Fox and Sky: what happens to media plurality now?
On June 29, the UK's Culture Secretary stated that she would submit to a full competition review the 21st Century Fox bid for 100% control of Sky plc. Will it happen?
1998:Film critic Barry Norman with the then Sky TV's General Manager Elisabeth Murdoch for the launch of three new channels for Sky Movies.Peter Jordan/Press Association. All rights reserved. For those people passionately opposed to Rupert Murdoch's media empire, the statement from Karen Bradley last Thursday would have aroused mixed emotions. One of the companies in which the Murdoch Family Trust (MFT) holds a controlling 39% interest – the US-based 21st Century Fox, hereafter "Fox" – is trying to buy all of Sky plc, in which Fox itself already holds a controlling 39% share.
After a 10-week review by the media regulator, Ofcom, Bradley was advised that there were presently no grounds for barring the transaction for reasons to do with commitment to broadcasting quality and upholding Ofcom standards: the "fit and proper test" had been passed. Anyone who thought that the still-running phone-hacking scandal or the recent reports of sexual harassment at Fox News in America might derail the bid, would have been disappointed.
If so, they had misunderstood the "fit and proper" assessment process: this is continuous, and not confined to the period when a transaction is in process. If Fox had been vulnerable on that front, it would have lost its licences already – and that would have been all 54 of Sky's Ofcom licences, not just that for Sky News. So tremendous an upheaval would have required long-term and persistent breaches of the broadcasting code and Ofcom licence conditions, with removal of licences only happening after clear public warnings.
Indeed, there have been far bigger scandals at News Corp (the publishing company that owns Murdoch's UK newspapers and radio stations, in which the MFT also owns 39%) than those that have afflicted Fox News (another Fox broadcaster), including the vast overpayment by News Corp for Shine, the production company created by Murdoch's daughter, Elisabeth (the board of News Corp paid out hundreds of millions of dollars in damages to disgruntled shareholders, and promised to amend its governance practices).
What continues to block the Sky deal is the verdict from Ofcom that there is a media plurality problem, in that 100% ownership of Sky would "reduce media plurality" and may provide an opportunity for the MFT to align the editorial approach of Sky News with that of the newspapers that the MFT controls through its 39% shareholding in News Corp.
The Ofcom report emphasises that the threshold for recommending a full inquiry by the Competition and Markets Authority (CMA), which could run for six months, is low. "The Secretary of State would need to hold the reasonable belief that it may be the case that the transaction may operate or may be expected to operate against the public interest" (their emphasis).
Essentially, the Ofcom assessment of this risk is based on an analysis of the media market, on the position of Sky News and News Corp in that market, and the change that might flow from the enlargement of Fox's shareholding in Sky.
Perhaps to the disappointment of the Murdochs, the alteration in the structure of their empire three years ago – whereby they clearly separated the news and publishing company (News Corp) from the entertainment company (21st Century Fox), with each having a different set of outside 61% shareholders – has been effectively ignored by Ofcom.
The regulator took the view that the MFT's position in each of Fox and News Corp gave it material control, and so the two companies should be looked at together (which the Ofcom report says conforms with previous regulatory practice).
So in terms of media market share, the newspapers, the radio interests and Sky News have been lumped together. However, this approach raises a rather more subtle issue. If the combined share of news consumption of the two different companies is a problem, why was News Corp two years ago allowed to purchase the radio stations in the Wireless Group (TalkRadio, TalkSport and Virgin Radio), thereby unquestionably increasing the Murdoch companies' share of news consumption?
Indeed, where were all the objectors then? There was no inquiry, and no calling in of Ofcom to investigate, even though there was measurably a reduction in the number of media owners (just as there was when Northern and Shell bought Channel 5 – again, with no Ofcom inquiry).
The Ofcom report this time concludes that buying out the 61% (a mass of small shareholders, with no collective identity as a "media enterprise owner") would result in a reduction in the number of those with control of media enterprises – quite a stretch, as the 61%, despite the presence of independent directors on the Sky board, had always allowed Fox (and before the corporate split, News Corp) full operational control of the business. Whether Fox's lawyers bother to challenge this ruling remains to be seen.
But the important issue is really what difference a shift from 39% control to 100% control might mean in terms of one small part of the Sky business: the editorial policies of Sky News.
To the best of my knowledge – and I was Head of Programming at Sky for four years – no-one from the MFT has ever tried to influence the output of Sky News during the 29 years it has been on air. Indeed, the only influence the MFT has brought to bear is an insistence on Sky accepting the on-going financial losses incurred by Sky News (cumulatively, at least £500 million).
Remarkably, when the Murdochs made a similar bid for Sky in 2010 (that time through News Corp), the independent directors suggested closing Sky News if that would allow the deal to go through. It was the Murdochs who spent six months trying to find a structure for Sky News that would satisfy Ofcom and keep the service alive: only for the Milly Dowler revelation to force News Corp to withdraw the bid.
There is no doubt that Rupert Murdoch from time to time influences the output of the newspapers in which the MFT holds a 39% shareholding: he can, and he does – but as far as I am aware, that is not the case where Sky News is concerned. During the last bid, Ofcom claimed that 100% control would allow Murdoch to select the editor of Sky News – I pointed out at the time that he had always had approval of such appointments, even with 39% control, and Ofcom has not advanced that argument this time.
Instead, it explores the possibility of aligning the editorial policy of Sky News with that of the Murdoch newspapers. Ofcom acknowledges that this could not be explicit, as it would breach the broadcasting code, but suggests that omission of certain news stories, managing the news agenda and offering screen time to News Corp newspaper writers might achieve a similar effect.
No doubt Fox will challenge this hypothesis: given the intense scrutiny to which all Murdoch activity is subjected, half a dozen university media departments would surely be tabulating every bulletin, every screen appearance, and every "missing" story.
That such a policy might be implemented but not detected is scarcely likely, Fox might argue: more importantly, if such subtle tweaks could not be detected, how could it be shown they were influencing public opinion? Indeed, would such tweaking be permissible within the terms of the Sky News licence from Ofcom, or not? If not, then the licence could be revoked. If permissible, surely the problem lies with the licence terms, not with the Murdochs (assuming Ofcom is willing and able to flex its regulatory muscles). It would be very curious to block the transaction for fear of changes to Sky News which were perfectly proper within the broadcasting code and licence terms.
Perhaps Ofcom might have created a simple matrix to frame its analysis. Absent the transaction, is there any evidence of the MFT trying to align Sky News with the editorial line in its newspapers? If not, is that because the Murdochs have no such intention, or because something is stopping them? Is the 39% ownership an inhibiting factor, where 100% would allow the alignment? That is clearly not the case with the newspapers: 39% seems more than sufficient to allow editorial interference. So why exactly is it assumed that 100% ownership of Sky would enable this supposed desire for editorial alignment to take place? After all, News Corp owned 100% of Sky during the first two years of Sky News, but there was no alignment then.
Ofcom is entitled to speculate as to possible outcomes of the transaction: but it goes further in its report, talking of "concern". Yet it seems not to have asked itself how the MFT would go about executing the nefarious plan it is "concerned" about. Which "commentators", and from which News Corp newspapers (The Sun, The Times, The Sunday Times and The Wall Street Journal all have somewhat different political slants and approaches)? If from The Times stable, would it be the fiercely anti-Brexit Oliver Kamm, or the strongly pro-Brexit Dominic Lawson?
If there were even a glimmer of such an ambition within the MFT, why is it the absence of News Corp writers appearing on Sky News that currently seems most notable? Why is the most regular duo of commentators on Sky News Kevin Maguire of the Daily Mirror and Andrew Pierce of the Daily Mail – two newspapers deeply hostile to News Corp and all its activities? Indeed, why is there no marketing co-operation between News Corp and Sky, or joint advertising sales arrangements, both of which would be perfectly legal under current ownership?
Sometimes, one wonders whether the problem with Ofcom is not so much that it is a poor regulator, but that it lacks the confidence in its ability to do its job, or even in whether the job is do-able. Sometimes, one wonders whether the problem with Ofcom is not so much that it is a poor regulator, but that it lacks the confidence in its ability to do its job, or even in whether the job is do-able. To recommend a full investigation of a transaction because it might either result in editorial changes at Sky News that were entirely permissible under the broadcasting code and the terms of the Sky News licence, or alternatively result in changes that were not permissible but which Ofcom did not feel able to prevent, tells us much more about Ofcom than about the MFT.
For reasons best known to itself, Ofcom has chosen to echo the fears of the most committed opponents of the Murdochs, without expressing an iota of scepticism about them. Its report quotes repeatedly from a submission from Ed Miliband and Sir Vince Cable, without mentioning their extraordinary article in The Guardian claiming that Sky and News Corp were responsible for 45% of all radio news consumption in the UK (true figure: less than 2%), or Cable's deplorable abuse of his then position as Secretary of State for Business in the coalition, when he was required to act in a quasi-judicial fashion in dealing with the original bid for Sky in 2010, only later to admit that he had referred the bid to Ofcom as part of his "war on the Murdochs" (an admission that forced his resignation).
In some circumstances, absence of evidence is not deemed sufficient to constitute evidence of absence. In the case of Sky News, however, the CMA (the Competition and Markets Authority, which would examine the deal if Karen Bradley goes ahead with her provisional decision to refer the bid) would surely look for some clue from 29 years of experience as to why it should be assumed that the MFT nurses the ambitions that the Ofcom report warns against, and why Ofcom seems so lacking in confidence in its ability to handle the outcome.
The threshold for concern at the first stage of inquiry was, as Ofcom concedes, low: for the CMA, the threshold will be a quantum higher – a probability rather than a possibility of the transaction proving contrary to the public interest.
Having strongly criticised the previous Ofcom report on the 2010 News Corp bid for Sky, I must express some relief at the improvement in the quality of analysis this time. A series of errors I identified last time have, gratifyingly, not been repeated this time. There is no attempt to create a specific reach figure for radio news, and the attribution entirely to Sky News of a (wrong) total for commercial radio news consumption is withdrawn, with full acknowledgement of the role of Independent Radio News and the commercial stations themselves in managing news content and assembling the news bulletins.
The role of broadcaster is now recognized as far as ITN's supply of news to ITV, Channel 4 and Five is concerned: last time, all of Five's news audience was attributed to the supplier (Sky News at the time), as if the broadcaster had no say in the matter – something that I, as a former CEO of Five, knew was a mistake.
Nor does Ofcom come up with a grossly exaggerated figure for consumption of news supplied by Fox/News Corp: unlike its 2010 report, where it calculated the Sky/News Corp share of all news consumption at 23% (and the BBC at 43%), when the correct figures were 10% and 61% respectively. Nor does Ofcom come up with a grossly exaggerated figure for consumption of news supplied by Fox/News Corp: unlike its 2010 report, where it calculated the Sky/News Corp share of all news consumption at 23% (and the BBC at 43%), when the correct figures were 10% and 61% respectively. I am sure the Fox lawyers will note, however, that the reason such errors have been avoided is that there are virtually no estimates of actual news consumption in the entire 134 page report. Instead, Ofcom is highly reliant on research into "resonance" and calculations of "reach" – something I pointed out last time as being both elusive and potentially misleading as a measure.
Let me explain. Reach is a calculation of the number of individuals who have sampled a product (newspaper, TV channel, radio service, online website). The sampling threshold is variable – a minimum of three continuous minutes, or five minutes, or longer; or the qualifying minutes might be non-continuous; these minutes could be consumed within the same day, or over six days, or even over a month; or the reach might be based on "recall" and ""diaries" rather than metered usage. Trying to combine different measures of reach within a multi-platform assessment is difficult; indeed, questionable. Moreover, reach figures do not distinguish multiple uses of the same source within the measurement period from a single use. Most important of all, reach figures tell you nothing helpful about the real issue: actual consumption of news and current affairs.
For instance, a news source might have a 100% reach, but the average usage might be just three minutes per individual. Another news source might have only 50% reach, but usage is 60 minutes per individual. In reach terms, the first source seems twice as important as the second; but in actual usage, the second source is ten times more important.
It is not clear why Ofcom is so reluctant to use well-established industry figures for actual news and current affairs consumption. We have very accurate numbers for television from the industry research body; details of actual time spend reading print newspapers; highly reliable estimates of radio listening that can be matched to schedules to deliver good estimates of consumption of radio news and current affairs; only with online consumption of news do we lack robust figures (though some plausible estimates are available).
Ofcom had commissioned last year another round of bespoke research by Kantar, which regularly asks a large sample of respondents what news sources they use "nowadays", how they rate them, and which they rely upon. The latest findings have been published as part of the merger assessment process, and are useful, but do not of course act as a substitute for actual consumption statistics. Indeed, Kantar does not even refer to the radio research body, RAJAR, as a source, though it claims that it is delivering "quantitative data".
It is in the area of online news provision and consumption that the Fox team will puzzle longest over the Kantar findings. The most frequently cited measuring tool here, ComScore, shows The Sun doing modestly well online, but that excludes mobile devices, which Ofcom believes are increasingly used to capture news stories.
Even as official newspaper industry statistics show sales of The Sun falling off a cliff (down 50% in just six years, part of the decline in newspaper circulation from 9 million sales a day to 5.6 million), Kantar suggests that there are 27 million mobile users who sample The Sun. Even as official newspaper industry statistics show sales of The Sun falling off a cliff (down 50% in just six years, part of the decline in newspaper circulation from 9 million sales a day to 5.6 million), Kantar suggests that there are 27 million mobile users who sample The Sun. What that means in terms of actual consumption is hard to say – assuming it is true – but what is certain is that News Corp would willingly swap that 27 million for just half a million real customers, whose purchasing of print copies would stem the heavy losses the title now incurs (it was once the cash cow of the Murdoch UK empire).
The Fox submission to Ofcom had argued that, even if the Sky and News Corp news provision figures were combined (ignoring the fact that the two providers were owned by separate companies), the huge expansion of online news provision from scores of suppliers meant that "those in control of media enterprises" (the legal definition of who was caught by the public interest provisions of the legislation) were steadily losing influence.
Ofcom disagrees: it judges that traditional media owners have not lost their influence as a result of the online surge – even that papers like The Sun might be disproportionate beneficiaries (clearly not true of The Times, which charges for online access to its news). The Ofcom judgment may not be robustly supported by data, but it remains a key element in the way Karen Bradley is required to assess the proposed deal.
Where the Fox lawyers might have more leverage, should the issue be referred to the CMA, is in attacking some of Ofcom's weaker conclusions. For instance, Ofcom makes considerable play of the "unique" position of Sky/News Corp in being distributed across "all four" news platforms – TV, radio, newspapers and online: "the strength of this position may allow Fox/Sky and News Corp to exercise a greater degree of influence over public opinion than might otherwise be the case absent the transaction" (Ofcom Report para 8.27 – though how the transaction itself might enlarge that influence is not explained).
Karen Bradley specifically cited this "unique" dimension in her Commons statement. Yet when we check the Kantar report, we discover that, of those consuming news from "all four" platforms, just 1% use newspapers in addition to TV, radio and online as a news source: that is the trivial margin of benefit of this "unique" advantage, as compared with media distributed over only three of these platforms.
In fact, restricting the definition of platforms to "all four" is possibly misleading. Kantar tells us that 20% of news consumers obtain news from local and regional outlets – TV, radio and newspapers. These three sectors are all distinct, with 600 newspaper titles, scores of radio services and regional TV provision from the BBC and ITV. The circulation of Metro alone increases the reported "reach" of the Daily Mail Group from 10% to 15%.
Neither Sky nor News Corp feature in these sectors – News Corp has no local papers or radio stations, and the Sky News supply of content to the commercial radio stations is just of national and international stories: the stations, or station groups, themselves assemble the bulletins, which only very rarely are solely dependent on the Sky supply.
If the definition of platforms is revised from four to seven, we can see that the BBC uses five, News Corp three, Sky three (four between them) and everyone else at least two (three in the case of Trinity Mirror, Daily Mail Group and the Lebedev Foundation) : so, not that "unique". And the Fox lawyers will also query the Ofcom claim (para 6.42) that the transaction will "close" the gap between Sky/News Corp and ITV in "cross-platform reach" (a metric itself open to query) and widen the gap to the Daily Mail Group, where the relevant charts in the Ofcom document show that the reported reach of the combined group is actually less than the sum of the two separate groups (which is not surprising). The Ofcom judgment may not be robustly supported by data, but it remains a key element in the way Karen Bradley is required to assess the proposed deal.
Another query will relate to Ofcom's "share of references" data – a specific piece of bespoke research tracking the possible influence of different news providers. Puzzlingly, the "share of references" for the combined group is greater than that for the Sky News and News Corp separately: no explanation is provided for this counter-intuitive finding.
What the Fox team will also home in on is the "saliency" aspect of Ofcom's research. Before the explosion in online news consumption, the combined share of news consumption attributable to Sky News and News Corp was about 10% – well below the level that competition authorities usually regard as a basis for intervention. Ofcom believes that the online surge will not have weakened the position of traditional suppliers: but clearly the share attributable directly to TV, radio and newspapers will have declined, so it is unlikely that the Sky/News Corp position is above 10%.
The Ofcom report declines to offer any hard metrics on consumption, but figure 4.6 tells us a great deal about saliency: a chart showing "the single most important news provider", as reported by consumers of news, and tabulated in the Kantar survey. Sky News comes in at 5% (a decline from 2014's 7%) and News Corp at 2% (down from 2014's 3%). The BBC tops the table at 49%.
Of course, the BBC is ascribed special status in all these media plurality assessments. It is seen as having no political agenda, and being dedicated to impartiality (though the Ofcom research shows that the public views it as no more impartial than Sky News or ITN, and – oddly enough – much less so than Channel 4 News). Yet the legislation is designed to prevent any one voice having too much influence: it does not say influence for good or ill, just influence.
In the United States, a 60% share of news consumption would be completely unacceptable, however admired and respected the supplier. In the US, the BBC would have had its many news distribution arms (BBC One, BBC News Channel, Radio 5 Live, Radio 1, Radio 4, national and regional news, local news) split up long ago. The BBC is ascribed special status in all these media plurality assessments... seen as having no political agenda, and being dedicated to impartiality (though the Ofcom research shows that the public views it as no more impartial than Sky News or ITN, and – oddly enough – much less so than Channel 4 News).
Three years ago, Ofcom itself asked the BBC Trust how it was proposing to introduce internal plurality to mitigate its dominant position. Answer came there none. Now, Ofcom has replaced the BBC Trust as the BBC's regulator. The issue of internal plurality at the BBC merits no mention in this latest report.
Labour Party deputy leader Tom Watson responds to Culture Secretary Karen Bradley in the House of Commons. Press Association. All rights reserved. Despite – or perhaps because of – the many repetitions in the Ofcom report of its concerns about the potential interference by the MFT in the editorial policy of Sky News – and the additional fear that politicians will perceive damage to the public interest from such potential interference whether it occurs or not – the offer of undertakings from Fox seems, towards the end of the 134 pages, effectively to solve the problem (or "mitigate" it, in legal language).
Essentially, Fox proposes the creation of an editorial board for Sky News, with an independent majority and an independent chairman, which would prevent any attempt by the MFT to interfere in the service, whilst guaranteeing the maintenance of the Sky News budget at its present level for five years. Ofcom's provisional view was that it would be preferable to lengthen the period of financial guarantee, and to tighten up the procedure for editorial board appointments: but it would be up to the Secretary of State to decide whether the offer was acceptable.
Unsurprisingly, Karen Bradley declined the offer, on the reasonable assumption that Fox had "more to give". But there was an unexpected element in her Commons statement and letter to the merger parties.
In 2010/11, Ofcom and News Corp had wrangled for months over the nature of the undertakings on offer, with the solution being a structure whereby News Corp would buy all of Sky apart from Sky News, leaving in place a quoted company, operating independently, and with financial support from News Corp.
At the time, I criticised this solution as putting Sky News seriously at risk of decline and eventual demise: cut off from the "mother ship", it would lose access to crucial content, marketing support, capital investment and the benefit of Murdoch's personal commitment. Ofcom seems to have absorbed that criticism, arguing in its report this time that a structural solution is too risky for Sky News, which might suffer terminal damage if left to its own devices.
Instead, Ofcom embraced what is called a "behavioural" solution, whilst leaving Sky News within the Fox business. The many critics of Murdoch (not just all political parties on the left, but some Tories, too) have warned against relying upon undertakings from Murdoch, citing previous bad experience at The Times and The Wall Street Journal, where promises to the vendors of those titles did not last long. The Ofcom report discounted those episodes, regarding legally binding undertakings (whether to Ofcom or the Secretary of State) as enforceable, on penalty of closing the service (by the way, many would argue that both The Times and The Wall Street Journal are significantly improved journalistic ventures under News Corp ownership, perhaps because of the broken promises).
The joker in the pack turned out to be the CMA itself: its standard guidance is that behavioural undertakings, after a first stage competition review, are not normally acceptable. Yet its preference for a structural solution has already been undercut by Ofcom's stated objections. How can this be resolved?
Both Ofcom and Karen Bradley have something to lose from just allowing the CMA investigation to proceed: a more hard-headed methodological approach to media plurality by the CMA could embarrass Ofcom, and if the resulting assessment established that the risks to plurality from the transaction were too minor to justify intervention, then even the undertakings currently on offer would fall away, leaving Bradley exposed politically.
Fox has less to fear from a CMA process, though it would have to honour a pledge to make a large payment to the non-Fox shareholders in Sky if it failed to complete the transaction by the year end: a typical 24-week CMA process would almost certainly run past that deadline. On the other hand, if the CMA could be persuaded that a beefed up version of the current undertakings would satisfy Ofcom, and was preferable to any possible structural change, then Karen Bradley would be off the hook.
Although theoretically she has absolute discretion, acting in a quasi-judicial capacity (that is, not as a member of the government, or according to party preferences), her position in practice almost certainly means adopting the advice provided by the competition authorities themselves. That is how Jeremy Hunt, in the same position last time, narrowly defined his discretionary space.
The unusual aspect of this dilemma is that a structural solution to a competition issue usually involves requiring the bidder to spin off a valuable asset. In this case, Sky News has negative commercial value (though Fox recognizes its considerable brand value). In effect, there is no prospect of anyone else buying the loss-making service, so – as Ofcom now acknowledges – separation is a high risk option.
Fox is unlikely to say so openly during the consultation period that Karen Bradley has launched (ending on July 14), but if there is a stand-off between Ofcom and the CMA, the nuclear option is simply to announce the planned closure of Sky News, thus removing all the media plurality issues in one ironic stroke.
So the likelihood is that Karen Bradley will see how far Ofcom and Fox can get with UILs ("undertakings in lieu" of a reference to the CMA), and whether the CMA will sign off on them: then she can with a clear conscience approve the deal (no doubt to the non-assuaged anger of Murdoch's many critics). But Fox will not want to spend months on such a process when it can be highly confident that a CMA investigation would clear the bid, requiring as it does a much higher threshold – probability, not possibility, as with Ofcom – of evidence that the transaction would damage the public interest. Tom Watson, for the Labour Party, in responding to Bradley's Commons statement saying she was minded to refer the bid to the CMA, predicted that her seemingly strong words would quickly be overtaken by an improved offer from Fox.
Tom Watson, for the Labour Party, in responding to Bradley's Commons statement saying she was minded to refer the bid to the CMA, predicted that her seemingly strong words would quickly be overtaken by an improved offer from Fox. Bradley protested at his cynicism: but the dynamics of the situation suggest that his forecast is likely to be proved correct.
If so, then we must all at least hope that the independence and financing of Sky News are robustly established in any undertakings given, and then properly enforced. To lose or damage Sky News, especially by miscalculation on the part of those responsible for monitoring the bid, would be a serious setback: not least to the need for media plurality, which is ostensibly the issue at stake.


Machine learning (ML) becomes a very promising research field in recent years. ML is a sub-area of artificial intelligence that teaches computers to learn. Current successful applications of ML include medicine, social networking, driverless care, autonomous robot, and a lot of image recognition tasks. Computer vision and pattern recognition are two very related fields/courses with ML.
In this course the instructor will teach a selected topics of ML, such as linear regression, logistic regression, SVM(support vector machine), neural networks, and deep learning. Some deep learning models such as CNN and R-CNN will be introduced. This course will focus more on the understanding of those ML topics, but not mathematical foundations of those ML topics.
Evaluation of student's performance is based on a multitude of metrics, including reading reports, oral presentation, programming results, group collaboration, and peer review. Programming skills including Matlab/C/C++ is necessary to practice and implement the deep learning method. Some topics in the course will be presented by students. Interactive forms of in-classroom activities will be planned in the course. A project will be assigned with paper reading, program coding, oral presentation and report writing. Project can be done by individuals or with team work. Some presentations and reports are evaluated by peer review.
Skill: Python, Matlab, or C/C++.
👍 Introduction to Machine Learning, 3rd, E. Alpaydin, MIT Press, 2014.
C.M. Bishop, N.M. Nasrabadi. Pattern Recognition and Machine Learning. New York: springer, 2006.


#1 position on iTunes. He has since released two other New York Times Best Selling Books, NOT TACO BELL MATERIAL and PRESIDENT ME.
Adam also recently completed three feature length films: ROAD HARD, a crowd-funded comedy that raised over a million dollars in a single month, the inspirational documentary WINNING: THE RACING LIFE OF PAUL NEWMAN, and THE 24 HOUR WAR, which documents the battle between Ford and Ferrari in the 1960's for racing dominance at Le Mans.
In 2018, Adam is touring his new one-man show, "NOT TACO BELL MATERIAL", based on his autographical book, to audiences all over the nation.


text:
Discover The History and Future of St John's Church
September 3, 2012 September 3, 2012 christoff3000carl chinn, sedgley local history society, st johns church1 Comment
Sedgley Local History Society will be hosting a talk dedicated to the preservation of St John's Church, Kates Hill, Dudley
Sedgley residents can learn about the fascinating history of St John's Church,Kates Hill, Dudley, as the new season of talks from Sedgley Local History Society gets underway.
During the event, held on Thursday, September 13 at 7.30pm in St. Andrew's Church, Bilston Street, Sedgley, guests will be treated to an illustrated talk from group chair, Deb Brownlee and Chris Smith, editor of community magazine, Village Voice. The speakers will be highlighting the efforts of the St
John's Church Preservation Group and how their project to restore this 1840s building to its former glory depends on community support in order to succeed.
The Significance of St John's Church
St John's Church is a hugely significant building in local history. Not only is it the burial place of William Perry, the famous Tipton Slasher, it's also the last resting place of Julia Hanson, who is synonymous with Hanson's beer and brewery. Marion Richardson, the famous art and handwriting teacher, is also buried there, and Reverend Edward Noot, who served at St John's for 63 years until his death in 1905, was related to Edward Jenner, the famous physician who discovered vaccination and thereby helped eradicate smallpox in the UK.
This Grade II listed building was closed in 2002 due to health and safety concerns but, since forming in 2007, the St John's Preservation Group has worked hard to demonstrate how it can be fully restored and take its rightful place in the community once more.
Numerous local celebrities have lent their support to the campaign, including former cycling champion Geoff Hill and Professor Carl Chinn – himself a descendent of the famous Tipton Slasher, William Perry.
Sedgley residents are warmly invited to the talk, which costs £1 to attend. Annual membership of the Society costs £5 and gives free access to the programme of talks.
Visit the St John's Preservation Group's website for more details on this important local landmark.

redpajama_set_name: RedPajamaCommonCrawl


Blue Paint Colors on the Chair. Blue is a kind of favorite colour in this world. Blue contains with many kind of type that may be used to paint your chair like dark blue, light blue blue, Mozart blue, Covington blue, Wythe blue, Palladian blue, navy blue, breath of refreshing air, turquoise, Aquarius, blue wave, blue blue, and many more.
This name of The Modern 2 Person Recliner Chair Property Plan is gallery form 2 Person Recliner Chair. Wish you can find your picture here as inspiration from our gallery here.


Forbidding peaks, resourceful commandos, beautiful spies, nonstop action, and neck-snapping plot twists make this the classic adventure thriller—the kind of page-turner that readers actually will find impossible to put down. A team of British Special Forces commandos parachutes into the high peaks of the Austrian Alps with the mission of stealing into an invulnerable alpine castle—accessible only by aerial gondola—the headquarters of Nazi intelligence. Supposedly sent in to rescue one of their own, their real mission turns out to be a lot more complicated—and the tension climbs as team members start to die off, one by one. Written by Alistair Maclean, author of the Guns of Navarone, this is the novel that set the pace for the modern action thriller (the film version, with Richard Burton and Clint Eastwood, also helped), and it still packs twice the punch of most contemporary best-selling thrillers. What's more, the cast of spooks, turncoats, and commandos who drive this story are more relevant than ever in our new era of special forces, black ops, and unpredictable alliances.


This Leatherette L Shaped sectional Sofa is perfect for your urban homes. Made to be the focal point of the room it is placed in, it comes with a detachable side chaise, making it a multi-utility model fit for modern homes. Clean lines, handcrafted framework and short square legs buids its contemporary look. Hand finished taut upholstery and lavish loose fit cushioning gives a comfortable seating experience.


So this morning i notice that my servers dont appear in the server list , since i can see that everything seems fine on the server side i though maybe there was an update.
So i went to the AA forums (i.e here) and i checked News and announcments and nothing was posted.
So then i though the problem was at my end so i started troubleshooting restarting hardware and software checking ports ect ect losing a lot of time and effort.
Maybe we can work out a system where server owners get some form of notification when there is an update? maybe an email to the Email used to registrer the server?
Why only the server admins, its simple really.. If you update a game, post those notes on the forums. Like every other game dev team does, that way you don't get 20 people reinstalling their servers thinking there is a problem on their side.
But, this has been going on for couple years now.. They never learn.
Yeah ,i honestly dont understand, it seems very unprofessional.


text:
Die Bue Shpata (auch Bua Spata) waren eine albanische Adelsfamilie. Die älteste schriftliche Quelle, in der ein Angehöriger dieses Geschlechts genannt wird, datiert auf das Jahr 1304. Die Bue Shpata – oder auch nur Bue oder nur Shpata genannt – gelangten unter der Regierung des serbischen Zaren Stefan Dušan (1346–1355) zu größerer Macht und Ansehen. Nikola Bue Shpata war 1345 bis 1349 Protovestarios des serbischen Zaren. Sein Sohn Pjetër I. wurde 1354 als Herr von Delvino und Angelokastron (seit 2011 ein Ortsteil von Agrinio) im Despotat Epirus genannt.

Der mächtigste Vertreter des Geschlechts war Gjin Bua Shpata (auch Ghin Bua Spata) († 1400). Seit 1358 ist er als Despot von Angelokastron und Acheloos bezeugt. Nach 1375 herrschte er auch über Arta und Lepanto.

Pjetër II. beteiligte sich unter Skanderbeg (1405–1468) an der Liga von Lezha gegen die Osmanen.

Die anderen in der zweiten Hälfte des 15. Jahrhunderts lebenden Familienmitglieder verkauften ihre kleinen Herrschaften in Griechenland und Albanien zumeist an die Republik Venedig, einige wurden auch Vasallen der Osmanen.

Weblinks
 Genealogie der Bue Shpata

Albanisches Adelsgeschlecht

redpajama_set_name: RedPajamaWikipedia


text:
\section{Coamoebas}\label{Coamoebas}
As above and for the remainder of this article, let $P(z)=\sum _{i=1,\ldots,n} \varepsilon_i c_{i} z^{\alpha^i} + c_0$, where $\varepsilon=(\varepsilon_1, \ldots, \varepsilon_n) \in \{1,-1 \}^n$ and $c_i \in {\mathbb R}_{>0}$  for all $i$,  be a simplicial real polynomial (we can suppose without loss of generality that the constant term is positive). Let $X:=X_P \subset ({\mathbb C}^*)^n$ be the associated simplicial real algebraic hypersurface, and let $\Delta_P$ be the Newton polytope of $P$, and $A$ be the $n \times n$ matrix whose rows correspond to the vertices of $\Delta _ P$ and the monomials of $P$, \textit{i.e.} $A^i = \alpha^i$.

\subsection{Definition and description of ${\mathcal C}_X$}

The \textit{coamoeba} ${\mathcal C}_X \subset T^n$ of $X$ is the closure in $T^n$ of the image $Arg({\mathbb C} X)$. The conjugation $c$ acts as $-\id $ on $T^n$; if we fix a representation of $T^n$ as $[0,2\pi]^n$ with its boundary quotiented as needed, which we do from now on, $c$ is the central symmetry.

We will use the convenient description of ${\mathcal C}_X$ given by Kerr and Zharkov in the linear case, \textit{i.e.} when the coamoeba is given by an affine hyperplane, as an intermediate step to get to the general case.
Let us introduce the simplicial polynomials $P_L^+(z)= \sum _{i=1,\ldots,n} c_{i} z_i + c_0$ and $P_L(z)=\sum _{i=1,\ldots,n}  \varepsilon_i c_{i} z_i +  c_0$, where the coefficients $c_i$ and $\varepsilon_i$ are the same as in $P$. Name ${\mathcal C}_{X_{P_L^+}}$ and ${\mathcal C}_{X_{P_L}}$ the associated coamoebas.

Consider $T:=T^1={\mathbb R} /2\pi {\mathbb Z} $ and identify it with the unit circle via the map $E:[\theta] \mapsto \exp(i\theta)$.
We say that  $(\theta _1, \ldots, \theta _ n) \in T^n$ are in an \textit{allowed configuration} if there is no open half-circle in $T$ containing all the $\theta_i$ as well as the point $1=E[0]$ ( which corresponds to the constant term).

 A \textit{zonotope} is the Minkowski sum in ${\mathbb R}^n$ of a finite collection of segments. Now consider such a zonotope $\tilde{Z}$; if the quotient map ${\mathbb R}^n \longrightarrow T^n= ({\mathbb R} / 2\pi {\mathbb Z} )^n$ restricted to the interior of $\tilde{Z}$ is an embedding, we call in the rest of this article,  by extension, the image $Z$ of the interior of $\tilde{Z}$ an (open) zonotope of $T^n$.

 \begin{lemma}\label{CoamoebaLinearPlusCase}
 The points of ${\mathcal C}_{X_{P_L^+}}$ are exactly the n-tuples $(\theta _1, \ldots, \theta _ n) \in T^n$ in allowed configurations. There is an $n$-dimensional zonotope $ \tilde{Z}$ in ${\mathbb R}^n$, generated by $n+1$ segments, such that the quotient map ${\mathbb R}^n \longrightarrow T^n= ({\mathbb R} / 2\pi {\mathbb Z} )^n$ restricted to the interior of $\tilde{Z}$ is an embedding, and such that the open zonotope $Z$ of $T^n$ which is the image of the interior of $\tilde{Z}$ is the complement of the points in allowed configurations (see Figure \ref{Zonotope}).
 \end{lemma}

 The proof is almost trivial - see \cite{KerrZharkov} for related details. Note that $Z$ is contractible in $T^n$, hence $(T^n,{\mathcal C}_{X_{P_L^+}})$ is of the same homotopy type as $( T^n,T^n \backslash \{ \star \})$.

\begin{figure}
\includegraphics[scale=0.2]{Zonotope.png}
\caption{
In black, the coamoeba ${\mathcal C} _{X _{P_L^+}}$. In white, the open zonotope $Z$}
\label{Zonotope}
\end{figure}

We now consider the slightly more general case of ${\mathcal C}_{X_{P_L}}$. Let $\delta=(\delta_1, \ldots,\delta_n):=\delta(\varepsilon_1, \ldots, \varepsilon_n)\in {\mathbb Z}_2^n$ be as in Section \ref{Definitions}.

\begin{lemma}\label{CoamoebaLinearCase}
The coamoeba ${\mathcal C}_{X_{P_L}}$ is a translate of ${\mathcal C}_{X_{P_L^+}}$ by $\pi \cdot \delta \in T^n$, \textit{i.e.} ${\mathcal C}_{X_{P_L}}={\mathcal C}_{X_{P_L^+}}+\pi \cdot \delta \subset T^n$.
\end{lemma}

The zonotope of forbidden configurations from Lemma \ref{CoamoebaLinearPlusCase} has simply been translated; the proof is once again almost trivial.

Let us now consider the general case of ${\mathcal C}_X$.

\begin{lemma}\label{CoamoebaGeneralCase}
Using the same notations as above, the coamoeba ${\mathcal C} _X $ is the preimage of the coamoeba ${\mathcal C}_{X_{P_L}}$ by the map $A \cdot : T^n \longrightarrow T^n$, \textit{i.e.} $ {\mathcal C}_X = A^{-1}({\mathcal C}_{X_{P_L}})$.
\end{lemma}
\begin{proof}
First, let us show that $Arg({\mathbb C} X) = A^{-1}(Arg({\mathbb C} X_{P_L}))$.

Indeed, if $\theta =(\theta_1, \ldots, \theta_n) \in Arg({\mathbb C} X)$, by definition there exists $r=(r_1, \ldots, r_n) \in ({\mathbb R} _{>0})^n$ such that  $z:= (r_1\theta_1,\ldots, r_n \theta _n)$ belongs to ${\mathbb C} X$, \textit{i.e.} $P(r_1\theta_1,\ldots, r_n \theta _n)=\sum _{i=1,\ldots,n} \varepsilon_i c_{i} r^{A^i}\theta^{A^i}+ c_0= P_L(r^{A^1}\theta^{A^1}, \ldots, r^{A^n}\theta^{A^n})=0$. Hence $(r^{A^1}\theta^{A^1}, \ldots, r^{A^n}\theta^{A^n})\in {\mathbb C} X_{P_L}$ and $\theta ^A = A \cdot \theta$ belongs to $Arg({\mathbb C} X_{P_L})$.

Conversely, suppose that $A \cdot \theta \in Arg({\mathbb C} X_{P_L}).$ Then by definition there exists $s=(s_1, \ldots, s_n) \in ({\mathbb R} _{>0}) ^n$ such that $P_L(s_1 \theta^{A_1}, \ldots, s_n \theta^{A_n})=\sum _{i=1,\ldots,n} \varepsilon_i c_{i} s_i \theta^{A^i}+ c_0= 0$. If there exists $r=(r_1, \ldots, r_n) \in ({\mathbb R} _{>0})^n$ such that $r^A=s$, then $z:=(r_1\theta_1,\ldots, r_n \theta _n)$ is such that $P(z) = 0$ and we can conclude that $\theta \in Arg({\mathbb C} X)$.
Now consider the cofactor matrix $B$ of $A$, and
\begin{align*}
     ({\mathbb R}_{>0})^{n}  & \xrightarrow{\Phi_B} ({\mathbb R}_{>0})^{n}   \xrightarrow{\Phi_A}  ({\mathbb R}_{>0})^{n} \\
     r =(r_1, \ldots, r_n) & \longmapsto r^B  \longmapsto  (r^B)^A=r^{det(A)} = (r_1^{det(A)}, \ldots, r_n^{det(A)}).
\end{align*}
This clearly shows that $\Phi_A$ is surjective from $ ({\mathbb R}_{>0})^{n} $ onto itself; thence we have shown that  $Arg({\mathbb C} X) = A^{-1}(Arg({\mathbb C} X_{P_L}))$.

As $A \cdot : T^n \longrightarrow T^n$ is continuous, we immediately have that ${\mathcal C} _X = \overline{Arg({\mathbb C} X)} \subset A^{-1}(\overline{Arg({\mathbb C} X_{P_L})}) =A^{-1}({\mathcal C}_{X_{P_L}})$.
Following the same reasoning as in Remark \ref{Smoothness}, the map  $A \cdot : T^n \longrightarrow T^n$ is a local diffeomorphism, because $(AB) \cdot = (det(A) \cdot Id)\cdot  : T^n \longrightarrow T^n$ is a local diffeomorphism, where $B$ is the comatrix of $A$.
Consider $\theta \in A^{-1}(\overline{Arg({\mathbb C} X_{P_L})})$, an open neighborhood $U \subset T^n$ of $\theta$ such that $(A \cdot)|_U$ is an embedding, and some open neighborhood $V \subset T^n$ of $\theta$. Then $A\cdot(U \cap V) $ is an open neighborhood of $A\cdot \theta \in \overline{Arg({\mathbb C} X_{P_L})}$, hence there exists $\rho \in A\cdot(U \cap V)\cap  Arg({\mathbb C} X_{P_L})$. Then $((A \cdot)|_U)^{-1}(\rho) \in V \cap  A^{-1}(Arg({\mathbb C} X_{P_L})) = V \cap Arg({\mathbb C} X) $, which shows that $\theta \in \overline{Arg({\mathbb C} X)} $.
\end{proof}

\subsection{A more explicit description of ${\mathcal C}_X$}\label{explicitDescription}

It is well-known (for example, using the theorem of structure of finitely generated abelian groups) that there exists two (non-uniquely defined) invertible matrices $G, H \in M_{n\times n}({\mathbb Z})$ such that
$$G\cdot A\cdot H=D=\begin{bmatrix}
    d_{1} & & 0 \\
    & \ddots & \\
    0 & & d_{n}
  \end{bmatrix}, $$
where $d_1|d_2|\ldots|d_n\in {\mathbb Z}$ and $d_1 d_2\ldots d_k$ is the greatest common divisor of the non-trivial $k$-minors of $A$ (for $k=1, \ldots,n$).

Consider $\delta^G = G\cdot \delta \in {\mathbb Z}_2^n$, where $\delta= \delta(\varepsilon_1, \ldots, \varepsilon_n) \in {\mathbb Z}_2^n$ is as above.

We partition $\{1,\ldots,n\}$ in the following way:
$$I^{0,0}:=\{i \in \{1,\ldots,n\} \quad | \quad \delta^G_i \equiv [0]_2, d_i \equiv [0]_2 \},$$
$$I^{1,0}:=\{i \in \{1,\ldots,n\} \quad | \quad \delta^G_i \equiv [1]_2, d_i \equiv [0]_2\},$$
$$I^{0,1}:=\{i \in \{1,\ldots,n\} \quad | \quad \delta^G_i \equiv [0]_2, d_i \equiv [1]_2 \},$$
$$I^{1,1}:=\{i \in \{1,\ldots,n\} \quad | \quad \delta^G_i \equiv [1]_2, d_i \equiv [1]_2\}.$$

For the remainder of this text, we fix two such matrices $G$ and $H$. Moreover, as observed in section \ref{Definitions}, we can consider $A$ up to right-multiplication by invertible matrices with integer coefficients: hence we can, and do, assume that $H$ is the identity matrix.
Hence from now on we have

$$
A = G^{-1}\cdot D.
$$

Then ${\mathcal C}_X$ lends itself to the following description: define the set of indices $\Omega := \{0,\ldots,d_1 -1\} \times \ldots \times \{0,\ldots, d_n -1 \}$ and let $\widetilde{\delta^G} \in \{0,1\}^n$ be the unique lifting of $\delta^G$. Let also $\widetilde{G\cdot {\mathcal C}_{X_{P_L}}}$ be the preimage of $G\cdot {\mathcal C}_{X_{P_L}}$ by the quotient map $[0,2\pi]^n \longrightarrow T^n$ (see the left part of Figure \ref{FigureGeneralCoamoeba}).

Cover  $[0,2\pi]^n$ with $d_1 \ldots d_n$ hyperrectangular cells $C_\alpha :=[\alpha_1\frac{2\pi}{d_1}, (\alpha_1 +1)\frac{2\pi}{d_1} ]\times \ldots \times [\alpha_n\frac{2\pi}{d_n},(\alpha_n+1)\frac{2\pi}{d_n} ]$ for $\alpha=(\alpha_1, \ldots, \alpha_n) \in \Omega$. Define $\widetilde{{\mathcal C}_X} \subset [0,2\pi]^n$ as the set such that the pair $(C_\alpha, C_\alpha \cap \widetilde{{\mathcal C}_X})$ is mapped to the pair $([0,2\pi]^n ,\widetilde{G\cdot {\mathcal C}_{X_{P_L}}})$ by  $(z_1, \ldots,z_n) \mapsto (d_1z_1 - \alpha_1 2\pi,\ldots, d_n z_n - \alpha_n 2\pi )$ (see the right part of Figure \ref{FigureGeneralCoamoeba}). Then

\begin{proposition}\label{DescriptionCoamoeba}
The pair $(T^n, {\mathcal C} _  X)$ is the image of $([0,2\pi]^n ,\tilde{{\mathcal C}_X})$ by the quotient map $[0,2\pi]^n \longrightarrow T^n$.

The coamoeba ${\mathcal C}_X$ is the complement in $T^n$ of $d_1 \ldots d_n$ open zonotopes indexed by the set of indices $\Omega$, such that the center of zonotope $Z_\alpha$, for $\alpha= (\alpha_1,\ldots, \alpha_n)\in  \Omega)$,  is in $\pi \cdot (\widetilde{\delta^G}_1 / d_1,\ldots, \widetilde{\delta^G}_n /d_n)+2\pi \cdot (\alpha_1/d_1, \ldots, \alpha_n/d_n) \in T^n$. Those zonotopes are translate of each other.
\end{proposition}

\begin{proof}
From \ref{CoamoebaGeneralCase}, we know that $ {\mathcal C} _  X= A^{-1}({\mathcal C}_{ X_{P_L}}) = D^{-1}(G\cdot {\mathcal C}_{ X_{P_L}})$.

As explained in Lemmas \ref{CoamoebaLinearPlusCase} and \ref{CoamoebaLinearCase}, the coamoeba ${\mathcal C}_{X_{P_L^+}}$ is the complement in $T^n$ of an open zonotope centered in $0$, and ${\mathcal C}_{X_{P_L}}$ is the complement in $T^n$ of the same zonotope translated and now centered in $(\delta_1 \pi, \ldots, \delta _ n \pi)=\pi \cdot \delta \in T^n$.

The linear isomorphism $G$ is then applied to it, so that $ G\cdot {\mathcal C}_{ X_{P_L}}$ is the complement in $T^n$ of an open zonotope (geometrically the starting zonotope deformed by $G$) centered in $\pi\cdot \delta^G \in T^n$. The last step is to take the preimage by the covering map $D\cdot : T^n \longrightarrow T^n$.

\end{proof}

\begin{figure}
\includegraphics[scale=0.5]{CoamoebaPartial.png}
\includegraphics[scale=0.5]{Coamoeba.png}
\caption{
On the left, $\widetilde{G\cdot {\mathcal C}_{X_{P_L}}}$. On the right:  the coamoeba ${\mathcal C} _X$ in black and the zonotopes in white.}
\label{FigureGeneralCoamoeba}
\end{figure}

\section*{Acknowledgement}
The author is very grateful to Ilia Itenberg, Oleg Viro and Ilia Zharkov for helpful discussions.

\input{Introduction.tex}
\input{Definitions.tex}
\input{Coamoebas.tex}

\input{HomologicalComputations.tex}
\input{GaloisMaximality.tex}

\bibliographystyle{alpha}

\section{Definitions and notations}\label{Definitions}

We denote the $n$-dimensional torus $({\mathbb R}  /2\pi {\mathbb Z}  )^n$ (seen as the product of $n$ unit circles) by $T^n$, and use either additive or multiplicative notations, based on context, to describe its natural group law. In particular, we frequently apply matrices with integer coefficients to elements of $T^n$.

Given a vector of signs $(\varepsilon_1, \ldots,\varepsilon_n) \in \{1,-1 \}^n $, we define $\delta(\varepsilon)=(\delta_1, \ldots,\delta_n)\in {\mathbb Z}_2^n$ by the relation $\varepsilon_i = (-1) ^{\delta_i}$.\\
Given a finite set $S$, let $|S|$ denote the cardinality of $S$. \\
We define
$Arg: ({\mathbb C}^*)^n \longrightarrow T^n,\quad (z_1, \ldots, z_n) \longmapsto (arg(z_1),\ldots,arg(z_n)).$

Throughout this text, we use the following conventions: for any $z=(z_1, \ldots, z_n) \in {\mathbb C}^n$, any matrix $G\in M_{n\times n}({\mathbb Z}) $ and any vector $v=(v_1,\ldots, v_n) \in {\mathbb Z}^n$, we define $z^v:= z_1^{v_1} z_2^{v_2} \ldots z_n^{v_n} \in {\mathbb C}$ and $z^G := (z^{G^1}, \ldots, z^{G^n}) \in {\mathbb C}^n$, where $G^i$ is the $i$-th \textbf{row} of $G$. Choosing rows instead of columns has the advantage of allowing us to write $Arg(z^G) = G\cdot Arg(z)$, and the disadvantage that for another matrix $H\in M_{n\times n}({\mathbb Z})$, we have ${(z^G)}^H=z^{H\cdot G}$.

Consider as above a simplicial polynomial $P(z) =\sum _{\alpha\in \Delta_P} c_\alpha z^\alpha$, for some coefficients $c_\alpha \in {\mathbb R}^*$, and the associated simplicial real algebraic hypersurface $X_P$. Up to multiplication by a non-trivial Laurent monomial (which doesn't change $X_P$), the polynomial $P$ can be chosen so that $\Delta_P$ has $0\in {\mathbb Z} ^n$ as one of its vertices. We assume this to be the case from now on. Moreover, let us denote the non-null vertices of $\Delta_P$ by $\alpha^1_P,\ldots,\alpha^n_P \in {\mathbb Z} ^n$, and define $A_P \in M_{n\times n}({\mathbb Z})$ as the matrix whose $i$-th \textbf{row} is $\alpha^i_P$ for $i=1, \ldots ,n$. If we also define $\alpha^0_P:=0\in {\mathbb Z} ^n$, we can write indifferently $P(z) =\sum _{\alpha\in \Delta_P} c_\alpha z^\alpha = \sum _{i=0,\ldots,n} c_{\alpha_P^i} z^{\alpha_P^i}  = \sum _{i=1,\ldots,n} c_{\alpha_P^i} z^{A_P^i}+c_0.$

For any $G\in M_{n\times n}({\mathbb Z})$, we define the algebraic morphism
\begin{align*}
   \Phi _G : \quad  ({\mathbb C}^*)^{n}  & \to  ({\mathbb C}^*)^{n} \\
    z & \mapsto  z^G.
\end{align*}

If $G$ is invertible, $\Phi_G$ is an algebraic isomorphism that sends $X_{\tilde{P}} = \{ \tilde{P}(z)=\sum _{i=1,\ldots,n} c_{\alpha_P^i} z^{(A_P \cdot G)^i}+c_0= 0  \}$ to $X_P = \{ P(z)=\sum _{i=1,\ldots,n} c_{\alpha_P^i} z^{A_P^i}+c_0= 0  \}$ and respects the real structure. This means that up to such isomorphisms, we can consider the matrix $A_P$ defining the simplex $\Delta_P$ up to right-multiplication by invertible matrices with integer coefficients.

\begin{remark}\label{Smoothness}
The complex part ${\mathbb C} X_P \subset ({\mathbb C}^*)^n$ of a simplicial real algebraic hypersurface $X_P$ is smooth.
\end{remark}
Let $X_P$ be given as above by $P(z) = \sum _{i=1,\ldots,n}  c_{i} z^{\alpha_i} +c_0$, for some coefficients $c_i \in {\mathbb R}^*$ and a simplex $\Delta_P$. Let $A$ be the $n \times n$ matrix whose rows correspond to the vertices $\alpha_i$ of $\Delta _ P$ and the monomials of $P$ (for $i=1,\ldots,n$). We also introduce $B$, the cofactor matrix of $A$, and $P_L(z) = \sum _{i=1,\ldots,n}  c_{i} z_i + c_0$ (the $L$ stands for "linear").

Then $P(z)=P_L\circ \Phi_A(z)$, thus $\Phi_A:({\mathbb C}^*)^{n}   \to  ({\mathbb C}^*)^{n}$ maps ${\mathbb C} X_P$ to the hyperplane ${\mathbb C} X_{P_L} :=\{P_L(z) = 0 \}$. But $\Phi_A \circ \Phi_B = \Phi_{det(A) \cdot Id}: (z_1,\ldots,z_n) \mapsto (z_1^{det(A)},\ldots,z_n ^{det(A)})$ is a local diffeomorphism, which implies that so is $\Phi_A$ by a dimensional argument. Hence the smoothness of ${\mathbb C} X_P$ can be deduced from the smoothness of ${\mathbb C} X_{P_L} $.

\section{Galois maximality}\label{GaloisMaximality}

We are still using the notations of the previous sections.

To prove Theorem \ref{MainResult}, we need to compare $\dim_{{\mathbb Z} _2}\left(\frac{Ker(1+c_*)}{Im(1+c_*)}\right)$ (where $1+c_*: H_{*}({\mathcal C}_ X)\longrightarrow H_{*}({\mathcal C} _X)$) and $ \dim_{{\mathbb Z} _2}(H_*({\mathbb R} X))$. We consider $H_*({\mathbb R} X)$ first, with the following lemma.

\begin{lemma}\label{LemmaRealPartHomology}

The real part ${\mathbb R} X \subset ({\mathbb R}^ *)^n$ consists of $2^n$ contractible connected components if $\delta = \delta(\varepsilon) \in {{\mathbb Z}_2}^n$ is not in the image of $A\cdot:{{\mathbb Z}_2}^n \longrightarrow {{\mathbb Z}_2}^n$, and of $2^n - 2 ^{n-\rank (A)}$ contractible connected components if it is. Thus $H_*({\mathbb R} X)$ is of ${\mathbb Z}_2$-dimension $2^n$ if $\delta \notin Im(A)$ and of ${\mathbb Z}_2$-dimension $2^n - 2 ^{n-\rank (A)}$ if $\delta \in Im(A)$.
\end{lemma}

\begin{proof}
Consider the $2^n$ quadrants of $({\mathbb R}^ *)^n$, and let $\gamma=(\gamma_1,\ldots, \gamma_n ) \in \{1,-1 \}^n$ index the quadrant $Q_\gamma := \{(x_1,\ldots,x_n) \in  ({\mathbb R}^ *)^n | x_1 \gamma_1 >0, \ldots, x_n \gamma_n >0 \}$. Consider as in Section \ref{Coamoebas} the real polynomial  $P_L(z) = \sum _{i=1,\ldots,n}  \varepsilon_i c_{i} z_i + c_0$ associated to $P(z) = \sum _{i=1,\ldots,n}  \varepsilon_i c_{i} z^{A_i} +  c_0$. Then the map
\begin{align*}
   \Phi _{A^{-1}} : \quad  Q_{(1,\ldots,1)}  & \to  Q_{(1,\ldots,1)} \\
    z & \mapsto  z^{A^{-1}},
\end{align*}
where we extend the notation to include rational exponents, is a well defined homeomorphism that maps ${\mathbb R} X_{P_L} \cap  Q_{(1,\ldots,1)}$ to  ${\mathbb R} X \cap  Q_{(1,\ldots,1)}$. In particular, ${\mathbb R} X \cap  Q_{(1,\ldots,1)}$ is empty if $\varepsilon_i=1$ for all $i\in \{1,\ldots,n \}$, and is non-empty and contractible otherwise.
Now observe that ${\mathbb R} X \cap  Q_{\gamma}$ is isomorphic to ${\mathbb R} X_{P_ \gamma} \cap  Q_{(1,\ldots,1)}$, where $P_\gamma (z) := \sum _{i=1,\ldots,n}  \varepsilon_i \gamma^{A_i} c_{i} z^{A_i} +  c_0$. Since $ \delta (\varepsilon_1 \gamma^{A_1}, \ldots, \varepsilon_n \gamma^{A_n})= \delta(\varepsilon) + A\cdot \delta(\gamma)$, we see that ${\mathbb R} X \cap  Q_{\gamma}$ is empty if $ \delta(\varepsilon) = A\cdot \delta(\gamma)$, and non-empty and contractible otherwise.

\end{proof}

The next lemma links the conditions of Theorem \ref{MainResult}, Lemma \ref{LemmaImageHomology} and Proposition \ref{PropositionRank}.
\begin{lemma}\label{LemmaEquivalenceConditions}
 The real part ${\mathbb R} X$ intersects non-trivially each quadrant of $({\mathbb R}^*)^n$ if and only if  $\delta \notin Im(A)$ if and only if $|I^{1,0}| \neq 0$.
\end{lemma}
\begin{proof}
The first equivalence comes from Lemma \ref{LemmaRealPartHomology}. Then
\begin{align*}
\delta \in Im(A) \iff \exists x \in {{\mathbb Z}_2}^n \: s.t. \: A\cdot x=G^{-1}\cdot D \cdot x=\delta \iff  \\
\exists x \in {{\mathbb Z}_2}^n \: s.t. \: D\cdot x=G\cdot \delta= \delta ^G
\iff \delta ^G \in Im(D) \iff \\
(\delta ^G)_i = 0 \: \forall i \: s.t. \: d_i= [0]_2
\iff |I^{1,0}| = 0.
\end{align*}
\end{proof}

We are now ready to prove the main result.
\begin{proof}[Proof of Theorem \ref{MainResult}]

As mentioned at the beginning of Subsection \ref{SubsectionHomologyCoamoeba}, $1+c_*: H_{k}({\mathcal C}_ X)\longrightarrow H_{k}({\mathcal C} _X)$ is trivial for $k\neq n-1$.
Moreover, we know that $H_k({\mathcal C}_X)$ is isomorphic to $H_k(T^n)$ for $k\neq n-1$, and that $dim(H_{n-1}({\mathcal C}_X))=n + d_1\ldots d_n -1$ (as is shown by Lemma \ref{descriptionHomologie}).

Therefore, from Proposition \ref{PropositionRank} and Lemmas \ref{LemmaRealPartHomology} and \ref{LemmaEquivalenceConditions}, we see that if ${\mathbb R} X$ intersects non-trivially each quadrant of $({\mathbb R}^*)^n$,
\begin{align*}
\dim \left(\frac{Ker(1+c_*)}{Im(1+c_*)}\right) = \dim(H_*({\mathbb C} X))-2\dim (Im(1+c_*)) =\\
2^n-1+(d_1\ldots d_n -1) -2( \frac{d_1\ldots d_n} {2}-1)=2^n= \dim(H_*({\mathbb R} X)).
\end{align*}
Hence $X$ is Galois maximal.

Otherwise, we have by definition of the sets $I^{\pm 1, \pm 1}$ in Section \ref{Coamoebas} that $\rank _ {{\mathbb Z} _2}(\Delta _ P) = \rank _ {{\mathbb Z} _2}(A)= |I^{0,1}|+|I^{1,1}| = n- |I^{0,0}|$ (since $|I^{1,0}| = 0)$ and
\begin{align*}
\dim \left(\frac{Ker(1+c_*)}{Im(1+c_*)}\right) = \dim(H_*({\mathbb C} X))-2\dim (Im(1+c_*)) =\\
2^n-1+(d_1\ldots d_n -1) -2( |I^{0,0}| +\frac{d_1\ldots d_n - 2^{|I^{0,0}|}} {2})=\\
2^n-2-2(n-\rank _{{\mathbb Z}_2}(A)) + 2^{n-\rank _{{\mathbb Z}_2}(A)},
\end{align*}
hence
\begin{flalign*}
& \dim \left(\frac{Ker(1+c_*)}{Im(1+c_*)}\right) -\dim (H_*({\mathbb R} X)) = & \\
 & 2^n-2-2(n-\rank _{{\mathbb Z}_2}(A)) + 2^{n-\rank _{{\mathbb Z}_2}(A)} -(2^n - 2 ^{n-\rank (A)}) = &\\
& 2(2^{n-\rank _{{\mathbb Z}_2}(A)}-1- (n-\rank _{{\mathbb Z}_2}(A))) &.
\end{flalign*}

This ends the proof.
\end{proof}

\section{Homological computations}\label{HomologicalComputations}
Let us now study the action of the conjugation $c$ on ${\mathcal C}_ X$, using the description from Proposition \ref{DescriptionCoamoeba}. As pointed out earlier, $c$ simply acts as the central symmetry on $T^n$ seen as an appropriate quotient of $[0,2\pi]^n$. It maps ${\mathcal C}_ X$ to itself, and zonotopes to one another. Let us describe that action more precisely.

\subsection{Action of the conjugation on the zonotopes}\label{SubsectionActionIndices}

Let $\alpha =(\alpha_1, \ldots, \alpha_n) \in \Omega =  \{0,\ldots,d_1 -1\} \times \ldots \times \{0,\ldots, d_n -1\}$. We define $c(\alpha) \in \Omega$ as the index such that $Z_{c(\alpha)}=c(Z_\alpha)$. Then:

If $i\in I^{0,0}$, then $c(\alpha)_i$ is the unique lifting to $\{0,\ldots,d_i -1 \}$ of $[d_i - \alpha_i]_{d_i}$. In particular, notice that $c(\alpha)_i = \alpha_i$ if and only if $\alpha_i \in \{ 0, d_i / 2\}$.

If $i\in I^{0,1}$ , then $c(\alpha)_i$ is the unique lifting to $\{0,\ldots,d_i -1 \}$ of $[d_i - \alpha_i]_{d_i}$. In particular, notice that $c(\alpha)_i = \alpha_i$ if and only if $\alpha_i=0$.

If $i\in I^{1,0}$, then $c(\alpha)_i = (d_i -1)- \alpha_i$. In particular, notice that $c(\alpha)_i$ is never equal to $\alpha_i$.

If $i\in I^{1,1}$, then $c(\alpha)_i = (d_i -1)- \alpha_i$. In particular, notice that $c(\alpha)_i = \alpha_i$ if and only if $\alpha_i=(d_i-1)/2$.

Denote by $F:=\{\alpha \in \Omega | c(\alpha) = \alpha \}$ the set of fixed points of $c$. If $|I^{1,0}| \neq 0$, then $F=\emptyset$. If $|I^{1,0}| = 0$, there are $2^{|I^{0,0}|}$ fixed points: the indices $\beta \in \Omega$ such that $\beta_i \in \{ 0, d_i / 2\}$ if $i\in I^{0,0}$, $\beta_i=0$ if $i\in I^{0,1}$ and $\beta_i=(d_i-1)/2$ if $i\in I^{1,1}$.

\subsection{Homology of ${\mathcal C}_X$}\label{SubsectionHomologyCoamoeba}

Denote by $\Gamma_1, \ldots, \Gamma_n \in H_{n-1}(T^n)$ the $n$ homology classes induced by the $(n-1)$-dimensional tori $\{[\theta_i] =0\} \subset T^n $. They form a basis of $H_{n-1}(T^n)$.

Using a Mayer-Vietoris long exact sequence, it is easy to see that
$$ H_k({\mathcal C}_X) \xrightarrow{in_*} H_k(T^n),$$
where $in_ *$ is induced by the inclusion, is an isomorphism for $k=0,\ldots,n-2$, and that $ H_k({\mathcal C}_X)=0$ for $k\geq n$.
We also get a short exact sequence
$$ 0 \longrightarrow H_n(T^n) \longrightarrow H_{n-1}(\bigsqcup_{\alpha\in \Omega} \partial Z_\alpha) \longrightarrow H_{n-1}({\mathcal C} _X) \longrightarrow H_{n-1}(T^n) \longrightarrow 0 ,$$
where  $\partial Z_ \alpha$ is the border of zonotope $Z_\alpha$, which is homeomorphic to a $(n-1)$-sphere. From it, we deduce the following Lemma:

\begin{lemma}\label{descriptionHomologie}
Given $B_1, \ldots B_n \in H_{n-1}({\mathcal C} _X)$ such that $in_*(B_i) =\Gamma_ i $ for all $i$, we have an isomorphism
\begin{equation*}
    H_{n-1}({\mathcal C} _X) \cong \left(\bigoplus_{i=1,\ldots,n} {\mathbb Z}_2 \cdot B_i \right) \oplus \left( \frac{\bigoplus_{\alpha\in \Omega} {\mathbb Z}_2 \cdot [\partial Z_\alpha]}{{\mathbb Z}_2 \cdot \sum_{\alpha\in \Omega} [\partial Z_\alpha]}\right).
\end{equation*}
\end{lemma}

We already know that $c_*([\partial Z_ \alpha]) = [\partial Z_ {c(\alpha)}]$ (as we consider homology with coefficients in ${\mathbb Z}_2$, we need not concern ourselves with orientation), where $c(\alpha)$ is as above - we only need to find suitable classes $B_i$ and describe how $c_*$ acts on them.

Let us define $B_i$, for $i\in \{ 1,\ldots,n\}$. Consider the $(n-1)$-dimensional torus $\{[\theta_i] =0\} \subset T^n $, and the set $I_i$ of intersections between $\{[\theta_i] =0\}$ and the open zonotopes of the complement of ${\mathcal C} _X$ in $T^n$. Notice that a given zonotope $Z_\alpha$ can intersect several times $\{[\theta_i] =0\}$ - each of these intersections appears as a distinct element of $I_i$. Call $N_i(\alpha)$ the number of such intersections. We would like to define $B_i$ as the class of $\{[\theta_i] =0\}$, with a modification (since $\{[\theta_i] =0\}$ is in general not included in ${\mathcal C}_X$).

For each intersection $\gamma \in I_i$, let $\alpha(\gamma)\subset \Omega$ be the index of the zonotope corresponding to $\gamma$. The intersection $\gamma$ of $\{[\theta_i] =0\}$ and $\overline{Z_{\alpha(\gamma)}}$ cuts $\partial Z_{\alpha(\gamma)}$ in two (topological) half-spheres of dimension $n-1$. Call $S_\gamma^+$ the half-sphere that lies in the direction $e_i$ of $\{[\theta_i] =0\}$, and $S_\gamma^-$ the one lying in direction $-e_i$ .

We start our construction with $\{[\theta_i] =0\}$. For each $\gamma \in I_i$, we remove the intersection $\{[\theta_i] =0\}\bigcap Z_{\alpha(\gamma)}$, and we add $S_\gamma^+$. Thus we obtain a $(n-1)$-cycle - name it $\widetilde{B_i}$ and name its class $B_i$.
Observe that parts of the border of a given zonotope $Z_\alpha$ can appear several times in $\widetilde{B_i}$: in other words, there can be several $S_\gamma^+$ that are subsets of the same $\overline{Z_\alpha}$ (with non-trivial intersections).
Notice also that $S_\gamma^+$ can be homotopically contracted in $T^n$ to the intersection $\{[\theta_i] =0\}\bigcap \overline{Z_{\alpha(\gamma)}}$. Thus $B_i$ is a lift of $\Gamma_i$, as required. This construction is illustrated in Figure \ref{CycleConstruction}.

\begin{figure}
\includegraphics[scale=0.6]{Cycles.png}
\caption{
As a black broken line, the cycle $\widetilde{B_1}$. In that example, $I_1$ is of cardinal $4$.}
\label{CycleConstruction}
\end{figure}

\subsection{Image of $1+c_*$}

We want to compute the image of $1+c_*:H_{n-1}({\mathcal C} _ X) \longrightarrow H_{n-1}({\mathcal C} _ X)$, and in particular its dimension. Using the description in Lemma \ref{descriptionHomologie}, we see that
\begin{equation}\label{descriptionImage}
   Im(1+c_*)= \sum_{i=1,\ldots,n}{\mathbb Z}_2 \cdot ((1+c_*)(B_i)) + \sum_{\alpha \in \Omega} {\mathbb Z}_2 \cdot [\partial Z_\alpha + \partial Z_{c(\alpha)}]
\end{equation}

Let us make some observations about the image of $B_i$ under $c_*$. It is the class of the mirror image by central symmetry of $\widetilde{B_i}$. This means that it consists of $\{[\theta_i] =0\}$, without the intersections $\{[\theta_i] =0\}\bigcap Z_{\alpha(\gamma)}$ (for $\gamma \in I_i$), and with all the $S_\gamma^-$, i.e. the half-spheres lying in the direction $-e_i$.
Thus $B_i+c_*(B_i) = \sum_{\gamma \in I_i}[\partial Z_{\alpha (\gamma)}]$ (since by definition $S_\gamma^+ +S_\gamma^- = \partial Z_\alpha$) - each $\partial Z_{\alpha}$ appears $N_i(\alpha)$ times.

It would be possible to compute $N_i(\alpha)$ exactly. However, we only want to compute the image of $1+c_*$ - the following observations suffice for our purpose.

Since we are considering coefficients in ${\mathbb Z} _2$, we only have to pay attention to the parity of $N_i(\alpha)$.
Let us define $J_i : =\{ \alpha \in \Omega \quad | \quad N_i(\alpha) \equiv [1]_2\}$. The number of intersections $N_i(\alpha)$ depends only on the length of the zonotope $Z_\alpha$ in the direction $e_i$ (which is the same for all $\alpha$) and on the $i$-th coordinate of the center of $Z_\alpha$, \textit{i.e.} on $\alpha_i$.

As we only have to consider the coordinate $\alpha_i$ and because of the symmetry of the situation, it is clear that if $\beta \in J_i$, then $c(\beta)\in J_i$. Consider the description (\ref{descriptionImage}) of $Im(1+c_*)$; if $\beta \neq c(\beta)$, we have
$$\partial Z_\beta + \partial Z_{c(\beta)} \in \sum_{\alpha\in\Omega} {\mathbb Z}_2 \cdot [\partial Z_\alpha + \partial Z_{c(\alpha)}]. $$
Thus for any $i\in \{1,\ldots,n \}$ we get

\begin{align*}
{\mathbb Z}_2 \cdot ((1+c_*)(B_i)) + \sum_{\alpha\in \Omega} {\mathbb Z}_2 \cdot [\partial Z_\alpha + \partial Z_{c(\alpha)}] &=\\
{\mathbb Z}_2 \cdot \sum_{\beta\in J_i}[\partial Z_\beta] + \sum_{\alpha\in \Omega} {\mathbb Z}_2 \cdot [\partial Z_\alpha + \partial Z_{c(\alpha)}]= \\
{\mathbb Z}_2 \cdot \sum_{\beta \in J_i \cap F}[\partial Z_{\beta}] + \sum_{\alpha\in \Omega} {\mathbb Z}_2 \cdot [\partial Z_\alpha + \partial Z_{c(\alpha)}],
\end{align*}
where as above $F$ is the set of fixed elements $\beta \in \Omega$.

We need to determine for all $i\in \{1,\ldots,n \}$ the set $J_i \cap F$. Doing so will allow us to prove the following Lemma.

\begin{lemma}\label{LemmaImageHomology}

If $|I^{1,0}| \neq 0$, the image of $1+c_*$ is $\sum_{\alpha\in \Omega} {\mathbb Z}_2 \cdot [\partial Z_\alpha + \partial Z_{c(\alpha)}]$.

If $|I^{1,0}| = 0$,
$$    Im(1+c_*) = \sum_{i\in I^{0,0}}{\mathbb Z}_2 \cdot ((1+c_*)(B_i)) + \sum_{\alpha\in \Omega} {\mathbb Z}_2 \cdot [\partial Z_\alpha + \partial Z_{c(\alpha)}].$$
\end{lemma}

\begin{proof}
Indeed, as noted in Subsection \ref{SubsectionActionIndices}, there are no fixed points if $|I^{1,0}| \neq 0$.

Let us now consider the case $|I^{1,0}| = 0$.

If $i\in I^{0,0}$,  $\beta\in J_i$ for all $\beta$ such that $\beta_i = 0$, and $\beta\notin J_i$ for all $\beta$ such that $\beta_i =(d_i -1)/2$. In particular, $J_i \cap F $ is the set of cardinality $2^{|I^{0,0}|-1}$ of all $\beta \in \Omega$ such that $\beta_i=0$, $\beta_j \in \{ 0, d_j / 2\}$ if $j\in I^{0,0} - \{i\}$, $\beta_j=0$ if $j\in I^{0,1}$ and $\beta_j=(d_j-1)/2$ if $j\in I^{1,1}$.

If $i\in I^{0,1}$, $\beta\in J_i$ for all $\beta$ such that $\beta_i = 0$. In particular, $J_i \cap F = F$. Remember description (\ref{descriptionHomologie}); we have $\sum_{\alpha \in \Omega}[\partial Z_{\alpha}] =0$ in $H_{n-1}({\mathcal C}_X)$.
Then,
$$  \sum_{\alpha \in F}[\partial Z_{\alpha}] = \sum_{\alpha \in F}[\partial Z_{\alpha}] +  \sum_{\alpha \in \Omega}[\partial Z_{\alpha}] \subset \sum_{\alpha\in \Omega} [\partial Z_\alpha + \partial Z_{c(\alpha)}] .$$
This implies that ${\mathbb Z}_2 \cdot ((1+c_*)(B_i)) \subset \sum_{\alpha\in \Omega} {\mathbb Z}_2 \cdot [\partial Z_\alpha + \partial Z_{c(\alpha)}]$.

If $i\in I^{1,1}$, $\beta\notin J_i$ for all $\beta$ such that $\beta_i =(d_i -1)/2$. In particular, $J_i \cap F = \emptyset$ .

Using description (\ref{descriptionImage}) , we can conclude.
\end{proof}

\subsection{Rank of $1+c_*$}

Let us compute the rank of $1+c_*$. We get the following proposition.
\begin{proposition}\label{PropositionRank}
If $|I^{1,0}| \neq 0$,
$$\dim({Im(1+c_*)}) = \frac{d_1\ldots d_n} {2}-1. $$
If $|I^{1,0}| = 0$,
$$\dim({Im(1+c_*)})= |I^{0,0}|+ \frac{d_1\ldots d_n-2^{|I^{0,0}|}}{2}.$$

\end{proposition}

\begin{proof}
We will first compute the dimension of $$\sum_{i=1,\ldots,n}{\mathbb Z}_2 \cdot ((1+c_*)(B_i)) + \sum_{\alpha\in \Omega} {\mathbb Z}_2 \cdot [\partial Z_\alpha + \partial Z_{c(\alpha)}]$$ in
$$\left(\bigoplus_{i=1,\ldots,n} {\mathbb Z}_2 \cdot B_i\right) \oplus \left(\bigoplus_{\alpha\in \Omega} {\mathbb Z}_2 \cdot [\partial Z_\alpha]\right),$$
then quotient by $Z_2 \cdot \sum_{\alpha\in \Omega} [\partial Z_\alpha]$ to get the dimension in $H_{n-1}({\mathcal C} _ X)$ (see Lemma \ref{descriptionHomologie}).

Denote by $\tilde{\Omega}\subset \Omega$ a set defined as such: for each pair of elements $\{\alpha, c(\alpha)\}$, where $\alpha \in \Omega \backslash F$, choose exactly one element to be included in $\tilde{\Omega}$.
Thus $\sum_{\alpha\in\Omega} {\mathbb Z}_2 \cdot [\partial Z_\alpha + \partial Z_{c(\alpha)}] = \sum_{\alpha\in \tilde{\Omega}} {\mathbb Z}_2 \cdot [\partial Z_\alpha + \partial Z_{c(\alpha)}]$.
This sum is clearly a direct sum before quotienting.

If $|I^{1,0}| \neq 0$, then $Im(1+c_*) =\sum_{\alpha\in \Omega} {\mathbb Z}_2 \cdot [\partial Z_\alpha + \partial Z_{c(\alpha)}]= \sum_{\alpha \in \tilde{\Omega}} {\mathbb Z}_2 \cdot [\partial Z_\alpha + \partial Z_{c(\alpha)}]$. When we quotient by $Z_2 \cdot \sum_{\alpha\in \Omega} [\partial Z_\alpha]$, the dimension decreases by exactly one, as $\sum_{\alpha\in \Omega} [\partial Z_\alpha] \in \sum_{\alpha \in \tilde{\Omega}} {\mathbb Z}_2 \cdot [\partial Z_\alpha + \partial Z_{c(\alpha)}]$ (since there is no fixed point). Hence
$$\dim({Im(1+c_*)})= |\tilde{\Omega}|-1 = \frac{d_1\ldots d_n} {2}-1. $$

If $|I^{1,0}| = 0$, we can see that
$$\sum_{i\in I^{0,0}}{\mathbb Z}_2 \cdot ((1+c_*)(B_i)) + \sum_{\alpha\in \tilde{\Omega}} {\mathbb Z}_2 \cdot [\partial Z_\alpha + \partial Z_{c(\alpha)}]$$
is actually a free sum before quotienting. Indeed, suppose that for each $ \alpha \in \tilde{\Omega} $ (respectively, $i\in I^{0,0}$) there is $\lambda_\alpha \in {\mathbb Z}_2$ (respectively, $\lambda_i \in {\mathbb Z}_2$), not all $0$, such that
\begin{equation}\label{PreuveIndep}
    \sum_{i\in I^{0,0}}\lambda_i  (1+c_*(B_i)) + \sum_{\alpha\in \tilde{\Omega}} \lambda_\alpha [\partial Z_\alpha + \partial Z_{c(\alpha)}] = 0.
\end{equation}
For $i\in I^{0,0}$, consider $\beta$ defined by $\beta_i=0$, $\beta_j =d_j / 2$ if $j\in I^{0,0} - \{i\}$, $\beta_j=0$ if $j\in I^{0,1}$ and $\beta_j=(d_j-1)/2$ if $j\in I^{1,1}$. As explained in the description of $J_i\cap F$ in the proof of Lemma \ref{LemmaImageHomology}, the class $[\partial Z_\beta]$ only appears in $((1+c_*)(B_i))$ among all the terms of equation (\ref{PreuveIndep}). Thus $\lambda_i$ is necessarily $0$. We can then conclude from the independence of the family $\{[\partial Z_\alpha + \partial Z_{c(\alpha)}]\}_{\alpha \in \tilde{\Omega}}$ before the quotient that all $\lambda_\alpha$ are $0$, which proves our point.

Consider $\beta \in \Omega $ defined by $\beta_j =d_j / 2$ if $j\in I^{0,0}$, $\beta_j=0$ if $j\in I^{0,1}$ and $\beta_j=(d_j-1)/2$ if $j\in I^{1,1}$. Since $[\partial Z_\beta]$ appears in $\sum_{\alpha\in \Omega} [\partial Z_\alpha]$ but not in $\sum_{i\in I^{0,0}}{\mathbb Z}_2 \cdot ((1+c_*)(B_i)) + \sum_{\alpha\in \tilde{\Omega}} {\mathbb Z}_2 \cdot [\partial Z_\alpha + \partial Z_{c(\alpha)}]$ (once again going back to the proof of Lemma \ref{LemmaImageHomology}), we see that
$\sum_{\alpha\in \Omega} [\partial Z_\alpha] \notin \sum_{i\in I^{0,0}}{\mathbb Z}_2 \cdot ((1+c_*)(B_i)) + \sum_{\alpha\in \tilde{\Omega}} {\mathbb Z}_2 \cdot [\partial Z_\alpha + \partial Z_{c(\alpha)}]$. This means that the dimension of $\sum_{i\in I^{0,0}}{\mathbb Z}_2 \cdot ((1+c_*)(B_i)) + \sum_{\alpha\in \tilde{\Omega}} {\mathbb Z}_2 \cdot [\partial Z_\alpha + \partial Z_{c(\alpha)}]$ does not decrease when we quotient.
Thus
$$\dim({Im(1+c_*)})=|I^{0,0}| +|\tilde{\Omega}| = |I^{0,0}|+ \frac{|\Omega|-|F|}{2}= |I^{0,0}|+ \frac{d_1\ldots d_n-2^{|I^{0,0}|}}{2},$$
as stated.
\end{proof}

\section{Introduction}

Any non-constant real Laurent polynomial $P$ in $n$ variables defines a real algebraic hypersurface $X_P$ in the complex torus $({\mathbb C}^*) ^n$. Given such a polynomial $P(z)=\sum_{\lambda = ( \lambda_1,\ldots,\lambda_n) \in \Lambda } c_\lambda z_1^{\lambda_1},\ldots, z_n^{\lambda_n} $, where $\Lambda$ is a finite subset of ${\mathbb Z} ^n$ and $c_\lambda \in {\mathbb R}^*$ for all $\lambda \in \Lambda$, we call the convex hull in ${\mathbb R}^n$ of $\Lambda$ the \textit{Newton polytope} of $P$.

If $P$ has exactly $n+1$ monomials with non-zero coefficients and if the Newton polytope $\Delta _P$ of $P$ is a non-degenerate $n$-dimensional simplex, we call $P$ a \textit{simplicial} real polynomial and $X_P$ a \textit{simplicial} real algebraic hypersurface. In particular, the monomials of $P$ and the vertices of $\Delta_P$ are in bijection.

Let $X$ be a simplicial real algebraic hypersurface.
We denote the real (respectively, complex) points of $X$ in $({\mathbb R}^*) ^n$ (respectively, $({\mathbb C}^*)^n$) by ${\mathbb R} X$ (respectively, ${\mathbb C} X$).
The complex conjugation $c$ on the complex torus naturally acts on ${\mathbb C} X$. Our goal here is to better understand this action and the induced action $c_*$ on the homology $H_*({\mathbb C} X)$ of ${\mathbb C} X$ with coefficients in ${\mathbb Z}_2:={\mathbb Z}/2{\mathbb Z}$ (unless otherwise specified, we always consider homology with coefficients in ${\mathbb Z}_2$).

Simplicial real algebraic hypersurfaces play a key role in Viro's Patchworking method, in particular in the combinatorial case, where varieties with prescribed topology are obtained by gluing together simplicial real algebraic hypersurfaces. Viro's method is one of the main tools in the study of topology of real algebraic varieties, and is deeply connected to tropical geometry. It was used by Viro to disprove the \textit{Ragsdale conjecture} regarding real algebraic curves in \cite{Viro80}, and later in \cite{Itenberg93} by I. Itenberg to show that the conjecture was also asymptotically wrong.  A complete exposition of the method can be found in \cite{ViroPatchworking} .

Hence, in addition to improving our understanding of simplicial real algebraic hypersurfaces, which are among the most natural and simple examples of real algebraic varieties, our results might also lead to a better comprehension of the varieties obtained using the combinatorial case of Viro's Patchworking.

In particular, A. Renaudineau and K. Shaw proved in \cite{RS} a bound conjectured by Itenberg, which states that in the case of a real algebraic projective hypersurface $X \subset {\mathbb R} \mathbb{P}^n$ obtained via \textit{primitive} Patchworking,
\begin{equation}\label{BorneKristinArthur1}
    \dim_{{\mathbb Z} _2}(H_q({\mathbb R} X)) \leq h^{q,n-1-q}({\mathbb C} X) +1
\end{equation}
if $q\in \{0,\ldots,n\}$ and $q\neq \frac{n-1}{2}$, and
\begin{equation}\label{BorneKristinArthur2}
    \dim_{{\mathbb Z} _2}(H_\frac{n-1}{2}({\mathbb R} X)) \leq h^{\frac{n-1}{2},\frac{n-1}{2}}({\mathbb C} X)
\end{equation}
if $n-1$ is even, where $h^{q,p}({\mathbb C} X)$ is the $(q,p)$-th Hodge number of ${\mathbb C} X$. This can be compared to the much looser \textit{Smith-Thom inequality}, valid for all real algebraic varieties, which states that
\begin{equation}\label{SmithThom}
    \dim_{{\mathbb Z} _2}(H_*({\mathbb R} X)) \leq \dim_{{\mathbb Z} _2}( H_*({\mathbb C} X) ).
\end{equation}
Taking the sum of inequalities (\ref{BorneKristinArthur1}) and (\ref{BorneKristinArthur2}) over all $q\in \{0,\ldots,n\}$ yields the Smith-Thom inequality. A real algebraic variety for which (\ref{SmithThom}) is an equality is called \textit{Smith-Thom maximal}, or simply \textit{maximal}. They are objects of particular interest.

A primitive Patchwork is a Patchwork obtained by gluing together simplicial real algebraic hypersurfaces whose Newton polytopes are of Euclidean volume $\frac{1}{n!}$ (more details can be found in \cite{RS}). Such hypersurfaces are the simplest possible objects from which to construct more complicated hypersurfaces; in particular, they are maximal, which is a key ingredient in the proof of inequalities (\ref{BorneKristinArthur1}) and (\ref{BorneKristinArthur2}).

The \textit{Borel-Swan inequality}, which implies the Smith-Thom inequality,  states that
\begin{equation*}
    \dim_{{\mathbb Z} _2}(H_*({\mathbb R} X)) \leq \dim_{{\mathbb Z} _2}\left( \frac{Ker(1+c_*)}{Im(1+c_*)} \right),
\end{equation*} where $1+c_* : H_*({\mathbb C} X) \longrightarrow H_*({\mathbb C} X)$  (from now on, for any ${\mathbb Z}_2$-vector space $V$, we write indifferently $\dim_{{\mathbb Z}_2}(V)$ or $\dim (V)$ to denote the dimension over ${\mathbb Z}_2$ of $V$). When this inequality is an equality, $X$ is called \textit{Galois maximal} (see  \cite{RealEnrique} for additional details). The notion has been considered of interest in itself; see for example Krasnov \cite{Krasnov}, where the Galois maximality of various families of varieties is proved.

Galois maximality is equivalent to a condition of degeneration on the second page of a certain associated spectral sequence, called \textit {Kalinin spectral sequence}, which was introduced by Kalinin in \cite{Kalinin2} - further explanations can also be found in \cite{StiefelOrientations}. Smith-Thom maximality is equivalent to degeneration on the first page of this spectral sequence,  which plays an important role in the arguments used in \cite{RS}.

Therefore, one possible way to relax the primitivity condition in order to generalize the results from \cite{RS} could be to consider Patchworks obtained from gluing together simplicial real algebraic hypersurfaces that are merely Galois maximal, rather than primitive (and thus Smith-Thom maximal). Note that we already know that inequalities (\ref{BorneKristinArthur1}) and (\ref{BorneKristinArthur2}) do not hold as they are under such hypotheses - additional work is needed.

 Hence our interest in the action of the conjugation on the homology of simplicial real algebraic hypersurfaces, and in particular in the rank of the map $1+c_*$, in order to determine the conditions under which such a hypersurface is Galois maximal.

In  \cite{KerrZharkov}, G. Kerr and I. Zharkov have in particular showed that the complex part ${\mathbb C} X$ of a simplicial real algebraic hypersurface is homeomorphic to the associated \textit{phase tropical variety} $\mathcal{T}X$.
Moreover, it is easy to see that the phase tropical variety $\mathcal{T}X$ retracts by deformation to the \textit{coamoeba} ${\mathcal C}_X$ of $X $. Hence we get a homotopy equivalence ${\mathbb C} X \longrightarrow {\mathcal C} _X$.
Private conversations with Zharkov have led us to believe that this homotopy equivalence satisfies the condition of the following conjecture (which we hope to prove soon):

\begin{conj}\label{ConjectureConjugationZharkov}
Let $X$ be a simplicial real algebraic hypersurface. There is a homotopy equivalence $\phi: {\mathbb C} X \longrightarrow {\mathcal C}_X$ such that the following diagram commutes for all $i\geq 0$:
\begin{center}
\begin{tikzcd}
H_i({\mathbb C} X) \arrow[r, "\phi_*"] \arrow[d, "c_*"]
& H_i({\mathcal C}_X) \arrow[d, "c_*"] \\
H_i({\mathbb C} X) \arrow[r, "\phi_*"]
& H_i({\mathcal C}_X)
\end{tikzcd}
\end{center}
where $c_*$ is induced by the complex conjugation on either ${\mathbb C} X$ or ${\mathcal C}_X$.
\end{conj}
This would immediately imply
\begin{align*}
\dim (Im(1+c_* :H_{i}({\mathbb C} X)\rightarrow H_{i}({\mathbb C} X) )) =\\ \dim (Im(1+c_* :H_{i}({\mathcal C}_ X)\rightarrow H_{i}({\mathcal C}_ X) )),
\end{align*}
where $1$ is the identity.

Our main focus in the rest of this text is to prove the following result, though the details of the proof might be of interest in themselves when considering other related questions.
Consider a simplicial real algebraic hypersurface $X$, the Newton polytope $\Delta$ of the simplicial real Laurent polynomial $P$ that defines $X$, and pick a vertex $O$ of $\Delta$. The edges of $\Delta$ containing $O$ define $n$ integer vectors (choosing $O$ as their initial point). Define $A\in M_{n\times n}({\mathbb Z}_2)$ as the matrix whose rows are these $n$ vectors modulo $2$. Then:

\begin{thm}\label{MainResult}
If ${\mathbb R} X$ intersects non-trivially each quadrant of $({\mathbb R}^*)^n$, then
$$\dim_{{\mathbb Z} _2}(H_*({\mathbb R} X))  =  \dim_{{\mathbb Z} _2}\left( \frac{Ker(1+c_*)}{Im(1+c_*)} \right),$$
 where $1+c_* : H_*({\mathcal C}_X) \longrightarrow H_*({\mathcal C}_ X)$.

Otherwise, we have
\begin{align*}
\dim_{{\mathbb Z} _2} \left( \frac{Ker(1+c_*)}{Im(1+c_*)}\right) - \dim_{{\mathbb Z} _2}( H_*({\mathbb R} X))  = 2(2^{n-\rank _{{\mathbb Z}_2}(A)}-1- (n-\rank _{{\mathbb Z}_2}(A))).
\end{align*}
\end{thm}
The condition on the intersection of ${\mathbb R} X$ with the quadrants of  $({\mathbb R}^*)^n$ is equivalent to a condition on the matrix $A$ and the signs of the monomials of $P$, as proved in Lemma \ref{LemmaRealPartHomology}.

If Conjecture \ref{ConjectureConjugationZharkov} holds, then Theorem \ref{MainResult} immediately implies
\begin{thm}\label{TheoremConjugationMainResultConsequence}
If ${\mathbb R} X$ intersects non-trivially each quadrant of $({\mathbb R}^*)^n$, then $X$ is always Galois maximal.

Otherwise, we have
\begin{align*}
\dim_{{\mathbb Z} _2} \left( \frac{Ker(1+c_*)}{Im(1+c_*)}\right) - \dim_{{\mathbb Z} _2}( H_*({\mathbb R} X))  = 2(2^{n-\rank _{{\mathbb Z}_2}(A)}-1- (n-\rank _{{\mathbb Z}_2}(A))),
\end{align*}
where $1+c_* : H_*({\mathbb C} X) \longrightarrow H_*({\mathbb C} X)$,
and $X$ is Galois maximal if and only if $n-\rank _{{\mathbb Z}_2}(A) \in \{0,1\}$.
\end{thm}

The rest of this article is organized as follows:
in Section \ref{Definitions}, we go over some definitions and notations.
In Section \ref{Coamoebas}, the coamoeba ${\mathcal C}_X$ associated to $X$ is introduced and described in a way suited to computations. As the coamoeba ${\mathcal C}_X$ is homotopically equivalent to ${\mathbb C} X$, we can study the action of the conjugation on ${\mathcal C} _ X$ and deduce from it its action on ${\mathbb C} X$ (assuming Conjecture \ref{ConjectureConjugationZharkov}).
In Section \ref{HomologicalComputations}, the action of the conjugation on the homology of ${\mathcal C}_ X$ is described. In particular, the rank of $1+c_ *$ is computed.
Finally, Theorem \ref{MainResult} is proved in Section \ref{GaloisMaximality}.

redpajama_set_name: RedPajamaArXiv


VVat makes a Pontiac so eager to go .9 Tiis Catalina ]as more l,orsepower per pound t,lan any otler automatic transmission car made in America. (As much as one horse per 10.65 pounds.) Tlat's an ideal balance between weight and po-wer. Gives you hustle wl,en you need it.. Improves gas mileaRe, too. Eager to go Go see your fine Pontiac dealer. THE CATALINA VISTA FOR 1961 t THE ONLY WIDE-TRACK CAR! Pontiac :=as the widest track of any car. Body width trimmed to reduce side overhang. More weight balanced between the wheels for sure-footed driving stabihty PONTIAC MOTOR DlV., GENERAL MOTORS CORP.


Super cute, well maintained updated home on the west side of town. Newer roof on home and garage. Above ground pool and covered patio area! There is time left to swim and enjoy the out of doors! Great layout and generously sized rooms. Hardwood floors, main level laundry center island kitchen. You don't want to miss this deal! Cul-de-sac neighborhood. This home is really a great place to call home! Come in a see for yourself!
One level living with one car garage under. Large (32x22) partially finished lower level with walkout to back yard could easily be finished for additional 600+ sq. ft. of living area. 6 Rooms, 3 bedrooms, 2 full baths, kitchen with breakfast bar and dining area, parquet floors throughout, new furnace, washer and dryer included. Level back yard.
Eastside 3 bedroom colonial. Spacious eat in kitchen with breakfast bar and slider to back deck and patio. Formal dining room, front to back living room, laundry is located in the half bath on the first floor, master bedroom has a walk in closet and office area. New gravel driveway and drainage system. Expansion potential in the lower level.


With the current growth in sales of wood burning stoves and boilers a wood store could be a very handy asset for a garden. The interest comes from two areas, cost and environmental.
With gas and electricity prices soaring in recent years, a wood burning stove can provide spot heat in a lounge and for the room above. This is a good alternative to putting the central heating on for many houses, saving the full central heating for the coldest days. A wood stove with a back boiler can feed heat into the central heating and hot water system as well.
From the ecological point of view wood burning is considered a renewable source that is also carbon neutral. The carbon released into the air when wood is burnt is roughly equivalent to the carbon trapped by the tree as it was grown. There is also less pollution than with fossil fuels; as there's virtually no sulphur dioxide and low levels of nitrous oxides so burning wood doesn't promote acid rain.
But Where are You Going to Put All that Wood?
So far so good, until you get your first truck-load of wood delivered from your local supplier and realise that you don't know what to do with it. And if it's not seasoned, it needs somewhere to dry out for a couple of years, depending on the wood.
Step forward the garden wood store, which at its simplest can be a tarpaulin pegged over the wood pile and at the other end of the market can be a wooden barn-type construction costing over a thousand pounds. Or you can make a usable and attractive wood store yourself.
The most vital part of a wood store is a roof as it's important to keep the rain off the wood. Sides are less important, you can move the outer logs inside as you get through the stack. So a roof (of any material that comes to hand) supported on fours poles securely anchored in the ground will do fine.
Another simple wood store idea is to fix a roof to a wall somewhere at the back or side of the house, perhaps on a garage wall. This simple canopy will protect the wood from the worst of the rain. Go for an angle of at least 15 degrees on a wood store roof to make sure that rain and snow fall off properly.
If the wind tends to drive rain from a certain direction most of the time then you could put a side on to protect the wood there. In fact there's nothing to stop you putting two sides and a back on a wood store and it will probably make it look better too. But you must leave gaps between the panels so that air can get in and around the wood to help the drying process.
Living gaps between the panels of a wood store is most important if you have to season the wood, not just store it. With green wood you need to chop it and split it into logs before storing it so that each log dries out. Stack them roughly so that there are plenty of gaps around them.
If you are buying wood that you are certain is already properly seasoned then you can store it in a wood store that has proper sides, front and doors, like a small shed. There are plastic or metal wood stores available from large DIY stores and garden centres that will store seasoned wood very well.


I realize the reality of investing psycologically in a mythology which,by pattern and practice, induces ,encourages and demands behaviors which results in murder,suicide,patholgical hatred, and paranoia. The reaction to that reality of attacking as "Islamaphobia" or " hate for Mu'slams, etc is an indicator of the disorder itself. I would recommend a really good therapist, trained in the special needs of personality disorders. No amount of discussion of the mythology you obviously have internalized will resolve your conflict . An understanding of the mythology's effect on the human person, and the destructiveness to the individual personality needs to be addressed. Good luck.


text:
Q: Android NDK: ndk-build script fails - script "check-cygwin-make.mk" not found I'm new to development with the NDK of Android.
As I'm developing on Win7 I installed Cygwin in order to build the native binaries.
When running ndk-build from the project folder /home/simon/ndk/hello-neon I get the error
ERROR: You are using a non-Cygwin compatible Make program.
Currently using: C:/Programs/cygwin/bin/make

To solve the issue, follow these steps:

1. Ensure that the Cygwin 'make' package is installed.
   NOTE: You will need GNU Make 3.81 or later!

2. Define the GNUMAKE environment variable to point to it, as in:

     export GNUMAKE=/usr/bin/make

3. Call 'ndk-build' again.

The problem that stops me now is that the ndk-build script seems not to be able to call the check-cygwin-make.mk script in the following lines:
GNUMAKE=`cygpath -u $GNUMAKE`
PROGDIR_MIXED=`cygpath -m $PROGDIR`
CYGWIN_GNUMAKE=`$GNUMAKE -f "$PROGDIR_MIXED/build/core/check-cygwin-make.mk" 2>&1`

When calling
echo $CYGWIN_GNUMAKE

I get the reply:
make: C:/Programs/cygwin/home/simon/build/core/check-cygwin-make.mk: No such fil
e or directory make: AndroidManifest.xml build.properties default.properties jni
 res src No rule to make target `C:/Programs/cygwin/home/simon/build/core/check-
cygwin-make.mk'. Stop.

Now I'm lost because I don't understand the role of the check-cygwin-make.mk script and I could not find it anywhere on my drive.
I'm aware of the problem with spaces in the C:/Program Files/ folder name, and I think this is not problem now.
I hope someone can help my to solve this issue,
Regards,
Simon

A: Install android-ndk and cygwin(with "make" support command) then set the varaibles like this
Right-click My Computer, and then click Properties.
Click the Advanced tab.
Click Environment variables.
then edit"PATH" add this line

C:\cygwin\bin like this
%SystemRoot%\system32;%SystemRoot%;%SystemRoot%\System32\Wbem;C:\Program Files......;C:\cygwin\bin
then try with your built steps

A: 'check-cygwin-make.mk' is in '\android-ndk-r6b\build\core', and you should have installed Android NDK to a directory without spaces in the path.
Next you should be running ndk-build either from Cygwin-shell, or with the command 'bash -c ndk-build'.
And of course the necessary paths both to Cygwin binaries (as in the previous answer), and to ndk-build should be specified in PATH.

redpajama_set_name: RedPajamaStackExchange