
\end{eqnarray}
Setting $g=1$ and again identifying the squashing parameter $v=1/(2s)$, notice this is identical
 to our original equation (\ref{newks}), but where we have replaced $A^{(3)}\rightarrow -\mathrm{i} A^{(3)}$. Of course,
 given the relative difference in Wick rotations of the gauge field in two the cases, this was
 precisely to be expected. In fact, comparing the $A^{(3)}$ (\ref{globalpart}) in this section with its counterpart
 (\ref{A3}) in the previous section, we see that equation (\ref{3dkse2}) is in fact \emph{identical} to (\ref{newks}), due to the
 factor of $\mathrm{i}$ difference in (\ref{globalpart}), (\ref{A3}).

The solution to (\ref{3dkse2}) is therefore given by an appropriate analytic continuation of the solution presented in \cite{Imamura:2011wg},
and reads
\begin{eqnarray}\label{3dspin}
\chi &=& \mathrm{e}^{\eta\tau_3/2} \mathtt{g}^{-1}\chi_0~,
\end{eqnarray}
where $\mathtt{g}\in SU(2)$, $\chi_0$ is a constant two-component spinor, and
\begin{eqnarray}
v &=& \frac{1}{\cosh \eta}~,
\end{eqnarray}
where $v=1/(2s)$. In terms of Euler angles $(\psi,\theta,\varphi)$, recall that
\begin{eqnarray}\label{su2}
\mathtt{g} &=& \left(\begin{array}{cc}\cos\frac{\theta}{2}\mathrm{e}^{\mathrm{i}(\psi+\varphi)/2} & \sin\frac{\theta}{2}\mathrm{e}^{-\mathrm{i}(\psi-\varphi)/2}
\\ -\sin\frac{\theta}{2}\mathrm{e}^{\mathrm{i}(\psi-\varphi)/2} & \cos\frac{\theta}{2}\mathrm{e}^{-\mathrm{i}(\psi+\varphi)/2}\end{array}\right)~.
\end{eqnarray}

\subsection*{Regularity of the metric}

We must again consider regularity of the metric (\ref{TNAdS}). A complete metric will necessarily close off at the largest root
$r_0$ of $\Omega(r)$, which must satisfy $r_0\geq s$. From (\ref{roots}) we see that either $r_0=r_+$ or $r_0=r_-$, where it is convenient to
define
\begin{eqnarray}\label{rpm}
r_+ & \equiv & r_4~, \qquad r_- \ \equiv \ r_2~.
\end{eqnarray}
{\it A priori} the coordinate $\psi$ must have period $2\pi/n$, for some positive integer $n$, so that
the surfaces of constant $r$ are Lens spaces $S^3/\Z_n$.
Assuming that $r_0>s$ is strict, then the metric (\ref{TNAdS}) will have the topology of
a complex line bundle $M_4=\mathcal{O}(-n)\rightarrow S^2$ over $S^2$, where $r-r_0$ is the radial direction away from the zero section.

Regularity of the metric near to the $S^2$ zero section at $r=r_0$ requires
\begin{eqnarray}\label{regularity}
\left|\frac{r_0^2-s^2}{s\Omega'(r_0)}\right| &=& \frac{2}{n}~.
\end{eqnarray}
This conditon ensures that near to $r=r_0$ the metric (\ref{TNAdS}) takes the form
\begin{eqnarray}
\diff s^2_4 &\approx & \diff\rho^2 + \rho^2\left[\diff \left(\frac{n\psi}{2}\right)+\frac{n}{2}\cos\theta\diff\varphi\right]^2 + (r_0^2-s^2)(\diff\theta^2+\sin^2\theta\diff\varphi^2)~,
\end{eqnarray}
near to $\rho=0$. Here note that $n\psi/2$ has period $2\pi$.
Imposing (\ref{regularity}) at $r_0=r_\pm$ gives
\begin{eqnarray}\label{Qvalues}
Q &=& Q_\pm(\s) \ \equiv \ \mp \frac{128g^4s^4-16g^2s^2-n^2}{64g^3\s^2}~.
\end{eqnarray}
In turn, substituting $Q=Q_\pm(s)$ into (\ref{rpm}) one then finds
\begin{eqnarray}\label{rootschoice}
r_\pm(Q_\pm(\s)) &=& \frac{1}{8g}\left[\frac{n}{g\s}\pm 4\sqrt{4g^2\s^2-1}\right]~.
\end{eqnarray}
Recall that in order to have a smooth metric, we require $r_0>s$. Imposing this for $r_0=r_\pm(Q_\pm(\s))$ gives
\begin{eqnarray}
r_\pm(Q_\pm(\s)) - s &=& \frac{1}{2g}\, f_n^\pm (2gs)~,
\end{eqnarray}
where the function
\begin{eqnarray}
f_n^\pm(x) &\equiv & \frac{n}{2x}-x \pm\sqrt{x^2-1}
\end{eqnarray}
is required to be \emph{positive} for a smooth metric with $s=x/(2g)$. Notice here that $s\geq 1/(2g)$ implies $x\geq 1$.
It is straightforward to show that $f_n^-(x)$ is monotonic decreasing on $x\in [1,\infty)$.
For simplicity here we will restrict our attention to $n\leq 2$.\footnote{In the first version of this paper
it was argued that $n>2$ breaks supersymmetry; however, this is incorrect.}
The analysis then splits into the cases $\{n=1\}$, $\{n=2\}$, which have a qualitatively different behaviour:

\subsubsection*{$n=1$}

It is easy to see that $f_1^\pm(x)< 0$ on $x\in [1,\infty)$, and thus the metric
(\ref{TNAdS}) cannot be made regular in this case. Specifically, $f_1^\pm(1)=-1/2$:
since $f_1^-(x)$ is monotonic decreasing, this rules out taking $r_0=r_-(Q_-(s))$ given by (\ref{rootschoice});
on the other hand $f_1^+(x)$ monotonically increases to zero
from below as $x\rightarrow\infty$, and we thus also rule out $r_0=r_+(Q_+(s))$ in (\ref{rootschoice}).

\subsubsection*{$n=2$}

It is easy to see that $f_2^-(x)< 0$ for $x\in (1,\infty)$,
while $f_2^+(x)>0$ on the same domain, which means we must set
\begin{eqnarray}\label{Qset}
Q &\equiv & Q_+(s) \ = \  - \frac{(4g^2\s^2-1)(1+8g^2\s^2)}{16g^3\s^2}~,
\end{eqnarray}
and
\begin{eqnarray}\label{r0}
r_0(s) & = & \frac{1}{4g}\left[\frac{1}{g\s}+ 2\sqrt{4g^2\s^2-1}\right]~,
\end{eqnarray}
may then be shown to be the largest root of $\Omega(r)$, for all $s\geq 1/(2g)$.
In particular,
this involves showing that $r_0(s)-r_-(Q_+(s))>0$  for all $s\geq 1/(2g)$, which follows since
\begin{eqnarray}
r_0(s)-r_-(Q_+(s)) &=& \frac{1}{2g}\, h(2gs)~,
\end{eqnarray}
where we have defined
\begin{eqnarray}
h(x) & \equiv & \frac{1}{x} + 2\sqrt{x^2-1}-\sqrt{4x^2-2-\frac{1}{x^2}}~.
\end{eqnarray}
It is a simple exercise to prove that $h(x)>0$ on $x\in (1,\infty)$.

\

After this slightly involved analysis, for $n=2$ we end up with a smooth complete metric on $M_4=T^*S^2$, given by
(\ref{TNAdS}), (\ref{unlikely}) with $Q=Q_+(s)$ given by (\ref{Qset}),  for all $s> 1/(2g)$.
The $S^2$ zero section is at $r=r_0(s)$ given by (\ref{r0}). The metric is thus of
Eguchi-Hanson-AdS type, although we stress that it is \emph{not} Einstein
for any $s>1/(2g)$. The large $r$ behaviour is again given by (\ref{approxmetric}),
so that the conformal boundary is a squashed $S^3/\Z_2$. The $s=1/2g$
limit gives a round $S^3/\Z_2$  at infinity
with the bulk metric being the singular AdS$_4/\Z_2$,  albeit with a non-trivial torsion gauge field, as we shall
see momentarily.

It follows that another interesting difference to the Taub-NUT-AdS solution of the previous section is that
the gauge field (\ref{noninstanton}) no longer has (anti)-self-dual field strength $F=\diff A$; moreover,
the latter has a non-trivial flux. Indeed,
although the gauge potential in (\ref{noninstanton}) is \emph{singular} on the
$S^2$ at $r=r_0$,  one can easily see that the field strength $F=\diff A$ is a
globally defined smooth two-form on our manifold. One
 computes the period of this through the $S^2$ at $r_0(s)$ to be
\begin{eqnarray}\label{fluxy}
\frac{g}{2\pi}\int_{S^2} F &=& -\frac{2gs}{r_0(s)^2-s^2}\left[-2Q_+(s)r_0(s)-(r_0(s)^2+s^2)\sqrt{4g^2\s^2-1}\right]\nonumber\\
&=& 1~,
\end{eqnarray}
the last line simply being a remarkable identity satisfied by the largest root $r_0(s)$. Setting $g=1$, we thus
see that we have precisely one unit of flux through the $S^2$! It follows that the gauge field $A$ is
 a connection on the non-trivial line bundle
$\mathcal{O}(1)\rightarrow T^*S^2$. The corresponding first Chern class $c_1=[F/2\pi]\in H^2(T^*S^2;\Z)\cong \Z$
is the generator of this group. Moreover, the map
$H^2(T^*S^2;\Z)\rightarrow H^2(S^3/\Z_2;\Z)\cong\Z_2$ that restricts the gauge
field to the conformal boundary is reduction modulo $2$. Hence
at infinity the background gauge field is more precisely given by the
global one-form (\ref{globalpart}) \emph{plus} the flat
non-trivial Wilson line that represents the element $1 \in H^2(S^3/\Z_2;\Z)\cong H_1(S^3/\Z_2;\Z)\cong\Z_2$.
One would be able to see this explicitly by writing the gauge field $A$ as a one-form that is locally well-defined
 in coordinate patches, and undergoes appropriate gauge transformations between these coordinate patches.
It follows that the gauge field at infinity is more precisely a connection on the non-trivial
torsion line bundle over $S^3/\Z_2$.

\subsection*{The holographic free energy}

Although we will not pursue the holographic interpretation of this solution in the present paper, below we will
compute its holographic free energy using  standard formulas.
Since the gauge field here is real, the relevant action is the Euclidean action with standard signs
\begin{eqnarray}
S_{\mathrm{Euclidean}} &=& -\frac{1}{16\pi G_4}\int \diff^4x\sqrt{\det g_{\mu\nu}}\left(R + 6g^2 - F^2\right)~.
\end{eqnarray}
Notice that upon taking the trace of the Einstein equation, we see that \emph{all} solutions
(supersymmetric or not) of $d=4$ gauged supergravity are metrics with  constant scalar curvature
$R = - 12 g^2$. Using this,
a straightforward calculation then gives for the total (bulk plus boundary) gravity part a finite result, after
sending the cut-off $r=R\to \infty$. Namely, after setting $g=1$ we get
\begin{eqnarray}
I^{\mathrm{grav}}_\mathrm{tot} \  =\  I^{\mathrm{grav}}_{\mathrm{bulk}}  +I_\mathrm{bdry}^\mathrm{grav}  + I_\mathrm{ct}^\mathrm{grav} & =&  \frac{\left(1-12 s^2\right) \pi }{32 G_4 s^2}\nonumber \\
& - &   \frac{ \sqrt{4 s^2-1} \left(1+4 s^2 \left(-3+8 s^2\right)\right) \pi }{16 G_4  s}~,
\end{eqnarray}
where we note that the contribution on the second line comes entirely from the boundary terms.
Although the gauge field is not (anti-)-self-dual, it is straightforward to compute its on-shell action,
which is still finite, namely we get
\begin{eqnarray}
 I^F_{\mathrm{bulk}} &=& \frac{(1+4  s^2)\pi}{32 G_4  s^2} -\frac{\sqrt{4 s^2-1} \left(1+4 s^2 \left(1-8 s^2\right)\right) \pi }{16 G_4  s}~.
\end{eqnarray}
Therefore for  the total on-shell action
we obtain
\begin{eqnarray}
I & = & \frac{\pi}{2 G_4} + \left(s^2 - \tfrac{1}{4}\right)^{3/2}\frac{\pi }{G_4 s }~.
\label{predict}
\end{eqnarray}
 Notice this  makes sense for any $s>1/2$. Moreover, in the $s\to 1/2$ limit the second term vanishes and we are left with
a result that is the same as that for the round three-sphere $S^3$. This might seem a contradiction, but in fact if we
look back at where this result comes from, we see that in this limit
\begin{eqnarray}
\lim_{s\to 1/2} I^{\mathrm{grav}}_\mathrm{tot} &  = & \frac{\pi}{4 G_4}~,
\end{eqnarray}
which is the correct contribution expected from the (singular) AdS$_4/\Z_2$ solution with round $S^3/\Z_2$ boundary. However, we get
an equal non-zero contribution from the gauge field action
\begin{eqnarray}
\lim_{s\to 1/2} I^F_{\mathrm{bulk}} &  =& \frac{\pi}{4 G_4}~,
\end{eqnarray}
despite the fact that the gauge field curvature $F\to 0$ in this limit. The calculation captures correctly the contribution from the flat torsion gauge field, which indeed cannot be
turned off continuously since in the bulk has one unit of flux through the
 vanishing $S^2$ at the $\Z_2$ singularity.
More precisely, the complement of the singular point
has toplogy $\R_+\times S^3/\Z_2$, and the gauge field is a flat connection on the non-trivial
torsion line bundle over this.

\section{Discussion}
\label{discussione}

In this letter we have extended the results of \cite{Martelli:2011fu}, discussing a new
class of supersymmetric solutions of $d=4$, $\mathcal{N}=2$ gauged supergravity, which in turn
uplift to solutions of eleven-dimensional supergravity. The solutions in section \ref{nuts} provide the holographic duals
to ${\cal N}=2$  supersymmetric gauge theories on the background of a biaxially squashed three-sphere and a $U(1)$ gauge field,
whose localized partition function was recently computed in \cite{Imamura:2011wg}. In particular, as in \cite{Martelli:2011fu}, we have shown that the
bulk metric, gauge field, and Killing spinors reduce precisely to their field theory counterparts on the boundary.
Moreover, the holographic free energy is identical to the leading large $N$ contribution to the field theoretic free energy computed from
the quiver matrix model. The solution is a special case of the general class of supersymmetric
Plebanski-Demianski solutions \cite{AlonsoAlberca:2000cs}, but it differs from the solution discussed in \cite{Martelli:2011fu} in various respects.
The graviphoton field is again an instanton, hence  the bulk metric is Einstein, but it is not now diffeomorphic to AdS$_4$.
The results of \cite{Martelli:2011fu}, and of this letter,  suggest that the AdS/CFT correspondence is a useful setting
for studying supersymmetric gauge theories on curved backgrounds.

We conclude  noting that although the results presented here share a number of similarities with those in
\cite{Chamblin:1998pz,Hawking:1998ct}, there are some crucial differences that are worth summarizing. In contrast to the solutions we have discussed,
the AdS-Taub-NUT and AdS-Taub-Bolt solutions in \cite{Chamblin:1998pz,Hawking:1998ct}  are \emph{not} supersymmetric, and moreover no
gauge field was turned on. In addition, while those solutions have the same biaxially squashed $S^3$  boundary, the boundary of our  Eguchi-Hanson-AdS
solution has the different topology $S^3/\Z_2$. Therefore, although we have computed the free energy for both families, it does not make
sense to compare them along the lines of \cite{Chamblin:1998pz,Hawking:1998ct}.
It would be very interesting to understand the precise
field theory dual interpretation of the Eguchi-Hanson-AdS solution discussed here.

\subsection*{Acknowledgments}
\noindent
We thank Jan Gutowski for a useful discussion. D.~M. is supported by an EPSRC Advanced
 Fellowship EP/D07150X/3 and J.~F.~S. by a Royal Society University Research Fellowship.

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