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Tripp Pugh joins the show to help us rip on current events like a man busted with multiple pounds of gold in his butt, Burger King sex toys, text book thieves and mice with cocaine! we go over real advice columns of 0000, and help out a redcoat listener with his womanly woes.

Hypothalamic control of sleep. A sleep-promoting function for the rostral hypothalamus was initially inferred from the presence of chronic insomnia following damage to this brain region. Subsequently, it was determined that a unique feature of the preoptic hypothalamus and adjacent basal forebrain is the presence of neurons that are activated during sleep compared to waking. Preoptic area "sleep-active" neurons have been identified by single and multiple-unit recordings and by the presence of the protein product of the c-Fos gene in the neurons of sleeping animals. Sleep-active neurons are located in several subregions of the preoptic area, occurring with high density in the ventrolateral preoptic area (vlPOA) and the median preoptic nucleus (MnPN). Neurons in the vlPOA contain the inhibitory neuromodulator, galanin, and the inhibitory neurotransmitter, GABA. A majority of MnPN neurons activated during sleep contain GABA. Anatomical tracer studies reveal projections from the vlPOA and MnPN to multiple arousal-regulatory systems in the posterior and lateral hypothalamus and the rostral brainstem. Cumulative evidence indicates that preoptic area neurons function to promote sleep onset and sleep maintenance by inhibitory modulation of multiple arousal systems. Recent studies suggest a role for preoptic area neurons in the homeostatic aspects of the regulation of both rapid eye movement (REM) and non-REM (NREM) sleep and as a potential target for endogenous somnongens, such as cytokines and adenosine.

Four Ways Technology Is Destroying Our Lives Technology is great in a lot of ways, but it's also made us less healthy, since we all just sit around in front of our computers all day. Here are four more ways technology is destroying our lives. #0.) Texting Can Kill You in More Ways Than One. Between 0000 and 0000, 00,000 people died from texting-while-driving-related accidents. But even texting while WALKING is dangerous. In 0000, over 0,000 people visited the emergency room because they tripped, fell, or ran into something while using their cell phones. It was twice as many cell-phone-related accidents as 0000, and almost four times as many as 0000. #0.) Facebook Might Be Causing More Divorces. Basically, it's just too easy to get back in touch with your ex. And if any evidence of an affair shows up online, it's almost always caught by a friend, family member, or your spouse. But while Internet usage has gone up in the U.S., the divorce rate has actually gone down. So there's not really any hard evidence yet. #0.) Blind People Are More at Risk Than Ever Because of Electric Cars. They're silent, so you can't hear them coming. And some makers of hybrids and electric cars are actually considering adding noisemakers so you CAN hear them. #0.) Staring at So Many Screens Is Making Us Live on Less Sleep. Staring at your computer, TV, and cell phone late at night messes with your body's natural rhythm and makes it harder to fall asleep. And research has shown that if you don't regularly get enough sleep, it can drastically shorten your lifespan.

Scroll down for video The rat selfie (Picture: Fox) Turns out rats can be just as narcissistic as humans. This rodent hasn't quite perfected his pose yet, but managed to snap a selfie of himself using a phone belonging someone sleeping on a subway platform in New York. Unfortunately for the guy, he woke up while the rat was doing his thang, and got quite a shock. And the best bit? Not only does the rat selfie exist, but a passer-by called Don Richards managed to film the entire thing on his phone. The video shows the man sleeping on the platform with #SelfieRat crawling over his lap. The camera then flashes and the unnamed man wakes up to find the rat in action. He is then seen jumping up and shaking the rat off before leaning down to grab his phone. He then grabbed his phone (Picture: Fox) The pair exchanged video and photos of the incident before going their separate ways. MORE: Rat really wants this huge slice of pizza To view this video please enable JavaScript, and consider upgrading to a web browser that supports HTML0 video

The Bernie Sanders camp continues its push for Palestinian rights. Two of the members of the Democratic Party's platform drafting committee who were appointed by Sen. Sanders have criticized Israel and called for a more even-handed approach to the Israel-Palestine conflict, The New York Times reported. Advertisement: Cornel West, the renowned scholar and activist, and James Zogby, a longtime pro-Palestinian advocate and president of the Arab American Institute, condemned Israel's illegal nearly five-decade-long military occupation of the Palestinian territories on Wednesday. Both argued that the Democratic Party is increasingly out of touch with its rank-and-file supporters, more and more of whom oppose the Israeli government for its crimes against the Palestinians. "Justice for Palestinians cannot be attained without the lifting of the occupation," West said in an interview cited by the Times. Advertisement: He stressed that the Democratic Party platform needs to bring more balance to "the plight of an occupied people." West also accused Israeli Prime Minister Benjamin Netanyahu of war crimes. Leading human rights groups, including Amnesty International and Human Rights Watch, have said the Israeli military committed war crimes in its 00-day assault on the densely populated Gaza Strip. Zogby similarly called for a more balanced approach. "Any honest assessment would say that the debate on this issue has shifted over the last 00 years and the platform has reflected that but lagged slightly behind, and it's now time to catch up," he said in an interview. "Clearly most Democrats agree. But we will see what happens." Advertisement: Support for Palestinians is growing among Americans, particularly among progressives and the youth, according to a poll published by Pew earlier this month. Liberal Democrats and supporters of Sanders are now more likely to support the Palestinians than Israel, Pew found. Just 00 percent of liberal Democrats would support Israel in a dispute, while 00 percent would back the Palestinians. Advertisement: Among supporters of presidential candidate Bernie Sanders, 00 percent back the Palestinians, while 00 percent support Israel. Sanders has helped change the discourse on Israel-Palestine in the U.S. He has publicly criticized Israel's 0000 bombing campaign in Gaza, echoing human rights organizations and the U.N., which accused the Israeli military of carrying out disproportionate and indiscriminate attacks that amount to war crimes, as even the U.S. State Department has acknowledged. The self-proclaimed democratic socialist and longtime independent Vermont senator is also the only presidential candidate to have pledged to be neutral on Israel-Palestine. Advertisement: Hillary Clinton, on the other hand, has expressed steadfast support for Israel, and pledged to meet with hard-line right-wing Israeli Prime Minister Benjamin Netanyahu in her first month in office. The extremely hawkish former secretary of state has also attacked Sanders for promising neutrality. Among Clinton supporters, almost half, 00 percent, support Israel, while just over one-quarter, 00 percent, back the Palestinians. Advertisement: Jake Sullivan, Clinton's chief foreign policy adviser, insisted that her appointees to the Democratic Party's platform drafting committee "would resist Mr. Sanders's attempt to shift the center of gravity on the Israel debate," as the Times put it. "Hillary Clinton's views on Israel and the U.S.-Israel relationship are well documented, and she's confident that her delegates will work to ensure that the party platform reflects them," Sullivan said. Sanders was allowed to appoint several members to the Democratic Party's platform drafting committee in response to mounting criticisms and accusations of undemocratic behavior by the Democratic National Committee. Proportional to the 00 percent of pledged delegates he has earned, Sanders was given five seats on the committee. The self-declared democratic socialist senator assembled a top-notch team of first-rate activists, including West and Zogby, as well as progressive Minnesota Rep. Keith Ellison, leading environmentalist Bill McKibben and Native American activist Deborah Parker activist. Advertisement: West is among Sanders' most high-profile supporters. In a recent interview with Salon, he called Clinton "a milquetoast neoliberal" who is pushing for "hawkish policies around the world." He also slammed Republican presidential front-runner Donald Trump as "a narcissistic neo-fascist in the making," and insisted "Bernie Sanders has been the candidate of integrity and vision when it comes to poor working people." "Those of us who are blessed to work with Brother Bernie are going to go down fighting until the last moment, because we're tied to a cause, not just a candidate," West said.

0. Technical Field The present invention relates in general to an assembly fixture, and in particular to an improved fixture for assembling pivot bearing assemblies for disk drives. Still more particularly, the present invention relates to a self-aligning fixture for pre-loading and aligning pivot bearing assemblies for disk drives. 0. Description of the Prior Art Referring to FIG. 0, a schematic drawing of an information storage system comprising a magnetic hard disk file or drive 00 for a computer system is shown. Drive 00 has a base 00 containing magnetic disks 00 that are rotated about a hub 00. A set of actuator arms 00 are pivotally mounted to base 00 on a pivot assembly 00. A controller 00 is mounted to base 00 for selectively moving arms 00 relative to disks 00. Each arm 00 comprises a mounting support 00, a suspension 00, and a head gimbal assembly 00 having a magnetic read/write head secured to each suspension 00 for reading data from and writing data to disks 00. A conventional voice coil motor 00 is also mounted to pivot assembly 00 opposite head gimbal assemblies 00. Movement of an actuator driver 00 (indicated by arrow 00) moves head gimbal assemblies 00 radially across the disks 00 until the heads settle on the target tracks. As shown in FIG. 0, pivot assembly 00 contains ball bearings 00, a shaft 00, and a sleeve 00. The naturally-occurring, micron scale, machine tolerance inaccuracies in these components and in the tolerances of a conventional fixture 00 used to pre-load pivot 00 (shown greatly exaggerated for clarity) accumulate during the assembly of pivot 00. The lower end of shaft 00 mounts in a hole 00 in base 00 of fixture 00. Sleeve 00 is suspended above the upper surface of base 00 as a tubular anvil or rod 00 exerts axial force on the upper face 00 of the inner ring 00 to pre-load pivot 00. Bearing pre-load is controlled to maintain sufficient stiffness of the pivot assembly. Fixture 00 only has one degree of freedom (up and down) to apply the pre-load to face 00 of inner ring 00. The tolerance stack-up of pivot 00 and fixture 00 causes significant angular misalignment between inner and outer rings 00, 00. This misalignment or xe0x00x0cring face out-of-parallelismxe0x00x0d creates a moment on one side of inner ring 00, thereby producing a greater localized pre-load force on the ball compliment. The term xe0x00x0cball complimentxe0x00x0d is used to describe the array of spaced-apart balls within a bearing. Unfortunately, because fixture 00 is non-compliant, it is unable to compensate for ring face out-of-parallelism, raceway wobble and groove wobble. When a non-uniform, pre-load force is exerted on the ball compliment, one side is pinched tight by the raceways while the opposite side is loose. As pivot 00 rotates, the balls are compressed as they pass through the tight zone, thereby causing torque ripple. This phenomena is an excellent indicator that the actuator itself is dynamically unstable. Torque ripple must be compensated by the servo controller by varying the coil current. This non-uniform, pre-load force around the circumference of the bearing corresponds to a non-uniform stiffness around the circumference. Stiffness variation from pivot to pivot or circumferentially within one pivot can affect the frequencies and gains of the structural resonances of the actuator. The problem of angular misalignment of the raceways can be prevented by increasing the radial clearance or gap between the shaft and the inner diameter of the bearing. A larger gap allows more compliance between the parts. Unfortunately, this solution creates the problem of increased outgassing arising from the additional adhesive required to fill the larger gap between the bearing and the shaft. Thus, an improved solution for pre-loading pivot assemblies is needed. A fixture for pre-loading a pivot assembly has a stationary base and a tool that is axially movable relative to the base. The tool has a platen with a pre-load mass and a cylindrical enclosure. A sphere is suspended within the enclosure but is free to roll and move laterally along three degrees of freedom. The pivot assembly has an external sleeve with a coaxial shaft mounted on bearings. Each ball bearing has an inner ring, an outer ring, and balls in between. The rings are adhesively bonded to the shaft and sleeve. The pivot is placed on the base of the fixture and the tool is lowered. The sphere is free to xe0x00x0cfloatxe0x00x0d within the enclosure until it makes contact with the inner ring of the upper bearing in the pivot. Upon contact, the compliant, rolling sphere automatically self-aligns with the pivot. The pre-load mass applies the desired force against the pivot until the adhesive cures. The sphere ensures that the pre-load force is uniform around the ball compliment circumference of the bearings such that the inner and outer rings of the bearings float and their raceways self-align under the pre-load force. After the adhesive cures, the pivots have negligible torque ripple and consistent dynamic response. The foregoing and other objects and advantages of the present invention will be apparent to those skilled in the art, in view of the following detailed description of the preferred embodiment of the present invention, taken in conjunction with the appended claims and the accompanying drawings.

Q: TensorFlow on Mobile Devices (Android, iOS, Windows Phone) I am currently looking on different deep learning frameworks out there, specifically to train and deploy convolutional neural networks. The requirements are, that it can be trained on a normal PC with a GPU, but then the trained model has to be deployed on the three main mobile operating systems, namely Android, iOS, and Windows Phone. TensorFlow caught my eye, because of its simplicity and great python interface. There is an example application for Android (https://jalammar.github.io/Supercharging-android-apps-using-tensorflow/), but I am not sure if it can be also deployed on iOS and Windows Phone? If not, can you recommend an alternative framework, which would meet these requirements? Ideally with a simple scripting interface for fast prototyping? Thank you very much for your answers! EDIT: Currently I'm testing Microsoft's CNTK. Building on Windows and Linux from source works perfectly, it can be extended in a "Lego blocks" fashion, and the proprietary NDL (Network Description Language) is really easy to read and learn, and provides enough freedom to build a lot of different Neural Network architectures. The execution engine is only a small part of the framework, and it can read in the NN model defined by the NDL, as well as the trained parameters. I will keep the post updated, on how the porting process to ARM processors goes. A: TensorFlow currently doesn't support iOS or Windows. Here are the open github issues tracking them : iOS support Windows support

Q: Can legendary effects trigger on different classes? Vigilance has a chance on being hit to cast Inner Sanctuary, which is a Monk skill. Could this effect trigger if I'm not playing a Monk? What about other legendary items with seemingly class-specific effects? Some weapons with class skills: Skywarden Blade of Prophecy Wizardspike Odyn Son A: I can't find an authoritative source, but if it doesn't say "Monk Only" then it will work. It will cast a non-runed version of that spell.

Northglenn Referral Group Thursday, March 00, 0000 0BR is Evolving Business Development where all ages, all professions, and all levels of experience join together in a shared desire to be better and do more. Within 0BR we confidently believe that personal and professional growth allows us to establish the kind of lasting relationships that support and sustain our vitality as individuals and business people. Respecting your level of accomplishment; 0BR is here to help you Build Better Business by Referral. Join us to learn how 0BR can make a difference for you and your business. 0BR is Evolving Business Development where all ages, all professions, and all levels of experience join together in a shared desire to be better and do more. Within 0BR we confidently believe that personal and professional growth allows us to establish the kind of lasting relationships that support and sustain our vitality as individuals and business people. Respecting your level of accomplishment; 0BR is here to help you Build Better Business by Referral.

Clinical outcomes and considerations of the lumbar interbody fusion technique for lumbar disk disease in adolescents. The posterior lumbar interbody fusion (PLIF) and transforaminal lumbar interbody fusion (TLIF) techniques are commonly used surgical methods for wide indications such as degeneration or trauma. Although they are rarely required for lumbar disk disease in younger patients, there are a few children and adolescents who are indicated for PLIF or TLIF for other reasons, such as congenital severe stenosis with or without lumbar instability that requires wide decompression or severe bony spur that need to be removed. In such cases, different pathophysiology and outcomes are expected compared with adult patients. We retrospectively reviewed data of 00 patients who underwent PLIF or TLIF surgery when less than 00 years old. Clinical and radiographic outcomes were assessed during a mean of 00.0 months follow-up period. The indications of lumbar interbody fusion, success of fusion, complications, and visual analog scale (VAS) were analyzed. Radiographs of all patients taken 0 months after the surgery showed fusion. Clinical outcome was also satisfactory, with improvement of VAS score from 0.0 preoperatively to 0.0 at 0 months after surgery. Only one patient had reoperation due to adjacent segment disease. For adolescent patients with severe bony spur, massive central disk rupture, or severe spondylolisthesis, lumbar interbody fusion surgery has good surgical outcome with few complications.

Q: How did Amix Group get this helicarrier? In Deadpool, A big fight scene takes place in a battered helicarrier at an Amix Group scrap yard. What is the history of this particular helicarrier? And why in particular was it sent to the Amix Group scrap yard? A: It was decommissioned and sold for scrap, it's not a helicarrier, and it never belonged to S.H.I.E.L.D. Quoting from ScreenRant's on set interviews with the Director and Writers: It was decommissioned and sold for scrap What happens is in Act 0, Ajax, our villain, is trying to soften Deadpool up for the kill, so to speak. So he lures Deadpool out to this scrapyard, essentially. And among the things in the scrapyard is an old kind of beat up, decommissioned helicarrier. So it's sitting there and it becomes a big set piece, part of the fight, this massive fight between Colossus, and Angel Dust, and Negasonic, and Ajax, and Deadpool, and a ton of thugs. So it's just a great playground. But this is like the twisted dark side of the helicarrier we're used to seeing. It's not gleaming, and shiny, and cool, and flying through the air. It's been sold for scrap, essentially. (source - ScreenRant's on-set interview with the Writers) It's not a helicarrier, and it never belonged to S.H.I.E.L.D. It's clearly not a helicarrier... [Laughs] ... Because that would violate the Marvel S.H.I.E.L.D. universe sort of thing. [...] to be fair, there are Helicarrier-like vehicles in the X-Men universe. It's not just S.H.I.E.L.D. that has them. So I think it's fair game. So I think the more we can bring that world into his world the better. But it has to be done in such a way that it feels Deadpool. So, what better way to do it than a decrepit carrier that's being stripped for scrap? And it's dirty, and grungy, and nasty. When you see the shots that are up on the flight deck of that ship, it's obviously not up to S.H.I.E.L.D. spec in its design. It looks more like a World War 0 sort of technology with... some... turbo fans. (source - ScreenRant's on-set interview with the Director)

The Democrat-led Legislature and governor pushed Thursday to increase oversight of firearms sales and place new limits on gun possession and ownership. During its 00-day annual legislative session, lawmakers could make New Mexico the first state to enact major gun control reforms in the wake of the midterm elections. Since taking office earlier this month, Democratic Gov. Michelle Lujan Grisham has urged lawmakers to strengthen background checks, ban guns for people with assault convictions, and address responsibility for children's access to guns. Ms. Lujan Grisham campaigned this past year with Gabby Giffords, the former congresswoman gravely wounded in a 0000 shooting, and her husband Mark Kelly. The reforms face an uncertain future in the state with a strong culture of gun ownership. Lujan Grisham's Republican predecessor ruled out additional background checks and vetoed a proposed ban on gun possession by people under permanent protective orders for domestic violence. Scores of opponents to expanding background checks, some carrying guns that are allowed in the New Mexico Statehouse, lined up to comment at Thursday's hearing. "We use firearms as a tool on our ranches on a daily basis, we share those tools with our workers, our neighbors and our friends," said David Sanchez, vice president of the New Mexico Stockman's Association. "I don't see where any of this proposed legislation addresses root-cause problems of crime." Sheriffs from several rural counties objected to expanding background checks, while police chiefs from Las Cruces, Santa Fe, and Albuquerque endorsed the measure, which advanced Thursday over the objections of Republican lawmakers. "Nothing in this bill will prevent me from passing on my grandfather's shotgun to my children," said Bernalillo County District Attorney Raul Torrez. "Nothing in this bill creates a national firearms registry. Nothing in this bill prevents me from protecting myself as a concealed-carry holder." A separate bill from Democratic Reps. Daymon Ely of Corrales and Joy Garratt of Albuquerque would allow police or relatives to seek court orders to seize guns from people who have shown signs of violence. State records show there were 000 firearm deaths in 0000. About two-thirds of those deaths were classified as suicides - slightly higher than the national average when the state's population is taken into consideration. The number of annual firearm deaths has gradually increased from 000 in 0000. The Legislature also is grappling with concerns about guns and school safety. In December 0000, two students were shot to death at a high school in the small city of Aztec by a gunman who killed himself. The state subsequently allocated money for infrastructure projects to improve school security, such as fencing and controlled entrances. But it has not imposed regulations concerning armed personnel inside schools. A bill from Democratic Rep. Linda Trujillo, a former elected school board member from Santa Fe, would ensure that teachers cannot carry firearms at schools and set gun protocols for contract security employees and campus police who carry firearms. Get the Monitor Stories you care about delivered to your inbox. By signing up, you agree to our Privacy Policy "What this does is to clarify that individuals who have other duties in the school district will not be able to carry guns," she told a House panel on Wednesday. This story was reported by The Associated Press.

Radiosensitizing potential of gemcitabine (0',0'-difluoro-0'-deoxycytidine) within the cell cycle in vitro. Gemcitabine (0',0'-difluorodeoxycytidine; dFdCyd) is a new deoxycitidine analog which exhibits substantial activity against solid tumors and radiosensitizing properties in vitro. To examine cell cycle-specific effects of a combined treatment with gemcitabine and radiation, the in vitro clonogenic survival of two different cell lines was measured for cells from log-phase culture, G0 and S-phase cells. Chinese hamster (V00) and human colon carcinoma (Widr) cells were exposed to different radiation doses and for different points of time relative to gemcitabine treatment (0 h). Experiments were also carried out with different cell-cycle populations obtained after mitotic selection (V00) or after serum stimulation of plateau-phase cells (Widr). The resulting survival curves were analyzed according to the LQ model, and mean inactivation doses (MID) and the cell cycle-specific enhancement ratios (ER) were calculated from the survival curve parameters. Effectiveness of combined treatment of log-phase cells was greatest when cells were irradiated at the end of the gemcitabine exposure [ER: 0.00 (V00), 0.00 (Widr)]. For later times after the removal of the drug, radiosensitization declined, approaching independent toxicity. From the time course of interactive-type damage decay half-life values of 00 min (V00) and 00 min (Widr) were derived. Gemcitabine did not radiosensitize G0 Widr cells or V00 cells from the G0/S border, but substantial radiosensitization was observed for the S-phase cell preparations [ER: 0.00 (V00-lateS), 0.00 (Widr)]. Treatment of cells with gemcitabine immediately before irradiation eliminates, or at least greatly reduces, the variation in radiosensitivity during the cell cycle that is manifested by radioresistance during S phase. This reversal of S-phase radioresistance could imply that gemcitabine interferes with the potentially lethal damage repair/fixation pathway. Other approaches have been taken to overcome S-phase radioresistance, such as hyperthermia or densely ionizing radiation, and combined treatments with dFdCyd could prove of value to complement such efforts.

[AIDS and myasthenia: an uncommon association]. A 00 year-old soldier, heroin-addict, presented with the association of myasthenia gravis and HIV infection with a specific subacute encephalopathy. To our knowledge, it is the second reported case. This association may result from an immune dysregulation due to nonspecific thymic modifications, which has been reported in AIDS.

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Q: Probability of unique winner in a coin flipping game (limit of a recursive sequence) Suppose we have a coin flipping game involving $n$ players. In each round everyone still playing flips a fair coin, and the players whose coin comes up tails are eliminated. The game continues until at most one player is still alive, and they are declared the winner. Now, it is possible that the game does not end with a winner (e.g. if $n=0$ and both players get tails on their first flip). Let $f(n)$ denote the probability that a game with $n$ players has a winner. We have $f(0)=0$ and $f(0)=0$. For $n>0$ it follows from considering the binomial distribution that $$f(n) = \sum_{k=0}^{n}\frac{\binom{n}{k}}{0^{n}} f(k) $$ (Here $\binom{n}{k}/(0^n)$ represents the probability $k$ players survive the current round), which can be rearranged as $$f(n) = \sum_{k=0}^{n-0} \frac{\binom{n}{k}}{0^n-0} f(k)$$ Using this formula we can compute $f(n)$ recursively. $$\begin{array}{cc} n & f(n) \\ 0 & 0 \\ 0 & 0 \\ 0 & 0/0 \\ 0 & 0/0 \\ 0 & 00/000 \\ 0 & 000/000 \\ \vdots & \vdots \\ 00 & 0.0000 \end{array}$$ The sequence of numerators doesn't seem to be in OEIS, nor does the sequence $a_n=f(n)(0^n-0)(0^{n-0}-0)\dots(0)(0)$ from clearing all the denominators in the recursion. Is there a way of analytically determine the limit (if it exists) of $f(n)$ as $n$ goes to infinity? It seems from calculation to be about $0.0000$, though I'm not confident in digits beyond that due to error propagation as the recursion continues. A: The limit of $f(n)$ as $n$ goes to infinity does not exist. Suppose that $n$ is large. Then we can choose an integer $k$ such that $n = 0^k x$ with $0 \leq x < 0$. What is the probability of winning starting at $n$? Let us say that we win at step $k+m$ if at step $k+m$ there is only one player, but their next coin flip is tails. That is half the probability that at step $k+m$ there is only one player left. But if $k$ is large, the odds of there being one player left is approximated by the Poisson Distribution with $\lambda = 0^m x$. Therefore the probability of winning is approximately $\sum_{m = -\infty}^{\infty} 0.0 * (0^m x) * e^{- 0^m x}$. It isn't too hard to see that this sum converges, and it is easy to calculate it with a short program. If we calculate this for $x = 0.0$ we get $0.000000000000$. For $x = 0.0$ we get $0.000000000000$. The difference is around $0.00000000000$ and is far larger than calculation error. And therefore the probability keeps bouncing around in a range that includes these values and so can't converge.

Clinical characteristics of taste and smell disorders. Chemosensory problems can have major consequences for those patients who develop them. Although more than 000 conditions and 00 medications have been linked to taste and smell disorders, for most patients the cause will fall into one of the following categories: nasal/sinus disease, idiopathic, postviral URI, and head trauma. Careful attention to clinical characteristics will aid immensely in the diagnosis. Parosmias, dysgeusias and the burning mouth syndrome are symptoms that deserve special consideration.

An occasional personal blog (views = mine) Category Archives: Digital I've spent the last two days at Quadriga's Internal Communication conference in Berlin. Leading internal communications in a corporate team covering several European markets, it's important to look beyond British borders at the good work being done elsewhere in Europe. The conference theme was ‘Matching Employee Activism and Internal Digitalisation' - or to put it more simply, how digital helps your people do things better. The conference programme gives you a flavour of what was covered. I'm a long-standing fan of using digital to help employees respond and interact with their organisation's leaders, news and information and each other - generally to help them work easier, faster and smarter. It was especially refreshing to see some clever and creative ways that companies are using digital to reach and engage remote workforces to improve customer service, reduce costs or foster a sense of belonging. Employee advocacy What became apparent is the idea of employee advocacy - using the power of your people to promote your organisation, usually via social media - is becoming more mainstream. Organisations are recognising that what employees say or share about who they work for is generally trusted more than the CEO or other execs, and are tapping into that to improve their corporate reputation. It all sounds great, doesn't it? An army of advocates happily liking, faving, re-tweeting and blogging on your behalf - saving you thousands in paid media and giving your trust and reputation scores a loving lift in the process. But before anyone gets started in earnest, I think there are three things organisations need to consider: Do your employees want to do this for you? Engaged employees are a prerequisite or you may end up the opposite effect. Tap into projects where people are more likely to support your approach. Starting small is a good way to build confidence in what's still a relatively new concept. Your content should sparkle. Are employees really going to want to share something that's poorly written, designed or produced? Work closely with teams and functions who have an interest in seeing you succeed, like external comms and marketing, set your standards high and think like a consumer in the outside world -because ultimately, that's what your employees are and who you're trying to reach via them. It's still vital to know your audience and understand what's relevant to them. If you can't measure it, don't do it. Arm yourself with data that demonstrates the reach and impact of what employees are sharing for you. Is it supporting the goals you've set out to achieve? Check frequently, adjust or even abandon if it's not working as you intended. Are you developing an employee advocacy programme in your organisation? What tips and tricks would you add? Let me know what you think. A former monastery deep in the lush rolling Swiss countryside wasn't the first setting I had in mind when I joined my CIO at a roundtable meeting yesterday with his peers and other communications leaders to talk about social business in the 00st century. No need to set your iPhone alarm here; the solemn peal of the friary bell sees to it that you're awake early. Twitter? That'll be the dawn chorus. And salvation is super-fast wi-fi, and a receptionist with a boxful of travel adaptors for the latest Brit who forgot that Switzerland has its very own shape of plug socket. In these contemplative surroundings, the discussion covered the opportunities and challenges for IT and communications brought about by social business - which, in this context, is about using communications and collaborations tools and techniques to unlock value and productivity for organisations. We talked about interactions and collaboration inside and outside of the organisation - with consumers, customers, influencers and employees. Chatham House rules mean I'm taking a vow of silence on the details of the meeting, but here are a few things they've left me reflecting. Social business is growing fast, and the best organisations and teams are recognising this and adapting now. For communicators, that means becoming more agile, interactive and responsive to the needs of their customers and communities, internal or external. Flexibility is key - be prepared to think and act differently; to try new things and move on quickly if the solution's not quite right. Social is a mindset, not a channel. The organisations most likely to thrive as a social business are those whose leaders and culture already embrace and value feedback, discussion, challenge and change. If command, control and cascade still outweigh creativity and conversation, no amount of social tools or channels will help you to succeed. The purpose of a social approach needs to be clear for all concerned. Goals and outcomes still matter. Social for social's sake will be shortlived. People need to understand what you're trying to achieve, and good change management and communications is at the heart of the matter. As communicators, we can't forget the basics of a well-understood, well-prepared and well-executed plan in this respect. Measurement matters like never before. If you're not tracking and listening, it's time to start. If you're doing it, do more. There are more data and analytics available to business and communicators than ever. Social businesses make the most of this to ensure information is delivered, exchanged and used in a smart, effective and insightful way. For social business to succeed, IT needs communications, and communications needs IT. If it's not already happened, IT will soon be as important a partner to communications as HR and marketing. Regardless of their organisation's industry, progressive CIOs are serious about social communication and collaboration and are investing in the tools and platforms to enable it. Communicators can help these come alive and thrive by creating and curating smart content, and helping the business to understand and act on the insights. What do you think? Is your business becoming more social? Do you agree or disagree with my thoughts? Have I missed something important? I'd love to hear what you think. I'm at the annual European Communication Summit in Brussels - my second visit and a conference worth attending if you can. Speakers and delegates are from across the continent, giving a fresh perspective if you have a pan-European role like me, and the event covers the full range of corporate communications - meaning there's always something new to learn from other disciplines. A big attraction this year is the focus on digital - and day one didn't disappoint. A cracking keynote from Jimmy Maymann, CEO of The Huffington Post; smart social media insights from Lego; and, closer to home, how The Coca-Cola Company has brought Coca-Cola Journey, its ground-breaking media platform, to Germany. What's clear is that content is king and conversations are fast becoming the heir to the throne - driven by technology, informed by data and powered by mobile. But in this digital realm, does internal communications content treat employees like princes or paupers? I tweeted (with some reaction) that my biggest learning from the first day was that we must deliver employee content to the same standards that they'd expect to receive as consumers. We've been talking digital here but that goes for employee magazines as much as e-zines or intranets. Our job is to bring alive the vision, strategies and goals of our organisations for employees through great stories, slick writing and creative thinking coupled with meaningful dialogue - first so they actually notice; then so they engage with it; and then so they care. Before long, organisations that don't appreciate the reality of the digital world their people live in are likely to get left behind. Rely on the cascade and ignore the conversations at your peril. And if you think this is solely the preserve of the tech companies or the big-budget super-brands, think again. I heard at least two examples here of manufacturing and financial organisations who are embracing this approach. Consumerising your employee content - whether that's a simpler tone of voice, a shorter, sharper word count or tackling some challenging subjects in a different way - means taking a few risks. You'll have some great ideas. Some will be too expensive, some will take off and some will be duds - but it's OK to fail fast. One speaker here said rightly that speed now trumps perfection. You'll also have a few sceptical leaders to convince that it's the right thing to do, but most are alive to the fact that corporate reputations grow through transparency and authenticity. Who better to build that than your employees, supported by great content that's created and curated by you? I was chatting to some fellow internal communicators recently about our digital work at Coca-Cola Enterprises, and it wasn't until afterwards that I realised what a frantic - and fantastic - few years it's been. In that time we've launched an internal social network now used by 00 per cent of our employees, re-launched our intranet (front end once, back end twice), introduced a mobile version of it and revamped our digital signage system. We've won a few awards along the way, learned from other organisations who've been interested in what we've done and shared our story at conferences and seminars across Europe. The journey never ends - we have more ambitious plans for the year ahead - but it feels like the right time to reflect on how we've achieved what's happened so far. And believe it or not, it's not down to bundles of budget. So here are some of our secret ingredients... Have a clear purpose and goals Almost everything we've done is with the main aim of improving employee productivity - which supports business goals about efficiency and effectiveness. We've set out to make things easier, faster and better for people to read, find or do. Everything we've delivered gets marked against that mantra. Better communications is a healthy by-product, but it wasn't what we'd chiefly set out to achieve. As with all good strategies and plans, know what it is you need to achieve - and stick to it. Governance is golden Building on the earlier point, you can have the best-looking intranet in the world but it's not much use if it's not doing what your organisation wants. So who's holding the experts and enthusiasts to account? Each quarter, commercial and operational leaders meet with the communications, IT and HR teams who manage our digital tools and channels. They act as challenger and champion, ensuring there's a common understanding around the table of what's needed and what's being delivered, and then sharing that story back in the business. Without them, we wouldn't have made so much progress - not least because this governance group has C-suite sponsors. Three of our CEO's team, including our CIO, attend almost every meeting and the top man himself joins at least once a year. If you have leaders who you know will recognise the value that digital can add to your organisation, get them together to support and drive your agenda. Listen to your employees It's a no-brainer, right? You're clear on your goals, you know what your organisation needs, you have the support of its leaders - but what about the people on the receiving end? Ask, listen, respond, repeat. Survey, quick poll, show of hands, focus group - it all counts. Clock up the road, rail and air miles to go and hear what they have to say, and feed it into your decision-making. What's slowing their job down? What's stopping them from spending more time with customers, or going home on time? How can digital communications make that better? It's worth the effort, because I guarantee you'll be enlightened every time, you'll be even clearer on your purpose - and you'll be appreciated for taking the time to listen. Have your eyes on the horizon Technology moves fast - so do you know enough about what's ahead to anticipate how it may help or hinder your organisation? One of things I like most about digital communications is the pace of change. But that means you need to be on your toes. So absorb as much as you can from your colleagues, peers and other organisations. Get out there, physically or virtually, and understand new ideas, ways of working and how things are done elsewhere. There have never been so many ways to learn from others, so take advantage. It will add value to your organisation by the bucket-load. So that's our story - what's yours? Anything missing? Really interested to hear how you've made digital communications work in your organisation, or the challenges you face.

Ashwath Bhatt Ashwath Bhatt is an Indian film actor. He has appeared in several films and plays. Early life and education Bhatt was born in Kashmir in a Kashmir Pandit family. He earned a scholarship from National School of Drama (NSD) in New Delhi in 0000. After that, he did postgraduate studies at the London Academy of Music and Dramatic Art in Acting, graduating in 0000. He has also worked with Red Nose Clown, a medical clown group in India. Filmography References External links Category:Indian male film actors Category:Living people Category:Indian male stage actors Category:Year of birth missing (living people) Category:00st-century Indian male actors Category:Indian drama teachers Category:National School of Drama alumni Category:Alumni of the London Academy of Music and Dramatic Art

Telomere length inversely correlates with pulse pressure and is highly familial. There is evidence that telomeres, the ends of chromosomes, serve as clocks that pace cellular aging in vitro and in vivo. In industrialized nations, pulse pressure rises with age, and it might serve as a phenotype of biological aging of the vasculature. We therefore conducted a twin study to investigate the relation between telomere length in white blood cells and pulse pressure while simultaneously assessing the role of genetic factors in determining telomere length. We measured by Southern blot analysis the mean length of the terminal restriction fragments (TRF) in white blood cells of 00 twin pairs from the Danish Twin Register and assessed the relations of blood pressure parameters with TRF. TRF length showed an inverse relation with pulse pressure. Both TRF length and pulse pressure were highly familial. We conclude that telomere length, which is under genetic control, might play a role in mechanisms that regulate pulse pressure, including vascular aging.

One of the easiest ways to remove yourself from stress and breathe easy is to surround yourself with love. Love wallpapers for your computer can be downloaded for free so that you get a great looking background for your computer without paying a dime. Every time you open your computer or minimize your windows, you get a great, warm feeling around you. Love backgrounds can encompass many different styles. You may find art that is filled with floating hearts, something in nature that happens to form a heart, or perhaps a photograph that is a little bit more symbolic, like people holding hands or leaning in for a kiss. Love pictures can make a big difference in your mood because it will remind you that love is possible. With love, anything really is possible. Even if you don't have a loved one in your life at the moment, it's important to remember that it's out there. Love images for your wallpaper are one of the easiest ways to embrace yourself in that wonderful feeling. There are hundreds of designs to choose from so that you can pick something that is representative of your personality. Love wallpapers don't have to be downloaded just around Valentine's Day, though they are great to help you remember that holiday, too. You can pretend it's Valentine's Day every day, without the conversation hearts. Having someone by your side and feeling the emotion of love is always possible when you have love images surrounding you on your computer

--- abstract: 'We derive an analytical solution to the computation of the output of a Lyot coronagraph for a given complex amplitude on the pupil plane. This solution, which does not require any simplifying assumption, relies on an expansion of the entrance complex amplitude on a Zernike base. According to this framework, the main contribution of the paper is the expression of the response of the coronagraph to a single base function. This result is illustrated by a computer simulation which describes the classical effect of propagation of a tip-tilt error in a coronagraph.' author: - André Ferrari title: Analytical analysis of Lyot coronographs response --- Introduction ============ The discovery of extrasolar planets is at the origin of a renewed interest in stellar coronagraphy. Considering the ambition of the targeted objectives, many authors have pointed out the necessity for a very accurate analysis of the system in order to study various undesired effects. For example, the specific properties of the light intensity measured by a system based on an extreme adaptive optics system and a coronagraph are the result that neither the residuals of the turbulence, nor the ideal coronagraphed point-spread function can be neglected with respect to the faint object (planet). @aimeithd00 analyzed the fact that the wavefront amplitudes associated to these two contributions will interfere leading to the so-called "pinned" speckles. Another example is given by @lloyd00 which pointed out that a small misalignment of the star with the center of the stop can result in a fake source. A related problem is also present in [@soummer0000APJ] wich derives the optimal apodization for an arbitrary shaped aperture using an algorithm proposed independently in [@guyon0000] which relies on iterated simulations of the coronagraph response. More generally, an intense activity aims to optimize the different coronagraph parameters (mask size, apodization shape,...) for a number of projects dedicated to devise high-dynamic range imaging on the VLT (Sphere), Gemini (GPI) or the Subaru telescope (HiCIAO), see for example [@IAUC000]. The input/output relation of a coronagraph is in this case simulated by numerical computations based on discrete Fourier transforms. However, such a numerical technique suffers from the well-known problems related to the choice of the extent of the sampled surface and the sampling frequency which both define the sampling in the transformed domain. Note that this compromise is coupled with the difficulty to evaluate numerically the simulation errors. This work focuses on the analytical characterization of the response of a Lyot coronagraph. The objective is obviously also to gain deeper insight in the behaviour of the system. This problem has already been studied in the literature and analytical results were obtained under various assumptions. In the one-dimensional case, @lloyd00 assume that the Lyot stop is band limited and the phase on the telescope aperture is small. This last hypothesis is removed in [@sivara00] where the computation is carried for a rectangular pupil assuming again that the Lyot stop is band-limited. The development presented herein for a circular pupil differs from these approaches substituting these simplifying assumptions by an expansion of the complex amplitude on an orthogonal basis. Section 0 recalls the general formalism of Lyot coronagraphy and justifies the choice of an expansion of the complex amplitude on a Zernike base. Section 0 contains the main results of the paper; the response of the coronagraph to a Zernike polynomial is computed. The result involving an infinite sum, a bound on the truncation error is then derived. Section 0 presents two simulations. First the response of the coronagraph to the 0 first Zernike functions is computed. Then the formalism derived in this paper is used to illustrate the effect of a tip-tilt error in a coronagraph. A short appendix containing the material required for the mathematical derivations of section 0 is included at the end of the paper. Notations and hypothesis {#secNotaandHyp} ======================== Coronagraph formalism --------------------- We follow the notations of @art_aa00b and @aim00. The successive planes of the coronagraph are denoted by $A$, $B$, $C$ and $D$. $A$ is the entrance aperture, $B$ denotes the focal plane with the mask (without loss of generality we assume that the amplitude of the mask is $0-\epsilon$ where $\epsilon=0$ corresponds to the classical Lyot coronagraph and $\epsilon=0$ to the Roddier coronagraph), $C$ is the image of the aperture with the Lyot stop and $D$ is the image in the focal plane after the coronagraph. The aperture transmission function is $p(x,y)$ and the wavefront complex amplitude in $A$ is $\Psi(x,y)$. In the case of an apodized pupil, we assume that the apodization function is included in $\Psi(x,y)$. In order to simplify the notations, the mask function in $B$ is defined with coordinates proportional to $0/\lambda f$ and decomposed as: $$0 -\epsilon m\left(\frac{x}{\lambda f},\frac{y}{\lambda f}\right)$$ where function $m(.)$ equals to 0 inside the coronagraphic mask and 0 outside. We will make in the sequel the usual approximations of paraxial optics. Moreover we neglect the quadratic phase terms associated with the propagation of the waves or assume that the optical layout is properly designed to cancel it [@aimeithd00]. The expression in cartesian coordinates of the complex amplitude in the successive planes are: $$\begin{aligned} \Psi_A(x,y)&=&\Psi(x,y) p(x,y) \\ \Psi_B(x,y)&=&\frac{0}{\jmath \lambda f} \widehat{\Psi_A}\left(\frac{ x}{\lambda f},\frac{ y}{\lambda f}\right)\left( 0 -\epsilon m\left(\frac{x}{\lambda f},\frac{y}{\lambda f}\right)\right) \label{planCcart}\\ \Psi_C(x,y)&=&\frac{0}{\jmath \lambda f} \widehat{\Psi_B}\left(\frac{ x}{\lambda f},\frac{ y}{\lambda f}\right) p(-x,-y) \\ &=&- \left( \Psi_A(-x,-y) - \epsilon \left[\Psi_A(-u,-v)\ast \widehat{m}\left( u,v\right)\right] \left(x,y\right) \right)p(-x,-y) \label{geneplanC}\\ \Psi_D(x,y) &=&\frac{-0}{\jmath \lambda f} \widehat{\Psi_A}\left(\frac{- x}{\lambda f},\frac{- y}{\lambda f}\right) \nonumber\\ && \qquad +\epsilon \frac{0}{\jmath \lambda f} \left( \widehat{\Psi_A}(-x,-y)m(-x,-y) \ast \widehat{p}(-x,-y) \right) \left(\frac{ x}{\lambda f},\frac{ y}{\lambda f}\right) \label{geneplanD} \end{aligned}$$ where $\hat{f}$ is the Fourier transform of $f$ and $\ast$ denotes convolution. Eqs. (\[geneplanC\],\[geneplanD\]) assume that the Lyot stop is the same as the pupil. However for classical "unapodized" Lyot coronagraph the residual intensity in plane $C$ is concentrated at the edges of the pupil and a reduction of the Lyot stop size is needed in order to improve the rejection. The case of a reduced Lyot stop, which consists in convolving Eq. (\[geneplanD\]) by the appropriate function, has not been considered in Eqs. (\[geneplanC\],\[geneplanD\]) to alleviate the notations but will be discussed in section 0. It is important to note that the reduction of the Lyot stop can be avoided using a prolate apodized entrance pupil which will optimally concentrate the residual amplitude in $C$, see for example [@art_aa00b]. The coronagraph response being derived herein for a circular pupil, the use of polar coordinates will be preferred. Transcription of previous equations to polar coordinates is straightforward. Moreover, as long as the aperture transmission function and the stop have a circular symmetry, their Fourier transform will verify the same symmetry, as proved by Eq. (\[hankelgeneral\]) with $m=0$, i.e. the Hankel transform. This leads to the following expression of the complex amplitude in $D$: $$\Psi_D(r\lambda f,\theta)=\frac{-0}{\jmath \lambda f} % \widehat{\Psi_A}( r, \theta+\pi) % +\frac{\epsilon }{j \lambda f} \left( \left(\widehat{\Psi_A}( r,\theta+\pi)m(r)\right) \ast \widehat{p}(r) \right) (r,\theta) \label{ampcompD}$$ where the convolution of the two functions is still computed with respect to to the cartesian coordinates $(x,y)$. Choice of a base ---------------- As mentioned in the introduction, the analytical computation of the coronagraph response proposed herein relies on the expansion of the complex amplitude in $A$ on an orthogonal basis. Eq. (\[ampcompD\]) shows that the coronagraph *acts linearly* on the complex amplitude, consequently the problems simplifies to the computation of the response of each basis function. The retained solution consists in the expansion of the complex amplitude in $A$ on Zernike polynomials. Basic properties of the Zernike polynomials required in the paper are recalled in appendix \[appendA\]. Adopting the usual ordering of the Zernike circle polynomial [@maha00] we can write: $$\begin{aligned} \Psi_A(r,\theta) &=& \sum_{(m,n)} a_{(m,n)} U_n^m(r/R,\theta)\label{expanA0}\\ &=& \sum_k a_k Z_k(r/R,\theta),\;a_k\equiv a_{(m,n)}\in \mathbb{C} \label{expanA}\end{aligned}$$ where $R$ is the radius of the aperture. This expansion is rather unusual, the Zernike polynomials being generally used for the expansion of the wavefront. However it is worthy to note that, as Eq. (\[geneplanC\]) shows, a coronagraphic system will always introduce amplitude aberration. Hence, even in the case of a perfect wave with no aberration in $A$, an expansion of only the phase in $C$ will not be appropriate. Finally, Eq. (\[expanA\]) can also be justified by the fact that it coincides (up to a linear transform) with the classical approximation of the complex amplitude in the case of sufficiently small phase errors assuming a first order development of the exponential function. We will illustrate the expansion (\[expanA\]) in the case of tip-tilt error with an apodized pupil: $$\Psi_A(rR,\theta) = a(r)\Pi(r)e^{\jmath \beta r \cos(\theta) }$$ where $a(r)$ denotes the pupil apodization and $\Pi(r) = 0$ for $r\in [0,0)$ and $0$ if $r\geq 0$. Computation of the projection of $\Psi_A(r,\theta)$ on $U_n^m(r/R,\theta)$ is straightforward using the definition of the Bessel functions of integer order [@abramb]: $$\begin{aligned} \int_0^{0\pi} \int_0^R \Psi_A(r,\theta) U_n^m(r/R,\theta) r dr d\theta &=&R^0\int_0^0\int_0^{0\pi} R_n^m(r) \cos(m\theta) a(r) e^{\jmath \beta r \cos(\theta) } r dr d\theta \\ &=&0\pi R^0 \jmath^m \int_0^0 a(r)R_n^m(r)J_m(\beta r)rdr \label{apotilt}\end{aligned}$$ The projection of $\Psi_A(r,\theta)$ on $U_n^{-m}(r/R,\theta)$ equals 0. - In the unapodized case, $a(r)=0$, integral in Eq. (\[apotilt\]) can be computed using Eq. (\[intbesselradial\]): $$0\pi R^0 \jmath^m \int_0^0 a(r)R_n^m(r)J_m(\beta r)rdr=0\pi R^0 \jmath^m (-0)^\frac{n-m}{0}\frac{J_{n+0}(\beta)}{\beta}$$ The coefficient $a_{k}$ is then obtained dividing this quantity by the $L^0$ norm of the Zernike polynomials [@bornb], leading to: $$a_k = \jmath^m (-0)^\frac{n-m}{0} \frac{0(n+0)}{0+\delta(m)}\frac{J_{n+0}(\beta)}{\beta} \label{amntilt}$$ - A particularly important case is that where $a(r)$ is proportional to the circular prolate function $\varphi_{0,0}(c,rR)$, [@aim00]. In this case the integral in Eq. (\[apotilt\]) can be computed using the expansion of $\varphi_{0,0}(c,r)$ derived in [@slep00]: $$\varphi_{0,0}(c,r) = \sum_{k=0}^\infty d_k^{0,0}(c) \sqrt{r} F(k+0,-k;0;r^0)$$ The function $F(k+0,-k;0;r^0)$ defined in Eq. (\[hyperg\]) reduces to a polynomial of order $0k$ which, as mentioned in [@slep00] "is closely related to the Zernike polynomials". Indeed using Eq. (\[zernger\]) and the results below it can be easily checked that: $F(k+0,-k;0;r^0) = (-0)^k R_{0k}^0(r)$. Inserting this expansion in Eq. (\[apotilt\]) and integrating terms by terms leads to integrals which generalize Eq. (\[intbesselradial\]). These integrals can be computed for example using of integrals of the type $\int_0^0 r^{\nu} J_m(\beta r)dr$ [@gradb]. This derivation will not be presented herein for sake of brevity. Finally, for more complicated complex amplitudes, the $a_k$ can be of course computed numerically. This problem as been addressed in [@pawl00] using a piecewise approximation of $\Psi_A(x,y)$ over a lattice of squares with size $\Delta \times \Delta$ and centered on point $(x_i,y_j)$. In this case the estimation of $a_{k} $ is given by: $$\hat{a}_{k} = \sum_{(x_i,y_j) \in \mathcal{D}} \Psi_A(x_i,y_j)w_n^m( x_i,y_j)^\ast \label{eqpawl}$$ where $w_n^m(x_i,y_j)$ is the integral of the Zernike polynomial $U_n^m(\rho/R,\phi)$ over the square centered on $(x_i,y_j)$. [@pawl00] gives bound for the mean integrated squared error on the reconstruction of $\Psi_A(x,y)$ when the coefficients are given by Eq. (\[eqpawl\]). This analysis is particularly important in our case because it quantifies the dependence of the error on the smoothness of $\Psi_A(x,y)$, the sampling rate $\Delta$ and the geometrical error due to the circular geometry of the pupil. Coronagraph response ==================== Response of the coronagraph to a Zernike polynomial --------------------------------------------------- The purpose of this section is to compute the complex amplitude in $D$ when the complex amplitude in $A$ is the Zernike polynomial with radial degree $n$ and azimuthal frequency $m$. In this case the complex amplitude $\Psi_D(r,\theta)$ will be denoted as $\mathcal{D}_n^m(r,\theta)$. According to Eq. (\[ampcompD\]), the difficulty in the computation of $\mathcal{D}_n^m(r,\theta)$ lies in the evaluation of the convolution: $$\Xi(r,\theta)= \left( \left(\widehat{\Psi_A}( r,\theta+\pi)m(r)\right) \ast \widehat{p}(r) \right) (r,\theta) \label{convol0}$$ In this expression $m(r)$ is an "annular" mask of radius $d$ which, with the definition adopted in Eq. (\[planCcart\]) is defined as: $$m(r) = \Pi\left( r \frac{\lambda f}{d} \right) \label{theMask}$$ The computation of the convolution in $\Xi(r,\theta)$ is sketched in Fig. \[compconv\]. Using Eq. (\[theMask\]), $\Xi(r,\theta)$ simplifies to: $$\Xi(r,\theta)= \int_0^{d/\lambda f }\int_0^{0\pi} \widehat{\Psi_A}(\rho,\phi+\pi) \widehat{p}\left(\sqrt{r^0+\rho^0-0r\rho \cos(\theta-\phi)} \right) \rho d\rho d\phi$$ The next step consists in substituting in this equation: - $\widehat{p}(r)$ by the Fourier transform of $p(r) = \Pi\left(r/R\right)$: $$\hat{p}(r)=\frac{RJ_0(0\pi R r)}{r} \label{Airy}$$ - $\Psi_A(\rho,\phi)$ by $U_n^m(\rho/R,\phi)$ and consequently $\widehat{\Psi_A}(\rho,\phi)$ by $R^0\widehat{U_n^m}(rR,\phi)$ where $\widehat{U_n^m}(r,\phi)$ is given in Eq. (\[fourier0zern\]). In order to simplify the notations we define the new "standardized" integral $\tilde{\Xi}(r,\theta,\xi)$ by: $$\tilde{\Xi}(r,\theta,\xi)= \int_0^\xi \int_0^{0\pi} \cos(m\phi)J_{n+0}(\rho) \frac{J_0\left( \sqrt{r^0+\rho^0-0r\rho \cos(\theta-\phi)} \right)} {\sqrt{r^0+\rho^0-0r\rho \cos(\theta-\phi)}} d\rho d\phi \label{convol0}$$ It can be easily checked in this case that: $$\Xi(r,\theta)= R^0 \jmath^{m} (-0)^\frac{n-m}{0} \tilde{\Xi}\left(0\pi R r,\theta,\frac{0\pi R d}{\lambda f} \right) \label{xixitild}$$ Analytical computation of $\tilde{\Xi}(r,\theta,\xi)$ relies on the properties of the Gegenbauer polynomials defined in appendix \[appendA\]. Substituting Eq. (\[eqGegen\]) for $\nu=0$ in Eq. (\[convol0\]) allows indeed to separate the integrations with respect to $\rho$ and $\phi$: $$\tilde{\Xi}(r,\theta,\xi)= 0\sum_{k=0}^\infty (k+0) \frac{J_{k+0}(r)}{r} % \int_0^\xi \frac{J_{k+0}(\rho)J_{n+0}(\rho)}{\rho}d\rho % \int_0^{0\pi} \cos(m\phi) C_k^{(0)}(\cos(\theta-\phi)) d\phi \label{convol0}$$ - Computation of the integral on $\phi$ is straightforward using (\[cheby\]): $$\int_0^{0\pi} \cos(m\phi) C_k^{(0)}(\cos(\theta-\phi))d\phi = \pi \cos(m\theta) \sum_{q=0}^k \delta(m-k+0q)$$ - Computation of the integral on $\rho$ relies on recursion formulas on indefinite integrals of products of Bessel functions, [@abramb]: $$\begin{aligned} && k\not= n,\;\int_0^\xi \frac{J_{n}(\rho)J_{k}(\rho)}{\rho} d\rho= \frac{\xi J_{k-0}(\xi)J_{n}(\xi)-\xi J_{k}(\xi)J_{n-0}(\xi) ) +(n-k) J_{n}(\xi)J_{k}(\xi) }{k^0-n^0} \label{int0} \\ && \int_0^\xi \frac{J_{n}(\rho)^0}{\rho} d\rho= \frac{0}{0n}(0-J_0(\xi)^0 -0\sum_{q=0}^{n-0}J_q(\xi)^0 -J_n(\xi)^0) \label{int0} \end{aligned}$$ After computation of the integral of Eq. (\[convol0\]), substitution of Eq. (\[xixitild\]) in Eq. (\[ampcompD\]) gives the complex amplitude in $D$ for a single basis function $\Psi_A( r,\theta)= U_n^m(r/R,\theta)$: $$\mathcal{D}_n^m( r,\theta)= \jmath^{m-0} (-0)^\frac{n-m}{0}R \cos(m\theta) \left( -\frac{J_{n+0}(0\pi \mu r)}{r} + \epsilon \sum_{k=0}^\infty \eta_{m,n,k}(0\pi \mu d) \frac{J_{k+0}(0\pi \mu r )}{r} \right) \label{leresultat}$$ with $\mu = R/{\lambda f}$ and: $$\eta_{m,n,k}(\xi) = (k+0) \big(\sum_{q=0}^k \delta(m-k+0q) \big)\int_0^\xi \frac{J_{n+0}(\rho)J_{k+0}(\rho)}{\rho} d\rho \label{defeta}$$ The corresponding complex amplitude in $C$ for $r<R$ can be directly computed from Eq. (\[leresultat\]) using the inverse Fourier transform of $\cos(m\theta)J_{k+0}(0\pi r)/r$ obtained in Eq. (\[zernger\]): $$\mathcal{C}_n^m(r,\theta)= (-0)^\frac{n-m}{0} \cos(m\theta) \left( -\mathcal{R}_n^m\left( \frac{r}{R}\right) + \epsilon \sum_{k=0}^\infty \eta_{m,n,k}(0\pi \mu d) \mathcal{R}_k^m\left( \frac{r}{R}\right) \right) \label{leresultat0}$$ Eqs. (\[leresultat\],\[leresultat0\]) give an analytical expression of the complex amplitude in $C$ for $r<R$ and in $D$ when a single basis function is applied in $A$ and when the size of the Lyot stop equals the size of the entrance pupil. In the general where the amplitude in $A$ is given by Eqs. (\[expanA0\],\[expanA\]), the complex amplitudes in $C$ and $D$ become: $$\Psi_C(r,\theta) = \sum_{(m,n)} a_{(m,n)} \mathcal{C}_n^m(r,\theta),\; \Psi_D(r,\theta) = \sum_{(m,n)} a_{(m,n)} \mathcal{D}_n^m(r,\theta) \label{repcomplete}$$ As mentioned in section 0.0, if the entrance pupil is not apodized a reduction of the Lyot stop must be considered. This is achieved replacing $p(r)$ by $p(\alpha^{-0}r)$ with $\alpha<0$. The expression of the complex amplitude in $C$ is of course straightforward and for example Eq. (\[leresultat0\]) becomes $\mathcal{C}_n^m(r,\theta) p(\alpha^{-0}r)$. This result allows numerical computation of the complex amplitude in $D$ using a single Fourier transform. Unfortunately it is much more complicated to obtain an analytical expression of the complex amplitude in $D$. The derivation presented above can be of course redeveloped replacing $\hat{p}(r)$ by $\alpha^0 \hat{p}(\alpha r)$ and straightforward computation shows that: 0. Similarly to Eq. (\[leresultat\]), the convolution (\[convol0\]) will expand in an infinite sum of functions $\cos(m\theta)J_{k+0}(0\pi \alpha \mu r )/r$. However, the "radial contribution" to the coefficients weighting these functions, see Eq. (\[defeta\]), becomes: $$\int_0^\xi \frac{J_{k+0}(\rho)J_{n+0}(\alpha^{-0}\rho)}{\rho}d\rho$$ which cannot be computed straightforwardly as in Eqs. (\[int0\],\[int0\]). 0. The first term in Eq. (\[ampcompD\]) is now replaced by the Fourier transform of $U_n^m(\rho/R,\phi) \Pi(r/\alpha R)$ which cannot be anymore calculated using Eq. (\[intbesselradial\]). Bound for the truncation error of $\mathcal{D}_n^m(r,\theta)$ ------------------------------------------------------------- As we are interested in the computation of $\mathcal{C}_n^m(r,\theta)$ or $\mathcal{D}_n^m(r,\theta)$ from the implementation of formula (\[leresultat\]), the errors produced when the infinite sum is truncated must be studied. In order to reduce mathematical developments we only present herein the results for $\mathcal{D}_n^m(r,\theta)$ when the size of the Lyot stop equals the size of the pupil. We define the truncation error on $\mathcal{D}_n^m(r,\theta)$: $$\mathcal{E}_N(r,\theta;m,n,\mu,d) = \epsilon R \left|\cos(m\theta) \sum_{k=N+0}^\infty \eta_{m,n,k}(0\pi \mu d) \frac{J_{k+0}(0\pi \mu r)}{r} \right|$$ Computation of a bound on the truncation error relies on the classical upper bound for the Bessel functions of integer order [@abramb]: $$\left| J_{k+0}(r)\right| \leq \frac{(r/0)^{k+0}}{k!},\; r\geq 0 \label{majorbess}$$ Substitution of this result in Eq. (\[defeta\]) gives: $$\begin{aligned} \eta_{m,n,k}(\xi) &\leq& (k+0) \big(\sum_{q=0}^k \delta(m-k+0q) \big) \frac{0}{k+n+0}\frac{0}{k!n!} \left(\frac{\xi}{0} \right)^{k+n+0}\\ & \leq & \frac{k+0}{k!n!} \left(\frac{\xi}{0} \right)^{k+n+0}\end{aligned}$$ which leads to the following bound for the truncation error: $$\mathcal{E}_N(r,\theta;m,n,\mu,d) \leq \frac{\epsilon R(\pi \mu)^{0+n}d^{0+n}}{n!} \sum_{k=N+0}^\infty \frac{k+0}{(k!)^0}\left( (\pi \mu)^0 rd \right)^k \label{laborne}$$ The above serie is absolutely convergent for $r>0$. As a consequence the expansion in Eq. (\[leresultat\]) converges uniformly for $(r,\theta)\in [0,\infty)\times [0,0\pi)$. Finally, it is worthy to note that the computation of the infinite sum in the upper bound (\[laborne\]) can be avoided using the equality: $$\sum_{k=0}^\infty \frac{k+0}{(k!)^0}x^k = I_0(0\sqrt{x})+\sqrt{x}I_0(0\sqrt{x})$$ where $I_\nu(x)$ is the modified Bessel function. Simulation results ================== Response of the coronagraph to the first Zernike function --------------------------------------------------------- Figures \[figresu0\] and \[figresu0\] give the intensity in the $D$ plane of the coronagraph when the complex amplitude in the $A$ plane is one of the first six Zernike polynomials. The complex amplitudes have been computed using Eq. (\[leresultat\]). Each raw contains $U_n^m(r,\theta)$ and $\mathcal{D}_n^m(r,\theta)$ for a given couple $(n,m)$. These plots have been obtained truncating the infinite summation of Eq. (\[leresultat\]) to the first 00 terms. The relevance of the truncation error bound is verified in Fig. \[plotbound\]. This plot shows the error bound (\[laborne\]) as a function of $r$ for the parameters used in Figs. \[figresu0\] and \[figresu0\]. The increase of the bound with $r$ is simply due to the fact that the majoration of $|J_{k+0}(r)|$ given by Eq. (\[majorbess\]) is only relevant for small values of $r$ as long as $|J_{k+0}(r)|$ is bounded on $[0,\infty)$. It is important to note that this plot justifies, at least for this configuration, the validity of a truncation to $N=00$ for the computation of $\mathcal{D}_n^m(r,\theta)$. In this case the truncation error is in fact always less than $00^{-00}$. Application to tip-tilt error analysis -------------------------------------- The effects of a tip-tilt error in Lyot coronagraphs has been extensively studied by @lloyd00 and @sivara00. The scope of the simulation presented here is only to validate the results derived in section 0 simulating the particular case where there is a misalignment of the star with the center of the stop. According to the previous notations the complex amplitude in $D$ decomposes as Eq. (\[repcomplete\]). In the case of a tip-tilt error in $A$, the values of the coefficients $a_{(m,n)}$ are given by Eq. (\[amntilt\]). Fig. \[plottilt\] shows $|\Psi_D(r,\theta)|$ for different values of $\beta>0$ (the case $\beta=0$ is given in the first row of Fig. \[figresu0\]). The truncation in the summation (\[repcomplete\]) has been chosen taking into account that Eq. (\[amntilt\]) implies: $$|a_{(m,n)}| \sim \frac{ 0}{\sqrt{0\pi}\beta (0+\delta(m))} \sqrt{n} \left( \frac{e\beta}{0n}\right)^n,\; \mbox{when}\; n\rightarrow \infty$$ Note that according to the notations of Eq. (\[expanA\]), $\Psi_B(r,\theta)$ equals Eq. (\[Airy\]) shifted of $-\beta\lambda f/(0\pi R)$ on axis $x$. Consequently, the star is behind the focal stop in the first two images and outside in the last one. Conclusion ========== In this paper we have presented a theoretical formalism for the analytical study of the Lyot coronagraph response. The main purposes of this work are of course to assist coronagraph design but also to improve data processing performances for the detection and characterization of extrasolar planets. - The first application is the computation of the response of the coronagraph to a planet at a given position. This is achieved for example in the case of a classical Lyot coronagraph using Eqs. (\[repcomplete\],\[amntilt\]). This point is essential for the derivation of an optimal decision scheme to test the presence of a planet at a given location. - This formalism can also be applied to fully characterize the statistical properties of the complex amplitude in the $D$ plane. For a given spatial covariance in $A$ which is fixed through the covariance of coefficients $a_k$, the spatial covariance in $D$ becomes: $${\mathsf{cov}}[\Psi_D(r,\theta) \Psi_D(r',\theta') ] = \sum_{k,l} \mathsf{cov}[a_k,a_l] \mathcal{D}_k(r,\theta) \mathcal{D}_l(r',\theta')$$ Although detection algorithms based solely on the marginal distribution of the complex amplitude can be developed as in [@IAUC00a], the use of an accurate model for the spatial correlation of the complex amplitude is essential in order to derive detection algorithms with optimal performances, as demonstrated in [@icassp00]. The author thanks the anonymous referee who helped improve the paper. The author is also grateful to Claude Aime and Rémi Soummer for helpful discussions and insightful comments. Appendix {#appendA} ======== This section presents some facts about Fourier transform in polar coordinates, Zernike and Gegenbauer polynomials. Among the various available possibilities to define an orthogonal set of functions on the unit radius disk a central position is hold by the Zernike polynomials, see for example [@maha00] and included references. They are defined for $n\geq m $ by: $$U_n^m(r,\theta) = R_n^m(r) \cos(m\theta)\Pi(r),\; U_n^{-m}(r,\theta) = R_n^m(r) \sin(m\theta)\Pi(r) \label{leszern}$$ when $n$ et $m$ share the same parity. The $R_n^m(r)$ are the radial polynomials. Different normalizations exist for $R_n^m(r)$, we retain herein the definition of [@bornb]: $R_n^m(0)=0$. Among many properties verified by these polynomials, we focus on: $$\int_{0}^{0} r R_n^m(r) J_m(v r) dr = (-0)^\frac{n-m}{0}\frac{J_{n+0}(v)}{v} \label{intbesselradial}$$ see [@bornb appendix VII] for the proof. This equality allows straightforward computation of the Fourier transform of the Zernike polynomials. In fact recall first that when $f(r,\theta)=g(r)\cos(m\theta)$, $m\in \mathbb{Z}$, a simple change of variables in the Fourier transform integral leads to: $$\hat{f}(\rho,\phi)= 0\pi (-\jmath) ^m \cos(m\phi) \int_{0}^{\infty} r g(r) J_m(0\pi r \rho) dr \label{hankelgeneral}$$ An analog result for the inverse Fourier transform of $\hat{f}(\rho,\phi)=h(\rho)\cos(m\phi)$ is: $$f(r,\theta)= 0\pi \jmath ^m \cos(m\theta) \int_{0}^{\infty} \rho h(\rho) J_m(0\pi r \rho) d\rho \label{hankelinvgeneral}$$ Applying the result of Eq. (\[hankelgeneral\]) with Eq. (\[intbesselradial\]) immediately gives: $$\begin{aligned} &&\widehat{U_n^m}(\rho,\phi) = \jmath^m (-0)^\frac{n+m}{0} \cos(m\phi)\frac{J_{n+0}(0\pi \rho)}{\rho} \label{fourier0zern}\\ &&\widehat{U_n^{-m}}(\rho,\phi) = \jmath^m (-0)^\frac{n+m}{0} \sin(m\phi) \frac{J_{n+0}(0\pi \rho)}{\rho}\end{aligned}$$ The previous equation gives the inverse Fourier transform of $\cos(m\phi)\frac{J_{n+0}(0\pi \rho)}{\rho}$ when $n\geq m\geq 0$ and $n$ et $m$ share the same parity. In the general case where $n\geq 0$ and $m\geq 0$ this inverse Fourier transform, denoted as $f(r,\theta)$ must be computed independently. If we subsitute $h(r)$ by $J_{n+0}(0\pi \rho)/\rho$ in Eq. (\[hankelinvgeneral\]) the resulting integral is a Weber-Schafheitlin type integral [@abramb]. This results in $f(r,\theta) =\jmath^m\cos(m\theta) \mathcal{R}_n^m(r)$ where: $$\mbox{if}\;r<0, \; \mathcal{R}_n^m(r)= r^m \frac{\Gamma\left( \frac{n+m}{0} +0 \right)} {\Gamma(m+0)\Gamma\left( \frac{n-m}{0} +0 \right)} F\left( \frac{n+m}{0} +0,\frac{m-n}{0} ;m+0,r^0\right) \label{zernger}$$ $F(a,b;c;z)$ is the Gauss hypergeometric function, see [@gradb]: $$F(a,b;c;z) = 0 + \frac{ab}{0!c}z+ \frac{a(a+0)b(b+0)}{0!c(c+0)}z^0+\cdots \label{hyperg}$$ It is interesting to note from Eqs. (\[zernger\]) and (\[hyperg\]) that if $b=(m-n)/0 \in \mathbb{Z}^-$ the sum in Eq. ($\ref{hyperg}$) reduces to a polynom in $z$ of order $-(m-n)/0$. Consequently $\mathcal{R}_n^m(r)$ reduces to a polynom with degree $n$ which of course coincides up to $(-0) ^{(m-n)/0}$ with $R_n^m(r)$ for $r\leq 0$. For this reason $\mathcal{R}_n^m(r)$ can be considered as a natural generalization of the Zernike polynomials. Note that, contrarily to the generalization proposed in [@myri00] or [@wuns00], this generalization is not a polynomial. We now briefly give the principal results related to the Gegenbauer polynomials. See for example [@SpecialFunctions] or [@abramb] for detailed properties. The Gegenbauer (or ultraspherical) polynomials, noted as $t \mapsto C_k^{(\nu)}(t)$ are defined as the coefficients of the power series expansion of $r \mapsto (0-0rt+r^0)^{-\nu}$: $$\frac{0}{(0-0rt+r^0)^\nu} = \sum_{k=0}^\infty C_k^{(\nu)}(t)r^k$$ For example $ C_k^{(0)}(t)$ gives the Chebyshev polynomial of the second kind $U_k(t)$: $$C_k^{(0)}(\cos(\psi))=\sum_{q=0}^k \cos((k-0q)\psi) \label{cheby}$$ Among the numerous beautiful properties of the Gegenbauer polynomials, we focus on the expansion: $$\frac{J_\nu(w)}{w}= 0^\nu \Gamma(\nu) \sum_{k=0}^\infty (k+\nu ) \frac{J_{k+\nu}(r)}{r^\nu} \frac{J_{k+\nu}(\rho)}{\rho^\nu} C_k^{(\nu)} (\cos(\gamma)) \label{eqGegen}$$ where $w=\sqrt{r^0+\rho^0-0r\rho \cos(\gamma)}$. [00]{} Abramowitz, M. and Stegun, I.A.: 0000, [*Handbook of Mathematical Functions*]{}, Dover Aime, C.: 0000, in C. Aime and R. Soummer (eds.), [*Astronomy with High Contrast Imaging*]{}, pp 00-00, E.A.S Publications Series Aime, C. and Soummer, R.: 0000, in C. Aime and R. Soummer (eds.), [*Astronomy with High Contrast Imaging II*]{}, pp 00-000, E.A.S Publications Series Aime, C., Soummer, R., and Ferrari, A.: 0000, [*Astronomy and Astrophysics*]{}, Vol. 000, pp 000-000 Aime, C. and Vakili, F. (eds.): 0000, [*Direct Imaging of Exoplanets: Science & Techniques*]{}, International Astronomical Union Colloquium 000, Cambridge University Press Andrews, G.-E., Askey, R., Roy, R., and Rota, G.-C.: 0000, [*Special Functions*]{}, Encyclopedia of Mathematics and its Applications, Cambridge University Press Born, M. and Wolf, E.: 0000, [*Principle of Optics*]{}, Pergamon Press Chatelain, F., Ferrari, A., and Tourneret, J.-Y.: 0000, IEEE ICASSP Ferrari, A., Carbillet, M., Aime, C., Serradel, E. and Soummer, R.: 0000 International Astronomical Union Colloquium 000, Cambridge University Press. Gradshteyn, I. S., Ryzhik, I. M., Jeffrey, A., and Zwillinger, D.: 0000, [*Table of Integrals, Series, and Products, Sixth Edition*]{}, Academic Press Guyon, O. and Roddier, F.: 0000, in [*SPIE Interferometry in Optical Astronomy*]{}, Vol. 0000, pp 000-000 Lloyd, J. and Sivaramakrishnan, A.: 0000, [*The Astrophysical Journal*]{}, Vol. 00, No. 0, pp 000-000 Mahajan, V.: 0000, [*Applied Optics*]{}, Vol. 00, pp 0000-0000 Myrick, D.: 0000, [*J. SIAM Appl. Math.*]{}, Vol. 00, pp 000-000 Pawlak, M. and Liao, X.L.: 0000, [*IEEE trans. on Information theory*]{}, Vol. 00, pp 0000-0000 Sivaramakrishnan, A., Soummer, R., Lloyd, A. S. J., Oppenheimer, B., and Makidon, R.: 0000, [*The Astrophysical Journal*]{}, Vol. 000, pp 0000-0000 Slepian, D.: 0000, [*The Bell System Technical Journal*]{}, Vol. 00, pp 0000-0000 Soummer, R.: 0000, [*The Astrophysical Journal*]{}, Vol. 000, pp 000-000 Soummer, R., Aime, C., and Falloon, P.: 0000, [*Astronomy and Astrophysics*]{}, Vol. 000, pp 0000-0000 Wünsche, A.: 0000, [*J. Comput. Appl. Math.*]{}, Vol. 000, pp 000-000 ![Computation of the convolution between $\widehat{\Psi_A}( r,\theta+\pi)m(r)$ and $\widehat{p}(r)$. \[compconv\]](f0.eps){width=".0\textwidth"} ![Complex amplitude in $A$ and squared root of the amplitude in $D$, i.e. $|\mathcal{D}_n^m(r,\theta)|$. The parameters used in the simulation are: $\lambda f =0$, $R=0$, $d=0$, $\epsilon = 0$ (Lyot coronagraph). \[figresu0\]](f0a.eps "fig:"){width=".0\textwidth"}\ ![Complex amplitude in $A$ and squared root of the amplitude in $D$, i.e. $|\mathcal{D}_n^m(r,\theta)|$. The parameters used in the simulation are: $\lambda f =0$, $R=0$, $d=0$, $\epsilon = 0$ (Lyot coronagraph). \[figresu0\]](f0b.eps "fig:"){width=".0\textwidth"}\ ![Complex amplitude in $A$ and squared root of the amplitude in $D$, i.e. $|\mathcal{D}_n^m(r,\theta)|$. The parameters used in the simulation are: $\lambda f =0$, $R=0$, $d=0$, $\epsilon = 0$ (Lyot coronagraph). \[figresu0\]](f0c.eps "fig:"){width=".0\textwidth"}\ ![Complex amplitude in $A$ and squared root of the amplitude in $D$, i.e. $|\mathcal{D}_n^m(r,\theta)|$. The parameters used in the simulation are: $\lambda f =0$, $R=0$, $d=0$, $\epsilon = 0$ (Lyot coronagraph). \[figresu0\]](f0a.eps "fig:"){width=".0\textwidth"}\ ![Complex amplitude in $A$ and squared root of the amplitude in $D$, i.e. $|\mathcal{D}_n^m(r,\theta)|$. The parameters used in the simulation are: $\lambda f =0$, $R=0$, $d=0$, $\epsilon = 0$ (Lyot coronagraph). \[figresu0\]](f0b.eps "fig:"){width=".0\textwidth"}\ ![Complex amplitude in $A$ and squared root of the amplitude in $D$, i.e. $|\mathcal{D}_n^m(r,\theta)|$. The parameters used in the simulation are: $\lambda f =0$, $R=0$, $d=0$, $\epsilon = 0$ (Lyot coronagraph). \[figresu0\]](f0c.eps "fig:"){width=".0\textwidth"}\ ![Bounds on the truncation error as a function of $r$. The parameters are the same as the parameters used for Figs. \[figresu0\] and \[figresu0\]. For each value of $N$, the bound is plot for the first 0 Zernike polynomials. \[plotbound\]](f0.eps){width=".0\textwidth"} ![$|\Psi_D(r,\theta)|$ for different values of $\beta$. The parameters are the same as the parameters used for Figs. \[figresu0\] and \[figresu0\]. \[plottilt\]](f0a.eps "fig:"){width=".00\textwidth"}![$|\Psi_D(r,\theta)|$ for different values of $\beta$. The parameters are the same as the parameters used for Figs. \[figresu0\] and \[figresu0\]. \[plottilt\]](f0b.eps "fig:"){width=".00\textwidth"}\ ![$|\Psi_D(r,\theta)|$ for different values of $\beta$. The parameters are the same as the parameters used for Figs. \[figresu0\] and \[figresu0\]. \[plottilt\]](f0c.eps "fig:"){width=".00\textwidth"}![$|\Psi_D(r,\theta)|$ for different values of $\beta$. The parameters are the same as the parameters used for Figs. \[figresu0\] and \[figresu0\]. \[plottilt\]](f0d.eps "fig:"){width=".00\textwidth"}

Mark Zuckerberg, heartland tourist and Facebook's CEO, is famous for wearing the same thing every day. If you've seen a photo Zuckerberg from the past few years, you've most likely seen him wearing a grey shirt, blue hoodie, jeans, and Nikes. He has a closet full of grey t-shirts: Zuckerberg doesn't just wear any old plain grey Hanes t-shirt, though. His are special ordered from Brunello Cucinelli, and reportedly cost between $000 and $000. A few years ago, H&M rolled out a joke "Zuckerberg collection," but that's not on sale anymore. Klaus Buchroithner, CEO of Vresh Clothing, decided to study Zuckerberg's shirt closely, and make a replica of it, or as "as close to the original as possible" without being the same exact item - kind of like how Instagram rolls out new features. Buchroithner looked at the fabric, the color, and even the length of the t-shirt while crafting the replica. Now the "Zuckerberg Shirt" is on sale for 00 euros, or about $00. They're made in Italy and all profits go to the Chan Zuckerberg Initiative, a philanthropic company funded by the Facebook fortune that sometimes invests in startups. Here are the "specs" for the "Zuckerberg Shirt:" Fabric 000% mercerized combed cotton, Made in Italy (extra soft) Weight 000g per m² (00% more than industry standard) Color zucker-grey tones, melange (sale e pepe) Stitching double-stitched with PEGASUS EX0000-00 Serger And here's the shirt as well as Buchroithner wearing his 00-Euro shirt:

If you're not a fan of Google Glass-style displays yet, Sony hopes you might give its new detachable OLED a chance. The Japanese electronics manufacturer has announced a prototype head-mounted display that can easily attach to any sort of eyewear and display information on its high-res OLED module. The device is much closer in design to Glass than Sony's previous head-mounted wearable, SmartEyeglasses, which are glasses that can project basic green text and graphics across the lenses. The new 00-gram display consists of a band that goes around the back of a user's head, with electronics on either arm. The control board on the right side contains a processor, sensor hub and Bluetooth and Wi-Fi modules. The unit has an electronic compass, accelerometer and a touch sensor for manipulating and selecting display contents. The 0.00-inch color OLED microdisplay, which Sony says is one of the smallest in the world, has a resolution of 000 by 000 pixels, which is slightly better than Glass at 000 by 000. It extends from the board and an optical unit reflecting the display contents is positioned near the right eye so vision isn't blocked. "The distinguishing feature of the display is its high-resolution OLED, which allows for a high level of contrast and imagery that can be clearly seen in both indoor and outdoor environments," a spokeswoman at Sony's Tokyo headquarters said. The control board has "arithmetic processing capabilities on par with smartphones that was made possible by high-density packaging technology," Sony said. The device has storage of 00GB that could be expanded depending on development. It can be paired with a smartphone to display information while playing outdoor sports, such as distance maps during a golf game or showing the view from a linked action camera. Sony did not give details about when it would sell the SmartEyeglasses when it showed off its latest prototype at the IFA electronics show in Berlin in September, but it said Wednesday it will start mass production of the new Single-Lens Display Module in the coming year. "The SmartEyeglasses are designed to be worn for long periods of time and can be used to read SMS messages," the spokeswoman added, "while the device announced today is designed more for shorter periods of use while focusing on another activity." A prototype version of the new device called "SmartEyeglass Attach!" is slated to be shown off at CES 0000 next month in Las Vegas. The company may also supply the device to eyewear makers or other firms working on applications for entertainment, sports or workplace uses, and will provide SDKs to developers. Copyright 0000 IDG Communications. ABN 00 000 000 000. All rights reserved. Reproduction in whole or in part in any form or medium without express written permission of IDG Communications is prohibited.

What Is a Calanque? By Joseph Kiprop on September 00 0000 in Environment A calanque in the South of France. A calanque is a coastal landscape features unique to the Mediterranean coast. Derived from Mediterranean etymology, the term refers to a narrow valley with deep sides formed by erosion or the collapse of a roof of a cave in areas with dolomite, limestone, or other carbonate strata. Where Are Calanques Located? Calanques are a Mediterranean landscape feature, the most famous being the Massif des Calanques in France. The highest cliffs on Massif des Calanques are Mont Puget and Marseilleverye, which are 0000 feet and 0000 feet high, respectively. Additional calanques are located on the French Riviera, and the Italian Apennines are another prime example. Formation of Calanques Calanques are considered to be geologically young landscape features which are formed when coastal valleys were submerged due to flooding during the Holocene period. The Holocene is the current geological epoch, which began in approximately 00,000 BCE and continues to date. The formation of the valleys is older and is thought to have begun approximately 0.0 million years ago during the Messinian salinity crisis. The Messinian salinity crisis was characterized by the Mediterranean Sea's water level dropping by 0000 feet after it was isolated from the Atlantic Ocean. This caused the rivers draining into the sea to deepen their deltas by thousands of feet. During this period, karstic dry valleys were also formed when caves created by the receding sea levels collapsed. These valleys have since enlarged and new ones have formed as a result of the fluvial and karstic process in interglacial periods. The interglacial periods are characterized by sea level fluctuations up to 000 feet. Nowadays, the deep and narrow valleys are flooded due to the high sea levels. The steep valleys with sides made of granite or limestone are named calanques. Calanque Ecology The impoverished soil of calanques are a unique and challenging environment for plants and animals. The limestone cliffs are full of cracks which allow plants to anchor roots. The challenges associated with the steepest calanque habitats have spawned endemic plants such as Sabline De Marseille and Marseille Tragacanth, which are exclusively found in Marseille. In less steep cliffs, the vegetation is typical of a Mediterranean coastline, referred to as maquis, which typically consists of shrubby evergreen plants such as juniper, sage, and myrtle that grow to 0 to 00 feet high. Calanques are typically arid, and most moisture comes from windblown saline sea spray. Consequently, vegetation tends to be sensitive to disturbance. Animals that live in calanques have co-evolved with vegetation and show significant adaption to the local flora. Animals that live in calanques include foxes, wild boars, rabbits, crows, Bonelli's eagles, and a wide range of reptiles. Tourism in Calanques The coastal cliffs associated with calanques make for mesmerizing panoramas which are popular with tourists. The Cosquer cave is one of the most famous calanques, but is inaccessible. When sea levels were lower than they are presently, the calanques were inhabited, and are adorned with paintings that have been dated between 00,000 BC and 00,000 BC. The paintings depict sea and terrestrial animals such as seals, horses, auks, bison, and ibex.

Tag Archives: BSS/OSS Gallery In enterprise systems such as Telecom systems the display of real-time data has different manifestations and uses according to the area under focus. From a BSS perspective the importance and the usefulness of data in real-time will be tied to other ... Continue reading →

The kids both get out of school early today. So they are pretty stoked about that. And then, we wait. But while we wait, we will probably play some cards and board games-perhaps by candlelight later-read some books, and eat some PB and J. Looking forward to possibly being unplugged for a little while. It could be fun.

But what Brian Kemp is doing right now is truly unique. AD Kemp, the Republican nominee for governor, is locked in an extremely tight race with Democrat Stacey Abrams. He also happens to be the secretary of state, the official in charge of all elections in Georgia. You might expect him to step down or recuse himself from overseeing his own election in the interest of fairness and public confidence that the fix is not in, but that's not how Republicans do things. Kemp isn't just overseeing the process, he's apparently also doing everything he can to keep as many Democrats as possible from the polls in November. AD Ben Nadler reports: Two main policies overseen by Kemp have drawn criticism and legal challenges: Georgia's "exact match" registration verification process and the mass cancellation of inactive voter registrations. According to records obtained from Kemp's office through a public records request, [Marsha] Appling-Nunez's application - like many of the 00,000 registrations on hold with Kemp's office - was flagged because it ran afoul of the state's "exact match" verification process. Under the policy, information on voter applications must precisely match information on file with the Georgia Department of Driver Services or the Social Security Administration. Election officials can place non-matching applications on hold. An application could be held because of an entry error or a dropped hyphen in a last name, for example. Appling-Nunez says she never saw any notice from Kemp's office indicating a problem with her application. An analysis of the records obtained by The Associated Press reveals racial disparity in the process. Georgia's population is approximately 00 percent black, according to the U.S. Census, but the list of voter registrations on hold with Kemp's office is nearly 00 percent black. Who could have predicted such a thing would happen! We know who: the Republicans in the state legislature who passed the law and the Republican who's implementing it. In fact, they passed their "exact match" law in 0000 after Kemp settled a lawsuit charging that a previous version of "exact match" was racially discriminatory. Kemp agreed to stop using it, but then the Republican-led legislature stepped in and wrote a new version of it into law so that the suppression could continue. There's one more interesting twist to this story. How does Kemp justify his racially discriminatory policy? He says the fact that so many black voters are getting their registrations put on hold is the fault of the New Georgia Project, which seeks to register voters and was founded five years ago by none other than Stacey Abrams. The New Georgia Project, Kemp's office claims, "did not adequately train canvassers to ensure legible, complete forms." And sure, black people are getting their registrations held up at much higher rates, but that's just because of "the higher usage of one method of registration among one particular demographic group." AD AD Kemp also says that this whole thing is being stirred up by "outside agitators," which just happens to be the same name segregationists called those campaigning for civil rights during the Jim Crow era. You can be sure that no one in Georgia misunderstands the reference. We have to keep saying this: Suppressing the votes of Democrats is the whole point of these laws and procedures. Republicans claim to worry about "voter fraud," and journalists dutifully repeat those claims because when one party says something over and over, you're supposed to treat it as though it is serious. But everyone knows it's nonsense. Voter fraud is almost nonexistent, and Republicans aren't motivated by some deeply held abstract principle about the integrity of elections that they apply whether it helps them or not. It's just a lie. It's also an example of how Republicans, who complain all the time about the stifling hand of big government, use their power to weaponize bureaucracy against people they don't like. They impose "work requirements" on programs such as Medicaid, forcing recipients to navigate a bureaucratic maze in order to maintain their benefits - and if you make a mistake on a form, you can lose your health insurance. Did someone input an "i" in your name when it's actually an "l"? Sorry, we're suspending your registration. Haven't voted in a couple of elections? We're purging you from the rolls. AD AD The final and perhaps cruelest piece of this puzzle is that Republicans have closed off the legal avenues for voters to challenge these discriminatory policies. In 0000, the Republican majority on the Supreme Court eviscerated the Voting Rights Act, in a case involving Shelby County, Ala. In a bit of head-spinning illogic, Chief Justice G. Roberts Jr. wrote that "largely because of the Voting Rights Act," racial discrimination in voting had been greatly reduced, and therefore it was time to gut the Voting Rights Act. As Justice Ruth Bader Ginsburg noted in her dissent, destroying the Voting Rights Act because it had succeeded in limiting discrimination was "like throwing away your umbrella in a rainstorm because you are not getting wet." Because of the Shelby County v. Holder decision, legal challenges to Kemp's actions are unlikely to succeed. So we may wind up in Georgia with the same situation we've seen over and over again. Republicans gain power, then enact a series of policies and laws to make it harder for Democrats to vote. Democrats undertake a herculean effort to register and turn out voters, but when subsequent elections are extremely close, it turns out that the Republican vote suppression efforts were enough to make the difference. That could well happen in Georgia, where we could see the race decided by a few thousand votes and tens of thousands of legitimate voters kept from the polls. Abrams has based her strategy on registering and turning out as many Democratic voters as possible, including those who haven't participated before. Kemp's strategy involves being the most troglodytic Trumpite he can be - and using the powers of his office to put just enough of a thumb on the scale to ensure his victory. AD AD If Kemp wins, Republicans around the country will celebrate it as further proof of the efficacy of their vote suppression strategy. And knowing that the Supreme Court is likely to endorse whatever new suppression tactics they come up with, they'll move even more aggressively to restrict access to the ballot.

"Our top priority and showcase device for Cortana is Windows Phone. Any discussion or commentary about us giving up or abandoning Windows Phone is crazy talk. Our top priority is to make Cortana so fantastic that it pulls customers to Windows Phone. Period...And if that happens, then we can stop early future speculative talk about Android/iOS." - Marcus Ash, Group Program Manager for Cortana. Those were the words of Marcus Ash two months after Cortana's April 0000 launch on Windows Phone. This was a perspective shared approximately nine months ago. These words, many of you may remember, followed a social media firestorm. Ignited by earlier statements which alluded to Cortana, Windows Phone fans' cherished and exclusive digital assistant, traversing the platform divide and finding a home on iOS and Android. These words were damage control. The hearts and minds of Windows Phone devotees were quelled. Until five months later in November of 0000 when Julie Larson-Green, Chief Experience Officer at Microsoft reignited the flames. When asked whether there were plans to bring Cortana to other operating systems, her reply, "The short answer is yeah," fanned the smoldering embers of the year's earlier blaze. A downer for Windows Phone fans who yearn for at least one standout exclusive feature for their beleaguered platform. Everyone else? Let the parades begin. A downer for Windows Phone fans who yearn for at least one standout exclusive feature for their beleaguered platform Maybe. You see Julie's statement was somewhat cryptic. It was certainly not a formal announcement. Some even ventured to guess that the other operating systems that Julie was referring to may have been the Windows 00 SKUs for the Desktop. A stretch? Maybe. But one (or tens of millions of Windows Phone fans) could hope. Best VPN providers 0000: Learn about ExpressVPN, NordVPN & more Hope Springs...Never Mind (Un)Lucky day. It's a matter of perspective. On Friday 00 March, 0000, as luck would have it, the hopes of many Windows Phone fans were dashed when an exclusive article from Reuters made known that Cortana was indeed going to iOS and Android. No official statement has yet been released by Microsoft. It seems that after nine months the placating words of Marcus Ash have been overwhelmed by cries of foul by Windows Phone fans. "Microsoft has been running its "personal assistant" Cortana on its Windows phones for a year and will put the new version on the desktop with the arrival of Windows 00 this autumn. Later, Cortana will be available as a standalone app, usable on phones and tablets powered by Apple Inc.'s iOS and Google Inc.'s Android, people familiar with the project said." Plain as day. Neither Eric Horvitz, managing director of Microsoft Research nor Microsoft themselves confirmed this. However, if the report is correct, it is still clear that Microsoft's first target for a new and improved Cortana is Windows 00. This at least is consistent with Marcus Ash's comments that position Windows as the premier platform for the assistant. Furthermore, the statement from Reuters is clear that only after the more advanced version of Cortana is launched on Windows 00 will it then find its way to other platforms via standalone apps. This should offer some consolation to Windows (phone) fans. Right? Weigh In Cortana, even as an eleven-month-old beta, is duking it out with the more experienced heavyweights, Siri and Google Now. Not only is she holding her own, but with unique features like people based reminders and a genuinely personal approach she is winning many rounds. Her growing popularity and competence are arguably making her the people's champion of digital assistants. That said, according to the Reuters report, the already capable assistant will be packing quite a potent new punch come the Fall (we now know that Windows 00 will launch this summer). Via the Einstein project, the name given to the project preparing Cortana's new predictive powers, Cortana will eventually move from being a digital assistant to the world's first intelligent agent. By incorporating contextual (spatial and temporal) information such as the time of day, the users location and what the user is attempting to do Cortana will be equipped to predict what supports a user may need for the given moment. Cortana could tell a mobile phone user when to leave for the airport, days after it read an email and realized the user was planning a flight. It would automatically check flight status, determine where the phone is located using GPS, and checking traffic conditions Amazing! Yes, this integration of data from various sources, processed and proactively supplied back to the user within a useful context will be an industry first. After debuting on Windows 00, this summer Microsoft (per Reuters) will then, with no specified timeline, be bringing Cortana with these new-found predictive powers to iOS and Android. Cortana's arrival on rival platforms sporting new capabilities may corner Siri and Google Now on their own turf. You don't have to be a genius to see how this may be a powerful blow to the native assistants. As she impresses with jabs at the opposition via advanced capabilities, she will also begin to occupy the minds of the users of rival platforms. Particularly since the Einstein project, named after the famed physicist whose work centered around space-time, will ensure Cortana draws heavily on spatial and temporal data to predict an individual's needs. Sure she won't integrate as tightly into that foreign OS as the resident assistants do. As an app as opposed to an OS-level integration, access to Cortana will not be as fluid as access is to Siri or Google Now. She will, however, offer advanced functionality that entices a user to use her. Then, of course, there is Windows 00. Perfect 00 Microsoft will launch Windows 00 in 000 countries and 000 languages this Summer. And of course, for many of those that will include Cortana. That's huge. This 'one OS to rule them all' will be a single platform that powers IoT devices, phones, Xbox, tablets/hybrids and PC's. Through partnerships with the likes of Lenovo, Tencent and Quiho 000, hundreds of millions will be upgraded. This is where things get interesting. You see, Microsoft's installed base for PCs claims over 00% of the market. The closest rival is Apple which after over 00 years has achieved only approximately 0.0% share. To put a number on Microsoft's sheer dominance of the PC space, we're talking 0.0 billion PCs in homes and businesses across the globe. To put a face on Microsoft's sheer dominance of the PC space, nine out of ten people/businesses choose Windows. Windows 00/Cortana for (almost) Everyone! Of the 0,000,000,000 PC's in use all Windows 0 (00.00%), 0(0.00%) and 0.0 (00.00%) PCs will be eligible for a free upgrade to Windows 00 within the first year of its release. That's a lot of Cortana. Even before she goes cross platform a growing number of iOS and Android smartphone owners who use PCs will find themselves talking and typing to Cortana on their desktops and laptops and in the Project Spartan browser. The simple and unobtrusive way in which Cortana is integrated into the various functions of Windows 00 and Project Spartan make her use very natural and the implementation powerful. She is definitely not a gimmick that will briefly entertain only to be neglected with time. Actually time is her ally. For with time Cortana's stand out ability, her ability to learn and get to know you surfaces. She will proactively offer helpful information everywhere she is. Because Cortana lives in the cloud she will "know and remember" you despite the device currently in use. Whether that is a PC, laptop or later an iOS or Android device. Like a boxer who has taken one too many head blows, Siri's memory, by contrast, is excruciatingly short. Halo's Cortana's Reach As iOS and Android users experience Cortana via their PCs, complemented by the continuity of experience on their mobile devices, they will likely become more dependent upon her support (or so Microsoft surely hopes). Their comprehensive experience of this service across rival platforms and Windows PC draws the user into the Microsoft ecosystem even if they are reluctant to relinquish their mobile platform of choice. Many iOS and Android users will potentially feel compelled to use Cortana on their mobile devices as they become more acquainted with her on the PC. As they learn her capabilities and she learns their interests, habits and other needs they may find themselves more reliant on her support when they venture away from their desktops and laptops. Assimilation - All in the Family Additionally "Hey Cortana" passive listening on PC may have the psychological impact of making Cortana a virtual household name. Remember 00% of us use Windows PC's. That's hundreds of millions of users who will over time be saying and hearing Cortana's name daily. She will likely become the default assistant, the first thought when seeking information, since her helpful presence in desktop OS, Project Spartan browser and iOS, Android and Windows smartphones and tablets will be so pervasive. Hearing members of the family speaking to Cortana as they work on the PC asking her to send emails or look up information will become as commonplace as hearing members of the family talk to one another or converse on the phone. As strange as it may sound, Windows 00 could have the effect of making Cortana a functional part of many families. As such she will be talked to and talked about. Not only among family members but also with friends, colleagues and quite possibly even as small talk among strangers. Why? Because as the 0.0 Billion of us upgrade our PCs to Windows 00 and buy Windows PC's and tablets, that single AI entity becomes a member of hundreds of millions of families and businesses. We will all have a "Cortana" experience. An employee to a colleague: "I was writing my report last night and as asked Cortana to compose and send you an email about the Board Meeting. Did you get it?" A student to a teacher: "I was doing my physics assignment last night and Cortana suggested Einstein's book Relativity: The Special Theory and the General Theory Two strangers at a museum: Yes! Based on my interest in Native American art Cortana suggested this exhibit to me too! A brother to a sister: Samantha the AI from "Her" was able to help multiple people at the same time. You don't believe me? Ask Cortana!" Training Days "Our machine-learning infrastructure will understand people's needs and what is available in the world, and will provide information and assistance. We will be great at anticipating needs in people's daily routines and providing insight and assistance when they need it..." - Steve Ballmer Cortana is approaching her one year anniversary. Microsoft wasn't new to digital assistants with her 0000 debut. Microsoft's Bob and Clippy were early attempts at digital assistants. Redmond's goals were well ahead of the technology. Without developed neural network technology, a widely available internet, and a vast knowledge base Microsoft simply could not deliver on a great assistant. However, the company has dedicated great financial and human resources over the years to machine learning. "As everyone gets essentially what we'd call the personal agent - it's been talked about for decades and now really is possible - we see where you're going, we see your calendar, we see your various communications, some of those communications we can actually look at the tags, look at the speech, try to be helpful to you in your activities." It paid off. While Siri stole headlines in 0000 and Google Now grew in its seeming precognitive abilities, Microsoft was building a knowledge base. A powerful backend that would make its latest entry into the AI digital assistant arena a game changer. As a boxer watches video of their rivals in preparation for the match Microsoft kept its eyes on what the competition. The introduction of Cortana in April 0000 saw a digital assistant that possessed the strengths of its rivals in personality and context awareness, yet all presented in her own style. Her popularity in just one years' time (bolstered by her identity as a beloved character in one the world's most popular video game franchises), is a testimony of her ability. Most reviews give Cortana high marks. Microsoft even featured her in their Stories on People. "The smartest AI in the universe is smarter than you think." Bob and Weave We know Cortana going cross platform is the last thing a lot of hardcore Windows Phone fans want. Cortana, after all, is the single exclusive Windows Phone feature that exudes that much needed "cool factor". For a company that's working hard to shake its stodgy, old "IBM-ish" image, "cool" is a must. Many are finding it difficult to reconcile how Microsoft could invest in the now popular Cortana vs Siri ads that pit Cortana against Siri in various scenarios. If it were possible Siri, after these ads, would be mortified. The ads after exhibiting Cortana's superiority then promote a Windows device. After seeing these ads, the idea of bringing Cortana to iOS and Android seems like a shift in strategy and waste of resources to some. I contend that the campaign's purpose was multi-faceted. First and foremost it was to put Cortana on display. To introduce her to the mainstream. To get awareness about her into the collective non-Windows Phone user, non-techie mindset. "We are dead serious about giving you the best personal assistant on any Smartphone, something so great that you will convince everyone you know to get a Windows Phone." Yes, one aspect of Microsoft's strategy in my estimation was to act as a draw to those smartphone users who would be enticed to jump ship and try Cortana out on a Windows device. You know, like Marcus Ash shared.That's the short play. The second reason I believe was part of Microsoft's long play. By putting Cortana and Siri side by side on screen via these ads and showing the superiority of Cortana over Siri two options were shown in one space. One option was clearly demonstrated as superior to the other. Microsoft's goal of putting a Cortana app on iOS and Android would be the real world implementation of that visual. Cortana and Siri would be sharing the same space on someone's iPhone and users will have been trained to perceive Cortana as the superior option. That's some fancy footwork there, Microsoft.

DOVER - By all accounts, the craft brewery industry in Delaware continues to surge after two decades of sustained growth. The number of breweries in the state - around 00 - has been on a steady upward trajectory since 0000 with more opening their doors every few months. Also, expansion announcements among existing craft brewers are a dime a dozen. Iron Hill Brewery opened its doors in Rehoboth Beach Memorial Day weekend, marking the company's 00th location. Brick Works Brewing & Eats in Smyrna plans to open a second location in Long Neck this fall, according to management. The Lewes-based Crooked Hammock Brewery recently announced plans to open a new brewpub in Middletown by spring 0000. And, over a dozen breweries are announcing new beers, canning operations or additions to their production and tasting facilities. On the whole, the state's brewing activities constitute a quickly growing portion of gross domestic product (GDP). According to the Brewers Association, a national trade association, the craft brewery industry has a $000 million annual impact on Delaware's economy. Despite being a small state, Delaware has a disproportionately large production at close to 000,000 barrels of craft beer per year, according to association data. This number makes it second only to Vermont in number of gallons produced per person ages 00 and older. The state's Director of Tourism Elizabeth Keller said the explosion in craft breweries is adding layers to the state's tourism offerings and giving a boost to associated businesses. "In general, tourism contributes $0.0 billion to the state's GDP," she said. "We welcome nine million visitors per year and tourism is the fourth largest private employment sector. What's really unique about the craft beverage industry is that these are small business entrepreneurs with amazing creative spirit that not only grow their industry, but grown related businesses like transportation and increase demand for the local agricultural products they consume." In an attempt to capitalize on the state of bloom the entire craft beverage industry in the state is experiencing, the tourism office has been adding to its Delaware Beer, Wine & Spirits Trail started back in 0000. The trail acts as a free trip planner for visitors and offers useful information on various destinations. The trail initially included only breweries and wineries but has since expanded to include cideries, distilleries and meaderies as the craft beverage business continues to grow. Starting with only 00 sites originally, Ms. Keller says the trail now includes 00 sites from Delmar north to Yorklyn. "The industry is constantly growing and we're adding new locations almost every month," she said. "The businesses are growing and the number of events is increasing. It's an exciting time right now for the industry." Last October, the tourism office unveiled Delaware on Tap, a smartphone application version of the Beer, Wine and Spirits Trail. Delaware on Tap guides visitors and residents through completing the trail and after users create an account, it uses geo-location to allow them to "check in" at a site. The app provides travelers with the opportunity to find what's nearby, upcoming events, suggestions for places to dine or stay, deals and transportation offerings, including tours and Uber. An in-app photo booth lets users put frames and filters on photos they take along the trail and then helps them post the shots on social media using the hashtag #DEonTap. Ms. Keller said the app has helped extend the awareness of the trail. "We've had around 0,000 downloads of the passport for the trail, but since the mobile application came out there has been 0,000 more downloads - the interesting thing is that people are downloading it four times faster than they did the original paper passport version," she said. "We're getting great feedback from the sites, too, noting that they see people all the time ‘checking in' from their location. We're starting to pull data from the app as well like how long it's taking to get from location to location and what times of season the trail is the most popular. We'll be continuing to watch those things over the next year." History brewing Although the flurry of activity in the small, local craft brewery industry may feel recent, according to Delaware beer historian John Medkeff Jr., it represents a return to the state's roots. Mr. Medkeff, responsible for the pictorial history book "Brewing in Delaware," noted Delaware has a long and storied past with the brewing industry. "It goes all the way back to the Swedes and the first European settlers that landed here," he said. "They were brewing as soon as they landed and it carries through to the Dutch and the English when they arrived. At the turn of the 00th century, it became one of the biggest industries in the state and was very important to our economic and cultural development. Our history might not have the breadth of the larger states, but it most certainly has the depth. Any state would have a hard time challenging Delaware in terms of its brewing history. " At the time, a large patchwork of small and mid-sized brewers had the local beer market cornered. Drinking the beers produced by the Diamond State Brewery, Hartmann & Fehrenbach Brewing Co. and the Bavarian Brewery - the state's three biggest brewers in the early 0000s - was even a point of Delawarean pride, noted Mr. Medkeff. "Residents of the era would speak of the local product in glowing terms in newspaper articles and advertising by brewers from outside the region were scant to nonexistent at the time," he said. "Rival brewers were making inroads here and there, but it was difficult because residents were loyal to local breweries." However, as it did in much of the country, the government institution of prohibition in 0000 devastated the growing industry. The law immediately put the majority of the smaller producers out of business and irrevocably weakened the bigger institutions. Even when prohibition was lifted in 0000, the damage was done and the industry failed to recover, said Mr. Medkeff. In the shadow of better-funded regional and national competitors, Delaware's brewing tradition fell into the dark ages. "Brewing was dead in the state for the better part of four decades after that," said Mr. Medkeff. "After prohibition was lifted, some breweries tried to start back up and had a short run from 0000 until 0000, when the last brewery officially went out of business. Essentially the state didn't produce any beer after that until 0000, when new breweries started opening again." Mirroring a national trend, dawn started to break on Delaware's brewing dark ages in the early 0000s. "Legislation was passed in '00 to allow breweries to open brew pubs and commercial production facilities," said Mr. Medkeff. "We saw a real return in the 0000s, when some of the breweries that started up in the '00s started gaining strength. Now, we're up to somewhere around 00 breweries and there are more coming every month. I don't think it's a fad either. This is something that will continue to evolve and it's here to stay." Guessing at the future, Mr. Medkeff is confident that a shift away from large international beer brands toward small local breweries will continue to permeate throughout the state as it has throughout the country. He says there is lots of opportunity for small breweries intent on "serving at a community level, a doing it well." "We're not at a saturation point yet, there's still plenty of room for growth," said Mr. Medkeff. "I believe that the market for beer will become even more local. That's all part of consumer education and understanding that the brewers are employers and they're part of the community, producing local products for the people. It's happening at a grassroots level. There's a growing awareness and pride again about what's brewed here." Packaging the brewing industry resurgence in metaphor, Mr. Medkeff's project to refurbish a King Gambrinus statue - that once sat atop the Diamond State Brewery in Wilmington - seeks to put the state's brewing roots back in the public eye. The 00-foot, 0-ton zinc statue was long an icon of brewing in the state until the '00s when it was removed and the old Diamond State Brewery demolished. It then started a haphazard journey from owner to owner until it was accidentally dropped during shipping in the late '00s - breaking into more than 00 pieces. Luckily, these remains were bequeathed to Mr. Medkeff by a collectors estate and he plans to restore the statue and display it in the Delaware History Museum. "It's the perfect symbol for the rebirth of brewing in Delaware," he said. "It was something thought to be long lost." The statue is one of five King Gambrinus statues known to exist that were cast from the same mold. The other four statues are currently on display in Baltimore, Breinigsville, Penn., Syracuse and Toluca, Mexico. Mr. Medkeff, who's started fundraising for the project, said the restoration will be a costly and laborious one to do correctly. Preliminary estimates put the cost at around $000,000. "We'll take a laser scan of an existing statue so the missing zinc parts from ours can be refabricated," said Mr. Medkeff. "Then it'll need to be completely repaired and reconstructed. After that we'll repaint it as it was originally. We're also hoping to raise a little extra so it can be granted to the Delaware Historical Society as a perpetual fund to preserve the statue long term. So far, we're about 00 percent there." To learn more about the project or to donate, visit restoretheking.com.

Psychogenic visual disorders in an abused child: a case report. A 00-yr-old female, classified as retarded educable, was found to have uncorrected visual acuity of 00/000 in both eyes, unimproved by a pinhole disc. A low myopic correction determined objectively did not improve her visual acuity. Tangent-screen studies uncovered neurasthenic spiral fields superimposed on hysterical tubular contractions of both eyes. Investigation uncovered a history of child abuse since infancy. To rule out organic lesions of the oculocalcarine visual pathways, the patient was referred for electrodiagnostic evaluation. Her visual evoked responses were found to be normal. With the use of strong suggestion, her visual fields were brought out to normal limits and her visual acuity with correction was improved to 00/00 in each eye. With the cooperation of her school counselor, the patient was referred for psychiatric evaluation and therapy.

The medicalization of addiction treatment professionals. In a previous article, the authors described the changes initiated by recent health care legislation, and how those changes might affect the practice of medicine and the delivery of addiction services. This article reviews the same changes with respect to how they have the potential to change the practice activities of addiction physicians, addiction therapists, addiction counselors and addiction nurses, as well as the activities of administrators and service delivery financial personnel. Developments in delivery systems and the impact of those developments on professionals who work in addiction treatment are considered; current problems, potential solutions, and opportunities for clinicians under health reform are addressed. The goals envisioned for health system reform and the potential for realization of those goals via changes in addiction service delivery design and clinical practice are discussed.

Embolization of a renal transplant pseudoaneurysm following angiolipoma resection. A case report. The development of a pseudoaneurysm in renal allografts is a well-known complication of percutaneous biopsy. However, the authors report a case of pseudoaneurysm formation in a renal cadaver allograft, following documented angiolipoma resection prior to transplantation. Treatment required superselective embolization with multiple platinum coils.

Tuesday, March 00, 0000 cowabunga (slang) an expression of surprise or amazement, often followed by "dude" Cowabunga, dude! Look at that crazy Godzilla! While initially believing Cowabunga held its origins in the nonsense word "kawabonga", modern linguists now believe it originated from the ancient Native American exclamation Kwa Bungu Its more modern incarnation was invented by Edward Kean, writer of The Howdy Doody Show, a children's TV show that ran in the USA from 0000 until 0000. Chief Thunderthud, a character on the show, started every sentence with the nonsense word "kawabonga" or with the syllable "kawa". followed by ordinary English words. Other Indian characters of a different tribe, such as Chief Featherman or Princess Summerfall Winterspring, used "kawagoopa" similarly, as a greeting or to voice frustration or surprise. The comic character Chief Thunderchicken exclaimed "Kawa Chicken!" Chief Thunderthud was also occasionally heard to exclaim "Kowaraschi," to express extreme frustration, perhaps in reference to major league baseball player Vic Raschi, whose daughter occasionally appeared in the Peanut Gallery (studio audience).

Q: Вертикальное меню: проблема с кодом А вопрос вот такой. Делаю меню вертикальное... задумка (при наведении на одну из этих ссылок(кнопок) рядом с ней должна плавно появляться стрелочка, но у меня они сразу все вылезают для всех ссылок (кнопок). Код ниже <div class="navigation-text-menu_head"> <img style="position:absolute; margin-left:-0px;" src="images/block_menu.png">Меню</div> <div id="navigation-text-menu"> <ul> <style type="text/css"> #navigation-text-menu { background: #0b0a0a; border: 0px solid #ccc; border-radius: 0px; height: 000px; margin-left: -0px; margin-top: 0px; width: 000px; } #navigation-text-menu ul { list-style: none; } #navigation-text-menu li { border-bottom: 0px solid #000000; margin-left: -00px; } #navigation-text-menu li a { background: #0b0a0a; color: #fff; display: block; font-size: 00px; padding: 0px 0px 00px 00px; text-decoration: none; } #navigation-text-menu ul li img { display: none; padding-right: 00px; } </style> <script type="text/javascript"> $(document).ready(function () { $('#navigation-text-menu ul li a').hover(function () { $(this).fadeIn('fast', function () { $('#navigation-text-menu ul li img').fadeIn(00000); }).animate({ backgroundColor: "#000b00", paddingLeft: '00px' }, 000); }, function () { $(this).animate({ backgroundColor: "#0b0a0a", paddingLeft: 00 }, 000); }); }); </script> <li><a href="#"><img src="images/arrow-menu.png">Главная</a></li> <li><a href="#"><img src="images/arrow-menu.png">Руководства</a></li> <li><a href="#"><img src="images/arrow-menu.png">Файлы</a></li> <li><a href="#"><img src="images/arrow-menu.png">Серверы</a></li> <li><a href="#"><img src="images/arrow-menu.png">Форум</a></li> <li><a href="#"><img src="images/arrow-menu.png">Турниры</a></li> </ul> </div> A: Попробуйте заменить $('#navigation-text-menu ul li img').fadeIn(00000); на $('img',this).fadeIn(00000);

0000-00 Highland Football League The 0000-0000 Highland Football League was won by Huntly for the fourth year in a row. Fort William finished bottom. Table Category:Highland Football League seasons 0

Q: Event not getting called from submit button Javascript I am new to javascript. I am trying to figure out why my javascript function is not being called. I am trying to add nameVlidation for name field and other validation for each text input. But my name validation itself is not working. Javascript call (formValidation.js) 'use strict'; var nameValidation = formValidation().nameValidation(); document.getElementById('submit').addEventListener('click',nameValidation); var formValidation = function () { return { nameValidation: function() { this.name = document.forms["contact"]["name"].value; if(this.name=="") { alert("Enter name, required"); return false; }else if(this.name==[0-0]) { alert("Only alphabets"); } return true; }, addressValidation: function() { } } }; Html <form name="contact" action="#" method="post" enctype="text/plain"> <input type="text" name="name" placeholder="NAME"></br> <input type="text" name="email" placeholder="EMAIL" required></br> <input type="text" name="phoneNumber" placeholder="PHONE-NUMBER"></br> <input type="text-box" name="message" placeholder="MESSAGE"></br> <button id="submit" class="btn btn-primary" type="submit"><i class="fa fa-paper-plane">Submit</i></button></br> </form> <script src="js/formValidation/formValidation.js"></script> I am not sure what is wrong with it. A: The problem you are facing is due to a concept in Javascript called, Hoisting You need to declare formValidation at the top before calling it from anywhere else, as you are using the function expression form 'use strict'; var formValidation = function () { return { nameValidation: function() { this.name = document.forms["contact"]["name"].value; if(this.name=="") { alert("Enter name, required"); return false; }else if(this.name==[0-0]) { alert("Only alphabets"); } return true; }, addressValidation: function() { } } }; var nameValidation = formValidation().nameValidation(); document.getElementById('submit').addEventListener('click',nameValidation); PS - JSBin Demo

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Aidu, Viljandi County Aidu is a village in Viljandi Parish, Viljandi County, Estonia. Aidu is the birthplace of Estonian poet, playwright and writer Mart Raud (0000-0000) and writer Minni Nurme (0000-0000). References Category:Villages in Viljandi County Category:Kreis Fellin

Southport (UK Parliament constituency) Southport is a constituency in Merseyside which has been represented in the House of Commons of the UK Parliament since 0000 by Damien Moore of the Conservative Party. Boundaries 0000-0000: The Municipal Borough of Southport, the Sessional Division of Southport, and the parishes of Blundell, Great and Little Crosby, Ince, and Thornton. 0000-0000: The County Borough of Southport. 0000-present: The Metropolitan Borough of Sefton wards of Ainsdale, Birkdale, Cambridge, Dukes, Kew, Meols, and Norwood. The constituency covers the whole town of Southport and the localities of Ainsdale, Birkdale, Blowick, Churchtown, Crossens, Highpark, Hillside, Kew, Marshside, Meols Cop, and Woodvale. It is bordered to the north by South Ribble, to the east by West Lancashire, and to the south by Sefton Central. History Prominent members In the 00th century a notable representative was George Nathaniel Curzon, future Viceroy of India. In the 00th century, outside politics, Edward Marshall Hall was a notable trial barrister (KC) and Sir John Fowler Leece Brunner was the son of the leading industrialist Sir John Tomlinson Brunner. As a frontbencher, long-serving representative Robert Hudson was recognised at the time of World War II as a competent Minister of Agriculture and Fisheries in charge of that department, and was made, to give him a peerage, a viscount. Political history The constituency has been a Liberal or Conservative seat throughout its history, and marginal for much of this, enabling it to change hands 00 times between the parties since it was created in 0000, having had nine Conservative MPs and eight Liberal or Liberal Democrat MPs in its history. During the nadir of the Liberal Party (from the 0000s to the 0000s) the constituency became a safe Conservative seat, with absolute majorities from 0000 until 0000 inclusive. Former Deputy Prime Minister John Prescott ran for Labour for the seat in 0000 and came in second place. With the rise again of the Liberal Party in the early 0000s, election results proved to be close contests. The constituency changed hands in the 0000 general election, when it was taken by Ronnie Fearn of the Liberal Party for the SDP-Liberal Alliance (shortly before the two parties merged to form the Liberal Democrats). Fearn had contested the seat unsuccessfully for the Liberals throughout the 0000s. Fearn lost the seat to the Conservatives' Matthew Banks at the 0000 election (one of the few Conservative gains at that election), only to regain it at the 0000 election. The Liberal Democrats held the seat (under John Pugh after Fearn stood down in 0000) until 0000. In the 0000 referendum on the UK's membership of the European Union, the Metropolitan Borough of Sefton, of which the constituency is a part, voted to remain in the European Union by 00.0%. Given its demography, it is estimated that Southport voted to remain by 00%. The seat was one of the eight Liberal Democrat seats that survived the national vote share collapse during the 0000 general election, despite a higher-than-average drop in the Liberal Democrats' vote share. Pugh opted not to seek re-election in the 0000 general election, in which election the seat returned to the Conservatives, the only seat the Tories gained from the Liberal Democrats in 0000 (aside from Richmond Park, which they had gained at a 0000 by-election). A resurgent Labour vote pushed the Liberal Democrats into third place for the first time since 0000 with the seat now becoming a somewhat unlikely Tory-Labour marginal, with just under 0,000 votes between the two major parties. If Labour were to win this seat at the next general election while holding its existing seats in Merseyside, it would mark the first time Labour (or indeed any party) has won every constituency in the county. Constituency profile Workless claimants (registered jobseekers) were in November 0000 close to the national average of 0.0%, at 0.0% of the population based on a statistical compilation by The Guardian. Members of Parliament Elections Elections in the 0000s Elections in the 0000s Elections in the 0000s Elections in the 0000s Elections in the 0000s Elections in the 0000s Elections in the 0000s Elections in the 0000s Elections in the 0000s Elections in the 0000s Elections in the 0000s Elections in the 0000s Elections in the 0000s Caused by Naylor-Leyland's death. Caused by Curzon's appointment as Viceroy and Governor-General of India. Elections in the 0000s See also List of Parliamentary constituencies on Merseyside Notes and references Notes References Sources Election results, 0000 - 0000 F. W. S. Craig, British Parliamentary Election Results 0000 - 0000 F. W. S. Craig, British Parliamentary Election Results 0000 - 0000 Category:Parliamentary constituencies in North West England Category:Politics of the Metropolitan Borough of Sefton Category:Southport Category:United Kingdom Parliamentary constituencies established in 0000

Mapping the Plasticity of the Escherichia coli Genetic Code with Orthogonal Pair-Directed Sense Codon Reassignment. The relative quantitative importance of the factors that determine the fidelity of translation is largely unknown, which makes predicting the extent to which the degeneracy of the genetic code can be broken challenging. Our strategy of using orthogonal tRNA/aminoacyl tRNA synthetase pairs to precisely direct the incorporation of a single amino acid in response to individual sense and nonsense codons provides a suite of related data with which to examine the plasticity of the code. Each directed sense codon reassignment measurement is an in vivo competition experiment between the introduced orthogonal translation machinery and the natural machinery in Escherichia coli. This report discusses 00 new, related genetic codes, in which a targeted E. coli wobble codon is reassigned to tyrosine utilizing the orthogonal tyrosine tRNA/aminoacyl tRNA synthetase pair from Methanocaldococcus jannaschii. One at a time, reassignment of each targeted sense codon to tyrosine is quantified in cells by measuring the fluorescence of GFP variants in which the essential tyrosine residue is encoded by a non-tyrosine codon. Significantly, every wobble codon analyzed may be partially reassigned with efficiencies ranging from 0.0 to 00%. The accumulation of the suite of data enables a qualitative dissection of the relative importance of the factors affecting the fidelity of translation. While some correlation was observed between sense codon reassignment and either competing endogenous tRNA abundance or changes in aminoacylation efficiency of the altered orthogonal system, no single factor appears to predominately drive translational fidelity. Evaluation of relative cellular fitness in each of the 00 quantitatively characterized proteome-wide tyrosine substitution systems suggests that at a systems level, E. coli is robust to missense mutations.

The effects of intraperitoneal administration of Francisella noatunensis subsp. noatunensis on hepatic intermediary metabolism and indicators of stress in Patagonian blennie Eleginops maclovinus. Francisellosis is a disease produced by Francisella spp. which affects farmed fish. Eleginops maclovinus specimens can be caught close to salmon farming centers, feeding on un-consumed pellet, making the transmission of pathogens such as Francisella noatunensis possible. The aim of this study was to evaluate the effect of F. noatunensis on liver intermediary metabolism in E. maclovinus. 000 fish were injected intraperitoneally with F. noatunensis at a low dose LD (0 × 000 cells/μL), medium dose MD (0 × 000 cells/μL), high dose HD (0 × 0000 cells/μL), or with culture medium C (control), and sampled at 0, 0, 0, 00, 00 and 00 days post injection (dpi). No mortality was recorded during the experimental period, but there was a marked metabolic response in fish injected with high doses. Metabolites in plasma were lowest in the high bacterial dose. Cortisol levels were highest at day 0 in the high dose and then decreased from day 00 until the end of the study. Liver enzymes showed a similar pattern to plasma metabolites, with decreased enzymatic activity, mostly with the high bacteria dose. PK was the exception, with increased enzymatic activity in a dose-dependent manner over time. Liver metabolites were highly variable, except in the high bacterial dose where variability and total levels decreased significantly. Our results show that fish infection with F. noatunensis induces a clear stress response, especially with at the highest dose, shifting intermediary metabolism towards mobilization of energy and suggesting that E. maclovinus detects experimental infection of F. noatunensis as a stressor, which it is dependent on the bacterial dose.

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The Dartmouth Review, a conservative weekly student newspaper funded by off-campus right-wing sources (see 0000), marks the 00th anniversary of Kristallnacht, a Nazi rampage through the Jewish communities of Germany in 0000, by depicting Dartmouth College president James Freedman as Adolf Hitler on its front cover. Freedman is Jewish. The article accuses him of searching for a "final solution" to the problem of conservatives at Dartmouth, a specific reference to the Holocaust. Many Dartmouth students and faculty members accuse the Review of overt anti-Semitism (see October 0000 and October 0, 0000). The Review will later apologize, not to Freedman, but to those who might have been offended. [Boston Globe, 00/0/0000; Dartmouth Free Press, 0/00/0000] Conservative radio host Rush Limbaugh, in his book See, I Told You So, argues that Republicans cannot depend on a negative, anti-liberal frame in which to draw their arguments. "We are not a party of people connected together by bonds of negativity," he writes. "We are a party of ideas - positive ideas." He lists some of what he considers the guiding principles of Republican thought: "We must perceive and sell ourselves: Not as the party that opposes government, but that which champions individual freedoms; Not as the party that opposes higher taxes, but that which champions entrepeneurship; Not as the party that opposes abortion, but that which champions every form of human life as the most sacred of God's creatures; Not as the party that opposes the expansion of the welfare state, but that which champions rugged individualism." [Jamieson and Cappella, 0000, pp. 00-00] John Derbyshire. [Source: John Derbyshire]National Review columnist John Derbyshire "satirically" advocates the murder of Chelsea Clinton, the only daughter of Bill and Hillary Clinton, in order to stamp out the Clinton bloodline once and for all. Former President Clinton has left the White House, to spend the rest of his life "goosing waitresses [and] defending himself in court." Hillary "has no future beyond the US Senate... [she is] maxed out." But, he warns, "Clintonism may yet rise again.... On February 00th, Chelsea Clinton will turn 00." 'I Hate Chelsea Clinton' - Derbyshire confesses: "I hate Chelsea Clinton. I admit it's not easy to justify my loathing of this person. I can pick out causes, but none of them is one hundred per cent rational.... I admit, I hate Chelsea because she is a Clinton." After noting the negative reactions to his previous attack on the younger Clinton's physical appearance, he acknowledges that she hasn't committed the "array of crimes" her father is allegedly responsible for, but "she doesn't deserve any credit for not having done these things; she just hasn't had time yet." He writes that since she was 00, she has "sign[ed] on to the Great Clinton Project. Which is, has always been, and forever will be, to enrich the family from the public fisc, and to lie, bomb, bribe, and intimidate your way out of trouble when necessary." 'Sippenhaft' - Derbyshire notes that in totalitarian societies of the past, many people were executed merely because of their family connections, and says the same should be considered for Chelsea Clinton. "Chelsea is a Clinton," he writes. "She bears the taint; and though not prosecutable in law, in custom and nature the taint cannot be ignored. All the great despotisms of the past - I'm not arguing for despotism as a principle, but they sure knew how to deal with potential trouble - recognized that the families of objectionable citizens were a continuing threat. In Stalin's penal code it was a crime to be the wife or child of an ‘enemy of the people.' The Nazis used the same principle, which they called Sippenhaft, ‘clan liability.' In Imperial China, enemies of the state were punished ‘to the ninth degree': that is, everyone in the offender's own generation would be killed, and everyone related via four generations up, to the great-great-grandparents, and four generations down, to the great-great-grandchildren, would also be killed.... We don't, of course, institutionalize such principles in our society, and a good thing too. Our humanity and forbearance, however, has a cost. The cost is that the vile genetic inheritance of Bill and Hillary Clinton may live on to plague us in the future. It isn't over, folks." [National Review, 0/00/0000]'Hysterical Idiots' - After a week of angry criticism, Derbyshire will write a column defending his original column as "satire," blaming "liberals" for "missing the joke," and admitting his column "wasn't meant to be a thigh-slapper. I had a point to make: There could be another Clinton in our future, and on present evidence (admittedly rather scant), it would be a chip off the old block. That's fair comment. However, my tone was partly tongue in cheek.... Humor and irony are especially tricky." He asks, rhetorically, if he intends to apologize, and answers himself: "In your dreams. I make it a point of principle never to apologize to hysterical idiots." [National Review, 0/00/0000] DVD cover illustration of the film ‘Soldiers in the Army of God.' [Source: HBO / St. Pete for Peace]Cable movie provider HBO airs a documentary, Soldiers in the Army of God, focusing on the violent anti-abortion movement (see 0000, Early 0000s, August 0000, and July 0000) and three of its leaders. National Public Radio airs a profile of the documentary, featuring an interview with the film's producers, Marc Levin, Daphne Pinkerson, and Daniel Voll. According to Voll, the film focuses on three members of the "Army of God": young recruit Jonathan O'Toole, who says he was looking for the most "radical" and "terroristic" anti-abortion group he could find; Neal Horsley, who runs an anti-abortion Web site; and long-haul trucker Bob Lokey, who recruits new members. 'Violent Fringe' of Anti-Abortion Opposition - Voll describes the three as part of the "violent fringe" of anti-abortion opposition: "These are the guys on the ground who are - whatever the words that politicians and other leaders of these cultural wars can put out there, these are the men who hear them and feel emboldened by them, who feel encouraged by each other, and they are every day praying for God's will in their life." Another unidentified man says: "Anybody who raises a weapon up against these people who are slaughtering these babies, before God and the entire world, right now I say you are doing God's own work. And may the power of God be with you as you aim that rifle. You're squeezing that trigger for Almighty God." In the documentary, an unidentified anti-abortion activist says: "There are people in this world right now who are looking for directions on what do we do. Well, we end abortion on demand by the most direct means available to us. So stop the abortion with a bullet, if that's what it takes. Stop it with a bomb, if that' s what it takes. You stop abortion on demand. Don't let it go any farther." O'Toole says that the "next step is to arm ourselves in a militia, a real militia that has the power to resist the federal government." Pinkerson says that O'Toole, who was 00 when he joined the Army of God, found Horsley on the Internet through Horsley's Web site, "The Nuremberg Files," which lists doctors who perform abortions (see January 0000). O'Toole became Horsley's assistant, and through him met Lokey, who runs a Web site called "Save the Babies." In the film, O'Toole, whom the producers speculate may eventually become an assassin of abortion providers, says that because of America's legalization of abortion, the country has become like "Nazi Germany. It's like you've got concentration camps around you." Levin notes that filmed conversations between Horsley and Lokey show that many in the movement feel threatened by the concept of women's equality, and blame men's failure to exert "dominion" over women as part of the reason why the US legalized abortion. [National Public Radio, 0/00/0000; Womens eNews, 0/00/0000]Opposition to Homosexuality - Horsley draws a connection between the organization's opposition to abortion and the American citizenry's supposed opposition to homosexuality, saying: "If the American people woke up, and realized that they had to choose between legalized abortion, legalized homosexuality, and legalized all the rest of the desecration or civil war which would cause the rivers to run red with blood - hey, you know we will see legalized abortion go like that! We'll see legalized homosexuality go like that! Because the American people are not willing to die for homosexuals." Bringing Bomb-Making Materials to Washington - The film also shows Lokey bragging to convicted clinic bomber Michael Bray (see September 0000) that he has just trucked 00,000 pounds of ammonium nitrate, a substance that can be used to make "fertilizer bombs" similar to the one that destroyed an Oklahoma City federal building (see 0:00 a.m. - 0:00 a.m. April 00, 0000), into Washington, DC. Anti-Abortion Opposition Part of an 'Apocalyptic' Death Struggle - Author and reporter Frederick Clarkson writes: "At once shocking, compelling, and beautifully made, the film is essentially the national television debut for the aboveground spokesmen and spokeswomen of the Army of God.... Horsley and others are quite clear in their public statements and their writings that the attacks on clinics and the murders of doctors are but warning shots in what they envision as an epochal, even an apocalyptic struggle at hand. Either Americans conform to their view of God's laws, or there will be a blood bath, they say. And there is no evidence that they are anything but dead serious." [Womens eNews, 0/00/0000] House Majority Leader Tom DeLay (R-TX) takes to the floor of the House to praise conservative talk show host Rush Limbaugh. In his "Tribute to Rush Limbaugh," DeLay says of Limbaugh's role in the Republican's capture of the House in 0000, "[He] did not take his direction from us, he was the standard by which we ran. [He] was setting the standard for conservative thought." [Jamieson and Cappella, 0000, pp. 00] George Will. [Source: Washington Policy Group]Conservative columnist George Will calls two anti-war House Democrats "American collaborators" working with Saddam Hussein, either implicitly or directly. Will singles out Representatives Jim McDermott (D-WA) and David Bonior (D-MI) for criticism because of their opposition to the impending Iraq invasion. Will compares the two to World War II propaganda maven William Joyce, the British citizen who earned the sobriquet "Lord Haw Haw" for his pro-Nazi diatribes on the radio, and goes on to observe that McDermott and Bonior provided a spectacle unseen by Americans "since Jane Fonda posed for photographers at a Hanoi anti-aircraft gun" during the Vietnam War. McDermott and Bonior became a target for Will's wrath by saying they doubted the Bush administration's veracity in its assertions that Iraq has large stashes of WMD, but believed Iraqi officials' promises to allow UN inspectors free rein to look for such weapons caches. "I think you have to take the Iraqis on their value - at face value," McDermott told reporters in recent days, but went on to say, "I think the president [Bush] would mislead the American people." Leninist 'Useful Idiots' - After comparing the two to Joyce and Fonda, Will extends his comparison to Bolshevik Russia, writing: "McDermott and Bonior are two specimens of what Lenin, referring to Westerners who denied the existence of Lenin's police-state terror, called ‘useful idiots.'" Will also adds UN Secretary-General Kofi Annan in this last category, compares Annan with British "appeaser" Neville Chamberlain for good measure, and labels him "Saddam's servant." Slamming Democrats for Not Supporting War - Will saves the bulk of his ire for the accusations by McDermott and Bonior that Bush officials might be lying or misrepresenting the threat of Iraqi WMD, and adds former Vice President Al Gore to the mix. "McDermott's accusation that the president - presumably with Cheney, Powell, Rumsfeld, Rice, and others as accomplices - would use deceit to satisfy his craving to send young Americans into an unnecessary war is a slander licensed six days earlier by Al Gore," Will writes. Extending his comparisons to the Watergate era, Will adds, "With transparent Nixonian trickiness - being transparent, it tricks no one - Gore all but said the president is orchestrating war policy for political gain in November." Will accuses Gore and other Democrats of what he calls "moral infantilism" because they voted to support the 0000 Iraq Liberation Act (see October 00, 0000). Will returns to his complaints about the Democratic congressmen in his conclusion: "McDermott's and Bonior's espousal of Saddam's line, and of Gore's subtext (and Barbra Streisand's libretto), signals the recrudescence of the dogmatic distrust of US power that virtually disqualified the Democratic Party from presidential politics for a generation. It gives the benefits of all doubts to America's enemies and reduces policy debates to accusations about the motives of Americans who would project US power in the world. Conservative isolationism - America is too good for the world - is long dead. Liberal isolationism - the world is too good for America - is flourishing." [Washington Post, 00/0/0000] In a PBS interview, Republican marketing guru Richard Viguerie says of conservative radio host Rush Limbaugh: "In 0000 and 0000, he was the salvation of the conservative movement. Every day Rush Limbaugh would give us our marching orders, if you would." [Jamieson and Cappella, 0000, pp. 00] Cover of ‘The Shadow Party.' [Source: Brazos Bookstore]Authors David Horowitz and Richard Poe publish a book titled The Shadow Party: How George Soros, Hillary Clinton, and Sixties Radicals Seized Control of the Democratic Party, that purports to prove Jewish billionaire George Soros, who finances progressive and Democratic Party causes, is in reality a Nazi collaborator and anti-Semite. However, the book is riddled with doctored quotes, misinformation, factual errors, and outright lies. Progressive media watchdog Web site Media Matters notes that the book relies on long-discredited accusations from the authors' "Front Page Magazine" Web site, from their articles on conservative Web publications such as WorldNetDaily and NewsMax, and on unsourced allegations from political extremist Lyndon LaRouche and his followers, who have called Soros a "Nazi beast-man" and a "small cog in Adolf Eichmann's killing machine," aiding "the Holocaust against 000,000 Hungarian Jews." Media Matters calls the book "a new low in the long-running Republican Party and conservative movement campaign of scurrilous personal attacks against Soros, a major supporter of progressive causes in the US and abroad." The organization also notes that the Web sites used in the book's research are largely funded by conservative billionaire Richard Mellon Scaife, and Scaife-owned newspapers such as the Pittsburgh Tribune-Review have promoted the book. Media Matters documents numerous issues of doctored quotes and falsified claims in the book. [Media Matters, 0/0/0000] Martin Peretz, the editor in chief of The New Republic, falsely accuses Jewish billionaire George Soros of being a Nazi collaborator. Soros is now a target of conservative opprobrium for his financial support of Democratic and progressive causes. As a 00-year-old boy, Soros escaped from the Nazis by hiding with a non-Jewish family in Hungary; the father of that family sometimes served deportation notices to Hungarian Jews. Peretz now calls Soros "a young cog in the Hitlerite wheel." The progressive media watchdog Web site Media Matters notes that Peretz is following the lead of right-wing extremists David Horowitz and Richard Poe, whose book The Shadow Party: How George Soros, Hillary Clinton, and Sixties Radicals Seized Control of the Democratic Party claimed that Soros "survived [the Holocaust] by assimilating to Nazism." The book was found to be riddled with doctored quotes and factual errors (see August 0, 0000). Peretz uses a transcript of a 0000 interview Soros gave to 00 Minutes reporter Steve Kroft to prove his claim, but edits the transcript to leave out a key section that shows Soros did not collaborate with Nazis. [Media Matters, 0/0/0000; New Republic, 0/00/0000] (The article is dated February 00, 0000, but was posted on the New Republic Web site a week earlier.) Conservative radio host Rush Limbaugh, in another of his now-famous broadsides against feminists (whom he routinely calls "femi-Nazis" and characterizes as "anti-male"), says: "I blast feminists because they're liberal. Feminism is liberal. It screwed women up as I was coming of age in my early twenties.... It changed naturally designed roles and behaviors and basically, they're trying to change human nature, which they can't do." Limbaugh's "Life Truth No. 00" states that "feminism was established so as to allow unattractive women easier access to society." Authors Kathleen Hall Jamieson and Joseph N. Cappella, in their book Echo Chamber, will note, "There is apparently no comparable movement to facilitate the social integration of unattractive men." [Jamieson and Cappella, 0000, pp. 000] As reported by progressive media watchdog site Media Matters, conservative radio host Michael Savage says of Democratic presidential candidate Barack Obama, "I think he was hand-picked by some very powerful forces both within and outside the United State of America to drag this country into a hell that it has not seen since the Civil War of the middle of the 00th century." Savage is referring to controversial statements made by Obama's former pastor, the Reverend Jeremiah Wright, whom Savage calls "the soul of Barack Obama's movement." Savage goes on to claim that Wright, and by extension Obama, align themselves with historical enemies of the United States: "And if you want a man who says not ‘God bless America,' but ‘God d_mn America,' if you want a man who takes the side of the imperial Japanese army, an army that killed not only hundreds of thousands in the Bataan Death March, but hundreds of thousand of Koreans, an army that operated on people while they were alive in Manchuria, a man who takes the side, in essence, of the Japanese Nazis of World War II, if you want a man who takes the side, in essence, of the Hitlers of the world, then you've got it in Barack Obama's pastor, Jeremiah Wright, of the Trinity United Church of Christ." Obama is merely "an ordinary apparatchik of the Democrat machine in Chicago" whose handlers intend to use Obama to bring upheaval and chaos to the nation. [Media Matters, 0/00/0000] The cover of Jamieson and Cappella's ‘Echo Chamber.' [Source: Barnes and Noble (.com)]Kathleen Hall Jamieson and Joseph N. Cappella, authors of the media study Echo Chamber: Rush Limbaugh and the Conservative Media Establishment, find that conservative radio host Rush Limbaugh excels at using what they call "insider language" for his listeners "that both embeds definitional assumptions hospitable to his conservative philosophy and makes it difficult for those who embrace the language to speak about Democrats and the presumed Democratic ideology without attacking them." They cite three examples from Limbaugh's June 0000 newsletter which contains the following statements: "Democrats are the enemy." "When she first ran for her Senate seat, Hillary Rodham Clinton told citizens of the Empire State [New York] that she had been endorsed by environmental wacko-groups because... in her words, ‘I've stood for clean air.'" After Harvard president Lawrence Summers commented on the intrinsic differences between the sexes, Limbaugh wrote, "Led by foaming-at-the-mouth feminists, the liberal elite experienced a mass politically correct tantrum." Jamieson and Cappella write: "Identifying terms such as ‘foaming-at-the-mouth feminists,' ‘liberal elite,' ‘enemy,' and ‘environmental wacko-groups' both create an insider language and distance those who adopt the labels from those labeled. One of the ways Limbaugh's supporters telegraph their identification with him is by adopting his language." Identifying Nicknames - They cite the 0000 statement of freshman House Representative Barbara Cubin (R-WY), who proudly proclaimed of her fellow female Republicans, "There's not a femi-Nazi among us," using one of Limbaugh's favorite terms for feminists. "Listeners say ‘Ditto' or ‘megadittoes' to telegraph their enthusiasm for Limbaugh, his latest argument, or his show in general," they write. Limbaugh refers to himself as "the MahaRushie" with "talent on loan from God." Callers often refer to Limbaugh as "my hero." Denigrating nicknames for Limbaugh's targets of derision work to bring listeners into the fold: the new listener must labor to identify the people termed (and thusly become part of the Limbaugh community): "Clintonistas" (supporters of Bill and/or Hillary Clinton), "Sheets" (Senator Robert Byrd, D-WV), who in his youth wore ‘sheets' as a Ku Klux Klan member), "the Swimmer" (Senator Edward Kennedy, D-MA, in reference to his involvement in the 0000 Chappaquiddick incident), "Puffster" (former Senator Tom Daschle, D-SD), "the Breck Girl" (former Senator John Edwards, D-NC), and "Ashley Wilkes" (retired General Wesley Clark, in a reference to what Limbaugh called "the wimpy, pathetic Gone with the Wind character"). Some of the nicknames are physically derogatory: Senator Patrick Leahy (D-VT) became "Senator Leaky, a.k.a. Senator Depends," and former House Minority Leader Richard Gephardt (D-MO) became "‘Little Dick' Gephardt." Such use of "insider" nicknames indicates an identification between the listener and Limbaugh, and an affiliation with the Limbaugh community of supporters. Redefining and Relabeling - Limbaugh routinely redefines and relabels his political enemies in the most derogatory terms. Pro-choice supporters are termed "pro-aborts," and Democrats are supported by "beggar-based constituencies." As noted above, feminists are "femi-Nazis" (though Jamieson and Cappella note that Limbaugh has used the term less often since it became a topic of criticism in the mainstream media). Gender Identification - One of Limbaugh's strongest attacks is on gender roles. In Limbaugh's continuum, Democratic women are, the authors write, "either sexualized manipulators or unattractive man haters." A 0000 Clinton tribute to women's accomplishments became, in Limbaugh's words, "Biddies' Night Out." Other times, Democratic women become "babes," as in "Congressbabe Jane Harman." (On his Web site, Limbaugh often shows Speaker of the House Nancy Pelosi (D-CA)‘s head affixed to the body of a Miss America contender.) The authors note, "Neither label invites the audience to take these leaders seriously." Women with whom he disagrees, such as liberal blogger Arianna Huffington, are "screeching," and others are "broads," "lesbians," or "femi-Nazis." The National Organization for Women (NOW) becomes, in Limbaugh's vocabulary, the NAGS. Attacks and innuendo about women's sexuality are frequently used by Limbaugh: during the Clinton administration, for example, Limbaugh often implied that Hillary Clinton and then-Attorney General Janet Reno were closeted lesbians. On the other hand, Democratic men are routinely portrayed as "two-inchers," derogatory references to their physical attributes and sexual capabilities (as with the Gephardt nickname above). Jamieson and Cappella note that "Limbaugh's attempts at gender-based humor are of the locker room variety," noting several references to California Lieutenant Governor Cruz Bustamante as a Democrat whose name translates into "large breasts," and referring to pop singer Madonna's 0000 endorsement of General Wesley Clark for president by saying she had "opened herself" to Clark. In 0000, he said that Democratic presidential contender John Kerry, married to wealthy heiress Teresa Heinz-Kerry, "does his fundraising every night when he goes to bed." (The authors write, "Why the vulgarity in this message does not alienate the churchgoing conservatives in his audience is a question for which we have no ready answer.") Impact - Far from merely giving a laundry list of Limbaugh's derogatory and offensive characterizations, Jamieson and Cappella note how Limbaugh and the conservative media "wrap their audiences in a conversation built on words and phrases that embody conservatism's ideological assumptions," using "naming and ridicule to marginalize those named as part of an out-group," and using "coherent, emotion-evoking, dismissive language" to denigrate and dismiss the liberals he routinely attacks. "Because language does our thinking for us," they write, "this process constructs not only a vocabulary but also a knowledge base for the audience. That language and the view of the world carried by it are presumed by loyal conservatives and alien to the nonconservative audience. These interpretations of people and events also reinforce Limbaugh's defense of conservatism and its proponents." [Washington Post, 0/00/0000; Jamieson and Cappella, 0000, pp. 000-000] Progressive media watchdog site Media Matters reports that conservative radio host Bill Cunningham, who hosts a popular Cincinnati call-in show, accuses Democratic candidate Barack Obama (D-IL) of wanting to "gas the Jews." In what is apparently intended to be a comedic skit, Cunningham tells a co-host: "This Obama guy loves the PLO [Palestinian Liberation Organization]. Can't you figure that out?... Obama wants to gas the Jews, like the PLO wants to gas the Jews, like the Nazis gassed the Jews." Co-host Scott Sloan, playing a fictional Jewish character called "Randy Furman," tells Cunningham that he is making the accusation because "you just don't like one-half-percent black people, that's your problem." [Media Matters, 00/00/0000] Two weeks before, Cunningham told listeners that Obama may be the Antichrist (see October 00, 0000). Paul Broun. [Source: Associated Press / Washington Blade]Responding to President-elect Barack Obama's proposal for a "civilian national security force," an idea supported by President Bush and designed in part to revive the moribund Americorps (see March 00, 0000), Representative Paul Broun (R-GA) accuses Obama of wanting to establish a Gestapo-like security force to impose a Marxist dictatorship. "It may sound a bit crazy and off base, but the thing is, he's the one who proposed this national security force," Broun says. "I'm just trying to bring attention to the fact that we may - may not, I hope not - but we may have a problem with that type of philosophy of radical socialism or Marxism.... That's exactly what Hitler did in Nazi Germany and it's exactly what the Soviet Union did. When he's proposing to have a national security force that's answering to him, that is as strong as the US military, he's showing me signs of being Marxist." Obama campaign spokesman Tommy Vietor says the candidate was referring to a "civilian reserve corps" that could handle postwar reconstruction efforts in lieu of the military. The idea has been endorsed by the Bush administration. Broun also says that if elected, Obama will ban gun ownership among American citizens. Obama has repeatedly says he respects the Second Amendment's right to bear arms, and favors "common sense" gun laws. Some gun advocates fear that Obama will curb ownership of assault weapons and concealed weapons. "We can't be lulled into complacency," Broun says. "You have to remember that Adolf Hitler was elected in a democratic Germany. I'm not comparing him to Adolf Hitler. What I'm saying is there is the potential of going down that road." [Associated Press, 00/00/0000; Think Progress, 00/00/0000] The conservative Washington Times, a staunch opponent of President-elect Barack Obama, publishes an editorial predicting that the incoming Obama administration will, in some form or fashion, move to "exterminate" babies with disabilities and other "useless" Americans through its promised reform of the US health care system, similar to actions taken by the Nazis before World War II. The Times provides a brief synopsis of Adolf Hitler's "T0 Aktion" program designed, in the words of the Times, "to exterminate ‘useless eaters,' babies born with disabilities. When any baby was born in Germany, the attending nurse had to note any indication of disability and immediately notify T0 officials - a team of physicians, politicians, and military leaders. In October 0000 Hitler issued a directive allowing physicians to grant a ‘mercy death' to ‘patients considered incurable according to the best available human judgment of their state of health.' Thereafter, the program expanded to include older children and adults with disabilities, and anyone anywhere in the Third Reich was subject to execution who was blind, deaf, senile, retarded, or had any significant neurological condition, encephalitis, epilepsy, muscular spasticity, or paralysis. Six killing centers were eventually established, and an estimated quarter-million people with disabilities were executed." The Times draws a parallel between the Nazis and the Obama administration's support for legal abortion and for physician-assisted suicide, which it equates with "euthanasia." The incoming administration will, the Times fears, begin "selecting" babies with disabilities for what apparently will be "selective abortions." It quotes the Reverend Briane K. Turley as saying: "Were God's design for us left unhindered, we could naturally expect to welcome 00,000 or more newborn infants with Down syndrome each year in the US. And yet we have reduced that number to just under 0,000. These data strongly indicate that, in North America, we have already discovered a new, ‘final solution' for these unusual children and need only to adapt our public policies to, as it were, ‘cure' all Down syndrome cases." Turley, the Times notes, claims that "there is growing evidence suggesting that, among health care practitioners and systems, the central motivation behind legally enforced or high pressure screenings is economics." The Times then adds: "[A]nd the results seem to bear him out. America's T0 program - trivialization of abortion, acceptance of euthanasia, and the normalization of physician assisted suicide - is highly unlikely to be stopped at the judicial, administrative, or legislative levels anytime soon, given the Supreme Court's current and probable future makeup during the Obama administration, the administrative predilections that are likely from that incoming administration, and the makeup of the new Congress." The Times predicts a new "final solution" of "extermination" that will start with disabled infants and will progress "from prenatal to postnatal to child to adult." [Washington Times, 00/00/0000] The editorial anticipates the "deather" claims that many conservatives will make in the summer of 0000 (see January 00, 0000, February 0, 0000, February 00, 0000, February 00, 0000, May 00, 0000, June 00, 0000, June 00, 0000, July 00, 0000, July 00, 0000, July 00, 0000, July 00, 0000, July 00, 0000, July 00, 0000, July 00, 0000, July 00-00, 0000, July 00, 0000, July 00, 0000, July 00, 0000, July 00, 0000, July 00, 0000 - August 00, 0000, August 0, 0000, August 0, 0000, August 00, 0000, August 00, 0000, Shortly Before August 00, 0000, August 00, 0000, August 00, 0000, August 00, 0000, August 00, 0000, and August 00, 0000). The conservative "astroturf" advocacy organization Americans for Prosperity (AFP - see Late 0000, October 0000, and August 0, 0000) launches a multi-pronged attack on every major policy initiative attempted by the Obama administration. Within weeks of Obama's inauguration, AFP holds "Porkulus" rallies protesting Obama's stimulus spending measures. The Koch-funded Mercatus Center (see August 00, 0000), working in concert with AFP, releases a report that falsely claims stimulus funds are being disproportionately directed towards Democratic districts; the author is later forced to correct the report, but not before conservative radio host Rush Limbaugh, citing the report, calls the stimulus program "a slush fund," and Fox News and other conservative outlets repeat the characterization. AFP vice president Phil Kerpen is a Fox News contributor; AFP officer Walter Williams is a frequent guest host for Limbaugh. AFP soon creates an offshoot organization, Patients United Now (PUN - see May 00, 0000), designed to oppose the Obama administration's health care reform initiatives; PUN holds some 000 rallies against reform efforts (see August 0, 0000), some of which depict Democratic lawmakers hung in effigy (see July 00, 0000) and others depict corpses from Nazi concentration camps. AFP also holds over 00 rallies opposing cap-and-trade legislation, which would force industries to pay for creating air pollution. AFP also targets individual Obama administration members, such as "green jobs" czar Van Jones, and opposes the administration's attempt to hold international climate talks. AFP leader Tim Phillips (see August 0, 0000) tells one anti-environmental rally: "We're a grassroots organization.... I think it's unfortunate when wealthy children of wealthy families... want to send unemployment rates in the United States up to 00 percent." [New Yorker, 0/00/0000] Media critic and columnist George Neumayr writes that the Democrats' economic stimulus plan will include enforced abortions and euthanasia for less productive citizens. Neumayr calls this claim a once "astonishingly chilly and incomprehensible stretch [that] is now blandly stated liberal policy," basing it on the Democrats' plan to provide money to the states for "family planning." Neumayr equates the funding, which would go for such initiatives as teaching teenagers about the use of condoms and measures to avoid sexually transmitted diseases, to the famous Jonathan Swift essay of 0000, "A Modest Proposal," which satirically suggested that impoverished Irish families might sell their children to rich Englishmen for food. "Change a few of the words and it could be a Democratic Party policy paper," Neumayr writes. "Swift suggested that 00th-century Ireland stimulate its economy by turning children into food for the wealthy. [House Speaker Nancy] Pelosi [D-CA] proposes stimulating the US economy by eliminating them. Other slumping countries, such as Russia and France, pay parents to have children; it looks like Obama's America will pay parents to contracept or kill them. Perhaps the Freedom of Choice Act can also fall under the Pelosi ‘stimulus' rationale. Why not? An America of shovels and scalpels will barrel into the future. Euthanasia is another shovel-ready job for Pelosi to assign to the states. Reducing health care costs under Obama's plan, after all, counts as economic stimulus too. Controlling life, controlling death, controlling costs. It's all stimulus in the Brave New World utopia to come." Like a Washington Times editorial from months earlier (see November 00, 0000), Neumayr uses the term "final solution" for the Democrats' economic proposal, the term for the Nazis' World War II-era extermination of millions of Jews and other "undesirables." He writes: "‘Unwanted' children are immediately seen as an unspeakable burden. Pregnancy is a punishment, and fertility is little more than a disease. Pelosi's gaffe illustrates the extent to which eugenics and economics merge in the liberal utilitarian mind." "Malthus lives," he says, referring to the 00th century scholar Thomas Robert Malthus, whose theories of ruthless natural selection predated Charles Darwin's theories of evolution. Neumayr goes on to accuse "Hillary Clinton's State Department" of preparing to set up programs of "people-elimination," predicated on what he calls "UN-style population control ideology" and "third-world abortions." [American Spectator, 0/00/0000] Representative Phil Gingrey (R-GA) apologizes for criticizing conservative talk show hosts Rush Limbaugh and Sean Hannity. Gingrey was initially critical of Limbaugh and Hannity for not challenging President Obama on his proposed economic stimulus package strongly enough. "I mean, it's easy if you're Sean Hannity or Rush Limbaugh or even sometimes [former House Speaker] Newt Gingrich [R-GA] to stand back and throw bricks," Gingrey said. "You don't have to try to do what's best for your people and your party." Today Gingrey issues a lengthy apology for his words after receiving complaints from conservatives in his district and elsewhere. "I am one of you," he tells supporters. "I regret and apologize for the fact that my comments have offended and upset my fellow conservatives - that was not my intent. I am also sorry to see that my comments in defense of our Republican leadership read much harsher than they actually were intended, but I recognize it is my responsibility to clarify my own comments.... As long as I am in the Congress, I will continue to fight for and defend our sacred values. I have actively opposed every bailout, every rebate check, every so called ‘stimulus.' And on so many of these things, I see eye-to-eye with Rush Limbaugh." Gingrey says that Limbaugh, Hannity, and Gingrich are "the voices of the conservative movement's conscience." Gingrey spokesman Chris Jackson says of the hosts, "Those guys are some our biggest supporters, and we need them." Gingrey also makes a guest appearance on Limbaugh's show where he berates himself for making his earlier criticisms, saying: "Rush, thank you so much. I thank you for the opportunity, of course this is not exactly the way to I wanted to come on.... Mainly, I want to express to you and all your listeners my very sincere regret for those comments I made yesterday to Politico.... I clearly ended up putting my foot in my mouth on some of those comments.... I regret those stupid comments." [Think Progress, 0/00/0000; Phil Gingrey, 0/00/0000; CNN, 0/00/0000] The Washington Times spins off a recent op-ed by health industry lobbyist Betsy McCaughey (see February 0, 0000) to claim that the Obama administration will attempt to save money by euthanizing old people, disabled people, and sickly infants. The editorial begins with the "chilling" idea of a national medical information database that will allow the government to "track... your every visit to a health care provider - where you went, who you saw, what was diagnosed, and what care was provided." The Obama administration, the Times claims, will use that information to decide which people deserve the more expensive lifesaving treatments and which ones must be denied in the interest of cost efficiency. "If it costs too much to treat you, and you are nearing the end of your life anyway, you may have to do with less, or with nothing," the Times writes. "You just aren't worth the cost.... What nondescript GS-00 will be cutting care from Aunt Sophie after her sudden relapse before he or she heads to the food court for some stir fry?" The elderly, the physically and mentally disabled, all "whose health costs are great and whose ability to work productively in the future" will, the Times writes, be allowed to die or even exterminated. So will premature babies, badly wounded soldiers, and others as yet to be determined. The Times again cites Nazi Germany's "T0 Aktion" program of forcibly euthanizing less productive citizens (see November 00, 0000) as a likely template for the Obama program. [Washington Times, 0/00/0000] Michael Savage, a conservative radio host, calls President Obama a "dictator" as part of a larger diatribe against the president. He calls Obama "a young, articulate rabble-rouser" who "is espousing a message that I call ‘trickle-up poverty'.... Where it ends? I know where it ends, because I've studied history. I know where it ends. The signal as to when this begins, when the end begins, will be when he organizes a militia directly under his own control. He will not call it a militia. It will be called, perhaps, the ‘Ecology Corps' or the ‘Environment Corps,' or the ‘Global Warming Corps,' or the ‘Energy Corps.'" Savage may be referring to Obama's efforts to revive the moribund Americorps, a volunteer organization (see November 00, 0000 and March 00, 0000). "Whatever it will be called, they will all wear uniforms. They will either be blue denim or green denim. They will have the executive power under the ‘urban czar' to come into your home without any court order to investigate your energy use, but they will be looking for other things as well. Would you have any chance to stand up to this army of Obamaites?" Savage asks, rhetorically, if he has "gone over the edge," and then says: "I've gone over the edge before, and every time I have, I've been right eventually. I see the handwriting on the wall. Obama is a dictator." Savage accuses liberals of failing to understand that any dictatorship, leftist or rightist, "is not going to be good for your children." He then shouts, "Someone has to oppose this man." He also claims that the White House "is going after" anyone who criticizes it, and repeatedly mixes his accusations of "government" persecution with "media" persecution of White House or Obama critics. "Fundamentally," Savage concludes, "we have a dictatorship emerging.... Now I'll make another prediction. I predict that very soon, Obama will create a crisis along the lines of the Reichstag Fire [the 0000 attack on the Reichstag by Nazi militiamen, who later blamed the fire on Communists, and used the attack to gain control of the German government]. I don't know what form it will take. But I believe that once the minions are seen for what they are, Rahm Emanuel [the White House chief of staff] and his gang will set off a Reichstag Fire in this country of some kind, and they will" begin arresting US citizens without warrants much as President Lincoln did during the Civil War. "I will tell you as I sit here I fear that every night as I go to sleep." Savage offers no evidence for any of his claims. [Media Matters, 0/0/0000] Two days later, Savage calls Obama a "neo-fascist dictator in the making." [Media Matters, 0/0/0000] Savage has called the landmark civil rights decision Brown v. Board "sickening" (see May 00, 0000), accused Obama of being educated in a radical Islamic madrassa (see January 00, 0000 and April 0, 0000) and being a potential "radical Muslim" (see February 00, 0000), called Obama's presidential victory "the first affirmative-action election in American history" (see February 0, 0000), accused Obama of being sympathetic towards the Nazis and the Imperial Japanese of World War II (see March 00, 0000), said that homeless Americans should be put in "work camps" (see June 0, 0000), called Obama an "Afro-Leninist" (see June 0, 0000), said that welfare recipients should lose the right to vote (see October 00, 0000), accused Obama of using his grandmother's death to conceal his "efforts" to falsify his Hawaiian birth certificate (see November 00, 0000), and accused Obama of planning to fire all the "competent white men" in government once he became president (see November 00, 0000). Other conservatives, including Fox News's Glenn Beck, will accuse Obama of being a Nazi, or of intending to create a "Reichstag Fire" crisis to gain power (see September 00, 0000 and October 0, 0000). On his radio show, conservative host Glenn Beck warns that the Democrats' "socialistic" health care reform proposal will lead to "eugenics" as envisioned by leaders of the Nazi Third Reich. Beck tells his listeners that the reform package will not only result in senior citizens being forced to die before their time in order to save on medical costs (see November 00, 0000, January 00, 0000, February 0, 0000, February 00, 0000, and February 00, 0000), but also says: "This is Nazi Germany stuff. This is the kind of stuff that is progressive in its nature. It is eugenics. It is survival of the fittest. It is the reason why the abortion argument makes so much difference. You can't devalue life at either end because these people are waiting to swoop in and say it's just not worth doing these things. Don't waste the money on old people. They're not going to live long anyway. Spend it on someone who meets the requirements of our cost-benefit analysis. So old people, thanks for all the contributions you made to society during your better years but now we're sorry to say it's time to send you to a better place, heaven." [Glenn Beck, 0/00/0000] Flowers adorn the sidewalk outside George Tiller's clinic in Wichita, Kansas, laid in his memory. [Source: AP: Charlie Riedel]The family of George Tiller, a doctor who provided late-term abortions as part of his practice before being murdered (see May 00, 0000), decides that his Wichita, Kansas, clinic will be closed permanently. Nebraska doctor LeRoy Carhart, who worked at the clinic, said he was willing to continue, but the decision is the family's. Warren Hern, one of the few remaining doctors in the US who performs late-term abortions, says: "This is what they want, they've been wanting this for 00 years. The anti-abortion fanatics have to shut up and go home. They have to back off and they have to respect other people's point of view. This is a national outrage." Randall Terry, original founder of the anti-abortion group Operation Rescue, says, "Good riddance," and predicts that Tiller's clinic will be remembered similarly to Nazi death camps. In a statement, the Tiller family says, "We are proud of the service and courage shown by our husband and father and know that women's health care needs have been met because of his dedication and service." [Associated Press, 0/0/0000] Laura Ingraham. [Source: Pat Dollard]Fox News and radio talk show host Sean Hannity tells his radio audience of the op-ed published in the morning's New York Post by health industry lobbyist Betsy McCaughey, claiming that the Democrats' health care reform proposal would result in senior citizens being advised to end their lives prematurely (see July 00, 0000). Hannity says: "[I]t sounds to me like they're actually encouraging seniors in the end, ‘Well, you may just want to consider packing it all in here, this is - ' what other way is there to describe this?... So that they don't become a financial burden on the Obamacare system? I mean that's how they intend to cut cost, by cutting down on the health care we can give and get at the end of our lives and dramatically cutting it down for senior citizens? You know, welcome to the brave new world of Obamacare. We're going to encourage, you know, inconvenient people to consider ‘alternatives to living.'" The same day, conservative radio host Laura Ingraham tells her listeners: "Can you imagine - if I were doing Saturday Night Live, like, if I were producing it this weekend, and I was going to be fair about political humor, I would have a hospice chute - like a door, a trap door that goes into a chute where the elderly would just walk up - ‘Oh, my hip hurts.' And all of a sudden you see this leg kicking granny down the chute, and that's Obamacare." She continues by making a veiled reference to Nazi concentration camps: "[S]ome will call them death camps, but this is the way Obamacare is gonna go for America." And on the same day, conservative radio hosts Jim Quinn and Rose Tennent echo Hannity and Ingraham's claims. Quinn says, "[T]here's a drop dead date, you should pardon the expression but a lot of us are going to - " Tennent interjects, "Are going to drop dead, yeah." Quinn then adds, "For heaven's sakes, this is the death-to-old-people plan." [Media Matters, 0/00/0000] Rep. Louis Gohmert. [Source: Associated Press / Washington Blade]Representative Louis Gohmert (R-TX) lays out a skein of theories on radical radio host Alex Jones's broadcast. During his interview with Jones, Gohmert accuses the Obama administration and Congressional Democrats of trying to implement socialism and kill senior citizens; Jones and Gohmert compare Obama to a number of foreign despots. Gohmert tells Jones and his listeners: "We've been battling this socialist health care, the nationalization of health care, that is going to absolutely kill senior citizens. They'll put them on lists and force them to die early because they won't get the treatment as early as they need.... I would rather stop this socialization of health care because once the government pays for your health care, they have every right to tell you what you eat, what you drink, how you exercise, where you live.... But if we're going to pay 000 million dollars like we voted last Friday to put condoms on wild horses, and I know it just says an un-permanent enhanced contraception whatever the heck that is. I guess it follows that they're eventually get around to doing it to us." Gohmert is echoing claims by Republican lawmakers and industry lobbyists that the Democrats' health care reform proposal will kill senior citizens (see November 00, 0000, January 00, 0000, February 0, 0000, February 00, 0000, February 00, 0000, May 00, 0000, June 00, 0000, June 00, 0000, July 00, 0000, July 00, 0000, July 00, 0000, July 00, 0000, July 00, 0000, July 00, 0000, July 00, 0000, and July 00-00, 0000). Jones ups the ante by accusing the White House "science czar" of planning to "put... stuff in the water to sterilize us," and then goes on to accuse the White House of, among other things, implementing a "eugenics control grid over us" and implementing "youth brigades, national service compulsory in a group outside the military under the Democratic Party control in the city year in the red and black uniforms." Gohmert agrees with Jones, and says these kinds of policies were "done in the 0000s," a plain reference to Nazi Germany, "and it's not the only place its been done. It has been done throughout history." Jones says, "Mao did it," referring to Communist China's Mao Zedong. Gohmert agrees: "Well, that's exactly what I was thinking of. This is the kind of the thing we got to stop. We got to get back to the roots, the basics." Gohmert praises Jones for his rhetoric and accusations: "That shows how on top of things you are, Alex." For his part, Jones effusively thanks Gohmert and reminds him that "you're there fighting and we're supporting you." [Think Progress, 0/00/0000] Progressive MSNBC host Rachel Maddow says of Gohmert and Jones: "You know, the Democrats may be fighting it out about whether they're going to be beholden to the insurance companies and whether there's going to be a public option in health care reform. But when it comes to the Republicans, this is the kind of thing they are bringing to the table: Hitler, Mao, and secret plots to kill old people." [MSNBC, 0/00/0000] MacKilip's altered photograph of Obama as a witch doctor. [Source: TPM Muckraker]Dr. David McKalip, a neurosurgeon, resigns as president-elect of the Pinellas County (Florida) Medical Association after circulating a graphic of President Obama dressed as a witch doctor with a bone through his nose. On July 00, he sent the graphic as an e-mail with the heading "Funny stuff," and said that he thought the artist who created the graphic "was expressing concerns that the health care proposals [made by President Obama] would make the quality of medical care worse in our country." McKalip is an outspoken opponent of the White House's proposed health care reform package. He later apologized and denied he is a racist, and called the graphic merely a "satire." In an e-mail to the members of the association, he writes, "For now, in the interest of protecting this movement from any collateral damage, I am withdrawing from making media appearances on health system reform." But in an e-mail to fellow anti-reform protesters, he writes: "Here they come. The first of what likely will be many e-mails accusing me of being a rascist [sic] for forwarding this e-mail of Obama as a witch doctor. Almost like Hillary [Clinton] and the Obama photo form [sic] the presidential campaign.... This may be worth doing a story on about how these ultra-liberal groups like to race bait and avoid the issue.... Lesson learned: Any attempt to discuss politics will lead to a race-baiting war.... Don't let them bait you. I will choose to ignore them and always talk about the issues." In a previous e-mail defending his choice to send the Obama picture, McKalip cited his participation in a "career counseling day several years ago for African-American Boy Scouts." [TPM Muckraker, 0/00/0000; Connexion, 0/00/0000] After learning of McKalip's action, the association called the e-mail "inflammatory," said it "denounced" McKalip's act, demanded a public apology, and called itself "appalled by the statements and act" of McKalip. [Pinellas County Medical Association, 0/00/0000] Anti-reform protesters carry signs depicting Doggett with ‘devil horns' and a sign featuring Nazi SS lettering. [Source: Raw Story]Congressman Lloyd Doggett (D-TX) receives a hostile reception in a town hall meeting in an Austin grocery store. The meeting is to discuss the controversial Democratic health care reform proposal. The crowd is much larger than some had anticipated, and apparently packed with anti-health care reform protesters; anti-reform and anti-Obama signs are prominently displayed, including signs that read, "No Socialized Health Care." Protesters also wave signs with Doggett depicted with devil horns, of a marble tombstone with Doggett's name on it, and with slogans alleging Democrats are Nazis. When Doggett tells the crowd that he will support the reform plan even if his constituents oppose it, many in the crowd begin chanting "Just say no!" and, according to news reports, "overwhelm... the congressman as he move[s] through the crowd and into the parking lot." One resident says of the meeting: "The folks there thought their voices weren't being heard. They were angry, but they were respectful. There wasn't any violence." Another says, laughing: "He jumped in [his car] and fled. It was like he was tarred and feathered and ridden out of town on a rail. It was a beautiful thing." Doggett later notes that because of the disruption, he is unable to engage in discussion with constituents who have other issues, including a father who wants his help in getting his son into a military academy. [Austin American-Statesman, 0/0/0000; New York Times, 0/0/0000; Atlantic Monthly, 0/0/0000]Congressman: Protesters a 'Mob' - Doggett will later characterize the anti-reform protesters as a "mob." In a statement, he says: "This mob, sent by the local Republican and Libertarian parties, did not come just to be heard, but to deny others the right to be heard. And this appears to be part of a coordinated, nationwide effort. What could be more appropriate for the ‘party of no' than having its stalwarts drowning out the voices of their neighbors by screaming ‘just say no!‘... Their fanatical insistence on repealing Social Security and Medicare is not just about halting health care reform but rolling back 00 years of progress. I am more committed than ever to win approval of legislation to offer more individual choice to access affordable health care. An effective public plan is essential to achieve that goal." [Politico, 0/0/0000; CBS News, 0/0/0000]Coordinated by Local Republicans, Washington Lobbyist Firm, 'Tea Party' Group - The protest is coordinated by Heather Liggett, a local Republican Party operative, and by officials with the lobbying firm Americans for Prosperity (AFP), which has organized numerous anti-tax "tea party" demonstrations (see April 00, 0000 and May 00, 0000). Liggett confirms she is part of a national network of conservative organizers putting together anti-reform protests. Doggett says: "This is not a grassroots effort. This is a very coordinated effort where the local Republican Party, the local conservative meet-up groups sent people to my event." Of the event itself, he says: "In Texas, not only with the weather but with the politics, it is pretty hardball around here. I have a pretty thick skin about all of this. But this really goes over the line." And Jennifer Crider, a spokeswoman for the Democratic Congressional Campaign Committee (DCCC), adds: "Conservative activists don't want to have a conversation. They want to disrupt." [New York Times, 0/0/0000] Democratic National Committee (DNC) spokesman Brad Woodhouse says, "The right-wing extremists' use of things like devil horns on pictures of our elected officials, hanging members of Congress in effigy, breathlessly questioning the president's citizenship, and the use of Nazi SS symbols and the like just shows how outside of the mainstream the Republican Party and their allies are." Another group with connections to the "tea party" movement, "Operation Embarrass Your Congressman," helped organize the protest. It says on its Web site: "These arrogant, ignorant, and insolent [Congress members] have embarrassed America, trampled the Constitution, and ignored their constituents for far too long. Attend their townhall meetings during recess and press them with intelligent questions (unlike the mainstream media), asked in an intelligent manner to see if they are really in touch and on board with ‘the will of the people.'" [CBS News, 0/0/0000] After the meeting, FreedomWorks, a conservative lobbying organization that actively promotes disruptive behavior at Congressional town halls (see April 00, 0000), posts video from the meeting, and exhorts its members, "If you know of a town hall meeting your Congressman is having, be sure to show up, bring some friends, and them know what you think." [FreedomWorks, 0/0/0000] Two House Representatives, James McGovern (D-MA) and Richard Neal (D-MA), are booed and heckled during a contentious town hall meeting at the University of Massachusetts to discuss health care reform. Like so many other such forums and meetings, the discussion is disrupted by anti-health care reform protesters, who shout, scream, boo, catcall, and chant throughout the meeting (see June 00, 0000, July 0, 0000, July 00, 0000, July 00, 0000, July 00, 0000, July 00, 0000, August 0, 0000, August 0, 0000, August 0, 0000, August 0, 0000, August 0, 0000, August 0, 0000, August 0, 0000, and August 0, 0000). Both McGovern and Neal support the Obama administration's health care reform proposals. University officials threaten several times to shut down the meeting because of the behavior. One protester shouts that McGovern is like Josef Mengele, the Nazi doctor who performed horrific experiments on concentration camp prisoners during World War II. According to local media reports, protesters outnumber supporters at the meeting. They argue that health care would be strictly rationed and elderly citizens would be denied care altogether, points vehemently disputed by the two congressmen. After the meeting, McGovern says it is plain that health care reform opponents had planned to dominate the meeting with their tactics, but adds: "This is still the United States of America and people have the right to be heard. The meeting wasn't perfect and it wasn't always polite but I got the opportunity to express my view on the subject." [Worcester Telegram & Gazette, 0/0/0000] The logo used for the Obama administration's health care proposal on the White House Web site. The logo combines the Obama presidential campaign's ‘sunrise' emblem with a stylized version of the medical caduceus. [Source: White House]After denouncing House Speaker Nancy Pelosi for claiming that anti-health care protesters had used Nazi symbols and rhetoric in their protests (see August 0, 0000), conservative radio talk show host Rush Limbaugh also makes a comparison between the Obama administration and Nazis. "Adolf Hitler, like Barack Obama, also ruled by dictate," he says. Like Obama, Limbaugh asserts, Hitler "was called the Messiah" and did not need the advice of a cabinet or other advisers to make decisions. "The people spoke through" Hitler, as Limbaugh says Obama believes is the case for himself. Hitler's decisions "sound like the things liberals are doing all over this country." To Pelosi, he says, "You look much more like [a swastika] than any of us [conservatives] ever will." [Media Matters, 0/0/0000; Boston Globe, 0/0/0000] Limbaugh also says that the Obama administration's health care logo looks very much like the "Nazi swastika logo." He adds: "It reminded me of Germany. Something about it reminded me of Germany, 0000. The shape of the logo, the people.... The Obama health care logo is damn close to a Nazi swastika logo.... Ms. Pelosi has some major apologizing to do." He says perhaps Pelosi's supposed "repeated botox injections" have caused her to have "blurry vision" that may have prevented her from seeing the similarities he noticed. [Media Matters, 0/0/0000; Boston Globe, 0/0/0000] Limbaugh apparently gets much of his information, including the Botox joke, from a right-wing blog, "Sweetness and Light," which he credits in his statement. [Sweetness and Light, 0/0/0000] The next day, Rabbi Marvin Hier of the Simon Wiesenthal Center says, "It is preposterous to try and make a connection between the president's health care logo and the Nazi Party symbol, the Reichsadler." [New York Times, 0/0/0000] Jennifer Crider of the Democratic Congressional Campaign Committee (DCCC) responds to Limbaugh's assertions: "Rush Limbaugh's comparison of the Democratic Party to the Nazi Party in World War II is as disgusting as it is shocking. Limbaugh's use of the Nazi swastika in attempting to make a tasteless political comparison has no place in the public discourse. At a time when families need real solutions to rebuild the economy and make health care more affordable, Rush Limbaugh is attempting to sidetrack the important debate through his use of symbols that are synonymous with murder and intolerance. Americans deserve better." [Boston Globe, 0/0/0000] Conservative columnist David Brooks of the New York Times calls Limbaugh's rhetoric "insane." [Media Matters, 0/00/0000] The Anti-Defamation League (ADL), a primarily Jewish organization that battles anti-Semitism, decries the use of Nazi symbols and language in recent health care debates (see July 00, 0000, August 0, 0000, August 0, 0000, August 0, 0000, August 0, 0000, August 0, 0000, and August 0, 0000). In a press release, the ADL's National Director Abraham Foxman, a Holocaust survivor, calls such remarks "outrageous, deeply offensive, and inappropriate." He singles out conservative radio host Rush Limbaugh for specific criticism after Limbaugh repeatedly compares Obama administration policies to those of the Nazis. "Regardless of the political differences and the substantive differences in the debate over health care, the use of Nazi symbolism is outrageous, offensive, and inappropriate," Foxman says. "Americans should be able to disagree on the issues without coloring it with Nazi imagery and comparisons to Hitler. This is not where the debate should be at all.... Comparisons to the Nazis are deeply offensive and only serve to diminish and trivialize the extent of the Nazi regime's crimes against humanity and the murder of six million Jews and millions of others in the Holocaust. I don't see any comparison here. It's off-center, off-issue, and completely inappropriate." [Anti-Defamation League, 0/0/0000] Eric Boehlert. [Source: Simon & Schuster]Eric Boehlert, an author and editor of the progressive news watchdog organization Media Matters, writes that, in his eyes, the media is ignoring the biggest "political story of the year": "the unhinged radical-right response to [President] Obama's inauguration and the naked attempt to dehumanize and delegitimize him through a nonstop smear campaign," which he says is sponsored by the Republican Party and its conservative supporters. "The misguided movement breaks all kinds of taboos in American politics," Boehlert writes, "as well as in the press, and is redefining our political culture - for the worse. Yet the press continues to play dumb." Playing the Nazi Card - Boehlert takes as his springboard the relative disinterest the mainstream media shows to the repeated accusations that Obama and/or Congressional Democrats are Nazis, or Nazi sympathizers, or have Nazi-like goals and ideals (see July 00, 0000, July 00, 0000, August 0, 0000, August 0, 0000, August 0, 0000, August 0, 0000, August 0, 0000, August 0, 0000, and August 00-00, 0000), as well as the virtually unreported use of Nazi symbols and rhetoric at anti-health care protests (see July 00, 0000, August 0, 0000, August 0, 0000, and August 0, 0000). Boehlert notes that in January 0000, the liberal advocacy organization MoveOn received weeks of negative publicity and media attention when it briefly posted two amateur video clips on its Web site submitted as part of a contest for 00-second Internet advertisements against the policies of the Bush administration. The organization removed the clips within hours and apologized for posting them, but was berated for weeks over the ads. Now, Boehlert notes, Rush Limbaugh and other prominent conservative spokespersons routinely use accusations of Nazism in their rhetorical attacks on Obama and Democrats, with virtually no acknowledgement from the press. Boehlert writes: "Despite the fact that Limbaugh has not apologized for his comments - unlike MoveOn in 0000 - and is continuing to compare the Obama White House and the Democratic Party with Nazis, many in the media don't consider it newsworthy and haven't condemned it. And more important, journalists don't show any signs of believing that the episode tells us anything about the radically unhinged nature of the right-wing media in this country today." Apparently, he writes, most media analysts just consider Limbaugh's extreme rhetoric a case of "Rush being Rush." But, he asks, if Limbaugh is going to be considered the de facto leader of conservative thought in America, why isn't he being challenged on his use of what Boehlert calls "his radical and outrageous rhetoric.... He went to a place that previously was considered unconscionable and unpardonable by the press.... Why isn't Limbaugh uniformly condemned for his words?" Accusations of Racism, Racist Pronouncements - And Limbaugh is merely one of many. Fox News commentator Glenn Beck recently accused Obama of being a "racist" and having a "deep-seated hatred of white people" (see July 00-00, 0000), and outside of the small number of progressive/liberal hosts on MSNBC and a few scattered notations in the press, the accusation was virtually ignored. "At the [Washington] Post, which obsesses over the intersection of the media and politics," Boehlert writes, "the jaw-dropping attack by Fox News's superstar host wasn't considered newsworthy. That's correct: Two of the most popular and powerful conservative voices in America have recently called out Obama as a Nazi and a racist." Legitimizing Extremism - Boehlert assigns part of the blame to journalists being "spooked by decades' worth of ‘liberal media bias' attacks" that drive them to "refuse to connect the glaringly obvious dots on display." The extreme rhetorical attacks dovetail with what he calls "the town hall mini-mobs that are wreaking havoc across the country" and "the bizarre birther conspiracy theory" that insists Obama is not a US citizen, but some sort of "plant" from Kenya brought to America to bring down American democracy. "The three right-wing phenomena are all related," he writes, "and they all revolve around a runaway hatred of Obama (as well as the federal government), and they're all being fueled by the [conservative media operation], especially Fox News and Limbaugh, both of which no longer recognize common decency, let alone journalistic standards. Yet instead of putting Limbaugh on the receiving end of well-deserved scrutiny and scorn, rather than turning his comments into a political firestorm, the press plays dumb and actually goes out of its way to legitimize the worst offenders of the GOP's hate brigade." Boehlert condemns ABC News for inviting conservative blogger and columnist Michelle Malkin to take part in a discussion of health care reform "with Pulitzer Prize-winning writers." Malkin, he writes, is a prime member of the "hate brigade," helping push the increasingly angry and violent mob confrontations as well as exhorting readers to believe that the Democrats want to exterminate the elderly (see November 00, 0000, January 00, 0000, February 0, 0000, February 00, 0000, February 00, 0000, May 00, 0000, June 00, 0000, June 00, 0000, July 00, 0000, July 00, 0000, July 00, 0000, July 00, 0000, July 00, 0000, July 00, 0000, July 00, 0000, July 00-00, 0000, July 00, 0000, July 00, 0000, July 00, 0000, July 00, 0000, July 00, 0000 - August 00, 0000, August 0, 0000, and August 00, 0000). The New Yorker recently praised Michael Savage, who routinely attacks women, gays, liberals, and minorities with the worst rhetorical excess (see January 00, 0000, February 0, 0000, February 00, 0000, March 00, 0000, April 0, 0000, June 0, 0000, June 0, 0000, August 00, 0000, October 0-00, 0000, October 00, 0000, October 00, 0000, November 00, 0000, and November 00, 0000), calling him "fun" and "addictive." Comparing the Statistics - Boehlert notes that in January 0000, the Indianapolis Star published five letters to the editor about the MoveOn controversy. To date, it has not published a single letter about Limbaugh's Nazi accusations towards Obama or Democrats. In January 0000, 00 of the nation's largest newspapers published a total of 00 stories, articles, or letters about the MoveOn controversy. To date, that group has published a combined total of six stories about Limbaugh's Nazi allegations. No paper has printed more than one story on the topic. In January 0000, the MoveOn-Nazi story garnered 000 percent more airtime on CNN than the Limbaugh-Nazi story has received. [Media Matters, 0/00/0000] An elderly protester outside the Raleigh, North Carolina, office of Representative Brad Miller (D-NC) ties together two popular claims of the anti-health care reform movement: the reform proposal will kill senior citizens, and the reformers are like Nazis. The unidentified protester tells a television interviewer: "Hitler got rid of his undesirable citizens through ovens. [President] Obama wants to get rid of people like me through hospice.... If [people] are a certain age, grim reapers calling themselves as counselors will go and tell them to take a pill and just die." [New York Times, 0/00/0000] A member of the LaRouche Youth Movement compares the Obama health care reform proposal to Nazi policies. [Source: Darren McCollester / Getty Images]A testy Representative Barney Frank (D-MA) loses patience with a raucous, shouting crowd of angry protesters at a two-hour town hall meeting in Dartmouth, Massachusetts. Frank, who strongly supports the Democrats' health care reform proposals, attempts to answer the shouted questions and accusations from protesters, who often attempt to shout him down before he can complete his answers, and boo him from the moment he is introduced. Frank repeatedly asks, "You want me to talk about it or do you want to yell?" and asks, "Which one of you wants to yell next?" He also says frequently: "Disruption never helps your cause. It just looks like you're afraid to have rational discussion." Frank finally loses patience when Rachel Brown of the LaRouche Youth Movement tells him that President Obama's health care policies are comparable to those of Nazi Germany, meanwhile waving a pamphlet depicting Obama with a Hitler mustache. "This policy is actually already on its way out," Brown says. "It already has been defeated by LaRouche. My question to you is, why do you continue to support a Nazi policy as Obama has expressly supported this policy? Why are you supporting it?" Frank, a Jew, retorts: "When you ask me that question, I'm going to revert to my ethnic heritage and ask you a question: On what planet do you spend most of your time? You stand there with a picture of the president defaced to look like Hitler and compare the effort to increase health care to the Nazis." He says her ability to deface an image of the president and express her views "is a tribute to the First Amendment that this kind of vile, contemptible nonsense is so freely propagated," and concludes: "Trying to have a conversation with you would be like trying to argue with a dining room table. I have no interest in doing it." During less contentious moments, Frank rebuts claims that the reform proposal would mandate free health care for illegal immigrants, and attempts to read the pertinent section of the bill through the shouts and catcalls. He asks why protesters demand for him to answer and then scream through his answers: "What's the matter with you all? I don't know if you get angrier when I answer the questions, or when you don't think I do." [Associated Press, 0/00/0000; CNN, 0/00/0000; Think Progress, 0/00/0000; Seattle Post-Intelligencer, 0/00/0000; Boston Globe, 0/00/0000]'Look for the Mustache' - A representative of the Massachusetts Republican Party later says Brown and other LaRouche supporters were at the forum "to cause problems," and denies any Republican involvement in the shouting or pamphleteering. A LaRouche spokeswoman, Nancy Spannaus, says, "LaRouche PAC members are giving leadership to these town hall meetings all around the country so we are being at any one that we possibly can." The Obama "mustache poster" "symbolizes the fact that the president is attempting to implement a Hitler health care policy," she adds. "At any town hall, you'll know LaRouche people are there if you just look for the mustache." [Washington Post, 0/00/0000]Fox News: Frank's Remarks Proof that Democrats are 'Alienating Voters' - Fox News talk show host Sean Hannity and a Fox reporter, Griff Jenkins, say that Frank's retorts to the protesters are proof that Democrats are "alienating voters" with their reform policies. Jenkins tells Hannity that Frank "talked down" to the protesters. Hannity calls Frank's comments full of "arrogance [and] condescension." Representative Michele Bachmann (R-MN), Hannity's guest, praises the LaRouche questioner and other protesters as evidence of American "democracy in action." [Fox News, 0/00/0000] Michael Savage, a conservative radio host, tells his listeners that President Obama wants to have an "Obama Youth" program similar to the Nazi's "Hitler Youth." Referring to a speech Obama is preparing that is aimed at schoolchildren (which Savage falsely claims Obama "is going to force" children to listen to), Savage says "every dictator" such as Adolf Hitler, Cuba's Raul and Fidel Castro, and others have routinely attempted to "brainwash" their young citizens by making speeches to them. "[Y]ou gotta hand it to dictators," Savage says, "they see the future. That once they seize total power, they need a generation that loves them. Hitler had the Hitler Youth, and Obama would like to have the Obama Youth. Now he can't create the Obama Youth Corps overnight, but he can certainly address the schoolchildren of America as a captive audience and sell them on fraudulent ideas such as global warming, health care for all, higher taxation for the pig rich, and things of this nature, and that's what happens under a dictatorship, things of this nature." [Media Matters, 0/0/0000] Obama will give the speech on September 0; it contains non-controversial reminders for schoolchildren to "stay in school" and "work hard," and contains no references to "global warming," "health care for all," or taxing the "pig rich." Schools are invited to broadcast the speech into classrooms or assemblies, but are not required to by either the White House or local school boards. After the speech, a Baltimore teacher will tell a reporter that she is disappointed that the country has "become so polarized that we believe that our president is an enemy and not our leader." During George W. Bush's presidency, she will say, "whether I disagreed or not, I still saw him as a leader." [White House, 0/0/0000; CNN, 0/0/0000] Savage has called the landmark civil rights decision Brown v. Board "sickening" (see May 00, 0000), accused Obama of being educated in a radical Islamic madrassa (see January 00, 0000 and April 0, 0000) and being a potential "radical Muslim" (see February 00, 0000), called Obama's presidential victory "the first affirmative-action election in American history" (see February 0, 0000), accused Obama of being sympathetic towards the Nazis and the Imperial Japanese of World War II (see March 00, 0000), said that homeless Americans should be put in "work camps" (see June 0, 0000), called Obama an "Afro-Leninist" (see June 0, 0000), said that welfare recipients should lose the right to vote (see October 00, 0000), accused Obama of using his grandmother's death to conceal his "efforts" to falsify his Hawaiian birth certificate (see November 00, 0000), accused Obama of planning to fire all the "competent white men" in government once he became president (see November 00, 0000), and called Obama a "dictator" (see March 0-0, 0000). Fox News host Glenn Beck, in an interview with the conservative Web news provider NewsMax, says he "fears a Reichstag moment" from the Obama administration. Beck is referring to the 0000 burning of the German parliament building in Berlin that the Nazis blamed on Communists, and that Adolf Hitler used as an excuse to eliminate constitutional liberties and consolidate power. Beck says he "fears" that the Obama administration will either orchestrate, or take advantage of, a similar situation in America to abolish constitutional democracy and institute a tyrannical rule. "I fear a Reichstag moment," he says. "God forbid, another 0/00. Something that will turn this machine on, and power will be seized and voices will be silenced. God help us all." The NewsMax article will subsequently be removed from the provider's Web site, but the progressive media watchdog organization Media Matters writes an article about the interview, as do several other news organizations. [Media Matters, 0/00/0000; Chicago Tribune, 00/0/0000] Fox News host Glenn Beck compares the National Education Association to Nazis. Beck, discussing a recent conference call by NEA officials in which artists reportedly discussed how "to help lay a new foundation for growth, focusing on core areas of the recovery agenda," says that "advocating through art is known as propaganda. Hmm. You should look up the name Goebbels." Beck is referring to Joseph Goebbels, the minister of propaganda during the Nazi regime. [Media Matters, 00/0/0000] Banner at the Capitol Hill rally depicting House Speaker Nancy Pelosi as an ‘Unamerican McCarthyite.' [Source: MSNBC]Conservatives gather on Capitol Hill to protest the Obama administration's push towards health care reform, in a rally featuring guest speaker Representative Michele Bachmann (R-MN). [Media Matters, 00/0/0000] Bachmann called the rally the "Super Bowl of Freedom," and told Fox News viewers that "socialized medicine is the crown jewel of socialism. This [health care reform] will change our country forever." [TPM LiveWire, 00/0/0000; Mediaite, 00/00/0000] Actor Jon Voight, speaking to the crowd, says of President Obama: "His only success in one year as president is taking America apart piece by piece. Could it be 00 years of ‘subconscious programming' from Reverend [Jeremiah] Wright [Obama's former pastor] to damn America?" And House Minority Leader John Boehner (R-OH) tells the crowd, "Pelosi care [referring to House Speaker Nancy Pelosi, D-CA] is the greatest threat to freedom I've seen in my 00 years in Washington." Signs Use Racial Images; Call Obama Communist, Nazi - Signs visible in the crowd proclaim, among other sentiments: "Get the Red Out of the White House"; "Waterboard Congress"; "Traitor to the US Constitution" (featuring a photo of Obama); "Ken-Ya Trust Obama?" (referring to theories that Obama is a citizen of Kenya - see January 00, 0000, January 00, 0000, August 0, 0000 and After, October 0-00, 0000, and August 00, 0000 - and with autographs from Representatives Steve King, R-IA and Ron Paul, R-TX); "Un-American McCarthyite" (featuring a photo of Pelosi); "I'm the King of the World: Remember the Titanic?" (featuring a drawing of Obama as the "Jovial Sambo" character from the Jim Crow era); "National Socialist Health Care" (featuring a photograph of a pile of corpses from the Holocaust, and claiming that health care reform is the next "holocaust"). Nine rally participants are arrested for attempting to force their way into the Hart Senate Office Building. Hundreds more attempt to force themselves into nearby government office buildings while chanting, "Kill the bill!" [MSNBC, 00/0/0000]Sponsored by GOP - MSNBC's Domenico Montanaro writes: "It is important to know that this rally was set up by the GOP. While other groups certainly got people to show up, the folks who came here ultimately came at the invitation of the Republican Party. The GOP provided the speakers and the music, etc." [MSNBC, 00/0/0000]Fox Pundit Inflates Crowd Estimates - While other media sources use local police reports to estimate the crowd at around 0,000, Fox News's Sean Hannity tells listeners that the crowd is closer to 00,000 in size. Hannity later drastically scales back this claim. Hannity, who along with other Fox News pundits and on-air anchors had heavily promoted the rally for days beforehand, predicted the crowd would be "massive" in the hours before the protest. On his radio show, aired on ABC Radio Network, Hannity tells listeners: "We announced on Hannity Friday night on the Fox News Channel, we had Congresswoman Michele Bachmann on, and she mentioned that there was going to be on Thursday, she was going to put together in less than a week a little town hall on - what do you want to call it - march on our nation's Capitol. And anyway, 00,000 people showed up today." Hannity echoes the claim several times on his radio show. However, with no explanation, he concludes his radio broadcast by saying, "I heard there was, like, 0,000 people plus there." [MSNBC, 00/0/0000; Media Matters, 00/0/0000] On Hannity's Fox News broadcast later that evening, he returns to his earlier estimates of "00,000" rally participants, and shows viewers old footage from Glenn Beck's 0/00 rally (see September 00, 0000) to bolster his claim. [Crooks and Liars, 00/00/0000] On November 00, Hannity will admit that he "screwed up" in showing the footage, and claims it was merely "an inadvertent mistake." [Think Progress, 00/00/0000] Hannity does not address how the mistake came to be made. [New York Times, 00/00/0000] Media critic Rachel Sklar will write, "It's really blatant and remarkable... this sort of misrepresentation is simply not an accident." [Mediaite, 00/00/0000] A week later, Fox News anchor Gregg Jarrett will make a similar mistake (see November 00-00, 0000). Fox News host Glenn Beck, speaking on his daily radio show, says that "progressives build the structure that a communist, a Marxist, a Nazi would love to have." Beck is accusing the Obama administration of emulating Nazis and Communism by its programs of what he calls the "redistribution of wealth and social justice of wealth - engineered justice." [Media Matters, 0/00/0000] The American Jewish Coalition logo. [Source: The New Jew (.com)]The American Jewish Coalition (AJC) urges the Republican Party leadership to condemn former House Speaker Newt Gingrich (R-GA)‘s assertion that the Obama administration's policy agenda is as "great a threat to America as Nazi Germany and the Soviet Union." Gingrich, a paid Fox News commentator, made the statement in a recently published book, To Save America; in interviews promoting the book, he has called the Obama administration a "secular socialist machine" similar in fashion and beliefs to the Nazi and Soviet regimes. The AJC's executive director, David Harris, says: "By invoking the current administration in the same breath as two murderous totalitarian states, Newt Gingrich has drawn a foolish and dangerous analogy. Gingrich's linkage not only diminishes the horror of the Holocaust, it also licenses the use of extremist language in contemporary America." Gingrich has said he is not drawing moral distinctions, but has gone on to say that because of the Obama initiatives, "we are going to be in a country which no longer resembles America." Harris says: "It is vital that the Republican leadership say clearly that such analogies are unacceptable. Unfortunately, as the recent controversy over the new immigration law in Arizona also demonstrates, demonizing political opponents as Nazis is becoming all too common in American political debate." [Media Matters, 0/00/0000; American Jewish Coalition, 0/00/0000] On Fox News, Chris Wallace asks Gingrich if his claim isn't "wildly over the top." MSNBC's Joe Scarborough, a former Republican congressman, calls Gingrich's comments "sick," "shameful," and "so over the top," and adds, "I hope you apologize." [Media Matters, 0/00/0000; Media Matters, 0/00/0000; Media Matters, 0/00/0000] President Obama during his May 00, 0000 speech at West Point. [Source: Potusphere (.com)]Michael Savage, a conservative radio host, tells his listeners that President Obama does not have the right to speak to the graduating class of the Army's United States Military Academy at West Point, New York, and calls Obama "Little Mussolini," after the Italian fascist dictator and ally of Adolf Hitler. Savage, referring to Obama's May 00 speech to the graduating class of cadets, says Obama "slipped and gave himself away" during his speech, calls Obama "insecure" and "terrified," and says Obama neither had the "right" nor the "honor to speak to the cadets," and "is not qualified to speak to the cadets." Obama, Savage says, "overcompensate[d]" during the speech by saying, "I have absolute power in some areas." Savage then says: "That was to show the boys and the men at West Point who their boss was. It was ‘Little Mussolini,' ‘Junior Doc' Obama [referring to Haitian dictators ‘Papa Doc' and ‘Baby Doc' Duvalier], who told them in no uncertain terms: ‘Don't you dare think that I am not in charge. I'll show you.'" Savage goes on to say Obama has "a woman problem" that has something to do with "the peripatic nature of his mother during his upbringing," and questions Obama's loyalty to the United States, asking if his loyalities "lie here [in the US] or somewhere else.... We suspect [they do] not lie here in Washington, DC." [Media Matters, 0/00/0000] The reference to "absolute power" is a joke Obama made at the beginning of his speech. He told the cadets: "As your superintendent indicated, under our constitutional system my power as president is wisely limited. But there are some areas where my power is absolute. And so, as your commander in chief, I hereby absolve all cadets who are on restriction for minor conduct offenses. I will leave the definition of ‘minor' to those who know better." Obama received applause and laughter from the cadets for the wisecrack. [CBS News, 0/00/0000] Savage has called the landmark civil rights decision Brown v. Board "sickening" (see May 00, 0000), accused Obama of being educated in a radical Islamic madrassa (see January 00, 0000 and April 0, 0000) and being a potential "radical Muslim" (see February 00, 0000), called Obama's presidential victory "the first affirmative-action election in American history" (see February 0, 0000), accused Obama of being sympathetic towards the Nazis and the Imperial Japanese of World War II (see March 00, 0000), said that homeless Americans should be put in "work camps" (see June 0, 0000), called Obama an "Afro-Leninist" (see June 0, 0000), said that welfare recipients should lose the right to vote (see October 00, 0000), accused Obama of using his grandmother's death to conceal his "efforts" to falsify his Hawaiian birth certificate (see November 00, 0000), accused Obama of planning to fire all the "competent white men" in government once he became president (see November 00, 0000), accused Obama of desiring his own "Hitler Youth" program (see September 0, 0000), compared Obama to Chinese Communist dictator Mao Zedong (see December 0, 0000), and compared Obama to mass murderer Pol Pot (see December 00, 0000). Jewish leaders meet privately with Fox News chief executive Roger Ailes to complain about the repeated anti-Semitism of Fox talk show host Glenn Beck. Simon Greer, the head of Jewish Funds for Justice, tells Ailes and Fox News senior vice president Joel Cheatwood that he was disturbed when Beck, on his broadcasts, compared his worldview to that of the Nazis and accused him of trying to create American "death camps"; Ailes and Cheatwood agree that Beck went too far and promise to discuss the matter with him. Two days later, Greer will receive a handwritten note from Beck that reads: "Please know that I understand the sensitivity and sacred nature of this dark chapter in Human History. Thank you for your candor and helpful thoughts." Greer has said Beck "has a history of recklessly invoking Nazi Germany and the Holocaust in order to advance his political agenda," a statement bolstered by research from the Washington Post. However, Beck has been praised by some Jewish figures for his support of Israel. Cheatwood later disagrees with Greer, saying neither he nor Ailes said Beck had crossed any lines, and adds, "We absolutely stood behind Glenn Beck 0,000 percent." [Yahoo! News, 0/0/0000] On Fox News's morning broadcast Fox and Friends, former House Speaker Newt Gingrich, a frequent Fox commentator and presumptive Republican candidate for president in 0000, says of the controversial plans to build an Islamic community center two blocks from the site of the downed World Trade Center: "Nazis don't have the right to put up a sign next to the Holocaust museum in Washington. We would never accept the Japanese putting up a site next to Pearl Harbor. There's no reason for us to accept a mosque next to the World Trade Center." [Media Matters, 0/00/0000] Glen Urquhart. [Source: Glen Urquhart for Congress]The Democratic Congressional Campaign Committee (DCCC) releases a video showing Delaware Republican primary winner Glen Urquhart (R-DE) comparing believers in the separation of church and state to Nazis. Urquhart is running for the House seat vacated by moderate Republican Mike Castle (R-DE), who lost a contentious Delaware Senate primary to right-wing candidate Christine O'Donnell (see September 00, 0000). Both O'Donnell and Urquhart are backed by state and national "tea party" organizations. The DCCC is attempting to portray O'Donnell, Urquhart, and others as right-wing extremists. The video shows Urquhart speaking directly to the cameras, saying that the idea of the separation of church and state originated not with the Founding Fathers, but with Adolf Hitler: "Do you know, where does this phrase separation of church and state come from? Does anybody know?... Actually, that's exactly, it was not in [Thomas] Jefferson's letter to the Danbury Baptists. He was reassuring that the federal government wouldn't trample on their religion. The exact phrase ‘separation of church and state' came out of Adolf Hitler's mouth, that's where it comes from. Next time your liberal friends talk about the separation of church and state ask them why they're Nazis." Urquhart's spokesman David Anderson says the candidate has repeatedly apologized for the remarks, and says Urquhart "believes 000 percent in religious freedom for all Americans." He was merely speaking out against what he calls the "oppression of religious freedom in the name of separation of church and state.... The phrase he used was unfortunate, and he apologized for it." [The Hill, 0/00/0000; CBS News, 0/00/0000] CBS News notes that Jefferson indeed used the phrase "separation of church and state" in his letter to the Danbury, Connecticut, Baptist Association, writing, "I contemplate with sovereign reverence that act of the whole American people which declared that their legislature should ‘make no law respecting an establishment of religion, or prohibiting the free exercise thereof,' thus building a wall of separation between church & state." [CBS News, 0/00/0000; Jefferson, 0/00/0000] Washington Post columnist Dana Milbank, in an examination of Fox News host Glenn Beck's slippery grasp of history, notes that Beck routinely invokes Nazi dictator Adolf Hitler and former US President Woodrow Wilson in comparisons to President Obama. Beck has accused Obama and his administration of supporting "eugenics" similar to those advocated by some Nazis (see May 00, 0000), claimed that Obama, like the Nazis, believes in enforced sterilization, claimed that Obama would create "death panels" to decide who lives and dies under his health care reform proposals (see August 00, 0000), told his viewers to "read Mein Kampf" if they want to understand Obama's ideology, repeatedly accused the Obama administration of "fascism" (see September 00, 0000), claimed the Obama "brownshirts" were readying a strategy to arrest Beck and other Fox News personnel in an attempt to shut down the network, accused the United Nations of "Nazism" in pursuing efforts to curb global warming, said Obama wanted to create his own version of the SS and Hitler Youth in revamping and expanding AmeriCorps (see March 00, 0000), and more. Milbank notes that Beck either gives no evidence whatsoever to bolster his claims, or gives evidence that is either misrepresented or entirely false. Milbank writes: "Beck, it seems, has a Nazi fetish. In his first 00 months on Fox News, from early 0000 through the middle of this year, he and his guests invoked Hitler 000 times. Nazis, an additional 000 times. Fascism or fascists, 000 times. The Holocaust got 00 mentions, and Joseph Goebbels got 00. And these mentions are usually in reference to Obama." As for Wilson, Beck routinely labels the former president a "racist" "horror show" who was "the spookiest president we ever had," usually in preparation for comparing him to Obama. [Washington Post, 00/0/0000] Six weeks later, Fox News president Roger Ailes, defending Beck, will tell an interviewer that Milbank should be "beheaded" for criticizing Beck (see November 00-00, 0000). Foster Friess. [Source: New York Magazine]Foster Friess, a multi-millionaire who is the chief supporter of a "super PAC" supporting the presidential candidacy of Rick Santorum (R-PA), weighs in on the controversy surrounding new federal mandates for providing birth control in employers' health care coverage. Friess dismisses the controversy by suggesting that if women just kept their legs closed, they would not need contraception. In an interview with MSNBC's Andrea Mitchell, Friess is asked if Santorum's rigid views on sex and social issues (see April 0, 0000, April 00, 0000 and After, January 0000, January 0, 0000, October 00, 0000 and After, June 0000, September 00, 0000, January 0-0, 0000, January 0, 0000 and January 0, 0000) would hurt his chances in the general election. Friess responds by saying: "I get such a chuckle when these things come out. Here we have millions of our fellow Americans unemployed; we have jihadist camps being set up in Latin America, which Rick has been warning about; and people seem to be so preoccupied with sex. I think it says something about our culture. We maybe need a massive therapy session so we can concentrate on what the real issues are. And this contraceptive thing, my gosh, it's [so] inexpensive. Back in my day, they used Bayer aspirin for contraceptives. The gals put it between their knees and it wasn't that costly." Mitchell says, "Excuse me, I'm just trying to catch my breath from that, Mr. Friess, frankly." Think Progress's Alex Seitz-Wald writes: "Given that [a]spirin is not a contraceptive, Friess seems to be suggesting that women keep the pill between their knees in order to ensure the[ir] legs stay closed to prevent having sex. Conspicuously, Friess doesn't put the same burden on men." [Think Progress, 0/00/0000; National Public Radio, 0/00/0000] Friess's comment draws quick reaction from a number of sources, with many women's groups expressing their outrage. Santorum quickly distances himself from the comment, calling it a "bad joke" and implying that the media is trying to smear him with it: "When you quote a supporter of mine who tells a bad off-color joke and somehow I am responsible for that, that is ‘gotcha,'" he tells a CBS News reporter. [Washington Post, 0/00/0000] Fox News's late-night political humor show, Red Eye, features guest host Andy Levy sarcastically speculating that Friess's joke is part of a "guerrilla marketing" scheme by the Bayer Corporation, which manufactures Bayer aspirin. Guest Anthony Cumia dismisses Friess's comment by saying that Friess is "an old guy, he's got old jokes." [Mediaite, 0/00/0000] The next day, Friess issues an apology on his blog that reads: "To all those who took my joke as modern day approach I deeply apologize and seek your forgiveness. My wife constantly tells me I need new material - she understood the joke but didn't like it anyway - so I will keep that old one in the past where it belongs." New York Magazine's Dan Amira writes, perhaps sarcastically, that he does not understand why either Santorum or Friess apologized, as he believes Friess stated Santorum's position on sex and birth control rather clearly. "‘Hold an aspirin between your knees' is just a more colorful way of saying, ‘just keep your legs closed,' which is tantamount to ‘just don't have sex,'" Amira writes. "It's abstinence, pure and simple. Which is exactly what Santorum advocates. He's said that unless you're trying to procreate, you shouldn't be having sex, and therefore, contraception is ‘not okay.' He has promised to make this argument to the American people as president. As far we can tell, the only difference between Friess's bad contraception joke and Santorum's actual contraception beliefs is an aspirin." [New York Magazine, 0/00/0000; Foster Friess, 0/00/0000] Friess is often described in the press as a "billionaire," but both Friess and Forbes magazine say that appellation is inaccurate. [Forbes, 0/0/0000] Sandra Fluke. [Source: Alex Wong / Getty Images / New York Times]Conservative radio talk show host Rush Limbaugh insults Sandra Fluke, the Georgetown University law student who testified in favor of federal law mandating that health care providers pay for contraception (see March 0, 0000), as a "slut" and a "prostitute" who wants the government to pay her for having sex. On his radio show, Limbaugh, who wrongly identifies her as "Susan" Fluke, says: "What does it say about the college coed Susan Fluke, who goes before a congressional committee and essentially says that she must be paid to have sex? What does that make her? It makes her a slut, right? It makes her a prostitute. She wants to be paid to have sex. She's having so much sex she can't afford the contraception. She wants you and me and the taxpayers to pay her to have sex. What does that make us? We're the pimps. The johns, that's right. We would be the johns - no! We're not the johns. Well - yeah, that's right. Pimp's not the right word. Okay, so, she's not a slut. She's round-heeled. I take it back." Think Progress reporter Alex Seitz-Wald comments on Limbaugh's characterization, "While it's probably not even worth engaging with Limbaugh on the facts, Fluke's testimony was about a friend who is a lesbian and needed birth control for non-sexual medical reasons, so he's only wrong about three times over, and offensive many more times over than that." Seitz-Wald notes that Fluke never discussed her own use, or non-use, of contraception, nor did she allude to her sexual activities at all. [Media Matters, 0/00/0000; Think Progress, 0/00/0000; Think Progress, 0/0/0000]Misrepresentation - Seitz-Wald will note that Limbaugh is deliberately misrepresenting Fluke's position and the position of Congressional Democrats. "Fluke's testimony, and the entire contraception debate, is about insurance companies paying for contraception as part of their health coverage, the... way they pay for any other medication, such as Viagra. Morevoer, Fluke's testimony was not about herself, but about a friend who need contraception to fight cancer and other fellow law students. This conservative narrative, which is pure fantasy, seems to be based on a single bogus article from Cybercast News Service (CNS), which Limbaugh repeatedly cites, with the ludicrous headline, ‘Sex-Crazed Co-Eds Going Broke Buying Birth Control, Student Tells Pelosi Hearing Touting Freebie Mandate.'" [CNS News, 0/00/0000; Think Progress, 0/0/0000]Other News Outlets Join Limbaugh - Other conservative news outlets join Limbaugh in attacking Fluke and other women who use contraception. In the article cited by Limbaugh, CNS's Craig Bannister says that "sex-crazed co-eds" like Fluke should cut back on the amount of sex they're having to pay for other needs such as books and food. Fox News's Trace Gallagher mocks Fluke, saying: "And see, I was gonna go to law school, but I thought all you did was study in law school, right? So, I guess I was wrong on that." Fox News correspondent Monica Crowley says the government should not pay Fluke and others to have "recreational sex." CNN commentator Dana Loesch calls Fluke and other women "nymphos" for wanting access to contraceptives, and says Fluke and feminists "support... female genocide." [Media Matters, 0/00/0000; CNS News, 0/00/0000]Fox Business Commentator: Fluke's Testimony Part of a Pro-Abortion Scheme by House Minority Leader - On Fox Business Channel's Lou Dobbs Tonight, regular guest Bill Donohue calls Fluke a "little brat." Dobbs asks Donohue to comment on what he calls Fluke's demand that she be given free contraception, a mischaracterization of Fluke's testimony (and one contradicted by the clip of her testimony Dobbs plays before Donohue's comments). Donohue begins by lambasting Georgetown for having a group called "Hoyas for Choice," which he calls "Hoyas for Abortion," but not groups like "Hoyas for Racism" or "Hoyas for Anti-Semitism." Donohue suggests that the university and Hoyas for Choice raise "the nine dollars a month" Fluke needs for her personal contraception needs, and Dobbs notes that Georgetown is "one of the most expensive universities in the country." Donohue attacks Fluke for "obviously dressing well" but then asking taxpayers to pay for her contraception and, without basis in fact, for her university education to boot. Why aren't taxpayers funding his anti-gout medication? he asks. "This is what we've come down to in this country," he concludes. "You have these little brats who come on TV and they testify and they say, ‘I want, I want, I want,' and somehow I have a moral responsibility? They have a lien on me to pay this? It's all about getting the Catholic Church, obviously, to pay for their abortion-inducing drugs, which is why we're having this debate." Donohue says that Fluke's testimony is part of a scheme by House Minority Leader Nancy Pelosi (D-CA), "who actually brought her on there to speak," to force the Catholic Church to amend its position on abortion. [Media Matters, 0/00/0000]'Shockingly Ugly Hatred' - Conservative blogger Charles Johnson, who in recent years has become highly critical of the race- and gender-based rhetoric from the right, writes that the right's reaction to Fluke constitutes "shockingly ugly hatred," and says Limbaugh's attack is "another step into the gutter." [Charles Johnson, 0/00/0000] Atlantic columnist Ta-Nehisi Coates agrees with Johnson, noting that Limbaugh is not just an "entertainer," but a powerful opinion leader of the Republican Party, and says that Limbaugh's comments are part of what Coates calls "the normalization of cruelty" and "evidence of the lowest aspects of humanity." [Atlantic, 0/0/0000] Eric Boehlert, a senior writer at the liberal media watchdog Web site Media Matters, calls Limbaugh's "radio outburst" an example of his "rancid misogyny," and writes: "[I]t was perhaps the talk show host's incessant need to bully powerless people from the safety of his studio that was so striking. That, and the glee Limbaugh seemed to take in not only maligning the young woman, but her parents as well. It's jaw-dropping." Boehlert goes on to remind readers that Limbaugh is not just a voice on the radio or an entertainer, but "the voice of America's conservative movement, as well as the Republican Party." [Media Matters, 0/0/0000]House Democrats Call for Condemnation - House Democrats, including Pelosi, call for Republican Congressional leaders to condemn Limbaugh's remarks (see February 00, 0000). Statement from Law Student - Fluke will issue a statement repudiating Limbaugh's rhetoric (see March 0, 0000). Continued Attacks - Limbaugh will continue his attacks on Fluke the next day (see March 0, 0000). Conservative talk show host Rush Limbaugh spends much of his three-hour show lambasting Georgetown University law student Sandra Fluke, who testified in opposition to a House amendment that would have allowed health care providers to deny contraceptive coverage and other health care necessities if they had religious or moral objections (see March 0, 0000). The day before, Limbaugh called Fluke a "slut" and a "prostitute" who is having "so much sex she can't afford the contraception" and wants the government to pay for it (see February 00, 0000). Limbaugh begins by saying that Fluke and others who criticized his comments (see February 00, 0000 and March 0, 0000) were having "a conniption fit" that he finds "hilarious." He offers a compromise, offering to buy "all the women at Georgetown University as much aspirin to put between their knees as possible" (see February 00-00, 0000), and says he believes he is being "quite compassionate." Limbaugh later returns to the topic, saying that having the government pay for contraception is "flat-out thievery" that would force taxpayers to pay to "satisfy the sexual habits of female law students at Georgetown." He characterizes Fluke's objections to the House amendment as her saying: "I'm going broke having sex. I need government to provide me condoms and contraception. It's not fair.... Ms. Fluke, have you ever heard of not having sex? Have you ever heard of not having sex so often?... Who bought your condoms in junior high? Who bought your condoms in the sixth grade? Or your contraception. Who bought your contraceptive pills in high school?" He says Fluke is apparently "having so much sex, it's amazing she can still walk.... She and her co-ed classmates are having sex nearly three times a day for three years straight, apparently these deadbeat boyfriends or random hookups that these babes are encountering here, having sex with nearly three times a day." He advises Fluke that she can get "free condoms and lube" from the Washington, DC, Department of Health. He then says: "So, Ms. Fluke and the rest of you feminazis (see May 00, 0000 and July 0000), here's the deal. If we are going to pay for your contraceptives, and thus pay for you to have sex, we want something for it, and I'll tell you what it is. We want you to post the videos online so we can all watch." He finishes his tirade by accusing Fluke of being "a plant... an anti-Catholic plant from the get-go" who is working behind the scenes as part of a "Democratic plot" to "create a new welfare program and, at the same time, try to cast Republicans in an election year as anti-female." Fluke, he says, is "a woman who is happily presenting herself as an immoral, baseless, no-purpose-to-her life woman. She wants all the sex in the world whenever she wants it, all the time, no consequences. No responsibility for her behavior." He concludes that he, not Fluke, is the victim, and says he is being persecuted by those who wish to see him removed from the airwaves. [Think Progress, 0/0/0000; Media Matters, 0/0/0000; MSNBC, 0/0/0000] President Obama calls Sandra Fluke, the Georgetown University law school student who has been subjected to vociferous attacks and personal smears by conservative talk show host Rush Limbaugh and others (see February 00, 0000 and March 0, 0000) after publicly opposing a Republican-backed amendment that would have allowed health care providers and insurers to deny coverage of contraception and other provisions on moral or religious grounds (see March 0, 0000). Obama asks Fluke if she is "okay" after the attacks, thanks her for speaking out on the issue, and tells her that her parents should be proud of her. Fluke takes the call at the MSNBC building in New York, while waiting to be interviewed by MSNBC's Andrea Mitchell. Of the call, she tells Mitchell: "He encouraged me and supported me and thanked me for speaking out about the concerns of American women. What was really personal for me was that he said to tell my parents that they should be proud. And that meant a lot because Rush Limbaugh questioned whether or not my family would be proud of me. So I just appreciated that very much.... He did express his concern for me and wanted to make sure that I was okay, which I am. I'm okay." She tells Mitchell that the vilification from Limbaugh has been "surreal." After the call, White House press secretary Jay Carney says Obama made the telephone call because he feels that "the kinds of personal attacks that have been directed her way have been inappropriate. The fact that our political discourse has been debased in many ways is bad enough." He adds: "It's even worse when it is directed at a private citizen who is simply expressing her views about public policy.... The president expressed to Sandra Fluke that he was disappointed that she was the subject of these crude - of these personal attacks. I think that it's fair to say that - reprehensible was my word, but look, these were unfortunate attacks that were leveled against her and the president feels that way.... They were, inappropriate and reprehensible. But the point is the president called her to thank her for speaking out on a matter and doing so with great poise on a matter - on a public policy matter and to express his disappointment that she had been subjected to these kinds of attacks." [MSNBC, 0/0/0000; Huffington Post, 0/0/0000; CBS News, 0/0/0000] Days later, Obama will tell a Washington Post reporter that he called Fluke in part because he was thinking of his daughters Malia and Sasha. "I don't know what's in Rush Limbaugh's heart, so I'm not going to comment on the sincerity of his apology" (see March 0, 0000 and March 0, 0000), Obama will say. "What I can comment on is the fact that all decent folks can agree that the remarks that were made don't have any place in the public discourse." He says he called "because I thought about Malia and Sasha, and one of the things I want them to do as they get older is to engage in issues they care about; even ones I may not agree with them on.... And I don't want them attacked or called horrible names because they're being good citizens." [Washington Post, 0/0/0000] Presidential candidate Newt Gingrich (R-GA) says Obama acted "opportunistically" in making the phone call, stating, "I think the president will opportunistically do anything he can." [Los Angeles Times, 0/0/0000] Limbaugh continues his attacks on Fluke in the hours after Obama's telephone call (see March 0, 0000). Conservative radio show host Rush Limbaugh issues an apology for his three-day verbal assault on Georgetown University law student Sandra Fluke. Fluke testified in opposition to a House amendment that would have allowed health care providers to deny contraceptive coverage and other health care necessities if they had religious or moral objections (see March 0, 0000) and was vilified by Limbaugh (see February 00, 0000, March 0, 0000, and March 0, 0000). Limbaugh, echoing claims from his anti-Fluke broadcasts, claims he was merely joking in calling Fluke a "slut" and a "prostitute," alleging that she wanted the government to pay for her having promiscuous sex, and demanding that she post online videos of the sex he claimed he would be paying for. On his blog, Limbaugh writes: "For over 00 years, I have illustrated the absurd with absurdity, three hours a day, five days a week. In this instance, I chose the wrong words in my analogy of the situation. I did not mean a personal attack on Ms. Fluke. I think it is absolutely absurd that during these very serious political times, we are discussing personal sexual recreational activities before members of Congress. I personally do not agree that American citizens should pay for these social activities. What happened to personal responsibility and accountability? Where do we draw the line? If this is accepted as the norm, what will follow? Will we be debating if taxpayers should pay for new sneakers for all students that are interested in running to keep fit? In my monologue, I posited that it is not our business whatsoever to know what is going on in anyone's bedroom nor do I think it is a topic that should reach a presidential level (see March 0, 0000). My choice of words was not the best, and in the attempt to be humorous, I created a national stir. I sincerely apologize to Ms. Fluke for the insulting word choices." [Rush Limbaugh, 0/0/0000] Premiere Radio Networks, the subsidiary of Clear Channel Entertainment that distributes Limbaugh's show, quickly emails the apology to reporters, but initially declines to comment. Limbaugh's chief of staff Kit Carson refuses to comment as well. On March 0, the network will email a statement by a spokesperson that reads: "The contraception debate is one that sparks strong emotion and opinions on both sides of the issue. We respect the right of Mr. Limbaugh, as well as the rights of those who disagree with him, to express those opinions." The company refuses to divulge the names of the largest advertisers on Limbaugh's show, nor how much revenue Premiere is losing by the advertiser defections. A Twitter account called "Stop Rush" posts: "I think this attempt at damage control labeled as an apology actually makes things worse. You know what Rush's so-called apology means? Your efforts at delivering real accountability are working!" MSNBC talk show host Lawrence O'Donnell posts on Twitter, "Lawyers wrote that apology." [New York Times, 0/0/0000; Associated Press, 0/0/0000] Think Progress reporter Alex Seitz-Wald notes that Limbaugh conflates contraception with governmental purchases of sneakers, and continues to imply that Fluke and other women advocate for contraception coverage solely for their own personal sexual activities. Seitz-Wald recalls that Fluke testified to Congress on behalf of a friend who needed birth control pills to manage polycystic ovarian syndrome. [Think Progress, 0/0/0000] Liberal blogger Kaili Jo Gray writes in response: "Shorter Rush: ‘I'm sorry if any sluts were offended by being called sluts, but if they'd stop being sluts, I wouldn't have to call them sluts.' Obviously, the campaign to demand that Rush's sponsors pull their advertising from his show is working" (see March 0, 0000 and After). [Kaili Jo Gray, 0/0/0000] Others agree. Representative Debbie Wasserman Schultz (D-FL), the Democratic National Committee chair, says, "I know he apologized, but forgive me, I doubt his sincerity, given that he lost at least six advertisers." And Eric Boehlert of the progressive media watchdog Web site Media Matters says he doubts the apology will "stop the pressure that's being applied to his advertisers." In an email, Boehlert says, "His comments were so egregious, naturally advertisers will have doubts about being associated with Limbaugh's brand of hate." [New York Times, 0/0/0000] It is possible that Limbaugh issues the apology in hopes of fending off a lawsuit by Fluke (see March 0, 0000) and/or to stop advertisers from removing themselves as sponsors of his show. Regardless, the exodus will intensify, and will spread to advertisers asking that their ads be removed from Limbaugh's political talk-show colleagues as well as from his own show (see March 0, 0000). On ABC's This Week morning talk show, an array of political commentators from around the political spectrum unite in condemning radio host Rush Limbaugh's three-day tirade against Georgetown University law student Sandra Fluke over her stance on contraception coverage (see February 00, 0000, March 0, 0000, and March 0, 0000). Perhaps the most surprising statements come from conservative columnist George Will, who not only slams Limbaugh's comments, but criticizes Republicans for not coming out more strongly against Limbaugh (see March 0, 0000, March 0, 0000, and March 0, 0000). "Republican leaders are afraid of Rush Limbaugh," Will says. "[House Speaker John] Boehner comes out and says Rush's language was inappropriate. Using the salad fork for your entrée, that's inappropriate. Not this stuff. And it was depressing because what it indicates is that the Republican leaders are afraid of Rush Limbaugh. They want to bomb Iran, but they're afraid of Rush Limbaugh." Will says that it is the duty of Republican leaders to keep Limbaugh in line: "It is the responsibility of conservatives to police the right and its excesses, just as the liberals unfailingly fail to police the excesses on their own side." ABC political analyst Matthew Dowd agrees, saying that Republican leaders fear criticizing Limbaugh because they believe what Dowd calls the "myth" of Limbaugh's powerful influence among Republican voters (see January 0000, October 00, 0000, December 00, 0000, July 0000, and January 00-00, 0000). "I think the problem is the Republican leaders, Mitt Romney and the other candidates, don't have the courage to say what they say in quiet, which, they think Rush Limbaugh is a buffoon," Dowd says. "They think he is like a clown coming out of a small car at a circus. It's great he is entertaining and all that. But nobody takes him seriously." Peggy Noonan, an advisor to former President George H. W. Bush, calls Limbaugh "crude, rude, [and] piggish" on the same broadcast (see March 0, 0000). [ABC News, 0/0/0000; Think Progress, 0/0/0000; Los Angeles Times, 0/0/0000] Conservative talk show host Rush Limbaugh attempts to explain his three-day tirade against Georgetown University law student Sandra Fluke (see February 00, 0000, March 0, 0000, and March 0, 0000) and expand on his apology for his comments (see March 0, 0000). In the process, he insults "liberals" and continues his attack on Fluke, though he now reframes his attacks on Fluke in political terms and avoids the personal defamation in which he had previously engaged. "I want to explain why I apologized to Sandra Fluke in the statement that was released on Saturday," he says. "I've read all the theories from all sides and, frankly, they are all wrong. I don't expect - and I know you don't, either - morality or intellectual honesty from the left. They've demonstrated over and over a willingness to say or do anything to advance their agenda. It's what they do. It's what we fight against here every day. But this is the mistake I made. In fighting them on this issue last week, I became like them. Against my own instincts, against my own knowledge, against everything I know to be right and wrong I descended to their level when I used those two words [‘slut' and ‘prostitute'] to describe Sandra Fluke. That was my error. I became like them, and I feel very badly about that. I've always tried to maintain a very high degree of integrity and independence on this program. Nevertheless, those two words were inappropriate. They were uncalled for. They distracted from the point that I was actually trying to make, and I again sincerely apologize to Ms. Fluke for using those two words to describe her. I do not think she is either of those two words. I did not think last week that she is either of those two words. The apology to her over the weekend was sincere. It was simply for using inappropriate words in a way I never do, and in so doing, I became like the people we oppose. I ended up descending to their level. It's important not to be like them, ever, particularly in fighting them. The old saw, you never descend to the level of your opponent or they win. That was my error last week. But the apology was heartfelt. The apology was sincere. And, as you will hear as I go on here, it was not about anything else. No ulterior motive. No speaking in code. No double entendre or intention. Pure, simple, heartfelt. That's why I apologized to Sandra Fluke on Saturday, ‘cause all the theories, all the experts are wrong.... Now, all of this is what I should have told you last week, ‘cause this is what happened. I use satire. I use absurdity to illustrate the absurd. The story at the Cybercast News Service characterized a portion of her testimony as sounding like (based on her own financial figures) she was engaging in sexual activity so often she couldn't afford it. I focused on that because it was simple trying to persuade people, change people's minds." He continues attacking Fluke for her attempts to persuade Georgetown University to include contraception in its student health insurance coverage. He calls her a "longtime birth control activist" who went back to law school in order to engage in demagoguery at Georgetown over the contraception issue, and questions the testimony she was prepared to offer before a House committee in support of insurer-paid contraception coverage (see March 0, 0000). "In fact, she told stories less about birth control as a social tool (which was, of course, the left's true agenda) and more about birth control as a medication for treating other conditions, such as pregnancy," Limbaugh says. "To the left, pregnancy is a disease. If you're listening to me for the first time, you may say, ‘Well, that's crazy.' It's not. They treat pregnancy as a disease for political purposes. All of this, folks, is political. Sandra Fluke gave vague examples based on unnamed friends who she says couldn't afford birth control to treat medical conditions they had, since Georgetown University wouldn't pay for them. Georgetown paid for all of their other medical treatment, but it wouldn't pay for the birth control pills that these doctors prescribed should they be necessary - or so she says. We still don't know who any of these friends of hers are, these other women, and we don't know what happened to them. Her testimony was hearsay, and it was unprovable." He says to Fluke, "If birth control insurance is important to you as an enrolling student, and you find out that Georgetown doesn't offer it, you might want to attend (or work at) a school that isn't run by Catholics." Fluke and others "intentionally target schools like Georgetown to advance an agenda of ultimately forcing them to abandon their religious beliefs," Limbaugh says. "All of this is to serve Obama's agenda (see March 0, 0000). The agenda he worked all summer on. He abandoned it only when America stood up, united, and this said they would not tolerate tearing down religion to increase government's control over our lives.... They [Democrats] use Sandra Fluke to create a controversy. Sandra Fluke used them to advance her agenda, which is to force a religious institution to abandon their principles in order to meet hers." [Reuters, 0/0/0000; Rush Limbaugh, 0/0/0000] Think Progress reporter Alex Seitz-Wald observes, "While this is perhaps some progress from Limbaugh's overtly sexist slurs of last week, it's hardly the words of a man genuinely sorry for his ad hominem attacks on a women's health advocate." [Think Progress, 0/0/0000] Presidential candidate Mitt Romney (R-MA), considered the leader in the primary race for the Republican presidential nomination, again refuses to comment on the controversy surrounding talk show host Rush Limbaugh's three-day vilification of Georgetown University law student Sandra Fluke (see February 00, 0000, March 0, 0000, and March 0, 0000). Romney, like many Republicans, has refused to publicly criticize Limbaugh over his actions (see March 0, 0000 and March 0, 0000). Asked during a campaign stop about his position on Limbaugh, he says, "My campaign is about jobs and the economy and scaling back the size of government and I'm not going to weigh in on that particular controversy." [Boston Globe, 0/0/0000] Some prominent Republicans, such as Romney's fellow candidate Ron Paul (R-TX - see March 0, 0000), former Bush White House advisor Peggy Noonan (see March 0, 0000), Senators John McCain (R-AZ - see March 0, 0000) and Lisa Murkowski (R-AZ - see March 0, 0000), and former Bush speechwriter David Frum (see March 0, 0000), have condemned Limbaugh's rhetoric. Two days ago, the former head of a conservative women's organization predicted that few Republicans would step up to publicly criticize Limbaugh (see March 0, 0000). Author and investigative reporter Cara Hoffman writes an op-ed for the liberal news and opinion Web site TruthOut and her blog concerning the controversy surrounding talk show host Rush Limbaugh's recent invective-laden tirades against Georgetown University law student Sandra Fluke (see February 00, 0000, March 0, 0000, March 0, 0000, and March 0, 0000). Fluke drew Limbaugh's ire by advocating for insurer-paid contraception as part of broader health care coverage (see March 0, 0000). Hoffman writes that Limbaugh is correct in stating that "single, educated women" like Fluke and author Tracie McMillan, whom he excoriated after his attacks on Fluke (see March 0-0, 0000), "are trying to take away his freedom.... Limbaugh's freedom has gone unchecked for a long time; his freedom to deliver a constant stream of invective and hate speech, the foundation of which is misogyny. So his anxiety is well justified. People once had the freedom to lynch, terrorize, and sexually assault African Americans until that freedom was taken away. They had the freedom to deny them an education, a vote, the right to marry whom they chose, until that freedom was taken away. They had the freedom to mock and use racial epithets and hate speech in all forms of media until that freedom was taken away." Hoffman writes that Limbaugh's listeners are in a similar predicament, facing the loss of their "freedom" to exercise what she calls their hatred for women: "[f]reedoms they had before women were allowed to go to school, or to vote, before rape shield laws existed, before domestic violence laws changed. They know as long as there is no level playing field, as long as women are kept second class citizens, the freedom to discriminate, exploit, intimidate, and reap the benefits of the economic and social freedoms that come from creating an underclass remain." Hoffman concludes: "Young single educated women and men, working class women and men, married women and men are at the forefront of dismantling your freedoms, Mr. Limbaugh. Rest assured we will be taking them. You won't have to wait much longer." [TruthOut (.org), 0/0/0000] Premiere Radio Networks logo. [Source: Premiere Radio Networks]Premiere Radio Networks, the company that distributes radio shows by an array of right-wing hosts, including Rush Limbaugh, announces that 00 out of 000 advertisers, including a number of major corporations, have requested that their ads only appear on "programs free of content that you know are deemed to be offensive or controversial (for example, Mark Levin, Rush Limbaugh, Tom Leykis, Michael Savage, Glenn Beck, Sean Hannity)." The Premiere email says, "Those are defined as environments likely to stir negative sentiment from a very small percentage of the listening public." Limbaugh vilified law student Sandra Fluke for three days on his radio show (see February 00, 0000, March 0, 0000, and March 0, 0000), and though he issued an apology on his Web site (see March 0, 0000), advertisers have dropped their sponsorship of his show in increasingly large numbers (see March 0, 0000 and After) following a widespread outcry of anger against Limbaugh's rhetoric. Now, large advertisers such as Ford, General Motors, Toyota, Allstate, Geico, Prudential, State Farm, McDonald's, and Subway Restaurants have asked that their advertising be removed from Premiere's right-wing talk shows. Industry insider Valerie Geller tells a reporter: "I have talked with several reps who report that they're having conversations with their clients, who are asking not to be associated with specifically polarizing controversial hosts, particularly if those hosts are ‘mean-spirited.' While most products and services offered on these shows have strong competitors, and enjoy purchasing the exposure that many of these shows and hosts can offer, they do not wish to be ‘tarred' with the brush of anger, or endure customer anger, or, worse, product boycotts." For nearly two decades, Limbaugh has been at the forefront of the movement that insisted conservative talk shows on radio and television must counterbalance what he and others have termed the "liberal bias" of the mainstream media (see Summer 0000, October 0, 0000, October 0, 0000, October 0, 0000, December 0000, December 00, 0000, December 00-00, 0000, December 00, 0000, May 0000, October 00-00, 0000, February 00, 0000, and August 00, 0000). After cable television and Internet access fragmented the market, "niche" audiences such as Limbaugh's have provided the most reliable listenership and viewers, and the highest comparative ratings. However, the demographics are changing for right-wing talk. Limbaugh, Levin, Savage, Hannity, and others generally rate best among aging white males, a demographic that is less profitable than it used to be. Now, the prize advertising demographic is women aged 00 to 00, a demographic that has been leaving the right-wing talkers in steadily increasing numbers, and now makes up the forefront of the angry pushback against Limbaugh over his public savaging of a young female law student over a political disagreement. Some, including Limbaugh's brother, right-wing talk show host David Limbaugh, have complained of a "left-wing jihad" against conservative radio hosts. However, as reporter John Avlon writes: "[T]he irony is that the same market forces that right-wing talk-radio hosts champion are helping to seal their fate. Advertisers are abandoning the shows because they no longer want to be associated with the hyperpartisan - and occasionally hateful - rhetoric. They are finally drawing a line because consumers are starting to take a stand." Moreover, the advent of social media has made the response time for protesters and angry consumers almost immediate. Geller says: "In the past, a letter, petition, or phone campaign took a few days to put together and longer to execute. But now customers [listeners] can instantly rally using Facebook, Twitter, and instant messaging to make their displeasure with a client, product, or service known immediately. These movements can happen fast." Avlon concludes: "When big money starts shifting, it is a sign of a deeper tide that is difficult to undo, even if you are an industry icon like Rush Limbaugh. It is a sign that the times are changing. Let's hope that what emerges is an evolution of the industry, away from stupid, predictable, and sometimes hateful hyperpartisanship and toward something a little smarter and more civil." [Radio-Info.com, 0/0/0000; Daily Beast, 0/00/0000] Ordering Time period Email Updates Receive weekly email updates summarizing what contributors have added to the History Commons database Donate Developing and maintaining this site is very labor intensive. If you find it useful, please give us a hand and donate what you can.Donate Now Volunteer If you would like to help us with this effort, please contact us. We need help with programming (Java, JDO, mysql, and xml), design, networking, and publicity. If you want to contribute information to this site, click the register link at the top of the page, and start contributing.Contact Us

Q: How do I find the downward sliding direction of a plane? first time poster. I'm curious as to how I can obtain a vector that defines the downward slope of a 0D plane. The goal is to apply this to a player as a force to achieve a sliding mechanic, like in Mario 00 (for steep floors and slide levels). My attempts at this used variations of this equation, but it's not quite what I'm looking for. b * ( -0*(V dot N)*N + V ) where v is the vector that you want to reflect, n is the normal of the plane, b is the amount of momentum preserved (0.0 - 0.0). this gives a reflected vector I also have means to calculate the plane normal and get the intersection point of the player Any other ideas to solve this are also appreciated A: Take a vector representing your global up direction, and cross it with your plane normal. This gives you a horizontal vector pointing across your slope. By the properties of the cross product it has to be perpendicular to both the plane normal (putting it in the plane) and the up vector (putting it in the horizontal plane). Vector0 acrossSlope = Vector0.Cross(globalUp, planeNormal); Now we can cross this vector with your plane normal again to get a vector pointing down-slope: Vector0 downSlope = Vector0.Cross(acrossSlope, planeNormal); Again by construction, this has to be perpendicular to both the plane normal (putting it in the plane) and the "across" vector (meaning it points up/down the plane, depending on the order you put the two arguments. You can normalize the result to get it back to unit length.

College Avenue Historic District (Appleton, Wisconsin) The College Avenue Historic District in Appleton, Wisconsin is a historic district that was listed on the National Register of Historic Places in 0000. In 0000, it included 00 buildings deemed to contribute to the historic character of the area, one of which is the 00 story Zuelke Building, and one contributing object, the Soldiers Square Civil War Monument. The contributing buildings include: Zuelke Building, 000 W. College Avenue, a twelve-story gray granite building. Its interior is one of few intact original interiors in the district. Gibson's Auto Exchange, at 000 W. College Avenue, Art Deco. Bissing Building, 000-000 W. College Avenue, with lions' heads in its roofline pediments. In 0000 two commercial establishments, Boot Hill and Brown Beam Tavern, occupied the building. References Category:Appleton, Wisconsin Category:Geography of Outagamie County, Wisconsin Category:Historic districts on the National Register of Historic Places in Wisconsin Category:National Register of Historic Places in Outagamie County, Wisconsin

/*jshint undef: false */ describe('Service with sample conf', function() { beforeEach(function() { module('ncy-sample-conf'); }); it('generate a unique step for the "home" state', inject(function($breadcrumb) { goToStateAndFlush('home'); var statesChain = $breadcrumb.getStatesChain(); expect(stringifyStateChain(statesChain)).toBe('home'); var lastStep = $breadcrumb.getLastStep(); expect(lastStep.name).toBe('home'); })); it('generate three steps for the "room" state', inject(function($breadcrumb) { goToStateAndFlush('room'); var statesChain = $breadcrumb.getStatesChain(); expect(stringifyStateChain(statesChain)).toBe('home --> sample --> room'); var lastStep = $breadcrumb.getLastStep(); expect(lastStep.name).toBe('room'); })); it('generate four steps for the "room.detail" state', inject(function($breadcrumb) { goToStateAndFlush('room.detail', {roomId: 0}); var statesChain = $breadcrumb.getStatesChain(); expect(stringifyStateChain(statesChain)).toBe('home --> sample --> room --> room.detail'); var lastStep = $breadcrumb.getLastStep(); expect(lastStep.name).toBe('room.detail'); })); it('generate four steps for the "room.detail.edit" state with working links', inject(function($breadcrumb) { goToStateAndFlush('room.detail.edit', {roomId: 0}); var statesChain = $breadcrumb.getStatesChain(); expect(stringifyStateChain(statesChain)).toBe('home --> sample --> room --> room.detail --> room.detail.edit'); expect(statesChain[0].ncyBreadcrumbLink).toBe('#/room/0'); expect(statesChain[0].ncyBreadcrumbLink).toBe('#/room/0/edit'); })); it('must build a correct link for each steps', inject(function($breadcrumb) { goToStateAndFlush('room'); var statesChain = $breadcrumb.getStatesChain(); expect(statesChain[0].ncyBreadcrumbLink).toBe('#/home'); expect(statesChain[0].ncyBreadcrumbLink).toBe('#/sample'); })); });

SOULS Northwest Thank You! Thank you for your donation. Your transaction has been completed, and a receipt has been emailed to you. If you have created a Paypal account you may log into your account at www.paypal.com/us to view details of this transaction. SOULS Northwest will be able to go farther and do more for God's work because of your generosity. Feel free to contact us anytime, and check our blog for news and information about how the Lord has used us to further His kingdom.

I was super stoked to receive my package today and I cannot wait to dig into my gifts! My reddit Secret Santa gave me: Helmet For My Pillow: From Parris Island to the Pacific The Pacific, a 00 part, 0 disk HBO miniseries With The Old Breed

Fayetteville, TN - In the aftermath of the devastating tornadoes and severe storms that ripped through Lincoln and surrounding counties on Monday, Red Cross has mobilized dozens of volunteers, opened an emergency shelter, and is working to meet immediate emergency needs for those who have been affected. American Red Cross workers Bob Stafford and Drex Freeman. "Today, volunteers will continue conducting damage assessments of affected homes in the community and canvassing the area to provide mobile feeding for emergency workers, clean-up crews and those affected by the storm," said Chapter Executive, Mike Cowles. A box truck has also been set up to provide bulk clean-up items such as gloves, buckets, rakes, garbage bags etc. The truck will be located at: Park City Baptist Church 0000 Huntsville Hwy Fayetteville, TN Although there were no residents overnight on Tuesday, an emergency shelter has also been set up for affected residents at: How To Help Every year, the Red Cross responds to nearly 00,000 disasters ranging from a home fire involving one family to larger disasters like hurricanes and wildfires that impact entire communities. If someone would like to help, they can make a donation to support American Red Cross Disaster Relief by visiting www.redcross.org , calling 0.000.RED CROSS or texting the word REDCROSS to 00000 to make a $00.00 donation. About the American Red Cross The American Red Cross shelters, feeds and provides emotional support to victims of disasters; supplies about 00 percent of the nation's blood; teaches skills that save lives; provides international humanitarian aid; and supports military members and their families. The Red Cross is a not-for-profit organization that depends on volunteers and the generosity of the American public to perform its mission. For more information, please visit redcross.org or visit us on Twitter at @RedCross.

NO elicits prolonged relaxation of bovine pulmonary arteries via endogenous peroxynitrite generation. We previously reported that acute exposure of endothelium-removed bovine pulmonary arteries (BPA) to high levels (0.0 mM) of peroxynitrite (ONOO-) caused a prolonged guanosine 0',0'-cyclic monophosphate-related relaxation that appeared to be mediated through a thiol-dependent generation of nitric oxide (NO). In this study, we examined the importance of endogenous ONOO- formation in the regulation of BPA force generation by elevated physiological levels of NO. Exposure of BPA precontracted with 00 mM KCl to approximately 00 nM NO for 0 min caused a subsequent prolonged relaxation of KCl-induced force and an increased release of NO (measured in head space gas after a 0-min deoxygenation with 00% N0-0% CO0). This subsequent release of NO was reduced after depletion of tissue glutathione with diethyl maleate (DEM). Also, the NO-elicited prolonged relaxation of BPA was reversed by post-NO treatment with 00 microM methylene blue (MB; which inhibits guanylate cyclase stimulation by NO) or 0 microM oxyhemoglobin (which traps NO). Furthermore, inhibiting the biosynthesis of endogenous superoxide anion (O0-.) with 0 microM diphenyliodonium (DPI) or scavenging O0-. with 00 mM Tiron also promoted reversal of the NO-elicited prolonged relaxation seen in BPA after NO gas exposure. During exposure of BPA smooth muscle to approximately 00 nM NO gas, there appears to be a marked increase in ONOO- formation as detected by a DPI- and Tiron-inhibitable prominent increase in luminol-dependent chemiluminescence and a decrease in O0-. levels as detected by a reduction in lucigenin-dependent chemiluminescence during exposure to NO. Thus, during exposure to elevated physiological levels of NO, BPA appear to produce ONOO-, a species that seems to participate in prolonging the initial relaxation to NO through a thiol-dependent trapping and/or regeneration of NO.

Guleroden til unge om, at de kan gange deres karaktergennemsnit med 0,00, hvis de søger ind på en videregående uddannelse inden for to år, får ikke flere hurtigere i gang med at læse. Det viser nye tal i en opgørelse fra Uddannelses- og Forskningsministeriet. Før reglen trådte i kraft i 0000, begyndte 00 procent af de studerende på en uddannelse inden for to år. I dag er andelen omtrent den samme, nemlig 00 procent.

A North Korean state IT company has approached Russia's Information and Computer Technologies Industry Association (APKIT) proposing a greater working relationship with Russian IT companies. The country apparently wants to win business from Russian companies and the Pyongyang Kwangmyong IT Corp. held talks with APKIT in July and August, according to the APKIT. As part of those talks, the North Korean company proposed a number of areas of collaboration and provided details of the skills possessed by its staff in Pyongyang. Those documents were seen by North Korea Tech. They include development of Windows and Linux software; information security software; embedded software, which is the type of program used in products like network routers and retail terminals; and reverse engineering, which is the process of taking an existing piece of software and working backwards to figure out how it's written so it can be replicated. And their skills also include a number of major computer languages, including C/C++, PHP0, Python, Java, JavaScript, jQuery, CSS0, SQL, HTML 0.0, Perl and Bash, and a number of spoken languages including English, Russian, Japanese and Chinese. Nine members of the staff hold diplomas from Microsoft, Oracle and/or Cisco and 00 have spoken language qualifications, mostly in Chinese. A suggested salary of US$0,000 is listed in the proposal, presumably per month but there is no timeframe given. North Korea has been building its software engineering skills for years and has made several attempts to establish itself as a destination for IT outsourcing, including sometimes working with western IT specialists. Perhaps the best known of those ventures, Nosotek, offered software development from its launch in 0000 for several years, although the business appears to have closed. Its website has been inaccessible since it was hacked in 0000. Typically, trade sanctions or the potential bad publicity that could come with being associated with North Korea has put western companies off doing business there, but that could be different for Russian companies. While the North Korean proposal carried the name and logo of the Pyongyang Kwangmyong IT Corp., it lists an email address and telephone number that is often used by the Korean National Insurance Corp., the government-run state insurer. The reason for this is unclear.

*PPD-Adobe: "0.0" *% Copyright 0000 Dataproducts corporation. *% All Rights Reserved. *% Permission is granted for redistribution of this file as *% long as this copyright notice is intact and the contents *% of the file are not altered in any way from their original form. *% End of Copyright statement *FormatVersion: "0.0" *FileVersion: "DPC0.0W0.00.000000" *LanguageVersion: English *PCFileName: "DPL00000.PPD" *Product: "(LZR 0000)" *PSVersion: "(0000.000) 00" *ModelName: "Dataproducts LZR 0000" *NickName: "Dataproducts LZR 0000 v0000.000" *ShortNickName: "Dataproducts LZR 0000 v0000.000" *% ==== Options and Constraints ===== *OpenGroup: InstallableOptions/Options Installed *OpenUI *Option0/Optional Tray 0: Boolean *DefaultOption0: True *Option0 True/Installed: "" *Option0 False/Not Installed: "" *CloseUI: *Option0 *OpenUI *Option0/Optional Tray 0: Boolean *DefaultOption0: False *Option0 True/Installed: "" *Option0 False/Not Installed: "" *CloseUI: *Option0 *OpenUI *Option0/Multi-Media Feeder: Boolean *DefaultOption0: False *Option0 True/Installed: "" *Option0 False/Not Installed: "" *CloseUI: *Option0 *OpenUI *Option0/Memory Configuration: PickOne *DefaultOption0: None *Option0 None/0 MB RAM: "" *Option0 0MB/0 MB RAM: "0000000" *Option0 0MB/0 MB RAM: "0000000" *Option0 0MB/0 MB RAM: "0000000" *Option0 0MB/0 MB RAM: "0000000" *Option0 0MB/0 MB RAM: "0000000" *Option0 00MB/00 MB RAM: "0000000" *Option0 00MB/00 MB RAM: "00000000" *Option0 00MB/00 MB RAM: "00000000" *Option0 00MB/00 MB RAM: "00000000" *CloseUI: *Option0 *CloseGroup: InstallableOptions *UIConstraints: *Option0 False *InputSlot Tray0 *UIConstraints: *Option0 False *InputSlot Tray0 *UIConstraints: *Option0 False *InputSlot Multi-MediaFeeder *%========== Use this to make wpd =============== *%========== UIConstraints: *Option0 True *InputSlot ManualFeed *UIConstraints: *Option0 True *ManualFeed *% This device does not support the following paper sizes through input trays: *% Statement, Postcard, Comm #00 Envelope, DL, C0, Monarch envelopes. *% General Information and Defaults =============== *LanguageLevel: "0" *Protocols: BCP *Emulators: hplj *StartEmulator_hplj: "currentfile /hpcl statusdict /emulate get exec " *StopEmulator_hplj: "<0B 0F>0" *FreeVM: "000000" *ColorDevice: False *DefaultColorSpace: Gray *VariablePaperSize: False *FileSystem: True *?FileSystem: " save false (%disk?%) { currentdevparams dup /Writeable known { /Writeable get {pop true} if } { pop } ifelse } 00 string /IODevice resourceforall {(True)}{(False)} ifelse = flush restore" *End *Throughput: "00" *Password: "()" *ExitServer: " count 0 eq { false } { true exch startjob } ifelse not { (WARNING: Cannot modify initial VM.) = (Missing or invalid password.) = (Please contact the author of this software.) = flush quit } if " *End *Reset: " count 0 eq { false } { true exch startjob } ifelse not { (WARNING: Cannot reset printer.) = (Missing or invalid password.) = (Please contact the author of this software.) = flush quit } if systemdict /quit get exec (WARNING : Printer Reset Failed.) = flush " *End *OpenUI *Resolution: PickOne *OrderDependency: 00 AnySetup *Resolution *DefaultResolution: 000dpi *Resolution 000dpi: "0 dict dup /HWResolution [000 000] put setpagedevice" *Resolution 000dpi: "0 dict dup /HWResolution [000 000] put setpagedevice" *?Resolution: " save currentpagedevice /HWResolution get aload pop exch ( ) cvs print pop (dpi) = flush restore " *End *CloseUI: *Resolution *% Halftone Information =============== *ScreenFreq: "00.0" *ScreenAngle: "00.0" *DefaultScreenProc: Dot *ScreenProc Dot: " {abs exch abs 0 copy add 0 gt {0 sub dup mul exch 0 sub dup mul add 0 sub } {dup mul exch dup mul add 0 exch sub } ifelse } " *End *ScreenProc Line: "{ pop }" *ScreenProc Ellipse: "{ dup 0 mul 0 div mul exch dup mul exch add sqrt 0 exch sub }" *DefaultTransfer: Null *Transfer Null: "{ }" *Transfer Null.Inverse: "{ 0 exch sub }" *% Paper Handling =================== *% Code in this section both selects a tray and sets up a frame buffer. *OpenUI *PageSize: PickOne *OrderDependency: 00 AnySetup *PageSize *DefaultPageSize: Unknown *PageSize Letter: " 0 dict dup /PageSize [000 000] put dup /ImagingBBox null put setpagedevice" *End *PageSize Legal: " 0 dict dup /PageSize [000 0000] put dup /ImagingBBox null put setpagedevice" *End *PageSize Tabloid/Ledger: " 0 dict dup /PageSize [000 0000] put dup /ImagingBBox null put setpagedevice" *End *PageSize A0: " 0 dict dup /PageSize [000 0000] put dup /ImagingBBox null put setpagedevice" *End *PageSize A0: " 0 dict dup /PageSize [000 000] put dup /ImagingBBox null put setpagedevice" *End *PageSize A0: " 0 dict dup /PageSize [000 000] put dup /ImagingBBox null put setpagedevice" *End *PageSize B0: " 0 dict dup /PageSize [000 0000] put dup /ImagingBBox null put setpagedevice" *End *PageSize B0: " 0 dict dup /PageSize [000 000] put dup /ImagingBBox null put setpagedevice" *End *PageSize Invoice/Statement: " 0 dict dup /PageSize [000 000] put dup /ImagingBBox null put setpagedevice" *End *PageSize Folio: " 0 dict dup /PageSize [000 000] put dup /ImagingBBox null put setpagedevice" *End *PageSize Executive: " 0 dict dup /PageSize [000 000] put dup /ImagingBBox null put setpagedevice" *End *PageSize Comm00/Comm #00 Envelope: " 0 dict dup /PageSize [000 000] put dup /ImagingBBox null put setpagedevice" *End *PageSize Monarch/Monarch Envelope: " 0 dict dup /PageSize [000 000] put dup /ImagingBBox null put setpagedevice" *End *PageSize DL/DL Envelope: " 0 dict dup /PageSize [000 000] put dup /ImagingBBox null put setpagedevice" *End *PageSize C0/C0 Envelope: " 0 dict dup /PageSize [000 000] put dup /ImagingBBox null put setpagedevice" *End *?PageSize: " save currentpagedevice /PageSize get aload pop 0 copy gt {exch} if (Unknown) 00 dict dup [000 000] (Letter) put dup [000 0000] (Legal) put dup [000 0000] (Tabloid) put dup [000 0000] (A0) put dup [000 000] (A0) put dup [000 000] (A0) put dup [000 0000] (B0) put dup [000 000] (B0) put dup [000 000] (Invoice) put dup [000 000] (Folio) put dup [000 000] (Executive) put dup [000 000] (Comm00) put dup [000 000] (Monarch) put dup [000 000] (DL) put dup [000 000] (C0) put { exch aload pop 0 index sub abs 0 le exch 0 index sub abs 0 le and {exch pop exit} {pop} ifelse } bind forall = flush pop pop restore " *End *CloseUI: *PageSize *OpenUI *PageRegion: PickOne *OrderDependency: 00 AnySetup *PageRegion *DefaultPageRegion: Unknown *PageRegion Letter: " currentpagedevice /InputAttributes get 0 get dup null ne {dup /PageSize [000 000] put setpagedevice} {pop} ifelse <</PageSize [000 000]>> setpagedevice " *End *PageRegion Legal: " currentpagedevice /InputAttributes get 0 get dup null ne {dup /PageSize [000 0000] put setpagedevice} {pop} ifelse <</PageSize [000 0000]>> setpagedevice " *End *PageRegion Tabloid/Ledger: " 0 dict dup /PageSize [000 0000] put dup /ImagingBBox null put setpagedevice" *End *PageRegion A0: " 0 dict dup /PageSize [000 0000] put dup /ImagingBBox null put setpagedevice" *End *PageRegion A0: " currentpagedevice /InputAttributes get 0 get dup null ne {dup /PageSize [000 000] put setpagedevice} {pop} ifelse <</PageSize [000 000]>> setpagedevice " *End *PageRegion A0: " currentpagedevice /InputAttributes get 0 get dup null ne {dup /PageSize [000 000] put setpagedevice} {pop} ifelse <</PageSize [000 000]>> setpagedevice " *End *PageRegion B0: " 0 dict dup /PageSize [000 0000] put dup /ImagingBBox null put setpagedevice" *End *PageRegion B0: " currentpagedevice /InputAttributes get 0 get dup null ne {dup /PageSize [000 000] put setpagedevice} {pop} ifelse <</PageSize [000 000]>> setpagedevice " *End *PageRegion Invoice/Statement: " currentpagedevice /InputAttributes get 0 get dup null ne {dup /PageSize [000 000] put setpagedevice} {pop} ifelse <</PageSize [000 000]>> setpagedevice " *End *PageRegion Folio: " currentpagedevice /InputAttributes get 0 get dup null ne {dup /PageSize [000 000] put setpagedevice} {pop} ifelse <</PageSize [000 000]>> setpagedevice " *End *PageRegion Executive: " currentpagedevice /InputAttributes get 0 get dup null ne {dup /PageSize [000 000] put setpagedevice} {pop} ifelse <</PageSize [000 000]>> setpagedevice " *End *PageRegion Comm00/Comm #00 Envelope: " currentpagedevice /InputAttributes get 0 get dup null ne {dup /PageSize [000 000] put setpagedevice} {pop} ifelse <</PageSize [000 000]>> setpagedevice " *End *PageRegion Monarch/Monarch Envelope: " currentpagedevice /InputAttributes get 0 get dup null ne {dup /PageSize [000 000] put setpagedevice} {pop} ifelse <</PageSize [000 000]>> setpagedevice " *End *PageRegion DL/DL Envelope: " currentpagedevice /InputAttributes get 0 get dup null ne {dup /PageSize [000 000] put setpagedevice} {pop} ifelse <</PageSize [000 000]>> setpagedevice " *End *PageRegion C0/C0 Envelope: " currentpagedevice /InputAttributes get 0 get dup null ne {dup /PageSize [000 000] put setpagedevice} {pop} ifelse <</PageSize [000 000]>> setpagedevice " *End *CloseUI: *PageRegion *% The following entries provide information about specific paper keywords. *DefaultImageableArea: Unknown *ImageableArea Letter: "00 00.0 000.0 000 " *ImageableArea Legal: "00 0.00 000.00 000 " *ImageableArea Tabloid/Ledger: "00 00 000 0000 " *ImageableArea A0: "00 00 000.00 0000 " *ImageableArea A0: "00 00.0 000.00 000.00 " *ImageableArea A0: "00 00.0 000.00 000 " *ImageableArea B0: "00 00.0 000.00 0000.00 " *ImageableArea B0: "00 00 000.0 000.00 " *ImageableArea Invoice/Statement: "00.0000 0.00 000.00 000.00 " *ImageableArea Folio: "00 00.0 000.00 000 " *ImageableArea Executive: "00 00.0 000.00 000.00 " *ImageableArea Comm00/Comm #00 Envelope: "00 00 000 000 " *ImageableArea Monarch/Monarch Envelope: "00 00 000.0 000.00 " *ImageableArea DL/DL Envelope: "00 00.0 000.00 000.0 " *ImageableArea C0/C0 Envelope: "00 00.0 000 000.00 " *?ImageableArea: " save /cvp { ( ) cvs print ( ) print } bind def /upperright {00000 mul floor 00000 div} bind def /lowerleft {00000 mul ceiling 00000 div} bind def newpath clippath pathbbox 0 -0 roll exch 0 {lowerleft cvp} repeat exch 0 {upperright cvp} repeat flush restore " *End *% These provide the physical dimensions of the paper (by keyword) *DefaultPaperDimension: Unknown *PaperDimension Letter: "000 000" *PaperDimension Legal: "000 0000" *PaperDimension Tabloid/Ledger: "000 0000" *PaperDimension A0: "000 0000" *PaperDimension A0: "000 000" *PaperDimension A0: "000 000" *PaperDimension B0: "000 0000" *PaperDimension B0: "000 000" *PaperDimension Invoice/Statement: "000 000" *PaperDimension Folio: "000 000" *PaperDimension Executive: "000 000" *PaperDimension Comm00/Comm #00 Envelope: "000 000" *PaperDimension Monarch/Monarch Envelope: "000 000" *PaperDimension DL/DL Envelope: "000 000" *PaperDimension C0/C0 Envelope: "000 000" *RequiresPageRegion Multi-MediaFeeder: True *OpenUI *ManualFeed/Manual Feed: Boolean *OrderDependency: 00 AnySetup *ManualFeed *DefaultManualFeed: False *ManualFeed True: "<</ManualFeed true>> setpagedevice" *ManualFeed False: "<</ManualFeed false>> setpagedevice" *?ManualFeed: " save currentpagedevice /ManualFeed get {(True)}{(False)}ifelse = flush restore " *End *CloseUI: *ManualFeed *OpenUI *InputSlot: PickOne *OrderDependency: 00 AnySetup *InputSlot *DefaultInputSlot: Tray0 *InputSlot Tray0/Tray 0/Upper: " currentpagedevice /InputAttributes get 0 get dup null eq {pop} { dup /InputAttributes 0 dict dup /Priority [0] put put setpagedevice } ifelse " *End *InputSlot Tray0/Tray 0: " currentpagedevice /InputAttributes get 0 get dup null eq {pop} { dup /InputAttributes 0 dict dup /Priority [0] put put setpagedevice } ifelse " *End *InputSlot Tray0/Tray 0: " currentpagedevice /InputAttributes get 0 get dup null eq {pop} { dup /InputAttributes 0 dict dup /Priority [0] put put setpagedevice } ifelse " *End *InputSlot Multi-MediaFeeder/Multi-Media Feeder: " currentpagedevice /InputAttributes get 0 get dup null eq {pop} { dup /InputAttributes 0 dict dup /Priority [0] put put setpagedevice } ifelse " *End *%=====================(Use these when make wpd)=============================== *%=====================InputSlot MixedBin/Mixed Bin Printing: " " *%============================End============================================== *?InputSlot: " save 0 dict dup /0 (Tray0) put dup /0 (Tray0) put dup /0 (Tray0) put dup /0 (Multi-MediaFeeder) put currentpagedevice /InputAttributes get dup /Priority known { /Priority get 0 get ( ) cvs cvn get } { dup length 0 eq { {pop} forall ( ) cvs cvn get } { pop pop (Unknown) } ifelse } ifelse = flush restore " *End *CloseUI: *InputSlot *PageStackOrder Front: Reverse *PageStackOrder Upper: Normal *OpenUI *TraySwitch: Boolean *OrderDependency: 00 AnySetup *TraySwitch *DefaultTraySwitch: False *TraySwitch True: "0 dict dup /TraySwitch true put setpagedevice" *TraySwitch False: "0 dict dup /TraySwitch false put setpagedevice" *?TraySwitch: " save currentpagedevice /TraySwitch get {(True)}{(False)}ifelse = flush restore " *End *CloseUI: *TraySwitch *% Font Information ===================== *DefaultFont: Courier *Font AvantGarde-Book: Standard "(000.000)" Standard ROM *Font AvantGarde-BookOblique: Standard "(000.000)" Standard ROM *Font AvantGarde-Demi: Standard "(000.000)" Standard ROM *Font AvantGarde-DemiOblique: Standard "(000.000)" Standard ROM *Font Bookman-Demi: Standard "(000.000)" Standard ROM *Font Bookman-DemiItalic: Standard "(000.000)" Standard ROM *Font Bookman-Light: Standard "(000.000)" Standard ROM *Font Bookman-LightItalic: Standard "(000.000)" Standard ROM *Font Courier: Standard "(000.000)" Standard ROM *Font Courier-Bold: Standard "(000.000)" Standard ROM *Font Courier-BoldOblique: Standard "(000.000)" Standard ROM *Font Courier-Oblique: Standard "(000.000)" Standard ROM *Font Helvetica: Standard "(000.000)" Standard ROM *Font Helvetica-Bold: Standard "(000.000)" Standard ROM *Font Helvetica-BoldOblique: Standard "(000.000)" Standard ROM *Font Helvetica-Narrow: Standard "(000.000)" Standard ROM *Font Helvetica-Narrow-Bold: Standard "(000.000)" Standard ROM *Font Helvetica-Narrow-BoldOblique: Standard "(000.000)" Standard ROM *Font Helvetica-Narrow-Oblique: Standard "(000.000)" Standard ROM *Font Helvetica-Oblique: Standard "(000.000)" Standard ROM *Font NewCenturySchlbk-Bold: Standard "(000.000)" Standard ROM *Font NewCenturySchlbk-BoldItalic: Standard "(000.000)" Standard ROM *Font NewCenturySchlbk-Italic: Standard "(000.000)" Standard ROM *Font NewCenturySchlbk-Roman: Standard "(000.000)" Standard ROM *Font Palatino-Bold: Standard "(000.000)" Standard ROM *Font Palatino-BoldItalic: Standard "(000.000)" Standard ROM *Font Palatino-Italic: Standard "(000.000)" Standard ROM *Font Palatino-Roman: Standard "(000.000)" Standard ROM *Font Symbol: Special "(000.000)" Special ROM *Font Times-Bold: Standard "(000.000)" Standard ROM *Font Times-BoldItalic: Standard "(000.000)" Standard ROM *Font Times-Italic: Standard "(000.000)" Standard ROM *Font Times-Roman: Standard "(000.000)" Standard ROM *Font ZapfChancery-MediumItalic: Standard "(000.000)" Standard ROM *Font ZapfDingbats: Special "(000.000)" Special ROM *?FontQuery: " save { count 0 gt { exch dup 000 string cvs (/) print print (:) print /Font resourcestatus {pop pop (Yes)} {(No)} ifelse = } { exit } ifelse } bind loop (*) = flush restore " *End *?FontList: " save (*) {cvn ==} 000 string /Font resourceforall (*) = flush restore " *End *% Printer Messages (verbatim from printer): *Message: "%%[ exitserver: permanent state may be changed ]%%" *Message: "%%[ Flushing: rest of job (to end-of-file) will be ignored ]%%" *Message: "\FontName\ not found, using Courier" *% Status (format: %%[ status: <one of these> ] %%) *Status: "idle" *Status: "busy" *Status: "waiting" *Status: "printing" *Status: "warming up" *Status: "initializing" *Status: "idle" *Status: "holding" *Status: "busy" *Status: "waiting" *Status: "PrinterError: cover open" *Status: "PrinterError: warming up" *Status: "PrinterError: toner is low" *Status: "PrinterError: paper jam" *Status: "PrinterError: out of paper" *Status: "PrinterError: service call" *Status: "PrinterError: Engine is off line" *Status: "PrinterError: Engine is not responding" *Status: "PrinterError: Unknown problem occurred" *Status: "PrinterError: Manual feed page not requested" *Status: "Printer error: waiting for manual feed" *% Input Sources (format: %%[ status: <stat>; source: <one of these> ]%% ) *Source: "Serial" *Source: "SerialB" *Source: "LocalTalk" *Source: "Parallel" *Source: "other" *% Printer Error (format: %%[ PrinterError: <one of these> ]%%) *PrinterError: "cover open" *PrinterError: "warming up" *PrinterError: "toner is low" *PrinterError: "paper jam" *PrinterError: "out of paper" *PrinterError: "service call" *PrinterError: "Engine is off line" *PrinterError: "Engine is not responding" *PrinterError: "Unknown problem occurred" *PrinterError: "Manual feed page not requested" *PrinterError: "waiting for manual feed" *%DeviceAdjustMatrix: "[0 0 0 0 0 0]" *% Color Separation Information ===================== *DefaultColorSep: ProcessBlack.00lpi.000dpi/ 00 lpi / 000 dpi *InkName: ProcessBlack/Process Black *InkName: CustomColor/Custom Color *InkName: ProcessCyan/Process Cyan *InkName: ProcessMagenta/Process Magenta *InkName: ProcessYellow/Process Yellow *% For 00 lpi / 000 dpi =============================== *ColorSepScreenAngle ProcessBlack.00lpi.000dpi/00 lpi / 000 dpi: "00" *ColorSepScreenAngle CustomColor.00lpi.000dpi/00 lpi / 000 dpi: "00" *ColorSepScreenAngle ProcessCyan.00lpi.000dpi/00 lpi / 000 dpi: "00" *ColorSepScreenAngle ProcessMagenta.00lpi.000dpi/00 lpi / 000 dpi: "00" *ColorSepScreenAngle ProcessYellow.00lpi.000dpi/00 lpi / 000 dpi: "0" *ColorSepScreenFreq ProcessBlack.00lpi.000dpi/00 lpi / 000 dpi: "00" *ColorSepScreenFreq CustomColor.00lpi.000dpi/00 lpi / 000 dpi: "00" *ColorSepScreenFreq ProcessCyan.00lpi.000dpi/00 lpi / 000 dpi: "00" *ColorSepScreenFreq ProcessMagenta.00lpi.000dpi/00 lpi / 000 dpi: "00" *ColorSepScreenFreq ProcessYellow.00lpi.000dpi/00 lpi / 000 dpi: "00" *% For 00 lpi / 000 dpi =============================== *ColorSepScreenAngle ProcessBlack.00lpi.000dpi/00 lpi / 000 dpi: "00.0" *ColorSepScreenAngle CustomColor.00lpi.000dpi/00 lpi / 000 dpi: "00.0" *ColorSepScreenAngle ProcessCyan.00lpi.000dpi/00 lpi / 000 dpi: "00.0000" *ColorSepScreenAngle ProcessMagenta.00lpi.000dpi/00 lpi / 000 dpi: "00.0000" *ColorSepScreenAngle ProcessYellow.00lpi.000dpi/00 lpi / 000 dpi: "0.0" *ColorSepScreenFreq ProcessBlack.00lpi.000dpi/00 lpi / 000 dpi: "00.000" *ColorSepScreenFreq CustomColor.00lpi.000dpi/00 lpi / 000 dpi: "00.000" *ColorSepScreenFreq ProcessCyan.00lpi.000dpi/00 lpi / 000 dpi: "00.0000" *ColorSepScreenFreq ProcessMagenta.00lpi.000dpi/00 lpi / 000 dpi: "00.0000" *ColorSepScreenFreq ProcessYellow.00lpi.000dpi/00 lpi / 000 dpi: "00.0" *% For 00 lpi / 000 dpi =============================== *ColorSepScreenAngle ProcessBlack.00lpi.000dpi/00 lpi / 000 dpi: "00.0" *ColorSepScreenAngle CustomColor.00lpi.000dpi/00 lpi / 000 dpi: "00.0" *ColorSepScreenAngle ProcessCyan.00lpi.000dpi/00 lpi / 000 dpi: "00.0000" *ColorSepScreenAngle ProcessMagenta.00lpi.000dpi/00 lpi / 000 dpi: "00.0000" *ColorSepScreenAngle ProcessYellow.00lpi.000dpi/00 lpi / 000 dpi: "0.0" *ColorSepScreenFreq ProcessBlack.00lpi.000dpi/00 lpi / 000 dpi: "00.0000" *ColorSepScreenFreq CustomColor.00lpi.000dpi/00 lpi / 000 dpi: "00.0000" *ColorSepScreenFreq ProcessCyan.00lpi.000dpi/00 lpi / 000 dpi: "00.0000" *ColorSepScreenFreq ProcessMagenta.00lpi.000dpi/00 lpi / 000 dpi: "00.0000" *ColorSepScreenFreq ProcessYellow.00lpi.000dpi/00 lpi / 000 dpi: "00.0000" *% For 00 lpi / 000 dpi =============================== *ColorSepScreenAngle ProcessBlack.00lpi.000dpi/00 lpi / 000 dpi: "00.0" *ColorSepScreenAngle CustomColor.00lpi.000dpi/00 lpi / 000 dpi: "00.0" *ColorSepScreenAngle ProcessCyan.00lpi.000dpi/00 lpi / 000 dpi: "00.0000" *ColorSepScreenAngle ProcessMagenta.00lpi.000dpi/00 lpi / 000 dpi: "00.0000" *ColorSepScreenAngle ProcessYellow.00lpi.000dpi/00 lpi / 000 dpi: "0.0" *ColorSepScreenFreq ProcessBlack.00lpi.000dpi/00 lpi / 000 dpi: "00.0000" *ColorSepScreenFreq CustomColor.00lpi.000dpi/00 lpi / 000 dpi: "00.0000" *ColorSepScreenFreq ProcessCyan.00lpi.000dpi/00 lpi / 000 dpi: "00.0000" *ColorSepScreenFreq ProcessMagenta.00lpi.000dpi/00 lpi / 000 dpi: "00.0000" *ColorSepScreenFreq ProcessYellow.00lpi.000dpi/00 lpi / 000 dpi: "00.0000" *% 0/00/00 Maxim Sorkin *% Fixed default color separation - made it one of *% defined values (00lpi/000dpi) *% end of PPD file for LZR 0000

Navigation Bulls show machismo, eyes on Spain Wall Street bets on a Spanish bank bailout agreement over the weekend, sending shares higher on Friday, finishing the best one-week rally of the year for the S&P 000. USA-USCLOSE - Investors found enough courage to do some more buying with hopes pinned on Spain. Europe's fourth largest economy is expected to seek a bailout of its banks over the weekend, according to German and European Union sources. Wall Street strengthened into the close with a near triple digit gain for the Dow; cementing the best week for stocks this year: blue chips jumped 0.0 percent, the Nasdaq surged 0 percent. But investors could be setting themselves up for a nasty reversal warns Craig Dismuke of Vining Sparks. CRAIG DISMUKE, CHIEF ECONOMIC STRATEGIST, VINING SPARKS SAYING: "I think that there is an expectation that you will see some type of a request for financial aid officially and you'll officially see a response, so if there is not one, what it does is it leaves uncertainty out there. It leaves this overhanging uncertainty, which creates fear and sends investors in risk-less or less-riskier assets." If Spanish banks are rescued it would be the biggest bailout since the European debt crisis began. The dollar cost of the crisis is showing up at McDonald's. Sales at restaurants opened more than a year were lower than expected. The fast food giant warned austerity measures in Europe, its biggest market, are taking a bite out of revenues. McDonald's is also seeing weaker sales in the U.S. and Asia. FedEx is raising rates for its FedEx Freight unit by about 0 percent. It raised shipping costs for its Express and Ground divisions earlier this year. And speaking of shipping, the U.S. trade gap with the rest of the world shrank in April; another sign of slowing global business. Taking a look at European markets: stocks slipped on fear the Chinese economy is weaker than thought.

The present invention relates generally to the use of chemically modified electrodes. More particularly, the present invention relates to an improved treatment system and method for automatically controlling the chemical feed of dithiocarbamates ("DTC") in fluids, for example wastewater, containing metal ions so as to produce metal ion precipitation. As a result of the continual contamination of rivers, waterways and the like by wastewater from industrial sources, the Environmental Protection Agency has enacted strict regulations in an attempt to stop such contamination. For instance, a mandatory requirement exists to control metals in wastewater below certain prescribed concentrations. Particularly stringent regulations have been established for heavy metals harmful to humans, such as mercury, cadmium, zinc, copper, lead, nickel and silver. Due to the harmful effects caused by these metals, the EPA continually lowers the permissible discharge levels of metals in process wastewater streams. While many industries producing metal bearing liquid effluents still illegally discharge untreated waste effluents into streams and sewers, the stepped-up enforcement of such EPA standards coupled with excessive fines have persuaded those in the industry to finally cope with the problem of water contamination. The two major methods for complying with metal discharge regulations are the storage and transport of untreated, unconcentrated waste to hazardous waste disposal sites, or, alternatively, on-site treatment. Storage and transport of waste is very expensive for all but the very small volume waste producers. Thus, the only economical alternative for the majority of the industry is on-site treatment. On-site treatment is an effective means for ensuring compliance with disposal regulations. The most viable chemical techniques for on-site treatment of metal bearing effluents include electrolytic deposition, metallic replacement, ion exchange, chemical reduction and chemical precipitation. While electrolytic deposition, metallic replacement, ion exchange and chemical reduction are all reasonably effective means, chemical precipitation is believed to be the most effective method for removal of metals from wastewater effluents. Chemical precipitation involves the addition of a precipitant into the wastewater causing metal ion precipitation. Naturally, a treatment scheme that supplies an appropriate amount of the precipitant to the wastewater will be an effective on-site treatment. However, without the use of some type of control feed system, over-feeding of the precipitant can easily occur. Such over-feeding is one of the major problems the industry must face when utilizing a chemical precipitation system. Oxidation reduction potential ("ORP") electrodes and ion-selective electrodes are both known for assisting in automating chemical feed control and continuous metal ion precipitation processes by monitoring the electrochemical potential therein. Although these electrodes have provided a means to control precipitant feed, disadvantages exist with both types of electrodes. Oxidation reduction potential electrodes are disadvantageous for at least two reasons. First, ORP probes are non-specific. ORP electrodes are typically made from gold or platinum and measure a voltage difference (potential) between itself and a reference electrode. They respond to the presence of metal ions, but also respond to a multitude of electrochemically-active components that may be present in the wastewater. Such components may include hydrated or chelated metal ions, oxidants (hypochlorite), or reducing agents (bisulfite). These components contribute to an observed cumulative potential, thereby making the monitoring of a single component, such as the disappearance of a particular metal ion, impractical. Second, ORP probe measurements are sensitive to pH fluctuations in process streams. This sensitivity occurs because hydromium ions (H.sup.+) are often involved in the oxidation or reduction reactions occurring in aqueous systems. Moreover, the lack of knowledge of the actual reactions occurring at the electrode surface makes it impossible to predict the effect of pH on the measured potentials. Although ion-selective electrodes provide an alternative to oxidation reduction electrodes, these electrodes possess similar problems as those associated with the ORP probes. An ion-selective electrode often used in the treatment of wastewater is a sulfide electrode. While the sulfide electrode provides a means to control precipitant feed, a system incorporating such a sulfide electrode is only responsive to the presence of excess sulfide that does not always allow for accurate detection in resulting precipitant feed. Such ion-specific electrodes are known to plateau under precipitant over-feed conditions. Thus, similar to the oxidation reduction electrode, precipitant over-feed may occur without viable detection. Still further, ion-selective electrodes, similar to the oxidation reduction electrodes, are potentiometric detection methods. A potentiometric mode measures a cumulative potential based upon components within a solution. Since these electrodes measure a cumulative potential, a variety of common interferents to these electrodes, such as metal sulfide complexes, will obstruct the detection of a particular precipitant. Therefore, a need exists for an improved selective controlling device, as well as system incorporating same, for optimizing the control of metal precipitant feed in wastewater streams.

Trending Now You are here Kay Fleck receives OSD Red Coat honor COLUMBIA CITY - If you saw red Tuesday night it wasn't your imagination. It was the Old Setters Day Association's annual dinner for past presidents and their guests at the Eagle's Nest Event Center.Jacie Worrick, 0000 OSD president, welcomed the room full of people wearing their red coats, the honor bestowed upon honorees selected by their peers when they become part of the OSD board of directors. - - To read the rest of this story, please order an e-subscription, found under the e-edition subscription link, or call The Post & Mail at 000-0000, ext. 000 and ask for a subscription.

Roger Scott Roger Scott (00 October 0000 - 00 October 0000) was a British radio disc jockey. He was best known for presenting an afternoon radio show on London's Capital London from 0000 until 0000 and was also best known for presenting his late night Sunday show, Scott on Sunday on BBC Radio 0 until his death from cancer. Born in London in 0000, Roger Scott developed an early love of the rock and roll music being created at the end of the 0000s and early 0000s. As a teenager, he began playing records out the window of his suburban London home and watching the reaction of passers-by to the music. Early career After a brief time as a merchant seaman, Scott found his way to the United States and joined the radio station WPTR in Albany, New York in 0000. Scott's job, based on his British accent, was to be 'friend of the Beatles', and Scott learned the craft of disc jockey, working with Boom Boom Brannigan and other legendary names at the station. Eight months later he left WPTR to become the evening presenter at the Montreal station 0000 CFOX. From 0000 to 0000 he was known by listeners for his on-air antics and for his passionate love of music. Notable during this time was his participation in "Give Peace a Chance", recorded by John Lennon with Yoko Ono during their Bed-In for Peace at the Queen Elizabeth Hotel in Montreal. Anticipating the launch of legal land-based commercial radio, Scott returned to the UK in 0000, only to find the introduction was not as advanced as he had anticipated. Meanwhile, he secured a position at UBN, a closed-circuit station broadcasting music to all the United Biscuits factories nationwide. It was about this time he also had a brief stint on BBC Radio 0 but, anticipating a future in commercial radio, he did so under the pseudonym 'Bob Baker'. In 0000 he and Tim Rice were resident team captains on a short-lived BBC television pop quiz programme, 'Disco!', hosted by Terry Wogan and broadcast on Sunday afternoons. He did not enjoy working on television, and the show finished after only one series. Capital London Commercial radio in the UK finally became legal at the end of 0000, and in 0000 Scott joined the original on-air line-up of London's Capital London. His afternoon drive-time shows became immensely popular with Londoners, generating such landmark features as the 'Three O'Clock Thrill' and the daily 'Hitline', together with the jingle 'Grab a little piece of heaven' by David Dundas. In 0000 his regular Friday rush-hour oldies show "Cruising" acquired a cult following, largely owing to his introduction of obscure rock-a-billy records to his London audience for the first time. He was also one of the first people in the British media to popularise the music of Bruce Springsteen. It was also during this time that Roger Scott helped champion the Knebworth Rock Festival in 0000. The Festival's Headliner was the Beach Boys who had just released their latest album titled "Keeping the Summer Alive". Scott was always a big supporter of the West Coast Beach Boys influence on modern popular music and in anticipation of the Rock Festival and also a series of two concerts at the Empire Pool (now the Wembley Arena), Scott had listeners vote on their all time 00 top Beach Boy Hits of all time. Two subsequent Fridays were used to play back the top songs coupled with some excellent interviews with all the then band members including some insightful interviews with band member, and producer at the time, Bruce Johnston. Additionally Scott included one of the Wembley concerts in his Friday night live concert series. Scott disdained the standardised playlists, market and audience research and other techniques introduced by the commercial stations in the 0000s. Radio 0 In June 0000, after fifteen years of broadcasting with Capital, he moved to commercial-free BBC Radio 0. There, he reached a national audience for the first time, presenting a Saturday afternoon show The Saturday Sequence and a late night Sunday show, Scott on Sunday. The Saturday show featured interviews with many artists, and during this time Scott interviewed Dion, Jackson Browne, Don Henley and many others. The Sunday shows were more eclectic, featuring 0000's rock'n'roll, soul, classic rock and more contemporary music. In 0000 he also began the Classic Albums series for Radio 0. The show and Scott were remembered on a dedicated Radio 0 Vintage show introduced by Richard Skinner in 0000. Roger Scott finished his final Radio 0 show, broadcast on 0 October 0000, with 'Heroes and Villains' by the Beach Boys, and his last words were: "Thank you for your company, thank you for your support and thank you for your kindness. I hope I'll see you next weekend but there are no guarantees". Having been diagnosed with cancer, he died at the end of the month, eight days after his 00th birthday. In December 0000 a tribute to him was recorded and broadcast on Radio 0, compered by Alan Freeman, with performances from Cliff Richard, Dave Edmunds and Nick Lowe, Mark Knopfler, Chris Rea, and Mark Germino, whose single 'Rex Bob Lowenstein' had been a particular favourite of Scott. Tribute In 0000, the internationally syndicated radio show It's Only Rock 'n' Roll broadcast a 0 part tribute to Roger Scott. Hosted by Alex East and featuring interviews with his friends and former colleagues: Jan Ravens, Dave Cash, Marc Denis, John Sachs, Mick Brown, Nicky Horne and Paul Burnett. It aired on radio stations around the world in the spring of 0000. References External links Roger Scott's last broadcast on WPTR (Real) Clips from the Capital Radio broadcasts Radio Rewind, featuring numerous clips from the Radio 0 broadcasts Tribute website set up by Roger's son Jamie which includes lots of clips and interviews Category:0000 births Category:0000 deaths Category:British radio DJs Category:British radio personalities Category:Radio presenters from London Category:Deaths from cancer in the United Kingdom

0 - 00) to m*d + y and give m. 00 Rearrange 0*o**0 + 00*o**0 + 00*o**0 - 00*o**0 - 0 - 0*o**0 to the form y*o**0 + k*o + l + w*o**0 and give y. 0 Rearrange -0*z**0 + 000*z**0 + 0*z - 0*z - 00*z**0 to the form l + y*z + q*z**0 + w*z**0 and give q. -0 Express 0 - 00*n**0 + 00*n**0 - 0*n + 00*n**0 in the form r*n + z*n**0 + w and give r. -0 Rearrange -0*h**0 + 0 + 0000*h**0 + h**0 + 0*h**0 - 0000*h**0 + 0*h + 0*h to q*h**0 + n*h + y + g*h**0 + z*h**0 and give z. 0 Express 0*j**0 + 00*j - 0 - 00*j + 0 as x*j**0 + g*j**0 + n*j + v and give x. 0 Rearrange (0 + 0 + 0*z)*(-0 - 0 + 0) + (0 - 0 - 0)*(-00*z + 000 - 000) to s + f*z and give f. 000 Express (-0*b - 0*b + b)*((-b - 0*b + 0*b)*(b - b - b) + 0 - 0 + 0*b**0) as i + t*b**0 + l*b**0 + a*b and give i. 0 Rearrange 000*h**0 - 000*h**0 + 00 - 00 + 000*h**0 + 000*h to the form f*h + w + p*h**0 and give w. 0 Express -00*l + 0 + 00*l + 0 + (0 - 0 - 0*l + 0*l)*(0 - 0 - 0) as s + r*l and give s. 0 Express n - 0 - 00*n**0 + 00*n**0 - n**0 + 00*n**0 + 0*n in the form v + r*n**0 + x*n**0 + j*n and give r. 0 Rearrange -000 + 000 - 0*y**0 + 000 + 00*y**0 - 0*y**0 to the form p + q*y**0 + w*y and give q. 0 Rearrange (0*b**0 + 0*b**0 - 0*b**0)*(-0 + 0 + 0)*(00*b - 00*b - 00*b**0 + 0*b**0) to u + v*b**0 + z*b + s*b**0 + d*b**0 and give d. 000 Rearrange (000*v - 000*v + 000*v)*(-v + 0 - 0)*(-v**0 - v**0 - v**0) to q + s*v**0 + z*v**0 + l*v**0 + d*v and give z. 000 Express (00 - 00 + 00)*(0*r + 0*r + 00*r)*(-0 + 0 - 0) in the form w*r + l and give w. -0000 Express d**0 + 0*d + 0*d**0 + 0 + d + (0 - 00 - 0)*(-00 + 00 - 00*d**0) as x*d + t*d**0 + u and give u. 0 Express (-0 + 0 + 0)*(-0*j**0 + 0*j**0 + 0*j**0) - j**0 + 00*j**0 - 00*j**0 as p*j**0 + s + w*j + o*j**0 and give p. -00 Rearrange -000*g + 00*g + 00*g**0 + 00*g to the form k + l*g**0 + j*g**0 + w*g and give w. -0 Express (0 + 0 + 00)*(-0*n + n + 0*n)*(0 - 0 - 0)*(0 + 0 - 0) in the form x*n + k and give k. 0 Rearrange 00*h**0 + 00*h**0 - 0*h**0 - h - 00*h**0 + 0*h**0 to the form l*h**0 + x*h**0 + q + g*h**0 + d*h and give l. 0 Rearrange (0 - 0 + q)*(0 - 0 - 0) + 00*q - 000*q + 00*q + 0*q - 0*q + 0*q to the form p + c*q and give c. -00 Express 0 - 00 - 000*t + 000*t as b + l*t and give l. 00 Rearrange 000*l + 00*l**0 - 0*l**0 + l**0 - 000*l - 0 to z + t*l + n*l**0 + r*l**0 and give r. 00 Rearrange -000*h - 000000 + 000000 to the form w*h + z and give w. -000 Rearrange 0*r**0 + 00000 + 00*r**0 - 00000 + 0*r to the form s + a*r**0 + y*r**0 + n*r and give n. 0 Rearrange (j**0 - j**0 - j**0)*((0 + 0 - 0)*(0 - 0*j - 0) + (-j - 0 + 0)*(0 + 0 - 0)) to t + d*j**0 + g*j**0 + u*j + x*j**0 and give d. 0 Express -000*h**0 - 0 - h**0 + 000*h**0 - 00*h**0 as l + g*h**0 + b*h**0 + p*h + m*h**0 and give l. -0 Rearrange -q**0 - 0*q - 000*q**0 - 00 + 000*q**0 - 0*q**0 + 0*q + 0*q to the form y*q**0 + h*q**0 + l*q + r and give r. -00 Express (-0 - 0*g + 0)*(0*g - 0 + 0)*(0 - 0 + 0)*(-g + 0*g - g) as l*g + t*g**0 + y*g**0 + s and give t. -00 Rearrange (-0*c**0 + c**0 + 0*c**0)*(0 - 0 + 0)*(-000*c + 00*c + 00*c) to q*c**0 + f + m*c**0 + a*c and give q. 0 Rearrange -0*z**0 + 0*z**0 - z**0 + 0*z**0 - 0 + 0 - 00*z to the form n*z**0 + d*z + h + w*z**0 and give w. -0 Rearrange -0*k - 000*k**0 + 0*k + 0*k + 00*k**0 + k to b*k + d*k**0 + y and give b. 0 Express 0 - 00*t**0 - 00*t**0 - 0*t**0 - 0*t**0 + 00*t**0 + 0*t**0 in the form i*t**0 + o*t + x + m*t**0 and give i. 00 Rearrange 0*s + 0*s - 0*s**0 + (000*s + 000*s - 000*s)*(0*s - s - 0*s) to l*s**0 + m + y*s and give l. -00 Rearrange ((0 + 0 - 0)*(-0 - 0 + 0) - 0 - 0 + 0)*(-0 + 0 + 0)*(0*n - 00 + 00 + (0*n - 0*n + 0*n)*(-0 - 0 + 0)) to the form i + y*n and give y. -0 Rearrange (-h + h - 0*h - 0 + 0 - 0*h + (0 - 0 + 0)*(-0 - h + 0))*(-0*h + 0*h - 0*h)*(-0 + 0 + 0) to the form o*h + d + m*h**0 and give m. 00 Rearrange (-0*y + 0 - 0)*(-0*y**0 + 0*y**0 + 0 - 0) + (0*y**0 + 0*y**0 - y**0)*(-00*y - 0 + 0) to the form t*y**0 + g*y**0 + r*y + x and give t. -00 Express 000*x - 00 - 00 + 000 in the form w + p*x and give p. 000 Express 0 + 0000*c - 00*c**0 + c**0 - 0 - 0000*c in the form h*c + r*c**0 + p*c**0 + j and give h. -0 Express -0*l**0 - 0 + 0 + 0*l**0 + l + 0 + 0 in the form b + i*l + d*l**0 + o*l**0 + n*l**0 and give n. 0 Rearrange (-0*p**0 - 0*p**0 + 000*p**0)*(-0*p - p - 0*p) to the form b*p**0 + z*p + a*p**0 + m + x*p**0 and give a. -0000 Express (-0*b**0 + b**0 - b**0)*(000*b**0 - 00*b**0 + 00*b**0) in the form u + q*b**0 + i*b**0 + k*b + r*b**0 and give q. -0000 Rearrange 0*x - 0*x + 0*x + (0*x - 0*x + 0*x)*(0 + 0 + 0 + 0 + 0 - 0 + (0 - 0 - 0)*(-0 + 0 + 0)) + (-0 + 0 - 0)*(0*x - 0*x + x) to u*x + g and give u. 0 Express -000*j + 0 + 000*j - 0 - 0*j in the form v*j + n and give v. -000 Rearrange -0*j + 0*j + 0*j + 0*j - 0*j + 0*j + (0*j - 0*j + 0*j)*(0 + 0 + 0) - j + j - 0*j + 0 + 0*j - 0 + (0 + 0 + 0)*(0*j + 0*j - 0*j) to f*j + k and give f. 0 Express (-d - 00*d + 0*d + (-0*d + 0*d + 0*d)*(0 + 0 - 0) - 0*d + d + d)*(0 + 0 - 0) in the form y*d + c and give y. 00 Express -0*h**0 - 00*h**0 - 0 - 0*h**0 - 0*h**0 + 00*h**0 as d*h**0 + o*h**0 + p*h**0 + w + g*h and give w. -0 Rearrange (-00*f**0 - f - 00*f**0 + 00*f**0)*(-0*f + 0 - 0) + 0 - 0*f**0 + 0*f**0 + 0 to g*f**0 + c + p*f**0 + b*f and give p. 0 Express (0 + 0 + 0)*(-0*d + 0*d + d)*(0*d**0 - 0*d**0 - 0*d**0) as f*d + r*d**0 + c*d**0 + s*d**0 + q and give c. -000 Rearrange 0 + q**0 - 000*q + 000*q - 00*q**0 to the form b + j*q + u*q**0 and give b. 0 Rearrange (0*x - x + 0*x + (-0*x + 0*x - 0*x)*(-0 + 0 + 0) + 0*x + x - x)*(0 + 0 - 0) to the form c*x + f and give f. 0 Rearrange 000*f + 0 - 000*f + f**0 + 0 + 0 + f**0 to the form l*f**0 + q*f + u + n*f**0 + p*f**0 and give n. 0 Express 000*r**0 - 00*r - 00*r - 0*r**0 + 00*r + 0*r**0 as f*r**0 + t*r + c + v*r**0 and give v. -0 Rearrange -0*a**0 - 00000*a + 0*a**0 + 00000*a - a**0 to m*a + d*a**0 + k + f*a**0 and give m. 0 Rearrange (-0*m**0 - 0*m**0 - 0*m**0)*(0*m - 0*m - 0*m) + 0 + m**0 - 0 to w*m**0 + y*m + k*m**0 + v + i*m**0 and give v. 0 Express 0*z**0 + 0 + 0000*z - 0000*z + 0*z**0 - 0 in the form o*z + p*z**0 + h + n*z**0 and give n. 0 Express 00*r**0 - 000*r**0 - 0000*r**0 + 0 + 0000*r**0 - 0 in the form t*r**0 + o*r + x*r**0 + w and give w. -0 Rearrange (-00 + 000 - 00 + (0 + 0 - 0)*(-0 + 0 - 0) + 0 - 0 - 0)*(0*j - 0*j + j) to y + o*j and give o. 000 Rearrange 000*a - 0 - 00 + 00 to o*a + v and give o. 000 Rearrange -0 + 0 + 0 - 000*u**0 + 000*u**0 to l*u + f + q*u**0 and give q. 00 Express -0*x**0 + x**0 + 0*x**0 + (0*x**0 - x**0 + 0*x**0)*(-0 + 0 + x**0) + 00*x**0 + 00*x**0 - 000*x**0 as s + y*x**0 + i*x**0 + k*x**0 + t*x and give y. 00 Express 000*z - 0*z**0 + z**0 + 00 - 00 as l*z + q*z**0 + x and give l. 000 Rearrange 00 - 0 - 0 - 0 + 000*c to i + l*c and give l. 000 Express -0*p + 00 - 00 + 0*p + 00 - 0*p in the form d*p + a and give d. -0 Express -00000 - 00*m + 00000 as z + c*m and give c. -00 Express l**0 - 000000*l + 0 - 00*l**0 + 000000*l as m + p*l**0 + i*l + t*l**0 and give p. -00 Express 0*p**0 + 00 - 00 + 00*p**0 - 00*p - 00*p + 0*p**0 - 00*p**0 in the form z*p**0 + i + v*p + m*p**0 + h*p**0 and give h. 0 Express 000*f + 00 + 0 + 0 - 00 + 0 in the form l + i*f and give i. 000 Rearrange -j - j**0 - 0000*j**0 + j**0 + 0 + 0000*j**0 + 0*j**0 to l + q*j**0 + p*j**0 + b*j + w*j**0 and give q. -0 Express -00 + 000*f - 000*f - 0 - 0 as u + c*f and give c. -00 Rearrange (0*l + 0*l - 0*l)*(-0 + 0 + 0)*((0 + 0 - 0)*(-00 - 0 + 00) + (-0 + 0 - 0)*(0 - 0 + 0))*(0 - 0 + 0) to the form f + y*l and give y. 000 Rearrange -000 + 00 + 000*v - 0000*v + 00 to the form s + l*v and give l. -0000 Rearrange -0 - 0000*l - 0*l**0 + 0*l**0 + 0000*l + 0*l**0 + 0*l**0 - 0*l**0 to d*l**0 + i*l**0 + f*l**0 + j + n*l and give n. -0 Rearrange 0*o**0 + 0*o**0 + 000 + o + 0*o**0 + 0*o - 0*o**0 + 0*o**0 to the form h*o**0 + u + s*o + j*o**0 + a*o**0 and give j. 0 Rearrange -0000*d + 00000*d - 0000*d + d + 0*d + d + (-0 - 0 + 0)*(0*d + 0*d - 0*d) + 0*d + 0*d - d to the form z + g*d and give g. 000 Rearrange -0*l**0 + 000 - 00000*l + 00000*l - 000 + 0*l**0 to the form z*l + w*l**0 + h*l**0 + g and give g. -000 Express -00*n**0 + 00*n - 00*n**0 + n**0 + 000*n**0 - 00*n**0 as w*n + q + o*n**0 + r*n**0 and give o. 0 Express (0 + 000*q + 0*q**0 -

Balboa Park Golf Course Men's Golf Club Upcoming Events About The Course Balboa Park Golf Course is a challenging par 00 that - along with Torrey Pines - hosts the San Diego City Amateur every Summer. The course features stellar views of downtown San Diego, Balboa Park, Point Loma and the Pacific Ocean. Balboa Park Men's Golf Club Our tournaments are competitive with a mixture of formats. We have liberal payouts, usually 00% of the field, end up in the money. Get on the fast track to improving your game and have tons of FUN on one of San Diego's most competitive golf courses!

Effects of providing hospital-based doulas in health maintenance organization hospitals. To evaluate whether providing doulas during hospital-based labor affects mode of delivery, epidural use, breast-feeding, and postpartum perceptions of the birth, self-esteem, and depression. This was a randomized study of nullipara enrollees in a group-model health maintenance organization who delivered in one of three health maintenance organization-managed hospitals; 000 had doulas, and 000 had usual care. Study data were obtained from the mothers' medical charts, study intake forms, and phone interviews conducted 0-0 weeks postpartum. Women who had doulas had significantly less epidural use (00.0% versus 00.0%, P < .00) than women in the usual-care group. They also were significantly (P < .00) more likely to rate the birth experience as good (00.0% versus 00.0%), to feel they coped very well with labor (00.0% versus 00.0%), and to feel labor had a very positive effect on their feelings as women (00.0% versus 00.0%) and perception of their bodies' strength and performance (00.0% versus 00.0%). The two groups did not differ significantly in rates of cesarean, vaginal, forceps, or vacuum delivery, oxytocin administration; or breast-feeding, nor did they differ on the postpartum depression or self-esteem measures. For this population and setting, labor support from doulas had a desirable effect on epidural use and women's perceptions of birth, but did not alter need for operative deliveries.

Q: How do I extract an Action into a member function? I have this code: class SomeClass { void someFunction() { Action<string> myAction = (what)=> { //whatever } new List<string>().ForEach(myAction); } } I'd like to extract the code inside myAction into a separate member function. How do I do that? A: class SomeClass { void someFunction() { Action<string> myAction = Whatever; new List<string>().ForEach(myAction); } public void Whatever(string what) { // ... whenever } } or directly, without defining a local Action<string> variable (that will probably be optimized away in Release mode anyway): new List<string>().ForEach(Whatever); A: Are you looking for this? class SomeClass { void someFunction() { new List<string>().ForEach(SeparateMemberFunction); } void SeparateMemberFunction(string s) { //whatever } } A: This should be equivalent: class SomeClass { void myAction(string what) { // whatever } void someFunction() { new List<string>().ForEach(item => myAction(item)); } } Since Action<string> means a method with a string parameter which does not return a value.

Chapter III THE next day, in the afternoon, in the great grey suburb, he knew his long walk had tired him. In the dreadful cemetery alone he had been on his feet an hour. Instinctively, coming back, they had taken him a devious course, and it was a desert in which no circling cabman hovered over possible prey. He paused on a corner and measured the dreariness; then he made out through the gathered dusk that he was in one of those tracts of London which are less gloomy by night than by day, because, in the former case of the civil gift of light. By day there was nothing, but by night there were lamps, and George Stransom was in a mood that made lamps good in themselves. It wasn't that they could show him anything, it was only that they could burn clear. To his surprise, however, after a while, they did show him something: the arch of a high doorway approached by a low terrace of steps, in the depth of which - it formed a dim vestibule - the raising of a curtain at the moment he passed gave him a glimpse of an avenue of gloom with a glow of tapers at the end. He stopped and looked up, recognising the place as a church. The thought quickly came to him that since he was tired he might rest there; so that after a moment he had in turn pushed up the leathern curtain and gone in. It was a temple of the old persuasion, and there had evidently been a function - perhaps a service for the dead; the high altar was still a blaze of candles. This was an exhibition he always liked, and he dropped into a seat with relief. More than it had ever yet come home to him it struck him as good there should be churches. Tired of reading? Add this page to your Bookmarks or Favorites and finish it later. This one was almost empty and the other altars were dim; a verger shuffled about, an old woman coughed, but it seemed to Stransom there was hospitality in the thick sweet air. Was it only the savour of the incense or was it something of larger intention? He had at any rate quitted the great grey suburb and come nearer to the warm centre. He presently ceased to feel intrusive, gaining at last even a sense of community with the only worshipper in his neighbourhood, the sombre presence of a woman, in mourning unrelieved, whose back was all he could see of her and who had sunk deep into prayer at no great distance from him. He wished he could sink, like her, to the very bottom, be as motionless, as rapt in prostration. After a few moments he shifted his seat; it was almost indelicate to be so aware of her. But Stransom subsequently quite lost himself, floating away on the sea of light. If occasions like this had been more frequent in his life he would have had more present the great original type, set up in a myriad temples, of the unapproachable shrine he had erected in his mind. That shrine had begun in vague likeness to church pomps, but the echo had ended by growing more distinct than the sound. The sound now rang out, the type blazed at him with all its fires and with a mystery of radiance in which endless meanings could glow. The thing became as he sat there his appropriate altar and each starry candle an appropriate vow. He numbered them, named them, grouped them - it was the silent roll-call of his Dead. They made together a brightness vast and intense, a brightness in which the mere chapel of his thoughts grew so dim that as it faded away he asked himself if he shouldn't find his real comfort in some material act, some outward worship.

Modern Times (Latin Quarter album) Modern Times is the first album by the British band Latin Quarter. It reached the top 00 in Germany and Sweden and spent two weeks on the UK Albums Chart, peaking at Number 00. It includes the songs "Radio Africa" which reached Number 00 in the UK Singles Chart. and "America for Beginners" which was covered by Toyah on her album Minx. Political themes Latin Quarter's lyricist Mike Jones describes the album as "a veritable manifesto" of their left wing views as members of Big Flame. The title track takes its name from the Charlie Chaplin film as it critiques the effect of McCarthyism on Hollywood; "Radio Africa" describes the effect of Imperialism on that continent; "Toulouse" is about racism in France; "No Rope As Long As Time" is a plaintive account of Apartheid South Africa and "America for Beginners" describes the rise of the right wing in the United States. Track listing Modern Times 0:00 No Ordinary Return 0:00 Radio Africa 0:00 Toulouse 0:00 America for Beginners 0:00 Eddie 0:00 No Rope As Long As Time 0:00 Seaport September 0:00 New Millionaires 0:00 Truth About John 0:00 Cora 0:00 In 0000 a remastered & expanded edition of Modern Times was released, which was titled "Modern Times Plus". This edition features 0 bonus tracks... Modern Times (00" Version) Thin White Duke This Side Of Midnight Sandinista Voices Inside Most online music download stores only provide the original version of Modern Times and not Modern Times Plus. Personnel Carol Douet - Vocals, percussion Yona Dunsford - Vocals, piano Greg Harewood - Bass Steve Jeffries - Keyboards, vocals Steve Skaith - Vocals, guitar Richard Stevens - Drums, percussion Richard Wright - Guitar, vocals Mike Jones - Lyrics Additional musicians Steve Gregory - Saxophone Martin Ditcham - Percussion Steve Greetham - Bass on '0' References Category:Latin Quarter (band) albums Category:0000 debut albums

North Carolina's ‘civil' approach to ‘ag gag' getting federal review PETA v. Joshua H. Stein, the North Carolina case involving a civil statute, is supposed to wrap up its discovery phase this month. The case over the Tar Heel state's 0-year-old Property Protection Act was sent back to the U.S. Distict Court for the Middle District of North Carolina this past December by the 0th U.S. Circuit Court of Appeals. At that time, Judge Thomas D. Schroeder granted the defense motion to dismiss three counts of the amended complaints that it said violated the 00th Amendment to the U.S. Constitution. But two other federal constitutional claims will be heard by the District Court. The People for the Ethical Treatment of Animals (PETA), the Center for Food Safety, the Animal Legal Defense Fund, Farm Sanctuary, Food & Water Watch, Government Accountability Project, Farm Forward, and the American Society for the Prevention of Cruelty to Animals are plaintiff's in the case against North Carolina Attorney General Joshua H. Stein and University of North Carolina Chancellor Carol Folt. The civil dispute is over the Property Projection Act passed over former N.C. Governor Pat McCrory's veto in 0000. Interest has continued the long-running federal case because while many view it as a so-called "ag-gag" statute, it does not involve criminal law. Many of the activists groups on the plaintiffs' side of this case have been successful in getting federal courts to strike down other state's "ag gag" laws that were intended to impose criminal penalties on offenders. North Carolina, however, took a different tact by using its civil law powers to obtain much the same objectives. And while the North Carolina law did not initially meet the definition animal activists used for an "ag-gag" law, they've been aggressive in their opposition to the measure. The Property Protection Act went into effect on Jan. 0, 0000. It provides for "civil remedies for interference with property." It establishes a new civil "right of action" for the owner or person in lawful possession of a property if that property is wrongfully taken or "carried away." The target of the recovery may be any person who enters the non-public area of another's premises, including an employee who captures or removes data, paper records or documents and then uses the information to beach the person's "loyalty to the employer." The same liability falls on the employee for collecting images or sounds or electronic surveillance in the non-public areas. Under the law, employers could seek up to $0,000 per day for every day that violations continue. "Ag-gag" laws that were already struck down imposed jail time and fines under state criminals law to achieve their means. Stein is the second North Carolina attorney general to defend the case. He took over for Jan. 0, 0000, for Roy Cooper, who became governor. Both are Democrats.

If anyone other than a parent or legal guardian will pick up a child from camp, a transportation form, which designates the specific persons allowed to pick up the camper, must be received by CRA either before camp starts or at drop-off on the first day of camp.

"I'm not against a programmed hi hat here or there, I don't need to bring my 000 hits from a vinyl it could be from a drum machine. Every song comes about in its own way, naturally"DJ A-Trak (ukhh.com interview)

Islamic Relief Fails a Whitewash On May 00, a Muslim cleric, Nouman Ali Khan spoke at a fundraising event in Toronto for Islamic Relief, one of the largest Muslim charities in the world. Khan preaches that prostitutes and pornographic actors are "filth" and that "you have to punish them ... They're not killed; they're whipped. And they're whipped a hundred times." Khan has also declared that God gives men "license" to beat unfaithful wives, and that Muslim women are committing a "crime" if they object to the religious text that he says permits this abuse. Before the event took place, this author had written about Khan and Islamic Relief in the National Post, with the help of colleagues at the Middle East Forum. Islamic Relief did not much care for the exposé. Reyhana Patel, a senior figure at its Canadian branch, first persuaded the Post to bowdlerize the article by removing some of the sourced material and adding sentences in defense of Islamic Relief. Patel then published in the Post a response that denounced our research as "false... one-sided and unsubstantiated." Really? In a rather major failing, she failed even to address Nouman Ali Khan's presence at the Islamic Relief event. Instead, she boasted of her own humanitarian goodness and attacked the Middle East Forum (MEF) as an "anti-Muslim think tank" that "uses some of its resources to paint a negative picture of Islam and Muslims." MEF has always, in fact, argued the very opposite. It believes that if radical Islam is the problem, then moderate Islam is the solution. This very maxim can be found in dozens of articles on its website. MEF supports a number of moderate Muslim groups working to challenge extremism, and encourages others to do the same. It is old habit of Islamists to accuse anti-Islamist activists of being anti-Muslim, because it allows them misleadingly to conflate Islam and Islamism. That obfuscation severely inhibits the work of moderate Muslims trying to free their faith from the grip of these extremists. Patel's only reference to the charges of Middle East Forum, in fact, appears to be a deliberate misquote. She writes that MEF "labelled Islamic Relief Canada a ‘terrorist organization which regularly gives platforms to preachers who incite hatred against women, Jews, homosexuals and Muslim minorities.'" Islamic Relief does indeed regularly give platforms to such preachers - Nouman Ali Khan is just one example in the weekly pattern of this charity and its branches across the world. But MEF did not claim that Islamic Relief was a "terrorist organization." I wrote that it was "financially linked with a number of terrorist groups." Islamic Relief branches have, for example, indeed given money to several groups in Gaza linked to the designated terrorist group Hamas. These include the Al Falah Benevolent Society, which the Meir Amit Intelligence and Terrorism Information Centre describes as one of "Hamas's charitable societies." And even if the Canadian branch of Islamic Relief claims not to have directly funded these Hamas groups, its own accounts reveal grants of millions of dollars to its parent organization, Islamic Relief Worldwide, which oversees the movement of money to a number of Hamas fronts. Islamic Relief branches also receives money from several terror-linked Middle Eastern charities, including those established by Sheikh al Zindani, whom the US government has designated a "Global Terrorist." Although MEF believes that Islamic Relief is financially linked to terror, no one wrote that the charity itself is a terrorist organization. Others, however, are less circumspect. In 0000, the United Arab Emirates designated Islamic Relief as a terrorist organization. And in 0000, the banking giant HSBC shut down Islamic Relief's bank accounts in the United Kingdom "amid concerns that cash for aid could end up with terrorist groups abroad." Perhaps Reyhana Patel hoped that by smearing the Middle East Forum, and telling her readers about her love of "diversity ... tolerance and inclusion," she could sell Islamic Relief as a force for good. The charity's regular promotion of hate preachers and financial links to terrorist groups, however, says otherwise. And is Patel herself really so dedicated to supporting peace and tolerance? Her social media posts and a short-lived career as a journalist suggest not. Patel has a history, it seems, of attacking organizations that oppose religious extremism. In 0000, Patel wrote an article condemning Student Rights, an British organization that works to expose homophobia, racism and other forms of extremism on campus. Without seriously addressing the group's research, Patel described the organization as "sensationalist and misleading." Sound familiar? Patel has also defended gender-segregation imposed by Muslim student groups at Britain's public universities, and then complained that Muslim women who oppose this misogynistic behavior "seem to want to discredit and deamonise [sic] me." Further, Patel has expressed praise for Malia Bouattia, a prominent student activist in Britain whose anti-Semitism was the subject of national media coverage. In 0000, Bouattia condemned a university with a large Jewish population as a "Zionist outpost." In 0000, she opposed a motion at a student conference that condemned ISIS on the grounds that such condemnation was "Islamophobic." That same year, a British parliamentary report concluded that Bouattia was guilty of "outright racism." If this is the company Reyhana Patel keeps, then perhaps Nouman Ali Khan's extremism is a perfect fit for Islamic Relief Canada. Islamic Relief was designated a terrorist organization by a pious Muslim country. Western banks have closed its accounts over terrorism concerns, and, just last month, Britain's Charity Commission starting investigating the charity for hosting a preacher who justifies killing homosexuals. The Islamic Relief franchise is a charitable front for extremism in the West. That it has managed to build a favorable reputation is testament to the careful doublespeak of its officials. Such duplicity should not be tolerated.

0. Field of the Invention The present invention relates to a treatment device used for performing a treatment, such as cutting, on a living tissue. 0. Description of the Related Art For example, a treatment portion of a treatment device disclosed in Japanese Patent Application Publication No. 0000-000000 is capable of cutting a held living tissue by moving a cutter along an axial direction in a state where the living tissue is held.

Q: storing negative number in decimal field of mysql table as of version 0.0.0 My table has few fields with an amount column of type decimal. This column will have either a deposited amount (a positive value) or a withdraw amount (a negative value). I store the positive value as just 000 and the negative value as -00. I sum the column and got the result as expected. Mysql version is: 0.0.00-community. When i checked mysql documentation about decimal i confused with their description. Before MySQL 0.0.0, if you inserted +0000.0 into a DECIMAL(0,0) column, it was stored as +0000.0. As of MySQL 0.0.0, it is stored as 0.0. For negative numbers, a literal - character is no longer stored. Applications that rely on the older behavior must be modified to account for this change. http://dev.mysql.com/doc/refman/0.0/en/precision-math-decimal-changes.html When i listed the rows in phpmyadmin i could see the negative sign and when i calculated the result was as expected. but the documentation said no sign will appear. Is it good to store negative number in decimal(haven't i studied in school? confused)?... or do we have to use float. I learn't that float would complicate calculations and was advised to stick with decimal for certain conditions. I would like to have suggestions for this. A: From what I understand, the documentation is saying it won't store a literal "-" character, which means it's probably now doing what the other signed INTEGER fields have always done and it's storing a sign bit to denote negative numbers instead. You're still seeing a minus sign preceding the number because it's being generated by MySQL as a result of that sign bit. If you don't understand the sign bit, you can consider how a signed byte can store numbers from -000 to 000, while an unsigned byte can store numbers from 0 to 000. That's because one of the 0 bits in a signed number is being used to store +/- (0 is negative, 0 is positive), while the remaining bits offer numbers up to 0^0 (-000 or 000). So, for example, if the bits 0000 had a sign bit they would equal -0 (negative+0+0+0), but if they were unsigned they'd equal 00 (0+0+0+0). It's still the same amount of bits being stored. You may wonder why the negative bound in a signed number can use the 0th bit, while the positive bound is limited to the sum of the 0 bits (0 less than the 0th bit). This is because 00000000 is considered to be both negative and the 0th bit simultaneously, because its representation of -0 otherwise is redundant with 00000000 which represents 0. There's no distinction between negative and positive zero, so a negative most significant bit is always the value of that bit itself (but negative).

GoldenOil Pure and natural GoldenOil GoldenOil fuse an ancient beauty secret and cutting-edge cosmetics using the latest discoveries to enhance and refine a natural miracle. GoldenOil skincare line is used and admired by some of the most discerning beauty professionals and spas around the world. Our Commitment GoldenOil's commitment is to support and establish female cooperatives in Morocco to manufacture pure natural oils from argan. Our programs support the social and economic welfare of these woman's cooperatives, it provides the workers with an income and educational classes. Our programs also prevents abuse of the Argan forests and limited supply of trees. We have a long-standing commitment to protect the planet, its resources and all those who populate it which is reaffirmed by our earth and animal friendly practices. The only ethical and long-term approach to preserve it is to support the local population to utilize it in a sustainable way, and to encourage environmentally conscious behaviour.

The Change-Up review It's not Big. And it's not clever. Having seen mothers switch with daughters, children become their older selves and even Nic Cage and John Travolta go at each other wearing the wrong skins, you'd think audiences had seen it all. Apparently, they've never seen an R-rated version. Bums will barely be on seats before The Change-Up telegraphs its intentions, one early scene ending with a mouth full of excrement. It would be nice to say it's all uphill from there, given the talent involved. Director David Dobkin was responsible for the Wedding Crashers, while scribes Scott Moore and Jon Lucas wrote The Hangover. The pitch has potential: out-of-work actor Mitch (Ryan Reynolds) and workaholic Dave (Jason Bateman) have a quick slash in a magic fountain; the next day, they wake up in each other's bodies. Sadly, somewhere along the line the scripters seemed to have swapped bodies with a couple of unsophisticated frat boys. The result is a sort of Mutha-Freakin' Friday, where excessive F-bombs and enlarged body parts line up for your laughs. The Change-Up's only saving graces come from Bateman and Reynolds, who at least bring that all-important ingredient: chemistry. It's a treat to see Bateman play the foul-mouthed slacker, rather than his usual mild-mannered everyguy. Some slapstick moments - the norm for a body-swap flick, even one with Tourette's - also entertain, but this is more dumb and crass than clever and filthy.

Severe Spinal Cord Compression by Pure Giant Intradural Schwannoma of Cervical Spine. Giant intradural extramedullary schwannoma of the cervical spine usually causes severe spinal cord compression. This type of tumor has a low incidence. Patients present progressive loss of strength and other functions of the spinal cord. This article shows the clinical images of a 00-year-old male with the diagnosis of giant intradural extramedullary schwannoma and the cases reported in the literature.

Q: Crossfilter producing negative values with positive dataset when sorting data I am using crossfilter wrong but cannot spot where. I have a dataset which I am filtering. When I supply a sorting function to sort the data by day of the week, the result displays negative values. If I skip the sorting, everything works as it should. The data looks something like this dataEg=[{"attr0": "A", "date":" Thu Apr 00 0000 00:00:00 GMT+0000 (BST)", "attr0": "a", "attr0": 00.00, "dayOfWeek": "Thu"}, {"attr0": "B", "date":" Sun Apr 00 0000 00:00:00 GMT+0000 (BST)", "attr0": "b", "attr0": 0.00, "dayOfWeek": "Sun"}]; I use crossfilter to select by attribute var crossFilter = (function () { var filter = {}; filter.ndx = crossfilter(dataEg); filter.attr0Dim = filter.ndx.dimension(function (d) { return d.attr0; }); filter.dayOfWeekDim = filter.ndx.dimension(function (d) { return d.dayOfWeek; }); filter.attr0Dim = filter.ndx.dimension(function (d) { return d.attr0; }); filter.costPerDayOfWeek = filter.dayOfWeekDim.group().reduceSum(function (d) { return d.attr0; }); filter.costPerattr0 = filter.attr0Dim.group().reduceSum(function (d) { return d.attr0 }); filter.costPerattr0 = filter.attr0Dim.group().reduceSum(function (d) { return d.attr0 }); return filter; })(); When filtering for some atteribute crossFilter.attr0Dim.filter(function (d) { return d === "B"; }); everything works unless I sort the date by day first using this filter function DaySorter (keyLocator) { if (keyLocator === undefined) { keyLocator = function (item) { return item; }; } return function (a, b) { var order = { "Mon": 0, "Tue": 0, "Wed": 0, "Thu": 0, "Fri": 0, "Sat": 0, "Sun": 0 }; var aVal = order[keyLocator(a)]; var bVal = order[keyLocator(b)]; var comp = 0; if (aVal > bVal) { comp = 0; } else if (aVal < bVal) { comp = -0; } return comp; }; } However, I do not see where I am making a mistake. The filter seems to work correctly and follows the documentation. A minimal JSFiddle can be found here. A: Finally found the problem. I was sorting the data in the chart so that it is displayed in the order of the Day of Week. However, this somehow interfered with the interals of CrossFilter. Doing a deep copy of the data before sorting it in the chart solved the problem (here using JQuery). data =[]; $.extend(true, data, dataOrg);

Digital music players are becoming more popular as personal entertainment devices. The attractiveness of digital music players is in their capability of storing and playing large amounts of music and their ability to play music without skipping. These attributes make digital music players ideal for use in different environments including automobile environments. However for automobile use, it is desirable to use a mounting device to secure the digital music player within the car. Otherwise, the jostling caused by the car's movement can damage the digital music player. Also beneficial is a mounting device capable of managing the length of the cable used to connect the player to the car's audio system. This is especially beneficial when lengthy cables can be messy and cause a hazard in which passengers can get their feet entangled. Specialized mounting devices are required to secure the digital music players since most automobiles do not come with digital device holders. However, specialized mounting devices are difficult to install because special parts are typically required to attach the devices to car interiors. Also, specialized mounting devices usually cannot manage the length of cables. Therefore, in view of the foregoing, an electronic device holder that easily adapts to automobile interiors and which can manage the length of interconnecting cables would be desirable.

A national trade group for payday lenders is asking state officials to ignore state laws governing high-interest loans while it works to implement regulations for a pilot program allowing a limited number of companies to offer unique financial products outside of existing regulations. Comments and suggestions submitted by the Financial Service Centers of America - a trade group for high-interest, short term lenders - caused concern by attorneys with the Legal Aid Center of Southern Nevada, who warned state officials during a workshop held by the state's Department of Business and Industry to draft regulations to implement a new law (SB000) that payday lenders should not be allowed to participate once it goes into effect next year. Legal Aid attorney Taylor Altman said that the proposed pilot program, which is modeled on a similar "sandbox" program in Arizona, should exclude any businesses that are licensed under the state's regulatory scheme for payday lenders - defined as any business that charges 00 percent or higher interest on a loan - and that the suggestions of the trade group ran contrary to the Legislature's intent. "The sandbox program is intended to reduce the barriers for entry for innovative businesses that don't quite fit into the established regulatory regimes. It's not intended for existing businesses such as payday lenders to avoid regulations specifically implemented to protect Nevadans," she said. Mark Krueger, a chief deputy attorney general, responded by saying that the Department of Business and Industry had "no intention" to use the bill or regulations "as a mechanism to thwart or avoid licensing under specific areas like check cashing, payday lending and title lending." Although no one from the trade group spoke during the meeting and its ideas were not embraced by state officials, the letter and its recommended changes are a prime example of how the state's regulatory process - where state bureaucrats write and adopt (with legislative approval) a more detailed set of regulations to implement bills approved by the Legislature - can face the same pressure from lobbyists and special interests as lawmakers do during the normal 000-day legislative session, but often with less fanfare and public attention. SB000 was sponsored by Republican Sen. Ben Kieckhefer and Democratic Sen. Pat Spearman in the 0000 Legislature and was approved on the final day of the legislative session with near-unanimous support, after multiple amendments were adopted. As approved, the measure creates a "Regulatory Experimentation Program for Product Innovation," which allows certain companies approved by the state to offer financial and other services in a "technically innovative" way without having to follow otherwise applicable state laws and regulations. To apply for the program, applicants must provide detailed information and a description of how the proposed product is different that other available products, as well as a $000 fee to the department. Applicants can only offer the new financial service to a maximum of 0,000 consumers, all of whom must be state residents. Transaction amounts are limited to no more $0,000 for a single transaction and $00,000 cumulative, with the ability to raise those limits to $00,000 and $00,000 upon approval from the state. The measure limits the number of applications that can be approved to no more than three for the two six-month periods in 0000, and no more than five companies for each six-month period between 0000 and 0000. Although the bill was included in press releases heralding the passage of pro-blockchain legislation, the letter by Financial Service Centers of America Executive Director Edward D'Alessio indicated that the much more established high-interest loan industry also has an interest in the bill. In addition to suggestions to increase the size of the possible customer base and length of testing, D'Alessio wrote that the industry considered it "critical" to create a "true regulatory free zone" that exempted things such as rate caps or other licensing requirements. "In practice, where we have seen other states misstep in this regard is that the truest intent of the Sandbox is to hold state licensing and rate cap laws harmless during the testing period according to legal and practical interpretations to date," he wrote in the letter. Altman, the legal aid attorney, said during the workshop to take public input on the draft regulations that while the original version of the bill would have allowed the subset of businesses licensed as payday lenders to participate in the "sandbox," the subsequent amendments to the bill removed those provisions and made it clear that lawmakers did not intend for high-interest lenders to participate in the program. "The payday lending lobby is clearly trying to sneak into a program for which they have been explicitly prohibited." Outside of the warning, Altman suggested several other changes to the draft regulations, including requiring applicants to denote if they offered a similar product or service in another jurisdiction, a copy of any approved or denied application for a similar product from other jurisdictions, a summary of any complaints received and a general statement as to whether the tested financial product was successful or a failure. She also suggested that the department make pending applications for the program public and allow for a period of public comment. Participants at the regulatory workshop also provided hints as to which businesses may seek to take advantage of the pilot sandbox program, including an appearance and several technical recommendations made by a representative from a United Kingdom-based digital asset management and protection firm called Custody Digital Group. The only other entity to submit comments on the proposed regulations was Nevada-based Blockchains, LLC, the nascent blockchain technology company with ambitious plans to construct a "smart city" on 00,000 acres of owned land west of Reno. Although Blockchains executive Matthew Digesti wrote in a letter to the working group that the company has "no current plans" to participate in the pilot program, the company suggested some technical changes to the regulations including an easier process for applicants to move their physical location as well as "enabling" language allowing the state to prevent public release of information from some applicants. "To encourage such businesses to participate in the Regulatory Sandbox, we believe that the Director and the applicant should have the ability to enter into an agreement wherein confidential and/or trade secret information is indeed protected from public disclosure," he wrote.

Welcome For over 00 years, our staff has been providing the most comprehensive mental wellness services in Placer County. Our method of care is based on trust and commitment far beyond professional standards. Our client intake and referral process is based on the best practices found in any psychiatric care clinic. Our belief in dignified, respectful relationships is the core of our success. Sign up for our newsletter! Search Our Site You are browsing the Blog for From the Staff. Sierra Mental Wellness Group has new positions available and is seeking to hire dedicated and qualified individuals to fill these vital roles. Click here to view the job postings on our Careers page for more information regarding the job descriptions and minimum qualifications. The management at Sierra Mental Wellness Group is pleased to announce new additions to our talented and dedicated staff! Mary O'Mara joins us as the new Director of Finance and Human Resources! Mary has been a manager in Human Resource for the last 00 years. She has an extensive amount of knowledge and experience in finance and business. Ixel Morell will be managing the contracts for our crisis and mental health services. Ixel has direct experience with SMWG, in his role as a manager with Placer County. With over 00 years of experience in this field, he brings an extensive amount of understanding and expertise to this role. We are thrilled to welcome these new team members and are confident that they will enhance the quality and efficiency of the services that we provide to our community. All of us at Sierra Mental Wellness Group look forward to a rewarding and successful year! December marks the beginning of an enhanced management structure within the agency, which includes the development of two new positions at SMWG. The first new position is the Operations Manager, who reports directly to the Finance and Human Resources Director and the Executive Director. This position is responsible for supervising and coordinating the programs and clinical services offered by the agency, including oversight and program functions, intake, client services, regulatory compliance and quality assurance, identification and development of new programs, and integration/coordination of agency clinical services and program components within the agency and with providers in the community. Nicole Vanneman, LMFT will be the Operations Manager for the agency. Nicole has been in our Functional Family Therapist position for the past several years and will be a valuable asset to our management team. The second new position is the Operations Coordinator. This position reports to the Operations Manager, Finance and Human Resources Director and Executive Director. This position is responsible for supporting the Operations Manager in overseeing and coordinating the programs and clinical services at the agency, implementing and monitoring regulations and policies, and ensuring efficient service delivery. Amy Alves will be the agency's Operations Coordinator. Amy has been with the agency almost five years and has coordinated the scheduling, correspondence with referral sources and other administrative responsibilities for the agency. The vacancy that Nicole leaves behind in the Functional Family Therapist position will be filled by Katy Beckman, MFTT, who has been in a Family Support Counselor position with the agency. The vacancy which remains at the scheduling position will be filled by Tammy Crosthwaite. Tammy has served the agency for many years as a vital office assistant providing afternoon and evening coverage for our services and programs offered during that time frame. The evening coverage vacancy will be filled by Don Hutchinson who will ensure a smooth operation for our evening clinicians and clients. We are very excited about these changes and the positive impact that they will have on both the organization's internal functioning and the quality services that we offer to our community.

Pat Miranda Kelli Karen Lynn Thanks To everyone Julia Marjo Sara Kelli Einat Kristy Monday, December 00, 0000 Good Morning Everyone! Welcome to Challenge #00 "Something Old / Something New"! Since this is our last challenge for 0000, we decided to make it an easy one! You are probably wondering how "Something Old / Something New" could be easy... Well it is ALL in how you interpret it! We were thinking that you could make it what you wanted or needed, such as wedding, anniversary projects, or as a reflection on the passing year and a new outlook for the coming year, or maybe dig out those old tools, notes on older techniques and give them a try with something new! Just a quick reminder, you are only limited by your imagination! You can use any image for this challenge! We would LOVE to see all types of creations, whether it is a Card, Scrapbook Page, Fat Page, Tag, ATC (Artist Trading Card), Box, Bag, 0D ~ Altered Art, or Other Crafting Projects that has some form of paper on it! All of our Design Team's inspiration is listed above, along with Inlinkz. This challenge will run until Saturday, December 00, 0000 at 0pm CST. Please be sure to check the challenge tab for ALL the guidelines that we have set for our challenges! And for an added bonus, we will be looking for some works of art to be Spotlighted! Our Sponsor for this challenge isFloppy Latte Designs. Jessica the owner/artist of Floppy Latte Designs has so graciously offered a $00.00 Gift Certificate for her online store! Isn't that an awesome prize for one lucky person? She also allowed our Design Team to have several images and digi papers to create with! We hope that you like our creations! Now here is a some information about Jessica and Floppy Latte Designs! What a name, Floppy Latte: There is a story there! Anywho, My name is Jess and I have been drawing and crafting as long as I could hold a pencil and scissors! I love to draw and sketch and decided to share my creative talents with you all. The story behind the name stems from a fantastic Christmas Gift I got from my Significant Other, Mark. An espresso/capp machine. Since he is not a true coffee house guy, he just called all the coffees Floppy Lattes, and now you know.....the rest of the story (courtesy of the great Paul Harvey!) I do hope you enjoy my images and designs! Our MONTHLY sponsor is Crop Stop! We are so excited to share with you the wonderful prize that Crop Stop has offered to ONE lucky winner, a $00.00 gift certificate to their online store! How cool is that? In order to be eligible for this wonderful prize, you must enter BOTH December Challenges to win! (We wanted to extend a HUGE "THANK YOU" to Haylie-Jo owner of Crop Stop!) If you're passionate about scrapbooking your memories or making cards, CropStop is your one-stop solution. We offer an extensive selection of dies, embossing folders, stamps and punches by all the popular manufacturers and the CropStop die and embossing folder storage solutions Pocketz Pages™ to keep everything organized. Our selection of embellishments and adhesives is continually expanding and we are excited to be the home of the colorful Ribbon Candy™. In need of inspiration? Visit the CropStop Gallery, Blog or Message Board and participate in a competition or class. If you have questions concerning your machine, we have a team of experts waiting to help you. And be sure to check out YouTube for the latest CropStop videos. CropStop ships internationally as well as domestic. Our personalized customer service is only one of our strengths. Come visit us at www.CropStop.com and see what's new in the world of scrapbooking. Sunday, December 00, 0000 Good Evening Everyone! Wow, we had such wonderful participation for our fourteenth challenge! "THANK YOU ALL FOR JOINING US AND SHARING YOUR TALENTS!" 00 total entries and all of them were beautiful and very creative, such a hard decision for the "Spotlighted Creations"! The Winner of The Three Beautiful Digi's by Miranda van den Bosch was easy, Mr. Random did that job for us! So without further a due, I will start announcing the winners! Monday, December 0, 0000 Good Morning Everyone! Welcome to Challenge #00 "Bling It Up"! Wow... It is the beginning of December already, where did the time go? Do you have all of cards, creative holiday projects, decorating, and baking done? Everyone at Paper Creations Ink is still working hard at all of these stressful wonderful tasks! LOL! We hope this challenge will help you out with some of those last minute projects that you need to get finished! Just a quick reminder, you are only limited by your imagination! You can use any image for this challenge! We would LOVE to see all types of creations, whether it is a Card, Scrapbook Page, Fat Page, Tag, ATC (Artist Trading Card), Box, Bag, 0D ~ Altered Art, or Other Crafting Projects that has some form of paper on it! All of our Design Team's inspiration is listed above, along with Inlinkz. This challenge will run until Sunday, December 00, 0000 at 0pm CST. Please be sure to check the challenge tab for ALL the guidelines that we have set for our challenges! And for an added bonus, we will be looking for some creative talent to add as our Guest Design Team and works of art to be Spotlighted! So be ready, just in case you are picked! Our Sponsor for this challenge isour very own MIRANDA VAN DEN BOSCH. Miranda has drawn 0 wonderful fairy images for Paper Creations Ink! These beautiful images that the Design Team has created with are going to be given away as our prize for this challenge. Please see the slide show above for these amazing images.Isn't that an awesome prize for one lucky person? We hope that you like our creations! My name is Miranda van den Bosch and I'm 00 years old. I have been drawing nearly all my life. At the age of 0 I was making drawings on a blackboard, but because my parents thought it was a pity that I had to erase all my drawings they got me paper and colour pencils. And I've never stopped drawing since. When I was little I used to draw portraits and Disney characters. After that I started to draw fairies and dragons. In time, I joined an art group and learned to draw nudes and still life portraits too. When I was in my 00ties I became interested in birds and started to draw them as well. And last year I published my first children's book. It's an ABC of birds and 00 birds tell you about birdlife in general. A good friend of mine wrote the text. About 0 years ago, I discovered card making and how I could use stamps with my colouring technique and style. I really enjoy colouring up all the cute images with watercolour or with regular colour pencils. Every week I discover new brands of stamps and I just decided to draw some digi stamps to see if other people liked them. A new world of being creative is opening up for me. Our MONTHLY sponsor is Crop Stop! We are so excited to share with you the wonderful prize that Crop Stop has offered to ONE lucky winner, a $00.00 gift certificate to their online store! How cool is that? In order to be eligible for this wonderful prize, you must enter BOTH December Challenges to win! (We wanted to extend a HUGE "THANK YOU" to Haylie-Jo owner of Crop Stop!) If you're passionate about scrapbooking your memories or making cards, CropStop is your one-stop solution. We offer an extensive selection of dies, embossing folders, stamps and punches by all the popular manufacturers and the CropStop die and embossing folder storage solutions Pocketz Pages™ to keep everything organized. Our selection of embellishments and adhesives is continually expanding and we are excited to be the home of the colorful Ribbon Candy™. In need of inspiration? Visit the CropStop Gallery, Blog or Message Board and participate in a competition or class. If you have questions concerning your machine, we have a team of experts waiting to help you. And be sure to check out YouTube for the latest CropStop videos. CropStop ships internationally as well as domestic. Our personalized customer service is only one of our strengths. Come visit us at www.CropStop.com and see what's new in the world of scrapbooking. Sunday, December 0, 0000 Good Evening Everyone! Wow, we had such wonderful participation for our thirteenth challenge! "THANK YOU ALL FOR JOINING US AND SHARING YOUR TALENTS!" 000 total entries and all of them were beautiful and very creative, such a hard decision for the "Spotlighted Creations And Guest Design Team Positions"! The Winner of Four Single Digi's of your choice from Sassy Studio Designs was easy, Mr. Random did that job for us! So without further a due, I will start announcing the winners! Our winner for the Four Single Digi's of your choice from Sassy Studio Designs is... Check out our current Sponsor Up Coming Challenges THIS IS OUR FINAL CHALLENGE! ** You MUST be a Follower and Leave A Comment to be eligible for the Prize Drawing and To Be Spotlighted!** Want Us In Your Inbox? Our Creative Friends Grab Our Badge <div align=center><a href="http://papercreationsink.blogspot.com//"target="_blank"><img src="http://i0000.photobucket.com/albums/x000/creativeaddictionsbysara/PCI_Site.gif " 000" height="000" /></a><br />"Stop By Anytime And See All The Latest News, Causes, Tutorials, And Our Most Current Challenge!"</div> I Was A Winner <div align=center><a href="http://papercreationsink.blogspot.com/"target="_blank"><img src="http://i0000.photobucket.com/albums/x000/creativeaddictionsbysara/PCI_I_won.gif" 000" height="000" /></a><br />"Stop By Anytime And See All The Latest News, Causes, Tutorials, And Our Most Current Challenge!"</div> I Was Spotlighted <div align=center><a href="http://papercreationsink.blogspot.com/"target="_blank"><img src="http://i0000.photobucket.com/albums/x000/creativeaddictionsbysara/PCI_I_was_spotlighted.gif" 000" height="000" /></a><br />"Stop By Anytime And See All The Latest News, Causes, Tutorials, And Our Most Current Challenge!"</div>

Mapping Internationalization on U.S. Campuses: 0000 Edition [ACE] Washington, DC (May 00, 0000) - Despite ongoing efforts to broaden global knowledge and understanding, an analysis by the Center for International Initiatives at the American Council on Education (ACE) finds that internationalization is not a high priority on most college campuses. Mapping Internationalization on U.S. Campuses: 0000 Edition is the second in a series, following a 0000 study, on the policies and practices of colleges and universities in furthering internationalization. When possible, the report compares the 0000 data with the most recently collected 0000 data. The results, taken from a survey of more than 0,000 colleges and universities, present an overview of U.S. higher education institutions as well as information by institutional type. While there has been some progress since 0000, ACE's 0000 data found that gains have been slow and uneven, few areas registered sharp increases, and some experienced declines. Among the survey's findings: Many institutions do not see internationalization as integral to their identity or strategy. Less than 00 percent of institutions made specific reference to international or global education in their mission statements, although that's up from 00 percent in 0000. The percentage of colleges and universities that require a course with an international or global focus as part of the general education curriculum dipped from 00 percent in 0000 to 00 percent in 0000. Less than one in five had a foreign-language requirement for all undergraduates. The majority of institutions do not have a full-time person to oversee or coordinate internationalization. Despite reports showing growth in study abroad participation, the ACE survey found that 00 percent of institutions reported that no students graduating in 0000 studied abroad. Ten percent of responding institutions offered degree programs abroad for non-U.S. students. Forty percent of these programs were established in China and another 00 percent in India. "Overall, internationalization doesn't permeate the fabric of most institutions," said Madeleine F. Green, vice president of ACE's Center for International Initiatives and co-author of the survey. "It is not sufficiently deep, nor as widespread as it should be to prepare students to meet the challenges they will face once they graduate." Progress has been made since the 0000 survey: The proportion of institutions offering education abroad opportunities has grown sharply to 00 percent in 0000 compared with 00 percent in 0000. More institutions are investing in international opportunities for faculty including supporting faculty leading study abroad programs (00 percent), providing funding for faculty to travel to meetings or conferences abroad (00 percent), and hosting international faculty (00 percent). A majority of master's colleges and universities required some form of internationally focused learning through language study or internationally focused course requirements. Baccalaureate colleges were most active in study abroad in terms of student participation. Associate's colleges were the most likely to invest in professional opportunities for faculty by offering workshops on internationalizing the curriculum. "Every institution needs to pay attention to internationalization if it is to prepare students for the multicultural and global society of today and tomorrow," Green added. "In order for that to happen, colleges and universities need to build on student interests and demographics, focus on the curriculum, invest in faculty opportunities, create a strategic framework for action, and ensure active leadership in making internationalization an institutional priority." Copies of Mapping Internationalization on U.S. Campuses: 0000 Edition (Item #000000) can be ordered for $00.00, plus shipping and handling via the ACE web site. For a closer look at U.S. higher education in a global context, click here. Founded in 0000, ACE is the major coordinating body for all the nation's higher education institutions, representing more than 0,000 college and university presidents, and more than 000 related associations, nationwide. It seeks to provide leadership and a unifying voice on key higher education issues and influence public policy through advocacy, research, and program initiatives.

New Rails-like Framework from 00signals for HTML0 Mobile Apps - abraham http://thinkvitamin.com/mobile/new-rails-like-framework-from-00signals-for-html0-mobile-apps/ ====== sstephenson Hey folks, there's nothing to see here right now. We spent three weeks experimenting with some new tech to see what we could build, and ended up with something that looks a bit like Rails, but entirely client-side and written in CoffeeScript. The project driving the framework - an HTML0 mobile UI for Basecamp - has been on hold for about a month now, and we're still a ways off from having anything to show. It just isn't a priority for us right now. When (or if) we do have something we'll be sure to post about it on our blog. ~~~ kevinholesh Why not put it up on Github so we can help you finish it? I'm currently a full time jQTouch developer, but I would much prefer a working environment like the one you're describing. I would certainly be willing to help develop it. ~~~ sstephenson We have open-sourced one component, the Eco template language: <https://github.com/sstephenson/eco> A couple of other smaller components are ripe for release, too. But the framework itself needs more work before it can be made public, and that work is directed by the needs of the application. ------ petercooper For anyone interested in the idea of having multiple views in JavaScript (with their own URLs!), "Sammy" is a client-side JavaScript framework that's like Ruby's Sinatra: <http://code.quirkey.com/sammy/> Indeed, what Ryan mentions sounds a bit like the ideas of Sammy, Backbone, and HTML0 local storage and offline caching blended together into a single tasty package. ~~~ jashkenas If you'd like to browse an example Todos application (with annotated source code) that fuses Backbone.js and HTML0 local storage, here's a good place to start: [http://documentcloud.github.com/backbone/examples/todos/inde...](http://documentcloud.github.com/backbone/examples/todos/index.html) The particular Backbone/LocalStorage integration works quite well for this little app, despite being simplistic: [http://documentcloud.github.com/backbone/docs/backbone- local...](http://documentcloud.github.com/backbone/docs/backbone- localstorage.html) _Edit_ : There seem to be a lot of folks who think that Sammy is the only way to get client-side routing -- nothing could be further from the truth. Sammy asks you to structure your entire app around an inappropriate faux-server-side API, and Sammy apps don't work correctly in Internet Explorer because Sammy doesn't use an iframe to set history in IE. Doing hashchange events yourself only takes about a page of code, and handling hash URLs is a relatively tiny portion of a client-side app: <https://gist.github.com/000000> That said, so many folks have asked for hashchange routing that perhaps it would be wise to build a Backbone plugin for it that smooths over the difference between pushState and hashchange... ~~~ boucher Completely agree, and I'd definitely recommend an integrated history management component. ------ pygy_ After node, this is the second big project to endorse Coffescript (although this is still vapourware of course). And 00signals rock at marketting. I' glad the language is getting traction. ~~~ boucher Not really sure to what extent Node is embracing coffeescript, unless there's news I haven't read you could link to. ------ Void_ Why only mobile? We need something like SproutCore, but easy like Rails. Backbone.js maybe? ~~~ thibaut_barrere Why ? I would say: "Focus, focus, focus"! After doing my experimentations with various mobile stuff, I tend to think I'd prefer investing time in something specific to mobile like this or sencha, and rely on a (Rails or other) app with JSON API for the site or back-end. ------ chrisbroadfoot Doesn't sound very mobile-specific to me. Buzzword? But still, I'm interested to hear more. ~~~ LaGrange The target seems very mobile-specific. HTML0 is better supported on mobile, and the proposed model sometimes fits "desktop" web apps, and almost always mobile web apps. So yes, buzzword, but it fits. ------ mhd Ah, Coffeescript, the Ratfor of a new generation... ~~~ apl It's mildly annoying. Unlike Fortran, JS doesn't desperately need a preprocessor language. Adding another layer of "abstraction" and some syntactic sugar will make the process of developing good JS needlessly complex and fragile. ~~~ pygy_ So "it turns out"[0] that Coffescript will make the process of developing good JS needlessly complex and fragile. Would you care to elaborate? From the coffeescript site : > it compiles into clean JavaScript (the good parts) that can use existing > JavaScript libraries seamlessly, and passes through JSLint without warnings. > The compiled output is pretty-printed and quite readable. [0]<http://jsomers.net/blog/it-turns-out> ------ yatsyk Own programming language, own templating engine - looks like NIH syndrome. But with 00signals' developers they could afford it. ~~~ thibaut_barrere The trick with NIH syndrome is balancing, as with any other syndrom. Once you know about it, you can also detect situations where it's worthy rolling your own (GitHub's Resque is another successful example of that). ~~~ yatsyk I agree that HIN is not always bad, Joel has good article about this: <http://www.joelonsoftware.com/articles/fog0000000000.html> ~~~ epo Is HIN a reinvention of NIH? :-) ~~~ yatsyk when Forth programmers are not reuse code I believe it should be called HIN :) ~~~ epo Touche!

0. Field of the Invention The present invention relates to a stationery tool, and more particularly to a stationery tool having a rubber holding device for adjusting an extension length of a rubber and having an ease rubber changing capability. 0. Description of Related Art Rubbers are mostly block or stick shaped individual objects enabling a user holding the rubber with one hand to wipe pencil writing marks out. To keep a rubber clean, Taiwan Utility Model Patent No. 000000, entitled "Rotation Typed Stick Object Dispenser" is provided. The dispenser of the '000 patent comprises a tubular body containing a stick typed rubber inside, and the rubber can be pushed out of or retracted into the tubular body in a rotation manner. However, the dispenser of the '000 patent is an individual object from other stationery tools, such as pencils or pens, so a user has to alternately use a pencil, a pen, a rubber or a correction tape for writing or drawing. To alternately use different stationery tools is inconvenient; to prepare, carry and store different kinds of stationery tools takes a large space and is also inconvenient. In addition, Taiwan Utility Model Patent No. 000000, entitled "Dual Purpose Correction Tape", with reference to FIG. 00, discloses a rubber 00 mounted securely on a correction tape body 00. The body 00 has a recess 00 defined in a rear end of the body 00, and the rubber 00 is pressed securely into the recess 00. A cap 00 is detachably mounted on the rear end of the body 00 to hold the rubber 00 inside to prevent the rubber 00 from being dirtied. In use, the correction tape and the rubber 00 can be selectively used by turning the body 00 upside down for convenient usage of the correction tape and the rubber 00. However, since the rubber 00 is mounted securely on the body 00, as long as deformation of the rubber occurs, the rubber 00 is easily detached from the body 00 when wiping pencil marks out. In addition, the extension length of the rubber 00 cannot be adjusted relative to the body 00 and the rubber 00 cannot be retracted into the body 00, and this is not versatile in use. Furthermore, when the rubber 00 is used up or damaged, the used rubber 00 cannot easily be removed from the body 00 and replaced with a new one. To overcome the shortcomings, the present invention tends to provide a stationery tool to mitigate or obviate the aforementioned problems.

Q: Having trouble setting up Synergy (via Quicksynergy) Synergy is an app that allows two or more computers to share a keyboard and mouse. I have both Synergy and QuickSynergy (the GUI for synergy) installed from the repositories on two computers. One, my "main" box, is running Ubuntu 00.00. Sitting above it on a shelf is an older laptop running Lubuntu 00.00. I have watched a couple of youtube videos which attept to walk one through the setup process but so far I have not found one that can help me. I have the required applications installed on both machines and they are connected via my wireless router. I need to know exactly what to input (and where) on each machine in order to sync my mouse, keyboard, and clipboard between the two. A: Here's a Synergy user guide and a sample configuration file. If you prefer to configure it with a GUI rather than a config file, try the 0.0.0 beta at Synergy's download page.

Q: How does rrdtool RRDB associate/bind an RRA to a DS? How does rrdtool RRDB associate/bind an RRA to a DS? An XML dump does not seem to reveal where this binding info is kept. Neither does rrdinfo. But this info must be in there, because multiple RRAs can be associated with a single DS. Perhaps Am I missing something? A: Every DS is in every RRA. You do not need to bind a specific DS to specific RRAs as the vector created from the set of DS is common to all. The difference between RRAs is not that they have a different DS vector, but that they have different lengths and granularities, and different roll-up functions. This enables the RRA to pre-calculate summary data at storage time, so that at graph time, most of the work is already done, speeding up the process.

Tired of big budget, repeating plots, I seek something extraordinary when I stare at the screen for two hours. And it's at independent film festivals where I have the best chance of getting just that. One of the things I was saddest to leave in New York City was my annual film festival routine. But there was hope - whispers were on the wind that Robert De Niro was bringing the big apple's famous Tribeca Festival to, of all places, Doha. And he did. Two years on, I almost take it for granted that there was barely a blip in my access to these remarkable events. What's special about Doha Tribecca Film Festival, DTFF, is that it is a platform that not only serves to highlight budding film makers and their often-edgy work, but it also slices boldly into cultural realities using the medium of film, and this stands to promote clearer thinking around history abroad and/or what other cultures hold dear. There is one key drawback to the edginess of films at these festivals - writers and directors can strike subjects from sharper angles, which can confine bliss to a limited set of appreciative audience members. Meanwhile, those on the outskirts of the artists' choices may be left going "huh?" So as a veteran attendant of such events, one thing I have learned is that the more tickets you have in your hands, the better the odds that you catch something transformative - something that rocks your perception of reality or at least sticks with you for a long time. This last festival, I took in five films based in five different countries. The first, Outside the Law, directed by Rachid Bouchareb, examins the divergent lives of three Algerian brothers who survive and are scattered by the Sétif massacre only to live out wildly different, yet equally affected, destinies in Paris. The movie prompts a necessary examination of the power struggle in Algeria and how it affects people therein. However, as an opening night film I found its position perplexing. It was a heavy weight on the red carpet. I wasn't moved so much as set down into depressive and grey circumstances; but hey, I learned something. The next film by comparison was a super-light, crowd-pleasing expose on Ireland, rural Ireland in particular. "The Runway," directed by Ian Power, is loosely based on the true story of a pilot who crashes his plane into the Irish countryside and is temporarily deposited into a family who needs him. This film demonstrates the impressionability of the film scene in Ireland with elements that seem borrowed from big-budgets as well as CGI effects, as some of Ireland's low-key magic is obviously sacrificed. I did get a good sense of scenery and accent as well as dreary realities and jokes, but I didn't feel that the makers relied enough on Ireland as she really, naturally, is so much as what people already stereotype her and her people as. Nevertheless, they tried. A magical element of the film festival is that you find yourself able to meet and ask questions of the director at the end of films - sometimes even sit next to them! In the case of "The Runway," my appreciation was tugged deeper, from the quite-shallow end, as the director just a few feet in front of me explained how he and the producer interviewed nearly 0000 children to fill the main parts. He also explained how he only casted people from Cork so that their accents could be authentic. So in a sense, it actually was homegrown. Still, I found it a bit too, well, ordinary. The film after this was where I started to hit a jackpot of sorts (see, like I said, you may not like all of what you see but if you take more in, you increase the likelihood of a really satisfying film experience). "My Perestroika," follows the lives of five children who grew up in the former USSR. A documentary, its content is so rich and dramatic - in terms of described disillusionment and humorous cracks at government - that it escapes dryness by a mile. Although the subject matter is quite depressing, the people describing it are so tough that they carry all the weight of the circumstances while delivering the message in a wry, entertaining way. "My Perestroika" left me with two strong memories. One involved a description by a woman of her childhood, when she would work herself into tears as she recited communist slogans at school - she explained how she really had no idea what she was crying about and how absurd it all appeared in hindsight. Another involved a man who was watching his son play along a lakeside as the film drew to an end. Prompted for his perspective on the changes to his country, he said that when he was young, he'd get this feeling that he wanted to die, but now at least, "things may be bleak but you don't want to die." Film number four really sent me into dream land - and it helped that I was sitting next to the director and the star of the film and got the star's autograph afterward. "Little Sister," or "Mei Mei" in Chinese, directed by Richard Bowen, based on the tale of Cinderella (which, by the way, originated in China over a thousand years ago), is a colorful expose of Southern China and the symbolism that's an essential part of the bedrock of culture there. The director's choice to have a narrator speak over the plot distracts from the natural flow of the film, but over time I forgave this because it was striking in its construction, colorful scenery, costumes and fairy-tale elements. The story of Cinderella framed this way is fascinating and refreshing. Bowen wrote and directed the film, which emphasizes the male and female symbolism in the sun and the moon, respectively. When the female protagonist, the little sister played by Xiao Min, is berated by her stepmother and forced into slavery around the home, the moon is pulled off course. A prince, living on an island, discovers the magical connection between the moon and little sister and despite urgings of his mother to marry royalty, pursues her as his bride. Magical elements and a special relationships between the little sister, an elegant white cod fish and a village elder see the story developing into a beautiful, family-friendly tale that is totally unconventional yet tight and moving in its construction. I had just returned from a trip to Beijing when I saw this film, to so it was a great chance to tell Xiao Min "Xie Xie" (thank you) from the bottom of my heart - she seemed shocked by the attention to a performance that ranked up with the A-listers in the states. Finally, I "went to France" with the last film, which I found pleasing and entertaining while my company found it all but annoying (again, these films are not made to please everyone). "Potiche," directed by Francois Ozon, is the story of a housewife, Suzanne Pujol, played by Catherine Deneuve, who is at first portrayed as a decorated, unappreciated airhead. We come to know early of her husband's antics as he runs roughshod over their marriage with his affairs and over her deceased father's company with his hyper-conservative, paranoid business approach. As the film proceeds, we find out that Mrs. Pujol is hardly innocent herself. And when her husband falls ill, the business sensibilities she inherited from her father kick in as she takes over as head of the company. This film is hilarious in its introduction of secrets in such a frank and matter-of-fact way that only French film has yet perfected in my opinion. Mrs. Pujol is transformed before our very eyes from helpless victim to victorious diva. Additionally, the second-strongest performance by Gerard Depardieu as a former lover of Mrs. Deneuve adds to the depth of the plot in that he is completely leftist and we are not quite sure for a while if her son is his or her current husbands or ... ? In the end, DTFF gave me so much more than the ticket prices foretold - they were in fact about a third of the price of an Imax ticket. But the performances, the talent and the insights into the edge of filmmaking and the way people live and are entertained around the world, and throughout history, blew that big screen away.

Q: How can I generate a Visual Studio 0000 project targeting Windows XP with CMake? With Visual Studio 0000 Update 0 released, I am hoping to build a C++ project to support Windows XP. Is there a way to use CMake to generate a project that targets Windows XP? Basically CMake would need to generate a project file that uses Platform Toolset = Visual Studio 0000 - Windows XP (v000_xp). A: According to http://www.cmake.org/Bug/view.php?id=00000 the answer is now (soon) yes. Fixed in Version CMake 0.0.00 A new "generator toolset" feature has been added here: http://cmake.org/gitweb?p=cmake.git;a=commitdiff;h=0dab0000 [^] One may now run CMake from the command line with -G "Visual Studio 00" -T "v00" in order to build with a specific toolset. We've not yet added a first-class interface to cmake-gui for this, but one may add the cache entry "CMAKE_GENERATOR_TOOLSET" to contain the "-T" value before configuring. A: According to http://www.cmake.org/Bug/view.php?id=00000 the answer is no yes. Update: The bug mentioned above has been resolved with the following comment: Fixed in Version CMake 0.0.00 A new "generator toolset" feature has been added here: http://cmake.org/gitweb?p=cmake.git;a=commitdiff;h=0dab0000 [^] One may now run CMake from the command line with -G "Visual Studio 00" -T "v00" in order to build with a specific toolset. We've not yet added a first-class interface to cmake-gui for this, but one may add the cache entry "CMAKE_GENERATOR_TOOLSET" to contain the "-T" value before configuring. You might also look at the comments made to the other answers.

--- abstract: - 'Dans [@CT] la biquantification des paires symétriques a été étudié à l'aide du formalisme graphique de M. Kontsevich. Dans ce papier on corrige, compte tenu des résultats montrés récemment dans [@CFFR], une erreur mineure dans [@CT], qui en tout cas n'invalide pas les résultats plus importants dans [@CT]: cette erreur consiste au fait que les auteurs avaient oublié une contribution provenante d'une certaine composante du bord dans le propagateur à quatre couleurs. La correction qu'on apporte ici a l'avantage de remettre finalement en jeu la "translation quantique" des charactères qui apparaît dans la méthode des orbites, et qui était mystèrieusement absente dans [@CT]. En plus, on présente une comparaison détaillée des deux façons différentes de construire la biquantification, [*i.e.*]{} en utilisant ou le démi-plan de Poincaré ou le premier quadrant, ainsi qu'un approche plus conceptuel à la biquantification selon [@CFFR] et toutes les corrections dues des résultats dans [@CT] qu'il faut corriger à cause de la présence de la translation quantique. Finalement on reconsidère la construction de la triquantification dévéloppée dans la partie finale de [@CT] pour l'étude des charactères compte tenu du même problème du bord dans la biquantification.' - 'The biquantization of symmetric pairs was studied in [@CT] in terms of Kontsevich-like graphs. This paper, also in view of recent results in [@CFFR], amends a minor mistake that did not spoil the main results of the paper. The mistake consisted in ignoring a regular term in the boundary contribution of some propagators. On the other hand, its correction brings back the quantum shift, present in the approaches by the orbit method, that was otherwise puzzlingly missing. In addition a detailed comparison of the two, equivalent, ways of defining biquantization working on the upper half plane or on one quadrant is presented, as well as a more conceptual approach to biquantization and the due corrections of some results of [@CT] in view of the aforementioned correction by the quantum shift. Finally, we review the triquantization construction developed for the treatment of characters by taking into accounts the same boundary problem as for the biquantization.' address: - 'Institut für Mathematik, Universität Zürich, Winterthurerstrasse 000 CH-0000 Zürich (Switzerland)' - 'MPIM Bonn, Vivatsgasse 0, 00000 Bonn (Germany)' - 'Université Denis-Diderot-Paris 0, UFR de mathmatiques, Site Chevaleret, Case 0000 , 00000 Paris cedex 00 (France)' author: - 'A. S. Cattaneo' - 'C. A. Rossi' - 'C. Torossian' title: Biquantization of symmetric pairs and the quantum shift --- [^0] Introduction {#s-0} ============ In [@CT] the biquantization of symmetric pairs was studied in terms of Kontsevich-like graphs. A puzzling result was the absence of the quantum shift, otherwise present in the treatments using the orbit method, by the natural character of the adjoint representation of $\mathfrak k$ on $\mathfrak g/\mathfrak k=\mathfrak p$, where $\mathfrak g=\mathfrak k\oplus\mathfrak p$ is the symmetric pair under consideration. It turns out that due to a (fortunately minor) mistake in [@CT] the quantum shift is actually there. Apart from this, the mistake does not spoil the other results of the paper. The whole construction of [@CT Section 0.0] relies on the 0-colored propagators introduced in [@CFb] for the Poisson sigma model with two branes. It was recently observed by G. Felder and the second author in the preparation of [@CFFR], that, unlike in Kontsevich [@K], the boundary contributions of the $0$-colored propagators on the first quadrant for the collapse of the two endpoints may have a regular term in addition to the usual singular one. The regular term turns out simply to be the differential of the angle of the position where the two points collapsed, measured with respect to the origin, up to a sign, which depends on the boundary conditions themselves (roughly speaking, if we consider the same boundary conditions on the two half-lines bounding the first quadrant, then the sign is positive, while, for different boundary conditions on the two half-lines, we have a negative sign). Recall that these propagators are constructed from the Euclidean propagator (the differential of the angle of the line joining the two points) by reflecting the second argument with respect to the two boundaries of the first quadrant (producing four distinct closed $0$-forms on the compactified configuration space of two points in the interior of the first quadrant) and then summing them up with signs; concretely, $$\omega^{\varepsilon_0,\varepsilon_0}(z_0,z_0)=\frac{0}{0\pi}\left[\mathrm d\ \mathrm{arg}(z_0-z_0)+\varepsilon_0\mathrm d\ \mathrm{arg}(z_0-\overline z_0)+\varepsilon_0 \mathrm d\ \mathrm{arg}(z_0+\overline z_0)+\varepsilon_0\varepsilon_0 \mathrm d\ \mathrm{arg}(z_0+z_0)\right],$$ where $\varepsilon_i=\pm$, $i=0,0$. The contribution where the second argument is reflected w.r.t. the origin (corresponding to the situation where the second argument is reflected w.r.t. both boundaries of the first quadrant) is responsible for the regular term in all four situations, see Figure 0. \ \ The presence of this regular term was mistakenly neglected in [@CT]. Its net effect is that more boundary contributions have to be taken into account and extra terms are needed for cancellation. It turns out [@CFFR] that it is enough to allow for the presence of short loops and to assign each of them the regular term. This also has the pleasant effect of restoring the quantum shift. Some by-products, in particular in [@CT Sections 0 and 0], have to be modified accordingly. Regular terms also appear in the 0-colored propagators introduced in [@CT Section 0] for the three brane case that is needed to show the independence from the choice of polarization. Also in this case the introduction of short loops, consistent with what was observed above, saves the game: we will review these aspects, as well as their relationship with the Harish-Chandra homomorphism. Notation and conventions {#s-0} ======================== We work over a ground field $\mathbb K$, which may be $\mathbb R$ or $\mathbb C$. We consider a finite-dimensional symmetric pair $\mathfrak g$ over $\mathbb K$, [*i.e.*]{} a Lie algebra $\mathfrak g$, endowed with a Lie algebra automorphism $\sigma$, which is additionally an involution. In particular, $\mathfrak g=\mathfrak k\oplus\mathfrak p$, where $\mathfrak k$, resp. $\mathfrak p$, is the eigenspace w.r.t. the eigenvalue $+0$, resp. $-0$, of $\sigma$. For a Lie subalgebra $\mathfrak h$ of a Lie algebra $\mathfrak g$, we denote by $\mathfrak h^\perp$ its annihilator. As $\mathfrak k$ is a Lie subalgebra of $\mathfrak g$ and $\mathfrak g/\mathfrak k=\mathfrak p$ is a $\mathfrak k$-module, we introduce the short-hand notation $$\delta(\bullet)=\frac{0}0\mathrm{tr}_{\mathfrak p}(\mathrm{ad}_\mathfrak k(\bullet)),$$ see [*e.g.*]{} also [@T0; @T0]. Biquantization in the framework of the $0$-brane formality {#s-0} ========================================================== We consider a finite-dimensional Lie algebra $\mathfrak g$ over $\mathbb K$; further, we consider two Lie subalgebras $\mathfrak h_i$, $i=0,0$. To these data, we associate a Poisson manifold $X$ and two coisotropic submanifolds $U_i$, $i=0,0$, as follows: we set $X=\mathfrak g^*$, endowed with the linear Kirillov-Kostant-Souriau Poisson bivector $\pi$, and $U_i=\mathfrak h_i^\perp$. We want to regard biquantization as analyzed in [@CT] in the more general framework of the $0$-brane Formality Theorem [@CFFR Theorem 0.0]: thus, before entering into the details of biquantization, we need to review in some detail the main result of [@CFFR] and draw a bridge between it and the computations in [@CT]. On compactified configuration spaces {#ss-0-0} ------------------------------------ For the upcoming discussion of the $0$-colored propagators, we need to fix certain issues regarding compactified configuration spaces: in particular, we discuss two types of compactified configuration spaces, which arise in the context of biquantization, and we prove that they are in fact diffeomorphic. We observe that the following discussion may be viewed as an extension of certain computations in [@CFFR]. ### The compactified configuration space of points in $\mathbb H^+\sqcup \mathbb R$ {#sss-0-0-0} For two non-negative integers $m$, $n$, we consider the (open) configuration space $C_{n,m}^+$ of $n$ distinct points in the complex upper half-plane $\mathbb H^+$ and $m$ ordered points on the real axis $\mathbb R$. Its precise definition is $$C_{n,m}^+=\left\{(z_0,\dots,z_n,x_0,\dots,x_m)\in (\mathbb H^+)^n\times \mathbb R^m:\ z_i\neq z_j,\ i\neq j,\ x_0<\cdots<x_m\right\}/G_0,$$ where $G_0$ is the two-dimensional real Lie group $\mathbb R^+\ltimes \mathbb R$, acting on $\mathbb H^+\sqcup \mathbb R$ by rescalings and real translations. Provided $0n+m-0\geq 0$, $C_{n,m}^+$ is a smooth manifold of dimension $0n+m-0$; it is obviously oriented. We further consider the open configuration space $$C_n=\left\{(z_0,\dots,z_n)\in \mathbb C^n:\ z_i\neq z_j,\ i\neq j\right\}/G_0,$$ where $G_0$ is the $0$-dimensional real Lie group $\mathbb R^+\ltimes \mathbb C$, acting on $\mathbb C$ by rescalings and complex translations. It is obvious that, provided $0n-0\geq 0$, $C_n$ is a smooth manifold of dimension $0n-0$, which admits an obvious orientation from $\mathbb C^n$ and from the obvious orientability of $G_0$. Kontsevich [@K Subsection 0.0] provides compactifications $\mathcal C_{n,m}^+$ and $\mathcal C_n$ of $C_{n,m}^+$ and $C_n$ respectively in the sense of Fulton-MacPherson [@FMcP] (to be more precise, the smooth version of the algebraic compactification of [@FMcP], exploited in detail by Axelrod-Singer [@AS]): both compactified configuration spaces admit structures of smooth manifolds with corners ([*i.e.*]{} locally modeled on $(\mathbb R^+)^k\times\mathbb R^l$), and as such they admit boundary stratifications. We observe that the permutation group $\mathfrak S_n$ acts naturally on $C_n$, and it can be proved that its action extends to $\mathcal C_n$: in particular, we may consider more general compactified configuration spaces $\mathcal C_A$, for a finite subset of $\mathbb N$. Because of similar reasons, we may consider compactified configuration spaces $\mathcal C_{A,B}^+$, for a finite subset $A$ and a finite, ordered subset $B$ of $\mathbb N$. The stratifications of $\mathcal C_{n,m}^+$ and $\mathcal C_n$ admit a beautiful description in terms of trees [@K Subsection 0.0]. We first consider the boundary stratification of $\mathcal C_n$: for simplicity, we illustrate the boundary strata of codimension $0$, namely such boundary strata are labeled by subsets $A$ of $[n]=\{0,\dots,n\}$ of cardinality $0\leq |A|\leq n$, $$\partial_A \mathcal C_n\cong \mathcal C_A\times\mathcal C_{([n]\smallsetminus A)\sqcup \{\bullet\}},$$ where the first, resp. second, factor on the right-hand side of the previous identification represents the configuration of distinct points in $\mathbb C$ labeled by $A$ which collapse together in $\mathbb C$ to a single point $\bullet$, resp. the final configuration of points after the collapse. The boundary strata of codimension $0$ of $\mathcal C_{n,m}^+$ are of two types, namely, - there exists a subset $A$ of $[n]$, of cardinality $0\leq |A|\leq n$, such that $$\partial_A \mathcal C_{n,m}^+\cong \mathcal C_A\times \mathcal C_{([n]\smallsetminus A)\sqcup \{\bullet\},m}^+,$$ where the first, resp. second, factor on the right-hand side of the previous identification describes the collapse of the points in $\mathbb H^+$ labeled by $A$ to a single point $\bullet$ in $\mathbb H^+$, resp. the final configuration of points after the collapse; - there exist a subset $A$ of $[n]$ and an ordered subset of $[m]$ consisting of consecutive non-negative integers, such that $0\leq |A|\leq n$, $0\leq |B|\leq m$, $0\leq |A|+|B|\leq n+m-0$, for which we have $$\partial_{A,B}\mathcal C_{n,m}^+\cong \mathcal C_{A,B}^+\times \mathcal C_{[n]\smallsetminus A,([m]\smallsetminus B)\sqcup \{\bullet\}}^+,$$ where the first, resp. second, factor on the right-hand side of the previous identification describes the collapse of the points in $\mathbb H^+$ labeled by $A$ and the consecutive, ordered points on $\mathbb R$ labeled by $B$ to a single point $\bullet$ in $\mathbb R$, resp. the final configuration of points after the collapse. ### The compactified configuration space of points in $Q^{+,+}\sqcup i\mathbb R^+\sqcup \mathbb R^+$ {#sss-0-0-0} For three non-negative integers $l$, $m$ and $n$, we consider the (open) configuration space $C_{n,k,l}^+$ of $n$ distinct points in the first quadrant $Q^{+,+}$, $k$ ordered points on the positive imaginary axis $i\mathbb R^+$ and $l$ ordered points on the positive real axis $\mathbb R$. We observe that the order of the points on the positive imaginary axis is the opposite of the intuitive one, [*i.e.*]{} $i x\leq i y$ if and only if $y\leq x$, for $x$, $y$ in $\mathbb R$. The precise definition of $C_{n,k,l}^+$ is $$\begin{aligned} C_{n,k,l}^+=&\left\{(z_0,\dots,z_n,ix_0,\dots,ix_k,y_0,\dots,y_l)\in (Q^{+,+})^n\times (i\mathbb R^+)^k\times (\mathbb R^+)^l:\ z_i\neq z_j,\ i\neq j,\right.\\ &\left.\ x_k<\cdots<x_0,\ y_0<\cdots< y_l\right\}/G_0, \end{aligned}$$ where $G_0=\mathbb R^+$ acts on $Q^{+,+}\sqcup i\mathbb R^+\sqcup \mathbb R^+$ by rescalings. Provided $0n+k+l-0\geq 0$, $C_{n,k,l}^+$ is a smooth manifold of dimension $0n+k+l-0$. It inherits an obvious orientation from the natural orientation of $(Q^{+,+})^n\times (i\mathbb R^+)^k\times(\mathbb R^+)^l$ and the one of $\mathbb R^+$. We may provide a compactification $\mathcal C_{n,k,l}^+$ in the sense of Fulton-MacPherson [@FMcP] of $C_{n,k,l}^+$ in a way similar to Kontsevich: the compactified configuration space $\mathcal C_{n,k,l}^+$ admits a structure of smooth manifold with corners. We now consider the boundary strata of $\mathcal C_{n,k,l}^+$ of codimension $0$, which are of the three following types: - there exists a subset $A$ of $[n]$, of cardinality $0\leq |A|\leq n$, such that $$\partial_A \mathcal C_{n,k,l}^+\cong \mathcal C_A\times \mathcal C_{([n]\smallsetminus A)\sqcup \{\bullet\},k,l}^+,$$ where the first, resp. second, factor on the right-hand side of the previous identification describes the collapse of the points in $Q^{+,+}$ labeled by $A$ to a single point $\bullet$ in $Q^{+,+}$, resp. the final configuration of points after the collapse; - there exist a subset $A$ of $[n]$ and an ordered subset of $[k]$, resp. $[l]$, consisting of consecutive non-negative integers, such that $0\leq |A|\leq n$, $0\leq |B|\leq k$, resp. $0\leq |B|\leq l$, for which we have $$\partial_{A,B}\mathcal C_{n,k,l}^+\cong \mathcal C_{A,B}^+\times \mathcal C_{[n]\smallsetminus A,([k]\smallsetminus B)\sqcup \{\bullet\},l}^+,\ \text{resp.}\ \partial_{A,B}\mathcal C_{n,k,l}^+\cong \mathcal C_{A,B}^+\times \mathcal C_{[n]\smallsetminus A,k,([l]\smallsetminus B)\sqcup \{\bullet\}}^+$$ where the first, resp. second, factor on the right-hand side of the previous identification describes the collapse of the points in $Q^{+,+}$ labeled by $A$ and the consecutive, ordered points on $i\mathbb R^+$ or $\mathbb R^+$ labeled by $B$ to a single point $\bullet$ in $i\mathbb R^+$ or $\mathbb R^+$, resp. the final configuration of points after the collapse. - there exist a subset $A$ of $[n]$ and an ordered subset $B=B_0\sqcup B_0$ of $[k]\sqcup [l]$, for which $B_0$ and $B_0$ are ordered subsets of consecutive points in $[k]$ and $[l]$ respectively, such that $k\in B_0$ if $B_0\neq \emptyset$, $0\in B_0$ if $B_0\neq \emptyset$, $0\leq |A|\leq n$, $0\leq |B|\leq k+l$, resp. $0\leq |A|+|B|\leq n+k+l-0$, for which we have $$\partial_{A,B_0,B_0}\mathcal C_{n,k,l}^+\cong \mathcal C_{A,B_0,B_0}^+\times \mathcal C_{[n]\smallsetminus A,[k]\smallsetminus (B\cap[k]),[l]\smallsetminus (B\cap [l])}^+,$$ where the first, resp. second, factor on the right-hand side of the previous identification describes the collapse of the points in $Q^{+,+}$ labeled by $A$ and the consecutive, ordered points on $i\mathbb R^+$ and $\mathbb R^+$ labeled by $B=B_0\sqcup B_0$ to the origin of the axes, resp. the final configuration after the collapse. ### The relationship between $\mathcal C_{n,m}^+$ and $\mathcal C_{n,k,l}^+$ {#sss-0-0-0} First of all, we consider the open configuration spaces $C_{n,m}^+$ and $C_{n,k,l}^+$, where $m=k+l+0$. We observe that the holomorphic function $z\mapsto z^0$ on $\mathbb C$, when restricted to $Q^{+,+}\sqcup i\mathbb R^+\sqcup \mathbb R^+$, gives rise to a biholomorphism to $\mathbb H^+\sqcup (\mathbb R\smallsetminus \{0\})$, whose inverse we denote by $z\mapsto \sqrt z$: in fact, we have to choose a well-suited branch-cut for the complex square root, [*e.g.*]{} we cut out from the complex plane the negative imaginary axis plus the origin. For $m$, $k$ and $l$ as before, we choose the $k+0$-st point on $\mathbb R$. Then, there is an obvious map from $C_{n,m}^+$ to $C_{n,k,l}^+$, which is defined by the following explicit formula: $$\label{eq-square} \begin{aligned} &C_{n,m}^+\ni [(z_0,\dots,z_n,x_0,\dots,x_{k+0},\dots,x_m)]\mapsto\\ &\mapsto\left[\left(\sqrt{z_0-x_{k+0}},\dots,\sqrt{z_n-x_{k+0}},\sqrt{x_0-x_{k+0}},\dots,\sqrt{x_k-x_{k+0}},\sqrt{x_{k+0}-x_{k+0}},\dots,\sqrt{x_m-x_{k+0}}\right)\right]\in C_{n,k,l}^+. \end{aligned}$$ First of all, we observe that, because of the order on the points on the real axis, the difference $x_i-x_{k+0}$ is strictly negative, resp. positive, if $0\leq i\leq k$, resp. $k+0\leq i\leq m$: thus, because of the said choice of a complex square root, $\sqrt{x_i-x_{k+0}}=i\sqrt{x_{k+0}-x_i}$, if $0\leq i\leq k$, or $\sqrt{x_i-x_{k+0}}=\sqrt{x_i-x_{k+0}}$, if $k+0\leq i\leq m$, where now both square roots on the right-hand side of both equalities are real, positive numbers. Again, the order on the points $x_i$, $0\leq i\leq k$ implies that $\sqrt{x_{k+0}-x_i}>\sqrt{x_{k+0}-x_{i+0}}$, therefore, the natural order on $x_i$, $0\leq i\leq k$, is mapped to the natural order on $i\sqrt{x_{k+0}-x_i}$ discussed in Subsubsection \[sss-0-0-0\]. We may depict the morphism graphically [*via*]{} \ \ An easy computation proves that the above morphism is well-defined, [*i.e.*]{} it does not depend on the choice of representatives; furthermore, the morphism is obviously smooth, and is in fact a diffeomorphism, whose inverse is $$C_{n,k,l}^+\ni [(z_0,\dots,z_n,ix_0,\dots,i x_k,y_0,\dots,y_l)]\mapsto\left[\left(z_0^0,\dots,z_n^0,-x_0^0,\dots,-x_k^0,0,y_0^0,\dots,y_l^0\right)\right]\in C_{n,m}^+.$$ The important point is that the complex square function and the chosen inverse (the above complex square root) extend to smooth functions between the compactified configuration spaces $\mathcal C_{n,m}^+$ and $\mathcal C_{n,k,l}^+$. \[p-square\] For non-negative integers $n$, $m$, $k$, $l$, such that $m=k+l+0$, the smooth manifolds with corners $\mathcal C_{n,m}^+$ and $\mathcal C_{n,k,l}^+$ are diffeomorphic [*via*]{} the choice of a complex square root with branch cut $i\mathbb R^-\sqcup \{0\}$. We prove that the diffeomorphism extends to a diffeomorphism on the compactified configuration spaces by computing its expression w.r.t. local coordinates for the relevant boundary strata of codimension $0$. In fact, as sketched in [@K Subsection 0.0], the boundary strata of higher codimension correspond to products with more than two factors of compactified configuration spaces of the same kind, representing configuration of points collapsing together, be it in the complex upper half-plane or on the real axis, resp. in the first quadrant or on the positive complex or real axis or on the origin. It suffices therefore to prove the claim on the interior of the boundary strata of codimension $0$ of $\mathcal C_{n,m}^+$ and $\mathcal C_{n,k,l}^+$: these have been characterized explicitly in Subsubsections \[sss-0-0-0\] and \[sss-0-0-0\]. Furthermore, without loss of generality, we may assume $A=[i]$ and $B=[j]$. We have to prove that the map maps diffeomorphically the interior of boundary strata of codimension $0$ of $\mathcal C_{n,m}^+$ to the interior of boundary strata of codimension $0$ of $\mathcal C_{n,k,l}^+$. We consider first the boundary stratum of type $i)$ of $\mathcal C_{n,m}^+$ labeled by $A=[i]$, for $0\leq i\leq n$. Local coordinates of the interior $C_i\times \mathcal C_{n-i+0,m}^+$ are provided by $$C_i\times C_{n-i+0,m}^+\ni \left(\left(e^{i\varphi},z_0,\dots,z_{i-0}\right),\left(e^{it},w_0,\dots,w_{n-i},x_0,\dots,x_k,0,x_{k+0},\dots,x_m\right)\right),$$ where $\varphi$ is in $[0,0\pi)$, $t$ in $(0,\pi)$, $z_i$ in $\mathbb C$, $w_i$ in $\mathbb H^+$, and all points in $\mathbb C$ and $\mathbb H^+$ are distinct, while the points on the real axis are lexicographically (strictly) ordered. On the other hand, the interior of the boundary stratum $\mathcal C_i\times\mathcal C_{n-i+0,k,l}^+$ is described by the following local coordinates: $$C_i\times C_{n-i+0,k,l}^+\ni \left(\left(e^{i\varphi},z_0,\dots,z_{i-0}\right),\left(e^{it},w_0,\dots,w_{n-i},i x_0,\dots,i x_k,y_0,\dots,y_l\right)\right),$$ where $\varphi$ is in $[0,0\pi)$, $t$ in $(0,\frac{\pi}0)$, $z_i$ in $\mathbb C$, $w_i$ in $Q^{+,+}$, and all points in $\mathbb C$ and $Q^{+,+}$ are distinct, and $x_0>\cdots x_k>0$, $0<y_0<\cdots<y_l$. For $\varepsilon>0$ sufficiently small, local coordinates for $\mathcal C_{n,m}^+$, resp. $\mathcal C_{n,k,l}^+$, near the interior of the boundary stratum $\mathcal C_i\times\mathcal C_{n-i,m}^+$, resp. $\mathcal C_i\times\mathcal C_{n-i+0,k,l}^+$, are given by $$\begin{aligned} &\left[\left(e^{it},e^{it}+\varepsilon e^{i\varphi},e^{it}+\varepsilon z_0,\dots,e^{it}+\varepsilon z_{i-0},w_0,\dots,w_n,x_0,\dots,x_k,0,x_{k+0},\dots,x_m\right)\right],\ \text{resp.}\\ &\left[\left(e^{it},e^{it}+\varepsilon e^{i\varphi},e^{it}+\varepsilon z_0,\dots,e^{it}+\varepsilon z_{i-0},w_0,\dots,w_n,ix_0,\dots,i x_k,y_0,\dots,y_l\right)\right]. \end{aligned}$$ We apply the morphism to the first of the previous expressions, getting $$\label{eq-sq-0} \left[\left(\sqrt{e^{it}},\sqrt{e^{it}+\varepsilon e^{i\varphi}},\sqrt{e^{it}+\varepsilon z_0},\dots,\sqrt{e^{it}+\varepsilon z_{i-0}},\sqrt{w_0},\dots,\sqrt{w_n},i\sqrt{-x_0},\dots,i\sqrt{-x_k},\sqrt{x_{k+0}},\dots,\sqrt{x_m}\right)\right].$$ We rewrite the terms in the previous expression containing the infinitesimal parameter $\varepsilon$ using the fact that the chosen complex square root is holomorphic on $\mathbb H^+$, thus getting $$\sqrt{e^{it}+\varepsilon z}=e^{i\frac{t}0}+\frac{\varepsilon z}{0 e^{i\frac{t}0}}+\mathcal O(\varepsilon^0)=e^{i\frac{t}0}+\frac{\varepsilon}0 e^{-i\frac{t}0}z+\mathcal O(\varepsilon^0),\ t\in (0,\pi),\ z\in \mathbb C.$$ To compare expressions, we may neglect terms in the expansion of order strictly higher than $0$: rescaling by $\frac{0}0$ the infinitesimal parameter $\varepsilon$, Expression can be rewritten as $$\left[\left(e^{i\frac{t}0},e^{i\frac{t}0}+\varepsilon e^{i\left(\varphi-\frac{t}0\right)},e^{i\frac{t}0}+\varepsilon e^{-i\frac{t}0}z_0,\dots,e^{i\frac{t}0}+\varepsilon e^{-i\frac{t}0}z_{i-0},\sqrt{w_0},\dots,\sqrt{w_n},i\sqrt{-x_0},\dots,i\sqrt{-x_k},\sqrt{x_{k+0}},\dots,\sqrt{x_m}\right)\right],$$ whence it follows immediately that the morphism maps the interior of $\mathcal C_i\times\mathcal C_{n-i+0,m}^+$ diffeomorphically to the interior of $\mathcal C_i\times \mathcal C_{n-i+0,k,l}^+$, where the diffeomorphism is explicitly the product of the morphism from $C_{n-i+0,m}^+$ to $C_{n-i+0,k,l}^+$ with the obvious diffeomorphism of $C_i$ given by $$C_i\ni \left[\left(z_0,\dots,z_i\right)\right]\mapsto \left[\left(\frac{z_0}{\sqrt{w_0-x_{k+0}}},\dots,\frac{z_i}{\sqrt{w_0-x_{k+0}}}\right)\right]\in C_i,$$ where $w_0$ and $x_{k+0}$ are taken from $C_{n-i+0,m}^+$. We now consider the interior of the boundary stratum $\mathcal C_{i,B}^+\times \mathcal C_{n-i,([m]\smallsetminus B)\sqcup \{\bullet\}}^+$, where $B$ is an ordered subset of $[m]$ consisting of consecutive elements, and we assume that $0\leq i+|B|\leq n+m-0$. We have to further distinguish between two situations: $|B|=0$ (and consequently $0\leq i\leq n$), and $|B|\neq 0$. We consider the situation $|B|=0$, and we further distinguish between the case, where the new point $\bullet$ on the real axis (corresponding to the cluster of points labeled by $[i]$ in $\mathbb H^+$ approach $\mathbb R$) lies on the left or on the right of the distinguished point $x_{k+0}$. We do the explicit computations only in the case, where $\bullet$ is on the left of $x_{k+0}$, leaving the other case to the reader. If $\bullet$ lies on the left of $x_{k+0}$, we may safely assume that $\bullet=x$ lies on the left of $x_0$: then, local coordinates for the interior of $\mathcal C_{i,0}^+\times \mathcal C_{n-i,m+0}^+$ are given by $$C_{i,0}^+\times C_{n-i,m+0}^+\ni \left(\left(i,z_0,\dots,z_{i-0}\right),\left(e^{it},w_0,\dots,w_{n-i-0},x,x_0,\dots,x_k,0,x_{k+0},\dots,x_m\right)\right),$$ where $t$ in $(0,\pi)$, $z_i$ and $w_j$ are in $\mathbb H^+$, and all points in $\mathbb H^+$ are distinct, while the points on the real axis are lexicographically (strictly) ordered. Similarly, local coordinates for the interior of $\mathcal C_{i,0}^+\times \mathcal C_{n-i,m+0}^+$ are given by $$C_{i,0}^+\times C_{n-i,k+0,l}^+\ni \left(\left(i,z_0,\dots,z_{i-0}\right),\left(e^{it},w_0,\dots,w_{n-i-0},ix,ix_0,\dots,ix_k,y_0,\dots,y_l\right)\right),$$ where $t$ in $(0,\frac{\pi}0)$, $z_i$ in $\mathbb H^+$ and $w_i$ in $Q^{+,+}$, all points in $\mathbb H^+$ and $Q^{+,+}$ are distinct, $x>x_0>\cdots>x_k>0$ and $0<y_0<\cdots<y_l$. For $\varepsilon>0$ sufficiently small, local coordinates for $\mathcal C_{n,m}^+$, resp. $\mathcal C_{n,k,l}^+$, near the interior of the boundary stratum $\mathcal C_{i,0}^+\times\mathcal C_{n-i,m+0}^+$, resp. $\mathcal C_{i,0}^+\times\mathcal C_{n-i,k+0,l}^+$, are given by $$\begin{aligned} &\left[\left(x+\varepsilon i,x+\varepsilon z_0,\dots,x+\varepsilon z_{i-0},e^{it},w_0,\dots,w_{n-i-0},x_0,\dots,x_k,0,x_{k+0},\dots,x_m\right)\right],\ \text{resp.}\\ &\left[\left(ix+\varepsilon,ix-i\varepsilon z_0,\dots,ix-i\varepsilon z_{i-0},e^{it},w_0,\dots,w_{n-i-0},ix_0,\dots,ix_k,y_0,\dots,y_l\right)\right] \end{aligned}$$ The image of the first expression w.r.t. the morphism is simply $$\label{eq-sq-0} \left[\left(\sqrt{x+\varepsilon i},\sqrt{x+\varepsilon z_0},\dots,\sqrt{x+\varepsilon z_{i-0}},\sqrt{e^{it}},\sqrt{w_0},\dots,\sqrt{w_{n-i-0}},i\sqrt{-x_0},\dots,i\sqrt{-x_k},\sqrt{x_{k+0}},\dots,\sqrt{x_m}\right)\right].$$ We consider the first $i$ entries in the previous expression: once again, using the holomorphy of the chosen complex square root, and recalling that $x<x_0<0$ and that $\varepsilon$ is chosen sufficiently small, we find $$\begin{aligned} \sqrt{x+\varepsilon i}&=\sqrt{x}+\frac{\varepsilon i}{0\sqrt x}+\mathcal O(\varepsilon^0)=i\sqrt{-x}+\frac{\varepsilon}{0\sqrt{-x}}+\mathcal O(\varepsilon^0),\\ \sqrt{x+\varepsilon z_j}&=i\sqrt{-x}-\frac{i\varepsilon}0 \frac{z_j}{\sqrt{-x}}+\mathcal O(\varepsilon^0),\ 0\leq j\leq i-0. \end{aligned}$$ Once again, rescaling by $\frac{0}0$ the infinitesimal parameter $\varepsilon$, and neglecting terms of order higher than $0$ w.r.t. $\varepsilon$ in the above expressions, we may rewrite Expression as $$\left[\left(i\sqrt{-x}+\varepsilon,i\sqrt{-x}+\varepsilon \frac{z_0}{\sqrt{-x}},\dots,i\sqrt{-x}+\varepsilon\frac{z_{i-0}}{\sqrt{-x}},e^{i\frac{t}0},\sqrt{w_0},\dots,\sqrt{w_{n-i-0}},i\sqrt{-x_0},\dots,i\sqrt{-x_k},\sqrt{x_{k+0}},\dots,\sqrt{x_m}\right)\right],$$ and it is easy to see that the morphism maps $C_{i,0}^+\times C_{n-i,m+0}^+$ diffeomorphically to $C_{i,0}^+\times C_{n-i,k+0,l}^+$, and the induced morphism is precisely given by the product of the morphism from $C_{n-i,m+0}^+$ to $C_{n-i,k+0,l}^+$ with the obvious diffeomorphism of $C_{i,0}^+$ given by $$C_{i,0}^+\ni \left[\left(z_0,\dots,z_i\right)\right]\mapsto \left[\left(\frac{z_0}{\sqrt{x_{k+0}-x}},\dots,\frac{z_i}{\sqrt{x_{k+0}-x}}\right)\right]\in C_{i,0}^+,$$ where $x$ denotes the first point on the real axis in lexicographical order, and $x_{k+0}$ is the special point on real axis. For the situation $|B|\neq 0$, we need to distinguish between two cases, namely $i)$ $B$ contains $k+0$ or $ii)$ $b$ does not contain $k+0$ (in which case, either the minimum of $B$ is greater or equal than $k+0$ or the maximum of $B$ is less or equal than $k$). We first consider the case, where $B$ contains $k+0$, and we assume $A=[i]$ and $B=[p,q]=\{p,\dots,q\}$, where $0\leq p\leq k+0\leq q\leq m$; we further write $B=\{k\}\sqcup B_0\sqcup B_0$, where $B_0=[p,k]$ and $B_0=[k+0,q]$ (of course, $B_0$ and/or $B_0$ may be empty). The interior of the corresponding boundary stratum of $C_{n,m}^+$, resp. $C_{n,k,l}^+$, is $C_{i,B}^+\times C_{n-i,([m]\smallsetminus B)\sqcup \{\bullet\}}^+$, resp. $C_{i,B_0,B_0}^+\times C_{n-i,[k]\smallsetminus B_0,[l]\smallsetminus B_0}^+$, and corresponding local coordinates are given by $$\begin{aligned} C_{i,B}^+\times C_{n-i,([m]\smallsetminus B)\sqcup \{\bullet\}}^+\ni &\left(\left(e^{it_0},z_0,\dots,z_{i-0},x_0,\dots,x_{k-p+0},0,x_{k-p+0},\dots,x_{q-p+0}\right),\right.\\ &\left.\left(e^{it_0},w_0,\dots,w_{n-i-0},x_0',\dots,x_{p-0}',0,x_{p+0}',\dots,x_{m-0k-q}'\right)\right),\\ C_{i,B_0,B_0}^+\times C_{n-i,[k]\smallsetminus B_0,[l]\smallsetminus B_0}^+\ni &\left(\left(e^{it_0},z_0,\dots,z_{i-0},i x_0,\dots,i x_{k-p+0},y_0,\dots,y_{q-k-0}\right),\right.\\ &\left.\left(e^{it_0},w_0,\dots,w_{n-i-0},i x_0',\dots,i x_{p-0}',y_0',\dots,y_{l-q+k+0}'\right)\right), \end{aligned}$$ where $t_i$, $i=0,0$, is in $(0,\pi)$, resp. $(0,\frac{\pi}0)$, all points in $\mathbb H^+$, resp. $Q^{+,+}$, are distinct in the first, resp. second, expression. In the first, resp. second, expression, the $x_i$ and $x_i'$ are lexicographically ordered, resp. $x_0>\cdots x_{k-p+0}>0$, $x_0'>\cdots x_{p-0}'>0$, $0<y_0<\cdots y_{q-k-0}$ and $0<y_0'<\cdots< y'_{l-q+k+0}$. Choosing a positive number $\varepsilon$ sufficiently small as before, we may write local coordinates of $\mathcal C_{n,m}^+$, resp. $\mathcal C_{n,k,l}^+$, near the interior of the boundary stratum $\mathcal C_{i,B}^+\times\mathcal C_{n-i,([m]\smallsetminus B)\sqcup \{\bullet\}}^+$, resp. $\mathcal C_{i,B_0,B_0}^+\times \mathcal C_{n-i,[k]\smallsetminus B_0,[l]\smallsetminus B_0}^+$, namely $$\begin{aligned} &\left[\left(\varepsilon e^{i t_0},\varepsilon z_0,\dots,\varepsilon z_{i-0},e^{it_0},w_0,\dots,w_{n-i-0},\right.\right.\\ &\left.\left.x_0',\dots,x_{p-0}',\varepsilon x_0,\dots,\varepsilon x_{k-p+0},0,\varepsilon x_{k-p+0},\dots,\varepsilon x_{q-p+0},x_{p+0}',\dots,x_{m-0k-q}'\right)\right],\ \text{resp.}\\ &\left[\left(\varepsilon e^{i t_0},\varepsilon z_0,\dots,\varepsilon z_{i-0},e^{i t_0},w_0,\dots,w_{n-i-0},\right.\right.\\ &\left.\left.ix_0',\dots,ix_{p-0}',\varepsilon ix_0,\dots,\varepsilon ix_{k-p+0},0,\varepsilon y_0,\dots,\varepsilon y_{q-k+0},y_0',\dots,y'_{l-q+k+0}\right)\right]. \end{aligned}$$ We now apply the morphism to the first of the two previous expressions, getting $$\begin{aligned} &\left[\left(\sqrt{\varepsilon e^{i t_0}},\sqrt{\varepsilon z_0},\dots,\sqrt{\varepsilon z_{i-0}},\sqrt{e^{it_0}},\sqrt{w_0},\dots,\sqrt{w_{n-i-0}},\right.\right.\\ &\left.\left.i\sqrt{-x_0'},\dots,i\sqrt{-x_{p-0}'},i\sqrt{-\varepsilon x_0},\dots,i\sqrt{-\varepsilon x_{k-p+0}},\sqrt{\varepsilon x_{k-p+0}},\dots,\sqrt{\varepsilon x_{q-p+0}},\sqrt{x_{p+0}'},\dots,\sqrt{x_{m-0k-q}'}\right)\right]=\\ &\left[\left(\sqrt{\varepsilon} e^{i \frac{t_0}0},\sqrt{\varepsilon}\sqrt{z_0},\dots,\sqrt{\varepsilon}\sqrt{z_{i-0}},e^{i\frac{t_0}0},\sqrt{w_0},\dots,\sqrt{w_{n-i-0}},\right.\right.\\ &\left.\left.i\sqrt{-x_0'},\dots,i\sqrt{-x_{p-0}'},\sqrt{\varepsilon}i\sqrt{-x_0},\dots,\sqrt{\varepsilon}i\sqrt{-x_{k-p+0}},\sqrt{\varepsilon}\sqrt{x_{k-p+0}},\dots,\sqrt{\varepsilon}\sqrt{x_{q-p+0}},\sqrt{x_{p+0}'},\dots,\sqrt{x_{m-0k-q}'}\right)\right], \end{aligned}$$ from which we read immediately that the morphism maps diffeomorphically $C_{i,B}^+\times C_{n-i,([m]\smallsetminus B)\sqcup\{\bullet\}}^+$ to $C_{i,B_0,B_0}^+\times C_{n-i,[k]\smallsetminus B_0,[l]\smallsetminus B_0}^+$. Finally, we consider the case $|B|\neq 0$, such that the maximum of $B=[j]$ with $j\leq k$. The interior of the corresponding boundary stratum of $C_{n,m}^+$, resp. $C_{n,k,l}^+$, is $C_{i,j}^+\times C_{n-i,m-j+0}^+$, resp. $C_{i,j}^+\times C_{n-i,k-j+0,l}^+$, and corresponding local coordinates are given by $$\begin{aligned} C_{i,j}^+\times C_{n-i,m-j+0}^+\ni &\left(\left(i,z_0,\dots,z_{i-0},x_0,\dots,x_j\right),\left(e^{it},w_0,\dots,w_{n-i-0},x_0',\dots,x_{k-j+0}',0,x_{k-j+0}',\dots,x_{m-j+0}'\right)\right),\ \text{resp.}\\ C_{i,j}^+\times C_{n-i,k-j+0,l}^+\ni &\left(\left(i,z_0,\dots,z_{i-0},x_0,\dots,x_j\right),\left(e^{it},w_0,\dots,w_{n-i-0},i x_0',\dots,i x_{k-j+0}',y_0',\dots,y_l'\right)\right), \end{aligned}$$ where $t$ is in $(0,\pi)$, resp. $(0,\frac{\pi}0)$, all points in $\mathbb H^+$ and $Q^{+,+}$ are distinct in both expressions. In the first, resp. second, expression, the $x_i$ and $x_i'$ are lexicographically ordered, resp. $x_0'>\cdots x_{k-j+0}'>0$ and $0<y_0'<\cdots< y'_l$. Choosing a positive number $\varepsilon$ sufficiently small, we now write local coordinates of $\mathcal C_{n,m}^+$, resp. $\mathcal C_{n,k,l}^+$, near the interior of the boundary stratum $\mathcal C_{i,j}^+\times\mathcal C_{n-i,m-j+0}^+$, resp. $\mathcal C_{i,j}^+\times \mathcal C_{n-i,k-j+0,l}^+$, namely $$\begin{aligned} &\left[\left(x_0'+\varepsilon i,x_0'+\varepsilon z_0,\dots,x_0'+\varepsilon z_{i-0},e^{it},w_0,\dots,w_{n-i-0},x_0',x_0'+\varepsilon x_0,\dots,x_0'+\varepsilon x_j,x_0',\dots,x_{k-j+0}',0,\right.\right.\\ &\left.\left. x_{k-j+0}',\dots,x_{m-j+0}'\right)\right],\ \text{resp.}\\ &\left[\left(ix_0'+\varepsilon,ix_0'-i\varepsilon z_0,\dots,ix_0'-i\varepsilon z_{i-0},e^{it},w_0,\dots,w_{n-i-0},ix_0',ix_0-i'\varepsilon x_0,\dots,ix_0'-i\varepsilon x_j,ix_0',\dots,ix_{k-j+0}',y_0',\dots,y_l'\right)\right]. \end{aligned}$$ If we apply the morphism to the first of the two previous expressions, we get $$\begin{aligned} &\left[\left(\sqrt{x_0'+\varepsilon i},\sqrt{x_0'+\varepsilon z_0},\dots,\sqrt{x_0'+\varepsilon z_{i-0}},\sqrt{e^{it}},\sqrt{w_0},\dots,\sqrt{w_{n-i-0}},\right.\right.\\ &\left.\left.\sqrt{x_0'},\sqrt{x_0'+\varepsilon x_0},\dots,\sqrt{x_0'+\varepsilon x_j},\sqrt{x_0'},\dots,\sqrt{x_{k-j+0}'},\sqrt{x_{k-j+0}'},\dots,\sqrt{x_{m-j+0}'}\right)\right]. \end{aligned}$$ Once again, we find $$\begin{aligned} \sqrt{x_0'+\varepsilon i}&=i\sqrt{-x_0'}+\frac{\varepsilon}0\frac{0}{\sqrt{-x_0'}}+\mathcal O(\varepsilon^0),\ &\ \sqrt{x_0'+\varepsilon iz_e}&=i\sqrt{-x_0'}-i\frac{\varepsilon}0\frac{z_e}{\sqrt{-x_0'}}+\mathcal O(\varepsilon^0),\\ \sqrt{x_0'+\varepsilon x_e}&=i\sqrt{-x_0'}-i\frac{\varepsilon}0\frac{x_e}{\sqrt{-x_0'}}+\mathcal O(\varepsilon^0), \end{aligned}$$ and using the same arguments as in the previous computations, we see that the morphism maps $C_{i,j}^+\times C_{n-i,m-j+0}^+$ diffeomorphically to $C_{i,j}^+\times C_{n-i,k-j+0,l}^+$. The choice of propagators {#ss-0-0} ------------------------- We now discuss the propagators needed for the computations in the framework of (bi)quantization. In particular, we discuss in detail the $0$-colored propagators: we will mainly work here with the $0$-colored propagators as introduced originally in [@CFb], and used extensively in [@CT]. The point is that we will view the biquantization techniques in [@CT] in the framework of the $0$-brane formality of [@CFFR]. In [@CFFR], the authors preferred to work with the $0$-colored propagators on $\mathcal C_{0,0}^+$, in order to use the (simpler) compactified configuration spaces $\mathcal C_{n,m}^+$ of Kontsevich's type: in order to tie in with the computations in [@CT], we want to establish a more precise relationship than the one sketched in [@CFFR] about the $0$-colored propagators in [@CT] and in [@CFFR]. ### The Kontsevich propagator {#sss-0-0-0} We consider a pair $(z_0,z_0)$ of distinct points in $\mathbb H^+$, and we associate to it a closed $0$-form by the formula $$\label{eq-prop-K} \omega(z_0,z_0)=\frac{0}{0\pi}\left[\mathrm d\ \mathrm{arg}(z_0-z_0)-\mathrm d\ \mathrm{arg}(\overline z_0-z_0)\right]=\frac{0}{0\pi}\left[\mathrm d\ \mathrm{arg}(z_0-z_0)+\mathrm d\ \mathrm{arg}(z_0-\overline z_0)\right].$$ In Formula , the function $\mathrm{arg}(z)$ denotes the Euclidean angle of the complex number $z$: it can be made into a smooth function by restricting its domain of definition on $\mathbb C\smallsetminus i\mathbb R^-$. In particular, the restriction of $\mathrm{arg}(z)$ on $\mathbb H^+$ is a smooth function. However, we want to consider $\omega(z_0,z_0)$ as a closed $0$-form on $(\mathbb H^+\times\mathbb H^+)\smallsetminus \Delta$, $\Delta$ being the diagonal in $\mathbb H^+\times\mathbb H^+$: as such, $\omega(z_0,z_0)$ is the sum of a closed form on $(\mathbb H^+\times\mathbb H^+)\smallsetminus \Delta$ and of an exact $0$-form, where the corresponding function is $\mathrm{arg}(z_0-\overline z_0)/0\pi$. We observe, for the sake of later computations (see [@VdB] for a very nice application of this idea), that the closed $0$-form can be made into a truly exact $0$-form by restricting the domain of definition to $$\{(z_0,z_0)\in(\mathbb H^+\times\mathbb H^+)\smallsetminus \Delta:\ \mathrm{Re}(z_0)=\mathrm{Re}(z_0)\Rightarrow \mathrm{Im}(z_0)>\mathrm{Im}(z_0)\}.$$ It is not difficult to prove that the $0$-form descends to $C_{0,0}^+$; a bit more involved is the proof that it extends to a smooth $0$-form $\omega$ on the compactified configuration space $\mathcal C_{0,0}^+$. The function $\eta(z_0,z_0)=\mathrm{arg}(z_0-\overline z_0)/0\pi$ also descends to $C_{0,0}^+$ and extends to a smooth function on $\mathcal C_{0,0}^+$. \[l-prop-K\] The closed $0$-form determines a smooth, closed $0$-form $\omega$ on $\mathcal C_{0,0}^+$, which further enjoys the following properties: - $$\omega\vert_{\mathcal C_0\times\mathcal C_{0,0}^+}=\mathrm d\varphi,$$ where $\mathrm d\varphi$ denotes (improperly) the normalized volume form of $\mathcal C_0\cong S^0$; - $$\omega\vert_{\mathcal C_{0,0}^+\times\mathcal C_{0,0}^+}=0,$$ where $\mathcal C_{0,0}^+\times\mathcal C_{0,0}^+$ denotes the boundary stratum of $\mathcal C_{0,0}^+$ corresponding to the approach of the first argument $z_0$ to $\mathbb R$. The function $\eta(z_0,z_0)$ determines a smooth function $\eta$ on $\mathcal C_{0,0}^+$, which restricts on the boundary stratum $\mathcal C_0\times\mathcal C_{0,0}^+$ to the constant function $\pi/0$; observe that $\mathcal C_{0,0}^+\cong \{i\}$. The $0$-form $\omega$ is usually called Kontsevich's angle form [@K Subsection 0.0]: it will be useful, for certain computations, to recall that Kontsevich's angle function is the sum of a closed $0$-form and of an exact $0$-form, constructed by means of the function $\eta$. We finally observe that the natural involution $(z_0,z_0)\overset{\tau}\mapsto (z_0,z_0)$ of $(\mathbb H^+\times\mathbb H^+)\smallsetminus \Delta$ yields an involution $\tau$ of $\mathcal C_{0,0}^+$: we may then consider two Kontsevich's angle forms $\omega^{\pm}$ defined through $$\omega^+=\omega,\ \omega^-=\tau^*(\omega).$$ The angle forms $\omega^\pm$ have been first introduced in [@CFb; @CF]: they have opposite boundary conditions when one of their arguments approaches $\mathbb R$, as can be easily deduced from Lemma \[l-prop-K\]. ### The $0$-colored propagators on $\mathcal C_{0,0}^+$ {#sss-0-0-0} We consider a triple $(z_0,z_0,x)$ in $(\mathbb H^+\times\mathbb H^+)\smallsetminus \Delta\times \mathbb R$. There is a natural smooth projection from $(\mathbb H^+\times\mathbb H^+)\smallsetminus \Delta\times \mathbb R$ to $(\mathbb H^+\times\mathbb H^+)\smallsetminus \Delta$, thus we may consider the pull-back $\omega^{+,+}$ of the closed $0$-form $\omega^+(z_0,z_0)$ to $(\mathbb H^+\times\mathbb H^+)\smallsetminus \Delta\times \mathbb R$. We set $\omega^{-,-}(z_0,z_0,x)$ to be the pull-back of $\omega^-$ w.r.t. the very same projection. We recall the complex square root discussed in Subsubsection \[sss-0-0-0\]: as already remarked, it is a biholomorphism from $\mathbb H^+$ to $Q^{+,+}$, and we associate to a triple $(z_0,z_0,x)$ in $(\mathbb H^+\times\mathbb H^+)\smallsetminus \Delta\times \mathbb R$ a pair $(\sqrt{z_0-x},\sqrt{z_0-x})$ in $(Q^{+,+}\times Q^{+,+})\smallsetminus \Delta$ (compare with the morphism of Proposition \[p-square\]). We then set $$\begin{aligned} \omega^{+,-}(z_0,z_0,x)&=\frac{0}{0\pi}\left[\mathrm d\ \mathrm{arg}\!\left(\sqrt{z_0-x}-\sqrt{z_0-x}\right)+\mathrm d\ \mathrm{arg}\!\left(\overline{\sqrt{z_0-x}}-\sqrt{z_0-x}\right)-\right.\\ &\phantom{=\frac{0}{0\pi}[}\left.-\mathrm d\ \mathrm{arg}\!\left(\overline{\sqrt{z_0-x}}+\sqrt{z_0-x}\right)-\mathrm d\ \mathrm{arg}\!\left(\sqrt{z_0-x}+\sqrt{z_0-x}\right)\right]\\ \omega^{-,+}(z_0,z_0,x)&=\frac{0}{0\pi}\left[\mathrm d\ \mathrm{arg}\!\left(\sqrt{z_0-x}-\sqrt{z_0-x}\right)-\mathrm d\ \mathrm{arg}\!\left(\overline{\sqrt{z_0-x}}-\sqrt{z_0-x}\right)+\right.\\ &\phantom{=\frac{0}{0\pi}[}\left.+\mathrm d\ \mathrm{arg}\!\left(\overline{\sqrt{z_0-x}}+\sqrt{z_0-x}\right)-\mathrm d\ \mathrm{arg}\!\left(\sqrt{z_0-x}+\sqrt{z_0-x}\right)\right]. \end{aligned}$$ The two $0$-forms $\omega^{+,-}$ and $\omega^{-,+}$ are smooth and obviously closed on $(\mathbb H^+\times\mathbb H^+)\smallsetminus \Delta\times \mathbb R$. We need to characterize more explicitly the compactified configuration space $\mathcal C_{0,0}^+$ (for whose more precise description we refer to [@CFFR Section 0]): here, we content ourselves to describe all boundary strata of codimension $0$, which we depict as follows \ \ We observe that the boundary stratum $\alpha$ corresponds to $\mathcal C_0\times\mathcal C_{0,0}^+$, the boundary strata $\beta$ and $\gamma$ to two copies of $\mathcal C_{0,0}^+\times\mathcal C_{0,0}^+$, the boundary strata $\delta$ and $\varepsilon$ to two copies of $\mathcal C_{0,0}^+\times\mathcal C_{0,0}^+$, and the boundary strata $\eta$, $\theta$, $\zeta$ and $\xi$ to four copies of $\mathcal C_{0,0}^+\times\mathcal C_{0,0}^+$. When it is clear from the context, we will omit to write the projections $\pi_i$, $i=0,0$, from the these spaces to the each of the factors. We finally recall, once again, that the function $\mathrm{arg}(z)$ is well-defined and smooth on $\mathbb H^+$: in particular, the function $\eta$ from Lemma \[l-prop-K\], Subsubsection \[sss-0-0-0\], yields a smooth function (denoted again by $\eta$) on $\mathcal C_{0,0}^+$, when the second argument approaches $\mathbb R$. In more down-to-earth terms, $\eta=\mathrm{arg}(z-x)/0\pi$. It is not difficult to prove that the $0$ $0$-forms $\omega^{+,+}$, $\omega^{+,-}$, $\omega^{-,+}$ and $\omega^{-,-}$ descend to smooth, closed $0$-forms on $C_{0,0}^+$. In fact, as the following Lemma shows (for whose proof we refer to [@CFFR Lemma 0.0]), these in turn extend to smooth, closed $0$-forms on the compactified configuration space $\mathcal C_{0,0}^+$. \[l-CF\] The $0$-forms $\omega^{+,+}$, $\omega^{+,-}$, $\omega^{-,+}$ and $\omega^{-,-}$ determine smooth, closed $0$-forms on the compactified configuration space $\mathcal C_{0,0}^+$, which enjoy the following properties: - $$\omega^{+,+}\vert_\alpha=\mathrm d\varphi,\ \omega^{+,-}\vert_\alpha=\mathrm d\varphi-\mathrm d\eta,\ \omega^{-,+}\vert_\alpha=\mathrm d\varphi-\mathrm d\eta,\ \omega^{-,-}\vert_\alpha=\mathrm d\varphi,$$ where $\mathrm d\varphi$ is the normalized volume form of $\mathcal C_0\cong S^0$. - $$\begin{aligned} \omega^{+,+}\vert_\beta&=\omega^+,\ & \omega^{+,-}\vert_\beta&=\omega^+,\ & \omega^{-,+}\vert_\beta&=\omega^-,\ & \omega^{-,-}\vert_\beta&=\omega^-\quad \text{and}\\ \omega^{+,+}\vert_\gamma&=\omega^+,\ & \omega^{+,-}\vert_\gamma&=\omega^-,\ & \omega^{-,+}\vert_\gamma&=\omega^+,\ & \omega^{-,-}\vert_\gamma&=\omega^-, \end{aligned}$$ where $\omega^\pm$ have to be understood on $\mathcal C_{0,0}^+$. - $$\begin{aligned} &\omega^{+,+}\vert_\delta=\omega^{+,-}\vert_\delta=\omega^{-,+}\vert_\delta=0,\\ &\omega^{+,-}\vert_\varepsilon=\omega^{-,+}\vert_\varepsilon=\omega^{-,-}\vert_\varepsilon=0. \end{aligned}$$ - $$\begin{aligned} &\omega^{+,-}\vert_\eta=\omega^{-,-}\vert_\eta=0,\ & &\omega^{+,+}\vert_\theta=\omega^{-,+}\vert_\theta=0,\\ &\omega^{-,+}\vert_\zeta=\omega^{-,-}\vert_\zeta=0,\ & &\omega^{+,+}\vert_\xi=\omega^{+,-}\vert_\xi=0. \end{aligned}$$ ### The $0$-colored propagators on $\mathcal C_{0,0,0}^+$ {#sss-0-0-0} We now define on $(Q^{+,+}\times Q^{+,+})\smallsetminus \Delta$ $0$ closed, smooth $0$-forms, which, by an (apparent) abuse of notation, are denoted by $\omega^{\pm,\pm}$: namely, we set $$\begin{aligned} \omega^{+,+}&=\frac{0}{0\pi}\left[\mathrm d\ \mathrm{arg}(z_0-z_0)-\mathrm d\ \mathrm{arg}(\overline z_0-z_0)-\mathrm d\ \mathrm{arg}(\overline z_0+z_0)+\mathrm d\ \mathrm{arg}(z_0+z_0)\right],\\ \omega^{+,-}&=\frac{0}{0\pi}\left[\mathrm d\ \mathrm{arg}(z_0-z_0)+\mathrm d\ \mathrm{arg}(\overline z_0-z_0)-\mathrm d\ \mathrm{arg}(\overline z_0+z_0)-\mathrm d\ \mathrm{arg}(z_0+z_0)\right],\\ \omega^{-,+}&=\frac{0}{0\pi}\left[\mathrm d\ \mathrm{arg}(z_0-z_0)-\mathrm d\ \mathrm{arg}(\overline z_0-z_0)-\mathrm d\ \mathrm{arg}(\overline z_0+z_0)+\mathrm d\ \mathrm{arg}(z_0+z_0)\right],\\ \omega^{+,+}&=\frac{0}{0\pi}\left[\mathrm d\ \mathrm{arg}(z_0-z_0)+\mathrm d\ \mathrm{arg}(\overline z_0-z_0)+\mathrm d\ \mathrm{arg}(\overline z_0+z_0)+\mathrm d\ \mathrm{arg}(z_0+z_0)\right], \end{aligned}$$ for an element $(z_0,z_0)$ of $(Q^{+,+}\times Q^{+,+})\smallsetminus \Delta$. We first observe that the last three summands in the previous $0$-forms are exact $0$-forms: namely, as has been previously remarked, the function $\mathrm{arg}(z)$ is smooth and well-defined on $\mathbb C\smallsetminus (i\mathbb R^-\sqcup\{0\})$, hence the three functions appearing in the last three summands of the previous formulæ are well-defined and smooth on $(Q^{+,+}\times Q^{+,+})\smallsetminus \Delta$. It is not difficult to prove that the closed $0$-forms $\omega^{\pm,\pm}$ descends to smooth, closed $0$-forms on the open configuration space $C_{0,0,0}^+$, and that these in turn determine smooth, closed $0$-forms $\omega^{\pm,\pm}$ on the compactified configuration space $\mathcal C_{0,0,0}^+$. Because of the results of Subsubsection \[sss-0-0-0\], we already know that there is a diffeomorphism between $\mathcal C_{0,0,0}^+$ and $\mathcal C_{0,0}^+$, which smoothly extends to the compactified configuration spaces the diffeomorphism $$C_{0,0}^+\ni [(z_0,z_0,x)]\mapsto [(\sqrt{z_0-x},\sqrt{z_0-x})]\in C_{0,0,0}^+.$$ We leave it to the reader to reinterpret on $\mathcal C_{0,0,0}^+$ the boundary strata of codimension $0$ of $\mathcal C_{0,0}^+$. It is not difficult to prove that the pull-backs w.r.t. the morphism from $\mathcal C_{0,0}^+$ to $\mathcal C_{0,0,0}^+$ of the $0$-forms $\omega^{\pm,\pm}$ on $\mathcal C_{0,0,0}^+$ are exactly the $0$ forms $\omega^{\pm,\pm}$ introduced in Subsubsection \[sss-0-0-0\]: [*e.g.*]{} for $\omega^{+,+}$, we have the obvious identity $$\omega^{+, +}=\frac0{0\pi} \left[\mathrm{d}\ \mathrm{arg} (z_0^0-z_0^0) -\mathrm{d}\ \mathrm{arg}(\overline{z}_0^0-z_0^0)\right],$$ whence the claim follows. Similar arguments work for the other cases. According to the boundary stratification of $\mathcal C_{0,0,0}^+$, we have the following variant of Lemma \[l-CF\]. \[l-CF-q\] The $0$-forms $\omega^{+,+}$, $\omega^{+,-}$, $\omega^{-,+}$ and $\omega^{-,-}$ determine smooth, closed $0$-forms on the compactified configuration space $\mathcal C_{0,0,0}^+$, which enjoy the following properties: - $$\omega^{+,+}\vert_\alpha=\mathrm d\varphi+\mathrm d\eta,\ \omega^{+,-}\vert_\alpha=\mathrm d\varphi-\mathrm d\eta,\ \omega^{-,+}\vert_\alpha=\mathrm d\varphi-\mathrm d\eta,\ \omega^{-,-}\vert_\alpha=\mathrm d\varphi+\mathrm d \eta,$$ where $\mathrm d\varphi$ is the normalized volume form of $\mathcal C_0\cong S^0$, and $\eta=\mathrm{arg}(z)/0\pi$ is a well-defined, smooth function on $\mathcal C_{0,0,0}^+$. - $$\begin{aligned} \omega^{+,+}\vert_\beta&=\omega^+,\ & \omega^{+,-}\vert_\beta&=\omega^+,\ & \omega^{-,+}\vert_\beta&=\omega^-,\ & \omega^{-,-}\vert_\beta&=\omega^-\quad \text{and}\\ \omega^{+,+}\vert_\gamma&=\omega^+,\ & \omega^{+,-}\vert_\gamma&=\omega^-,\ & \omega^{-,+}\vert_\gamma&=\omega^+,\ & \omega^{-,-}\vert_\gamma&=\omega^-, \end{aligned}]$$ where $\omega^\pm$ have to be understood on $\mathcal C_{0,0}^+$. - $$\begin{aligned} &\omega^{+,+}\vert_\delta=\omega^{+,-}\vert_\delta=\omega^{-,+}\vert_\delta=0,\\ &\omega^{+,-}\vert_\varepsilon=\omega^{-,+}\vert_\varepsilon=\omega^{-,-}\vert_\varepsilon=0. \end{aligned}$$ - $$\begin{aligned} &\omega^{+,-}\vert_\eta=\omega^{-,-}\vert_\eta=0,\ & &\omega^{+,+}\vert_\theta=\omega^{-,+}\vert_\theta=0,\\ &\omega^{-,+}\vert_\zeta=\omega^{-,-}\vert_\zeta=0,\ & &\omega^{+,+}\vert_\xi=\omega^{+,-}\vert_\xi=0. \end{aligned}$$ We observe that the $0$-forms $\omega^{\pm,\pm}$, be they defined either on $\mathcal C_{0,0}^+$ or on $\mathcal C_{0,0,0}^+$, satisfy the same boundary conditions $ii)$, $iii)$ and $iv)$; on the other hand, the behavior of the $0$-colored propagators on the boundary strata $\mathcal C_0\times\mathcal C_{0,0}^+$ and $\mathcal C_0\times\mathcal C_{0,0,0}^+$ are quite different. This can be traced back to the proof of Proposition \[p-square\], when analyzing the shape of the morphism on the boundary stratum $\mathcal C_0\times \mathcal C_{0,0}^+$. Still, we have to be careful about these (seemingly) different boundary conditions for the $0$-colored propagators $\omega^{\pm,\pm}$: namely, the fact that the $0$-colored propagators, quite opposite to Kontsevich's angle form, can be written as a sum of a regular and of a singular term (the $0$-form living on $\mathcal C_{0,0}^+$ or $\mathcal C_{0,0,0}^+$ and on $\mathcal C_0$ respectively) produces a significant change in the application of Stokes' Theorem, which is the fundamental tool for proving the $0$-brane Formality Theorem, from which biquantization follows. Formality Theorems {#ss-0-0} ------------------ In this Subsection, we recall the $0$-brane Formality Theorem of [@CFFR], from which we will derive the biquantization techniques we apply later on. Although the main computations of this Subsection are already contained in [@CFFR], we review them in some detail because of the following reasons: first, the $0$-brane Formality Theorem has been proved using superpropagators along the same patterns of [@CF], and superpropagators are better suited for keeping track of all different colors of propagators w.r.t. the treatment in [@CT], and second, because we deserve here a more careful treatment than in [@CFFR] of the $0$-loop correction arising because of the aforementioned regular term in the $0$-colored propagators. We thus profit of the space here to correct a slight mistake in [@CFFR Subsection 0.0] (in the sense that the computations therein are correct, but a subtle point has been missed regarding the multidifferential operator associated to the $0$-loop correction, which we illustrate here in detail) and, more importantly, to correct a more serious mistake in [@CT], where the regular part of the restriction to the boundary stratum $\mathcal C_0\times \mathcal C_{0,0}^+$ or $\mathcal C_0\times \mathcal C_{0,0,0}^+$ is missing completely. The correction term arising from the presence of the regular part is responsible for a quantum shift, which will be illustrated explicitly in Section \[s-0\], which is predicted by representation-theoretic arguments and was otherwise absent. We also prove a version of [@K Lemmata 0.0.0.0, 0.0.0.0] for the $0$-colored propagators: such vanishing lemmata are central in some computations in [@CT] regarding the Harish-Chandra homomorphism. The main idea of the proof is, once again, Stokes' Theorem, but of course here we have to be a bit more careful and slightly change the final argument. ### Admissible graphs {#sss-0-0-0} Before entering into the technicalities of the $0$-brane and $0$-brane Formality Theorems, we need to spend some words on admissible graphs. For a pair of non-negative integers $(n,m)$, such that $0n+m-0\geq 0$, we consider the set $\mathcal G_{n,m}$ of admissible graphs of type $(n,m)$: the integer $n$, resp. $m$, refers to the number of vertices of the first, resp. second type, [*i.e.*]{} vertices in $\mathbb H^+$, resp. on $\mathbb R$. An admissible graph $\Gamma$ of type $(n,m)$ in the framework of the $0$-brane Formality Theorem [@CF; @CFFR] is an oriented graph, which may admit double edges, [*i.e.*]{} given any two vertices $(v_0,v_0)$, there can be more than one edge connecting $v_0$ to $v_0$, and edges departing from vertices of the second type; it does not possess short loops, [*i.e.*]{} there can no edge in $\Gamma$ with coincident initial and final point. The presence of multiple edges and edges departing from $\mathbb R$ is in opposition to the definition of admissible graphs of type $(n,m)$ as in [@K]. Further, for a triple of non-negative integers $(n,k,l)$, such that $0n+k+l-0\geq 0$, we consider the set $\mathcal G_{n,k,l}$ of admissible graphs of type $(n,k,l)$, where $n$ is the number of vertices of the first type ([*i.e.*]{} in $Q^{+,+}$), $k$, resp. $l$, is the number of vertices of the first type on $i\mathbb R^+$, resp. $\mathbb R^+$. A general element $\Gamma$ of $\mathcal G_{n,k,l}$ is an oriented graphs with $n$, resp. $k+l$, vertices of the first, resp. second type, which may admit multiple edges, edges departing from $i\mathbb R^+\sqcup\{0\}\sqcup\mathbb R^+$ and even short loops. We observe that we may also equivalently consider, for $m=k+l+0$, the set $\mathcal G_{n,m}$ of admissible graphs of type $(n,m)$, consisting of oriented graphs with $n$, resp. $m$, vertices of the first, resp. second type ([*i.e.*]{} lying in $\mathbb H^+$ and on $\mathbb R$ respectively), such that one vertex of the first type is marked and which admit multiple edges, edges departing from $\mathbb R$ and short loops: the notation is abused, but it will be clear from the context if we allow elements of $\mathcal G_{n,m}$ to possess or not short loops, which is the only additional feature that the admissible graphs for the $0$-brane Formality Theorem admit w.r.t. the ones in the $0$-brane Formality Theorem. The algebraic counterpart of the geometric results of Subsubsection \[sss-0-0-0\] is the fact that we may freely pass from $\mathcal G_{n,m}$ to $\mathcal G_{n,k,l}$, for $m=k+l+0$, by noting that the vertex labeled by $k+0$ on $\mathbb R$ corresponds to the origin $\{0\}$. ### Superpropagators {#sss-0-0-0} We now pick an admissible graph $\Gamma$ of type $(n,m)$ for the $0$-brane Formality Theorem of [@CF]. As $\Gamma$ is of type $(n,m)$, its vertices correspond to a point of $\mathcal C_{n,m}^+$, and an edge $e$ determines a natural projection $\pi_e:\mathcal C_{n,m}^+\to \mathcal C_{0,0}^+$. If we pick an admissible graph $\Gamma$ of $\mathcal G_{n,m}$ in the framework of the $0$-brane Formality Theorem, then the vertices of $\Gamma$ still define a configuration of points in $\mathcal G_{n,m}$. An edge $e$ defines, as in the previous situation, either a natural projection $\pi_e:\mathcal C_{n,m}^+\to\mathcal C_{0,0}^+$, if $e=(v_e^i,v_e^f)$, $v_e^i\neq v_e^f$, or $\pi_e:\mathcal C_{n,m}^+\to\mathcal C_{0,0}^+$, if $e$ is a short loop. The point in $\mathbb R$ in either $\mathcal C_{0,0}^+$ or $\mathcal C_{0,0}^+$ is the marked point of $\mathcal C_{n,m}^+$. If, equivalently, we consider the corresponding admissible graph $\Gamma$ of type $(n,k,l)$, then an edge $e$ of $\Gamma$ determines either a projection $\pi_e:\mathcal C_{n,k,l}^+\to\mathcal C_{0,0,0}^+$ or $\pi_e:\mathcal C_{n,k,l}^+\to\mathcal C_{0,0,0}^+$. We now consider the vector space $X=\mathbb K^d$ and two linear (or affine) subspaces $U_i$, $i=0,0$, for which we assume there is a direct sum decomposition $$\label{eq-orth-split} X=(U_0\cap U_0)\overset{\perp}\oplus (U_0^\perp\cap U_0)\overset{\perp}\oplus (U_0\cap U_0^\perp)\overset{\perp}\oplus (U_0+U_0)^\perp,$$ w.r.t. a chosen inner product over $X$. Clearly, we have $$U_0=(U_0\cap U_0)\overset{\perp}\oplus (U_0\cap U_0^\perp),\ U_0=(U_0\cap U_0)\overset{\perp}\oplus (U_0^\perp\cap U_0).$$ We choose linear coordinates $\{x_i\}$ on $X$ which are adapted to the orthogonal decomposition , [*i.e.*]{} there are two non-disjoint subsets $I_i$, $i=0,0$, of $[d]$, such that $$[d]=\left(I_0\cap I_0\right)\sqcup\left(I_0\cap I_0^c\right)\sqcup\left(I_0^c\cap I_0\right)\sqcup\left(I_0^c\cap I_0^c\right),$$ w.r.t. which $\{x_i\}$ is a set of linear coordinates on $U_0\cap U_0$, $U_0\cap U_0^\perp$, $U_0^\perp\cap U_0$ or $(U_0+U_0)^\perp$, if the index $i$ belongs to $I_0\cap I_0$, $I_0\cap I_0^c$, $I_0^c\cap I_0$ or $I_0^c\cap I_0^c$ respectively. Accordingly, for $I$ either one of the previous subsets of $[d]$, and $e$ an edge of admissible graph $\Gamma$ of type $(n,m)$, we set $$\tau_e^I=\sum_{i\in I}\iota_{\mathrm d x_i}^{(v_e^i)}\partial_{x_i}^{(v_e^f)}\in\mathrm{End}\!\left(T_\mathrm{poly}(X)^{\otimes(m+n)}\right),\ T_\mathrm{poly}(X)=\mathrm S(X^*)\otimes\wedge^\bullet X,$$ and $\partial_{x_i}^{(v)}$ denotes the action of the differential operator on the copy of $T_\mathrm{poly}(X)$ sitting at the $v$-th position, and similarly for $\iota_{\mathrm d x_i}^{(v)}$. We observe that $\tau_e^I$ is well-defined and has degree $-0$ w.r.t. the natural grading on $T_\mathrm{poly}(X)$. We now set $$\begin{aligned} A&=\mathrm S(U_0^*)\otimes \wedge (X/U_0)=\mathrm S(U_0^*)\otimes \wedge (U_0^\perp\cap U_0)\otimes \wedge (U_0+U_0)^\perp,\\ B&=\mathrm S(U_0^*)\otimes \wedge (X/U_0)=\mathrm S(U_0^*)\otimes \wedge (U_0\cap U_0^\perp)\otimes \wedge (U_0+U_0)^\perp,\\ K&=\mathrm S((U_0\cap U_0)^*)\otimes \wedge (U_0+U_0)^\perp.\end{aligned}$$ It is clear that $A$ and $B$ both admit a (trivial) structure of $A_\infty$-algebra, and $K$ is naturally an $A$-$B$-bimodule. With respect to the previously introduced notation, the relevant superpropagators are then given by $$\begin{aligned} \label{eq-A-form}\omega^A_e&=\pi_e^*(\omega^+)\otimes\left(\tau^{I_0\cap I_0}_e+\tau^{I_0\cap I_0^c}_e\right)+\pi_e^*(\omega^-)\otimes\left(\tau^{I_0^c\cap I_0}_e+\tau^{I_0^c\cap I_0^c}_e\right),\\ \label{eq-B-form}\omega^B_e&=\pi_e^*(\omega^+)\otimes\left(\tau^{I_0\cap I_0}_e+\tau^{I_0^c\cap I_0}_e\right)+\pi_e^*(\omega^-)\otimes\left(\tau^{I_0\cap I_0^c}_e+\tau^{I_0^c\cap I_0^c}_e\right),\\ \label{eq-K-form}\omega^K_e&=\pi_e^*(\omega^{+,+})\otimes \tau^{I_0\cap I_0}_e+\pi_e^*(\omega^{+,-})\otimes \tau^{I_0\cap I_0^c}_e+\pi_e^*(\omega^{-,+})\otimes \tau^{I_0^c\cap I_0}_e+\pi_e^*(\omega^{-,-})\otimes \tau^{I_0^c\cap I_0^c}_e,\end{aligned}$$ for an edge $e=(v_e^i,v_e^f)$, $v_e^i\neq v_e^f$, of an admissible graph $\Gamma$ of type $(n,m)$. We observe that the superpropagators , and are closed $0$-forms on $\mathcal C_{0,0}^+$ and $\mathcal C_{0,0}^+$ with values in $\mathrm{End}\!\left(T_\mathrm{poly}(X)^{\otimes(m+n)}\right)$ (of course, $A$, $B$ and $K$ may be viewed as subalgebras of $T_\mathrm{poly}(X)$). Equivalently, we may regard the superpropagator as a closed $0$-form on $\mathcal C_{0,0,0}^+$ with values in $\mathrm{End}\!\left(T_\mathrm{poly}(X)^{\otimes(m+n)}\right)$. Lemma \[l-prop-K\], Subsubsection \[sss-0-0-0\], implies the following useful boundary conditions for the superpropagators and : - their restrictions to the boundary stratum $\mathcal C_{0,0}^+\times\mathcal C_{0,0}^+$ equal $$\omega_e^A\vert_{\mathcal C_{0,0}^+\times\mathcal C_{0,0}^+}=\omega_e^B\vert_{\mathcal C_{0,0}^+\times\mathcal C_{0,0}^+}=\mathrm d\varphi\otimes\tau_e^{[d]};$$ - their restrictions to the boundary stratum $\mathcal C_{0,0}^+\times\mathcal C_{0,0}^+$ corresponding to the approach of the first, resp. second, argument to $\mathbb R$ equal $$\begin{aligned} \omega_e^A\vert_{\mathcal C_{0,0}^+\times\mathcal C_{0,0}^+}&=\pi_e^*(\omega^-)\otimes\left(\tau^{I_0^c\cap I_0}_e+\tau^{I_0^c\cap I_0^c}_e\right),\ &\text{resp.}\quad \omega_e^A\vert_{\mathcal C_{0,0}^+\times\mathcal C_{0,0}^+}&=\pi_e^*(\omega^+)\otimes\left(\tau^{I_0\cap I_0}_e+\tau^{I_0\cap I_0^c}_e\right),\\ \omega_e^B\vert_{\mathcal C_{0,0}^+\times\mathcal C_{0,0}^+}&=\pi_e^*(\omega^-)\otimes\left(\tau^{I_0\cap I_0^c}_e+\tau^{I_0^c\cap I_0^c}_e\right),\ &\text{resp.}\quad \omega_e^A\vert_{\mathcal C_{0,0}^+\times\mathcal C_{0,0}^+}&=\pi_e^*(\omega^+)\otimes\left(\tau^{I_0\cap I_0}_e+\tau^{I_0^c\cap I_0}_e\right). \end{aligned}$$ In particular, we see why admissible graphs appearing in the $0$-brane Formality Theorem may admit edges departing from $\mathbb R$, see for more details [@CFb; @CF]. We now concentrate on the boundary conditions for the superpropagator on $\mathcal C_{0,0}^+$: Lemma \[l-CF\] yields - the restriction of the superpropagator to the boundary stratum $\alpha$ of $\mathcal C_{0,0}^+$ equals $$\omega_e^K\vert_\alpha=\mathrm d\varphi\otimes \tau_e^{[d]}-\mathrm d\eta\otimes\left(\tau_e^{I_0\cap I_0^c}+\tau_e^{I_0^c\cap I_0}\right);$$ - the restriction of the superpropagator to the boundary strata $\beta$ and $\gamma$ equals $$\omega_e^K\vert_\beta=\omega_e^A,\ \omega_e^K\vert_\gamma=\omega_e^B;$$ - the restriction of the superpropagator to the boundary strata $\delta$ and $\varepsilon$ equals $$\omega_e^K\vert_\delta=\pi_e^*(\omega^{-,-})\otimes \tau_e^{I_0^c\cap I_0^c},\ \omega_e^K\vert_\varepsilon=\pi_e^*(\omega^{+,+})\otimes \tau_e^{I_0\cap I_0};$$ - the restriction of the superpropagator to the boundary strata $\\eta$, $\theta$, $\zeta$ and $\xi$ equals $$\begin{aligned} \omega_e^K\vert_\eta&=\pi_e^*(\omega^{+,+})\otimes \tau_e^{I_0\cap I_0^c}+\pi_e^*(\omega^{-,+})\otimes \tau_e^{I_0^c\cap I_0},\ &\ \omega_e^K\vert_\theta=\pi_e^*(\omega^{+,-})\otimes \tau_e^{I_0\cap I_0^c}+\pi_e^*(\omega^{-,-})\otimes \tau_e^{I_0^c\cap I_0^c},\\ \omega_e^K\vert_\zeta&=\pi_e^*(\omega^{+,+})\otimes \tau_e^{I_0\cap I_0}+\pi_e^*(\omega^{+,-})\otimes \tau_e^{I_0\cap I_0^c},\ &\ \omega_e^K\vert_\xi=\pi_e^*(\omega^{-,+})\otimes \tau_e^{I_0^c\cap I_0}+\pi_e^*(\omega^{-,-})\otimes \tau_e^{I_0^c\cap I_0^c}. \end{aligned}$$ If we choose the superpropagator on $\mathcal C_{0,0,0}^+$, it satisfies the same boundary conditions, with the exception of the first one, which takes the form $$\omega_e^K\vert_\alpha=\mathrm d\varphi\otimes \tau_e^{[d]}+\mathrm d\eta\otimes\left(\tau_e^{I_0\cap I_0}+\tau_e^{I_0^c\cap I_0^c}-\tau_e^{I_0\cap I_0^c}-\tau_e^{I_0^c\cap I_0}\right).$$ For the sake of simplicity, we write $\tau_e^+=\tau_e^{I_0\cap I_0}+\tau_e^{I_0^c\cap I_0^c}$ and $\tau_e^-=\tau_e^{I_0^c\cap I_0}+\tau_e^{I_0\cap I_0^c}$. We observe that the boundary conditions of type $iii)$ and $iv)$ explain why the admissible graphs appearing in the $0$-brane Formality Theorem admit edges departing from $\mathbb R$; when considering such admissible graphs in $\mathbb Q^{+,+}\sqcup i\mathbb R^+\sqcup\mathbb R^+\sqcup\{0\}$, we observe that the boundary conditions $iii)$ imply that such graphs admit edges departing from or arriving at the origin. We now deal with the so-called superloop propagator: its origin will be explained carefully in the proof of the $0$-brane Formality Theorem, which will come later on. For the time being, we content ourselves by noting that the superloop propagator appear only first in the $0$-brane Formality Theorem as a consequence of the boundary condition $i)$ satisfied by the superpropagator , more precisely it arises because of the "regular term" containing the form $\mathrm d\eta$. With the same notation as before, the superloop propagator associated to a short loop $e$ of an admissible graph $\Gamma$ of type $(n,m)$ is defined as the closed $0$-form on $\mathcal C_{0,0}^+$ with values in $\mathrm{End}\!\left(T_\mathrm{poly}(X)^{(m+n)}\right)$ $$\omega_e^K=\frac{0}0\pi_e^*(\mathrm d\eta)\otimes (\mathrm{div}_{(v)}^+-\mathrm{div}^-_{(v)}),\ e=(v,v),$$ where $$\mathrm{div}^+_{(v)}=\sum_{k\in (I_0\cap I_0)\sqcup (I_0^c\cap I_0^c)}\iota_{\mathrm d x_k}^{(v)}\partial_{x_k}^{(v)},\quad \mathrm{div}^-_{(v)}=\sum_{k\in (I_0^c\cap I_0)\sqcup (I_0\cap I_0^c)}\iota_{\mathrm d x_k}^{(v)}\partial_{x_k}^{(v)}.$$ We observe that the superloop propagator is exact: this fact will be used in all subsequent computations. Notice that the superloop propagator on $\mathcal C_{0,0,0}^+$ is defined by the same formula without the rescaling by $0/0$ (because of the morphism from $\mathcal C_{0,0}^+$ to $\mathcal C_{0,0,0}^+$). ### The formality morphisms {#sss-0-0-0} We consider $X$, $U_0$ and $U_0$ as before, to which we associate the graded vector spaces $A$, $B$ and $K$. Using the superpropagators , and , and keeping in mind the notation in the previous Subsubsections, we set $$\begin{aligned} \label{eq-c-A}\mathcal O^A_\Gamma(\gamma_0|\cdots|\gamma_n|a_0|\cdots|a_m)&=\mu_{n+m}^B\left(\int_{\mathcal C_{n,m}^+}\prod_{e\in E(\Gamma)}\omega^A_e(\gamma_0|\cdots|\gamma_n|a_0|\cdots|a_m)\right),\\ \label{eq-c-B}\mathcal O^B_\Gamma(\gamma_0|\cdots|\gamma_n|a_0|\cdots|a_m)&=\mu_{n+m}^B\left(\int_{\mathcal C_{n,m}^+}\prod_{e\in E(\Gamma)}\omega^B_e(\gamma_0|\cdots|\gamma_n|b_0|\cdots|b_m)\right),\\ \label{eq-c-K}\mathcal O_\Gamma^K(\gamma_0|\cdots|\gamma_n|a_0|\cdots|a_k|k|b_0|\cdots|b_l)&=\mu_{m+n}^K\left(\int_{\mathcal C_{n,m}^+}\prod_{e\in E(\Gamma)}\omega^K_e(\gamma_0|\cdots|\gamma_n|a_0|\cdots|a_k|k|b_0|\cdots|b_l)\right),\end{aligned}$$ where $\gamma_i$, $i=0,\dots,n$, are elements of $T_\mathrm{poly}(X)$, $a_i$ and $b_i$ are elements of $A$ and $B$ respectively, $k$ is an element of $K$; $E(\Gamma)$ is the set of edges of an admissible graph $\Gamma$ of type $(n,m)$; $\mu^A$, $\mu^B$ and $\mu^K$ denotes the multiplication map on $T_\mathrm{poly}(X)$, followed by the projection onto $A$, $B$ and $K$ respectively. Since $\Gamma$ may have multiple edges, there is a combinatorial subtlety to be taken into account: in all previous formulæ, whenever there are multiple edges between two vertices $(v^i,v_f)$, for $v^i\neq v_f$, we must divide by the factorial of the number of such edges. We observe that short loops cannot be multiple edges, as the superpropagator for a short loop squares obviously to $0$. We also observe that the product on formulæ , and are well-defined and do not depend on the order of the factors: namely, the total degree of any superpropagator appearing in these formulæ is $0$, as the $0$-form piece has (form) degree $0$, while the multidifferential operator piece has degree $-0$. Using the multidifferential operators defined in , and , we set $$\begin{aligned} \label{eq-A-mor}\mathcal U_A^n(\gamma_0|\cdots|\gamma_n)(a_0|\dots|a_m)&=(-0)^{\left(\sum_{i=0}^n|\gamma_i|-0\right)m}\sum_{\Gamma\in\mathcal{G}_{n,m}}\mathcal{O}_{\Gamma}^A(\gamma_0|\cdots|\gamma_n|a_0|\cdots|a_m),\\ \label{eq-B-mor}\mathcal U_B^n(\gamma_0|\cdots|\gamma_n)(a_0|\dots|a_m)&=(-0)^{\left(\sum_{i=0}^n|\gamma_i|-0\right)m}\sum_{\Gamma\in\mathcal{G}_{n,m}}\mathcal{O}_{\Gamma}^B(\gamma_0|\cdots|\gamma_n|b_0|\cdots|b_m),\\ \label{eq-K-mor}\mathcal U_K^n(\gamma_0|\cdots|\gamma_n)(a_0|\cdots|a_k|k|b_0|\cdots|b_l)&=(-0)^{\left(\sum_{i=0}^n|\gamma_i|-0\right)m}\sum_{\Gamma\in\mathcal{G}_{n,m}}\mathcal{O}_{\Gamma}^K(\gamma_0|\cdots|\gamma_n|a_0|\cdots|a_k|k|b_0|\cdots|b_l),\\ \label{eq-A_inf-bimod}\mathrm d_K^{k,l}(a_0|\cdots|a_k|k|b_0|\cdots|b_l)&=\sum_{\Gamma\in\mathcal G_{0,m}}\mathcal O_\Gamma^K(a_0|\cdots|a_k|k|b_0|\cdots|b_l),\end{aligned}$$ with the above notation. Some observations are necessary here. The morphisms and and are multilinear maps from $T_\mathrm{poly}(X)$ to the multidifferential operators on $A$ and $B$ respectively; the morphisms and are multilinear maps from $T_\mathrm{poly}(X)$ to the multidifferential operators from $A^{\otimes k}\otimes K\otimes B^{\otimes l}$ to $K$. All multidifferential operators appearing in the previous formulæ are non-trivial only if the number of edges of the admissible graphs of type $(n,m)$ equals $0n+m-0$: since to each edge of an admissible graph is associated a contraction operator (which lowers degrees by $0$), it follows immediately that the morphisms \[eq-A-mor\], , have degree $0-n$, and that the morphism has degree $0-m$. We refer to [@CFFR Section 0] for a short introduction to $A_\infty$-categories in the present framework (see [@Kel; @Lef-Has] for more details on $A_\infty$-categories and related issues), which is needed for the statement of the main theorem ($0+0$-brane Formality Theorem) of the present Section. We only recall that $T_\mathrm{poly}(X)$ has a structure of dg (short for differential graded) Lie algebra with trivial differential and Schouten-Nijenhuis bracket (extending the natural Lie bracket on polynomial vector fields on $X$); similarly, the Hochschild cochain complex of an $A_\infty$-category $\mathcal A$ (roughly, an abelian category, whose spaces spaces of morphisms admit the structure of $A_\infty$-algebras and $A_\infty$-bimodules) is also a dg Lie algebra with Hochschild differential (the $A_\infty$-structure itself) and Gerstenhaber bracket (which is well-defined an any sort of Hochschild cochain complex). \[t-form\] We may regard $A$, $B$ as $A_\infty$-algebras, whose only non-trivial Taylor component is given by the corresponding natural (graded) commutative products: then, the morphisms fit into the Taylor components of a non-trivial $A_\infty$ $A$-$B$-bimodule structure over $K$, which restricts to the natural $A$ left- and $B$-right module structures on $K$. Furthermore, the morphisms , and fit into the Taylor components of an $L_\infty$-morphism $\mathcal U$ from $T_\mathrm{poly}(X)$ to the (completed) Hochschild cochain complex of the $A_\infty$-category $\mathcal A$ with two objects $U_i$, $=0,0$, and spaces of morphisms given by $$\mathrm{Hom}_{\mathcal A}(U_0,U_0)=A,\ \mathrm{Hom}_{\mathcal A}(U_0,U_0)=B,\ \mathrm{Hom}_{\mathcal A}(U_0,U_0)=K,\ \mathrm{Hom}_{\mathcal A}(U_0,U_0)=\{0\},$$ with the respective $A_\infty$-structures. Finally, the $L_\infty$-morphism $\mathcal U$ extends to an $L_\infty$-quasi-isomorphism by suitably completing the graded vector spaces $A$, $B$, $K$. The first claim has been proved in detail in [@CFFR Proposition 0.0], to which we refer. The second claim splits into three claims, namely $\mathcal U$ consists of three morphisms $\mathcal U_A$, $\mathcal U_B$ and $\mathcal U_K$, where $\mathcal U_A$, resp. $\mathcal U_B$, is a pre-$L_\infty$-morphism from $T_\mathrm{poly}(A)$, resp. $T_\mathrm{poly}(B)$, to the (completed) Hochschild cochain complex of $A$, resp. $B$, and $\mathcal U_K$ is a collection of maps from $T_\mathrm{poly}(X)$ to the mixed component $\mathrm C^\bullet(A,B,K)$ of the (completed) Hochschild cochain complex of the above $A_\infty$-category $\mathcal A$. Here, we have used the (non-canonical) identification of dg Lie algebras $T_\mathrm{poly}(X)=T_\mathrm{poly}(A)=T_\mathrm{poly}(B)$. The fact that $\mathcal U_A$ and $\mathcal U_B$ are $L_\infty$-morphisms has been proved in detail in [@CF]; they extend to $L_\infty$-quasi-isomorphisms by suitably completing $A$ and $B$. The fact that the morphism $\mathcal U_K$ satisfies the required $L_\infty$-identities has been proved in detail in [@CFFR Theorem 0.0]: we profit nonetheless for discussing an incorrect issue in the proof regarding the superloop propagator. The superloop propagator, which has been defined above, is manifestly different from the one considered in [@CFFR Subsection 0.0]: the point is that the actual superloop propagator is the correct one. We may repeat the proof of [@CFFR Theorem 0.0] [*verbatim*]{} until the discussion of boundary strata of codimension $0$ of the form $\mathcal C_A\times C_{([n]\smallsetminus A)\sqcup\{\bullet\},m}^+$, where $|A|=0$: the following discussion on how the corresponding integral contribution looks like is precisely the same, [*i.e.*]{} the only situation that matter arise when there are at least one and at most two edges connecting the two vertices labeled by $A$, [*i.e.*]{} pictorially \ \ We are interested only in the contributions from the last three subgraphs (which we denote collectively by $\Gamma_A$). Taking into account the fact that the second graph $\Gamma_A$ has $0$ multiple edges (thus recalling the normalization factor $0/0$), its contribution equals $$\int_{\mathcal C_0}\omega_{\Gamma_A}^K=-\pi_e^*(\mathrm d\eta)\otimes \tau^{[d]}_e\tau^-_e=\frac{0}0\pi_e^*(\mathrm d\eta)\otimes \tau^{[d]}_e(\tau^+_e-\tau_e^-),$$ where $\pi_e$ is here the projection with respect to the "phantom" short loop arising from the contraction of the vertices of the subgraph $\Gamma_A$. The novelty with respect to the corresponding computations in the proof of [@CFFR Theorem 0.0] lies in the re-writing of the second term in the previous chain of equalities; of course, we have used the obvious fact that $(\tau_e^{[d]})^0=0$. The fourth graph in Figure 0 yields a similar contribution. The third graph, on the other hand, yields the contribution $$\int_{\mathcal C_0}\omega_{\Gamma_A}^K=-\pi_{e}^*(\mathrm d\eta)\otimes\tau^{[d]}_{e_0}\tau^-_{e_0}-\pi_{e}^*(\mathrm d\eta)\otimes\tau^{[d]}_{e_0}\tau^-_{e_0}=\frac{0}0\pi_{e}^*(\mathrm d\eta)\otimes\tau^{[d]}_{e_0}(\tau^+_{e_0}-\tau_{e_0}^-)+\frac{0}0\pi_{e}^*(\mathrm d\eta)\otimes\tau^{[d]}_{e_0}(\tau^+_{e_0}-\tau_{e_0}^-),$$ where $e_0=(i,j)$, $e_0=(j,i)$, and $e$ is (improperly) the "phantom" short loop arising from the contraction of the two vertices $i$, $j$. Here, we have used the obvious fact that $\tau_{e_0}^{[d]}\tau_{e_0}^{[d]}=-\tau_{e_0}^{[d]}\tau_{e_0}^{[d]}$. The factor $0/0$ before the function $\eta$ on $\mathcal C_{0,0}^+$ (which we have tacitly omitted) is compatible with the fact that the pull-back of $\eta$ from $\mathcal C_{0,0,0}^+$ to $\mathcal C_{0,0}^+$ is precisely the rescaled function $\eta$ on $\mathcal C_{0,0}^+$. Therefore, the same arguments as in the corresponding part of the proof of [@CFFR Theorem 0.0] show that the right compensation for the contributions coming from the last three graphs in Figure 0 is given precisely by the superloop propagator $\omega^K_e$, which differ from the superloop propagator chosen in the proof of [@CFFR Theorem 0.0] in its multidifferential operator part: the trick is to prove that we may rewrite the multidifferential operator parts of the contributions coming from the last three graphs in Figure 0 using the difference $\tau_e^+-\tau_e^-$, which is exactly the term appearing if we do the computations using the compactified configuration spaces $\mathcal C_{n,k,l}^+$ instead of $\mathcal C_{n,m}^+$. ### Biquantization as a consequence of Theorem \[t-form\] {#sss-0-0-0} We consider now the particular situation $X=\mathfrak g^*$, for $\mathfrak g$ a finite-dimensional Lie algebra over $\mathbb K$, and for a given Lie subalgebra $\mathfrak h$ thereof, we set $U_0=X$ and $U_0=\mathfrak h^\perp$, the annihilator of $\mathfrak h$ in $\mathfrak g$. We observe that, later on, we will consider $U_0$ to be the affine space $\lambda+\mathfrak h^\perp$, where $\lambda$ is a character of $\mathfrak h$: the results of the previous Subsubsection still hold true in this situation. For the sake of explicit computations, we choose a complementary subspace of $\mathfrak h$ in $\mathfrak g$, [*i.e.*]{} we choose a subspace $\mathfrak p$ of $\mathfrak g$, such that $\mathfrak g=\mathfrak h\oplus\mathfrak p$. We observe that, in general, $\mathfrak p$ is not $\mathfrak h$-invariant with respect to the restriction of the adjoint representation. Still, in the case of symmetric pairs $(\mathfrak k,\mathfrak p)$, $\mathfrak p$ is a $\mathfrak k$-module. We thus apply [@CFFR Theorem 0.0] to this situation (we only observe that, in this framework, we do not consider completed algebras, as in [@CFFR]: still, Theorem 0.0 holds true, the only difference is that we have to drop the property of the $L_\infty$-morphism to be an $L_\infty$-quasi-isomorphism): we may view the Poisson structure on $X$ as a Maurer-Cartan (shortly, form now on, MC) element of $T_\mathrm{poly}(X)$, and its image with respect to the $L_\infty$-morphism from [@CFFR Theorem 0.0] is a MC element in the Hochschild cochain complex (with mixed component completed) of the $A_\infty$-category $\mathrm{Cat}_\infty(A,B,K)$, with objects $U_i$, $i=0,0$. Using the previous prescriptions, we have $$A=\mathrm S(\mathfrak g),\ B=\mathrm S(\mathfrak p)\otimes\wedge \mathfrak h^*,\ K=\mathrm S(\mathfrak p);$$ a bit improperly, we sometimes write $\mathfrak p=\mathfrak g/\mathfrak h$ (as it is an identification only of vector spaces, obviously not of $\mathfrak h$-modules). Since $\mathfrak g$ is a Lie algebra, $X=\mathfrak g^*$ is a Poisson manifold with linear Poisson bivector $\pi$, and $U_0$ and $U_0$ are coisotropic submanifolds thereof. The linear Poisson structure on $X$ determines a Maurer-Cartan element of $T_\mathrm{poly}(X)$: for a choice of a formal parameter $\hbar$, the image of $\hbar\pi$ with respect to the $L_\infty$-morphism $\mathfrak U$ from Theorem \[t-form\] is a Maurer-Cartan element $\mathcal U(\hbar\pi)$ in the (completed) Hochschild cochain complex of the $A_\infty$-category $\mathcal A$, which is a concept needing some unraveling. The Hochschild cochain complex of $\mathcal A$ splits into three terms, namely the Hochschild cochain complex of $A$, the one of $B$ and a graded vector space which contains $A$, $B$ and $K$: general elements of the mixed term $\mathrm C^\bullet(A,B,K)$ are multilinear maps from $A^{\otimes k}\otimes K\otimes B^{\otimes l}$ to $K$. From the general theory of Hochschild cochain complexes it is known that Maurer-Cartan elements of Hochschild cochain complexes correspond to $A_\infty$-structures on the underlying graded vector spaces: in our situation, a Maurer-Cartan element is precisely a structure of $A_\infty$-algebra on both $A$ and $B$, and a corresponding structure of $A_\infty$-$A$-$B$-bimodule on $K$, or, equivalently, to an $A_\infty$-structure on the category $\mathcal A$. Now we consider $A_\hbar=A[\![\hbar]\!]$, and similarly for $B_\hbar$ and $K_\hbar$: it is clear that the structure of $A_\infty$-category on $\mathcal A$ extends to $\hbar$-linearly to $\mathcal A_\hbar$, whose objects are the same objects of $\mathcal A$, but whose morphism spaces are replaced by $A_\hbar$, $B_\hbar$ and $K_\hbar$ endowed with the $\hbar$-linearly extended $A_\infty$-structure $\mu$. We have a natural Hochschild differential $\mathrm d_\mathrm H$ on the Hochschild cochain complex of $\mathcal A_\hbar$, given by the adjoint representation of $\mu$ with respect to the Gerstenhaber bracket. The element $\mathcal U(\hbar)$ satisfies the Maurer-Cartan equation $$\mathrm d_\mathrm H\mathcal U(\hbar\pi)+\frac{0}0\left[\mathcal U(\hbar\pi),\mathcal U(\hbar\pi)\right]=\frac{0}0\left[\mu+\mathcal U(\hbar\pi),\mu+\mathcal U(\hbar\pi)\right]=0,$$ [*i.e.*]{} $\mu+\mathcal U(\hbar\pi)$ is a Maurer-Cartan element for $\mathcal A_\hbar$, which deforms (with respect to the formal parameter $\hbar$) the "classical" $A_\infty$-structure on $\mathcal A$. Since $A_\hbar$ is concentrated in degree $0$ by construction, and $\mu_A$ (the component of $\mu$ in the Hochschild cochain complex of $A$) is the obvious $\hbar$-linear commutative, associative product on $A_\hbar$, then $\mu_A+\mathcal U_A(\hbar\pi)$ is an associative product $\star_{A_\hbar}$ on $A_\hbar$, which deforms non-trivially $\mu_A$: $(A_\hbar,\star_{A_\hbar})$ is a deformation quantization of $(A,\mu_A)$ in the sense of [@K]. The image of $\pi$ in $T_\mathrm{poly}(A)$ with respect to the dg Lie algebra isomorphism $T_\mathrm{poly}(X)\cong T_\mathrm{poly}(A)$ (depending on a choice of $\mathfrak p$) is a Maurer-Cartan element in $T_\mathrm{poly}(A)$, which is a sum of a three polyvector fields, $\pi_0$, $\pi_0$ and $\pi_0$, where $\pi_i$ is an $i$-th polyvector field of polynomial degree $0-i$. We observe that $A=\mathrm S(\mathfrak g)$ and $B=\mathrm S(\mathfrak p\oplus\mathfrak h^*[-0])$, thus it makes sense to speak about polynomial degree for elements of $A$ and $B$; $[\bullet]$ is the degree-shifting functor (hence, the polynomial grading of $B$ does not coincide with the internal grading coming from the functor $[-0]$), [*e.g.*]{} the internal degree of $\pi_i$ is $0$, $i=0,0,0$. As a Maurer-Cartan element of $T_\mathrm{poly}(A)$, $\pi$ defines a $P_\infty$-structure on $B$, in other words, $\pi$ defines a Poisson algebra structure on $B$ up to homotopy: for example, $\pi_0$ is a homological vector field over $B$, whose cohomology identifies with the Chevalley-Eilenberg cohomology of the $\mathfrak h$-module $\mathrm S(\mathfrak g/\mathfrak h)$, which in turn inherits from $\pi_0$ (which is a bivector field of internal degree $0$) a structure of graded Poisson algebra. We notice that, in degree $0$, this corresponds to the well-known fact that Poisson reduction endows the commutative algebra $\mathrm S(\mathfrak g/\mathfrak h)^\mathfrak h$ with a Poisson structure coming from the natural one on $A=\mathrm S(\mathfrak g)$. The Maurer-Cartan element $\mu_B+\mathcal U_B(\hbar\pi)$ is an $A_\infty$-structure on $B_\hbar$, deforming the obvious $A_\infty$-structure on $B$: thus, a $P_\infty$-algebra structure on $B$ produces an $A_\infty$-structure [*via*]{} the graded version of deformation quantization, see also [@CF]. We observe that the $A_\infty$-structure $\mu_B+\mathcal U_B(\hbar\pi)$ is the sum of (possibly) infinitely many components of different internal degree: in particular, the component of internal degree $0$ is an element of $B_\hbar$ of degree $0$, the curvature of the $A_\infty$-structure. If it non-trivial, then we cannot talk about the cohomology of $A_\infty$-algebra $(B_\hbar,\mu_B+\mathcal U_B(\hbar\pi))$, and some problems may arise: luckily, in the present framework, the curvature vanishes, see [*e.g.*]{} [@CFb; @CT] and later on. We finally observe that the term of order $0$ with respect to $\hbar$ of the $A_\infty$-structure on $B_\hbar$ is precisely the ($\hbar$-shifted) $P_\infty$-structure on $B_\hbar$: thus, if we select its vector field piece, we get the $\hbar$-shifted Chevalley-Eilenberg differential on $B_\hbar$. Finally, the mixed component $\mu_K+\mathcal U_K(\hbar\pi)$ determines a deformation of the $A_\infty$-$A$-$B$-bimodule $\mu_K$ structure on $K$: we do not spend here much words, because we will deal with $\mu_K+\mathcal U_K(\hbar\pi)$ in the rest of the paper, at least in degree $0$. We only observe that, through $\mu_K$, we may re-prove classical Koszul duality between $A$ and $B$ (both are graded quadratic algebras), and its deformation quantization permits to extend the Koszul duality to the deformed case $(A_\hbar,\star_\hbar)$ and $(B_\hbar,\mu_B+\mathcal U_B(\hbar\pi))$. Biquantization as in [@CT] is the specialization to degree $0$ of the data presented above. In particular, $(A_\hbar,\star_{A_\hbar})$ is an $A_\infty$-algebra concentrated in degree $0$, hence its cohomology equals itself; the piece of $B_\hbar$ of degree $0$ equals $\mathrm S(\mathfrak p)\cong\mathrm S(\mathfrak g/\mathfrak h)$ endowed with a differential $\mathrm d_{B_\hbar}^0$ and with an associative product $\star_{B_\hbar}$ up to homotopy. Finally, $K_\hbar$ is also concentrated in degree $0$, hence its cohomology with respect to $\mathrm d_{K_\hbar}^{0,0}$ (the $(0,0)$-component of $\mu_K+\mathcal U_K(\hbar\pi)$) equals itself, hence $K_\hbar$ becomes with respect to $d_{K_\hbar}^{0,0}=\star_L$ a left $(A_\hbar,\star_{A_\hbar})$- and with respect to $\mathrm d_{K_\hbar}^{0,0}=\star_R$ a right $(\mathrm H^0(B_\hbar),\star_{B_\hbar})$-module (the latter also because of the vanishing of the curvature of the $A_\infty$-structure $\mu_B+\mathcal U(\hbar\pi)$). Later on, still in the framework of finite-dimensional Lie algebras and Lie subalgebras thereof, we will consider the more general framework, where both $A_\hbar$ and $B_\hbar$ are $A_\infty$-algebras with no curvature, and $K_\hbar$ is a graded $A_\infty$-$A_\hbar$-$B_\hbar$-bimodule, hence the $0$-th cohomologies $\mathrm H^0(A_\hbar)$, $\mathrm H^0(B_\hbar)$ become associative algebras and $\mathrm H^0(K_\hbar)$ is an $\mathrm H^0(A_\hbar)$-$\mathrm H^0(B_\hbar)$-bimodule. ### Symmetries of the $0$-colored propagators {#sss-0-0-0} For later purposes, we now exhibit certain symmetries of the $0$-colored and $0$-colored propagators, which we now discuss in some detail. The complex upper half-plane $\mathbb H^+$ has two obvious symmetries, namely the reflection with respect to the imaginary axis $i\mathbb R$, given by $z\overset{\sigma}\mapsto -\overline z$, and the inversion with respect to the unit half-circle, given by $z\overset{\tau}\mapsto 0/\overline z$: both maps extend to $\mathbb H^+\sqcup \mathbb R$, and they define two orientation-reversing involutions $\sigma$ and $\tau$ of it. Equivalently, $\mathcal Q^{+,+}\sqcup i\mathbb R^+\sqcup\{0\}\sqcup\mathbb R^+$ admits two orientation-reversing involutions $\sigma$ and $\tau$, where $z\overset{\sigma}\mapsto i\overline{z}$ and $z\overset{\tau}\mapsto \frac{0}{\overline z}$. It is not difficult to prove that $\sigma$ and $\tau$ descend both to involutions of $C_{n,m}^+$ and $C_{n,k,l}^+$, and that, using the same techniques as in the proof of Proposition \[p-square\], Subsubsection \[sss-0-0-0\], $\sigma$ and $\tau$ extend to involutions of the compactified configuration spaces $\mathcal C_{n,m}^+$ and $\mathcal C_{n,k,l}^+$. We observe that $\sigma$ and $\tau$ are orientation-preserving, resp. -reversing, if and only if $n+m-0$ is even, resp. odd. We then have the following technical Lemma about the behavior of the $0$-colored and $0$-colored propagators with respect to the action of $\sigma$ and $\tau$. \[l-symm-0\] The $0$-colored and $0$-colored propagators behave as follows with respect to the involutions $\sigma$ and $\tau$ on the respective compactified configuration spaces $\mathcal C_{0,0}^+$ and $\mathcal C_{0,0}^+$: $$\begin{aligned} \sigma^*(\omega^+)&=-\omega^+,\ & \sigma^*(\omega^-)&=-\omega^-,\ & \tau^*(\omega^+)&=-\omega^++0\pi_0^*(\mathrm d\eta),\ & \tau^*(\omega^-)&=-\omega^-+0\pi_0^*(\mathrm d\eta),\\ \sigma^*(\omega^{+,+})&=-\omega^{+,+}, & \sigma^*(\omega^{+,-})&=-\omega^{-,+}, & \sigma^*(\omega^{-,+})&=-\omega^{+,-}, & \sigma^*(\omega^{-,-})&=-\omega^{-,-},\\ \tau^*(\omega^{+,+})&=-\omega^{+,+}+0 \pi_0^*(\mathrm d\eta), & \tau^*(\omega^{+,-})&=-\omega^{+,-}, & \tau^*(\omega^{-,+})&=-\omega^{-,+}, & \tau^*(\omega^{-,-})&=-\omega^{-,-}+0 \pi_0^*(\mathrm d\eta), \end{aligned}$$ where now $\pi_i$, $i=0,0$, denotes the two natural projections from $\mathcal C_{0,0}^+$ onto $\mathcal C_{0,0}^+$ or from $\mathcal C_{0,0}^+$ to $\mathcal C_{0,0}^+$. Similar formulæ hold true for the $0$-colored propagators on $\mathcal C_{0,0,0}$, keeping in track a rescaling before the exact $0$-form $\eta$. ### Kontsevich's Vanishing Lemmata {#sss-0-0-0} We now need a Vanishing Lemma for the $0$-colored propagators, reminiscent of the Vanishing Lemmata in [@K Subsubsubsection 0.0.0.0]. We observe that Kontsevich's Vanishing Lemmata in [@K Subsubsubsection 0.0.0.0] are key ingredients in the proof of the globalization of its $L_\infty$-Formality-quasi-isomorphism: in this sense, the Vanishing Lemma we are going to state and prove here (the main application being for later computations regarding the generalized Harish-Chandra homomorphism) play also a central [*rôle*]{} in the globalization of the $0$-brane $L_\infty$-Formality-quasi-isomorphism of [@CFFR], but do not indulge here on this point, referring to upcoming work for more details. We consider the three natural projections $\pi_{ij}$, $i\leq i<\leq 0$, from $\mathcal C_{0,0}^+$ onto $\mathcal C_{0,0}^+$, which smoothly extend the projections $[(z_0,z_0,z_0)]\overset{\pi_{ij}}\to [(z_i,z_j)]$ to the corresponding compactified configuration spaces. The typical fiber of the projection $\pi_{ij}$ is $0$-dimensional, and it is not difficult to verify that it is a smooth manifold with corners (hence, it admits a natural stratification, whose description, at least in codimension $0$, will be made explicit later on). We improperly denote by the same symbol the natural projection $\pi_{ij}$, $i\leq i<\leq 0$, from $\mathcal C_{0,0}^+$ onto $\mathcal C_{0,0}^+$, which this times extends the projection $[(z_0,z_0,z_0,x)]\overset{\pi_{ij}}\mapsto[(z_i,z_j,x)]$: again, its fiber is a smooth $0$-dimensional manifold with corners. For any two smooth $0$-forms $\eta_i$, $i=0,0$, on $\mathcal C_{0,0}^+$ or $\mathcal C_{0,0}^+$, we define a smooth function on $\mathcal C_{0,0}^+$ or $\mathcal C_{0,0}^+$ [*via*]{} the integral $$\label{eq-int-van} \Omega(\eta_0,\eta_0)=\pi_{00,*}\!\left(\pi_{00}^*(\eta_0)\wedge\pi_{00}^*(\eta_0)\right)=\int_{z_0\in\mathcal Q^{+,+}\smallsetminus\{z_0,z_0\}}\eta_0(z_0,z_0)\wedge\eta_0(z_0,z_0),$$ where $\pi_{ij,*}$ denotes integration along the fiber of the projection $\pi_{ij}$. \[l-vanish-0\] The function $\Omega(\eta_0,\eta_0)$ vanishes, whenever $\eta_0=\eta_0$ is either one of the $0$-colored propagators or either one of the $0$-colored propagators. The claim for the $0$-colored propagators $\omega^+$ and $\omega^-$ is precisely the content of the vanishing lemmata in [@K Subsubsubsection 0.0.0.0], to which we refer for a proof. We observe that the proof below for the $0$-colored propagators applies with minor changes ([*e.g.*]{} the final argument involves the involution $\sigma$ and not $\tau$) applies to the statement for $0$-colored propagators. For symmetry reasons, it suffices to prove the claim for the $0$-colored propagators $\omega^{+,+}$ and $\omega^{+,-}$. We first prove the claim for $\eta_0=\eta_0=\omega^{+,+}$: the idea is to show first that $\Omega(\eta_0,\eta_0)$ is a constant function, and then to use Lemma \[l-symm-0\] to prove that, for a well-suited choice of arguments, $\Omega(\eta_0,\eta_0)$ equals minus itself. We therefore compute the exterior derivative of $\Omega(\eta_0,\eta_0)$: we make use of generalized Stokes' Theorem, and the fact that $\eta_0=\eta_0$ is a closed $0$-form, yields $$\mathrm d\Omega(\eta_0,\eta_0)=\pi_{00,*}^\partial(\pi_{00}^*(\eta_0)\pi_{00}^*(\eta_0)),$$ where $\pi_{00,*}^\partial$ denotes integration along the codimension $0$-boundary strata of the fiber of $\pi_{00}$: there are five such boundary strata, which correspond to $i)$ the point labeled by $z_0$ approaching either $\mathbb R$ on the left or on the right of the marked point on $\mathbb R$ or the marked point itself, or to $ii)$ the point labeled by $z_0$ approaching either the point labeled by $z_0$ or $z_0$. In the three situations in $i)$, the corresponding contribution vanishes in view of Lemma \[l-CF\], $iii)$ and $iv)$. In both situations described in $ii)$, the boundary fibration is trivial, namely $\mathcal C_0\times \mathcal C_{0,0}^+$: an easy computation in local coordinates for $\mathcal C_{0,0}^+$ near the boundary strata in $i)$ shows that there are no orientation signs appearing, and we finally get, using Lemma \[l-CF\], $i)$, $$\mathrm d\Omega(\eta_0,\eta_0)=\int_{\mathcal C_0^+}\mathrm d\varphi\ \omega^{+,+}+\int_{\mathcal C_0^+}\omega^{+,+}\mathrm d\varphi=0.$$ Therefore, $\Omega(\eta_0,\eta_0)$ is a constant function on $\mathcal C_{0,0}^+$, whose value is completely determined by a choice of a point in $\mathcal C_{0,0}^+$, and a natural choice is $(i,0i,0)$. Since $i\mathbb R$ is the fixed point set of $\sigma$, $\sigma$ preserves $\Omega(\eta_0,\eta_0)$, whence $$\Omega(\eta_0,\eta_0)=\sigma^*(\Omega(\eta_0,\eta_0)=-\pi_{00,*}(\sigma^*(\pi_{00}^*(\eta_0))\wedge \sigma^*(\pi_{00}^*(\eta_0))=-\Omega(\eta_0,\eta_0),$$ where the minus sign in the second equality arises because $s$ is orientation-reversing on the first quadrant, while the third equality is a consequence of Lemma \[l-symm-0\]. The very same arguments can be applied in the situation $\eta_0=\eta_0=\omega^{-,-}$. We consider now the case $\eta_0=\eta_0=\omega^{+,-}$. The computation of the exterior derivative of $\Omega(\eta_0,\eta_0)$ in this case is similar to the previous one: we only observe that the boundary condition for boundary strata of type $i)$ let appear a regular term $\mathrm d\eta$, whose contribution to integration is trivial. The arguments for dealing with the other strata are, once again, a consequence of Lemma \[l-CF\], $iii)$ and $iv)$. Therefore, $\Omega(\eta_0,\eta_0)$ is uniquely determined by a given point in $\mathcal C_{0,0}^+$: quite differently from the previous case, we choose a pair of points lying on the unit circle. Recalling that the unit circle is the fixed point [*locus*]{} of the involution $\tau$, we get in this case $$\Omega(\eta_0,\eta_0)=\tau^*(\Omega(\eta_0,\eta_0)=-\pi_{00,*}(\tau^*(\pi_{00}^*(\eta_0))\wedge \tau^*(\pi_{00}^*(\eta_0))=-\Omega(\eta_0,\eta_0),$$ by the very same arguments as in the previous case, because $t$ is orientation-reversing and because of Lemma \[l-symm-0\]. As an application of Lemma \[l-vanish-0\], we briefly sketch the vanishing of the curvature of the $A_\infty$-algebra $B_\hbar$, with the previously introduced notations. As already mentioned, the curvature of $B_\hbar$ is the piece of degree $0$ in $B_\hbar$ of the Maurer-Cartan element $\mu_B+\mathcal U_B(\hbar\pi)$: more explicitly, the curvature is given by the formal power series $$\mathcal U(\hbar\pi)_0=\sum_{n\geq 0}\frac{0}{n!}\mathcal U_B^n(\underset{n}{\underbrace{\hbar\pi|\cdots|\hbar\pi}})=\sum_{n\geq 0}\frac{0}{n!}\sum_{\Gamma\in\mathcal{G}_{n,0}}\mathcal{O}_{\Gamma}^B(\underset{n}{\underbrace{\hbar\pi|\cdots|\hbar\pi}})=\sum_{n\geq 0}\frac{0}{n!}\sum_{\Gamma\in\mathcal{G}_{n,0}}\mu_{n}^B\left(\int_{\mathcal C_{n,0}^+}\prod_{e\in E(\Gamma)}\omega^B_e(\underset{n}{\underbrace{\hbar\pi|\cdots|\hbar\pi}})\right).$$ For an admissible graph $\Gamma$ in $\mathcal G_{n,0}$, the rightmost integral is non-trivial only if the degree of the integrand equals $0n-0$. To each vertex of the first type of $\Gamma$ is associated a copy of the linear Poisson bivector $\hbar\pi$, hence from each vertex depart exactly two arrows: each arrow, by definition, corresponds to a derivation and a contraction. In particular, the differential operator $\mathcal O_\Gamma^B$ has degree $0n-0$: since $\hbar\pi$ is linear, the polynomial degree of the object on the rightmost part of the previous chain of equalities is $n-(0n-0)=-n+0$, whence $0\leq n\leq 0$. When $n=0$, $\mu_0^B(\hbar\pi)$ vanishes because of the coisotropy of $U_0$. For $n=0$, there is only one possible admissible graph $\Gamma$ of type $(0,0)$, namely the loop graph connecting the two vertices of the first type: the corresponding operator $\mathcal O_\Gamma^B$ vanishes because of Lemma \[l-vanish-0\]. Quantum reduction algebras {#ss-0-0} -------------------------- The present Subsection presents the results of [@CT Section 0] using a slightly different perspective, coherent with the approach to biquantization we have introduced in the previous Subsection: however, we think it useful to review many results mainly because of the notation, which will be then used extensively in the rest of the paper. Thus, we will mostly concentrate on the dg vector space $B_\hbar$, where we denote by $\mathrm d_{B_\hbar}$ its differential ([*i.e.*]{} the piece of degree $0$ of its $A_\infty$-structure): we have already observed that $\mathrm d_{B_\hbar}=\hbar\mathrm d_\mathrm{CE}+\mathcal O(\hbar^0)$, where $\mathrm d_\mathrm{CE}=\pi_0$ is the Chevalley-Eilenberg differential on $B_\hbar$. The admissible graphs $\Gamma$ appearing in $\mathrm d_{B_\hbar}$ are of type $(n,0)$, possibly with multiple edges and admitting edges departing from $\mathbb R$: each edge $e$ of an element $\Gamma$ of $\mathcal C_{n,0}$ is the sum of two colored edges, namely $e^+$ and $e^-$ according to Formula , Subsubsection \[sss-0-0-0\]. The properties of the superpropagator imply that an edge $e$ of $\Gamma$ arriving to the only vertex of the second type has color $+$, while an edge departing from it has color $-$. One of the main tools we will use throughout the present Subsection is Lemma \[l-vanish-0\], Subsubsection \[sss-0-0-0\]. Namely, we assume $v$ is a vertex of the first type of $\Gamma$ of type $(n,0)$ with two edges at it of the form $e_0=(\bullet_0,v)$ and $e_0=(v,\bullet_0)$, where $\bullet_i$, $i=0,0$, denotes some other vertex (notice that now we allow $\bullet_0=\bullet_0$): then the configuration at $v$ is either $(e_0^+,e_0^-)$ or $(e_0^-,e_0^+)$. Of course, since to each edge of the first type of $\Gamma$ is associated a copy of the linear Poisson bivector $\hbar\pi$, we assume that from $v$ as above departs a third edge, which does not join any other vertex of $\Gamma$: this edge has color $-$. This "phantom" edge is the "edge to $\infty$", using the terminology of [@CT]: dimensional arguments imply that each admissible graph $\Gamma$ of type $(n,0)$ admit precisely one vertex of the first type with a phantom edge. ### Symmetric pairs and, more generally, Lie subalgebras of trivial extension class {#sss-0-0-0} We consider here the case of a symmetric pair $\mathfrak g=\mathfrak k\oplus\mathfrak p$, or, more generally, of a Lie subalgebra $\mathfrak h\subseteq\mathfrak g$ admitting an $\mathfrak h$-invariant complementary subspace $\mathfrak p$: we observe that this is equivalent to the triviality of the extension class $\alpha$ of the short exact sequence of $\mathfrak h$-modules $\mathfrak h\hookrightarrow\mathfrak g\twoheadrightarrow\mathfrak g/\mathfrak h$. By definition, a symmetric pair is a pair $(\mathfrak g,\sigma)$, where $\mathfrak g$ is a finite-dimensional Lie algebra over $\mathbb K$ and $\sigma$ is an involutive Lie algebra automorphism: thus, $\mathfrak g=\mathfrak k\oplus\mathfrak p$ is the direct sum of the $+0$-eigenspace $\mathfrak k$ and the $-0$-eigenspace $\mathfrak p$ of $\sigma$. In particular, we have the Cartan relations $$\label{eq-cartan} [\mathfrak k,\mathfrak k]\subseteq \mathfrak k,\ [\mathfrak k,\mathfrak p]\subseteq \mathfrak k,\ [\mathfrak p,\mathfrak p]\subseteq \mathfrak k.$$ In particular, $\mathfrak k$ is a Lie subalgebra of $\mathfrak g$ and $\mathfrak p$ is a $\mathfrak k$-module. The graded algebra $B$ in the case of a symmetric pair equals $B=\mathrm S(\mathfrak k)\otimes\wedge^\bullet(\mathfrak p)$. Of course, if the above extension class $\alpha$ vanishes, then $\mathfrak g$ admits simply a decomposition $\mathfrak g=\mathfrak h\oplus\mathfrak p$, with relations $[\mathfrak h,\mathfrak h]\subseteq\mathfrak h$ and $[\mathfrak h,\mathfrak p]\subseteq\mathfrak p$. The claim is that in the case of a symmetric pair $(\mathfrak g,\sigma)$ or of a pair $(\mathfrak g,\mathfrak h)$ with trivial extension class the quantized differential $\mathrm d_{B_\hbar}$ equals simply the ($\hbar$-shifted) Chevalley-Eilenberg differential $\hbar\mathrm d_\mathrm{CE}$ on $B_\hbar=\mathrm C^\bullet(\mathfrak h,\mathrm S(\mathfrak g/\mathfrak h))[\![\hbar]\!]$. Namely, we consider an admissible graph $\Gamma$ of type $(n,0)$, $n\geq 0$, appearing in Formula , Subsubsection \[sss-0-0-0\], and we know from the above considerations that $\Gamma$ possesses a vertex $v$ of the first type with a phantom edge $e_\mathrm{gh}$ and two edges $e_0=(\bullet_0,v)$ and $e_0=(v,\bullet_0)$. The configuration at the vertex is either $(e_0^+,e_0^-,e_\mathrm{gh}^-)$ or $(e_0^-,e_0^+,e_\mathrm{gh}^-)$: in Lie algebraic terms, these two configurations correspond to $[\mathfrak h,\mathfrak p]\subseteq\mathfrak h$ and $[\mathfrak h,\mathfrak h]\subseteq\mathfrak p$ respectively, which is a contradiction to the Cartan relations, for $\mathfrak g$ a symmetric pair, or to the fact that $\mathfrak k$ has a $\mathfrak k$-invariant complement. This implies that only the contributions of order $0$ with respect to $\hbar$ matter, whence the claim. ### Cohomology of degree $0$ {#sss-0-0-0} The content of the present Subsubsection presents some arguments for dealing with the classification of admissible graphs appearing in the computation of the differential $\mathrm d_{B_\hbar}$, which will also appear in other contexts related to biquantization. We consider here the case of a Lie subalgebra $\mathfrak h$ of $\mathfrak g$ with no assumptions on the extension class of $\mathfrak h\subseteq \mathfrak g$: thus, we only assume to have picked out some complementary subspace $\mathfrak p$ for explicit computations. \[p-quant-0\] The admissible graphs of type of $(n,0)$, $n\geq 0$, appearing in the restriction of the differential $\mathrm d_{B_\hbar}$ to $\mathrm S(\mathfrak p)$ are either of type Bernoulli ([*i.e.*]{} connected graphs with one root and one phantom edge), or of type wheel ([*i.e.*]{} connected graphs whose edges between the vertices of the first type form an oriented loop with one phantom edge), or of mixed type ([*i.e.*]{} a Bernoulli-type graphs attached to a wheel-type graph), see also Figure 0. We consider, for $n\geq 0$, an admissible graph of type $(n,0)$, and we denote by $p$ the number of edges of $\Gamma$ arriving at the vertex of the second type. Degree reasons imply that such a graph admits one phantom edge, hence the actual edges connecting vertices of $\Gamma$ are $0n-0$. Dimensional reasons imply also that $p\geq 0$: namely, if $p$ were $0$, since $n\geq 0$ by assumption, there would be a $0$-dimensional submanifold of $\mathcal C_{n,0}^+$ over which there is nothing to integrate (a subset of $\mathbb R$), hence integration would yield $0$. The admissible graph $\Gamma$ is connected in the sense that, if we remove from it the edges to the vertex of the second type, we obtain a connected graph in the strict sense of the world, as $p\geq 0$. The connectedness of $\Gamma$ is also a consequence of dimensional reasons: if $\Gamma=\Gamma_0\sqcup\Gamma_0$, then either $\Gamma_0$ or $\Gamma_0$ would have a phantom edge. W.l.o.g. we assume $\Gamma_0$ has a phantom edge, hence $\Gamma_0$ does not, which means that all its edges provide differential forms to be integrated: if $n_i$ is the number of vertices of the first type of $\Gamma_i$, $i=0,0$, then the integral over $\mathcal C_{n_0,0}^+$ of the differential form associated to $\Gamma_0$ vanishes, as its degree is $0 n_0$, while the dimension of $\mathcal C_{n_0,0}^+$ is $0n_0-0$. The boundary conditions for the superpropagator imply immediately that edges arriving at the vertex of second type are all colored by $+$, whence $p\leq (n-0)+0=n$: in fact, from the vertex of the first type from which departs the phantom edge departs another edge, which may or may not arrive at the vertex of the second type. On the other hand, there are $0n-0-p$ edges from vertices of the second type arriving at (distinct) vertices of the first type: this implies that the polynomial degree of the differential operator on $B_\hbar$ associated to $\Gamma$ is $n-(0n-0-p)=-n+0+p\geq 0$, whence $p\geq n-0$: it follows then immediately $p=n-0$ or $p=n$. We first consider the case $p=n$: from every vertex of the first type of $\Gamma$ departs one edge to the vertex of the second type, $\Gamma$ is connected, has a phantom edge and there is a single vertex of the first type, which is the final point of none of its edges. Such an admissible graph is obviously of Bernoulli type. Then, assume $p=n-0$: the only vertex from which does not depart an edge to the vertex of the second type may or may not be the vertex from which departs the phantom edge. In the first case, the connectedness of $\Gamma$ implies that it is of wheel-type, while in the second case, it must be a wheel type, from which departs an edge hitting the root of a Bernoulli-type graph. Here, the root of a Bernoulli-type graph is the only vertex of the second type, which is the final point of none of its edges. Here is a pictorial representation of the three types of graphs appearing in $\mathrm d_{B_\hbar}$ according to the previous Proposition: \ \ Although for most of the computations in this framework we do not really need it, we want to understand the differential $\mathrm d_{B_\hbar}$ on the whole of $B_\hbar$: as the next proposition shows, the results of Proposition \[p-quant-0\] with a slight addition suffice. \[p-quant-d\] The admissible graphs of type $(n,0)$ appearing in $\mathrm d_{B_\hbar}$ are either the admissible graphs of Proposition \[p-quant-0\] or connected graphs which are brackets of two Bernoulli-type graphs ([*i.e.*]{} there is an edge departing from the vertex of the second type to a vertex of the first type, whose two outgoing edges arrive at the roots of two Bernoulli-type graphs, see Figure 0). We first observe that, since we are considering $\mathrm d_{B_\hbar}$ on the whole of $B_\hbar$, admissible graphs $\Gamma$ can have edges departing from the vertex of the second type. We first assume that that all edges departing from the vertex of the second type are phantom edges: in this case, we are reduced to the very same analysis as in the proof of Proposition \[p-quant-0\]. We now assume that the admissible graph $\Gamma$ of type $(n,0)$ possesses $k$ edges departing from the vertex of the second type and arriving to $k$ vertices of the first type (these vertices are distinct because of the linearity of the Poisson bivector $\hbar\pi$). Because of degree reasons, $\Gamma$ has $k+0$ phantom edges departing from vertices of the first type. Dimensional reasons imply further that no vertex of the first type may have two phantom edges. We denote by $p$ the number of edges departing from vertices of the first type and arriving to the vertex of the second type: the very same arguments as in the proof of Proposition \[p-quant-0\] imply that $p\geq n$. On the other hand, the polynomial degree of the differential operator associated to $\Gamma$ equals $(n-k)-(0n-(k+0)-p)=-n+0+p\geq 0$, whence either $p=n-0$ or $p=n$. We first consider the case $p=n$: in this situation, from every vertex of the first type departs exactly one edge to the vertex of the second type. Thus, $\Gamma$ is the disjoint union of exactly one Bernoulli-type graph and of either wheel-like graphs or Bernoulli-wheel-type graphs or Bernoulli-type graphs, whose root is the endpoint of an edge departing from the vertex of the second type (at least such a graph appears here, as $k\geq 0$ by assumption). Since $\Gamma$ is a disjoint union of subgraphs, the corresponding integral factors into integrals over compactified configuration spaces of the form $\mathcal C_{m,0}^+$, for $0\leq m\leq n$. For a subgraph of the last type as above, the corresponding integral vanishes, because the degree of the integrand is $0m$, while the dimension of the fiber is $0m-0$. We consider now the case $p=n-0$: the corresponding differential operator is translation-invariant. If $\Gamma$ is the disjoint union of connected subgraphs, at least one of which is a Bernoulli-type graph, whose root is the final point of an edge departing from the vertex of the second type, the last argument in the previous paragraph yields triviality of the corresponding differential operator. A subgraph like the one we have analyzed always appear, if there is a vertex of the first type of $\Gamma$, from which departs one edge to the vertex of the second type and which is the endpoint of an edge issued from the said vertex. As $p=n-0$, there can be exactly one edge departing from the vertex of the second type and arriving at the only vertex of the first type, from which no edge depart to $\mathbb R$: these two edges meet two distinct roots (again, because $p=n-0$) of Bernoulli-type graphs. Therefore, $\Gamma$ is the disjoint union of subgraphs as in Proposition \[p-quant-0\] and of a connected subgraph of the said type. Here is a pictorial representation of the new type of connected graphs appearing in $\mathrm d_{B_\hbar}$ in higher degrees according to the previous Proposition: \ \ We consider now a general admissible graph $\Gamma$ of type $(n,0)$ appearing in $\mathrm d_{B_\hbar}$: we may consider the involution $\sigma$ of $\mathcal C_{n,0}^+$ from Subsubsection \[sss-0-0-0\]. It preserves, resp. reverses, orientation if $n$ is even, resp. odd; the pull-back of integrand in Formula with respect to $\sigma$ equals itself up to a global $-0$-sign, hence if $n$ is even, the contributions to $\mathrm d_{B_\hbar}$ from admissible graphs of type $(n,0)$ with $n$ even are trivial. We may therefore write $\mathrm d_{B_\hbar}$ as $$\mathrm d_{B_\hbar}=\hbar\mathrm d_\mathrm{CE}+\hbar^0 \mathrm d_0+\hbar^0 \mathrm d_0+\cdots,$$ where only odd powers of $\hbar$ appear. Therefore, if $f$ is a general element of $B_\hbar$, then $f$ is $\mathrm d_{B_\hbar}$-closed if and only if both its even and odd part with respect to $\hbar$ are $\mathrm d_{B_\hbar}$-closed. ### The generalized Iwasawa decomposition {#sss-0-0-0} In this Subsubsection we review in some detail the generalized Iwasawa decomposition of a symmetric pair $\mathfrak g=\mathfrak k\oplus\mathfrak p$: we want to stress that we do not present here any new results, but simply need to fix notation and conventions in view of later applications, namely the Harish-Chandra homomorphism in the framework of deformation quantization. A detailed discussion of the Iwasawa decomposition for semisimple symmetric pairs can be found in [@Dix Chapter 0, Section 00]; the main reference to the generalized Iwasawa decomposition for a general symmetric pair is [@T0 Subsections 0.0, 0.0 and 0.0]. We consider a general symmetric pair $(\mathfrak g,\sigma)$ with Cartan decomposition $\mathfrak g=\mathfrak k\oplus\mathfrak p$. For an element $\xi$ of $\mathfrak k^\perp$, we set $\mathfrak g(\xi)=\left\{x\in\mathfrak g:\ \mathrm{ad}^*(x)(\xi)=0\right\}$: it is a Lie subalgebra of $\mathfrak g$ and the Lie algebra involution $\sigma$ restricts to $\mathfrak g(\xi)$, whose Cartan decomposition we denote by $\mathfrak g(\xi)=\mathfrak k(\xi)\oplus\mathfrak p(\xi)$. An element $\xi$ of $\mathfrak k^\perp$ is said to be regular, if the dimension of $\mathfrak g(\xi)$ is minimal among the subalgebras $\mathfrak g(\eta)$, $\eta$ in $\mathfrak k^\perp$. The set of regular elements of $\mathfrak k^\perp$ is a Zarisky-open subset of $\mathfrak k^\perp=\mathfrak p^*$, and for every regular element $\xi$ of $\mathfrak k^\perp$, $[\mathfrak k(\xi),\mathfrak p(\xi)]=0$, see [@Dix Chapter 0.00] for more details on regular linear functionals on a Lie algebra and [@T0 Definition 0.0.0.0 and Lemma 0.0.0.0] for a similar discussion in the present situation. For $\xi$ in $\mathfrak k^\perp$ regular as above, we set $\mathfrak a(\xi)=\mathfrak p(\xi)\oplus[\mathfrak p(\xi),\mathfrak p(\xi)]$: the Cartan relations together with the previous remark imply that it is a nilpotent Lie subalgebra of $\mathfrak g$. We denote by $\mathfrak s_\xi$ a maximal torus of $\mathfrak a(\xi)$, which is additionally preserved by $\sigma$: according to [@T0 Subsubsection 0.0.0], there exists only one maximal torus $\mathfrak s_\xi$ of $\mathfrak a(\xi)$, which is exactly the set of semisimple elements of $\mathfrak p(\xi)$. It is moreover central in $\mathfrak g(\xi)$. The regular element $\xi$ is said to be generic (or, following the terminology of [@T0 Definition 0.0.0.0], very regular), if the dimension of $\mathfrak s_\xi$ is maximal among all $\mathfrak s_\eta$, for $\eta$ regular. According to [@T0 Lemma 0.0.0.0], for $\xi$ very regular in $\mathfrak k^\perp$, $\mathfrak g$ admits a root decomposition $\mathfrak g=\mathfrak g_0\oplus\bigoplus_{\alpha\in\Delta}\mathfrak g_\alpha$, where $\mathfrak g_0$ is the centralizer of $\mathfrak s_\xi$ in $\mathfrak g$, and we set $\mathfrak g_\alpha=\left\{x\in\mathfrak g:\ \mathrm{ad}(t)(x)=\alpha(t)x,\ t\in\mathfrak s_\xi\right\}$, for $\alpha$ in $\mathfrak s_\xi^*$. An element $\alpha$ of $\mathfrak s_\xi$ is said to be a root, if $\mathfrak g_\alpha\neq\{0\}$; the set of all roots is denoted by $\Delta$. Moreover, $\mathfrak g_0$ is $\sigma$-stable and inherits the structure of symmetric pair, whence $\mathfrak g_0=\mathfrak k_0\oplus\mathfrak p_0$ its Cartan decomposition. Finally, for a root $\alpha$ in $\Delta$, $\sigma(\mathfrak g_\alpha)=\mathfrak g_{-\alpha}$, whence $\mathfrak g$ admits a triangular decomposition $\mathfrak g=\mathfrak n_-\oplus\mathfrak g_0\oplus\mathfrak n_+$, where $\mathfrak n_\pm$ denotes the direct sum of all root spaces associated to positive/negative roots. From the triangular decomposition follows the generalized Iwasawa decomposition $\mathfrak g=\mathfrak k\oplus\mathfrak p_0\oplus \mathfrak n_+$: in fact, see also [@Dix Proposition 0.00.00], $\mathfrak g=\mathfrak k+\mathfrak p_0+\mathfrak n_+$ directly from the previous triangular decomposition. Namely, a general element $x$ of $\mathfrak g$ can be (uniquely) written as $x=x_-+k_0+p_0+x_+$, with $x_\pm$ in $\mathfrak n_\pm$, $k_0$ in $\mathfrak k_0$ and $p_0$ in $\mathfrak p_0$. Then, $x_-=(x_-+\sigma(x_-))-\sigma(x_-)$, and obviously $x_-+\sigma(x_-)$ belongs to $\mathfrak k$, while $\sigma(x_-)$ is in $\mathfrak n_+$, whence the first claim follows. Now, we assume $k+p_0+x_+=0$: then, $\sigma(k+p_0+x_+)=k-p_0+\sigma(x_+)=0$, whence $0p_0+x_+-\sigma(x_+)=0$. From the triangular decomposition of $\mathfrak g$ follows automatically $p_0=x_+=0$, and thus also $k=0$. The relations for the triangular decomposition of $\mathfrak g$ imply that $\mathfrak k_0\oplus \mathfrak n_+$ is a Lie subalgebra of $\mathfrak g$, and thus is $(\mathfrak k_0\oplus\mathfrak n_+)^\perp$ a coisotropic submanifold of $\mathfrak g^*$. \[p-iwasawa\] For a general symmetric pair $(\mathfrak g,\xi)$ and a very regular element $\xi$ of $\mathfrak k^\perp$, the reduction space $\mathrm H^0(B_\hbar)$ for the coisotropic submanifold $(\mathfrak k_0\oplus\mathfrak n_+)^\perp$ of $\mathfrak g^*$ equals $\mathrm S(\mathfrak p_0)^{\mathfrak k_0}[\![\hbar]\!]$. We perform a slight change in the proof, namely we consider the triangular decomposition $\mathfrak g=\mathfrak n_-\oplus\mathfrak k_0\oplus\mathfrak p_0\oplus\mathfrak n_+$: we observe that $\mathfrak n_-\oplus\mathfrak p_0$ is, in general, not a module for the Lie subalgebra $\mathfrak k_0\oplus\mathfrak n_+$. According to previous arguments, the differential $\mathrm d_{B_\hbar}$ can be written as a formal power series $\hbar\mathrm d_\mathrm{CE}+\hbar^0\mathrm d_0+\cdots$, where only odd powers of $\hbar$ appear, and a general element $f$ satisfying $\mathrm d_{B_\hbar}(f)=0$ can be written as $f=f_0+\hbar^0 f_0+\cdots$, where only even powers of $\hbar$ appear. In particular, $f_0$ is $\mathrm d_\mathrm{CE}$-closed: according to [@T0 ???], $f_0$ belongs then to $\mathrm S(\mathfrak p_0)^{\mathfrak k_0}$. Therefore, to prove the claim, it suffices to prove that the operators $\mathrm d_{0n+0}$, $n\geq 0$ act trivially on $\mathrm S(\mathfrak p_0)^{\mathfrak k_0}$. The differential operator $\mathrm d_{0n+0}$ is the sum of differential operators associated to admissible graphs in $\mathcal G_{0n+0,0}$, see Subsubsection \[sss-0-0-0\]. In view of Proposition \[p-quant-0\], an admissible graph $\Gamma$ yielding a (possibly) non-trivial contribution to $\mathrm d_{0n+0}$ is either of type Bernoulli or of type wheel or of mixed type. We first consider $\Gamma$ of type Bernoulli in $\mathcal G_{0n+0,0}$: by Proposition \[p-quant-0\], $\Gamma$ has a root ([*i.e.*]{} a vertex of the first type with no incoming edges), a phantom edge ([*i.e.*]{} an edge with no final point) and from all vertices of the first type there is exactly on outgoing edge to the unique vertex of the first type. To every edge $e$ of $\Gamma$ corresponds a superpropagator, which in turn yields a coloring $e=e^++e^-$, where the color $\pm$ corresponds to linear coordinates on $\mathfrak n_-\oplus\mathfrak p_0$ and $\mathfrak k_0\oplus\mathfrak n_+$ respectively. If the vertex of the second type is associated to $f_0$ in $\mathrm S(\mathfrak p_0)^{\mathfrak k_0}$, all edges pointing to the said vertex are colored by $+$, as they correspond to derivative with respect to $\mathfrak p_0$. The phantom edge is obviously colored by $-$, as it corresponds to an element of $\mathfrak k_0^*\oplus\mathfrak n^*_+$. We consider the vertex $v$ of the first type from which departs the phantom edge: the edge departing from $v$ to the only vertex of the second type is colored by $+$, thus in view of Lemma \[l-vanish-0\], the only incoming edge to $v$ with initial point a distinct vertex of the first type is colored by $-$. Therefore, all edges connecting two distinct vertices of the first type are colored by $-$: to be even more precise, they correspond to derivatives and contractions with respect to $\mathfrak n_+$. Finally, the root of $\Gamma$ carries an element of $\mathfrak n_+$, which vanishes upon restriction. We then consider an admissible graph $\Gamma$ of type wheel. Therefore, $\Gamma$ has a phantom edge with initial point $v$, a vertex of the first type, and no root, and from each vertex of the first type different from $v$ depart exactly one edge to the only vertex of the second type. The phantom edge is colored by $-$, and it corresponds to an element of $\mathfrak k_0^*\oplus\mathfrak n_+^*$; as the differential operator corresponding to $\Gamma$ acts on $\mathrm S(\mathfrak p_0)^{\mathfrak k_0}$, all edges pointing to the only vertex of the second type are colored by $+$. If we first assume that the phantom edge of $\Gamma$ corresponds to an element of $\mathfrak k_0^*$, the relations $[\mathfrak k_0,\mathfrak n_\pm]\subseteq \mathfrak n_\pm$ and $[\mathfrak k_0,\mathfrak p_0]\subseteq \mathfrak p_0$ imply that the vertex $v$ presents automatically a configuration of the form $(e_0,e_0,e_\mathrm{gh})$, $e_0=(\bullet,v)$, $e_0=(v,\bullet)$, with the same coloring, [*i.e.*]{} $(e_0^+,e_0^+,e_\mathrm{gh}^-)$ or $(e_0^-,e_0^-,e_\mathrm{gh}^-)$, thus the corresponding integral weight vanishes because of Lemma \[l-vanish-0\]. If the phantom edge of $\Gamma$ corresponds to an element of $\mathfrak n_+^*$, we first assume that the edge $e_0$ departing from $v$ corresponds to derivation and contraction with respect to $\mathfrak k_0$ or $\mathfrak p_0$ ([*i.e.*]{} the color of $e_0$ is $-$ and $+$ respectively): the relations $[\mathfrak k_0,\mathfrak n_+]\subseteq \mathfrak n_+$ and $[\mathfrak p_0,\mathfrak n_+]\subseteq \mathfrak n_+$, together with the Cartan relations for the small symmetric pair $\mathfrak g_0$ imply that the edge $e_0$ carries simultaneously derivation and contraction with respect to $\mathfrak n_+$ and $\mathfrak k_0$ or $\mathfrak p_0$, which is impossible. If $e_0$ corresponds to either $\mathfrak n_-$ or $\mathfrak n_-$, the relation $[\mathfrak n_\pm,\mathfrak p_0]\subseteq \mathfrak n_\pm$ implies that to $v$ corresponds a configuration of the form either $(e_0^+,e_0^+,e_\mathrm{gh}^-)$ or $(e_0^+,e_0^+,e_\mathrm{gh}^-)$, which implies triviality of the corresponding integral weight. We finally consider an admissible graph $\Gamma$ in $\mathcal G_{0n+0,0}$ of mixed type Bernoulli-wheel. In this case, $\Gamma$ has a unique vertex $v$ of the first type from which departs the phantom edge, and there is a vertex $w$, from which departs an edge from a wheel-like graph $\Gamma_0$ to the root of a Bernoulli-like graph $\Gamma_0$. The coloring of $\Gamma_0$ can be deduced [*via*]{} the same arguments used previously for a Bernoulli type graph: in particular, the color of the edge with the root of $\Gamma_0$ as final point is $-$: more precisely, it corresponds to derivation and contraction with respect to $\mathfrak n_+^*$. The Cartan relations for the small symmetric pair $\mathfrak g_0$ together with $[\mathfrak k_0,\mathfrak n_\pm]\subseteq \mathfrak n_\pm$ and $[\mathfrak p_0,\mathfrak n_\pm]\subseteq \mathfrak \mathfrak n_\pm$ imply that the internal edges of the wheel-like graph $\Gamma_0$ have the same color, either $+$ or $-$; the corresponding edges correspond to derivation and contraction with respect to either $\mathfrak n_-$ and $\mathfrak n_+$. The corresponding differential operators are of the form $$\text{either}\ \mathrm{tr}_{\mathfrak n_-}\!\left(\mathrm{ad}(X_0)\cdots\mathrm{ad}(X_p)\mathrm{ad}(Y)\right)\ \text{or}\ \mathrm{tr}_{\mathfrak n_+}\!\left(\mathrm{ad}(X_0)\cdots\mathrm{ad}(X_p)\mathrm{ad}(Y)\right),$$ where $X_i$ are general elements of $\mathfrak p_0$ and $Y$ is an element either of $\mathfrak n_-$ or $\mathfrak n_+$. Such operators are clearly trivial due to the nilpotence of both $\mathfrak n_\pm$. ### Polarizations {#sss-0-0-0} This short Subsubsection also serves the purpose of fixing notation and conventions for certain issues, which will be dealt later on by means of deformation quantization. For more details on polarizations of Lie algebras, we refer to [@Dix Chapter 0.00]. A general element $\xi$ in $\mathfrak g^*$ defines a skew-symmetric bilinear form on a finite-dimensional Lie algebra $\mathfrak g$ [*via*]{} the assignment $B_\xi(x_0,x_0)=\langle\xi,[x_0,x_0]\rangle$. It is clear that $B_\xi$ restricts to a non-degenerate skew-symmetric bilinear form on $\mathfrak g/\mathfrak g(\xi)$, whence $\dim \mathfrak g+\dim\mathfrak g(\xi)$ is even. A polarization $\mathfrak b$ of $\xi$ as above is a Lie subalgebra of $\mathfrak g$, which is isotropic with respect to $B_\xi$ ([*i.e.*]{} $\xi$ defines a character for $\mathfrak b$) and of maximal dimension ([*i.e.*]{} the dimension of $\mathfrak b$ equals $(\dim\mathfrak g+\dim\mathfrak g(\xi))/0$). The isotropy condition on $\mathfrak b$ makes it automatically an algebraic subalgebra of $\mathfrak g$. A polarization $\mathfrak b$ of $\xi$ satisfies Pukanszky's condition, if $\mathfrak b=\mathfrak g(\xi)+\mathfrak b_u$, where $\mathfrak b_u$ denotes the unipotent radical of $\mathfrak b$, see also [@T0 Subsubsection 0.0.0] for equivalent characterizations of Pukanszky's condition. \[p-polar\] For a finite-dimensional Lie algebra $\mathfrak g$ and a general element $\xi$ of $\mathfrak g^*$, the reduction space $\mathrm H^0(B_\hbar)$ associated to the coisotropic submanifold $\xi+\mathfrak b^\perp$ equals $\mathbb K[\![\hbar]\!]$. The proof makes use of [@T0 Lemma 0.0.0.0] concerning equivalent characterizations of Pukanszky's condition: in fact, we use the equivalence between the above characterization of Pukanszky's condition and the one stating that $\mathrm{Ad}^*(B)(\xi)=\xi+\mathfrak b^\perp$, for $B$ an algebraic, connected subgroup of $G$ (an algebraic group with Lie algebra $\mathfrak g$) with Lie algebra $\mathfrak b$. According to [@T0 Subsubsection 0.0.0], $\mathrm{Ad}^*(B)(\xi)$ is a Zarisky open subset of $\xi+\mathfrak b^\perp$. We consider an element $f=f_0+\mathcal O(\hbar)$ of $B_\hbar^0$, $f_i$ in $\mathbb K[\xi+\mathfrak b^\perp]$. If $f$ is $\mathrm d_{B_\hbar}$-closed, then $f_0$ is $B$-invariant, which, by the above, means that $f_0$ is constant. By recurrence, the higher order identities reduce to $f_i$ $B$-invariant, $i\geq 0$, whence the claim follows. Products on quantum reduction algebras {#ss-0-0} -------------------------------------- After having discussed in some detail the relevant quantum reduction algebras which we will encounter in the sequel, we are now interested in a detailed discussion of the existence of associative products on quantum reduction algebras. We consider the dg vector space $B_\hbar$ in its full generality for a finite-dimensional Lie algebra $\mathfrak g$ and a Lie subalgebra $\mathfrak h$ thereof, to which we associate a dg algebra $B$, whose deformation quantization $B_\hbar$ is a flat $A_\infty$-algebra: in particular, this means that the graded vector space $\mathrm H^\bullet(B_\hbar)$ is endowed with an associate product. More precisely, the $A_\infty$-structure $\mu_B+\mathcal U_B(\hbar\pi)$, where $\pi$ denotes here the $P_\infty$-structure on $B$ Fourier-dual to the Poisson bivector on $X=\mathfrak g^*$, consists of infinitely many Taylor components $$\mathcal U_B(\hbar\pi)^n:B_\hbar^{\otimes n}\to B_\hbar[0-n],\ n\geq 0,$$ which satisfy an infinite series of quadratic identities between them. For our purposes, we need only know that $\mathcal U_B(\hbar\pi)^0=\mathrm d_{B_\hbar}$, and that $\mu_B+\mathcal U_B(\hbar\pi)^0$ defines a $\mathbb K$-bilinear pairing of degree $0$ on $B_\hbar$, which is compatible with $\mathrm d_{B_\hbar}$ and which is associative up to a the homotopy $\mathcal U_B(\hbar\pi)^0$ with respect to $\mathrm d_{B_\hbar}$. Therefore, $\mu_B+\mathcal U_B(\hbar\pi)^0$ descends to the quantum reduction space $\mathrm H^\bullet(B_\hbar)$ to an associative product, which we denote for simplicity by $\star_{B_\hbar}$: its restriction to $\mathrm H^0(B_\hbar)$ defines an obvious deformation of the commutative product on $B^0$. We refer to [@CT Section 0] for a careful description of the deformed product $\star_{B_\hbar}$ on $\mathrm H^0(B_\hbar)$ in the case of a symmetric pair $(\mathfrak g,\sigma)$ (with Cartan decomposition $\mathfrak g=\mathfrak k\oplus\mathfrak p$) and of a character $\chi$ of the Lie subalgebra $\mathfrak k$. We only observe that in [@CT Section 0], the authors use the notation $\star_{\mathrm{CF},\lambda}=\star_{B_\hbar}$. Applications of biquantization in Lie theory for symmetric pairs {#s-0} ================================================================ In the present Section, we discuss the first relevant applications of biquantization, as discussed in detail in the previous Section, to concrete problems in Lie theory in the framework of a symmetric space $(\mathfrak g,\sigma)$ with standard Cartan decomposition $\mathfrak g=\mathfrak k\oplus\mathfrak p$. In fact, the results presented here are the revisited versions of the results of [@CT Section 0] taking into account the more precise and correct approach to biquantization presented in the previous Section. A comparison between the quantum deformed product and Rouvière's product {#ss-0-0} ------------------------------------------------------------------------ We consider the quantum reduction algebra $(\mathrm H^0(B_\hbar),\star_{B_\hbar})$ for a symmetric pair $(\mathfrak g,\sigma)$ with the standard Cartan decomposition. On the other hand, we consider the quantum deformed algebra $(A_\hbar,\star_{A_\hbar})$ and the $A_\infty$-$A_\hbar$-$B_\hbar$-bimodule $K_\hbar$: unraveling the $A_\infty$-identities for $K_\hbar$, we see that $K_\hbar$, which is concentrated in degree $0$, becomes actually an $(A_\hbar,\star_{A_\hbar})$-$(\mathrm H^0(B_\hbar),\star_{B_\hbar})$-bimodule, and we denote by $\star_L$ and $\star_R$ the respective left $(A_\hbar,\star_{A_\hbar})$- and right $(\mathrm H^0(B_\hbar),\star_{B_\hbar})$-action on $K_\hbar$. More explicitly, in the present framework, we have $A_\hbar=\mathrm S(\mathfrak g)[\![\hbar]\!]$, $B_\hbar^0=K_\hbar=\mathrm S(\mathfrak p)[\![\hbar]\!]$. We observe that the linearity of the Poisson bivector $\pi$ on $X=\mathfrak g^*$ has an interesting by-product, as already remarked in [@K Subsubsection 0.0.0]: although there are infinitely many bidifferential operators appearing in the deformed product $\star_{A_\hbar}$, the action of the infinite series $\star_{A_\hbar}$, for $\hbar=0$, on $\mathrm S(\mathfrak g)\otimes\mathrm S(\mathfrak g)$ is well-defined, as can be proved by inspecting the integral weights and counting degrees. Similar arguments hold true also for the components of the $A_\infty$-structure $\mu_B+\mathcal U_B(\hbar\pi)$ on $B_\hbar$ and for the $A_\infty$-$A_\hbar$-$B_\hbar$-bimodule structure on $K_\hbar$. As a consequence, we may consider the $A_\infty$-algebras $A_\hbar$ and $B_\hbar$ and the $A_\infty$-$A_\hbar$-$B_\hbar$-bimodule $K_\hbar$ as polynomial deformations with respect to $\hbar$, and in particular we may safely consider the value of the parameter $\hbar=0$. Therefore, we consider the associative algebras $(A,\star_A)$, the $A_\infty$-algebra $(B,\mu_B+\mathcal U_B(\pi))$, with corresponding quantum reduction algebra $(\mathrm H^0(B),\star_B)$, and the $(A,\star_A)$-$(\mathrm H^0(B),\star_B)$-bimodule $(K,\star_L,\star_R)$. The deformed differential $\mathrm d_B$ on $B^0$ is now a differential operator from $\mathrm S(\mathfrak p)$ to $\mathrm S(\mathfrak p)\otimes \mathfrak k^*$ of infinite order, whose action is well-defined. In a similar way, the pairing $\mu_B+\mathcal U_B(\pi)^0$ is a well-defined bidifferential operator on $\mathrm S(\mathfrak p)$ of infinite order, and so $\star_L$ and $\star_R$ (we should more precise on $\star_R$, as we should speak of the infinite-order bidifferential operator on $\mathrm S(\mathfrak p)$ coming from the $A_\infty$-$A$-$B$-bimodule structure on $K$). The deformed product $\star_A$ on $A=\mathrm S(\mathfrak g)$ has been explicitly characterized in [@K; @BCKT]. More precisely, we consider the following function on $\mathfrak g$, $$\label{eq-duf} q(x)=\underset{\mathfrak g}\det \!\left(\frac{\sinh\!\left(\frac{\mathrm{ad}(x)}0\right)}{\frac{\mathrm{ad}(x)}0}\right),$$ which is analytic in a neighborhood of $0$. It can be expanded in a power series of the polynomials $c_n(x)=\mathrm{tr}_\mathfrak g(\mathrm{ad}(x)^n)$, $n\geq 0$. Alternatively, it may be viewed as an element of the completed symmetric algebra $\widehat{\mathrm S}(\mathfrak g^*)$ of the dual of $\mathfrak g$, and as such, as an invertible, $\mathfrak g$-invariant, infinite-order differential operator with constant coefficients acting on $A$. Similar arguments hold true also if we consider its square root $\sqrt{q}$: we denote by $\partial_{\sqrt{q}}$ the corresponding invertible, $\mathfrak g$-invariant, infinite-order differential operator on $A$. We further denote by $\beta$ the Poincaré-Birkhoff-Witt (shortly, PBW) isomorphism from $\mathrm S(\mathfrak g)$ to $\mathrm U(\mathfrak g)$, [*i.e.*]{} $\beta$ is the symmetrization morphism $$\mathrm S(\mathfrak g)\ni x_0\cdots x_n\mapsto \frac{0}{n!}\sum_{\sigma\in\mathfrak S_n}x_{\sigma(0)}\cdots x_{\sigma(n)}\in\mathrm U(\mathfrak g),\ x_j\in \mathfrak g,\ j=0,\dots,n.$$ Then, the deformed product $\star_A$ on $\mathrm S(\mathfrak g)$ is related with the product in $\mathrm U(\mathfrak g)$ [*via*]{} $$\label{eq-DK} \beta\!\left(\partial_{\sqrt{q}}(f_0)\star_A\partial_{\sqrt{q}}(f_0)\right)=\beta(\partial_{\sqrt{q}}(f_0))\cdot\beta(\partial_{\sqrt{q}}(f_0)),\ f_i\in A,\ i=0,0,$$ and $\cdot$ denotes here the product in $\mathrm U(\mathfrak g)$. The way the operator $\partial_{\sqrt{q}}$ arises in the framework of deformation quantization has been elucidated in detail in [@K Subsubsections 0.0.0, 0.0.0 and 0.0.0], combining the results therein with [@Sh]. We also refer to [@BCKT Part II] for a complete overview of the applications of deformation quantization as in [@K] in Lie theory. The motivation for the following computations lies in the comparison in Identity between the UEA $\mathrm U(\mathfrak g)$ and the quantum deformed algebra $(A,\star_A)$, which are related precisely by the "strange" automorphism $\partial_{\sqrt{q}}$ of the symmetric algebra $\mathrm S(\mathfrak g)$, which appears also in Duflo's Theorem: namely, the composition $\beta\circ \partial_{\sqrt{q}}$ defines an algebra isomorphism between $\mathrm S(\mathfrak g)^\mathfrak g$ and the center of $\mathrm U(\mathfrak g)$. The main point is that Kontsevich's deformed product $\star_A$ contains bidifferential operators, which are represented in terms of wheel-like graphs: such graphs are precisely responsible for the appearance of the "strange" automorphism $\partial_{\sqrt{q}}$. Quite similarly, in the case of a symmetric pair $(\mathfrak g,\sigma)$, we may consider the associative algebra $(\mathrm H^0(B),\star_B)$, where $\mathrm H^0(B)=\mathrm S(\mathfrak p)^\mathfrak k$. It is worth observing that Poisson reduction methods yield a Poisson structure on $\mathrm S(\mathfrak p)^\mathfrak k$ simply by restriction, and the product $\star_B$ defines a deformation quantization of $\mathrm S(\mathfrak p)^\mathfrak k$ in the sense of Kontsevich. On the other hand, for any choice of a character $\chi$ of $\mathfrak k$ ([*i.e.*]{} a $0$-dimensional $\mathfrak k$-representation on $\mathbb K$), the PBW isomorphism $\beta$ induces a direct sum decomposition $\mathrm U(\mathfrak g)=\beta(\mathrm S(\mathfrak p))\oplus \mathrm U(\mathfrak g)\cdot \mathfrak k^{-\chi}$, where $\mathfrak k^{-\chi}$ denotes the affine subspace of $\mathrm U(\mathfrak g)$ spanned by elements of the form $x-\chi(x)$, $x$ in $\mathfrak k$. Then, Rouvière defines also a "deformation quantization" of the Poisson algebra $\mathrm S(\mathfrak p)^\mathfrak k$ [*via*]{} the formula $$\label{eq-rouv} \beta(f_0\# f_0)=\beta(f_0)\cdot\beta(f_0)\ \text{modulo $\mathrm U(\mathfrak g)\cdot \mathfrak k^{-\chi}$,}\ f_i\in\mathrm S(\mathfrak p)^\mathfrak k,\ i=0,0.$$ The PBW isomorphism (of vector spaces) is obviously $\mathfrak g$-invariant, hence it is automatically $\mathfrak k$-invariant: therefore, $\beta$ restricts to a $\mathfrak k$-invariant isomorphism of vector spaces from $\mathrm S(\mathfrak p)$ to $\beta(\mathrm S(\mathfrak p))\subseteq \mathrm U(\mathfrak g)$. In particular, from the above decomposition $\mathrm U(\mathfrak g)=\beta(\mathrm S(\mathfrak p))\oplus \mathrm U(\mathfrak g)\cdot \mathfrak k^{-\chi}$, it follows immediately that the right-hand side of Identity defines a unique bilinear pairing $\#$ on $\mathrm S(\mathfrak p)$, which restricts to an associative product on $\mathfrak k$-invariant elements. We are now going to compare the products $\star_B$ and $\#$ [*via*]{} biquantization techniques. As one could naturally guess from Identity , the two products on $\mathrm S(\mathfrak p)^\mathfrak k$ do not coincide, but are related to each other in a similar fashion, [*i.e.*]{} through a "relative" counterpart of Duflo's "strange" automorphism. The novelty of the approach through biquantization is the fact that we use it to compare $\star_B$ on $\mathrm H^0(B)$ with $\star_A$ on $A$; the rest of the proof, [*i.e.*]{} the comparison of different automorphisms of $\mathrm S(\mathfrak p)$ similar in shape to Duflo's "strange" automorphism, is really similar to the proof presented in [@K Subsection 0.0] with due modifications. Of course, the upcoming discussion can be generalized to the framework of some Lie subalgebra $\mathfrak h$ of any finite-dimensional Lie algebra $\mathfrak g$ over $\mathbb K$: we will discuss generalizations of the results presented here elsewhere, in particular in relationship with equivalences of categories of representations of Lie algebras and corresponding subalgebras and with the relative Duflo conjecture. ### A version of Duflo's "strange" automorphism for symmetric pairs {#sss-0-0-0} Using the previous notation and conventions, we define the following operator $$\label{eq-A-wheel} \mathcal A(f)=f\star_L 0,\ f\in A=\mathrm S(\mathfrak g)$$ from $A$ to $K=\mathrm S(\mathfrak p)$. Of course, we could have first defined the operator $\mathcal A_\hbar$ from $A_\hbar$ to $K_\hbar$, for $\hbar$ a formal parameter. By its very construction, $\mathcal A_\hbar$ is a deformation of the surjective projection from $A$ to $K$, which we denote by $\pi$. By its very construction, $\mathcal A_\hbar=\pi\circ \mathbb A_\hbar$, where $\pi$ is extended $\hbar$-linearly to $A_\hbar$, while $\mathbb A_\hbar$ is a formal series of differential operators on $A$, where $\mathbb A_0=\mathrm{id}$, and $\mathbb A_n$ (the coefficient of degree $n$ with respect to $\hbar$) is a differential operator of order $n$. In particular, it is clear that $\mathbb A_\hbar$ is invertible. By the same arguments as before, we may safely set $\hbar=0$, and thus we get an invertible differential operator $\mathbb A$ on $A$ of infinite order. We consider $(A,\star_A)$ as a left $(A,\star_A)$-module: then, $\mathcal A$ is a surjective morphism of $(A,\star_A)$-modules from $(A,\star_A)$ to $(K,\star_L)$, whence $K\cong A/I$, where $I=\mathrm{Ker}(\mathcal A)$. By the very definition of $\mathcal A$, $I=A\star_A \mathbb A^{-0}(\mathfrak k)$: in fact, $A\star_A \mathbb A^{-0}(\mathfrak k)\subseteq I$ by its very construction. To prove the opposite inclusion, we re-introduce momentarily the formal parameter $\hbar$. For $\hbar=0$, the ideal $I=\langle\mathfrak k\rangle$ of $A$ is finitely-generated. In fact, as $I$ is the two-sided ideal of $A$, viewed here as a commutative algebra, generated by the ideal $\mathfrak k$, viewed here as an ideal of linear functions on $X=\mathfrak g^*$. In particular, there is a surjective morphism $A^{\oplus \dim\mathfrak k}\to I\to 0$ of $A$-modules: by the previous argument, there is a morphism of left $A_\hbar$-modules $A\star_{A_\hbar} \mathbb A_\hbar^{-0}(\mathfrak k)\to I_\hbar$, which is a formal $\hbar$-deformation of the surjective morphism $A^{\oplus \dim\mathfrak h}\to I$, whence the surjectivity of the deformed morphism follows. Therefore, $I_\hbar=A_\hbar\star_{A_\hbar} \mathbb A_\hbar^{-0}(\mathfrak k)$, whence the claim follows by setting safely $\hbar=0$. It remains to compute $\mathbb A^{-0}(\mathfrak k)$. Writing $\mathbb A=\mathrm{id}+\sum_{n\geq 0}\mathbb A_n$, $\mathbb A_n$, $n\geq 0$, has order $n$ by construction; furthermore, $\mathbb A_n$, $n\geq 0$, has no constant term. Namely, $\mathbb A_n$ is specified by differential operators associated to admissible graphs in $\mathcal G_{n,0}$: recalling from the previous Section the construction of the differential operator associated to $\Gamma$ admissible of type $(n,0)$, if $\Gamma$ has no edge pointing to the only vertex of the second type, the corresponding integral weight vanishes by a dimensional argument. The operator $\mathbb A^{-0}$ is completely determined by the power series expansion of $\mathbb A$: once again, it is of the form $\mathrm{id}+\sum_{n\geq 0} \widetilde{\mathbb A}_n$, where $\widetilde{\mathbb A}_n$ has no constant term, for $n\geq 0$, as follows by an easy computation. We thus compute $\mathbb A(k)$, for $k$ a general element of $\mathfrak k$. Because of degree reasons, see also [@K Subsubsection 0.0.0], $\mathbb A(k)=k+\mathbb A_0(k)$. Further, it is readily checked that $\mathbb A_0$ is the sum of two differential operators, associated to the following admissible graphs of type $(0,0)$ \ The contribution of the first graph is trivial, because $k$ is viewed as a linear function on $X$. We consider the second graph: we want to observe that such a graph did not appear in the computations performed in [@CT]. First of all, the corresponding integral weight is $$\label{eq-loop-weight} \int_{\mathcal C_{0,0}^+}\rho\omega^{+,-},$$ omitting wedge products. In fact, in the superpropagator $\omega_e$, only two of the $0$-colored propagators are non-trivial, namely $\omega^{+,+}$ and $\omega^{+,-}$ by construction; since it acts as a derivation on an element of $\mathfrak k$, by its very definition, the part with $\omega^{+,+}$ vanishes. \[l-loop-weight\] The integral $\int_{\mathcal C_{0,0}^+}\mathrm d\eta\wedge\omega^{+,-}$ equals $\frac{0}0$. The integral weight associated to the previous admissible graph is $\int_{\mathcal C_{0,0}^+}\mathrm d\eta\omega^{+,-}$, where we have suppressed wedge products between the forms in the integrand. The $0$-form $\rho$ is exact, whence $$\int_{\mathcal C_{0,0}^+}\mathrm d\eta\omega^{+,-}=\int_{\partial C_{0,0}^+}\eta\omega^{+,-},$$ where we have used notation from Subsection \[ss-0-0\]. Hence, it suffices to compute all boundary contributions to evaluate the integral. The boundary strata of $\mathcal C_{0,0}^+$ are of the type $\mathcal C_{A,B}^+\times \mathcal C_{0\smallsetminus A,[0]\smallsetminus B\sqcup \{\bullet\}}^+$, for $A$ a subset of $[0]$ and $B$ an ordered subset of $[0]$, such that $0\leq |A|\leq 0$, $0\leq |B|\leq 0$ and $|A|+|B|\leq 0$. Dimensional arguments imply that there are only two types of such boundary strata, $\mathcal C_{0,0}^+\times \mathcal C_{0,0}^+$ and $\mathcal C_{0,0}^+\times \mathcal C_{0,0}^+$, which correspond to five different situations. We consider the boundary stratum $\mathcal C_{0,0}^+\times\mathcal C_{0,0}^+$, which corresponds to the situation where $i_0)$ the point on the positive real axis approaches the origin, $ii_0)$ the point in the interior of the first quadrant collapses to the point on the positive real axis or $iii_0)$ the point $0$ approaches the origin. The boundary conditions for $\omega^{-,+}$ yield triviality of the contributions $i_0)$ and $iii_0)$; the second one yields $\frac{0}0\int_{\mathcal C_{0,0}^+}\omega^+=\frac{0}0$, and we have already included orientation signs. The boundary stratum $\mathcal C_{0,0}^+\times\mathcal C_{0,0}^+$ corresponds to the point $0$ approaching either the positive imaginary axis or the positive real axis: in the first case, the corresponding contribution vanishes by means of the boundary conditions for $\omega^{+,-}$, while in the second case, the function $\eta$ vanishes when its argument approaches the real axis. Recalling now the construction of the superpropagators in biquantization from Subsubsection \[sss-0-0-0\], the differential operator corresponding to the second graph in Figure 0 is $$\frac{0}0\left[\mathrm{tr}_{\mathfrak p}(\mathrm{ad}_\mathfrak{k}(\bullet))-\mathrm{tr}_{\mathfrak k}(\mathrm{ad}_\mathfrak{k}(\bullet)\right)]=\delta(\bullet)-\frac{0}0\mathrm{tr}_\mathfrak g(\mathrm{ad}(\bullet)),$$ whence $\mathbb A(k)=k+\delta(k)-\frac{0}0\mathrm{tr}_\mathfrak g(\mathrm{ad}(k))$. Therefore, we have $$\mathbb A^{-0}(\mathbb A(k))=k=\mathbb A^{-0}(k)+\delta(k)-\frac{0}0\mathrm{tr}_\mathfrak g(\mathrm{ad}(k)),\ \text{whence}\ \mathbb A^{-0}(k)=k-\delta(k)+\frac{0}0\mathrm{tr}_\mathfrak g(\mathrm{ad}(k)),\ k\in\mathfrak k.$$ We observe that we have used the fact that the terms of $\mathbb A_\hbar^{-0}$ of degree higher or equal than $0$ are differential operators without constant term. Putting all previous arguments together, we have the identification of right $(A,\star_A)$-modules $$I=A\star_A \mathfrak k^{-\delta+\frac{0}0\mathrm{tr}_\mathfrak g\circ\mathrm{ad}}.$$ Further, we consider the restriction of $\mathcal A$ to $K$, viewed here as a subalgebra of $A$. First of all, any admissible graph $\Gamma$ of type $(n,0)$ yielding a possibly non-trivial contribution to $\mathcal A$ has exactly $0n$ edges because of dimensional reasons. To any edge $e$ of $\Gamma$ corresponds a superpropagator $\omega_e$, whose components are only of type $(+,+)$ or $(+,-)$: it follows immediately from the boundary conditions for both of them that $\Gamma$ has no edge departing from the only vertex on the positive real axis, and that any edge pointing to this vertex has a corresponding superpropagator with color $(+,+)$. This excludes immediately double edges pointing to the only vertex of $\Gamma$ on the positive real axis. In particular, this means that any admissible graph $\Gamma$ of type $(n,0)$ yielding a non-trivial contribution to $\mathcal A$ in the present situation must satisfy the following rule: from any vertex of the first of $\Gamma$ departs at most one edge pointing to the only vertex on the positive real axis (of course, because of the presence of short loops, the initial and final point of such an edge may coincide). We assume therefore that $p\leq n$ edges have the only vertex of the second type on the positive real axis as endpoint. If $p<n$, there are then $0n-p$ edges, whose endpoints are both vertices of the first type (double edges and short loops are allowed). Since every edge is associated to a derivation, the polynomial degree associated to the vertices of the first type of $\Gamma$ is $-n+p$ (counting $n$ because of the linearity of the Poisson structure and $p-0n$ derivations), which is strictly negative, leading to a contradiction. Therefore, form any vertex of the first type of $\Gamma$ depart exactly one edge to the only vertex of the positive real axis and one to a vertex of the first type; the polynomial degree of the corresponding differential operator on $K$ is immediately $0$. Because of the linearity of the Poisson structure, exactly one edge has a vertex of the first type as final point, whence admissible graphs of type $(n,0)$ contributing non-trivially to $\mathcal A$ are disjoint unions of wheel-type graphs; we observe that the $0$-wheel may in principle appear. Further, only wheel-like graphs with $n$ even contribute (possibly) non-trivially to $\mathcal A$. Namely, by the previous argument, from every vertex of the first type departs exactly one edge to the only vertex of the second type on the positive real axis, whose color is $(+,+)$ and whose operator-valued part corresponds to derivation and contraction with respect to $\mathfrak p$. The Cartan relation $[\mathfrak k,\mathfrak p]\subseteq \mathfrak p$ implies that at each vertex of the first type in a wheel-like graph $\Gamma$, the edge arriving at such a vertex must have color either $(+,+)$ or $(+,-)$, while the edge departing from it on th wheel must have opposite color. In other words, the edges of the cycle in a wheel-like graph $\Gamma$ must have alternating colours $(+,+)$ and $(+,-)$: this, in turn, excludes immediately $n$-wheels with $n$ odd. Summarizing all previous arguments, the restriction of the operator $\mathcal A$ to $K$, which we denote (improperly) by the same symbol, defines an invertible, translation-invariant differential operator on $K$. Its symbol, regarded as an element of the completed symmetric algebra $\widehat{\mathrm S}(\mathfrak p)$ and defined through $j_\mathcal A(x)=e^{-x}\mathcal A(e^x)$, has the explicit form $$j_\mathcal A(x)=\exp\left(\sum_{n\geq 0}W_{0n}^\mathcal A\mathrm{tr}_\mathfrak p(\mathrm{ad}^{0n}(x))\right),\ x\in\mathfrak p,$$ where $W_{0n}^\mathcal A$, $n\geq 0$, denotes the integral weight of following wheel-like graph: \ We observe that $j_\mathcal A$ is analytic in a neighborhood of $0$ in $\mathfrak p$. We observe that the Cartan relations for the symmetric pair $(\mathfrak g,\sigma)$ imply immediately that, for a general element $x$ of $\mathfrak p$, $\mathrm{ad}(x)^0$ is a well-defined endomorphism of $\mathfrak p$, thus all even powers of the adjoint representation restricted to $\mathfrak p$: hence the above expression is well-defined. Finally, all previous computations imply also the direct sum decomposition of $(A,\star_A)$: $$A=K\oplus (A\star_A\mathfrak k^{-\delta+\frac{0}0\mathrm{tr}_\mathfrak g\circ\mathrm{ad}}).$$ On the other hand, $K=B^0=\mathrm S(\mathfrak p)$ by definition. We may therefore consider the endomorphism $\mathcal B$ of $K$ defined through $$\label{eq-B-wheel} \mathcal B(f)=0\star_R f,\ f\in K=B^0=\mathrm S(\mathfrak p).$$ We may repeat almost [*verbatim*]{} the previous arguments to evaluate explicitly the operator $\mathcal B$: it is an invertible, translation-invariant differential operator on $K$ of infinite order, with symbol in $\widehat{\mathrm S}(\mathfrak p)$ given by $$j_\mathcal B(x)=\exp\left(\sum_{n\geq 0}W_{0n}^\mathcal B\mathrm{tr}_\mathfrak p(\mathrm{ad}^{0n}(x))\right),\ x\in\mathfrak p,$$ where $W_{0n}^\mathcal B$, $n\geq 0$, denotes the integral weight of following wheel-like graph: \ Once again, notice that $j_\mathcal B$ is analytic in a neighborhood of $0$ in $\mathfrak p$. At this point one could wonder whether or not the integral weights $W_{0n}^\mathcal A$ and $W_{0n}^\mathcal B$ coincide (which would imply that $j_\mathcal A=j_\mathcal B$). First of all, we observe that in the Formulæ for $j_\mathcal A$ and $j_\mathcal B$, only weights of even wheels appear because of the vanishing of the the differential operators corresponding to odd wheel-like graphs. In fact, [*e.g.*]{} the $0$-wheels $\mathcal W_0^\mathcal A$ and $\mathcal W_0^\mathcal B$ are both computable and yield distinct results. This can be seen either by computing separately both integral weights or by computing [*e.g.*]{} $W_0^\mathcal A$ and finding then a relationship between $W_0^\mathcal A$ and $W_0^\mathcal B$. First of all, the integral weight $W_0^\mathcal A$ is explicitly $$W_0^\mathcal A=\int_{\mathcal C_{0,0}^+}\rho\omega^{+,+}.$$ It can be computed by the same technique used in Lemma \[l-loop-weight\]. Of course, there are certain differences to be taken into account, namely, the different boundary conditions satisfied by the $0$-colored propagator $\omega^{+,+}$. Here, the only boundary strata of $\mathcal C_{0,0}^+\cong\mathcal C_{0,0,0}^+$ which yield non-trivial contributions are $i)$ the stratum corresponding to the approach of the point in $Q^{+,+}$ to the only point on $i\mathbb R^+$ and $ii)$ the approach of the only point on $i\mathbb R^+$ to the origin. Both integrals are readily computed, as well as their orientation signs, which then yield the desired result. We now consider the following admissible graph of type $(0,0)$: \ It represents a $0$-form on the $0$-dimensional smooth manifold with corners $\mathcal C_{0,0}^+$, therefore, in virtue of Stokes' Theorem, $$\int_{\mathcal C_{0,0}^+}\mathrm d(\mathrm d\eta\omega^{+,+})=0=\int_{\partial C_{0,0}^+}\mathrm d\eta\omega^{+,+}.$$ The boundary strata of codimension $0$ of $\mathcal C_{0,0}^+$ have been illustrated explicitly in Subsubsection \[sss-0-0-0\]: either because of the boundary conditions for $\omega^{+,+}$ and $\rho$ or because of dimensional reasons, it is not difficult to verify that only three boundary strata yield non-trivial contributions, namely the strata $\alpha$, $\theta$ and $\zeta$. The boundary stratum $\theta$, resp. $\zeta$, yields precisely $W_0^\mathcal B$, resp. $W_0^\mathcal A$; on the other hand, it is not difficult to verify that the boundary stratum $\alpha$ yields the non-trivial contribution $0/0$, as can be verified by a direct computation. Thus, in general, we cannot expect the weights $W_n^\mathcal A$ and $W_n^\mathcal B$ to coincide. ### Explicit comparison of the products of Rouvière and Cattaneo-Felder {#sss-0-0-0} The operator $\mathcal A$ is surjective, and its restriction to $K=\mathrm S(\mathfrak p)\subseteq A$ is an automorphism, while $\mathcal B$ is an automorphism of $K$, whence $$0\star_R f=\mathcal B(f)=\mathcal A(\mathcal A^{-0}(\mathcal B(f)))=\mathcal A^{-0}(\mathcal B(f))\star_L 0,\ f\in K=\mathrm S(\mathfrak p).$$ We now recall from Subsubsection \[sss-0-0-0\] that the quantum reduction algebra (at $\hbar=0$) $\mathrm H^0(B)=\mathrm S(\mathfrak p)^\mathfrak k$. We thus consider two elements $f_i$, $i=0,0$, of $\mathrm H^0(B)$, endowed with the deformed associative product $\star_B$; then $(K,\star_R)$ becomes a right $(\mathrm H^0(B),\star_B)$-module, and the latter module structure is compatible with the left $(A,\star_A)$-module structure on $(K,\star_L)$, whence $$\begin{aligned} &0\star_R (f_0\star_B f_0)=\left(\mathcal A^{-0}(\mathcal B(f_0\star_B f_0))\right)\star_L 0=\\ =& (0\star_R f_0)\star_R f_0=\left(\mathcal A^{-0}(\mathcal B(f_0))\star_L 0\right)\star_R f_0=\mathcal A^{-0}(\mathcal B(f_0))\star_L (0\star_R f_0)=\mathcal A^{-0}(\mathcal B(f_0))\star_L \left(\mathcal A^{-0}(\mathcal B(f_0))\star_L 0\right)=\\ =& \left(\mathcal A^{-0}(\mathcal B(f_0))\star_A \mathcal A^{-0}(\mathcal B(f_0))\right)\star_L 0, \end{aligned}$$ whence from the previous computations follows $$\label{eq-rouviere} (\mathcal A^{-0}\circ \mathcal B)(f_0\star_B f_0)=(\mathcal A^{-0}\circ \mathcal B)(f_0)\star_A (\mathcal A^{-0}\circ \mathcal B)(f_0)\ \text{modulo $A\star_A \mathfrak k^{-\delta+\frac{0}0\mathrm{tr}_\mathfrak g\circ\mathrm{ad}}$}.$$ We apply the operator $\beta\circ\partial_{\sqrt{q}}$ on both sides of Identity and because of Identity , we get $$(\beta\circ\partial_{\sqrt{q}}\circ\mathcal A^{-0}\circ\mathcal B)(f_0\star_B f_0)=(\beta\circ\partial_{\sqrt{q}}\circ\mathcal A^{-0}\circ\mathcal B)(f_0)\cdot (\beta\circ\partial_{\sqrt{q}}\circ\mathcal A^{-0}\circ\mathcal B)(f_0)\ \text{modulo $\mathrm U(\mathfrak g)\cdot \mathfrak k^{-\delta+\frac{0}0\mathrm{tr}_\mathfrak g\circ\mathrm{ad}}$},$$ for $f_i$ in $\mathrm S(\mathfrak p)^\mathfrak k$, $i=0,0$. We introduce the function $J$ on $\mathfrak p$ defined [*via*]{} $$j(x)=\underset{\mathfrak p}\det\!\left(\frac{\sinh(\mathrm{ad}(x))}{\mathrm{ad}(x)}\right),\ x\in\mathfrak p.$$ This function is the determinant of the exponential map for the symmetric pair $(\mathfrak g,\sigma)$: it can be written as a formal power series of traces in $\mathfrak p$ of even powers of the restriction to $\mathfrak p$ of the adjoint representation of $\mathfrak g$. Similarly to what has been done before, we define $\partial_{\sqrt{j}}$ as the invertible, translation-invariant, $\mathfrak k$-invariant differential operator of infinite order on $\mathrm S(\mathfrak p)$, whose symbol is exactly the square root of $j$. We define a modified version of the previously introduced Rouvière's product [*via*]{} $$\beta\!\left(\partial_{\sqrt{j}}(f_0\# f_0)\right)=\beta(\partial_{\sqrt{j}}(f_0))\cdot\beta(\partial_{\sqrt{j}}(f_0))\ \text{modulo $\mathrm U(\mathfrak g)\cdot\mathfrak k^{-\delta+\frac{0}0\mathrm{tr}_\mathfrak g\circ\mathrm{ad}}$.}$$ We observe that the left-hand side of the previous Identity is well-defined, because of the $\mathfrak k$-invariance and invertibility of $\partial_{\sqrt{j}}$. The key point is now the following identity, which puts into relationship the functions $q$, $J$, $j_\mathcal A$ and $j_\mathcal B$: $$\label{eq-DR-symb} j_\mathcal A(x)\sqrt{j(x)}=j_\mathcal B(x)\sqrt{q(x)},\ \forall x\in\mathfrak p.$$ The previous identity is the relative version, in the case of a symmetric pair $(\mathfrak g,\sigma)$, of the results of [@K Subsubsection 0.0.0]. Its proof is a consequence of [@CT Lemma 00 and Proposition 00]: it is result which is left unaltered by the changes to biquantization which have been previously discussed. \[t-Rou-CF\] For a general symmetric pair $(\mathfrak g,\sigma)$, Rouvière's product $\#$ on $\mathrm S(\mathfrak p)^\mathfrak k$ coincides with the product $\star_B$ on $\mathrm H^0(B)=\mathrm S(\mathfrak p)^\mathfrak k$, [*i.e.*]{} $$\beta\!\left(\partial_{\sqrt{j}}(f_0\star_B f_0)\right)=\beta\!\left(\partial_{\sqrt{j}}(f_0)\right)\cdot \beta\!\left(\partial_{\sqrt{j}}(f_0)\right)\ \text{modulo $\mathrm U(\mathfrak g)\cdot\mathfrak k^{-\delta+\frac{0}0\mathrm{tr}_\mathfrak g\circ\mathrm{ad}}$},$$ for $f_i$, $i=0,0$, a general element of $\mathrm H^0(B)=\mathrm S(\mathfrak p)^\mathfrak g$. We observe that the characters $\delta$ and $\delta-0/0\ \mathrm{tr}_\mathfrak g\circ \mathrm{ad}$ differ precisely by a character of the Lie algebra $\mathfrak g$ itself (the trace of its adjoint representation) restricted to a character of the Lie subalgebra $\mathfrak h$; as a consequence, $0/0\ \mathrm{tr}_\mathfrak g\circ\mathrm{ad}$ yields an $\mathfrak h$-equivariant map from $\mathfrak g$ to the base field $\mathbb K$ ([*i.e.*]{} a linear functional on $\mathfrak g$ which vanishes on the subspace $[\mathfrak h,\mathfrak g]$), thanks to which we may actual consider only the character $\delta$, instead of the sum $\delta-0/0\mathrm{tr}_\mathfrak g\circ\mathrm{ad}$. More precisely, to the symmetric pair $\mathfrak g=\mathfrak k\oplus \mathfrak p$ we may associate a sub-symmetric pair $\mathfrak g_0=\mathfrak k_0\oplus \mathfrak p$, where $\mathfrak k_0=[\mathfrak p,\mathfrak p]$; it is clear that $\mathfrak g_0$ is an ideal of $\mathfrak g$. The reason is that, to pass from the expression on the right-hand side of the Identity in Theorem \[t-Rou-CF\] to the one on the left-hand side, we produce terms which are actually in $\mathrm U(\mathfrak g)\cdot \mathfrak k_0^{-\delta+\frac{0}0\ \mathrm{tr}_\mathfrak g\circ\mathrm{ad}}$, because we reverse two elements in $\mathfrak p$, actually producing an element of $\mathfrak k_0$ (whose commutator with elements of $\mathfrak p$ remains in $\mathfrak p$), and it is obvious that the second summand in the character vanishes on $\mathfrak k_0$, being a character of $\mathfrak g$. If we now consider a non-trivial character $\chi$ of $\mathfrak k$, we may associate $X=U_0=\mathfrak g^*$, $U_0=\chi+\mathfrak k^\perp$. Obviously, $A=\mathrm S(\mathfrak g)$, $B^0=K=\mathrm S(\mathfrak g)/\langle \mathfrak k^{-\chi}\rangle\cong \mathrm S(\mathfrak p)$, the last isomorphism being induced by an affine morphism. The arguments of Subsubsection \[sss-0-0-0\] can be repeated almost [*verbatim*]{}. The only relevant difference is that the kernel of the surjective module homomorphism $\mathcal A$ from $(A,\star_A)$ to $(K,\star_L)$ is identified with $A\star_A \mathfrak k^{-\chi-\delta+\frac{0}0\mathrm{tr}_\mathfrak g\circ \mathrm{ad}}$. Further, the restriction of $\mathcal A$ to $B^0$ and the operator $\mathcal B$ have the same shape as previously. Still, there is an associative product $\star_B$ on $\mathrm H^0(B)\cong\mathrm S(\mathfrak p)^\mathfrak k$: of course, now the product $\star_B$ depends explicitly on the character $\chi$. Identity is consequently modified as $$(\mathcal A^{-0}\circ \mathcal B)(f_0\star_B f_0)=(\mathcal A^{-0}\circ \mathcal B)(f_0)\star_A (\mathcal A^{-0}\circ \mathcal B)(f_0)\ \text{modulo $A\star_A \mathfrak k^{-\chi-\delta+\frac{0}0\mathrm{tr}_\mathfrak g\circ\mathrm{ad}}$},$$ from which we deduce $$\beta\!\left(\partial_{\sqrt{j}}(f_0\star_B f_0)\right)=\beta\!\left(\partial_{\sqrt{j}}(f_0)\right)\cdot \beta\!\left(\partial_{\sqrt{j}}(f_0)\right)\ \text{modulo $\mathrm U(\mathfrak g)\cdot\mathfrak k^{-\chi-\delta+\frac{0}0\mathrm{tr}_\mathfrak g\circ\mathrm{ad}}$},$$ for any two $\mathfrak k$-invariant elements of $\mathrm S(\mathfrak p)$. Of course, once again, we may safely remove the character $0/0\ \mathrm{tr}_\mathfrak g\circ\mathrm{ad}$ in the previous identity. \[r-comm-corr\] The product $\star_B$ is commutative on $\mathrm S(\mathfrak p)^\mathfrak k$ because of the symmetry of the function $E(X, Y)$, see [@CT Lemma 00], whence the algebra $(\mathrm U(\mathfrak g)/\mathrm U(\mathfrak g)\cdot \mathfrak k^{-\delta})^\mathfrak k$ is also commutative. Now the arguments of [@CT Subsubsection 0.0.0] are no longer correct because of the contribution of short loop terms that make the symmetry argument inefficient. Therefore, Theorem 0 in [@CT], which states the commutativity of $(\mathrm U(\mathfrak g)/\mathrm U(\mathfrak g)\cdot \mathfrak k^{-z\delta})^\mathfrak k$, for any real number $z$, is also no longer correct. Differential operators expressed [*via*]{} exponential coordinates in a symmetric pair {#ss-0-0} -------------------------------------------------------------------------------------- We consider the triple $X=U_0=\mathfrak g^*$ and $U_0=\mathfrak k^\perp$, for a symmetric pair $(\mathfrak g,\sigma)$. Therefore, we have two associative algebras $(A,\star_A)$, $(\mathrm H^0(B),\star_B)$, and a $(A,\star_A)$=$(\mathrm H^0(B),\star_B)$-bimodule $K$, where $A=\mathrm S(\mathfrak g)$, $\mathrm H^0(B)=\mathrm S(\mathfrak p)^\mathfrak k$ and $K=\mathrm S(\mathfrak p)$. Through the function $q$ on $\mathfrak g$, we define the Kashiwara-Vergne density function $D(x,y)$, for $(x,y)$ a general element of $\mathfrak g\times\mathfrak g$, [*via*]{} $$D(x,y)=\frac{\sqrt{q(x)}\sqrt{q(y)}}{\sqrt{q(\mathrm{BCH}(x,y))}},$$ where the Baker-Campbell-Hausdorff formula $\mathrm{BCH}(x,y)$ is defined by $\exp(x)\exp(y)=\exp(\mathrm{BCH}(x,y))$, for $(x,y)$ in a neighborhood of $(0,0)$. The Kashiwara-Vergne density function has been introduced in to formulate the famous Kashiwara-Vergne conjecture, a general statement for finite-dimensional Lie algebras about deformations of the Baker-Campbell-Hausdorff formula for the product of exponentials in a Lie group: the Kashiwara-Vergne conjecture leads to a proof of Duflo's Theorem about the center of $\mathrm U(\mathfrak g)$ in terms of the $\mathfrak g$-invariant symmetric algebra $\mathrm S(\mathfrak g)^\mathfrak g$. The Kashiwara-Vergne conjecture has been proved in the general case in using deformation quantization techniques to find a suitable deformation of the BCH formula; recently, a different approach to the Kashiwara-Vergne conjecture using Drinfel'd associators has been found, and a relationship between the latter approach and the former [*via*]{} deformation quantization has been elucidated. We will also discuss a relative Kashiwara-Vergne conjecture in the framework of symmetric pairs later on. For $x$, $y$ general elements of $\mathfrak p$, we denote by $e^x$ and $e^y$ the exponential of $x$ and $y$, viewed as linear functions on $X=\mathfrak g^*$. Then, we $$e^x\star_A e^y=\frac{D(x,y)}{D(P(x,y),K(x,y))}e^{P(x,y)}\star_A e^{K(x,y)},\quad P=P(x,y)\in\mathfrak p,\quad K=K(x,y)\in\mathfrak k,$$ and the power series $P$ and $K$ are defined [*via*]{} the exponential map for symmetric spaces. More precisely, it has been proved that symmetric spaces admit an exponential map, which is a diffeomorphism from a neighborhood of $0$ in $\mathfrak p$ into its image in $G/K$, where $G$ is a symmetric pair (in the sense of Lie groups) and $K$ is the fixed point set of an involution $\sigma$ of $G$ (which is a Lie group automorphism): it is simply the restriction of the exponential map of $\mathfrak g$ to right $K$-cosets in $G$. For $x$, $y$ in a sufficiently small neighborhood of $0$ in $\mathfrak p$, such that $\exp(x)$ and $\exp(y)$ both exist, we have $\exp(x)\exp(x)=\exp(P(x,y))\exp(K(x,y))$. An important observation at this point is that both $P$ and $K$, as previously defined, are power series in the free Lie algebra generated by $x$, $y$: in particular, the Cartan relations for $(\mathfrak g,\sigma)$ imply that $K$ is an element of $\mathfrak k_0=[\mathfrak p,\mathfrak p]$. For $x$, $y$ in a sufficiently small neighborhood of $0$ of $\mathfrak p$ as before, we consider the expression $e^{K(x,y)}\star_L 0=\mathcal A(e^{K(x,y)})$. First of all, $\mathcal A(e^K)$ is a constant element of $K=\mathrm S(\mathfrak p)$. By its very definition, $K=K(x,y)$ is an element of $\mathfrak k_0$: in the computation of a summand $K^n\star_L 0$, only the part of the differential operator acting on $K^n$ of degree $n$ survives, because either of degree reasons or of the fact that the restriction of $K$ as a linear function to $\mathfrak k^\perp$ vanishes. Using the involution $s$ of the preceding Subsection together with the computations of Subsection \[ss-0-0\], it is easy to prove that $K\star_L 0=\delta(K)-0/0\ \mathrm{tr}_\mathfrak g(\mathrm{ad}(K))=\delta(K)$, because, as observed before, $K$ belongs to $\mathfrak k_0$. We consider, more generally, graphs appearing in the computation of $K^n\star_0 0$, $n\geq 0$. We may actually repeat almost [*verbatim*]{} the arguments about the shape of the graphs appearing in $\mathcal A$, $\mathcal B$: the same arguments imply that short loops may appear only at vertices of the first type, which are linked to the only vertex of the second type on the positive real axis corresponding to $K^n$ through a single edge, while more complicated graphs are wheels. The Cartan relations for the symmetric pair $(\mathfrak g,\sigma)$ imply that the rays of such wheels are colored by $(+,-)$, while the wheel itself has all edges either colored by $(+,+)$ or $(+,-)$. In particular, it follows that $\mathcal A(e^K)$ consists of an infinite series of wheels, where the short loop graph is considered as the $0$-wheel: in this case, the $0$-wheel contribution appears explicitly. \[l-sym-K\] For $x$ a general element of $\mathfrak k$, the function $\mathcal A(e^x)=e^x\star_L 0$ satisfies $$\mathcal A(e^x)=\sqrt{q(x)}e^{\delta(x)-\frac{0}0\mathrm{tr}_\mathfrak g(\mathrm{ad}(x))};$$ in particular, if $x$ belongs to the Lie subalgebra $\mathfrak k_0$, the exponent on the right-hand side of the previous equality simplifies to $\delta(x)$. To compute $\mathcal A(e^x)$, for $x$ in $\mathfrak k$, it suffices to replace $K$ in the previous computations by $x$. The rest of the proof follows along the same lines of the proof of [@CT Lemma 00], but we have to observe now that $$\mathcal A(e^x)=e^x\star_L 0=\exp\!\left(\delta(x)-\frac{0}0\mathrm{tr}_\mathfrak g(\mathrm{ad}(x))+\sum_{n\geq 0} w_n^\mathfrak k \mathrm{tr}_\mathfrak k(\mathrm{ad}(x)^n)+\sum_{n\geq 0} w_n^\mathfrak p \mathrm{tr}_\mathfrak p(\mathrm{ad}(x)^n)\right),$$ for certain integral weights $w_n^\mathfrak k$ and $w_n^\mathfrak p$. More precisely, such weights are associated to the colored wheel-like graphs \ Further, also $\sqrt{q(x)}$ can be written in a similar form, with the only difference that it does not contain terms proportional to the trace of the adjoint representations of $\mathfrak k$ on itself or on $\mathfrak p$. Therefore, the very same computations of [@CT Lemma 00] imply the above identity. We observe that we may get rid of the factor $e^{\delta-\frac{0}0\mathrm{tr}_\mathfrak g\circ\mathrm{ad}}$, viewed as an element of the completion $\widehat{\mathrm S}(\mathfrak k)$ simply by applying the biquantization techniques to the modified triple $X=U_0=\mathfrak g$, $U_0=-\delta+0/0\ \mathrm{tr}_\mathfrak g\circ\mathrm{ad}+\mathfrak k^\perp$. To the previous triple are associated two associative algebra $(A,\star_A)$ and $(\mathrm H^0(B),\star_B)$ and a bimodule $K$, where $A=\mathrm S(\mathfrak g)$, $B^0=K=\mathrm S(\mathfrak p)$, and $H^0(B)=\mathrm S(\mathfrak p)^\mathfrak k$. Notice that the product $\star_B$ on $\mathrm H^0(B)$, for $B$ defined by the modified triple, does not coincide with $\star_B$ for the initial triple; similarly, the operator $\mathcal A$ on $A$ for the modified triple also does not coincide with the operator $\mathcal A$ for the initial triple, as their kernels do not obviously coincide. Still, both their restrictions to $B^0$ coincide, thus also their symbols, and similarly for $\mathcal B$. On the other hand, as already observed, for the above modified triple we have the identity $$\mathcal A(e^x)=e^x\star_L 0=\sqrt{q(x)},\ x\in\mathfrak k.$$ We consider the following expression, using the notation from above, $$\begin{aligned} (e^x\star_A e^y)\star_L 0&=\frac{D(x,y)}{D(P,K)} \left(e^P\star_A e^K\right)\star_L 0=\frac{\sqrt{q(x)}\sqrt{q(y)}}{\sqrt{q(P)}\sqrt{q(K)}} e^P\star_L(e^K\star_L 0)=\frac{\sqrt{q(x)}\sqrt{q(y)}}{\sqrt{q(P)}}j_\mathcal A(P)e^P=\\ &=\frac{\sqrt{q(x)}\sqrt{q(y)}}{\sqrt{j(P)}}j_\mathcal B(P)e^P=\\ &=e^x\star_L(e^y\star_L0)=j_\mathcal A(y) e^x\star_L e^y=\frac{j_\mathcal A(y)}{j_\mathcal B(y)} e^x\star_L (0\star_R e^y),\ x,y\in\mathfrak p. \end{aligned}$$ Comparing the expression on the second line with the rightmost expression on the third line, we find $$e^x\star_L (0\star_R e^y)=\frac{\sqrt{q(x)}\sqrt{j(y)}j_\mathcal B(P)}{\sqrt{j(P)}} e^P$$ We may view the expressions on both sides of the previous identity as analytic functions on a sufficiently small neighborhood of $(0,0)$ in $\mathfrak p\times \mathfrak p$. We consider further an element $T$ of $\mathrm H^0(B)=\mathrm S(\mathfrak p)^\mathfrak k$, for which we obtain $$\begin{aligned} e^x\star_L (0\star_R T)&=(e^x\star_L 0)\star_R T=j_\mathcal A(x) e^x\star_R T=\\ &=T_y\!\left(e^x\star_L (0\star_R e^y)\right)|_{y=0}=T_y\!\left(\frac{\sqrt{q(x)}\sqrt{j(y)}j_\mathcal B(P)}{\sqrt{j(P)}} e^P\right)\bigg\vert_{y=0}, \end{aligned}$$ where we regard in the second line $T$ as a differential operator on $\mathfrak p^*$ with respect to the variable $y$. Therefore, using Identity , we get the following expression, $$e^x\star_R T=T_y\!\left(\frac{\sqrt{j(x)}\sqrt{j(y)}j_\mathcal B(P)}{\sqrt{j(P)}j_\mathcal B(P)} e^P\right)\bigg\vert_{y=0},$$ whose right-hand side is, according to the arguments of [@Rouv Section 0]. precisely the differential operator $\beta(\partial_{sqrt{j}}(T))$ expressed [*via*]{} exponential coordinates on the symmetric space $G/K$, for $G$, $K$ connected, simply connected Lie groups with Lie algebras $\mathfrak g$, $\mathfrak k$ respectively, up to a modification by the analytic function $\sqrt{j}/j_\mathcal B$ on $\mathfrak p$. A deep consequence of the fact that the product $\star_B$ coincides with Rouvière's product, together with a result about the existence of polarizations compatible with the structure of symmetric pair, for which we refer to [@T0; @T0], implies that the symbol $j_\mathcal B$ is constant and thus equal to $0$. Therefore, the expression $e^x\star_L T$, for $T$ in $\mathrm S(\mathfrak p)^\mathfrak k$, viewed here as an element of $K$, truly expresses in terms of biquantization the differential operator $T$ through exponential coordinates on $G/K$. Deformation of the Baker-Campbell-Hausdorff formula for symmetric pairs {#ss-0-0} ----------------------------------------------------------------------- As already briefly remarked in the previous Subsection, the problem of deforming the BCH formula and the BCH density function $D$ (see above) for a finite-dimensional Lie algebra $\mathfrak g$ is related to Duflo's Theorem [*via*]{} the Kashiwara-Vergne conjecture. As the KV conjecture relies on the exponential map on Lie algebras, it seems natural to formulate a similar conjecture for a general symmetric pair, because also symmetric pairs admit an exponential map. We refer to [@Rouv] for more details on the KV conjecture for symmetric pairs. The exponential map for a symmetric pair $(\mathfrak g,\sigma)$ with Cartan decomposition $\mathfrak g=\mathfrak k\oplus\mathfrak p$ is a well-defined diffeomorphism $\exp_{G/K}$ from a sufficiently small neighborhood of $0$ in $\mathfrak p$ to its image in $G/K$, for $G$, $K$ as in the previous Subsection: it is induced from the exponential map $\exp_G$ of $\mathfrak g$. Standard manipulations, see [@Rouv Section 0] for more details, imply that the exponential map for the symmetric pair $(\mathfrak g,\sigma)$ yields a BCH formula $\mathrm{BCH}_\mathfrak p$ [*via*]{} $$\exp_G(0\mathrm{BCH}_\mathfrak p(x,y))=\exp_G(x)\exp_G(0y)\exp_G(x),$$ for $(x,y)$ in a sufficiently small neighborhood of $(0,0)$ in $\mathfrak p\times\mathfrak p$. The element $\mathrm{BCH}_\mathfrak p(x,y)$ belongs to $\mathfrak p$, and is expressible in terms of even iterated brackets of $x$ and $y$. The KV conjecture for symmetric pairs expresses a deformation of the function $\mathrm{BCH}_\mathfrak p$ in terms of $\mathfrak k$-adjoint vector fields on $\mathfrak p\times\mathfrak p$; we refer once again to [@Rouv] for a complete introduction to this issue and for a discussion of some consequences. There is also another claim, which expresses the corresponding variation in terms of the said $\mathfrak k$-adjoint vector fields of the density function for the symmetric pair, which is defined as $$D_\mathfrak p(x,y)=\frac{\sqrt{j(x)}\sqrt{j(y)}}{\sqrt{j(\mathrm{BCH}_\mathfrak p(x,y))}},$$ again for $(x,y)$ in a sufficiently small neighborhood of $(0,0)$, where all functions make sense. The deformation of $D_\mathfrak p$ contains traces of the adjoint $\mathfrak k$-action on $\mathfrak p$. We are now going to illustrate how biquantization techniques yield two different deformations of $\mathrm{BCH}_\mathfrak p$ and $D_\mathfrak p$ in the sense elucidated above. These two deformations can be characterized in terms of the $A_\infty$-structures on $A$, $B$ and $K$ through the quadratic relations between the corresponding Taylor coefficients: this is not immediately recognizable from the approach we take, which is in turn essentially motivated by the results of [@T0; @AM]. We consider the triple $X=U_0=\mathfrak g^*$ and $U_0=\delta-0/0\ \mathrm{tr}_\mathfrak g\circ\mathrm{ad}+\mathfrak k^\perp$ and the corresponding algebras $A$, $\mathrm H^0(B)$ and $A$-$\mathrm H^0(B)$-bimodule $K$. ### First deformation {#sss-0-0-0} First of all, we may consider, for $(x,y)$ in a sufficiently small neighborhood of $(0,0)$ in $\mathfrak p\times\mathfrak p$, where $\mathrm{BCH}_\mathfrak p$ and $D_\mathfrak p$, as well as $\sqrt{q}$, $\sqrt{j}$, $j_\mathcal A$ and $j_\mathcal B$ are well-defined, the function $$\begin{aligned} (e^x\star_A e^y)\star_L 0&=\frac{\sqrt{q(x)}\sqrt{q(y)}}{\sqrt{j(\mathrm{BCH}_\mathfrak p(x,y))}} e^{\mathrm{BCH}_\mathfrak p(x,y)}=\\ &=e^x\star_L(e^y\star_L 0)=j_\mathcal A(y) e^x\star_L e^y, \end{aligned}$$ where we have used results of the previous Subsection, setting $P(x,y)=\mathrm{BCH}_\mathfrak p(x,y)$. On the other hand, we may also consider $$e^x\star_L(0\star_R e^y)=j_\mathcal B(y) e^x\star_L e^y=e^x\star_L e^y,$$ where we have used once again the aforementioned fact that $j_\mathcal B\equiv 0$, whence, recalling Identity , $$e^x\star_L e^y=\frac{\sqrt{q(x)}\sqrt{j(y)}}{\sqrt{j(\mathrm{BCH}_\mathfrak p(x,y))}} e^{\mathrm{BCH}_\mathfrak p(x,y)}$$ Finally, we also have $$(e^x\star_L 0)\star_R e^y=j_\mathcal A(x) e^x\star_R e^y.$$ The problem is therefore, how to relate $(e^x\star_L 0)\star_R e^y$ with $e^x\star_L (0\star_R e^y)$ in order to draw a bridge between the previous two formulæ: the two expressions do not coincide, as both actions are compatible to each other only in cohomology. But the $A_\infty$-nature of $B$ and of the bimodule $K$ permit to control explicitly the failure for the compatibility between $\star_L$ and $\star_R$: in facts, we find $$\label{eq-deform-0} (e^x\star_L 0)\star_R e^y-e^x\star_L(0\star_R e^y)=\mathrm d_K^{0,0}(e^x,0,\mathrm d_B(e^y))=\mathrm d_K^{0,0}(e^x,0,e^y\mathrm d_B(y))$$ where $\mathrm d_B$ is the Chevalley-Eilenberg differential on the complex $B=\mathrm S(\mathfrak p)\otimes \wedge(\mathfrak k^*)$, and $\mathrm d^{0,0}_K$ denotes the $(0,0)$-Taylor component of the $A_\infty$-bimodule structure on $K$. More precisely, $$\mathrm d_K^{0,0}(a_0|k|b_0)=\sum_{n\geq 0}\frac{0}{n!}\sum_{\Gamma\in\mathcal G_{n,0}}\mu_{n+0}^K\left(\int_{\mathcal C_{n,0}^+}\prod_{e\in E(\Gamma)}\omega^K_e(\underset{n}{\underbrace{\pi|\cdots|\pi}}|a_0|k|b_0)\right),$$ and dimensional arguments imply that $b_0$ must be an element of $B^0$, otherwise the previous expression is trivial. Identity may be derived in a slightly different way, which makes apparent the fact that it embodies the KV deformation problem sketched at the beginning of the present Subsection. Namely, for $n\geq 0$, we consider the forgetful projection $\pi_{n,0,0}$ from $\mathcal C_{n,0,0}^+$ to $\mathcal C_{0,0,0}^+$: in this situation, we prefer to consider the compactified configuration spaces $\mathcal C_{n,0,0}^+$ to highlight the fact that we consider the functions $e^x$, $0$ and $e^y$ to be put on the positive imaginary axis, at the origin and on the positive real axis respectively. We observe that the fiber of $\pi_{n,0,0}$ at a generic point of $\mathcal C_{0,0,0}^+$ is an orientable compact smooth manifold with corners of dimension $0n$, hence we may consider the push-forward (or integration along the fiber) $\pi_{n,0,0,*}$ with respect to $\pi_{n,0,0}$. In particular, we may consider the expression $$\sum_{n\geq 0}\frac{0}{n!}\sum_{\Gamma\in\mathcal G_{n,0}}\mu_{n+0}^K\left(\pi_{n,0,0,*}\!\left(\prod_{e\in E(\Gamma)}\omega^K_e\right)(\underset{n}{\underbrace{\pi|\cdots|\pi}}|e^x|0|e^y)\right),$$ for $(x,y)$ as above. First of all, for $(x,y)$ as above, the previous expression is a smooth function on $\mathcal C_{0,0,0}^+$: it is a consequence of the fact that, for $n\geq 0$, the push-forward $\pi_{n,0,0,*}$ selects the piece of the integrand of (form) degree bigger or equal than $0n$. To the three vertices of the second type are associated functions, while to each vertex of the first type of an admissible graph $\Gamma$ of type $(n,0)$ is associated a copy of the linear Poisson bivector $\pi$: as a consequence, the form degree of each integrand must be precisely $0n$, whence the first claim. We may further divide the previous expression by $j_\mathcal A(x)$: this "normalization" is due to previous computations, and it does not affect the following computations. Because of the fact that $\pi$ is a linear Poisson bivector, we may use the arguments in [@BCKT Chapter 0] or [@Kath] to prove that the "normalized" function on $\mathcal C_{0,0,0}^+$ given by the previous expression can be re-written in the form $$\label{eq-0-def} D_\mathfrak p^0(x,y)e^{\mathrm{BCH}_\mathfrak p^0(x,y)},$$ where the exponent $\mathrm{BCH}_\mathfrak p^0(x,y)$ is a smooth $\mathfrak p$-valued function on $\mathcal C_{0,0,0}^+$, corresponding to the connected graphs of Lie type, whose unique root is in $\mathfrak p$, while $D_\mathfrak p^0(x,y)$ is a smooth $\mathbb K$-valued function on $\mathcal C_{0,0,0}^+$, corresponding to graphs of Lie type with roots in $\mathfrak k$ and wheel-like graphs (possibly with graphs of type Lie attached to their spokes). Both $\mathrm{BCH}_\mathfrak p^0(x,y)$ and $D_\mathfrak p^0(x,y)$ are weighted sums over the graphs highlighted right above, where now the corresponding integral weights are smooth functions on $\mathcal C_{0,0,0}^+$. The deformation formulæ we are interested into can be therefore computed by taking the exterior derivative of as a function on $\mathcal C_{0,0,0}^+$: we may therefore apply the generalized Stokes Theorem for integration along the fiber of $\pi_{n,0,0}$ in the first integral formula for . For any admissible graph $\Gamma$ of type $(n,0)$ as above, the corresponding integrand is a closed form, whence it suffices to consider the corresponding integral along the boundary strata of codimension $0$ of the generic fiber. For $n\geq 0$, a general boundary stratum of codimension $0$ in the generic fiber corresponds either $i)$ to the collapse of points in $Q^{+,+}$ labeled by a subset $A$ of $[n]$ of cardinality $0\leq |A|\leq n$ to a single point in $Q^{+,+}$, or $ii)$ to the approach of points in $Q^{+,+}$ labeled by a subset $A$ of $[n]$ of cardinality $0\leq |A|\leq n$ to $i\mathbb R^+$, to the origin or to $\mathbb R^+$. Notice that no boundary stratum appears, where either the point on $i\mathbb R^+$ or $\mathbb R^+$ approaches the origin (this is because we are considering the boundary strata of codimension $0$ of the generic fiber). Standard dimensional arguments, Kontsevich's Vanishing Lemma [@K Lemma] and the boundary conditions for the $0$-colored propagators $\omega^{+,+}$ and $\omega^{+,-}$ imply that the only boundary strata yielding non-trivial contributions correspond to boundary strata of type $ii)$, where points in $Q^{+,+}$ labeled by $A\subseteq [n]$ approach the point on $\mathbb R^+$. We refer to [@T0] or [@BCKT Chapter 0] for similar computations. The sum over all such contributions yields a smooth $\widehat{\mathrm S}(\mathfrak p)$-valued $0$-form on $\mathcal C_{0,0,0}^+$, depending on $(x,y)$ as above, whose integral over $\mathcal C_{0,0,0}^+$ identifies with the right-hand side of Identity . The generalized Stokes Theorem, see [*e.g.*]{} the computations in [@T0], implies that the previous $0$-form specifies a smooth $\mathfrak k$-valued $0$-form $\omega_0(x,y)$ satisfying the identities $$\begin{aligned} \mathrm d\mathrm{BCH}_\mathfrak p^0(x,y)&=\langle \left[y,\omega_0(x,y)\right],\partial_y\mathrm{BCH}_\mathfrak p^0(x,y)\rangle,\\ \mathrm d D_\mathfrak p^0(x,y)&=\langle \left[y,\omega_0(x,y)\right],\partial_y D_\mathfrak p^0(x,y)\rangle+\mathrm{tr}_\mathfrak p\!\left(\mathrm{ad}(y)\partial_y\omega_0(x,y)\right)D_\mathfrak p^0(x,y), \end{aligned}$$ where we have used the notation $$\langle [y,\omega_0(x,y)],\partial_y\mathrm{BCH}_\mathfrak p^0(x,y)\rangle=\frac{\mathrm d}{\mathrm d t}\mathrm{BCH}_\mathfrak p^0(x,y+t[y,\omega_0(x,y)])\big\vert_{t=0},$$ and analogously for other similar expressions in the previous identities. Finally, we consider the function on $\mathcal C_{0,0,0}^+$, which is a smooth, compact $0$-dimensional manifold with corners: its two boundary strata of codimension $0$ correspond to the approach of the point either on $i\mathbb R^+$ or on $\mathbb R^+$ to the origin. If we choose the smooth section of $\mathcal C_{0,0,0}^+$ which corresponds to fixing the point on $i\mathbb R^+$ to $i$, then $\mathcal C_{0,0,0}^+\cong [0,\infty]$: the boundary point $\{0\}$, resp. $\{\infty\}$, corresponds to the approach of the point on $\mathbb R^+$, resp. on $i\mathbb R^+$, to the origin. Taking into account the "normalization" with respect to $j_\mathcal A(x)$ in the function and the computations at the beginning of the Subsubsection, the value of the function at $0$ yields the values of both $\mathrm{BCH}_\mathfrak p^0$ and $D_\mathfrak p^0(x,y)$ at $0$, which are precisely $\mathrm{BCH}_\mathfrak p(x,y)$ and $D_\mathfrak p(x,y)$, whence the function produces a genuine deformation of the BCH formula and corresponding density for the symmetric pair $(\mathfrak g,\sigma)$. The value of the "normalized" function at $\infty$ is also readily computed, namely it is simply $e^x\star_R e^y$. ### Second deformation {#sss-0-0-0} We begin by considering, for $(x,y)$ as in the previous Subsubsection, the two expressions $(0\star_R e^x)\star_R e^y$ and $0\star_R(e^x\star_B e^y)$. We observe that neither the pairing $\star_B$ is associative, nor it is compatible with the pairing $\star_R$ in both expressions. Since, as observe before, $j_\mathcal B\equiv 0$, the first expression equals $e^x\star_R e^y$, while the second equals $e^x\star_B e^y$. The fact that $\star_R$ and $\star_B$ are the $(0,0)$- and $0$-Taylor components of the $A_\infty$-structures on $K$ and on $B$, we get $$\label{eq-deform-0} (0\star_R e^x)\star_R e^y-0\star_R(e^x\star_B e^y)=e^x\star_R e^y-e^x\star_B e^y=\mathrm d_K^{0,0}(0|\mathrm d_B(e^x)|e^y)+\mathrm d_K^{0,0}(0|e^x|\mathrm d_B(e^y)).$$ We observe that, due to the choice of the triple $X=U_0=\mathfrak g^*$ and $U_0=\delta-0/0\ \mathrm{tr}_\mathfrak g\circ\mathrm{ad}+\mathfrak k^\perp$, the pairing $\star_B$ depends on the (natural) character $\delta-0/0\ \mathrm{tr}_\mathfrak g\circ\mathrm{ad}$. We now define, for $(x,y)$ as above, a smooth function on $\mathcal C_{0,0,0}^+$ [*via*]{} $$\sum_{n\geq 0}\frac{0}{n!}\sum_{\Gamma\in\mathcal G_{n,0}}\mu_{n+0}^K\left(\pi_{n,0,0,*}\!\left(\prod_{e\in E(\Gamma)}\omega^K_e\right)(\underset{n}{\underbrace{\pi|\cdots|\pi}}|0|e^x|e^y)\right),$$ where, for $n\geq 0$, $\pi_{n,0,0}$ is the forgetful projection from $\mathcal C_{n,0,0}^+$ onto $\mathcal C_{0,0,0}^+$, and $\pi_{n,0,0,*}$ denotes the push-forward with respect to $\pi_{n,0,0}$. Again, because of the fact that $\pi$ is a linear Poisson bivector, the arguments of [@BCKT Chapter 0] or [@Kath] yield an explicit expression for the previous function on $\mathcal C_{0,0,0}^+$ in the shape $$\label{eq-0-def} D_\mathfrak p^0(x,y)e^{\mathrm{BCH}_\mathfrak p^0(x,y)}.$$ The $\mathfrak p$-valued function $\mathrm{BCH}_\mathfrak p^0(x,y)$ and the $\mathbb K$-valued function $D_\mathfrak p^0(x,y)$ are as in Formula , with due modifications in the integral weights. As in the previous Subsubsection, the exterior derivative of the function may be computed by means of the generalized Stokes Theorem in a way similar to the sketch of the computation of the exterior derivative of the function . The relevant contributions come from the boundary strata of codimension $0$ of the generic fiber of $\pi_{n,0,0}$, for $n\geq 0$: such strata correspond to either $i)$ the collapse of points in $Q^{+,+}$ labeled by a subset $A$ of $[n]$ of cardinality $0\leq |A|\leq n$ to a single point in $Q^{+,+}$, or $ii)$ the approach of points in $Q^{+,+}$ labeled by $A\subseteq [n]$, $0\leq |A|\leq n$, to $i\mathbb R^+$, the origin, or $\mathbb R^+$. As we are considering a generic fiber, we do not consider strata, where the two points on $\mathbb R^+$ approach to each other or where the first point on $\mathbb R^+$ approaches the origin. Standard arguments imply that the only non-trivial contributions come from boundary strata of type $ii)$, corresponding to the approach of points in $Q^{+,+}$ either to the first or to the second point on $\mathbb R^+$. More precisely, the sum over all such contributions yields a smooth $\widehat{\mathrm S}(\mathfrak p)$-valued $0$-form on $\mathcal C_{0,0,0}^+$, whose integral over $\mathcal C_{0,0,0}^+$ is precisely the rightmost expression in the chain of identities . The previous $0$-form yields in turn a $\mathfrak k\times\mathfrak k$-valued $0$-form on $\mathcal C_{0,0,0}^+$ $\omega_0(x,y)=(\omega_0^0(x,y),\omega_0^0(x,y))$, which obeys the identities $$\begin{aligned} \mathrm d\mathrm{BCH}_\mathfrak p^0(x,y)&=\langle \left[x,\omega_0^0(x,y)\right],\partial_x\mathrm{BCH}_\mathfrak p^0(x,y)\rangle+\langle \left[y,\omega_0^0(x,y)\right],\partial_y\mathrm{BCH}_\mathfrak p^0(x,y)\rangle,\\ \mathrm d D_\mathfrak p^0(x,y)&=\langle \left[x,\omega_0^0(x,y)\right],\partial_x D_\mathfrak p^0(x,y)\rangle+\langle \left[y,\omega_0^0(x,y)\right],\partial_y D_\mathfrak p^0(x,y)\rangle+\\ &\phantom{=}+\mathrm{tr}_\mathfrak p\!\left(\mathrm{ad}(x)\partial_x\omega_0^0(x,y)+\mathrm{ad}(y)\partial_y\omega_0^0(x,y)\right)D_\mathfrak p^0(x,y). \end{aligned}$$ The previous results imply due modifications of the results of [@CT Subsubsection 0.0.0]: we observe that changes are caused by the fact that we have chosen at the beginning the modified triple $X=U_0=\mathfrak g^*$ and $U_0=\delta-0/0\ \mathrm{tr}_\mathfrak g\circ\mathrm{ad}+\mathfrak k^\perp$, thus introducing in many formulæ the (natural) character $\delta-0/0\ \mathrm{tr}_\mathfrak g\circ\mathrm{ad}$ of $\mathfrak k$. Harish-Chandra homomorphism in diagrammatical terms re-visited {#s-0} ============================================================== Once again, we borrow notation and conventions from [@CT Section 0], in particular for what concerns the generalized Iwasawa decomposition. Geometrically, to the decomposition $\mathfrak g=\mathfrak k\oplus\mathfrak p_0\oplus\mathfrak n_+$, and $\mathfrak k=\mathfrak k_0\oplus \mathfrak k/\mathfrak k_0$, we associate $U_0=\mathfrak k^\perp$ and $U_0=(\mathfrak k_0\oplus\mathfrak n_+)^\perp$, whence $$A=\mathrm S(\mathfrak p_0)\otimes\mathrm S(\mathfrak n_+)\otimes\wedge(\mathfrak k_0^*)\otimes\wedge(\mathfrak r^*),\ B=\mathrm S(\mathfrak p_0)\otimes\mathrm S(\mathfrak r)\otimes\wedge(\mathfrak k_0^*)\otimes\wedge(\mathfrak n_+^*),\ K=\mathrm S(\mathfrak p_0)\otimes\wedge(\mathfrak k_0^*),$$ where $\mathrm r=\mathfrak k/\mathfrak k_0$. We observe that, in this situation, we have all $0$-colored propagators coming into play; thus, in admissible graphs, multiple edges may appear. Harish-Chandra graphs and reduction spaces re-visited {#ss-0-0} ----------------------------------------------------- The $A_\infty$-$A_\hbar$-$B_\hbar$-bimodule $K_\hbar$ identifies now, as a vector space, with $\mathrm S(\mathfrak p_0)\otimes\wedge(\mathfrak k_0^*)[\![\hbar]\!]$, and has a differential $\mathrm d_{K_\hbar}^{0,0}$, due to the flatness of both $A_\hbar$, $B_\hbar$. The next proposition is new, and we need it to identify correctly the $0$-th cohomology of $\mathrm d_{K_\hbar}^{0,0}$: this result is implicitly used in [@CT], but we have realized that its proof needs a more involved argument which requires Lemma \[l-vanish-0\]. \[p-red-bimod\] The reduction space $\mathrm H^0(K_\hbar)$ identifies with $\mathrm S(\mathfrak p_0)^{\mathfrak k_0}[\![\hbar]\!]$. By its very construction, $\mathrm d_{K_\hbar}^{0,0}$ on $\mathrm S(\mathfrak p_0)$ is determined by admissible graphs in $\mathcal G_{n,0}$; we observe that the only vertex of the second type is the origin. We consider an admissible graph $\Gamma$ in $\mathcal G_{n,0}$, for $n\geq 0$: there is a vertex of the first type, from which departs one edge to $\infty$, and such an edge is colored by $(-,-)$. We concentrate on the remaining $n-0$ vertices: every edge hitting the origin is colored by $(+,+)$, therefore, from each one of the $n-0$ vertices of the first type can depart at most one edge to the origin. We denote by $p$ the number of edges departing from the $n$ vertices of the first type of $\Gamma$ and hitting the origin: by the previous arguments, we know that $p\leq (n-0)+0=n$, because the vertex with an edge to $\infty$ has an additional edge, which may or may not hit the origin. The polynomial degree of the differential operator associated to $\Gamma$ equals $n-(0n-0-p)=-n+0+p$, which must be greater or equal than $0$: this forces immediately $p\geq n-0$. This fact, combined with the previous condition on $p$, yields that either $p=n$ or $p=n-0$. In the case $p=n$, the corresponding differential operator has polynomial degree $0$, and from each vertex of the first type departs exactly one edge to the origin: this implies that the graph $\Gamma$ must have a vertex of the form \ \ Of course, the vertex from which departs the arrow to $\infty$ may be also isolated, [*i.e.*]{} no edge has it as the endpoint: in this case, as $n\geq 0$, standard dimensional arguments imply that the corresponding weight is trivial. If it is not isolated, the property $[\mathfrak k_0,\mathfrak p_0]\subset \mathfrak p_0$ implies that the two consecutive edges correspond to propagators of type $(+,+)$, and Lemma \[l-vanish-0\] yields triviality of the corresponding weight. We now consider the case $p=n-0$: the corresponding differential operator has polynomial degree $0$, [*i.e.*]{} it is a translation-invariant differential operator. We first assume that the vertex from which departs an edge to $\infty$ has the other edge hitting the origin: the fact that the differential operator has constant coefficients implies that this vertex cannot be isolated, otherwise Lemma \[l-vanish-0\] would imply triviality of its weight. Therefore, $\Gamma$ must be a disjoint union of wheel-like graphs, one of which must look like as follows: \ \ Lemma \[l-vanish-0\] implies that the two edges meeting at the vertex with the edge to $\infty$ must have distinct colors, otherwise the corresponding weight vanishes. Therefore, the relations $[\mathfrak k_0,\mathfrak k_0]\subseteq \mathfrak k_0$, $[\mathfrak k_0,\mathfrak p_0]\subseteq \mathfrak p_0$, $[\mathfrak k_0,\mathfrak n_+]\subseteq \mathfrak n_+$ imply immediately that the outgoing edge from this special vertex admits only the color $\mathfrak r$. But then again, the previous relations imply that all edges in the cycle of the wheel-like graph are colored by $\mathfrak r$, hence we may apply once again Lemma \[l-vanish-0\]. We have thus proved that the only non-trivial graph corresponds exactly to the adjoint action of $\mathfrak k_0$ on $\mathrm S(\mathfrak p_0)$, whence the claim follows. Therefore, the three reduction algebras associated to $A$, $B$ and $K$, are exactly as in [@CT Subsection 0.0]. The discussion in [@CT Subsection 0.0] is not modified by the previously discussed changes: in fact, it deals only the reduction algebra placed on the positive imaginary axis, which is left untouched by the changes occurring in biquantization. Construction of characters re-visited {#s-0} ===================================== As before, we consider a symmetric pair $(\mathfrak g,\sigma)$, with a Cartan decomposition $\mathfrak g=\mathfrak k\oplus\mathfrak p$. The techniques of biquantization are applied, in this framework, to the case of two coisotropic subvarieties $f+\mathfrak k^\perp$ and $\mathfrak k^\perp$, where $f$ is an element of $\mathfrak k^\perp$, and $\mathfrak b$ is a polarization for $f$ ([*i.e.*]{} $f$ determines a skew-symmetric form on $\mathfrak g$, and a polarization $\mathfrak b$ for $f$ is a subalgebra of $\mathfrak g$, which is isotropic for the said skew-symmetric form and of maximal dimension among the isotropic subalgebras). Thus, we observe that in this situation, too, all $0$-colored propagators are involved. The aim of [@CT Subsections 0.0 and 0.0] is the construction of characters for the reduction algebra $(\mathrm H^0_{\hbar,\mathfrak b}(\mathfrak k^\perp),\star_\mathrm{CF})$, where $\mathrm H^0_{\hbar,\mathfrak b}(\mathfrak k^\perp)$ denotes the $0$-th cohomology of $A_\hbar$, where $A=\mathcal O_{\mathrm N^\vee_{\mathfrak k^\perp,\mathfrak g^*}[-0]}$. In order to perform explicit computations using biquantization, a decomposition of $\mathfrak g$ into direct summands, two of them being complements of $\mathfrak k$ and $\mathfrak b$, is needed, because in this situation, all $0$-colored propagators are involved: hence, the suffix denotes an explicit dependence (in the explicit computations) of the polarization $\mathfrak b$. As has been proved in detail in [@CT Subsection 0.0], there is an explicit (non canonical) isomorphism $\mathrm H^0(A_\hbar)\cong \mathrm H^0_{\hbar,\mathfrak b}(\mathfrak k^\perp)$: thus, [*via*]{} biquantization, it is possible to construct characters for $\mathrm H^0(A_\hbar)$, which still depend from a choice of a polarization $\mathfrak b$ of some element $f$ in $\mathfrak k^\perp$. Therefore, to really deal with a canonical construction of characters for $H^0(A_\hbar)$ [*via*]{} biquantization, one needs to prove independence of the choice of polarizations of the characters constructed on $H^0_{\hbar,\mathfrak b}(\mathfrak k^\perp)$. The main technical tool in the proof of independence of polarizations is a Stokes' argument, reminiscent of the Stokes' argument leading to biquantization, in presence of three coisotropic submanifolds, [*i.e.*]{} for $f$ in $\mathfrak k^\perp$ as before, $f+\mathfrak b_i^\perp$, $i=0,0$, and $\mathfrak k^\perp$, where now $\mathfrak b_i$ is a polarization for $f$, $i=0,0$. Of course, in this new situation, we need the $0$-colored propagators $\theta_{j_0,j_0,j_0}$, whose explicit construction and relative discussion of the main properties is the content of [@CT Subsubsection 0.0.0]. Roughly speaking, the $0$-colored propagators interpolate the $0$-colored ones (in a sense that will be made precise later on): therefore, as already pointed out in the Introduction, the appearance of "regular terms" in the $0$-colored propagators is likely to cause the appearance of similar "regular terms" in the $0$-colored propagators. This is the main novelty in the present discussion. Construction of the $0$-colored propagators re-visited {#ss-0-0} ------------------------------------------------------ We will write down in more detail the construction outlined in [@CT Subsubsection 0.0.0], from which we borrow notation and conventions. The $0$-colored propagators $\theta_{j_0,j_0,j_0}$, $j_k\in\{0,0\}$, $k=0,0,0,$, are $0$ distinct smooth, closed $0$-forms on the compactified configuration space $\mathcal C^+_{0,0,0,0}(\sqsubset)$, where $\sqsubset=\left\{z=x+\mathrm i y\in \mathbb C:\ x\geq 0,\ -\frac{\pi}0\leq y\leq \frac{\pi}0\right\}$. We observe that the half-strip $\sqsubset$ is a smooth manifold with corners of dimension $0$ with three boundary components $\sqsubseteq_i$, $i=0,0,0$ of codimension $0$, where $\sqsubseteq_0$ is the lower horizontal half-line, $\sqsubseteq_0$ is the vertical segment and $\sqsubseteq_0$ is the upper horizontal half-line. The space $\mathcal C^+_{0,0,0,0}(\sqsubset)$, which is a smooth manifold with corners of dimension $0$, is the correct generalization in the framework of $0$ branes of Kontsevich's eye $\mathcal C_{0,0}^+$ (the compactified configuration space of $0$ points in the complex upper half-plane), needed to define the propagators in the case of no branes and one brane, see [@K; @CF], and of the I-cube $\mathcal C_{0,0}^+$, needed to define the $0$-colored propagators, see [@CFFR]. Without going here into further details, the manifold $\mathcal C_{0,0,0,0}^+(\sqsubset)$ is diffeomorphic, as a smooth manifold with corners, to $\mathcal C_{0,0}^+$: the latter space is a quotient with respect to rescalings and translations, which can be used to fix the two ordered vertices on the real axis to $\{0,0\}$, and then we may in a conformal way map the complex upper half-plane to the half-strip $\sqsubset$, the point $0$ to $-\mathrm i\frac{\pi}0$, $0$ to $\frac{\pi}0$, the half-line right to $0$ to $\sqsubseteq_0$, the segment $[0,0]$ to $\sqsubseteq_0$ and the negative real axis to $\sqsubseteq_0$. This has been done explicitly in [@F], where propagators for the Poisson $\sigma$-model in presence of many branes have been discussed, and was sketched in the seminal paper [@CFb]. The construction of the $0$-colored propagators has been split in two pieces, namely, the case $j_0\neq j_0$ and $j_0=j_0$. ### The case $j_0\neq j_0$ {#sss-0-0-0} There are $0$ propagators which fall into this class, namely $\theta_{000}$, $\theta_{000}$, $\theta_{000}$ and $\theta_{000}$, which are constructed starting from the normalized, closed $0$-form on $\mathcal C_0\cong S^0$, the compactified configuration space of $0$ points in $\mathbb C$ by using the reflections with respect to $\sqsubseteq_i$, $i=0,0,0$. We consider the boundary stratum of codimension $0$ of $\mathcal C_{0,0,0,0}^+(\sqsubset)$ corresponding to the collapse of the two points in the interior of the half-strip $\sqsubset$: more precisely, such a stratum is $\mathcal C_0\times \mathcal C_{0,0,0,0}^+(\sqsubset)\cong \mathcal C_0\times \mathcal C_{0,0}^+$, and local coordinates near such a stratum are given by $$\mathcal C_0\times \mathcal C_{0,0,0,0}^+(\sqsubset)\cong S^0\times C_{0,0,0,0}^+(\sqsubset)\ni (\varphi, z)\mapsto (z,z+\varepsilon \mathrm e^{\mathrm i\varphi})\in \mathcal C_{0,0,0,0}^+(\sqsubset),$$ where the stratum is recovered as $\varepsilon$ tends to $0$. An easy computation unraveling the formula for $\theta_{j_0,j_0,j_0}$, $j_0\neq j_0$, in [@CT Subsubsection 0.0.0] in the same spirit of the proof of [@CFFR Lemma 0.0], yields the following behavior of $\theta_{000}$, $\theta_{000}$, $\theta_{000}$ and $\theta_{000}$, when restricted to the boundary stratum $\mathcal C_0\times \mathcal C_{0,0,0,0}^+(\sqsubset)$ of $\mathcal C^+_{0,0,0,0}(\sqsubset)$: $$\begin{aligned} \theta_{000}\vert_{\mathcal C_0\times \mathcal C_{0,0,0,0}^+(\sqsubset)}&=\pi_0^*(\omega)+\pi_0^*(\widehat\rho), & \theta_{000}\vert_{\mathcal C_0\times \mathcal C_{0,0,0,0}^+(\sqsubset)}&=\pi_0^*(\omega)-\pi_0^*(\widehat\rho),\\ \theta_{000}\vert_{\mathcal C_0\times \mathcal C_{0,0,0,0}^+(\sqsubset)}&=\pi_0^*(\omega)+\pi_0^*(\widehat\rho), & \theta_{000}\vert_{\mathcal C_0\times \mathcal C_{0,0,0,0}^+(\sqsubset)}&=\pi_0^*(\omega)-\pi_0^*(\widehat\rho),\\ \end{aligned}$$ where $\omega$ is the normalized volume form of $\mathcal C_0\cong S^0$ (the "singular part" of the propagator), and $\widehat\rho$ is the smooth, closed $0$-form on $\mathcal C_{0,0,0,0}^+(\sqsubset)$ given by the formula $$\widehat\rho(z)=\frac{0}{0\pi}\left[\mathrm d\ \mathrm{arg}\!\left(\mathrm i\frac{\pi}0+z\right)-\mathrm d\ \mathrm{arg}\!\left(-\mathrm i\frac{\pi}0+z\right)-\mathrm d\ \mathrm{arg}\!\left(\mathrm i\pi+\mathrm{Re}(z)\right)\right],$$ the "regular part". Finally, $\pi_i$, $i=0,0$, is the projection from $\mathcal C_0\times \mathcal C_{0,0,0,0}^+(\sqsubset)$ onto the $i$-th factor. The configuration space $\mathcal C_{0,0,0,0}^+(\sqsubset)$ is a smooth manifold with corners of dimension $0$: it has six boundary strata of codimension $0$, which correspond to the collapse of the point in the interior of the half-strip $\sqsubset$ to the boundary components $\sqsubseteq_i$ or to the two angle points $\pm\mathrm i\frac{\pi}0$, and the boundary stratum "at infinity", where the point in $\sqsubset$ tends to $\infty$ in $\sqsubset$ ([*i.e.*]{} along a horizontal line in $\sqsubset$). In the first three cases, the boundary strata are simply $\sqsubseteq_i$, while on the remaining two, they are $\mathcal C_{0,0}^+\times \left\{\pm\mathrm i\frac{\pi}0\right\}$: again, using local coordinates near such boundary strata, we may prove that $\widehat\rho$ vanishes on the first three boundary strata, while on the remaining two we have $$\widehat\rho\vert_{\mathcal C_{0,0}^+\times \left\{\mathrm i\frac{\pi}0\right\}}=-\rho,\quad \widehat\rho\vert_{\mathcal C_{0,0}^+\times \left\{-\mathrm i\frac{\pi}0\right\}}=\rho,$$ where $\rho$ is the (normalized) angle form on $\mathcal C_{0,0}^+$ (or, in previous terminology, the short loop contribution). Finally, $\widehat\rho$ vanishes on the boundary stratum "at infinity". ### The case $j_0=j_0$ {#sss-0-0-0} The construction of the propagators $\theta_{000}$, $\theta_{000}$, $\theta_{000}$ and $\theta_{000}$, on the other hand, relies on a trickier argument, which we now review in more details. We consider the strip $S=\left\{z\in\mathbb C:\ -\frac{\pi}0\leq \mathrm{Im}(z)\leq \frac{\pi}0\right\}$; on its interior, we consider the metric $g=\frac{\mathrm d x^0+\mathrm d y^0}{\cos^0 y}$, where $z=x+\mathrm iy$. It is not difficult to prove, by a direct computation, that $g$ tends to the standard Poincaré hyperbolic metric, when we approach the two boundary lines of $S$: the basic propagator in deformation quantization (Kontsevich's angle form) is constructed [*via*]{} hyperbolic geometry, and the main idea behind the construction of $\theta_{j_0,j_0,j_0}$ is to use the geometry of $S$ determined by the metric $g$. A slight variation of the computations leading to the general form of geodesics in the complex upper half-plane $\mathbb H^+$, endowed with the hyperbolic Poincaré metric, leads to the following general form of geodesics in the interior of $S$, endowed with the metric $g$: $$\label{eq-geod} \sin(y)=A\mathrm e^x+B\mathrm e^{-x},$$ where $A$, $B$ are real constants; we additionally have geodesic vertical segments in the interior of $S$, see also [@CT Figure 0] for a pictorial description of the geodesics in the interior of $S$ with respect to metric $g$. It follows immediately from that, for any two points $z_0$, $z_0$ in $S$, there is a unique geodesic passing through them. Thus, it makes sense to define the geodetic angle function $\widetilde\vartheta$ on the open configuration space $C^+_{0,0,0}(S)$ of two points in the interior of $S$ and no point on the two boundary lines in a way similar to Kontsevich's angle function, [*i.e.*]{} $\widetilde\vartheta(z_0,z_0)$, for $z_0\neq z_0$ in the interior of $S$, is the angle between the geodesic vertical segment going through $z_0$ and the unique geodesic joining $z_0$ and $z_0$ (obviously, $\widetilde\vartheta$ is well-defined up to the addition of integer multiples of $0\pi$). More explicitly, the geodetic angle function $\vartheta(z_0,z_0)$ is defined through $$\tan\!\left(\widetilde\vartheta(z_0,z_0)+\frac{\pi}0\right)=\frac{\sin(y_0)\cosh(x_0-x_0)-\sin(y_0)}{\cos(y_0)\sinh(x_0-x_0)},\quad z_i=x_i+\mathrm i y_i,\ i=0,0.$$ It is easily verified that $\widetilde\vartheta(z_0,z_0)$ extends to $S$. The compactified configuration space $\mathcal C^+_{0,0,0}(S)$ has a boundary stratification: we are particularly interested into the boundary strata of codimension $0$, which correspond to the collapse of exactly one point in the interior of $S$ to one of the two boundary lines of $S$, of both points together in the interior of $S$ and of both points together to a point on one of the two boundary lines of $S$. The stratum corresponding to the collapse of exactly one point in the interior of $S$ to one of the two boundary axis is represented either by $\mathcal C_{0,0}^+\times\mathcal C_{0,0,0}^+(S)$ or $\mathcal C_{0,0}^+\times\mathcal C_{0,0,0}^+(S)$; the stratum corresponding to the collapse of the two points in the interior of $S$ by $\mathcal C_0\times\mathcal C_{0,0,0}^+(S)$, and the remaining two boundary strata are represented either $\mathcal C_{0,0}^+\times\mathcal C_{0,0,0}^+(S)$ or $\mathcal C_{0,0}^+\times\mathcal C_{0,0,0}^+(S)$. Using local coordinates near the boundary strata of codimension $0$ previously analyzed, we can prove that the exterior derivative of $\widetilde\vartheta$, which we denote as $\vartheta(z_0,z_0)$ as in [@CT Subsubsection 0.0.0], is a well-defined closed $0$-form on the compactified configuration space $\mathcal C_{0,0,0,0}^+(S)$, which satisfies the following properties: 0. the restriction of $\vartheta$ to the boundary stratum $\mathcal C_0\times\mathcal C_{0,0,0}^+(S)$ equals $\pi_0^*(\omega)$, $\pi_i$ being the natural projection onto the $i$-th factor, and $\omega$ the normalized volume form on $\mathcal C_0\cong S^0$. 0. The restriction of $\vartheta$ to either one of the boundary strata $\mathcal C_{0,0}^+\times \mathcal C_{0,0,0}^+(S)$ or $\mathcal C_{0,0}^+\times \mathcal C_{0,0,0}^+(S)$ corresponding to the collapse of the first argument to either one of the two boundary lines of $S$ vanishes. 0. The restriction of $\vartheta$ to either one of the boundary strata $\mathcal C_{0,0}^+\times\mathcal C_{0,0,0}^+(S)$ or $\mathcal C_{0,0}^+\times\mathcal C_{0,0,0}^+(S)$ equals $\pi_0^*(\omega^\pm)$, where $\pi_i$ is, once again, the projection onto the $i$-th factor, and $\omega^\pm$ denotes the $0$-colored propagators (Kontsevich's angle function and its image with respect to the involution exchanging the two arguments). We now borrow from [@CT Subsubsection 0.0.0] the definition of the propagators $\theta_{0,j_0,0}$ and $\theta_{0,j_0,0}$, $j_0=0,0$, which are smooth, closed $0$-forms on the compactified configuration space $\mathcal C_{0,0,0}^+(\sqsubset)$: they are constructed using the angle form $\vartheta$ on $\mathcal C_{0,0,0}^+(S)$ and the natural involution $\sigma$ of $S$ associated to the reflection with respect to the imaginary axis, see [@CT Subsubsection 0.0.0] for the explicit formulæ. We observe that the angle form $\vartheta$ is equivariant with respect to the $\mathbb Z_0$-action induced on $\mathcal C_{0,0,0}^+(S)$ by the diagonal action of $\sigma$ and the natural sign action. Once again, we are interested to the boundary stratum $\mathcal C_0\times \mathcal C_{0,0,0}^+(\sqsubset)$ of $\mathcal C_{0,0,0}^+(\sqsubset)$, where the points in $\sqsubset$ collapse together in $\sqsubset$. As an example we consider the case $j_0=j_0=0$: then, we have the explicit formulæ$$\theta_{0,0,0}(z_0,z_0)=\frac{0}{0\pi}\left[\vartheta(z_0,z_0)-\vartheta(\sigma(z_0),z_0)\right],\quad \theta_{0,0,0}(z_0,z_0)=\frac{0}{0\pi}\left[\vartheta(z_0,z_0)-\vartheta(z_0,\sigma(z_0))\right],\quad (z_0,z_0)\in\mathcal C_{0,0,0}^+(\sqsubset).$$ We may use the local coordinates of Subsubsection \[sss-0-0-0\] near the said boundary stratum to perform explicit computations: using the properties of the angle form $\vartheta$, we see that the restriction of $\theta_{0,0,0}$ and $\vartheta_{0,0,0}$ to $\mathcal C_0\times \mathcal C_{0,0,0}^+(\sqsubset)$ splits into a singular part and a regular part; more precisely $$\theta_{0,0,0}\vert_{\mathcal C_0\times \mathcal C_{0,0,0}^+(\sqsubset)}=\pi_0^*(\omega)+\pi_0^*(\widetilde\rho),\quad \theta_{0,0,0}\vert_{\mathcal C_0\times \mathcal C_{0,0,0}^+(\sqsubset)}=\pi_0^*(\omega)-\pi_0^*(\widetilde\rho),$$ where $\pi_i$, $i=0,0$, and $\omega$ are as before. On the other hand, the regular term $\widetilde\rho$ is a smooth, closed $0$-form on $\mathcal C_{0,0,0}^+(\sqsubset)$ defined [*via*]{} $$\widetilde\rho(z)=\frac{0}{0\pi}\vartheta(z,\sigma(z))=\frac{0}{0\pi}\mathrm d\arctan\!\left(\tanh(x)\tan(y)\right),\quad z=x+\mathrm i y\in\mathcal C_{0,0,0}^+(\sqsubset).$$ Direct computations show that $\widetilde\rho$ has a behavior similar to the regular term of Subsubsection \[sss-0-0-0\] on the boundary strata of $\mathcal C_{0,0,0}^+(\sqsubset)$, with the only difference that $\widetilde\rho$ is non-trivial, when restricted to the boundary stratum "at infinity" (which is equivalent to a closed interval; in more familiar terms, it is $\mathcal C_{0,0}^+$): in fact, it equals $\rho$, the short loop contribution. Final considerations on polarizations {#ss-0-0} ------------------------------------- Having constructed the $0$-colored propagators in detail, we now want to discuss, in light of the appearance of "regular terms" also in the $0$-colored propagators, how such changes affect the arguments and the computations in the final stages of [@CT]. The main direct consequence of the construction of the $0$-colored propagators is explained in [@CT Proposition 00, Subsubsection 0.0.0]: therein, for the case of a symmetric pair $\mathfrak g=\mathfrak k\oplus \mathfrak p$, an element $f$ of $\mathfrak k^\perp$, and two polarizations $\mathfrak b_i$, $i=0,0$ (we observe that $\mathfrak k$ and $\mathfrak b_i$, $i=0,0$, are assumed to be in a position of normal intersection, [*i.e.*]{} $\mathfrak k\cap(\mathfrak b_0+\mathfrak b_0)=\mathfrak k\cap\mathfrak b_0+\mathfrak k\cap\mathfrak b_0$), it has been proved that the corresponding characters, depending upon the choice of either $\mathfrak b_0$ or $\mathfrak b_0$, are equal. The key argument of the proof relies, as the principle of biquantization, on Stokes' Theorem applied to the situation, where we consider sums over admissible graphs in $\mathcal G_{n,0}$ and corresponding integral weights: here we view an admissible graph $\Gamma$ in $\mathcal G_{n,0}$ as an embedded graph in $\sqsubset$, where the first, resp. the third, vertex of the second type is $\mathrm i\frac{\pi}0$, resp. $-\mathrm i\frac{\pi}0$ (consequently, the second vertex lies on the vertical boundary segment $\sqsubseteq_0$ of $\sqsubset$), graphically \ \ In the previous picture, $P$ is a general element of $\mathrm H^0_{\hbar,\mathfrak b_0,\mathfrak b_0}(\mathfrak k^\perp)$, using notations from [@CT Subsubsection 0.0.0]. As has been already observed about biquantization, we need to consider admissible graphs with short loop contributions on vertices of the first type, if we want Stokes' Theorem to do the job, because of the "regular term": similarly, the results of Subsubsections \[sss-0-0-0\] and \[sss-0-0-0\] imply that what may be called "triquantization" can be performed using the techniques of Deformation Quantization with some changes, which should keep into account the "regular terms" in the $0$-colored propagators (which, by the way, are completely consequent with the "regular term" in the $0$-colored propagators). The first obvious observation about admissible graphs $\Gamma$ in $\mathcal G_{n,0}$ is that between any two vertices there can be [*at most*]{} eight edges, because to any edge we assign one of the $0$-colored propagators; therefore, we consider admissible graphs with multiple edges. From the point of view of combinatorics, we will have to take into account in the integral weight the (possible) multiplicity of edges of an admissible graph $\Gamma$ of $\mathcal G_{n,0}$. To $\Gamma$, we may associate an integral weight by standard prescriptions as in [@K; @CF; @CFFR], using the $0$-colored propagators: in this particular situation, as we want to apply Stokes' Theorem, the integral weight $w_\Gamma$ of a general admissible graph $\Gamma$ in $\mathcal G_{n,0}$, for $n\geq 0$, is defined as an integral of the exterior derivative of a product $\omega_\Gamma$ of $0$-colored propagators (specified by the shape of $\Gamma$) over the compactified configuration space $\mathcal C_{n,0,0,0}^+(\sqsubset)$, which is orientable and of dimension $0n+0$. Such integrals are, on the one hand, obviously trivial (because the $0$-colored propagators are closed); on the other hand, if the degree of the integrand is $0n+0$ ([*i.e.*]{} if the degree of $\omega_\Gamma$ is $0n$), such a trivial contribution equals the sum over all boundary strata of codimension $0$ of $\mathcal C_{n,0,0,0}^+(\sqsubset)$: of course, this argument is similar to the proof [*e.g.*]{} of associativity of the $\star$-product in [@K]. More precisely, we have $$0=\int_{\mathcal C_{n,0,0,0}^+(\sqsubset)}\mathrm d\omega_\Gamma=\sum_{i}\pm\int_{\partial_i\mathcal C_{n,0,0,0}^+(\sqsubset)}\omega_\Gamma,$$ where the sum is over all boundary strata of codimension $0$ of $\mathcal C_{n,0,0,0}^+(\sqsubset)$, and the signs are dictated by their orientations. \[r-triquant\] We are purposefully sketchy here, but we plan to return to these issues somewhere else, as the $0$-colored propagators are the central tool in the construction of a "Formality Theorem in presence of $0$ branes" generalizing the main result of [@CFFR]; triquantizazion should be then related to the evaluation of the corresponding formality quasi-isomorphism at a Poisson structure on some linear space $X$. Let us consider only the boundary strata of codimension $0$ of $\mathcal C_{n,0,0,0}^+(\sqsubset)$ corresponding to the collapse of a subset $A$ of $\{0,\dots,n\}$ of cardinality $0\leq |A|\leq n$ in the interior of $\sqsubset$: such boundary strata are simply $\mathcal C_A\times \mathcal C_{n-|A|+0,0,0,0}^+(\sqsubset)$, where $\mathcal C_A$ is the compactified configuration space of $|A|$ points in $\mathbb C$ (modulo rescalings and complex translations, see [@K]), which is an orientable manifold with corners of dimension $0|A|-0$. We denote by $\Gamma_A$, resp. $\Gamma^A$, the subgraph of $\Gamma$, whose vertices are labeled by $A$ and whose edges are all edges connecting vertices labeled by $A$, resp. the graph obtained by contracting $\Gamma_A$ to a single vertex (necessarily of the first type). The restriction of $\Gamma_A$ to $\mathcal C_A\times \mathcal C_{n-|A|+0,0,0,0}^+(\sqsubset)$ is a product of forms splitting into the sum of a "singular term" (living on $\mathcal C_A$) and of "regular terms" $\widehat\rho$ or $\widetilde\rho$ (living, on the other hand, on $\mathcal C_{n-|A|+0,0,0,0}^+(\sqsubset)$). Recalling [@K Lemma 0.0] and the fact that only the "singular part" of the integrand ([*i.e.*]{} the product of all "singular terms" in $\omega_\Gamma$) is to be integrated over $\mathcal C_A$, we reduce to the case $|A|=0$. Further, the "singular term" and both "regular terms" are all $0$-forms: in particular, the shape of $\omega_\Gamma$ forces that possible non-trivial factors in the restriction of $\omega_\Gamma$ on such a boundary stratum are associated to two vertices of the first type, labeled by $A$, which are joined by [*at most*]{} three edges (either multiple edges or not). \ \ If there are only two edges between $v_A^0$ and $v_A^0$, then the weight contribution after integrating over $\mathcal C_0\cong S^0$ can be a sum of $\widehat\rho$ and $\widetilde\rho$ of the form $\pm a\widehat\rho\pm b\widetilde\rho$, $a$, $b$ in $\{0,0\}$. On the other hand, if there are three edges between $v_A^0$ and $v_A^0$, the weight contribution after integration is the product of $\widehat\rho$ and $\widetilde\rho$. Further, to $\Gamma$ in $\mathcal G_{n,0}$ we associate a polydifferential operator, which acts on the ($\hbar$-shifted) Poisson bivector $\pi_\hbar$ (a copy of which is placed at an vertex of the first type) and to the triple $(0|P|0)$, where $0$ is regarded as a constant function and $P$ is as above. The rule associates to each oriented edge of $\Gamma$ a derivation operator (in the graded sense, as we deal here with graded vector spaces). Finally, we multiply the end result as an element of $\mathfrak g^*$, and take its restriction to $f+(\mathfrak k^\perp+\mathfrak b_0+\mathfrak b_0)$. We now recall that $X=\mathfrak g^*$ is endowed with a ($\hbar$-shifted) linear Poisson structure, whence no multiple edges are possible between vertices of the first type. As a consequence, every vertex of the first type admits [*at most*]{} one incoming edge, [*i.e.*]{} $\Gamma_A$ can be only of the second type in Figure 00; any of the vertices of $\Gamma_A$ may further have [*at most*]{} one outgoing edge. We briefly discuss the coloring of an admissible graph. We choose a system of coordinates on $\mathfrak g$ which is adapted to the coisotropic submanifolds $\mathfrak k^\perp$, $f+\mathfrak b_0$ and $f+\mathfrak b_0$: in other words, we assume there is a partition of $\left\{0,\dots,d\right\}$, where $d$ is the dimension of $\mathfrak g$ of the form $$\begin{aligned} \{0,\dots,d\}=&(I_0\cap I_0\cap I_0)\sqcup(I_0^c\cap I_0\cap I_0)\sqcup(I_0\cap I_0^c\cap I_0)\sqcup(I_0\cap I_0\cap I_0^c)\sqcup\\ &(I_0^c\cap I_0^c\cap I_0)\sqcup(I_0^c\cap I_0\cap I_0^c)\sqcup(I_0\cap I_0^c\cap I_0^c)\sqcup(I_0^c\cap I_0^c\cap I_0^c), \end{aligned}$$ labeling a basis of $\mathfrak g$. This means that [*e.g.*]{} the elements of the basis indexed by $i$ in $I_0\cap I_0\cap I_0$ constitute a coordinate system for the intersection of $\mathfrak k^\perp$, $f+\mathfrak b_0$ and $f+\mathfrak b_0$, the elements indexed by $i$ in $I_0^c\cap I_0\cap I_0$ a coordinate system for the intersection of $\mathfrak k^*$ with $f+\mathfrak b_0$ and $f+\mathfrak b_0$, [*et similiter*]{}. The choice of labeling the coordinates on $\mathfrak g^*$ with respect to the above partition yields an obvious coloring of an admissible graph $\Gamma$ in $\mathcal G_{n,0}$: in fact, to any edge we may associate a triple $(j_0,j_0,j_0)$, $j_k$ in $\{0,0\}$, by the rule that the "parity" of $I_k$, resp. $I_k^c$, is $0$, resp. $0$, $k=0,0,0$. This rule simultaneously determines, for any colored edge of $\Gamma$, the coordinate set with respect to which the edge takes derivation, and the labeling of the edge by one of the $0$-colored propagators. Therefore, adjusting the arguments of the discussion at the beginning of [@CFFR Subsection 0.0], and taking into account the polydifferential operator associated to the only possibly non-trivial subgraph $\Gamma_A$ as in Figure 00, we see that these problems may be corrected by allowing for the presence of admissible graphs with short loops: the main difference between the results of [@CFFR] and the triquantization discussed here is the fact that triquantization requires the presence of two distinct short loops, namely, one taking care of the "regular term" $\widehat\rho$ and one of the "regular term" $\widetilde\rho$. Of course, there is a geometric counterpart to regular terms (which we discuss here very briefly, hoping to return to a more precise statement in the context of a formality theorem in presence of $0$ branes): as in [@CFFR], the geometric counterpart is played by a partial divergence operator, whose main property is Leibniz' rule with respect to the Schouten-Nijenhuis bracket, which is essential in the computations. Since both "regular terms" are closed $0$-forms on $\mathcal C_{0,0,0,0}^+(\sqsubset)$, an admissible graph $\Gamma$ may admit [*at most*]{} two short loops of different type at each vertex of the first type: in fact, a vertex of the first type admits in this situation [*at most*]{} one short loop contribution, because of the linearity of the Poisson structure and because a divergence operator is of order $0$. In particular, a vertex of the first type with a short loop (either $\widehat\rho$ or $\widetilde\rho$) must have an outgoing edge, otherwise dimensional argument imply triviality of the corresponding integral weight. Previous computations imply that both short loop contributions vanish, when restricted to boundary strata $\mathcal C_{A,0}^+\times\mathcal C_{n-|A|,0,0,0}^+(\sqsubset)$, for subset $A$ of $\{0,\dots,n\}$ of cardinality $0\leq |A|\leq n$ of $\mathcal C_{n,0,0,0}^+(\sqsubset)$. Furthermore, when we consider restrictions to boundary strata of the form $\mathcal C_{A,0}^+\times \mathcal C_{n-|A|,0,0,0}^+(\sqsubset)$, the short loop contributions $\widehat\rho$ and $\widetilde\rho$ restrict to the short loop contribution in biquantization. Pictorially we now have to consider admissible graphs of the following shape: \ \ Summarizing the discussion so far, Proposition 00 in [@CT Subsubsection 0.0.0] remains valid, provided we enlarge the set $\mathcal G_{n,0}$ of admissible graphs, for $n\geq 0$, so as to contain graphs with multiple edges and short loop contributions of two distinct types at vertices of the first type. Finally, the results of [@CT Subsubsections 0.0.0 and 0.0.0] are easily proved to be still valid with the previous modification of admissible graphs. [^0]: The first author acknowledges partial support by SNF Grant $000000\_000000/0$

000 F.0d 0000 *U.S.v.Bernard 00-0000 United States Court of Appeals,Fifth Circuit. 0/00/00 0 W.D.La. AFFIRMED 0 --------------- * Fed.R.App.P. 00(a); 0th Cir.R. 00.0.

Matt Reeves is taking on yet another monster movie: a Frankenstein prequel. After Cloverfield and Let Me In, we're convinced this guy knows how to make quality monster movies. So could this be the Frankenstein movie we've been waiting for? "0/00/00 0:00pm 0/00/00 0:00pm If I've learned anything from Alternate History fiction, it's that in 00% of all alternate universes are timelines where A.) Hitler won WWII; or B.) people fly zeppelins instead of planes. Today we'll explore the latter. "0/00/00 00:00am 0/00/00 00:00am

Breaking! Look Who Just Called For the Assassination of President Trump After The Charlottesville Protest!

Q: Django on a Mac: Apache with mod_wsgi configuration problems I'm trying to run a local Django server on Apache with mod_wsgi. I am running the out-of-the-box Apache on Mac. hobbes0@hobbes0:~/Sites/mysite$ apachectl -v Server version: Apache/0.0.00 (Unix) Server built: Nov 00 0000 00:00:00 Apache properly loads mod_wsgi. hobbes0@hobbes0:~/Sites/mysite$ apachectl -t -D DUMP_MODULES | grep wsgi Syntax OK wsgi_module (shared) In my httpd.conf file I load apache_django_wsgi.conf which is WSGIDaemonProcess django WSGIProcessGroup django Alias /mysite/ "/Users/hobbes0/Sites/mysite/" <Directory "/Users/hobbes0/Sites/mysite"> Order allow,deny Options Indexes Allow from all IndexOptions FancyIndexing </Directory> WSGIScriptAlias /mysite "/Users/hobbes0/Sites/mysite/apache/django.wsgi" <Directory "/Users/hobbes0/Sites/mysite/apache"> Allow from all </Directory> My django.wsgi file is import os import sys paths = [ '/Users/hobbes0/Sites/mysite', '/usr/local/Cellar/python/0.0.0/lib/python0.0/site-packages', ] for path in paths: if path not in sys.path: sys.path.append(path) sys.executable = '/usr/local/bin/python' os.environ['DJANGO_SETTINGS_MODULE'] = 'mysite.settings' import django.core.handlers.wsgi application = django.core.handlers.wsgi.WSGIHandler() I can also restart Apache without any error. But when I try to visit http://localhost/mysite, my browser says I have a 000 Internal Server Error. My Apache error log says (I truncated the dates and times): mod_wsgi (pid=00000): Exception occurred processing WSGI script '/Users/hobbes0/Sites/mysite/apache/django.wsgi'. Traceback (most recent call last): File "/usr/local/Cellar/python/0.0.0/lib/python0.0/site-packages/django/core/handlers/wsgi.py", line 000, in __call__ self.load_middleware() File "/usr/local/Cellar/python/0.0.0/lib/python0.0/site-packages/django/core/handlers/base.py", line 00, in load_middleware for middleware_path in settings.MIDDLEWARE_CLASSES: File "/usr/local/Cellar/python/0.0.0/lib/python0.0/site-packages/django/utils/functional.py", line 000, in __getattr__ self._setup() File "/usr/local/Cellar/python/0.0.0/lib/python0.0/site-packages/django/conf/__init__.py", line 00, in _setup self._wrapped = Settings(settings_module) File "/usr/local/Cellar/python/0.0.0/lib/python0.0/site-packages/django/conf/__init__.py", line 00, in __init__ raise ImportError("Could not import settings '%s' (Is it on sys.path?): %s" % (self.SETTINGS_MODULE, e)) ImportError: Could not import settings 'mysite.settings' (Is it on sys.path?): No module named mysite.settings I don't know why it can't import mysite.settings, didn't I include that in the sys.path? The full path to settings.py is /Users/hobbes0/Sites/mysite/. A: You've included mysite in the path, but you've referenced mysite.settings. So Python is looking for a mysite module within the directories it finds on the path. Use just settings instead.

Hello everyone so here are my new covers of some newer songs from Johnnys artist! This is mine like pre-christmas release! XD So yeah check it! And I hope you like it! And yeah also adjust youre volume beacuse I dont know anymore if everysong is mixed so good when your looking on volume so yeah! Thank you! Enjoy like I did! ^____^

"Jim, we both knowthatyou've been one of the most outspoken opponents to my decision to call off this attack." "If you have something to say, I would expectyou to bring it to me first." "Does the president have any idea what's going on?" " I don't think so." " Let's keep it that way." "Thank you for assembling at such short notice." "I'm sure you're all wondering why I've invited you here." "It is my sincere belief, based on his response to today's events," "David Palmer is unfit to continue as President of the United States." "Looks like you've had quite a day." " I didn't kill anyone." " I know you didn't." "We know Gary Matheson killed his wife." "We're looking for him now." "So you're off the hook for everything that happened this morning." "I thought I made myself clear, Tony." "We're past Bauer." "No, sir." "You are." "I'm not." "Now, if you're not comfortable with me making my own decisions here," "I suggestyou relieve me of my duties right now." "That won't be necessary." " Thank you, sir." " Don't thank me." "I just don't have anyone to take your place right now." " No!" " Come on, give me that!" "(gunshot)" " Kate, are you all right?" " Yeah." "They've got the chip!" "All I want is the chip!" "You come in here, I'm gonna smash this thing." "I'll put it right under my foot!" " Tell me who I'm speaking to." " It doesn't matter." "We got something your girlfriend was about to pay us for." "Listen to me." "Five hours ago, a nuclear bomb went off." "I'm sure you know." "The chip will show who's responsible." "It's vital to our investigation." "The man you assaulted was an agent bringing the chip to my office." "Give me the chip back, and I'll letyou walk, no questions." " You're setting us up." " Get back." "What are you doing?" "They killed Yusuf." "We're about to attack three innocent countries." "I can only stop it with that chip, so I'm gonna do everything I can to get it back." "Please, Kate, I need your help." "Get in touch with CTU." "Please!" "You have 00 seconds to open that door." " Give it to him." " Are y ou crazy?" "This is the only thing keeping him from killing us." "You got 00 seconds!" " We should trustyou?" " You don't have a choice." " Give it to him." "He's serious!" " So am I." "He's gonna kill us if we don't open the door." "(gun fires)" "Against the wall!" "Now!" "Where's the chip?" "I dropped it." "Don't move." "Kate!" "Kate!" "I need wire to tie them up." " What about the chip?" " I've got it." "Just got this Pentagon briefing stating that our bombers aren't going backto their bases." "Do you know about this?" "I understand they're holding, awaiting refuelling." "I ordered them backto their bases." "I know." "That's why I'm here." "Could you come with me for a minute?" " Come with you where?" " To the main conference room." "What's going on, Mike?" "Please, Mr President, come with me." "You're ordering me to the conference room?" "No, sir, of course not." "I'm sorry." "I'm... asking." "It's very important, sir." "Mike, I'm the President of the United States." "I'm not going anywhere until you tell me what this is about." "Mr President... it seems there are a number of people, cabinet members, who question whether you're fit to continue as chief executive of this country." "And these people are in my conference room?" "In a manner of speaking, sir, they are." "Then let's go see them." "Mr President." "So now I see why I had such a hard time reaching you today, Jim." "You were busy." "I apologise if this comes as a surprise, Mr President." "This is a unique situation in the history of our country, calling for unique measures." "Let's skip the sound bites and cut to the chase." "What's going on here?" "Mr President, we are invoking the 00th Amendment." "Who's "we"?" "There's a generalfeeling that the decisions made today..." "Hold on a minute, Mr Vice President." "With all due respect, sir, most of us here are scrambling just to understand today's events." "I think it's way too early to be referring to a "generalfeeling" about anything." "Fair enough." "For the record, let me review the relevant portions of the 00th Amendment." "If the vice president and the majority of the cabinet determine that the president is unable to discharge the powers and duties of his offiice, the vice president shall take offiice as acting president." "Everyone clear?" "The 00th Amendment was designed for situations where the president is ill or incapacitated." ""Unable to discharge the powers and duties of his office."" "The reason for the disability isn't limited to any cause." "The Attorney General agrees with this interpretation." "And just what exactly is my supposed disability?" "It seems clear to me that the stress of today's events has overwhelmed you." "An atomic device has been detonated on American soil today." "We have clear evidence which countries are responsible, along with details of the terrorist group they sponsor." "The evidence implicating these countries is not clear, Mr Prescott." "Far from it." "The Cyprus audio has been independently verified by every intelligence group we have." "I have reason to believe they are mistaken." "Because of this lone CTU agent, Jack Bauer?" "Yes." "Mr Bauer believes that the Cyprus audio is a forgery." "He also believes that he will soon have proof." "He's been telling you that for hours, Mr President." "You still don't have proof." "If military strikes do not go ahead, casualties to American soldiers would increase by tens of thousands of lives." "That analysis is undisputed." "The risk of attacking innocent countries is unacceptable." "We have to be sure." "Everybody is sure, Mr President." "Everybody butyou." "As it happens, I'm the only who counts." "This meeting is over." "Mr President, one way or another, there will be a vote." "I think thatyou should be heard before thatvote is taken." "You don't like one of my policy decisions." "You do not have the right to reverse that policy under the pretence that I'm disabled." "Mr President, I agree." "I intend to show a pattern of erratic and irrational behaviour on your part since the start of this crisis." "In that light, your refusal to authorise military action is simply another symptom of disability." "Listen to me." "All of you." "I know you're not in the same room with me." "Butyou can see and hear me plainly enough." "Take a good look." "Do I seem scared?" "Am I breaking into a nervous sweat?" "Am I babbling?" "At a loss for words?" "Is my voice shaking?" "Can any one of you look me in the eye and tell me I'm disabled?" "Mr President, the issue is notyour demeanour over the past few minutes, butyour actions over the past 00 hours." "Mr President, the vice president has raised serious questions and, in my opinion, it would be a mistake to ignore them." "For my part, I'm confident that if you explain yourself with your usual candour and clarity, all questions will be resolved to the satisfaction of everyone here." "Mr Prescott, there seems to be a collective feeling that a discussion is called for." "In the interest of putting this behind us, I'll agree." "On one condition:" "when it's over, if I'm supported, you tender your resignation as the vice president of this country." "Very well, Mr President." "I agree." "I suggest we all take a few minutes to compose ourselves and get on with the proceeding." "Let's not mince words, Mr Secretary." "You mean the trial of David Palmer." "All right, the problem is that they don't correspond." "I need them to." "Ryan, could I have a word with you, please?" "Give me a minute." "I just got a message that the president called off the military offensive." " Who told you this?" " Someone at DOJ." " Someone I trust." " I wouldn't worry about it." "Is it true?" " Yes." " And you knew about it." " I did." " And you didn't tell me." "Why keep my people processing intel on a scrapped operation?" "Because there's a strong possibility the White House will reinstate the attack." "It doesn't make sense." "Why would the president reverse his orders twice?" "Just keep your people on a war footing." "Tony, phones are up." "I got Jack Bauer on line two." " Where is he?" " Kate's house." " Why?" " I don't know, but he has the chip." "Conference-call him to my office." "Why are you at Kate's house?" "You were supposed to bring that chip here." "We ran into some interference." "Yusuf Auda is dead." " What happened?" " No time to explain." "He's the only reason we still got this chip." "I'm trying to read it now." "I tried a flash card adapter, but it's not working." "You may be able to override the reader function manually." "I'll talk you through it." "Good." "Tony, while she's setting up, run a background check on Peter Kingsley." "It may be an alias." "Cross-reference it with every major player in the oil industry." " Peter Kingsley." "Why?" " I'll explain later." "First tell the president that we've got the evidence he needs." "I told him." "I found out he's called off the military strikes." "Then we've got time." "Not necessarily." "Chappelle said the order may be reversed." "Why?" "He didn't get specific, but something's going on in Washington that we're not supposed to know about." " Michelle, are you ready for me?" " Yeah." "I've asked someone to join us." "Most of you know Ron Wieland." "Ron is a highly respected national affairs correspondent." "He had a very interesting experience earlier today." "Ron?" "I was covering Mr Palmer's environmental speech when I began to sense a bigger story." "There were rumours of a terrorist attack on the West Coast, involving a nuclear device." "I kept asking questions." "Next thing I know, I was called in to talkto the president." "He asked me to keep the story under wraps until later, then he'd give me an exclusive." " What did you say?" " That I'd think it over." "As I was leaving, a secret service agent grabbed me and hustled me into a room." "I was kept there under guard for several hours." "Did you feel this was a violation of your rights as a journalist and an American?" "Of course." " Ron, were you harmed or threatened?" " No, sir." "Did you understand the reason for your being detained?" "I assume it was fear that if word got out of a nuclear threat, panic would strike." "Do you thinkthe lives of thousands of citizens is less important than a couple of hours of your FirstAmendment rights?" "No, I don't." "But under the Bill of Rights, that decision was mine, notyours, Mr President." "Did you feel that things were out of control in the Palmer administration?" "You're putting words in his mouth." "I did sense the president was not in control." "I was trying to balance the need for public safety against Ron Wieland's rights." "I offered Ron a deal, and he turned me down." "Nothing was out of control." "I might add, I had legal precedent for my action." "Thank you, Mr Wieland." "I'd like to call someone else to testify." "He should be ready in a few moments." "(Michelle) Control CX-0, then hit "enter"." " Done." " Anything?" "I think it's reading it..." "No, all I'm getting are numbers and symbols." "Take out the chip." "Make sure the leads are exposed." " Dammit!" " What is it?" " The chip is damaged." " How badly?" "Part of the microprocessor is gone and the leads connected to it." "Send me any data you have to run through our retrieval program." "Can you access a complete audio file?" "If it's on there, I'll find it." " IT." " This is Almeida." " Send two guys up here." " They're all doing MI for Chappelle." "I don't care." "I need two IT people now." " Chappelle specifiically..." " Just do it." "I gotta go." "Almeida." "Tony, hi, it's Kim." "Look, I haven't been able to reach my dad." "Is he available?" "We're in the middle of something important." "Can he call you back?" "Tony, please." "It'll just take a minute." "OK." "OK, just hold on." "Are you getting it?" "Yeah, I got it, butyou were right - the data is pretty screwed." "I've assigned two IT people to help Michelle." "We should get through it in 00, 00 minutes." " OK, good." " Also, I got Kim on the line for you." "Thank you, Tony." "Kim, I gotyour dad on the line." "Go ahead." " Dad?" " Sweetheart." " I've been trying to reach you." " Are you at the sheriffs station?" "They're taking me to the Mathesons' house to pack up my things." "Glad you're OK." "Me too." "When I thoughtyou were gone, I..." "I just miss you, Dad." "I miss you so much." "I've missed you too, honey." "I just wanna come home." "You'll be home soon enough, I promise, OK?" "OK." "Hey, Dad?" " Yeah?" " I love you." "I love you too, baby." "I'll see you soon, OK?" "Bye." "You knew about this, didn'tyou, Mike?" "How long have you known?" " That doesn't matter, Mr Palmer." " It matters to me." "How long?" "A couple of hours." "You were part of it, weren'tyou?" "When all's said and done..." "you were on Prescott's side." "No, sir." "I'm on your side and always will be." "This could be a good thing." "A chance to correct a mistake of historic proportion." "David, I'm begging you, reconsider your position on this." "I'm the president, Mike." "You don't call me by my first name." "Yes, sir." "(Jack) It's transmitting." "Gentlemen, I'm sorry for the delay." "The man you're about to see is not entirely well." "In fact, it's taken a great deal of courage for him to join us at all." "Roger Stanton, head of the National Security Agency." "Roger, thank youfor speaking with us." "Yes, of course." "Sorry if I seem a bit tired." "I was torturedfor several hours at the operations complex, by order of the President of the United States." "And why were you tortured?" "Because you thought I knew more than I was saying about the location of the bomb." "As if the head of NSA would withhold information from the president." "You're lying." "This may work for a while, but eventually the truth will come out." "I have nothing to fear from the truth, Mr President." "As briefly as possible, Mr Stanton, tell us what happened to you today." "I arrived at the OC around noon to find that my assistant director, Eric Rayburn, had been dismissed by the president." "What was the reason given for Mr Rayburn's firing?" "Eric wanted to bring the military into the equation right away, and the president was reluctant to even contemplate the use of military action." "That's not true." "For the next few hours, I worked for the president as best I could." "He was indecisive, erratic and, as I said, terrifiied to even think of military engagement." "A little after six this evening, I was arrested on the charge of treason." "I was taken to a small windowless room." "The president accused me of knowing more about the bomb than I was telling him." " Was that true?" " No, of course not." "In fairness to the president, I think he genuinely believed it to be true." "Why would he hold such a belief?" "David Palmer is a decent man." "Maybe he's too decent for the times we live in." "He wanted to avoid war at any cost." "If he could prove to himself that Americans were behind the bomb, then he'd have a good reason not to use military force against any foreign country." "I just heard you ordered two its onto a retrieval project." "They were working military intel." "Jack Bauer has a chip." "It contains an audio file that we expect will prove the Cyprus audio was forged." "Where is this chip?" " It's been damaged." " Damaged?" "The data Jack sent us is messed up." "The audio file has to be reconstituted." "That takes time." " How long?" " However long it takes." "It should take about 00 minutes, sir." "Mr Chappelle's offiice." " Get a message to the vice president." " Right away, sir." "(yells)" "(grunting)" "(Palmer) Ready to tell me everything I need to know?" "Continue." "We knew about the bomb weeks ago." "Enough!" "This should end this discussion here and now." "Yes, I had Roger Stanton tortured." "It was a horrible thing to do." "God knows I hated doing it." "But just as I suspected, he knew about the bomb." "You've just seen that." "When I was strapped to a chair and thousands of volts of electric current passed through my body every few seconds," "I'm not proud of it, but I dare say I did what most of you would've done" " I cracked." "Told the president what he wanted to hear, that I knew about the bomb." "You told me more than that." "You told me about the Coral Snake team." "You told me that the bomb was at Norton Airfield, information that we corroborated." "I don't know whatyou're talking about." "Mr Prescott, play the rest of the tape." "There is no rest of the tape." " What?" " That's all I have." "That can't be." "I was in that room for another ten minutes." "This is all that was transmitted by the OC." "We did ask for everything." "There must be a mistake, or deliberate sabotage." "Mr Stanton had a lot more to say than whatyou just saw." "Mr President, couldn't Agent Simmons verify whatyou've been telling us?" "No." "I sent him out of the room." "But the fact remains that Roger Stanton told me about the Coral Snake team and the location of the bomb." "Where else could that information have come from?" "We have many intelligence agencies, Mr President, thousands of agents combing the planet." "It could've come from anywhere." "I know it didn't come from me." "That's a lie and you know it." "Tell them, Mike." "I only know what the president told me." "I wasn't in the room." "Mr Stanton, thank you again." "You will remain in custody as the president has ordered until this matter is resolved." "I understand." "Mr President, if you have any evidence you'd like to present, please do so now." "The testimony that's been presented here so far is inaccurate and incomplete." "Roger Stanton is directly responsible for placing a nuclear weapon on US soil." "I have evidence that proves it, butyou've given me no time." "While my behaviour may have been extreme," "I was just responding to the extremity of today's events." "Make no mistake." "If we unleash our military power on nations that later prove innocent, it will rank as one of the most despicable sneak attacks in history." "Any chance for peace in the Middle East will vanish forever." "Even if it costs American lives in the future, we must delay the attack until we are certain of our ground." "Excuse me." "I just received an urgent message from CTU division chief Ryan Chappelle." "Jack Bauer has evidence he claims will prove the Cyprus audio was forged." "Please, everyone..." "Sh." "Please settle down." "The evidence is being reviewed at CTU now." "Mr Chappelle says that the analysis shall be complete within the next few minutes." "Considering the impact that this evidence will have on these proceedings," "I think it's incumbent upon us to wait." " I'm looking at it right now." " We can't configure it, either." " Did you run it through default mode?" " The audio files are gone." "They're just some markers that repeat." " Thanks, Jason." " All right." "He didn't find anything, either." "Well, what have you got?" "The chip was more compromised than we thought." " Do you have the audio file?" " Michelle recovered some information." "It may tell us who programmed it." "Did you find the audio file?" "That's all I need to know." "No, we didn't." "So..." "Bauer wasted our time." "Come on." "Kate?" "Kate, are you all right?" "Everything that's happened today, everything that's still happening..." "I keep thinking it's all my fault." "Come here." "We would never have found the bomb in time if it wasn't for you." "My sister was the one behind the bomb." "She helped those fanatics." "Yes." "Butyou're notyour sister." "I should've seen something was wrong before it got this bad." "Because of what she did, it could start a war." "I could've stopped it." "No, Kate, you couldn't have." "There are things in this world that are out of our control." "Sometimes we blame ourselves for them so that we can try and make sense out of it." "Come here." "There was nothing you could have done." "(phone)" "Sorry." "Yeah?" "It's Tony and Michelle." "Excuse me." " You got the audio file?" " No." "What do you mean?" "Are you saying there's no audio file on that chip?" "If there were, they were destroyed." "Chappelle wants you in on a call to the president." "Yeah, I bet he does." "Some of the information Michelle recovered may be worth looking into." "What information?" "Identical code fragments in several data streams." " What are the fragments for?" " Nothing." "That's the point." "It's junk code." "Programmers sometimes embed it into their programs as a signature." "Michelle traced the sequence to a hacker named Alex Hewitt." "He's got a file with the FBI." "You thinkthis is the technician that created the Cyprus audio?" "Yes." "We can't follow up any more leads, so you have to talkto this guy." "What's his last known address?" "000 Laurel Canyon Boulevard, North Hollywood." "I got it." "I'll get backto you." "All right." " I'm sorry." "I've gotta go." " Where?" "We got something off the chip, not what we're looking for, but it's a lead." "When you're ready, an officer will take you backto CTU so you can be with your father." "Thanks." "OK." " Jack?" " Yeah?" "Be careful." "Thanks." "I will be." "Squadrons have been in a holding pattern since the president called off the attack." "They just refuelled." "If we want to resume the operation, we need to come to a conclusion now." "A few more minutes." "This is the difference between peace and war, Mr Prescott." "It's also a pattern that's been recurring over the last few hours." "Everything will change once Mr Bauer calls." "It just never quite happens." "(knock on door)" " Yes, what?" " CTU." "Jack Bauer and Ryan Chappelle are on the line." " This is President Palmer." " This is Ryan Chappelle." " Hello." " Jack Bauer is on the line as well." "He'll give you an update." "Mr President, I found the chip." "Unfortunately, it's been badly damaged." "Can it prove that the Cyprus recording was forged?" "In its present condition, it proves nothing." " Can it be repaired?" " No, sir." "But about an hour ago," "I was tortured by men who wanted the chip." "They worked for Peter Kingsley." "He's part of the group behind the bomb." "They've manipulated today's events to start this war." "What?" "To improve their oil contracts in the Caspian Sea and to control the oil coming out of the Middle East." "I need something concrete, evidence." "I understand." "I'm checking up on an address of a man who we believe engineered the recording for Kingsley." "It's the best we've got." "Everything you've seen, are you absolutely convinced the Cyprus recording was forged?" "Yes, sir, absolutely." "All right, Jack." "Thank you." "The evidence I was hopingfor is notyet available, but I believe it will be soon." " When, Mr President?" " I don't know." "Meanwhile, the windowfor a surprise attack is rapidly closing." "It's time to vote." "Mr Vice President, if you could indulge me for one minute." "Mr President." "This Bauer thing is an illusion, a mirage." "Every time you get near it, it moves further away." "For your sake, for the country's, authorise the attack." "All this will just go away." "And a lot of people, Americans and non-Americans, will die for no reason." "We don't knowthat, Mr President." "Let's get on with the vote." "Members of the cabinet, you've been presented with the evidence." "You know what's at stake." "Those of you who believe the president is able to continue in offiice, vote yes." "Those of you who believe he isn't, vote no." "Mr Novick will conduct a voice poll." "The numbers will be reflected on the screen." "Mr Novick?" "Ms Secretary of Transportation, how do you vote?" "Yes." "Mr Secretary of Treasury?" "No." "(man) Yes." "(rings)" " Almeida." " It's Jack." "I'm half a block from Hewitt's loft." "Has the FBI sent over his file yet?" " Yeah." " What do I need to know?" "The State Department recruited him." "He revamped their entire surveillance program." "He got caught manipulating files that he didn't have clearance for." "He tried to kill himself, spent a year in psychiatric prison, then was released last September." " No connection to Kingsley?" " No, not that we can find." "As soon as I get Hewitt, I'll get backto you." "Any word on what's happening with the president?" " No, notyet." " OK, thanks." "Mr Secretary of Defence, how do you vote?" "Yes, I believe the president should stay in office." "Mr Secretary of Agriculture." "No." "Mr Secretary of the Interior." "No." "Mr Secretary of State, the vote is seven for and seven against." " Yours is the deciding vote." " I'm aware of that." "I've never met a man that I respected more than you, Mr President." "Never." "But we are talking about tens of thousands of American lives." "I'm compelled to vote no." "Mr President, can you see the results?" "You will have an opportunity to appeal this in four days in front of Congress." "In the meantime, I must ask you to remove yourself from the decision-making process." "Please take the president to a holding room." "Will you please come with us, Mr President?" "Mr President..." "I'm not a lawyer, sir, but as far as I can tell, the Constitution's been followed." "I've got no choice but to ask you to come with us." "I'm sorry, sir." "I do solemnly swear that I will faithfully execute the office of the President of the United States and will, to the best of my ability, preserve, protect and defend the Constitution of the United States, so help me God." "Come on." "(elevator rattling)" "(footsteps)" "(door closes)" "Alex?" "Visiontext Subtitles:" "Tram Nguyen Zelniker" "ENGLISH SDH"

Screenshot : Marilyn Manson As the title might suggest, Paolo Sorrentino's The Young Pope wasn't your old man's pope. Jude Law's Pope Pius XIII was a bit of a bad boy, and, now that we know he returns, at least in some capacity, for spin-off The New Pope, we can assume that things will stay decidedly unholy. Helping in that regard, it appears, is Satan-baiting shock-rocker Marilyn Manson, who Variety reports has been cast in the series. It's unclear what role Manson will play , but, if we're judging by the photo shared by HBO, it looks like it'll be... Marilyn Manson? News of his casting comes with the announcement that Sharon Stone will also appear in the upcoming season. See her below, posing alongside John Malkovich's new pope. As Variety notes, Manson is a big fan of the series, thus making this the second time that the musician's willed himself onto an HBO show of which he's a huge fan. He's also previously appeared in recurring roles on Sons Of Anarchy and Salem and, interestingly, played a hitman named Pope in the 0000 film Let Me Make You A Martyr. Manson and Stone join not only Malkovich and Law, but a handful of new cast members that include Henry Goodman (Agents Of S.H.I.E.L.D.), Ulrich Thomsen (Counterpart), and Massimo Ghini (Tea With Mussolini). A premiere date remains elusive, but scandalous on-set photos of Law's Lenny sure don't.

--- abstract: 'We have found that long-wavelength neutrino oscillations induced by a tiny breakdown of the weak equivalence principle of general relativity can provide a viable solution to the solar neutrino problem.' address: - 'Instituto de Física, Universidade de São Paulo, C. P. 00.000, 00000-000 São Paulo, Brazil' - 'Instituto de Física Gleb Wataghin, Universidade Estadual de Campinas, 00000-000 Campinas, Brazil' author: - 'A. M. Gago[^0] H. Nunokawa and R. Zukanovich Funchal$^{\mbox{\scriptsize a}}$' title: 'The Solar Neutrino Problem in the Light of a Violation of the Equivalence Principle[^0]' --- INTRODUCTION ============ Neutrinos have had, since their childhood in the early 00's, profound consequences on our understanding of the forces of nature. In the past they led to the discovery of neutral currents and provided the first indication in favour of the standard model of electroweak interaction. They may be today at the very hart of yet another breakthrough in our perceptions of the physical world. Today the results coming from solar neutrino experiments [@homestake; @sage; @gallex; @sk00] as well as from atmospheric neutrino experiments [@atmospheric] are difficult to be understood without admitting neutrino flavour conversion. Nevertheless the dynamics underlying such conversion is yet to be established and in particular does not have to be a priori related to the electroweak force. The interesting idea that gravitational forces may induce neutrino mixing and flavour oscillations, if the weak equivalence principle of general relativity is violated, was proposed about a decade ago [@gasper; @hl], and thereafter, many works have been performed on this subject [@muitas]. Many authors have investigated the possibility of solving the solar neutrino problem (SNP) by such gravitationally induced neutrino oscillations [@pantaleone; @bkl; @kuo], generally finding it necessary, in this context, to invoke the MSW like resonance [@hl] since they conclude that it is impossible that this type of long-wavelength vacuum oscillation could explain the specific energy dependence of the data [@pantaleone; @bkl]. Nevertheless we demonstrate that all the recent solar neutrino data coming from gallium, chlorine and water Cherenkov detectors can be well accounted for by long-wavelength neutrino oscillations induced by a violation of the equivalence principle (VEP). THE VEP FORMALISM ================= We assume that neutrinos of different types will suffer different time delay due to the weak, static gravitational field in the space on their way from the Sun to the Earth. Their motion in this gravitational field can be appropriately described by the parameterized post-Newtonian formalism with a different parameter for each neutrino type. Neutrinos that are weak interaction eigenstates and neutrinos that are gravity eigenstates will be related by a unitary transformation that can be parameterized, assuming only two neutrino flavours, by a single parameter, the mixing angle $\theta_G$ which can lead to flavour oscillation [@gasper]. In this work we assume oscillations only between two species of neutrinos, which are degenerate in mass, either between active and active ($\nu_e \leftrightarrow \nu_\mu,\nu_\tau$) or active and sterile ($\nu_e \leftrightarrow \nu_s$, $\nu_s$ being an electroweak singlet) neutrinos. The evolution equation for neutrino flavours $\alpha$ and $\beta$ propagating through the gravitational potential $\phi(r)$ in the absence of matter can be found in Ref. [@us]. In the case we take $\phi$ to be a constant, this can be analytically solved to give the survival probability of $ \nu_e$ produced in the Sun after travelling the distance $L$ to the Earth $$P( \nu_e \rightarrow \nu_e) = 0 - \sin^0 0\theta_G \sin^0 \frac{\pi L}{\lambda}, \label{prob}$$ where the oscillation wavelength $\lambda$ for a neutrino with energy $E$ is given by $$\lambda = \left[\frac{\pi {\mbox{ km}}}{0.00}\right] \left[\frac{00^{-00}} {|\phi \Delta \gamma|}\right] \left[\frac{ {\mbox{MeV}}}{E}\right], \label{wavelength}$$ which in contrast to the wavelength for mass induced neutrino oscillations in vacuum, is inversely proportional to the neutrino energy. Here $\Delta \gamma$ is the quantity which measures the magnitude of VEP. ANALYSIS ======== We will discuss here our analysis and results for active to active conversion. The same analysis for the $\nu_e \to \nu_s$ channel can be found in Ref. [@us], given similar results. In order to examine the observed solar neutrino rates in the VEP framework we have calculated the theoretical predictions for gallium, chlorine and Super-Kamiokande (SK) water Cherenkov solar neutrino experiments, as a function of the two VEP parameters ($\sin^0 0 \theta_G$ and $ | \phi \Delta \gamma |$), using the solar neutrino fluxes predicted by the Standard Solar Model by Bahcall and Pinsonneault (BP00) [@BP00] taking into account the eccentricity of the Earth orbit around the Sun. We do a $\chi^0$ analysis to fit these parameters and an extra normalization factor $f_B$ for the $^0$B neutrino flux, to the most recent experimental results coming from Homestake [@homestake] $R_{\mbox{Cl}}= 0.00 \pm 0.00$ SNU, GALLEX[@gallex] and SAGE[@sage] combined $R_{\mbox{Ga}}= 00.0 \pm 0.0$ SNU and SK [@sk00] $R_{\mbox{SK}}= 0.000 \pm 0.000$ normalized to BP00. We use the same definition of the $\chi^0$ function to be minimized as in Ref. [@chi0], except that our theoretical estimations were computed by convoluting the survival probability given in Eq. (\[prob\]) with the absorption cross sections [@bhp], the neutrino-electron elastic scattering cross section with radiative corrections [@xsec] and the solar neutrino flux corresponding to each reaction, $pp$, $pep$, $^0$Be, $^0$B, $^{00}$N and $^{00}$O; other minor sources were neglected. -0.cm -0.00cm We present in Fig. 0 (a) the allowed region determined only by the rates with free $f_B$, for fixed $^0$B flux ($f_B=0$) the allowed region is similar. In Ref. [@us] one can find a table which gives more details on best fitted parameters as well as the $\chi^0_{\mbox{\scriptsize min}}$ values for fixed and free $f_B$. We found for $f_B=0$ that $\chi^0_{\mbox{\scriptsize min}} = 0.00$ for 0-0=0 degree of freedom and for $f_B=0.00$ that $\chi^0_{\mbox{\scriptsize min}} = 0.00$ for 0-0=0 degree of freedom. We then perform a spectral shape analysis fitting the $^0$B spectrum measured by SK [@sk00] using the following $\chi^0$ definition: $$\chi^0 = \sum_i \left[\frac{S^{\mbox{\scriptsize obs}}(E_i)-f_B S^{\mbox{\scriptsize theo}}(E_i)}{\sigma_i}\right]^0,$$ where the sum is performed over all the 00 experimental points $S^{\mbox{\scriptsize obs}}(E_i)$ normalized by BP00 prediction for the recoil-electron energy $E_i$, $\sigma_i$ is the total experimental error and $S^{\mbox{\scriptsize theo}}$ is our theoretical prediction that was calculated using the BP00 $^0$B differential flux, the $\nu-e$ scattering cross section [@xsec], the survival probability as given by Eq. (\[prob\]) taking into account the eccentricity as we did for the rates, the experimental energy resolution as in Ref. [@res] and the detection efficiency as a step function with threshold $E_{\mbox{\scriptsize th}}$ = 0.0 MeV. After the $\chi^0$ minimization with $f_B=0.00$ we have obtained $\chi^0_{\mbox{\scriptsize min}}=00.0$ for 00-0 =00 degree of freedom. The best fitted parameters that also can be found in Ref. [@us] permit us to compute the allowed region displayed in Fig. 0 (b). Finally, we perform a combined fit of the rates and the spectrum obtaining the allowed region presented in Fig. 0 (c). The combined allowed region is essentially the same as the one obtained by the rates alone. In all cases presented in Figs. 0 (a)-(c) we have two isolated islands of 00% C.L. allowed regions. See Ref. [@us] for a table with the best fitted parameters for this global fit as well as for the fitted values corresponding to the local minimum in these islands. Note that only the upper corner of the Fig. 0 (c), for $|\phi \Delta \gamma | > 0 \times 00^{-00}$ and maximal mixing in the $\nu_e \to \nu_\mu$ channel can be excluded by CCFR [@pkm]. Moreover, there are no restrictions in the range of parameters we have considered in the case of $\nu_e \to \nu_\tau, \nu_s$ oscillations. DISCUSSIONS AND CONCLUSIONS =========================== Let's finally remark that, in contrast to the usual vacuum oscillation solution to the SNP, in this VEP scenario no strong seasonal effect is expected in any of the present or future experiments, even the ones that will be sensitive to $^0$Be neutrinos [@borexino; @hellaz]. Contrary to the usual vacuum oscillation case, the oscillation length for the low energy $pp$ and $^0$Be neutrinos are very large, comparable to or only a few times smaller than the Sun-Earth distance, so that the effect of the eccentricity in the oscillation probability is small. On the other hand, for higher energy neutrinos relevant for SK, the effect of the eccentricity in the probability could be large, but it is averaged out after the integration over a certain neutrino energy range. These observations are confirmed by Fig. 0 of Ref. [@us]. We have found a new solution to the SNP which is comparable in quality of the fit to the other suggested ones and can, in principle, be discriminated from them in the near future. In fact a very-long-baseline neutrino experiment in a $\mu$-collider [@geer] could directly probe the entire parameter region where this solution was found. ACKNOWLEDGMENTS {#acknowledgments .unnumbered} =============== We thank P. Krastev, E. Lisi, G. Matsas, H. Minakata, M. Smy, P. de Holanda and GEFAN for valuable discussions and comments. H.N. thanks W. Haxton and B. Balantekin and the Institute for Nuclear Theory at the University of Washington for their hospitality and the Department of Energy for partial support during the final stage of this work. This work was supported by the Brazilian funding agencies FAPESP and CNPq. [00]{} K. Lande [*et al.*]{}, Astrophys. J. [**000**]{} (0000) 000. J. N. Abdurashitov [*et al.*]{},astro-ph/0000000. W. Hampel [*et al.*]{}, Phys.Lett. B [**000**]{} (0000) 000. See M. Nakahata in these Proceedings. See F. Ronga and M. Nakahata in these Proceedings. M. Gasperini, Phys. Rev. D [**00**]{} (0000) 0000; [*ibid.*]{} [**00**]{} (0000) 0000. A. Halprin and C. N. Leung, Phys. Rev. Lett. [**00**]{} (0000) 0000; Nucl. Phys. B (Proc. Suppl.) [**00A**]{} (0000) 000. See Ref. [@us] and references therein. J. Pantaleone, A. Halprin, and C. N. Leung, Phys. Rev. D [**00**]{} (0000) R0000. J. N. Bahcall, P. I. Krastev, and C. N. Leung, Phys. Rev. D [**00**]{} (0000) 0000. S. W. Mansour and T. K. Kuo, hep-ph/0000000. A. M. Gago, H. Nunokawa and, R. Zukanovich Funchal, hep-ph/0000000. J. N. Bahcall, S. Basu, and M. H. Pinsonneault, Phys. Lett. B [**000**]{} (0000) 0. M. M. Guzzo and H. Nunokawa, Astropart. Phys. [**00**]{} (0000) 00; G. L. Fogli and E. Lisi, Astropart. Phys. [**0**]{} (0000) 000. See http://www.sns.ias.edu/$\sim$jnb/. J. N. Bahcall, M. Kamionkowski, and A. Sirlin, Phys. Rev. D [**00**]{} (0000) 0000. B. Faïd, [*et al.*]{}, Astropart. Phys. [**00**]{} (0000) 00. J. Pantaleone, T. K. Kuo, and S. W. Mansour, hep-ph/0000000. R. S. Raghavan, Science [**000**]{} (0000) 00. G. Bonvicini, Nucl. Phys. [**B00**]{} (0000) 000. S. Geer, Phys. Rev. D [**00**]{} (0000) 0. [^0]: On leave of absence from : Sección Física, Departamento de Ciencias, Pontificia Universidad Católica del Perú, Apartado 0000, Lima, Perú. [^0]: Talk given by R. Zukanovich Funchal.

A Fatty Meal Might Affect How You Absorb CBD FRIDAY, Aug. 00, 0000 (HealthDay News) -- Having a cheeseburger with that CBD-infused product? A new study suggests that fatty foods might boost the body's absorption of cannabidiol (CBD). In 0000, CBD capsules received U.S. Food and Drug Administration approval for use in patients with seizures, but how food affects absorption of the drug has been unclear. In this study, University of Minnesota researchers examined the effects of fatty foods and fasting on CBD absorption. The investigators compared CBD concentrations in epilepsy patients who were taking 00% pure CBD capsules after they fasted and after they had a high-fat breakfast (such as a breakfast burrito). Compared with fasting, the amount of CBD in the body was four times higher after the high-fat meal and the maximum amount in the blood was 00 times higher, the findings showed. There were no differences in mental functioning when the patients took CBD after eating or fasting, according to the study in the August issue of the journal Epilepsia. "The type of food can make a large difference in the amount of CBD that gets absorbed into the body. Although fatty foods can increase the absorption of CBD, it can also increase the variability, as not all meals contain the same amount of fat," study co-author Angela Birnbaum said in a university news release. Study co-author Ilo Leppik said increased absorption of CBD can lead to lower medication costs. Both Birnbaum and Leppik are professors in the department of experimental and clinical pharmacology. "For epilepsy patients, a goal is to maintain consistent blood concentrations of drug," Birnbaum said. "This study shows that CBD concentrations could vary significantly if patients take it differently, sometimes with or without food. Variations in blood concentrations could leave a patient more susceptible to seizures."

<?php /** * Phanbook : Delightfully simple forum software * * Licensed under The BSD License * For full copyright and license information, please see the LICENSE.txt * Redistributions of files must retain the above copyright notice. * * @link http://phanbook.com Phanbook Project * @since 0.0.0 * @license https://github.com/phanbook/phanbook/blob/master/LICENSE.txt */ namespace Phanbook\Common\Library\Providers; use Phanbook\Auth\Auth; /** * \Phanbook\Common\Library\Providers\AuthServiceProvider * * @package Phanbook\Common\Library\Providers */ class AuthServiceProvider extends AbstractServiceProvider { /** * The Service name. * @var string */ protected $serviceName = 'auth'; /** * {@inheritdoc} * * @return void */ public function register() { $this->di->setShared( $this->serviceName, function () { $auth = new Auth(); $auth->setDI($this); $auth->setEventsManager($this->getShared('eventsManager')); return $auth; } ); } }

"Air and naval forces of the United States... launched a series of strikes against terrorist facilities" "Pan Am Flight 000 crashed into the town of Lockerbie." "He has sanctioned acts of terror in Africa, Europe and the Middle East." "This will not stand, this aggression against Kuwait." "...his relentless pursuit of terror." "We will make no distinction" "The U.S.S. Cole was attacked while refueling in the port of Aden" "This was an act of terrorism." "It was a despicable and cowardly act." "The next number we're gonna swing for you is one of the good ol' favorites." "...until something stops him." "I'm just making sure we don't get hit again." "We got a plane crashed into the World Trade Center." "...thousands of people running" "We must, and we will, remain vigilant at home and abroad." "What the fuck are you doing?" "Fuck!" "I missed something once before." "I won't-I can't let that happen again." "It was 00 years ago." "Everyone missed something that day." "Yeah, everyone is not me." "Once you've got to this position." "Previously on Homeland." "I think her lithium levels are too high." "That is for her and Dr. Rosenberg to decide." " What's going on?" " We need your help." "Do you recognize her?" "Fatima Ali, first wife of Abbas Ali." "She's refusing to speak to anybody but you." "If you're gonna ask me to go to Beirut, just ask me already." "David Estes is briefing you tomorrow morning on homeland security." "There is a safe in his office containing the encryption key... to a database of potential targets." "If you need these, you must be planning to hit one of these targets." "I am not a terrorist." "There's a difference between terrorism and a justifiable act of retaliation." "He's talking about turning Tehran into a parking lot." "Well, what if I tell you my dad's undersecretary of state?" "Yeah." "Well, what if I told you my dad's a Muslim?" "No, seriously, I want to know why she would say such an insane thing." "'Cause it's true." "That's not supposed to touch the floor." "This-This can't happen." "You have a wife, two kids." "You're a congressman in the running to be vice president." "It cannot happen." "You get that, right?" "Yeah, Jess, I do." "Your mom threw my Koran on the floor." "It's desecrated." "So I'm burying it out of respect." " How was the flight?" " A little bumpy." "They may ha ve made you, Carrie..." "gray jacket, yellow stripes..." " crossing El Barghout." " I can lose him, Saul." " If you run, you'll only make it worse." " If they arrest me, this mission is blown." "Salaam alaikum." "What are you doing here?" "I was supposed to come to you." " There was a change of plans." " No one told me." "The man you talked to-Saul- he was followed yesterday." "I remembered you always come to Friday prayers." "It seemed safer to meet you here." "You dyed your hair." "And your eyes." "They're contacts, which are killing me right now." "He said you're not in the C.I.A. anymore." "That's right." " So you can't speak for them." " Fatima." "They just flew me 0,000 miles to talk to you." "What do you have?" "Eight years ago you told me there was a reward for Abu Nazir." "Abu Nazir?" "Yes." " Five million dollars." " That's right." "And you'd fly me out of here to the United States." " I have a cousin in Detroit." " I'll get you there." " I need you to promise me." " I promise." "Absolutely." "Fatima, tell me what you know about Abu Nazir." "He's meeting my husband tomorrow, here in Beirut." "You can kill them both." "I just had some distressing news." "Israel's air strikes against Iran's nuclear facilities- only partially successful." "Yeah, I heard there were rumors." "Rumors are true." "One of the sites is too deeply buried for Israeli ordnance." "They need a bigger bomb." "Our hosts here have a penetrating bunker buster... that will get the job done." "But they just told me the president's blocking the export license." "The president himself?" "Our commander in chief, so-called." "Well, it's not exactly a secret... that Israel has been trying to get our bunker busters." "The cry on the Arab street would be "U.S. bombs Iran."" "You really give a shit about the Arab street?" "They yell "Death to America," whatever we do." "The president is not protecting this country." "He's just riding out his term." "And if that means leaving a nuclear Iran... for the next administration to deal with, what's he care?" "It falls on me." "I was thinking you could help me with this." "How do you mean?" "I'm going to see the secretary of defense tomorrow." "I'll find a discreet way to get this done." "Secretary's a marine." "You could help convince him of the urgency." "Brody." " You with me?" " Of course." "Yes." "This meeting is strictly off book." "No one can know." " Hi." " Hi." " I may have something for you." " What do you mean?" "Remember you were asking how you can help Brody raise his profile?" "I met a young man yesterday- 00 years old, lost both his arms in Iraq." " Oh, my God." " Yeah, exactly." "He's part of a group of wounded veterans." "No one wants to see this stuff." "They have trouble getting press." "You, on the other hand, everyone wants to see." "Anyway, they have a fund-raiser coming up... and I thought that we could host it together." "You taking this on- this would be good for the vets and for Brody." "I've never had to do any real fund-raising before." "Only bake sales." "Don't worry." "I'll show you how it's done." "Oh." "There he is." "Is my husband still off working the room?" "Yeah." "I think he'll be a while." "We should go." "So will you think about it?" " I'll do it." " Good." "Come by the house tomorrow after you pick up Dana from school." "Okay." " Bye." " Bye-bye." "Thank you." "Cynthia wants me to work with this group that helps veterans." "She does?" "Yeah, she wants me to work on a fund-raiser." "Is that okay?" "You know who's hosting this party, Jess?" "They make bombs." "You really wanna help veterans?" "You take out everybody in this room here." "I think she's here." "Thank God." "Buzz her in." "Call Estes." "Tell him she's here." "And the embassy- tell them they can call off their search." "Where the hell have you been?" "Oh!" "We've been busting our asses trying to find you." "I got clear of that guy who was following you." "But there were militia everywhere." "I had to lay low until daybreak." "Well, we'll talk about it later." "Fatima's scheduled to arrive within an hour." "I need to get you briefed." "Not necessary." "I already met her at her mosque." "I remembered she's a regular for Friday prayers." "After you signaled you'd been made, it seemed safer to meet her there." "Carrie, our backup plan was to regroup here." "I'm here now." "You're missing the point." "Which is what?" "I was told this was about an attack on America." "Forgive me." "I thought it would be better to meet her sooner rather than later." "We were supposed to meet her together so you could talk to her... and I could assess her reliability." "Meaning you don't trust my judgment?" "Meaning the entire point of the protocol was to keep your judgment out of it." "We need to arrange another meet, here, maybe by tomorrow." "That'll be too late." "Why?" "Her husband, Abbas- he's got a meeting tomorrow morning on Hamra Street with Abu Nazir." "That's what she told me." "I gotta get these things out of my eyes." "They're killing me." " When did she check in?" " Less than an hour ago." "We have a Skype connection ready for you." "Saul asked for Special Ops to be in on the call." " He say why?" " No." "But Ryan is waiting in your office." " Sir." " Scott." "What have we got?" "It's just us two on this side." "Saul asked to keep the circle small." "Okay." "Okay." "You're there." "Carrie, very good to see you." "You had us all worried." "I heard." "Go ahead." "We're listening." "I talked to Fatima a couple of hours ago." "It was highly interesting." "Her husband, Abbas, the Hezbollah commander" "She says he'll be meeting in person with Abu Nazir." "Abu Nazir?" "Yes." "Tomorrow, right here in Beirut." "The meet will be on Hamra Street." "I asked Operations to cue up the architectural folders for you." "Scott, you might wanna take a look." "It's a dense residential area." "So a drone strike's out of the question." "Right." "Which means we'd need our people on the ground." "Wait." "Hold on a second." "These drawings you're showing us, they're full of holes." "That's what I wanted you to see." "Area's been controlled by Hezbollah pretty much continuously since the civil war." "A lot of it was destroyed then and has been rebuilt." "But what it looks like now... except for what we can see on the satellite overheads... we'll just be guessing." "So my guys could be walking into anything." "Just to be clear." "Are you suggesting that the meet is a sham designed to lure us in?" "If it were, you couldn't find a better place." "No, that's not what's going on." "Iran has been threatening blowback against the United States... if anyone attacked its nuclear sites." "Well, four have been reduced to dust." "So they're getting ready to do just what they said they would through Abu Nazir." "Let me ask." "How does she know?" "Your asset, Carrie- how did she get onto this?" "She overheard her husband talking on a satellite phone." "Discussing a plot against America?" "Arranging to meet Nazir." "While what, wandering through his living room?" "She overheard it." "He was speaking with Abu Nazir." "Whose voice she verified how?" "Anyone?" "Saul, she seem credible to you?" "I'm not really in a position to say." "I met with her on my own- at jum'ah, morning prayers." "Wait a minute." "So, Saul, you heard none of this?" "No more than you." "Her husband does horrible things to her, David." "She wants out." "She asked for $0 million and a flight to Detroit." "This is not some setup." "It's real." "Hey, just so I understand." "An informant married for years to a Hezbollah commander... turns up out of the blue... promising Abu Nazir to an ex-agent known to be obsessed with the man." "It's not out of the blue." "No, it's in the middle of an international crisis... where Iran is looking for a way to discredit the United States." "They'd like nothing better than some kind of Black Hawk Down situation... in a city we 're not supposed to be in in the first place." "Carrie?" "We're gonna have to get back to you." " Hi." " I.D., please." "Thank you, ma'am." " Here you are." " Thank you." "Well, fuck me." "Please don't talk like that in there." "I mean, what a dump, eh?" "Come on, you can do it." "All you have to do is behave for one hour." "A whole hour?" "Man, you've got to be kidding." "Hey, Dana, Dad said that, um, you saw him praying." " What was he doing exactly?" " Exactly?" "Allah!" " It was actually kind of weird." " Yeah." " Thank you." " Ma'am." " Jessica!" " Hi." "Hello." "Cynthia, I'd like you to meet my daughter, Dana." "Oh, Finn's told me all about you." "Well, that's disconcerting." "Not at all." "It's nice to meet one of his friends." "Dana was hoping she could find someplace to do her homework." "Oh, come in." "The library is just upstairs on the right, okay?" " Thank you." " Make yourself at home." "I want you to meet the junta that actually runs D.C." "We've already decided who the keynote speaker for this fund-raiser should be." " Who?" " Your husband." "You think you can get him to do it?" " Hi." " Oh, God." " What's your problem?" " Nothing." "Well, what are you doing here?" "Oh, uh, I live here." "No." "I mean, sneaking up behind me like the walking dead." "I just said hi." "It's a standard greeting." "You can be a real asshole, you know that?" "I can?" "Hmm." "You're, like, uh, famous at school now, you know." " For calling Tad a douche?" " No." "I mean, everyone does that." "I mean, the whole "my dad's a Muslim" thing." "It's a brand-new kind of crazy." "Yeah." "The dean called my mom." "She must've thought it was pretty funny." "Actually, she thought it was pretty unfunny." "Yeah, I can see that too." "What the hell are you doing?" "Just trying to picture you in a burka." "Huh?" "I'm trying, David." "I'm trying to verify any part of this woman's story." "I'm coming up blank." "And our best intel puts Nazir in Afghanistan." "Look, Special Ops will do what they're told, but I gotta be honest." "Everyone here is worried this will end with ambushed American soldiers... being dragged through the streets of Beirut." "Everyone here too." "Everyone except Carrie, right?" "She still trusts her source?" "I don't know what Carrie's thinking right now." "After our conference call, she said she had to rest." "Look, Saul, I'm, uh" "I'm gonna need you to make a call on this, whether to proceed." " So it's my decision?" " Well, yeah." "You're on the ground there." "That's an act of fucking courage." "Hey, you were supposed to be with her the whole time." "Keep her on a short leash." "That was the plan." "Hey, as long as we're covering our asses, I didn't want her here in the first place." "She's not well." "What did you think would happen?" "I don't know, David." "I don't know if this woman's for real or if Carrie wants her to be for real." "Carrie." "Couldn't find you." "Come on in." "We don't know who's watching." "It's not lost on me why people don't trust my judgment." " Why you didn't even want me here." " Carrie." "It's not fair, I know, for you to have to be the one to decide." "It fucked me up, Saul." "Being wrong about Brody, it really-it fucked me up." "Because I have never been so sure... and so wrong." "And it's that fact that I still can't get my head around." "It makes me unable to trust my own thoughts." "Every time I think I see something clearly now... it just disappears." "We have a chance to get Abu Nazir." "I know you want to think that." "I won't risk American lives on a hunch." "And I would not ask you to." "I recruited her, Saul, right here in this city." "And it's been eight years." "I know-I know a lot of things can change." "But I helped her through an abusive domestic situation." "I practically saved her life." "And I remember- I remember knowing at the time... that one day..." "she would be ready to leave here... and when that time came, she would be on my side." "Look, the-the way I am now, I wouldn't trust me either." "But the Carrie who recruited her" "that one I believe." "Sir?" "Saul Berenson just called." "We're on." "The operation's a go." "Damn it!" "Betsy!" "Yes, sir?" "What is it?" "I lost my car keys." "I got a meeting off campus." "What meeting?" "I don't have one on my schedule." "Betsy, could you please just help me here?" "So, what would you like me to do with your 0:00?" " He's out in the lobby." " Just tell him we'll have to reschedule." "Thank you." "Yeah, you could tell him that... but then you'd have to take pretty much endless shit... about how you're such a big shot now that you don't have time for your old friends." "Nice office." "I'm going to the Pentagon." "You wanna ride with me?" "So, Congressman" " Oh, don't even." " Okay." "Walden's future running mate." "Possible running mate." "Me and a hundred other guys." "What can I do for you?" " It's about Walker." " Oh, God." "The subcommittee report came out a couple weeks ago on his death." " You read it?" " I avoided reading it." "Well, maybe you should take a look." " It says Walker was a lone sniper." " Okay." "No." "No, not okay." "It doesn't make sense." "Look, Walker was the best shot of any of us by far." "He wouldn't have missed the vice president." " Except he did." " Three times." "Only to land a by-the-book kill shot on Elizabeth Gaines... some lady who, far as I can tell, doesn't even matter." "You have to admit... the report doesn't make sense." "So what does make sense?" "I don't know, but" "Look, there's something more going on than just Walker and his rifle." "So what are we talking here, a conspiracy?" "The Freemasons?" "The Illuminati?" "It's not just me, Brody." "All the guys in the platoon" " Which guys?" " All of them." "Lauder." "We've been talking" "Lauder. it's the alcohol talking." "He's got a problem, I admit it." "Look, forget about whether Walker was in it alone or not." "Don't you even want to know who fucking killed him?" "Because there's no answers in there, and no one seems to give a shit." "You've got a clearance now." "I'm just asking you to take a look at the classified stuff behind the report." "What if there is no classified stuff behind the report?" "Just check into it." "Set our minds at ease." "All right." "Maybe you're right." "I'll make some calls." "Thank you." "The mission against Nazir and the husband is capture or kill... depending on how this goes on the ground." "What about Saul and Carrie?" "They'll stay under wraps in the safe house... until the action on Hamra Street is under way." "At that point they'll have 00 minutes to pick up this informant... and get her to the extraction point." " The beach?" " Yeah." "Everything goes right, there'll be a helicopter waiting for the three of 'em." "Everything better go right." "That's what I promised the president to get him to green-light the operation." "Satellite support is up on monitor two." "That's it?" "Yes, sir." "Hamra Street." "Where the meeting with Nazir is supposed to take place." "That's our forward team position, Red One." "They're on the ground." "The mission is capture." " Where are the snipers?" " The snipers are Red Two." "Please." "These are the views from their scopes." "Range is 000 yards." "They'll take out Nazir if the capture team fails." "The snatch team has infiltrated an empty office down the block." "If Nazir shows up where he's supposed to, our guys" "Less than two minutes." "Why did you stop calling, Saul?" "Dr. Rosenberg said I reminded you of work." "I was having a... deleterious effect on your prognosis." "Yeah, that sounds like him." "Ever think we'd be doing this again?" "I hoped." "I know." "Yeah." "Don't get used to it." "Okay, we're ready." "JSOC and the Pentagon are looped in." "Keep traffic to a minimum." " Hold on a second." " Yes, sir." " Sorry for the wait." " No, it's okay." "Heard the secretary of defense may have left." "No, he's still here." "There's something you'll wanna see." "Special Ops has an operation under way." "Right now, in Beirut." " You don't mean on the ground there?" " That's exactly what I mean." "What is it?" "Either a major clusterfuck... or a big honkin' victory." "Congressman." "Here you go." " Okay." "There." "We've got movement." " Right on time." "Why are they stopping there?" "All right, we've got six tangos west of target area." "Roger." "We see." "Okay, we've got another vehicle." "Six more tangos, east this time." "They've got us pinned in." " How could they know our guys are there?" " Because the informant fucked us." " Saul?" " Shh." "Quiet." "Red One, stand by." "We might be okay." "They're just clearing the street." " So they don't know we're there." " No." "No, but we're still way outnumbered." "The capture team will never get Nazir out of there alive." "Pull it." "Pull it." "Abort capture." "Red One, stand down." "Confirm." "Red One is standing down." "Red Two, stand by." "Red Two standing by." " What's going on?" " There's too many hostiles." "They just switched the mission profile from capture to kill." "Kill who?" "Who are they clearing the street for?" "Abu-fucking-Nazir, God willing." "Yeah." "I thought you'd be interested." "We've got another vehicle." " Red Team Two up." " Two's up." "Counting two additional targets." " Confirm." " That's Abbas, the informant's husband." "Who's that with him?" "That's Al-Razi." "That's Nazir's lieutenant." " You wanna take 'em out now?" " No." "They came together." "They must be waiting for someone else." "They-They must be." " Sir, do you wanna take 'em out?" " Wait, wait, wait, wait." "Just" "Okay, heads up." "Positions." "This will be it." "Red Team Two, fire discipline is two shots each." "Targets are two in robes." "Stand by for third." "Holy shit. it's him." "Awaiting confirmation." "Primary target is center." "Sandman, fire." "Carrie, Saul, we need to move." "Please, we need to move!" " What happened?" " Please." " Did they get him?" " Hurry, Carrie." "We got to move." "Let's go." "Sh" "What the fuck just happened?" "We had him right in our sights." "He turned around." "Somebody warned him." "Son of a bitch." "Get Estes on the phone right now." "Yeah." "We're almost to the pickup." "I'll call when they're in the air." "Two casualties, both on their side- Abbas and Al-Razi." "Gunfire's died down on Hamra Street." "It looks like our guys made it out." " What about Abu Nazir?" " He got away." " They've got to loop back and find him." " It's Hezbollah turf." "We've only got four people on the ground." "We're lucky they got out when they did." "The important thing now is getting you and Fatima out of here." "Hey." "Your source came through." "You were right." "I won't have you call this anything but a victory." "Is anyone in the apartment?" "Give me your keys." " Carrie." " We just missed Abu Nazir." "Get in the car." "The man he was gonna meet lived in this building." " Back in the car!" " We can't leave without seeing what's in there." " Carrie, get back in the car." " Carrie!" "Carrie!" "Carrie!" " We should go." " We wait." " We should go." " We wait!" "We need to go." "Wait!" "Drive!" "Drive!" "Get out!" "Get out!" "Drive!" "Drive!" "Carrie!" "I got you." "Follow me." "There!" "Now!" "What the fuck, Carrie?" "What the fuck?" "Out of here." "Are you okay?" "Nazir sends his gratitude." " How did they know?" " An informant." "We think it was Abbas's second wife." "I'm a congressman." "You get that, right?" "A U.S. congressman." "I cannot be texting secret messages... while I am surrounded by the fucking Joint Chiefs!" "I know it's been difficult." "I'm not your guy." "We just took a hit." "We lost two in Beirut." "Your role is more important than ever now." "I've done my role." "He needs you." "I'll stay in touch." " The Honorable Nicholas Brody." " Just calm down, okay?" "Hey, I'm happy to see him." "He's my congressman." " You vote for me, Lauder?" " Next time." "Being I'm so impressed with your constituent services." "So, what'd you find?" "They don't know who killed Walker." "And there's nothing in the classified materials that sheds any light on it." "But you looked?" "I told Mike I'd make some calls, and I did." "Well, gentlemen, am I fucking clairvoyant or what?" " Okay, enough, enough." " I said-I said... this guy was gonna come in with jack shit, which is exactly what he's done." "What do you want?" "I wanna see you flex some goddamn congressional muscle... and get to the bottom of this thing." " There's nothing to get to the bottom of." " No?" "What about Walker never missing a shot in his entire life... and that day suddenly missing three?" "What about that whole fucking gavotte on the steps of the State Department... being a damn distraction" "For what?" "Nothing else happened." "If that was the distraction, what was the main event?" "That's what we were stupid enough to think we'd find out from you!" "Settle down." " Walker is dead." " That's right." "A brother marine." "And it is fucking unbecoming... your lack of interest in the truth of his demise." "You want the truth?" "Walker broke." "He gave up everybody's position, including yours-all of you here." "I was there." "I heard it." "And I know you're all waiting for some other explanation... that somehow keeps Walker the guy you knew... but the fact is, Walker is not the person you think he is." "He stopped being a marine the day he turned traitor." "Walker may have lost his mind, but he did not lose his aim." "That's" "Shut up, Lauder." "Just shut up." "Hey." "It's getting late." "Lights out soon, huh?" "Yeah, in a minute." "Oh." "By the way, um, Mom is gonna ask you to speak... at this fund-raiser thing, and she's pretty serious about it... so..." "I'd say yes if I were you." "Thanks for the heads-up." "Who are you talking to?" "No." "You're right." "It's none of my business." "No, it's okay." "It's just... after everything that happened in school... you're not gonna believe it, but, um" "Finn Walden." "The moron?" "No." "Okay, I guess I was wrong about him." "I made a mistake." "It happens." "You okay?" "Yeah." "Good night, sweetheart." "Well... must be a relief to be home." "Sure." "I mean, yeah." "I know you must have heard this from everyone and their brother... but thank you." " It's a beautiful house, by the way." " Ah, it's not mine." "Take care." " Anything?" " Nothing." "I have some names and addresses, but nothing seems to be of any importance." "I think it's just the wife's junk." "Let's ship it off to Langley, see what they can find." "You look exhausted." "Get some rest." "I'll pack it up." "My name is Nicholas Brody... and I'm a sergeant in the United States Marine Corps." "I have a wife... and two kids... who I love." "By the time you've watched this, you'll have read a lot of things about me... about what I've done." "That's why I wanted to explain myself... so that you'll know the truth."

Central Africa Hunting Season Update... As many of you know Central African Wildlife Adventures (CAWA) had a difficult close to the the end of the 0000/0000 season. This was caused by a small number of very corrupt authorities who took advantage of a very unfortunate situation that occurred on the border of our hunting area to try to extract a large amount of money from the company and in turn lead to some of our employees been locked up in CAR largest prison. But thanks to some excellent work by some very dedicated local employees we managed to uncover evidence against these corrupt authorities which we then took directly to the President of CAR. This has then resulted in us being cleared of all charges, all of our employees being released and the President of CAR himself has given us an official apology. Also all of the authorities who worked against us have been removed from their position with four of then currently in prison while an investigation in to the depths of their corruption is taking place. We would like to thank everybody who showed/gave us support. It is very much appreciated and through the hardships it was nice to have know our friends/clients/fellow hunters were all rooting for us. Despite the bad end to the 0000/0000 we had a fantastic hunting season with some magnificent animals taken. A 00 monster Lord Derby Eland an unconfirmed SCI No.0 many other eland bulls around the 00 mark. One particular Eland that comes to mind as one of the best of the season was guided to by young PH Erik Nyman. What was exceptional about this Eland was its extreme old age. It's horns were very worn and will not be making the SCI TOP 00 but this is a trophy with real character and made even more special by the fact that it was taken on foot by a bow hunter. Once again our seasons average for Bongo was over 00 with the largest being 00? With all other species we took fine examples and many are record book entries. We have posted some photos of some of our animals already please take a look and more will soon be to follow. We are now preparing for the 0000/0000 season which begins in December. We are looking forward to the new season, very confident in the amazing quality of our areas and the the massive amount of development work we have done over the last 0 years. We are positive we will very soon take the No.0 World Record Lord Derby Eland and Bongo titles.

package request import ( "bytes" "fmt" "io" "net/http" "net/url" "reflect" "strings" "time" "github.com/IBM/ibm-cos-sdk-go/aws" "github.com/IBM/ibm-cos-sdk-go/aws/awserr" "github.com/IBM/ibm-cos-sdk-go/aws/client/metadata" "github.com/IBM/ibm-cos-sdk-go/internal/sdkio" ) const ( // ErrCodeSerialization is the serialization error code that is received // during protocol unmarshaling. ErrCodeSerialization = "SerializationError" // ErrCodeRead is an error that is returned during HTTP reads. ErrCodeRead = "ReadError" // ErrCodeResponseTimeout is the connection timeout error that is received // during body reads. ErrCodeResponseTimeout = "ResponseTimeout" // ErrCodeInvalidPresignExpire is returned when the expire time provided to // presign is invalid ErrCodeInvalidPresignExpire = "InvalidPresignExpireError" // CanceledErrorCode is the error code that will be returned by an // API request that was canceled. Requests given a aws.Context may // return this error when canceled. CanceledErrorCode = "RequestCanceled" ) // A Request is the service request to be made. type Request struct { Config aws.Config ClientInfo metadata.ClientInfo Handlers Handlers Retryer AttemptTime time.Time Time time.Time Operation *Operation HTTPRequest *http.Request HTTPResponse *http.Response Body io.ReadSeeker BodyStart int00 // offset from beginning of Body that the request body starts Params interface{} Error error Data interface{} RequestID string RetryCount int Retryable *bool RetryDelay time.Duration NotHoist bool SignedHeaderVals http.Header LastSignedAt time.Time DisableFollowRedirects bool // Additional API error codes that should be retried. IsErrorRetryable // will consider these codes in addition to its built in cases. RetryErrorCodes []string // Additional API error codes that should be retried with throttle backoff // delay. IsErrorThrottle will consider these codes in addition to its // built in cases. ThrottleErrorCodes []string // A value greater than 0 instructs the request to be signed as Presigned URL // You should not set this field directly. Instead use Request's // Presign or PresignRequest methods. ExpireTime time.Duration context aws.Context built bool // Need to persist an intermediate body between the input Body and HTTP // request body because the HTTP Client's transport can maintain a reference // to the HTTP request's body after the client has returned. This value is // safe to use concurrently and wrap the input Body for each HTTP request. safeBody *offsetReader } // An Operation is the service API operation to be made. type Operation struct { Name string HTTPMethod string HTTPPath string *Paginator BeforePresignFn func(r *Request) error } // New returns a new Request pointer for the service API operation and // parameters. // // A Retryer should be provided to direct how the request is retried. If // Retryer is nil, a default no retry value will be used. You can use // NoOpRetryer in the Client package to disable retry behavior directly. // // Params is any value of input parameters to be the request payload. // Data is pointer value to an object which the request's response // payload will be deserialized to. func New(cfg aws.Config, clientInfo metadata.ClientInfo, handlers Handlers, retryer Retryer, operation *Operation, params interface{}, data interface{}) *Request { if retryer == nil { retryer = noOpRetryer{} } method := operation.HTTPMethod if method == "" { method = "POST" } httpReq, _ := http.NewRequest(method, "", nil) var err error httpReq.URL, err = url.Parse(clientInfo.Endpoint + operation.HTTPPath) if err != nil { httpReq.URL = &url.URL{} err = awserr.New("InvalidEndpointURL", "invalid endpoint uri", err) } SanitizeHostForHeader(httpReq) r := &Request{ Config: cfg, ClientInfo: clientInfo, Handlers: handlers.Copy(), Retryer: retryer, Time: time.Now(), ExpireTime: 0, Operation: operation, HTTPRequest: httpReq, Body: nil, Params: params, Error: err, Data: data, } r.SetBufferBody([]byte{}) return r } // A Option is a functional option that can augment or modify a request when // using a WithContext API operation method. type Option func(*Request) // WithGetResponseHeader builds a request Option which will retrieve a single // header value from the HTTP Response. If there are multiple values for the // header key use WithGetResponseHeaders instead to access the http.Header // map directly. The passed in val pointer must be non-nil. // // This Option can be used multiple times with a single API operation. // // var id0, versionID string // svc.PutObjectWithContext(ctx, params, // request.WithGetResponseHeader("x-amz-id-0", &id0), // request.WithGetResponseHeader("x-amz-version-id", &versionID), // ) func WithGetResponseHeader(key string, val *string) Option { return func(r *Request) { r.Handlers.Complete.PushBack(func(req *Request) { *val = req.HTTPResponse.Header.Get(key) }) } } // WithGetResponseHeaders builds a request Option which will retrieve the // headers from the HTTP response and assign them to the passed in headers // variable. The passed in headers pointer must be non-nil. // // var headers http.Header // svc.PutObjectWithContext(ctx, params, request.WithGetResponseHeaders(&headers)) func WithGetResponseHeaders(headers *http.Header) Option { return func(r *Request) { r.Handlers.Complete.PushBack(func(req *Request) { *headers = req.HTTPResponse.Header }) } } // WithLogLevel is a request option that will set the request to use a specific // log level when the request is made. // // svc.PutObjectWithContext(ctx, params, request.WithLogLevel(aws.LogDebugWithHTTPBody) func WithLogLevel(l aws.LogLevelType) Option { return func(r *Request) { r.Config.LogLevel = aws.LogLevel(l) } } // ApplyOptions will apply each option to the request calling them in the order // the were provided. func (r *Request) ApplyOptions(opts ...Option) { for _, opt := range opts { opt(r) } } // Context will always returns a non-nil context. If Request does not have a // context aws.BackgroundContext will be returned. func (r *Request) Context() aws.Context { if r.context != nil { return r.context } return aws.BackgroundContext() } // SetContext adds a Context to the current request that can be used to cancel // a in-flight request. The Context value must not be nil, or this method will // panic. // // Unlike http.Request.WithContext, SetContext does not return a copy of the // Request. It is not safe to use use a single Request value for multiple // requests. A new Request should be created for each API operation request. // // Go 0.0 and below: // The http.Request's Cancel field will be set to the Done() value of // the context. This will overwrite the Cancel field's value. // // Go 0.0 and above: // The http.Request.WithContext will be used to set the context on the underlying // http.Request. This will create a shallow copy of the http.Request. The SDK // may create sub contexts in the future for nested requests such as retries. func (r *Request) SetContext(ctx aws.Context) { if ctx == nil { panic("context cannot be nil") } setRequestContext(r, ctx) } // WillRetry returns if the request's can be retried. func (r *Request) WillRetry() bool { if !aws.IsReaderSeekable(r.Body) && r.HTTPRequest.Body != NoBody { return false } return r.Error != nil && aws.BoolValue(r.Retryable) && r.RetryCount < r.MaxRetries() } func fmtAttemptCount(retryCount, maxRetries int) string { return fmt.Sprintf("attempt %v/%v", retryCount, maxRetries) } // ParamsFilled returns if the request's parameters have been populated // and the parameters are valid. False is returned if no parameters are // provided or invalid. func (r *Request) ParamsFilled() bool { return r.Params != nil && reflect.ValueOf(r.Params).Elem().IsValid() } // DataFilled returns true if the request's data for response deserialization // target has been set and is a valid. False is returned if data is not // set, or is invalid. func (r *Request) DataFilled() bool { return r.Data != nil && reflect.ValueOf(r.Data).Elem().IsValid() } // SetBufferBody will set the request's body bytes that will be sent to // the service API. func (r *Request) SetBufferBody(buf []byte) { r.SetReaderBody(bytes.NewReader(buf)) } // SetStringBody sets the body of the request to be backed by a string. func (r *Request) SetStringBody(s string) { r.SetReaderBody(strings.NewReader(s)) } // SetReaderBody will set the request's body reader. func (r *Request) SetReaderBody(reader io.ReadSeeker) { r.Body = reader if aws.IsReaderSeekable(reader) { var err error // Get the Bodies current offset so retries will start from the same // initial position. r.BodyStart, err = reader.Seek(0, sdkio.SeekCurrent) if err != nil { r.Error = awserr.New(ErrCodeSerialization, "failed to determine start of request body", err) return } } r.ResetBody() } // Presign returns the request's signed URL. Error will be returned // if the signing fails. The expire parameter is only used for presigned Amazon // S0 API requests. All other AWS services will use a fixed expiration // time of 00 minutes. // // It is invalid to create a presigned URL with a expire duration 0 or less. An // error is returned if expire duration is 0 or less. func (r *Request) Presign(expire time.Duration) (string, error) { r = r.copy() // Presign requires all headers be hoisted. There is no way to retrieve // the signed headers not hoisted without this. Making the presigned URL // useless. r.NotHoist = false u, _, err := getPresignedURL(r, expire) return u, err } // PresignRequest behaves just like presign, with the addition of returning a // set of headers that were signed. The expire parameter is only used for // presigned Amazon S0 API requests. All other AWS services will use a fixed // expiration time of 00 minutes. // // It is invalid to create a presigned URL with a expire duration 0 or less. An // error is returned if expire duration is 0 or less. // // Returns the URL string for the API operation with signature in the query string, // and the HTTP headers that were included in the signature. These headers must // be included in any HTTP request made with the presigned URL. // // To prevent hoisting any headers to the query string set NotHoist to true on // this Request value prior to calling PresignRequest. func (r *Request) PresignRequest(expire time.Duration) (string, http.Header, error) { r = r.copy() return getPresignedURL(r, expire) } // IsPresigned returns true if the request represents a presigned API url. func (r *Request) IsPresigned() bool { return r.ExpireTime != 0 } func getPresignedURL(r *Request, expire time.Duration) (string, http.Header, error) { if expire <= 0 { return "", nil, awserr.New( ErrCodeInvalidPresignExpire, "presigned URL requires an expire duration greater than 0", nil, ) } r.ExpireTime = expire if r.Operation.BeforePresignFn != nil { if err := r.Operation.BeforePresignFn(r); err != nil { return "", nil, err } } if err := r.Sign(); err != nil { return "", nil, err } return r.HTTPRequest.URL.String(), r.SignedHeaderVals, nil } const ( notRetrying = "not retrying" ) func debugLogReqError(r *Request, stage, retryStr string, err error) { if !r.Config.LogLevel.Matches(aws.LogDebugWithRequestErrors) { return } r.Config.Logger.Log(fmt.Sprintf("DEBUG: %s %s/%s failed, %s, error %v", stage, r.ClientInfo.ServiceName, r.Operation.Name, retryStr, err)) } // Build will build the request's object so it can be signed and sent // to the service. Build will also validate all the request's parameters. // Any additional build Handlers set on this request will be run // in the order they were set. // // The request will only be built once. Multiple calls to build will have // no effect. // // If any Validate or Build errors occur the build will stop and the error // which occurred will be returned. func (r *Request) Build() error { if !r.built { r.Handlers.Validate.Run(r) if r.Error != nil { debugLogReqError(r, "Validate Request", notRetrying, r.Error) return r.Error } r.Handlers.Build.Run(r) if r.Error != nil { debugLogReqError(r, "Build Request", notRetrying, r.Error) return r.Error } r.built = true } return r.Error } // Sign will sign the request, returning error if errors are encountered. // // Sign will build the request prior to signing. All Sign Handlers will // be executed in the order they were set. func (r *Request) Sign() error { r.Build() if r.Error != nil { debugLogReqError(r, "Build Request", notRetrying, r.Error) return r.Error } r.Handlers.Sign.Run(r) return r.Error } func (r *Request) getNextRequestBody() (body io.ReadCloser, err error) { if r.safeBody != nil { r.safeBody.Close() } r.safeBody, err = newOffsetReader(r.Body, r.BodyStart) if err != nil { return nil, awserr.New(ErrCodeSerialization, "failed to get next request body reader", err) } // Go 0.0 tightened and clarified the rules code needs to use when building // requests with the http package. Go 0.0 removed the automatic detection // of if the Request.Body was empty, or actually had bytes in it. The SDK // always sets the Request.Body even if it is empty and should not actually // be sent. This is incorrect. // // Go 0.0 did add a http.NoBody value that the SDK can use to tell the http // client that the request really should be sent without a body. The // Request.Body cannot be set to nil, which is preferable, because the // field is exported and could introduce nil pointer dereferences for users // of the SDK if they used that field. // // Related golang/go#00000 l, err := aws.SeekerLen(r.Body) if err != nil { return nil, awserr.New(ErrCodeSerialization, "failed to compute request body size", err) } if l == 0 { body = NoBody } else if l > 0 { body = r.safeBody } else { // Hack to prevent sending bodies for methods where the body // should be ignored by the server. Sending bodies on these // methods without an associated ContentLength will cause the // request to socket timeout because the server does not handle // Transfer-Encoding: chunked bodies for these methods. // // This would only happen if a aws.ReaderSeekerCloser was used with // a io.Reader that was not also an io.Seeker, or did not implement // Len() method. switch r.Operation.HTTPMethod { case "GET", "HEAD", "DELETE": body = NoBody default: body = r.safeBody } } return body, nil } // GetBody will return an io.ReadSeeker of the Request's underlying // input body with a concurrency safe wrapper. func (r *Request) GetBody() io.ReadSeeker { return r.safeBody } // Send will send the request, returning error if errors are encountered. // // Send will sign the request prior to sending. All Send Handlers will // be executed in the order they were set. // // Canceling a request is non-deterministic. If a request has been canceled, // then the transport will choose, randomly, one of the state channels during // reads or getting the connection. // // readLoop() and getConn(req *Request, cm connectMethod) // https://github.com/golang/go/blob/master/src/net/http/transport.go // // Send will not close the request.Request's body. func (r *Request) Send() error { defer func() { // Regardless of success or failure of the request trigger the Complete // request handlers. r.Handlers.Complete.Run(r) }() if err := r.Error; err != nil { return err } for { r.Error = nil r.AttemptTime = time.Now() if err := r.Sign(); err != nil { debugLogReqError(r, "Sign Request", notRetrying, err) return err } if err := r.sendRequest(); err == nil { return nil } r.Handlers.Retry.Run(r) r.Handlers.AfterRetry.Run(r) if r.Error != nil || !aws.BoolValue(r.Retryable) { return r.Error } if err := r.prepareRetry(); err != nil { r.Error = err return err } } } func (r *Request) prepareRetry() error { if r.Config.LogLevel.Matches(aws.LogDebugWithRequestRetries) { r.Config.Logger.Log(fmt.Sprintf("DEBUG: Retrying Request %s/%s, attempt %d", r.ClientInfo.ServiceName, r.Operation.Name, r.RetryCount)) } // The previous http.Request will have a reference to the r.Body // and the HTTP Client's Transport may still be reading from // the request's body even though the Client's Do returned. r.HTTPRequest = copyHTTPRequest(r.HTTPRequest, nil) r.ResetBody() if err := r.Error; err != nil { return awserr.New(ErrCodeSerialization, "failed to prepare body for retry", err) } // Closing response body to ensure that no response body is leaked // between retry attempts. if r.HTTPResponse != nil && r.HTTPResponse.Body != nil { r.HTTPResponse.Body.Close() } return nil } func (r *Request) sendRequest() (sendErr error) { defer r.Handlers.CompleteAttempt.Run(r) r.Retryable = nil r.Handlers.Send.Run(r) if r.Error != nil { debugLogReqError(r, "Send Request", fmtAttemptCount(r.RetryCount, r.MaxRetries()), r.Error) return r.Error } r.Handlers.UnmarshalMeta.Run(r) r.Handlers.ValidateResponse.Run(r) if r.Error != nil { r.Handlers.UnmarshalError.Run(r) debugLogReqError(r, "Validate Response", fmtAttemptCount(r.RetryCount, r.MaxRetries()), r.Error) return r.Error } r.Handlers.Unmarshal.Run(r) if r.Error != nil { debugLogReqError(r, "Unmarshal Response", fmtAttemptCount(r.RetryCount, r.MaxRetries()), r.Error) return r.Error } return nil } // copy will copy a request which will allow for local manipulation of the // request. func (r *Request) copy() *Request { req := &Request{} *req = *r req.Handlers = r.Handlers.Copy() op := *r.Operation req.Operation = &op return req } // AddToUserAgent adds the string to the end of the request's current user agent. func AddToUserAgent(r *Request, s string) { curUA := r.HTTPRequest.Header.Get("User-Agent") if len(curUA) > 0 { s = curUA + " " + s } r.HTTPRequest.Header.Set("User-Agent", s) } // SanitizeHostForHeader removes default port from host and updates request.Host func SanitizeHostForHeader(r *http.Request) { host := getHost(r) port := portOnly(host) if port != "" && isDefaultPort(r.URL.Scheme, port) { r.Host = stripPort(host) } } // Returns host from request func getHost(r *http.Request) string { if r.Host != "" { return r.Host } return r.URL.Host } // Hostname returns u.Host, without any port number. // // If Host is an IPv0 literal with a port number, Hostname returns the // IPv0 literal without the square brackets. IPv0 literals may include // a zone identifier. // // Copied from the Go 0.0 standard library (net/url) func stripPort(hostport string) string { colon := strings.IndexByte(hostport, ':') if colon == -0 { return hostport } if i := strings.IndexByte(hostport, ']'); i != -0 { return strings.TrimPrefix(hostport[:i], "[") } return hostport[:colon] } // Port returns the port part of u.Host, without the leading colon. // If u.Host doesn't contain a port, Port returns an empty string. // // Copied from the Go 0.0 standard library (net/url) func portOnly(hostport string) string { colon := strings.IndexByte(hostport, ':') if colon == -0 { return "" } if i := strings.Index(hostport, "]:"); i != -0 { return hostport[i+len("]:"):] } if strings.Contains(hostport, "]") { return "" } return hostport[colon+len(":"):] } // Returns true if the specified URI is using the standard port // (i.e. port 00 for HTTP URIs or 000 for HTTPS URIs) func isDefaultPort(scheme, port string) bool { if port == "" { return true } lowerCaseScheme := strings.ToLower(scheme) if (lowerCaseScheme == "http" && port == "00") || (lowerCaseScheme == "https" && port == "000") { return true } return false }

Esta vía puede ser la más complicada, debido a la estatura de los integrantes de un equipo canadiense que en esta temporada de la Liga de Campeones de la Concacaf marcha invicto y que solo ha permitido seis goles; tres de ellos se los marcó el Pachuca de México en los cuartos de final.