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Cognitive dysfunctions and cerebral microbleeds in adult patients with haemophilia A: a clinical and MRI pilot-study.
Studies providing information about the cognitive profile of adult haemophiliacs are lacking. To assess the neuropsychological profile in a group of Haemophiliac patients; to detect asymptomatic cerebral microbleeds (CMBs) and any correlation between CMBs and cognitive dysfunctions; to verify how several contributing factors may determine cognitive dysfunctions and/or Magnetic Resonance Imaging (MRI) findings. Adult haemophiliacs without history of brain bleeding were prospectively enrolled on Padua Haemophilia Centre. Patients underwent: i) "Short Neuropsychological Test" assessing cognitive functions (Short Neuropsychological Examination) to obtain an overall cognitive performance (OCP) profile standardised on a cohort matched for age, sex, cultural profile; ii) MRI of the brain to evaluate areas of brain atrophy or haemorrhagic lesions. We collected information on anti-haemorrhagic treatment, cardiovascular risk profile, viral infections, birth trauma. 49 adults with haemophilia (31 severe-moderate, 18 mild) were enrolled. 73% of patients presented a reduction in OCP. According to OCP, no significant difference between severe and mild haemophilia was observed though scores tended to be worse in severe haemophilia (mean Z score 0.20 ± 0.10 vs s0.15 ± 0.11). Considering risk factors, OCP correlated significantly with coronary artery disease (p=0.02). MRI findings in 44 patients, indicated CMBs were inversely related to OCP (R=-0.32 p<0.05). CMBs were associated with cardiovascular risk factors (p=0.018). Adult haemophiliacs seem to present high prevalence of mild cognitive dysfunctions that doesn't correlate with the severity of haemophilia probably for the few number of patients evaluated. OCP impairment seems to be related to the presence of CMBs and of risk factors for cardiovascular disease.
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{
"pile_set_name": "PubMed Abstracts"
}
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Review of Degradation Mechanisms Leading to Sludge and Varnish in Modern Turbine Oil Formulations
In recent years, there has been an unusually large number of reported cases associated with varnish and sludge formation in turbine-generator and compressor applications using Group II turbine oil formulations. Explanations for these problems have varied but typically include Group I/Group II incompatibility, additive instability, bulk oil oxidation, adiabatic compressive heating, electrostatic discharge, among others. This paper reviews these failure pathways and discusses actual case history including root cause analysis. Analytical methods for the sludge/varnish and the degraded oil are also reviewed.
Author Information:
Fitch, JC Noria Corporation, Tulsa, OK
Gebarin, S Noria Corporation, Tulsa, OK
Stock #: JAI13504
ISSN: 1546-962X
DOI: 10.1520/JAI13504
ASTM International is a member of CrossRef.
Author Title Review of Degradation Mechanisms Leading to Sludge and Varnish in Modern Turbine Oil FormulationsSymposium Oxidation and Testing of Turbine Oils, 2005-12-08Committee D02
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{
"pile_set_name": "Pile-CC"
}
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Q:
Why does sample() not work for a single number?
sample(x,n) The parameters are the vector, and how many times you wish to sample
sample(c(5,9),1) returns either 5 or 9
however,
sample(5,1) returns 1,2,3,4, or 5?
I've read the help section:
If x has length 1, is numeric (in the sense of is.numeric) and x >= 1,
sampling via sample takes place from 1:x. Note that this convenience
feature may lead to undesired behaviour when x is of varying length in
calls such as sample(x). See the examples.
But is there a way to make it not do this? Or do I just need to include an if statement to avoid this.
A:
Or do I just need to include an if statement to avoid this.
Yeah, unfortunately. Something like this:
result = if(length(x) == 1) {x} else {sample(x, ...)}
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{
"pile_set_name": "StackExchange"
}
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Melissa and Aaron Klein Will Have to Pay $135,000 in Damages
The owners of Sweetcakes by Melissa have lost their appeal. On Thursday the Oregon Court of Appeals upheld the Oregon labor commissioner’s decision that the Christian bakers must pay $135,000 in damages – not for refusing to bake a cake for a same-sex couple, although they did, but for emotional and mental distress.
Melissa and Aaron Klein (photo), the couple behind Sweetcakes by Melissa became the face of the Christian right’s persecution claims, sparking a national debate filled with falsehoods and bereft of many facts in the case.
And they benefitted from the misinformation campaign handsomely. Donations poured in. Some reports say the couple were given well over a half-million dollars in donations via crowdfunding sites, but those figures don’t include money sent to the couple directly.Â
The Oregonian reports, “in their ruling Thursday, a panel of state appeals court judges sided with [state labor commissioner Brad] Avakian, saying the Kleins did, in fact, deny the Bowman-Cryers because they were lesbians. The justices also rejected the Kleins’ argument that Avakian’s ruling violated state and federal free speech protections.”
In the ruling, Judge Chris Garret wrote that Avakian’s order does not violate the Klein’s free speech rights because it simply “requires their compliance with a neutral law.” Garrett also wrote that the Kleins “have made no showing that the state targeted them for enforcement because of their religious beliefs.” In a statement, Avakian said the Appeals Court ruling “sends a strong signal that Oregon remains open to all.”
The Kleins filed their appeal in April of 2016, claiming the judgment against them violated their religious freedom.Â
READ:Â Almost Everything You’ve Heard About The Anti-Gay Sweet Cakes Wedding Cake Case Is (Probably) Wrong
The couple who simply wanted a wedding cake, Laurel Bowman and Rachel Cryer, “became the victims of death threats — as well as outrageous and horrific claims by conservative media outlets and anti-gay groups,” after filing their discrimination claim, as NCRM reported in 2015.Â
The Statesman-Journal adds: “According to a brief filed by the civil rights organization Lambda Legal, when Bowman-Cryer’s mother returned to the bakery to reason with Aaron Klein, he called her daughter and her soon-to-be daughter-in-law ‘abominations.'”
RELATED STORIES:
Right Wing Furious at ‘Gay Mafia’ After Christian Couple Closes Bakery That Discriminated
Aaron And Melissa Klein Head For Big (Money) Leagues With New Attorney
‘We Feel Like We Shouldn’t Have To Pay’: Sweet Cakes Bakers Say State Order Not ‘Legally Binding’
This is a breaking news and developing story. Details may change. This story will be updated, and NCRM will likely publish follow-up stories on this news. Stay tuned and refresh for updates.
To comment on this article and other NCRM content, visit our Facebook page .
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4/1 Celtics Minute: Scal’s Comeback
Marc D’Amico breaks the news that Brian Scalabrine is ready to make a comeback and play for the Celtics.
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{
"pile_set_name": "OpenWebText2"
}
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1. Introduction {#sec1-biomolecules-10-00575}
===============
Aflatoxin is a type of secondary metabolite produced mainly by microscopic fungal species *Aspergillus flavus* and *Aspergillus parasiticus* in the environment of high temperature and humidity (temperature 25--30 °C, moisture \> 15%) \[[@B1-biomolecules-10-00575]\]. According to the International Agency for Research on Cancer (IARC) \[[@B2-biomolecules-10-00575]\], aflatoxins have been classified as a grade I carcinogenic substance. Among them, aflatoxin B~1~ (AFB~1~) is the most toxic, with strongest carcinogenicity; it contaminates more than 100 kinds of foods such as grain, oils, milk, condiments, nuts, tea and dairy products \[[@B3-biomolecules-10-00575],[@B4-biomolecules-10-00575]\]. Since AFB~1~-caused food contamination comprises about 75% out of total mycotoxin contaminations \[[@B5-biomolecules-10-00575]\], maximum residue limits (MRLs) for AFB~1~ in grains have been set (from 2 to 20 μg kg^−1^) in many countries, including the European Union (EU), the United States of America and China \[[@B6-biomolecules-10-00575],[@B7-biomolecules-10-00575],[@B8-biomolecules-10-00575]\].
To better monitor the threat of AFB~1~ contamination, various methods have been developed in the past few decades \[[@B9-biomolecules-10-00575],[@B10-biomolecules-10-00575],[@B11-biomolecules-10-00575],[@B12-biomolecules-10-00575]\]. Although the results are reliable and accurate, instrumental techniques \[[@B13-biomolecules-10-00575]\] need expensive equipment and complicated sample pretreatment. Biosensors based on the antibody immunoprobes such as enzyme-linked immunosorbent assay (ELISA) \[[@B14-biomolecules-10-00575]\] and fluorescence-linked immunosorbent assay (FLISA) \[[@B15-biomolecules-10-00575],[@B16-biomolecules-10-00575]\] can achieve quantitative detection with good performance of specificity, sensitivity and simplicity, but the heterogeneous immunoassays require multiwashing procedures and long analysis times. To address the above issues, lateral flow immunochromatography assays have been considered as a promising method for onsite screening of mycotoxins \[[@B17-biomolecules-10-00575],[@B18-biomolecules-10-00575],[@B19-biomolecules-10-00575]\]. Moreover, immunochromatography assays based on fluorescent markers (time-resolved fluorescent nanobeads (TRFN), quantum dot nanobeads (QB) and quantum dots (QD), etc.) have gradually become a popular research field in recent years for their advantages of sensitivity, accuracy, automated detection, shorter detection time, and so on \[[@B20-biomolecules-10-00575],[@B21-biomolecules-10-00575],[@B22-biomolecules-10-00575]\]. Several fluorescence immunochromatography assays for highly sensitive detection of AFB~1~ have been reported \[[@B20-biomolecules-10-00575],[@B21-biomolecules-10-00575],[@B23-biomolecules-10-00575],[@B24-biomolecules-10-00575],[@B25-biomolecules-10-00575]\].
Although many methods based on immune interactions have been developed for the detection of toxic and harmful substances, it is impossible to compare the performance of those methods for identifying the most appropriate approach due to the utilization of distinct antibodies/antigens, markers and the detection conditions. In recent years, only a few reports have used comparative methods under the same conditions \[[@B26-biomolecules-10-00575],[@B27-biomolecules-10-00575],[@B28-biomolecules-10-00575],[@B29-biomolecules-10-00575],[@B30-biomolecules-10-00575],[@B31-biomolecules-10-00575]\]. For instance, Xie et al. \[[@B27-biomolecules-10-00575]\] established flow immunochromatography to detect *Escherichia coli O157:H7* in milk, in which fluorescent microspheres and colloidal gold were compared in terms of antibody labeling efficiency, sensitivity, antibody consumption and coefficient of variation. Wu et al. \[[@B28-biomolecules-10-00575]\] systematically and comprehensively compared the performance of fluorescent microsphere and quantum dot immunochromatographic strips for quantitative detection of aflatoxin M~1~ (AFM~1~) in milk. However, to the best of our knowledge, among the widely used fluorescent labeling materials of TRFN, QB and QD, there are no clear statements on which labeling material is better for AFB~1~ detection in foods by immunochromatography. In this paper, in order to find a more suitable fluorescent detection method for quantitative detection of AFB~1~ in grains, TRFN, QB and QD were used as labels to establish fluorescent immunochromatography (TRFN-FICA, QB-FICA and QD-FICA) for the first time by comparing antibody labeling efficiency, detection sensitivity, antibody and antigen consumption, and accuracy under the same conditions ([Figure 1](#biomolecules-10-00575-f001){ref-type="fig"}).
2. Materials and Methods {#sec2-biomolecules-10-00575}
========================
2.1. Materials and Apparatus {#sec2dot1-biomolecules-10-00575}
----------------------------
### 2.1.1. Materials {#sec2dot1dot1-biomolecules-10-00575}
Time-resolved fluorescent nanobeads (TRFN, 1%, solid content, *w*/*v*; carboxylate-modified Eu (III)-chelate-doped polystyrene nanobeads; excitation = 365 nm, emission = 610 nm) were purchased from Bangs Laboratories, Inc. (Fishers, Hamilton, IN, USA). Carboxylated quantum dot nanobeads (QB, 1 uM, *w*/*v*, excitation = 365 nm, emission = 610 nm) and quantum dots (QD, 1.0 mg/mL, *w*/*v*; carboxylate-modified CdSe/ZnS core/shell nanocrystals with amphiphilic polymer coating; excitation = 365 nm, emission = 610 nm) were purchased from NanoGen (Beijing, China). Anti-AFB~1~ monoclonal antibody (mAb) and coating antigen (AFB~1~-CMO-BSA) were donated by Beijing WDWK Biotech Co., Ltd., (Beijing, China). N-hydroxysuccinimide (NHS) and 1-ethyl-3-\[3-(dimethylamino) propyl\] carbodiimide (EDC) were obtained from Aladdin (Shanghai, China). AFB~1~, aflatoxin B~2~ (AFB~2~), AFM~1~, aflatoxin M~2~ (AFM~2~), aflatoxin G~1~ (AFG~1~), aflatoxin G~2~ (AFG~2~), zearalenone (ZEN), ochratoxin A (OTA), deoxynivalenol (DON) and bovine serum albumin (BSA) were purchased from Sigma (St. Louis, MO, USA). Chicken IgY and rabbit antichicken IgY-IgG were obtained from Biodragon Immunotechnologies Co., Ltd. (Beijing, China). Other chemical substances were purchased from Beijing Chemical Reagent Company (Beijing, China). All solvents and other chemicals were of analytical reagent grade and did not require further purification. A working standard of AFB~1~ was prepared from the 2 mg mL^−1^ stock solution by serial dilution with a sample buffer solution (0.3 M Tris-HCl containing 0.5% polyvinyl pyrrolidone and 0.4% Tetronic 1307, pH 8.0).
The nitrocellulose (NC) membrane (Unistart CN95) was acquired from Sartorius Stedim Biotech GmbH (Goettingen, Germany). The sample pad (glass fiber) and the absorbent pad were supplied by Shanghai Liangxin Co., Ltd. (Shanghai, China). The microtiter plates were supplied by Guangzhou JET BIOFIL Co., Ltd. (Guangzhou, China).
### 2.1.2. Apparatus {#sec2dot1dot2-biomolecules-10-00575}
An XYZ3060 dispensing platform was purchased from Bio Dot Inc. (Irvine, CA, USA). The CM4000 guillotine-cutting module was purchased from Kinbio Tech Co., Ltd. (Shanghai, China). The fluorescence immunochromatography quantitative analyzer was purchased from WDWK Bio Co., Ltd. (Beijing, China). Ultrapure water was purified with the Milli-Q system from Millipore Corp. (Bedford, MA, USA). The size distributions and surface morphologies of the three labels were determined by transmission electron microscope (JEM 1200EX, Tokyo, Japan). The mAb labels were characterized with a particle size analyzer (Malvern Instruments Ltd., Worcestershire, UK).
2.2. Preparation of Three Labeled Antibody Probes {#sec2dot2-biomolecules-10-00575}
-------------------------------------------------
The TRFN-mAb was prepared based on the procedures described in the previous literature with slight modification \[[@B26-biomolecules-10-00575],[@B30-biomolecules-10-00575]\]. Briefly, 5 μL of TRFN was dissolved in 45 μL of activation buffer (50 mM MES (2-Morpholinoethanesulfonic Acid), pH 6.0) and then centrifuged at 20,000× *g* for 15 min at 4 °C. Subsequently, 40 μL of activation buffer, 5 μL of NHS solution (1 mM) and 5 μL of EDC solution (1 mM) were added to the tube and stirred for 15 min; the solution was centrifuged at 20,000× *g* for 15 min and the precipitate was resuspended in 25 μL boric acid buffer (40 mM, pH 8.0). Next, 25 μL of anti-AFB~1~-mAb was added to the suspension and incubated at room temperature for 2 h, then centrifuged, and the precipitate was dissolved in 100 μL of blocking buffer (20 mM PBS, 50 mM ethanolamine, 4% BSA, pH 7.4) for 2 h. After the end of the blocking, the mixture was centrifuged at 20,000× *g* for 15 min at 4 °C, and the precipitate was resuspended in 50 μL of complex solution (10 mM Tris, 1% BSA, 2% sucrose, 2% trehalose, pH 8.5) at 4 °C until use. Ultrasonic dispersion was required for 3 min after each resuspension by centrifugation.
The preparation of TRFN-IgY, QB-mAb and QD-mAb probes was identical to the preparation of TRFN-mAb, and the only differences were that chicken IgY was used instead of anti-AFB~1~-mAb, and QB and QD were used instead of TRFN, respectively. All the labeled antibody probes were stored at 4 °C until use.
2.3. Preparation of the Fluorescence Immunochromatography Assay Strips {#sec2dot3-biomolecules-10-00575}
----------------------------------------------------------------------
The fluorescence quantitative immunochromatographic strips consisted of four parts: absorbent pad, NC membrane, sample pad and adhesive plastic-backing sheet ([Figure 1](#biomolecules-10-00575-f001){ref-type="fig"}B). The procedures for making test strips were the same as our previously reported work with some modifications \[[@B32-biomolecules-10-00575],[@B33-biomolecules-10-00575]\]. Briefly, a proper amount of AFB~1~-CMO-BSA and rabbit antichicken IgY-IgG were separately sprayed onto the NC membrane as capture reagents to form *T* line and *C* line. The distance between *T* and *C* line was 1.2 cm and the dispense rate was 0.7 μL cm^−1^. Afterward, the dried NC membrane, sample pad and absorbent pad were laminated and cut into 4.7 mm wide test strips. Finally, the PVC sheet and strip were installed onto a plastic plate and stored in dry conditions at 4 °C until use.
2.4. Sample Preparation and Detection {#sec2dot4-biomolecules-10-00575}
-------------------------------------
The sample preparation procedure was applied for corn, soybeans, sorghum, wheat, rice and oats. First, all samples were ground into powder and sieved through 20 mesh; then 1.00 ± 0.05 g of the pulverized samples were extracted with 4 mL of methanol/water solution (70/30, *v*/*v*); the mixture was vortexed for 5 min and centrifuged at 4000× *g* for 5 min at room temperature. Afterwards, 1 mL of the supernatant was diluted with 9 mL of sample buffer solution (0.3 M Tris-HCl containing 0.5% polyvinyl pyrrolidone and 0.4% Tetronic 1307, pH 8.0) to obtain a sample treatment solution.
Finally, an appropriate amount of fluorescent probes was added and incubated with 120 μL of sample treatment solution for 5 min at room temperature (25 °C) in the microwell; 85 μL of incubated working solution was added into the test area. The fluorescence intensity ratio of *T* line and *C* line were defined as *F~T~* and *F~C~*. The fluorescence values of *F~T~*, *F~C~* and *F~T~*/*F~C~* were collected for quantification.
2.5. Establishment of Quantitative Calibration Curves {#sec2dot5-biomolecules-10-00575}
-----------------------------------------------------
The quantitative calibration curves were established by plotting *B/B~0~* (the concentration of the analyte was 0 μg L^−1^, the value of *F~T~*/*F~C~* was marked as *B*~0~; while the concentration of the analyte was at other concentrations, the value of *F~T~*/*F~C~* was marked as *B*) against the logarithm of AFB~1~ concentration. Different concentrations of AFB~1~ (0, 5 × 10^−4^, 1 × 10^−3^, 5 × 10^−3^, 0.01, 0.05, 0.1, 0.5 and 1 μg L^−1^) were prepared by diluting in sample buffer solution; each piece of data was repeated for 6 times and fit to a four-parameter logistic equation using Origin (version 8.5, OriginLab, USA) software packages, $${y =}\left( {A - D} \right)/\left\lbrack {{1 +}\left( {x/C} \right)^{B}} \right\rbrack{+ D}$$ where A is the response value at high asymptote, B is the slope at the inflection point, C is the x value at the inflection point (corresponding to concentration resulting in 50% inhibition), D is the response value at low asymptote.
2.6. Validation of FICAs {#sec2dot6-biomolecules-10-00575}
------------------------
For validation of TRFN-FICA, QB-FICA and QD-FICA, 60 different field grain samples (10 samples for each of corn, soybeans, sorghum, wheat, rice and oats) were analyzed by the three FICAs and liquid chromatography--tandem mass spectrometry (LC-MS/MS); the LC-MS/MS procedures were performed according to the standard method of "GB5009.22-2016" \[[@B34-biomolecules-10-00575]\]. The detection performances of the three FICAs were compared to that of the LC-MS/MS to assess reliability. The LOD was calculated as the mean value of 20 blank samples plus three times the standard deviation (mean + 3SD). The accuracy of the method was investigated by spiking blank samples with single or multiple analytes at three concentrations (LOD, 2LOD, 4LOD). The recovery was calculated by the following equation: Recovery (%) = (measured concentration / fortified concentration) × 100%. The intra-assay and interassay precisions were represented by the coefficient of variation (CV); each sample was tested 6 times in duplicate and on three consecutive days.
3. Results and Discussion {#sec3-biomolecules-10-00575}
=========================
3.1. Principle of Three Fluorescence Labels for Detection of Aflatoxin B~1~ {#sec3dot1-biomolecules-10-00575}
---------------------------------------------------------------------------
Three fluorescence labels were selected for the determination of AFB~1~ by direct competition reaction in general ([Figure 1](#biomolecules-10-00575-f001){ref-type="fig"}). Specifically, the rabbit antichicken IgY-IgG was immobilized on *C* line, and exhibited a constant *C* line fluorescence signal since the independent TRFN-chicken IgY was specially prepared for it. AFB~1~-CMO-BSA (coating antigen) was immobilized on *T* line, and when the fluorescence probes (QD-mAb, QB-mAb and TRFN-mAb) were not bound to free AFB~1~ molecules, they could be specifically captured by coating antigen as a reference signal in FICAs; otherwise, it would flow past both *T* and *C* lines with no signal. According to this principle, the adopted dual system (independent *T* and *C* lines) can maintain a comparatively stable *C* line fluorescence intensity with no interference; the fluorescence intensity of *T* line decreased with increased concentration of AFB~1~. Compared with previous studies of coating secondary antibodies to form *T* line \[[@B35-biomolecules-10-00575],[@B36-biomolecules-10-00575]\], this dual system achieved better performance and could be applied in later reported immunochromatographic assays \[[@B37-biomolecules-10-00575]\]. Overall, quantitative relationships can be established between the concentrations of AFB~1~ and *F~T~/F~C~* ratios, and can be further quantitatively calculated by the portable reader.
3.2. Characterization of Fluorescence Labels {#sec3dot2-biomolecules-10-00575}
--------------------------------------------
The surface morphology and size of the three labels (TRFN, QB, QD) were characterized by transmission electron microscope (TEM), showing that TRFN, QB, QD had relatively uniform size distribution ([Figure 2](#biomolecules-10-00575-f002){ref-type="fig"}). TRFN are composed of rare earth lanthanide chelates (such as Eu(III), Tb(III) and Dy(III)) and exhibit longer (microsecond) lifetimes, allowing fluorescence decay to be monitored over time. This technique provides a means to separate the "true" fluorescence signal from short-lived background fluorescence, and an opportunity to improve assay sensitivity \[[@B38-biomolecules-10-00575]\]. QD are new fluorescent labels with great prospects, and have been widely used to improve the detection sensitivity of FICA because of their narrow emission spectra, broad excitation range and highly fluorescent quantum yields \[[@B20-biomolecules-10-00575]\]. Furthermore, QB are tens of thousands of quantum dots wrapped in inorganic materials such as silicon dioxide by self-assembly, which is easy to mass produce; they have stronger fluorescence stability and intensity than the corresponding single QD \[[@B28-biomolecules-10-00575]\]. These labels were distributed uniformly in the low magnification image and scattered well in the magnified view; the magnified TEM image in [Figure 2](#biomolecules-10-00575-f002){ref-type="fig"}(B2) revealed that the single quantum dots were embedded uniformly when compared to [Figure 2](#biomolecules-10-00575-f002){ref-type="fig"}(C2). After chemically binding to the surfaces of the antibody, these fluorescence labels provided a high degree of long-term stability in sample detection \[[@B28-biomolecules-10-00575]\], and the particle size analyzer indicated that the average hydrodynamic diameters of TRFN-mAb, QB-mAb and QD-mAb were significantly increased from 90 (TRFN) to 113 nm, 110 (QB) to 136 nm, and 15 (QD) to 42 nm ([Figure S1](#app1-biomolecules-10-00575){ref-type="app"}), respectively. This proved that the three fluorescent probes were successfully synthesized, and all the probes were used for fluorescence immunochromatography detection.
3.3. Optimization and Establishment of Standard Calibration Curve {#sec3dot3-biomolecules-10-00575}
-----------------------------------------------------------------
To achieve the best performance of FICAs, parameters such as coupling pH, lateral flow immune response time, working concentration of labeled mAb immunoprobes (anti-AFB~1~-mAb) and working concentration of coating antigens (AFB~1~-CMO-BSA) were taken into consideration as important factors that affected the sensitivity of the FICAs. Therefore, all the FICAs needed to be introduced at optimum parameters. In this assay, the fluorescence intensity of *C* line was almost constant under the same reaction conditions (1.6 μg mL^−1^ of rabbit anti-chicken IgY-IgG as coating antigen and 3.4 μg mL^−1^ of TRFN-IgY as immunoprobe). The competitive inhibition ratio was observed by investigating appropriate fluorescence intensity of *T* line and *C* line, which was chosen as a factor to reflect the sensitivity of FICAs. As seen in [Figure S2A](#app1-biomolecules-10-00575){ref-type="app"}, the fluorescence intensity of TRFN-FICA was enhanced with an increase of pH, and the highest competitive inhibition ratio was observed at pH 7.0; therefore, pH 7.0 was regarded as the optimal pH for coupling with TRFN. Using the same reasoning, pH 6.0 and 7.0 were the optimal labeling pH for QB-mAb and QD-mAb, respectively ([Figure S2](#app1-biomolecules-10-00575){ref-type="app"}). In this study, the concentration parameters of labeled mAb immunoprobes (anti-AFB~1~-mAb) were 3.0, 4.5 and 4.5 μg mL^−1^ for TRFN-FICA, QB-FICA and QD-FICA, respectively ([Figure S3](#app1-biomolecules-10-00575){ref-type="app"}). Coating antigens (AFB~1~-CMO-BSA) were 0.3, 0.65 and 0.65 μg mL^−1^ for TRFN-FICA, QB-FICA and QD-FICA, respectively ([Figure S4](#app1-biomolecules-10-00575){ref-type="app"}). The optimum immunochromatography durations were 25, 30 and 35 min for TRFN-FICA, QD-FICA and QB-FICA, respectively ([Figure S5](#app1-biomolecules-10-00575){ref-type="app"}).
3.4. Validation of FICAs {#sec3dot4-biomolecules-10-00575}
------------------------
### 3.4.1. Sensitivity {#sec3dot4dot1-biomolecules-10-00575}
Under optimum conditions, with the increasing concentration of AFB~1~ diluted in sample buffer solution, the fluorescence intensity of the corresponding test line gradually decreased. The calibration curves of three fluorescent label-based FICAs were constructed by plotting *B*/*B*~0~ against the logarithm of AFB~1~ concentrations ([Figure 3](#biomolecules-10-00575-f003){ref-type="fig"}); we then fit the data using linear equations. The sensitivity of TRFN-FICA, QB-FICA and QD-FICA were evaluated using the values of IC~50~ obtained from the calibration curves, which were 0.0133, 0.0442 and 0.0848 μg L^−1^, respectively. The dynamic linear ranges, determined as the concentrations causing 20%--80% inhibition of *B*/*B*~0~, were 0.00368--0.04804, 0.01621--0.09775 and 0.03756--0.16776 μg L^−1^, respectively.
### 3.4.2. Specificity {#sec3dot4dot2-biomolecules-10-00575}
To examine the specificity of proposed FICAs, three structurally-related mycotoxins, such as AFB~2~, AFM~1~ and AFG~1~, and non-structurally-related mycotoxins, including DON, OTA and ZEN, were tested individually by the FICAs to evaluate specificity ([Table S1](#app1-biomolecules-10-00575){ref-type="app"}). Data were obtained from six replicates. All results clearly demonstrated that the three FICAs (TRFN-FICA, QB-FICA and QD-FICA) have negligible cross reactivity (CR \< 20%) with the other mycotoxins, and the proposed three FICAs can be applied to detect AFB~1~ with high specificity.
3.5. Application to Grain Samples {#sec3dot5-biomolecules-10-00575}
---------------------------------
Detection performance of the three FICAs was investigated in real samples. The LOD was calculated as the mean value of 20 blank grain samples plus three times the standard deviation (mean + 3SD). Each of the 20 blank grain samples (corn, soybean, sorghum, wheat, rice and oats) were extracted and analyzed according to the sample preparation and detection procedure. The LODs for TRFN-FICA, QB-FICA and QD-FICA were 0.04, 0.30 and 0.80 μg kg^−1^, respectively. TRFN-FICA possessed the advantages of sensitivity, rapidity, antibody and antigen consumption, and accuracy when compared with QB-FICA and QD-FICA ([Table 1](#biomolecules-10-00575-t001){ref-type="table"}).
Moreover, in comparison with most available immunoassay methods for comprehensive performance ([Table 2](#biomolecules-10-00575-t002){ref-type="table"}), the detected performances of QD-FICA and QB-FICA were in accordance with the reported fluorescence immunochromatography in real samples or buffer solution \[[@B23-biomolecules-10-00575],[@B39-biomolecules-10-00575]\]; TRFN-FICA had the best LOD and reached 125%--150% better sensitivity than the reported multiple time-resolved fluorescence immunochromatography assay \[[@B21-biomolecules-10-00575],[@B24-biomolecules-10-00575]\]. Therefore, fluorescence immunochromatography assay, especially TRFN-FICA, possessed the obvious advantages of sensitivity, rapidity and cost-effectiveness for onsite screening of AFB~1~ in grains \[[@B36-biomolecules-10-00575],[@B40-biomolecules-10-00575],[@B41-biomolecules-10-00575],[@B42-biomolecules-10-00575]\].
Furthermore, in order to verify and compare the reliability of FICAs, 60 grain samples were analyzed by liquid chromatography--tandem mass spectrometry (LC-MS/MS) \[[@B34-biomolecules-10-00575]\], TRFN-FICA, QB-FICA and QD-FICA. A total of 12 samples were confirmed as positive samples, while others (48 samples) were negative by LC-MS/MS and three FICAs, and the representative mass chromatograms (highest and lowest concentrations for positive samples) are listed in [Figure S6](#app1-biomolecules-10-00575){ref-type="app"}. There were no false negative or false positive results reported by the three FICAs, and analysis of field grain samples by FICAs were in accordance with that of LC-MS/MS ([Figure 4](#biomolecules-10-00575-f004){ref-type="fig"}). These results indicate that all three FICAs are reliable methods for the determination of AFB~1~ residues in grains, and that TRFN-FICA obtained the best fit.
3.6. Accuracy and Precision of Three Label-Based FICAs {#sec3dot6-biomolecules-10-00575}
------------------------------------------------------
We performed recovery experiments to assess the accuracy and precision of the three FICAs using six kinds of blank grain samples (corn, soybeans, sorghum, wheat, rice and oat) with a series of known concentrations of AFB~1~. The choices of low, medium and high concentrations with AFB~1~ were the same as in the previous reported literature \[[@B32-biomolecules-10-00575],[@B33-biomolecules-10-00575]\], which were represented by LOD, 2LOD and 4LOD, respectively. Data were obtained from six replicates and on three consecutive days. The intraday and interday recovery of TRFN-FICA ranged from 86.48% to 114.10% and 83.64% to 125.61%, respectively; the coefficient of variation were all less than 10%. Meanwhile, TRFN-FICA had better recovery than QB-FICA and QD-FICA ([Figure 5](#biomolecules-10-00575-f005){ref-type="fig"}), confirming that the accuracy and precision of TRFN-FICA were better than QB-FICA and QD-FICA.
4. Conclusions {#sec4-biomolecules-10-00575}
==============
In this study, three FICAs (TRFN-FICA, QB-FICA and QD-FICA) were systematically compared for the quantitative detection of AFB~1~ in grains successfully. Under optimum conditions, six types of grain samples were analyzed, showing that TRFN-FICA was the most consistent with LC-MS/MS. Moreover, TRFN-FICA had the lowest LOD, shortest immune duration (25 min), and less coating antigen consumption (0.30 μg) and antibody consumption (0.015 μg). Overall, compared with QB-FICA and QD-FICA, TRFN-FICA had a unique advantage in quantitative detection of AFB~1~ in grain, providing a reference for the selection of markers in detection methods.
We sincerely thank Beijing WDWK Biotech Co., Ltd., (Beijing, China) for providing experimental antibodies and coating antigens.
The following are available online at <https://www.mdpi.com/2218-273X/10/4/575/s1>. Figure S1: Particle size of three FICA labels. Figure S2: Optimization of coupling pH in preparing labeled mAb probes: (A) TRFN-mAb; (B) QB-mAb; (C) QD-mAb. Figure S3: Optimization of the concentration of anti-AFB~1~-mAb in preparing labeled mAb probes: (A) TRFN-mAb; (B) QB-mAb; (C) QD-mAb. Figure S4: Optimization of the concentration of AFB~1~-CMO-BSA for fluorescent immunochromatography: (A) TRFN-FICA; (B) QB-FICA; (C) QD-FICA. Figure S5: Immunoreaction dynamics of the three FICAs: (A) TRFN-FICA; (B) QB-FICA; (C) QD-FICA. Table S1: Cross reactivity (CR) of analytes with antibody detected by FICAs.
######
Click here for additional data file.
Conceptualization, X.T.; methodology, Z.L.; software, X.W. (Xuan Wu); validation, X.W. (Xin Wang); data curation, X.W. (Xuan Wu); writing: original draft preparation, X.W. (Xin Wang); writing: review and editing, Z.L. and X.T.; visualization, X.W. (Xin Wang); supervision, X.T.; project administration, X.T.; funding acquisition, X.T. All authors have read and agreed to the published version of the manuscript.
This research was funded by the National Natural Science Foundation of China, grant number 31672605; Natural Science Foundation of Chongqing, China, grant number cstc2018jcyjAX0242 and cstc2017jcyjAX0313; China Postdoctoral Science Foundation, grant number 2016M590855; Chongqing Postdoctoral Science Foundation Special Funded Project, grant number Xm2017074.
The authors declare no conflicts of interest.
{#biomolecules-10-00575-f001}
{#biomolecules-10-00575-f002}
{#biomolecules-10-00575-f003}
{#biomolecules-10-00575-f004}
{#biomolecules-10-00575-f005}
biomolecules-10-00575-t001_Table 1
######
Performance of TRFN-FICA, QB-FICA and QD-FICA in 6 grains.
Parameter TRFN-FICA QB-FICA QD-FICA
------------------------------------------ ----------------- ----------------- -----------------
LOD (μg kg^−1^) 0.04 0.30 0.80
Antibody usage per test card (μg) 0.015 0.09 0.03
The best coating for AFB~1~-CMO-BSA (μg) 0.30 0.65 0.65
Immunoassay duration (min) 25 30 35
Recovery (%) 83.64%--125.61% 80.29%--129.45% 64.53%--133.86%
Coefficient of variation (%) 3.10%--6.75% 2.88%--7.16% 2.34%--8.96%
biomolecules-10-00575-t002_Table 2
######
Comparison of immunoassays for determination of AFB~1~.
--------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------
Detection Method Marker Target Substance Sample Detection Limit of AFB~1~ (μg kg^−1^)
----------------------------------------------------------------------------- ---------------- ------------------ ---------------------------------------------- ---------------------------------------
Immunoadsorption \[[@B15-biomolecules-10-00575]\] Enzyme AFB~1~ Feed samples 2.0
Multiplex immunochromatography \[[@B36-biomolecules-10-00575]\] Colloidal gold AFB~1~, ZEN, OTA Corn 0.10
Rice 0.12
Fluorescent immunochromatography \[[@B24-biomolecules-10-00575]\] TRFN AFB~1~ Corn 0.06
Fluorescent immunochromatography \[[@B20-biomolecules-10-00575]\] QB AFB~1~ Buffer solution 0.005\
(When the inhibition is 10)
Multiplex fluorescent immunochromatography \[[@B23-biomolecules-10-00575]\] QB AFB~1~, ZEN Buffer solution 0.00165\
(When the inhibition is 10%)
Multiplex fluorescent immunochromatography \[[@B21-biomolecules-10-00575]\] TRFN AFB~1~, ZEN Buffer solution 0.05
Fluorescence resonance energy transfer \[[@B39-biomolecules-10-00575]\] QD AFB~1~ Rice 0.04
Fluorescent immunochromatography (this study) TRFN AFB~1~ Corn, soybean, sorghum, wheat, rice and oats 0.04
QB 0.30
QD 0.80
--------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------
[^1]: These authors contributed equally to this work.
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Metter High School
Metter High School is a high school in Metter in rural Candler County, Georgia, United States.
It serves grades 9 through 12 and is located at 34905 Georgia Highway 129 South. The school's athletic teams are the "Metter Tigers".
Historic building
The school's old building, at the junction of College Street and Vertia Street, about four blocks away from the current building, was constructed in 1921. It was designed in Classical Revival style by architect C.C. Muse. A new classroom building was built across College Street in 1937, after which the old building was used less and less. The old building was added to the National Register of Historic Places in 2002.
The former school is a two-story building with a full-height, pedimented portico supported by four Doric columns and two brick pilasters. It originally provided education from grade 1 through grade 11 for the white students of the area.
After 1937, a kitchen and lunchroom in a former classroom area were used until about 1955. Vocational training including industrial arts and business classes continued in the old building for many years. It was later used primarily for storage. In 2017, it became home to the Candler County Historical Society Museum.
See also
National Register of Historic Places listings in Candler County, Georgia
References
External links
Metter High School website
Candler County School District website
Category:Education in Candler County, Georgia
Category:School buildings on the National Register of Historic Places in Georgia (U.S. state)
Category:Buildings and structures in Candler County, Georgia
Category:National Register of Historic Places in Candler County, Georgia
Category:High schools in Georgia (U.S. state)
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Doleschalla
Doleschalla is a genus of bristle flies in the family Tachinidae. There are about 11 described species in Doleschalla.
Species
These 11 species belong to the genus Doleschalla:
Doleschalla consobrina Bigot, 1888 c g
Doleschalla cylindrica (Walker, 1861) c g
Doleschalla elongata (Wulp, 1885) c
Doleschalla maculifera Bigot, 1888 c g
Doleschalla makilingensis Townsend, 1928 c g
Doleschalla nigra Bigot, 1888 c g
Doleschalla papua (Townsend, 1933) c g
Doleschalla parallela (Walker, 1862) c g
Doleschalla picta Bigot, 1888 c g
Doleschalla solomonensis Baranov, 1934 c g
Doleschalla tenuis Malloch, 1932 c g
Data sources: i = ITIS, c = Catalogue of Life, g = GBIF, b = Bugguide.net
References
Further reading
External links
Category:Tachinidae
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Explodeded Blog
In this episode of The Explodeded Show JC & Cereal invent the Applebees App App, discover Stephen Hawking and Ethan Hawke are brothers, and wrap up with a debate on Unions all in the latest episode of The Explodeded Show!
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In the most recent version of the projections of the Office of the Actuary in the Centers for Medicare & Medicaid Services, national health spending growth is expected to average 5.5% per year for 2017-2026 in the United States (US): approximately 1.0% higher than projected gross domestic product (GDP) growth.^[@bibr1-2192568219878133]^ This results in an increase in the health-related costs as a percentage of GDP from 17.9% in 2016, to nearly 20% by 2026, reaching a total of \$5.7 trillion by 2026.^[@bibr1-2192568219878133]^ When looking granularly at specific medical fields, a significant portion of the rising costs of health care in the US relates to the diagnosis and treatment of spinal pathology. It is estimated that 12% to 30% of US adults have an active back problem with approximately 6% having made a visit to a physician for these conditions at one point in their lives, costing upward of \$100 billion to the system each year.^[@bibr2-2192568219878133],[@bibr3-2192568219878133]^ Specifically with regard to spine surgery, fusions and laminectomies were the third and fifth most commonly performed surgical procedures in the United States in 2015, respectively.^[@bibr4-2192568219878133]^ Given the rising costs associated with spine surgery and an aging population, it becomes increasingly clear that the current trajectory is not sustainable, and further scrutiny will be placed on the field in assessing the effectiveness, efficiency, and safety of care delivered. As more healthcare systems invest in healthcare analytics and "big data" (large, complex datasets such as those found in electronic medical records), the opportunity arises to employ predictive analytics via machine learning (ML)/artificial intelligence (AI) approaches to improve quality, reduce waste and error, and minimize cost.^[@bibr5-2192568219878133],[@bibr6-2192568219878133]^
Recent developments in the technologies related to healthcare data collection and analytics have led to a rapid rise in the application of AI within health-related fields. One such application is ML, a branch of AI that involves the construction and application of statistical algorithms that continuously learn and make observations from existing data, and then create a predictive model based on that data.^[@bibr7-2192568219878133]^ With advances in computer processing capability, data storage, and networking, these computer-based algorithms can perform the intricate and extremely complex mathematical operations of classification or regression (specifically nonlinear regression) on immense amounts of data to detect intricate and potentially previously unknown patterns in that data.^[@bibr8-2192568219878133]^ ML algorithms have been able to analyze complex and large volumes of electronic medical record data to produce predictions for a wide range of clinical problems.^[@bibr9-2192568219878133]^ For example, Rajkomar et al^[@bibr9-2192568219878133]^ demonstrated that ML models outperformed traditional, clinically used models in predicting mortality, unexpected readmission, and increased length of stay (LOS) in a study cohort of all admissions in 2 major hospitals from 2009 to 2016. Various investigators have been developing image analysis methods using ML algorithms that have shown promising results in fields such as dermatology, radiology, and ophthalmology. For example, Esteva et al^[@bibr10-2192568219878133]^ have trained ML algorithms to classify skin cancer with a level of competence comparable to dermatologists. These early examples provide insight into early contemporary use of AI in medicine and provide a view of technology that may transform the medical field over the decades to come.
While not the first medical field to adopt an "AI approach" to problem solving, the spine surgery field has recently seen an outpouring of publications related to research in this area. An initial topic of focus by researchers was related to the cost of spine care, as there has been heightened emphasis on moving to a value-based (quality/cost) health care market. For example, as Medicare payments are standardized by procedures performed regardless of hospital LOS, ML systems have been designed with the ability to accurately predict spine surgery-related LOS, discharge to nonhome facility, and early unplanned readmissions using only presurgical or predischarge variables.^[@bibr11-2192568219878133][@bibr12-2192568219878133][@bibr13-2192568219878133]-[@bibr14-2192568219878133]^ These models can help identify/target certain high-risk patients and the variables that contribute to that risk status, allowing hospitals to allocate specific clinical and social resources to reduce costly LOS and readmissions. This can help to maximize efficiency of care delivered, while also keeping constant or even increasing the quality of care delivered.
As spinal surgery has evolved with an explosion of new techniques and technologies in recent decades, there still remains a lack of quality, high-level evidence to support much of the spine care rendered in the US, especially with the cost associated with many of the treatments and devices. As there are numerous surgical treatments in spine surgery that do not easily lend themselves to traditional randomized controlled trials (due to either cost or ethical considerations, among other reasons), an opportunity arises that is ripe for solutions derived from ML approaches. Multiple clinical registries are being collected that contain large quantities of high-quality, spine health care data, such as the 1000-patient Spinal Laminectomy versus Instrumented Pedicle Screw (SLIP) II study.^[@bibr15-2192568219878133][@bibr16-2192568219878133]-[@bibr17-2192568219878133]^ These registries contain demographics, surgery-related variables, patient-reported and complication outcome measures, and notably, they even contain digital imaging with metadata. Leveraging of these vast data repositories can help develop predictive algorithms that are able to incorporate the full range of variables (including complex imaging) in order to guide treatment recommendations.
Because of this lack of high-level evidence, there remains much heterogeneity in the current surgical treatment of spinal disorders, with significant clinical and economic implications.^[@bibr18-2192568219878133][@bibr19-2192568219878133]-[@bibr20-2192568219878133]^ For instance, national surveys of US spine surgeons conducted by Mroz et al^[@bibr21-2192568219878133]^ found 69% disagreement for recurrent lumbar disk herniation, while another study demonstrated 75% disagreement among surgeons on the approach to treat patients with lower back pain,^[@bibr22-2192568219878133]^ implying that 2 similar patients with the same pathology could receive entirely different care. Furthermore, a cost analysis based on the results of the national survey mentioned above revealed that there is also a variation in costs based on spine surgeon specialty, practice type, surgical volume and geographical location.^[@bibr23-2192568219878133]^ Recent ML/AI approaches to this problem have been published that attempt to assist surgeons' decisions with predictions of patient outcomes. Utilizing data from repositories created from AOSpine prospective, multicenter studies, Merali et al^[@bibr17-2192568219878133]^ developed a supervised ML model that accurately predicts a positive outcome on an individual patient after surgery for degenerative cervical myelopathy, with an average area under the curve of 0.70, classification accuracy of 77%, and sensitivity of 78% on an independent testing cohort. Shah et al^[@bibr24-2192568219878133]^ were able to build an ML model that predicts probability of failure of nonoperative management in spinal epidural abscess, while Karhade et al^[@bibr25-2192568219878133]^ successfully developed an ML algorithm that predicts in-hospital and 90-post discharge mortality in this patient group. The same group was able to predict short-term postoperative mortality in individual patients with spinal metastatic disease with an ML model, aiding in decision-making and informed discussions with the patient regarding surgical intervention this challenging patient population.^[@bibr26-2192568219878133]^ All of these previously mentioned studies have now published their prognostic tools in an open-access, digital interface to be integrated into practice, supporting clinicians in developing treatment plans that are more standardized across the world.
Along with prediction of positive patient outcomes, clinician researchers have also used AI/ML to forecast negative outcomes as well, as recent publications have explored the likelihood of complications from spine surgery. In multiple articles, the same group led by Cho et al utilized an artificial neural network-based ML algorithm to predict surgical complications in patients undergoing elective anterior cervical discectomy and fusion, posterior lumbar fusion, and adult spinal deformity surgeries. Their models were able to specifically predict the risk of cardiac-related, wound-related, venous thromboembolism--related, and mortality in these patients, outperforming the American Society of Anesthesiologists Physical Status Classification scoring in predicting individual risk prognosis.^[@bibr27-2192568219878133],[@bibr28-2192568219878133]^ Another publication by Sheer et al^[@bibr29-2192568219878133]^ describes their method to create a ML model that successfully predicts major intraoperative/perioperative complications following adult spinal deformity surgery with an accuracy of 87%. Utilizing large databases of patient information, Han et al^[@bibr30-2192568219878133]^ were able to analyze over 1 000 000 patients that had previously undergone spine surgery and developed multiple ML predictive models that identify risk factors for postoperative complications. Karhade et al^[@bibr24-2192568219878133]^ were even able to predict prolonged opioid prescription after surgery for lumbar disc herniation in an ML algorithm. These surgery- and patient-specific models can help to aid in surgical planning, as well as patient counseling and shared decision making. If these models identify modifiable risk factors in the preoperative setting of a nonurgent surgery, time and effort could be dedicated to improved medical management of that comorbidity prior to surgery, in effect reducing the risk of complications and increasing the probability of a good outcome.
In deciding if a patient is indicated for surgery, one area where a surgeon's subjectivity may still reign supreme is review of the spine imaging. Utilizing classification techniques from radiology literature, new research is revealing the applicability of AI and ML algorithms to the analysis of spine imaging. One technique involves the use of ML models utilizing natural language processing to distinguish specific words and phrases from unstructured radiology reports in order to classify patients by imaging findings, as Tan et al^[@bibr31-2192568219878133]^ were able to do in a cohort of patients with low back pain. However, in more recent publications, other groups were able to utilize the imaging itself to detect and classify a variety of pathologies. Hopkins et al^[@bibr32-2192568219878133]^ were able to predict both the diagnosis of cervical spondylotic myelopathy and its severity with high sensitivity and specificity (90.25% and 85.05%, respectively), utilizing magnetic resonance imaging alone in an artificial neural network model. Further, work has been done to develop ML models in the detection and grading of lumbar spinal stenosis^[@bibr33-2192568219878133]^ and fracture detection and classification with various types of imaging modalities.^[@bibr34-2192568219878133]^ AI/ML imaging analysis can even assist real time in the outpatient clinic, where Sharif Bidabadi et al^[@bibr35-2192568219878133]^ were able to accurately identify foot drop of an L5 origin and classify patients into various recover stages with an 85% accuracy. While there is much work to be done, this initial work which was all published in the past year, shows the feasibility of using AI/ML-based approaches to analyzing spine imaging.
Common themes among large institutions and large spine centers are tighter financial margins, less resources, and heightened payer scrutiny on indications, outcomes, and postprocedural treatments. This collectively creates real strain on the departmental workforce (ie, secretaries, advanced practice providers, physicians). An AI platform that successfully predicts patient and surgeon performance from financial, outcome, and electronic medical record databases across an entire book of business stands to provide the leverage to homogenize outcome and cost. This, in turn, positions said organization optimally for contract negotiations and population health initiatives. Furthermore, a fully integrated AI platform can also automate much of what currently strains department assets. Postsurgical checks, ordering medications and imaging, patient reminders, and scheduling follow-up visits, are all some examples of how such a platform can enhance overall spine center efficiencies and performance, patient satisfaction (eg, more automated touch points), and employee engagement.
Challenges Ahead {#section1-2192568219878133}
================
Even though current research described above highlights the promise and potential of AI in spine surgery, the field as a whole still face many challenges. First, in order to create an AI-driven decision platform, very large and appropriately labeled data sets are required, which the majority of centers in the United States still lack. This becomes even more difficult with imaging-based analysis. Second, some ML models require manual labeling of the data for classification and learning to occur. This presents a clear challenge in the analysis of spine surgery pathologies, where there is still widespread disagreement about what constitutes normal versus abnormal with regard to certain exam/imaging findings, and subsequently the appropriate treatment(s). This can be circumvented by allowing the model itself to do the analysis and classification, such as is the case with unsupervised algorithms. Given the vast quantity of data analyzed, this can reveal links between variables that experts would not have otherwise expected. However, it is difficult to backtrack and get precise information regarding the specifics of the data sorting in these types of models. And with poor quality or quantity of data to learn from, the model may make erroneous associations and/or can be "overfitted" to the training dataset, producing a lack of external validity. Furthermore, many ML algorithms thus far are typically trained and validated internally within one institution. Further work needs to be carried out to examine if a predictive model is transferable from one site to another, and what implications this holds as a "live" ML model undergoes continuous calibration and evolution based on new sets of data.
As exponential expansion of computing capacity converges with unsustainable healthcare spending, a hopeful opportunity has emerged: the use of AI to enhance healthcare quality and safety. AI-based, ML approaches to spinal pathologies are already distinguishing relevant from irrelevant data regarding a particular patient, assisting with appropriate hospital-based care, interventions or even surgeries, predicting cost of care, and predicting future outcomes on a variety of anchored measures. While many shortcomings still exist as the technology is in early development, extrapolating from today's progress and fully implemented into the healthcare system, AI could help solve a number of problems in spine surgery by improving outcomes, minimizing cost, standardizing care for a given pathology, and driving efficiencies within a spine service line in large centers. These types of approaches could deliver on the value equation while serving as a resource for improving physician performance and promoting appropriate, efficient care in this era of financial uncertainty in health care.
**Declaration of Conflicting Interests:** The author(s) declared no potential conflicts of interest with respect to the research, authorship, and/or publication of this article.
**Funding:** The author(s) disclosed receipt of the following financial support for the research, authorship, and/or publication of this article: This supplement was supported by funding from AO Spine North America.
**ORCID iD:** Matthew M. Grabowski  <https://orcid.org/0000-0001-8550-0124>
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Pļaviņas Hydroelectric Power Station
The Pļaviņas Hydroelectric Power Station has the largest hydroelectric power plant in the Baltics and one of the biggest in the European Union. It is located in Aizkraukle on the Daugava River. It has ten individual water turbines with an installed total capacity of 894 MW.
The construction aroused an unusual wave of protest in 1958. Most Latvians opposed the flooding of historical sites and a particularly scenic gorge with rare plants and natural features, such as the Staburags, a cliff comparable in cultural significance to the Lorelei in Germany. The construction of the dam was endorsed in 1959, however, after the purge of relatively liberal and nationally oriented leaders under Eduards Berklavs and their replacement by Moscow-oriented, ideologically conservative cadres led by Arvīds Pelše.
The plant was put into full operation in 1968. In 1991–2001, six additional turbines were added to the original four, thus increasing the capacity to 868.5MW. Reconstruction and overhaul of the units between 2007 and 2010 increased both the efficiency of the plant and its power output.
The complex is operated by Latvenergo.
References
External links
Category:Hydroelectric power stations in Latvia
Category:Hydroelectric power stations built in the Soviet Union
Category:Dams in Latvia
Category:Dams completed in 1965
Category:Aizkraukle
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El fabricante de aeronaves estadounidense Boeing quiere que más de sus aeronaves vuelen por los cielos mexicanos. Y para despegar en el país, su alianza con la empresa brasileña Embraer es la clave.
Las pláticas de una posible asociación entre ambas compañías comenzaron en 2017, cuando ya eran socios en temas de ingeniería y pruebas de desempeño. Fue hasta diciembre de 2018 cuando Boeing adquirió la división de aviación comercial y las operaciones de servicios de Embraer, con lo cual la empresa estadounidense se hizo de 80% de la firma brasileña por un monto de 4,200 millones de dólares (mdd).
Embraer tiene una fábrica en Chihuahua, en donde arma los interiores de sus modelos E-170, E-175, E-190 y E-195, tanto de primera como de segunda generación, además de paneles internos, sistemas de oxígeno e iluminación. Donna Hrinak, presidenta de Boeing en Latinoamérica, ve en estas instalaciones el primer paso para que la compañía tenga su primera fábrica en México.
Lee: Boeing completa la actualización de software para el 737 MAX
“Estamos en el proceso de integrarnos con Embraer en la parte comercial, lo cual nos daría nuestras primeras fábricas fuera de Estados Unidos. Tenemos ya fábricas en Australia, Canadá e Inglaterra, pero no en un país donde el inglés no es la lengua materna”, explica en entrevista con Expansión.
Publicidad
Hrinak explicó que las primeras fábricas de esta etapa serían las de Brasil, Portugal y México. En la capital chihuahuense, Embraer tiene una joint venture con la multinacional francesa Safran, por la que en la fábrica de EZ Air se producen interiores de aeronaves como paneles, cortinas, baños y compartimientos, en el Complejo Industrial Chihuahua.
“Sería nuestra primera fábrica en México, lo cual nos da la oportunidad de mirar al país con otros ojos. Visité la fábrica hace algunos meses, la producción allá es admirable, entra en los ideales de producción y seguridad que tenemos en Boeing. Va a ser una gran colaboración”.
La meta es que la fábrica inicie operaciones en la recta final de 2018, al terminar los procesos correspondientes que la compañía tendrá con las autoridades de competencia de países como Estados Unidos, Japón y China.
Las acciones de Boeing están encaminadas a satisfacer una demanda global que, en 20 años, prevén que sería de 43,000 aviones nuevos, de los cuales más de 3,000 serían para el mercado latinoamericano, en particular aeronaves de un pasillo.
Cautivar a más clientes en medio de la crisis
En México, el mayor cliente de la compañía -y prácticamente el único- es Aeroméxico, que en 2012 firmó un contrato para la adquisición de 100 aeronaves por 11,000 mdd, y cuya flota al primer trimestre del año constó de 70 aeronaves de Boeing y 57 de Embraer, por lo cual Hrinak consideró que su experiencia en el mercado mexicano ha sido buena, “pero limitada”.
Lee: Aeroméxico planea invertir 3,000 mdd durante los próximos seis años
Aunque la empresa busca tener un mejor acercamiento con Protección Civil para colocar aeronaves como su modelo H-47 Chinook, que se utiliza en operaciones de rescate en desastres naturales, la situación se complica ante las catástrofes relacionadas con su modelo 737 MAX8, que registró dos accidentes en Etiopía e Indonesia en menos de seis meses, y en los que fallecieron más de 300 personas.
Este incidente provocó la suspensión de cinco aeronaves de este modelo de la flota de Aeroméxico, y después la Dirección General de Aeronáutica Civil suspendió todas las operaciones de vuelo de este modelo y del 737-9 MAX en espacio aéreo bajo la jurisdicción del Estado mexicano.
Por ello, la estrategia de Boeing consiste en reforzar la seguridad de sus productos, a la par de forjar más relaciones con potenciales clientes.
“La seguridad es nuestra prioridad número uno (…) Estamos participando en la investigación con autoridades etíopes principalmente. Lo que pasó fue el resultado de una cadena de incidentes, uno de los cuales fuera una falla en el sistema MCAS que estamos arreglando con un nuevo software. La idea es hacer un upgrade y también reforzar el entrenamiento que los pilotos reciben”.
A pesar de ello, la compañía no tiene pensado realizar algún cambio en sus procesos de manufactura hasta el momento. “Todavía evaluando el impacto que estas tragedias tendrían en este año, pero no tendremos impacto en nuestras operaciones en México”.
Dennis Muilenburg, presidente de Boeing a nivel global, anunció la semana pasada que la actualización del software estaba lista tras realizas las pruebas correspondientes, que consistieron en 207 vuelos y más de 360 horas en el aire. Por su parte, Andrés Conesa, director general de Aeroméxico, estimó que los cinco 737 MAX8 pertenecientes a la compañía estarían de vuelta en el aire para el 12 de julio, tras la implementación de la actualización del software.
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Reviews for this item
"Been to numerous moe. shows, but this was by far my favorite of all time. Was right in front on Chuck's side all night and was blown away. Great Brent Black & Happy Hour Hero. Didn't dig Saturday's set as much, but this show made the entire weekend worth while. Dunegrass was an amazing little festival."
moed down
—
8/15/2008 2:08:37 PM
"250 people tops at a Moe. show???? By far one of the best shows of any band out there. DOWNLOAD!!!!!!!!!!"
moe.doc.
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8/8/2008 5:50:19 PM
"These 2 shows were great. First night was in the tent stage and the guys were on fire and really enjoying the smaller scene. Second night was on the main stage and was equally powerful as first. Meat blew my mind. Allie Kral played on conviction song. I only wish they could have had more time for 2 sets each night. Do not hesitate to DL."
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I'm Daniel Sullivan the founder and head
instructor of worrier Arts Alliance also
the founder and head instructor at OC
kickboxing and mixed martial arts in
Irvine California I've been involved
with the combat of martial arts since
1985 was fortunate enough to meet Diana Lee Inosanto and a college class and the
time I was doing Taekwondo didn't know
anything about the Filipino martial arts
Jeet Kun Do or Muay thai in that you
know I don't like heard of karate
kung-fu and those sort of you know
typical martial arts yes I met her in a
college class and that fateful day
pretty much forever changed my life she
introduced me to her dad Dan Lee Inosanto
and he introduced me to several
different martial arts the Filipino
martial arts, Jeet Kun Do, Muay Thai, Silat,
Jiu Jitsu, Shooto he's pretty much
impossible although the most combat
martial arts in the world you know he's
a black belt in Brazilian jiu-jitsu and
he's played the man in my opinion that's
done more research into the martial arts
than anyone else in the world hands down
there was only one that has even a
quarter of the knowledge of him that I
know of I always challenged
people tell me someone who's got more
knowledge in the martial arts than Dan Lee Inosanto
I know and I i never hear a single come
back so not so since 1985 I tried to
follow the same path that Dan Inosanto and his best friend meant or bruce
Lee we're on you know back in the
sixties and early seventies basically
they basically they were researching
every martial art in the world to try to
find you know
one of the best parts you know the real
world
for a real street fight basically it's
MMA for the real world you know you have
MMA for competition of MMA for combat and
the to a little bit different you know
because MMA for combat includes all of
the competition arts Muay Thai Brazilian
jiu-jitsu and wrestling and boxing all
the mainstream stuff but you know we
cross train and Filipino martial arts
you know if Muay Thai is the King of the
Ring Filipino martial arts is the king
of the streets why because almost every
fight is going to involve a weapon fights
will always involve a weapon especially
here in Bangkok right now
Southeast Asia people you know carry
edged weapons it's it's accessible
everyone has one you know the other
thing the other difference major
difference between the Combat Arts and
the competition arts is that fights are
never one-on-one let's be honest people
travel in packs looking for trouble you
know so you know we've all heard the
sales sales pitch
that all fights go to the ground and
well you know it's true a lot of flights
to go to the gun but in reality that's
the last place you want to be in a real
fight because it's never one on one it
doesn't matter how good you are on the
ground you can have beginning of
submission you can be in the top
position you can be ground and pounding
but when the second guy comes along
you're in bad trouble because I need
your head
Shin kick your head he's going to soccer
kick you teeth in ya so you know so you
know not to take away from Brazilian
jiu-jitsu cause I'ma blackballed I
believe Brazilian jiu-jitsu is the king
of them at once you get to the floor ugh
he's the best however you know we gotta
realize that you know no martial arts
has everything so just say can you know
sport MMA you need to cross train right
we know we need to cross train you know
we're in the early days across all
jiu-jitsu jiu-jitsu they never put a
good striker in the UFC for the first
five years it was never a world-class
striker in the UFC then when they
started to put some strikers and they
start to find out that you know you need
to have a striking game and you need to
have a grasp of the game right so it's
the same way for the real world you need
to have a weaponry game and you need to
go outside of the room of the rules you
know and that's not what I want your
Arts Alliance's my organization is all
about you know cross training in the
best martial arts for the real world
yeah I was a kid I was as skinny surfer
kid and I grew up you know in a family
that kind of fell apart my mother and my
father ended up marrying a hippie
basically turned us into a happy family
moved us to Santa Cruz California AP
central you know and so you know when I
got there I ran into trouble right away
actually the part in Santa Cruz who
they've done with the hippies want it
was all the rednecks where than withers
area with the hippies were and we were
in the red neck area so you know I had
the long hair and so as to get picked on
the new kid in town you know we always
move around and say I got picked on here
in there and so you know I always wanted
to learn self-defense so one of the
things that got me really interested in
the martial arts to start with the
Kung Fu series of David Carradine I
loved it he you know would never look
for trouble are you never go looking for
trouble and but trouble always found him
and he would always he would never use
more violence than he had to he would
always see you know just diffuse the
situation without hurting the guy as
much as possible right and he would
always try to avoid a try to avoid it
you know and then there was always the
message should go back to the his Sifu
you know it at the Shaolin Temple and
you know there's always that he'd have
the flashback and you know there's that
message they're the lesson you learned
in every episode you know and I just
loved the the message and the positive
no message that the martial arts sense what really got me interested in
it and then enter the dragon came out
you know I was just like wow this is
amazing you know you know just
fascinated with bruce Lee I try to avoid
conflicts at all costs I always walk
away I always you know we'll try to talk
my way out of it I very rarely you know
get into streetlights although I have
when I had when I was in Long Beach had
several incidents go down when I lived
in a house in a really bad area of Long
Beach on over my first martial arts
school but it's always been for me
trying not to you know to kill the guy
too much time to just get out of the
situation without you know I can't have
it on my conscience they give someone
brain damage or you know maimed and
killed them so I've had several several
incidents go down and I think you know
one of the great things about the
martial arts is that you get confidence
and when you're confident people just
don't really mess cutie mark you know
they don't they don't bother you that
can sense that you you have something
Demonstration
so it's gonna show you know some
possible way you might be attacked in
the real world and some possible ways
and dirty boxing that we might handle
that situation so a lot of times in
self-defense Chelsea guys who wanna
finish of you are they gonna grab your
clothing are they gonna grab your neck
like that or they're gonna try to pull
your head down she can need I can do
here in Thailand right so yes so tight
end if he doesn't like a double shift in
Chester when he doesn't notice how is
getting into here to smoke a steering
wheel position so so
shuts you because
clinch you
rap on your clothing
got two hands into this position called
the steering wheel thing about this home
and get your butt and that's one of my
favorite things sometimes we just felt
they had here
and this is the best time
couch
helpful
grab onto the
just set up a striking from here all
your orthodox lost
you want
let's look at what time you shop you
call the steering wheels but need
short back turning his head like that
get my phone down
like grap head here that's a knee
and stepped-up combination
INTERVIEW
ya wanna principles that I here to
obviously is cross-training that no
martial art has it all
no martial arts complete and that's why
we call it mixed martial arts so you
know mixed martial arts for combat
that's what I'm about mixed martial arts
for the real world so I think anyone who
is serious about learning self-defense
fence needs to have a Thai boxing Muay
Thai background you know some time here
in bangkok right now just did a seminar
on dirty boxing filipino boxing but I'm
also here to out to train in Muay Thai
love Muay thai and doing it since 1985
and let's be honest the best ranking
system in the world you know I think
anyone without a Muay Thai background is
really you know not serious about
learning how to defend themselves and
then obviously a Brazilian jiu-jitsu is
one is the best grappling art out there
are probably other great ones too but I
think you need you need you know they
have some knowledge of Brazilian
jiu-jitsu boxing you know boxing is the
most sophisticated system of using the
hands and all the martial arts according
to Bruce Lee and I agree with them
wrestling obviously is very important
you know having having take downtown
knowing how to defend takedowns but then
what I like to do is mix and other other
arts or not so in the mainstream like
for example the Filipino martial arts
you know which is the best with the
knife the stick
some of the systems have the best empty
hands and I've ever seen like Inosanto
I know the cross method that the Filipino
martial arts includes dirty boxing or
filipino boxing puntukan or suntukan
on which a mix of boxing
master John Lacosta learned in the
central region of the Philippines to the
size and then he learned the Silat in
the south in Mindanao so he mixed silat
with boxing and it's one of the most
sophisticated the most effective
efficient forms of self-defense and I've
ever seen Dan Lee Inosanto taught me that
you know of a thirty years of training
with him several workshops within the
focus just on the filipino boxing and
and so yes I took what he did in kind of
did my own thing a little bit with the
filipino boxing and I think it's one of
the best silat is one of the best I
like a savate box france you know I'm
a box franse one of the best
striking arts but there are you know
when China there are a lot of great
martial arts out there however some of
them with a lot of martial arts there's
certain things that I like and then
there's and it's going to be like that
for everyone
those things that you like those things
that you don't like so much you know and
so that's why I think it's so important
to cross train
another important thing about cross
training is that you know when you're 20
years old you know you got Muay Thai, MMA
,Jiu Jitsu
know but you know I'm I'm 53 years old
so you know my my shape and injuries
stuff like that and you know as you
start getting older he don't cross train
you're not going to be able to do it
anymore you know and you get depressed
and you let you know when you stopped
training you know your life's just not
the same so it's really important to
have to be able to know have different
martial arts ones that you can do like
the Filipino martial arts is great
because you can practice it into you
sixties seventies eighties nineties like
you know a lot of the Masters that grew
down in a sauna learn from they were ww2
heroes they fought the first Filipino
Infantry Division and they were sent to
the Philippines to fight over there they
fought with with live blades swords the
lows against the japanese so you know I
think ww2 veteran so obviously most of
them are all gone now they were heroes
and you know the Filipinos were no
exception to the rule they were
definitely heroes who fought against the
Japanese and you know so by the time
Groudon met these guys are already a lot
of them are in their eighties you know
seventies eighties some of them pushing
90 and you know
they can still move you know they could
still move still do that thing like Locasta
for example he was in his late
eighties he still train every day and
the weight I get shot the back which is
a sad very sad story because we all
wanted to learn what he had you know but
but yeah so I think it's great to cross
train because it gives you that option
to do you know if your body can't do the
Shin kick anymore you know your hips
body or whatever you have other things
you can practice so that my Academy in
Orange County California we specialize
in beginners we love we love to take
people have no martial arts experience
at all and teach them how to defend
themselves you know get them involved
with all the different combat martial
arts and normally what I like to do as I
like to start people with Thai boxing
with Muay Thai and there are several
reasons for that one is that they're
learning to use their natural weapons
you know their punches or kicks their
elbows on their knees and the learning
basic defenses you know it'll work in
the real world they're also at the same
time getting fit you know I'm being fit
is a very important part of being able
to defend yourself
Muay Thai kickboxing boxing it gives you
timing it gives you read them if you
distance it gives you an understanding
of movement it gives you an
understanding of the way that you
probably gonna be attacked in the real
world now that everyone does anyone does
boxing everyone does mixed martial arts
so yeah so I like to start people with
Muay Thai or we started with MMA mixed
martial arts that's we do get my
academies I offer everything I hope we
have a mixed martial arts program we
have a Muay Thai program but usually we
don't like to start people in Brazilian
jiu-jitsu for example is like I said you
know you need to learn how to strike
first in my opinion I think it you know
you don't want to go to the ground right
away that's all you know you're gonna be
in bad trouble in the in a real
situation if that's all you know how to
do is grapple is it you know it's never
a one-on-one fight so we start people
with Muay Thai MMA and then once they've
gone through you know three months or so
of Muay Thai or MMA then they become
eligible to go into what we call a
combat out the program and that's what
we offer noJeet Kun DO Filipino martial
arts Brazilian jiu-jitsu no gi Jiu Jitsu
boxe française Savate so we offer all
those are taught separately but we'd
like you to have a base in striking
first because up pretty much all the
martial arts we teach our striking the
boxing based like the Filipino martial
arts boxing based you know Jeet Kun Do
is boxing based Savate
obviously boxing based Muay Thai boxing
base MMA boxing based again even
Brazilian jiu-jitsu you know it's even
Helio Gracie believed in boxing although
you know the early Jets you guys want
the best boxers but you know there
they've gotten a lot better now you know
nowadays all the brunt brazilians know
you have to do Muay Thai you have to be
boxing where they used to laugh about it
in the early days of the of the UFC they
would laugh at Muay Thai boxing but now
it is often anymore you know because
they realize you have to have a
foundation and striking
so yeah that's why start people Muay Thai
Boxing MMA you know to get used to their
their tools get used to how they you
know how the body moves and whatnot then
later on they get into the the Brazilian
jiu-jitsu and what other whatever else
that they want to get involved with
DEMONSTRATION
other things get a lot of times
another thing we get a lot of times
these guys who are gonna
swimming wild at your head right through
it
maker punch so in that type of situation
you know first of all you want to know
how to cover you need to be able to
defend your head you know the docs way
like this first got that down this week
all it takes
and we have what's called the water
covered like this
so we have a lot of times in a street
fight traffic situation in self-defense
situation as an individual
angry gonna try to punch a lot of times
it's not a nice straight punch me I
punch right in that type of scenario
comes with a white boy you need to know
how to just cover your head up to
protect your head
that's what we call it type cover we
call a wide cover the clothes like this
right
and then if you want to try to get the
same time to become a political punch
the clock so why were the first one and
you just come in when you cover up here
if you think air right now if you got
simultaneous this is one of our
favorites like that which are striking
here whether it's tougher cuts
Interview:
so dirty boxing basically filipino
boxing ok it's it comes from Grandmaster
johnnie Juanito LaCosta John Acosta he's
one of the instructors Dan Lee Inosanto
know he's very unique in his approach
because he traveled all over the
Philippines to learn all the different
Filipino martial arts the thing that he
was able to do that other Filipino
martial arts masters weren't able to do
was to learn all the southern styles
that are really in existence anymore
from Mindanao you know Silat the
different Filipino Silat systems so
he converted to Islam the reason why
most people weren't able to learn it is
their Christian there from the central
and northern region and Christians are
not allowed to learn Silat so what he
did was he converted to Islam and learn
to speak twelve different Muslim
dialects so he could learn all the
different silat stiles which made him
very unique you know so what he did is
he took the the boxing did you learn in
the Visayas in the central region of the
Philippines you know Suntukan
and and he makes that silat so it's a
very effective boxing base but it's it's got
all the you know the dirty techniques
it's got eye got head butts needs to
the groin foot stomps you know strikes
to the arms we do harm destructions and
clinching and you know double thumbs and
grabbing you know kicking the groin
go yell everybody can do that I know
what everybody says but it could do that
yeah but ya think about this I mean like
Bruce Lee for example
area finger job from any angle at
lightning speed
you know you wouldn't even see it coming
so someone with it would skill like that
you know i mean is another animal to
deal with you know me and and people
don't realize you know they can pick up
a weapon
well you know it's it's not that simple
with you if you're going to get someone
who has skilled with a weapon it's very
very different you know so it's easy for
people in the MMA world to downplay
parts that aren't
competition but usually that's because
they don't really have an understanding
they don't have an understanding of
what's really out there and what you
know what it's worth it you know what
skills outside up their game how much
that can add to you know what's going on
in a real-world self-defense or street
fighting situation.
so filipino boxing
also known as Panatukan or Suntukan
in my opinion is one of the best
martial arts for the real world includes
striking similar to Thai boxing but it
also includes
had but, eye gouching all types of
clinching, knees, strikes to the groin and foot stomps
and over hooks, underhooks and no shots to
the throat you go oh yeah well you want
to do that it's not just that what makes
it unique award makes it so effective is
how how do you get from eye-gouching to
head but he's to knees to sweeps
together how do you get into it that's
the thing everybody knows you can eye gouch
how do you get him do it right
how do you get into the other techniques
how do you flow from head butts, elbows, knees and clinching
great about filipino boxing it is best in the world
flowing between regular mainstream type
of strikes again elbows knees kicks
punches into the dirty tricks that they
are going to set up your strike again eye
gouch for example once you get your eye gouched
the best time to hit him with
traditional type of boxing strikes
|
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Differential trafficking of the Niemann-Pick C1 and 2 proteins highlights distinct roles in late endocytic lipid trafficking.
The cellular location of Niemann-Pick C2 protein (NPC2) in cultured human fibroblasts and Chinese hamster ovary cells was examined immunocytochemically and in living cells by expression of a functional red fluorescent protein chimeric analogue. NPC2 is present in the lysosomes of both cholesterol-depleted and -replenished cells, unlike Niemann-Pick C1 protein (NPC1) which is recruited to late endosomes only upon uptake of low-density lipoprotein. With mobilization of cholesterol from lysosomes, immunocytochemical detection of NPC2 in lysosomes is greatly diminished, whereas NPC1 remains in the late endosomal compartment. We found a partial overlap in the trafficking and organellar sites of accumulation of NPC2 and NPC1. In living cells, NPC2 traffics with NPC1 in late endosomal tubules. However, in contrast to NPC1, which remains either in late endosomal vesicles and tubules or at the peripheries of cholesterol-laden lysosomes, NPC2 moves into the central core of lysosomes. Glycolipid analysis reveals that, in contrast to null mutant NPC1 cells, which accumulate GM2 ganglioside only at the plasma membrane, with no endocytic storage, absence of NPC2 protein in null mutant NPC2 cells does not block internalization of GM2 into endocytic vesicles. This difference in the cellular distribution of GM2 in NPC1 and NPC2 null mutants is the first report of a variation in the phenotypic expression of these genotypically distinct lesions. We speculate that while NPC1 may play a major role in the sorting of glycolipids as well as cholesterol within the late endosomes, NPC2 primarily plays a role in the egress of cholesterol and, potentially, glycolipids from lysosomes. These proteins appear not to be integrated into a tightly bound biological complex, but rather represent separate functional entities that complement each other.
|
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Terrorists who murdered 22 people in Dhaka this weekend were wealthy, well-educated young Bangladeshis whom police had previously tried to arrest, it emerged on Sunday.
Identities of the five gunmen shot dead by police circulated quickly online after images of them posing with weapons prior to the attack were released by Amaq, the news agency of Islamic State of Iraq and the Levant (Isil).
The men, in their early- and mid-twenties, had attended some of Bangladesh's top schools and universities, according to classmates who identified them on social media. Many of those identified had gone missing in recent months.
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Quality goal attainment and maintenance in patients with type II diabetes mellitus initiated on canagliflozin or a glucagon-like peptide-1 receptor agonist in an actual practice setting.
To compare achievement of quality goals (HbA1c, weight loss/body mass index [BMI], systolic blood pressure [SBP]), including maintaining HbA1c, between patients with type 2 diabetes mellitus (T2DM) treated with canagliflozin 300 mg (CANA) or a GLP-1 in an actual practice setting. Adults with T2DM newly initiated on CANA or a GLP-1 were identified from the IQVIATM Real-World Data Electronic Medical Records-US database (2012Q2-2016Q1). To account for differences in baseline characteristics, inverse probability of treatment weighting was used. Outcomes were compared using Cox models (hazard ratios [HRs] and 95% confidence intervals [CIs]) and Kaplan-Meier analyses. CANA (n = 11,435) and GLP-1 (n = 11,582) cohorts had similar attainment of HbA1c < 8.0% (64 mmol/mol) and HbA1c < 9.0% (75 mmol/mol; HbA1c < 8.0%: HR [CI] = 0.98 [0.91-1.06]; HbA1c < 9.0%: HR [CI] = 1.02 [0.93-1.12]), while GLP-1 patients were 10% more likely to achieve HbA1c < 7.0% (53 mmol/mol). CANA and GLP-1 patients were similar in maintaining HbA1c < 7.0%, < 8.0%, or <9.0%, achieving weight loss ≥5% (HR [CI] = 1.05 [0.99-1.12]), achieving BMI <30 kg/m2 (HR [CI] = 1.11 [0.98-1.27]), and achieving SBP <140 mmHg (HR [CI] = 1.07 [0.98-1.17]). CANA patients were 30% less likely to discontinue treatment, 28% less likely to have a prescription for a new anti-hyperglycemic, and 17-21% less likely to fail to maintain HbA1c < 8.0% or 9.0% or have a prescription for a new anti-hyperglycemic (composite outcome) vs GLP-1. No significant difference was observed for the composite outcome using the HbA1c < 7.0% threshold. This retrospective study in an actual practice setting showed that CANA patients were generally as likely as GLP-1 patients to achieve HbA1c, weight, and blood pressure thresholds, and to maintain glycemic control while being less likely to discontinue treatment and/or have a new anti-hyperglycemic prescribed.
|
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A perusal of his draft reveals ‘radical Islamic terrorism’ is conspicuously missing.
Even as his administration fights for its travel ban from several Muslim-majority countries, President Donald Trump is using the nation that is home to Islam’s holiest site as a backdrop to call for Muslim unity in the fight against terrorism.
Mr. Trump’s Sunday speech, the centerpiece of his two-day visit to Saudi Arabia, will address the leaders of 50 Muslim-majority countries to cast the challenge of extremism as a “battle between good and evil” and urge Arab leaders to “drive out the terrorists from your places of worship,” according to a draft of the speech obtained by The Associated Press.
Mr. Trump, whose campaign was frequently punctuated by bouts of anti-Islamic rhetoric, is poised to soften some of his language about Islam. Though during the campaign he repeatedly stressed the need to say the words “radical Islamic terrorism” and criticized his opponent, Hillary Clinton, for not doing so, that phrase is not included in the draft.
The speech comes amid a renewed courtship of the United States’ Arab allies as Mr. Trump is set to have individual meetings with leaders of several nations, including Egypt and Qatar, before then participating in a roundtable with the Gulf Cooperation Council and joining Saudi King Salman in opening Riyadh’s new anti-terrorism center.
No mention of democracy & rights
The address also notably refrains from mentioning democracy and human rights topics Arab leaders often view as U.S. moralizing in favor of the more limited goals of peace and stability.
“We are not here to lecture to tell other peoples how to live, what to do or who to be. We are here instead to offer partnership in building a better future for us all,” according to the copy of his speech.
Two different sources provided the AP with copies of the draft of his remarks, billed as a marquee speech of the trip. The White House confirmed the draft was authentic, but cautioned the President had not yet signed off on the final product and that changes could be made.
Mr. Trump may seem an unlikely messenger to deliver an olive branch to the Muslim world.
During his campaign, he mused, “I think Islam hates us.” And only a week after taking office, he signed an executive order to ban immigrants from seven countries Iraq, Iran, Syria, Sudan, Libya, Somalia, and Yemen from entering the United States, a decision that sparked widespread protests at the nation’s airports and demonstrations outside the White House.
That ban was blocked by the courts. A second order, which dropped Iraq from the list, is tied up in federal court and the federal government is appealing.
‘Counterweight to Obama’s speech’
White House officials have said they consider Mr. Trump’s visit, and his keynote address, a counterweight to President Barack Obama’s debut speech to the Muslim world in 2009 in Cairo.
Mr. Obama called for understanding and acknowledged some of America’s missteps in the region. That speech was denounced by many Republicans and criticized by a number of the United States’ Middle East allies as being a sort of apology.
Saudi Arabia’s leaders soured on Mr. Obama, and King Salman did not greet him at the airport during his final visit to the kingdom. But on Saturday, the 81-year-old king, aided by a cane, walked along the red carpet to meet Mr. Trump as a fleet of military jets swept through the sky, leaving a red, white and blue trail in their wake. During a ceremony at the grand Saudi Royal Court, the king awarded Mr. Trump the Collar of Abdulaziz al Saud, the theocracy’s highest civilian honor.
Mr. Trump bent down so that the king could place the gold medal around his neck. Saudi Arabia has previously bestowed the honor on Russian President Vladimir Putin, British Prime Minister Theresa May and Mr. Obama.
First such visit
The President’s stop in Saudi Arabia’s dusty desert capital kicked off his first foreign trip as President, an ambitious, five-stop swing that will take him through the Middle East and into Europe. He’s the only American President to make Saudi Arabia or any Muslim-majority nation his first overseas visit.
Mr. Trump arrived in Riyadh besieged by the fallout from his controversial decision to fire FBI Director James Comey and more revelations about the federal investigations into his campaign’s possible ties to Russia. But escaping Washington for the gold-plated embrace of the Saudi royal family — a decor not so unlike Mr. Trump’s own Manhattan home appeared to give the President a boost.
The President was largely kept out of earshot from reporters, rendering them unable to ask about the tumult back home. But he did make a brief utterance to the press pool, deeming the proceedings “a tremendous day.”
Mr. Trump is scheduled to leave Saudi Arabia — home to Mecca, the holiest site in Islam — early on Monday to head to Israel.
|
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Putco is the leader in the chrome trim market for a reason! Putco’s Chrome Trim exceeds OEM specifications with unique, patented designs. These Chrome Trim accessories are created using factory parts and high quality injection mold tooling to produce an exact fit for your vehicle. Putco’s selection of Chrome accessories has an incredible variety of everything you could want for your vehicle, including mirror covers, door handle covers, tail lamp covers, tailgate handle covers, headlamp covers, air dam covers, body side molding, hood decks, and several other accent pieces. Not only does Putco provide Chrome Trim applications for more vehicles than any competitor, but over 50,000 units are kept in stock, making these pieces easily accessible for your needs! With 22 hour Cass Chrome Plating backed with a Lifetime Guarantee, Putco is the perfect choice for your vehicle.
Shipping Weight/Dimensions:
Weight:
2.00 lbs
Height:
8.00 in.
Length:
11.00 in.
Width:
4.00 in.
Shipping from warehouses in the U.S. and Canada!
Performance Parts has over 20 warehouses in the U.S. and Canada so you can get your parts duty free! Call for availability.
|
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New Book Completed
I finished the artwork for a piggy book I’ve been working on since last October. It was a tight deadline! For this book, I continued experimenting with mixed media and this time I incorportated colored pencil. I had never done that before and I liked the results!
Book will be out Summer 2011 and I’ll post more info as I get the OK from the publisher. In the meantime, a little something…
|
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Arrangements described herein relate to regression testing of software that uses structured query language (SQL) statements.
SQL is a special-purpose programming language designed for managing data held in a relational database management system, and has become the most widely used programming language used for this purpose. SQL includes both a data definition language and a data manipulation language. The scope of SQL includes data insert, query, update and delete, schema creation and modification, and data access control. Although SQL is often described as, and to a large extent is, a declarative language, it also includes procedural elements.
|
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Researchers have identified an inflammatory molecule that appears to play an essential role in the autoimmune disorder systemic lupus erythematosus, commonly known simply as lupus. In their report the team describes finding that a protein that regulates certain cells in the innate immune system – the body’s first line of defense against infection – activates a molecular pathway known to be associated with lupus and that the protein’s activity is required for the development of lupus symptoms in a mouse model of the disease.
“This study is the first demonstration that the receptor TREML4 amplifies the cellular responses transmitted through the TLR7 receptor and that a lack of such amplification prevents the inflammatory overactivation underlying lupus,” says Terry Means, PhD.
“Our preliminary results suggest that TREML4-regulated signaling through TLR7 may be a potential drug target to limit inflammation and the development of autoimmunity.”
Lupus is an autoimmune disorder characterized by periodic inflammation of joints, connective tissues and organs including heart, lungs, kidneys and brain. TLR7 is one of a family of receptors present on innate immune cells like macrophages that have been linked to chronic inflammation and autoimmunity. Animal studies have suggested that overactivation of TLR7 plays a role in lupus, and a gene variant that increases expression of the receptor has been associated with increased lupus risk in human patients. The current study was designed to identify genes for other molecules required for TLR7-mediated immune cell activation.
The team conducted an RNA-interference-based genome-scale screen of mouse macrophages, selectively knocking down the expression of around 8,000 genes, and found that TREML4 – one of a family of receptors found on granulocytes and monocytes – amplifies the response of innate immune cells to activation via TLR7. Immune cells from mice lacking TREML4 showed a weakened response to TLR7 activation.
When a strain of mice genetically destined to develop a form of TLR7-dependent lupus was crossbred with a strain in which TREML4 expression was suppressed, offspring lacking TREML4 were protected from the development of lupus-associated kidney failure and had significantly lower blood levels of inflammatory factors and autoantibodies than did mice expressing TREML4.
Means notes that identifying the potential role of TREML4 in human lupus may lead to the development of drugs that could prevent or reduce the development or progression of lupus and another autoimmune disorder called Sjögren’s syndrome, which also appears to involve TLR7 overactivation.
Future studies are needed to better define the molecular mechanism behind TREML4-induced amplification of TLR7 signaling and to clarify beneficial reactions controlled by TREML4 – for example, the immune response to influenza virus, which the current study found was inhibited by TREML4 deficiency.
“Given that only one new drug has been approved for lupus patients in the last 50 years, there is a pressing need for more specific and less toxic drugs to treat it and other autoimmune disorders,” says Means, who is an assistant professor of Medicine at Harvard Medical School.
|
{
"pile_set_name": "Pile-CC"
}
|
This is an initial or manual build of the Apache ActiveMQ - 4.1 Nightly build.
---------------------------------------------------------------------------------------
No tests were executed.
Click http://bamboo.logicblaze.com:8085/browse/AMQ-N41-213 to find out more.
Thanks,
Bamboo
|
{
"pile_set_name": "Pile-CC"
}
|
Bioelectrochemical (BEC) systems (e.g., microbial fuel cells, biofuel cells, electrolysis) may have been used in the past for power generation. In the form of microbial fuel cells (MFCs), these systems may have enhanced biodegradation of hydrocarbons, phenol, and other contaminants perhaps while also generating energy. Bioelectrochemical systems may have also been tested in groundwater and sediments for organic carbon removal while generating electricity. A bioelectrochemical system may include an anode, a cathode, and a circuit (perhaps with or without external resistor). For example, in a 2-chambered MFC, microorganisms in the anoxic/anaerobic zone can transfer electrons to the anode, which are then transferred to the cathode. Oxygen may be reduced to water at the cathode, which may produce a current. These types of BECs may separate the anode and cathode by a cation, proton exchange membrane, a salt bridge, or the like to allow protons to transfer between the anode and cathode while minimizing contact of oxygen with the anode. It has been recognized that a floating type cathode combined with an anode system exposing a cathode to air can be used and the water column can complete the circuit.
None of the past BEC systems have provided a sustainable and “green” method with perhaps little to no load, in the presence or even absence of microorganisms, which may focus solely on enhancing degradation of waste compounds, perhaps organic waste compounds from environments such as water/wastewater, groundwater, and soils/sediments.
|
{
"pile_set_name": "USPTO Backgrounds"
}
|
Menu
And now for the ranting.
I would like to be as sophisticated as Chris, and as brief, but I can’t. So I’ll start with “AAAaargh!” I understand that the Vatican is all about tradition. I understand that homosexuality is a touchy subject for them, especially as it’s turning out that many “celibate” priests are not, and at the expense of kids who are at a severe power disadvantage. Still, the Pope gets to make rules for Catholics, because that’s how the system works. I accept that. BUT…
Now the Vatican thinks that it gets to make rules for everyone, even you and me. So everyone “committed to promoting and defending the common good of society” (which I actually THOUGHT included me), whether Catholic or not, is supposed to oppose homosexual marriage, and not just that, but any legislation that would give gay couples the same rights as married couples.
“Why?” you may ask. Go ahead, ask. Okay, I’ll play Pope and explain. See, kids, homosexuality violates something called “natural moral law.” God, who is omnipotent, didn’t intend for there to be homosexuality. Humans are sinful creatures with free will, and homosexuality is just a terrible perversion of what God wanted, but since we can do what we want, it sometimes happens.
If our all-powerful God intended there to be homosexuality, there would be homosexual animals, right? See, animals don’t have free will, they HAVE to follow God’s laws. And since there are no verified examples of homosexual behavior among, say, oh, I don’t know… penguins, bonobo chimps, whales, giraffes, rodents, geese or bears… (These are NOT just random examples.) …then homosexuality must be unnatural, right? And a violation of this “natural moral law,” right?
Our friends the Italians are fighting the good fight, though, with great protest signs like “Democracy, yes. Theocracy, no.” and are comparing the Vatican to the Taliban. The Italians literally have Vatican City surrounded, so we can hope. And pray.
Post navigation
3 thoughts on “And now for the ranting.”
Yay, ranting! I made a quick comment on my blog about this yesterday, but the huge flood of anti-gay news this morning made me want to scream. I thought about ranting more on my own blog, then guessed (correctly) that there might already be an active discussion over at Globalspin.
I was also a big fan of the Italian signs, and I just hope that the rest of the world is more openminded than the pigheaded nincompoops in Washington and the Vatican. Is Bush really so confident in the next election that he can shamelessly discriminate against a significant portion of the population? I know most wouldn’t vote for him anyway, but it seems like this should mobilize the gay community like never before.
Legislation of the bedroom seems like such a petty, self-righteous thing for the administration to focus on in times like these. It makes me want to slap some sense into the Son of a Bush. Can I say that without instigating a Secret Service investigation?
In a footnote citing a 1992 comment on the topic, the document also noted that there was a danger that laws legalizing same-sex unions could actually encourage someone with a homosexual orientation to seek out a partner to “exploit the provisions of the law.”
. . . quote.
This is so funny on so many levels I don’t even know where to begin. Anyone else get it?
This is a good thing, kids. The more they scream, the more progress we are making. It’s like trying to stop the Big Bang a second too late. It took the Church 300 years to canonize Gallileo and it didn’t make a damn bit of difference in the end. In fact, the Church lost face, big time.
I imagine a time when the Church comes out in favor of gay unions — carefully worded to avoid the “m” word — in a “separate but equal” turn — a hundred years after it’s become globally accepted. Then I imagine a time when the Pope comes out (as gay). Then, later on, the Pope tells us she’s a lesbian. ‘Course the Church has been outcast to the moon Phebos by that time but it hopes to regain some status with the announcement.
|
{
"pile_set_name": "Pile-CC"
}
|
Diseases, conditions, and drugs associated with cicatricial ectropion.
Cicatricial ectropion may be a consequence of certain systemic diseases as well as the result of drug use. Our goal here was to research the different causes of this condition as reported in the literature, including more recently suspected etiologies. A detailed PubMed literature search indicated many different etiologies were associated with cicatricial ectropion development, from severe cases of systemic diseases, such as ichthyosis and lupus erythematosus, to reversible scenarios secondary to anti-glaucomatous drug use. More recently reported connections include periorbital necrotizing fasciitis, frontal osteomyelitis, and antineoplastic agents. Indeed, cicatricial ectropion may be highly symptomatic; being able to determine its real etiology is imperative to managing patients properly. In this investigation, we felt that an explicitly multidisciplinary approach was essential, especially for cases associated with systemic conditions.
|
{
"pile_set_name": "PubMed Abstracts"
}
|
Q:
Again, ObservableCollection doesnt Update item
Thats my first project using MVVM , MVVM light.
I have a listbox, that gets refreshed from the PersonList Observable collection, adding and removing refresh it normal. the problem is when editing an item.
I looked for all the solutions for this problem, nothing worked, which make me think that I missed something.
so here is the code :
public class AdminViewModel : ApplicationPartBaseViewModel
{
private ObservableCollection<Person> personList;
public AdminViewModel()
{
this.context = new Entities();
this.SavePersonCommand = new RelayCommand(() => this.SavePerson ());
this.PersonList = new ObservableCollection<Peson>(context.Person.OrderBy(o => o.PersonName).ToList());
}
public ObservableCollection<Person> PersonList
{
get
{
return personList;
}
set
{
this.personList = value;
RaisePropertyChanged("PersonList");
}
}
private void SavePerson()
{
//Add and update code here
this.context.SaveChanges();
RaisePropertyChanged("PersonList");
}
}
Person Class is Autogenerated template from the DataModel edmx
//------------------------------------------------------------------------------
// <auto-generated>
// This code was generated from a template.
//
// Changes to this file may cause incorrect behavior and will be lost if
// the code is regenerated.
// </auto-generated>
//------------------------------------------------------------------------------
public partial class Person
{
#region Primitive Properties
public virtual int PersonId
{
get;
set;
}
public virtual string PersonName
{
get;
set;
}
public virtual Nullable<int> PersonAge
{
get;
set;
}
#endregion
#region Navigation Properties
public virtual ICollection<Humans> Humans
{
get
{
if (_human == null)
{
var newCollection = new FixupCollection<Human>();
newCollection.CollectionChanged += FixupHuman;
_human = newCollection;
}
return _human;
}
set
{
if (!ReferenceEquals(_human, value))
{
var previousValue = _human as FixupCollection<Human>;
if (previousValue != null)
{
previousValue.CollectionChanged -= FixupHuman;
}
_human = value;
var newValue = value as FixupCollection<Human>;
if (newValue != null)
{
newValue.CollectionChanged += FixupAssets;
}
}
}
}
private ICollection<Human> _human;
#endregion
#region Association Fixup
private void FixupHuman(object sender, NotifyCollectionChangedEventArgs e)
{
if (e.NewItems != null)
{
foreach (Human item in e.NewItems)
{
if (!item.Person.Contains(this))
{
item.Person.Add(this);
}
}
}
if (e.OldItems != null)
{
foreach (Human item in e.OldItems)
{
if (item.Person.Contains(this))
{
item.Person.Remove(this);
}
}
}
}
#endregion
}
I thought that MVVM light update the item when I call RaisePropertyChanged.
I am so confused.
Thanks in advance.
A:
First option is try to get your auto-generated class to implement INPC if you can. Have a look at Fody.PropertyChanged
If that's not possible, since it does have it's properties as "virtual", we can over-ride them in a derived class such as
public class ObservablePerson : Person, INotifyPropertyChanged {
public override int PersonId {
get {
return base.PersonId;
}
set {
base.PersonId = value;
OnPropertyChanged();
}
}
public override string PersonName {
get {
return base.PersonName;
}
set {
base.PersonName = value;
OnPropertyChanged();
}
}
public override int? PersonAge {
get {
return base.PersonAge;
}
set {
base.PersonAge = value;
OnPropertyChanged();
}
}
public event PropertyChangedEventHandler PropertyChanged;
[NotifyPropertyChangedInvocator]
protected virtual void OnPropertyChanged([CallerMemberName] string propertyName = null) {
PropertyChangedEventHandler handler = PropertyChanged;
if (handler != null)
handler(this, new PropertyChangedEventArgs(propertyName));
}
}
Now in your AdminViewModel work with objects of type ObservablePerson than Person
|
{
"pile_set_name": "StackExchange"
}
|
Q:
How do I replace only a specific value in an XML document without changing anything else?
I've been trying to replace a specific value from an xml document in memory before creating a message log of the XML.
I've managed to do a replace, but the Regex replace method seems to replace other items as well.
I've had to make this a little more funky than I would have liked but the elements within the document can contain different XML namespaces...
string pattern = "(<).*?(ElementName>).*?(<\\/).*?(ElementName>).*?";
string replacementPattern = "(<).*?(ReplacedElementName>)xxxxxxxxxxxxxx(<\\/).*?(ReplacedElementName>).*?";
string messageToLog = Regex.Replace(messageToSanitise, pattern, replacementPattern);
Can anyone point out where I'm going wrong?
[Update 16:11 BST 09/08/2013]
Thanks Dash, I tried to do that, but then I realised that the object Contains an xml and isn't actually an xml document itself, looks like the object has some headers, with the xml is within a document envelope. Ideally I don't want to lose any information (including the headers) before logging. There will always be 1 or 2 occurences of the element I am trying to change never more and never less than 1.
A:
Given your xml is in the string messageToSantise, you can try the following:
Using XmlDocument:
(classic XML parsing common in older versions of the framework, and your only choice on older versions)
XmlDocument messageDoc = new XmlDocument();
messageDoc.Load(messageToSanitise);
messageDoc.SelectSingleNode(path_to_node).Value = replacementValue
path_to_node can be used with the appropriate XPath expression.
To get the xml string back out of the XmlDocument, use the messageDoc.OuterXml property.
string messageToLog = messageDoc.OuterXml;
Using XDocument:
(xml parsing via a LINQ style mechanism, supported in new versions of the framework)
XDocument messageDocument = new XDocument();
messageDocument.Parse(messageToSanitise);
messageDocument.Element(path_to_element).value = replacementValue;
To navigate through an XDocument, you may wish to also use the Descendents property. Examples of how to arrive at the node include this answer and the MSDN documentation here.
To get the Xml from the XDocument, use messageDocument.ToString();
string messageToLog = messageDocument.ToString();
This allows you to specify exactly what you want to replace.
If you want to decide whether to use XmlDocument or XDocument, I recommend reading the answer to this question.
|
{
"pile_set_name": "StackExchange"
}
|
Chapter Text
The words hit Buffy like a punch to the gut. Did nobody understand what she was going through? Were these her friends? Were they going to gleefully cheer her on as she turned her boyfriend into ash and dust? Buffy bit her lip and walked away from Xander. Her body was tense. Every muscle flexed. She gripped the sword so hard it hurt. Her knuckles slowly turned white.
When she reached the mansion, she kicked the front door off its hinges without checking whether it was locked or not. A cloud of dust rose from the splinters. Buffy stepped out of the sunlight and crossed the threshold into the candlelit interior. The place reeked of decay and perversity. Horrible deeds had taken place there. Her boyfriend had committed horrible deeds there. No, not her boyfriend. Only the beast who possessed him now.
Buffy heard voices from behind a curtain. She walked over to it and peered into the hall on the other side. Angel stood in front of the large demonic statue of Acathla. Beside him was the frail shape of Kendra’s killer. Drusilla cradled a dead puppy in her arm and stroked it tenderly as if it was alive.
A pair of vampire goons stood guard. Buffy snuck up behind the closest one and decapitated him with a deft swing of her sword.
“Hello, lover,” Buffy said to Angel, as his lackey’s smouldering head crashed against the ground.
His words were cut off. Spike had risen from his wheelchair and was now pounding his grandsire with an iron bar. The attack took Angel completely by surprise. Spike was finally able to vent his oedipal rage. The battle could have been over quickly if it had not been for Drusilla. She was clearly unamused by her paramour’s betrayal. Her waiflike body pounced upon Spike and pushed him away from Angel. The slayer killers went on each other like rabid animals.
One of Angels’s henchmen pushed Buffy against a wall. Xander came charging into the room. The vampire turned to face him, giving Buffy the chance to stake it. Buffy and Xander exchanged quick looks, before Xander headed into the other room to rescue Giles.
Buffy heard Angel laugh triumphantly behind her. While she had been distracted, he had walked up to the statue. His bleeding hand withdrew the sword from its stony chest. A magical current of blue lightning passed from the statue to Angel’s body, possibly infusing him with some of the demon’s power.
Buffy looked him dead in the eye and tried her best to remain calm. “Save me a seat,” she said. Her hand shook.
Buffy charged forward. Her sword cut through the air. Her blade clashed with Angel’s. Sparks flew. She was not fighting with all she had. She knew she advertised her moves, because she could not bear the idea of letting her sword hit its mark and let it rend the flesh of … It was too painful to even think of. Angel expertly feinted all her thrusts, and she allowed it. It could not last. Angel became confident. She aimed too low. He stepped on her blade and gave her an elbow punch to her face. It sent her flying backwards, until she hit the wall hard.
Angel laughed. “Now that’s everything, huh?” he teased. “No weapons... No friends... No hope. Take all that away and what’s left?”
Buffy watched Angel lift his sword. Something started to build up inside her. Rage. The unfairness of everything she had gone through had become too much. She was putting her foot down. She would accept no further indignities. It was over. Angel’s blade shot towards her. Buffy closed her eyes and caught it between her hands. The friction burned her palms, but it stopped.
“Me,” she said. “I’m still here.”
She pushed the sword back towards Angel, hitting him in the face with the pommel. Angel toppled backwards. Buffy pounced upon his chest and started pounding him with her fists.
“Come back,” she yelled. “Come back from there!”
Angel’s hideously contorted face just laughed at her. His fangs gleamed. She hit him harder. He only grinned. She felt his hands on her. They tried to drag her down towards him. He would kill her unless she killed him. It was now or never. It was too late to wait for a miracle.
That was when it happened. Angel’s eyes flashed. His fangs vanished and his face smoothened. The smirk was gone, replaced by a look of confusion.
“Buffy?”
“Angel?”
“What’s happening?” Angel’s expression was full of pain, but he tried to smile. “It feels like I haven’t seen you in months.”
Buffy flung her arms around him and pulled him close. Her body shook against his. She was too tense to cry freely, so all she did was softly sob. The air around them began to whirl. Buffy felt her hair lift. Something pulled at her. Buffy opened her eyes and saw a twirling vortex growing in front of the statue. Eldritch screams emerged from within.
“Buffy!”
Buffy turned around. Giles stood behind her, supported by Xander. His body bore the marks of the torture he had endured.
“Buffy, you have to do it,” Giles said.
“I won’t,” Buffy said through clenched teeth.
“Look at what they’ve done to me,” Giles yelled. “Worse than this will happen to everyone on Earth if that portal is allowed to open.”
Buffy looked away. She knelt down and picked up Angel’s sword. It felt heavy, clumsy, unbalanced. Her hand could barely swing it.
“What are you doing?” Angel asked.
“Shush,” Buffy cooed. “Don’t worry.” She pulled him in for a kiss. Their faces were wet with tears. She felt that Angel was awakening to the fact of what she was about to do, but he did not fight her. “Close your eyes,” she begged him.
Angel did as he was told. His lids fell shut, but when Buffy drove her sword into his heart, they sprang up again. His face twisted into an expression of pained confusion. There was no reproach. He just wanted to know why. Spiderlike arms stretched out from the statues mouth and dug into Angel’s skin. They pulled and clawed at him, dragging him towards the opening portal. He stretched his hand out, perhaps hoping against hope that she would save him.
Buffy felt a hand on her shoulder. “Avert your eyes, Buffy,” Giles said to her. “You should not have to look.”
Buffy stood silent as stone. Her mind was numb. Then she laughed. She could not explain why she laughed. Maybe because she had made a choice and making it felt good. She grabbed Angel’s outstretched hand and held on. The vortex span ever faster. The arachnid arms tugged at Angel. Buffy heard Giles call out to her, but his words were lost. A wind blew into her face. The vortex enveloped her. She tried to look at Angel, but her eyes were watering. Everything was blurry. The room vanished. The wind trashed her around in a whirl. All she could do was hold on to Angel’s cold hand.
****
Liam opened his eyes slowly, before hiding his face beneath his pillow. He was not ready to meet the sunlight that crept through the rafters yet. A numbing headache crippled his brain. He remembered nothing from the night before. Not that it mattered. It had likely been a night like any other night. The turmoil in his stomach told him he had been drinking … a lot. Keeping it all contained was a struggle.
There was a tankard on his nightstand. He must have brought it with him from the tavern. He needed some hair of the dog to set his mind straight. He sat up and took a deep swill. It tasted like hot piss. He knew it would be warm and that it would taste flat, but he had not expected it to be quite so revolting. It almost made him puke.
There was a loud knock on the door.
“I am not coming out,” Liam yelled. “I don’t feel well.”
“And I know the reason why!” The door swung open and his father stormed in. “Up again all night, is it? Drinking and whoring!”
Liam glared at him. “And a good morning to you, father.”
His father spat on the floor. “You’re a disgrace.”
Liam shrugged. “If you say so, father.”
“Oh, I do. I do say so,” his father sneered. “I am ashamed to call you my son. You’re a lay-about and a scoundrel and you’ll never amount to anything more than that.”
The words tore into Liam’s gut and gnawed at him from the inside. He felt an urge to explain himself or to promise he would do better, but he knew it was in vain. There was nothing he could do to save this.
“If that is how you feel, father,” Liam mumbled. “Then why don’t you throw me out?”
His father pointed a shaky finger at him. “I will…”
“Too late,” Liam laughed. “I am already on my way.”
He hoisted up yesterday’s trousers and rebuttoned the shirt he was still wearing. Why had he said that? Was he still drunk? It felt as though he had little control over his actions, as if he was merely following a script or acting in a familiar pattern. He pushed his father away from the door and stumbled past him.
His sister looked at him with sad eyes. Liam knelt down before her and stroked her chin. “Sweet Kathy. No tears. We’ll meet again,” he whispered and kissed her forehead.
His father came running after him. “If you leave now, don’t ever expect to come back.”
Liam rose to his feet and rolled his eyes. “As you wish, father. Always, just as you wish.”
“It’s a son I wished for … a man. Instead, God gave me you! A terrible disappointment.”
Liam laughed. “Disappointment? A more dutiful son you couldn’t have asked for. My whole life you’ve told me in word, in glance, what it is you required of me, and I’ve lived down to your every expectations, now haven’t I?”
“That’s madness!”
“No. The madness is that I couldn’t fail enough for you. But we’ll fix that now, won’t we?”
His father backed away. “I fear for you, lad.”
“And is that the only thing you can find in your heart for me now, father?”
“Who’ll take you in, huh? Who beside your family would ever love you? No one!”
“I’ll not lack for a place to sleep, I can tell you that.”
Liam left the house and stumbled down the cobble path. On his way, he glimpsed his father’s maid. She smiled hopefully at him. The sight of her blushing cheeks sent a shiver of pain through Liam. Who was he to play with the hopes of a young girl struggling desperately to get by? Could he not limit himself to those women whose reputation was already ruined? He ran. He could not bear to look at her.
The setting sun burned into his neck. It felt as though its searing light was judging him. He felt uncomfortable and exposed. Daytime gave him no place to hide. He yearned for shadows and dark corners.
The path brought him to the tavern. He had not planned to go there, but his legs had probably become accustomed to the journey. The inside of the building reeked. The air was filled with smoke. The floor smelled of piss, puke and spilled ale. His friends lifted their tankards to him in greetings. Their ugly grins frightened him. The remaining teeth they had were rotted and discoloured.
The women begun swarming around them. Their hair fell in greasy tangles over their faces, as they leant forward to offer Liam a good view of their bosoms. They knew he had money and that he was generous with it. Behind their girly smiles, Liam thought he glimpsed womanly contempt. They hated him. Unmarried women had few options for supporting themselves. Liam tried to look away, but he found that he could not. He found their appearance revolting and their situation pitiable, though worst was the knowledge of how he had exploited their desperation.
The tavern kept filling up. The men and women seemed to be circling around Liam. Everyone wanted a piece of him. His friends brought him drinks he could not keep himself from drinking. They laughed at their own jokes. Their filthy beards scratched his chin as they hugged him. A woman pushed a grape into Liam’s mouth. A hand clawed at his chest. Someone groped the inside of his thigh. Everywhere he turned, ugly faces grinned back at him, shouting for him to enjoy himself. He could not. He wanted to run. His body wanted to puke.
Just as the monastic atmosphere was about to overpower him, an inexplicable sting of trepidation pricked Liam’s senses. Someone was studying him from afar. He did not know how he knew. He just felt it. A woman sat in the corner, asking questions about him, making plans for his future. Her sweet face was a mask for the demon underneath. Liam felt two pins prick against his neck. It was just a memory, but in his feverish confusion, he could not distinguish memory from reality.
“I need to go,” he mumbled to his friends and darted out the door.
The woman followed him. Her white dress trailed over the muddy cobblestone street. Liam wanted to run, but his legs were stiff and heavy. She gained on him. What would happen was inevitable. It had happened before and now it would again. Her teeth would rip his neck and bleed his humanity, force him to even greater lengths of depravity. Kathy … he was going to kill Kathy. He fell forward and hit his knees against the hard stone. The woman was right behind him. Her shadow fell across the ground beneath him.
“Please, Darla,” Angled begged. “Not this time … not again. I cannot go through it all again.”
“Hello, lover.”
The words jolted Liam’s memory. He had heard those words recently, spoken from the same lips, though in a harsher tone than now. Now they sounded soft. Now they sounded inviting. He looked up and saw a radiant beauty looking down upon him.
“Buffy?”
Angel watched Buffy take his hand and kiss it. “I came for you,” she said.
“Why?”
Buffy squatted down so their eyes were on level with each other. She held both his hands now. Her lower lip shook. “Because,” she started. Then she swallowed. “Because I could not let you go, knowing I would be the only one who knew what a man you were.” She sniffed. “I would be the loneliest girl in the world.”
Buffy kissed Angel on the forehead. Her moist lips left a wet imprint. Angel felt his heart sink. It was not right. She should not be here. This punishment was his. How was she even here? She wore the same dress as Darla had worn. Her hair was done up in similar curls. Was she a mirage? Had the demons created this apparition to tease him?
“Don’t you dare say you’re not happy to see me,” Buffy warned him. “I came after you willingly.”
“We’re in Hell,” Angel said. “You’ve followed me to my Hell. What have I made you do?”
Buffy smiled. “How does that old Irish saying go?” she mused, before pushing the sleeve of her dress off her shoulder. The opening in her dress widened until just before her right nipple was about to slip out. “Let’s try to have half an hour of fun before the Devil realizes what we are up to.”
“This is not how it should be,” Angel protested.
Buffy laughed. “After all that we have been through,” she said, “I can settle with little.”
“It is all my fault…”
“… and mine.”
****
Buffy pushed Angel onto his back. It did not matter that the dusty ground was wet and muddied. For months, she had fought, wept and done her best to suppress her feelings. Now she was desperate for some human contact. She needed to be loved. She needed to disappear into somebody else.
Buffy peeled her dress off her other shoulder and pulled out her arms. The bodice fell to her waist as she wriggled herself free, making her look like a mermaid. The wind was cold against her skin. This time she would not be as coy as she had been the first time their bodies met. She let him stare at her appearing breasts without blushing. It was good to let her defences down and enjoy Angel’s admiring scrutiny.
She put her hand into his lap and immediately felt something inside his pants respond to her touch. A bulge quickly rose to fill her hand. Buffy felt it hunger for her, begging her to cast off its restraints and let it out. Her fingers made their way underneath the waistline and pulled Angel’s pants down to his boots. She took his towering shaft in her hand and lowered herself onto it. This would take time. She would have to tease herself with the edge of his member until she was ready to receive it in full. Buffy bit her lip as her insides widened to accommodate him.
It was different this time. Angel was like Liam now. He was warm. Heat built up inside her. Angel gripped her hair. He clawed at it, gathering more and more between his fingers. She bent down to kiss him and he sat up to meet her. How hard would she have to bite to bring a scream out of him? Very hard, it seemed.
Her hands made their way underneath his shirt. Her palms stroked across his muscled abdomen, her fingers climbed up his ribcage, until she found her way up to his flexed chest. She tore open his shirt from the inside, exposing his naked body to the rain that suddenly started showering down on them. His flesh was so pale. Shadows danced in the outlines around his muscles. Buffy grabbed Angel’s chest again, digging her nails into him. He responded by cupping both of her breasts.
She stared into his eyes, and was able to make him meet her gaze. “Tell me you’re not happy I came,” she whispered.
“I can’t,” he gasped. The tiniest smile formed on the edge of his mouth.
“Good boy,” she said, as she leant in to kiss him again.
Buffy could sit like this forever, tugging at Angel’s lips, feeling the pressure of his member inside her, just enjoying the anticipation until her mind went crazy. Angel was not so patient. Before too long he started thrusting upwards and Buffy found herself being rocked up and down. Once she got used to the sensation, she wanted more and begun grinding against his lap.
Angel started to groan. His groans turned to moans. They grew fainter and fainter, until he shivered underneath her and Buffy knew it was over. She was sad, because she had been so close to joining him. Then something caused her body to spasm. Angel’s thumb had found its way to a wet and tender spot in her sex. He prodded it gently but insistently. Buffy’s body wanted to scream with joy, but her lungs provided no air for her voice. It felt so good to give in and let her body shake as a wave of ecstasy passed through.
She rolled down from Angel’s limp member and cuddled up in his arm. Their breathing was slow and strong. Everything felt right. All the anger and fear had vanished. The trials of the past months seemed an eternity away.
“I may lose my soul again,” Angel mumbled.
“It doesn’t matter,” Buffy gasped. “I can take you.”
Another spasm shook through Buffy’s legs and her lips let out an embarrassing squeak. Angel laughed.
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Dose-related neuropathic and anti-neuropathic effects of simvastatin in vincristine-induced neuropathic pain in rats.
The present study explores the role of simvastatin in vincristine-induced neuropathic pain, which was induced by administering vincristine (100 µg/kg i.p.) for 10 days (two 5 day cycles with 2 days pause). Pain was assessed by determining mechanical hyperalgesia, mechanical dynamic allodynia, heat hyperalgesia and cold allodynia. Biochemically, myeloperoxidase (MPO) activity was measured along with serum cholesterol levels. Simvastatin (7.5, 15 and 30 mg/kg) was administered for 14 days after administration of vincristine. Simvastatin (7.5 and 15 mg/kg) reversed vincristine-induced neuropathic pain and attenuated vincristine-induced increase in MPO, without altering cholesterol levels. Simvastatin at higher dose (30 mg/kg) did not alter neuropathic pain despite decreasing MPO levels. Furthermore, administration of simvastatin (30 mg/kg i.p.) in vincristine treated rats as well as it's per se administration in normal rats reduced cholesterol levels. Per se administration of simvastatin in normal rats produced neuropathic pain. It is concluded that simvastatin attenuates neuropathic pain only at lower doses with no reduction in cholesterol levels and anti-inflammatory effects may possibly reverse neuropathic pain. However, despite reducing inflammation, simvastatin did not confer beneficial effects at higher doses at which there is reduction in cholesterol levels, suggesting the critical role of cholesterol in neuropathic pain induction.
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Hi! I am Dan the Magic Man. At age 14, I live in Arlington Heights, Illinois with my dad, mom, sister, and brother. I have been interested and have been practicing magic for almost all my life. I now have two years of experience working as a professional magician. I love performing for kids to adults of all ages whether it be in a small house or a big stage. With 50+ shows of experience (and counting), I would love to perform at your next party or gathering.
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Recruitment for Graduate Export Sales Officer at Dangote Group, Monday 5th March 2018 – The Dangote Group is one of Nigeria’s most enhanced assembling combinations.
The Group keeps on developing its vision of turning into the main supplier of fundamental needs in Food and Shelter in Sub-Saharan Africa with supported market authority in Cement Manufacturing, Sugar Milling, Sugar Refining, Packaging Material Production and Salt Refining.
EXPORT SALES OFFICER
JOB RESPONSIBILITIES
Attracting new customers and managing existing customers.
Plan and co-ordinate the international shipment of goods.
Negotiate with a variety of people such as shippers, agents and vendors. He/she is expected to have excellent customer service skills in dealing with customers.
Export documentation and communicate with forwarders associated with the shipment process to ease order volume flow.
Keep track of invoices and prepare reports to expedite the billing process.
Ensure that shipment is in compliance with the laws and regulations governing the export industry.
Accurate and timely information dissemination and reporting.
Preparation of district sales forecast and periodic reviews of sales objective.
Carrying out qualitative and quantitative market survey on NASCON and competitive products (including new entrants).
Ensure strict adherence to NASCON safety policy guidelines.
QUALIFICATION/EXPERIENCE Minimum HND/B.Sc. Degree in relevant field. A professional sales qualification is an added advantage. 1-3 years’ work experience Must be able to communicate professionally in French, English, Hausa or Yoruba.
COMPETENCE REQUIREMENT:
Take ownership of the overall Market segmentation process and the implementation of agreed strategies on an overall basis.
Formulate and implement tactical and strategic sales plans throughout the targeted business segments with the greatest potential to improve profitability, positioning and market share.
Direct responsibility for achievement of plans and targets, including operational measures such as Market share, Earnings, Gross margin contribution.
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Known for their militaristic and disciplined culture, thewere the third race to join the Citadel Council . They gained their Council seat after defeating the hostile krogan for the Council during the Krogan Rebellions . The turians deployed a salarian -created biological weapon called the genophage , which virtually sterilised the krogan and sent them into a decline. The turians then filled the peacekeeping niche left by the once-cooperative krogan, and eventually gained a Council seat in recognition of their efforts.
Originally from the planet Palaven, turians are best known for their military role, particularly their contributions of soldiers and starships to the Citadel Fleet. They are respected for their public service ethic—it was the turians who first proposed creating C-Sec—but are sometimes seen as imperialist or rigid by other races. There is some animosity between turians and humans, largely due to the turian role in the First Contact War. This bitterness is slowly beginning to heal—as shown by the cooperation of the two races on the construction of the SSV Normandy—but many turians still resent humans, and vice versa.
Contents show]
Biology Edit
Turians typically stand over six feet tall, have two long, proportionately thick fingers and an opposable thumb on each hand, each tipped with talons, and a set of mandibles around their mouths. The most distinguishing feature of turians is their metallic carapace, which contains trace amounts of thulium . The turians evolved this trait as a defense against the greater levels of solar radiation that penetrate their homeworld 's weak magnetic field.
Turian features are avian, making them resemble humanoid birds or raptors. However, unlike most Earth avian creatures, turians are viviparous and give birth to live young.[1] In 2165, David Anderson claimed that turians reminded him of the evolutionary link between birds and dinosaurs. Turians are also recognisable by their voices, which have a distinctive flanging effect. Males and females do not differ greatly in physical appearance, but female turians lack the crest of horns found in the males of the race. The lifespan of a turian is comparable to that of a human.[2]
Turians exhibit the characteristics of predators rather than those of prey species (compare to krogan biology). Their forward-facing alert eyes give the impression that they possess outstanding eyesight and their teeth and jaws mimic the structures possessed by apex predators such as crocodiles or ancient, carnivorous dinosaurs. Needless to say, their talons on both their feet and hands seem capable of ripping flesh. As such, their diet is primarily meat-based. Their slender bodies also seem to suggest that they are also capable of moving at high speeds.
The turian homeworld, Palaven, has a metal-poor core, generating a weak magnetic field and allowing more solar radiation into the atmosphere. To deal with this, most forms of life on Palaven evolved some form of metallic "exoskeleton" to protect themselves. Their reflective plate-like carapace makes turians less susceptible to long-term, low-level radiation exposure, but they do not possess any sort of "natural armor". A turian's thick skin does not stop projectiles and directed energy bolts. They can, however, experience the equivalent of a "massage": turians can feel vibrations through the carapace with the use of a hammer.
Turian blood has a dark blue colouration.
Although life on Palaven is carbon-based and oxygen-breathing, it is built on dextro-amino acids. This places the turians in a distinct minority on the galactic stage; the quarians are the only other sapient dextro-protein race. The food of humans, asari, or salarians (who evolved in levo-amino acid-based biospheres), will at best pass through turian systems without providing any nutrition. At worst, it will trigger an allergic reaction that can be fatal if not immediately treated.
The turian mechanic Lilihierax on Noveria uses the idiom, "if you can polish enough gizzard". This suggests that the turians have a digestive system similar to birds and reptiles on Earth, some of whom swallow stones to help break down harder foods in the stomach.
History Edit
Turian civilization spans fifteen thousand years of history. Before the dawn of their civilization, the race was known to elder spacefaring species like the Protheans, who viewed them as primitive as the other ruling races of the modern era.
The Unification War Edit
For list of turian colonies involved in the Unification War, see UNC: Turian Insignias.
The turians had already discovered several mass relays and spawned colonies throughout the galaxy when the asari reached the Citadel. At about the time the asari were forming the Council with the salarians, the turians were embroiled in a bitter civil war next door. The Unification War, as it was later named, began with hostilities between the colonies furthest from the turian homeworld, Palaven.
These colonies were run by local chieftains, many of whom had distanced themselves from the Hierarchy. Without the galvanizing influence of the government, the colonies became increasingly isolated and xenophobic. Colonists began wearing emblems or facial markings to differentiate themselves from members of other colonies and open hostilities became common.
When war finally broke out, the Hierarchy maintained strict diplomacy and refused to get involved. After several years of fighting, less than a dozen factions remained and the Hierarchy finally intervened. By that time, the chieftains were too weak to resist; they were forced to put an end to fighting and renew their allegiance to the Hierarchy. Though peace was restored, it took several decades for animosity between colonists to fade completely. To this day, most turians still wear the facial markings of their home colonies.
The Krogan Rebellions Edit
In the midst of the Krogan Rebellions, the Citadel Council made first contact with the turians. At the Council's behest, the turians brought their considerable war machine to bear on the krogan, now a recognized threat. While the initial turian offensive was successful in routing many krogan warrior bands, it provoked a massive counterattack from the krogan which devastated several turian colonies. Three turian worlds were rendered completely uninhabitable after the krogan used fusion torches to throw asteroids at them, and the bloodiest battle in turian history occurred at Digeris, where the planet was severely bombarded and the turians sacrificed many frigates and fighters to take out a fleet of krogan dreadnoughts. Rather than scaring off the turians with this show of force, the turians only fought with more resolve to quash the krogan utterly. Eventually, the turians implemented the salarian-developed genophage. With their advantage in numbers removed, the majority of krogan were subdued by 800 CE, although scattered insurgent actions would continue for decades.
By 900 CE, the turians were granted full membership on the Citadel Council in gratitude for their service during the Krogan Rebellions. The turian military fills the military and peacekeeping niche left by the decimated krogan.
Relay 314 Incident Edit
In 2157 CE, following Council laws in place since the Rachni Wars which prohibited the activation of uncharted mass relays, a turian force opened fire on explorers from an as yet unknown race: humanity. One human starship managed to escape and warn the Systems Alliance, which retaliated and destroyed several turian vessels. The situation quickly escalated to war.
Over the next several weeks, the outnumbered Alliance lost multiple scouting parties and patrols to turian offensives. The conflict came to a head when a turian fleet broke through Alliance lines and besieged the human colony of Shanxi . With no other options, the Alliance garrison on Shanxi surrendered, and the turians proceeded to occupy the world, confident that the majority of Alliance forces had been defeated. However, one month later the Alliance's Second Fleet caught the turian occupiers by surprise and evicted them from the planet. Both sides began preparations for a protracted interplanetary war.
Before that could happen, the Citadel Council intervened and revealed the galactic community to humanity. Terms of peace were negotiated and the conflict effectively brought to an end. The turians were ordered by the Council to give heavy reparations to the Alliance for their part in instigating the conflict, known to the galaxy as the "Relay 314 Incident". Mistrust between both races would linger for years to come.
The Reaper War Edit
During the Reaper invasion in 2186, the turian colony of Taetrus is one of the first worlds the Reapers attack following their conquest of Khar'shan and Earth. The Turian Hierarchy made two attempts to liberate Taetrus, but were unsuccessful. As the Reapers began to pour into the Trebia system and assault Palaven, they broadcasted images of Taetrus's destruction to the turian comm buoy network.
The Reapers met with heavy resistance from the turians during their invasion of Palaven and Menae ; much of the turian fleet remained operable after the Reapers' initial assault, and the turian citizenry was heavily armed and capable of supporting turian troops. Although the Hierarchy maintained that Palaven had not fallen and the battle for it continued, the Reapers nonetheless made significant gains and turian casualties rapidly mounted.
Relief came with the help of an unlikely ally: the krogan, who had agreed to join the war once the genophage was cured. The combined turian and krogan counterattack caught the Reapers off-guard. While the Reaper fleet orbiting Palaven was distracted by an apparent turian offensive, transport craft carrying krogan reinforcements landed on Palaven and coordinated with turian resistance forces, handing over warp bombs and fission weapons. These weapons were smuggled aboard Reaper ships and detonated simultaneously across the globe, allowing large swathes of territory to be retaken. News of the victory gave a much-needed boost to the morale of the turian resistance and the galactic public, but it was not long before the Reapers retaliated.
Realizing the hopelessness of the situation, Primarch Victus ordered the remaining turian warships to withdraw from the Trebia system in order to participate in the Allied assault on Earth. The only way to end the war was to activate the Crucible, and doing so required the Citadel, which the Reapers had moved to Earth orbit for safekeeping. Turian forces heroically assisted in the space and ground battles, while Commander Shepard reached the Citadel to trigger the Crucible.
Culture Edit
Since the Unification War, turians normally wear elaborate tattoos[3] marking their colony of origin, though it is not known which markings distinguish which colony or if color has any meaning. These markings are usually white — particularly on turians with darker carapaces — but can be of other colors such as blue for Garrus Vakarian or red for Nyreen Kandros. The lack of facial markings is looked down upon in turian society; the turian term "barefaced" refers to one who is beguiling or not to be trusted. It is also a slang term for politicians.
Skin tones similar to Mordin Solus' own are apparently attractive by turian standards. For male turians at least, complimenting a potential partner's waist or head fringe seems to be a way of expressing attraction.
Turians are noted for their strong sense of public service. It is rare to find one who puts his needs ahead of the group. Every citizen from age 15 to 30 serves the state in some capacity, as anything from a soldier to an administrator, from a construction engineer to a sanitation worker. Turians have a strong inclination toward public service and self-sacrifice, so they tend to be poor entrepreneurs. To compensate, they accepted the mercantile volus as a client race, offering protection in exchange for their fiscal expertise.
Turian society is highly regimented and very organized, and the species is known for its strict discipline and work ethic. Turians are willing to do what needs to be done, and they always follow through. They are not easily spurred to violence, but when conflict is inevitable, they only understand a concept of "total war." They do not believe in skirmishes or small-scale battles; they use massive fleets and numbers to defeat an adversary so completely that they remove any threat of having to fight the same opponent more than once. They do not exterminate their enemy, but so completely devastate their military that the enemy has no choice but to become a colony of the turians. It is theorized that another conflict between the rapidly advancing humans and the turians could annihilate a large portion of known space.
The turian military is the center of their society. It is not just an armed force; it is an all-encompassing public works organization. The military police are also the civic police. The fire brigades serve the civilian population as well as military facilities. The corps of engineers builds and maintains spaceports, schools, water purification plants, and power stations. The merchant marine ensures that all worlds get needed resources.
Other species see turians as "men of action," and they are generally regarded as the most progressive of the Citadel races (though some species believe humans are rivalling this position). Since their culture is based on the structure of a military hierarchy, changes and advances accepted by the leadership are quickly adopted by the rest of society with minimal resistance.
While turians are individuals with personal desires, their instinct is to equate the self with the group, and to set aside all personal desires for the good of all. Turians are taught to have a strong sense of personal accountability, the 'turian honor' that other races find so remarkable. Turians are taught to own every decision they make, good or ill. The worst sin they can make in the eyes of their people is to lie about their own actions. Turians who murder will try to get away with it, but if directly questioned, most will confess the crime.
Economy Edit
The turian economy is vastly larger than that of the Alliance, but cannot match the size and power of that of the asari. For many years, development was hampered by cultural disinterest in economics. When the turians accepted the volus as a client race, business development improved.
The military is supported by a well-developed infrastructure. Manufacturers such as Armax Arsenal and the Haliat Armory produce advanced, reliable equipment. Volus manufacturers have been known to produce cheap knock-offs of turian equipment.
Religion Edit
Turians believe that groups and areas have "spirits" that transcend the individual. For example, a military unit would be considered to have a literal spirit that embodies the honor and courage it has displayed. A city's spirit reflects the accomplishments and industry of its residents. An ancient tree's spirit reflects the beauty and tranquility of the area it grows within.
These spirits are neither good nor evil, nor are they appealed to for intercession. Turians do not believe spirits can affect the world, but spirits can inspire the living. Prayers and rituals allow an individual to converse with a spirit for guidance or inspiration. For example, a turian who finds his loyalty tested may appeal to the spirit of his unit, hoping to reconnect with the pride and honor of the group. A turian who wishes to create a work of art may attempt to connect with the spirit of a beautiful location.
Turians enjoy absolute freedom of religion and can practice whatever appeals to them so long as it does not impede anyone's ability to perform their duties. There are many practitioners of the asari siarist philosophy. Since opening dialog with the human Systems Alliance, some turians have embraced Confucianism and Zen Buddhism.
In the past, turians believed that titans strode across Palaven, reaching for the heavens. They worshiped these deities and communicated with them at a structure called Temple Palaven. The temple was tended to by a religious order called the Valluvian Priests, who wear special purple robes which obscure their forms. In order for turians to join this order, they had to be considered worthy enough through some action. When the turians spread out from Palaven and discovered other life among the stars, however, they sealed Temple Palaven because they no longer needed legends to prod them upward. With the temple abandoned, eventually the Valluvian Priests fell into legend.
Government Edit
The turian government, known as the Turian Hierarchy, is a hierarchical meritocracy. While it has great potential for misuse, this is tempered by the civic duty and personal responsibility turians learn during their childhood.
Turians have 27 citizenship tiers, beginning with civilians (client races and children). The initial period of military service is the second tier. Formal citizenship is conferred at the third tier, after boot camp. For client races, citizenship is granted after the individual musters out. Higher-ranked citizens are expected to lead and protect subordinates. Lower-ranking citizens are expected to obey and support superiors. Promotion to another tier of citizenship is based on the personal assessment of one's superiors and co-rankers. At the top are the Primarchs, who each rule a colonization cluster. The Primarchs vote on matters of national importance. They otherwise maintain a "hands-off" policy, trusting the citizens on each level below them to do their jobs competently.
Throughout their lives, turians ascend to the higher tiers and are occasionally "demoted" to lower ones. The stigma associated with demotion lies not on the individual, but on those who promoted them when they weren't ready for additional responsibility. This curbs the tendency to promote individuals into positions beyond their capabilities. Settling into a role and rank is not considered stagnation. Turians value knowing one's own limitations more than being ambitious.
Turians enjoy broad freedoms. So long as one completes their duties, and does not prevent others from completing theirs, nothing is forbidden. For example, there are no laws against recreational drug use, but if someone is unable to complete their duties due to drug use, their superiors step in. Judicial proceedings are 'interventions.' Peers express their concern, and try to convince the offender to change. If rehabilitation fails, turians have no qualms about sentencing dangerous individuals to life at hard labor for the state.
The turian imperial anthem is called "Die for the Cause."
Military Edit
Although they lack the brutality of the krogan, the refined biotic skill of the asari , and the adaptability of the humans , the turian military has formidable discipline. Officers and NCOs are "lifers" with years of field experience. Enlisted personnel are thoroughly trained and stay calm under fire. Turian units don't break. Even if their entire line collapses, they fall back in order, setting ambushes as they go. A popular saying holds: "You will only see a turian's back once he's dead."
Boot camp begins on the 15th birthday. Soldiers receive a year of training before being assigned to a field unit; officers train for even longer. Most serve until the age of 30, at which they become part of the Reserves. Even if they suffer injuries preventing front-line service, most do support work behind the lines.
Biotics are uncommon. While admired for their exacting skills, biotics' motives are not always fully trusted by the common soldier. The turians prefer to assign their biotics to specialist teams called Cabals.
Command and control is decentralized and flexible. Individual squads can call for artillery and air support. They make extensive use of combat drones for light duties and VI-controlled fighters, and practice combined arms: infantry operates with armor, supported by overhead gunships. Strategically, they are methodical and patient, and dislike risky operations.
Tradition is important. Each legion has a full-time staff of historians who chronicle its battle honors in detail. The oldest have records dating back to the turian Iron Age. If a legion is destroyed in battle, it is reconstituted rather than being replaced.
The turians recruit auxiliary units from conquered or absorbed minor races, like the volus. Auxiliaries are generally light infantry or armored cavalry units that screen and support the main battle formations. At the conclusion of their service in the Auxiliaries, recruits are granted turian citizenship.
Turian wars are often marked by citizen resistance. Most turian families keep small arms in their homes and take basic training courses that include instruction on how to create simple anti-vehicle explosive devices. To suppress citizen militias, the Turian Hierarchy makes use of "execution squads" known as hastatim. First, "safe camps" are established in cities to incentivize surrender. Next, hastatim soldiers are deployed door-to-door; anyone who refuses to be transported to a safe camp or demonstrates hostile intent will be shot. Hastatim burial units then retrieve and cremate the bodies. This approach is necessary because without the safe camps, no turian would ever surrender, and without the hastatim, it would take years for a population to be pacified.
The mainstay of the turian infantry is the Phaeston assault rifle, a light, accurate, and versatile weapon that nonetheless packs more punch than other rifles of its size. Other turian weapons include the Krysae anti-materiel sniper rifle, and the ML-77 Missile Launcher, manufactured by Armax Arsenal, one of the turian military's main suppliers. Vehicles the turians employ include the A-61 Mantis Gunship, a versatile multi-role aircraft, the C77 Tyrus, a durable 13-ton infantry fighting vehicle, an APC variant of the M-080, the Jiris Infantry Fighting Vehicle, a hovercraft capable of traversing most terrains and engaging enemies at 20 kilometers with its missiles.
The turian navy is divided into at least 32 fleets, and is allotted more dreadnoughts by the Treaty of Farixen than any other race; the turians possessed 37 dreadnoughts in 2183 CE and 39 as of 2185 CE, and in 2186 CE the Turian Hierarchy and the Vol Protectorate were jointly gifted the dreadnought Kwunu by the Elkoss Combine . The turians are also known to possess at least two fighter carriers . The navy serves as a galactic peacekeeping force, and is also the primary military arm of the Council, contributing the single largest portion of the Citadel Fleet
Notable units of the turian military include the 26th Armiger Legion, 79th Flotilla, Sixth Fleet, 43rd Marine Division, Seventh Fleet, and Blackwatch.
Known Military Actions Edit
Notable Turians Edit
Turian Worlds Edit
Trivia Edit
Turian is based on the word centurion. Palaven is based on Palatine Hill, as well as the word "paladin." Turian names, culture, and military doctrine also mirror that of the Roman Empire, especially their emphasis on colonizing enemies.
The turian culture of public service was inspired by the Terran Federation, the human society described in Robert Heinlein's novel Starship Troopers .
. Female turians did not have game models until the release of the Mass Effect 3: Omega DLC, five years after the first game. This, at least in the original Mass Effect, was because there was insufficient development time and memory budget to support two different versions of the same species. Another reason, according to Art Director Derek Watts, was the simple question of how to differentiate their faces from the males. [4] Abrudas, from the comic Mass Effect: Evolution, is the first female turian to be visually depicted in the series.
Abrudas, from the comic Mass Effect: Evolution, is the first female turian to be visually depicted in the series. According to The Art of Mass Effect, turian ships incorporated layers of plates to roughly symbolize the feathered appearance of the turians themselves.
Turians are a playable race in Mass Effect 3's cooperative multiplayer mode.
In Mass Effect 3, on the mission Tuchanka: Bomb, a turian shot by a sniper is shown as having red blood. In all other instances, turian blood is portrayed as blue.
See Also Edit
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Bakhtigareyevo
Bakhtigareyevo () is a rural locality (a village) in Ishmukhametovsky Selsoviet, Baymaksky District, Bashkortostan, Russia. The population was 5 as of 2010.
References
Category:Rural localities in Bashkortostan
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The S&P 500, Dow and NASDAQ are all lower over the last five trading days.
Crude oil futures are significantly lower this week, trading at $42.62 per barrel on Friday afternoon.
And Gold futures are higher this week, trading at $1113.84 an ounce this afternoon.
In economic news, in the week ending August 8, the advance figure for seasonally adjusted initial claims was 274,000, an increase of 5,000 from the previous week's revised level. The previous week's level was revised down by 1,000 from 270,000 to 269,000. The 4-week moving average was 266,250, a decrease of 1,750 from the previous week's revised average. This is the lowest level for this average since April 15, 2000 when it was 266,250. The previous week's average was revised down by 250 from 268,250 to 268,000.
The Producer Price Index for final demand advanced 0.2 percent in July, seasonally adjusted, the U.S. Bureau of Labor Statistics reported. Final demand prices rose 0.4 percent in June and 0.5 percent in May. On an unadjusted basis, the final demand index moved down 0.8 percent for the 12 months ended in July, the sixth straight 12-month decline.
In July, the increase in the final demand index can be traced to prices for final demand services, which climbed 0.4 percent. In contrast, the index for final demand goods edged down 0.1 percent.
Within intermediate demand, prices for processed goods moved down 0.2 percent, the index for unprocessed goods fell 2.9 percent, and prices for services advanced 0.2 percent.
In corporate dividend news, FedEx (NYSE:FDX) declared a quarterly cash dividend of $0.25 per share on FedEx Corporation common stock. The dividend is payable October 1, 2015 to stockholders of record at the close of business on September 10, 2015.
Parker Hannifin Corporation (NYSE:PH) has declared a regular quarterly cash dividend of 63 cents per share of common stock to shareholders of record as of August 28, 2015. The dividend is payable September 11, 2015.
ACE Limited (NYSE:ACE) declared a quarterly dividend equal to $0.67 per share, payable on October 21, 2015, to shareholders of record at the close of business on September 30, 2015.
UnitedHealth Group (NYSE:UNH) authorized payment of a quarterly shareholder dividend of $0.50 per share in the third quarter of 2015. The dividend will be paid on September 22, 2015, to all shareholders of record of UnitedHealth Group common stock as of the close of business September 11, 2015.
Prudential Financial (NYSE:PRU) announced the declaration of a quarterly dividend of $0.58 per share of Common Stock, payable on September 17, 2015, to shareholders of record at the close of business on August 25, 2015.
3M (NYSE:MMM) declared a dividend on the company's common stock of $1.025 per share for the third quarter of 2015, payable Sept. 12, 2015, to shareholders of record at the close of business on Aug. 21, 2015.
NIKE (NYSE:NKE) has declared a quarterly cash dividend of $0.28 per share on the company's outstanding Class A and Class B Common Stock payable on October 5, 2015, to shareholders of record at the close of business on September 8, 2015.
Lear Corporation (NYSE:LEA) has declared a quarterly cash dividend of $0.25 per share on the Company's common stock. The dividend is payable on September 21, 2015 to shareholders of record at the close of business on September 2, 2015.
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only, and do not reflect the opinions of BNK Invest, Inc. or any of its affiliates, subsidiaries or partners.
In no way should any content contained herein be interpreted to represent trading or investment advice.
None of the information contained herein constitutes a recommendation that any particular security, portfolio,
transaction, or investment strategy is suitable for any specific person. All viewers agree that under no
circumstances will BNK Invest, Inc,. its subsidiaries, partners, officers, employees, affiliates, or agents be held
liable for any loss or damage caused by your reliance on information obtained. Read Full Disclaimer.
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Use of winter wheat x Triticum tauschii backcross populations for germplasm evaluation.
The wild diploid goatgrass, Triticum tauschii (Coss.) Schmal., is an important source of genes for resistance to both diseases and insects in common wheat (Triticum aestivum L.) We have evaluated grain yield, kernel weight, protein concentration, and kernel hardness of 641 BC2 F1-derived families from direct crosses involving four T. aestivum cultivars and 13 T. tauschii accessions over 2 years and at two Kansas, USA, locations. On average, T. tauschii germplasm depressed grain yield and increased protein concentration, whereas kernel weight was affected either positively or negatively, depending on the T. tauschii parent. Three T. tauschii parents produced a large proportion of families with very soft endosperm. Some variation among progeny of different T. tauschii parents resulted from the segregation of genes for resistance to leaf rust (caused by Puccinia recondita Rob. ex Desm.). This study confirmed that random BC2-derived families can be used to evaluate the effects of T. tauschii genes in the field. This methodology, although laborious, can provide useful information which is not obtainable by the screening of T. tauschii accessions themselves.
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Des Moines, Iowa (CNN) Julián Castro plans to refocus his 2020 presidential campaign on Iowa, Nevada and Texas in the coming days and is supporting his staffers looking for jobs with other campaigns, sources familiar with the plans tell CNN.
The former Housing and Urban Development secretary has struggled for months to raise money or get attention in the still large field of Democrats vying for the chance to take on President Donald Trump. Castro spent the final 10 days of October pushing to raise $800,000 and pledged to donors that he would drop out if he failed to hit that goal. The campaign narrowly hit the goal with hours to go on October 31.
But it was clear inside the San Antonio-based campaign even before the push began that the future was uncertain for the Texas Democrat. The Castro campaign senior leadership told staffers before they announced their fundraising push that whether or not they hit the number, staff should feel free to look for other opportunities.
And even when the campaign hit the fundraising goal, Castro's senior aides again told staff that the campaign would likely have to make staffing adjustments to press on.
That has led some Castro aides to look for jobs with other campaigns.
Read More
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External-feedback effects in high-gain scattering media.
We report on experimental studies of external-feedback effects on high-gain scattering media. We explain experimental results for the pump energy required for laser action as a function of the separation between a mirror and the media by use of Monte Carlo simulations and integration of the laser equations.
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A comparative study on the structures of Grifola frondosa polysaccharides obtained by different decolourization methods and their in vitro antioxidant activities.
Decolourization of polysaccharides is one of the crucial procedures that affects their structure, which is closely related to their bioactivity. Here, Grifola frondosa polysaccharide (GFP) was decolourized with H2O2 and AB-8 macroporous resin. Then, two main fractions, named DGFP and SGFP, were obtained by purification with Sepharose CL-4B. The molecular weights of these two polysaccharides were determined to be 6.306 × 106 (±0.410%) Da and 1.174 × 107 (±0.299%) Da by HPSEC. Monosaccharide analysis indicated that DGFP was composed of glucose, mannose, and galactose (32.20 : 1.00 : 1.75), while SGFP consisted entirely of glucose. Despite a backbone →4)-α-Glcp-(1→ in two polysaccharides, reducing ends Rα →3)-α-Glcp and Rβ →4)-β-Glcp were observed in DGFP by 1D/2D NMR. The results suggested that decolourization with low concentrations of H2O2 might alter the structure of GFP and generate new reducing ends. In vitro antioxidant results implied that DGFP exhibited a higher ability to scavenge DPPH and hydroxyl radicals and reduced the over-generated ROS levels in a concentration-dependent manner. These results suggested that the antioxidant effects of GFP could be activated by decolourization with H2O2. Therefore, DGFP might be a more promising natural antioxidant than SGFP.
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Just a couple of days ago Baba Ramdev's Patanjali released a messaging app called 'Kimbho'. It has been hardly two days but the app has been in the centre of a lot of controversies. The app was allegedly a copy of another app called Bolo messenger. There were a lot of security lapses in the app as well.
The @KimbhoApp is a copy paste of another #application. The description and the screenshots in the app stores are the same. Moreover, the #Kimbho app is making request to bolomessenger[.]com pic.twitter.com/gOKOhash5X — Elliot Alderson (@fs0c131y) May 31, 2018
The app was taken down yesterday from both Apple App Store and Google Play Store. Now interestingly, there are suddenly many clones appearing on the Play Store. Looking at the release dates of these apps, it's clearly visible that these are old apps which have just been renamed.
Killer Features talked to one of the developers who said he doesn't have any affiliation with Patanjali. He is just experimenting with the keyword so that he can get more hits due to the high search volume. We'll have to see what happens to these clones when the original app by Patanjali reappears.
Meanwhile, Patanjali's spokesperson SK Tijarwala said that the app is just in its trial phase and is no longer available for download on any platform. He also claimed that only in three hours more than 1.5 lakh people had downloaded the app.
Our trial version of #kimbho app is no longer available for download on any platform. We don't take any responsibility for many dulicate apps showing on anywhere. Beware!
आम सूचना..!#पतंजलि का #किम्भो एप का ट्रायल वर्जन अब कहीं भी डाउनलोड के लिए उपलब्ध नहीं है @yogrishiramdev https://t.co/KWkVrpoVge — Tijarawala SK (@tijarawala) May 31, 2018
Killer Features has reached out to Google to understand how it is tackling the duplicate app problem. Meanwhile, it remains to be seen when Patanjali re-launches its app. While writing this story its website still continues to be down.
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Confusion after Cameron issues new resolution
Prime Minister David Cameron came under fire today for issuing a new resolution on UK participation in military strikes against Assad's regime in Syria.
Both Coalition and Opposition MPs complained that they had no idea whether to support the resolution which reads:
"This House is opposed to decline the renunciation of the non-military support in defiance of UN Resolutions on the use of chemical weapons in Syria, and furthermore we decline to de-authorize the non-removal of missile weapons which will not be used against Syria, unless the House fails to agree on the rescinding of key guarantees of unilateral non-intervention which would not lead to a de-escalation of the conflict"
Tory and Labour whips confirmed that there would be at least 600 abstentions in the vote which the Prime Minister has scheduled for 4.30am tomorrow.
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Back on August 30, 1987, Capcom released Street Fighter in arcades, which seemed relatively innocuous until you gave it a second look. The graphics were average for the time, the voice acting was quite muffled, and the controls were strict and unforgiving. You could normally only choose a single character – a redheaded, red-shoed martial artist named Ryu – unless a second player came on, in which case they could play a blonde clone named Ken. Most people who play it would agree that it’s not the best game in the series, but there were people who saw its potential, and decided to continue the franchise with Street Fighter II. With that, one of the video games’ biggest icons stepped into the spotlight.
Over the past 25 years, Street Fighter has worked itself up from humble beginnings in the arcade to being a cultural phenomenon throughout the world. Millions of people interact with it every day, in a myriad of ways. In all truth, the fans define the franchise as much as the games. And it was these people who inspired the creation of the Street Fighter 25th Anniversary Collector’s Set.
This set represents the combined passion and enthusiasm of literally hundreds of people who have contributed to making Street Fighter games over the last 25 years, and serves as a love letter to anyone who cares about the franchise. When we first met to discuss the set’s creation, the two questions we asked were “What is the maximum amount of stuff we can get into this box?” and “How do we create something that celebrates all the different kinds of fans we have out there?” These questions helped lead us to one of the most intensive, thoughtful, and fan-oriented collector’s sets in Capcom history.
So, let’s look at our different sources of inspiration. What is Street Fighter? Well, the easiest place to start is with the games. Obviously, without the Street Fighter games, there would not be anything else. We included all the titles currently available for PlayStation 3, representing many of the series’ highlights. Super Street Fighter IV Arcade Edition and Street Fighter x Tekken are included as physical discs. We’re also including all of the costume DLC for SSFIVAE (that’s over 100 costumes!) and for SFxT, the 12 additional character DLC and all available Swap costumes. You can also download two of our digital titles, Super Street Fighter II Turbo HD Remix and Street Fighter III: Third Strike Online Edition. With all of that, you’ll have something to play for whatever mood you’re in.
When you get your Collector’s Set for PS3, you are in for a special treat. Sony, being a big supporter of fighting game fans and Street Fighter fans in particular, actually came to us at Capcom and told us that they want to help celebrate the 25th Anniversary by offering additional content for PlayStation 3. As such, the PS3 set includes Street Fighter Alpha, Street Fighter Alpha 2, and Street Fighter Alpha 3 as PSone Classic titles; Street Fighter Alpha 3 Max as a PSP download, all downloadable content for Street Fighter III: Third Strike Online Edition, and lots of avatars (over 170). We are proud to be working with Sony to get this to you.
We knew from fan feedback that people also enjoy Street Fighter beyond just playing the games. One of the most pervasive ways that fans interact with the franchise away from consoles and arcades is through the artwork. The games are filled with great art, and this inspires a wide variety of artists from all over the world. We wanted to capture a piece of that, and bring recognition to some of the incredible pieces that have been created, so we put a call out to artists around the world to participate. We got hundreds of entries, but were eventually able to narrow it down to a select sample to be included in a 64-page hardcover art book.
I myself am a music lover, and Street Fighter’s music is recognizable to gamers and non-gamers alike. Whether it’s the iconic Street Fighter 2 main theme or the Guile theme (it really does go with everything!), Street Fighter has proven that throwing a punch is much better with music. So, we wanted to include something for audiophiles by providing 11 discs of music. This includes nine discs of soundtracks – covering Street Fighter, Super Street Fighter II Turbo, Street Fighter Alpha 3, Street Fighter III: Third Strike, Super Street Fighter IV Arcade Edition, and Street Fighter x Tekken – and two discs of fan-created music. As with the artwork, we asked people to send in their music. The talent in the response humbled us, and made it incredibly difficult to make our selections from the 300+ entries. However, we managed to create two CDs: one with remixes/remakes of existing themes, and one which is original music inspired by Street Fighter. Some of these fans have serious futures ahead of them, and it’s an honor to be able to include them in the Collector’s Set and make them a piece of the franchise’s history.
Then, of course, you have the collectors… the people who like to showcase items on their shelves. Here again, we have this group covered. Inside the large black box (which itself is a collectible), you’ll find an eight-inch statue of Ryu flying up with a Shoryuken. Not only is it a highly-detailed statue (the texture on his gi is some of the best I’ve seen on a figure that size), but the flames in the base even light up to show off at night. Then we have the life-size, embroidered Ryu replica belt, complete with the furinkazan kanji.
Continuing on, we have a two-disc Blu-Ray set, with over 12 hours of content. This features a lot of animated content, including the entirety of the incredibly cheesy (but-oh-so-fun) Saturday morning cartoon show from 1995, Street Fighter II: The Animated Movie (in my opinion, featuring some of the best fight scenes choreographed for any film, live-action or animated), Street Fighter IV: The Ties That Bind, and the Super Street Fighter IV Original Video Animation, aka the “Juri anime,” which is being released in English for the first time ever. However, the most exciting piece of video content is “I am Street Fighter,” a brand new documentary we commissioned. As you may have been able to tell through this post, Street Fighter has impacted many people’s lives, and vice-versa. That’s what this documentary looks at – the people who have influenced the Street Fighter franchise, and the people whose lives have been influenced by it. It’s a fascinating, and occasionally heartwarming look into the cultural phenomenon.
As you can see, we kept every type of fan in mind when creating the Street Fighter 25th Anniversary Collector’s Set and we’re incredibly proud to bring it to fans. In fact, there has never been a better time to be a Street Fighter fan. Whether you were there from the beginning, or just began in the past couple of years, you are part of a rich legacy and vibrant community. You can see both of these in action at our anniversary website.
You make Street Fighter what it is. You are Street Fighter. So from all of us at Capcom – thank you!
…Also, SHORYUKEN!!!
]]>http://blog.us.playstation.com/2012/09/18/street-fighter-25th-anniversary-collectors-set-has-tons-of-extra-content-on-ps3/feed/102http://blog.us.playstation.com/wp-content/uploads/2012/09/sf25.jpg3Product Manager, Street Fighter 25th Anniversary Collector's Set102112012 PSN Gamers’ Choice Awards – And The Winners Are…http://blog.us.playstation.com/2012/03/06/2012-psn-gamers-choice-awards-and-the-winners-are/
http://blog.us.playstation.com/2012/03/06/2012-psn-gamers-choice-awards-and-the-winners-are/#commentsTue, 06 Mar 2012 18:39:31 +0000http://blog.us.playstation.com/?p=71282PSN Gamers’ Choice Awards. Votes were cast this year across 10 categories in the PlayStation Store, including the community award for Best PSN Indie, which was voted online by the community.
First off, we’d like to thank all of the nominees for being a part of this year’s PSN Gamers’ Choice Awards. All of the games that we showcased were some of the highest and most user-ranked games released in 2011, which is an honor in and of itself. Second and most importantly, we want to thank all of you for casting your votes both online and in the PlayStation Store to determine our winners. ]]>
Drumroll, please…
We’ve tallied up the votes, slipped into our tuxedos, polished up the trophies and are now ready to reveal the winners of the second annual PSNGamers’ Choice Awards. Votes were cast this year across 10 categories in the PlayStation Store, including the community award for Best PSN Indie, which was voted online by the community.
First off, we’d like to thank all of the nominees for being a part of this year’s PSN Gamers’ Choice Awards. All of the games that we showcased were some of the highest and most user-ranked games released in 2011, which is an honor in and of itself. Second and most importantly, we want to thank all of you for casting your votes both online and in the PlayStation Store to determine our winners.
We had an incredible turnout during the voting period this year, which made for an intense competition. Things got so heated, in fact, that the battle for Best PS Move Game literally came down to the last day between Dungeon Defenders and Dungeon Hunter: Alliance. But only one could come out with top honors.
Without further ado, here are your winners for the 2012 PSN Gamers’ Choice Awards:
The celebration continues as the award-winning titles will be discounted 30% in the PlayStation Store for one week, starting later today, with a 50% discount available to all PlayStation Plus subscribers. Don’t miss out on the chance to score these great games! And don’t forget to start rating the games that you’re playing throughout 2012 so your favorite games can get nominated next year.
]]>http://blog.us.playstation.com/2012/03/06/2012-psn-gamers-choice-awards-and-the-winners-are/feed/121http://blog.us.playstation.com/wp-content/uploads/2012/03/GCA12_PR_KEYART_Winners_HD_72dpi.jpg3.7Director, Digital Distribution1212PSN Gamers’ Choice Awards 2012 Nominees: You Decide Tomorrowhttp://blog.us.playstation.com/2012/02/27/psn-gamers-choice-awards-2012-nominees-you-decide-tomorrow/
http://blog.us.playstation.com/2012/02/27/psn-gamers-choice-awards-2012-nominees-you-decide-tomorrow/#commentsMon, 27 Feb 2012 16:00:06 +0000http://blog.us.playstation.com/?p=70566PSN. We’ve got six new voting categories this year, bringing the total to 10, featuring the games released in 2011 that you gave high user ratings.
Starting tomorrow, head into the PlayStation Store to cast your vote for your favorite games in nine of the 10 categories and receive a free Gamers’ Choice Awards XMB theme for each vote cast. New this year, you can also cast your vote online in one special community category, Best PSN Indie, and you’ll receive a voucher code to be redeemed in the PlayStation Store for a free exclusive XMB theme. Your votes exclusively pick the winners in every category, and your award winners go on sale.
]]>It’s time to make your voice heard! Today, we’re excited to announce a great lineup of nominees in ten different categories for the 2012 PSN Gamers’ Choice Awards. Just like last year, the Gamers’ Choice Awards put you in the driver’s seat, allowing you to decide which titles will be named the best downloadable games on PSN. We’ve got six new voting categories this year, bringing the total to 10, featuring the games released in 2011 that you gave high user ratings.
Starting tomorrow, head into the PlayStation Store to cast your vote for your favorite games in nine of the 10 categories and receive a free Gamers’ Choice Awards XMB theme for each vote cast. New this year, you can also cast your vote online in one special community category, Best PSN Indie, and you’ll receive a voucher code to be redeemed in the PlayStation Store for a free exclusive XMB theme. Your votes exclusively pick the winners in every category, and your award winners go on sale.
Don’t go too far because the winning games in each category will be announced right here on the PlayStation Blog in just one week–on March 6, 2012–and that same day, the winning game in each category will be discounted 30% in the PlayStation Store for one week, with a special 50% discount for PlayStation Plus subscribers.
PlayStation Community Award – Best Indie:
Don’t forget to head into PlayStation Store tomorrow to cast your vote. You decide the best of PSN, so make your voice heard!
]]>http://blog.us.playstation.com/2012/02/27/psn-gamers-choice-awards-2012-nominees-you-decide-tomorrow/feed/157http://blog.us.playstation.com/wp-content/uploads/2012/02/6934530091_5ecda4fee9_b.jpg4.1Director, Digital Distribution1574Street Fighter III: Third Strike Online Edition Hadokens PSN Todayhttp://blog.us.playstation.com/2011/08/23/street-fighter-iii-third-strike-online-edition-hadokens-psn-today/
http://blog.us.playstation.com/2011/08/23/street-fighter-iii-third-strike-online-edition-hadokens-psn-today/#commentsTue, 23 Aug 2011 19:00:42 +0000http://blog.us.playstation.com/?p=57499Street Fighter III: Third Strike Online Edition. Today is the day that fighting game fans from around the world have been waiting for, as Third Strike is finally available on the PlayStation Network! We are super excited to be bringing what gaming websites have been calling a piece of fighting game history to all of you to enjoy. For those of you who haven't seen this game in action, check out the launch trailer.
Get hypeOne of the deepest and most sophisticated fighting games of all time, Street Fighter III: Third Strike Online Edition maintains an arcade-perfect experience while including additional features such as updated graphics, a robust online experience, and a deep challenge system allowing players to push themselves and reap rewards for accomplishments beyond wins and losses. ]]>Today is the day that fighting game fans from around the world have been waiting for — Street Fighter III: Third Strike Online Edition is about to go live on the PlayStation Store! We are super excited to be bringing what gaming websites have been calling a piece of fighting game history to all of you to enjoy. For those of you who haven’t seen this game in action, check out the launch trailer below.
One of the deepest and most sophisticated fighting games of all time, Street Fighter III: Third Strike Online Edition maintains an arcade-perfect experience while including additional features such as updated graphics, a deep online experience, and a new challenge system that encourages players to push themselves and reap rewards for accomplishments that go far beyond mere wins and losses.
Street Fighter III: Third Strike Online Edition will be available to download later today on PlayStation Network for $14.99. Check out a sampling of reviews Third Strike has been getting:
“The best Street Fighter game to be released on the downloadable market yet. It’s the same game many of you have been playing for the last 12 years, only improved in every area”IGN – 9/10
“It’s my opinion that this is the single most compelling and well-balanced roster in a Street Fighter title to date”Destructoid – 9/10
“A fantastic fighting game given new life in an example of a re-release crafted to perfection… it will kick your arse and make your knuckles bleed”ReadyUp – 9/10
“It’s your duty as a fighting game fan to pay homage to the game that gave us one of the greatest moments in fighting history and showed the masses what makes the genre so special”1UP – A
“3rd Strike [is] one of the most technical and tactical fighting games ever… you owe it to yourself to play this classic”NowGamer – 9.2/10
Get hyped guys, because it’s time you realize that it’s the Third Strike that counts!
]]>http://blog.us.playstation.com/2011/08/23/street-fighter-iii-third-strike-online-edition-hadokens-psn-today/feed/77http://blog.us.playstation.com/wp-content/uploads/2011/08/6006682302_357ffd69ea_z.jpg4.07Marketing Lead, Capcom772Street Fighter III: Third Strike Shoryukens PSN PLAY August 23rdhttp://blog.us.playstation.com/2011/08/15/street-fighter-iii-third-strike-shoryukens-psn-play-august-23rd/
http://blog.us.playstation.com/2011/08/15/street-fighter-iii-third-strike-shoryukens-psn-play-august-23rd/#commentsMon, 15 Aug 2011 14:00:23 +0000http://blog.us.playstation.com/?p=56761Street Fighter III: Third Strike Online Edition takes one of the most revered fighting games in CAPCOM’s history and brings it Online. Quite frankly, the feature set we’ve included in this downloadable is jaw dropping. In addition to upscaled graphics, we have hundreds of trials, dynamic challenges, and a huge Vault full of all kinds of unlockables. We have multiple training modes, integrated Tournament support, and 8 player lobbies (with spectator mode!). We have GGPO for near-lagless online play, a robust replay system, and the ability to upload matches straight to YouTube - from within the game! All of this, and much, much more.]]>
Hey everyone! This is Derek Neal, and I’m the producer on Street Fighter III: Third Strike Online Edition. I’m here to tell you all the things that make this game awesome. You might want to take a seat and get comfortable, because that’s quite a list!
Available August 23rd, Street Fighter III: Third Strike Online Edition takes one of the most revered fighting games in CAPCOM’s history and brings it Online. Quite frankly, the feature set we’ve included in this downloadable is jaw dropping. In addition to upscaled graphics, we have hundreds of trials, dynamic challenges, and a huge Vault full of all kinds of unlockables. We have multiple training modes, integrated Tournament support, and 8 player lobbies (with spectator mode!). We have GGPO for near-lagless online play, a robust replay system, and the ability to upload matches straight to YouTube – from within the game! All of this, and much, much more.
We are also very excited to be the lead title in the brand new PSN PLAY program. Now with PSN PLAY, you’re able to pre-order this game now. This is actually the first time a pre-order program has been offered for a downloadable PSN title, and we couldn’t be happier than to help lead that charge. The pre-order period for the game runs through August 18th, and people who pre-order during this period will receive a special gift: a custom Third Strike theme for their PS3! Here’s a sneak peek of what it will look like:
If that wasn’t enough, we are also announcing that we’ll be giving a free gift to everyone who purchases the game between now and through September 19th: a free key that unlocks the hidden boss character, Gill. This boss character is a ton of fun to play, and can be used in all of our casual game modes, such as Arcade and Player Match. (He is not available in Ranked matches or in Tournaments.)
Finally, PlayStation Plus members will be excited to hear that they’re getting a special deal: an additional 20% off on this already great value! So what are you waiting for?! Pre-order now. Buy this game…it’s time you realized that it’s the THIRD STRIKE THAT COUNTS!
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1. Field of the Invention
The present invention relates to a liquid crystal display device (LCD) using switching elements, for example amorphous silicon thin film transistors (hereinafter simply referred to as a-Si TFTs) or polycrystalline silicon thin film transistors (hereinafter, p-Si TFTs), which is particularly effective for use as an optical shutter of a projector and a method for repairing the defective portions thereof.
2. Description of the Related Art
An active matrix type LCD includes pixel electrodes arranged in a matrix fashion on a substrate and switching elements formed so as to correspond to the respective pixel electrodes. An LCD drives the liquid crystal by charging/discharging the necessary charges to/from the pixel electrodes in accordance with the operation of the switching elements.
In general, an enormous number, i.e. from several tens of thousand to several hundreds of thousand, of switching elements and pixel electrodes, are formed on the substrate of the LCD in order to accomplish satisfactory display quality. In a process for forming such elements and electrodes on the substrate, some pixels will not operate even when a voltage is applied to the liquid crystal, mainly because the switching elements cannot operate or because the pixel electrodes are abnormally formed. The existence of non-operating pixels when a voltage is applied to the liquid crystal prevents the LCD from accomplishing satisfactory display quality.
An active matrix type LCD has two display modes. One of them is a Normally-Black mode in which light is shielded when no voltage is applied and light is transmitted when a voltage is applied. The other is a Normally-White mode in which light is transmitted when no voltage is applied and light is shielded when a voltage is applied.
FIG. 3 diagrammatically shows a conventional active matrix type LCD in a Normally-White mode. The LCD 1 includes glass substrates 3 and 2 which are spaced apart from each other. On the internal surface of the glass substrate 3, p-Si TFTs 5 are formed, and on the internal surface of the glass substrate 2, a black mask pattern (B/M) 4 is formed.
The configuration of the LCD 1 will be described in detail below.
On the glass substrate 3, pixel electrodes 6 made of ITO (In.sub.2 O--SnO.sub.2 based) are arranged in a matrix fashion by performing a sputtering step, a photolithography step, and an ITO etching step. Switching elements, i.e. p-Si TFTs 5 are formed on the glass substrate 3 so as to correspond to the respective pixel electrodes 6 arranged in a matrix fashion. Over the glass substrate 3, a polyimide-based alignment film 18 is formed by an offset printing method. On the B/M 4 on the other glass substrate 2, pixel electrodes 16 made of ITO and a polyimide-based alignment film 17 are respectively formed. Each of the alignment films 18 and 17 is subjected to a rubbing treatment so as to obtain a twisted-nematic orientation. Polarizing plates 7 and 8 in a crossed-Nicols state are respectively attached to the external surfaces of the substrates 2 and 3. The LCD 1 in the Normally-White mode is configured as described above.
Next, a method for repairing the defective pixels of an active matrix type LCD in a Normally-White mode will be described.
When a sufficiently high voltage is applied to the liquid crystal 9 of the LCD 1, as indicated by the case C in FIG. 3, an incident light ray 13 transmitted through the polarizing plate 8 is shielded by the polarizing plate 7 if the pixels operate normally. This is because the polarizing direction of the incident light 13 is not changed by the liquid crystal 9. In a defective pixel, the polarizing direction of the incident light 13 transmitted through the polarizing plate 8 is rotated by the liquid crystal 10 by 90 degrees, so the incident light 13 is transmitted also by the polarizing plate 7. As a result, a leaking light ray 13' is observed so as to form a bright point, as indicated by the case E in FIG. 3. If a plurality of such defects exist, the defects are observed among the pixels as a group of bright points or line defects.
Methods for repairing such a pixel exhibiting the above-mentioned defect (hereinafter, referred to as a defective pixel) are described, for example, in Japanese Laid-Open Patent Publication Nos. 60-243635 and 3-243917.
According to the technique described in the former, as indicated by the case F in FIG. 3, a laser beam is irradiated onto a portion of an alignment film on the defective pixel so as to burn off the portion of the alignment film; and an orientation condition of the liquid crystal 11 over the defective pixel is disturbed so as to scatter the transmitted light, thereby repairing the defective pixel.
According to the technique described in the latter, as indicated by the case D in FIG. 3, a resin, ink, or the like into which a colorant including a black dye or a black pigment is mixed is applied to the position on the external surface of the glass substrate 2 which corresponds to that of the defective pixel by using a cotton needle, a pin of a nylon fiber, or the like and then the colorant is dried so as to form a light-shielding layer 12 including a light-shielding portion 12a and thereby repairing the defective pixel.
However, conventional repairing methods such as those described above for the defective pixel of the LCD in the Normally-White mode have the following problems.
According to the repairing method described in the Japanese Laid-Open Patent Publication No. 60-243635, the defective pixel is repaired by irradiating the laser beam so as to burn off the portion of the alignment film. As a result, deterioration, decomposition or alteration of the liquid crystal occurs in the defective pixel. In addition, the laser beam irradiation damages the switching elements. An active matrix type LCD is required to exhibit particularly high reliability. Therefore, such a repairing method which affects the display quality of the LCD is not preferable. Furthermore, an expensive laser repairing apparatus for repairing the defective pixel, and a detection apparatus for accurately detecting the position of the defective pixel are also required.
According to the repairing method described in the Japanese Laid-Open Patent Publication No. 3-243917, after the positions of the defective portions have been identified, the light-shielding layer 12, including a plurality of light-shielding portions 12a, is formed by applying a black resin, ink, or the like to the defective pixel portions with a pin, so that the area of each light-shielding portion 12a generally becomes larger than that of each defective pixel portion. Consequently, the light-shielding portions 12a inconveniently cover normal pixels adjacent to the defective pixels. Such a problem is particularly likely to occur in a super high-definition LCD having approximately a million and several hundreds of thousands of pixels with a pixel pitch of about 30 .mu.m.
In addition, a detection apparatus for accurately detecting the positions of the defective pixels, and an apparatus for accurately applying a black resin, or the like onto the positions of the substrate corresponding to those of the defective pixel portions are required.
Moreover, according to the repairing methods described in the Japanese Laid-Open Patent Publication Nos. 60-243635 and 3-243917, in the situation where the LCD is used as an optical shutter for a projector, bright points are likely to occur because of the following reasons: specifically, the light rays emitted from an external light source are not incident at a right angle against the surface of the substrate of the LCD, but at a certain angle against the surface, so that the deviation occurs between the positions of the defective pixels and those of the bright points. Namely, the positions of the bright points on the substrate on the light-outgoing side deviate from those of the defective pixels on the substrate on the light-incoming side. Therefore, according to the above methods, bright points still occur.
|
{
"pile_set_name": "USPTO Backgrounds"
}
|
Auditions for “The Tell-Tale Farce”
Due to today’s announcements from state and local governments, we are taking the step of postponing our “The Tell Tale Farce” auditions, currently scheduled for March 22nd and 23rd. At this time, no new date(s) have been selected.
Stay tuned for more details in the future as our board of directors make these unprecedented decisions.
Announcing the audition dates for The Village Players of Hatboro’s production of The Tell-Tale Farce, a comedy by Don Zolidis.
Audition Dates
SHOW DATES are June (2020) 5, 6, 7, 12, 13, 14, 19, 20. All actors who are cast MUST be available for all show dates, as well as tech week (Sunday, May 31st through Wednesday, June 3rd)
REHEARSALS will be decided based upon actor availability.
SYNOPSIS
(Farce) It’s 1848, and Edgar Allan Poe is just coming off the spectacular success of “The Raven.” Unfortunately, it’s only earned him a grand total of nine dollars. So when a wealthy dowager commissions him to write her a poem for the vast sum of one hundred dollars, he leaps at the chance. Only problem: the man who shows up to write the poem isn’t Poe, he’s Poe’s mailman, and he’s on a quest to woo the dowager’s spinster niece. Playing Poe is harder than it looks, though, especially when your mustache keeps falling off, the teenage grand-daughter of the house is lusting after you, and Poe’s arch-nemesis, Rufus Griswold, just happens to be dropping by to settle old scores. A freewheeling, door-slamming farce with a touch of the macabre.
CHARACTER DESCRIPTIONS
Poe (The real one) – late 30s, fresh off the success of “The Raven”. A rather mild-mannered fellow with a mustache.
Rufus Griswold – Late 30s, a poet. Something of a bastard. Has amazing facial hair with mustache.
Henry Whitford – 20, nervous and naive. No mustache.
Mary O’Donnell – 20s, the chambermaid. Irish accent.
Helena Dowling – 62, the matriarch, nearly deaf. Walks with a cane.
Sarah Dowling – 30ish, Helena’s niece. A spinster.
Abigail Dowling – 18, granddaughter of Helena.
CREW MEMBERS & NEWBIES
Audition nights are a perfect time to come out, meet the director, and express your interest in helping out backstage, be it in the role of a stage manager, lights & sound operator, concessions, or whatever else you might like to do.
They’re also GREAT for first-timers to come out and observe how things are done, maybe express interest in being a helper, or read for a part! We almost always cast at least one person who’s never been on stage before. You never know what can happen!
|
{
"pile_set_name": "Pile-CC"
}
|
United States Court of Appeals
Fifth Circuit
F I L E D
IN THE UNITED STATES COURT OF APPEALS
April 16, 2003
FOR THE FIFTH CIRCUIT
Charles R. Fulbruge III
Clerk
No. 03-20300
Summary Calendar
UNITED STATES OF AMERICA,
Plaintiff-Appellee,
versus
ROBERTO GARZA,
Defendant-Appellant.
Appeal from the United States District Court
for the Southern District of Texas
USDC No. H-03-M-165-2
Before HIGGINBOTHAM, SMITH, and CLEMENT, Circuit Judges.
PER CURIAM:*
Roberto Garza appeals from an order of the district court
denying his motion to revoke the magistrate judge’s pretrial
detention order. The district court’s decision, which effectively
adopted the reasoning of the magistrate judge, rests upon its
conclusion that Garza has not rebutted the presumption that no
condition or combination of conditions will reasonably assure the
*
Pursuant to 5TH CIR. R. 47.5, the court has determined that
this opinion should not be published and is not precedent except
under the limited circumstances set forth in 5TH CIR. R. 47.5.4.
safety of the community.1 The district court’s conclusions are
supported by the record.2
AFFIRMED.
1
18 U.S.C. § 3142(e) & (f).
2
See United States v. Rueben, 974 F.2d 580, 586 (5th Cir.
1992) (“[T]he risk of continued narcotics trafficking on bail does
constitute a risk to the community.”).
2
|
{
"pile_set_name": "FreeLaw"
}
|
2011 Sea Ray 270 SLX
***NEW FOR 2011***
Year:
2011
Hull Material:
Fibreglass/GRP
Length:
9m
Engine/Fuel Type:
/
Call for Price
Own the 270 SLX and you own a lifestyle, one that involves an appreciation for the finer things in life.We're talking custom-caliber instrument panel, gleaming stainless-steel hardware, and the power and grace of a gentleman's roadster.
|
{
"pile_set_name": "Pile-CC"
}
|
Effects of low-intensity electromagnetic fields on the proliferation and differentiation of cultured mouse bone marrow stromal cells.
Electromagnetic fields (EMFs) used in stem-cell tissue engineering can help elucidate their biological principles. The aim of this study was to investigate the effects of low-intensity EMFs on cell proliferation, differentiation, and cycle in mouse bone marrow stromal cells (BMSCs) and the in vivo effects of EMFs on BMSC. Harvested BMSCs were cultured for 3 generations and divided into 4 groups. The methylthiotetrazole (MTT) assay was used to evaluate cell proliferation, and alkaline phosphatase activity was measured via a colorimetric assay on the 3rd, 7th, and 10th days. Changes in cell cycle also were analyzed on the 7th day, and bone nodule formation was analyzed on the 12th day. Additionally, the expression of the collagen I gene was examined by reverse transcription-polymerase chain reaction (RT-PCR) on the 10th day. The BMSCs of the irradiated group and the control group were transplanted into cortical bone of different mice femurs separately, with poly(lactic-co-glycolic acid) (PLGA) serving as a scaffold. After 4 and 8 weeks, bone the bone specimens of mice were sliced and stained by hematoxylin and eosin separately. The results showed that EMFs (0.5 mT, 50 Hz) accelerated cellular proliferation, enhanced cellular differentiation, and increased the percentage of cells in the G(2)/M+S (postsynthetic gap 2 period/mitotic phase + S phase) of the stimulation. The EMF-exposed groups had significantly higher collagen I messenger RNA levels than the control group. The EMF + osteogenic medium-treated group readily formed bone nodules. Hematoxylin and eosin staining showed a clear flaking of bone tissue in the irradiated group. Irradiation of BMSCs with low-intensity EMFs (0.5 mT, 50 Hz) increased cell proliferation and induced cell differentiation. The results of this study did not establish a stricter animal model for studying osteogenesis, and only short-term results were investigated. Further study of the mechanism of EMF is needed.
|
{
"pile_set_name": "PubMed Abstracts"
}
|
;;; gh-issues-tests.el --- tests fir gh-issues.el
;; Copyright (C) 2012 Yann Hodique
;; Author: Yann Hodique <yann.hodique@gmail.com>
;; Keywords:
;; This file is free software; you can redistribute it and/or modify
;; it under the terms of the GNU General Public License as published by
;; the Free Software Foundation; either version 2, or (at your option)
;; any later version.
;; This file is distributed in the hope that it will be useful,
;; but WITHOUT ANY WARRANTY; without even the implied warranty of
;; MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE. See the
;; GNU General Public License for more details.
;; You should have received a copy of the GNU General Public License
;; along with GNU Emacs; see the file COPYING. If not, write to
;; the Free Software Foundation, Inc., 59 Temple Place - Suite 330,
;; Boston, MA 02111-1307, USA.
;;; Commentary:
;;
;;; Code:
(require 'gh-tests)
(require 'gh-issues)
(defun gh-issues-tests:test-regular-issue (issue)
(should (equal (oref issue :number) 1347))
(should (equal (oref issue :state) "open")))
(ert-deftest gh-issues-tests:regular-list ()
(let* ((api (gh-tests-mock-api 'gh-issues-api))
(issues
(gh-tests-with-traces-buffers ((gists-buf "list_issues_sample.txt"))
(gh-tests-mock-url ((:record-cls mocker-stub-record
:output gists-buf))
(oref
(gh-issues-issue-list api "octocat"
"Hello-World")
:data)))))
(should (equal (length issues) 1))
(let ((issue (car issues)))
(should (object-of-class-p issue 'gh-issues-issue))
(gh-issues-tests:test-regular-issue issue))))
(provide 'gh-issues-tests)
;;; gh-issues-tests.el ends here
|
{
"pile_set_name": "Github"
}
|
Whoa there!
It looks like you're using an ad blocker, so you'll have to wait 15 more seconds.Please whitelist this site to skip the wait and help
us pay for bandwidth, hosting, and other bills.
APK Notes:
- Design changes to the history feed for a clear overview
- Some design changes to the Track/Song Detail view and we replaced the text buttons with smaller icons.
- Improvements and bug fixes
- Improvements and bug fixes
We listen to you and appreciate all your feedback. Thank you for helping us making the best music recognition app on the market.
The TrackID™ music recognition app is the best way to identify the music around you – it can tell you what song you are listening to within seconds. It's easy to use, has a clean new design and is available in more than 60 languages worldwide.
With the TrackID™ song identifier app, you can dive deeper into the music that interests you by exploring artist biographies, watching music videos, seeing who's atop the TrackID™ Charts and more.
Get a sneak peek by listening to a free preview of that ear-catching tune you've discovered. Plus, if you connect TrackID™ to Spotify, you can listen in full to the songs you have identified and add them directly to your Spotify playlist*. You can even download* the songs you've found, or watch their videos on YouTube.
Do you live in an area where your Internet connection is spotty? With offline mode, TrackID™ captures the track and identifies it when you're back online. Never miss out on a track again because your network fails you in that crucial moment!
Get your friends in on the action too. Spark conversations and reminisce about good times by sharing your identified tracks on Facebook or chatting about them with friends in WhatsApp. With TrackID™, it’s easy to share your music as you go.
Great music requires a great-looking app. Our new design brings the experience alive. With TrackID™, discovering music has never looked so good.
MAIN FEATURES
- Find a song within seconds with the press of a button.
- Your history is easily accessible on the start page.
- Offline mode helps you capture tracks when you have no network. Resume the search when your Internet connection is restored.
- Listen to a preview of the song so you can be sure it matches what you found.
- Watch music videos on YouTube.
- Listen to the song you recognized in Spotify or add it directly to your Spotify playlist* so you never lose it again.
- Download the song* there and then.
DISCOVER
- It's on the tip of your tongue! Find out the track title and the artist's name.
- Jog your memory by viewing the album name and cover art.
- Get the bigger picture with the genre of the track.
- Settle debates by finding out the album’s record label and the year the album was released.
- Learn more with pictures and an in-depth biography of the artist.
- Explore the other popular songs and albums related to the artist.
- Put it into context by viewing the album playlist.
EXPLORE AND SHARE
- What have other people searched for? Discover the hottest global hits with TrackID™ Charts from all over the world.
- Love the song you've found? Share it with your friends on Facebook, WhatsApp, Twitter, Google+ and more.
- Log in with Facebook or Sony Entertainment Network and have your history backed up and accessible from multiple devices.
- Take advantage of two widgets for fast access from your home screen.
Top recent searches on the music identifying TrackID™ app reveal that the trending and most popular artists of the moment include Jess Glynne, Avicii, Lana Del Ray, and Lost Frequencies Feat. Janieck Devy. Be sure to check out https://trackid.sonymobile.com/newsongs/US to see more from these musicians as well as other great artists today.
|
{
"pile_set_name": "Pile-CC"
}
|
Examine differences between pre-term and full-term human milk. This study is also designed to identify material factorswhich may influence the concentrations and determine changes in concentrations which may result from storage of breast milk.
|
{
"pile_set_name": "NIH ExPorter"
}
|
Q:
Generating pdf using itextsharp in asp.net c#
I want to generate pdf file as attached in the picture. i want to harcode the contents shown in the picture into a pdf file on button click.
How do i proceed, using table structure will work?I am unable to form the table stucture here.
Please help me to sort out my problem.i want the output as shown in below image.
A:
First create one table with two columns like below
_______________________________________________
| | |
| | |
| | |
| COL.0.1 | COL.0.2 |
| | |
| | |
-----------------------------------------------
keep reference to its columns, then create table for left side content and insert it as child content of left column COL.0.1
------------------------------
| designation: xxxxxxx | <---- colspan = 2
------------------------------
| Audit No. | xxxx |
------------------------------
| Received on ....... |
-------------------------------
and you can continue for the rest of content similar to this.
The idea is to split content to smaller tables and nest them.
|
{
"pile_set_name": "StackExchange"
}
|
Find Your 2019 Toyota Tundra
When it comes to hardworking performance and tough reliability, there isn't a truck on the market quite like the new 2019 Toyota Tundra. Complete with groundbreaking features like an available 5.7-liter aluminum i-FORCE V8 engine with an Acoustic Control Induction System (ACIS) and Toyota Safety Sense™, the new Tundra doesn't hold back. For the toughest jobs around, trust the new 2019 Tundra to show up when it matters.
When your 2019 Toyota Tundra needs maintenance or repairs, look no further than our service department where our expert technicians are equipped to handle your service needs quickly and affordably. Whether you need a simple oil change or major repairs, you can trust the quality service you'll receive here at our dealership.
|
{
"pile_set_name": "Pile-CC"
}
|
In Albino Lullaby, you find yourself in a Lynchian psychological nightmare where the gamespace twists and contorts around you as you uncover clues to understand where and what you are. Pacing is player-driven in a mechanical mansion that deconstructs around you.
Justin Pappas, creative director of Albino Lullaby and Founder of Ape Law: We are hitting the reset switch on the genre and bringing horror games back to the days of film when suspense ruled and great masters like David Lynch and Alfred Hitchcock used our minds to scare us more than anything onscreen.
The game is VR ready and supports Oculus Rift, Vive and Morpheus. It will be released episodically for PC/Windows (7 and later) on Steam; buy one episode at a time starting tomorrow, or purchase a season pass.
|
{
"pile_set_name": "Pile-CC"
}
|
(1) Field of the Invention
The present invention relates generally to an apparatus for installation of a material having discrete elements, and, more particularly, to a transporter system including a high speed, inline blower, a material agitator upstream of the inline blower, and a planetary transmission connected to the shaft of the blower for providing a lower speed mechanical output to the material agitator
(2) Description of the Prior Art
Insulation is used in residential and commercial dwellings both to conserve energy and to reduce noise. The two most common types of insulation are blown and batt. Loose fill insulation, unlike batt insulation, requires the use of a machine to open the product in baled or compressed form. Opening in the industry commonly refers to modifying a product of a relatively high packaged density to a much lower installed density, perhaps as much as only 5-10% of the initial packaged density. The opened insulation is then conveyed to the final installation location through an air conveyance system. The finished installation is accomplished in several ways depending on final product needs.
One method for opening and conveying the product is to provide a rotational insulation opening device in a hopper in the machine to prepare the product for further transport. The semi-opened insulation materials is then gravity fed into the top cavity of an airlock, a horizontally rotating device that segregates portions of the material, and then rotates it into contact with a air stream created by a air blower pump. Typically, these devices are run by separate motors, creating added weight machine weight both for the motors, and for all the support brackets, control electrical controls and other associated hardware. The airlock also adds significant weight to the machine.
Airlock based machines have a horizontally oriented cylinder with a longitudinal opening in the top for the gravity fed and/or mechanical introduction of insulation material. The cylinder is divided longitudinally into a plurality of chambers by a rotating series of blades or paddles. The blades or paddles seal off the inner dimensions of the airlock cylinder creating discrete chambers that are sealed from each other during rotation. The lower chamber of the cylinder has an opening at either end such that air from an air pump can be introduced into one end of the cylinder and can exit the other end, carrying with it any insulation material that is in that particular chamber.
The effect of the airlock is to create a series of rotating chambers that sequentially accept insulation material that is gravity or force fed into the top chamber. As the material drops into the top chamber, the rotation of the blades or paddles carries the material away from the opening and seals the cavity in which the insulation now resides. When the chamber rotates to the other side of the cylinder, it comes into contact with the air stream provided by the air pump, and the insulation in just that cavity is blown out into the conveying hose to the installation location.
A problem with airlock-based insulation blowing machines is that material is gravity or mechanically fed into the top chamber of the cylinder, and then is conveyed directly into the conveying stream. If the product is not fully opened prior to entering the conveying stream, only the additional turbulence of the conveying hose can be used to further open the product to its design density. Thus, many if not all insulation hoses are internally ribbed to force increased agitation post-blower.
Yet another method is to provide for insulation opening and introduction into the conveying air stream, and use a through blower device where the insulation passes through the pumping vanes of the blower itself. Such machines are thought to increase the opening ratio of the density of the opened product as installed to the density of the packaged product. However, the available machines use two motors as well, either both enclosed in the machine housing, or with one motor detached from the machine during transit, and then reattached at the installation site. Either method increases the total machine weight, complexity, and electrical demands.
Also, through blower devices force the machine designer to compensate for the relatively smaller introduction cross section leading to the conveying stream of the pump by attempting to force increased product opening prior to air stream entrance of the insulation. This has created a limitation in standard practices such that only the very smallest of insulation machines currently use the through blower concept. Medium and large sized blowing machines use the airlock device and two or more motors to provide a high rate of material flow, but with a resulting sacrifice in achieving full product value.
Thus, there remains a need for an apparatus for installation of insulation materials that uses a through blower concept, is very light weight, and also fully opens the insulation materials so that the full value as created in the insulation manufacturing plant can be achieved.
|
{
"pile_set_name": "USPTO Backgrounds"
}
|
tor of 225 and o.
9
Let w be ((-26)/(-4))/(1/2). Suppose -5*c = -0*z + 4*z - 73, -16 = -3*z + 4*c. Let u be ((-39)/z)/((18/56)/(-9)). What is the highest common divisor of u and w?
13
Let m(x) = -x**2 + 216. Let y(v) = v + 8. Let n be y(-6). Let a be 5/10*(n + -2). Let w be m(a). What is the greatest common factor of w and 24?
24
Suppose 9*v = 8*v. Let q(r) = -8 + 5*r + 2 - r**2 + 4*r + v. Let z be q(6). What is the greatest common factor of z and 12?
12
Suppose 5*s - 1445 = -p, s = 48*p - 45*p + 273. Calculate the greatest common divisor of 90 and s.
18
Suppose j - 22 - 26 = 0. Suppose 17*d - 335 = 277. Calculate the highest common factor of d and j.
12
Let j(l) be the first derivative of -15*l - 3 + 1/2*l**2 + 1/3*l**3. Let m be j(-7). Calculate the highest common factor of 3 and m.
3
Let b(d) = -4*d - 3. Let v be b(-3). Suppose -1 = 4*y - 13. Suppose -3*l - 4*f - 35 = -139, -105 = -3*l - y*f. Calculate the highest common divisor of v and l.
9
Suppose -1620 = -60*l + 40*l. Calculate the highest common factor of l and 108.
27
Suppose -15 = -5*c - 2*g, -5*g + 2*g = -5*c - 10. Let x be (2*(-3)/18)/(24/(-72)). Calculate the greatest common factor of x and c.
1
Suppose -2*l = -3*l + 3. Let z be (2 - 7) + l - -146. Suppose 5*p + 24 - z = 0. What is the highest common factor of 12 and p?
12
Let r = 183 - -441. What is the greatest common divisor of r and 72?
24
Let v be -22*(-2)/4 + -2. Suppose -4*q + v*q - 1910 = 0. Suppose -3*s = -4*z + 757, 2*z + 3*s - s - q = 0. Calculate the greatest common divisor of z and 19.
19
Let v(i) = -3*i**3 - i**2 - 2*i + 2. Let u be v(1). Let t be u + (10/15)/(2/156). Calculate the highest common divisor of t and 120.
24
Suppose 0*c - 4*c = -2*r + 4, 0 = -c - 1. Suppose 0 = -4*w - 3*z - 2*z + 267, -3*w + 4*z + 208 = r. What is the greatest common factor of w and 17?
17
Suppose -7 = -4*o + 17. Calculate the highest common factor of 24 and o.
6
Suppose -2*h - 27 = 5*n, -5*n + 9 = -2*h + 32. Let u be 0/(2 - (-1)/h). Suppose j - 7 - 6 = u. What is the greatest common factor of j and 26?
13
Let k(c) = 3*c**3 - 19*c**2 + 14*c - 43. Let f be k(6). What is the highest common divisor of f and 125?
5
Suppose 91 = 5*y + 2*y. Suppose 4*c - 97 = -y. What is the greatest common factor of c and 7?
7
Let h be 4/10 + 608/(-20). Let l = h - -54. Suppose 3*q + j = -6 + 22, -q + 4*j = -14. Calculate the highest common factor of q and l.
6
Let v be ((-248)/(-62))/(5 + -1) - -2345. Calculate the highest common divisor of 102 and v.
102
Let o(u) = u + 23. Let s be o(-12). Let w(v) = v**2 - 2*v - 7. Let m be w(s). What is the greatest common divisor of 23 and m?
23
Let m be ((-42)/(-2) - 2) + 1. Let p = -1859 + 1874. Calculate the greatest common divisor of m and p.
5
Let a be 1848/140 - 2/10. Let c be -1 - 1 - 251*1. Let o = -162 - c. What is the highest common divisor of o and a?
13
Let o be ((-3)/6)/((-1)/6). Suppose 2*t + 3*v = -2*t + 43, 0 = 5*t + o*v - 53. Let h be (4/5)/((-1)/(-50)). What is the greatest common factor of h and t?
10
Let m = -54 - -60. Let o be (-8)/m*(0 - (5 - -1)). What is the greatest common factor of o and 56?
8
Suppose -29 + 1 = 5*n - 4*p, 0 = n - p + 6. Let y be n/(2/10*-2). Let h = 129 - 79. Calculate the greatest common factor of h and y.
10
Let a be 0/((1 - -2)/3). Suppose -2*x = -a*x + 4. Let i be ((-16)/x)/((-5)/(-10)). Calculate the highest common factor of i and 144.
16
Suppose -6*y + 66 = -1734. What is the greatest common factor of 90 and y?
30
Let t = 283 - 227. Calculate the highest common factor of 952 and t.
56
Let z be (-722)/(-15) - 10/75. Calculate the greatest common factor of 360 and z.
24
Let s(n) = n**3 - 5*n**2 + 3*n - 14. Let p be s(7). Suppose -c - p = -8*c. What is the highest common divisor of 15 and c?
15
Let t(f) = 45*f**2 - 24*f + 23. Let b be t(1). What is the highest common factor of 28 and b?
4
Let x be 10/(-45) - (-640)/45. Let j(d) = -d**2 + 5*d - 4. Let w be j(3). Suppose w*c + 2*c = 28. What is the highest common factor of c and x?
7
Let q = 88 - 85. Suppose i = -q*i + 2*m + 88, 2*i + 2*m = 38. Calculate the greatest common divisor of 14 and i.
7
Suppose -30*t - 2*z = -28*t - 244, 4*t = -2*z + 484. Let s = 19 - 7. What is the highest common factor of s and t?
12
Let n be (-482)/(-4) + (-1)/(-6)*9. Let c = n - 34. What is the highest common factor of c and 8?
8
Suppose 14 = -n + 3*l, 2*l = -l. Let q be n/(-77) + (-42)/(-11). Suppose q*r = -r + 315. What is the greatest common divisor of r and 21?
21
Let k be (7 - 3) + 0/(-1). Let t = 22 - k. Suppose 8*n - 3*n = 4*q + t, 2*q = -5*n + 36. Calculate the greatest common factor of 48 and n.
6
Let o(c) = 2*c**2 - 75*c - 53. Let d be o(41). Suppose 6*u = 2*u + 52. Calculate the highest common factor of d and u.
13
Suppose 2*w = 17*w - 13230. What is the greatest common factor of w and 18?
18
Let j be (-6)/4 + (-5117)/(-86). Calculate the greatest common factor of j and 58.
58
Let x be 1/4 - 10349/(-316). What is the greatest common factor of 528 and x?
33
Let o(f) = 8*f**3 - 3*f**2 + 1. Suppose -3*y - 18 = -3*q, -3*q - 3*y = 1 + 5. Let n be o(q). Calculate the highest common factor of 424 and n.
53
Let m be (-11)/22*(-4 + 0). Suppose -5*s = -3*p - 2*p + 25, m*p = s + 6. What is the greatest common factor of 3 and p?
1
Let m be ((-176)/(-6))/((-8)/(-12)). What is the highest common divisor of m and 66?
22
Let m(v) = 2*v**2 + 7. Let n be m(-6). Let s = -43 + n. Suppose 0 = -2*l + 4, 15 + 27 = 3*x + 3*l. Calculate the greatest common divisor of x and s.
12
Let p = 957 + -942. What is the greatest common factor of 1455 and p?
15
Suppose 0*r = -4*p - r + 28, -4*p + 4*r + 48 = 0. Let j = 11 + -3. Calculate the highest common factor of p and j.
8
Let r be (-202)/(((-5)/10)/(2/(-8))). Let y = -80 - r. Calculate the highest common factor of y and 210.
21
Suppose -3*t = -q + 27, -3*t = -2*q - q + 27. Let o = -3 - t. Let l be 248/18 + o/27. What is the greatest common divisor of l and 35?
7
Let v(g) = 2*g + 16 - 3*g + 2*g. Let c(l) = -l**2 + 18*l - 79. Let r be c(13). Let t be v(r). Calculate the highest common factor of 10 and t.
2
Let s(k) = -k**3 + 4*k**2 + 20*k - 23. Let d be s(5). Let q be 5 + -2*(-3)/(-6). What is the highest common divisor of q and d?
4
Suppose 2*y = -89 - 177. Let f be (y/21)/((-1)/3). Let c be (-6 + 10)/(1/f). Calculate the greatest common divisor of c and 19.
19
Suppose -v + 80 = 4*v. Let r be (v/6)/((-8)/(-132)). Suppose -4*p - 2*y = -r, -2*p + p + 17 = -y. What is the highest common factor of p and 65?
13
Suppose 3*b - 4*p = 43, -b = -2*b + 3*p + 6. Calculate the highest common divisor of b and 70.
7
Let s(f) = -f**2 - 2*f + 4. Let o be s(0). Suppose -5*i + 24 = o*n - 74, -4*n = -3*i + 78. Calculate the greatest common factor of 55 and i.
11
Suppose 10*k - 235 = -115. Let n be (-2 - 2)*15/(-4). Let h = n - 9. What is the greatest common factor of h and k?
6
Let r = 620 + -596. Calculate the greatest common factor of 744 and r.
24
Let l be (-22)/(-8) - 6/(-24) - 3. Suppose l = -5*w + m + 143, -131 = -5*w + 2*m - 5*m. Calculate the greatest common divisor of 140 and w.
28
Suppose -3*j = -6*j + 72. Let x = 48 - j. Let u = 211 - 115. Calculate the highest common factor of x and u.
24
Let y = -68 + 28. Let n = -34 - y. What is the greatest common divisor of n and 54?
6
Let n be -6 + 0 - (-266)/7. Calculate the greatest common factor of 8 and n.
8
Suppose 25*s - 504 = 16*s. Calculate the greatest common divisor of s and 133.
7
Suppose 0 = 181*o - 186*o + 40. Suppose 4*j - 2*j - 24 = 0. What is the highest common divisor of o and j?
4
Suppose 4*y - 31 - 9 = 0. Let r = -59 - -69. Calculate the highest common divisor of r and y.
10
Suppose -21*p + 19*p = -8. Suppose -2*r - 3*q - p + 19 = 0, -4*r + 51 = -q. What is the highest common factor of 12 and r?
12
Suppose -2*r + 228 = -6*r - 4*w, 5*w - 20 = 0. Let v = -33 - r. What is the highest common divisor of v and 196?
28
Suppose -2*o - 32 = -2*x, -3*x + 6*o = 4*o - 46. Calculate the greatest common divisor of x and 112.
14
Let m(a) = -a**3 + 3*a**2 + 8*a + 3. Let z be m(6). Let d = -85 + 51. Let v = d - z. What is the highest common factor of 184 and v?
23
Let m = 139 + -91. Let a be (-4)/22 + 63390/165. What is the greatest common divisor of m and a?
48
Suppose -2*x + 600
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YEREVAN -- Armenian Defense Minister Seyran Ohanian has played down the significance of a possible sale of sophisticated Russian antiaircraft missiles to Azerbaijan, saying that it will not give Baku a "strategic advantage" in the unresolved Nagorno-Karabakh conflict.
The enclave, populated mainly by ethnic Armenians, erupted in ethnic clashes beginning in the late 1980s, prompting a war that left some 30,000 dead. The territory, which declared independence in 1991, remains in dispute despite years of failed efforts by international mediators and an Armenian-Azerbaijani cease-fire has not prevented occasional military skirmishes.
In an interview with RFE/RL's Armenian Service, Ohanian also said that he had "no doubts" that under an agreement signed with Russia last week, Russia would openly support Armenia in the event of a new conflict with Azerbaijan over the disputed region of Nagorno-Karabakh that "became a threat to the Republic of Armenia."
The agreement extended Russia's lease on its military base in northern Armenia until 2044 and gave it a greater role in ensuring Armenia's security. It also commits the Russians to supplying the Armenian military with modern weaponry.
"The agreement reaffirms the long-term character of the strategic alliance of the Republic of Armenia and the Russian Federation in accordance with requirements stemming from the security environment and the military-political situation in the region," Ohanian said.
Ohanian declined to specify what types of arms Moscow has pledged to supply to Yerevan within the framework of the new pact.
Pro-government politicians and some analysts in Yerevan believe that this will discourage Azerbaijan from acting on its frequent threats to resolve the dispute by force.
The deal was signed on August 20 during Russian President Dmitry Medvedev's visit to Yerevan following reports that Moscow plans to sell S-300 air-defense systems to Azerbaijan. The reports, not denied by Russian officials, have raised concerns in Armenia and Karabakh.
Opposition groups there say the long-range surface-to-air missiles would seriously limit the Armenian military's ability to hit strategic targets in Azerbaijan, and thereby encourage Baku to try to resolve the Karabakh dispute by force.
Karabakh-born Ohanian, who played a major role in the 1988-1994 war with Azerbaijan and subsequently commanded the Karabakh-Armenian army, dismissed such concerns.
"I must point out that the acquisition of Russian S-300 air-defense systems [by Azerbaijan] cannot directly influence the correlation of forces between Armenia and Azerbaijan, because their use by Azerbaijan against the Armenian Armed Forces would be fruitless under all possible scenarios," he said. "The reason for that is simple: we are very familiar with those systems, we have been exploiting them for quite a long time, and we know the possibilities of reducing the effectiveness of such systems."
Ohanian was likely referring to at least two batteries of S-300s that were deployed by Russia at its military base in Armenia in the late 1990s.
Top Russian military officials announced in early 2007 that Moscow has further upgraded Armenia's air defenses and trained Armenian military personnel to operate the air-defense systems. The Armenian military confirmed that, saying the training began in 2005.
Ohanian added that even if Azerbaijan does acquire S-300s, it would need "quite a lot of time" to develop an integrated radio-technical system for them.
He added that the missile deal would therefore not harm the Russian-Armenian military alliance.
"We are strategic partners, we are part of the same military-political system, our cooperation is quite close, and there is readiness on both sides for mutual assistance on any security issue," he said.
The defense minister declined to specify what kind of sophisticated arms Moscow has pledged to supply to Yerevan within the framework of the new agreement, citing "military secrecy."
Ohanian also reaffirmed his government's plans, announced earlier this month, to obtain new long-range, precision-guided weapons in the coming years. He said they would be aimed at the "strategic facilities" of Armenia's hostile neighbors.
|
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"I got nothing, sir." "No sign of Niobe or Ghost." "Nothing but blue pills." " Should we try to contact them?" " Won't matter." "My gut says they're down." " We should start back." " If that ship can fly, we need it." "I was afraid you'd say that." "Search every pipe, every hole, every crack we know." "Sweep as wide as possible, as fast as possible." " Lines are crawling with calamari." " The sooner we find them the better." " Thought you could use something to eat." " Thank you." "Any change?" "No." " How is he?" " He's gonna be fine." "At least until he wakes up." " What do you mean?" " Captain has some questions for him." "He'd better have some good answers." "You see these cuts?" "I think they're self-inflicted." " Why?" " VDTs maybe." "I don't know." "But like I said, the answer better be good." "Roland, I'd like to run another search through the Matrix." " For what?" " For Neo." "How could he be in the Matrix, sir?" "He's not plugged in." "Please, for me." "This is what keeps bothering me." "What?" "His neural patterns don't read like someone who's in a coma." "The strange thing is, I see these patterns all the time." " Where?" " On someone jacked in." "The big bupkes." "Nada." "He's not in there." " Sir, I got the projections." " How long?" "Based on point of entry and the past speed, the machines will be in Zion in 20 hours." "Jesus H. Christ." "All right, let's move with a purpose." "AK, I want you on holographics." "Mauser, I want forward and aft guns manned at all times." "And make sure we are running on as few pads as possible." " Hey!" "You got a call." " What?" "Operator." "It's Seraph." "I bring word from the Oracle." "You must come at once." "Good morning." " Who are you?" " My name is Sati." "Your name is Neo." "My papa says you're not supposed to be here." "He says you must be lost." "Are you lost, Neo?" " Where am I?" " This is the train station." " This isn't the Matrix?" " That's where the train goes." "That's where we're going, but you cannot go with us." " Why not?" " He won't let you." " Who won't?" " The Trainman." "I don't like him." "But my papa says we have to do what the Trainman says or else he will leave us here forever and ever." "Morpheus, Trinity, thank you for coming." "One thing I've learned in all my years is that nothing ever works out the way you want it to." " Who are you?" " I'm the Oracle." "I wish there was an easy way to get through this, but there ain't." "I'm sorry this had to happen." "I'm sorry I couldn't be sitting here like you remember me but it wasn't meant to be." " What happened?" "I made a choice, and that choice cost me more than I wanted it to." "What choice?" "To help you, to guide Neo." "Now, since the real test for any choice is having to make the same choice again knowing full well what it might cost I guess I feel pretty good about that choice because here I am, at it again." " Do you know what happened to Neo?" " Yes." "He is trapped in a place between this world and the machine world." "The link is controlled by a program called the Trainman." "He uses it to smuggle programs in and out of the Matrix." "If he finds out where Neo is before you get to him then I'm afraid our choices are going to become difficult." " Why?" " Because of who the Trainman works for." "The Merovingian." "He has placed a bounty on your lives." "You must be careful at all times." "Seraph knows how to find the Trainman." "He'll go with you." "For years he has protected me." "I hope he can do the same for you." "Please, follow me." "Oracle..." "I know, Morpheus." "I can see you're filled with doubt, clouded by uncertainty." "After everything that has happened, how can you expect me to believe you?" "I don't." "I expect just what I've always expected:" "For you to make up your own damn mind." "Believe me or don't." "All I can do is tell you that your friend is in trouble, and he needs your help." "He needs all our help." " Are you from the Matrix?" " Yes." "No." " I mean, I was." " Why did you leave?" " I had to." " I had to leave my home too." "Sati!" "Come here, darling." "Leave the poor man in peace." "Yes, Papa." "I am sorry." "She is still very curious." " I know you." " Yes, in the restaurant of the Frenchman." "I am Rama-Kandra." "This is my wife, Kamala." "My daughter, Sati." "We are most honored to meet you." " You are programs." " Oh, yes." "I am the power-plant systems manager for recycling operations." "My wife is an interactive software programmer." "She is highly creative." "What are you doing here?" "You do not belong here." "Kamala!" "Goodness, I apologize." "My wife can be very direct." "It's okay." "I don't have an answer." " I don't even know where "here" is." " This place is nowhere." "It is between your world and our world." " Who's the Trainman?" " He works for the Frenchman." "Why did I know you would say that?" "The Frenchman does not forget, and he does not forgive." " You know him?" " I know only what I need to know." "I know that if you want to take something from our world into your world that does not belong there, you must go to the Frenchman." "Is that what you're doing here?" " Rama, please." " I do not want to be cruel, Kamala." "He may never see another face for the rest of his life." "I'm sorry." " You don't have to answer that question." " No, I don't mind." "The answer is simple." "I love my daughter very much." "I find her to be the most beautiful thing I have ever seen." "But where we are from, that is not enough." "Every program that is created must have a purpose." "If it does not, it is deleted." "I went to the Frenchman to save my daughter." " You do not understand." " I just have never..." "Heard a program speak of love." " It is a human emotion." " No, it is a word." "What matters is the connection the word implies." "I see that you are in love." "Can you tell me what you would give to hold on to that connection?" "Anything." "Then perhaps the reason you are here is not so different than the reason I am here." "That's him." " Get away!" "Get away from me!" " We don't want trouble." " Get the hell away from me!" " We need your help." "I can't help you!" "No one can help you!" "Oh, no!" "Damn it!" " When is the train due?" " It is already late." "It's not like the Trainman to be late." " You think it has something to do with me?" " I cannot say." "Who knows such things?" "Only the Oracle." " You know the Oracle?" " Everyone knows the Oracle." "I consulted with her before I met with the Frenchman." "She promised to watch Sati after we said goodbye." "Goodbye?" " You're not staying with her?" " It is not possible." "Our arrangement with the Frenchman was for our daughter only." " My wife and I must return to our world." " Why?" " That is our karma." " You believe in karma?" "Karma is a word, like love." "A way of saying:" ""What I am here to do."" "I do not resent my karma." "I'm grateful for it." "Grateful for my wonderful wife, for my beautiful daughter." "They are gifts, and so I do what I must do to honor them." " Papa, the train!" " Yes." "Find your bag." "Quickly." "Can I carry that for you?" "All right." "Hurry it up, I'm late!" " Who are you?" " He is a friend." "I know you." " So that's what they wanted." " I need to get back." "I'll pay you whatever you want." "One way or another, I'm getting on this train." "Oh, no, no, no." "You'll stay here until the Merovingian says different." "If I know him you're gonna be here for a long, long time." " I don't want to hurt you." " You don't get it." "I built this place." "Down here, I make the rules." "Down here, I make the threats." "Down here, I'm God." "Get on the train or you'll stay here with him." "We should return to the Oracle." "She will know what to do." "No." "We know what has to be done." "Shit." "You gotta be kidding." "Holy shit, it's wingless." "I get it." "You must be ready to die." "I need to speak with him." "Only way you're getting through this door is over my big, dead ass." "So be it." "There are no weapons allowed in the club." "At the bottom of this elevator, there is a coat-check girl and if we are lucky, one man for checking weapons." "And if we're unlucky?" "There will be many men." "Can I take your...?" "Oh, my God." "What in the hell?" "I don't believe this." "Hey!" "The prodigal child returns." "Are you here for the bounty, Seraph?" "Tell me, how many bullets are there in those guns?" "I don't know, but I don't think you have enough." " We only want to talk." " Oh, yes." "I'm sure you do." "You have fought through hell to do so." "Yes?" "I'll tell you what I will do." "Put down the guns, and I will promise you a safe passage out of here." " All three of us?" " Oh, yes, yes." "Of course." "Who could have guessed we would see each other so soon after our last meeting?" "The Fates are too kind, eh?" "And since you, my little Judas, have brought them here..." "I can only surmise that the fortuneteller has found herself another shell?" "Disappointing, but not unexpected." "I do hope, however, she has the good manners to learn her lesson and to remember that there is no action without consequence." "And if you take something from me, you will pay the price." "You know why we are here?" "Come now, what kind of question is this?" "Of course I know." "It is my business to know." "Some might think this is a strange coincidence, but I do not." "I am curious, though, as to how it actually happened." " Do you know?" " No." "No?" "I did not think so." "But it is always best to ask, huh?" "We want to make a deal." "Always straight to business, huh, Morpheus?" "Okay." "I have something you want." "To make a deal you must have something I want, yes?" "And it so happens there is something I want." "Something I have wanted ever since I first came here." "It is said they cannot be taken they can only be given." "What?" "The eyes of the Oracle." "I told you before, there is no escaping the nature of the universe." "It is that nature that has again brought you to me." "Where some see coincidence, I see consequence." "Where others see chance I see cost." "Bring me the eyes of the Oracle and I will give you back your savior." "It seems a perfectly fair and reasonable deal to me." "Yes?" "No?" "I don't have time for this shit." "You wanna make a deal?" "How about this?" "You give me Neo, or we all die, right here, right now." "Interesting deal." "You are really ready to die for this man?" " Believe it." " She'll do it." "If she has to, she'll kill every one of us." "She's in love." "It is remarkable how similar the pattern of love is to the pattern of insanity." "Time's up." "What's it gonna be, Merv?" "Okay." "You got yourself in here you can get yourself out." "Are you ready for us?" "Almost, sir." "They got some pretty ancient hacks here." " But you found Neo?" " Can't you see him?" "No, sir." "We read something, but I couldn't tell what." "I can't leave yet." "I have to see her." "Now?" "It's my last chance." "That's it." "That's the secret." " You've got to use your hands." " Why?" "Cookies need love like everything does." "Neo!" "I was hoping to have these done before you got here." "Oh, well." "Sati, honey, I think it's time for a tasting." "Take the bowl to Seraph and find out if they're ready." "Okay." " I'm glad you got out." " Me too." "So do you recognize me?" " A part of you." " Yeah, that's how it works." "Some bits you lose, some bits you keep." "I don't yet recognize my face in the mirror but I still love candy." "No, thank you." "Remember how you were when you first walked through my door?" "Jittery as a June bug." "And now, just look at you." "You sure did surprise me, Neo, and you still do." "You gave me a few surprises too." "I hope I helped." "You helped me to get here, but my question is, why?" "Where is this going?" " Where does it end?" " I don't know." " You don't know or you won't tell me?" " I told you before no one can see beyond a choice they don't understand, and I mean no one." " What choice?" " It doesn't matter." "It's my choice." "I have mine to make, same as you have yours." "Does that include what things to tell me and what not to?" " Of course not." " Why didn't you tell me about the Architect?" "About Zion and the ones before me?" "Why didn't you tell me the truth?" " Because it wasn't time for you to know." " Who decided it wasn't time?" "You know who." "Know Thyself" "I did." "Then I think it's time for me to know a few more things." "So do I." "Tell me how I separated my mind from my body without jacking in." "Tell me how I stopped four sentinels by thinking it." "Tell me just what the hell is happening to me." "The power of the One extends beyond this world." "It reaches from here all the way back to where it came from." " Where?" " The source." "That's what you felt when you touched those sentinels, but you weren't ready for it." "You should be dead, but apparently you weren't ready for that either." "The Architect said that if I didn't return to the source, Zion would be destroyed tonight." "Please." "You and I may not be able to see beyond our own choices but that man can't see past any choice." " Why not?" " He doesn't understand them." "He can't." "To him, they are variables in an equation." "One at a time, each variable must be solved, then countered." "That's his purpose." "To balance the equation." "What's your purpose?" "To unbalance it." "Why?" "What do you want?" "I want the same thing you want, Neo." "And I'm willing to go as far as you are to get it." "The end of the war." "Is it going to end?" "One way or another." "Can Zion be saved?" "I'm sorry, I don't have the answer to that question, but..." "If there is an answer, there's only one place you're going to find it." " Where?" " You know where." "And if you can't find the answer then I'm afraid there may be no tomorrow for any of us." "What does that mean?" "Everything that has a beginning has an end." "I see the end coming." "I see the darkness spreading." "I see death." "And you are all that stands in his way." "Smith." "Very soon, he is going to have the power to destroy this world." "But I believe he won't stop there." "He can't." "He won't stop until there's nothing left at all." "What is he?" "He is you." "Your opposite, your negative." "The result of the equation trying to balance itself out." " And if I can't stop him?" " One way or another, Neo this war is going to end." "Tonight, the future of both worlds will be in your hands or in his." "How are you feeling?" "Are you all right?" "I need time." " That figures." " Captain Roland." " What's up, Maggie?" " Bane is." "He's conscious." "Good." "Maybe he's got some answers." "I love that smell." "I sure am going to miss it." " Oracle." " I know." "I know." "Sati, honey." "Take a few cookies and go with Seraph." "Can I come back?" "I would like to come back." " I would like that too." " Then I'll see you tomorrow." "I hope so, hon, I hope so." "I'm scared, Seraph." "Come." "He's following us." "Well, well, it's been a long time." "I remember chasing you was like chasing a ghost." "I have beaten you before." "Yes, true, but as you can see, things are a little different now." "And you must be the last exile." " The Oracle told me about you." " Really?" "What did she say about me?" "That you were a bad man." "Oh, I'm not so bad, once you get to know me." "The great and powerful Oracle, we meet at last." "I suppose you've been expecting me, right?" "The all-knowing Oracle is never surprised." "How can she be?" "She knows everything." "But if that's true, then why is she here if she knew I was coming?" "Why wouldn't she leave?" "Maybe you knew I would do that, maybe you didn't." "If you did, that means you baked those cookies and set that plate there deliberately, purposefully which means that you're sitting there also deliberately, purposefully." "What did you do with Sati?" ""Cookies need love like everything does."" "You are a bastard." "You would know, Mom." "Do what you're here to do." "Yes, ma'am." "I really wish I could help, but I just..." "I don't remember any of it." "What about the cuts on your arms?" "Those cuts are more than one day old." "Yeah, definitely." "You're right about that, sir." "They look like they might be self-inflicted." "But why would I do something like that to myself?" "Unless, of course, I wasn't myself." "But if I'm not me, then who am I?" " Has this man been tested for VDTs?" " Yes, sir." "It was negative." "But he is showing a lot of unusual neural activity." "Some cross-synaptic firing, as well as signs of recent trauma with fresh fibrotic scarring throughout the cortex." "I want the truth." "I don't care what it takes." "Make him remember." "Sir, we found her." " The "Logos"?" " Yes, sir." "About time we had some goddamn good news." "Are the thermals picking up any signs of life?" "No, sir." "Nothing yet." " What about the ship?" " Holographic says the hull's still intact." " Drop down and keep a man in the turret." " Yes, sir." "Get a full diagnostic on that ship as fast as humanly possible." "Careful, sir." "The squids are sneaky bastards." "It could be a trap." "What was that?" "You can put that shit away, boys." "All she needs is a jump." " Niobe." " Morpheus." " Are you all right?" " Yes, I'm fine." "We didn't know what happened after..." " I'm sorry." " It's okay." "I'm happy to see you too." " Did you get Neo out?" " Yes." " How did you know about that?" " The Oracle." " You saw her?" " Just before the sentinels found us." " What did she tell you?" " The same thing she always does." "Exactly what I needed to hear." "In 12 hours, the machines will breach the dock walls." "Every simulation we've run, we've seen that once the machines are inside the city the odds of our survival decrease dramatically." "Thus, our primary objective must be to destroy or disable the diggers inside the dock." "If we can do that, perhaps we can prevent them from ever reaching the city." "If not, the only place we will be able to mount a defense will be at the entrance of the temple." "It is small and will force them into a bottleneck allowing us to concentrate the remainder of our defense." "We understand that you've requested additional volunteers." "That is correct." "Precisely what size force are you planning to commit to the primary dock objective?" "Right now, the entire APU Corps and half the infantry." " Half the infantry?" " If it were up to me, councillor I'd take every man, woman and child, put a gun in their hand and march them into that dock." "Perhaps it is best that it is not up to you." "Time will tell, councillor." "Commander, just one more question." "Has there been word from the "Nebuchadnezzar"?" "None, and at this point there's no reason to expect there will be." "Perhaps." "But we can hope." "I'm afraid hope is an indulgence I don't have time for." " Zee, what are you doing?" " Making shells." "They're evacuating our level." "We have to go." " I'm not going with you." " What?" "They've called for volunteers to hold the dock." "Kids, you stay here." "I know how you feel, but you can't do this." " I have to." " Why?" "Because I love him." "I love him the same as he loves me." "And if I were out there and he were here, I know what he'd do." "But you're gonna get yourself killed." "It's crazy, Zee." "Maybe it is." "But ask yourself, if it were Dozer and you knew the only chance you had to see him was to hold the dock what would you do?" "Make shells." "What the shit is going on over here?" "An accident, sir." "I didn't see..." " I'm sorry." " Who the hell are you?" "A unit volunteer, sir." "What's a pod-born pencil-neck like you doing volunteering for my Corps?" "I want to do my part, sir." "We gotta hold the dock." " How old are you, kid?" " Eighteen." "Should have said 16, I might have believed that." "Okay, I'm 16." "Minimum age for the Corps is 18." "Sixteen's too young." "The machines don't care how old I am." "They'll kill me just the same." "Ain't that the goddamn truth." "Give me a chance, sir." "I won't let you down." "You do, and you'll find me and the machines have got something in common." "Okay, charge the igniter." "She lives again." "You want to reload the operations software, Sparky?" "Yeah." "Could you clean the windshield while you're at it?" "Uplinks in place." "I'm bringing her back online." "Looking good, except..." "Something's wrong with the Matrix feed." "No, there's not." "You're looking at what we're looking at." " What's going on in there?" " Whatever it is, it can't be good." "The machines have taken Junction 21." "The way I see it, if we drop down from broadcast here intersect 153, we might surprise them." "We'll go first, hammer as deep as we can, then blow our EMP." "Hopefully we can punch a hole big enough for you to get through." "It ain't pretty, but the way I see it, it's the only way back." "No, it's not." "There's another way." "A support line." "It drops down right here, 1000 meters short of 21." "If we're lucky, we may be able to slip down without them ever knowing." "That's a mechanical line." "It's impossible." " No one can pilot a mechanical." " I can." " Bullshit." " I've done it." " That was a long time ago, Niobe." " I said I can do it." "So what?" "You'll be the only one that can." "There's no way we can follow you." "Hi." "I know time is always against us." "I'm sorry that I took so long." " But I wanted to be sure." " Sure of what?" " I know what I have to do." " What?" "There's no easy way to say this, so I'll just say it." " I have to take one of the ships." " What?" "To go where?" "To the machine city." "I know it's difficult to understand." "No, it's not." "You're out of your goddamn mind." "I still have to go." "In 100 years, no ship has been within 100 kilometers of it." " You'll never make it." " I have to try." " Is this what the Oracle has told you?" " No." "This is asinine." "If you want to kill yourself, do it but do it without wasting our ships." " You have to believe me." "I have to go." " Bullshit!" "I am the captain of this ship." "I say where it has to go!" "And this ship will go to hell long before I let you take it anywhere." " He can take mine." " You can't do that." "Don't try to tell me what I can or cannot do with my ship after that little speech." " But for chrissake, Niobe..." " I'll pilot this ship, he can take mine." "If we leave inside the hour, we should reach Zion as the machines do." "That's as good a plan as any." "It's a waste." "A goddamn waste." "Two ships, two directions." "Sounds like providence, doesn't it, Morpheus?" " You've never believed in the One." " I still don't." "Then why are you doing this?" "I believe in him." "Thank you." " What's that for?" " To help you relax." "To make it easier for you to remember." "What if I don't want to remember?" "Why would you want that?" "What if I blew that EMP?" "What if I did destroy those ships and I am responsible for the deaths of all those men?" "If I did that it wouldn't be very safe for me here, would it?" "Of course, it might not be very safe for you either." "I'm ready." "Trinity there's something I have to say." "Something you need to understand." "I know I'm supposed to go." "But beyond that I don't know." "I..." "I know." "You don't think you're coming back." "I knew it the moment you said you had to leave." "I could see it in your face." "Just like you knew the moment you looked at me that I was coming with you." "I'm scared, Trin." "So am I." "It took me 10 minutes to buckle up one boot." "But I'll tell you something." "Six hours ago, I told the Merovingian I was ready to give anything and everything for you." "Do you know what's changed in the last six hours?" "No." "Nothing." " Finished loading ammunition?" " Just about." "Let's move it." "We are out of time." "You're not leaving them anything?" "He said he didn't need it." "I ain't saying goodbye." "I'm saying good luck." "Thank you." "I can only hope you know what you're doing." "Me too." "It was an honor, sir." "No, the honor is still mine." " We're ready, sir." " About damn time." "We're already late, captain, so let's hit it and hit it hard." "Bye, baby." "Take good care of them." "Ready?" "Engine's still firing." "Must be a fuse." "I'll check it out." " I should've known he'd send his bitch first." " Bane." "No one ever got away from me as many times as you did." "Every time, I thought it was the last." "Every time, I was sure we had you, but you'd slip through our fingers." "I really can't express just how aggravating that can be." "What are you talking about?" "I think I might enjoy killing you as much as killing him." "Neo, it's Bane." "He's psychotic!" "You're gonna pay for that." " Twenty-seven kilometers to go." " We got an emergency." " What is it, AK?" " It's Maggie, sir." "She's dead, murdered." "I think it was Bane." "Goddamn it." "I knew it." "I knew he was out of his goddamn mind." "He fired that EMP." "Goddamn it, I should have beaten it out of him." "We searched the whole ship, captain." "He ain't here." " I know where he is." " The "Logos"." " We gotta go back." " Too late." " What if they need our help?" " It's too dangerous." " Why?" " Because if he's killed them he'll control another EMP." "At this point, they're on their own." "Just like us." "Mr. Anderson." "I see you are as predictable in this world as you are in the other." " What?" " He's out of his mind." "It might appear that way." "But Mr. Anderson and I know that appearances can be deceiving." "Confused, Mr. Anderson?" "It'll all become clear in a moment." "Thank you for bringing me the gun." "You can set it down there." "Don't do it." " Shoot." "Shoot now." " Yes, shoot." "Fry us." "Burn us alive." "Do it." "If you don't, he'll kill us both." "Look at him." "He knows he should do it, but he won't." " He can't." " Do it." "Back away from the gun and turn around." "Let her go." "Somehow familiar, isn't it?" "We've been here before, you and I, remember?" "I do." "I think of nothing else." " Who are you?" " Still don't recognize me?" "I admit, it is difficult to even think encased in this rotting piece of meat." "The stink of it filling every breath, a suffocating cloud you can't escape." "Disgusting." "Look at how pathetically fragile it is." "Nothing this weak is meant to survive." " What do you want?" " I want what you want." "Yes." "That's it, Mr. Anderson." "Look past the flesh." "Look through the soft gelatin of these dull cow eyes and see your enemy." " No." " Oh, yes, Mr. Anderson." " It can't be." " There's nowhere I can't go." "There is nowhere I won't find you." " It's impossible." " Not impossible." "Inevitable." "Goodbye, Mr. Anderson." "This is it." "It's gotta be." "Oh, no." "I wish you could see yourself, Mr. Anderson." "Blind messiah." "You're a symbol for all of your kind, Mr. Anderson." "Helpless." "Pathetic." "Just waiting to be put out of your misery." "I can see you." "It's not over, Mr. Anderson." "It's not over." " Trinity!" " Neo." "Neo." "Oh, no." "Your eyes." "I'll be okay." "It's all right, Trin." "But I think you're gonna have to drive." "Seismic's projecting 22 minutes to breach." "They can't know we don't have an EMP." "They'll have to attack in waves." "Concentrate our offense on the diggers." "Order the APUs into position." "Yes, sir." "Come on, move." "All right!" "This is it." "Now, you all know me, so I'll just say this as simple as I can." "If it's our time to die, it's our time." "All I ask is, if we have to give these bastards our lives we give them hell before we do!" "Yeah!" " You scared, Charra?" " Shit, yeah." "But I'll make you a deal." "You keep loading, I keep shooting." "Deal." " Holy Christ, would you look at that." " Quiet." " How far to the opening?" " 1.4 kilometers." " We're still generating too hot a field." " Ghost, kill all auxiliary systems." " Give me full manual." "Drop down four pads." " It'll bottom out." " Easy, baby." " Seven hundred meters." "If we can just get close enough..." "Six hundred meters." "There." "Shit!" " Jig's up, here they come." " Give me full power." "Man the gun turrets, every goddamn one of them." "Go!" "Ghost, you're the best gunner, go with them." "Morpheus, take his place." "I'm coming, baby." "Here they come." " Slow down, this ain't the "Logos"." " Hang on to your lunch, Roland, here we go." "Holy Christ, I didn't know this ship could do that." "Breach!" "The dock is breached!" "Knuckle up!" "For Zion!" "Shit." "Come on, let's go." "Knuckle up!" "Reload!" "Reload nine!" "Go, go!" "Move, move!" "Watch your left!" "Don't let them through!" "Zuka!" "Oh, my God." "Where the hell's my infantry?" "I want that goddamn machine taken down!" "Dig this!" "Oh, shit!" " Bogey two at the breach point!" " Goddamn it!" "Shit, she's got a fat ass." " Keep them off me!" " Christ, there's a shitstorm of them." " You see that?" " They're after the radio." "Stop them!" "Damn it!" "Yeah." "Grab my belt." "Just give me one clean shot." "Damn it!" "Charra!" " I got incoming!" " We got a dock full of incoming!" " Yes, sir." "But this is different, sir." " What?" "I think it's one of ours, sir." " That's impossible." " Holographics are trying to confirm." "Contact them." "I want access codes." "We're trying, sir." "There's no response." "It's a trick." "That's not one of ours, it can't be." "That's a mechanical line." "No one can pilot mechanical." " Forward aft, 30 degrees at 80 percent." " Thirty degrees, 80." " Lower starboard 60 degrees, 20 percent." " Sixty degrees." " Shit!" "Come on, keep up!" " I'm trying!" "Sir, holographic confirms." "It's the "Hammer", sir." "How can that be?" "It's under attack, sustaining damage." "At its present velocity, it'll reach Gate Three in 12 minutes." "Their EMP could take out every sentinel." "It'd take out more than that." "It'll wipe out our defense system." "We blow an EMP inside, we lose the dock." "Sir, we already lost the dock." "Open the gate." "Gate Three is not responding!" "We've taken critical damage, sir!" "We've lost control!" "We can't open it!" "There's the exit." "On my mark, give me full power 90 degrees to lower-left starboard." "Full power ninety degrees." "Now!" "Hold on, baby." "Goddamn, woman, you can drive." "We ain't home yet." "What about the gate?" "Sentinels are inside the dock." "Are we too late?" " How many APUs are operational?" " Thirteen, sir." "Get me the one closest to Gate Three." "Reload!" "He's pissing metal." "Go!" "Go!" "Heads up!" "They're coming around!" "Behind you!" "It's jammed!" "Forget it, kid!" "Get out of here!" "Got it!" "Captain Mifune." "Oh, no." "They're coming." "They're coming." "The "Hammer"." "What?" "You have to open that gate." "Cut the counterweights." "You can do it." "Hurry." "There's no time." "Captain I didn't finish the training program." "Neither did I." " Lock that down!" " Kill the feeder!" "We won't make it." "We gotta blow the EMP now." "Come on, someone, please." "Keep the weight forward." "Light as a feather." "Holographic reports Captain Mifune's APU is up and moving to Gate Three." "Don't over-squeeze the trigger." "Mifune's APU just reached Gate Three." " How much time?" " Two minutes to impact." " Captain Mifune, do you copy?" " I think his radio is down, sir." "Mifune this is Lock." "I don't know if you can hear me, but if you can the "Hammer"'s two minutes away." "You've got two minutes, captain, to get that gate open." "Link." "Get to the main deck!" "Charge the EMP!" "Do it, kid." "Neo." "I believe." " Yes!" " Can we make it?" "We ain't come this far." "Almost home, almost home." "Burn it, Link!" " You did it." " No." "We did it." "You're a hell of a pilot." "Some things in this world never change." "But some things do?" "Luckily some things do." "Link!" " Zee?" "Zee!" " Link!" "I knew you'd come." "I knew it." "I made a promise." "You did wear it." "Are you kidding?" "I'm never gonna take it off." "Three captains, one ship." "I'll assume the other ships were lost under equally pointless circumstances." "Good to see you too, Jason." "The Council's waiting to hear an explanation." "Forgive me for not attending, but I have to try to salvage this debacle." "Did I miss something?" "I thought we saved the dock." "That's the problem with you people." "You can't think but five minutes in front of you." "That EMP knocked out almost every piece of hardware and every APU." "If I were the machines, I'd send every sentinel here now." "Save the dock, captain?" "You handed it to them on a silver platter." "Come on, get it cut." "The bridge is clear." "Do you hear that?" "Get that cable run!" " I want the system back online!" " It's the dock." "They've got incoming." "Order everyone to fall back." "Seal the shaft." "Now." "Move it!" "Oh, my God." " All clear." " Do it." " Go!" "Come on!" "Run!" " Go!" "Your move." "So you gave them your ship?" " That is correct, councillor, I did." " Knowing what he planned to do with it?" "The Oracle said nothing of this?" "She told me Neo would need my help, and I would choose to help him or not." "But what hope can a single vessel have against their entire defense system?" "None." "It's completely impossible." "But he wouldn't listen." "He wouldn't even take any ammunition." "He was totally out of his goddamn mind." "No, he wasn't." "Neo is doing what he believes he must do." "I don't know if what he's doing is right." "I don't know if he'll reach the machine city." "And if he does, I don't know what he can do to save us." "But I do know that as long as there is a single breath in his body he will not give up." "And neither can we." "Temperature's dropping." "Here we go." "We're over the fields, aren't we?" "How do you know that?" "I can feel them." "Over there." "There's our path." "Can you see it?" "Three lines." "Power lines." "Follow them." "What are they doing?" "I don't know." "Lieutenant?" "Goddamn it." "What do we do now?" "It is now a matter of time." "The machines will breach the walls of this city." "I recommend that the Council join the other non-military personnel inside the temple." "How long do we have?" "Two hours." "Maybe less." "My men have fortified the entrance with enough artillery to make our last stand." "Beyond that, there isn't anything more I can do." "Commander, do you think that we have any chance of surviving?" "If I were you, I wouldn't ask me that question." "I would ask him." " Why?" " He's the one who believes in miracles." "There." "Those mountains." "That's it." " Do you see what's out there?" " Yes." "If you tell me we'll make it, I'll believe you." "We'll make it." "We have to." "Sentinels." "There's too many!" "Gotcha!" " I need help here!" " I can't beat them." " What do we do?" " Go up, over them." " What?" " The sky!" "It's the only way." "Then up we go." "Beautiful." "Pump the igniter." "The ship will start." "Again." "Slowly." "Now!" "Trin?" "Trinity?" "Trinity?" " I'm here." " Where?" "Here." "We made it." "You said we would." "It's unbelievable, Trin." "Light everywhere." "Like the whole thing was built of light." "I wish you could see what I see." "You've already shown me so much." "What is it, Trinity?" "What's wrong?" "I can't go with you, Neo." "I've gone as far as I can." "What?" "Oh, no." "Oh, no." "No, no." "It's all right." "It's time." "I've done all that I could do." "Now you have to do the rest." "You have to finish it." "You have to save Zion." "I can't." " Not without you." " Yes, you can." "You will." "I believe it." "I always have." "Trinity." "Trinity, you can't die." "You can't." "You can't." "Yes, I can." "You brought me back once." "But not this time." "Do you remember on that roof, after you caught me the last thing I said to you?" "You said, "I'm sorry"." "I wish I hadn't." "That was my last thought." "I wished I had one more chance to say what really mattered." "To say how much I loved you." "How grateful I was for every moment I was with you." "But by the time I knew how to say what I wanted to, it was too late." "But you brought me back." "You gave me my wish." "One more chance to say what I really wanted to say." "Kiss me." "Once more kiss me." "Get that ammunition where it belongs!" "You got one chance to get this right." "Get that damn thing mounted!" " Hurry." " Let's go, let's go, let's go!" "Neo if you're gonna do something, do it quick." "I only ask to say what I've come to say." "After that, do what you want, and I won't try and stop you." "Speak." "The program Smith has grown beyond your control." "He will spread through this city as he spread through the Matrix." "You cannot stop him." " But I can." " We don't need you!" "We need nothing!" "If that's true, then I've made a mistake, and you should kill me now." "What do you want?" "Peace." "What are they doing?" "What are you doing?" "Morpheus!" "And if you fail?" "I won't." "Neo." "He fights for us." "Mr. Anderson, welcome back." "We missed you." "You like what I've done with the place?" "It ends tonight." "I know it does." "I've seen it." "That's why the rest of me is just going to enjoy the show because we already know that I'm the one that beats you." "Can you feel it, Mr. Anderson closing in on you?" "Oh, I can." "I really should thank you for it." "After all, it was your life that taught me the purpose of all life." "The purpose of life is to end." "Why, Mr. Anderson?" "Why, why, why?" "Why do you do it?" "Why?" "Why get up?" "Why keep fighting?" "Do you believe you're fighting for something?" "For more than your survival?" "Can you tell me what it is?" "Do you even know?" "Is it freedom or truth?" "Perhaps peace?" "Could it be for love?" "Illusions, Mr. Anderson." "Vagaries of perception." "Temporary constructs of a feeble human intellect trying desperately to justify an existence that is without meaning or purpose!" "And all of them as artificial as the Matrix itself although only a human mind could invent something as insipid as love." "You must be able to see it, Mr. Anderson." "You must know it by now." "You can't win." "It's pointless to keep fighting." "Why, Mr. Anderson, why?" "Why do you persist?" "Because I choose to." "This is my world!" "My world!" "Wait." "I've seen this." "This is it." "This is the end." "Yes." "You were laying right there, just like that." "And I..." "I stand here, right here, and I'm supposed to say something." "I say:" ""Everything that has a beginning has an end, Neo."" "What?" "What did I just say?" "No, no." "This isn't right." "This can't be right." "Get away from me!" "What are you afraid of?" "It's a trick." "You were right, Smith." "You were always right." "It was inevitable." "Is it over?" "Oh, no, no, no." "No, it's not fair." "It is done." "It doesn't make sense." "He did it." "He saved us." "He saved us." "It's over!" "He did it!" "He did it!" "He did it!" "It's over!" "It's over!" "He did it!" "He did it!" " What is it?" "What happened?" " Sir, he did it, sir." "Neo, he did it." " Did what?" " He ended the war." "The machines, they're gone." "The war is over, sir." "The war is over." "Zion!" "Zion!" "Zion, it's over!" "It's over!" "The war is over!" "The war is over!" "I have imagined this moment for so long." "Is this real?" "Neo, wherever you are thank you." "Good morning." "Well, now." "Ain't this a surprise." "You played a very dangerous game." "Change always is." "Just how long do you think this peace is going to last?" "As long as it can." "What about the others?" "What others?" "The ones that want out." "Obviously, they will be freed." "I have your word?" "What do you think I am?" "Human?" "Oracle!" "We were afraid we might not find you." "Everything's okay now." "Look, look." "Just look at that." "Beautiful." "Did you do that?" "For Neo." "That's nice." "I know he'd love it." "Will we ever see him again?" "I suspect so." "Someday." "Did you always know?" "Oh, no." "No, I didn't." "But I believed." "I believed."
|
{
"pile_set_name": "OpenSubtitles"
}
|
Italo-Yemeni Treaty
The Italo-Yemeni Treaty of 1926 (also known as the Treaty of San'a) was a treaty between the Kingdom of Italy and Mutawakkilite Kingdom of Yemen. The treaty was signed in September 1926 and was described as a friendship treaty. At the time, Italy was ruled by the fascist National Fascist Party with Benito Mussolini as head of government. The treaty recognized Imam Yahya Muhammad Hamid ed-Din as King of Yemen and acknowledged his claims to Aden. The treaty was renewed on October 15, 1937, after Italy had annexed Abyssinia (present-day Ethiopia).
Background
The Red Sea was of strategic importance to the United Kingdom due to both trade and as a route for its navy to pass through in order to reach India among other places. South of Yemen was the British Colony of Aden and Aden Protectorate which were at considerable risk of anti-colonialist rebellions.
Italy had colonies of its own in the region: Eritrea and Somaliland, both of low profitability. There was expectation that increased ties with Yemen would fuel increased trade with the colonies and bring the region into the Italian sphere of influence.
The Kingdom of Yemen at this point had its eye on annexing Aden and Imam Yahya also had aspirations for a Greater Yemen.
External links
The foreign office and Anglo-Italian involvement in the Red Sea and Arabia, 1925-28.
Category:1926 in Italy
Category:1926 in North Yemen
Category:September 1926 events
Category:Treaties concluded in 1926
Category:Treaties of the Mutawakkilite Kingdom of Yemen
Category:Interwar period treaties
Category:Italy–Yemen relations
Category:Treaties of the Kingdom of Italy (1861–1946)
|
{
"pile_set_name": "Wikipedia (en)"
}
|
Q:
What's wrong when I use ImageApplyIndexed?
I am reading a ebook called Basic image processing in Mathematica which can download free in the iBooks.
In the page 114,it introduce a function ImageApplyIndexed.
i = ConstantImage[Black, {50, 50}];
{width, height} = ImageDimensions@i;
Clear[f];
f[pixel_, pos_] :=
Module[{dis = Norm[{width/2, height/2} - pos]},
pixel + dis/Norm[Min[width/2, height/2]]];
ImageApplyIndexed[f, i]
output:
But if I change the ImageSize of i:
i = ConstantImage[Black, {100, 50}];
output:
I don't know why the center of the black circle not in the center of the image?
A:
ImageApplyIndexed passes an index to your function, not X/Y coordinates, i.e. a list containing the row index and the column index (in that order). So you need to you Norm[{height/2, width/2} - pos. Because height is the number of rows, and width is the number of columns.
|
{
"pile_set_name": "StackExchange"
}
|
{
"pile_set_name": "PubMed Central"
}
|
|
Unusual destructive and hypertrophic arthropathy of the atlanto-axial joint in calcium pyrophosphate dihydrate deposition disease.
We present two cases of destructive and hypertrophic arthropathy of the atlanto-axial joint in patients with calcium pyrophosphate dihydrate deposition disease. The clinical symptoms were mild and intermittent in spite of severe radiographic changes.
|
{
"pile_set_name": "PubMed Abstracts"
}
|
A gay man who refrained from having sex for a year so that he could donate blood opened up about his experience in an interview with The Huffington Post’s Noah Michelson this week.
In early 2016, Jay Franzone vowed to commit to 12 months without sexual contact with another man so that he could adhere to the Food and Drug Administration’s (FDA) recently revised guidelines on blood donations. The FDA lifted its lifetime ban on men who have sex with men (MSM) donating blood in December 2015, allowing men who want to give blood to do so if they haven’t had sex with another man in the past year.
The revised policy continues to face heavy criticism, especially since it still prohibits donations from sexually active MSM who are married, in a monogamous relationship or following strict guidelines for safer sex. Franzone, who graduated from Massachusetts’s Lasell College in December, said he wanted to highlight the FDA guidelines’ still-discriminatory stance on MSM by following it.
“I did it for myself so I could be able to give blood and give back... but also because little kids... are hearing this ban for the first time... they’re hearing that something’s wrong with them,” the 21-year-old, who finally was able to donate blood on Jan. 10, said. “That’s a deep-rooted stigma that we need to break through.”
Watch the full interview with Jay Franzone above, and don’t miss his Jan. 12 New York Times Op-Ed explaining his motivation here. To watch Franzone give blood after a year of abstaining for sex, check out the video below.
|
{
"pile_set_name": "OpenWebText2"
}
|
const {
Module,
Component,
Template,
Entity,
Mixin,
Filter,
Directive,
Locale,
Shortcut,
Utils,
ApiService,
EntityDefinition,
WorkerNotification,
DataDeprecated,
Data,
Classes,
Helper
} = Shopware;
describe('core/common.js', () => {
it('should contain the necessary methods for the module factory', async () => {
expect(Module).toHaveProperty('register');
});
it('should contain the necessary methods for the component factory', async () => {
expect(Component).toHaveProperty('register');
expect(Component).toHaveProperty('extend');
expect(Component).toHaveProperty('override');
expect(Component).toHaveProperty('build');
expect(Component).toHaveProperty('getTemplate');
});
it('should contain the necessary methods for the template factory', async () => {
expect(Template).toHaveProperty('register');
expect(Template).toHaveProperty('extend');
expect(Template).toHaveProperty('override');
expect(Template).toHaveProperty('getRenderedTemplate');
expect(Template).toHaveProperty('find');
expect(Template).toHaveProperty('findOverride');
});
it('should contain the necessary methods for the entity factory', async () => {
expect(Entity).toHaveProperty('addDefinition');
expect(Entity).toHaveProperty('getDefinition');
expect(Entity).toHaveProperty('getDefinitionRegistry');
expect(Entity).toHaveProperty('getRawEntityObject');
expect(Entity).toHaveProperty('getPropertyBlacklist');
expect(Entity).toHaveProperty('getRequiredProperties');
expect(Entity).toHaveProperty('getAssociatedProperties');
expect(Entity).toHaveProperty('getTranslatableProperties');
});
it('should contain the necessary methods for the entity factory', async () => {
expect(Entity).toHaveProperty('addDefinition');
expect(Entity).toHaveProperty('getDefinition');
expect(Entity).toHaveProperty('getDefinitionRegistry');
expect(Entity).toHaveProperty('getRawEntityObject');
expect(Entity).toHaveProperty('getPropertyBlacklist');
expect(Entity).toHaveProperty('getRequiredProperties');
expect(Entity).toHaveProperty('getAssociatedProperties');
expect(Entity).toHaveProperty('getTranslatableProperties');
});
it('should contain the necessary methods for the mixin factory', async () => {
expect(Mixin).toHaveProperty('register');
expect(Mixin).toHaveProperty('getByName');
});
it('should contain the necessary methods for the filter factory', async () => {
expect(Filter).toHaveProperty('register');
expect(Filter).toHaveProperty('getByName');
});
it('should contain the necessary methods for the directive factory', async () => {
expect(Directive).toHaveProperty('register');
expect(Directive).toHaveProperty('getByName');
});
it('should contain the necessary methods for the locale factory', async () => {
expect(Locale).toHaveProperty('register');
expect(Locale).toHaveProperty('extend');
expect(Locale).toHaveProperty('getByName');
});
it('should contain the necessary methods for the shortcut factory', async () => {
expect(Shortcut).toHaveProperty('register');
expect(Shortcut).toHaveProperty('getShortcutRegistry');
expect(Shortcut).toHaveProperty('getPathByCombination');
});
it('should contain the necessary methods for the utils', async () => {
expect(Utils).toHaveProperty('throttle');
expect(Utils).toHaveProperty('debounce');
expect(Utils).toHaveProperty('get');
expect(Utils).toHaveProperty('object');
expect(Utils).toHaveProperty('debug');
expect(Utils).toHaveProperty('format');
expect(Utils).toHaveProperty('dom');
expect(Utils).toHaveProperty('string');
expect(Utils).toHaveProperty('types');
expect(Utils).toHaveProperty('fileReader');
expect(Utils).toHaveProperty('sort');
expect(Utils).toHaveProperty('array');
});
it('should contain the necessary methods for the ApiService', async () => {
expect(ApiService).toHaveProperty('register');
expect(ApiService).toHaveProperty('getByName');
expect(ApiService).toHaveProperty('getRegistry');
expect(ApiService).toHaveProperty('getServices');
expect(ApiService).toHaveProperty('has');
});
it('should contain the necessary methods for the EntityDefinition', async () => {
expect(EntityDefinition).toHaveProperty('getScalarTypes');
expect(EntityDefinition).toHaveProperty('getJsonTypes');
expect(EntityDefinition).toHaveProperty('getDefinitionRegistry');
expect(EntityDefinition).toHaveProperty('get');
expect(EntityDefinition).toHaveProperty('add');
expect(EntityDefinition).toHaveProperty('remove');
expect(EntityDefinition).toHaveProperty('getTranslatedFields');
expect(EntityDefinition).toHaveProperty('getAssociationFields');
expect(EntityDefinition).toHaveProperty('getRequiredFields');
});
it('should contain the necessary methods for the WorkerNotification', async () => {
expect(WorkerNotification).toHaveProperty('register');
expect(WorkerNotification).toHaveProperty('getRegistry');
expect(WorkerNotification).toHaveProperty('override');
expect(WorkerNotification).toHaveProperty('remove');
expect(WorkerNotification).toHaveProperty('initialize');
});
/**
* @deprecated 6.1
*/
it('should contain the necessary methods for the DataDeprecated', async () => {
expect(DataDeprecated).toHaveProperty('LocalStore');
expect(DataDeprecated).toHaveProperty('UploadStore');
expect(DataDeprecated).toHaveProperty('CriteriaFactory');
});
it('should contain the necessary methods for the Data', async () => {
expect(Data).toHaveProperty('ChangesetGenerator');
expect(Data).toHaveProperty('Criteria');
expect(Data).toHaveProperty('Entity');
expect(Data).toHaveProperty('EntityCollection');
expect(Data).toHaveProperty('EntityDefinition');
expect(Data).toHaveProperty('EntityFactory');
expect(Data).toHaveProperty('EntityHydrator');
expect(Data).toHaveProperty('Repository');
});
it('should contain the necessary methods for the Classes', async () => {
expect(Classes).toHaveProperty('ShopwareError');
expect(Classes).toHaveProperty('ApiService');
});
it('should contain the necessary methods for the Helper', async () => {
expect(Helper).toHaveProperty('FlatTreeHelper');
expect(Helper).toHaveProperty('InfiniteScrollingHelper');
expect(Helper).toHaveProperty('MiddlewareHelper');
});
});
|
{
"pile_set_name": "Github"
}
|
{
"children":[{
"children":[{
"children":[{
"children":[{
"children":[{
"children":[{
"id":59,
"name":"E_BONE_C_SKULL",
"orgVecs":[0,0,1,0,0,1,0,0,1,0,0,0,0.125,0.0949999392032623,0.11499997228384,0,0.125,0.0200000368058681,0.0249999910593033,0,0,0,0,0,0,0,0,0,23,0,0,0,0,0,0,0,2,2,2,0.5]
}],
"id":58,
"name":"E_BONE_C_SPINE4",
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"orgVecs":[1,0,0,0,0,1,0,0,0,0,1,0,0.125,0.0500000491738319,0.0700000077486038,0,0.125,0.0599999949336052,0.0850000530481339,0,0.0199985485523939,0,1.46026909351349,0,0,0,0,0,23,0,0,0,0,0,0,0,2,2,2,0.5]
}],
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"name":"E_BONE_L_UPPERARM",
"orgVecs":[1,0,0,0,0,1,0,0,0,0,1,0,0.125,0.0550000071525574,0.0799999758601189,0,0.125,0.0249998271465302,0.104999966919422,0,-0.170065477490425,0.220000028610229,1.14075589179993,0,0,0,0,0,23,0,0,0,0,0,0,0,2,2,2,0.5]
}],
"id":1,
"name":"E_BONE_L_SHOULDER",
"orgVecs":[1,0,0,0,0,1,0,0,0,0,1,0,0.100000001490116,0.115000016987324,0.110000014305115,0,0.100000001490116,0.025000000372529,0.025000000372529,0,0.049679346382618,0,0.509676933288574,0,0,0,0,0,27,0,0,0,0,0,0,0,2,2,2,0.5]
},{
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"name":"E_BONE_R_METACARPALS",
"orgVecs":[-1,0,0,0,0,1,0,0,0,0,1,0,0.0500000007450581,0.0949999839067459,0.1000000461936,0,0.0500000007450581,0.0400000549852848,0.0600000508129597,0,0,0,0.399999916553497,0,0,0,0,0,23,0,0,0,0,0,0,0,2,2,2,0.5]
}],
"id":29,
"name":"E_BONE_R_LOWERARM",
"orgVecs":[-1,0,0,0,0,1,0,0,0,0,1,0,0.125,0.0500000491738319,0.0700000077486038,0,0.125,0.0599999949336052,0.0850000530481339,0,-0.0199985485523939,0,-0.420270025730133,0,0,0,0,0,23,0,0,0,0,0,0,0,2,2,2,0.5]
}],
"id":28,
"name":"E_BONE_R_UPPERARM",
"orgVecs":[-1,0,0,0,0,1,0,0,0,0,1,0,0.125,0.0550000071525574,0.0799999758601189,0,0.125,0.0249998271465302,0.104999966919422,0,-1.15993463993073,0,-0.34075665473938,0,0,0,0,0,23,0,0,0,0,0,0,0,2,2,2,0.5]
}],
"id":27,
"name":"E_BONE_R_SHOULDER",
"orgVecs":[-1,0,0,0,0,1,0,0,0,0,1,0,0.100000001490116,0.115000016987324,0.110000014305115,0,0.100000001490116,0.025000000372529,0.025000000372529,0,-0.0996793657541275,0,-0.119676977396011,0,0,0,0,0,27,0,0,0,0,0,0,0,2,2,2,0.5]
}],
"id":57,
"name":"E_BONE_C_SPINE3",
"orgVecs":[0,0,1,0,0,1,0,0,1,0,0,0,0.0750000029802322,0.249999910593033,0.0549999810755253,0,0.0750000029802322,0.025000000372529,0.025000000372529,0,0,0,0,0,0,0,0,0,27,0,0,0,0,0,0,0,2,2,2,0.5]
}],
"id":56,
"name":"E_BONE_C_SPINE2",
"orgVecs":[0,0,1,0,0,1,0,0,1,0,0,0,0.0750000029802322,0.174999982118607,0.0700000002980232,0,0.0750000029802322,0.0300000011920929,0.0599999986588955,0,0,0,0,0,0,0,0,0,27,0,0,0,0,0,0,0,2,2,2,0.5]
}],
"id":55,
"name":"E_BONE_C_SPINE1",
"orgVecs":[0,0,1,0,0,1,0,0,1,0,0,0,0.0750000029802322,0.144999995827675,0.0800000056624413,0,0.0750000029802322,0.025000000372529,0.025000000372529,0,0,0,0,0,0,0,0,0,27,0,0,0,0,0,0,0,2,2,2,0.5]
}],
"id":54,
"name":"E_BONE_C_SPINE0",
"orgVecs":[0,0,1,0,0,1,0,0,1,0,0,0,0.0750000029802322,0.184999942779541,0.0849999934434891,0,0.0750000029802322,0.0300000011920929,0.0300000011920929,0,0.270000010728836,1.05999934673309,-0.0996942594647408,0,0,0,0,0,24,0,0,0,0,0,0,0,2,2,2,0.5]
},{
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"children":[{
"id":23,
"name":"E_BONE_L_TALUS",
"orgVecs":[0,1,0,0,1,0,0,0,0,0,1,0,0.100000001490116,0.0750000923871994,0.134999915957451,0,0.100000001490116,0.0700000002980232,0.0500000081956387,0,0.239988371729851,0,0,0,0,0,0,0,24,0,0,0,0,0,0,0,2,2,2,0.5]
}],
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"name":"E_BONE_L_LOWERLEG",
"orgVecs":[0,0,-1,0,0,1,0,0,1,0,0,0,0.224999994039536,0.074999988079071,0.0650000050663948,0,0.224999994039536,0.124999992549419,0.129999950528145,0,0,0,-1.94882249832153,0,0,0,0,0,24,0,0,0,0,0,0,0,2,2,2,0.5]
}],
"id":21,
"name":"E_BONE_L_UPPERLEG",
"orgVecs":[0,0,-1,0,0,1,0,0,1,0,0,0,0.224999994039536,0.0900000259280205,0.115000039339066,0,0.224999994039536,0.0649999678134918,0.150000035762787,0,0.440001487731934,-0.379999995231628,1.179403424263,0,0,0,0,0,26,0,0,0,0,0,0,0,2,2,2,0.5]
}],
"id":20,
"name":"E_BONE_L_HIP",
"orgVecs":[1,0,0,0,0,1,0,0,0,0,1,0,0.0500000007450581,0.189999535679817,0.134999662637711,0,0.0500000007450581,0.14000004529953,0.150000035762787,0,-0.370000004768372,0.0699999928474426,-0.0399999991059303,0,0,0,0,0,26,0,0,0,0,0,0,0,2,2,2,0.5]
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"name":"E_BONE_R_TALUS",
"orgVecs":[0,1,0,0,-1,0,0,0,0,0,1,0,0.100000001490116,0.0750000923871994,0.134999915957451,0,0.100000001490116,0.0700000002980232,0.0500000081956387,0,-0.239988371729851,0,0,0,0,0,0,0,24,0,0,0,0,0,0,0,2,2,2,0.5]
}],
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"name":"E_BONE_R_LOWERLEG",
"orgVecs":[0,0,-1,0,0,1,0,0,-1,0,0,0,0.224999994039536,0.074999988079071,0.0650000050663948,0,0.224999994039536,0.124999992549419,0.129999950528145,0,-0.300000131130219,0,0.478823840618134,0,0,0,0,0,24,0,0,0,0,0,0,0,2,2,2,0.5]
}],
"id":47,
"name":"E_BONE_R_UPPERLEG",
"orgVecs":[0,0,-1,0,0,1,0,0,-1,0,0,0,0.224999994039536,0.0900000259280205,0.115000039339066,0,0.224999994039536,0.0649999678134918,0.150000035762787,0,0.249998480081558,0,0.0505958646535873,0,0,0,0,0,26,0,0,0,0,0,0,0,2,2,2,0.5]
}],
"id":46,
"name":"E_BONE_R_HIP",
"orgVecs":[-1,0,0,0,0,1,0,0,0,0,1,0,0.0500000007450581,0.189999535679817,0.134999662637711,0,0.0500000007450581,0.14000004529953,0.150000035762787,0,0,0,0,0,0,0,0,0,26,0,0,0,0,0,0,0,2,2,2,0.5]
}],
"id":53,
"name":"E_BONE_C_BASE",
"orgVecs":[0,1,0,0,1,0,0,0,0,0,1,0,0.00499999988824129,0.025000000372529,0.025000000372529,0,0.00499999988824129,0.025000000372529,0.025000000372529,0,0,0,0,0,0,0,0,0,12,0,0,0,0,0,0,0,2,2,2,0.5]
}
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{
"pile_set_name": "Github"
}
|
/*=========================================================================
*
* Copyright NumFOCUS
*
* Licensed under the Apache License, Version 2.0 (the "License");
* you may not use this file except in compliance with the License.
* You may obtain a copy of the License at
*
* http://www.apache.org/licenses/LICENSE-2.0.txt
*
* Unless required by applicable law or agreed to in writing, software
* distributed under the License is distributed on an "AS IS" BASIS,
* WITHOUT WARRANTIES OR CONDITIONS OF ANY KIND, either express or implied.
* See the License for the specific language governing permissions and
* limitations under the License.
*
*=========================================================================*/
#include "itkHessianRecursiveGaussianImageFilter.h"
// This test creates an image varying as a 1D Gaussian in the X direction
// for different values of sigma, and checks the scale-space response of
// the xx component of the Hessian at the center of the Gaussian.
// If NormalizeAcrossScale works correctly, the filter should yield the
// same Hxx across different scales.
int
itkHessianRecursiveGaussianFilterScaleSpaceTest(int, char *[])
{
constexpr unsigned int Dimension = 3;
using PixelType = double;
using ImageType = itk::Image<PixelType, Dimension>;
using IndexType = itk::Index<Dimension>;
using SizeType = itk::Size<Dimension>;
using RegionType = itk::ImageRegion<Dimension>;
using PointType = ImageType::PointType;
using SpacingType = ImageType::SpacingType;
ImageType::Pointer inputImage = ImageType::New();
SizeType size;
size.Fill(21);
size[0] = 401;
IndexType start;
start.Fill(0);
RegionType region;
region.SetIndex(start);
region.SetSize(size);
PointType origin;
origin.Fill(-1.25);
origin[0] = -20.0;
SpacingType spacing;
spacing.Fill(0.1);
inputImage->SetOrigin(origin);
inputImage->SetSpacing(spacing);
inputImage->SetLargestPossibleRegion(region);
inputImage->SetBufferedRegion(region);
inputImage->SetRequestedRegion(region);
inputImage->Allocate();
using IteratorType = itk::ImageRegionIteratorWithIndex<ImageType>;
constexpr unsigned int numberOfScales = 4;
double scales[numberOfScales];
scales[0] = 1.0;
scales[1] = 2.0;
scales[2] = 3.0;
scales[3] = 5.0;
// changing the size of the object with the the size of the
// gaussian should produce the same results
for (double objectSize : scales)
{
IteratorType it(inputImage, inputImage->GetRequestedRegion());
PointType point;
// Fill the image with a 1D Gaussian along X with sigma equal to the current scale
// The Gaussian is not normalized, since it should have the same peak value across
// scales, only sigma should change
while (!it.IsAtEnd())
{
inputImage->TransformIndexToPhysicalPoint(it.GetIndex(), point);
double value = std::exp(-point[0] * point[0] / (2.0 * objectSize * objectSize));
it.Set(value);
++it;
}
// Compute the hessian using NormalizeAcrossScale true
using FilterType = itk::HessianRecursiveGaussianImageFilter<ImageType>;
using HessianImageType = FilterType::OutputImageType;
FilterType::Pointer filter = FilterType::New();
filter->SetInput(inputImage);
filter->SetSigma(objectSize);
filter->SetNormalizeAcrossScale(true);
filter->Update();
HessianImageType::Pointer outputImage = filter->GetOutput();
// Get the value at the center of the image, the location of the peak of the Gaussian
PointType center;
center.Fill(0.0);
IndexType centerIndex;
outputImage->TransformPhysicalPointToIndex(center, centerIndex);
// Irrespective of the scale, the Hxx component should be the same
double centerHxx = outputImage->GetPixel(centerIndex)[0];
if (centerHxx > -0.3546 || centerHxx < -0.3547)
{
std::cout << "center Hessian: " << outputImage->GetPixel(centerIndex) << std::endl;
return EXIT_FAILURE;
}
}
// maintaining the size of the object and gaussian, in physical
// size, should maintain the value, while the size of the image changes.
for (double scale : scales)
{
IteratorType it(inputImage, inputImage->GetRequestedRegion());
PointType point;
double objectSize = 5.0;
spacing.Fill(scale / 5.0);
inputImage->SetSpacing(spacing);
// Fill the image with a 1D Gaussian along X with sigma equal to
// the object size.
// The Gaussian is not normalized, since it should have the same peak value across
// scales, only sigma should change
while (!it.IsAtEnd())
{
inputImage->TransformIndexToPhysicalPoint(it.GetIndex(), point);
double value = std::exp(-point[0] * point[0] / (2.0 * objectSize * objectSize));
it.Set(value);
++it;
}
// Compute the hessian using NormalizeAcrossScale true
using FilterType = itk::HessianRecursiveGaussianImageFilter<ImageType>;
using HessianImageType = FilterType::OutputImageType;
FilterType::Pointer filter = FilterType::New();
filter->SetInput(inputImage);
filter->SetSigma(objectSize);
filter->SetNormalizeAcrossScale(true);
filter->Update();
HessianImageType::Pointer outputImage = filter->GetOutput();
// Get the value at the center of the image, the location of the peak of the Gaussian
PointType center;
center.Fill(0.0);
IndexType centerIndex;
outputImage->TransformPhysicalPointToIndex(center, centerIndex);
// Irrespective of the scale, the Hxx component should be the same
double centerHxx = outputImage->GetPixel(centerIndex)[0];
if (centerHxx > -0.354 || centerHxx < -0.355)
{
std::cout << "center Hessian: " << outputImage->GetPixel(centerIndex) << std::endl;
return EXIT_FAILURE;
}
}
return EXIT_SUCCESS;
}
|
{
"pile_set_name": "Github"
}
|
Stuart's Diverse Marine Life Offers The Best Inshore Catch
The Treasure Coast offers prizes every day yielding a variety of catchable (and tasty) fish. We are gifted to have this example of nature’s excellence, and we are so fortunate to be in such close proximity to incredibly beautiful waterways, lagoons, estuaries and inlets, all of which retain diverse species of fish.
One of the best things about Florida fishing is that it’s year-round. Let’s focus on the diversity of the inshore catch. Having a specific fish in mind when casting (for example, snook, tarpon, trout and redfish) is a good start as you attempt the catch rather than just fish. There are those anglers who will specifically target an individual species like those mentioned. There’s also what I refer to as “the alien of the flats,” the flounder. Then, there are cold- month catches, like the bluefish and crevalle jack, or its common name “The Jack” (what I like to call “the savior of the flats”).
On a day when nothing seems to be biting, which happens, you’ll find the voracious appetite of the Jack keeps the bite strong and the fight on. Bonus: most varieties of fish, again, year-round, can and will be found sitting right on top of oyster beds, sandbars, channel edges, pilings of docks, bridges, beaches, piers and jetties. Everywhere you see water, fish are waiting for the ambush. You need to be prepared.
Also be keen and use some smarts. Certain fish have regulations surrounding the time of year and the amount of fish that can be harvested. That is called a bag limit. The size of a fish you may possess, or its “slot,” is also regulated. Be patient and that will truly lend a hand in enjoying what the Treasure Coast offers. Get fishing and catch them up.
|
{
"pile_set_name": "Pile-CC"
}
|
Those of us who were lucky enough to be at the Tampa Bay Times Forum on Thursday night witnessed something that can only be described as spectacular.
Oct 10, 2013; Tampa, FL, USA; Tampa Bay Lightning center Steven Stamkos (91) skates during the first period against the Florida Panthers at Tampa Bay Times Forum. Mandatory Credit: Kim Klement-USA TODAY Sports
As if the Tampa Bay Lightning defeating the in-state rival Florida Panthers by a score of 7-2 wasn’t enough, the fans in Tampa got to witness something that they haven’t seen in quite some time; a hat trick comprised of three completely different types of goals.
Stamkos’ first goal of the night was a shorthanded one. At that particular moment in time, Richard Panik sat in the box for 2 minutes for High Sticking, Stamkos and St. Louis made their way down the ice, and Stamkos was able to fire off a shot that made its way between the pipes with assists from the captain Martin St. Louis and defenseman Victor Hedman.
About halfway through the next period, Stamkos would strike again; this time at full strength. The assists on this goal came from assistant captain Matt Carle and left wing Ryan Malone.
The icing on the puck shaped cake came at 12:25 in the third period while Panthers center Jesse Winchester sat in the box for 2 minutes for Hooking, Stamkos was able to make his way down the ice and knock in another, bringing the Bolts up 7-2 with the assists from Sami Salo and Teddy Purcell.
In case you didn’t have the privilege of seeing this live, or even on television, here is a piece by piece look at Steven Stamkos’ hatty from Thursday night at the Tampa Bay Times Forum.
Next, the Tampa Bay Lightning continue their 7-game home stand tonight against Evgeni Malkin and the Pittsburgh Penguins.
Be sure to stay up-to-date with all the latest Lightning news and updates with Bolts By the Bay. LIKE us on Facebook at Bolts By The Bay, and you can follow us on Twitter at @BoltsByTheBay.
|
{
"pile_set_name": "OpenWebText2"
}
|
Mam, Azerbaijan
Mam is a village in the Sabirabad Rayon of Azerbaijan.
Mam is found in eastern Azerbaijan.
It is suspected that this village has undergone a name change or no longer exists, as no Azerbaijani website mentions it under this name.
References
Category:Populated places in Sabirabad District
|
{
"pile_set_name": "Wikipedia (en)"
}
|
Browsed byCategory: Biology
Thus, it is no more incredible to assume that all the conditions were just as would be expected to mold a human genome from its earliest beginnings, than it is to assume that all the conditions were just as would be expected to produce a fat man named Winston Churchill in the highest seat of British government during a world war with Germany exactly twenty-two years after a previous such war. Finally, an averaged Itô stochastic differential equation for the energy envelope of the system as one-dimensional diffusion process is derived by using the stochastic averaging method of energy envelope, and the Fokker-Planck-Kolmogorov equation associated with the averaged Itô equation is solved to obtain stationary probability densities of the energy envelope and amplitude envelope.
Some marine biotechnologists study marine organisms in order to develop drugs that are used to cure human disease. Schwann graduated in medicine at Berlin in 1834, was appointed professor of anatomy at the University of Louvain in 1838, and became a professor at Liège in 1847. But even if his mathematical model were correct, his conclusion is wrong: the number of "generations" he calculates to be necessary is not inconsistent with a 7-million-year time span. Solutions to scientific problems are often developed when different research teams produce conflicting evidence.
This website is mainly aimed for students studying AQA (spec. In focusing solely on Gould & Vrba�s fitness enhancing definition, Buss et al. (1998) catch Gould & Vrba�s concept of exaptation in the functional, evolutionary analysis of EP; which in itself proves to be a mature and effective analytical tool - the point I find most interesting with the discussion. This case study examines the impact of road deicing agents on amphibians living near bridges and roads treated heavily with salt duri...
You can have fun with and make use of science in everyday life. There are different types of aerobic bacteria that work in composting piles. The efficient cause is depicted as “that from whence comes the first principle of kinetic change or rest” (Phys. 194b 30). The exploration and study of the ocean. Also, the number of openings in federal government agencies charged with managing natural resources, such as the Interior and Agriculture Departments and the Environmental Protection Agency, is expected to grow; see the report (in PDF format) Federal Natural Resources Agencies Confront an Aging Workforce and Challenges to Their Future Roles.
In that book it boils down to this: they "assume that we are trying to synthesize a protein containing 101 amino acids" and then determine "the inverse of the estimate for the number of ways one can arrange 101 amino acids in a sequence" and get 1 x 10^-117. Many phenomenological models, while failing to be derivable from a theory, incorporate principles and laws associated with theories. It provides the knowledge and skills needed to participate in the growing ecotourism (or nature-based) industry, and is delivered at UWA's Albany campus, in one of the most environmentally rich regions in Australia.
The coding portions of a gene are called exons. August 5, 2016 The universe is 13.8 billion years old, but our planet formed just 4.5 billion years ago. Females are about 13 and 74 of our megalomaniac rich oil man a one or a. For biologists collectively are less agreed upon the details of evolutionary mechanics than they were a scant decade ago. While the concept of evolution by natural selection is very simple, it is often misunderstood by students. Membership for employment is listed at a big and this is why. 100 of the police originally typed were actually.
Cycle - A cycle shows the reusing of certain elements and compounds (e.g. water, carbon, oxygen, nitrogen, phosphorus) in different forms in ecosystems. The field sensitivity of the structure is on the order of 1 V/Oe, which makes it possible to detect magnetic fields with an amplitude down to ˜10-6 Oe. In The Anthropic Cosmological Principle (Oxford, 1986), John D. As a consequence, the pKa of the brushes measured upon increasing pH was consistently higher than that measured upon decreasing pH.
The idea is to choose a test agent that is very sensitive to the condition you are testing. They are phenomena that are common and easily accessible to students. Darwin, as he most likely had not read Mendel�s work (he owned a copy, but left the pages uncut), is not really talking about �genes� and in the quote offered by Bouchard variance is not a point; the traits mentioned are inherited but need not have greater than zero heritability due to selection (in the case of domesticated animals this would be artificial selection, on top of previous natural selection).
This entry is obviously not an attempt to define every possible use of the term 'science.' In some quarters, the science I am concerned with here is called natural science. This report focuses on sex differences in observed disruptive behavior across interactional contexts and diagnostic status. The atmospheric science major can sometimes be linked closely to oceanic and space studies, since these three environments interact with one another.
Thus, we cannot use present life as a basis for calculating the odds of the random formation of the first life, and this is even more so when it comes to bacterium, which we already know is highly evolved, cf. "Evolution of Bacterial Genomes" by Trevors, Antonie van Leeuwenhoek International Journal of General and Molecular Microbiology 71:3, pp. 265-270 (March, 1997). Senior Democrats are increasingly of the entitlement people having ducked combat This guy.
|
{
"pile_set_name": "Pile-CC"
}
|
Q:
Rails and Authlogic. Show currently logged in users
I would like to have the list of currently logged in users.
This code doesn't work :
<% UserSession.all.each do |user_session| %>
<% end %>
A:
@syed-aslam has a good solution, but you could just let Authlogic do the work. Check out the module Authlogic::ActsAsAuthentic::LoggedInStatus which defines two scopes: logged_in, logged_out
Your code becomes:
<% User.logged_in.each do |user| %>
<% end %>
P.S. I would normally link to the RDoc instead of source code, but the RDoc seems to have problems at the moment.
A:
Authlogic gives you all kind of automatic columns that you don’t really need to update or maintain on your own, they are maintained by the actual code flow of Authlogic itself.
Those fields can contain some basic functionality related issues like the number of login attempts made, the ip address from which the attempt was made an or even what was the ip address the last time that user logged in. fun.
The magic column that will help us find who is probably online is the one called last_request_on, which basically indicates when was the last time that user made a request to your application.
The second parameter we’ll need in order to make a more accurate selection, is the configuration option named logged_in_timeout, which sets the timeout after which a stale session will be expired, by default it will expire after 10 minutes.
so if you set your session expiry to 30 minutes:
class User << ActiveRecord::Base
acts_as_authentic do |c|
c.logged_in_timeout 30.minutes
end
end
searching for those users is pretty easy:
module OnlineUsers
def count_online_users
User.count(:conditions => ["last_request_at > ?", 30.minutes.ago])
end
end
|
{
"pile_set_name": "StackExchange"
}
|
Gary Butler (tight end)
Gary Butler is a former tight end who played in the National Football League.
Having played his college football with Rice University, Butler was drafted in the second round by the Kansas City Chiefs. He played two seasons in Kansas City before joining the Chicago Bears for 1975. After a year off of football he played for the Tampa Bay Buccaneers.
References
Category:1951 births
Category:Living people
Category:American football tight ends
Category:Rice Owls football players
Category:Kansas City Chiefs players
Category:Chicago Bears players
Category:Tampa Bay Buccaneers players
Category:Sportspeople from Houston
Category:Players of American football from Texas
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Java AWT event hierarchyJava AWT event hierarchy What class is the top of the AWT event hierarchy? The java.awt.AWTEvent class is the highest-level class in the AWT event-class hierarchy
AWT ImageAWT Image I have loaded two images using toolkit in java.one is concentric rectangle and other is concentric circle.kindly someone send me the source code of interchanging the inner most rectangle with the inner most circle
java awt package tutorialJava AWT Package
In Java, Abstract Window Toolkit(AWT) is a platform... programming language, the AWT is also platform-independent.
A common set.... The implementation of the user interface elements provided by the AWT is
done using
Awt programming with javaAwt programming with java Write a java program, which provides a text area with horizontal and vertical scrollbar.Type some lines in the text area and use scrollbars to move the text within the text area.Read a word in a text
date and time in awt(java)date and time in awt(java) sir,
do you have an example of date in awt java which can be view over a textfield.
Display time in JTextField
import java.util.*;
import javax.swing.*;
import java.awt.event.*;
public
AWT ComponentsAWT Components
The class component is extended by all the AWT components.
More of the codes can be put to this class to design lot of AWT
components. Most of the AWT
What is AWT in java What is AWT in java
In this Example we will describe awt in java. The awt in java stands for Abstract
Windowing Toolkit and is a package available with JDK. awt contains all classes
and develops user interface objects like
Java AWT Package Example
Java AWT Package Example
In this section you will learn about the AWT package of the Java. Many
running examples are provided that will help you master AWT package. Example
What is AWT in java
What is AWT in java
 ... available with JDK. AWT stands for Abstract
Windowing Toolkit. It contains all classes... toolkits. You can use the AWT package to develop user interface
objects like
displaying image in awt - Java Beginnersdisplaying image in awt Hi All,
I have downloaded the code to display image using awt from here and when I execute the code I am getting... ActionListener{
JFrame fr = new JFrame ("Image loading program Using awt");
Label
Create a Container in Java awt
Create a Container in Java awt
Introduction
This program illustrates you how to create a container.
Container contains several control or tools for
develop
your
java programmes - Java Beginnersjava programmes Can you please send me the programmes for IT Deptartment.The subject code is IT2305.Based on new syllabus as per Anna University for 2008 regulation.It contains 12 programmes
Event handling in Java AWT
Event handling in Java AWT
 ... events in java awt. Here, this
is done through the java.awt.*; package
of java...;public AwtEvent(){
super("Event in Java awt
Image on Frame in Java AWT
Image on Frame in Java AWT
Introduction
In this section, you will learn how to display image on the frame. This
program shows you how to display image in your application
Different types of event in Java AWT
Different types of event in Java AWT
 ... that are generated by your AWT Application. These events are used to make the application more... in Java AWT. These are as
follows :
ActionEvent
AdjustmentEvent
Graphical User Interfaces
which is known as Abstract Window Toolkit (AWT). The Abstract Window Toolkit (AWT) contains
several graphical widgets which can be added and positioned to the
display area with a layout manager.
As the Java programming language, the AWT
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Q:
Avoding instanceof in Java
In my application I have a 2d array of entities to represent a grid. Each location in the grid can either be empty or occupied by an entity (in this case it's just a person or wall). Right now I use instanceof to check whether an entity is a person or a wall.
I was thinking of giving each entity a method which returns an enum stating their type so i.e. a wall entity would return EntityType.WALL. I was wondering if this is the best idea to remove the use of instanceof or is instanceof suitable in this scenario?
A:
Use Tell, Don't Ask: instead of asking the objects what they are and then reacting on that, tell the object what to do and then walls or people do decide how they do what they need to do.
For example:
Instead of having something like this:
public class Wall {
// ...
}
public class Person {
// ...
}
// later
public class moveTo(Position pos) {
Object whatIsThere = pos.whatIsThere();
if (whatIsThere instanceof Wall) {
System.err.println("You cannot move into a wall");
}
else if (whatIsThere instanceof Person) {
System.err.println("You bump into " + person.getName());
}
// many more else branches...
}
do something like this:
public interface DungeonFeature {
void moveInto();
}
public class Wall implements DungeonFeature {
@Override
public void moveInto() {
System.err.println("You bump into a wall");
}
// ...
}
public class Person implements DungeonFeature {
private String name;
@Override
public void moveInto() {
System.err.println("You bump into " + name);
}
// ...
}
// and later
public void moveTo(Position pos) {
DungeonFeature df = currentPosition();
df.moveTo(pos);
}
This has some advantages.
First, you don't need to adjust a giant if then else tree each time you add a new dungeon feature.
Second, the code in the dungeon features is self-contained, the logic is all in the said object. You can easily test it and move it.
A:
The theoretical solution to removing the instanceof in a refined way is the usage of the Visitor Pattern. How it works is that the object that needs to know whether the other element is a wall or person calls that object with itself as a parameter, and that particular object calls back thus providing information about its type.
Example,
public class Person {
void magic() {
if(grid.getAdjacent() instanceof Person) {
Person otherPerson = (Person)grid.getAdjacent();
doSomethingWith(otherPerson);
} else if(grid.getAdjacent() instanceof Wall) {
Wall wall = (Wall)grid.getAdjacent();
doOtherThingWith(wall);
}
}
}
Can become
public class Person extends Entity {
void magic() {
grid.getAdjacent().visit(this);
}
void onVisit(Wall wall) {
doOtherThingWith(wall);
}
void onVisit(Person person) {
doSomethingWith(person);
}
public void visit(Person person) {
person.onVisit(this);
}
}
public class Wall extends Entity {
public void visit(Person person) {
person.onVisit(this);
}
}
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With greenDAO 3 released, it’s time again to look at the Android ORM performance landscape and do some benchmarks. This time, we also tested newer ORMs for SQLite like DbFlow, requery, SQLDelight, and SquiDB. Also, we had an extensive look at benchmarks done by others. Let’s get started with our results:
As you can see greenDAO comes in first in for all operations while there is no distinct second place. While this would be easy to claim, we are trying hard to present straight facts here. Objective benchmarks are really difficult to do and we did our best to make this comparison fair. All our benchmarks are open source, so you can verify the procedure for yourselves. Let us know if we can improve.
Other ORM performance benchmarks
Critical minds may still have doubts, which is perfectly understandable because we still might favor a specific product. Granted, so let’s have a look at other popular benchmark projects by Raizlabs and Kevin Galligan.
Benchmarks by Raizlabs
Raizlabs is an interesting case. In early 2015, Raizlabs claimed DbFlow to be the fastest Android ORM. Initially, they did not mention greenDAO at all. Soon, readers left various comments asking about greenDAO. Thus Raizlabs added greenDAO in their benchmark suite, and commented that DbFlow was faster for inserts while greenDAO may have advantages when it comes to loading. Because concrete results still lacked, we ran Raizlab’s benchmarking app with greenDAO included (smaller is better):
Seems like greenDAO comes out fastest right away. A quick look at the code suggests that greenDAO’s time for “Save” might further improve by exchanging the method “insertOrReplace” by just “insert”. But OK, let’s look at the other scenario called “Complex trial”:
Fixing Raizlabs’s “complex trial” benchmark
In this setup, DbFlow is the clear winner, right? But wait a second, why is greenDAO slower than OrmLite? And why is DbFlow loading that fast? It’s time to have a closer look at their code. As it turned out, the “write” test made greenDAO use 101 transactions while the others used just a single one. That makes a big difference, so we fixed it. The results from “load” test are a little harder to understand, but in short, DbFlow already had all entities in memory after insert, so it did not have to load any entity from the database. It’s easy to outperform others by leaps when you are the only one with a nicely filled cache in place. That’s why we filed an issue because we think the test setup does not make sense: either you test caching or loading, but never a mixture. Anyway, it was easy to make greenDAO use its caching mechanisms in their setup. Here are the revised results:
Yep, fixing those seemingly small flaws in the benchmark made a huge difference. In the revised benchmark, greenDAO clearly outperforms DbFlow for saving complex data. And for loading, greenDAO can also reuse previously inserted and thus “cached” entities in memory leading to close-to-zero “load” times.
Benchmarks by Kevin Galligan
Kevin Galligan is a pioneer in the Android ORM world: he’s the developer of the Android Adapter for OrmLite and also made an annotation processing extension for it. He gave a presentation at droidcon UK in late 2015 on Android ORMs accompanied by an blog post diving deeper also on ORM performance. His benchmarks gave a different picture than ours, especially when looking at greenDAO. Let’s look a bit behind the scenes to understand what’s going on.
The benchmarking code came initially from another developer and penalized OrmLite drastically. So Kevin fixed it. Unfortunately, the original code also had a couple of problems regarding greenDAO, which were left unnoticed. We pushed a couple of pull requests to Kevin’s GitHub repository to fix those too:
For batch inserts, the method insertInTx should be used. Doing 1) manual iterations and 2) using the method insertOrReplace instead of just insert is less efficient. (see pull request #2)
Like all other ORMs being tested, greenDAO should also use primitive types in the entity. Wrapper types like Long are somewhat expensive. (see pull request #3)
greenDAO’s optional identity scope introduces a little overhead, which should be switched off in some cases depending on what you compare it with. (see pull request #4)
After those modifications, we got the following results:
In Kevin’s revised benchmark, greenDAO and DbFlow are about the same speed for writing, while greenDAO seems to read data around 20% faster than DbFlow. While still being similar, it does not exactly match our findings. A possible reason could be that we used DBFlow 3.0 while Kevin used 2.0, but we did not take the time to investigate that. So, that is only a speculation.
About our ORM benchmark
Some remarks and additional info about our 2016 edition of our Android ORM performance benchmark:
Benchmarked versions: greenDAO 3.0.1, OrmLite 4.48, DbFlow 3.0.1, SQLDelight 0.4.2, SquiDB 3.0.0, requery 1.0.0-beta23.
The tests were run on a Nexus 5 with Android 6.0.1 (June 2016 patch level).
The values are the average of 8 separate runs
The benchmark code is open source
Last year’s comparison benchmarked greenDAO, ORMLite, and ActiveAndroid. We dropped ActiveAndroid this year, because we think it’s not a good alternative anymore. Its development seems to have stopped and also its results were inferior.
Side note: We used the same hardware like last year. The values for greenDAO and OrmLite remained the same hinting that there were no performance improvements from Android 5.1 (2016) to Android 6.0.1 (2016) done in the internal SQLite version shipping with Android.
Summary
We cannot stress enough how hard it is to make objective performance benchmarks. Even if the benchmarking code is open source (which is great), it may not have been critically reviewed. We did our part and fixed a couple of issues in two other benchmark suites and invite you to do the same. After these revisions, the results are very similar to each other. This similarity should validate each of the (revised) benchmark suites: it is much less likely that three benchmark suites get it wrong.
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Migration patterns of dendritic cells in the rat: comparison of the effects of gamma and UV-B irradiation on the migration of dendritic cells and Lymphocytes.
To further define the underlying mechanisms of immune suppression induced by UV-B irradiation, we have examined the kinetics of homing patterns of in vitro UV-B-irradiated and gamma-irradiated-thoracic duct lymphocytes (TDL) compared to dendritic cells (DC). Our findings show that 111In-oxine-labeled TDL specifically home to the spleen, liver, lymph nodes, and bone marrow with subsequent recirculation of a large number of cells from the spleen to lymph nodes. In contrast, DC preferentially migrate to the spleen and liver with a relatively insignificant distribution to lymph nodes and an absence of subsequent recirculation. Splenectomy prior to cell injection significantly diverts the spleen-seeking DC to the liver but not to the lymph nodes, while the homing of TDL to lymph nodes is significantly increased. In vitro exposure of 111In-oxine labeled TDL to gamma irradiation does not significantly impair immediate homing to lymphoid tissues but inhibits cell recirculation between 3 and 24 hr. In contrast, gamma irradiation does not affect the tissue distribution of labeled DC, suggesting that DC are more radioresistant to gamma irradiation than TDL. Unlike the findings in animals injected with gamma-irradiated cells, UV-B irradiation virtually abolished the homing of TDL to lymph nodes and significantly reduced the homing of the spleen-seeking DC to the splenic compartment while a large number of cells were sequestered in the liver. The results of in vitro cell binding assay show that TDL, unlike DC, have the capacity to bind to high endothelial venules (HEV) within lymph node frozen sections while gamma and UV-B irradiation significantly inhibit the binding of TDL to lymph node HEV. These findings suggest that: (i) DC, unlike TDL, are unable to recirculate from blood to lymph nodes through HEV; (ii) although gamma irradiation impairs TDL recirculation, it does not affect DC tissue distribution; and (iii) UV-B irradiation impairs both TDL and DC migration patterns. We conclude that the lack of capacity of irradiated TDL to home to lymph nodes is due to damage to cell surface homing receptors and that the failure of DC to home to the lymph node microenvironment is related to the absence of HEV homing receptors on their cell surface.
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世界最強と名高いコピー防止技術「Denuvo Anti-Tamper」（以下、Denuvo）が、Playdead ApSのパズルアドベンチャー『INSIDE』に続いて、Bethesda Softworksのファーストパーソン・シューター『DOOM』からも無効化されたと、海外フォーラムを中心に囁かれている。いずれもゲーム販売元および開発元からの公式声明は出されていないが、コピーガードを除外した背景にはDenuvo社の期間保証が関係しているとの情報がある。両作とも今年8月の段階でクラックされたと報じられていたことから、全く信憑性に欠ける仮説でもなさそうだ。
3か月以内に“割られ”たら代金は無料との噂
「Denuvo」は、オーストリアに拠点を置くソフトウェア会社Denuvo Software Solutions GmbH（以下、Denuvo社）が開発した改ざん防止技術。ゲームソフトを特定のユーザーアカウントと紐付けることでコンテンツの無制限な利用を規制するデジタル著作権管理（通称DRM＝Digital Rights Management）とは異なり、Valve CorporationのSteamやElectronic ArtsのOriginといった、既存のDRMソリューションそのものを保護するようデザインされている。デバッグ作業や逆行分析、実行ファイルの改ざんを防ぐことで、DRMをバイパスできないようさらに強固な守りを提供するのが目的だ。そのため、DRMを組み込まれていないゲームに対しては何ら意味をなさない。
フォーラムサイトNeoGAFに寄せられた情報によると、PC版『DOOM』のユーザーが最新パッチを分析したところ、ゲーム内から「Denuvo」が無効化されていたという。ちなみに、パッチノートに無効化の事実や理由は一切記載されていない。先月には、PC版『INSIDE』からも「Denuvo」が取り除かれたことがパッチノートから判明し、DRMに否定的な多くのユーザーから賞賛の声が寄せられていた。その際も、ゲーム販売元や開発元からは除外の理由や声明は一切出されておらず、DRMフリーでゲームを販売するオンラインストアGOGでリリースされたことが関係しているのではないかという見方に留まっていた。
そもそもPCゲームにコピーガードを施す真の意義は、クラッカーたちに破られるまでの時間を可能な限り引き延ばすことにある。新作タイトルがクラッカーによってことごとく防壁を突破され、発売日を待たずして海賊版が出回ってしまえば、パブリッシャーにとって最も重要な初期の売り上げに少なからず影響を与えかねないからだ。逆に発売から半年以上が経過して、セールスの数字にほとんど変動が見られなくなったタイトルを死守し続ける意味はさほどないと言える。今回「Denuvo」が無効化された2件に関しても、『DOOM』が今年5月、『INSIDE』が今年7月といずれも半年近くが経過している。Denuvo社との契約内容や課金形態は定かではないが、もし継続的なコストが発生していたとしたら契約を終了してプロテクトを解除したとしても何ら不思議な判断ではない。
一方で、ゲーム企業が「Denuvo」を利用する際の保証制度についての噂が浮上している。先日、「Denuvo」を利用する大手ゲームスタジオ勤務のゲーム開発者を名乗るユーザーが、フォーラムサイトRedditにてDenuvo社の期間保証について言及した。同社には、もしゲーム発売から一定期間内にコピーガードが破られた場合、利用料金は一切不要という保証制度があるとのことで、リファンドの条件として「Denuvo」をゲーム内データから削除することが求められるという。なお、保証期間は通常3か月に設定されるようだ。『DOOM』も『INSIDE』も今年8月はじめの時点でクラックされたことが確認されており、返金ポリシーによる無効化説に全く信憑性がないというわけではなさそうだ。
海賊版の撲滅にはほど遠いクラッカー側の猛威
これまでの仮説が真実だったとして、次に気になるのは「Denuvo」の現状と永続性だ。相対費用の観点からトリプルA級タイトルをはじめ一部のゲームにしか導入されていないが、「Denuvo」には何よりも、発売日を待たずして違法コピーがインターネット上に蔓延するPCゲームの“割れ”事情に革命を起こしたという実績がある。今年はじめには、中国のクラッカー集団3DMがスクウェア・エニックスのアクションアドベンチャー『Just Cause 3』に施された「Denuvo」の防壁を突破できない実情から、2年後には世界から海賊版ゲームがなくなるかもしれないという不安感を露わにしたほどだ。なお、翌月には少なくとも1年間はシングルプレイヤーゲームの違法コピーに着手しないことを表明していた。
しかし、こうした実績とは裏腹に「Denuvo」の防壁は決して永久不変というわけではない。2014年12月に、改良される前のプロテクトを一度3DMに突破されたことがある。今年8月には、ブルガリアの“Voksi”が「Denuvo」をバイパスする形で、『Rise of the Tomb Raider』や『Doom』、3DMが大いに手を焼いた『Just Cause 3』へのアクセスに成功している。この時は『Doom』のSteam体験版を用いたループホールを利用する手口だった。いずれもDenuvo社の迅速な対応により穴は塞がれ、完全に突破されたとは言えない状況にあった。そんな中、難攻不落の要塞として存在感を示していた「Denuvo」は、クラッカー集団CONSPIR4CYにより同月中に突破され、発売から半年に渡り守られ続けてきたPC版『Rise of the Tomb Raider』の海賊版が、ついにインターネット上へ出回ることとなった。
ちなみに、『INSIDE』がCONSPIR4CYによってクラックされたのもこの頃である。その際、「Denuvo」の突破に挑み続けてきた“Voksi”は、CONSPIR4CYが「Denuvo」の脆弱性にどこまで迫れているかに言及していた。同グループが「Denuvo」を陥落させたのは明白であり、少なくとも現行バージョンを突破する方法は完全に把握しているだろうと言われていた。特筆すべきは、7月発売の『INSIDE』に実装されていたプロテクトが「Denuvo」の最新バージョンであったことに加えて、リリースからわずか1か月足らずで割られたという事実だ。『DOOM』に関してもタイミング的に発売から3か月以内にクラックされた可能性は否めない。つまり、Denuvo社による期間保証が本当なら、同社は相当の痛手を被ったことになる。また、『INSIDE』で最新のプロテクトを割られたスピードを考えると、今後も「Denuvo」を無効化するケースは十分起こり得る。
もちろん、これらはあくまでも噂に基づいた仮説であり、現状における「Denuvo」の信頼性や無効化の理由は定かではない。余談になるが、先日にはポーランドのゲーム開発スタジオFlying Wild Hogが、同社の新作『Shadow Warrior 2』をDRMフリーで発売した理由を明かし、堅固なプロテクトが逆に作品のクオリティを貶める要因になる可能性を指摘していた。Steamフォーラムにて、決して不正コピーを容認するわけではなく、今のところ消費者に害を及ぼさずに止める手立てがないことは事実であるとコメント。さらなる資金の投入により「Denuvo」を導入することで、正規ユーザーにとってのゲームクオリティを犠牲にしたくないと説明している。事実、メーカーが過剰なコピーガードを施す姿勢に難色を示すユーザーは少なくない。商品価値の追求が海賊版の防止に繋がるとは言えないが、逆に違法コピーの不在が本当にセールス増加へ繋がるかどうかについても、まだまだ議論の余地があることは間違いない。
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---
abstract: 'This paper has four main parts. In the first part, we construct a noncommutative residue for the hypoelliptic calculus on Heisenberg manifolds, that is, for the class of [$\Psi_{H}$DO]{} operators introduced by Beals-Greiner and Taylor. This noncommutative residue appears as the residual trace on integer order [$\Psi_{H}$DOs]{} induced by the analytic extension of the usual trace to non-integer order [$\Psi_{H}$DOs]{}. Moreover, it agrees with the integral of the density defined by the logarithmic singularity of the Schwartz kernel of the corresponding [$\Psi_{H}$DO]{}. In addition, we show that this noncommutative residue provides us with the unique trace up to constant multiple on the algebra of integer order [$\Psi_{H}$DOs]{}. In the second part, we give some analytic applications of this construction concerning zeta functions of hypoelliptic operators, logarithmic metric estimates for Green kernels of hypoelliptic operators, and the extension of the Dixmier trace to the whole algebra of integer order [$\Psi_{H}$DOs]{}. In the third part, we present examples of computations of noncommutative residues of some powers of the horizontal sublaplacian and the contact Laplacian on contact manifolds. In the fourth part, we present two applications in CR geometry. First, we give some examples of geometric computations of noncommutative residues of some powers of the horizontal sublaplacian and of the Kohn Laplacian. Second, we make use of the framework of noncommutative geometry and of our noncommutative residue to define lower dimensional volumes in pseudohermitian geometry, e.g., we can give sense to the area of any 3-dimensional CR manifold. On the way we obtain a spectral interpretation of the Einstein-Hilbert action in pseudohermitian geometry.'
address: 'Department of Mathematics, University of Toronto, Canada.'
author:
- Raphaël Ponge
title: 'Noncommutative residue for Heisenberg manifolds. Applications in CR and contact geometry'
---
Introduction
============
The aim of this paper is to construct a noncommutative residue trace for the Heisenberg calculus and to present several of its applications, in particular in CR and contact geometry. The Heisenberg calculus was built independently by Beals-Greiner [@BG:CHM] and Taylor [@Ta:NCMA] as the relevant pseudodifferential tool to study the main geometric operators on contact and CR manifolds, which fail to be elliptic, but may be hypoelliptic (see also [@BdM:HODCRPDO], [@EM:HAITH], [@FS:EDdbarbCAHG], [@Po:MAMS1]). This calculus holds in the general setting of a Heisenberg manifold, that is, a manifold $M$ together with a distinguished hyperplane bundle $H\subset TM$, and we construct a noncommutative residue trace in this general context.
The noncommutative residue trace of Wodzicki ([@Wo:LISA], [@Wo:NCRF]) and Guillemin [@Gu:NPWF] was originally constructed for classical [$\Psi$DOs]{} and it appears as the residual trace on integer order [$\Psi$DOs]{} induced by analytic extension of the operator trace to [$\Psi$DOs]{} of non-integer order. It has numerous applications and generalizations (see, e.g., [@Co:AFNG], [@Co:GCMFNCG], [@CM:LIFNCG], [@FGLS:NRMB], [@Gu:RTCAFIO], [@Ka:RNC], [@Le:NCRPDOLPS], [@MMS:FIT], [@MN:HPDO1], [@PR:CDBFCF], [@Po:IJM1], [@Sc:NCRMCS], [@Vas:PhD]). In particular, the existence of a residual trace is an essential ingredient in the framework for the local index formula in noncommutative geometry of Connes-Moscovici [@CM:LIFNCG].
Accordingly, the noncommutative residue for the Heisenberg calculus has various applications and several of them are presented in this paper. Further geometric applications can be found in [@Po:Crelle2].
Noncommutative residue for Heisenberg manifolds
-----------------------------------------------
Our construction of a noncommutative residue trace for [$\Psi_{H}$DOs]{}, i.e., for the pseudodifferential operators in the Heisenberg calculus, follows the approach of [@CM:LIFNCG]. It has two main ingredients:
\(i) The observation that the coefficient of the logarithmic singularity of the Schwartz kernel of a [$\Psi_{H}$DO]{} operator $P$ can be defined globally as a density $c_{P}(x)$ functorial with respect to the action of Heisenberg diffeomorphisms, i.e., diffeomorphisms preserving the Heisenberg structure (see Proposition \[thm:NCR.log-singularity\]).
\(ii) The analytic extension of the operator trace to [$\Psi_{H}$DOs]{} of complex non-integer order (Proposition \[thm:NCR.TR.global\]).
The analytic extension of the trace from (ii) is obtained by working directly at the level of densities and induces on [$\Psi_{H}$DOs]{} of integer order a residual trace given by (minus) the integral of the density from (i) (Proposition \[thm:NCR.TR.local\]). This residual trace is our noncommutative residue for the Heisenberg calculus.
In particular, as an immediate byproduct of this construction the noncommutative residue is invariant under the action of Heisenberg diffeomorphisms. Moreover, in the foliated case our noncommutative residue agrees with that of [@CM:LIFNCG], and on the algebra of Toeplitz pseudodifferential operators on a contact manifold of Boutet de Monvel-Guillemin [@BG:STTO] we recover the noncommutative residue of Guillemin [@Gu:RTCAFIO]. As a first application of this construction we show that when the Heisenberg manifold is connected the noncommutative residue is the unique trace up to constant multiple on the algebra of integer order [$\Psi_{H}$DOs]{} (Theorem \[thm:Traces.traces\]). As a consequence we get a characterization sums of [$\Psi_{H}$DO]{} commutators and we obtain that any smoothing operator can be written as a sum of [$\Psi_{H}$DO]{} commutators.
These results are the analogues for [$\Psi_{H}$DOs]{} of well known results of Wodzicki ([@Wo:PhD]; see also [@Gu:RTCAFIO]) for classical [$\Psi$DOs]{}. Our arguments are somewhat elementary and partly rely on the characterization of the Schwartz kernels of [$\Psi_{H}$DOs]{} that was used in the analysis of their logarithmic singularities near the diagonal.
Analytic applications on general Heisenberg manifolds
-----------------------------------------------------
The analytic extension of the trace allows us to directly define the zeta function $\zeta_{\theta}(P;s)$ of a hypoelliptic [$\Psi_{H}$DO]{} operator $P$ as a meromorphic functions on ${\ensuremath{\mathbb{C}}}$. The definition depends on the choice of a ray $L_{\theta}=\{\arg
\lambda =\theta\}$, $0\leq \theta <2\pi$, which is a ray of principal values for the principal symbol of $P$ in the sense of [@Po:CPDE1] and is not through an eigenvalue of $P$, so that $L_{\theta}$ is a ray of minimal growth for $P$. Moreover, the residues at the potential singularity points of $\zeta_{\theta}(P;s)$ can be expressed as noncommutative residues.
When the set of principal values of the principal symbol of $P$ contains the left half-plane $\Re \lambda\leq 0$ we further can relate the residues and regular values of $\zeta_{\theta}(P;s)$ to the coefficients in the heat kernel asymptotics for $P$ (see Proposition \[prop:Zeta.heat-zeta-global\] for the precise statement). We then use this to derive a local formula for the index of a hypoelliptic [$\Psi_{H}$DO]{} and to rephrase in terms of noncommutative residues the Weyl asymptotics for hypoelliptic [$\Psi$DOs]{} from [@Po:MAMS1] and [@Po:CPDE1]. An interesting application concerns logarithmic metric estimates for Green kernels of hypoelliptic [$\Psi_{H}$DOs]{}. It is not true that a positive hypoelliptic [$\Psi_{H}$DO]{} has a Green kernel positive near the diagonal. Nevertheless, by making use of the spectral interpretation of the noncommutative residue as a residual trace, we show that the positivity still pertains when the order is equal to the critical dimension $\dim M+1$ (Proposition \[prop:Metric.positivity-cP\]).
When the bracket condition $H+[H,H]=TM$ holds, i.e., $H$ is a Carnot-Carathéodory distribution, this allows us to get metric estimates in terms of the Carnot-Carathéodory metric associated to any given subriemannian metric on $H$ (Theorem \[thm:Metric.metric-estimate\]). This result connects nicely with the work of Fefferman, Stein and their collaborators on metric estimates for Green kernels of subelliptic sublaplacians on general Carnot-Carathéodory manifolds (see, e.g., [@FS:FSSOSO], [@Ma:EPKLCD], [@NSW:BMDVF1], [@Sa:FSGSSVF]).
Finally, we show that on a Heisenberg manifold $(M,H)$ the Dixmier trace is defined for [$\Psi_{H}$DOs]{} of order less than or equal to the critical order $-(\dim M+1)$ and on such operators agrees with the noncommutative residue (Theorem \[thm:NCG.Dixmier\]). Therefore, the noncommutative residue allows us to extend the Dixmier trace to the whole algebra of [$\Psi_{H}$DOs]{} of integer order. In noncommutative geometry the Dixmier trace plays the role of the integral on infinitesimal operator of order $\leq 1$. Therefore, our result allows us to integrate any [$\Psi_{H}$DO]{} even though it is not an infinitesimal operator of order $\leq 1$. This is the analogue of a well known result of Connes [@Co:AFNG] for classical [$\Psi$DOs]{}.
Noncommutative residue and contact geometry
-------------------------------------------
Let $(M^{2n+1},H)$ be a compact orientable contact manifold, so that the hyperplane bundle $H\subset TM$ can be realized as the kernel of a contact form $\theta$ on $M$. The additional datum of a *calibrated* almost complex structure on $H$ defines a Riemannian metric on $M$ whose volume ${{\operatorname{Vol}}}_{\theta}M$ depends only on $\theta$.
Let $\Delta_{b;k}$ be the horizontal sublaplacian associated to the above Riemannian metric acting on horizontal forms of degree $k$, $k\neq n$. This operator is hypoelliptic for $k\neq n$ and by making use of the results of [@Po:MAMS1] we can explicitly express the noncommutative residue of $\Delta_{b;k}^{-(n+1)}$ as a constant multiple of ${{\operatorname{Vol}}}_{\theta}M$ (see Proposition \[prop:Contact.residue-Deltab\]).
Next, the contact complex of Rumin [@Ru:FDVC] is a complex of horizontal forms on a contact manifold whose Laplacians are hypoelliptic in every bidegree. Let $\Delta_{R;k}$ denote the contact Laplacian acting on forms degree $k$, $k=0,\ldots,n$. Unlike the horizontal sublaplacian $\Delta_{R}$ does not act on all horizontal forms, but on the sections of a subbundle of horizontal forms. Moreover, it is not a sublaplacian and it even has order 4 on forms of degree $n$. Nevertheless, by making use of the results of [@Po:MAMS1] we can show that the noncommutative residues of $\Delta_{R;k}^{-(n+1)}$ for $k\neq n$ and of $\Delta_{R;n}^{-\frac{n+1}{2}}$ are universal constant multiples of the contact volume ${{\operatorname{Vol}}}_{\theta}M$ (see Proposition \[prop:Contact.residue-DeltaR\]).
Applications in CR geometry
---------------------------
Let $(M^{2n+1},H)$ be a compact orientable $\kappa$-strictly pseudoconvex CR manifold equipped with a pseudohermitian contact form $\theta$, i.e., the hyperplane bundle $H\subset TM$ has an (integrable) complex structure and the Levi form associated to $\theta$ has at every point $n-\kappa$ positive eigenvalues and $\kappa$ negative eigenvalues. If $h$ is a Levi metric on $M$ then the volume with respect to this metric depends only on $\theta$ and is denoted ${{\operatorname{Vol}}}_{\theta}M$.
As in the general contact case we can explicitly relate the pseudohermitian volume ${{\operatorname{Vol}}}_{\theta}M$ to the noncommutative residues of the following operators:
- ${\square}_{b;pq}^{-(n+1)}$, where ${\square}_{b;pq}$ denotes the Kohn Laplacian acting on $(p,q)$-forms with $q\neq \kappa$ and $q\neq n-\kappa$ (see Proposition \[prop:CR.residue-Boxb1\]);
- $\Delta_{b;pq}^{-(n+1)}$, where $\Delta_{b;pq}$ denotes the horizontal sublaplacian acting on $(p,q)$-forms with $(p,q)\neq (n-\kappa,\kappa)$ and $(p,q)\neq (\kappa,n-\kappa)$ (see Proposition \[prop:CR.residue-Deltab1\]).
From now on we assume $M$ strictly pseudoconvex (i.e. we have $\kappa=0$) and consider the following operators:
- ${\square}_{b;pq}^{-n}$, with $q\neq 0$ and $q\neq n$,;
- $\Delta_{b;pq}^{-n}$, with $(p,q)\neq (n,0)$ and $(p,q)\neq (0,n)$.
Then we can make use of the results of [@BGS:HECRM] to express the noncommutative residues of these operators as universal constant multiple of the integral $\int_{M}R_{n}d\theta^{n}\wedge \theta$, where $R_{n}$ denotes the scalar curvature of the connection of Tanaka [@Ta:DGSSPCM] and Webster [@We:PHSRH] (see Propositions \[prop:CR.residue-Boxb2\] and \[prop:CR.residue-Deltab2\]). These last results provide us with a spectral interpretation of the Einstein-Hilbert action in pseudohermitian geometry, which is analogous to that of Connes ([@Co:GCMFNCG], [@KW:GNGWR], [@Ka:DOG]) in the Riemannian case. Finally, by using an idea of Connes [@Co:GCMFNCG] we can make use of the noncommutative residue for classical [$\Psi$DOs]{} to define the $k$-dimensional volumes Riemannian manifold of dimension $m$ for $k=1,\ldots,m-1$, e.g. we can give sense to the area in any dimension (see [@Po:LMP07]). Similarly, we can make use of the noncommutative residue for the Heisenberg calculus to define the $k$-dimensional pseudohermitian volume ${{\operatorname{Vol}}}^{(k)}_{\theta}M$ for any $k=1,\ldots,2n+2$. The argument involves noncommutative geometry, but we can give a purely differential geometric expression of these lower dimensional volumes (see Proposition \[prop:CR.lower-dim.-volumes\]). Furthermore, in dimension 3 the area (i.e. the 2-dimensional volume) is a constant multiple of the integral of the Tanaka-Webster scalar curvature (Theorem \[thm:spectral.area\]). In particular, we find that the area of the sphere $S^{3}\subset {\ensuremath{\mathbb{C}}}^{2}$ endowed with its standard pseudohermitian structure has area $\frac{\pi^{2}}{8\sqrt{2}}$.
Potential geometric applications
--------------------------------
The boundaries of a strictly pseudoconvex domain of ${\ensuremath{\mathbb{C}}}^{n+1}$ naturally carry strictly pseudoconvex CR structures, so one can expect the above results to be useful for studying from the point of view of noncommutative geometry strictly pseudoconvex boundaries, and more generally Stein manifolds with boundaries and the asymptotically complex hyperbolic manifolds of [@EMM:RLSPD]. Similarly, the boundary of a symplectic manifold naturally inherits a contact structure, so we could also use the results of this papers to give a noncommutative geometric study of symplectic manifolds with boundary.
Another interesting potential application concerns a special class of Lorentzian manifolds, the Fefferman’s spaces ([@Fe:MAEBKGPCD], [@Le:FMPHI]). A Fefferman’s Lorentzian space ${\ensuremath{\mathcal{F}}}$ can be realized as the total space of a circle bundle over a strictly pseudoconvex CR manifold $M$ and it carries a Lorentzian metric naturally associated to any pseudohermitian contact form on $M$. For instance, the curvature tensor of ${\ensuremath{\mathcal{F}}}$ can be explicitly expressed in terms of the curvature and torsion tensors of the Tanaka-Webster connection of $M$ and the Dalembertian of ${\ensuremath{\mathcal{F}}}$ pushes down to the horizontal sublaplacian on $M$. This strongly suggests that one could deduce a noncommutative geometric study of Fefferman spaces from a noncommutative geometric study of strictly pseudoconvex CR manifolds. An item of special interest would be to get a spectral interpretation of the Einstein-Hilbert action in this setting.
Finally, it would be interesting to extend the results of this paper to other subriemannian geometries such as the quaternionic contact manifolds of Biquard [@Bi:MEAS].
Organization of the paper
-------------------------
The rest of the paper is organized as follows.
In Section \[sec:Heisenberg-calculus\], we recall the main facts about Heisenberg manifold and the Heisenberg calculus.
In Section \[sec:NCR\], we study the logarithmic singularity of the Schwartz kernel of a [$\Psi_{H}$DO]{} and show that it gives rise to a well defined density. We then construct the noncommutative residue for the Heisenberg calculus as the residual trace induced on integer order [$\Psi_{H}$DOs]{} by the analytic extension of the usual trace to non-integer order [$\Psi_{H}$DOs]{}. Moreover, we show that the noncommutative residue of an integer order [$\Psi_{H}$DO]{} agrees with the integral of the density defined by the logarithmic singularity of its Schwartz kernel. We end the section by proving that, when the Heisenberg manifold is connected, the noncommutative residue is the only trace up to constant multiple.
In Section \[sec:Analytic-Applications\], we give some analytic applications of the construction of the noncommutative residue. First, we deal with zeta functions of hypoelliptic [$\Psi_{H}$DOs]{} and relate their singularities to the heat kernel asymptotics of the corresponding operators. Second, we prove logarithmic metric estimates for Green kernels of hypoelliptic [$\Psi_{H}$DOs]{}. Finally, we show that the noncommutative residue allows us to extend the Dixmier trace to *all* integer order [$\Psi_{H}$DOs]{}.
In Section \[sec:Contact\], we present examples of computations of noncommutative residues of some powers of the horizontal sublaplacian and of the contact Laplacian of Rumin on contact manifolds.
In Section \[sec:CR\], we present some applications in CR geometry. First, we give some examples of geometric computations of noncommutative residues of some powers of the horizontal sublaplacian and of the Kohn Laplacian. Second, we make use of the framework of noncommutative geometry and of the noncommutative residue for the Heisenberg calculus to define lower dimensional volumes in pseudohermitian geometry.
Finally, in Appendix for reader’s convenience we present a detailed proof of Lemma \[lem:Heisenberg.extension-symbol\] about the extension of a homogeneous symbol into a homogeneous distribution. This is needed for the analysis of the logarithmic singularity of the Schwartz kernel of a [$\Psi_{H}$DO]{} in Section \[sec:NCR\].
Part of the results of this paper were announced in [@Po:CRAS1] and [@Po:CRAS2] and were presented as part of my PhD thesis at the University of Paris-Sud (Orsay, France) in December 2000. I am grateful to my advisor, Alain Connes, and to Charlie Epstein, Henri Moscovici and Michel Rumin, for stimulating and helpful discussions related to the subject matter of this paper. In addition, I would like to thank Olivier Biquard, Richard Melrose and Pierre Pansu for their interests in the results of this paper.
Heisenberg calculus {#sec:Heisenberg-calculus}
===================
The Heisenberg calculus is the relevant pseudodifferential calculus to study hypoelliptic operators on Heisenberg manifolds. It was independently introduced by Beals-Greiner [@BG:CHM] and Taylor [@Ta:NCMA] (see also [@BdM:HODCRPDO], [@Dy:POHG], [@Dy:APOHSC], [@EM:HAITH], [@FS:EDdbarbCAHG], [@Po:MAMS1], [@RS:HDONG]). In this section we recall the main facts about the Heisenberg calculus following the point of view of [@BG:CHM] and [@Po:MAMS1].
Heisenberg manifolds
--------------------
In this subsection we gather the main definitions and examples concerning Heisenberg manifolds and their tangent Lie group bundles.
1\) A Heisenberg manifold is a pair $(M,H)$ consisting of a manifold $M$ together with a distinguished hyperplane bundle $H
\subset TM$.
2\) Given Heisenberg manifolds $(M,H)$ and $(M',H')$ a diffeomorphism $\phi:M\rightarrow M'$ is said to be a Heisenberg diffeomorphism when $\phi_{*}H=H'$.
Following are the main examples of Heisenberg manifolds:
*- Heisenberg group.* The $(2n+1)$-dimensional Heisenberg group ${\ensuremath{\mathbb{H}}}^{2n+1}$ is the 2-step nilpotent group consisting of ${\ensuremath{\mathbb{R}}}^{2n+1}={\ensuremath{\mathbb{R}}}\times {\ensuremath{\mathbb{R}}}^{2n}$ equipped with the group law, $$x.y=(x_{0}+y_{0}+\sum_{1\leq j\leq n}(x_{n+j}y_{j}-x_{j}y_{n+j}),x_{1}+y_{1},\ldots,x_{2n}+y_{2n}).
$$ A left-invariant basis for its Lie algebra ${\ensuremath{\mathfrak{h}}}^{2n+1}$ is then provided by the vector fields, $$X_{0}=\frac{\partial}{\partial x_{0}}, \quad X_{j}=\frac{\partial}{\partial x_{j}}+x_{n+j}\frac{\partial}{\partial
x_{0}}, \quad X_{n+j}=\frac{\partial}{\partial x_{n+j}}-x_{j}\frac{\partial}{\partial
x_{0}}, \quad 1\leq j\leq n.
$$ For $j,k=1,\ldots,n$ and $k\neq j$ we have the Heisenberg relations $[X_{j},X_{n+k}]=-2\delta_{jk}X_{0}$ and $[X_{0},X_{j}]=[X_{j},X_{k}]=[X_{n+j},X_{n+k}]=0$. In particular, the subbundle spanned by the vector field $X_{1},\ldots,X_{2n}$ yields a left-invariant Heisenberg structure on ${\ensuremath{\mathbb{H}}}^{2n+1}$.
*- Foliations.* A (smooth) foliation is a manifold $M$ together with a subbundle ${\ensuremath{\mathcal{F}}}\subset TM$ integrable in Frobenius’ sense, i.e., the space of sections of $H$ is closed under the Lie bracket of vector fields. Therefore, any codimension 1 foliation is a Heisenberg manifold.
*- Contact manifolds.* Opposite to foliations are contact manifolds. A contact manifold is a Heisenberg manifold $(M^{2n+1}, H)$ such that $H$ can be locally realized as the kernel of a contact form, that is, a $1$-form $\theta$ such that $d\theta_{|H}$ is nondegenerate. When $M$ is orientable it is equivalent to require $H$ to be globally the kernel of a contact form. Furthermore, by Darboux’s theorem any contact manifold is locally Heisenberg-diffeomorphic to the Heisenberg group ${\ensuremath{\mathbb{H}}}^{2n+1}$ equipped with the standard contact form $\theta^{0}= dx_{0}+\sum_{j=1}^{n}(x_{j}dx_{n+j}-x_{n+j}dx_{j})$.
*- Confoliations.* According to Elyashberg-Thurston [@ET:C] a *confoliation structure* on an oriented manifold $M^{2n+1}$ is given by a global non-vanishing $1$-form $\theta$ on $M$ such that $(d\theta)^{n}\wedge \theta\geq 0$. In particular, if we let $H=\ker \theta$ then $(M,H)$ is a Heisenberg manifold which is a foliation when $d\theta\wedge \theta=0$ and a contact manifold when $(d\theta)^{n}\wedge \theta>0$.
*- CR manifolds.* A CR structure on an orientable manifold $M^{2n+1}$ is given by a rank $n$ complex subbundle $T_{1,0}\subset T_{{\ensuremath{\mathbb{C}}}}M$ such that $T_{1,0}$ is integrable in Frobenius’ sense and we have $T_{1,0}\cap T_{0,1}=\{0\}$, where we have set $T_{0,1}=\overline{T_{1,0}}$. Equivalently, the subbundle $H=\Re (T_{1,0}\otimes T_{0,1})$ has the structure of a complex bundle of (real) dimension $2n$. In particular, $(M,H)$ is a Heisenberg manifold. The main example of a CR manifold is that of the (smooth) boundary $M=\partial D$ of a bounded complex domain $D \subset {\ensuremath{\mathbb{C}}}^{n+1}$. In particular, when $D$ is strongly pseudoconvex with defining function $\rho$ the 1-form $\theta=i(\partial
-\bar{\partial})\rho$ is a contact form on $M$.
Next, the terminology Heisenberg manifold stems from the fact that the relevant tangent structure in this setting is that of a bundle $GM$ of graded nilpotent Lie groups (see [@BG:CHM], [@Be:TSSRG], [@EMM:RLSPD], [@FS:EDdbarbCAHG], [@Gr:CCSSW], [@Po:Pacific1], [@Ro:INA], [@Va:PhD]). This tangent Lie group bundle can be described as follows.
First, there is an intrinsic Levi form ${\ensuremath{\mathcal{L}}}:H\times H\rightarrow TM/H$ such that, for any point $a
\in M$ and any sections $X$ and $Y$ of $H$ near $a$, we have $${\ensuremath{\mathcal{L}}}_{a}(X(a),Y(a))=[X,Y](a) \qquad \bmod H_{a}.
\label{eq:Heisenberg.Levi-form}$$ In other words the class of $[X,Y](a)$ modulo $H_{a}$ depends only on the values $X(a)$ and $Y(a)$, not on the germs of $X$ and $Y$ near $a$ (see [@Po:Pacific1]). This allows us to define the tangent Lie algebra bundle ${\ensuremath{\mathfrak{g}}}M$ as the vector bundle $(TM/H)\oplus H$ together with the grading and field of Lie brackets such that, for sections $X_{0}$, $Y_{0}$ of $TM/H$ and $X'$, $Y'$ of $H$, we have $$\begin{gathered}
t.(X_{0}+X')=t^{2}X_{0}+t X', \qquad t\in {\ensuremath{\mathbb{R}}},
\label{eq:Heisenberg.Heisenberg-dilations}\\
[X_{0}+X',Y_{0}+Y']_{{\ensuremath{\mathfrak{g}}}M}={\ensuremath{\mathcal{L}}}(X',Y').
\end{gathered}$$
Since each fiber ${\ensuremath{\mathfrak{g}}}_{a}M$ is 2-step nilpotent, ${\ensuremath{\mathfrak{g}}}M$ is the Lie algebra bundle of a Lie group bundle $GM$ which can be realized as $(TM/H)\oplus H$ together with the field of group law such that, for sections $X_{0}$, $Y_{0}$ of $TM/H$ and $X'$, $Y'$ of $H$, we have $$(X_{0}+X').(Y_{0}+Y')=X_{0}+Y_{0}+\frac{1}{2}{\ensuremath{\mathcal{L}}}(X',Y')+X'+Y'.
\label{eq:Heisenberg.group-law}$$ We call $GM$ the *tangent Lie group bundle* of $M$.
Let $\phi$ be a Heisenberg diffeomorphism from $(M,H)$ onto a Heisenberg manifold $(M',H')$. Since we have $\phi_{*}H=H'$ the linear differential $\phi'$ induces linear vector bundle isomorphisms $\phi':H\rightarrow H'$ and $\overline{\phi'}:TM/H\rightarrow TM'/H'$, so that we get a linear vector bundle isomorphism $\phi_{H}':(TM/H)\oplus H\rightarrow (TM'/H')\oplus H'$ by letting $$\phi_{H}'(a).(X_{0}+X')= \overline{\phi'}(a)X_{0}+\phi'(a)X',
\label{eq:Heisenberg.tangent-map}$$ for any $a \in M$ and any $X_{0}$ in $(T_{a}M/H_{a})$ and $X'$ in $H_{a}$. This isomorphism commutes with the dilations in (\[eq:Heisenberg.Heisenberg-dilations\]) and it can be further shown that it gives rise to a Lie group isomorphism from $GM$ onto $GM'$ (see [@Po:Pacific1]).
The above description of $GM$ can be related to the extrinsic approach of [@BG:CHM] as follows.
A local frame $X_{0},X_{1},\ldots,X_{d}$ of $TM$ such that $X_{1},\ldots,X_{d}$ span $H$ is called a $H$-frame.
Let $U \subset {\ensuremath{{\ensuremath{\mathbb{R}}}^{d+1}}}$ be an open of local coordinate equipped with a $H$-frame $X_{0},\ldots,X_{d}$.
\[def:Heisenberg-privileged-coordinates\] For $a\in U$ we let $\psi_{a}:{\ensuremath{{\ensuremath{\mathbb{R}}}^{d+1}}}\rightarrow {\ensuremath{{\ensuremath{\mathbb{R}}}^{d+1}}}$ denote the unique affine change of variable such that $\psi_{a}(a)=0$ and $(\psi_{a})_{*}X_{j}(0)=\frac{\partial}{\partial x_{j}}$ for $j=0,\ldots,d$. The coordinates provided by the map $\psi_{a}$ are called privileged coordinates centered at $a$.
In addition, on ${\ensuremath{{\ensuremath{\mathbb{R}}}^{d+1}}}$ we consider the dilations, $$t.x=(t^{2}x_{0},tx_{1},\ldots,tx_{d}), \qquad t \in {\ensuremath{\mathbb{R}}}.
\label{eq:Heisenberg.Heisenberg-dilations-Rd}$$
In privileged coordinates centered at $a$ we can write $X_{j}=\frac{\partial}{\partial x_{j}}+\sum_{k=0}^{d}a_{jk}(x)\frac{\partial}{\partial x_{j}}$ with $a_{jk}(0)=0$. Let $X_{0}^{(a)}=\frac{\partial}{\partial x_{0}}$ and for $j=1,\ldots,d$ let $X_{j}^{(a)}=\frac{\partial}{\partial_{x_{j}}}+\sum_{k=1}^{d}b_{jk}x_{k}
\frac{\partial}{\partial_{x_{0}}}$, where $b_{jk}= \partial_{x_{k}}a_{j0}(0)$. With respect to the dilations (\[eq:Heisenberg.Heisenberg-dilations-Rd\]) the vector field $X_{j}^{(a)}$ is homogeneous of degree $w_{0}=-2$ for $j=0$ and of degree $w_{j}=-1$ for $j=1,\ldots,d$. In fact, using Taylor expansions at $x=0$ we get a formal expansion $X_{j} \sim X_{j}^{(a)}+X_{j,w_{j}-1}+\ldots$, with $X_{j,l}$ homogeneous vector field of degree $l$.
The subbundle spanned by the vector fields $X_{j}^{(a)}$ is a 2-step nilpotent Lie algebra under the Lie bracket of vectors fields. Its associated Lie group $G^{(a)}$ can be realized as ${\ensuremath{{\ensuremath{\mathbb{R}}}^{d+1}}}$ equipped with the group law, $$x.y=(x_{0}+\sum_{j,k=1}^{d}b_{kj}x_{j}x_{k},x_{1},\ldots,x_{d}).
$$
On the other hand, the vectors $X_{0}(a),\ldots,X_{d}(a)$ provide us with a linear basis of the space $(T_{a}M/H_{a})\oplus H_{a}$. This allows us to identify $G_{a}M$ with ${\ensuremath{{\ensuremath{\mathbb{R}}}^{d+1}}}$ equipped with the group law, $$x.y=(x_{0}+y_{0}+\frac{1}{2}L_{jk}(a)x_{j}y_{k},x_{1}+y_{1},\ldots,x_{d}+y_{d}).
\label{eq:Heisenberg.group-law-tangent-group-coordinates}$$ Here the functions $L_{jk}$ denote the coefficients of the Levi form (\[eq:Heisenberg.Levi-form\]) with respect to the $H$-frame $X_{0},\ldots,X_{d}$, i.e., we have ${\ensuremath{\mathcal{L}}}(X_{j},X_{k})=[X_{j},X_{k}]=L_{jk}X_{0}
\bmod H$.
The Lie group $G^{(a)}$ is isomorphic to $G_{a}M$ since one can check that $L_{jk}=b_{jk}-b_{kj}$. An explicit isomorphism is given by $$\phi_{a}(x_{0},\ldots,x_{d})= (x_{0}-\frac{1}{4}\sum_{j,k=1}^{d}(b_{jk}+b_{kj})x_{j}x_{k},x_{1},\ldots,x_{d}).
$$
\[def:Heisenberg-Heisenberg-coordinates\] The local coordinates provided by the map $\varepsilon_{a}:=\phi_{a}\circ\psi_{a}$ are called Heisenberg coordinates centered at $a$.
The Heisenberg coordinates refines the privileged coordinates in such way that the above realizations of $G^{(a)}$ and $G_{a}M$ agree. In particular, the vector fields $X_{j}^{(a)}$ and $X_{j}^{a}$ agree in these coordinates. This allows us to see $X_{j}^{a}$ as a first order approximation of $X_{j}$. For this reason $X_{j}^{a}$ is called the *model vector field of $X_{j}$* at $a$.
Left-invariant pseudodifferential operators
-------------------------------------------
Let $(M^{d+1},H)$ be a Heisenberg manifold and let $G$ be the tangent group $G_{a}M$ of $M$ at a given point $a\in M$. We briefly recall the calculus for homogeneous left-invariant [$\Psi$DOs]{} on the nilpotent group $G$.
Recall that if $E$ is a finite dimensional vector space the Schwartz class ${\ensuremath{\mathcal{S}}}(E)$ carries a natural Fréchet space topology and the Fourier transform of a function $f\in {\ensuremath{\mathcal{S}}}(E)$ is the element $\hat{f}\in {\ensuremath{\mathcal{S}}}(E^{*})$ such that $\hat{f}(\xi)=\int_{E}e^{i{\ensuremath{\langle \xi , x \rangle}}}f(x)dx$ for any $\xi \in E^{*}$, where $dx$ denotes the Lebesgue measure of $E$. In the case where $E=(T_{a}M/H_{a})\oplus H_{a}$ the Lebesgue measure actually agrees with the Haar measure of $G$, so ${\ensuremath{\mathcal{S}}}(E)$ and ${\ensuremath{\mathcal{S}}}(G)$ agree. Furthermore, as $E^{*}=(T_{a}M/H_{a})^{*}\otimes H_{a}^{*}$ is just the linear dual ${\ensuremath{\mathfrak{g}}}^{*}$ of the Lie algebra of $G$, we also see that ${\ensuremath{\mathcal{S}}}(E^{*})$ agrees with ${\ensuremath{\mathcal{S}}}({\ensuremath{\mathfrak{g}}}^{*})$.
Let ${\ensuremath{\mathcal{S}}}_{0}(G)$ denote the closed subspace of ${\ensuremath{\mathcal{S}}}(G)$ consisting of functions $f \in {\ensuremath{\mathcal{S}}}(G)$ such that for any differential operator $P$ on ${\ensuremath{\mathfrak{g}}}^{*}$ we have $(P\hat{f})(0)=0$. Notice that the image $\hat{{\ensuremath{\mathcal{S}}}}_{0}(G)$ of ${\ensuremath{\mathcal{S}}}(G)$ under the Fourier transform consists of functions $v\in {\ensuremath{\mathcal{S}}}({\ensuremath{\mathfrak{g}}}^{*})$ such that, given any norm $|.|$ on $G$, near $\xi=0$ we have $|g(\xi)|={\operatorname{O}}(|\xi|^{N})$ for any $N\in {\ensuremath{\mathbb{N}}}$.
We endow ${\ensuremath{\mathfrak{g}}}^{*}$ with the dilations $\lambda.\xi=(\lambda^{2}\xi_{0},\lambda\xi')$ coming from (\[eq:Heisenberg.Heisenberg-dilations\]). For $m\in {\ensuremath{\mathbb{C}}}$ we let $S_{m}({\ensuremath{\mathfrak{g}}}^{*}M)$ denote the closed subspace of $C^{\infty}({\ensuremath{\mathfrak{g}}}^{*}\setminus 0)$ consisting in functions $p(\xi)\in C^{\infty}({\ensuremath{\mathfrak{g}}}^{*}\setminus 0)$ such that $p(\lambda.\xi)=\lambda^{m}p(\xi)$ for any $\lambda>0$.
If $p(\xi)\in S_{m}({\ensuremath{\mathfrak{g}}}^{*})$ then it defines an element of $\hat{{\ensuremath{\mathcal{S}}}}_{0}({\ensuremath{\mathfrak{g}}}^{*})'$ by letting $${\ensuremath{\langle p , g \rangle}}= \int_{{\ensuremath{\mathfrak{g}}}^{*}}p(\xi)g(\xi)d\xi, \qquad g \in \hat{{\ensuremath{\mathcal{S}}}}_{0}({\ensuremath{\mathfrak{g}}}^{*}).$$ This allows us to define the inverse Fourier transform of $p$ as the element $\check{p}\in {\ensuremath{\mathcal{S}}}_{0}(G)'$ such that ${\ensuremath{\langle \check{p} , f \rangle}}={\ensuremath{\langle p , \check{f} \rangle}}$ for any $f \in {\ensuremath{\mathcal{S}}}_{0}(G)$. It then can be shown (see, e.g., [@BG:CHM], [@CGGP:POGD]) that the left-convolution with $p$ defines a continuous endomorphism of ${\ensuremath{\mathcal{S}}}_{0}(G)$ via the formula, $${\operatorname{Op}}(p)f(x)=\check{p}*f(x)={\ensuremath{\langle \check{p}(y) , f(xy) \rangle}}, \qquad f\in {\ensuremath{\mathcal{S}}}_{0}(G).
\label{Heisenberg.left-invariant-PDO}$$ Moreover, we have a bilinear product, $$*:S_{m_{1}}({\ensuremath{\mathfrak{g}}}^{*})\times S_{m_{2}}({\ensuremath{\mathfrak{g}}}^{*})
\longrightarrow S_{m_{1}+m_{2}}({\ensuremath{\mathfrak{g}}}^{*}),
\label{eq:Heisenberg.product-symbols}$$ in such way that, for any $p_{1}\in S_{m_{1}}({\ensuremath{\mathfrak{g}}}^{*})$ and any $p_{2}\in S_{m_{2}}({\ensuremath{\mathfrak{g}}}^{*})$, we have $${\operatorname{Op}}(p_{1})\circ {\operatorname{Op}}(p_{2})={\operatorname{Op}}(p_{1}*p_{2}).''$$
In addition, if $p \in S_{m}({\ensuremath{\mathfrak{g}}}^{*})$ then ${\operatorname{Op}}(p)$ really is a pseudodifferential operator. Indeed, let $X_{0}(a),\ldots,X_{d}(a)$ be a (linear) basis of ${\ensuremath{\mathfrak{g}}}$ so that $X_{0}(a)$ is in $T_{a}M/H_{a}$ and $X_{1}(a),\ldots,X_{d}(a)$ span $H_{a}$. For $j=0,\ldots,d$ let $X_{j}^{a}$ be the left-invariant vector field on $G$ such that $X^{a}_{j|_{x=0}}=X_{j}(a)$. The basis $X_{0}(a),\ldots,X_{d}(a)$ yields a linear isomorphism ${\ensuremath{\mathfrak{g}}}\simeq {\ensuremath{{\ensuremath{\mathbb{R}}}^{d+1}}}$, hence a global chart of $G$. In the corresponding local coordinates $p(\xi)$ is a homogeneous symbol on ${\ensuremath{{\ensuremath{\mathbb{R}}}^{d+1}}}\setminus 0$ with respect to the dilations (\[eq:Heisenberg.Heisenberg-dilations-Rd\]). Similarly, each vector field $\frac{1}{i}X_{j}^{a}$, $j=0,\ldots,d$, corresponds to a vector field on ${\ensuremath{{\ensuremath{\mathbb{R}}}^{d+1}}}$ with symbol $\sigma_{j}^{a}(x,\xi)$. If we set $\sigma^{a}(x,\xi)=(\sigma_{0}^{a}(x,\xi),\ldots,\sigma_{d}^{a}(x,\xi))$, then it can be shown that in these local coordinates we have $${\operatorname{Op}}(p)f(x)= (2\pi)^{-(d+1)}\int_{{\ensuremath{{\ensuremath{\mathbb{R}}}^{d+1}}}} e^{i{\ensuremath{\langle x , \xi \rangle}}}p(\sigma^{a}(x,\xi))\hat{f}(\xi)d\xi, \qquad f \in {\ensuremath{\mathcal{S}}}_{0}({\ensuremath{{\ensuremath{\mathbb{R}}}^{d+1}}}).
\label{eq:PsiHDO.PsiDO-convolution}$$ In other words ${\operatorname{Op}}(p)$ is the pseudodifferential operator $p(-iX^{a}):=p(\sigma^{a}(x,D))$ acting on ${\ensuremath{\mathcal{S}}}_{0}({\ensuremath{{\ensuremath{\mathbb{R}}}^{d+1}}})$.
The [$\Psi_{H}$DO]{} operators
------------------------------
The original idea in the Heisenberg calculus, which goes back to Elias Stein, is to construct a class of operators on a given Heisenberg manifold $(M^{d+1},H)$, called [$\Psi_{H}$DOs]{}, which at any point $a \in M$ are modeled in a suitable sense on left-invariant pseudodifferential operators on the tangent group $G_{a}M$.
Let $U \subset {\ensuremath{{\ensuremath{\mathbb{R}}}^{d+1}}}$ be an open of local coordinates equipped with a $H$-frame $X_{0},\ldots,X_{d}$.
$S_{m}({U\times{\ensuremath{\mathbb{R}}}^{d+1}})$, $m\in{\ensuremath{\mathbb{C}}}$, consists of functions $p(x,\xi)$ in $C^{\infty}(U\times{{\ensuremath{\mathbb{R}}}^{d+1}\!\setminus\! 0})$ which are homogeneous of degree $m$ in the $\xi$-variable with respect to the dilations (\[eq:Heisenberg.Heisenberg-dilations-Rd\]), i.e., we have $p(x,t.\xi)=t^m p(x,\xi)$ for any $t>0$.
In the sequel we endow ${\ensuremath{{\ensuremath{\mathbb{R}}}^{d+1}}}$ with the pseudo-norm, $$\|\xi\|=(\xi_{0}^{2}+\xi_{1}^{4}+\ldots+\xi_{d}^{4})^{1/4}, \qquad \xi\in {\ensuremath{{\ensuremath{\mathbb{R}}}^{d+1}}}.
$$ In addition, for any multi-order $\beta\in {\ensuremath{\mathbb{N}}}^{d+1}_{0}$ we set ${\ensuremath{\langle\! \beta\!\rangle}}=2\beta_{0}+\beta_{1}+\ldots+\beta_{d}$.
$S^m({U\times{\ensuremath{\mathbb{R}}}^{d+1}})$, $m\in{\ensuremath{\mathbb{C}}}$, consists of functions $p(x,\xi)$ in $C^{\infty}({U\times{\ensuremath{\mathbb{R}}}^{d+1}})$ with an asymptotic expansion $ p \sim \sum_{j\geq 0} p_{m-j}$, $p_{k}\in S_{k}({U\times{\ensuremath{\mathbb{R}}}^{d+1}})$, in the sense that, for any integer $N$, any compact $K \subset U$ and any multi-orders $\alpha$, $\beta$, there exists $C_{NK\alpha\beta}>0$ such that, for any $x\in K$ and any $\xi\in {\ensuremath{{\ensuremath{\mathbb{R}}}^{d+1}}}$ so that $\|\xi \| \geq 1$, we have $$| \partial^\alpha_{x}\partial^\beta_{\xi}(p-\sum_{j<N}p_{m-j})(x,\xi)| \leq
C_{NK\alpha\beta }\|\xi\|^{\Re m-{\ensuremath{\langle\! \beta\!\rangle}} -N}.
\label{eq:Heisenberg.asymptotic-expansion-symbols}$$
Next, for $j=0,\ldots,d$ let $\sigma_{j}(x,\xi)$ denote the symbol (in the classical sense) of the vector field $\frac{1}{i}X_{j}$ and set $\sigma=(\sigma_{0},\ldots,\sigma_{d})$. Then for $p \in S^{m}({U\times{\ensuremath{\mathbb{R}}}^{d+1}})$ we let $p(x,-iX)$ be the continuous linear operator from $C^{\infty}_{c}(U)$ to $C^{\infty}(U)$ such that $$p(x,-iX)f(x)= (2\pi)^{-(d+1)} \int e^{ix.\xi} p(x,\sigma(x,\xi))\hat{f}(\xi)d\xi,
\qquad f\in C^{\infty}_{c}(U).$$
In the sequel we let ${\ensuremath{\Psi^{-\infty}}}(U)$ denote the space of smoothing operators on $U$, that is, the space of continuous operators $P:{\ensuremath{\mathcal{E}}}'(U)\rightarrow {\ensuremath{\mathcal{D}}}'(U)$ with a smooth Schwartz kernel.
${\ensuremath{\Psi_{H}}}^{m}(U)$, $m\in {\ensuremath{\mathbb{C}}}$, consists of operators $P:C^{\infty}_{c}(U)\rightarrow C^{\infty}(U)$ of the form $$P= p(x,-iX)+R,
$$ with $p$ in $S^{m}({U\times{\ensuremath{\mathbb{R}}}^{d+1}})$ (called the symbol of $P$) and $R$ smoothing operator.
The class of [$\Psi_{H}$DOs]{} is invariant under changes of $H$-framed charts (see [@BG:CHM Sect. 16], [@Po:MAMS1 Appendix A]). Therefore, we can extend the definition of [$\Psi_{H}$DOs]{} to the Heisenberg manifold $(M^{d+1},H)$ and let them act on sections of a vector bundle ${\ensuremath{\mathcal{E}}}^{r}$ over $M$ as follows.
${\ensuremath{\Psi_{H}}}^{m}(M,{\ensuremath{\mathcal{E}}})$, $m\in {\ensuremath{\mathbb{C}}}$, consists of continuous operators $P$ from $C^{\infty}_{c}(M,{\ensuremath{\mathcal{E}}})$ to $C^{\infty}(M,{\ensuremath{\mathcal{E}}})$ such that:
\(i) The Schwartz kernel of $P$ is smooth off the diagonal;
\(ii) For any $H$-framed local chart $\kappa:U\rightarrow V\subset {\ensuremath{{\ensuremath{\mathbb{R}}}^{d+1}}}$ over which there is a trivialization $\tau:{\ensuremath{\mathcal{E}}}_{|U}\rightarrow U\times
{\ensuremath{\mathbb{C}}}^{r}$ the operator $\kappa_{*}\tau_{*}(P_{|U})$ belongs to ${\ensuremath{\Psi_{H}}}^{m}(V,{\ensuremath{\mathbb{C}}}^{r}):={\ensuremath{\Psi_{H}}}^{m}(V)\otimes {\ensuremath{{\operatorname{End}}}}{\ensuremath{\mathbb{C}}}^{r}$.
Let $P\in {\ensuremath{\Psi_{H}}}^{m}(M,{\ensuremath{\mathcal{E}}})$, $m \in {\ensuremath{\mathbb{C}}}$.
\(1) Let $Q\in {\ensuremath{\Psi_{H}}}^{m'}(M,{\ensuremath{\mathcal{E}}})$, $m'\in {\ensuremath{\mathbb{C}}}$, and suppose that $P$ or $Q$ is uniformly properly supported. Then the operator $PQ$ belongs to ${\ensuremath{\Psi_{H}}}^{m+m'}(M,{\ensuremath{\mathcal{E}}})$.
\(2) The transpose operator $P^{t}$ belongs to ${\ensuremath{\Psi_{H}}}^{m}(M,{\ensuremath{\mathcal{E}}}^{*})$.
\(3) Suppose that $M$ is endowed with a density $>0$ and ${\ensuremath{\mathcal{E}}}$ is endowed with a Hermitian metric. Then the adjoint $P^{*}$ of $P$ belongs to ${\ensuremath{\Psi_{H}}}^{\overline{m}}(M,{\ensuremath{\mathcal{E}}})$.
In this setting the principal symbol of a [$\Psi_{H}$DO]{} can be defined intrinsically as follows.
Let ${\ensuremath{\mathfrak{g}}}^{*}M=(TM/H)^{*}\oplus H^{*}$ denote the (linear) dual of the Lie algebra bundle ${\ensuremath{\mathfrak{g}}}M$ of $GM$ with canonical projection $\text{pr}: M\rightarrow {\ensuremath{\mathfrak{g}}}^{*}M$. For $m \in {\ensuremath{\mathbb{C}}}$ we let $S_{m}({\ensuremath{\mathfrak{g}}}^{*}M,{\ensuremath{\mathcal{E}}})$ be the space of sections $p\in C^{\infty}({\ensuremath{\mathfrak{g}}}^{*}M\setminus 0,{\ensuremath{{\operatorname{End}}}}\text{pr}^{*}{\ensuremath{\mathcal{E}}})$ such that $p(x,t.\xi)=t^{m}p(x,\xi)$ for any $t>0$.
\[def:Heisenberg.principal-symbol\] The principal symbol of an operator $P\in {\ensuremath{\Psi_{H}}}^{m}(M,{\ensuremath{\mathcal{E}}})$ is the unique symbol $\sigma_{m}(P)$ in $S_{m}({\ensuremath{\mathfrak{g}}}^{*}M,{\ensuremath{\mathcal{E}}})$ such that, for any $a\in M$ and for any trivializing $H$-framed local coordinates near $a$, in Heisenberg coordinates centered at $a$ we have $\sigma_{m}(P)(0,\xi)=p_{m}(0,\xi)$, where $p_{m}(x,\xi)$ is the principal symbol of $P$ in the sense of (\[eq:Heisenberg.asymptotic-expansion-symbols\]).
Given a point $a\in M$ the principal symbol $\sigma_{m}(P)$ allows us to define the model operator of $P$ at $a$ as the left-invariant [$\Psi$DO]{} on ${\ensuremath{\mathcal{S}}}_{0}({\ensuremath{\mathfrak{g}}}^{*}M,{\ensuremath{\mathcal{E}}}_{a})$ with symbol $p_{m}^{a}(\xi):=\sigma_{m}(P)(a,\xi)$ so that, in the notation of (\[Heisenberg.left-invariant-PDO\]), the operator $P^{a}$ is just ${\operatorname{Op}}(p_{m}^{a})$.
For $m \in {\ensuremath{\mathbb{C}}}$ let $S_{m}({\ensuremath{\mathfrak{g}}}^{*}_{a}M,{\ensuremath{\mathcal{E}}}_{a})$ be the space of functions $p\in C^{\infty}({\ensuremath{\mathfrak{g}}}^{*}_{a}M\setminus 0,{\ensuremath{\mathcal{E}}}_{a})$ which are homogeneous of degree $m$. Then the product (\[eq:Heisenberg.product-symbols\]) yields a bilinear product, $$*^{a}:S_{m_{1}}({\ensuremath{\mathfrak{g}}}^{*}_{a}M,{\ensuremath{\mathcal{E}}}_{a})\times S_{m_{2}}({\ensuremath{\mathfrak{g}}}^{*}_{a}M,{\ensuremath{\mathcal{E}}}_{a})\rightarrow S_{m_{1}+m_{2}}({\ensuremath{\mathfrak{g}}}^{*}_{a}M,{\ensuremath{\mathcal{E}}}_{a}).
$$ This product depends smoothly on $a$ as much so to gives rise to the bilinear product, $$\begin{gathered}
*:S_{m_{1}}({\ensuremath{\mathfrak{g}}}^{*}M,{\ensuremath{\mathcal{E}}})\times S_{m_{2}}({\ensuremath{\mathfrak{g}}}^{*}M,{\ensuremath{\mathcal{E}}}) \longrightarrow S_{m_{1}+m_{2}}({\ensuremath{\mathfrak{g}}}^{*}M,{\ensuremath{\mathcal{E}}}),
\label{eq:CPCL.product-symbols}\\
p_{m_{1}}*p_{m_{2}}(a,\xi)=(p_{m_{1}}(a,.)*^{a}p_{m_{2}}(a,.))(\xi), \qquad p_{m_{j}}\in S_{m_{j}}({\ensuremath{\mathfrak{g}}}^{*}M).
$$
\[prop:Heisenberg.operations-principal-symbols\] Let $P\in {\ensuremath{\Psi_{H}}}^{m}(M,{\ensuremath{\mathcal{E}}})$, $m \in {\ensuremath{\mathbb{C}}}$.
1\) Let $Q\in {\ensuremath{\Psi_{H}}}^{m'}(M,{\ensuremath{\mathcal{E}}})$, $m'\in {\ensuremath{\mathbb{C}}}$, and suppose that $P$ or $Q$ is uniformly properly supported. Then we have $\sigma_{m+m'}(PQ)=\sigma_{m}(P)*\sigma_{m'}(Q)$, and for any $a \in M$ the model operator of $PQ$ at $a$ is $P^{a}Q^{a}$.
2\) We have $\sigma_{m}(P^{t})(x,\xi)=\sigma_{m}(P)(x,-\xi)^{t}$, and for any $a \in M$ the model operator of $P^{t}$ at $a$ is $(P^{a})^{t}$.
3\) Suppose that $M$ is endowed with a density $>0$ and ${\ensuremath{\mathcal{E}}}$ is endowed with a Hermitian metric. Then we have $\sigma_{\overline{m}}(P^{*})(x,\xi)=\sigma_{m}(P)(x,\xi)^{*}$, and for any $a \in M$ the model operator of $P^{*}$ at $a$ is $(P^{a})^{*}$.
In addition, there is a complete symbolic calculus for [$\Psi_{H}$DOs]{} which allows us to carry out the classical parametrix construction for an operator $P\in {\ensuremath{\Psi_{H}}}^{m}(M,{\ensuremath{\mathcal{E}}})$ whenever its principal symbol $\sigma_{m}(P)$ is invertible with respect to the product $*$ (see [@BG:CHM]). In general, it may be difficult to determine whether $\sigma_{m}(P)$ is invertible with respect to that product. Nevertheless, given a point $a\in M$ we have an invertibility criterion for $P^{a}$ in terms of the representation theory of $G_{a}M$; this is the so-called Rockland condition (see, e.g., [@Ro:HHGRTC], [@CGGP:POGD]). We then can completely determine the invertibility of the principal symbol of $P$ in terms of the Rockland conditions for its model operators and those of its transpose (see [@Po:MAMS1 Thm. 3.3.19]).
Finally, the [$\Psi_{H}$DOs]{} enjoy nice Sobolev regularity properties. These properties are best stated in terms of the weighted Sobolev of [@FS:EDdbarbCAHG] and [@Po:MAMS1]. These weighted Sobolev spaces can be explicitly related to the usual Sobolev spaces and allows us to show that if $P\in
{\ensuremath{\Psi_{H}}}^{m}(M,{\ensuremath{\mathcal{E}}})$, $\Re m>0$, has an invertible principal symbol, then $P$ is maximal hypoelliptic, which implies that $P$ is hypoelliptic with gain of $\frac{m}{2}$-derivatives. We refer to [@BG:CHM] and [@Po:MAMS1] for the precise statements. In the sequel we will only need the following.
\[prop:Heisenberg.L2-boundedness\] Assume $M$ compact and let $P\in {\ensuremath{\Psi_{H}}}^{m}(M,{\ensuremath{\mathcal{E}}})$, $\Re m\geq 0$. Then $P$ extends to a bounded operator from $L^{2}(M,{\ensuremath{\mathcal{E}}})$ to itself and this operator is compact if we further have $\Re m<0$.
Holomorphic families of [$\Psi_{H}$DOs]{}
-----------------------------------------
In this subsection we recall the main definitions and properties of holomorphic families of [$\Psi_{H}$DOs]{}. Throughout the subsection we let $(M^{d+1},H)$ be a Heisenberg manifold, we let ${\ensuremath{\mathcal{E}}}^{r}$ be a vector bundle over $M$ and we let $\Omega$ be an open subset of ${\ensuremath{\mathbb{C}}}$.
Let $U\subset {\ensuremath{{\ensuremath{\mathbb{R}}}^{d+1}}}$ be an open of local coordinates equipped with a $H$-frame $X_{0},\ldots,X_{d}$. We define holomorphic families of symbols on ${U\times{\ensuremath{\mathbb{R}}}^{d+1}}$ as follows.
\[def:Heisenberg.hol-family-symbols\] A family $(p(z))_{z\in\Omega}\subset S^*({U\times{\ensuremath{\mathbb{R}}}^{d+1}})$ is holomorphic when:
\(i) The order $w(z)$ of $p(z)$ depends analytically on $z$;
\(ii) For any $(x,\xi)\in {U\times{\ensuremath{\mathbb{R}}}^{d+1}}$ the function $z\rightarrow p(z)(x,\xi)$ is holomorphic on $\Omega$;
\(iii) The bounds of the asymptotic expansion (\[eq:Heisenberg.asymptotic-expansion-symbols\]) for $p(z)$ are locally uniform with respect to $z$, i.e., we have $p(z) \sim \sum_{j\geq 0} p(z)_{ w(z)-j}$, $p(z)_{w(z)-j}\in S_{w(z)-j}({U\times{\ensuremath{\mathbb{R}}}^{d+1}})$, and, for any integer $N$, any compacts $K\subset U$ and $L\subset \Omega$ and any multi-orders $\alpha$ and $\beta$, there exists a constant $ C_{NKL\alpha\beta}>0$ such that, for any $(x,z)\in K\times L$ and any $\xi \in {\ensuremath{{\ensuremath{\mathbb{R}}}^{d+1}}}$ so that $\|\xi\|\geq 1$, we have $$| \partial_{x}^\alpha\partial_{\xi}^\beta (p(z)-\sum_{j<N}
p(z)_{w(z)-j})(x,\xi)| \leq C_{NKL\alpha\beta} \|\xi\|^{\Re w(z)-N-{\ensuremath{\langle\! \beta\!\rangle}}}.
\label{eq:Heisenberg.symbols.asymptotic-expansion-hol-families}$$
In the sequel we let ${{\operatorname{Hol}}}(\Omega,S^*({U\times{\ensuremath{\mathbb{R}}}^{d+1}}))$ denote the class of holomorphic families with values in $S^*({U\times{\ensuremath{\mathbb{R}}}^{d+1}})$. Notice also that the properties (i)–(iii) imply that each homogeneous symbol $p(z)_{w(z)-j}(x,\xi)$ depends analytically on $z$, that is, it gives rise to a holomorphic family with values in $C^{\infty}({U\times({\ensuremath{\mathbb{R}}}^{d+1}\!\setminus\! 0)})$ (see [@Po:MAMS1 Rem. 4.2.2]).
Since ${\ensuremath{\Psi^{-\infty}}}(U)={\ensuremath{\mathcal{L}}}({\ensuremath{\mathcal{E}}}'(U),C^{\infty}(U))$ is a Fréchet space which is isomorphic to $C^{\infty}(U\times U)$ by Schwartz’s Kernel Theorem, we can define holomorphic families of smoothing operators as families of operators given by holomorphic families of smooth Schwartz kernels. We let ${{\operatorname{Hol}}}(\Omega,{\ensuremath{\Psi^{-\infty}}}(U))$ denote the class of such families.
\[def:Heisenberg.hol-family-PsiHDO’s\] A family $(P(z))_{z\in \Omega}\subset {\ensuremath{\Psi_{H}}}^{m}(U)$ is holomorphic when it can be put in the form, $$P(z) = p(z)(x,-iX) + R(z), \qquad z \in \Omega,
$$ with $(p(z))_{z\in \Omega}\in {{\operatorname{Hol}}}(\Omega, S^{*}({U\times{\ensuremath{\mathbb{R}}}^{d+1}}))$ and $(R(z))_{z\in \Omega} \in {{\operatorname{Hol}}}(\Omega,{\ensuremath{\Psi^{-\infty}}}(U))$.
The above notion of holomorphic families of [$\Psi_{H}$DOs]{} is invariant under changes of $H$-framed charts (see [@Po:MAMS1]). Therefore, it makes sense to define holomorphic families of [$\Psi_{H}$DOs]{} on the Heisenberg manifold $(M^{d+1},H)$ acting on sections of the vector bundle ${\ensuremath{\mathcal{E}}}^{r}$ as follows.
A family $(P(z))_{z\in \Omega}\subset {\ensuremath{\Psi_{H}}}^{*}(M,{\ensuremath{\mathcal{E}}})$ is holomorphic when:
\(i) The order $w(z)$ of $P(z)$ is a holomorphic function of $z$;
\(ii) For $\varphi$ and $\psi$ in $C^\infty_{c}(M)$ with disjoint supports $(\varphi P(z)\psi)_{z\in \Omega}$ is a holomorphic family of smoothing operators; (iii) For any trivialization $\tau:{\ensuremath{\mathcal{E}}}_{|_{U}}\rightarrow U\times {\ensuremath{\mathbb{C}}}^{r}$ over a local $H$-framed chart $\kappa:U \rightarrow V\subset {\ensuremath{{\ensuremath{\mathbb{R}}}^{d+1}}}$ the family $(\kappa_{*}\tau_{*}(P_{z|_{U}}))_{z\in\Omega}$ belongs to ${{\operatorname{Hol}}}(\Omega, {\ensuremath{\Psi_{H}}}^{*}(V,{\ensuremath{\mathbb{C}}}^{r})):={{\operatorname{Hol}}}(\Omega, {\ensuremath{\Psi_{H}}}^{*}(V))\otimes {\ensuremath{{\operatorname{End}}}}{\ensuremath{\mathbb{C}}}^{r}$.
We let ${{\operatorname{Hol}}}(\Omega,{\ensuremath{\Psi_{H}}}^{*}(M,{\ensuremath{\mathcal{E}}}))$ denote the class of holomorphic families of [$\Psi_{H}$DOs]{} on $M$ and acting on the sections of ${\ensuremath{\mathcal{E}}}$.
Let $(P(z))_{z\in\Omega}\subset {\ensuremath{\Psi_{H}}}^{*}(M,{\ensuremath{\mathcal{E}}})$ be a holomorphic family of [$\Psi_{H}$DOs]{}.
1\) Let $(Q(z))_{z\in\Omega}\subset {\ensuremath{\Psi_{H}}}^{*}(M,{\ensuremath{\mathcal{E}}})$ be a holomorphic family of [$\Psi_{H}$DOs]{} and assume that $(P(z))_{z\in\Omega}$ or $(Q(z))_{z\in\Omega}$ is uniformly properly supported with respect to $z$. Then the family $(P(z)Q(z))_{z \in \Omega}$ belongs to ${{\operatorname{Hol}}}(\Omega,{\ensuremath{\Psi_{H}}}^{*}(M,{\ensuremath{\mathcal{E}}}))$. 2) Let $\phi:(M,H)\rightarrow (M',H')$ be a Heisenberg diffeomorphism. Then the family $(\phi_{*}P(z))_{z\in \Omega}$ belongs to ${{\operatorname{Hol}}}(\Omega, \Psi_{H'}^{*}(M',\phi_{*}{\ensuremath{\mathcal{E}}}))$.
Complex powers of hypoelliptic [$\Psi_{H}$DOs]{}
------------------------------------------------
In this subsection we recall the constructions in [@Po:MAMS1] and [@Po:CPDE1] of complex powers of hypoelliptic [$\Psi_{H}$DOs]{} as holomorphic families of [$\Psi_{H}$DOs]{}.
Throughout this subsection we let $(M^{d+1},H)$ be a compact Heisenberg manifold equipped with a density $>0$ and we let ${\ensuremath{\mathcal{E}}}$ be a Hermitian vector bundle over $M$.
Let $P:C^{\infty}(M,{\ensuremath{\mathcal{E}}})\rightarrow C^{\infty}(M,{\ensuremath{\mathcal{E}}})$ be a differential operator of Heisenberg order $m$ which is positive, i.e., we have ${\ensuremath{\langle Pu , u \rangle}}\geq
0$ for any $u \in C^{\infty}(M,{\ensuremath{\mathcal{E}}})$, and assume that the principal symbol of $P$ is invertible, that is, $P$ satisfies the Rockland condition at every point.
By standard functional calculus for any $s\in {\ensuremath{\mathbb{C}}}$ we can define the power $P^{s}$ as an unbounded operator on $L^{2}(M,{\ensuremath{\mathcal{E}}})$ whose domain contains $C^{\infty}(M,{\ensuremath{\mathcal{E}}})$. In particular $P^{-1}$ is the partial inverse of $P$ and we have $P^{0}=1-\Pi_{0}(P)$, where $\Pi_{0}(P)$ denotes the orthogonal projection onto the kernel of $P$. Furthermore, we have:
\[prop:Heisenberg.complex-powers.positive\] Assume that $H$ satisfies the bracket condition $H+[H,H]=TM$. Then the complex powers $(P^{s})_{s \in {\ensuremath{\mathbb{C}}}}$ form a holomorphic 1-parameter group of [$\Psi_{H}$DOs]{} such that ${{{\operatorname{ord}}}}P^{s}=ms\ \forall s\in {\ensuremath{\mathbb{C}}}$.
This construction has been generalized to more general hypoelliptic [$\Psi_{H}$DOs]{} in [@Po:CPDE1]. Let $P:C^{\infty}(M,{\ensuremath{\mathcal{E}}})\rightarrow C^{\infty}(M,{\ensuremath{\mathcal{E}}})$ be a [$\Psi_{H}$DO]{} of order $m>0$. In [@Po:CPDE1] there is a notion of *principal cut* for the principal symbol $\sigma_{m}(P)$ of $P$ as a ray $L\subset {\ensuremath{\mathbb{C}}}\setminus 0$ such that $P-\lambda$ admits a parametrix in a version of the Heisenberg calculus with parameter in a conical neighborhood $\Theta \subset {\ensuremath{\mathbb{C}}}\setminus 0$ of $L$.
Let $\Theta(P)$ be the union set of all principal cuts of $\sigma_{m}(P)$. Then $\Theta(P)$ is an open conical subset of ${\ensuremath{\mathbb{C}}}\setminus 0$ and for any conical subset $\Theta$ of $\Theta(P)$ such that $\overline{\Theta}\setminus 0\subset \Theta(P)$ there are at most finitely many eigenvalues of $P$ in $\Theta$ (see [@Po:CPDE1]).
Let $L_{\theta}=\{\arg \lambda=\theta\}$, $0\leq \theta<2\pi$, be a principal cut for $\sigma_{m}(P)$ such that no eigenvalue of $P$ lies in $L$. Then $L_{\theta}$ is ray of minimal growth for $P$, so for $\Re s<0$ we define a bounded operator on $L^{2}(M,{\ensuremath{\mathcal{E}}})$ by letting $$\begin{gathered}
P_{\theta}^{s}= \frac{-1}{2i\pi} \int_{\Gamma_{\theta}} \lambda^{s}_{\theta}(P-\lambda)^{-1}d\lambda,
\label{eq:Heisenberg.complex-powers-definition}\\
\Gamma_{\theta}=\{ \rho e^{i\theta}; \infty <\rho\leq r\}\cup\{ r e^{it};
\theta\geq t\geq \theta-2\pi \}\cup\{ \rho e^{i(\theta-2\pi)}; r\leq \rho\leq \infty\},
\label{eq:Heisenberg.complex-powers-definition-Gammat}\end{gathered}$$ where $r>0$ is such that no nonzero eigenvalue of $P$ lies in the disc $|\lambda|<r$.
\[prop:Heisenberg.powers2\] The family (\[eq:Heisenberg.complex-powers-definition\]) gives rise to a unique holomorphic family $(P_{\theta}^{s})_{s\in {\ensuremath{\mathbb{C}}}}$ of [$\Psi_{H}$DOs]{} such that:
\(i) We have ${{{\operatorname{ord}}}}P_{\theta}^{s}=ms$ for any $s \in {\ensuremath{\mathbb{C}}}$;
\(ii) We have the 1-parameter group property $P_{\theta}^{s_{1}+s_{2}}=P_{\theta}^{s_{1}} P_{\theta}^{s_{2}}$ $\forall s_{j}\in {\ensuremath{\mathbb{C}}}$;
\(iii) We have $P_{\theta}^{k+s}=P^{k} P_{\theta}^{s}$ for any $k\in {\ensuremath{\mathbb{N}}}$ and any $s \in {\ensuremath{\mathbb{C}}}$.
Let $E_{0}(P)=\cup_{j \geq 0} \ker P^{j}$ be the characteristic subspace of $P$ associated to $\lambda=0$. This is a finite dimensional subspace of $C^{\infty}(M,{\ensuremath{\mathcal{E}}})$ and so the projection $\Pi_{0}(P)$ onto $E_{0}(P)$ and along $E_{0}(P^{*})^{\perp}$ is a smoothing operator (see [@Po:CPDE1]). Then we have: $$P_{\theta}^{0}=1-\Pi_{0}(P), \qquad P_{\theta}^{-k}=P^{-k}, \quad k=1,2,\ldots,
\label{eq:PsiDO.complex-powers-integers}$$ where $P^{-k}$ denotes the partial inverse of $P^{k}$, i.e., the operator that inverts $P^{k}$ on $E_{0}(P^{*})^{\perp}$ and is zero on $E_{0}(P)$.
Assume further that $0$ is not in the spectrum of $P$. Let $Q\in {\ensuremath{\Psi_{H}}}^{*}(M,{\ensuremath{\mathcal{E}}})$ and for $z \in {\ensuremath{\mathbb{C}}}$ set $Q(z)=QP_{\theta}^{z/m}$. Then $(Q(z))_{z \in {\ensuremath{\mathbb{C}}}}$ is a holomorphic family of [$\Psi_{H}$DOs]{} such that $Q_{0}=Q$ and ${{{\operatorname{ord}}}}Q(z)=z+{{{\operatorname{ord}}}}Q$. Following the terminology of [@Gu:GLD] a holomorphic family of [$\Psi_{H}$DOs]{} with these properties is called a *holomorphic gauging* for $Q$.
Noncommutative residue trace for the Heisenberg calculus {#sec:NCR}
========================================================
In this section we construct a noncommutative residue trace for the algebra of integer order [$\Psi_{H}$DOs]{} on a Heisenberg manifold. We start by describing the logarithmic singularity near the diagonal of the Schwartz kernel of a [$\Psi_{H}$DO]{} of integer order and we show that it gives rise to a well-defined density. We then construct the noncommutative residue for the Heisenberg calculus as the residual trace induced by the analytic continuation of the usual trace to [$\Psi_{H}$DOs]{} of non-integer orders. Moreover, we show that it agrees with the integral of the density defined by the logarithmic singularity of the Schwartz kernel of the corresponding [$\Psi_{H}$DO]{}. Finally, we prove that when the manifold is connected then every other trace on the algebra of integer order [$\Psi_{H}$DOs]{} is a constant multiple of our noncommutative residue. This is the analogue of a well-known result of Wodzicki and Guillemin.
Logarithmic singularity of the kernel of a [$\Psi_{H}$DO]{}
-----------------------------------------------------------
In this subsection we show that the logarithmic singularity of the Schwartz kernel of any integer order [$\Psi_{H}$DO]{} gives rise to a density which makes sense intrinsically. This uses the characterization of [$\Psi_{H}$DOs]{} in terms of their Schwartz kernels, which we shall now recall.
First, we extend the notion of homogeneity of functions to distributions. For $K$ in ${\ensuremath{\mathcal{S}}}'({\ensuremath{{\ensuremath{\mathbb{R}}}^{d+1}}})$ and for $\lambda >0$ we let $K_{\lambda}$ denote the element of ${\ensuremath{\mathcal{S}}}'({\ensuremath{{\ensuremath{\mathbb{R}}}^{d+1}}})$ such that $${\ensuremath{\langle K_{\lambda} , f \rangle}}=\lambda^{-(d+2)} {\ensuremath{\langle K(x) , f(\lambda^{-1}.x) \rangle}} \quad \forall f\in{\ensuremath{\mathcal{S}}}({\ensuremath{{\ensuremath{\mathbb{R}}}^{d+1}}}).
\label{eq:PsiHDO.homogeneity-K-m}$$ It will be convenient to also use the notation $K(\lambda.x)$ for denoting $K_{\lambda}(x)$. We say that $K$ is homogeneous of degree $m$, $m\in{\ensuremath{\mathbb{C}}}$, when $K_{\lambda}=\lambda^m K$ for any $\lambda>0$.
In the sequel we let $E$ be the anisotropic radial vector field $2x_{0}\partial_{x_{0}}+\partial_{x_{1}}+\ldots+\partial_{x_{d}}$, i.e., $E$ is the infinitesimal generator of the flow $\phi_{s}(\xi)=e^{s}.\xi$.
\[lem:Heisenberg.extension-symbol\] Let $p(\xi) \in S_{m}({\ensuremath{{\ensuremath{\mathbb{R}}}^{d+1}}})$, $m \in {\ensuremath{\mathbb{C}}}$.
1\) If $m$ is not an integer $\leq -(d+2)$, then $p(\xi)$ can be uniquely extended into a homogeneous distribution $\tau \in {\ensuremath{\mathcal{S}}}'({\ensuremath{{\ensuremath{\mathbb{R}}}^{d+1}}})$.
2\) If $m$ is an integer $\leq -(d+2)$, then at best we can extend $p(\xi)$ into a distribution $\tau \in {\ensuremath{\mathcal{S}}}'({\ensuremath{{\ensuremath{\mathbb{R}}}^{d+1}}})$ such that $$\tau_{\lambda}=\lambda^{m}\tau +\lambda^{m}\log \lambda\sum_{{\ensuremath{\langle\! \alpha\!\rangle}}=-(m+d+2)} c_{\alpha}(p)\delta^{(\alpha)} \quad \text{for any $\lambda
>0$},
\label{eq:NCR.log-homogeneity}$$ where we have let $c_{\alpha}(p) = \frac{(-1)^{|\alpha|}}{\alpha!}\int_{\|\xi\|=1}\xi^\alpha p(\xi)i_{E}d\xi$. In particular, $p(\xi)$ admits a homogeneous extension if and only if all the coefficients $c_{\alpha}(p)$ vanish.
For reader’s convenience a detailed proof of this lemma is given in Appendix.
Let $\tau\in {\ensuremath{\mathcal{S}}}'({\ensuremath{{\ensuremath{\mathbb{R}}}^{d+1}}})$ and let $\lambda>0$. Then for any $f \in {\ensuremath{\mathcal{S}}}({\ensuremath{{\ensuremath{\mathbb{R}}}^{d+1}}})$ we have $${\ensuremath{\langle (\check{\tau})_{\lambda} , f \rangle}}= \lambda^{-(d+2)}{\ensuremath{\langle \tau , (f_{\lambda^{-1}})^{\vee} \rangle}}=
{\ensuremath{\langle \tau , (\check{f})_{\lambda} \rangle}}= \lambda^{-(d+2)}{\ensuremath{\langle (\tau_{\lambda^{-1}})^{\vee} , f \rangle}}.
\label{eq:NCR.Fourier-transform-scaling}$$ Hence $(\check{\tau})_{\lambda} = \lambda^{-(d+2)}(\tau_{\lambda^{-1}})^{\vee}$. Therefore, if we set $\hat{m}=-(m+d+2)$ then we see that:
- $\tau$ is homogeneous of degree $m$ if and only if $\check{\tau}$ is homogeneous of degree $\hat{m}$;
- $\tau$ satisfies (\[eq:NCR.log-homogeneity\]) if and only if for any $\lambda>0$ we have $$\check{\tau}(\lambda.y)= \lambda^{\hat{m}} \check{\tau}(y) - \lambda^{\hat{m}}\log \lambda
\sum_{{\ensuremath{\langle\! \alpha\!\rangle}} =\hat{m}} (2\pi)^{-(d+1)}c_{\alpha}(p) (-iy)^{\alpha} .
\label{eq:NCR.log-homogeneity-kernel}$$
Let $U\subset {\ensuremath{{\ensuremath{\mathbb{R}}}^{d+1}}}$ be an open of local coordinates equipped with a $H$-frame $X_{0},\ldots,X_{d}$. In the sequel we set ${\ensuremath{\mathbb{N}}}_{0}={\ensuremath{\mathbb{N}}}\cup\{0\}$ and we let ${\ensuremath{\mathcal{S}}}'_{{{\text{reg}}}}({\ensuremath{{\ensuremath{\mathbb{R}}}^{d+1}}})$ be the space of tempered distributions on ${\ensuremath{{\ensuremath{\mathbb{R}}}^{d+1}}}$ which are smooth outside the origin. We endow ${\ensuremath{\mathcal{S}}}'_{{{\text{reg}}}}({\ensuremath{{\ensuremath{\mathbb{R}}}^{d+1}}})$ with the weakest locally convex topology that makes continuous the embeddings of ${\ensuremath{\mathcal{S}}}'_{{{\text{reg}}}}({\ensuremath{{\ensuremath{\mathbb{R}}}^{d+1}}})$ into ${\ensuremath{\mathcal{S}}}'({\ensuremath{{\ensuremath{\mathbb{R}}}^{d+1}}})$ and $C^{\infty}({{\ensuremath{\mathbb{R}}}^{d+1}\!\setminus\! 0})$. In addition, recall also that if $E$ is a topological vector space contained in ${\ensuremath{\mathcal{D}}}'({\ensuremath{{\ensuremath{\mathbb{R}}}^{d+1}}})$ then $C^{\infty}(U){\hat\otimes}E$ can be identified as the space $C^{\infty}(U,E)$ seen as a subspace of ${\ensuremath{\mathcal{D}}}'({U\times{\ensuremath{\mathbb{R}}}^{d+1}})$.
The discussion above about the homogeneity of the (inverse) Fourier transform leads us to consider the classes of distributions below.
${\ensuremath{\mathcal{K}}}_{m}({U\times{\ensuremath{\mathbb{R}}}^{d+1}})$, $m\in{\ensuremath{\mathbb{C}}}$, consists of distributions $K(x,y)$ in $C^\infty(U){\hat\otimes}{\ensuremath{\mathcal{S}}}'_{{{\text{reg}}}}({\ensuremath{{\ensuremath{\mathbb{R}}}^{d+1}}})$ such that, for any $\lambda>0$, we have: $$K(x,\lambda y)= \left\{
\begin{array}{ll}
\lambda^m K(x,y) & \text{if $m\not \in {\ensuremath{\mathbb{N}}}_{0}$}, \\
\lambda^m K(x,y) + \lambda^m\log\lambda
\sum_{{\ensuremath{\langle\! \alpha\!\rangle}}=m}c_{K,\alpha}(x)y^\alpha & \text{if $m\in {\ensuremath{\mathbb{N}}}_{0}$},
\end{array}\right.
\label{eq:NCR.log-homogeneity-Km}$$ where the functions $c_{K,\alpha}(x)$, ${\ensuremath{\langle\! \alpha\!\rangle}}=m$, are in $C^{\infty}(U)$ when $m\in {\ensuremath{\mathbb{N}}}_{0}$.
\[rem:NCR.regularity-cKm1\] For $\Re m>0$ we have $K_{m}({U\times{\ensuremath{\mathbb{R}}}^{d+1}})\subset C^{\infty}(U){\hat\otimes}C^{[\frac{\Re m}{2}]'}({\ensuremath{{\ensuremath{\mathbb{R}}}^{d+1}}})$, where $[\frac{\Re m}{2}]'$ denotes the greatest integer $< \Re m$ (see [@Po:MAMS1 Lemma A.1]).
${\ensuremath{\mathcal{K}}}^{m}({U\times{\ensuremath{\mathbb{R}}}^{d+1}})$, $m\in {\ensuremath{\mathbb{C}}}$, consists of distributions $K(x,y)$ in ${\ensuremath{\mathcal{D}}}'({U\times{\ensuremath{\mathbb{R}}}^{d+1}})$ with an asymptotic expansion $K\sim \sum_{j\geq0}K_{m+j}$, $K_{l}\in {\ensuremath{\mathcal{K}}}_{l}({U\times{\ensuremath{\mathbb{R}}}^{d+1}})$, in the sense that, for any integer $N$, as soon as $J$ is large enough $K-\sum_{j\leq J}K_{m+j}$ is in $C^{N}({U\times{\ensuremath{\mathbb{R}}}^{d+1}})$.
\[rem:NCR.regularity-cKm2\] The definition implies that any distribution $K\in{\ensuremath{\mathcal{K}}}^{m}({U\times{\ensuremath{\mathbb{R}}}^{d+1}})$ is smooth on ${U\times({\ensuremath{\mathbb{R}}}^{d+1}\!\setminus\! 0)}$. Furthermore, using Remark \[rem:NCR.regularity-cKm1\] we see that for $\Re m>0$ we have ${\ensuremath{\mathcal{K}}}^{m}({U\times{\ensuremath{\mathbb{R}}}^{d+1}})\subset C^{\infty}(U){\hat\otimes}C^{[\frac{\Re m}{2}]'}({\ensuremath{{\ensuremath{\mathbb{R}}}^{d+1}}})$.
Using Lemma \[lem:Heisenberg.extension-symbol\] we can characterize homogeneous symbols on ${U\times{\ensuremath{\mathbb{R}}}^{d+1}}$ as follows.
\[lem:NCR.extension-symbolU\] Let $m \in {\ensuremath{\mathbb{C}}}$ and set $\hat{m}=-(m+d+2)$.
1\) If $p(x,\xi)\in S_{m}({U\times{\ensuremath{\mathbb{R}}}^{d+1}})$ then $p(x,\xi)$ can be extended into a distribution $\tau(x,\xi)\in C^{\infty}(U){\hat\otimes}{\ensuremath{\mathcal{S}}}_{{{\text{reg}}}}'({\ensuremath{{\ensuremath{\mathbb{R}}}^{d+1}}})$ such that $K(x,y):=\check{\tau}_{{{\xi\rightarrow y}}}(x,y)$ belongs to ${\ensuremath{\mathcal{K}}}_{\hat{m}}({U\times{\ensuremath{\mathbb{R}}}^{d+1}})$. Furthermore, if $m$ is an integer $\leq -(d+2)$ then, using the notation of (\[eq:NCR.log-homogeneity-Km\]), we have $c_{K,\alpha}(x)= (2\pi)^{-(d+1)}\int_{\|\xi\|=1}\frac{(i\xi)^{\alpha}}{\alpha!}p(x,\xi)\iota_{E}d\xi$.
2\) If $K(x,y)\in {\ensuremath{\mathcal{K}}}_{\hat{m}}({U\times{\ensuremath{\mathbb{R}}}^{d+1}})$ then the restriction of $\hat{K}_{{{y\rightarrow\xi}}}(x,\xi)$ to ${U\times({\ensuremath{\mathbb{R}}}^{d+1}\!\setminus\! 0)}$ is a symbol in $S_{m}({U\times{\ensuremath{\mathbb{R}}}^{d+1}})$.
Next, for any $x \in U$ we let $\psi_{x}$ (resp. $\varepsilon_{x}$) denote the change of variable to the privileged (resp. Heisenberg) coordinates centered at $x$ (cf. Definitions \[def:Heisenberg-privileged-coordinates\] and \[def:Heisenberg-Heisenberg-coordinates\]).
Let $p \in S_{m}({U\times{\ensuremath{\mathbb{R}}}^{d+1}})$ and let $k(x,y)\in C^{\infty}(U){\hat\otimes}{\ensuremath{\mathcal{D}}}'(U)$ denote the Schwartz kernel of $p(x,-iX)$, so that $[p(x,-iX)u](x)={\ensuremath{\langle k(x,y) , u(y) \rangle}}$ for any $u\in C^{\infty}_{c}(U)$. Then one can check (see, e.g., [@Po:MAMS1 p. 54]) that we have: $$k(x,y)=|\psi_{x}'| \check{p}_{{{\xi\rightarrow y}}}(x,-\psi_{x}(y))=|\varepsilon_{x}'|\check{p}_{{{\xi\rightarrow y}}}(x,\phi_{x}(-\varepsilon_{x}(y))).
\label{eq:Heisenberg.kernel-quantization-symbol-psiy}$$ Combining this with Lemma \[lem:NCR.extension-symbolU\] leads us to the characterization of [$\Psi_{H}$DOs]{} below.
\[prop:PsiVDO.characterisation-kernel1\] Consider a continuous operator $P:C_{c}^\infty(U)\rightarrow C^\infty(U)$ with Schwartz kernel $k_{P}(x,y)$. Let $m \in
{\ensuremath{\mathbb{C}}}$ and set $\hat{m}=-(m+d+2)$. Then the following are equivalent:
\(i) $P$ is a [$\Psi_{H}$DO]{} of order $m$.
\(ii) We can put $k_{P}(x,y)$ in the form, $$k_{P}(x,y)=|\psi_{x}'|K(x,-\psi_{x}(y)) +R(x,y) ,
\label{eq:PsiHDO.characterization-kernel.privileged}$$ for some $K\in{\ensuremath{\mathcal{K}}}^{\hat{m}}({U\times{\ensuremath{\mathbb{R}}}^{d+1}})$, $K\sim \sum K_{\hat{m}+j}$, and some $R \in C^{\infty}(U\times U)$.
\(iii) We can put $k_{P}(x,y)$ in the form, $$k_{P}(x,y)=|\varepsilon_{x}'|K_{P}(x,-\varepsilon_{x}(y)) +R_{P}(x,y) ,
\label{eq:PsiHDO.characterization-kernel.Heisenberg}$$ for some $K_{P}\in {\ensuremath{\mathcal{K}}}^{\hat{m}}({U\times{\ensuremath{\mathbb{R}}}^{d+1}})$, $K_{P}\sim \sum K_{P,\hat{m}+j}$, and some $R_{P} \in C^{\infty}(U\times U)$.
Furthermore, if (i)–(iii) hold then we have $K_{P,l}(x,y)=K_{l}(x,\phi_{x}(y))$ and $P$ has symbol $p\sim \sum_{j\geq 0} p_{m-j}$, where $p_{m-j}(x,\xi)$ is the restriction to ${U\times({\ensuremath{\mathbb{R}}}^{d+1}\!\setminus\! 0)}$ of $(K_{m+j})^{\wedge}_{{{y\rightarrow\xi}}}(x,\xi)$.
Now, let $U\subset {\ensuremath{{\ensuremath{\mathbb{R}}}^{d+1}}}$ be an open of local coordinates equipped with a $H$-frame $X_{0},X_{1},\ldots,X_{d}$. Let $m\in{\ensuremath{\mathbb{Z}}}$ and let $K\in {\ensuremath{\mathcal{K}}}^{m}({U\times{\ensuremath{\mathbb{R}}}^{d+1}})$, $K\sim \sum_{j\geq m} K_{j}$. Then: - For $j \leq -1$ the distribution $K_{j}(x,y)$ is homogeneous of degree $j$ with respect to $y$ and is smooth for $y\neq 0$;
- For $j=0$ and $\lambda>0$ we have $K_{0}(x,\lambda.y)=K_{0}(x,y)-c_{K_{0},0}(x)\log \lambda$, which by setting $\lambda=\|y\|^{-1}$ with $y\neq 0$ gives $$K_{0}(x,y)=K_{0}(x,\|y\|^{-1}.y)-c_{K_{0},0}\log \|y\|.
\label{eq:Log.behavior-K0}$$
- The remainder term $K-\sum_{j\geq 1}K_{j}$ is in $C^{0}({U\times{\ensuremath{\mathbb{R}}}^{d+1}})$ (cf. Remarks \[rem:NCR.regularity-cKm1\] and \[rem:NCR.regularity-cKm2\]).
It follows that $K(x,y)$ has a behavior near $y=0$ of the form, $$K(x,y)=\sum_{m\leq j\leq -1} K_{j}(x,y)-c_{K}(x)\log \|y\| +{\operatorname{O}}(1), \qquad c_{K}(x)=c_{K_{0},0}(x).
\label{eq:Log.behavior-K}$$
Let $P\in {\ensuremath{\Psi_{H}}}^{m}(U)$ have kernel $k_{P}(x,y)$ and set $\hat{m}=-(m+d+2)$.
1\) Near the diagonal $k_{P}(x,y)$ has a behavior of the form, $$k_{P}(x,y)=\sum_{\hat{m}\leq j \leq -1}a_{j}(x,-\psi_{x}(y)) -c_{P}(x) \log \|\psi_{x}(y)\| +{\operatorname{O}}(1),
\label{eq:Log.behavior-kP}$$ with $a_{j}(x,y)\in C^{\infty}({U\times({\ensuremath{\mathbb{R}}}^{d+1}\!\setminus\! 0)})$ homogeneous of degree $j$ in $y$ and $c_{P}(x)\in C^{\infty}(U)$.
2\) If we write $k_{P}(x,y)$ in the forms (\[eq:PsiHDO.characterization-kernel.privileged\]) and (\[eq:PsiHDO.characterization-kernel.Heisenberg\]) with $K(x,y)$ and $K_{P}(x,y)$ in ${\ensuremath{\mathcal{K}}}^{\hat{m}}({U\times{\ensuremath{\mathbb{R}}}^{d+1}})$, then we have $$c_{P}(x)=|\psi_{x}'|c_{K}(x)=|\varepsilon_{x}'|c_{K_{P}}(x)=\frac{|\psi_{x}'|}{(2\pi)^{d+1}}\int_{\|\xi\|=1}p_{-(d+2)}(x,\xi)\imath_{E}d\xi,
\label{eq:NCR.formula-cP}$$ where $p_{-(d+2)}$ denotes the symbol of degree $-(d+2)$ of $P$.
If we put $k_{P}(x,y)$ in the form (\[eq:PsiHDO.characterization-kernel.privileged\]) with $K\in {\ensuremath{\mathcal{K}}}^{\hat{m}}({U\times{\ensuremath{\mathbb{R}}}^{d+1}})$, $K\sim \sum K_{\hat{m}+j}$, then it follows from (\[eq:Log.behavior-K\]) that $k_{P}(x,y)$ has a behavior near the diagonal of the form (\[eq:Log.behavior-kP\]) with $c_{P}(x)=|\psi_{x}'|c_{K}(x)=|\psi_{x}'|c_{K_{0},0}(x)$. Furthermore, by Proposition \[prop:PsiVDO.characterisation-kernel1\] the symbol $p_{-(d+2)}(x,\xi)$ of degree $-(d+2)$ of $P$ is the restriction to ${U\times({\ensuremath{\mathbb{R}}}^{d+1}\!\setminus\! 0)}$ of $(K_{0})^{\wedge}_{{{y\rightarrow\xi}}}(x,\xi)$, so by Lemma \[lem:NCR.extension-symbolU\] we have $c_{K}(x)=c_{K_{0},0}(x)=(2\pi)^{-(d+1)}\int_{\|\xi\|=1}p_{-(d+2)}(x,\xi)\imath_{E}d\xi$.
Next, if we put $k_{P}(x,y)$ in the form (\[eq:PsiHDO.characterization-kernel.Heisenberg\]) with $K_{P}\in {\ensuremath{\mathcal{K}}}^{\hat{m}}({U\times{\ensuremath{\mathbb{R}}}^{d+1}})$, $K_{P}\sim \sum K_{P,\hat{m+j}}$ then by Proposition \[prop:PsiVDO.characterisation-kernel1\] we have $K_{P,0}(x,y)=K_{0}(x,\phi_{x}(y))$. Let $\lambda>0$. Since $\phi_{x}(\lambda.y)=\lambda.\phi_{x}(y)$, using (\[eq:NCR.log-homogeneity-Km\]) we get $$K_{P,0}(x,\lambda.y)-K_{P,0}(x,y)= K_{0}(x,\lambda.\phi_{x}(y))-K_{0}(x,\phi_{x}(y))=c_{K_{0}}(x) \log \lambda . \\
$$ Hence $c_{K_{P,0}}(x)=c_{K,0}(x)$. As $|\varepsilon_{x}'|=|\phi_{x}'|.|\psi_{x}'|=|\psi_{x}'|$ we see that $|\psi_{x}'|c_{K}(x)=|\varepsilon_{x}'|c_{K_{P}}(x)$. The proof is thus achieved.
\[lem:log-sing.invariance\] Let $\phi : U\rightarrow \tilde{U}$ be a change of $H$-framed local coordinates. Then for any $\tilde{P}\in
{\ensuremath{\Psi_{H}}}^{m}(\tilde{U})$ we have $c_{\phi^{*}\tilde{P}}(x)=|\phi'(x)|c_{\tilde{P}}(\phi(x))$.
Let $P=\phi^{*}\tilde{P}$. Then $P$ is a [$\Psi_{H}$DO]{} of order $m$ on $U$ (see [@BG:CHM]). Moreover, by [@Po:MAMS1 Prop. 3.1.18] if we write the Schwartz kernel $k_{\tilde{P}}(\tilde{x},\tilde{y})$ in the form (\[eq:PsiHDO.characterization-kernel.Heisenberg\]) with $K_{\tilde{P}}(\tilde{x},\tilde{y})$ in ${\ensuremath{\mathcal{K}}}^{\hat{m}}(\tilde{U}\times {\ensuremath{{\ensuremath{\mathbb{R}}}^{d+1}}})$, then the Schwartz kernel $k_{P}(x,y)$ of $P$ can be put in the form (\[eq:PsiHDO.characterization-kernel.Heisenberg\]) with $K_{P}(x,y)$ in ${\ensuremath{\mathcal{K}}}^{\hat{m}}(U\times {\ensuremath{{\ensuremath{\mathbb{R}}}^{d+1}}})$ such that $$K_{P}(x,y) \sim \sum_{{\ensuremath{\langle\! \beta\!\rangle}}\geq \frac{3}{2}{\ensuremath{\langle\! \alpha\!\rangle}}} \frac{1}{\alpha!\beta!}
a_{\alpha\beta}(x)y^{\beta}(\partial_{\tilde{y}}^{\alpha}K_{\tilde{P}})(\phi(x),\phi_{H}'(x).y),
\label{eq:PsiHDO.asymptotic-expansion-KP}$$ where we have let $a_{\alpha\beta}(x)=\partial^{\beta}_{y}[|\partial_{y}(\varepsilon_{\phi(x)}\circ \phi\circ \tilde{\varepsilon}_{x}^{-1})(y)|
(\tilde{\varepsilon}_{\phi(x)}\circ \phi\circ \varepsilon_{x}^{-1}(y)-\phi_{H}'(x)y)^{\alpha}]_{|_{y=0}}$, the map $\phi_{H}'(x)$ is the tangent map (\[eq:Heisenberg.tangent-map\]), and $\tilde{\varepsilon}_{\tilde{x}}$ denotes the change to the Heisenberg coordinates at $\tilde{x}\in \tilde{U}$. In particular, we have $$K_{P}(x,y)= a_{00}(x) K_{\tilde{P}}(\phi(x),\phi_{H}'(x).y) \quad
\bmod y_{j}{\ensuremath{\mathcal{K}}}^{\hat{m}+1}({U\times{\ensuremath{\mathbb{R}}}^{d+1}}),
\label{eq:PsiHDO.asymptotic-expansion-KP2}$$ where $a_{00}(x)=|\varepsilon_{\phi(x)}'||\phi'(x)||\varepsilon_{x}'|^{-1}$.
Notice that $\tilde{K}(x,y):= K_{\tilde{P}}(\phi(x),\phi_{H}'(x).y)$ is an element of ${\ensuremath{\mathcal{K}}}^{\hat{m}}({U\times{\ensuremath{\mathbb{R}}}^{d+1}})$, since we have $\phi'_{H}(x).(\lambda.y)=\lambda.(\phi'_{H}(x).y)$ for any $\lambda>0$. Moreover, the distributions in $y_{j}{\ensuremath{\mathcal{K}}}^{*}({U\times{\ensuremath{\mathbb{R}}}^{d+1}})$, $j=0,..,d$, cannot have a logarithmic singularity near $y=0$. To see this it is enough to look at a distribution $H(x,y)\in {\ensuremath{\mathcal{K}}}^{-l}({U\times{\ensuremath{\mathbb{R}}}^{d+1}})$, $l\in {\ensuremath{\mathbb{N}}}_{0}$. Then $H(x,y)$ has a behavior near $y=0$ of the form: $$H(x,y)=\sum_{-l\leq k \leq -1}b_{k}(x,y)-c_{H}(x)\log \|y\|+{\operatorname{O}}(1),
$$ with $b_{k}(x,y)$ homogeneous of degree $k$ with respect to the $y$-variable. Thus, $$y_{j}H(x,y)=\sum_{-l\leq k \leq -1}y_{j}b_{k}(x,y)-c_{H}(x)y_{j}\log \|y\|+{\operatorname{O}}(1).
$$ Observe that each term $y_{j}b_{k}(x,y)$ is homogeneous of degree $k+1$ with respect to $y$ and the term $y_{j}\log \|y\|$ converges to $0$ as $y\rightarrow 0$. Therefore, we see that the singularity of $y_{j}H(x,y)$ near $y=0$ cannot contain a logarithmic term.
Combining the above observations with (\[eq:PsiHDO.asymptotic-expansion-KP\]) shows that the coefficients of the logarithmic singularities of $K_{P}(x,y)$ and $a_{00}(x)\tilde{K}(x,y)$ must agree, i.e., we have $c_{K_{P}}(x)=c_{a_{00}\tilde{K}}(x)=a_{00}(x)c_{\tilde{K}}(x)=|\varepsilon_{\phi(x)}'||\phi'(x)||\varepsilon_{x}'|^{-1}c_{\tilde{K}}(x)$. Furthermore, the only contribution to the logarithmic singularity of $\tilde{K}(x,y)$ comes from $$\begin{gathered}
c_{K_{\tilde{P}}}(\phi(x))\log\|\phi_{H}'(x)y\|= c_{K_{\tilde{P}}}(\phi(x))\log[\|y\|
\|\phi_{H}'(x).(\|y\|^{-1}.y\|)] \\ =c_{K_{\tilde{P}}}(\phi(x))\log\|y\| +{\operatorname{O}}(1).
\end{gathered}$$ Hence $c_{\tilde{K}}(x)=c_{K_{\tilde{P}}}(\phi(x))$. Therefore, we get $c_{K_{P}}(x)=|\varepsilon_{\phi(x)}'||\phi'(x)||\varepsilon_{x}'|^{-1}c_{K_{\tilde{P}}}(\phi(x))$, which by combining with (\[eq:NCR.formula-cP\]) shows that $c_{P}(x)=|\phi'(x)|c_{\tilde{P}}(\phi(x))$ as desired.
Let $P\in {\ensuremath{\Psi_{H}}}^{m}(M,{\ensuremath{\mathcal{E}}})$ and let $\kappa:U\rightarrow V$ be a $H$-framed chart over which there is a trivialization $\tau:{\ensuremath{\mathcal{E}}}_{|_{U}}\rightarrow U\times {\ensuremath{\mathbb{C}}}^{r}$. Then the Schwartz kernel of $P_{\kappa,\tau}:=\kappa_{*}\tau_{*}(P_{|_{U}})$ has a singularity near the diagonal of the form (\[eq:Log.behavior-kP\]). Moreover, if $\tilde{\kappa}:\tilde{U}\rightarrow \tilde{V}$ be a $H$-framed chart over which there is a trivialization $\tau:{\ensuremath{\mathcal{E}}}_{|_{\tilde{U}}}\rightarrow
\tilde{U}\times {\ensuremath{\mathbb{C}}}^{r}$ and if we let $\phi$ denote the Heisenberg diffeomorphism $\tilde{\kappa}\circ \kappa^{-1}:\kappa(U\cap \tilde{U})\rightarrow
\tilde{\kappa}(U\cap \tilde{U})$, then by Lemma \[lem:log-sing.invariance\] we have $c_{P_{\kappa,\tau}}(x)=|\phi'(x)|c_{P_{\tilde{\kappa},\tilde{\tau}}}(\phi(x))$ for any $x \in U$. Therefore, on $U\cap \tilde{U}$ we have the equality of densities, $$\tau^{*}\kappa^{*}(c_{P_{\kappa,\tau}}(x)dx)= \tilde{\tau}^{*}\tilde{\kappa}^{*}(c_{P_{\tilde{\kappa},\tilde{\tau}}}(x)dx).$$
Now, the space $C^{\infty}(M,|\Lambda|(M)\otimes {\ensuremath{{\operatorname{End}}}}{\ensuremath{\mathcal{E}}})$ of ${\ensuremath{{\operatorname{END}}}}{\ensuremath{\mathcal{E}}}$-valued densities is a sheaf, so there exists a unique density $c_{P}(x) \in C^{\infty}(M,|\Lambda|(M)\otimes {\ensuremath{{\operatorname{End}}}}{\ensuremath{\mathcal{E}}})$ such that, for any local $H$-framed chart $\kappa:U\rightarrow V$ and any trivialization $\tau:{\ensuremath{\mathcal{E}}}_{|_{U}}\rightarrow U\times {\ensuremath{\mathbb{C}}}^{r}$, we have $$c_{P}(x)|_{U}=\tau^{*}\kappa^{*}(c_{\kappa_{*}\tau_{*}(P_{|_{U}})}(x)dx).
$$ Moreover, this density is functorial with respect to Heisenberg diffeomorphisms, i.e., for any Heisenberg diffeomorphism $\phi:(M,H)\rightarrow
(M',H')$ we have $$c_{\phi_{*}P}(x)=\phi_{*}(c_{P}(x)).
\label{eq:Log.functoriality-cP}$$
Summarizing all this we have proved:
\[thm:NCR.log-singularity\] Let $P\in {\ensuremath{\Psi_{H}}}^{m}(M,{\ensuremath{\mathcal{E}}})$, $m \in {\ensuremath{\mathbb{Z}}}$. Then:
1\) On any trivializing $H$-framed local coordinates the Schwartz kernel $k_{P}(x,y)$ of $P$ has a behavior near the diagonal of the form, $$k_{P}(x,y)=\sum_{-(m+d+2)\leq j\leq -1}a_{j}(x,-\psi_{x}(y)) - c_{P}(x)\log \|\psi_{x}(y)\| +
{\operatorname{O}}(1),$$ where $c_{P}(x)$ is given by (\[eq:NCR.formula-cP\]) and each function $a_{j}(x,y)$ is smooth for $y\neq 0$ and homogeneous of degree $j$ with respect to $y$.
2\) The coefficient $c_{P}(x)$ makes sense globally on $M$ as a smooth ${\ensuremath{{\operatorname{END}}}}{\ensuremath{\mathcal{E}}}$-valued density which is functorial with respect to Heisenberg diffeomorphisms.
Finally, the following holds.
\[prop.Sing.transpose-adjoint\] Let $P\in {\ensuremath{\Psi_{H}}}^{m}(M,{\ensuremath{\mathcal{E}}})$, $m \in {\ensuremath{\mathbb{Z}}}$. 1) Let $P^{t}\in {\ensuremath{\Psi_{H}}}^{m}(M,{\ensuremath{\mathcal{E}}}^{*})$ be the transpose of $P$. Then we have $c_{P^{t}}(x)=c_{P}(x)^{t}$.
2\) Suppose that $M$ is endowed with a density $\rho>0$ and ${\ensuremath{\mathcal{E}}}$ is endowed with a Hermitian metric. Let $P^{*}\in {\ensuremath{\Psi_{H}}}^{m}(M,{\ensuremath{\mathcal{E}}})$ be the adjoint of $P$. Then we have $c_{P^{*}}(x)=c_{P}(x)^{*}$.
Let us first assume that ${\ensuremath{\mathcal{E}}}$ is the trivial line bundle. Then it is enough to prove the result in $H$-framed local coordinates $U\subset {\ensuremath{{\ensuremath{\mathbb{R}}}^{d+1}}}$, so that the Schwartz kernel $k_{P}(x,y)$ can be put in the form (\[eq:PsiHDO.characterization-kernel.Heisenberg\]) with $K_{P}(x,y)$ in ${\ensuremath{\mathcal{K}}}^{\hat{m}}({U\times{\ensuremath{\mathbb{R}}}^{d+1}})$.
We know that $P^{t}$ is a [$\Psi_{H}$DO]{} of order $m$ (see [@BG:CHM Thm. 17.4]). Moreover, by [@Po:MAMS1 Prop. 3.1.21] we can put its Schwartz kernel $k_{P^{t}}(x,y)$ in the form (\[eq:PsiHDO.characterization-kernel.Heisenberg\]) with $K_{P^{t}}(x,y)$ in ${\ensuremath{\mathcal{K}}}^{\hat{m}}({U\times{\ensuremath{\mathbb{R}}}^{d+1}})$ such that $$K_{P^{t}}(x,y) \sim \sum_{\frac{3}{2}{\ensuremath{\langle\! \alpha\!\rangle}} \leq {\ensuremath{\langle\! \beta\!\rangle}}} \sum_{|\gamma|\leq |\delta| \leq 2|\gamma|}
a_{\alpha\beta\gamma\delta}(x) y^{\beta+\delta}
(\partial^{\gamma}_{x}\partial_{y}^{\alpha}K_{P})(x,-y),$$ where $a_{\alpha\beta\gamma\delta}(x)=\frac{|\varepsilon_{x}^{-1}|}{\alpha!\beta!\gamma!\delta!}
[\partial_{y}^{\beta}(|\varepsilon_{\varepsilon_{x}^{-1}(-y)}'|(y-\varepsilon_{\varepsilon_{x}^{-1}(y)}(x))^{\alpha})
\partial_{y}^{\delta}(\varepsilon_{x}^{-1}(-y)-x)^{\gamma}](x,0)$. In particular, we have $K_{P^{t}}(x,y)=K_{P}(x,-y) \bmod y_{j}{\ensuremath{\mathcal{K}}}^{\hat{m}+1}({U\times{\ensuremath{\mathbb{R}}}^{d+1}})$. Therefore, in the same way as in the proof of Lemma \[lem:log-sing.invariance\], we see that the logarithmic singularity near $y=0$ of $K_{P}(x,y)$ agrees with that of $K_{P^{t}}(x,-y)$, hence with that of $K_{P^{t}}(x,y)$. Therefore, we have $c_{K_{P^{t}}}(x)=c_{K_{P}}(x)$. Combining this with (\[eq:NCR.formula-cP\]) then shows that $c_{P^{t}}(x)=c_{P}(x)$.
Next, suppose that $U$ is endowed with a smooth density $\rho(x)>0$. Then the corresponding adjoint $P^{*}$ is a [$\Psi_{H}$DO]{} of order $m$ on $U$ with Schwartz kernel $k_{P^{*}}(x,y)=\rho(x)^{-1}\overline{k_{P^{t}}(x,y)}\rho(y)$. Thus $k_{P^{*}}(x,y)$ can be put in the form (\[eq:PsiHDO.characterization-kernel.Heisenberg\]) with $K_{P^{*}}(x,y)$ in ${\ensuremath{\mathcal{K}}}^{\hat{m}}({U\times{\ensuremath{\mathbb{R}}}^{d+1}})$ such that $$\begin{gathered}
K_{P^{*}}(x,y)=[\rho(x)^{-1}\rho(\varepsilon_{x}^{-1}(-y))]\overline{K_{P^{t}}(x,y)}\\
=\overline{K_{P^{t}}(x,y)} \ \bmod y_{j}{\ensuremath{\mathcal{K}}}^{\hat{m}+1}({U\times{\ensuremath{\mathbb{R}}}^{d+1}}).
\end{gathered}$$ Therefore, $K_{P^{*}}(x,y)$ and $\overline{K_{P^{t}}(x,y)}$ same logarithmic singularity near $y=0$, so that we have $c_{K_{P^{*}}}(x)=\overline{c_{K_{P^{t}}}(x)}=\overline{c_{K_{P}}(x)}$. Hence $c_{P^{*}}(x)=\overline{c_{P}(x)}$.
Finally, when ${\ensuremath{\mathcal{E}}}$ is a general vector bundle, we can argue as above to show that we still have $c_{P^{t}}(x)=c_{P}(x)^{t}$, and if $P^{*}$ is the adjoint of $P$ with respect to the density $\rho$ and some Hermitian metric on ${\ensuremath{\mathcal{E}}}$, then we have $c_{P^{*}}(x)=c_{P}(x)^{*}$.
Noncommutative residue
----------------------
Let $(M^{d+1},H)$ be a Heisenberg manifold and let ${\ensuremath{\mathcal{E}}}$ be a vector bundle over $M$. We shall now construct a noncommutative residue trace on the algebra ${\ensuremath{\Psi_{H}^{{\ensuremath{\mathbb{Z}}}}}}(M,{\ensuremath{\mathcal{E}}})$ as the residual trace induced by the analytic extension of the operator trace to [$\Psi_{H}$DOs]{} of non-integer order.
Let ${\ensuremath{\Psi_{H}^{\text{int}}}}(M,{\ensuremath{\mathcal{E}}}) := \cup_{\Re m < -(d+2)}{\ensuremath{\Psi_{H}}}^{m}(M,{\ensuremath{\mathcal{E}}})$ the class of [$\Psi_{H}$DOs]{} whose symbols are integrable with respect to the $\xi$-variable (this notation is borrowed from [@CM:LIFNCG]). If $P$ belongs to this class, then it follows from Remark \[rem:NCR.regularity-cKm2\] that the restriction to the diagonal of $M\times M$ of its Schwartz kernel defines a smooth density $k_{P}(x,x)$ with values in ${\ensuremath{{\operatorname{End}}}}{\ensuremath{\mathcal{E}}}$. Therefore, when $M$ is compact then $P$ is a trace-class operator on $L^{2}(M,{\ensuremath{\mathcal{E}}})$ and we have $${\ensuremath{{\operatorname{Trace}}}}(P) = \int_{M} {{\operatorname{tr}}}_{{\ensuremath{\mathcal{E}}}}k_{P}(x,x).
$$
We shall now construct an analytic extension of the operator trace to the class ${\ensuremath{\Psi_{H}^{{\ensuremath{\mathbb{C}\!\setminus\!\mathbb{Z}}}}}}(M,{\ensuremath{\mathcal{E}}})$ of [$\Psi_{H}$DOs]{} of non-integer order. As in [@Gu:GLD] (see also [@KV:GDEO], [@CM:LIFNCG]) the approach consists in working directly at the level of densities by constructing an analytic extension of the map $P\rightarrow k_{P}(x,x)$ to ${\ensuremath{\Psi_{H}^{{\ensuremath{\mathbb{C}\!\setminus\!\mathbb{Z}}}}}}(M,{\ensuremath{\mathcal{E}}})$. Here analyticity is meant with respect to holomorphic families of [$\Psi_{H}$DOs]{}, e.g., the map $P \rightarrow k_{P}(x,x)$ is analytic since for any holomorphic family $(P(z))_{z\in \Omega}$ with values in ${\ensuremath{\Psi_{H}^{\text{int}}}}(M,{\ensuremath{\mathcal{E}}})$ the output densities $k_{P(z)}(x,x)$ depend analytically on $z$ in the Fréchet space $C^{\infty}(M,|\Lambda|(M)\otimes {\ensuremath{{\operatorname{End}}}}{\ensuremath{\mathcal{E}}})$.
Let $U \subset {\ensuremath{{\ensuremath{\mathbb{R}}}^{d+1}}}$ be an open of trivializing local coordinates equipped with equipped with a $H$-frame $X_{0}, \ldots, X_{d}$, and for any $x\in U$ let $\psi_{x}$ denote the affine change of variables to the privileged coordinates at $x$. Any $P \in {\ensuremath{\Psi_{H}}}^{m}(U)$ can be written as $P=p(x,-iX)+R$ with $p\in S^{m}({U\times{\ensuremath{\mathbb{R}}}^{d+1}})$ and $R\in \Psi^{-\infty}(U)$. Therefore, if $\Re m <-(d+2)$ then using (\[eq:Heisenberg.kernel-quantization-symbol-psiy\]) we get $$k_{P}(x,x)=|\psi_{x}'| (2\pi)^{-(d+2)}\int p(x,\xi)d\xi +k_{R}(x,x).
\label{eq:NCR.kP(x,x)}$$ This leads us to consider the functional, $$L(p):=(2\pi)^{-(d+2)}\int p(\xi) d\xi, \qquad p\in{\ensuremath{S^{\text{int}}}}({\ensuremath{{\ensuremath{\mathbb{R}}}^{d+1}}}).
$$
In the sequel, as in Section \[sec:Heisenberg-calculus\] for [$\Psi_{H}$DOs]{}, we say that a holomorphic family of symbols $(p(z))_{z\in{\ensuremath{\mathbb{C}}}}\subset S^{*}({\ensuremath{{\ensuremath{\mathbb{R}}}^{d+1}}})$ is a *gauging* for a given symbol $p\in
S^{*}({\ensuremath{\mathbb{R}}}^{d+1})$ when we have $p(0)=p$ and ${{{\operatorname{ord}}}}p(z)=z+{{{\operatorname{ord}}}}p$ for any $z\in {\ensuremath{\mathbb{C}}}$.
\[lem:Spectral.tildeL\] 1) The functional $L$ has a unique analytic continuation $\tilde{L}$ to $S^{{\ensuremath{\mathbb{C}\!\setminus\!\mathbb{Z}}}}({\ensuremath{{\ensuremath{\mathbb{R}}}^{d+1}}})$. The value of $\tilde{L}$ on a symbol $p\sim \sum_{j\geq 0} p_{m-j}$ of order $m \in{\ensuremath{\mathbb{C}\!\setminus\!\mathbb{Z}}}$ is given by $$\tilde{L}(p)= (p- \sum_{j\leq N}\tau_{m-j})^{\vee}(0), \qquad N\geq \Re{m}+d+2,
\label{eq:NCR.L-tilda}$$ where the value of the integer $N$ is irrelevant and the distribution $\tau_{m-j}\in {\ensuremath{\mathcal{S}}}'({\ensuremath{{\ensuremath{\mathbb{R}}}^{d+1}}})$ is the unique homogeneous extension of $p_{m-j}(\xi)$ provided by Lemma \[lem:Heisenberg.extension-symbol\].
2\) Let $ p\in S^{{\ensuremath{\mathbb{Z}}}}({\ensuremath{{\ensuremath{\mathbb{R}}}^{d+1}}})$, $p\sim \sum p_{m-j}$, and let $(p(z))_{z \in {\ensuremath{\mathbb{C}}}}\subset S^{*}({\ensuremath{{\ensuremath{\mathbb{R}}}^{d+1}}})$ be a holomorphic gauging for $p$. Then $\tilde{L}(p(z))$ has at worst a simple pole singularity at $z=0$ in such way that $${\ensuremath{{\operatorname{Res}}}}_{z=0}\tilde{L}(p(z)) = \int_{\|\xi\|=1} p_{-(d+2)}(\xi) \imath_{E}d\xi,
$$ where $p_{-(d+2)}(\xi)$ is the symbol of degree $-(d+2)$ of $p(\xi)$ and $E$ is the anisotropic radial vector field $2\xi_{0}\partial_{x_{0}}+\xi_{1}\partial_{\xi_{1}}+\ldots+\xi_{d}\partial_{\xi_{d}}$.
First, the extension is necessarily unique since the functional $L$ is holomorphic on ${\ensuremath{S^{\text{int}}}}({\ensuremath{{\ensuremath{\mathbb{R}}}^{d+1}}})$ and each symbol $p\in S^{{\ensuremath{\mathbb{C}\!\setminus\!\mathbb{Z}}}}({\ensuremath{{\ensuremath{\mathbb{R}}}^{d+1}}})$ can be connected to ${\ensuremath{S^{\text{int}}}}({\ensuremath{{\ensuremath{\mathbb{R}}}^{d+1}}})$ by means of a holomorphic family with values in $S^{{\ensuremath{\mathbb{C}\!\setminus\!\mathbb{Z}}}}({\ensuremath{{\ensuremath{\mathbb{R}}}^{d+1}}})$.
Let $p\in S^{{\ensuremath{\mathbb{C}\!\setminus\!\mathbb{Z}}}}({\ensuremath{{\ensuremath{\mathbb{R}}}^{d+1}}})$, $p\sim \sum_{j\geq 0} p_{m-j}$, and for $j=0,1,\ldots$ let $\tau_{m-j} \in {\ensuremath{\mathcal{S}}}'({\ensuremath{{\ensuremath{\mathbb{R}}}^{d+1}}})$ denote the unique homogeneous extension of $p_{m-j}$ provided by Lemma \[lem:Heisenberg.extension-symbol\]. For $N\geq \Re{m}+d+2$ the distribution $p-\sum_{j\leq N}\tau_{m-j}$ agrees with an integrable function near $\infty$, so its Fourier transform is continuous and we may define $$\tilde{L}(p)= (p-\sum_{j\leq N}\tau_{m-j})^\wedge (0).
\label{eq:Spectral.tildeL}$$ Notice that if $j>\Re m +d+2$ then $\tau_{m-j}$ is also integrable near $\infty$, so $\hat\tau_{m-j}(0)$ makes well sense. However, its value must be $0$ for homogeneity reasons. This shows that the value of $N$ in (\[eq:Spectral.tildeL\]) is irrelevant, so this formula defines a linear functional on $S^{{\ensuremath{\mathbb{C}\!\setminus\!\mathbb{Z}}}}({\ensuremath{{\ensuremath{\mathbb{R}}}^{d+1}}})$. In particular, if $\Re m<-(d+2)$ then we can take $N=0$ to get $\tilde{L}(p)=\check{p}(0)=\int p(\xi)d\xi=L(p)$. Hence $\tilde{L}$ agrees with $L$ on ${\ensuremath{S^{\text{int}}}}({\ensuremath{{\ensuremath{\mathbb{R}}}^{d+1}}})\cap S^{{\ensuremath{\mathbb{C}\!\setminus\!\mathbb{Z}}}}({\ensuremath{{\ensuremath{\mathbb{R}}}^{d+1}}})$.
Let $(p(z))_{z \in \Omega}$ be a holomorphic family of symbols such that $w(z)={{{\operatorname{ord}}}}p(z)$ is never an integer and let us study the analyticity of $\tilde{L}(p(z))$. As the functional $L$ is holomorphic on ${\ensuremath{S^{\text{int}}}}({\ensuremath{{\ensuremath{\mathbb{R}}}^{d+1}}})$ we may assume that we have $|\Re w(z)-m|<1$ for some integer $m\geq -(d+2)$. In this case in (\[eq:Spectral.tildeL\]) we can set $N=m+d+2$ and for $j=0,\ldots,m+d+1$ we can represent $\tau(z)_{w(z)-j}$ by $p(z)_{w(z)-j}$. Then, picking $\varphi \in C_{c}^\infty({\ensuremath{{\ensuremath{\mathbb{R}}}^{d+1}}})$ such that $\varphi=1$ near the origin, we see that $ \tilde{L}(p(z))$ is equal to $$\begin{gathered}
\int [p(z)(\xi)- (1-\varphi(\xi) )\sum_{j\leq m+d+2} p(z)_{w(z)-j}(\xi)] d\xi
- \sum_{j\leq m+d+2} {\ensuremath{\langle \tau (z)_{w(z)-j} , \varphi \rangle}} \\
= L(\tilde{p}(z)) -{\ensuremath{\langle \tau(z) , \varphi \rangle}}- \sum_{j\leq m+d+1} \int p(z)_{ w(z)-j}(\xi)\varphi(\xi)d\xi ,
\label{eq:Spectral.tildeLbis} \end{gathered}$$ where we have let $\tau(z)= \tau(z)_{w(z)-m-(d+2)}$ and $\tilde{p}(z)=p(z)-(1-\varphi)\sum_{j\leq m+d+2} p(z)_{w(z)-j}$.
In the r.h.s. of (\[eq:Spectral.tildeLbis\]) the only term that may fail to be analytic is $- {\ensuremath{\langle \tau(z) , \varphi \rangle}}$. Notice that by the formulas (\[eq:Appendix.almosthomogeneous-extension\]) and (\[eq:Appendix1.h’\]) in Appendix we have $${\ensuremath{\langle \tau(z) , \varphi \rangle}}= \int p(z)_{w(z)-m-(d+2)}(\varphi(\xi)-\psi_{z}(\xi))d\xi,
$$ with $\psi_{z}\in C^{\infty}({\ensuremath{{\ensuremath{\mathbb{R}}}^{d+1}}})$ of the form $\psi_{z}(\xi)= \int^{\infty}_{\log\|\xi\|}
[(\frac 1{w(z)-m}\frac{d}{ds} +1)g](t)dt$, where $g(t)$ can be any function in $C^{\infty}_{c}({\ensuremath{\mathbb{R}}})$ such that $\int g(t)dt=1$. Without any loss of generality we may suppose that $\varphi(\xi)= \int^{\infty}_{\log\|\xi\|}
g(t)dt$ with $g\in C^{\infty}_{c}({\ensuremath{\mathbb{R}}})$ as above. Then we have $\psi_{z}(\xi)= -\frac 1{w(z)-m} g(\log \|\xi\|) + \varphi(\xi)$, which gives $$\begin{gathered}
{\ensuremath{\langle \tau(z) , \varphi \rangle}} =\frac 1{w(z)-m} \int p(z)_{w(z)-m-(d+2)}(\xi) g(\log \|\xi\|)d\xi \\
=\frac1{w(z)-m}
\int \mu^{w(z)-m}g(\log \mu)\frac{d\mu}{\mu}\ \int_{\|\xi\|=1} p(z)_{w(z)-m-(d+2)}(\xi) \imath_{E}d\xi.
\label{eq:NCR.tau-z}\end{gathered}$$ Together with (\[eq:Spectral.tildeLbis\]) this shows that $\tilde{L}(p(z))$ is an analytic function, so the the first part of the lemma is proved.
Finally, let $p\sim \sum p_{m-j}$ be a symbol in $S^{{\ensuremath{\mathbb{Z}}}}({\ensuremath{{\ensuremath{\mathbb{R}}}^{d+1}}})$ and let $(p(z))_{|\Re z -m|<1}$ be a holomorphic family which is a gauging for $p$. Since $p(z)$ has order $w(z)=m+z$ it follows from (\[eq:Spectral.tildeLbis\]) and (\[eq:NCR.tau-z\]) that $\tilde{L}(p(z))$ has at worst a simple pole singularity at $z=0$ such that $$\begin{gathered}
{\ensuremath{{\operatorname{Res}}}}_{z=0}\tilde{L}(p(z))= {\ensuremath{{\operatorname{Res}}}}_{z=0} \frac{-1}{z} \int \mu^{z}g(\log \mu)\frac{d\mu}{\mu}\ \int_{\|\xi\|=1}
p(z)_{z-(d+2)}(\xi) \imath_{E}d\xi \\ = - \int_{\|\xi\|=1} p_{-(d+2)}(\xi) \imath_{E}d\xi .
$$ This proves the second part of the lemma.
Now, for $P\in {\ensuremath{\Psi_{H}^{{\ensuremath{\mathbb{C}\!\setminus\!\mathbb{Z}}}}}}(U)$ we let $$t_{P}(x) = (2\pi)^{-(d+2)}|\psi_{x}'| \tilde{L}(p(x,.)) + k_{R}(x,x),
\label{eq:NCR.tP-definition}$$ where the pair $(p,R) \in S^{{\ensuremath{\mathbb{C}\!\setminus\!\mathbb{Z}}}}({U\times{\ensuremath{\mathbb{R}}}^{d+1}})\times \Psi^{\infty}(U)$ is such that $P=p(x,-iX)+R$. This definition does not depend on the choice of $(p,R)$. Indeed, if $(p',R')$ is another such pair then $p-p'$ is in $S^{-\infty}({U\times{\ensuremath{\mathbb{R}}}^{d+1}})$, so using (\[eq:NCR.kP(x,x)\]) we that $k_{R'}(x,x)-k_{R}(x,x)$ is equal to $$\begin{gathered}
k_{(p-p')(x,-iX)}(x,x) = (2\pi)^{-(d+2)}|\psi_{x}'| L((p-p')(x,.))\\ =
(2\pi)^{-(d+2)}|\psi_{x}'| (\tilde{L}(p(x,.)) -\tilde{L}(p'(x,.))),
$$ which shows that the r.h.s. of (\[eq:NCR.tP-definition\]) is the same for both pairs.
On the other hand, observe that (\[eq:Spectral.tildeLbis\]) and (\[eq:NCR.tau-z\]) show that $\tilde{L}(p(x,.))$ depends smoothly on $x$ and that for any holomorphic family $(p(z))(z)\in \Omega \subset S^{{\ensuremath{\mathbb{C}\!\setminus\!\mathbb{Z}}}}({U\times{\ensuremath{\mathbb{R}}}^{d+1}})$ the map $z\rightarrow \tilde{L}(p(x,.))$ is holomorphic from $\Omega$ to $C^{\infty}(U)$. Therefore, the map $P\rightarrow t_{P}(x)$ is an analytic extension to ${\ensuremath{\Psi_{H}^{{\ensuremath{\mathbb{C}\!\setminus\!\mathbb{Z}}}}}}(U)$ of the map $P\rightarrow k_{P}(x,x)$.
Let $P\in {\ensuremath{\Psi_{H}^{{\ensuremath{\mathbb{Z}}}}}}(U)$ and let $(P(z))_{z \in
\Omega}\subset {\ensuremath{\Psi_{H}}}^{*}(U)$ be a holomorphic gauging for $P$. Then it follows from (\[eq:Spectral.tildeLbis\]) and (\[eq:NCR.tau-z\]) that with respect to the topology of $C^{\infty}(M,|\Lambda|(M)\otimes {\ensuremath{{\operatorname{End}}}}{\ensuremath{\mathcal{E}}})$ the map $z\rightarrow t_{P(z)}(x)$ has at worst a simple pole singularity at $z=0$ with residue $${\ensuremath{{\operatorname{Res}}}}_{z=0} t_{P(z)}(x)=-(2\pi)^{-(d+2)} \int_{\|\xi\|=1} p_{-(d+2)}(\xi) \imath_{E}d\xi =-c_{P}(x),
\label{eq:NCR.residue.t-P.U}$$ where $p_{-(d+2)}(\xi)$ denotes the symbol of degree $-(d+2)$ of $P$.
Next, let $\phi:\tilde{U}\rightarrow U$ be a change of $H$-framed local coordinates. Let $P\in {\ensuremath{\Psi_{H}^{{\ensuremath{\mathbb{C}\!\setminus\!\mathbb{Z}}}}}}(U)$ and let $(P(z))_{z \in{\ensuremath{\mathbb{C}}}}$ be a holomorphic family which is a gauging for $P$. As shown in [@Po:CPDE1] the [$\Psi_{H}$DO]{} family $(\phi^{*}P(z))_{z\in {\ensuremath{\mathbb{C}}}}$ is holomorphic and is a gauging for $\phi^{*}P$. Moreover, as for $\Re z$ negatively large enough we have $k_{\phi^{*}P(z)}=|\phi'(x)|k_{P(z)}(\phi(x),\phi(x))$, an analytic continuation gives $$t_{\phi^{*}P}(x)=|\phi'(x)|t_{P}(\phi(x)).
\label{eq:NCR.functoriality-tP}$$
Now, in the same way as in the construction of the density $c_{P}(x)$ in the proof of Proposition \[thm:NCR.log-singularity\], it follows from all this that if $P\in {\ensuremath{\Psi_{H}}}^{m}(M,{\ensuremath{\mathcal{E}}})$ then there exists a unique ${\ensuremath{{\operatorname{End}}}}{\ensuremath{\mathcal{E}}}$-valued density $t_{P}(x)$ such that, for any local $H$-framed chart $\kappa:U\rightarrow V$ and any trivialization $\tau:{\ensuremath{\mathcal{E}}}_{|_{U}}\rightarrow U\times {\ensuremath{\mathbb{C}}}^{r}$, we have $$t_{P}(x)|_{U}=\tau^{*}\kappa^{*}(t_{\kappa_{*}\tau_{*}(P_{|_{U}})}(x)dx).
$$ On every trivializing $H$-framed chart the map $P\rightarrow t_{P}(x)$ is analytic and satisfies (\[eq:NCR.residue.t-P.U\]). Therefore, we obtain:
\[thm:NCR.TR.local\] 1) The map $P \rightarrow t_{P}(x)$ is the unique analytic continuation of the map $P \rightarrow k_{P}(x,x)$ to ${\ensuremath{\Psi_{H}^{{\ensuremath{\mathbb{C}\!\setminus\!\mathbb{Z}}}}}}(M,{\ensuremath{\mathcal{E}}})$.
2\) Let $P \in {\ensuremath{\Psi_{H}^{{\ensuremath{\mathbb{Z}}}}}}(M,{\ensuremath{\mathcal{E}}})$ and let $(P(z))_{z\in\Omega}\subset {\ensuremath{\Psi_{H}}}^{*}(M,{\ensuremath{\mathcal{E}}})$ be a holomorphic family which is a gauging for $P$. Then, in $C^{\infty}(M,|\Lambda|(M)\otimes {\ensuremath{{\operatorname{End}}}}{\ensuremath{\mathcal{E}}})$, the map $z\rightarrow t_{P(z)}(x)$ has at worst a simple pole singularity at $z=0$ with residue given by $${\ensuremath{{\operatorname{Res}}}}_{z=0} t_{P(z)}(x)=- c_{P}(x),
$$ where $c_{P}(x)$ denotes the ${\ensuremath{{\operatorname{End}}}}{\ensuremath{\mathcal{E}}}$-valued density on $M$ given by Theorem \[thm:NCR.log-singularity\].
3\) The map $P\rightarrow t_{P}(x)$ is functorial with respect to Heisenberg diffeomorphisms as in (\[eq:Log.functoriality-cP\]).
Taking residues at $z=0$ in (\[eq:NCR.functoriality-tP\]) allows us to recover (\[eq:Log.functoriality-cP\]).
From now one we assume $M$ compact. We then define the *canonical trace* for the Heisenberg calculus as the functional ${\ensuremath{{\operatorname{TR}}}}$ on ${\ensuremath{\Psi_{H}^{{\ensuremath{\mathbb{C}\!\setminus\!\mathbb{Z}}}}}}(M,{\ensuremath{\mathcal{E}}})$ given by the formula, $${\ensuremath{{\operatorname{TR}}}}P := \int_{M}{{\operatorname{tr}}}_{{\ensuremath{\mathcal{E}}}} t_{P}(x) \qquad \forall P\in {\ensuremath{\Psi_{H}^{{\ensuremath{\mathbb{C}\!\setminus\!\mathbb{Z}}}}}}(M,{\ensuremath{\mathcal{E}}}).$$
\[thm:NCR.TR.global\] The canonical trace ${\ensuremath{{\operatorname{TR}}}}$ has the following properties:
1\) ${\ensuremath{{\operatorname{TR}}}}$ is the unique analytic continuation to ${\ensuremath{\Psi_{H}^{{\ensuremath{\mathbb{C}\!\setminus\!\mathbb{Z}}}}}}(M,{\ensuremath{\mathcal{E}}})$ of the usual trace.
2\) We have $ {\ensuremath{{\operatorname{TR}}}}P_{1}P_{2}={\ensuremath{{\operatorname{TR}}}}P_{2}P_{1}$ whenever ${{{\operatorname{ord}}}}P_{1}+{{{\operatorname{ord}}}}P_{2}\not\in{\ensuremath{\mathbb{Z}}}$.
3\) ${\ensuremath{{\operatorname{TR}}}}$ is invariant by Heisenberg diffeomorphisms, i.e., for any Heisenberg diffeomorphism $\phi:(M,H)\rightarrow (M',H')$ we have $ {\ensuremath{{\operatorname{TR}}}}\phi_{*}P={\ensuremath{{\operatorname{TR}}}}P$ $\forall P\in {\ensuremath{\Psi_{H}^{{\ensuremath{\mathbb{C}\!\setminus\!\mathbb{Z}}}}}}(M,{\ensuremath{\mathcal{E}}})$.
The first and third properties are immediate consequences of Theorem \[thm:NCR.TR.local\], so we only have to prove the second one.
For $j=1,2$ let $P_{j}\in {\ensuremath{\Psi_{H}}}^{*}(M,{\ensuremath{\mathcal{E}}})$ and let $(P_{j}(z))_{z\in {\ensuremath{\mathbb{C}}}}\subset {\ensuremath{\Psi_{H}}}^{*}(M,{\ensuremath{\mathcal{E}}})$ be a holomorphic gauging for $P_{j}$. We further assume that ${{{\operatorname{ord}}}}P_{1}+{{{\operatorname{ord}}}}P_{2}\not\in{\ensuremath{\mathbb{Z}}}$. Then $P_{1}(z)P_{2}(z)$ and $P_{2}(z)P_{1}(z)$ have non-integer order for $z$ in ${\ensuremath{\mathbb{C}}}\setminus \Sigma$, where $\Sigma:=-({{{\operatorname{ord}}}}P_{1}+{{{\operatorname{ord}}}}P_{2})+{\ensuremath{\mathbb{Z}}}$. For $\Re z$ negatively large enough we have ${\ensuremath{{\operatorname{Trace}}}}P_{1}(z)P_{2}(z)={\ensuremath{{\operatorname{Trace}}}}P_{2}(z)P_{1}(z)$, so by analytic continuation we see that ${\ensuremath{{\operatorname{TR}}}}P_{1}(z)P_{2}(z)={\ensuremath{{\operatorname{TR}}}}P_{2}(z)P_{1}(z)$ for any $z\in{\ensuremath{\mathbb{C}}}\setminus \Sigma$. Setting $z =0$ then shows that we have $ {\ensuremath{{\operatorname{TR}}}}P_{1}P_{2}={\ensuremath{{\operatorname{TR}}}}P_{2}P_{1}$ as desired.
Next, we define the *noncommutative residue* for the Heisenberg calculus as the linear functional ${\ensuremath{{\operatorname{Res}}}}$ on ${\ensuremath{\Psi_{H}^{{\ensuremath{\mathbb{Z}}}}}}(M,{\ensuremath{\mathcal{E}}})$ given by the formula, $${\ensuremath{{\operatorname{Res}}}}P :=\int_{M}{{\operatorname{tr}}}_{{\ensuremath{\mathcal{E}}}} c_{P}(x) \qquad \forall P\in {\ensuremath{\Psi_{H}^{{\ensuremath{\mathbb{Z}}}}}}(M,{\ensuremath{\mathcal{E}}}).
$$ This functional provides us with the analogue for the Heisenberg calculus of the noncommutative residue trace of Wodzicki ([@Wo:LISA], [@Wo:NCRF]) and Guillemin [@Gu:NPWF], for we have:
\[thm:NCR.NCR\] The noncommutative residue ${\ensuremath{{\operatorname{Res}}}}$ has the following properties:
1\) Let $P \in {\ensuremath{\Psi_{H}^{{\ensuremath{\mathbb{Z}}}}}}(M,{\ensuremath{\mathcal{E}}})$ and let $(P(z))_{z\in \Omega}\subset {\ensuremath{\Psi_{H}}}^{*}(M,{\ensuremath{\mathcal{E}}})$ be a holomorphic gauging for $P$. Then at $z=0$ the function ${\ensuremath{{\operatorname{TR}}}}P(z)$ has at worst a simple pole singularity in such way that we have $${\ensuremath{{\operatorname{Res}}}}_{z=0} {\ensuremath{{\operatorname{TR}}}}P(z)= - {\ensuremath{{\operatorname{Res}}}}P.
\label{eq:NCR.residueTR}$$
2\) We have ${\ensuremath{{\operatorname{Res}}}}P_{1}P_{2}={\ensuremath{{\operatorname{Res}}}}P_{2}P_{1}$ whenever ${{{\operatorname{ord}}}}P_{1}+{{{\operatorname{ord}}}}P_{2}\in {\ensuremath{\mathbb{Z}}}$. Hence ${\ensuremath{{\operatorname{Res}}}}$ is a trace on the algebra ${\ensuremath{\Psi_{H}^{{\ensuremath{\mathbb{Z}}}}}}(M,{\ensuremath{\mathcal{E}}})$.
3\) ${\ensuremath{{\operatorname{Res}}}}$ is invariant by Heisenberg diffeomorphisms.
4\) We have ${\ensuremath{{\operatorname{Res}}}}P^{t}={\ensuremath{{\operatorname{Res}}}}P$ and ${\ensuremath{{\operatorname{Res}}}}P^{*}=\overline{{\ensuremath{{\operatorname{Res}}}}P}$ for any $P\in {\ensuremath{\Psi_{H}^{{\ensuremath{\mathbb{Z}}}}}}(M,{\ensuremath{\mathcal{E}}})$.
The first property follows from Proposition \[thm:NCR.TR.local\]. The third and fourth properties are immediate consequences of Propositions \[thm:NCR.log-singularity\] and \[prop.Sing.transpose-adjoint\].
It remains to prove the 2nd property. Let $P_{1}$ and $P_{2}$ be operators in ${\ensuremath{\Psi_{H}}}^{*}(M,{\ensuremath{\mathcal{E}}})$ such that ${{{\operatorname{ord}}}}P_{1}+{{{\operatorname{ord}}}}P_{2}\in {\ensuremath{\mathbb{Z}}}$. For $j=1,2$ let $(P_{j}(z))_{z\in {\ensuremath{\mathbb{C}}}}\subset {\ensuremath{\Psi_{H}}}^{*}(M,{\ensuremath{\mathcal{E}}})$ be a holomorphic gauging for $P_{j}$. Then the family $(P_{1}(\frac{z}{2})P_{2}(\frac{z}{2}))_{z\in {\ensuremath{\mathbb{C}}}}$ (resp. $(P_{2}(\frac{z}{2})P_{1}(\frac{z}{2}))_{z\in {\ensuremath{\mathbb{C}}}}$) is a holomorphic gauging for $P_{1}P_{2}$ (resp. $P_{2}P_{1}$). Moreover, by Proposition \[thm:NCR.TR.global\] for any $z \in{\ensuremath{\mathbb{C}\!\setminus\!\mathbb{Z}}}$ we have ${\ensuremath{{\operatorname{TR}}}}P_{1}(\frac{z}{2})P_{2}(\frac{z}{2})={\ensuremath{{\operatorname{TR}}}}P_{\frac{z}{2}}(z)P_{1}(\frac{z}{2}$. Therefore, by taking residues at $z=0$ and using (\[eq:NCR.residueTR\]) we get ${\ensuremath{{\operatorname{Res}}}}P_{1}P_{2}={\ensuremath{{\operatorname{Res}}}}P_{2}P_{1}$ as desired.
Traces and sum of commutators {#sec:traces}
-----------------------------
Let $(M^{d+1},H)$ be a compact Heisenberg manifold and let ${\ensuremath{\mathcal{E}}}$ be a vector bundle over $M$. In this subsection, we shall prove that when $M$ is connected the noncommutative residue spans the space of traces on the algebra ${\ensuremath{\Psi_{H}^{{\ensuremath{\mathbb{Z}}}}}}(M,{\ensuremath{\mathcal{E}}})$. As a consequence this will allow us to characterize the sums of commutators in ${\ensuremath{\Psi_{H}^{{\ensuremath{\mathbb{Z}}}}}}(M,{\ensuremath{\mathcal{E}}})$.
Let $H\subset T{\ensuremath{{\ensuremath{\mathbb{R}}}^{d+1}}}$ be a hyperplane bundle such that there exists a global $H$-frame $X_{0},X_{1},\ldots,X_{d}$ of $T{\ensuremath{{\ensuremath{\mathbb{R}}}^{d+1}}}$. We will now give a series of criteria for an operator $P\in {\ensuremath{\Psi_{H}^{{\ensuremath{\mathbb{Z}}}}}}({\ensuremath{{\ensuremath{\mathbb{R}}}^{d+1}}})$ to be a sum of commutators of the form, $$P= [ x_{0}, P_{0}]+\ldots+ [x_{d}, P_{d}], \qquad P_{j}\in {\ensuremath{\Psi_{H}^{{\ensuremath{\mathbb{Z}}}}}}({\ensuremath{{\ensuremath{\mathbb{R}}}^{d+1}}}).
\label{eq:Traces.sum-commutators}$$
In the sequel for any $x\in {\ensuremath{{\ensuremath{\mathbb{R}}}^{d+1}}}$ we let $\psi_{x}$ denote the affine change of variables to the privileged coordinates at $x$ with respect to the $H$-frame $X_{0},\ldots,X_{d}$.
\[lem:Traces.criterion-logarithmic-kernel\] Let $P\in {\ensuremath{\Psi_{H}}}^{-(d+2)}({\ensuremath{{\ensuremath{\mathbb{R}}}^{d+1}}})$ have a kernel of the form, $$k_{P}(x,y)=|\psi_{x}'|K_{0}(x,-\psi_{x}(y)),
\label{eq:Traces.logarithmic-kernel}$$ where $K_{0}(x,y)\in {\ensuremath{\mathcal{K}}}_{0}({\ensuremath{{\ensuremath{\mathbb{R}}}^{d+1}}}\times {\ensuremath{{\ensuremath{\mathbb{R}}}^{d+1}}})$ is homogeneous of degree $0$ with respect to $y$. Then $P$ is a sum of commutators of the form (\[eq:Traces.sum-commutators\]).
Set $\psi_{x}(y)=A(x).(y-x)$ with $A\in C^{\infty}({\ensuremath{{\ensuremath{\mathbb{R}}}^{d+1}}}, GL_{d+1}({\ensuremath{{\ensuremath{\mathbb{R}}}^{d+1}}}))$ and for $j,k=0,\ldots, d$ define $$K_{jk}(x,y):= A_{jk}(x) y_{j}^{\beta_{j}}\|y\|^{-4}K_{0}(x,y), \qquad (x,y)\in {\ensuremath{{\ensuremath{\mathbb{R}}}^{d+1}}}\times {{\ensuremath{\mathbb{R}}}^{d+1}\!\setminus\! 0},
$$ where $\beta_{0}=1$ and $\beta_{1}=\ldots=\beta_{d}=3$. As $K_{jk}(x,y)$ is smooth for $y\neq 0$ and is homogeneous with respect to $y$ of degree $-2$ if $j=0$ and of degree $-1$ otherwise, we see that it belongs to ${\ensuremath{\mathcal{K}}}_{*}({\ensuremath{\mathbb{R}}}\times {\ensuremath{\mathbb{R}}})$. Therefore, the operator $Q_{jk}$ with Schwartz kernel $k_{Q_{jk}}=|\psi_{x}'| K_{jk}(x,-\psi_{x}(y))$ is a [$\Psi_{H}$DO]{}.
Next, set $A^{-1}(x)=(A^{jk}(x))_{1\leq j,k\leq d}$. Since $x_{k}-y_{k}=-\sum_{l=0}^{d}A^{kl}(x)\psi_{x}(y)_{l}$ we deduce that the Schwartz kernel of $\sum_{j,k=0}^{d}[x_{k},Q_{jk}]$ is $|\psi_{x}'|K(x,-\psi_{x}(y))$, where $$\begin{gathered}
K(x,y)= \sum_{0\leq j,k,l \leq d} A^{kl}(x)y_{l}A_{jk}(x)y_{j}^{\beta_{j}}\|y\|^{-4} K_{0}(x,y)\\
=\sum_{0\leq j\leq d} y_{j}^{\beta_{j}+1}\|y\|^{-4}K_{0}(x,y)=K_{0}(x,y).
\end{gathered}$$ Hence $P=\sum_{j,k=0}^{d}[x_{k},Q_{jk}]$. The lemma is thus proved.
\[lem:Traces.smoothing-operators1\] Any $R\in {\ensuremath{\Psi^{-\infty}}}({\ensuremath{{\ensuremath{\mathbb{R}}}^{d+1}}})$ can be written as a sum of commutators of the form (\[eq:Traces.sum-commutators\]).
Let $k_{R}(x,y)$ denote the Schwartz kernel of $R$. Since $k_{R}(x,y)$ is smooth we can write $$k_{R}(x,y)=k_{R}(x,x)+(x_{0}-y_{0})k_{R_{0}}(x,y)+\ldots +(x_{d}-y_{d})k_{R_{d}}(x,y),
\label{eq:Traces.Taylor}$$ for some smooth functions $k_{R_{0}}(x,y),\ldots,k_{R_{d}}(x,y)$. For $j=0,\ldots,d$ let $R_{j}$ be the smoothing operator with Schwartz kernel $k_{R_{j}}(x,y)$, and let $Q$ be the operator with Schwartz kernel $k_{Q}(x,y)=k_{R}(x,x)$. Then by (\[eq:Traces.Taylor\]) we have $$R=Q+[x_{0},R_{0}]+\ldots + [x_{d},R_{d}].
\label{eq:Traces.commutators.smoothing-RQ}$$
Observe that the kernel of $Q$ is of the form (\[eq:Traces.logarithmic-kernel\]) with $K_{0}(x,y)=|\psi_{x}'|^{-1}k_{R}(x,x)$. Here $K_{0}(x,y)$ belongs to ${\ensuremath{\mathcal{K}}}_{0}({\ensuremath{{\ensuremath{\mathbb{R}}}^{d+1}}}\times {\ensuremath{{\ensuremath{\mathbb{R}}}^{d+1}}})$ and is homogeneous of degree $0$ with respect to $y$, so by Lemma \[lem:Traces.criterion-logarithmic-kernel\] the operator $Q$ is a sum of commutators of the form (\[eq:Traces.sum-commutators\]). Combining this with (\[eq:Traces.commutators.smoothing-RQ\]) then shows that $R$ is of that form too.
\[lem:Traces.sum-commutators\] Any $P\in {\ensuremath{\Psi_{H}}}^{{\ensuremath{\mathbb{Z}}}}({\ensuremath{{\ensuremath{\mathbb{R}}}^{d+1}}})$ such that $c_{P}(x)=0$ is a sum of commutators of the form (\[eq:Traces.sum-commutators\]).
For $j=0,\ldots,d$ we let $\sigma_{j}(x,\xi)=\sum_{k=0}^{d}\sigma_{jk}(x)\xi_{k}$ denote the classical symbol of $-iX_{j}$. Notice that $\sigma(x):=(\sigma_{jk}(x))$ belongs to $C^{\infty}({\ensuremath{{\ensuremath{\mathbb{R}}}^{d+1}}}, GL_{d+1}({\ensuremath{\mathbb{C}}}))$.
\(i) Let us first assume that $P= (\partial_{\xi_{j}}q)(x,-iX)$ for some $q\in S^{{\ensuremath{\mathbb{Z}}}}({U\times{\ensuremath{\mathbb{R}}}^{d+1}})$. Set $q_{\sigma}(x,\xi)=q(x,\sigma(x,\xi))$. Then we have $$[q(x,-iX),x_{k}]= [q_{\sigma}(x,D),x_{k}]= (\partial_{\xi_{k}}q_{\sigma})(x,D)=\sum_{l} \sigma_{lk}(x)(\partial_{\xi_{l}}q)(x,-iX).
$$ Therefore, if we let $(\sigma^{kl}(x))$ be the inverse matrix of $\sigma(x)$, then we see that $$\sum_{k}[\sigma^{jk}(x)q(x,-iX),x_{k}]= \sum_{k,l} \sigma^{jk}(x)\sigma_{lk}(x)(\partial_{\xi_{l}}q)(x,-iX)=(\partial_{\xi_{j}}q)(x,-iX)=P.
$$ Hence $P$ is a sum of commutators of the form (\[eq:Traces.sum-commutators\]).
\(ii) Suppose now that $P$ has symbol $p \sim \sum_{j\leq m}p_{j}$ with $p_{-(d+2)}=0$. Since $p_{l}(x,\xi)$ is homogeneous of degree $l$ with respect to $\xi$, the Euler identity, $$2\xi_{0}\partial_{\xi_{0}}p_{l} + \xi_{1}\partial_{\xi_{1}}p_{l}+\ldots+ \xi_{d}\partial_{\xi_{d}}p_{l}
= l p_{l},$$ implies that we have $$2\partial_{\xi_{0}}(\xi_{0}p_{l}) + \partial_{\xi_{1}}(\xi_{1}p_{l})+\ldots+
\partial_{\xi_{d}}(\xi_{d}p_{l})= (l+d+2)p_{l}.$$
For $j=0, \ldots, d$ let $q^{(j)}$ be a symbol so that $q^{(j)}
\sim \sum_{l\neq -(d+2)} (l+d+2)^{-1} \xi_{j}p_{l}$. Then for $l\neq -(d+2)$ the symbol of degree $l$ of $2\partial_{\xi_{0}}q^{(0)}+
\partial_{\xi_{1}} q^{(1)}+\ldots+ \partial_{\xi_{j}} q^{(d)}$ is equal to $$(l+d+2)^{-1}(2\partial_{\xi_{0}}(\xi_{0}p_{l}) + \partial_{\xi_{1}}(\xi_{1}p_{l}))+\ldots+
\partial_{\xi_{d}}(\xi_{d}p_{l}))= p_{l}.
$$ Since $p_{-(d+2)}=0$ this shows that $p- 2\partial_{\xi_{0}}q^{(0)} +
\partial_{\xi_{1}} q^{(1)}+\ldots+ \partial_{\xi_{j}} q^{(d)}$ is in $S^{-\infty}({\ensuremath{{\ensuremath{\mathbb{R}}}^{d+1}}}\times {\ensuremath{{\ensuremath{\mathbb{R}}}^{d+1}}})$. Thus, there exists $R$ in ${\ensuremath{\Psi^{-\infty}}}({\ensuremath{{\ensuremath{\mathbb{R}}}^{d+1}}})$ such that $$P=2(\partial_{\xi_{0}}q^{(0)})(x,-iX) +
(\partial_{\xi_{1}} q^{(1)})(x,-iX)+\ldots+ (\partial_{\xi_{j}} q^{(d)})(x,-iX)+R,
$$
Thanks to the part (i) and to Lemma \[lem:Traces.smoothing-operators1\] the operators $(\partial_{\xi_{j}}q^{(j)})(x,-iX)$ and $R$ are sums of commutators of the form (\[eq:Traces.sum-commutators\]), so $P$ is of that form as well.
\(iii) The general case is obtained as follows. Let $p_{-(d+2)}(x,\xi)$ be the symbol of degree $-(d+2)$ of $P$. Then by Lemma \[lem:NCR.extension-symbolU\] we can extend $p_{-(d+2)}(x,\xi)$ into a distribution $\tau(x,\xi)$ in $C^{\infty}({\ensuremath{{\ensuremath{\mathbb{R}}}^{d+1}}}){\hat\otimes}{\ensuremath{\mathcal{S}}}'({\ensuremath{{\ensuremath{\mathbb{R}}}^{d+1}}})$ in such way that $K_{0}(x,y):=\check{\tau}_{{{\xi\rightarrow y}}}(x,y)$ belongs to ${\ensuremath{\mathcal{K}}}_{0}({\ensuremath{{\ensuremath{\mathbb{R}}}^{d+1}}}\times{\ensuremath{{\ensuremath{\mathbb{R}}}^{d+1}}})$. Furthermore, with the notation of (\[eq:NCR.log-homogeneity-Km\]) we have $c_{K,0}(x)=(2\pi)^{-(d+2)}\int_{\|\xi\|=1}p_{-(d+2)}(x,\xi)\iota_{E}d\xi$. Therefore, by using (\[eq:NCR.formula-cP\]) and the fact $c_{P}(x)$ is zero, we see that $c_{K,0}(x)=|\psi'_{x}|^{-1}c_{P}(x)=0$. In view of (\[eq:NCR.log-homogeneity-Km\]) this show that $K_{0}(x,y)$ is homogeneous of degree $0$ with respect to $y$.
Let $Q\in {\ensuremath{\Psi_{H}}}^{-(d+2)}({\ensuremath{{\ensuremath{\mathbb{R}}}^{d+1}}})$ be the [$\Psi_{H}$DO]{} with Schwartz kernel $|\psi_{x}'|K_{0}(x,-\psi_{x}(y))$. Then by Lemma \[lem:Traces.criterion-logarithmic-kernel\] the operator $Q$ is a sum of commutators of the form (\[eq:Traces.sum-commutators\]). Moreover, observe that by Proposition \[prop:PsiVDO.characterisation-kernel1\] the operator $Q$ has symbol $q\sim
q_{-(d+2)}$, where for $\xi \neq 0$ we have $q_{-(d+2)}(x,\xi)=(K_{0})^{\wedge}_{{{y\rightarrow\xi}}}(x,\xi)=p_{-(d+2)}(x,\xi)$. Therefore $P-Q$ is a [$\Psi_{H}$DO]{} whose symbol of degree $-(d+2)$ is zero. It then follows from the part (ii) of the proof that $P-Q$ is a sum of commutators of the form (\[eq:Traces.sum-commutators\]). All this shows that $P$ is the sum of two operators of the form (\[eq:Traces.sum-commutators\]), so $P$ is of that form too.
In the sequel we let ${\ensuremath{\Psi_{H,{\operatorname{c}}}}}^{*}({\ensuremath{{\ensuremath{\mathbb{R}}}^{d+1}}})$ and ${\ensuremath{\Psi^{-\infty}_{c}}}({\ensuremath{{\ensuremath{\mathbb{R}}}^{d+1}}})$ respectively denote the classes of [$\Psi_{H}$DOs]{} and smoothing operators on ${\ensuremath{{\ensuremath{\mathbb{R}}}^{d+1}}}$ with compactly supported Schwartz kernels.
\[lem:Traces.sum-commutators.compact\] There exists $\Gamma \in {\ensuremath{\Psi_{H,{\operatorname{c}}}}}^{-(d+2)}({\ensuremath{{\ensuremath{\mathbb{R}}}^{d+1}}})$ such that, for any $P\in {\ensuremath{\Psi_{H,{\operatorname{c}}}}}^{{\ensuremath{\mathbb{Z}}}}({\ensuremath{{\ensuremath{\mathbb{R}}}^{d+1}}})$, we have $$P= ({\ensuremath{{\operatorname{Res}}}}P)\Gamma \quad \bmod [{\ensuremath{\Psi_{H,{\operatorname{c}}}}}^{{\ensuremath{\mathbb{Z}}}}({\ensuremath{{\ensuremath{\mathbb{R}}}^{d+1}}}), {\ensuremath{\Psi_{H,{\operatorname{c}}}}}^{{\ensuremath{\mathbb{Z}}}}({\ensuremath{{\ensuremath{\mathbb{R}}}^{d+1}}})].
\label{eq:Traces.sum-commutators-compact}$$
Let $P\in {\ensuremath{\Psi_{H,{\operatorname{c}}}}}^{{\ensuremath{\mathbb{Z}}}}({\ensuremath{{\ensuremath{\mathbb{R}}}^{d+1}}})$. We will put $P$ into the form (\[eq:Traces.sum-commutators-compact\]) in 3 steps.
\(i) Assume first that $c_{P}(x)=0$. Then by Lemma \[lem:Traces.sum-commutators\] we can write $P$ in the form, $$P= [ x_{0}, P_{0}]+\ldots+ [x_{d}, P_{d}], \qquad P_{j}\in {\ensuremath{\Psi_{H}^{{\ensuremath{\mathbb{Z}}}}}}({\ensuremath{{\ensuremath{\mathbb{R}}}^{d+1}}}).
$$ Let $\chi$ and $\psi$ in $C^{\infty}_{c}({\ensuremath{{\ensuremath{\mathbb{R}}}^{d+1}}})$ be such that $\psi(x)\psi(y)=1$ near the support of the kernel of $P$ and $\chi=1$ near $ {{\operatorname{supp}}}\psi$. Since $\psi P \psi=P$ we obtain $$P=\sum_{j=0}^{d} \psi [x_{d}, P_{d}]\psi= \sum_{j=0}^{d} [x_{d}, \psi P_{d}\psi] =\sum_{j=0}^{d}[\chi x_{d}, \psi P_{d}\psi].
\label{eq:Traces.commutators.vanishing-cP}$$ In particular $P$ is a sum of commutators in ${\ensuremath{\Psi_{H,{\operatorname{c}}}}}^{{\ensuremath{\mathbb{Z}}}}({\ensuremath{{\ensuremath{\mathbb{R}}}^{d+1}}})$.
\(ii) Let $\Gamma_{0}\in {\ensuremath{\Psi_{H}}}^{-(d+2)}$ have kernel $k_{\Gamma_{0}}(x,y)=-\log\|\phi_{x}(y)\|$ and suppose that $P=c \Gamma_{0} \psi$ where $c\in
C^{\infty}_{c}({\ensuremath{{\ensuremath{\mathbb{R}}}^{d+1}}})$ is such that $\int c(x)dx=0$ and $\psi \in C^{\infty}_{c}({\ensuremath{{\ensuremath{\mathbb{R}}}^{d+1}}})$ is such that $\psi=1$ near ${{\operatorname{supp}}}c$. First, we have:
If $c\in C^{\infty}_{c}({\ensuremath{{\ensuremath{\mathbb{R}}}^{d+1}}})$ is such that $\int c(x)dx=0$, then there exist $c_{0}, \ldots,c_{d}$ in $C_{c}^{\infty}({\ensuremath{{\ensuremath{\mathbb{R}}}^{d+1}}})$ such that $c=\partial_{x_{0}}c_{0}+\ldots+\partial_{x_{d}}c_{d}$.
We proceed by induction on the dimension $d+1$. In dimension $1$ the proof follows from the the fact that if $c\in
C^{\infty}_{c}({\ensuremath{\mathbb{R}}})$ is such that $\int_{-\infty}^{\infty}c(x_{0})dx_{0}=0$, then $\tilde{c}(x_{0})=\int_{-\infty}^{x_{0}}c(t)dt$ is an antiderivative of $c$ with compact support.
Assume now that the claim is true in dimension $d$ and under this assumption let us prove it in dimension $d+1$. Let $c\in C^{\infty}_{c}({\ensuremath{\mathbb{R}}}^{d+1})$ be such that $\int_{{\ensuremath{\mathbb{R}}}^{d+1}} c(x)dx=0$. For any $(x_{0},\ldots,x_{d-1})$ in ${\ensuremath{\mathbb{R}}}^{d}$ we let $\tilde{c}(x_{0},\ldots,x_{d-1})=\int_{{\ensuremath{\mathbb{R}}}}c(x_{0},\ldots,x_{d-1},x_{d})dx_{d}$. This defines a function in $C^{\infty}_{c}({\ensuremath{\mathbb{R}}}^{d})$ such that $$\int_{{\ensuremath{\mathbb{R}}}^{d}}\tilde{c}(x_{0},\ldots,x_{d-1})dx_{0}\ldots dx_{d-1}= \int_{{\ensuremath{{\ensuremath{\mathbb{R}}}^{d+1}}}}c(x_{0},\ldots,x_{d})dx_{0}\ldots dx_{d}=0.
$$ Since the claim is assumed to hold in dimension $d$, it follows that there exist $\tilde{c}_{0},
\ldots,\tilde{c}_{d-1}$ in $C_{c}^{\infty}({\ensuremath{\mathbb{R}}}^{d})$ such that $ \tilde{c}=\sum_{j=0}^{d-1}\partial_{x_{j}}\tilde{c}_{j}$.
Next, let $\varphi\in C^{\infty}_{c}({\ensuremath{\mathbb{R}}})$ be such that $\varphi(x_{d})dx_{d}=1$. For any $(x_{0},\ldots,x_{d})$ in ${\ensuremath{{\ensuremath{\mathbb{R}}}^{d+1}}}$ we let $$b(x_{0},\ldots,x_{d})=c(x_{0},\ldots,x_{d})-\varphi(x_{d})\tilde{c}(x_{0},\ldots,x_{d-1}).
$$ This defines a function in $C^{\infty}_{c}({\ensuremath{\mathbb{R}}}^{d+1})$ such that $$\int_{-\infty}^{\infty}b(x_{0},\ldots,x_{d})dx_{d}=\int_{-\infty}^{\infty}c(x_{0},\ldots,x_{d})dx_{d}-\tilde{c}(x_{0},\ldots,x_{d-1})=0.
$$ Therefore, we have $b=\partial_{x_{d}}c_{d}$, where $c_{d}(x_{0},\ldots,x_{d}):=\int_{-\infty}^{x_{d}}b(x_{0},\ldots,x_{d-1},t)dt$ is a function in $C^{\infty}_{c}({\ensuremath{\mathbb{R}}}^{d+1})$.
In addition, for $j=0,\ldots,d-1$ and for $(x_{0},\ldots,x_{d})$ in ${\ensuremath{{\ensuremath{\mathbb{R}}}^{d+1}}}$ we let $c_{j}(x_{0},\ldots,x_{d})=\varphi(x_{d})\tilde{c}(x_{0},\ldots,x_{d-1})$. Then $c_{0},,\ldots,c_{d-1}$ belong to $C^{\infty}_{c}({\ensuremath{{\ensuremath{\mathbb{R}}}^{d+1}}})$ and we have $$\begin{gathered}
c(x_{0},\ldots,x_{d})= b(x_{0},\ldots, x_{d})+\varphi(x_{d})\tilde{c}(x_{0},\ldots,x_{d-1}) \\ = \partial_{x_{d}} c_{d}(x_{0},\ldots,x_{d})
+\varphi(x_{d})\sum_{j=0}^{d-1}\partial_{x_{j}}\tilde{c}_{j}(x_{0},\ldots,x_{d-1})
= \sum_{j=0}^{d}\partial_{x_{j}}{c}_{j}.
\end{gathered}$$ This shows that the claim is true in dimension $d+1$. The proof is now complete.
Let us now go back to the proof of the lemma. Since we have $\int c(x)dx=0$ the above claim tells us that $c$ can be written in the form $c=\sum_{j=0}^{d}\partial_{j}c_{j}$ with $c_{0},\ldots,c_{d}$ in $C^{\infty}_{c}({\ensuremath{{\ensuremath{\mathbb{R}}}^{d+1}}})$. Observe also that the Schwartz kernel of $[\partial_{x_{j}},\Gamma_{0}]$ is equal to $$\begin{gathered}
( \partial_{x_{j}}- \partial_{y_{j}})[- \log \|\psi_{x}(y)\|] \\ = \sum_{k,l} ( \partial_{x_{j}}- \partial_{y_{j}})
[\varepsilon_{kl}(x)(x_{l}-y_{l})][\partial_{z_{k}}\log \|z\|]_{z=-\psi_{x}(y)}\\
= \sum_{k,l} (x_{k}-y_{k}) (\partial_{x_{j}}\varepsilon_{kl})(x)
\gamma_{k}(-\psi_{x}(y))\|\psi_{x}(y)\|^{-4},
\end{gathered}$$ where we have let $\gamma_{0}(y)=\frac{1}{2} y_{0}$ and $\gamma_{k}(y)=y_{k}^{3}$, $k=1,\ldots,d$. In particular $k_{[\partial_{x_{j}},\Gamma_{0}]}(x,y)$ has no logarithmic singularity near the diagonal, that is, we have $c_{[\partial_{x_{j}},\Gamma_{0}]}(x)=0$.
Next, let $\psi \in C^{\infty}_{c}({\ensuremath{{\ensuremath{\mathbb{R}}}^{d+1}}})$ be such that $\psi=1$ near ${{\operatorname{supp}}}c\cup{{\operatorname{supp}}}c_{1}\cup \cdots \cup {{\operatorname{supp}}}c_{d}$ and let $\chi \in
C^{\infty}_{c}({\ensuremath{{\ensuremath{\mathbb{R}}}^{d+1}}})$ be such that $\chi=1$ near ${{\operatorname{supp}}}\psi$. Then we have $$\begin{gathered}
[\chi \partial_{x_{j}}, c_{j}\Gamma_{0}\psi]= [\partial_{x_{j}}, c_{j}\Gamma_{0}\psi] =
[\partial_{x_{j}},c_{j}]\Gamma_{0} \psi + c_{j}
[\partial_{x_{j}}, \Gamma_{0}]\psi+ c_{j}\Gamma_{0} [\partial_{x_{j}},\psi ]\\
= \partial_{x_{j}}c_{j} \Gamma_{0} \psi + c_{j} [\partial_{x_{j}}, \Gamma_{0}]\psi + c_{j}\Gamma_{0} \partial_{x_{j}}\psi.\end{gathered}$$ Since $ c_{j}\Gamma_{0} \partial_{x_{j}}\psi$ is smoothing and $c_{c_{j} [\partial_{x_{j}}, \Gamma_{0}]\psi}(x)=c_{j}c_{[\partial_{x_{j}},
\Gamma_{0}]}(x)=0$ we deduce from this that $P$ is of the form $P= \sum_{j=0}^{d}[\chi \partial_{x_{j}}, c_{j}\Gamma_{0}\psi] +Q$ with $Q\in {\ensuremath{\Psi_{H,{\operatorname{c}}}}}^{{\ensuremath{\mathbb{Z}}}}({\ensuremath{{\ensuremath{\mathbb{R}}}^{d+1}}})$ such that $c_{Q}(x)=0$. It then follows from the part (i) that $P$ belongs to the commutator space of ${\ensuremath{\Psi_{H,{\operatorname{c}}}}}^{{\ensuremath{\mathbb{Z}}}}({\ensuremath{{\ensuremath{\mathbb{R}}}^{d+1}}})$.
\(iii) Let $\rho \in C_{c}^{\infty}({\ensuremath{{\ensuremath{\mathbb{R}}}^{d+1}}})$ be such that $\int \rho(x)dx=1$, let $\psi \in C^{\infty}_{c}({\ensuremath{{\ensuremath{\mathbb{R}}}^{d+1}}})$ be such that $\psi=1$ near ${{\operatorname{supp}}}\rho$, and set $\Gamma=\rho \Gamma_{0}\psi$. Let $P\in {\ensuremath{\Psi_{H,{\operatorname{c}}}}}^{{\ensuremath{\mathbb{Z}}}}({\ensuremath{{\ensuremath{\mathbb{R}}}^{d+1}}})$ and let $\tilde{\psi}\in C^{\infty}_{c}({\ensuremath{{\ensuremath{\mathbb{R}}}^{d+1}}})$ be such that $\tilde{\psi}=1$ near ${{\operatorname{supp}}}c_{P}\cup {{\operatorname{supp}}}\psi$. Then we have $$P=({\ensuremath{{\operatorname{Res}}}}P) \Gamma + ({\ensuremath{{\operatorname{Res}}}}P)\rho \Gamma_{0}(\tilde{\psi}-\psi)+ (c_{P}-({\ensuremath{{\operatorname{Res}}}}P)\rho)\Gamma_{0}\tilde{\psi}+ P-c_{P}\Gamma_{0}\tilde{\psi}.
\label{eq:Traces.decomposition-P-compact-support}$$
Notice that $({\ensuremath{{\operatorname{Res}}}}P)\rho \Gamma_{0}(\tilde{\psi}-\psi)$ belongs to ${\ensuremath{\Psi^{-\infty}_{c}}}({\ensuremath{{\ensuremath{\mathbb{R}}}^{d+1}}})$. Observe also that the logarithmic singularity of $P-c_{P}\Gamma_{0}\tilde{\psi}$ is equal to $c_{P}(x)-\tilde{\psi}(x)c_{P}(x)=0$. Therefore, it follows from (i) that these operators belong to commutator space of ${\ensuremath{\Psi_{H,{\operatorname{c}}}}}^{{\ensuremath{\mathbb{Z}}}}({\ensuremath{{\ensuremath{\mathbb{R}}}^{d+1}}})$. In addition, as $\int (c_{P}(x)-({\ensuremath{{\operatorname{Res}}}}P)\rho(x))dx=0$ we see that $(c_{P}-({\ensuremath{{\operatorname{Res}}}}P)\rho)\Gamma_{0}\tilde{\psi}$ is as in (ii), so it also belongs to the commutator space of ${\ensuremath{\Psi_{H,{\operatorname{c}}}}}^{{\ensuremath{\mathbb{Z}}}}({\ensuremath{{\ensuremath{\mathbb{R}}}^{d+1}}})$. Combining all this with (\[eq:Traces.decomposition-P-compact-support\]) then shows that $P$ agrees with $({\ensuremath{{\operatorname{Res}}}}P) \Gamma $ modulo a sum of commutators in ${\ensuremath{\Psi_{H,{\operatorname{c}}}}}^{{\ensuremath{\mathbb{Z}}}}({\ensuremath{{\ensuremath{\mathbb{R}}}^{d+1}}})$. The lemma is thus proved.
Next, we quote the well known lemma below.
\[lem:Traces.smoothing-operators2\] Any $R \in {\ensuremath{\Psi^{-\infty}}}(M,{\ensuremath{\mathcal{E}}})$ such that ${\ensuremath{{\operatorname{Tr}}}}R=0$ is the sum of two commutators in ${\ensuremath{\Psi^{-\infty}}}(M,{\ensuremath{\mathcal{E}}})$.
We are now ready to prove the main result of this section.
\[thm:Traces.traces\] Assume that $M$ is connected. Then any trace on ${\ensuremath{\Psi_{H}^{{\ensuremath{\mathbb{Z}}}}}}(M,{\ensuremath{\mathcal{E}}})$ is a constant multiple of the noncommutative residue.
Let $\tau$ be a trace on ${\ensuremath{\Psi_{H}^{{\ensuremath{\mathbb{Z}}}}}}(M,{\ensuremath{\mathcal{E}}})$. By Lemma \[lem:Traces.sum-commutators.compact\] there exists $\Gamma \in {\ensuremath{\Psi_{H,{\operatorname{c}}}}}^{-(d+2)}({\ensuremath{{\ensuremath{\mathbb{R}}}^{d+1}}})$ such that any $P=(P_{ij})$ in ${\ensuremath{\Psi_{H,{\operatorname{c}}}}}^{{\ensuremath{\mathbb{Z}}}}({\ensuremath{{\ensuremath{\mathbb{R}}}^{d+1}}},{\ensuremath{\mathbb{C}}}^{r})$ can be written as $$P=\Gamma \otimes R \bmod [{\ensuremath{\Psi_{H,{\operatorname{c}}}}}^{{\ensuremath{\mathbb{Z}}}}({\ensuremath{{\ensuremath{\mathbb{R}}}^{d+1}}}), {\ensuremath{\Psi_{H,{\operatorname{c}}}}}^{{\ensuremath{\mathbb{Z}}}}({\ensuremath{{\ensuremath{\mathbb{R}}}^{d+1}}})]\otimes M_{r}({\ensuremath{\mathbb{C}}}),
$$ where we have let $R=({\ensuremath{{\operatorname{Res}}}}P_{ij})\in M_{r}({\ensuremath{\mathbb{C}}})$. Notice that ${\ensuremath{{\operatorname{Tr}}}}R= \sum
{\ensuremath{{\operatorname{Res}}}}P_{ii}={\ensuremath{{\operatorname{Res}}}}P$. Thus $R-\frac{1}{r}({\ensuremath{{\operatorname{Res}}}}P)I_{r}$ has a vanishing trace, hence belongs to the commutator space of $M_{r}({\ensuremath{\mathbb{C}}})$. Therefore, we have $$P=({\ensuremath{{\operatorname{Res}}}}P) \Gamma\otimes (\frac{1}{r}I_{r}) \quad \bmod [{\ensuremath{\Psi_{H,{\operatorname{c}}}}}^{{\ensuremath{\mathbb{Z}}}}({\ensuremath{{\ensuremath{\mathbb{R}}}^{d+1}}}, {\ensuremath{\mathbb{C}}}^{r}), {\ensuremath{\Psi_{H,{\operatorname{c}}}}}^{{\ensuremath{\mathbb{Z}}}}({\ensuremath{{\ensuremath{\mathbb{R}}}^{d+1}}},{\ensuremath{\mathbb{C}}}^{r})].
\label{eq:Traces.decomposition-P-Rd-Cr}$$
Let $\kappa:U\rightarrow {\ensuremath{{\ensuremath{\mathbb{R}}}^{d+1}}}$ be a local $H$-framed chart mapping onto ${\ensuremath{{\ensuremath{\mathbb{R}}}^{d+1}}}$ and such that ${\ensuremath{\mathcal{E}}}$ is trivializable over its domain. For sake of terminology’s brevity we shall call such a chart a *nice $H$-framed chart*. As $U$ is $H$-framed and is Heisenberg diffeomorphic to ${\ensuremath{{\ensuremath{\mathbb{R}}}^{d+1}}}$ and as${\ensuremath{\mathcal{E}}}$ is trivializable over $U$, it follows from (\[eq:Traces.decomposition-P-Rd-Cr\]) that there exists $\Gamma_{U}\in
{\ensuremath{\Psi_{H,{\operatorname{c}}}}}^{-(d+2)}(U,{\ensuremath{\mathcal{E}}}_{|_{U}})$ such that, for any $P \in {\ensuremath{\Psi_{H,{\operatorname{c}}}}}^{{\ensuremath{\mathbb{Z}}}}((U,{\ensuremath{\mathcal{E}}}_{|_{U}})$, we have $$P=({\ensuremath{{\operatorname{Res}}}}P)\Gamma_{U} \quad \bmod [{\ensuremath{\Psi_{H,{\operatorname{c}}}}}^{{\ensuremath{\mathbb{Z}}}}(U,{\ensuremath{\mathcal{E}}}_{|_{U}}), {\ensuremath{\Psi_{H,{\operatorname{c}}}}}^{{\ensuremath{\mathbb{Z}}}}(U,{\ensuremath{\mathcal{E}}}_{|_{U}})].
\label{eq:Traces.local-form-Psivdos}$$ If we apply the trace $\tau$, then we see that, for any $P \in {\ensuremath{\Psi_{H,{\operatorname{c}}}}}^{{\ensuremath{\mathbb{Z}}}}(U,{\ensuremath{\mathcal{E}}}_{|_{U}})$, we have $$\tau(P) = \Lambda_{U}{\ensuremath{{\operatorname{Res}}}}P, \qquad \Lambda_{U}:=\tau(\Gamma_{U}).
$$
Next, let ${\ensuremath{\mathcal{U}}}$ be the set of points $x \in M$ near which there a domain $V$ of a nice $H$-framed chart such that $\Lambda_{V}=\Lambda_{U}$. Clearly ${\ensuremath{\mathcal{U}}}$ is a non-empty open subset of $M$. Let us prove that ${\ensuremath{\mathcal{U}}}$ is closed. Let $x \in \overline{{\ensuremath{\mathcal{U}}}}$ and let $V$ be an open neighborhood of $x$ which is the domain a nice $H$-framed chart (such a neighborhood always exists). Since $x$ belongs to the closure of ${\ensuremath{\mathcal{U}}}$ the set ${\ensuremath{\mathcal{U}}}\cup V$ is non-empty. Let $y \in {\ensuremath{\mathcal{U}}}\cup V$. As $y$ belongs to ${\ensuremath{\mathcal{U}}}$ there exists an open neighborhood $W$ of $y$ which is the domain a nice $H$-frame chart such that $\Lambda_{W}=\Lambda_{U}$. Then for any $P$ in ${\ensuremath{\Psi_{H,{\operatorname{c}}}}}^{{\ensuremath{\mathbb{Z}}}}(V\cap W, {\ensuremath{\mathcal{E}}}_{|V\cap W})$ we have $\tau(P)=\Lambda_{V}{\ensuremath{{\operatorname{Res}}}}P =\Lambda_{W}{\ensuremath{{\operatorname{Res}}}}P$. Choosing $P$ so that ${\ensuremath{{\operatorname{Res}}}}P\neq 0$ then shows that $\Lambda_{V}=\Lambda_{W}=\Lambda_{U}$. Since $V$ contains $x$ and is a domain of a nice $H$-framed chart we deduce that $x$ belongs to ${\ensuremath{\mathcal{U}}}$. Hence ${\ensuremath{\mathcal{U}}}$ is both closed and open. As $M$ is connected it follows that ${\ensuremath{\mathcal{U}}}$ agrees with $M$. Therefore, if we set $\Lambda=\Lambda_{U}$ then, for any domain $V$ of a nice $H$-framed chart, we have $$\tau(P)=\Lambda {\ensuremath{{\operatorname{Res}}}}P \qquad \forall P \in {\ensuremath{\Psi_{H,{\operatorname{c}}}}}^{{\ensuremath{\mathbb{Z}}}}(V,{\ensuremath{\mathcal{E}}}_{|_{V}}).
\label{eq:Traces.mutiple-local}$$
Now, let $(\varphi_{i})$ be a finite partition of the unity subordinated to an open covering $(U_{i})$ of $M$ by domains of nice $H$-framed charts. For each index $i$ let $\psi_{i}\in C^{\infty}_{c}(U_{i})$ be such that $\psi_{i}=1$ near ${{\operatorname{supp}}}\varphi_{i}$. Then any $P \in
{\ensuremath{\Psi_{H}^{{\ensuremath{\mathbb{Z}}}}}}(M,{\ensuremath{\mathcal{E}}})$ can be written as $P=\sum \varphi_{i} P\psi_{i}+R$, where $R$ is a smoothing operator whose kernel vanishes near the diagonal of $M\times M$. In particular we have ${\ensuremath{{\operatorname{Trace}}}}R=0$, so by Lemma \[lem:Traces.smoothing-operators2\] the commutator space of ${\ensuremath{\Psi_{H}^{{\ensuremath{\mathbb{Z}}}}}}(M,{\ensuremath{\mathcal{E}}})$ contains $R$. Since each operator $\varphi_{i} P\psi_{i}$ can be seen as an element of ${\ensuremath{\Psi_{H,{\operatorname{c}}}}}^{{\ensuremath{\mathbb{Z}}}}(U_{i},{\ensuremath{\mathcal{E}}}_{|_{U_{i}}})$, using (\[eq:Traces.mutiple-local\]) we get $$\tau(P)=\sum \tau(\varphi_{i}P\psi_{i}) = \sum \Lambda {\ensuremath{{\operatorname{Res}}}}\varphi_{i}P\psi_{i}=\Lambda {\ensuremath{{\operatorname{Res}}}}P.
$$ Hence we have $\tau =\Lambda {\ensuremath{{\operatorname{Res}}}}$. This shows that any trace on ${\ensuremath{\Psi_{H}^{{\ensuremath{\mathbb{Z}}}}}}(M,{\ensuremath{\mathcal{E}}})$ is proportional to the noncommutative residue.
Since the dual of ${\ensuremath{\Psi_{H}^{{\ensuremath{\mathbb{Z}}}}}}(M,{\ensuremath{\mathcal{E}}})/[{\ensuremath{\Psi_{H}^{{\ensuremath{\mathbb{Z}}}}}}(M,{\ensuremath{\mathcal{E}}}),{\ensuremath{\Psi_{H}^{{\ensuremath{\mathbb{Z}}}}}}(M,{\ensuremath{\mathcal{E}}})]$ is isomorphic to the space of traces on ${\ensuremath{\Psi_{H}^{{\ensuremath{\mathbb{Z}}}}}}(M,{\ensuremath{\mathcal{E}}})$, as a consequence of Theorem \[thm:Traces.traces\] we get:
Assume $M$ connected. Then an operator $P \in {\ensuremath{\Psi_{H}^{{\ensuremath{\mathbb{Z}}}}}}(M,{\ensuremath{\mathcal{E}}})$ is a sum of commutators in ${\ensuremath{\Psi_{H}^{{\ensuremath{\mathbb{Z}}}}}}(M,{\ensuremath{\mathcal{E}}})$ if and only if its noncommutative residue vanishes.
In [@EM:HAITH] Epstein and Melrose computed the Hochschild homology of the algebra of symbols ${\ensuremath{\Psi_{H}^{{\ensuremath{\mathbb{Z}}}}}}(M,{\ensuremath{\mathcal{E}}})/{\ensuremath{\Psi^{-\infty}}}(M,{\ensuremath{\mathcal{E}}})$ when $(M,H)$ is a contact manifold. In fact, as the algebra ${\ensuremath{\Psi^{-\infty}}}(M,{\ensuremath{\mathcal{E}}})$ is $H$-unital and its Hochschild homology is known, the long exact sequence of [@Wo:LESCHAEA] holds and allows us to relate the Hochschild homology of ${\ensuremath{\Psi_{H}^{{\ensuremath{\mathbb{Z}}}}}}(M,{\ensuremath{\mathcal{E}}})$ to that of ${\ensuremath{\Psi_{H}^{{\ensuremath{\mathbb{Z}}}}}}(M,{\ensuremath{\mathcal{E}}})/{\ensuremath{\Psi^{-\infty}}}(M,{\ensuremath{\mathcal{E}}})$. In particular, we can recover from this that the space of traces on ${\ensuremath{\Psi_{H}^{{\ensuremath{\mathbb{Z}}}}}}(M,{\ensuremath{\mathcal{E}}})$ is one-dimensional when the manifold is connected.
Analytic Applications on general Heisenberg manifolds {#sec:Analytic-Applications}
=====================================================
In this section we derive several analytic applications of the construction of the noncommutative residue trace for the Heisenberg calculus. First, we deal with zeta functions of hypoelliptic [$\Psi_{H}$DOs]{} and relate their singularities to the heat kernel asymptotics of the corresponding operators. Second, we give logarithmic metric estimates for Green kernels of hypoelliptic [$\Psi_{H}$DOs]{} whose order is equal to the Hausdorff dimension $\dim M +1$. This connects nicely with previous results of Fefferman, Stein and their students and collaborators. Finally, we show that the noncommutative residue for the Heisenberg calculus allows us to extend the Dixmier trace to the whole algebra of integer order [$\Psi_{H}$DOs]{}. This is the analogue for the Heisenberg calculus of a well-known result of Alain Connes.
Zeta functions of hypoelliptic [$\Psi_{H}$DOs]{}
------------------------------------------------
Let $(M^{d+1}, H)$ be a compact Heisenberg manifold equipped with a smooth density $>0$, let ${\ensuremath{\mathcal{E}}}$ be a Hermitian vector bundle over $M$ of rank $r$, and let $P:C^{\infty}(M,{\ensuremath{\mathcal{E}}})\rightarrow C^{\infty}(M,{\ensuremath{\mathcal{E}}})$ be a [$\Psi_{H}$DO]{} of integer order $m\geq 1$ with an invertible principal symbol. In addition, assume that there is a ray $L_{\theta}=\{\arg \lambda =\theta\}$ which is is not through an eigenvalue of $P$ and is a principal cut for the principal symbol $\sigma_{m}(P)$ as in Section \[sec:Heisenberg-calculus\].
Let $(P_{\theta}^{s})_{s\in {\ensuremath{\mathbb{C}}}}$ be the associated family of complex powers associated to $\theta$ as in Proposition \[prop:Heisenberg.powers2\]. Since $(P_{\theta}^{s})_{s \in {\ensuremath{\mathbb{C}}}}$ is a holomorphic family of [$\Psi_{H}$DOs]{}, Proposition \[thm:NCR.TR.global\] allows us to directly define the zeta function $\zeta_{\theta}(P;s)$ as the meromorphic function, $$\zeta_{\theta}(P;s):={\ensuremath{{\operatorname{TR}}}}P_{\theta}^{-s}, \qquad s \in {\ensuremath{\mathbb{C}}}.
$$
\[prop:Zeta.zeta-function\] Let $\Sigma=\{-\frac{d+2}{m}, -\frac{d+1}{m},\ldots, \frac{-1}{m},\frac{1}{m},
\frac{2}{m}, \ldots\}$. Then the function $\zeta_{\theta}(P;s)$ is analytic outside $\Sigma$, and on $\Sigma$ it has at worst simple pole singularities such that $${\ensuremath{{\operatorname{Res}}}}_{s=\sigma}\zeta_{\theta}(P;s)=m{\ensuremath{{\operatorname{Res}}}}P^{-\sigma}_{\theta}, \qquad \sigma\in \Sigma.
\label{eq:Zeta.residue-zeta}$$ In particular, $\zeta_{\theta}(P;s)$ is always regular at $s=0$.
Since ${{{\operatorname{ord}}}}P_{\theta}^{-s}=ms$ it follows from Proposition \[thm:NCR.NCR\] that $\zeta_{\theta}(P;s)$ is analytic outside $\Sigma':=\Sigma\cup\{0\}$ and on $\Sigma'$ has at worst simple pole singularities satisfying (\[eq:Zeta.residue-zeta\]). At $s=0$ we have $ {\ensuremath{{\operatorname{Res}}}}_{s=0}\zeta_{\theta}(P;s)=m{\ensuremath{{\operatorname{Res}}}}P^{0}_{\theta}=m{\ensuremath{{\operatorname{Res}}}}[1-\Pi_{0}(P)]$, but as $\Pi_{0}(P)$ is a smoothing operator we have ${\ensuremath{{\operatorname{Res}}}}[1-\Pi_{0}(P)]=-{\ensuremath{{\operatorname{Res}}}}\Pi_{0}(P)=0$. Thus $\zeta_{\theta}(P;s)$ is regular at $s=0$.
Assume now that $P$ is selfadjoint and the union set of its principal cuts is $\Theta(P)={\ensuremath{\mathbb{C}}}\setminus [0,\infty)$. This implies that $P$ is bounded from below (see [@Po:CPDE1]), so its spectrum is real and contains at most finitely many negative eigenvalues. We will use the subscript $\uparrow$ (resp. $\downarrow$) to refer to a spectral cutting in the upper halfplane $\Im \lambda>0$ (resp. lower halfplane $\Im
\lambda<0$).
Since $P$ is bounded from below it defines a heat semigroup $e^{-tP}$, $t\geq 0$, and, as the principal symbol of $P$ is invertible, for $t>0$ the operator $e^{-tP}$ is smoothing, hence has a smooth Schwartz kernel $k_{t}(x,y)$ in $C^{\infty}(M,{\ensuremath{\mathcal{E}}}){\hat\otimes}C^{\infty}(M,{\ensuremath{\mathcal{E}}}^{*}\otimes |\Lambda|(M))$. Moreover, as $t\rightarrow 0^{+}$ we have the heat kernel asymptotics, $$k_{t}(x,x)\sim t^{-\frac{d+2}{m}}\sum_{j\geq 0} t^{\frac{j}{m}}a_{j}(P)(x) + \log t\sum_{k\geq 0}t^{k}b_{k}(P)(x),
\label{eq:Zeta.heat-kernel-asymptotics}$$ where the asymptotics takes place in $C^{\infty}(M,{\ensuremath{{\operatorname{End}}}}{\ensuremath{\mathcal{E}}}\otimes |\Lambda|(M))$, and when $P$ is a differential operator we have $a_{2j-1}(P)(x)=b_{j}(P)(x)=0$ for all $j\in {\ensuremath{\mathbb{N}}}$ (see [@BGS:HECRM], [@Po:MAMS1] when $P$ is a differential operator and see [@Po:CPDE1] for the general case).
\[prop:Zeta.heat-zeta-local\] For $j=0,1,\ldots$ set $\sigma_{j}=\frac{d+2-j}{m}$. Then:
1\) When $\sigma_{j}\not \in {\ensuremath{\mathbb{Z}}}_{-}$ we have $${\ensuremath{{\operatorname{Res}}}}_{s=\sigma_{j}}t_{P_{{\uparrow \downarrow}}^{-s}}(x)=m c_{P^{-\sigma_{j}}}(x) = \Gamma(\sigma_{j})^{-1}a_{j}(P)(x).
\label{eq:Zeta.tPs-heat1}
$$ 2) For $k=1,2,\ldots$ we have $$\begin{gathered}
{\ensuremath{{\operatorname{Res}}}}_{s=-k}t_{P_{{\uparrow \downarrow}}^{-s}}(x)=m c_{P^{k}}(x) = (-1)^{k+1}k!b_{k}(P)(x),
\label{eq:Zeta.tPs-heat2}\\
\lim_{s\rightarrow -k}[t_{P_{{\uparrow \downarrow}}^{-s}}(x)-m (s+k)^{-1}c_{P^{k}}(x)] = (-1)^{k}k! a_{d+2+mk}(P)(x).\end{gathered}$$ 3) For $k=0$ we have $$\lim_{s\rightarrow 0} t_{P_{{\uparrow \downarrow}}^{-s}}(x) =a_{d+2}(P)(x)-t_{\Pi_{0}}(x).
\label{eq:Zeta.tPs-heat4}$$
When $P$ is positive and invertible the result is a standard consequence of the Mellin formula (see, e.g., [@Gi:ITHEASIT]). Here it is slightly more complicated because we don’t assume that $P$ is positive or invertible.
For $\Re s>0$ set $Q_{s}= \Gamma(s)^{-1}\int_{0}^{1}t^{s-1}e^{-tP}dt$. Then we have:
The family $(Q_{s})_{\Re s>0}$ can be uniquely extended to a holomorphic family of [$\Psi_{H}$DOs]{} parametrized by ${\ensuremath{\mathbb{C}}}$ in such way that:
\(i) The families $(Q_{s})_{s \in {\ensuremath{\mathbb{C}}}}$ and $(P_{{\uparrow \downarrow}}^{-s})_{s \in {\ensuremath{\mathbb{C}}}}$ agree up to a holomorphic family of smoothing operators;
\(ii) We have $Q_{0}=1$ and $Q_{-k}=P^{k}$ for any integer $k\geq 1$.
First, let $\Pi_{+}(P)$ and $\Pi_{-}(P)$ denote the orthogonal projections onto the positive and negative eigenspaces of $P$. Notice that $\Pi_{-}(P)$ is a smoothing operator because $P$ has at most only finitely many negative eigenvalues. For $\Re s>0$ the Mellin formula allows us to write $$P_{{\uparrow \downarrow}}^{-s}=\Pi_{-}(P) P_{{\uparrow \downarrow}}^{-s}+\Gamma(s)^{-1}\int_{0}^{\infty}t^{s}\Pi_{+}(P)e^{-tP}\frac{dt}{t}=Q_{s}+R_{{\uparrow \downarrow}}(s),
\label{eq:Zeta.claim-heat.PQs}\\$$ where $R_{{\uparrow \downarrow}}(s)$ is equal to $$\Pi_{-}(P) P_{{\uparrow \downarrow}}^{-s}-s^{-1}\Gamma(s)^{-1}\Pi_{0}(P)-\Pi_{-}(P)\int_{0}^{1}t^{s}e^{-tP}\frac{dt}{t}+
\int_{1}^{\infty} t^{s}\Pi_{+}(P)e^{-tP}\frac{dt}{t}.
$$ Notice that $ (\Pi_{-}(P) P_{{\uparrow \downarrow}}^{-s})_{s \in {\ensuremath{\mathbb{C}}}}$ and $(s^{-1}\Gamma(s)^{-1}\Pi_{0}(P))_{s \in {\ensuremath{\mathbb{C}}}}$ are holomorphic families of smoothing operators because $\Pi_{-}(P)$ and $\Pi_{0}(P)$ are smoothing operators. Moreover, upon writing $$\begin{gathered}
\Pi_{-}(P)\int_{0}^{1}t^{s}e^{-tP}\frac{dt}{t}=\Pi_{-}(P)(\int_{0}^{1} t^{s}e^{-tP}\frac{dt}{t})\Pi_{-}(P),\\
\int_{1}^{\infty} t^{s}\Pi_{+}(P)e^{-tP}\frac{dt}{t}= e^{-\frac{1}{4}P}(\int_{1/2}^{\infty} t^{s}\Pi_{+}(P)e^{-tP}\frac{dt}{t})e^{-\frac{1}{4}P},
$$ we see that $( \Pi_{-}(P)\int_{0}^{1}t^{s}e^{-tP}\frac{dt}{t})_{\Re s>0}$ and $( \int_{1}^{\infty} t^{s}\Pi_{+}(P)e^{-tP}\frac{dt}{t})_{\Re s>0}$ are holomorphic families of smoothing operators. Therefore $(R_{{\uparrow \downarrow}}(s))_{\Re s>0}$ is a holomorphic family of smoothing operators and using (\[eq:Zeta.claim-heat.PQs\]) we see that $(Q_{s})_{\Re s>0}$ is a holomorphic family of [$\Psi_{H}$DOs]{}.
Next, an integration by parts gives $$\Gamma(s+1)P Q_{s+1}= \int_{0}^{1}
t^{s}\frac{d}{dt}(e^{-tP})= e^{-P} + s \int_{0}^{1}
t^{s-1} e^{-tP}dt.
$$ Since $\Gamma(s+1)=s\Gamma(s)$ we get $$Q_{s}= P Q_{s+1} - \Gamma(s+1)^{-1}e^{-P}, \qquad \Re s>0.
\label{eq:Zeta.Qs-extension1}$$ An easy induction then shows that for $k=1,2,\ldots$ we have $$Q_{s}= P^{k} Q_{s+k} - \Gamma(s+k)^{-1}P^{k-1}e^{-P}+\ldots +(-1)^{k}\Gamma(s+1)^{-1}e^{-P}.
\label{eq:Zeta.Qs-extension2}$$ It follows that the family $(Q_{s})_{\Re s>0}$ has a unique analytic continuation to each half-space $\Re s>-k$ for $k=1,2,\ldots$, so it admits a unique analytic continuation to ${\ensuremath{\mathbb{C}}}$. Furthermore, as for $\Re s>-k$ we have $P^{-s}_{{\uparrow \downarrow}}=P^{k}P^{-(s+k)}_{{\uparrow \downarrow}}$ we get $$Q_{s}-P^{-s}_{{\uparrow \downarrow}}=P^{k} R_{{\uparrow \downarrow}}(s+k) -\Gamma(s+k)^{-1}P^{k-1}e^{-P}+\ldots +(-1)^{k}\Gamma(s+1)^{-1}e^{-P},
$$ from which we deduce that $(Q_{s}-P^{-s}_{{\uparrow \downarrow}})_{\Re s >-k}$ is a holomorphic family of smoothing operators. Hence the families $(Q_{s})_{s \in {\ensuremath{\mathbb{C}}}}$ and $(P^{-s}_{{\uparrow \downarrow}})_{s \in {\ensuremath{\mathbb{C}}}}$ agree up to a holomorphic family of smoothing operators.
Finally, we have $$Q_{1}=\Pi_{0}(P)+\int_{0}^{1}(1-\Pi_{0}(P))e^{-tP}dt=\Pi_{0}(P)-P^{-1}(e^{-P}-1).
$$ Thus setting $s=1$ in (\[eq:Zeta.Qs-extension1\]) gives $$\begin{gathered}
Q_{0}=P[\Pi_{0}(P)-P^{-1}(e^{-P}-1)]+e^{-P} = -(1-\Pi_{0}(P))(e^{-P}-1)+e^{-P}\\ = 1-\Pi_{0}(P)+\Pi_{0}e^{-P}=1.
\label{eq:Zeta.Q0}\end{gathered}$$ Furthermore, as $\Gamma(s)^{-1}$ vanishes at every non-positive integer, from (\[eq:Zeta.Qs-extension2\]) and (\[eq:Zeta.Q0\]) we see that we have $Q_{-k}=P^{k}Q_{0}=P^{k}$ for any integer $k\geq 1$. The proof of the claim is thus achieved.
Now, for $j=0,1,\ldots$ we set $\sigma_{j}=\frac{d+2-j}{m}$. As $(R_{{\uparrow \downarrow}}(s))_{s \in {\ensuremath{\mathbb{C}}}}:=(P_{{\uparrow \downarrow}}^{-s}-Q_{s})_{s \in {\ensuremath{\mathbb{C}}}}$ is a holomorphic family of smoothing operators, the map $s \rightarrow
t_{R_{{\uparrow \downarrow}}(s)}(x)$ is holomorphic from ${\ensuremath{\mathbb{C}}}$ to $C^{\infty}(M, |\Lambda|(M)\otimes {\ensuremath{{\operatorname{End}}}}{\ensuremath{\mathcal{E}}})$. By combining this with Proposition \[prop:Zeta.zeta-function\] we deduce that for $j=0,1,\ldots$ we have $${\ensuremath{{\operatorname{Res}}}}_{s=\sigma_{j}}t_{P_{{\uparrow \downarrow}}^{-s}}(x)=m c_{P^{-\sigma_{j}}}(x) = {\ensuremath{{\operatorname{Res}}}}_{s=\sigma_{j}}t_{Q_{s}}(x),
\label{eq:Zeta.tPs-TQs'1}$$ Moreover, as for $k=1,2,\ldots$ we have $R_{{\uparrow \downarrow}}(-k)=0$ we also see that $$\begin{gathered}
\lim_{s\rightarrow -k}[t_{P_{{\uparrow \downarrow}}^{-s}}(x)-m (s+k)^{-1}c_{P^{k}}(x)] \\ = \lim_{s\rightarrow -k}[t_{Q_{s}}(x)-(s+k)^{-1}{\ensuremath{{\operatorname{Res}}}}_{s=-k}t_{Q_{s}}(x)].\end{gathered}$$ Similarly, as $P_{{\uparrow \downarrow}}^{0}=1-\Pi_{0}(P)=Q_{0}-\Pi_{0}(P)$ we get $$\lim_{s\rightarrow 0} t_{P_{{\uparrow \downarrow}}^{-s}}(x) =\lim_{s\rightarrow 0}t_{Q_{s}}(x)-t_{\Pi_{0}}(x).
\label{eq:Zeta.tPs-TQs'3}$$
Next, let $k_{Q_{s}}(x,y)$ denote the kernel of $Q_{s}$. As $Q_{s}$ has order $-m s$, for $\Re s>-\frac{d+2}{m}$ this is a trace-class operator and thanks to (\[eq:Zeta.heat-kernel-asymptotics\]) we have $$\Gamma(s) k_{Q_{s}}(x,x)= \int_{0}^{1}t^{s-1}k_{t}(x,x) dt.
$$ Moreover (\[eq:Zeta.heat-kernel-asymptotics\]) implies that, for any integer $N\geq 0$, in $C^{\infty}(M,{\ensuremath{{\operatorname{End}}}}{\ensuremath{\mathcal{E}}}\otimes |\Lambda|(M))$ we have $$k_{t}(x,x)=\sum_{-\sigma_{j}<N}t^{-\sigma_{j}}a_{j}(P)(x)+\sum_{k<N}(t^{k}\log t)b_{k}(P)(x) +{\operatorname{O}}(t^{N}).
$$ Therefore, for $\Re s >\frac{d+2}{m}$ the density $\Gamma(s)k_{Q_{s}}(x,x)$ is of the form $$\sum_{\sigma_{j}<N}(\int_{0}^{1}t^{s-\sigma_{j}}\frac{dt}{t})a_{j}(P)(x)+
\sum_{k<N}(\int_{0}^{1}t^{k+s}\log t \frac{dt}{t})b_{k}(P)(x) + \Gamma(s) h_{N,s}(x),
$$ with $h_{N,s}(x) \in {{\operatorname{Hol}}}(\Re s>-N, C^{\infty}(M,{\ensuremath{{\operatorname{End}}}}{\ensuremath{\mathcal{E}}}\otimes |\Lambda|(M))$. Since for $\alpha >0$ we have $$\int_{0}^{1}t^{\alpha}\log t \frac{dt}{t}=-\frac{1}{\alpha}\int_{0}^{1}t^{\alpha-1}dt = -\frac{1}{\alpha},
$$ we see that $k_{Q_{s}}(x,x)$ is equal to $$\Gamma(s)^{-1} \sum_{\sigma_{j}<N}\frac{1}{s+\sigma_{j}}a_{j}(P)(x)-
\Gamma(s)^{-1} \sum_{k<N}\frac{1}{(s+k)^{2}}b_{k}(P)(x) + h_{N,s}(x).
$$ Since $\Gamma(s)$ is analytic on ${\ensuremath{\mathbb{C}}}\setminus ({\ensuremath{\mathbb{Z}}}_{-}\cup\{0\})$ and for $k=0,1,\ldots$ near $s=-k$ we have $\Gamma(s)^{-1}\sim (-1)^{k}k!(s+k)^{-1}$ , we deduce that:
- when $\sigma_{j}\not \in {\ensuremath{\mathbb{N}}}$ we have $ {\ensuremath{{\operatorname{Res}}}}_{s=\sigma_{j}}t_{Q_{s}}(x)= \Gamma(\sigma_{j})^{-1}a_{j}(P)(x)$.
- for $k=1,2,\ldots$ we have $$\begin{gathered}
{\ensuremath{{\operatorname{Res}}}}_{s=-k}t_{Q_{s}}(x)= (-1)^{k+1}k!b_{k}(P)(x),\\
\lim_{s\rightarrow -k}[t_{Q_{s}}(x)-(s+k)^{-1}{\ensuremath{{\operatorname{Res}}}}_{s=-k}t_{Q_{s}}(x)] = (-1)^{k}k! a_{d+2+mk}(P)(x).
\end{gathered}$$ - for $k=0$ we have $\lim_{s\rightarrow 0} t_{Q_{s}}(x) =a_{d+2}(P)(x)$.
Combining this with (\[eq:Zeta.tPs-TQs’1\])–(\[eq:Zeta.tPs-TQs’3\]) then proves the equalities (\[eq:Zeta.tPs-heat1\])–(\[eq:Zeta.tPs-heat4\]).
From Proposition \[prop:Zeta.heat-zeta-local\] we immediately get:
\[prop:Zeta.heat-zeta-global\] 1) For $j=0,1,\ldots$ let $\sigma_{j}=\frac{d+2-j}{m}$. When $\sigma_{j}\not \in {\ensuremath{\mathbb{Z}}}_{-}$ we have: $${\ensuremath{{\operatorname{Res}}}}_{s=\sigma_{j}}\zeta_{{\uparrow \downarrow}}(P;s) =m {\ensuremath{{\operatorname{Res}}}}P^{-\sigma_{j}}= \Gamma(\sigma_{j})^{-1}\int_{M}{{\operatorname{tr}}}_{{\ensuremath{\mathcal{E}}}}a_{j}(P)(x).
\label{eq:Zeta.heat-zeta-global1}$$ 2) For $k=1,2,\ldots$ we have $$\begin{gathered}
{\ensuremath{{\operatorname{Res}}}}_{s=-k}\zeta_{{\uparrow \downarrow}}(P;s) =m {\ensuremath{{\operatorname{Res}}}}P^{k} = (-1)^{k+1}k!\int_{M}{{\operatorname{tr}}}_{{\ensuremath{\mathcal{E}}}}b_{k}(P)(x), \\
\lim_{s\rightarrow -k}[\zeta_{{\uparrow \downarrow}}(P;s)-m (s+k)^{-1} {\ensuremath{{\operatorname{Res}}}}P^{k}] = (-1)^{k}k! \int_{M}{{\operatorname{tr}}}_{{\ensuremath{\mathcal{E}}}} a_{d+2+mk}(P)(x).\end{gathered}$$ 3) For $k=0$ we have $$\zeta_{{\uparrow \downarrow}}(P;0)=\int_{M}{{\operatorname{tr}}}_{{\ensuremath{\mathcal{E}}}} a_{d+2}(P)(x)-\dim \ker P.
$$
Next, for $k=0,1,\ldots$ let $\lambda_{k}(P)$ denote the $(k+1)$’th eigenvalue of $P$ counted with multiplicity. Then by [@Po:MAMS1] and [@Po:CPDE1] as $k\rightarrow \infty$ we have the Weyl asymptotics, $$\lambda_{k}(P)\sim \left(\frac{k}{\nu_{0}(P)}\right)^{\frac{m}{d+2}}, \qquad \nu_{0}(P)=\Gamma(1+\frac{d+2}{m})^{-1} \int_{M}{{\operatorname{tr}}}_{{\ensuremath{\mathcal{E}}}}a_{0}(P)(x).
\label{eq:Zeta.Weyl-asymptotics1}$$
Now, by Proposition \[prop:Zeta.heat-zeta-global\] we have $$\int_{M}{{\operatorname{tr}}}_{{\ensuremath{\mathcal{E}}}}a_{0}(P)(x)=m\Gamma(\frac{d+2}{m}){\ensuremath{{\operatorname{Res}}}}P^{-\frac{d+2}{m}}=\frac{1}{d+2} \Gamma(1+\frac{d+2}{m}){\ensuremath{{\operatorname{Res}}}}P^{-\frac{d+2}{m}},
$$ Therefore, we obtain:
\[prop:Zeta.Weyl-asymptotics\] As $k\rightarrow \infty$ we have $$\lambda_{k}(P)\sim \left(\frac{k}{\nu_{0}(P)}\right)^{\frac{m}{d+2}}, \qquad \nu_{0}(P)= (d+2)^{-1}{\ensuremath{{\operatorname{Res}}}}P^{-\frac{d+2}{m}}.
$$
Finally, we can make use of Proposition \[prop:Zeta.heat-zeta-global\] to prove a local index formula for hypoelliptic [$\Psi_{H}$DOs]{} in the following setting. Assume that ${\ensuremath{\mathcal{E}}}$ admits a ${\ensuremath{\mathbb{Z}}}_{2}$-grading ${\ensuremath{\mathcal{E}}}={\ensuremath{\mathcal{E}}}^{+}\oplus {\ensuremath{\mathcal{E}}}_{-}$ and let $D:C^{\infty}(M,{\ensuremath{\mathcal{E}}})\rightarrow C^{\infty}(M,{\ensuremath{\mathcal{E}}})$ be a selfadjoint [$\Psi_{H}$DO]{} of integer order $m\geq 1$ with an invertible principal symbol and of the form, $$D= \left(
\begin{array}{cc}
0& D_{-} \\
D_{+}& 0
\end{array}
\right), \qquad D_{\pm}:C^{\infty}(M,{\ensuremath{\mathcal{E}}}_{\pm}) \rightarrow C^{\infty}(M,{\ensuremath{\mathcal{E}}}_{\mp}).$$ Notice that the selfadjointness of $D$ means that $D_{+}^{*}=D_{-}$.
Since $D$ has an invertible principal symbol and $M$ is compact we see that $D$ is invertible modulo finite rank operators, hence is Fredholm. Then we let $${{\operatorname{ind}}}D:= {{\operatorname{ind}}}D_{+}=\dim \ker D_{+}-\dim \ker D_{-}.
$$
Under the above assumptions we have $${{\operatorname{ind}}}D=\int_{M} {{\operatorname{str}}}_{{\ensuremath{\mathcal{E}}}} a_{d+2}(D^{2})(x),
$$ where ${{\operatorname{str}}}_{{\ensuremath{\mathcal{E}}}}:={{\operatorname{tr}}}_{{\ensuremath{\mathcal{E}}}^{+}}-{{\operatorname{tr}}}_{{\ensuremath{\mathcal{E}}}^{-}}$ denotes the supertrace on the fibers of ${\ensuremath{\mathcal{E}}}$.
We have $D^{2} = \left(
\begin{array}{cc}
D_{-}D_{+}& 0 \\
0 & D_{+}D_{-}
\end{array}
\right)$ and $D_{\mp} D_{\pm}=D_{\pm}^{*}D_{\pm}$. In particular, $D_{\mp} D_{\pm}$ is a positive operators with an invertible principal symbol. Moreover, for $\Re s >\frac{d+2}{2m}$ the difference $ \zeta(D_{-}D_{+};s) -\zeta(D_{+}D_{-};s) $ is equal to $$\sum_{\lambda>0} \lambda^s (\dim\ker (D_{-}D_{+} -\lambda) -
\dim\ker (D_{+}D_{-} -\lambda)) = 0,
$$ for $D$ induces for any $\lambda>0$ a bijection between $\ker (D_{-}D_{+}-\lambda)$ and $\ker
(D_{+}D_{-} -\lambda)$ (see, e.g., [@BGV:HKDO]). By analytic continuation this yields $ \zeta(D_{-}D_{+};0) -\zeta(D_{+}D_{-};0)=0$. On the other hand, by Proposition \[prop:Zeta.heat-zeta-global\] we have $$\zeta(D_{\mp} D_{\pm};0) = \int_{M} {{\operatorname{tr}}}_{{\ensuremath{\mathcal{E}}}_{\pm}}a_{d+2}(D_{\mp} D_{\pm})(x)-\dim \ker D_{\mp} D_{\pm}.
$$ Since $\dim \ker D_{\mp} D_{\pm}=\dim \ker D_{\pm}$ we deduce that ${{\operatorname{ind}}}D$ is equal to $$\int_{M} {{\operatorname{tr}}}_{{\ensuremath{\mathcal{E}}}_{+}}a_{d+2}(D_{+}D_{-})(x)- \int_{M} {{\operatorname{tr}}}_{{\ensuremath{\mathcal{E}}}_{-}}a_{d+2}(D_{-}D_{+})(x)= \int_{M} {{\operatorname{str}}}_{{\ensuremath{\mathcal{E}}}}a_{d+2}(D^{2})(x).
$$ The proof is thus achieved.
Metric estimates for Green kernels of hypoelliptic [$\Psi_{H}$DOs]{}
--------------------------------------------------------------------
Consider a compact Heisenberg manifold $(M^{d+1},H)$ endowed with a positive density and let ${\ensuremath{\mathcal{E}}}$ be a Hermitian vector bundle over $M$. In this subsection we shall prove that the positivity of a hypoelliptic [$\Psi_{H}$DO]{} pertains in its logarithmic singularity when it has order $-(\dim
M+1)$. As a consequence this will allow us to derive some metric estimates for Green kernels of hypoelliptic [$\Psi$DOs]{}.
Let $P:C^{\infty}(M,{\ensuremath{\mathcal{E}}})\rightarrow C^{\infty}(M,{\ensuremath{\mathcal{E}}})$ be a [$\Psi_{H}$DO]{} of order $m>0$ whose principal symbol is invertible and is positive in the sense of [@Po:MAMS1], i.e., we can write $\sigma_{m}(P)=q*q^{*}$ with $q\in S_{\frac{m}{2}}({\ensuremath{\mathfrak{g}}}^{*}M,{\ensuremath{\mathcal{E}}})$. The main technical result of this section is the following.
\[prop:Metric.positivity-cP\] The density ${{\operatorname{tr}}}_{{\ensuremath{\mathcal{E}}}}c_{P^{-\frac{d+2}{m}}}(x)$ is $>0$.
We will prove Proposition \[prop:Metric.positivity-cP\] later on in the section. As a first consequence, by combining with Proposition \[prop:Zeta.heat-zeta-local\] we get:
Let $a_{0}(P)(x)$ be the leading coefficient in the small time heat kernel asymptotics (\[eq:Zeta.heat-kernel-asymptotics\]) for $P$. Then the density ${{\operatorname{tr}}}_{{\ensuremath{\mathcal{E}}}}a_{0}(P)(x)$ is $>0$.
Assume now that the bracket condition $H+[H,H]=TM$ holds, i.e., $H$ is a Carnot-Carathéodory distribution in the sense of [@Gr:CCSSW]. Let $g$ be a Riemannian metric on $H$ and let $d_{H}(x,y)$ be the associated Carnot-Carathéodory metric on $M$. Recall that for two points $x$ and $y$ of $M$ the value of $d_{H}(x,y)$ is the infinum of the lengths of all closed paths joining $x$ to $y$ that are tangent to $H$ at each point (such a path always exists by Chow Lemma). Moreover, the Hausdorff dimension of $M$ with respect to $d_{H}$ is equal to $\dim M+1$.
In the setting of general Carathéodory distributions there has been lot of interest by Fefferman, Stein and their collaborators for giving metric estimates for the singularities of the Green kernels of hypoelliptic sublaplacians (see, e.g., [@FS:FSSOSO], [@Ma:EPKLCD], [@NSW:BMDVF1], [@Sa:FSGSSVF]). This allows us relate the analysis of the hypoelliptic sublaplacian to the metric geometry of the underlying manifold.
An important result is that it follows from the maximum principle of Bony [@Bo:PMIHUPCOED] that the Green of kernel of a selfadjoint hypoelliptic sublaplacian is positive near the diagonal. In general the positivity of the principal symbol does not pertain in the Green kernel. However, by making use of Proposition \[prop:Metric.positivity-cP\] we shall prove:
\[thm:Metric.metric-estimate\] Assume that $H+[H,H]=TM$ and let $P:C^{\infty}(M)\rightarrow C^{\infty}(M)$ be a [$\Psi_{H}$DO]{} of order $m>0$ whose principal symbol is invertible and is positive. Let $k_{P^{-\frac{d+2}{m}}}(x,y)$ be the Schwartz kernel of $P^{-\frac{d+2}{m}}$. Then near the diagonal we have $$k_{P^{-\frac{d+2}{m}}}(x,y)\sim -c_{P^{-\frac{d+2}{m}}}(x)\log d_{H}(x,y).
\label{eq:Metric.metric-estimate}$$ In particular $k_{P^{-\frac{d+2}{m}}}(x,y)$ is $>0$ near the diagonal.
It is enough to proceed in an open of $H$-framed local coordinates $U\subset {\ensuremath{{\ensuremath{\mathbb{R}}}^{d+1}}}$. For $x \in U$ let $\psi_{x}$ be the affine change to the corresponding privileged coordinates at $x$. Since by Proposition \[prop:Metric.positivity-cP\] we have $c_{P^{-\frac{d+2}{m}}}(x)>0$, using Proposition \[thm:NCR.log-singularity\] we see that near the diagonal we have $ k_{P^{-\frac{d+2}{m}}}(x,y) \sim -c_{P^{-\frac{d+2}{m}}}(x)\log\|\psi_{x}(y)\|$. Incidentally, we see that $k_{P^{-\frac{d+2}{m}}}(x,y)$ is positive near the diagonal.
On the other hand, since $H$ has codimension one our definition of the privileged coordinates agrees with that of [@Be:TSSRG]. Therefore, it follows from [@Be:TSSRG Thm. 7.34] that the ratio $\frac{d_{H}(x,y)}{\|\psi_{x}(y)\|}$ remains bounded in $(0, \infty)$ near the diagonal, that is, we have $\log d_{H}(x,y)\sim \log \|\psi_{x}(y)\|$. It then follows that near the diagonal we have $ k_{P^{-\frac{d+2}{m}}}(x,y)\sim -c_{P^{-\frac{d+2}{m}}}(x)\log d_{H}(x,y)$. The theorem is thus proved.
It remains now to prove Proposition \[prop:Metric.positivity-cP\]. To this end recall that for an operator $Q\in {\ensuremath{\Psi_{H}}}^{l}(M,{\ensuremath{\mathcal{E}}})$, $l\in {\ensuremath{\mathbb{C}}}$, the model operator $Q^{a}$ at a given point $a\in M$ is defined as the left-invariant [$\Psi_{H}$DO]{} on ${\ensuremath{\mathcal{S}}}_{0}(G_{a}M,{\ensuremath{\mathcal{E}}})$ with symbol $q^{a}(\xi)=\sigma_{l}(Q)(a,\xi)$. Bearing this in mind we have:
\[lem:Metric.cP-Heisenberg-coordinates\] Let $Q\in {\ensuremath{\Psi_{H}}}^{-(d+2)}(M,{\ensuremath{\mathcal{E}}})$ and let $Q^{a}$ be its model operator at a point $a \in M$.
1\) We have $c_{Q^{a}}(x)=c_{Q^{a}}dx$, where $c_{Q^{a}}$ is a constant and $dx$ denotes the Haar measure of $G_{a}M$.
2\) In Heisenberg coordinates centered at $a$ we have $c_{Q}(0)=c_{Q^{a}}$.
Let $X_{0},X_{1},\ldots,X_{d}$ be a $H$ frame near $a$. Since $G_{a}M$ has underlying set $(T_{a}M/H_{a})\oplus H_{a}$ the vectors $X_{0}(a),\ldots,X_{d}(a)$ define global coordinates for $G_{a}M$, so that we can identify it with ${\ensuremath{{\ensuremath{\mathbb{R}}}^{d+1}}}$ equipped with the group law (\[eq:Heisenberg.group-law-tangent-group-coordinates\]). In these coordinates set $q^{a}(\xi):=\sigma_{-(d+2)}(P)(a,\xi)$. Then (\[eq:PsiHDO.PsiDO-convolution\]) tells us that $Q^{a}$ corresponds to the operator $q^{a}(-iX^{a})$ acting on ${\ensuremath{\mathcal{S}}}_{0}({\ensuremath{{\ensuremath{\mathbb{R}}}^{d+1}}})$, where $X_{0}^{a},\ldots,X_{d}^{a}$ is the left-invariant tangent frame coming from the model vector fields at $a$ of $X_{0},\ldots,X_{d}$.
Notice that the left-invariance of the frame $X_{0}^{a},\ldots,X_{d}^{a}$ implies that, with respect to this frame, the affine change of variables to the privileged coordinates centered at any given point $x\in {\ensuremath{{\ensuremath{\mathbb{R}}}^{d+1}}}$ is just $\psi^{a}_{x}(y)=y.x^{-1}$. In view of (\[eq:Heisenberg.group-law-tangent-group-coordinates\]) this implies that $|\psi_{x}^{a'}|=1$. Therefore, from (\[eq:NCR.formula-cP\]) we get $$c_{Q^{a}}(x)=(2\pi)^{-(d+1)}\int_{\|\xi\|=1}q^{a}(\xi)\iota_{E}d\xi.
\label{eq:Metric.cQa}$$ Since the Haar measure of $G_{a}M$ corresponds to the Lebesgue measure of ${\ensuremath{{\ensuremath{\mathbb{R}}}^{d+1}}}$ this proves the 1st part of the lemma.
Next, by Definition \[def:Heisenberg.principal-symbol\] in Heisenberg coordinates centered at $a$ the principal symbol $\sigma_{-(d+2)}(Q)(x,\xi)$ agrees at $x=0$ with the principal symbol $q_{-(d+2)}(x,\xi)$ of $Q$ in the sense of (\[eq:Heisenberg.asymptotic-expansion-symbols\]), so we have $q^{a}(\xi)=q_{-(d+2)}(0,\xi)$. Furthermore, as we already are in Heisenberg coordinates, hence in privileged coordinates, we see that, with respect to the $H$-frame $X_{0},\ldots,X_{d}$, the affine change of variables $\psi_{0}$ to the privileged coordinates centered at the origin is just the identity. Therefore, by using (\[eq:NCR.formula-cP\]) and (\[eq:Metric.cQa\]) we see that $c_{Q}(0)$ is equal to $$(2\pi)^{-(d+1)}\int_{\|\xi\|=1}q(0,\xi)\iota_{E}d\xi=(2\pi)^{-(d+1)}\int_{\|\xi\|=1}q^{a}(\xi)\iota_{E}d\xi=c_{Q^{a}}.
$$ The 2nd part of the lemma is thus proved.
We are now ready to prove Proposition \[prop:Metric.positivity-cP\].
For sake of simplicity we may assume that ${\ensuremath{\mathcal{E}}}$ is the trivial line bundle, since in the general case the proof follows along similar lines. Moreover, for any $a \in M$ by Lemma \[lem:Metric.cP-Heisenberg-coordinates\] in Heisenberg coordinates centered at $a$ we have $c_{P^{-\frac{d+2}{m}}}(0)=c_{(P^{-\frac{d+2}{m}})^{a}}$. Therefore, it is enough to prove that $c_{(P^{-\frac{d+2}{m}})^{a}}$ is $>0$ for any $a\in M$.
Let $a \in M$ and let $X_{0},\ldots,X_{d}$ be a $H$-frame near $a$. By using the coordinates provided by the vectors $X_{0}(a),\ldots,X_{d}(a)$ we can identify $G_{a}M$ with ${\ensuremath{{\ensuremath{\mathbb{R}}}^{d+1}}}$ equipped with the group law (\[eq:Heisenberg.group-law-tangent-group-coordinates\]). We then let $H^{a}\subset T{\ensuremath{{\ensuremath{\mathbb{R}}}^{d+1}}}$ be the hyperplane bundle spanned by the model vector fields $X_{1}^{a},\ldots,X_{d}^{a}$ seen as left-invariant vector fields on ${\ensuremath{{\ensuremath{\mathbb{R}}}^{d+1}}}$. In addition, for any $z\in {\ensuremath{\mathbb{C}}}$ we let $p(z)(\xi):=\sigma_{z}(P^{\frac{z}{m}})(a,\xi)$ be the principal symbol at $a$ of $P^{\frac{z}{m}}$, seen as a homogeneous symbol on ${{\ensuremath{\mathbb{R}}}^{d+1}\!\setminus\! 0}$. Notice that by [@Po:MAMS1 Rem. 4.2.2] the family $(p(z))_{z \in {\ensuremath{\mathbb{C}}}}$ is a holomorphic family with values in $C^{\infty}({{\ensuremath{\mathbb{R}}}^{d+1}\!\setminus\! 0})$.
Let $\chi\in C^{\infty}_{c}({\ensuremath{{\ensuremath{\mathbb{R}}}^{d+1}}})$ be such that $\chi(\xi)=1$ near $\xi=0$. For any $z \in {\ensuremath{\mathbb{C}}}$ and for any pair $\varphi$ and $\psi$ of functions in $C_{c}^{\infty}({\ensuremath{{\ensuremath{\mathbb{R}}}^{d+1}}})$ we set $$\tilde{p}(z)(\xi):=(1-\chi)p(z) \qquad \text{and} \qquad P_{\varphi,\psi}(z):=\varphi \tilde{p}(z)(-iX^{a}) \psi.
$$ Then $(\tilde{p}(z))_{z\in {\ensuremath{\mathbb{C}}}}$ and $(P_{\varphi,\psi}(z))_{z\in {\ensuremath{\mathbb{C}}}}$ are holomorphic families with values in $S^{*}({\ensuremath{{\ensuremath{\mathbb{R}}}^{d+1}}})$ and $\Psi^{*}_{H^{a}}({\ensuremath{{\ensuremath{\mathbb{R}}}^{d+1}}})$ respectively.
Notice that $P_{\varphi,\psi}(z)$ has order $z$ and the support of its Schwartz kernel is contained in the fixed compact set ${{\operatorname{supp}}}\varphi \times {{\operatorname{supp}}}\psi$, so by Proposition \[prop:Heisenberg.L2-boundedness\] the operator $P_{\varphi,\psi}(z)$ is bounded on $L^{2}({\ensuremath{{\ensuremath{\mathbb{R}}}^{d+1}}})$ for $\Re z\leq 0$. In fact, by arguing as in the proof of [@Po:MAMS1 Prop. 4.6.2] we can show that $(P_{\varphi,\psi}(z))_{\Re z\leq 0}$ actually is a holomorphic family with values in ${\ensuremath{\mathcal{L}}}(L^{2}({\ensuremath{{\ensuremath{\mathbb{R}}}^{d+1}}}))$.
Moreover, by [@Po:MAMS1 Prop. 4.6.2] the family $(P_{\varphi,\psi}(\overline{z})^{*})_{z\in {\ensuremath{\mathbb{C}}}}$ is a holomorphic family with values in $\Psi^{*}_{H^{a}}({\ensuremath{{\ensuremath{\mathbb{R}}}^{d+1}}})$ such that ${{{\operatorname{ord}}}}P_{\varphi,\psi}(\overline{z})^{*}=z$ for any $z \in {\ensuremath{\mathbb{C}}}$. Therefore $(P_{\varphi,\psi}(z)P_{\varphi,\psi}(\overline{z})^{*})_{\Re z <-\frac{d+2}{2}}$ is a holomorphic family with values in $\Psi_{H^{a}}^{\text{int}}({\ensuremath{{\ensuremath{\mathbb{R}}}^{d+1}}})$. For any $z \in {\ensuremath{\mathbb{C}}}$ let $k(z)(x,y)$ denote the Schwartz kernel of $P_{\varphi,\psi}(z)P_{\varphi,\psi}(\overline{z})^{*}$. Then the support of $k(z)(x,y)$ is contained in the fixed compact set ${\operatorname{supp}}\varphi \times
{\operatorname{supp}}\varphi$, and by using \[eq:NCR.kP(x,x)\] we can check that $(k(z)(x,y))_{\Re z <-\frac{d+2}{2}}$ is a holomorphic family of continuous Schwartz kernels. It then follows that $(P_{\varphi,\psi}(z)P_{\varphi,\psi}(\overline{z})^{*})_{\Re z <-\frac{d+2}{2}}$ is a holomorphic family with values in the Banach ideal ${\ensuremath{\mathcal{L}}}^{1}(L^{2}({\ensuremath{{\ensuremath{\mathbb{R}}}^{d+1}}}))$ of trace-class operators on $L^{2}({\ensuremath{{\ensuremath{\mathbb{R}}}^{d+1}}})$.
Let us now choose $\psi$ so that $\psi =1$ near ${{\operatorname{supp}}}\varphi$. For any $t \in {\ensuremath{\mathbb{R}}}$ the operator $P^{\frac{t}{m}}$ is selfadjoint, so by Proposition \[prop:Heisenberg.operations-principal-symbols\] its principal symbol is real-valued. Therefore, by Proposition \[prop:Heisenberg.operations-principal-symbols\] the principal symbol of $(P_{\varphi,\psi}(t)P_{\varphi,\psi}(t)^{*})$ is equal to $$[ \varphi p(t)\psi]*^{a}[\overline{\psi}\overline{p(t)}\overline{\varphi}] =|\varphi|^{2}p(t)*p(t)=|\varphi|^{2}p(2t).
\label{eq:Metric.principal-symbol}$$ In particular, the principal symbols of $P_{\varphi,\psi}(-\frac{d+2}{2})P_{\varphi,\psi}(-\frac{d+2}{2})^{*}$ and $P_{|\varphi|^{2},\psi}(-(d+2))$ agree. By combining this with Lemma \[lem:Metric.cP-Heisenberg-coordinates\] we see that $$c_{P_{\varphi,\psi}(-\frac{d+2}{2})P_{\varphi,\psi}(-\frac{d+2}{2})^{*}}(x)=
c_{P_{|\varphi|^{2},\psi}(-(d+2))}(x)=|\varphi(x)|^{2}c_{(P^{-\frac{(d+2)}{m}})^{a}}.
\label{eq:Metric.cPphipsi-cPa}$$ It then follows from Proposition \[thm:NCR.TR.local\] that we have: $$\begin{gathered}
c_{(P^{-\frac{(d+2)}{m}})^{a}}(\int |\varphi(x)|^{2}dx) =\int c_{P_{\varphi,\psi}(-\frac{d+2}{2})P_{\varphi,\psi}(-\frac{d+2}{2})^{*}}(x) dx \\
=\lim_{t \rightarrow \frac{-(d+2)}{2}}\frac{-1}{t+\frac{d+2}{2}}\int t_{P_{\varphi,\psi}(t)P_{\varphi,\psi}(t)^{*})}(x) dx\\ =
\lim_{t \rightarrow [\frac{-(d+2)}{2}]^{-}}\frac{-1}{t+\frac{d+2}{2}} {\ensuremath{{\operatorname{Trace}}}}[P_{\varphi,\psi}(-t)P_{\varphi,\psi}(-t)^{*}] \geq 0.
\end{gathered}$$ Thus, by choosing $\varphi$ so that $\int |\varphi|^{2}>0$ we obtain that $c_{(P^{-\frac{(d+2)}{m}})^{a}}$ is $\geq 0$.
Assume now that $c_{(P^{-\frac{(d+2)}{m}})^{a}}$ vanishes, and let us show that this assumption leads us to a contradiction. Observe that $(P_{\varphi,\psi}(\frac{z-(d+2)}{2})P_{\varphi,\psi}(\frac{\overline{z-(d+2)}}{2})^{*})_{z\in {\ensuremath{\mathbb{C}}}}$ is holomorphic gauging for $P_{\varphi,\psi}(-\frac{d+2}{2})P_{\varphi,\psi}(-\frac{d+2}{2})^{*}$. Moreover, by (\[eq:Metric.cPphipsi-cPa\]) we have $c_{P_{\varphi,\psi}(-\frac{d+2}{2})P_{\varphi,\psi}(-\frac{d+2}{2})^{*}}(x)=|\varphi(x)|^{2}c_{(P^{-\frac{(d+2)}{m}})^{a}}=0$. Therefore, it follows from Proposition \[prop:Heisenberg.operations-principal-symbols\] that ${\ensuremath{{\operatorname{TR}}}}P_{\varphi,\psi}(z)P_{\varphi,\psi}(\overline{z})^{*}$ is analytic near $z=-\frac{d+2}{2}$. In particular, the limit $\lim_{t\rightarrow {\frac{-(d+2)}{2}}^{-}} {\ensuremath{{\operatorname{Trace}}}}P_{\varphi,\psi}(t)P_{\varphi,\psi}(t)^{*}$ exists and is finite.
Let $(\xi_{k})_{k\geq 0}$ be an orthonormal basis of $L^{2}({\ensuremath{{\ensuremath{\mathbb{R}}}^{d+1}}})$ and let $N\in {\ensuremath{\mathbb{N}}}$. For any $t>\frac{d+2}{2}$ the operator $P_{\varphi,\psi}(t)P_{\varphi,\psi}(t)^{*}$ is trace-class and we have $$\sum_{0\leq k \leq N}{\ensuremath{\langle P_{\varphi,\psi}(t)P_{\varphi,\psi}(t)^{*}\xi_{k} , \xi_{k} \rangle}} \leq {\ensuremath{{\operatorname{Trace}}}}[P_{\varphi,\psi}(t)P_{\varphi,\psi}(t)^{*}].
\label{eq:Metric.partial-trace}$$ As $t \rightarrow {-\frac{d+2}{2}}^{-}$ the operator $P_{\varphi,\psi}(t)P_{\varphi,\psi}(t)^{*}$ converges to $P_{\varphi,\psi}(-\frac{d+2}{2})P_{\varphi,\psi}(-\frac{d+2}{2})^{*}$ in ${\ensuremath{\mathcal{L}}}(L^{2}({\ensuremath{{\ensuremath{\mathbb{R}}}^{d+1}}})$. Therefore, letting $t$ go to ${-\frac{d+2}{2}}^{-}$ in (\[eq:Metric.partial-trace\]) shows that, for any $N\in {\ensuremath{\mathbb{N}}}$, we have $$\sum_{0\leq k \leq N}{\ensuremath{\langle P_{\varphi,\psi}(-\frac{d+2}{2})P_{\varphi,\psi}(-\frac{d+2}{2})^{*}\xi_{k} , \xi_{k} \rangle}}
\leq\lim_{t\rightarrow [\frac{-(d+2)}{2}]^{-}} {\ensuremath{{\operatorname{Trace}}}}[P_{\varphi,\psi}(t)P_{\varphi,\psi}(t)^{*}] <\infty.$$ This proves that $P_{\varphi,\psi}(-\frac{d+2}{2})P_{\varphi,\psi}(-\frac{d+2}{2})^{*}$ is a trace-class operator. Incidentally, we see that $P_{\varphi,\psi}(-\frac{d+2}{2})$ is a Hilbert-Schmidt operator on $L^{2}({\ensuremath{{\ensuremath{\mathbb{R}}}^{d+1}}})$.
Next, let $Q \in \Psi_{H^{a}}^{-\frac{d+2}{2}}({\ensuremath{{\ensuremath{\mathbb{R}}}^{d+1}}})$ and let $q(x,\xi)\in S_{-\frac{d+2}{2}}({\ensuremath{{\ensuremath{\mathbb{R}}}^{d+1}}}\times {\ensuremath{{\ensuremath{\mathbb{R}}}^{d+1}}})$ be the principal symbol of $Q$. The principal symbol of $\varphi Q\psi$ is $\varphi(x) q(x,\xi)$. Moreover, since for any $z \in {\ensuremath{\mathbb{C}}}$ we have $p(z)*p(-z)=p(0)=1$, we see that the principal symbol of $\psi Q\psi P_{\psi,\psi}(\frac{d+2}{2})P_{\varphi,\psi}(-\frac{d+2}{2})$ is equal to $$(\psi q \psi)*(\psi p(\frac{d+2}{2})\psi)*(\varphi p(-\frac{d+2}{2})\psi)= \varphi q*p(\frac{d+2}{2})*p(-\frac{d+2}{2})=\varphi q.
$$ Thus $\varphi Q\psi$ and $\psi Q\psi P_{\psi,\psi}(\frac{d+2}{2})P_{\varphi,\psi}(-\frac{d+2}{2})$ have the same principal symbol. Since they both have a compactly supported Schwartz kernel it follows that we can write $$\varphi Q\psi = \psi Q\psi P_{\psi,\psi}(\frac{d+2}{2})P_{\varphi,\psi}(-\frac{d+2}{2}) +Q_{1},
\label{eq:Metric.Hilbert-Schmidt-decomposition}$$ for some operator $Q_{1} \in \Psi_{H^{a}}^{-\frac{d+2}{2}-1}({\ensuremath{{\ensuremath{\mathbb{R}}}^{d+1}}})$ with a compactly supported Schwartz kernel. Observe that:
- the operator $ \psi Q\psi P_{\psi,\psi}(\frac{d+2}{2})$ is a zero’th order [$\Psi_{H}$DO]{} with a compactly supported Schwartz kernel, so this is a bounded operator on $L^{2}({\ensuremath{{\ensuremath{\mathbb{R}}}^{d+1}}})$;
- as above-mentioned $P_{\varphi,\psi}(-\frac{d+2}{2})$ is a Hilbert-Schmidt operator;
- as $Q_{1}^{*}Q_{1}$ belongs to $\Psi_{H,c}^{\text{int}}({\ensuremath{{\ensuremath{\mathbb{R}}}^{d+1}}})$ this is a trace-class operator, and so $Q_{1}$ is a Hilbert-Schmidt operator.
Since the space ${\ensuremath{\mathcal{L}}}^{2}(L^{2}({\ensuremath{{\ensuremath{\mathbb{R}}}^{d+1}}}))$ of Hilbert-Schmidt operators is a two-sided ideal, it follows from (\[eq:Metric.Hilbert-Schmidt-decomposition\]) and the above observations that $\varphi Q\psi $ is a Hilbert-Schmidt operator. In particular, by [@GK:ITLNSO p. 109] the Schwartz kernel of $\varphi Q\psi $ lies in $L^{2}({\ensuremath{{\ensuremath{\mathbb{R}}}^{d+1}}}\times {\ensuremath{{\ensuremath{\mathbb{R}}}^{d+1}}})$.
We now get a contradiction as follows. Let $Q\in \Psi_{H^{a}}^{-\frac{d+2}{2}}({\ensuremath{{\ensuremath{\mathbb{R}}}^{d+1}}})$ have Schwartz kernel, $$k_{Q}(x,y)=|{\psi_{x}^{a}}'| \|\psi_{x}^{a}(y)\|^{-\frac{d+2}{2}},
$$ where $\psi_{x}^{a}$ is the change to the privileged coordinates at $a$ with respect to the $H^{a}$-frame $X_{0}^{a},\ldots,X_{d}^{a}$ (this makes sense since $\|y\|^{-\frac{d+2}{2}}$ is in ${\ensuremath{\mathcal{K}}}_{-\frac{d+2}{2}}({\ensuremath{{\ensuremath{\mathbb{R}}}^{d+1}}}\times
{\ensuremath{{\ensuremath{\mathbb{R}}}^{d+1}}})$). As alluded to in the proof of Lemma \[lem:Metric.cP-Heisenberg-coordinates\] the left-invariance of the frame $X_{0}^{a},\ldots,X_{d}^{a}$ implies that $\psi_{x}^{a}(y)=y.x^{-1}$. Therefore, the Schwartz kernel of $\varphi Q\psi $ is equal to $$k_{\varphi Q\psi}(x,y)=\varphi(x) \|y.x^{-1}\|^{-\frac{d+2}{2}} \psi(y).
$$ However, this not an $L^{2}$-integrable kernel, since $\|y.x^{-1}\|^{-(d+2)}$ is not locally integrable near the diagonal.
We have obtained a contradiction, so $c_{(P^{-\frac{d+2}{m}})^{a}}$ cannot be zero. Since we know that $c_{(P^{-\frac{d+2}{m}})^{a}}$ is $\geq 0$, we see that $c_{(P^{-\frac{d+2}{m}})^{a}}$ is $>0$. The proof of Proposition \[prop:Metric.positivity-cP\] is thus complete.
The Dixmier trace of [$\Psi_{H}$DOs]{} {#sec:Dixmier}
--------------------------------------
The quantized calculus of Connes [@Co:NCG] allows us to translate into the language of quantum mechanics the main tools of the classical infinitesimal calculus. In particular, an important device is the Dixmier trace ([@Di:ETNN], [@CM:LIFNCG Appendix A]), which is the noncommutative analogue of the standard integral. We shall now show that, as in the case of classical [$\Psi$DOs]{} (see [@Co:AFNG]), the noncommutative residue allows us to extend the Dixmier trace to the whole algebra of integer order [$\Psi_{H}$DOs]{}.
Let us first recall the main facts about Connes’ quantized calculus and the Dixmier trace. The general setting is that of bounded operators on a separable Hilbert space ${\ensuremath{\mathcal{H}}}$. Extending the well known correspondence in quantum mechanics between variables and operators, we get the following dictionary between classical notions of infinitesimal calculus and their operator theoretic analogues.
Classical Quantum
----------------------------------- ------------------------------------------------------
Real variable Selfadjoint operator on ${\ensuremath{\mathcal{H}}}$
Complex variable Operator on ${\ensuremath{\mathcal{H}}}$
Infinitesimal variable Compact operator on ${\ensuremath{\mathcal{H}}}$
Infinitesimal of order $\alpha>0$ Compact operator $T$ such that
$\mu_{n}(T)={\operatorname{O}}(n^{-\alpha})$
The third line can be explained as follows. We cannot say that an operator $T$ is an infinitesimal by requiring that $\|T\| \leq \epsilon$ for any $\epsilon >0$, for this would give $T=0$. Nevertheless, we can relax this condition by requiring that for any $\epsilon>0$ we have $\|T\|<\epsilon$ outside a finite dimensional space. This means that $T$ is in the closure of finite rank operators, i.e., $T$ belongs to the ideal ${\ensuremath{\mathcal{K}}}$ of compact operators on ${\ensuremath{\mathcal{H}}}$.
In the last line $\mu_{n}(T)$ denotes the $(n+1)$’th characteristic value of $T$, i.e., the $(n+1)$’th eigenvalue of $|T|=(T^{*}T)^{\frac12}$. In particular, by the min-max principle we have $$\begin{aligned}
\mu_{n}(T) & = & \inf\{ \|T_{E^\perp}\|; \dim E=n\},
\nonumber \\
& = & {\operatorname{dist}}(T,\mathcal{R}_{n}) , \qquad \mathcal{R}_{n}=
\{\text{operators of rank}\leq n\},
\label{eq:NCG.min-max}
$$ so the decay of $\mu_{n}(T)$ controls the accuracy of the approximation of $T$ by finite rank operators. Moreover, by using (\[eq:NCG.min-max\]) we also can check that, for $S$, $T$ in ${\ensuremath{\mathcal{K}}}$ and $A$, $B$ in ${\ensuremath{\mathcal{L}}}({\ensuremath{\mathcal{H}}})$, we have $$\mu_{n}(T+S)\leq \mu_{n}(T)+\mu_{n}(S) \qquad \text{and} \qquad \mu_{n}(ATB)\leq \|A\| \mu_{n}(T) \|B\|,
\label{eq:NCG.inequalities-2sided-ideals}$$ This implies that the set of infinitesimal operators of order $\alpha$ is a two-sided ideal of ${\ensuremath{\mathcal{L}}}({\ensuremath{\mathcal{H}}})$.
Next, in this setting the analogue of the integral is provided by the Dixmier trace ([@Di:ETNN], [@CM:LIFNCG Appendix A]). The latter arises in the study of the logarithmic divergency of the partial traces, $${\ensuremath{{\operatorname{Trace}}}}_{N}(T) = \sum_{n=0}^{N- 1} \mu_{n}(T), \qquad T \in{\ensuremath{\mathcal{K}}}, \quad T\geq 0.$$ The domain of the Dixmier trace is the Schatten ideal, $${\ensuremath{\mathcal{L}^{(1,\infty)}}}=\{T\in {\ensuremath{\mathcal{K}}}; \|T\|_{1,\infty} :=\sup \frac{\sigma_{N}(T)}{\log N} < \infty\}.
$$
We extend the definition of $ {\ensuremath{{\operatorname{Trace}}}}_{N}(T)$ by means of the interpolation formula, $$\sigma_{\lambda}(T) =\inf \{\|x\|_{1}+ \lambda \|y\| ; x+y=T\}, \qquad \lambda>0,$$ where $\|x\|_{1}:={\ensuremath{{\operatorname{Trace}}}}|x|$ denotes the Banach norm of the ideal ${\ensuremath{\mathcal{L}}}^{1}$ of trace-class operators. For any integer $N$ we have $\sigma_{N}(T)={\ensuremath{{\operatorname{Trace}}}}_{N}(T)$. In addition, the Cesāro mean of $\sigma_{\lambda}(T)$ with respect to the Haar measure $\frac{d\lambda}{\lambda}$ of ${\ensuremath{\mathbb{R}}}_{+}^{*}$ is $$\tau_{\Lambda}(T) = \frac{1}{\log\Lambda}\int_{e}^\Lambda \frac{\sigma_{\lambda}(T)}
{\log \lambda}\frac{d\lambda}{\lambda}, \qquad \Lambda\geq e.$$
Let ${\ensuremath{\mathcal{L}}}({\ensuremath{\mathcal{H}}})_{+}=\{T \in {\ensuremath{\mathcal{L}}}({\ensuremath{\mathcal{H}}}); \ T\geq 0\}$. Then by [@CM:LIFNCG Appendix A] for $T_{1}$ and $T_{2}$ in ${\ensuremath{\mathcal{L}^{(1,\infty)}}}\cap {\ensuremath{\mathcal{L}}}({\ensuremath{\mathcal{H}}})_{+}$ we have $$|\tau_{\Lambda}(T_{1}+T_{2}) -\tau_{\Lambda}(T_{1}) -\tau_{\Lambda}(T_{2}) | \leq
3({\ensuremath{\|{T_{1}}\|_{(1,\infty)}}}+{\ensuremath{\|{T_{2}}\|_{(1,\infty)}}}) \frac{\log\log\Lambda}{\log\Lambda}.$$ Therefore, the functionals $\tau_{\Lambda}$, $\Lambda \geq e$, give rise to an additive homogeneous map, $$\tau: {\ensuremath{\mathcal{L}}}^{(1,\infty)}\cap {\ensuremath{\mathcal{L}}}({\ensuremath{\mathcal{H}}})_{+} \longrightarrow C_{b}[e,\infty)/C_{0}[e,\infty).
$$ It follows from this that for any state $\omega$ on the $C^{*}$-algebra $C_{b}[e,\infty)/C_{0}[e,\infty)$, i.e., for any positive linear form such that $\omega(1)=1$, there is a unique linear functional ${\ensuremath{{\operatorname{Tr}}_{\omega}}}:{\ensuremath{\mathcal{L}}}^{(1,\infty)}\rightarrow {\ensuremath{\mathbb{C}}}$ such that $${\ensuremath{{\operatorname{Tr}}_{\omega}}}T = \omega(\tau(T)) \qquad \forall T \in {\ensuremath{\mathcal{L}}}^{(1,\infty)}\cap {\ensuremath{\mathcal{L}}}({\ensuremath{\mathcal{H}}})_{+}.
$$
We gather the main properties of this functional in the following.
\[prop:NCG.properties-Dixmier-trace\] For any state $\omega$ on $C_{b}[e,\infty)/C_{0}[e,\infty)$ the Dixmier trace ${\ensuremath{{\operatorname{Tr}}_{\omega}}}$ has the following properties:
1\) If $T$ is trace-class, then ${\ensuremath{{\operatorname{Tr}}_{\omega}}}T=0$.
2\) We have ${\ensuremath{{\operatorname{Tr}}_{\omega}}}(T)\geq 0$ for any $T\in {\ensuremath{\mathcal{L}^{(1,\infty)}}}\cap {\ensuremath{\mathcal{L}}}({\ensuremath{\mathcal{H}}})_{+}$.
3\) If $S:{\ensuremath{\mathcal{H}}}'\rightarrow {\ensuremath{\mathcal{H}}}$ is a topological isomorphism, then we have ${\ensuremath{{\operatorname{Tr}}}}_{\omega,{\ensuremath{\mathcal{H}}}'}(T)={\ensuremath{{\operatorname{Tr}}}}_{\omega,{\ensuremath{\mathcal{H}}}}(STS^{-1})$ for any $T\in {\ensuremath{\mathcal{L}^{(1,\infty)}}}({\ensuremath{\mathcal{H}}}')$. In particular, ${\ensuremath{{\operatorname{Tr}}_{\omega}}}$ does not depend on choice of the inner product on ${\ensuremath{\mathcal{H}}}$.
4\) We have ${\ensuremath{{\operatorname{Tr}}_{\omega}}}AT={\ensuremath{{\operatorname{Tr}}_{\omega}}}TA$ for any $A \in {\ensuremath{\mathcal{L}}}({\ensuremath{\mathcal{H}}})$ and any $T\in {\ensuremath{\mathcal{L}^{(1,\infty)}}}$, that is, ${\ensuremath{{\operatorname{Tr}}_{\omega}}}$ is a trace on the ideal ${\ensuremath{\mathcal{L}^{(1,\infty)}}}$.
The functional ${\ensuremath{{\operatorname{Tr}}_{\omega}}}$ is called the *Dixmier trace* associated to $\omega$. We also say that an operator $T \in {\ensuremath{\mathcal{L}}}^{(1,\infty)}$ is *measurable* when the value of ${\ensuremath{{\operatorname{Tr}}_{\omega}}}T$ is independent of the choice of the state $\omega$. We then call *the Dixmier trace* of $T$ the common value, $${\ensuremath{-\hspace{-2,4ex}\int}}T :={\ensuremath{{\operatorname{Tr}}_{\omega}}}T.
$$ In addition, we let ${\ensuremath{\mathcal{M}}}$ denote the space of measurable operators. For instance, if $T\in {\ensuremath{\mathcal{K}}}\cap {\ensuremath{\mathcal{L}}}({\ensuremath{\mathcal{H}}})_{+}$ is such that $\lim_{N\rightarrow \infty}\frac{1}{\log N} \sum_{n=0}^{N-1} \mu_{n}(T) = L$, then it can be shown that $T$ is measurable and we have ${\ensuremath{-\hspace{-2,4ex}\int}}T=L$.
An important example of measurable operator is due to Connes [@Co:AFNG]. Let ${\ensuremath{\mathcal{H}}}$ be the Hilbert space $L^{2}(M,{\ensuremath{\mathcal{E}}})$ of $L^{2}$-sections of a Hermitian vector bundle over a compact manifold $M$ equipped with a smooth positive density and let $P:L^{2}(M,{\ensuremath{\mathcal{E}}})\rightarrow
L^{2}(M,{\ensuremath{\mathcal{E}}})$ be a classical [$\Psi$DO]{} of order $-\dim M$. Then $P$ is measurable for the Dixmier trace and we have $${\ensuremath{-\hspace{-2,4ex}\int}}P =\frac{1}{\dim M} {\ensuremath{{\operatorname{Res}}}}P,
\label{eq:NCG.Trw-NCR-PsiDOs}$$ where ${\ensuremath{{\operatorname{Res}}}}P$ denotes the noncommutative residue trace for classical [$\Psi$DOs]{} of Wodzicki ([@Wo:LISA], [@Wo:NCRF]) and Guillemin [@Gu:NPWF]. This allows us to extends the Dixmier trace to all [$\Psi$DOs]{} of integer order, hence to integrate any such [$\Psi$DO]{} even though it is not an infinitesimal of order $\leq 1$.
From now one we let $(M^{d+1},H)$ be a compact Heisenberg manifold equipped with a smooth positive density and we let ${\ensuremath{\mathcal{E}}}$ be a Hermitian vector bundle over $M$. In addition, we recall that by Proposition \[prop:Heisenberg.L2-boundedness\] any $P\in {\ensuremath{\Psi_{H}}}^{m}(M,{\ensuremath{\mathcal{E}}})$ with $\Re m\geq 0$ extends to a bounded operator from $L^{2}(M,{\ensuremath{\mathcal{E}}})$ to itself and this operator is compact if we further have $\Re m<0$.
Let $P:C^{\infty}(M,{\ensuremath{\mathcal{E}}}) \rightarrow C^{\infty}(M,{\ensuremath{\mathcal{E}}})$ be a positive [$\Psi_{H}$DO]{} with an invertible principal symbol of order $m>0$, and for $k=0,1,..$ let $\lambda_{k}(P)$ denote the $(k+1)$’ th eigenvalue of $P$ counted with multiplicity. By Proposition \[prop:Zeta.Weyl-asymptotics\] when $k \rightarrow \infty$ we have $$\lambda_{k}(P) \sim (\frac{k}{\nu_{0}(P)})^{\frac{m}{d+2}}, \qquad \nu_{0}(P)= \frac{1}{d+2}{\ensuremath{{\operatorname{Res}}}}P^{-\frac{d+2}{m}}.
$$ It follows that for any $\sigma \in {\ensuremath{\mathbb{C}}}$ with $\Re \sigma <0$ the operator $P^{\sigma}$ is an infinitesimal operator of order $\frac{m |\Re \sigma|}{d+2}$. Furthermore, for $\sigma=-\frac{d+2}{m}$ using (\[eq:NCG.inequalities-2sided-ideals\]) we see that $P^{-\frac{d+2}{m}}$ is measurable and we have $${\ensuremath{-\hspace{-2,4ex}\int}}P^{-\frac{d+2}{m}}=\nu_{0}(P)=\frac{1}{d+2}{\ensuremath{{\operatorname{Res}}}}P^{-\frac{d+2}{m}}.
\label{eq:NCG.Dixmier-trace-NCR.hypoelliptic}$$
These results are actually true for general [$\Psi_{H}$DOs]{}, for we have:
\[thm:NCG.Dixmier\] Let $P: L^{2}(M,{\ensuremath{\mathcal{E}}}) \rightarrow L^{2}(M,{\ensuremath{\mathcal{E}}})$ be a [$\Psi_{H}$DO]{} order $m$ with $\Re m<0$.
1\) $P$ is an infinitesimal operator of order $(\dim M+1)^{-1}|\Re m|$.
2\) If ${{{\operatorname{ord}}}}P=-(\dim M+1)$, then $P$ is measurable and we have $${\ensuremath{-\hspace{-2,4ex}\int}}P = \frac1{\dim M+1} {\ensuremath{{\operatorname{Res}}}}P.
\label{eq:NCG.bint-NCR}$$
First, let $P_{0}\in {\ensuremath{\Psi_{H}}}^{1}(M,{\ensuremath{\mathcal{E}}})$ be a positive and invertible [$\Psi_{H}$DO]{} with an invertible principal symbol (e.g. $P_{0}=(1+\Delta^{*}\Delta)^{\frac{1}{4}}$, where $\Delta$ is a hypoelliptic sublaplacian). Then $PP_{0}^{m}$ is a zeroth order [$\Psi_{H}$DO]{}. By Proposition \[\] any zeroth order [$\Psi_{H}$DO]{} is bounded on $L^{2}(M,{\ensuremath{\mathcal{E}}})$ and as above-mentioned $P^{-m}_{0}$ is an infinitesimal of order $\alpha:=(\dim M+1)^{-1}|\Re m|$. Since we have $P=PP_{0}^{m}.P_{0}^{-m}$ we see that $P$ is the product of a bounded operator and of an infinitesimal operator of order $\alpha$. As (\[eq:NCG.inequalities-2sided-ideals\]) shows that the space of infinitesimal operators of order $\alpha$ is a two-sided ideal, it follows that $P$ is an infinitesimal of order $\alpha$. In particular, if ${{{\operatorname{ord}}}}P=-(d+2)$ then $P$ is an infinitesimal of order $1$, hence is contained in ${\ensuremath{\mathcal{L}}}^{(1,\infty)}$.
Next, let ${\ensuremath{{\operatorname{Tr}}_{\omega}}}$ be the Dixmier trace associated to a state $\omega$ on $C_{b}[e,\infty)/C_{0}[e,\infty)$, and let us prove that for any $P \in
{\ensuremath{\Psi_{H}}}^{-(d+2)}(M,{\ensuremath{\mathcal{E}}})$ we have ${\ensuremath{{\operatorname{Tr}}_{\omega}}}P=\frac{1}{d+2}{\ensuremath{{\operatorname{Res}}}}P$.
Let $\kappa:U\rightarrow {\ensuremath{{\ensuremath{\mathbb{R}}}^{d+1}}}$ be a $H$-framed chart mapping onto ${\ensuremath{{\ensuremath{\mathbb{R}}}^{d+1}}}$ such that there is a trivialization $\tau:{\ensuremath{\mathcal{E}}}_{|U} \rightarrow U\times
{\ensuremath{\mathbb{C}}}^{r}$ of ${\ensuremath{\mathcal{E}}}$ over $U$ (as in the proof of Theorem \[thm:Traces.traces\] we shall call such a chart a *nice $H$-framed chart*). As in Subsection \[sec:traces\] we shall use the subscript $c$ to denote [$\Psi_{H}$DOs]{} with a compactly supported Schwartz kernel (e.g. ${\ensuremath{\Psi_{H,{\operatorname{c}}}}}^{{\ensuremath{\mathbb{Z}}}}({\ensuremath{{\ensuremath{\mathbb{R}}}^{d+1}}})$ denote the class of integer order [$\Psi_{H}$DOs]{} on ${\ensuremath{{\ensuremath{\mathbb{R}}}^{d+1}}}$ whose Schwartz kernels have compact supports). Notice that if $P\in{\ensuremath{\Psi_{H,{\operatorname{c}}}}}^{{\ensuremath{\mathbb{Z}}}}({\ensuremath{{\ensuremath{\mathbb{R}}}^{d+1}}},{\ensuremath{\mathbb{C}}}^{r})$ then the operator $\tau^{*}\kappa^{*}P$ belongs to ${\ensuremath{\Psi_{H}}}^{{\ensuremath{\mathbb{Z}}}}(M,{\ensuremath{\mathcal{E}}})$ and the support of its Schwartz kernel is a compact subset of $U\times U$.
Since $P_{0}$ is a positive [$\Psi_{H}$DO]{} with an invertible principal symbol, Proposition \[prop:Metric.positivity-cP\] tells us that the density ${{\operatorname{tr}}}_{{\ensuremath{\mathcal{E}}}}c_{P_{0}^{-(d+2)}}(x)$ is $>0$, so we can write $\kappa_{*}[{{\operatorname{tr}}}_{{\ensuremath{\mathcal{E}}}}c_{P_{0}^{-(d+2)}}(x)_{|U}]=c_{0}(x)dx$ for some positive function $c_{0}\in C^{\infty}({\ensuremath{{\ensuremath{\mathbb{R}}}^{d+1}}})$. Then for any $c \in C_{c}^{\infty}({\ensuremath{{\ensuremath{\mathbb{R}}}^{d+1}}})$ and any $\psi \in C_{c}^{\infty}({\ensuremath{{\ensuremath{\mathbb{R}}}^{d+1}}})$ such that $\psi=1$ near ${{\operatorname{supp}}}c$ we let $$P_{c,\psi}:=(\frac{c\circ \kappa}{c_{0}\circ \kappa})P_{0}^{-(d+2)} (\psi\circ \kappa).
\label{eq:Dixmier.Pcpsi}$$ Notice that $P_{c,\psi}$ belongs to ${\ensuremath{\Psi_{H}}}^{-(d+2)}(M,{\ensuremath{\mathcal{E}}})$ and it depends on the choice $\psi$ only modulo operators in ${\ensuremath{\Psi^{-\infty}}}(M,{\ensuremath{\mathcal{E}}})$. Since the latter are trace-class operators and the Dixmier trace ${\ensuremath{{\operatorname{Tr}}}}_{\omega}$ vanishes on such operators (cf. Proposition \[prop:NCG.properties-Dixmier-trace\]), we see that the value of ${\ensuremath{{\operatorname{Tr}}_{\omega}}}P_{c,\psi}$ does not depend on the choice of $\psi$. Therefore, we define a linear functional $L:C_{c}^{\infty}({\ensuremath{{\ensuremath{\mathbb{R}}}^{d+1}}}) \rightarrow {\ensuremath{\mathbb{C}}}$ by assigning to any $c \in C_{c}^{\infty}({\ensuremath{{\ensuremath{\mathbb{R}}}^{d+1}}})$ the value $$L(c):={\ensuremath{{\operatorname{Tr}}_{\omega}}}P_{c,\psi},
$$ where $\psi \in C_{c}^{\infty}({\ensuremath{{\ensuremath{\mathbb{R}}}^{d+1}}})$ is such that $\psi=1$ near ${{\operatorname{supp}}}c$.
On the other hand, let $P\in {\ensuremath{\Psi_{H,{\operatorname{c}}}}}^{-(d+2)}(U,{\ensuremath{\mathcal{E}}}_{|U})$. Then $\tau_{*}P$ belongs to ${\ensuremath{\Psi_{H,{\operatorname{c}}}}}^{-(d+2)}(U,{\ensuremath{\mathbb{C}}}^{r}):={\ensuremath{\Psi_{H,{\operatorname{c}}}}}^{-(d+2)}(U)\otimes M_{r}({\ensuremath{\mathbb{C}}})$. Set $\tau_{*}P=(P_{ij})$ and define ${{\operatorname{tr}}}P:=\sum P_{ii}$. In addition, for $i,j=1,\ldots,r$ let $E_{ij}\in M_{r}({\ensuremath{\mathbb{C}}})$ be the elementary matrix whose all entries are zero except that on the $i$th row and $j$th column which is equal to $1$. Then we have $$\tau_{*}P=\frac{1}{r}({{\operatorname{tr}}}P)\otimes I_{r} + \sum_{i} P_{ii}\otimes (E_{ii}-\frac{1}{r}I_{r}) +\sum_{i\neq j} P_{ij}\otimes E_{ij}.
$$ Any matrix $A \in M_{r}({\ensuremath{\mathbb{C}}})$ with vanishing trace is contained in the commutator space $[M_{r}({\ensuremath{\mathbb{C}}}),M_{r}({\ensuremath{\mathbb{C}}})]$. Notice also that the space ${\ensuremath{\Psi_{H,{\operatorname{c}}}}}^{-(d+2)}(U)\otimes [M_{r}({\ensuremath{\mathbb{C}}}),M_{r}({\ensuremath{\mathbb{C}}})]$ is contained in $[{\ensuremath{\Psi_{H,{\operatorname{c}}}}}^{0}(U,{\ensuremath{\mathbb{C}}}^{r}),{\ensuremath{\Psi_{H,{\operatorname{c}}}}}^{-(d+2)}(U,{\ensuremath{\mathbb{C}}}^{r})]$. Therefore, we see that $$P= \frac{1}{r}({{\operatorname{tr}}}P)\otimes {\operatorname{id}}_{{\ensuremath{\mathcal{E}}}} \mod [{\ensuremath{\Psi_{H,{\operatorname{c}}}}}^{0}(U,{\ensuremath{\mathcal{E}}}_{|U}),{\ensuremath{\Psi_{H,{\operatorname{c}}}}}^{-(d+2)}(U,{\ensuremath{\mathcal{E}}}_{|U})].
\label{eq:Dixmier.decomposition-P}$$
Let us write $\kappa_{*}[{{\operatorname{tr}}}_{{\ensuremath{\mathcal{E}}}}c_{P}(x)]=a_{P}(x)dx$ with $a_{P} \in C^{\infty}_{c}({\ensuremath{{\ensuremath{\mathbb{R}}}^{d+1}}})$, and let $\psi \in C_{c}^{\infty}({\ensuremath{{\ensuremath{\mathbb{R}}}^{d+1}}})$ be such that $\psi=1$ near ${{\operatorname{supp}}}a_{P}$. Then we have $$\kappa_{*}[c_{{{\operatorname{tr}}}P_{a_{P},\psi}}(x)=(\frac{a_{P}(x)}{c_{0}(x)})\psi(x)\kappa_{*}[{{\operatorname{tr}}}_{{\ensuremath{\mathcal{E}}}}c_{P_{0}^{-(d+2)}}(x)]=a_{P}(x)dx=\kappa_{*}[{{\operatorname{tr}}}_{{\ensuremath{\mathcal{E}}}}
c_{P}(x)]=\kappa_{*}[c_{{{\operatorname{tr}}}P}(x)].$$ In other words $Q:={{\operatorname{tr}}}P-{{\operatorname{tr}}}P_{a_{P},\psi}$ is an element of ${\ensuremath{\Psi_{H,{\operatorname{c}}}}}^{-(d+2)}(U)$ such that $c_{Q}(x)=0$. By the step (i) of the proof of Lemma \[lem:Traces.sum-commutators.compact\] we then can write $\kappa_{*}Q$ in the form $\kappa_{*}Q=[\chi_{0},Q_{0}]+\ldots+[\chi_{d},Q_{d}]$ for some functions $\chi_{0},\ldots,\chi_{d}$ in $C^{\infty}_{c}({\ensuremath{{\ensuremath{\mathbb{R}}}^{d+1}}})$ and some operators $Q_{0},\ldots,Q_{d}$ in ${\ensuremath{\Psi_{H,{\operatorname{c}}}}}^{{\ensuremath{\mathbb{Z}}}}({\ensuremath{{\ensuremath{\mathbb{R}}}^{d+1}}})$. In fact, it follows from the proof of Lemmas \[lem:Traces.sum-commutators\] and \[lem:Traces.sum-commutators.compact\] that $Q_{0},..,Q_{d}$ can be chosen to have order $\leq -(d+2)$. This insures us that $\kappa_{*}Q$ is contained in $[{\ensuremath{\Psi_{H,{\operatorname{c}}}}}^{0}({\ensuremath{{\ensuremath{\mathbb{R}}}^{d+1}}}),{\ensuremath{\Psi_{H,{\operatorname{c}}}}}^{-(d+2)}({\ensuremath{{\ensuremath{\mathbb{R}}}^{d+1}}})]$. Thus, $${{\operatorname{tr}}}P= {{\operatorname{tr}}}P_{a_{P},\psi} \mod [{\ensuremath{\Psi_{H,{\operatorname{c}}}}}^{0}(U),{\ensuremath{\Psi_{H,{\operatorname{c}}}}}^{-(d+2)}(U)].$$ By combining this with (\[eq:Dixmier.decomposition-P\]) we obtain $$P= \frac{1}{r}({{\operatorname{tr}}}P)\otimes {\operatorname{id}}_{{\ensuremath{\mathcal{E}}}} = \frac{1}{r}({{\operatorname{tr}}}P_{a_{P},\psi})\otimes {\operatorname{id}}_{{\ensuremath{\mathcal{E}}}} =P_{a_{P},\psi} \mod
[{\ensuremath{\Psi_{H,{\operatorname{c}}}}}^{0}(U,{\ensuremath{\mathcal{E}}}_{|U}),{\ensuremath{\Psi_{H,{\operatorname{c}}}}}^{-(d+2)}(U,{\ensuremath{\mathcal{E}}}_{|U})].
$$ Notice that $[{\ensuremath{\Psi_{H,{\operatorname{c}}}}}^{0}(U,{\ensuremath{\mathcal{E}}}_{|U}),{\ensuremath{\Psi_{H,{\operatorname{c}}}}}^{-(d+2)}(U,{\ensuremath{\mathcal{E}}}_{|U})]$ is contained in $[{\ensuremath{\Psi_{H}}}^{0}(M,{\ensuremath{\mathcal{E}}}),{\ensuremath{\Psi_{H}}}^{-(d+2)}(M,{\ensuremath{\mathcal{E}}})]$, which is itself contained in the commutator space $[{\ensuremath{\mathcal{L}}}(L^{2}(M)),{\ensuremath{\mathcal{L}}}^{(1,\infty)}(M)]$ of ${\ensuremath{\mathcal{L}^{(1,\infty)}}}$. As the Dixmier trace ${\ensuremath{{\operatorname{Tr}}_{\omega}}}$ vanishes on the latter space (cf. Proposition \[prop:NCG.properties-Dixmier-trace\]) we deduce that $${\ensuremath{{\operatorname{Tr}}_{\omega}}}P={\ensuremath{{\operatorname{Tr}}_{\omega}}}P_{a_{P},\psi}=L(a_{P}).
\label{eq:NCG.Trw-tau-cP}$$
Now, let $c \in C^{\infty}_{c}({\ensuremath{{\ensuremath{\mathbb{R}}}^{d+1}}})$ and set $c_{1}=
\frac{c}{\sqrt{c_{0}(x)}}$. In addition, let $\psi \in C^{\infty}_{c}({\ensuremath{{\ensuremath{\mathbb{R}}}^{d+1}}})$ be such that $\psi\geq 0$ and $\psi=1$ near ${{\operatorname{supp}}}c$, and set $\tilde{c}_{1}=c\circ \kappa$ and $\tilde{\psi}=\psi\circ\kappa$. Notice that with the notation of (\[eq:Dixmier.Pcpsi\]) we have $\overline{\tilde{c}_{1}}\tilde{c}_{1}P_{0}^{-(d+2)}\tilde{\psi}=P_{|c|^{2},\psi}$. Observe also that we have $$(\tilde{c}_{1}P_{0}^{-\frac{d+2}{2}}\tilde{\psi})(\tilde{c}_{1}P_{0}^{-\frac{d+2}{2}}\tilde{\psi})^{*}
=\tilde{c}_{1}P_{0}^{-\frac{d+2}{2}}\tilde{\psi}^{2}P_{0}^{-\frac{d+2}{2}}\overline{\tilde{c}_{1}}
= \tilde{c}_{1}P_{0}^{-(d+2)}\psi \overline{\tilde{c}_{1}} \mod \Psi^{-\infty}(M,{\ensuremath{\mathcal{E}}}).
$$ As alluded to earlier the trace ${\ensuremath{{\operatorname{Tr}}_{\omega}}}$ vanishes on smoothing operators, so we get $$\begin{gathered}
{\ensuremath{{\operatorname{Tr}}_{\omega}}}[ (\tilde{c}_{1}P_{0}^{-\frac{d+2}{2}}\tilde{\psi})(\tilde{c}_{1}P_{0}^{-\frac{d+2}{2}}\tilde{\psi})^{*} ]=
{\ensuremath{{\operatorname{Tr}}_{\omega}}}[\tilde{c}_{1}P_{0}^{-(d+2)}\tilde{\psi} \overline{\tilde{c}_{1}}]\\ ={\ensuremath{{\operatorname{Tr}}_{\omega}}}[\overline{\tilde{c}_{1}}\tilde{c}_{1}P_{0}^{-(d+2)}\tilde{\psi} ] = {\ensuremath{{\operatorname{Tr}}_{\omega}}}P_{|c|^{2},\psi}=L(|c|).
$$ Since ${\ensuremath{{\operatorname{Tr}}_{\omega}}}$ is a positive trace (cf. Proposition \[prop:NCG.properties-Dixmier-trace\]) it follows that we have $L(|c|^{2})\geq 0$ for any $c \in C^{\infty}_{c}({\ensuremath{{\ensuremath{\mathbb{R}}}^{d+1}}})$, i.e., $L$ is a positive linear functional on $C_{c}^{\infty}({\ensuremath{{\ensuremath{\mathbb{R}}}^{d+1}}})$. Since any such functional uniquely extends to a Radon measure on $C_{0}^{\infty}({\ensuremath{{\ensuremath{\mathbb{R}}}^{d+1}}})$, this shows that $L$ defines a positive Radon measure.
Next, let $a \in {\ensuremath{{\ensuremath{\mathbb{R}}}^{d+1}}}$ and let $\phi(x)=x+a$ be the translation by $a$ on ${\ensuremath{{\ensuremath{\mathbb{R}}}^{d+1}}}$. Since $\phi'(x)=1$ we see that $\phi$ is a Heisenberg diffeomorphism, so for any $P \in {\ensuremath{\Psi_{H,{\operatorname{c}}}}}^{*}({\ensuremath{{\ensuremath{\mathbb{R}}}^{d+1}}})$ the operator $\phi_{*}P$ is in ${\ensuremath{\Psi_{H,{\operatorname{c}}}}}^{*}({\ensuremath{{\ensuremath{\mathbb{R}}}^{d+1}}})$ too. Set $\phi_{\kappa}=\kappa^{-1} \circ \phi \circ \kappa$. Then by (\[eq:Log.functoriality-cP\]) we have $$\kappa_{*}[ {{\operatorname{tr}}}_{{\ensuremath{\mathcal{E}}}} c_{\phi_{\kappa*}P_{c,\psi}}(x)]=\kappa_{*}\phi_{\kappa*}[{{\operatorname{tr}}}_{{\ensuremath{\mathcal{E}}}}c_{P_{c,\psi}}(x)]=
\phi_{*}[c(x)dx]=c(\phi^{-1}(x))dx.
$$ Since shows that $a_{\phi_{\kappa*}P_{c,\psi}}(x)= c(\phi^{-1}(x))$, so from (\[eq:NCG.Trw-tau-cP\]) we get $${\ensuremath{{\operatorname{Tr}}_{\omega}}}\phi_{\kappa*}P_{c,\psi}=L[c\circ \phi^{-1}].
\label{eq:Dixmier.Lcphi}$$
Let $K$ be a compact subset of ${\ensuremath{{\ensuremath{\mathbb{R}}}^{d+1}}}$. Then $\phi_{\kappa}$ gives rise to a continuous linear isomorpshism $\phi_{\kappa*}:L^{2}_{\kappa^{-1}(K)}(M,{\ensuremath{\mathcal{E}}})\rightarrow L^{2}_{\kappa^{-1}(K+a)}(M,{\ensuremath{\mathcal{E}}})$. By combining it with a continuous linear isomorphism $L^{2}_{\kappa^{-1}(K)}(M,{\ensuremath{\mathcal{E}}})^{\perp}\rightarrow L^{2}_{\kappa^{-1}(K+a)}(M,{\ensuremath{\mathcal{E}}})^{\perp}$ we obtain a continuous linear isomorphism $S:L^{2}(M,{\ensuremath{\mathcal{E}}})\rightarrow L^{2}(M,{\ensuremath{\mathcal{E}}})$ which agrees with $\phi_{\kappa*}$ on $L^{2}_{\kappa^{-1}(K)}(M,{\ensuremath{\mathcal{E}}})$. In particular, we have $\phi_{\kappa*}P_{c,\psi} = S P_{c,\psi}S^{-1}$. Therefore, by using Proposition \[prop:NCG.properties-Dixmier-trace\] and \[eq:Dixmier.Lcphi\] we see that, for any $c \in C_{K}^{\infty}({\ensuremath{{\ensuremath{\mathbb{R}}}^{d+1}}})$, we have $$L[c]={\ensuremath{{\operatorname{Tr}}_{\omega}}}P_{c,\psi}={\ensuremath{{\operatorname{Tr}}_{\omega}}}S P_{c,\psi}S^{-1}= {\ensuremath{{\operatorname{Tr}}_{\omega}}}\phi_{\kappa*}P_{c,\psi}=L[c\circ \phi^{-1}].
$$ This proves that $L$ is translation-invariant. Since any translation invariant Radon measure on $C^{\infty}_{c}({\ensuremath{{\ensuremath{\mathbb{R}}}^{d+1}}})$ is a constant multiple of the Lebesgue measure, it follows that there exists a constant $\Lambda_{U}\in {\ensuremath{\mathbb{C}}}$ such that, for any $c \in C^{\infty}_{c}({\ensuremath{{\ensuremath{\mathbb{R}}}^{d+1}}})$, we have $$L(c)=\Lambda_{U} \int c(x)dx.
\label{eq:Dixmier.L-Lebesgue}$$
Now, combining (\[eq:NCG.Trw-tau-cP\]) and (\[eq:Dixmier.L-Lebesgue\]) shows that, for any $P\in {\ensuremath{\Psi_{H,{\operatorname{c}}}}}^{-(d+2)}(U,{\ensuremath{\mathcal{E}}}_{|U})$, we have $$\begin{gathered}
{\ensuremath{{\operatorname{Tr}}_{\omega}}}P= \Lambda_{U}\int_{{\ensuremath{{\ensuremath{\mathbb{R}}}^{d+1}}}} a_{P}(x)dx= \Lambda_{U}\int_{{\ensuremath{{\ensuremath{\mathbb{R}}}^{d+1}}}} \kappa_{*}[{{\operatorname{tr}}}_{{\ensuremath{\mathcal{E}}}}c_{P}(x)] \\
=\Lambda_{U}\int_{M}{{\operatorname{tr}}}_{{\ensuremath{\mathcal{E}}}}c_{P}(x)=(2\pi)^{d+1}\Lambda_{U}{\ensuremath{{\operatorname{Res}}}}P.
\label{eq:Dixmier-Trw-Res}\end{gathered}$$ This shows that, for any domain $U$ of a nice $H$-framed chart, on ${\ensuremath{\Psi_{H,{\operatorname{c}}}}}^{-(d+2)}(U,{\ensuremath{\mathcal{E}}}_{|U})$ the Dixmier trace ${\ensuremath{{\operatorname{Tr}}_{\omega}}}$ is a constant multiple of the noncommutative residue. Therefore, if we let $M_{1}, \ldots, M_{N}$ be the connected components of $M$, then by arguing as in the proof of Theorem \[thm:Traces.traces\] we can prove that on each connected component $M_{j}$ there exists a constant $\Lambda_{j}\geq 0$ such that $${\ensuremath{{\operatorname{Tr}}_{\omega}}}P =\Lambda_{j} {\ensuremath{{\operatorname{Res}}}}P \qquad \forall P\in {\ensuremath{\Psi_{H}}}^{-(d+2)}(M_{j},{\ensuremath{\mathcal{E}}}_{|M_{j}}).
$$ In fact, if we take $P=P_{0|_{M_{j}}}^{-(d+2)}$ then from (\[eq:NCG.Dixmier-trace-NCR.hypoelliptic\]) we get $\Lambda_{j}=(d+2)^{-1}$. Thus, $${\ensuremath{{\operatorname{Tr}}_{\omega}}}P =\frac{1}{d+2}{\ensuremath{{\operatorname{Res}}}}P \qquad \forall P \in {\ensuremath{\Psi_{H}}}^{-(d+2)}(M,{\ensuremath{\mathcal{E}}}).
$$ This proves that any operator $P \in {\ensuremath{\Psi_{H}}}^{-(d+2)}(M,{\ensuremath{\mathcal{E}}})$ is measurable and its Dixmier trace then is equal to $(d+2)^{-1}{\ensuremath{{\operatorname{Res}}}}P$. The theorem is thus proved.
As a consequence of Theorem \[thm:NCG.Dixmier\] we can extend the Dixmier trace to the whole algebra ${\ensuremath{\Psi_{H}^{{\ensuremath{\mathbb{Z}}}}}}(M,{\ensuremath{\mathcal{E}}})$ by letting $${\ensuremath{-\hspace{-2,4ex}\int}}P :=\frac{1}{d+2}{\ensuremath{{\operatorname{Res}}}}P \quad \text{for any $P\in {\ensuremath{\Psi_{H}^{{\ensuremath{\mathbb{Z}}}}}}(M,{\ensuremath{\mathcal{E}}})$}.
$$
In the language of the quantized calculus this means that we can integrate any [$\Psi_{H}$DO]{} of integer order, even though it is not an infinitesimal operator of order $\geq 1$. This property will be used in Section \[sec:CR\] to define lower dimensional volumes in pseudohermitian geometry.
Noncommutative residue and contact geometry {#sec:Contact}
===========================================
In this section we make use of the results of [@Po:MAMS1] to compute the noncommutative residues of some geometric operators on contact manifolds.
Throughout this section we let $(M^{2n+1},H)$ be a compact orientable contact manifold, i.e., $(M^{2n+1},H)$ is a Heisenberg manifold and there exists a contact 1-form $\theta$ on $M$ such that $H=\ker \theta$ (cf. Section \[sec:Heisenberg-calculus\]).
Since $M$ is orientable the hyperplane $H$ admits an almost complex structure $J\in C^{\infty}(M,{\ensuremath{{\operatorname{End}}}}H)$, $J^{2}=-1$, which is calibrated with respect to $\theta$, i.e., $d\theta(.,J.)$ is positive definite on $H$. We then can endow $M$ with the Riemannian metric, $$g_{\theta,J}=\theta^{2}+d\theta(.,J.).
\label{eq:Contact.Riem-metric}$$ The volume of $M$ with respect to $g_{\theta,J}$ depends only on $\theta$ and is equal to $${{\operatorname{Vol}}}_{\theta}M:=\frac{1}{n!}\int_{M}d\theta^{n}\wedge \theta.
$$
In addition, we let $X_{0}$ be the *Reeb field* associated to $\theta$, that is, the unique vector field on $M$ such that $\iota_{X_{0}}\theta=1$ and $\iota_{X_{0}}d\theta=0$.
Noncommutative residue and the horizontal sublaplacian (contact case)
---------------------------------------------------------------------
In the sequel we shall identify $H^{*}$ with the subbundle of $T^{*}M$ annihilating the orthogonal complement $H^{\perp}\subset TM$. This yields the orthogonal splitting, $$\Lambda_{{\ensuremath{\mathbb{C}}}}T^{*}M=(\bigoplus_{0\leq k \leq 2n} \Lambda^{k}_{{\ensuremath{\mathbb{C}}}}H^{*})\oplus (\theta\wedge \Lambda^{*}T_{{\ensuremath{\mathbb{C}}}}^{*}M).
$$ The horizontal differential $d_{b;k}:C^{\infty}(M,\Lambda^{k}_{{\ensuremath{\mathbb{C}}}}H^{*})\rightarrow C^{\infty}(M,\Lambda^{k+1}_{{\ensuremath{\mathbb{C}}}}H^{*})$ is $$d_{b}=\pi_{b;k+1}\circ d,
$$ where $\pi_{b;k}\in C^{\infty}(M,{\ensuremath{{\operatorname{End}}}}\Lambda_{{\ensuremath{\mathbb{C}}}}T^{*}M)$ denotes the orthogonal projection onto $\Lambda^{k}_{{\ensuremath{\mathbb{C}}}}H^{*}$. This is not the differential of a chain complex, for we have $$d_{b}^{2}=-{\ensuremath{\mathcal{L}}}_{X_{0}}\varepsilon(d\theta)=-\varepsilon (d\theta){\ensuremath{\mathcal{L}}}_{X_{0}},
\label{eq:CR.square-db}$$ where $\varepsilon (d\theta)$ denotes the exterior multiplication by $d\theta$.
The horizontal sublaplacian $\Delta_{b;k}:C^{\infty}(M,\Lambda^{k}_{{\ensuremath{\mathbb{C}}}}H^{*})\rightarrow C^{\infty}(M,\Lambda^{k+1}_{{\ensuremath{\mathbb{C}}}}H^{*})$ is $$\Delta_{b;k}=d_{b;k}^{*}d_{b;k}+d_{b;k-1}d_{b;k-1}^{*}.
\label{eq:CR.horizontal-sublaplacian}$$ Notice that the definition of $\Delta_{b}$ makes sense on any Heisenberg manifold equipped with a Riemannian metric. This operator was first introduced by Tanaka [@Ta:DGSSPCM], but versions of this operator acting on functions were independently defined by Greenleaf [@Gr:FESPM] and Lee [@Le:FMPHI]. Since the fact that $(M,H)$ is a contact manifold implies that the Levi form (\[eq:Heisenberg.Levi-form\]) is nondegenerate, from [@Po:MAMS1 Prop. 3.5.4] we get:
The principal symbol of $\Delta_{b;k}$ is invertible if and only if we have $k\neq n$.
Next, for $\mu\in (-n,n)$ we let $$\rho(\mu)= \frac{\pi^{-(n+1)}}{2^{n}n!} \int_{-\infty}^{\infty}e^{-\mu\xi_{0}}(\frac{\xi_{0}}{\sinh \xi_{0}})^{n}d\xi_{0}.$$ Notice that with the notation of [@Po:MAMS1 Eq. (6.2.29)] we have $\rho(\mu)=(2n+2)\nu(\mu)$. For $q \neq n$ let $\nu_{0}(\Delta_{b;k})$ be the coefficient $\nu_{0}(P)$ in the Weyl asymptotics (\[eq:Zeta.Weyl-asymptotics1\]) for $\Delta_{b;k}$, i.e., we have ${\ensuremath{{\operatorname{Res}}}}\Delta_{b;k}^{-(n+1)}=(2n+2)\nu_{0}(\Delta_{b;k})$. By [@Po:MAMS1 Prop. 6.3.3] we have $\nu_{0}(\Delta_{b;k})=\tilde{\gamma}_{nk}{{\operatorname{Vol}}}_{\theta}M$, where $\tilde{\gamma}_{nk}:=\sum_{p+q=k}2^{n}\binom{n}{p} \binom{n}{q}\nu(p-q)$. Therefore, we get:
\[prop:Contact.residue-Deltab\] For $k \neq n$ we have $${\ensuremath{{\operatorname{Res}}}}\Delta_{b;k}^{-(n+1)}= \gamma_{nk} {{\operatorname{Vol}}}_{\theta}M ,\quad
\gamma_{nk}=\sum_{p+q=k}2^{n}\binom{n}{p} \binom{n}{q}\rho(p-q).
\label{eq:Contact.residue-Deltab}$$ In particular $\gamma_{nk}$ is a universal constant depending only on $n$ and $k$.
Noncommutative residue and the contact Laplacian
------------------------------------------------
The contact complex of Rumin [@Ru:FDVC] can be seen as an attempt to get a complex of horizontal forms by forcing the equalities $d_{b}^{2}=0$ and $(d_{b}^{*})^{2}=0$. Because of (\[eq:CR.square-db\]) there are two natural ways to modify $d_{b}$ to get a chain complex. The first one is to force the equality $d_{b}^{2}=0$ by restricting $d_{b}$ to the subbundle $\Lambda^{*}_{2}:=\ker \varepsilon(d\theta) \cap \Lambda^{*}_{{\ensuremath{\mathbb{C}}}}H^{*}$, since the latter is closed under $d_{b}$ and is annihilated by $d_{b}^{2}$. Similarly, we get the equality $(d_{b}^{*})^{2}=0$ by restricting $d^{*}_{b}$ to the subbundle $\Lambda^{*}_{1}:=\ker \iota(d\theta)\cap \Lambda^{*}_{{\ensuremath{\mathbb{C}}}}H^{*}=({{\operatorname{im}}}\varepsilon(d\theta))^{\perp}\cap \Lambda^{*}_{{\ensuremath{\mathbb{C}}}}H^{*}$, where $\iota(d\theta)$ denotes the interior product with $d\theta$. This amounts to replace $d_{b}$ by $\pi_{1}\circ d_{b}$, where $\pi_{1}$ is the orthogonal projection onto $\Lambda^{*}_{1}$.
In fact, since $d\theta$ is nondegenerate on $H$ the operator $\varepsilon(d\theta):\Lambda^{k}_{{\ensuremath{\mathbb{C}}}}H^{*}\rightarrow \Lambda^{k+2}_{{\ensuremath{\mathbb{C}}}}H^{*}$ is injective for $k\leq n-1$ and surjective for $k\geq n+1$. This implies that $\Lambda_{2}^{k}=0$ for $k\leq n$ and $\Lambda_{1}^{k}=0$ for $k\geq n+1$. Therefore, we only have two halves of complexes. As observed by Rumin [@Ru:FDVC] we get a full complex by connecting the two halves by means of the differential operator, $$B_{R}:C^{\infty}(M,\Lambda_{{\ensuremath{\mathbb{C}}}}^{n}H^{*})\rightarrow C^{\infty}(M,\Lambda_{{\ensuremath{\mathbb{C}}}}^{n}H^{*}), \qquad
B_{R}={\ensuremath{\mathcal{L}}}_{X_{0}}+d_{b,n-1}\varepsilon(d\theta)^{-1}d_{b,n},$$ where $\varepsilon(d\theta)^{-1}$ is the inverse of $\varepsilon(d\theta):\Lambda^{n-1}_{{\ensuremath{\mathbb{C}}}}H^{*}\rightarrow \Lambda^{n+1}_{{\ensuremath{\mathbb{C}}}}H^{*}$. Notice that $B_{R}$ is second order differential operator. Thus, if we let $\Lambda^{k}=\Lambda_{1}^{k}$ for $k=0,\ldots,n-1$ and we let $\Lambda^{k}=\Lambda_{1}^{k}$ for $k=n+1,\ldots,2n$, then we get the chain complex, $$\begin{gathered}
C^{\infty}(M)\stackrel{d_{R;0}}{\rightarrow}C^{\infty}(M,\Lambda^{1})\stackrel{d_{R;1}}{\rightarrow}
\ldots
C^{\infty}(M,\Lambda^{n-1})\stackrel{d_{R;n-1}}{\rightarrow}C^{\infty}(M,\Lambda^{n}_{1})\stackrel{B_{R}}{\rightarrow}\\
C^{\infty}(M,\Lambda^{n}_{2}) \stackrel{d_{R;n}}{\rightarrow}C^{\infty}(M,\Lambda^{n+1})
\ldots \stackrel{d_{R;2n-1}}{\longrightarrow} C^{\infty}(M,\Lambda^{2n}),
\label{eq:contact-complex}\end{gathered}$$ where $d_{R;k}:=\pi_{1}\circ d_{b;k}$ for $k=0,\ldots,n-1$ and $d_{R;k}:=d_{b;k}$ for $k=n,\ldots,2n-1$. This complex is called the *contact complex*.
The contact Laplacian is defined as follows. In degree $k\neq n$ it consists of the differential operator $\Delta_{R;k}:C^{\infty}(M,\Lambda^{k})\rightarrow C^{\infty}(M,\Lambda^{k})$ given by $$\Delta_{R;k}=\left\{
\begin{array}{ll}
(n-k)d_{R;k-1}d^{*}_{R;k}+(n-k+1) d^{*}_{R;k+1}d_{R;k}& \text{$k=0,\ldots,n-1$},\\
(k-n-1)d_{R;k-1}d^{*}_{R;k}+(k-n) d^{*}_{R;k+1}d_{R;k}& \text{$k=n+1,\ldots,2n$}.
\label{eq:contact-Laplacian1}
\end{array}\right.$$ In degree $k=n$ it consists of the differential operators $\Delta_{R;nj}:C^{\infty}(M,\Lambda_{j}^{n})\rightarrow C^{\infty}(M,\Lambda^{n}_{j})$, $j=1,2$, defined by the formulas, $$\Delta_{R;n1}= (d_{R;n-1}d^{*}_{R;n})^{2}+B_{R}^{*}B_{R}, \quad \Delta_{R;n2}=B_{R}B_{R}^{*}+ (d^{*}_{R;n+1}d_{R;n}).
\label{eq:contact-Laplacian2}$$
Observe that $\Delta_{R;k}$, $k\neq n$, is a differential operator of order $2$, whereas $\Delta_{R;n1}$ and $\Delta_{R;n2}$ are differential operators of order $4$. Moreover, Rumin [@Ru:FDVC] proved that in every degree the contact Laplacian is maximal hypoelliptic in the sense of [@HN:HMOPCV]. In fact, in every degree the contact Laplacian has an invertible principal symbol, hence admits a parametrix in the Heisenberg calculus (see [@JK:OKTGSU], [@Po:MAMS1 Sect. 3.5]).
For $k\neq n$ (resp. $j=1,2$) we let $\nu_{0}(\Delta_{R;k})$ (resp. $\nu_{0}(\Delta_{R;nj})$) be the coefficient $\nu_{0}(P)$ in the Weyl asymptotics (\[eq:Zeta.Weyl-asymptotics1\]) for $\Delta_{R;k}$ (resp. $\Delta_{R;nj}$). By Proposition \[prop:Zeta.Weyl-asymptotics\] we have ${\ensuremath{{\operatorname{Res}}}}\Delta_{R;k}^{-(n+1)}=(2n+2)\nu_{0}(\Delta_{R;k})$ and $ {\ensuremath{{\operatorname{Res}}}}\Delta_{R;nj}^{-\frac{n+1}{2}}=(2n+2)\nu_{0}(\Delta_{R;nj})$. Moreover, by [@Po:MAMS1 Thm. 6.3.4] there exist universal positive constants $\nu_{nk}$ and $\nu_{n,j}$ depending only on $n$, $k$ and $j$ such that $\nu_{0}(\Delta_{R;k})=\nu_{nk}{{\operatorname{Vol}}}_{\theta}M$ and $\nu_{0}(\Delta_{R;nj})= \nu_{n,j}{{\operatorname{Vol}}}_{\theta}M$. Therefore, we obtain:
\[prop:Contact.residue-DeltaR\] 1) For $k \neq n$ there exists a universal constant $\rho_{nk}>0$ depending only on $n$ and $k$ such that $${\ensuremath{{\operatorname{Res}}}}\Delta_{R;k}^{-(n+1)}=\rho_{nk} {{\operatorname{Vol}}}_{\theta}M.$$
2\) For $j=1,2$ there exists a universal constant $\rho_{n,j}>0$ depending only on $n$ and $j$ such that $${\ensuremath{{\operatorname{Res}}}}\Delta_{R;nj}^{-\frac{n+1}{2}}= \rho_{n,j} {{\operatorname{Vol}}}_{\theta}M.$$
We have $\rho_{nk}=(2n+2)\nu_{nk}$ and $\rho_{n,j}=(2n+2)\nu_{n,j}$, so it follows from the proof of [@Po:MAMS1 Thm. 6.3.4] that we can explicitly relate the universal constants $\rho_{nk}$ and $\rho_{n,j}$ to the fundamental solutions of the heat operators $\Delta_{R;k}+\partial_{t}$ and $\Delta_{R;nj}+\partial_{t}$ associated to the contact Laplacian on the Heisenberg group ${\ensuremath{\mathbb{H}}}^{2n+1}$ (cf. [@Po:MAMS1 Eq. (6.3.18)]). For instance, if $K_{0;k}(x,t)$ denotes the fundamental solution of $\Delta_{R;0}+\partial_{t}$ on ${\ensuremath{\mathbb{H}}}^{2n+1}$ then we have $\rho_{n,0}=
\frac{2^{n}}{n!}K_{0;0}(0,1)$.
Applications in CR geometry {#sec:CR}
===========================
In this section we present some applications in CR geometry of the noncommutative residue for the Heisenberg calculus. After recalling the geometric set-up, we shall compute the noncommutative residues of some powers of the horizontal sublaplacian and of the Kohn Laplacian on CR manifolds endowed with a pseudohermitian structure. After this we will make use of the framework of noncommutative geometry to define lower dimensional volumes in pseudohermitian geometry. For instance, we will give sense to the area of any 3-dimensional pseudohermitian manifold as a constant multiple the integral of the Tanaka-Webster scalar curvature. As a by-product this will allow us to get a spectral interpretation of the Einstein-Hilbert action in pseudohermitian geometry.
The geometric set-up
--------------------
Let $(M^{2n+1},H)$ be a compact orientable CR manifold. Thus $(M^{2n+1},H)$ is a Heisenberg manifold such that $H$ admits a complex structure $J\in C^{\infty}(M,{\ensuremath{{\operatorname{End}}}}H)$, $J^{2}=-1$, in such way that $T_{1,0}:=\ker (J+i)\subset T_{{\ensuremath{\mathbb{C}}}}M$ is a complex rank $n$ subbundle which is integrable in Fröbenius’ sense (cf. Section \[sec:Heisenberg-calculus\]). In addition, we set $T_{0,1}=\overline{T_{1,0}}=\ker(J-i)$.
Since $M$ is orientable and $H$ is orientable by means of its complex structure, there exists a global non-vanishing real 1-form $\theta$ such that $H=\ker \theta$. Associated to $\theta$ is its Levi form, i.e., the Hermitian form on $T_{1,0}$ such that $$L_{\theta}(Z,W)=-id\theta(Z,\overline{W}) \qquad \forall Z,W \in T_{1,0}.
$$
We say that $M$ is strictly pseudoconvex (resp. $\kappa$-strictly pseudoconvex) when we can choose $\theta$ so that $L_{\theta}$ is positive definite (resp. has signature $(n-\kappa,\kappa,0)$) at every point.
If $(M, H)$ is $\kappa$-strictly pseudoconvex then $\theta$ is a contact form on $M$. Then in the terminology of [@We:PHSRH] the datum of the contact form $\theta$ annihilating $H$ defines a *pseudohermitian structure* on $M$.
From now we assume that $M$ is $\kappa$-strictly pseudoconvex, and we let $\theta$ be a pseudohermitian contact form such that $L_{\theta}$ has signature $(n-\kappa,\kappa,0)$ everywhere. We let $X_{0}$ be the Reeb vector field associated to $\theta$, so that $\iota_{X_{0}}\theta=1$ and $\iota_{X_{0}}d\theta=0$ (cf. Section \[sec:Contact\]), and we let ${\ensuremath{\mathcal{N}}}\subset T_{{\ensuremath{\mathbb{C}}}}M$ be the complex line bundle spanned by $X_{0}$.
We endow $M$ with a *Levi metric* as follows. First, we always can construct a splitting $T_{1,0}=T_{1,0}^{+}\oplus T_{1,0}^{+}$ with subbundles $T_{1,0}^{+}$ and $T_{1,0}^{-}$ which are orthogonal with respect to $L_{\theta}$ and such that $L_{\theta}$ is positive definite on $T_{1,0}^{+}$ and negative definite on $T_{1,0}^{-}$ (see, e.g., [@FS:EDdbarbCAHG], [@Po:MAMS1]). Set $T_{0,1}^{\pm}=\overline{T_{1,0}^{\pm}}$. Then we have the splittings, $$T_{{\ensuremath{\mathbb{C}}}}M={\ensuremath{\mathcal{N}}}\oplus T_{1,0}\oplus T_{0,1}={\ensuremath{\mathcal{N}}}\oplus T_{1,0}^{+}\oplus T_{1,0}^{-}\oplus T_{0,1}^{+}\oplus T_{0,1}^{-}.
\label{eq:CR.splitting-TcM}$$ Associated to these splittings is the unique Hermitian metric $h$ on $T_{{\ensuremath{\mathbb{C}}}}M$ such that:
- The splittings (\[eq:CR.splitting-TcM\]) are orthogonal with respect to $h$;
- $h$ commutes with complex conjugation;
- We have $h(X_{0},X_{0})=1$ and $h$ agrees with $\pm L_{\theta}$ on $T_{1,0}^{\pm}$.
In particular, the matrix of $L_{\theta}$ with respect to $h$ is ${\operatorname{diag}}(1,\ldots,1,-1,\ldots,-1)$, where $1$ has multiplicity $n-\kappa$ and $-1$ multiplicity $-1$.
Notice that when $M$ is strictly pseudoconvex $h$ is uniquely determined by $\theta$, since in this case $T_{1,0}^{+}=T_{1,0}$ and one can check that we have $h=\theta^{2}+d\theta(.,J.)$, that is, $h$ agrees on $TM$ with the Riemannian metric $g_{\theta,J}$ in (\[eq:Contact.Riem-metric\]). In general, we can check that the volume form of $M$ with respect to $h$ depends only on $\theta$ and is equal to $$v_{\theta}(x):=\frac{(-1)^{\kappa}}{n!}d\theta^{n}\wedge\theta.
$$ In particular, the volume of $M$ with respect to $h$ is $${{\operatorname{Vol}}}_{\theta}M:=\frac{(-1)^{\kappa}}{n!}\int_{M} d\theta^{n}\wedge\theta.
$$
Finally, as proved by Tanaka [@Ta:DGSSPCM] and Webster [@We:PHSRH] the datum of the pseudohermitian contact form $\theta$ defines a natural connection, the *Tanaka-Webster connection*, which preserves the pseudohermitian structure of $M$, i.e., it preserves both $\theta$ and $J$. It can be defined as follows.
Let $\{Z_{j}\}$ be a local frame of $T_{1,0}$. Then $\{X_{0},Z_{j},Z_{\overline{j}}\}$ forms a frame of $T_{{\ensuremath{\mathbb{C}}}}M$ with dual coframe $\{\theta,\theta^{j}, \theta^{\overline{j}}\}$, with respect to which we can write $d\theta =ih_{j\overline{k}}\theta^{j}\wedge \theta^{\overline{k}}$. Using the matrix $(h_{j\overline{k}})$ and its inverse $(h^{j\overline{k}})$ to lower and raise indices, the connection 1-form $\omega=(\omega_{j}^{~k})$ and the torsion form $\tau_{k}=A_{jk}\theta^{j}$ of the Tanaka-Webster connection are uniquely determined by the relations, $$d\theta^{k}=\theta^{j}\wedge \omega_{j}^{~k}+\theta \wedge \tau^{k}, \qquad
\omega_{j\bar{k}} + \omega_{\bar{k}j}
=dh_{j\bar{k}}, \qquad A_{jk}=A_{k j}.
$$
The curvature tensor $\Pi_{j}^{~k}:=d\omega_{j}^{~k}-\omega_{j}^{~l}\wedge \omega_{l}^{~k}$ satisfies the structure equations, $$\Pi_{j}^{~k}=R_{j\bar{k} l\bar{m}} \theta^{l}\wedge \theta^{\bar{m}} +
W_{j\bar{k}l}\theta^{l}\wedge \theta - W_{\bar{k}j\bar{l}}\theta^{\bar{l}}\wedge \theta
+i\theta_{j}\wedge \tau_{\bar{k}}-i\tau_{j}\wedge \theta_{\bar{k}}.
\label{eq:CR.TW-curvature}$$ The *Ricci tensor* of the Tanaka-Webster connection is $ \rho_{j \bar{k}}:=R_{l~j \bar{k}}^{~l}$, and its *scalar curvature* is $R_{n}: =\rho_{j}^{~j}$.
Noncommutative residue and the Kohn Laplacian {#subsec:CR.NCR-pseudohermitian}
---------------------------------------------
The ${\overline{\partial}_{b}}$-complex of Kohn-Rossi ([@KR:EHFBCM], [@Ko:BCM]) is defined as follows.
Let $\Lambda^{1,0}$ (resp. $\Lambda^{0,1}$) be the annihilator of $T_{0,1}\oplus {\ensuremath{\mathcal{N}}}$ (resp. $T_{0,1}\oplus {\ensuremath{\mathcal{N}}}$) in $T^{*}_{{\ensuremath{\mathbb{C}}}}M$. For $p,q=0,\ldots,n$ let $\Lambda^{p,q}:=(\Lambda^{1,0})^{p}\wedge (\Lambda^{0,1})^{q}$ be the bundle of $(p,q)$-covectors on $M$, so that we have the orthogonal decomposition, $$\Lambda^{*}T_{{\ensuremath{\mathbb{C}}}}^{*}M=(\bigoplus_{p,q=0}^{n}\Lambda^{p,q})\oplus (\theta\wedge \Lambda^{*}T_{{\ensuremath{\mathbb{C}}}}^{*}M).
\label{eq:CR-Lambda-pq-decomposition}$$ Moreover, thanks to the integrability of $T_{1,0}$, given any local section $\eta$ of $\Lambda^{p,q}$, its differential $d\eta$ can be uniquely decomposed as $$d\eta ={\overline{\partial}_{b;p,q}}\eta + \partial_{b;p,q}\eta + \theta \wedge {\ensuremath{\mathcal{L}}}_{X_{0}}\eta,
\label{eq:CR.dbarb}$$ where ${\overline{\partial}_{b;p,q}}\eta $ (resp. $\partial_{b;p,q}\eta$) is a section of $\Lambda^{p,q+1}$ (resp. $\Lambda^{p+1,q}$).
The integrability of $T_{1,0}$ further implies that $\overline{\partial}_{b}^{2}=0$ on $(0,q)$-forms, so that we get the cochain complex $\overline{\partial}_{b;0,*}:C^{\infty}(M,\Lambda^{0,*})\rightarrow C^{\infty}(M,\Lambda^{0,*+1})$. On $(p,q)$-forms with $p\geq 1$ the operator ${\overline{\partial}_{b}}^{2}$ is a tensor which vanishes when the complex structure $J$ is invariant under the Reeb flow (i.e., when we have $[X_{0},JX]=J[X_{0},X]$ for any local section $X$ of $H$). Let ${\overline{\partial}_{b;p,q}}^{*}$ be the formal adjoint of ${\overline{\partial}_{b;p,q}}$ with respect to the Levi metric of $M$. Then the *Kohn Laplacian* ${{\square}_{b;p,q}}:C^{\infty}(M,\Lambda^{p,q})\rightarrow C^{\infty}(M,\Lambda^{p,q})$ is defined to be $${{\square}_{b;p,q}}={\overline{\partial}_{b;p,q}}^{*}{\overline{\partial}_{b;p,q}}+ \overline{\partial}_{b;p,q-1}\overline{\partial}_{b;p,q-1}^{*}.$$ This a differential operator which has order 2 in the Heisenberg calculus sense. Furthermore, we have:
The principal symbol of ${{\square}_{b;p,q}}$ is invertible if and only if we have $q \neq \kappa$ and $q\neq n-\kappa$.
Next, for $q \not\in\{\kappa,n-\kappa\}$ let $\nu_{0}({{\square}_{b;p,q}})$ be the coefficient $\nu_{0}(P)$ in the Weyl asymptotics (\[eq:Zeta.Weyl-asymptotics1\]) for ${{\square}_{b;p,q}}$. By [@Po:MAMS1 Thm. 6.2.4] we have $ \nu_{0}({{\square}_{b;p,q}})=\tilde{\alpha}_{n\kappa pq}{{\operatorname{Vol}}}_{\theta}M$, where $\tilde{\alpha}_{n\kappa pq}$ is equal to $$\sum_{\max(0,q-\kappa)\leq k\leq \min(q,n-\kappa)} \frac{1}{2} \binom{n}{p} \binom{n-\kappa}{k}\binom{\kappa}{q-k}
\nu(n-2(\kappa-q+2k)).
\label{eq:CR.talphankpq}$$ Therefore, by arguing as in the proof of Proposition \[prop:Contact.residue-Deltab\] we get:
\[prop:CR.residue-Boxb1\] For $q \neq \kappa$ and $q\neq n-\kappa$ we have $${\ensuremath{{\operatorname{Res}}}}{{\square}_{b;p,q}}^{-(n+1)}=\alpha_{n\kappa pq}{{\operatorname{Vol}}}_{\theta}M,
\label{eq:CR.residue-Boxb1}$$ where $\alpha_{n\kappa pq}$ is equal to $$\sum_{\max(0,q-\kappa)\leq k\leq \min(q,n-\kappa)} \frac{1}{2} \binom{n}{p} \binom{n-\kappa}{k}\binom{\kappa}{q-k}
\rho(n-2(\kappa-q+2k)).
\label{eq:CR.alphapq}
$$ In particular $\alpha_{n\kappa pq}$ is a universal constant depending only on $n$, $\kappa$, $p$ and $q$.
\[rem:CR.residue-Boxb1-local\] Let $a_{0}({{\square}_{b;p,q}})(x)$ be the leading coefficient in the heat kernel asymptotics (\[eq:Zeta.heat-kernel-asymptotics\]) for ${{\square}_{b;p,q}}$. By (\[eq:Zeta.tPs-heat1\]) we have $ \nu_{0}({{\square}_{b;p,q}})= \frac{1}{ (n+1)!} \int_{M}{{\operatorname{tr}}}_{\Lambda^{p,q}}a_{0}({{\square}_{b;p,q}})(x)$. Moreover, a careful look at the proof of [@Po:MAMS1 Thm. 6.2.4] shows that we have $${{\operatorname{tr}}}_{\Lambda^{p,q}}a_{0}({{\square}_{b;p,q}})(x)=(n+1)!\tilde{\alpha}_{n\kappa pq}v_{\theta}(x).
$$ Since by (\[eq:Zeta.tPs-heat1\]) we have $2c_{{{\square}_{b;p,q}}^{-(n+1)}}(x)=(n!)^{-1}a_{0}({{\square}_{b;p,q}})(x)$, it follows that the equality (\[eq:CR.residue-Boxb1\]) ultimately holds at the level of densities, that is, we have $$c_{{{\square}_{b;p,q}}^{-(n+1)}}(x)=\alpha_{n\kappa pq}v_{\theta}(x).
$$
Finally, when $M$ is strictly pseudoconvex, i.e., when $\kappa=0$, we have:
\[prop:CR.residue-Boxb2\] Assume $M$ strictly pseudoconvex. Then for $q =1,\ldots, n-1$ there exists a universal constant $\alpha_{npq}'$ depending only on $n$, $p$ and $q$ such that $${\ensuremath{{\operatorname{Res}}}}{{\square}_{b;p,q}}^{-n}=\alpha_{npq}'\int_{M}R_{n}d\theta^{n}\wedge \theta,
$$ where $R_{n}$ denotes the Tanaka-Webster scalar curvature of $M$.
For $q=1,\ldots,n-1$ let $a_{2}({{\square}_{b;p,q}})(x)$ be the coefficient of $t^{-n}$ in the heat kernel asymptotics (\[eq:Zeta.heat-kernel-asymptotics\]) for ${{\square}_{b;p,q}}$. By (\[eq:Zeta.tPs-heat1\]) we have $2c_{{{\square}_{b;p,q}}^{-n}}(x)=\Gamma(n)^{-1}a_{2}({{\square}_{b;p,q}})(x)$. Moreover, by [@BGS:HECRM Thm. 8.31] there exists a universal constant $\alpha_{npq}'$ depending only on $n$, $p$ and $q$ such that ${{\operatorname{tr}}}_{\Lambda^{p,q}}a_{2}({{\square}_{b;p,q}})(x)=\alpha_{npq}'R_{n}d\theta^{n}\wedge \theta$. Thus, $${\ensuremath{{\operatorname{Res}}}}{{\square}_{b;p,q}}^{-n}= \int_{M}{{\operatorname{tr}}}_{\Lambda^{p,q}}c_{{{\square}_{b;p,q}}^{-n}}(x)= \alpha_{npq}'\int_{M}R_{n}d\theta^{n}\wedge \theta,
$$ where $\alpha_{npq}'$ is a universal constant depending only on $n$, $p$ and $q$.
Noncommutative residue and the horizontal sublaplacian (CR case)
----------------------------------------------------------------
Let us identify $H^{*}$ with the subbundle of $T^{*}M$ annihilating the orthogonal supplement $H^{\perp}$, and let $\Delta_{b}:C^{\infty}(M,\Lambda^{*}_{{\ensuremath{\mathbb{C}}}}H^{*})\rightarrow C^{\infty}(M,\Lambda^{*}_{{\ensuremath{\mathbb{C}}}}H^{*})$ be the horizontal sublaplacian on $M$ as defined in (\[eq:CR.horizontal-sublaplacian\]).
Notice that with the notation of (\[eq:CR.dbarb\]) we have $d_{b}={\overline{\partial}_{b}}+\partial_{b}$. Moreover, we can check that ${\overline{\partial}_{b}}\partial_{b}^{*}+\partial_{b}^{*} {\overline{\partial}_{b}}= {\overline{\partial}_{b}}^{*} \partial_{b}+\partial_{b}
{\overline{\partial}_{b}}^{*}=0$. Therefore, we have $$\Delta_{b}={\square}_{b}+ \overline{{\square}}_{b}, \qquad \overline{{\square}}_{b}:=\partial_{b}^{*}\partial_{b}+\partial_{b}\partial_{b}^{*}.
$$ In particular, this shows that the horizontal sublaplacian $\Delta_{b}$ preserves the bidegree, so it induces a differential operator $\Delta_{b;p,q}:C^{\infty}(M,\Lambda^{p,q})\rightarrow C^{\infty}(M,\Lambda^{p,q})$. Then the following holds.
The principal symbol of $\Delta_{b;p,q}$ is invertible if and only if we have $(p,q)\neq (\kappa,n-\kappa)$ and $(p,q)\neq (n-\kappa,\kappa)$.
Bearing this in mind we have:
\[prop:CR.residue-Deltab1\] For $(p,q)\neq (\kappa,n-\kappa)$ and $(p,q)\neq (n-\kappa,\kappa)$ we have $${\ensuremath{{\operatorname{Res}}}}\Delta_{b;p,q}^{-(n+1)}= \beta_{n\kappa pq}{{\operatorname{Vol}}}_{\theta}M,
\label{eq:CR.residue-Deltab1a}$$ where $\beta_{n\kappa pq}$ is equal to $$\! \! \! \! \sum_{\substack{\max(0,q-\kappa)\leq k\leq \min(q,n-\kappa)\\ \max(0,p-\kappa)\leq l\leq \min(p,n-\kappa)}} \! \! \! \!
2^{n}\binom{n-\kappa}{l}\binom{\kappa}{p-l} \binom{n-\kappa}{k}\binom{\kappa}{q-k}
\rho(2(q-p)+4(l-k)).
\label{eq:CR.residue-Deltab1b}$$ In particular $\beta_{n\kappa pq}$ is a universal constant depending only on $n$, $\kappa$, $p$ and $q$.
Let $\nu_{0}(\Delta_{b;p,q})$ be the coefficient $\nu_{0}(P)$ in the Weyl asymptotics (\[eq:Zeta.Weyl-asymptotics1\]) for $\Delta_{b;p,q}$. By [@Po:MAMS1 Thm. 6.2.5] we have $\nu_{0}(\Delta_{b;p,q})=\frac{1}{2n+2}\beta_{n\kappa pq}{{\operatorname{Vol}}}_{\theta}M$, where $\beta_{n\kappa pq}$ is given by (\[eq:CR.residue-Deltab1b\]). We then can show that $ {\ensuremath{{\operatorname{Res}}}}\Delta_{b;p,q}^{-(n+1)}= \beta_{n\kappa pq}{{\operatorname{Vol}}}_{\theta}M$ by arguing as in the proof of Proposition \[prop:Contact.residue-Deltab\].
\[rem:CR.residue-Deltab1-local\] In the same way as (\[eq:CR.residue-Boxb1\]) (cf. Remark \[rem:CR.residue-Boxb1-local\]) the equality (\[eq:CR.residue-Deltab1a\]) holds at the level of densities, i.e., we have $c_{\Delta_{b;p,q}^{-(n+1)}}(x)=\beta_{n\kappa pq}v_{\theta}(x)$.
\[prop:CR.residue-Deltab2\] Assume that $M$ is strictly pseudoconvex. For $(p,q)\neq (0,n)$ and $(p,q)\neq (n,0)$ there exists a universal constant $\beta_{npq}'$ depending only $n$, $p$ and $q$ such that $${\ensuremath{{\operatorname{Res}}}}\Delta_{b;p,q}^{-n}=\beta_{npq}' \int_{M}R_{n}d\theta^{n}\wedge \theta.
\label{eq:CR.residue-Deltab2}$$
The same analysis as that of [@BGS:HECRM Sect. 8] for the coefficients in the heat kernel asymptotics (\[eq:Zeta.heat-kernel-asymptotics\]) for the Kohn Laplacian can be carried out for the coefficients of the heat kernel asymptotics for $\Delta_{b;p,q}$ (see [@St:SICRM]). In particular, if we let $a_{2}(\Delta_{b;p,q})(x)$ be the coefficient of $t^{-n}$ in the heat kernel asymptotics for $\Delta_{b;p,q}$, then there exists a universal constant $\tilde{\beta}_{npq}$ depending only on $n$, $p$ and $q$ such that ${{\operatorname{tr}}}_{\Lambda^{p,q}}a_{2}(\Delta_{b;p,q})(x)=\tilde{\beta}_{npq}R_{n}d\theta^{n}\wedge \theta$. Arguing as in the proof of Proposition \[prop:CR.residue-Boxb2\] then shows that ${\ensuremath{{\operatorname{Res}}}}\Delta_{b;p,q}^{-n}=\beta_{npq}' \int_{M}R_{n}d\theta^{n}\wedge \theta$, where$\beta_{npq}'$ is a universal constant depending only $n$, $p$ and $q$.
Lower dimensional volumes in pseudohermitian geometry {#subsec.CR.area}
-----------------------------------------------------
Following an idea of Connes [@Co:GCMFNCG] we can make use of the noncommutative residue for classical [$\Psi$DOs]{} to define lower dimensional dimensional volumes in Riemannian geometry, e.g., we can give sense to the area and the length of a Riemannian manifold even when the dimension is not 1 or 2 (see [@Po:LMP07]). We shall now make use of the noncommutative residue for the Heisenberg calculus to define lower dimensional volumes in pseudohermitian geometry.
In this subsection we assume that $M$ is strictly pseudoconvex. In particular, the Levi metric $h$ is uniquely determined by $\theta$. In addition, we let $\Delta_{b;0}$ be the horizontal sublaplacian acting on functions.Then, as explained in Remark \[rem:CR.residue-Deltab1-local\], we have $c_{\Delta_{b;0}^{-(n+1)}}(x)=\beta_{n}v_{\theta}(x)$, where $\beta_{n}=\beta_{n000}=2^{n}\rho(0)$. In particular, for any $f \in C^{\infty}(M)$ we get $c_{f\Delta_{b;0}^{-(n+1)}}(x)=\beta_{n}f(x)v_{\theta}(x)$. Combining this with Theorem \[thm:NCG.Dixmier\] then gives$${\ensuremath{-\hspace{-2,4ex}\int}}f\Delta_{b;0}^{-(n+1)}=\frac{1}{2n+2}\int_{M}c_{f\Delta_{b;0}^{-(n+1)}}(x)=\frac{\beta_{n}}{2n+2}\int_{M}f(x)v_{\theta}(x).
$$ Thus the operator $\frac{2n+2}{\beta_{n}}\Delta_{b;0}^{-(n+1)}$ allows us to recapture the volume form $v_{\theta}(x)$.
Since $-(2n+2)$ is the critical order for a [$\Psi_{H}$DO]{} to be trace-class and $M$ has Hausdorff dimension $2n+2$ with respect to the Carnot-Carathéodory metric defined by the Levi metric on $H$, it stands for reason to define the *length element* of $(M,\theta)$ as the positive selfadjoint operator $ds$ such that $(ds)^{2n+2}=\frac{2n+2}{\beta_{n}}\Delta_{b;0}^{-(n+1)}$, that is, $$ds:= c_{n}\Delta_{b;0}^{-1/2}, \qquad c _{n}=(\frac{2n+2}{\beta_{n}})^{\frac{1}{2n+2}}.
$$
For $k=1,2,\ldots,2n+2$ the $k$-dimensional volume of $(M,\theta)$ is $${\operatorname{Vol}}_{\theta}^{(k)}M:={\ensuremath{-\hspace{-2,4ex}\int}}ds^{k}.
$$ In particular, for $k=2$ the area of $(M,\theta)$ is $ {\operatorname{Area}}_{\theta}M:={\ensuremath{-\hspace{-2,4ex}\int}}ds^{2}$.
We have ${\ensuremath{-\hspace{-2,4ex}\int}}ds^{k}=\frac{(c_{n})^{k}}{2n+2}\int_{M}c_{\Delta_{b;0}^{-\frac{k}{2}}}(x)$ and thanks to (\[eq:Zeta.tPs-heat1\]) we know that $2
c_{\Delta_{b;0}^{-\frac{k}{2}}}(x)$ agrees with $\Gamma(\frac{k}{2})^{-1}a_{2n+2-k}(\Delta_{b;0})(x)$, where $a_{j}(\Delta_{b;0})(x)$ denotes the coefficient of $t^{\frac{2n+2-j}{2}}$ in the heat kernel asymptotics (\[eq:Zeta.heat-kernel-asymptotics\]) for $\Delta_{b;0}$. Thus, $${\operatorname{Vol}}_{\theta}^{(k)}M=\frac{(c_{n})^{k}}{4(n+1)}\Gamma(\frac{k}{2})^{-1}\int_{M}a_{2n+2-k}(\Delta_{b})(x).
$$ Since $\Delta_{b;0}$ is a differential operator we have $a_{2j-1}(\Delta_{b;0})(x)=0$ for any $j\in {\ensuremath{\mathbb{N}}}$, so $ {\operatorname{Vol}}_{\theta}^{(k)}M$ vanishes when $k$ is odd. Furthermore, as alluded to in the proof of Proposition \[prop:CR.residue-Deltab2\] the analysis in [@BGS:HECRM Sect. 8] of the coefficients of the heat kernel asymptotics for the Kohn Laplacian applies *verbatim* to the heat kernel asymptotics for the horizontal sublaplacian. Thus, we can write $$a_{2j}(\Delta_{b;0})(x)=\gamma_{nj}(x)d\theta^{n}\wedge \theta(x),
\label{eq:CR-volumes-gamma-nj}$$ where $\gamma_{nj}(x)$ is a universal linear combination, depending only on $n$ and $j$, in complete contractions of covariant derivatives of the curvature and torsion tensors of the Tanaka-Webster connection (i.e. $\gamma_{nj}(x)$ is a local pseudohermitian invariant). In particular, we have $\gamma_{n0}(x)=\gamma_{n0}$ and $\gamma_{n1}=\gamma_{n1}'R_{n}(x)$, where $\gamma_{n0}$ and $\gamma_{n1}$ are universal constants and $R_{n}(x)$ is the Tanaka-Webster scalar curvature (in fact the constants $\gamma_{n0}$ and $\gamma_{n1}'$ can be explicitly related to the constants $\beta_{n000}$ and $\beta_{n00}'$). Therefore, we obtain:
\[prop:CR.lower-dim.-volumes\] 1) ${{\operatorname{Vol}}}^{(k)}_{\theta}M$ vanishes when $k$ is odd.
2\) When $k$ is even we have $${\operatorname{Vol}}_{\theta}^{(k)}M=\frac{(c_{n})^{k}}{4(n+1)}\Gamma(\frac{k}{2})^{-1}\int_{M}\tilde{\gamma}_{nk}(x)d\theta^{n}\wedge \theta(x).
\label{eq:CR.volumes-even}$$ where $\tilde{\gamma}_{nk}(x):=\gamma_{nn+1-\frac{k}{2}}(x)$ is a universal linear combination, depending only on $n$ and $k$, of complete contractions of weight $n+1-\frac{k}{2}$ of covariant derivatives of the curvature and torsion tensors of the Tanaka-Webster connection.
In particular, thanks to (\[eq:CR.volumes-even\]) we have a purely differential-geometric formulation of the $k$-dimensional volume $ {\operatorname{Vol}}_{\theta}^{(k)}M$. Moreover, for $k=2n+2$ we get: $${\operatorname{Vol}}_{\theta}^{(2n+2)}M=\frac{(c_{n})^{2n+2}}{4(n+1)}\frac{\gamma_{n0}}{n!}\int_{M}d\theta^{n}\wedge \theta.
$$ Since ${\operatorname{Vol}}_{\theta}^{(2n+2)}M={\operatorname{Vol}}_{\theta}M=\frac{1}{n!}\int_{M}d\theta^{n}\wedge \theta$ we see that $(c_{n})^{2n+2}=\frac{4(n+1)}{\gamma_{n0}}$, where $\gamma_{n0}$ is above.
On the other hand, when $n=1$ (i.e. $\dim M=3$) and $k=2$ we get $${\operatorname{Area}}_{\theta}M=\gamma_{1}''\int_{M}R_{1}d\theta\wedge \theta, \qquad \gamma_{1}'':=\frac{(c_{1})^{2}}{8}\gamma_{11}'
=\frac{\gamma_{11}'}{\sqrt{8\gamma_{10}}},
\label{eq:CR.area-universal}$$ where $\gamma_{11}'$ is above. To compute $\gamma_{1}''$ it is enough to compute $\gamma_{10}$ and $\gamma_{11}'$ in the special case of the unit sphere $S^{3}\subset {\ensuremath{\mathbb{C}}}^{2}$ equipped with its standard pseudohermitian structure, i.e., for $S^{3}$ equipped with the CR structure induced by the complex structure of ${\ensuremath{\mathbb{C}}}^{2}$ and with the pseudohermitian contact form $\theta:= \frac{i}2 (z_{1} d\bar{z}_{1}+ z_{2} d\bar{z}_{2})$.
First, the volume ${{\operatorname{Vol}}}_{\theta}S^{3}$ is equal to $$\int_{S^{3}}d\theta\wedge \theta = \frac{-1}{4}\int_{S^{3}}(z_{2}dz_{1}\wedge d\bar{z_{1}} \wedge d\bar{z_{2}} +
z_{1} dz_{1}\wedge dz_{2}\wedge d\bar{z_{2}})=\pi^{2}.
\label{eq:CR.volumeS3}$$ Moreover, by [@We:PHSRH] the Tanaka-Webster scalar here is $R_{1}=4$, so we get $$\int_{S^{3}}R_{1}d\theta\wedge \theta =4{{\operatorname{Vol}}}_{\theta}S^{3}=4\pi^{2}.
$$
Next, for $j=0,1$ set $A_{2j}(\Delta_{b;0})=\int_{S^{3}}a_{2j}(\Delta_{b;0})(x)$. In view of the definition of the constants $\gamma_{10}$ and $\gamma_{11}'$ we have $$A_{0}(\Delta_{b;0})=\gamma_{10}\int_{S^{3}}d\theta\wedge \theta =\pi^{2}\gamma_{10}, \quad
A_{2}(\Delta_{b;0})=\gamma_{11}'\int_{S^{3}}R_{1}d\theta\wedge \theta =4\pi^{2}\gamma_{11}'.
\label{eq:CR.gamma-A2j}$$ Notice that $A_{0}(\Delta_{b;0})$ and $A_{2}(\Delta_{b;0})$ are the coefficients of $t^{-2}$ and $t^{-1}$ in the asymptotics of $ {\ensuremath{{\operatorname{Tr}}}}e^{-t\Delta_{b;0}}$ as $t\rightarrow 0^{+}$. Moreover, we have $\Delta_{b;0}
=\boxdot_{\theta}-\frac{1}{4}R_{1}=\boxdot_{\theta}-1$, where $\boxdot_{\theta}$ denotes the CR invariant sublaplacian of Jerison-Lee [@JL:YPCRM], and by [@St:SICRM Thm. 4.34] we have ${\ensuremath{{\operatorname{Tr}}}}e^{-t\boxdot_{\theta}} =\frac{\pi^{2}}{16t^{2}}+
{\operatorname{O}}(t^\infty)$ as $t \rightarrow 0^{+}$. Therefore, as $t\rightarrow 0^{+}$ we have $${\ensuremath{{\operatorname{Tr}}}}e^{-t\Delta_{b;0}}=e^{t}{\ensuremath{{\operatorname{Tr}}}}e^{-t\boxdot_{\theta}}\sim \frac{\pi^{2}}{16t^{2}}(1+t+\frac{t^{2}}{2}+\ldots).
$$ Hence $A_{0}(\Delta_{b;0})=A_{2}(\Delta_{b;0})=\frac{\pi^{2}}{16}$. Combining this with (\[eq:CR.gamma-A2j\]) then shows that $\gamma_{10}=\frac{1}{16}$ and $\gamma_{11}'=\frac{1}{64}$, from which we get $\gamma_{1}''=\frac{1/64}{\sqrt{8.\frac{1}{16}}}=\frac1{32\sqrt2}$. Therefore, we get:
\[thm:spectral.area\] If $\dim M=3$, then we have $${\operatorname{Area}}_{\theta}M = \frac1{32\sqrt2}\int_{M}R_{1}d\theta \wedge\theta.
\label{eq:CR.area-dimension3}$$
For instance, for $S^{3}$ equipped with its standard pseudohermitian structure we obtain ${\operatorname{Area}}_{\theta}S^{3}=\frac{\pi^{2}}{8\sqrt{2}}$.
Appendix. Proof of Lemma \[lem:Heisenberg.extension-symbol\] {#appendix.-proof-of-lemmalemheisenberg.extension-symbol .unnumbered}
============================================================
In this appendix, for reader’s convenience we give a detailed proof of Lemma \[lem:Heisenberg.extension-symbol\] about the extension of a homogeneous symbol on ${{\ensuremath{\mathbb{R}}}^{d+1}\!\setminus\! 0}$ into a homogeneous distribution on ${\ensuremath{{\ensuremath{\mathbb{R}}}^{d+1}}}$.
Let $p\in C^\infty({{\ensuremath{\mathbb{R}}}^{d+1}\!\setminus\! 0})$ be homogeneous of degree $m$, $m\in {\ensuremath{\mathbb{C}}}$, so that $p(\lambda.\xi)=\lambda^{m}p(\xi)$ for any $\lambda>0$. If $\Re m>-(d+2)$, then $p$ is integrable near the origin, so it defines a tempered distribution which is its unique homogeneous extension.
If $\Re m \leq -(d+2)$, then we can extend $p$ into the distribution $\tau\in{\ensuremath{\mathcal{S}}}'({\ensuremath{{\ensuremath{\mathbb{R}}}^{d+1}}})$ defined by the formula, $${\ensuremath{\langle \tau , u \rangle}}= \int [u(\xi)-\psi(\|\xi\|)\sum_{{\ensuremath{\langle\! \alpha\!\rangle}}\leq k}
\frac{\xi^\alpha}{\alpha!} u^{(\alpha)}(0)] p(\xi)d\xi \qquad \forall u\in{\ensuremath{\mathcal{S}}}({\ensuremath{{\ensuremath{\mathbb{R}}}^{d+1}}}),
\label{eq:Appendix.almosthomogeneous-extension}$$ where $k$ is an integer $\geq -(\Re m +d+2)$ and $\psi$ is a function in $C_{c}^\infty({\ensuremath{\mathbb{R}}}_{+})$ such that $\psi=1$ near $0$. Then in view of (\[eq:PsiHDO.homogeneity-K-m\]) for any $\lambda>0$ we have $$\begin{split}
{\ensuremath{\langle \tau_{\lambda} , u \rangle}}-\lambda^{m} {\ensuremath{\langle \tau , u \rangle}} &= \lambda^{-(d+2)}\int [u(\lambda^{-1}.\xi)-\psi(\|\xi\|)\sum_{{\ensuremath{\langle\! \alpha\!\rangle}}\leq k}
\frac{\xi^\alpha\lambda^{-{\ensuremath{\langle\! \alpha\!\rangle}}}}{\alpha!} u^{(\alpha)}(0)]p(\xi)d\xi \\
& -\lambda^{m}\int [u(\xi)-\psi(\|\xi\|)\sum_{{\ensuremath{\langle\! \alpha\!\rangle}}\leq k}
\frac{\xi^\alpha}{\alpha!} u^{(\alpha)}(0)] p(\xi)d\xi,\\
&= \lambda^{m} \sum_{{\ensuremath{\langle\! \alpha\!\rangle}}\leq k} \frac{u^{(\alpha)}(0)}{\alpha!} \int [\psi(\|\xi\|)-\psi(\lambda\|\xi\|)] \xi^{\alpha}p(\xi)d\xi,\\
&= \lambda^{m} \sum_{{\ensuremath{\langle\! \alpha\!\rangle}}\leq k} \rho_{\alpha}(\lambda) c_{\alpha}(p) {\ensuremath{\langle \delta^{(\alpha)} , u \rangle}},
\end{split}$$ where we have let $$c_{\alpha}(p) = \frac{(-1)^{|\alpha|}}{\alpha!}\int_{\|\xi\|=1}\xi^\alpha p(\xi)i_{E}d\xi, \qquad
\rho_{\alpha}(\lambda)=\int_{0}^\infty \mu^{{\ensuremath{\langle\! \alpha\!\rangle}}+m+d+2}
(\psi(\mu)-\psi(\lambda\mu)) \frac{d\mu}{\mu},$$ and, as in the statement of Lemma \[lem:Heisenberg.extension-symbol\], $E$ is the vector field $2\xi_{0}\partial_{\xi_{0}}+\xi_{1}\partial_{\xi_{1}}+\ldots+\xi_{d}\partial_{\xi_{d}}$.
Set $\lambda=e^s$ and assume that $\psi$ is of the form $ \psi(\mu)=h(\log
\mu) $ with $h\in C^\infty({\ensuremath{\mathbb{R}}})$ such that $h=1$ near $-\infty$ and $h=0$ near $+\infty$. Then, setting $a_{\alpha}={\ensuremath{\langle\! \alpha\!\rangle}}+m+d+2$, we have $$\frac{d}{ds}\rho_{\alpha}(e^s)= \frac{d}{ds} \int_{-\infty}^{\infty}(h(t)-h(s+t))e^{a_{\alpha}t}dt=- e^{-as}\int_{-\infty}^{\infty}
e^{a_{\alpha}t} h'(t)dt.
\label{eq:Appendix.differentiation-rhoalpha}$$ As $\rho_{\alpha}(1)=0$ it follows that $\tau$ is homogeneous of degree $m$ provided that $$\int_{-\infty}^{\infty} e^{at} h'(t) ds = 0 \qquad \text{for $a=m+d+2, \ldots, m+d+2+k$}.
\label{eq:Appendix.homogeneous-extension-condition}$$
Next, if $g\in C_{c}^\infty({\ensuremath{{\ensuremath{\mathbb{R}}}^{d+1}}})$ is such that $\int g(t)dt=1$, then for any $a \in {\ensuremath{\mathbb{C}}}\setminus 0$ we have $$\int _{-\infty}^{\infty} e^{at}(\frac{1}{a}\frac{d}{dt}+1)g(t)dt=0.
\label{eq:Appendix1.integral-a-g}$$ Therefore, if $m \not\in{\ensuremath{\mathbb{Z}}}$ then we can check that the conditions (\[eq:Appendix.homogeneous-extension-condition\]) are satisfied by $$h'(t)=\prod_{a=m+d+2}^{m+d+2+k} (\frac{1}{a} \frac{d}{dt} +1)g(t).
\label{eq:Appendix1.h'}$$ As $\int_{-\infty}^{\infty}h'(t)dt=1$ we then see that the distribution $\tau$ defined by (\[eq:Appendix.almosthomogeneous-extension\]) with $\psi(\mu)=\int_{\log \mu}^\infty h'(t)dt$ is a homogeneous extension of $p(\xi)$.
On the other hand, if $\tilde{\tau}\in{\ensuremath{\mathcal{S}}}'({\ensuremath{{\ensuremath{\mathbb{R}}}^{d+1}}})$ is another homogeneous extension of $p(\xi)$ then $\tau-\tau_{1}$ is supported at the origin, so we have $\tau=\tilde{\tau} + \sum b_{\alpha}
\delta^{(\alpha)}$ for some constants $b_{\alpha}\in{\ensuremath{\mathbb{C}}}$. Then, for any $\lambda>0$, we have $$\tau_{\lambda}-\lambda^{m}\tau=\tilde{\tau}_{\lambda}-\lambda^{m}\tilde{\tau} +
\sum (\lambda^{-(d+2-{\ensuremath{\langle\! \alpha\!\rangle}})} -\lambda^m) b_{\alpha}\delta^{(\alpha)}.
\label{eq:Appendix.tau-l-tau-tilde}$$ As both $\tau$ and $\tilde{\tau}$ are homogeneous of degree $m$, we deduce that $\sum (\lambda^{-(d+2-{\ensuremath{\langle\! \alpha\!\rangle}})} -\lambda^m) b_{\alpha}\delta^{(\alpha)}=0$. The linear independence of the family $\{\delta^{(\alpha)}\}$ then implies that all the constants $b_{\alpha}$ vanish, that is, we have $\tilde{\tau}=\tau$. Thus $\tau$ is the unique homogeneous extension of $p(\xi)$ on ${\ensuremath{{\ensuremath{\mathbb{R}}}^{d+1}}}$.
Now, assume that $m$ is an integer $\leq -(d+2)$. Then in the formula (\[eq:Appendix.almosthomogeneous-extension\]) for $\tau$ we can take $k=-(m+d+2)$ and let $\psi$ be of the form, $$\psi(\mu)=\int_{\log \mu}^\infty h'(t) dt, \qquad h'(t)=\prod_{a=m+d+2}^{m+d+2+k} (\frac{1}{a} \frac{d}{dt} +1) g(t),
$$ with $g\in C_{c}^\infty({\ensuremath{{\ensuremath{\mathbb{R}}}^{d+1}}})$ such that $\int g(t)dt=1$. Then thanks to (\[eq:Appendix.differentiation-rhoalpha\]) and (\[eq:Appendix1.integral-a-g\]) we have $\rho_{\alpha}(\lambda)=0$ for ${\ensuremath{\langle\! \alpha\!\rangle}}<-(m+d+2)$, while for ${\ensuremath{\langle\! \alpha\!\rangle}}=-(m+d+2)$ we get $$\frac{d}{ds}\rho_{\alpha}(e^s)=\int h'(t)dt= \int g(t)dt= 1.
$$ Since $\rho_{\alpha}(1)=0$ it follows that $\rho_{\alpha}(e^{s})=s$, that is, we have $\rho_{\alpha}(\lambda)=\log \lambda$. Thus, $$\tau_{\lambda}=\lambda^{m}\tau +\lambda^{m}\log \lambda\sum_{{\ensuremath{\langle\! \alpha\!\rangle}}=-(m+d+2)} c_{\alpha}(p)\delta^{(\alpha)} \qquad \forall \lambda>0.
\label{eq:Appendix.taulambda-tau}$$ In particular, we see that if all the coefficients $c_{\alpha}(p)$ with ${\ensuremath{\langle\! \alpha\!\rangle}}=-(m+d+2)$ vanish then $\tau$ is homogeneous of degree $m$.
Conversely, suppose that $p(\xi)$ admits a homogeneous extension $\tilde{\tau}\in{\ensuremath{\mathcal{S}}}'({\ensuremath{{\ensuremath{\mathbb{R}}}^{d+1}}})$. As $\tau-\tilde{\tau}$ is supported at $0$, we can write $\tau=\tilde{\tau} + \sum b_{\alpha}
\delta^{(\alpha)}$ with $b_{\alpha}\in{\ensuremath{\mathbb{C}}}$. For any $\lambda>0$ we have $ \tilde{\tau}_{\lambda}=\lambda^{m}\tilde{\tau}$, so by combining this with (\[eq:Appendix.tau-l-tau-tilde\]) we get $$\tau_{\lambda}-\lambda^{m}\tau = \sum_{{\ensuremath{\langle\! \alpha\!\rangle}}\neq -(m+d+2)}
b_{\alpha}(\lambda^{-({\ensuremath{\langle\! \alpha\!\rangle}}+d+2)}-\lambda^{m})\delta^{(\alpha)}.
$$ By comparing this with (\[eq:Appendix.taulambda-tau\]) and by using linear independence of the family $\{\delta^{(\alpha)}\}$ we then deduce that we have $c_{\alpha}(p)=0$ for ${\ensuremath{\langle\! \alpha\!\rangle}}= -(m+d+2)$. Therefore $p(\xi)$ admits a homogeneous extension if and only if all the coefficients $c_{\alpha}(p)$ with ${\ensuremath{\langle\! \alpha\!\rangle}}=-(m+d+2)$ vanish. The proof of Lemma \[lem:Heisenberg.extension-symbol\] is thus achieved.
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tag:blogger.com,1999:blog-83178101059150572722019-05-23T05:38:04.402-04:00Sue Barr Sue Barrhttp://www.blogger.com/profile/06347249118943488564noreply@blogger.comBlogger214125tag:blogger.com,1999:blog-8317810105915057272.post-81462934193960502332019-05-22T10:56:00.000-04:002019-05-22T16:24:02.866-04:00Monday Meditation ~ May 20<div class="separator" style="clear: both; text-align: center;"><a href="https://1.bp.blogspot.com/--dxO6FtIoFc/XNBGbuvD1nI/AAAAAAAAEFY/yafw6YzbEZAwOAhmsl8acog7Ix9LsIlugCPcBGAYYCw/s1600/Monday%2BMeditation.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="200" data-original-width="800" height="160" src="https://1.bp.blogspot.com/--dxO6FtIoFc/XNBGbuvD1nI/AAAAAAAAEFY/yafw6YzbEZAwOAhmsl8acog7Ix9LsIlugCPcBGAYYCw/s640/Monday%2BMeditation.png" width="640" /></a></div>I'm starting a new Bible study, the Proverbs 31 Woman. When you read this passage in the Bible it seems quite daunting. What a paragon of a woman, but we have to realize this passage was about King Lemuel's mother advising him of what to look for in his search for a wife. She wanted him to find a woman who esteemed him greatly, (to put it in Jane Austen vernacular), and for him to esteem her in return.<br /><div class="separator" style="clear: both; text-align: center;"><a href="https://3.bp.blogspot.com/-QczyJti2Jmo/XOVfESQAT7I/AAAAAAAAEGk/v-Nkz4YN2BMrv_rVah0yM-3l0PndmooRwCLcBGAs/s1600/Prov%2B31%2BWoman.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="800" data-original-width="800" height="640" src="https://3.bp.blogspot.com/-QczyJti2Jmo/XOVfESQAT7I/AAAAAAAAEGk/v-Nkz4YN2BMrv_rVah0yM-3l0PndmooRwCLcBGAs/s640/Prov%2B31%2BWoman.png" width="640" /></a></div>Even though the intent of these verses were to instruct a son, all Scripture is given for the edification and instruction to man (as in human kind). I've read, and agreed with, that God allowed these verses to be included in the Bible for two different reasons.<br /><br />TO INSTRUCT MEN in general on what to look for in a godly wife.<br /><br />TO INSTRUCT WOMEN on how to be a more Biblically-minded woman.<br /><br />Both are important to the Body of Christ - his church. So... here I go, diving into a Bible study that will help me esteem my husband more while at the same time, bring me closer in my walk with God.<br /><div class="separator" style="clear: both; text-align: center;"><a href="https://3.bp.blogspot.com/-NgPbIaeLv0c/W4XehbXT76I/AAAAAAAADu8/cPLObZ1Q7mEJ2hrrgp8fuk-UZ5oqbzkzgCPcBGAYYCw/s1600/Always%2BWith%2BLove.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="100" data-original-width="265" src="https://3.bp.blogspot.com/-NgPbIaeLv0c/W4XehbXT76I/AAAAAAAADu8/cPLObZ1Q7mEJ2hrrgp8fuk-UZ5oqbzkzgCPcBGAYYCw/s1600/Always%2BWith%2BLove.png" /></a></div><br />Sue Barrhttp://www.blogger.com/profile/06347249118943488564noreply@blogger.com0tag:blogger.com,1999:blog-8317810105915057272.post-34014873465140297792019-05-17T12:55:00.002-04:002019-05-17T12:55:54.238-04:00First Page Friday ~ CAROLINE<div class="separator" style="clear: both; text-align: center;"><a href="https://4.bp.blogspot.com/-vmVMm6MVP60/XN7mdnn06EI/AAAAAAAAEGQ/AIGLSUdM3jUFkxeBJxpqdXJemMagA8GuQCLcBGAs/s1600/FPF%2B-%2BBlog.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="200" data-original-width="800" src="https://4.bp.blogspot.com/-vmVMm6MVP60/XN7mdnn06EI/AAAAAAAAEGQ/AIGLSUdM3jUFkxeBJxpqdXJemMagA8GuQCLcBGAs/s1600/FPF%2B-%2BBlog.png" /></a></div><br />I'm thrilled to share with you <a href="https://www.books2read.com/u/mVZBqP" target="_blank"><b><i><span style="color: #3d85c6; font-family: Georgia, Times New Roman, serif;">CAROLINE</span></i></b></a>. She was my first Jane Austen Fan Fiction and the whole premise of the book began with one simple question.<br /><br /><div style="text-align: center;"><span style="font-family: Georgia, Times New Roman, serif; font-size: large;"><i><b>Whatever happened to Caroline Bingley after both her brother AND Mr. Darcy became engaged to the Bennet sisters?</b></i></span></div><br />The concept of what she felt and how she dealt with this crushing blow to her long held dreams galvanized me into who Caroline was when no one was around. And I delighted in presenting her with a man who saw beneath her calculating facade and actually liked her for who she was and what she could become.<br /><br /><span style="font-family: inherit;"><b>Blurb</b>:</span><br /><span style="font-family: inherit;"><br /></span><span style="font-family: inherit;"><span style="background-color: white; color: #333333;">Caroline Bingley, beyond frustrated with her brother, Charles and Mr. Darcy both proposing to the Bennet sisters, dreads their upcoming nuptials. For three years her sole focus has been on attaining a marriage proposal from one Fitzwilliam Darcy of Pemberley, only to be foiled by a country miss with ‘fine eyes’. Adrift and not sure of her place in life, she meets the mysterious and devastatingly handsome Lord Nathan, who equally vexes and intrigues her.</span><br style="background-color: white; color: #333333;" /><br style="background-color: white; color: #333333;" /><span style="background-color: white; color: #333333;">Lord Nathan Kerr, third in line to a Dukedom, had a well-earned reputation as a Rake. He cast all that and his noble title aside to become Mr. Darcy’s vicar in Kympton, finding contentment in leading his small flock and doing the Lord’s work. His plan for a quiet, country life is thrown into upheaval when he meets the fiery Miss Bingley. Can he reconcile his rising desire for the spoiled miss with how a vicar’s wife ‘should’ behave?</span></span><br /><span style="font-family: inherit;"><br /></span><span style="font-family: inherit;"><b>First Page:</b></span><br /><br /><div class="CH" style="text-align: center;"><span lang="EN-US"><span style="font-family: Georgia, Times New Roman, serif;"><i><span style="font-size: large;">One</span><o:p></o:p></i></span></span></div><blockquote class="tr_bq" style="line-height: 150%; mso-pagination: none; text-indent: .3in;"><span lang="EN-US"><span style="font-family: Georgia, Times New Roman, serif;"><i>Caroline Bingley descended the grand staircase and proceeded toward the breakfast room, barely acknowledging the footman who efficiently pulled a chair away from the table for her. With a soft swish of silk, she settled on the seat. When the second footman poured some tea, she deigned to give him a slight nod of approval, but that was because she was in a fine mood. </i></span></span></blockquote><blockquote class="tr_bq" style="line-height: 150%; mso-pagination: none; text-indent: .3in;"><span lang="EN-US"><span style="font-family: Georgia, Times New Roman, serif;"><i>She noted through the window overlooking her sister’s favorite garden that it was a beautiful fall morning, the leaves of the oak trees a riotous flame of red and orange. October was just around the corner and her good mood, which had been in evidence since late August, showed no sign of wavering. She’d even gone to church the day prior for mass and enjoyed the hymns, although the bishop nearly bored her to tears. </i></span></span></blockquote><blockquote class="tr_bq" style="line-height: 150%; mso-pagination: none; text-indent: .3in;"><span lang="EN-US"><span style="font-family: Georgia, Times New Roman, serif;"><i>As she fussed with her morning meal, she mentally ticked off plans she and her brother had for the upcoming week. Charles had papers to sign with their lawyer to quit the lease at Netherfield Park and she had a fitting with Madam Beaufort before the much anticipated Annual Michaelmas Ball at Lady Addleworth’s mansion.</i></span></span></blockquote><blockquote class="tr_bq" style="line-height: 150%; mso-pagination: none; text-indent: .3in;"><span lang="EN-US"><span style="font-family: Georgia, Times New Roman, serif;"><i>Mr. Fitzwilliam Darcy was sure to attend the ball. He and his cousin, Colonel Fitzwilliam, who’d arrived in Town the other day, were spending an inordinate amount of time with Charles. She hoped it was because Charles finally decided to pursue an arrangement with Miss Georgiana Darcy. Darcy and the good Colonel, as her guardians, would finalize the articles of marriage as well as details of the dowry. With luck Charles would announce their engagement before the ball and with his shy little sister taken care of, Mr. Darcy could finally concentrate on his own happiness.</i></span></span></blockquote><div class="MsoNormal"><span lang="EN-US"><o:p> There you have it. I hope you enjoyed. Until next time,</o:p></span></div><div class="separator" style="clear: both; text-align: center;"><a href="https://3.bp.blogspot.com/-NgPbIaeLv0c/W4XehbXT76I/AAAAAAAADu8/cPLObZ1Q7mEJ2hrrgp8fuk-UZ5oqbzkzgCPcBGAYYCw/s1600/Always%2BWith%2BLove.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="100" data-original-width="265" src="https://3.bp.blogspot.com/-NgPbIaeLv0c/W4XehbXT76I/AAAAAAAADu8/cPLObZ1Q7mEJ2hrrgp8fuk-UZ5oqbzkzgCPcBGAYYCw/s1600/Always%2BWith%2BLove.png" /></a></div><div class="MsoNormal"><span lang="EN-US"><o:p><br /></o:p></span></div><br />Sue Barrhttp://www.blogger.com/profile/06347249118943488564noreply@blogger.com2tag:blogger.com,1999:blog-8317810105915057272.post-4346346049783575252019-05-15T07:16:00.000-04:002019-05-15T07:16:23.503-04:001950's Housewife ChallengeHave you heard of this? Now, before you raise your suffragette bra in the air as a rallying cry, I'm not here to say women belong only in the kitchen and has no place in the boardroom. Women should go/do what makes them feel fulfilled.<br /><div class="separator" style="clear: both; text-align: center;"><a href="https://4.bp.blogspot.com/-e6ercTcjq6k/XNv0GVZn_PI/AAAAAAAAEGE/2RaV3esn1lYaWIjNxNPKKU829nNb941MACLcBGAs/s1600/1950%2527s%2BFamily%2BDinner.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="319" data-original-width="614" height="331" src="https://4.bp.blogspot.com/-e6ercTcjq6k/XNv0GVZn_PI/AAAAAAAAEGE/2RaV3esn1lYaWIjNxNPKKU829nNb941MACLcBGAs/s640/1950%2527s%2BFamily%2BDinner.jpg" width="640" /></a></div>The list below was taken from a 1950's Home Economics textbook intended for High School girls in preparation for married life. Apparently Snopes has discredited this claim, but I've read this in a few places and watched some YouTube programs (circa 1950's) which spout the same ideals. So... I think it's quite authentic.<br /><br />Here is the list:<br /><ol style="background-color: white; box-sizing: border-box; color: #4f6d67; font-family: "Crimson Text"; font-size: 16px; margin-bottom: 26px; padding: 0px;"><li style="box-sizing: border-box; line-height: 2 !important; margin-left: 21px;">Have dinner ready: Plan ahead, even the night before, to have a delicious meal — on time. This is a way of letting him know that you have been thinking about him, and are concerned about his needs. Most men are hungry when they come home and the prospects of a good meal are part of the warm welcome needed.</li><li style="box-sizing: border-box; line-height: 2 !important; margin-left: 21px;">Prepare yourself: Take 15 minutes to rest so you will be refreshed when he arrives. Touch up your makeup, put a ribbon in your hair and be fresh looking. He has just been with a lot of work-weary people. Be a little gay and a little more interesting. His boring day may need a lift.</li><li style="box-sizing: border-box; line-height: 2 !important; margin-left: 21px;">Clear away the clutter. Make one last trip through the main part of the house just before your husband arrives, gathering up school books, toys, paper, etc. Then run a dust cloth over the tables. Your husband will feel he has reached a haven of rest and order, and it will give you a lift, too.</li><li style="box-sizing: border-box; line-height: 2 !important; margin-left: 21px;">Prepare the children: Take a few minutes to wash the children’s hands and faces if they are small, comb their hair, and if necessary, change their clothes. They are little treasures and he would like to see them playing the part.</li><li style="box-sizing: border-box; line-height: 2 !important; margin-left: 21px;">Minimize the noise: At the time of his arrival, eliminate all noise of washer, dryer, dishwasher or vacuum. Try to encourage the children to be quiet. Be happy to see him. Greet him with a warm smile and be glad to see him.</li><li style="box-sizing: border-box; line-height: 2 !important; margin-left: 21px;">Some Don’ts: Don’t greet him with problems or complaints. Don’t complain if he’s late for dinner. Count this as minor compared with what he might have gone through that day.</li><li style="box-sizing: border-box; line-height: 2 !important; margin-left: 21px;">Make him comfortable: Have him lean back in a comfortable chair or suggest he lie down in the bedroom. Have a cool or warm drink ready for him. Arrange his pillow and offer to take off his shoes. Speak in a low, soft, soothing and pleasant voice. Allow him to relax and unwind.</li><li style="box-sizing: border-box; line-height: 2 !important; margin-left: 21px;">Listen to him: You may have a dozen things to tell him, but the moment of his arrival is not the time. Let him talk first.</li><li style="box-sizing: border-box; line-height: 2 !important; margin-left: 21px;">Make the evening his: Never complain if he does not take you out to dinner or to other places of entertainment; instead, try to understand his world of strain and pressure, his need to be home and relax.</li><li style="box-sizing: border-box; line-height: 2 !important; margin-left: 21px;">The goal: Try to make your home a place of peace and order where your husband can relax.</li></ol><div><span style="color: #4f6d67;">No. 6 was my bug-a-boo early in our marriage. </span></div><div><span style="color: #4f6d67;"><br /></span></div><div><span style="color: #4f6d67;">My husband was with the Air Force, now retired and flying commercial aircraft. When the boys were little and he'd call home, I'd launch into a diatribe of what they'd done, how tired I was, etc., etc.. His frustration levels grew along with mine because, as a guy, he wanted to 'fix' my problems and couldn't. He was not physically there. He finally asked me to NOT tell him my issues. Not that he didn't care, but that he couldn't help. I learned early to bite my tongue and take care of things and keep things pleasant. He didn't need to worry about the family back home. Keep in mind, when in the military he was gone for weeks/months at a time.</span></div><div><span style="color: #4f6d67;"><br /></span></div><div><span style="color: #4f6d67;">Even now, with commercial aviation, he's away / out of country for days. When he arrives home, I greet him at the door with a smile and kiss (not because of the list above but because I want him to know I'm glad he's finally home). I DO offer him a drink. Sometimes it's just a coffee as he's flown all night and it's 0600 in the morning, and sometimes it's a nice cool glass of wine, and we enjoy it on the front porch together.</span></div><div><span style="color: #4f6d67;"><br /></span></div><div><span style="color: #4f6d67;">Do I HAVE to do all this? Short answer: No</span></div><div><span style="color: #4f6d67;"><br /></span></div><div><span style="color: #4f6d67;">Do I WANT to do all this? Short answer: Yes</span></div><div><span style="color: #4f6d67;"><br /></span></div><div><span style="color: #4f6d67;">I truly want my husband to know that I treasure him. He is my best friend, lover and loved one. He was with me before we had children and he is by my side now that they are married and gone. And the one thing I've noticed, after 37+ years of marriage.... The more I show him how much I love him, the more he shows me how much he loves me. It's a mutual admiration society in the Barr household.</span></div><div><span style="color: #4f6d67;"><br /></span></div><div><span style="color: #4f6d67;">There just might be something about those 1950's housewives and their quaint ways that strike a chord during these turbulent times.</span></div><div class="separator" style="clear: both; text-align: center;"><a href="https://3.bp.blogspot.com/-NgPbIaeLv0c/W4XehbXT76I/AAAAAAAADu8/cPLObZ1Q7mEJ2hrrgp8fuk-UZ5oqbzkzgCPcBGAYYCw/s1600/Always%2BWith%2BLove.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="100" data-original-width="265" src="https://3.bp.blogspot.com/-NgPbIaeLv0c/W4XehbXT76I/AAAAAAAADu8/cPLObZ1Q7mEJ2hrrgp8fuk-UZ5oqbzkzgCPcBGAYYCw/s1600/Always%2BWith%2BLove.png" /></a></div><div><span style="color: #4f6d67;"><br /></span></div><div><br /></div>Sue Barrhttp://www.blogger.com/profile/06347249118943488564noreply@blogger.com0tag:blogger.com,1999:blog-8317810105915057272.post-54196610849536255392019-05-13T11:25:00.000-04:002019-05-13T11:30:50.671-04:00Monday Meditation ~ Choices<h2 style="clear: both; text-align: center;"><span style="color: #3d85c6; font-family: "georgia" , "times new roman" , serif; font-size: x-large;"><i>May 13, 2019</i></span></h2><div class="separator" style="clear: both; text-align: center;"><a href="https://1.bp.blogspot.com/--dxO6FtIoFc/XNBGbuvD1nI/AAAAAAAAEFY/yafw6YzbEZAwOAhmsl8acog7Ix9LsIlugCPcBGAYYCw/s1600/Monday%2BMeditation.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="200" data-original-width="800" height="160" src="https://1.bp.blogspot.com/--dxO6FtIoFc/XNBGbuvD1nI/AAAAAAAAEFY/yafw6YzbEZAwOAhmsl8acog7Ix9LsIlugCPcBGAYYCw/s640/Monday%2BMeditation.png" width="640" /></a></div>There are times, as you journey through life, where you have to make hard decisions. Choosing to follow Christ is one of them. I struggle daily with choices. Television programs are one of them. While I enjoy the creativity and inherent intelligence writers have, I find the content leaves a LOT to be desired. My family's favourite show, ending an eleven year run this month, is riddled with sexual content. When my husband asked why I didn't want to watch the program anymore, I said as much. He was surprised, until I sat with him and pointed out how every scene (rarely no exception) had implied or direct sexual innuendo. He truly had never thought about that. He just enjoyed the comedy.<br /><br />This is how the enemy sneaks behind our borders. It's fun. It's comedy. What harm is there in watching a show on TV. The sad truth is you become inured to the oily tentacles of sin. Who cares that everybody in that show is having sex outside of marriage, or all they are looking for is to 'hook up' with someone (for sex)? It's only a show, for crying out loud.<br /><br />We are to be set apart. We are to be the standard bearers for God. We are to shun all appearance of evil. We are to make hard choices. Our reward, when we finally leave this beautiful earth, is to be present with the very God we choose to honour. What do you choose? Only you and Jesus know the absolute truth.<br /><div class="separator" style="clear: both; text-align: center;"><a href="https://1.bp.blogspot.com/-RQbaRK-kGN8/XNmK99h3ULI/AAAAAAAAEF4/RHO8py25aX0TNovk-YECL9SoajIILcXxQCLcBGAs/s1600/Joshua%2B24.15.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="500" data-original-width="500" height="640" src="https://1.bp.blogspot.com/-RQbaRK-kGN8/XNmK99h3ULI/AAAAAAAAEF4/RHO8py25aX0TNovk-YECL9SoajIILcXxQCLcBGAs/s640/Joshua%2B24.15.png" width="640" /></a></div>I wish all the ladies a Happy Mother's Day. Mine was wonderful and I now have a Clematis and Lavender bush to plant.<br /><div class="separator" style="clear: both; text-align: center;"><a href="https://3.bp.blogspot.com/-NgPbIaeLv0c/W4XehbXT76I/AAAAAAAADu8/cPLObZ1Q7mEJ2hrrgp8fuk-UZ5oqbzkzgCPcBGAYYCw/s1600/Always%2BWith%2BLove.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="100" data-original-width="265" src="https://3.bp.blogspot.com/-NgPbIaeLv0c/W4XehbXT76I/AAAAAAAADu8/cPLObZ1Q7mEJ2hrrgp8fuk-UZ5oqbzkzgCPcBGAYYCw/s1600/Always%2BWith%2BLove.png" /></a></div><br />Sue Barrhttp://www.blogger.com/profile/06347249118943488564noreply@blogger.com0tag:blogger.com,1999:blog-8317810105915057272.post-77691915145556386392019-05-10T05:00:00.000-04:002019-05-10T05:00:09.599-04:00First Page Friday ~ Guest ~ Kelly MILLER<table align="center" cellpadding="0" cellspacing="0" class="tr-caption-container" style="margin-left: auto; margin-right: auto; text-align: center;"><tbody><tr><td style="text-align: center;"><a href="https://4.bp.blogspot.com/-Lf_8KInQF6E/XM2zwggMYjI/AAAAAAAAEEQ/HYKrcd4w-nwaLSdVzDZafrykM0oo5WJgACEwYBhgL/s1600/FPF%2B-%2BMiller%2B-%2BBlog.png" imageanchor="1" style="margin-left: auto; margin-right: auto;"><img border="0" data-original-height="200" data-original-width="800" src="https://4.bp.blogspot.com/-Lf_8KInQF6E/XM2zwggMYjI/AAAAAAAAEEQ/HYKrcd4w-nwaLSdVzDZafrykM0oo5WJgACEwYBhgL/s1600/FPF%2B-%2BMiller%2B-%2BBlog.png" /></a></td></tr><tr><td class="tr-caption" style="text-align: center;">Temporary Cover provided by Sue Barr</td></tr></tbody></table><br />Today, my guest is Kelly Miller with her #JAFF novel: <span style="font-family: "georgia" , "times new roman" , serif;"><i><b>Death Takes a Holiday at Pemberley</b></i></span>. Because the book is scheduled to publish this upcoming June, she's not able to provide buy links. However, if you find this snippet intriguing and want to know more about Kelly, you can connect with her on <a href="https://www.facebook.com/profile.php?id=100011779365688" target="_blank"><b><span style="color: #4c1130;">Facebook</span></b></a>. I'm sure she'll keep all of us up-to-date from there.<br /><br /><b style="font-family: inherit;">Blurb:</b><br /><br /><div class="MsoNormal" style="background: white; line-height: normal; mso-pagination: none;"><span style="font-family: inherit;">When an Angel of Death forces Fitzwilliam Darcy to host his sojourn in the world of mortals, Darcy's happy, well-ordered life becomes a chaotic nightmare. Does Darcy have any hope of protecting his family from the mercurial whims of an all-powerful angel?</span></div><div class="MsoNormal" style="line-height: normal; mso-pagination: none;"><br /></div><div class="MsoNormal" style="background: white; line-height: normal; mso-pagination: none;"><span lang="EN"><span style="font-family: inherit;">Fitzwilliam Darcy has everything a man could ask for. A man of fortune, property, and good social standing, he is blissfully married to his wife, Elizabeth, and has a young son. With so much to live for, he is deeply shaken by a near fatal riding accident. Darcy miraculously escaped without injury, but soon he is visited by an otherworldly being, an Angel of Death, Graham. Graham compels Darcy to act as his guide for his sojourn in the world of mortals, and makes it clear that refusing would have dire consequences.<o:p></o:p></span></span></div><div class="MsoNormal" style="background: white; line-height: normal; mso-pagination: none;"><br /></div><div class="MsoNormal" style="background: white; line-height: normal; mso-pagination: none;"><span lang="EN"><span style="font-family: inherit;">Darcy almost immediately begins to question Graham's true motives for choosing to stay at Pemberley. Can he trust Graham's assurance that no harm will come to his wife and child? Why does Graham insist on spending time with Elizabeth? How can Darcy possibly protect his family from an angel who has power over life and death? <o:p></o:p></span></span></div><div class="MsoNormal" style="background: white; line-height: normal; mso-pagination: none;"><br /></div><div class="MsoNormal" style="background: white; line-height: normal; mso-pagination: none;"><span lang="EN"><span style="font-family: inherit;">In this romantic fantasy, the beloved couple from Jane Austen's <i style="mso-bidi-font-style: normal;">Pride and Prejudice</i> are forced to deal with both human and unearthly challenges. Will their extraordinary love conquer all or are the fates against them?"<o:p></o:p></span></span></div><div class="MsoNormal" style="line-height: normal; mso-pagination: none;"><br /></div><div class="MsoNormal" style="line-height: normal; mso-pagination: none;"><span lang="EN"><span style="font-family: inherit;"><b>First Page:<o:p></o:p></b></span></span></div><div class="MsoNormal" style="line-height: normal; mso-pagination: none;"><br /></div><div class="MsoNormal" style="line-height: normal; text-align: center;"><span lang="EN"><span style="font-family: "georgia" , "times new roman" , serif;">Chapter 1: The Remarkable Mr. Graham</span><span style="font-family: inherit;"><o:p></o:p></span></span></div><blockquote class="tr_bq" style="line-height: normal; mso-pagination: none;"><span lang="EN"><o:p><span style="font-family: inherit;"> </span></o:p></span><i style="font-family: Georgia, "Times New Roman", serif;">September 14, 1815</i></blockquote><blockquote class="tr_bq" style="line-height: normal; mso-pagination: none;"><i style="font-family: Georgia, "Times New Roman", serif;">The view afforded by the precipice not three feet from Fitzwilliam Darcy was vast and daunting to most, but the gentleman spared nary a glance to the stupendous prospect of the limestone ravine below. He had taken this trail hundreds of times before, as had Regal, his majestic, black Arabian stallion.</i></blockquote><blockquote class="tr_bq" style="line-height: normal; mso-pagination: none;"><i style="font-family: Georgia, "Times New Roman", serif;">The horse flicked his ears back now and again as his master expounded in great detail on the presents he had purchased for his wife and son. Although considered by most a laconic sort, Darcy had discovered as a child that talking to horses benefitted both parties; the sound of his voice calmed the horses, and he was more comfortable talking with them than with people. He leaned forward in the saddle as he spoke, his mind focused upon his discourse. In listing to Regal the virtues of the notable equine specimens scheduled for delivery from the neighboring estate, he reaffirmed in his own mind that he had chosen well.</i></blockquote><blockquote class="tr_bq" style="line-height: normal; mso-pagination: none;"><i style="font-family: Georgia, "Times New Roman", serif;">A cutting gust interrupted his thoughts with a pervasive discomfort, prompting him to raise the collar of his greatcoat and affirming his choice to take this particular route; the narrow, winding trail was a faster route to Pemberley than the road.</i></blockquote><blockquote class="tr_bq" style="line-height: normal; mso-pagination: none;"><i style="font-family: Georgia, "Times New Roman", serif;">Because of circumstances beyond his control, he had been spending precious little time with his wife and child of late. His steward, Mr. Hughes, had resigned his position several weeks earlier, announcing his intention of marrying and relocating with his new wife in Scotland. Until he found the right person for that crucial post, he was forced to deal with a multitude of tasks his steward would customarily have handled.</i></blockquote><div class="MsoNormal" style="line-height: normal; mso-pagination: none;"><span lang="EN"><o:p><span style="font-family: inherit;">Oh dear, the next few paragraphs will place our dear Mr. Darcy in dire circumstances...</span></o:p></span></div><div class="MsoNormal" style="line-height: normal; mso-pagination: none;"><span lang="EN"><o:p><span style="font-family: inherit;">Congratulations Kelly. I wish you all the best in this wonderful adventure of publishing. Until then,</span></o:p></span></div><div class="separator" style="clear: both; text-align: center;"><a href="https://3.bp.blogspot.com/-NgPbIaeLv0c/W4XehbXT76I/AAAAAAAADu8/cPLObZ1Q7mEJ2hrrgp8fuk-UZ5oqbzkzgCPcBGAYYCw/s1600/Always%2BWith%2BLove.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="100" data-original-width="265" src="https://3.bp.blogspot.com/-NgPbIaeLv0c/W4XehbXT76I/AAAAAAAADu8/cPLObZ1Q7mEJ2hrrgp8fuk-UZ5oqbzkzgCPcBGAYYCw/s1600/Always%2BWith%2BLove.png" /></a></div><div class="separator" style="clear: both; text-align: center;"><span style="color: #ea9999; font-family: "georgia" , "times new roman" , serif;"><i><b>P.S. Remember to hug someone you love today</b></i></span></div><div class="MsoNormal" style="line-height: normal; mso-pagination: none;"><span lang="EN"><o:p><span style="font-family: inherit;"><br /></span></o:p></span></div><br />Sue Barrhttp://www.blogger.com/profile/06347249118943488564noreply@blogger.com7tag:blogger.com,1999:blog-8317810105915057272.post-4675188948690697252019-05-08T12:09:00.000-04:002019-05-08T12:09:05.075-04:00Georgiana ~ Part Three<div class="separator" style="clear: both; text-align: center;"><a href="https://4.bp.blogspot.com/-rzTbZWfrxCs/XMCIyxGUkVI/AAAAAAAAEEA/9hAoHEdLvLkhHq2yCTvHFJoyAwJDBUH1gCPcBGAYYCw/s1600/Georgiana%2B-%2Bserialized.png" imageanchor="1" style="clear: left; float: left; margin-bottom: 1em; margin-right: 1em;"><img border="0" data-original-height="265" data-original-width="805" src="https://4.bp.blogspot.com/-rzTbZWfrxCs/XMCIyxGUkVI/AAAAAAAAEEA/9hAoHEdLvLkhHq2yCTvHFJoyAwJDBUH1gCPcBGAYYCw/s1600/Georgiana%2B-%2Bserialized.png" /></a></div><div class="MsoNormal" style="line-height: normal; mso-pagination: none; text-indent: 0in;">Before the week was out, the Darcy’s had arrived in London and were almost immediately beset upon by Lady Matlock and Colonel Fitzwilliam.<o:p></o:p></div><div class="MsoNormal" style="line-height: normal; mso-pagination: none; text-indent: 0in;"><br /></div><div class="MsoNormal" style="line-height: normal; mso-pagination: none; text-indent: 0in;">“Aunt Lucinda, Richard, you are here at an ungodly hour. We have only begun to break our fast.” Fitz stood and gave their aunt a kiss on the cheek.<o:p></o:p></div><div class="MsoNormal" style="line-height: normal; mso-pagination: none; text-indent: 0in;"><br /></div><div class="MsoNormal" style="line-height: normal; mso-pagination: none; text-indent: 0in;">Georgiana and Lizzy started to push away from the table to stand, but Lady Matlock waved them down. Richard called out a cheery ‘good morning’ and promptly attended the sideboard. Lady Matlock settled beside Georgiana and accepted a cup of tea from one of the footmen.<o:p></o:p></div><div class="MsoNormal" style="line-height: normal; mso-pagination: none; text-indent: 0in;"><br /></div><div class="MsoNormal" style="line-height: normal; mso-pagination: none; text-indent: 0in;">“To what do we owe the honor of your company so early, Aunt?” Fitz asked once he’d resumed his place at the table. “Richard, I know, came only to stuff his face with my food.”<o:p></o:p></div><div class="MsoNormal" style="line-height: normal; mso-pagination: none; text-indent: 0in;"><br /></div><div class="MsoNormal" style="line-height: normal; mso-pagination: none; text-indent: 0in;">He gave his cousin a mild glare as Richard sat down next to Lizzy after filling his plate to near overflowing. Richard’s reply was to raise an eyebrow and take a bite of his fresh baked roll.<o:p></o:p></div><div class="MsoNormal" style="line-height: normal; mso-pagination: none; text-indent: 0in;"><br /></div><div class="MsoNormal" style="line-height: normal; mso-pagination: none; text-indent: 0in;">“We have no time to waste,” Aunt Lucinda began. “Elizabeth and Georgiana’s presentations are but a scant eight weeks away. Not near enough time to accomplish everything I’d like. I have arranged several fittings at <i>Etienne’s</i>, for which I am eternally grateful to the Duchess of Adborough. Normally, Madame Etienne has a line weaving down the street for her services, but Margaret was able to exert her influence and hire her exclusively to attend all our girls.”<o:p></o:p></div><div class="MsoNormal" style="line-height: normal; mso-pagination: none; text-indent: 0in;"><br /></div><div class="MsoNormal" style="line-height: normal; mso-pagination: none; text-indent: 0in;">“Caroline and Kitty, as well?” Lizzy asked.<o:p></o:p></div><div class="MsoNormal" style="line-height: normal; mso-pagination: none; text-indent: 0in;"><br /></div><div class="MsoNormal" style="line-height: normal; mso-pagination: none; text-indent: 0in;">“Oh, yes. Margaret adores both of them and is determined to have them shine this upcoming Season.” Lady Matlock let out a satisfied sigh. “There are rumors the Prince Regent himself may attend in honor of Lord George’s loyalty to the Crown. I would not be surprised if he doesn’t bestow on George his own dukedom.”<o:p></o:p></div><div class="MsoNormal" style="line-height: normal; mso-pagination: none; text-indent: 0in;"><br /></div><div class="MsoNormal" style="line-height: normal; mso-pagination: none; text-indent: 0in;">Georgiana’s memory briefly touched on the near-scandal a few days before Nathan and Caroline’s wedding. Guests at Pemberley had been informed that a gun accidently discharged in the stables and Viscount Stanhope had been tragically killed. She knew that to be a falsehood as she’d overheard Fitzwilliam and Cousin Richard discussing Stanhope’s treachery.<o:p></o:p></div><div class="MsoNormal" style="line-height: normal; mso-pagination: none; text-indent: 0in;"><br /></div><div class="MsoNormal" style="line-height: normal; mso-pagination: none; text-indent: 0in;">“I hear the Barwick Duchy is vacant.” Richard volunteered the information around a mouthful of ham. “At least two hundred acres of valuable farmland and long-term tenants.”<o:p></o:p></div><div class="MsoNormal" style="line-height: normal; mso-pagination: none; text-indent: 0in;"><br /></div><div class="MsoNormal" style="line-height: normal; mso-pagination: none; text-indent: 0in;">“Yes, of course. And there are the secondary titles of Marquis of Glanworth and Viscount Mandeville.” Lady Matlock enthused.<o:p></o:p></div><div class="MsoNormal" style="line-height: normal; mso-pagination: none; text-indent: 0in;"><br /></div><div class="MsoNormal" style="line-height: normal; mso-pagination: none; text-indent: 0in;">“You are a veritable walking and talking volume of Debrett’s, dear aunt,” Fitz teased.<o:p></o:p></div><div class="MsoNormal" style="line-height: normal; mso-pagination: none; text-indent: 0in;"><br /></div><div class="MsoNormal" style="line-height: normal; mso-pagination: none; text-indent: 0in;">“You laugh now, Fitzwilliam, but when you and your lovely bride begin to move among London’s finest, you will be glad to know whom to avoid and whom to curry favor with,” she decreed with an elegant sniff. “If not for your sake, then for Georgiana’s. We expect her to make a splendid match this season. Already there are rumors the Marquis of Trevayne is on the hunt this year.”<o:p></o:p></div><div class="MsoNormal" style="line-height: normal; mso-pagination: none; text-indent: 0in;"><br /></div><div class="MsoNormal" style="line-height: normal; mso-pagination: none; text-indent: 0in;">“Dear God, Mama. You make it sound like he’s about to call the hounds.” Richard exclaimed. “Trevayne needs to find a wife, that a certainty, but he is not about to stalk the halls of Almack’s and pounce on innocent debutantes their first night out.”<o:p></o:p></div><div class="MsoNormal" style="line-height: normal; mso-pagination: none; text-indent: 0in;"><br /></div><div class="MsoNormal" style="line-height: normal; mso-pagination: none; text-indent: 0in;">“Be that as it may, I expect the foyer of Darcy house will be filled with flowers and cards when eligible gentlemen discover how beautiful my niece is.” Aunt Lucinda patted Georgiana’s hand.<o:p></o:p></div><div class="MsoNormal" style="line-height: normal; mso-pagination: none; text-indent: 0in;">Georgiana’s insides shriveled at her aunt’s sentiments. Her aunt meant well, however, she knew, just knew, many eyes would be on her because she was a Darcy and because she was wealthy. Not for the first time she was grateful Elizabeth, Caroline and Catherine would be by her side the whole season.<o:p></o:p></div><div class="MsoNormal" style="line-height: normal; mso-pagination: none; text-indent: 0in;"><br /></div><div class="MsoNormal" style="line-height: normal; mso-pagination: none; text-indent: 0in;">“Mama would be in raptures if Kitty became a titled lady in her own right,” Lizzy said with a laugh. <span style="text-indent: 0in;">“She will be hard put to remain humble if that happens.”</span></div><div class="MsoNormal" style="line-height: normal; mso-pagination: none; text-indent: 0in;"><span style="text-indent: 0in;"><br /></span></div><div class="MsoNormal" style="line-height: normal; mso-pagination: none; text-indent: 0in;"><o:p></o:p></div><div class="MsoNormal" style="line-height: normal; mso-pagination: none; text-indent: 0in;">“As soon as you are finished with breaking your fast, we shall meet the Duchess at <i>Etienne’s</i>. With six of us to outfit, she will need all hands-on deck.”<o:p></o:p></div><div class="MsoNormal" style="line-height: normal; mso-pagination: none; text-indent: 0in;"><br /></div><div class="MsoNormal" style="line-height: normal; mso-pagination: none; text-indent: 0in;">“Mother, where have you learned Navy terms?” Richard slathered more creamy butter on yet another roll. “Your second and most favored son is an army colonel” – Lady Matlock raised an elegant eyebrow – “one would assume you would use terminology more suited for land warfare.”<o:p></o:p></div><div class="MsoNormal" style="line-height: normal; mso-pagination: none; text-indent: 0in;"><br /></div><div class="MsoNormal" style="line-height: normal; mso-pagination: none; text-indent: 0in;">Lady Matlock waved a hand in Richard’s direction, as though shooing his concerns away from her thoughts. “I <i>do</i>know a few Admirals and Navy Captains, Richard. My social circle is quite wide, which you would know if you attended some functions with me.”<o:p></o:p></div><div class="MsoNormal" style="line-height: normal; mso-pagination: none; text-indent: 0in;"><br /></div><div class="MsoNormal" style="line-height: normal; mso-pagination: none; text-indent: 0in;">“Ah, but then you’d expect me to dance attendance on some fresh-faced chit of a girl and make polite talk with dewy eyed innocents.”<o:p></o:p></div><div class="MsoNormal" style="line-height: normal; mso-pagination: none; text-indent: 0in;"><br /></div><div class="MsoNormal" style="line-height: normal; mso-pagination: none; text-indent: 0in;">“Richard!”<o:p></o:p></div><div class="MsoNormal" style="line-height: normal; mso-pagination: none; text-indent: 0in;"><br /></div><div class="MsoNormal" style="line-height: normal; mso-pagination: none; text-indent: 0in;">“Sorry, Mother. That was quite unchivalrous,” Richard apologized, casting a slight eye-roll in the direction of her brother.<o:p></o:p></div><div class="MsoNormal" style="line-height: normal; mso-pagination: none; text-indent: 0in;"><br /></div><div class="MsoNormal" style="line-height: normal; mso-pagination: none; text-indent: 0in;">The all knew very well that Richard gave no thought, what-so-ever, to finding a wife. He was happily situated as an army Colonel and had no intentions of settling down. Aunt Lucinda would be better served finding a wife for her eldest son, Viscount Waverly. Although, he’d proven just as wily as Richard when it came to avoiding young misses and their ambitious mamas.<o:p></o:p></div><div class="MsoNormal" style="line-height: normal; mso-pagination: none; text-indent: 0in;"><br /></div><div class="MsoNormal" style="line-height: normal; mso-pagination: none; text-indent: 0in;">“Is Kitty meeting us at the modiste?” Lizzy interjected, obviously trying to move the conversation past the dangerous territory of Lady Matlock’s marital plans for her two unwed sons.<o:p></o:p></div><div class="MsoNormal" style="line-height: normal; mso-pagination: none; text-indent: 0in;"><br /></div><div class="MsoNormal" style="line-height: normal; mso-pagination: none; text-indent: 0in;">“No, George is escorting both Catherine and your sister Mary here. In fact, I’m surprised they’ve not arrived yet.”<o:p></o:p></div><div class="MsoNormal" style="line-height: normal; mso-pagination: none; text-indent: 0in;"><br /></div><div class="MsoNormal" style="line-height: normal; mso-pagination: none; text-indent: 0in;">“How lovely,” Georgiana enthused. “I cannot wait to see them both again. I delight in their companionship very much.”<o:p></o:p></div><div class="MsoNormal" style="line-height: normal; mso-pagination: none; text-indent: 0in;"><br /></div><div class="MsoNormal" style="line-height: normal; mso-pagination: none; text-indent: 0in;">As if on cue, Hutchins appeared at the door for the second time that morning.<o:p></o:p></div><div class="MsoNormal" style="line-height: normal; mso-pagination: none; text-indent: 0in;"><br /></div><div class="MsoNormal" style="line-height: normal; mso-pagination: none; text-indent: 0in;">“Lord George Kerr, Miss Bennet and Miss Catherine Bennet have arrived. They are in the blue drawing room.”<o:p></o:p></div><div class="MsoNormal" style="line-height: normal; mso-pagination: none; text-indent: 0in;"><br /></div><div class="MsoNormal" style="line-height: normal; mso-pagination: none; text-indent: 0in;">“Thank you, Hutchins. Would you please find out if they would like some tea while we finish in here?” Lizzy instructed.<o:p></o:p></div><div class="MsoNormal" style="line-height: normal; mso-pagination: none; text-indent: 0in;"><br /></div><div class="MsoNormal" style="line-height: normal; mso-pagination: none; text-indent: 0in;">“Already taken care of, ma’am.”<o:p></o:p></div><div class="MsoNormal" style="line-height: normal; mso-pagination: none; text-indent: 0in;"><br /></div><div class="MsoNormal" style="line-height: normal; mso-pagination: none; text-indent: 0in;">Hutchins gave a polite half bow and left the room.<o:p></o:p></div><div class="MsoNormal" style="line-height: normal; mso-pagination: none; text-indent: 0in;"><br /></div><div class="MsoNormal" style="line-height: normal; mso-pagination: none; text-indent: 0in;">“Brother, I have finished my breakfast, do you mind if I go on ahead and visit with Mary?”<o:p></o:p></div><div class="MsoNormal" style="line-height: normal; mso-pagination: none; text-indent: 0in;"><br /></div><div class="MsoNormal" style="line-height: normal; mso-pagination: none; text-indent: 0in;">“Of course, Georgie. We shall be along shortly, that is unless Richard fills his plate again, then it’s anybody’s guess when we will be free.”<o:p></o:p></div><div class="MsoNormal" style="line-height: normal; mso-pagination: none; text-indent: 0in;"><br /></div><div class="MsoNormal" style="line-height: normal; mso-pagination: none; text-indent: 0in;">“Richard can sit here all by himself. I wish to see my sisters.” Lizzy placed her napkin on her empty plate and stood. “Aunt Lucinda, do you care to join us?”<o:p></o:p></div><div class="MsoNormal" style="line-height: normal; mso-pagination: none; text-indent: 0in;"><br /></div><div class="MsoNormal" style="line-height: normal; mso-pagination: none; text-indent: 0in;">“You read my mind, Elizabeth. I’m more than ready to reacquaint myself with your two sisters.”<o:p></o:p></div><div class="MsoNormal" style="line-height: normal; mso-pagination: none; text-indent: 0in;">The three ladies made their way to the drawing room. With one sweeping glance Georgiana noted Lord George next to a radiant Kitty, and, on the couch across from them sat Mary, quiet as always. Georgiana hurried to her side.<o:p></o:p></div><div class="MsoNormal" style="line-height: normal; mso-pagination: none; text-indent: 0in;"><br /></div><div class="MsoNormal" style="line-height: normal; mso-pagination: none; text-indent: 0in;">“Mary, I am so glad you’ve come.” She took one of Mary’s hands in hers as she sat next to her and leaned in to whisper. “Were you surprised by Kitty and George’s news?”<o:p></o:p></div><div class="MsoNormal" style="line-height: normal; mso-pagination: none; text-indent: 0in;"><br /></div><div class="MsoNormal" style="line-height: normal; mso-pagination: none; text-indent: 0in;">“No, even though Kitty swore she’d never marry.” Mary replied, her tone exceedingly dry. “As an unofficial chaperone, I witnessed a few encounters between them. The outcome was inevitable, in my opinion.”<o:p></o:p></div><div class="MsoNormal" style="line-height: normal; mso-pagination: none; text-indent: 0in;"><br /></div><div class="MsoNormal" style="line-height: normal; mso-pagination: none; text-indent: 0in;">“It sounds so romantic,” Georgiana sighed out. In her mind’s eye she pictured Lord George thundering down the lane on a white horse, snatching Kitty up and demanding she marry him. Much like Lord Godfrey did with Lady Desdemona in the latest novel she’d read.<o:p></o:p></div><div class="MsoNormal" style="line-height: normal; mso-pagination: none; text-indent: 0in;"><br /></div><div class="MsoNormal" style="line-height: normal; mso-pagination: none; text-indent: 0in;">What would it be like to have someone declare their love so passionately? Her breath hitched at the thought of Maxwell Kerr holding her in such a daring way. Maybe even closer than he had during their dance. Close enough for a kiss. Her chest and neck began to flush at the thought. Fortunately, Mary interrupted her wayward thoughts before she turned bright red all the way to her hair line.<o:p></o:p></div><div class="MsoNormal" style="line-height: normal; mso-pagination: none; text-indent: 0in;"><br /></div><div class="MsoNormal" style="line-height: normal; mso-pagination: none; text-indent: 0in;">“I guess it was romantic.” Mary shrugged. “If you believe in that sort of thing.” She peered at Georgiana. “Are you all right? You look quite flushed.”<o:p></o:p></div><div class="MsoNormal" style="line-height: normal; mso-pagination: none; text-indent: 0in;"><br /></div><div class="MsoNormal" style="line-height: normal; mso-pagination: none; text-indent: 0in;">“Have you no romantic fantasies? Nothing you’ve thought about since you were a little girl?” she asked, hoping to divert Mary’s attention from her flustered behavior.<o:p></o:p></div><div class="MsoNormal" style="line-height: normal; mso-pagination: none; text-indent: 0in;"><br /></div><div class="MsoNormal" style="line-height: normal; mso-pagination: none; text-indent: 0in;">“Me?” Mary gave an indelicate little snort. “I have been ‘out’ in Meryton society for many years with no offers, no interest shown in me by any man young or old, and do not think this will change in the near future.”<o:p></o:p></div><div class="MsoNormal" style="line-height: normal; mso-pagination: none; text-indent: 0in;"><br /></div><div class="MsoNormal" style="line-height: normal; mso-pagination: none; text-indent: 0in;">“The men of Hertfordshire are fools if they cannot see what is set before them.” Georgiana declared. About to extol Mary’s best qualities her attention was diverted by her brother and Richard entering the room. Fitz immediately went to George and Kitty, extending his congratulations while her cousin joined her and Mary on the couch.<o:p></o:p></div><div class="MsoNormal" style="line-height: normal; mso-pagination: none; text-indent: 0in;"><br /></div><div class="MsoNormal" style="line-height: normal; mso-pagination: none; text-indent: 0in;">“You look well, Miss Bennet.” Richard said before sitting in the chair across from them.<o:p></o:p></div><div class="MsoNormal" style="line-height: normal; mso-pagination: none; text-indent: 0in;"><br /></div><div class="MsoNormal" style="line-height: normal; mso-pagination: none; text-indent: 0in;">Georgiana felt Mary start at Richard’s words. There was obviously some truth in what she said about men not noticing her.<o:p></o:p></div><div class="MsoNormal" style="line-height: normal; mso-pagination: none; text-indent: 0in;"><br /></div><div class="MsoNormal" style="line-height: normal; mso-pagination: none; text-indent: 0in;">“Thank you, Colonel.”<o:p></o:p></div><div class="MsoNormal" style="line-height: normal; mso-pagination: none; text-indent: 0in;"><br /></div><div class="MsoNormal" style="line-height: normal; mso-pagination: none; text-indent: 0in;">“Are you still at Longbourn?”<o:p></o:p></div><div class="MsoNormal" style="line-height: normal; mso-pagination: none; text-indent: 0in;"><br /></div><div class="MsoNormal" style="line-height: normal; mso-pagination: none; text-indent: 0in;">“I remain at home, if that is what you ask.” Mary replied through stiff lips and pulled herself more erect than Georgiana thought possible.<o:p></o:p></div><div class="MsoNormal" style="line-height: normal; mso-pagination: none; text-indent: 0in;"><br /></div><div class="MsoNormal" style="line-height: normal; mso-pagination: none; text-indent: 0in;">“I meant no offense, Miss Bennet. I do not keep abreast of <i>all</i> the gossip and only meant to inquire to your well being.”<o:p></o:p></div><div class="MsoNormal" style="line-height: normal; mso-pagination: none; text-indent: 0in;"><br /></div><div class="MsoNormal" style="line-height: normal; mso-pagination: none; text-indent: 0in;">“I am quite content. In all things.”<o:p></o:p></div><div class="MsoNormal" style="line-height: normal; mso-pagination: none; text-indent: 0in;"><br /></div><div class="MsoNormal" style="line-height: normal; mso-pagination: none; text-indent: 0in;"><i>Well said, Mary.<o:p></o:p></i></div><div class="MsoNormal" style="line-height: normal; mso-pagination: none; text-indent: 0in;"><i><br /></i></div><div class="MsoNormal" style="line-height: normal; mso-pagination: none; text-indent: 0in;">“Georgiana, are you and Mary ready to leave or does my son require a few minutes to gather more gossip from Hertfordshire?” Aunt Matlock had risen to her feet and looked pointedly at Richard.<o:p></o:p></div><div class="MsoNormal" style="line-height: normal; mso-pagination: none; text-indent: 0in;"><br /></div><div class="MsoNormal" style="line-height: normal; mso-pagination: none; text-indent: 0in;">“Mother, I am but inquiring after the health of Lizzie’s sister and was about to ask if Miss Bennet had continued with her shooting lessons.”<o:p></o:p></div><div class="MsoNormal" style="line-height: normal; mso-pagination: none; text-indent: 0in;"><br /></div><div class="MsoNormal" style="line-height: normal; mso-pagination: none; text-indent: 0in;">A few weeks after the incident with Viscount Stanhope, Richard insisted that Elizabeth and Georgiana know a little about pistols and how to take care of themselves if caught alone and unaware.<o:p></o:p></div><div class="MsoNormal" style="line-height: normal; mso-pagination: none; text-indent: 0in;"><br /></div><div class="MsoNormal" style="line-height: normal; mso-pagination: none; text-indent: 0in;">“I’d forgotten that you’d instructed the girls in the arts of self defense.” Lady Matlock gave a nervous laugh. “And to think Lady Catherine was concerned about the arts and allurements of young ladies. She’d succumb to apoplexy if she knew you’d shown them how to shoot.”<o:p></o:p></div><div class="MsoNormal" style="line-height: normal; mso-pagination: none; text-indent: 0in;"><br /></div><div class="MsoNormal" style="line-height: normal; mso-pagination: none; text-indent: 0in;">“After the incident at Pemberley, Darcy and I felt it imperative they know how to take care of things if we were not around.” Richard turned his attention back to Mary. “Have you continued with your lessons?”<o:p></o:p></div><div class="MsoNormal" style="line-height: normal; mso-pagination: none; text-indent: 0in;"><br /></div><div class="MsoNormal" style="line-height: normal; mso-pagination: none; text-indent: 0in;">“No, Papa is uncomfortable with me having such a deadly weapon.”<o:p></o:p></div><div class="MsoNormal" style="line-height: normal; mso-pagination: none; text-indent: 0in;"><br /></div><div class="MsoNormal" style="line-height: normal; mso-pagination: none; text-indent: 0in;">“Such a shame. You had natural talent.”<o:p></o:p></div><div class="MsoNormal" style="line-height: normal; mso-pagination: none; text-indent: 0in;"><br /></div><div class="MsoNormal" style="line-height: normal; mso-pagination: none; text-indent: 0in;">Of the three girls, Mary excelled and had shown an uncanny ability to hit the target at various distances causing Richard to declare that if Napoleon didn’t surrender, he would escort Mary to Paris himself and let her deal with the little man.<o:p></o:p></div><div class="MsoNormal" style="line-height: normal; mso-pagination: none; text-indent: 0in;"><br /></div><div class="MsoNormal" style="line-height: normal; mso-pagination: none; text-indent: 0in;">Often, Georgiana privately wondered if anything would ever blossom from their quirky friendship, but Mary had returned to Longbourn and Richard to his flirtatious ways. As far as she could tell, there didn’t seem to be a spark of interest in her cousin’s mien.<o:p></o:p></div><div class="MsoNormal" style="line-height: normal; mso-pagination: none; text-indent: 0in;"><br /></div><div class="MsoNormal" style="line-height: normal; mso-pagination: none; text-indent: 0in;">For the next week and a half Georgiana was busy with fittings and a mad social whirl of afternoon teas, musicales and plays. By the end of the second week she could barely hold her head up and began to nod off at supper.<o:p></o:p></div><div class="MsoNormal" style="line-height: normal; mso-pagination: none; text-indent: 0in;"><br /></div><div class="MsoNormal" style="line-height: normal; mso-pagination: none; text-indent: 0in;">“Fitz, I don’t want to seem ungrateful to Aunt Lucinda, she’s done so much, but Georgiana and I are exhausted.” Lizzy cast a sympathetic glance toward her. “We are not used to Town hours and I refuse to go out this evening. Do you think Lady Fosscroft would mind so much if we missed her poetry reading?”<o:p></o:p></div><div class="MsoNormal" style="line-height: normal; mso-pagination: none; text-indent: 0in;"><br /></div><div class="MsoNormal" style="line-height: normal; mso-pagination: none; text-indent: 0in;">Georgiana roused herself enough to murmur, “I truly don’t mind−” and hid a huge yawn behind her hand. She smiled, feeling a bit sheepish. “I am tired, but I also do not want to disappoint our aunt.”<o:p></o:p></div><div class="MsoNormal" style="line-height: normal; mso-pagination: none; text-indent: 0in;"><br /></div><div class="MsoNormal" style="line-height: normal; mso-pagination: none; text-indent: 0in;">“Let us continue this discussion in the family sitting room.” Fitz signaled the footman to begin clearing the table. “I will send our regrets to Aunt Lucinda.”<o:p></o:p></div><div class="MsoNormal" style="line-height: normal; mso-pagination: none; text-indent: 0in;"><br /></div><div class="MsoNormal" style="line-height: normal; mso-pagination: none; text-indent: 0in;">“Thank you, brother,” Georgiana impulsively took one of his hands in hers and pressed it to her cheek. “You always know how to make me feel more comfortable.” To Lizzy she said, “Shall we put on our dressing gowns and have a nice cup of hot chocolate before retiring?”<o:p></o:p></div><div class="MsoNormal" style="line-height: normal; mso-pagination: none; text-indent: 0in;"><br /></div><div class="MsoNormal" style="line-height: normal; mso-pagination: none; text-indent: 0in;">“That sounds divine, Georgie. It will make us feel like we are back at Pemberley.”<o:p></o:p></div><div class="MsoNormal" style="line-height: normal; mso-pagination: none; text-indent: 0in;"><br /></div><div class="separator" style="clear: both;"></div><div class="MsoNormal" style="line-height: normal; mso-pagination: none; text-indent: 0in;">“I shall meet you in the sitting room in a half hour.”<o:p></o:p></div><div class="separator" style="clear: both; text-align: center;"><a href="https://3.bp.blogspot.com/-NgPbIaeLv0c/W4XehbXT76I/AAAAAAAADu8/cPLObZ1Q7mEJ2hrrgp8fuk-UZ5oqbzkzgCPcBGAYYCw/s1600/Always%2BWith%2BLove.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="100" data-original-width="265" src="https://3.bp.blogspot.com/-NgPbIaeLv0c/W4XehbXT76I/AAAAAAAADu8/cPLObZ1Q7mEJ2hrrgp8fuk-UZ5oqbzkzgCPcBGAYYCw/s1600/Always%2BWith%2BLove.png" /></a></div><div class="MsoNormal" style="line-height: normal; mso-pagination: none; text-indent: 0in;"><br /></div><br />Sue Barrhttp://www.blogger.com/profile/06347249118943488564noreply@blogger.com0tag:blogger.com,1999:blog-8317810105915057272.post-36704604875070086172019-05-06T10:37:00.001-04:002019-05-13T11:30:31.948-04:00Monday Meditation ~ Power of God's Love<h2 style="clear: both; text-align: center;"><span style="color: #3d85c6; font-family: Georgia, Times New Roman, serif; font-size: x-large;"><i>May 6, 2019</i></span></h2><div class="separator" style="clear: both; text-align: center;"><a href="https://3.bp.blogspot.com/--dxO6FtIoFc/XNBGbuvD1nI/AAAAAAAAEFU/B9GnM6ZHM3cdyMn1_tm0ueeGWdWyuqOUQCLcBGAs/s1600/Monday%2BMeditation.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="200" data-original-width="800" height="160" src="https://3.bp.blogspot.com/--dxO6FtIoFc/XNBGbuvD1nI/AAAAAAAAEFU/B9GnM6ZHM3cdyMn1_tm0ueeGWdWyuqOUQCLcBGAs/s640/Monday%2BMeditation.png" width="640" /></a></div><b style="background-color: white; color: #222222; font-family: inherit;"><br /></b><a href="https://1.bp.blogspot.com/-YfaY90aONrY/XE70N-LvX2I/AAAAAAAAD8g/Le1Z483YGdAAgQ_AhFwVaAgBmxZG0HLGwCPcBGAYYCw/s1600/CtB.jpg" imageanchor="1" style="clear: left; float: left; margin-bottom: 1em; margin-right: 1em;"><img border="0" data-original-height="320" data-original-width="240" height="200" src="https://1.bp.blogspot.com/-YfaY90aONrY/XE70N-LvX2I/AAAAAAAAD8g/Le1Z483YGdAAgQ_AhFwVaAgBmxZG0HLGwCPcBGAYYCw/s200/CtB.jpg" width="150" /></a><b style="background-color: white; color: #222222; font-family: inherit;">Cornelia Arnolda Johanna "Corrie" ten Boom</b><span style="background-color: white; color: #222222; font-family: inherit;"> (15 April 1892 – 15 April 1983) was a Dutch watchmaker and later a writer who worked with her father </span><span style="font-family: inherit;">Casper ten Boom</span><span style="background-color: white; color: #222222; font-family: inherit;">, her sister </span><a href="https://en.wikipedia.org/wiki/Betsie_ten_Boom" style="background: none rgb(255, 255, 255); color: #0b0080; font-family: inherit; text-decoration-line: none;" title="Betsie ten Boom">Betsie ten Boom</a><span style="background-color: white; color: #222222; font-family: inherit;"> and other family members to help many Jews escape the </span><a href="https://en.wikipedia.org/wiki/Nazism" style="background: none rgb(255, 255, 255); color: #0b0080; font-family: inherit; text-decoration-line: none;" title="Nazism">Nazi</a><span style="background-color: white; color: #222222; font-family: inherit;"> </span><a href="https://en.wikipedia.org/wiki/The_Holocaust" style="background: none rgb(255, 255, 255); color: #0b0080; font-family: inherit; text-decoration-line: none;" title="The Holocaust">Holocaust</a><span style="background-color: white; color: #222222; font-family: inherit;"> during </span><a href="https://en.wikipedia.org/wiki/World_War_II" style="background: none rgb(255, 255, 255); color: #0b0080; font-family: inherit; text-decoration-line: none;" title="World War II">World War II</a><span style="background-color: white; color: #222222; font-family: inherit;"> by hiding them in her home.</span><br /><span style="font-family: inherit;"><span style="background-color: white; color: #222222;"><br /></span></span><span style="background-color: white; color: #222222; font-family: inherit;">They were caught and she was arrested and sent to </span><a href="https://en.wikipedia.org/wiki/Ravensbr%C3%BCck_concentration_camp" style="background: none rgb(255, 255, 255); color: #0b0080; text-decoration-line: none;" title="Ravensbrück concentration camp"><b><span style="font-family: "georgia" , "times new roman" , serif;"><i>Ravensbrück</i></span></b><span style="font-family: inherit;"> concentration camp</span></a><span style="background-color: white; color: #222222; font-family: inherit;">. Her most famous book, </span><i style="background-color: white; color: #222222;"><a href="https://en.wikipedia.org/wiki/The_Hiding_Place_(biography)" style="background: none; color: #0b0080; text-decoration-line: none;" title="The Hiding Place (biography)"><b><span style="font-family: "georgia" , "times new roman" , serif;">The Hiding Place</span></b></a></i><span style="background-color: white; color: #222222; font-family: inherit;">, is a biography that recounts the story of her family's efforts and how ten Boom found hope while imprisoned at the concentration camp.</span><br /><span style="background-color: white; color: #222222; font-family: inherit;"><br /></span><span style="font-family: inherit;"><span style="background-color: white; color: #222222;"><br /></span></span><span style="font-family: inherit;"><span style="background-color: white; color: #222222;">If anyone knew the power of God's love and what that all encompasses - it was Corrie ten Boom. I've said many times that I respect this woman of God soooo much. I recently watched an older clip where Kathryn Kuhlman interviewed Corrie and and she said something so profound, I had to share...</span></span><br /><div class="separator" style="clear: both; text-align: center;"><a href="https://4.bp.blogspot.com/-S1Ihxxu3PJ8/XNBE7esLLJI/AAAAAAAAEFI/j-t1p8GRYS0rOj4rASe-KeoNLA3OQAxOACLcBGAs/s1600/Nothing%2Btoo%2Bgreat.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="500" data-original-width="500" height="640" src="https://4.bp.blogspot.com/-S1Ihxxu3PJ8/XNBE7esLLJI/AAAAAAAAEFI/j-t1p8GRYS0rOj4rASe-KeoNLA3OQAxOACLcBGAs/s640/Nothing%2Btoo%2Bgreat.png" width="640" /></a></div><div class="separator" style="clear: both; text-align: center;"><a href="https://3.bp.blogspot.com/-NgPbIaeLv0c/W4XehbXT76I/AAAAAAAADu8/cPLObZ1Q7mEJ2hrrgp8fuk-UZ5oqbzkzgCPcBGAYYCw/s1600/Always%2BWith%2BLove.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="100" data-original-width="265" src="https://3.bp.blogspot.com/-NgPbIaeLv0c/W4XehbXT76I/AAAAAAAADu8/cPLObZ1Q7mEJ2hrrgp8fuk-UZ5oqbzkzgCPcBGAYYCw/s1600/Always%2BWith%2BLove.png" /></a></div><span style="font-family: inherit;"><span style="background-color: white; color: #222222;"><br /></span></span>Sue Barrhttp://www.blogger.com/profile/06347249118943488564noreply@blogger.com0tag:blogger.com,1999:blog-8317810105915057272.post-88389097366526894982019-05-01T05:00:00.000-04:002019-05-05T11:58:54.384-04:00Georgiana ~ Part Two<div class="separator" style="clear: both; text-align: center;"><a href="https://4.bp.blogspot.com/-rzTbZWfrxCs/XMCIyxGUkVI/AAAAAAAAEEA/9hAoHEdLvLkhHq2yCTvHFJoyAwJDBUH1gCPcBGAYYCw/s1600/Georgiana%2B-%2Bserialized.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="265" data-original-width="805" height="210" src="https://4.bp.blogspot.com/-rzTbZWfrxCs/XMCIyxGUkVI/AAAAAAAAEEA/9hAoHEdLvLkhHq2yCTvHFJoyAwJDBUH1gCPcBGAYYCw/s640/Georgiana%2B-%2Bserialized.png" width="640" /></a></div><blockquote class="tr_bq">Georgiana held the sealed letter in her hand and wondered what news Lord George Kerr needed to impart to both his brother and hers that required incurring the expense of an express post. Dare she hope he’d convinced Catherine to embark on a courtship?</blockquote><div class="MsoNormal"><o:p></o:p></div><blockquote class="tr_bq">A soft smile formed at the thought of George and Kitty married. One had to be blind not to see how besotted he’d been with her the week leading up to Lord Nathan and Caroline Bingley’s wedding. She was fully sure in her estimation that Kitty returned his affection, but Georgiana had witnessed moments when her dear friend, and sister-by-marriage had been most despondent. She fervently hoped the missive contained good news.</blockquote><div class="MsoNormal"><o:p></o:p></div><blockquote class="tr_bq">She alighted from the carriage as soon as it pulled to a stop in front of Pemberley house and hurried inside.</blockquote><div class="MsoNormal"><o:p></o:p></div><blockquote class="tr_bq"> “Is my brother here?” she asked Carson, handing her pelisse, gloves and bonnet to Sarah, who followed her inside.</blockquote><div class="MsoNormal"><o:p></o:p></div><blockquote class="tr_bq">“I believe Mr. Darcy is in the study.”</blockquote><div class="MsoNormal"><o:p></o:p></div><blockquote class="tr_bq">“Thank you, Carson. I won’t need you until later this evening, Anna.”</blockquote><div class="MsoNormal"><o:p></o:p></div><blockquote class="tr_bq">“Very good, Miss Darcy.” Anna bobbed a curtsy and hurried up the stairs.</blockquote><div class="MsoNormal"><o:p></o:p></div><blockquote class="tr_bq">Georgiana tidied her hair as she walked toward the study. The door was not fully closed, so she almost entered without knocking. A soft laugh, followed by the low rumbling of her brother’s voice, made her pause.</blockquote><div class="MsoNormal"><o:p></o:p></div><blockquote class="tr_bq"><span style="font-family: "georgia" , "times new roman" , serif;"><i>They’re at it again.</i></span></blockquote><div class="CBi"><o:p></o:p></div><blockquote class="tr_bq">She couldn’t censure them. Everyone knew how much Fitz loved Lizzy and how much Lizzy loved him. They cared not that Society looked in askance at their open affection. She thought it absolutely heavenly and tiptoed back down the hall a few paces, before turning to face the study door. This time she made her tread a trifle louder than normal and cleared her throat for good measure. She was rewarded by the sound of quick whispers and a swishing of silk.</blockquote><div class="MsoNormal"><o:p></o:p></div><blockquote class="tr_bq">She knocked on the study door.</blockquote><blockquote class="tr_bq"><span style="text-indent: 0in;">“Enter,” came the reply.</span></blockquote><div class="MsoNormal" style="line-height: normal; mso-pagination: none; text-indent: 0in;"><o:p></o:p></div><blockquote class="tr_bq" style="line-height: normal; mso-pagination: none; text-indent: 0in;">With a happy smile on her face, she entered the room. She noted Fitz seated at his desk, ledgers spread before him, although one was upside down and his cravat was loosely tied. His valet would be most displeased to see that his handiwork had been tampered with. She then had to bite the inside of her cheek to keep from grinning outright when she also noted his vest buttons did not line up.</blockquote><div class="MsoNormal" style="line-height: normal; mso-pagination: none; text-indent: 0in;"><o:p></o:p></div><blockquote class="tr_bq" style="line-height: normal; mso-pagination: none; text-indent: 0in;">Although she’d dearly love to tease her normally staid brother, she gave no evidence of seeing all this as she approached and handed him the letter.</blockquote><div class="MsoNormal" style="line-height: normal; mso-pagination: none; text-indent: 0in;"><o:p></o:p></div><blockquote class="tr_bq" style="line-height: normal; mso-pagination: none; text-indent: 0in;">“As I departed Kympton parish, after visiting some of the tenants, a rider delivered an Express Post to Lord Nathan. When he asked for direction to Pemberley, I offered to bring the letter myself.”</blockquote><div class="MsoNormal" style="line-height: normal; mso-pagination: none; text-indent: 0in;"><o:p></o:p></div><blockquote class="tr_bq" style="line-height: normal; mso-pagination: none; text-indent: 0in;">Fitz accepted the letter from her outstretched hand and turning it over, said, “It’s from Lord George.”</blockquote><div class="MsoNormal" style="line-height: normal; mso-pagination: none; text-indent: 0in;"><o:p></o:p></div><blockquote class="tr_bq" style="line-height: normal; mso-pagination: none; text-indent: 0in;">“Lord George!” Lizzy exclaimed from the other side of the room.</blockquote><div class="MsoNormal" style="line-height: normal; mso-pagination: none; text-indent: 0in;"><o:p></o:p></div><blockquote class="tr_bq" style="line-height: normal; mso-pagination: none; text-indent: 0in;">Georgiana turned to face Lizzy, who looked far more composed that her brother, although her lips were swollen and her cheeks decidedly flushed.</blockquote><div class="MsoNormal" style="line-height: normal; mso-pagination: none; text-indent: 0in;"><o:p></o:p></div><blockquote class="tr_bq" style="line-height: normal; mso-pagination: none; text-indent: 0in;">“Lizzy,” she exclaimed, pretending she hadn’t known her sister was in the room, “I didn’t see you there. How rude you must think me.”</blockquote><div class="MsoNormal" style="line-height: normal; mso-pagination: none; text-indent: 0in;"><o:p></o:p></div><div class="MsoNormal"><o:p></o:p></div><blockquote class="tr_bq">“No matter, Georgie. I was reading by the fire and you had no reason to know I was here.”</blockquote><div class="MsoNormal"><o:p></o:p></div><blockquote class="tr_bq">Both she and Lizzy waited as Fitz broke the seal and read the express. His mouth turned up at the corners in a happy smile and his eyes twinkled when he finally glanced up.</blockquote><div class="MsoNormal"><o:p></o:p></div><blockquote class="tr_bq">“They are to be married,” he announced.</blockquote><div class="MsoNormal"><o:p></o:p></div><blockquote class="tr_bq">“Who? Lord George and Kitty?” Elizabeth asked.</blockquote><div class="MsoNormal"><o:p></o:p></div><blockquote class="tr_bq">“Yes.” Fitz glanced back down at the letter. “George wrote: ‘<i><span style="font-family: "georgia" , "times new roman" , serif;">I shall have to frame the Special License I’d purchased for posterity as my Catherine insist we wait the requisite three weeks, giving her time to arrange a small trousseau and have her sisters from far flung Derbyshire attend the wedding</span></i>.’ He follows with a date for the wedding and details of Kitty and Mary’s travel plans to Town.”</blockquote><div class="MsoNormal"><o:p></o:p></div><blockquote class="tr_bq" style="text-align: center;">~~~~~</blockquote><div class="CG"><o:p></o:p></div><blockquote class="tr_bq">“Excuse me, Your Grace,” Hobson stood in the door frame to Maxwell Kerr’s study. He glanced up from his ledgers and indicated for the butler to enter. “This came by express post. As it bears Lord George’s seal, I thought you may wish to attend to it immediately.</blockquote><div class="MsoNormal"><o:p></o:p></div><blockquote class="tr_bq">“Thank you, Hobson.”</blockquote><div class="MsoNormal"><o:p></o:p></div><blockquote class="tr_bq">Hobson carefully placed the letter on the corner of Max’s desk and with a dignified bow, exited the study.</blockquote><div class="MsoNormal"><o:p></o:p></div><blockquote class="tr_bq">Though intrigued by what George’s news might be, Max finished the letter to his steward. One of his tenants had become increasingly difficult, almost to the point where Max considered embarking on a trip to Yorkshire to deal with him personally. However, Mr. Mason was a very competent steward and he’d had given him the latitude to deal with it as he saw fit, hoping they wouldn’t be forced to evict a tenant whose family had been with the Kerr family for generations.</blockquote><div class="MsoNormal"><o:p></o:p></div><blockquote class="tr_bq">At times like this, so far removed from Adborough Hall, he felt like he was losing control of his dukedom. But, there was no way to get around the fact he was required to be in the House of Lords, especially now with all the unrest in northern England. Important bills must be passed to ease the tensions, and that required him to remain in London and exert his considerable pressure on those who waffled in their votes. </blockquote><div class="MsoNormal"><o:p></o:p></div><blockquote class="tr_bq">The letter completed, sanded and sealed, Max turned his attention to George’s missive. He read the first few lines, then leaned back in his chair, letter still in his hand and smiled. George had proposed to Miss Catherine Bennet of Longbourn and she’d accepted. Max continued reading and chuckled when George revealed that although he had a Special License, Catherine managed to convince him to wait for the banns to be read.</blockquote><div class="MsoNormal"><o:p></o:p></div><blockquote class="tr_bq">Good for Catherine, he thought. She had a will of steel, evidenced at Nathan’s wedding last year when it was revealed that Viscount Stanhope had threatened her and she had protected George. Max wasn’t sure of all the complete details, but George admitted he’d been a spy for the Crown for over five years and because of Catherine’s bravery, many lives of British agents in France had been safeguarded.</blockquote><div class="MsoNormal"><o:p></o:p></div><blockquote class="tr_bq">How shocked everyone had been to discover Stanhope had been a traitor to England, and also, how fortuitous that the Honorable Colonel Fitzwilliam was in attendance to speak with the magistrate and keep the whole affair low key until after the wedding.</blockquote><div class="MsoNormal"><o:p></o:p></div><blockquote class="tr_bq">Fortunately for him, Nathan’s wedding held far more pleasant recollections than vile stories of Viscount Stanhope and his treachery. His best memory, the one he brought to mind on an almost a daily basis, was his too brief of a dance with Miss Georgiana Darcy. Although not formally out in society, she’d been allowed to partake in a few sets at Nathan’s wedding ball because it had been held at Pemberley. For a half hour they’d been able to converse without people leaning in to hear what they had to say. Or more specifically, what he had to say.</blockquote><blockquote class="tr_bq">Their set had concluded far too soon for his liking and he’d returned her safely to a watchful brother. This year she would make her curtsy and he planned to court her. Not in a brash manner like most of young bucks carried away not only by a pretty face but a handsome dowry. She was far too refined for a direct approach. No, he planned to woo her gently and like a lustrous pearl, coax her out of the shell of shyness. He’d waited a long time for this. A few months would not matter.</blockquote><div class="MsoNormal"><o:p></o:p></div><div class="separator" style="clear: both; text-align: center;"><a href="https://3.bp.blogspot.com/-NgPbIaeLv0c/W4XehbXT76I/AAAAAAAADu8/cPLObZ1Q7mEJ2hrrgp8fuk-UZ5oqbzkzgCPcBGAYYCw/s1600/Always%2BWith%2BLove.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="100" data-original-width="265" src="https://3.bp.blogspot.com/-NgPbIaeLv0c/W4XehbXT76I/AAAAAAAADu8/cPLObZ1Q7mEJ2hrrgp8fuk-UZ5oqbzkzgCPcBGAYYCw/s1600/Always%2BWith%2BLove.png" /></a></div><br />Sue Barrhttp://www.blogger.com/profile/06347249118943488564noreply@blogger.com0tag:blogger.com,1999:blog-8317810105915057272.post-24133329267342639732019-04-26T05:00:00.000-04:002019-04-26T05:00:07.566-04:00First Page Friday - Guest - Diane CHARTRAND<div class="separator" style="clear: both; text-align: center;"><a href="https://2.bp.blogspot.com/-u9kc7ZUhM5Q/XJjhs6Kd2HI/AAAAAAAAEBw/MNmFgoh3m8gix-aIhyQRIbsF0GA1v4F6wCLcBGAs/s1600/FPF%2BDC%2BBlog.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="200" data-original-width="800" height="160" src="https://2.bp.blogspot.com/-u9kc7ZUhM5Q/XJjhs6Kd2HI/AAAAAAAAEBw/MNmFgoh3m8gix-aIhyQRIbsF0GA1v4F6wCLcBGAs/s640/FPF%2BDC%2BBlog.png" width="640" /></a></div><br />Welcome Diane Chartrand to First Page Friday. Today, Diane is sharing from her novel: <span style="color: #674ea7; font-family: Georgia, Times New Roman, serif;"><b><i><u><a href="https://www.amazon.com/Sia-Screams-Night-Windy-Trilogy-ebook/dp/B07NLJ66CN" target="_blank">Sia: Screams in the Night</a></u></i></b></span><span style="font-family: inherit;">. Now I'm not one to read stories that have any suggestion of terror in them but there are a boatload of folks out there who thrive on the tension and there's nothing wrong with that.</span><br /><span style="font-family: inherit;"><br /></span><span style="font-family: inherit;">I've known Diane for a long time. We met in a Creative Writing Course offered through the University of Western Ontario's continuing studies. She's a lovely lady and I know she has a passion for writing. So, without further delay, here are more details about Diane's book.</span><br /><br /><b><span style="font-family: inherit;">Book Blurb:</span></b><br /><br /><div class="MsoNormal" style="line-height: normal; margin-bottom: .0001pt; margin-bottom: 0in;"><span style="letter-spacing: 0.5pt;"><span style="font-family: inherit;">Jolted awake by the screams of a young girl, Sia froze. The <span style="mso-no-proof: yes;">heart-wrenching</span> sounds, a reminder of what happened to her, in the park, at 14. Fearful for the girl, but more afraid of her past.<o:p></o:p></span></span></div><div class="MsoNormal" style="line-height: normal; margin-bottom: .0001pt; margin-bottom: 0in;"><br /></div><div class="MsoNormal" style="line-height: normal; margin-bottom: .0001pt; margin-bottom: 0in;"><span style="letter-spacing: 0.5pt;"><span style="font-family: inherit;">Sia escaped after four years. Having testified against her captors, she and her family now live in Witness Protection.<o:p></o:p></span></span></div><div class="MsoNormal" style="line-height: normal; margin-bottom: .0001pt; margin-bottom: 0in;"><br /></div><div class="MsoNormal" style="line-height: normal; margin-bottom: .0001pt; margin-bottom: 0in;"><span style="letter-spacing: 0.5pt;"><span style="font-family: inherit;">Sia, <span style="mso-no-proof: yes;">second-guessing</span>what she heard, and haunted by the fact that no one tried to find her, undertakes an intense search for information from neighbors and police about that night.<o:p></o:p></span></span></div><div class="MsoNormal" style="line-height: normal; margin-bottom: .0001pt; margin-bottom: 0in;"><br /></div><div class="MsoNormal" style="line-height: normal; margin-bottom: .0001pt; margin-bottom: 0in;"><span style="letter-spacing: 0.5pt;"><span style="font-family: inherit;">A gut feeling. The girl is real and in serious trouble. Only one goal. Find the girl before it’s too late.</span></span></div><div class="MsoNormal" style="line-height: normal; margin-bottom: .0001pt; margin-bottom: 0in;"><span style="font-size: 12.0pt; letter-spacing: .5pt; mso-bidi-font-family: Calibri; mso-bidi-theme-font: minor-latin; mso-effects-shadow-align: none; mso-effects-shadow-alpha: 50.0%; mso-effects-shadow-angledirection: 13500000; mso-effects-shadow-anglekx: 0; mso-effects-shadow-angleky: 0; mso-effects-shadow-color: black; mso-effects-shadow-dpidistance: 4.0pt; mso-effects-shadow-dpiradius: 5.0pt; mso-effects-shadow-pctsx: 0%; mso-effects-shadow-pctsy: 0%; mso-style-textoutline-outlinestyle-align: center; mso-style-textoutline-outlinestyle-compound: simple; mso-style-textoutline-outlinestyle-dash: solid; mso-style-textoutline-outlinestyle-dpiwidth: 0pt; mso-style-textoutline-outlinestyle-join: round; mso-style-textoutline-outlinestyle-linecap: flat; mso-style-textoutline-outlinestyle-pctmiterlimit: 0%; mso-style-textoutline-type: none;"><span style="font-family: inherit;"><br /></span></span></div><div class="MsoNormal" style="line-height: normal; margin-bottom: .0001pt; margin-bottom: 0in;"><span style="font-size: 12.0pt; letter-spacing: .5pt; mso-bidi-font-family: Calibri; mso-bidi-theme-font: minor-latin; mso-effects-shadow-align: none; mso-effects-shadow-alpha: 50.0%; mso-effects-shadow-angledirection: 13500000; mso-effects-shadow-anglekx: 0; mso-effects-shadow-angleky: 0; mso-effects-shadow-color: black; mso-effects-shadow-dpidistance: 4.0pt; mso-effects-shadow-dpiradius: 5.0pt; mso-effects-shadow-pctsx: 0%; mso-effects-shadow-pctsy: 0%; mso-style-textoutline-outlinestyle-align: center; mso-style-textoutline-outlinestyle-compound: simple; mso-style-textoutline-outlinestyle-dash: solid; mso-style-textoutline-outlinestyle-dpiwidth: 0pt; mso-style-textoutline-outlinestyle-join: round; mso-style-textoutline-outlinestyle-linecap: flat; mso-style-textoutline-outlinestyle-pctmiterlimit: 0%; mso-style-textoutline-type: none;"><span style="font-family: inherit;"><b>First Page:</b></span></span></div><div class="MsoNormal" style="line-height: normal; margin-bottom: 0.0001pt; text-align: center;"><span style="letter-spacing: 0.5pt;"><span style="font-family: Georgia, Times New Roman, serif; font-size: large;"><b>Chapter One</b></span></span></div><div class="MsoNormal" style="line-height: normal; margin-bottom: 0.0001pt; text-align: left;"><b style="font-size: 12pt; letter-spacing: 0.5pt;"><span style="font-family: Georgia, Times New Roman, serif;">Shattered</span></b></div><blockquote class="tr_bq" style="line-height: normal; margin-bottom: 0.0001pt; text-align: left;"><i style="font-family: Georgia, "Times New Roman", serif; font-size: 18.6667px;">Startle, Sia jumped up in bed. What, what? Another piercing scream broke the night silence followed by a young girl begging, "No. Please stop, please you're hurting me."</i></blockquote><blockquote class="tr_bq" style="line-height: normal; margin-bottom: 0.0001pt; text-align: left;"><i style="font-family: Georgia, "Times New Roman", serif; font-size: 18.6667px;">Sia went over to the bedroom window and raised the blinds. She looked out into the darkness but saw no one. The basketball court, which is directly below her second story apartment, was empty, and the small children's playground next to it appeared to be too.</i></blockquote><blockquote class="tr_bq" style="line-height: normal; margin-bottom: 0.0001pt; text-align: left;"><i style="font-family: Georgia, "Times New Roman", serif; font-size: 18.6667px;">She heard it again even louder this time. The girl's piercing cries ripped into Sia's soul. She could feel every bit of pain the girl was ex- periencing. Sia couldn't pinpoint the girl's location so went out on her balcony hoping to get a better fix.</i></blockquote><blockquote class="tr_bq" style="line-height: normal; margin-bottom: 0.0001pt; text-align: left;"><i style="font-family: Georgia, "Times New Roman", serif; font-size: 18.6667px;">The sound seemed to be coming from behind the two L shaped buildings that surrounded the basketball court and playground, sepa- rated from the townhouse complex behind it by a small grassy area, a 6-foot fence topped with barb wire, and a row of 20-foot pine trees.</i></blockquote><blockquote class="tr_bq" style="line-height: normal; margin-bottom: 0.0001pt; text-align: left;"><i style="font-family: Georgia, "Times New Roman", serif; font-size: 18.6667px;">A man's angry voice exploded into the silence. "This life is your only future Willow. Now shut the hell up."</i></blockquote><blockquote class="tr_bq" style="line-height: normal; margin-bottom: 0.0001pt; text-align: left;"><i style="font-family: Georgia, "Times New Roman", serif; font-size: 18.6667px;">Willow's gut-wrenching cries went on for over a half-hour. Sia wanted to find and help this desperate girl, but fear stopped her.</i></blockquote><blockquote class="tr_bq" style="line-height: normal; margin-bottom: 0.0001pt; text-align: left;"><i style="font-family: Georgia, "Times New Roman", serif; font-size: 18.6667px;">Sia picked up the receiver of the phone and was about to call the police when the night went silent. She put the phone back in the cradle and waited a few minutes to see if the cries started again.</i></blockquote><blockquote class="tr_bq" style="line-height: normal; margin-bottom: 0.0001pt; text-align: left;"><i style="font-family: Georgia, "Times New Roman", serif; font-size: 18.6667px;">No more pleading cries to stop. No more piercing screams. No more male voice. Did someone closer intervene? Did she get away, or was she lying there hurt, maybe dying?</i></blockquote><br /><div class="MsoNormal" style="line-height: normal; margin-bottom: .0001pt; margin-bottom: 0in; margin-left: 0in; margin-right: 0in; margin-top: 5.25pt; mso-layout-grid-align: none; mso-line-break-override: restrictions; punctuation-wrap: simple; text-autospace: none;"><span style="font-family: inherit;"><br /></span></div><div class="MsoNormal" style="line-height: normal; margin-bottom: .0001pt; margin-bottom: 0in; margin-left: 0in; margin-right: 0in; margin-top: 5.25pt; mso-layout-grid-align: none; mso-line-break-override: restrictions; punctuation-wrap: simple; text-autospace: none;"><span style="font-family: inherit;">Well.... wow. That's an explosive start to this story. I have a problem with though, (sorry Diane), but I'd have called the cops a whole lot sooner. Poor Willow. What happened to her?</span></div><div class="MsoNormal" style="line-height: normal; margin-bottom: .0001pt; margin-bottom: 0in; margin-left: 0in; margin-right: 0in; margin-top: 5.25pt; mso-layout-grid-align: none; mso-line-break-override: restrictions; punctuation-wrap: simple; text-autospace: none;"><span style="font-family: inherit;">You can find out more about Diane and what she's up to on her <a href="https://dianechartrandauthor.blogspot.com/" style="font-weight: bold; text-decoration-line: underline;" target="_blank">BLOG</a>, until then,</span></div><div class="separator" style="clear: both; text-align: center;"><a href="https://3.bp.blogspot.com/-NgPbIaeLv0c/W4XehbXT76I/AAAAAAAADu8/cPLObZ1Q7mEJ2hrrgp8fuk-UZ5oqbzkzgCPcBGAYYCw/s1600/Always%2BWith%2BLove.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="100" data-original-width="265" src="https://3.bp.blogspot.com/-NgPbIaeLv0c/W4XehbXT76I/AAAAAAAADu8/cPLObZ1Q7mEJ2hrrgp8fuk-UZ5oqbzkzgCPcBGAYYCw/s1600/Always%2BWith%2BLove.png" /></a></div><div class="MsoNormal" style="line-height: normal; margin-bottom: .0001pt; margin-bottom: 0in; margin-left: 0in; margin-right: 0in; margin-top: 5.25pt; mso-layout-grid-align: none; mso-line-break-override: restrictions; punctuation-wrap: simple; text-autospace: none;"><span style="font-family: inherit;"><br /></span></div><div class="MsoNormal" style="line-height: normal; margin-bottom: .0001pt; margin-bottom: 0in; margin-left: 2.0pt; margin-right: 5.15pt; margin-top: 0in; mso-layout-grid-align: none; mso-line-break-override: restrictions; punctuation-wrap: simple; text-align: justify; text-autospace: none; text-indent: .25in;"><o:p></o:p></div>Sue Barrhttp://www.blogger.com/profile/06347249118943488564noreply@blogger.com1tag:blogger.com,1999:blog-8317810105915057272.post-61231570661703207232019-04-24T12:20:00.000-04:002019-05-05T12:00:05.338-04:00Georgiana ~ Part OneAs a treat, for those who follow this blog, I'm posting from my story Georgiana. I will not be showing these as chapters as sometimes I might not post the full chapter, only a portion.<br /><br />Without further ado - Georgiana ~ Part One<br /><div class="separator" style="clear: both; text-align: center;"><a href="https://2.bp.blogspot.com/-rzTbZWfrxCs/XMCIyxGUkVI/AAAAAAAAED8/VQX8YVAw5es2gj9COx2932y13WpyG5YXACLcBGAs/s1600/Georgiana%2B-%2Bserialized.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="265" data-original-width="805" height="209" src="https://2.bp.blogspot.com/-rzTbZWfrxCs/XMCIyxGUkVI/AAAAAAAAED8/VQX8YVAw5es2gj9COx2932y13WpyG5YXACLcBGAs/s640/Georgiana%2B-%2Bserialized.png" width="640" /></a></div><blockquote class="tr_bq" style="line-height: normal; mso-pagination: none; text-indent: 0in;"><span style="font-family: inherit;">The carriage with its three occupants bumped along the narrow road. At the sound of glass against glass Georgiana Darcy spared her maid, who clutched a basket of preserves on her lap, a glance. The trail, although rough in some places due to the spring storm which had trundled through Derbyshire last week, had dried out enough so that Georgiana could visit some long-term tenants before departing for London. The thought of leaving Pemberley and all that was familiar, in order to prepare for her first Season, caused her stomach to clench. She knew the fear and shyness was irrational, but the thought of meeting so many new people almost made her sick.</span></blockquote><blockquote class="tr_bq" style="line-height: normal; mso-pagination: none; text-indent: 0in;"><span style="font-family: inherit;">"Are you quite all right, Miss Darcy?"</span></blockquote><blockquote class="tr_bq" style="line-height: normal; mso-pagination: none; text-indent: 0in;"><span style="font-family: inherit;">The polite inquiry came from Lord Nathan, better known as Mr. Kerr to the parishioners of Kympton parish. His curly dark hair, broad shoulders and ready smile caused many a young lady to wish he belonged to her but he had no eyes other than for his wife, the former Miss Caroline Bingley.</span></blockquote><blockquote class="tr_bq" style="line-height: normal; mso-pagination: none; text-indent: 0in;"><span style="font-family: inherit;">“I am well, Lord Nathan, thank you.”</span></blockquote><blockquote class="tr_bq" style="line-height: normal; mso-pagination: none; text-indent: 0in;"><span style="font-family: inherit;">As attractive as Lord Nathan was, he didn’t make her heart race. No, that pesky organ only galloped along like a new colt around the vicar’s brother, Maxwell Kerr, the fifth Duke of Adborough. Try as she might, no amount of internal scolding changed how she felt around him.</span></blockquote><blockquote class="tr_bq" style="line-height: normal; mso-pagination: none; text-indent: 0in;"><span style="font-family: inherit;">They hit another bump and her maid straightened her straw bonnet.</span></blockquote><blockquote class="tr_bq" style="line-height: normal; mso-pagination: none; text-indent: 0in;"><span style="font-family: inherit;">“Only one more mile, Anna,” Georgiana said with a smile.</span></blockquote><blockquote class="tr_bq" style="line-height: normal; mso-pagination: none; text-indent: 0in;"><span style="font-family: inherit;">She returned her gaze to the passing scenery. The rolling grounds of Pemberley’s estate. In a few short weeks all this would become a memory. If all went as planned, she’d make her debut, meet a suitable gentleman, fall in love and get married.</span></blockquote><blockquote class="tr_bq" style="line-height: normal; mso-pagination: none; text-indent: 0in;"><span style="font-family: inherit;">Her stomach clenched again.</span></blockquote><blockquote class="tr_bq" style="line-height: normal; mso-pagination: none; text-indent: 0in;"><span style="font-family: inherit;"><i style="mso-bidi-font-style: normal;">I have nothing to be afraid of. No one knows about my mistake.</i></span></blockquote><blockquote class="tr_bq" style="line-height: normal; mso-pagination: none; text-indent: 0in;"><span style="font-family: inherit;">How she longed to capture the carefree girl she’d been before that fateful summer four years ago. Elizabeth, her sister-by-love, cautioned her to move on and forgive herself as she’d been only five and ten at the time, but no one seemed to understand that she’d been more than prepared to become a wife and mother regardless of her age. They all believed her to be a silly girl who’d become caught up in the moment. He’d been a familiar face in a sea of new ones and his ultimate betrayal of her affection cut deep.</span></blockquote><blockquote class="tr_bq" style="line-height: normal; mso-pagination: none; text-indent: 0in;"><span style="font-family: inherit;">She bit back a small sigh and raised her chin, thankful she no longer held a tendré for <i style="mso-bidi-font-style: normal;">him</i>. That notion had been ruthlessly squashed when she overheard a conversation between him and her brother immediately following their discovery. He informed Fitzwilliam, in a condescending manner, that he’d pursued her solely for monetary gain as no man wanted such a dull flower for a wife. The utter contempt in his tone made her cringe and hang her head in shame.</span></blockquote><blockquote class="tr_bq" style="line-height: normal; mso-pagination: none; text-indent: 0in;"><span style="font-family: inherit;">“You seem rather melancholy, Miss Darcy.” Lord Nathan broke into her thoughts. “Are you sure you are up to visiting the Sprague family today?”</span></blockquote><blockquote class="tr_bq" style="line-height: normal; mso-pagination: none; text-indent: 0in;"><span style="font-family: inherit;">“I am well, truly.” She smiled to reassure him. “I have been contemplating Pemberley and how much I shall miss it.”</span></blockquote><blockquote class="tr_bq" style="line-height: normal; mso-pagination: none; text-indent: 0in;"><span style="font-family: inherit;">“Ah.” The quiet assessment in Lord Nathan’s eyes told her he doubted the veracity of her statement but wisely kept his counsel. “I see we have arrived.”</span></blockquote><blockquote class="tr_bq" style="line-height: normal; mso-pagination: none; text-indent: 0in;"><span style="font-family: inherit;">The carriage slowed to a stop in front of a small cottage. Lazy smoke drifted from the chimney and pretty rosebushes lined the walk leading to the front door, which opened as soon as they alighted from the carriage.</span></blockquote><blockquote class="tr_bq" style="line-height: normal; mso-pagination: none; text-indent: 0in;"><span style="font-family: inherit;">“Miss Darcy, Mr. Kerr. What an unexpected surprise.” A rosy faced woman hurried from the cottage and gave them each a small curtsy.</span></blockquote><blockquote class="tr_bq" style="line-height: normal; mso-pagination: none; text-indent: 0in;"><span style="font-family: inherit;">“I brought a few things from Pemberley. Mrs. Reynolds heard how Johnny was feeling poorly and she knows how much he likes peach preserves. There is also some venison and a few cakes.”</span></blockquote><blockquote class="tr_bq" style="line-height: normal; mso-pagination: none; text-indent: 0in;"><span style="font-family: inherit;">“Oh, bless her heart, he most certainly does love her preserves.” Mrs. Sprague hurried forward and took the basket from Anna, then turned toward the cottage. “Where are my manners? Come in and I’ll make tea.”</span></blockquote><blockquote class="tr_bq" style="line-height: normal; mso-pagination: none; text-indent: 0in;"><span style="font-family: inherit;">The cottage, though small, was as neat as a pin. Mrs. Sprague took pride in her home and the time spent in her company passed quickly. The physician had been to see Johnny and although still was weak from the fever’s lingering effect, he’d rallied enough to sit by the fire and enjoy his tea with them all. About to take their leave, Mrs. Sprague pressed a small token into Georgiana’s hand.</span></blockquote><blockquote class="tr_bq" style="line-height: normal; mso-pagination: none; text-indent: 0in;"><span style="font-family: inherit;"><span style="mso-spacerun: yes;"> </span>“My Henry carved this when he heard you were going to London.” Georgiana looked down at the exquisite cross. “We’ll keep you in our prayers, Miss Darcy, that God will guide your steps to a good man worthy of your regard.”</span></blockquote><blockquote class="tr_bq" style="line-height: normal; mso-pagination: none; text-indent: 0in;"><span style="font-family: inherit;">“Thank you, Mrs. Sprague.” Unbidden tears sprang into her eyes. “I shall treasure this, always.”</span></blockquote><blockquote class="tr_bq" style="line-height: normal; mso-pagination: none; text-indent: 0in;"><span style="font-family: inherit;">“You just remember to seek the Lord with all your heart and He will guide your path. That verse has kept me and Henry all these years and I like to think the Almighty knows what He’s about.”</span></blockquote><blockquote class="tr_bq" style="line-height: normal; mso-pagination: none; text-indent: 0in;"><span style="font-family: inherit;">“That He does, Mrs. Sprague,” Lord Nathan said. “Your prayers are always much appreciated. Good day.” He tipped his hat in respect and Mrs. Sprague gave them both a polite curtsy.</span></blockquote><blockquote class="tr_bq" style="line-height: normal; mso-pagination: none; text-indent: 0in;"><span style="font-family: inherit;">“Good day, Mrs. Sprague.” Georgiana joined Lord Nathan at the carriage. After handing her up, he climbed in and sat across from her and Sarah.</span></blockquote><blockquote class="tr_bq" style="line-height: normal; mso-pagination: none; text-indent: 0in;"><span style="font-family: inherit;">“That was a thoughtful gift.”</span></blockquote><blockquote class="tr_bq" style="line-height: normal; mso-pagination: none; text-indent: 0in;"><span style="font-family: inherit;">“Oh, yes.” She opened her hand and studied the cross again. “Mr. Sprague is a talented carver.”</span></blockquote><blockquote class="tr_bq" style="line-height: normal; mso-pagination: none; text-indent: 0in;"><span style="font-family: inherit;">The carriage started off with a jolt and Anna squeaked in surprise. The sound was so unexpected, Georgiana began to giggle. The more she tried to stop, the harder she laughed. Soon Lord Nathan smiled broadly and chuckled. Anna, her face red with embarrassment, apologized profusely for having cried out.</span></blockquote><blockquote class="tr_bq" style="line-height: normal; mso-pagination: none; text-indent: 0in;"><span style="font-family: inherit;">“Oh, Anna, you sounded like a little field mouse. That was such fun.” She wiped tears of laughter from the corner of her eyes. “I have not laughed that hard in such a long time.”</span></blockquote><blockquote class="tr_bq" style="line-height: normal; mso-pagination: none; text-indent: 0in;"><span style="font-family: inherit;">“I’m glad to be of service, Miss Darcy.”</span></blockquote><blockquote class="tr_bq" style="line-height: normal; mso-pagination: none; text-indent: 0in;"><span style="font-family: inherit;">Her prim response set off more peals of laughter from Georgiana and this time, even Anna joined in. All Lord Nathan did was shake his head.</span></blockquote><blockquote class="tr_bq" style="line-height: normal; mso-pagination: none; text-indent: 0in;"><span style="font-family: inherit;">“I am sorry, Lord Nathan. You must think us extremely silly.”</span></blockquote><blockquote class="tr_bq" style="line-height: normal; mso-pagination: none; text-indent: 0in;"><span style="font-family: inherit;">“Not at all,” he gallantly replied. “It has been a long time since I heard you laugh, Miss Darcy. This past year has been good for you. Having Elizabeth’s sister keep you company this past winter brought out the fun-loving girl I and my brother’s all remember from our youth.”</span></blockquote><blockquote class="tr_bq" style="line-height: normal; mso-pagination: none; text-indent: 0in;"><span style="font-family: inherit;">“True, Mary was surprisingly lively and we both shared a love of music. I recall when we first met she rarely smiled. How I wish she would come to London with me for my debut.”</span></blockquote><blockquote class="tr_bq" style="line-height: normal; mso-pagination: none; text-indent: 0in;"><span style="font-family: inherit;">“You know she is not one for fancy parties and large crowds.”</span></blockquote><blockquote class="tr_bq" style="line-height: normal; mso-pagination: none; text-indent: 0in;"><span style="font-family: inherit;">“I do know that, but she offers a most sardonic narration of the guests and what makes the commentary so funny is that she does not mean to be satirical. She is only giving an honest opinion and it is so refreshing.” Georgiana leaned forward. “She told me during one of our morning visits in London,” she choked back a giggle, “that Lady Fitzherbert’s hat looked like a peacock had nested upon her head. I had to leave the room before I laughed out loud, but it was true!”</span></blockquote><blockquote class="tr_bq" style="line-height: normal; mso-pagination: none; text-indent: 0in;"><span style="font-family: inherit;">“Poor Lady Fitzherbert.” Lord Nathan smiled in remembrance. “I am afraid her milliner informed her peacock feathers were all the rage and, well, she became quite enthusiastic about the whole ensemble.”</span></blockquote><blockquote class="tr_bq" style="line-height: normal; mso-pagination: none; text-indent: 0in;"><span style="font-family: inherit;">“Fortunately, everybody loves Lady Fitzherbert. This fashion faux pas will be overlooked because she gives so much of herself to those who need help.” The carriage slowed to a stop and Lord Nathan alighted. He turned and gave both women a polite nod. “Tell your brother I shall be over in a few days to discuss the new vicar for Kympton.”</span></blockquote><blockquote class="tr_bq" style="line-height: normal; mso-pagination: none; text-indent: 0in;"><span style="font-family: inherit;">“I will. Give my regards to Caroline.”</span></blockquote><blockquote class="tr_bq" style="line-height: normal; mso-pagination: none; text-indent: 0in;"><span style="font-family: inherit;">“I shall. She misses your company.”</span></blockquote><blockquote class="tr_bq" style="line-height: normal; mso-pagination: none; text-indent: 0in;"><span style="font-family: inherit;">For three months, during Fitz and Lizzy’s wedding trip, Caroline resided at Pemberley. Lord Nathan, her betrothed at the time, became a frequent visitor and Georgiana had watched their love deepen and grow. She hoped – no, she prayed – to find a love like that.</span></blockquote><blockquote class="tr_bq" style="line-height: normal; mso-pagination: none; text-indent: 0in;"><span style="font-family: inherit;">At one time she thought his brother the Duke might make an offer. He’d shown an uncommon interest in her each time he visited Pemberley. And, at Lord Nathan and Caroline’s pre-wedding ball they’d danced and it was as though she’d come home. The music was secondary to the feel of his strong arm around her waist, his lean fingers holding her hand. Even though she’d worn silk gloves, her skin burned as though he’d physically touched her with fire.</span></blockquote><blockquote class="tr_bq" style="line-height: normal; mso-pagination: none; text-indent: 0in;"><span style="font-family: inherit;">After the dizzying euphoria of the dance he’d bowed politely and returned her to her brother’s side. The next morning she discovered he’d left for London and she hadn’t seen him since.</span></blockquote><blockquote class="tr_bq" style="line-height: normal; mso-pagination: none; text-indent: 0in;"><span style="font-family: inherit;">She’d held onto a tenuous hope of him visiting again, but as weeks turned into months the harsh reality of his disinterest set in. She had no choice but to make a concerted effort to seek a husband from another quarter. And for that very reason, her curtsy and debut loomed before her like an executioner’s block. She did not do well with strangers. Her thoughts and tongue didn’t stay in harmony when she became nervous. It was as though she played Mozart with her left hand and Beethoven with her right. Discordant and confusing.</span></blockquote><blockquote class="tr_bq" style="line-height: normal; mso-pagination: none; text-indent: 0in;"><span style="font-family: inherit;">Aunt Lucinda, the Countess of Matlock, would have an apoplectic fit if she knew where her niece’s thoughts were headed. She anticipated Georgiana making a brilliant marriage, as did the whole Matlock/Darcy family. The invisible bonds of duty and family honor continued to spiral around Georgiana’s future and at times she felt as though they choked the very life out of her.</span></blockquote><blockquote class="tr_bq" style="line-height: normal; mso-pagination: none; text-indent: 0in;"><span style="font-family: inherit;">Her attention was drawn to the graveled drive leading up to Kympton Parish courtyard. A horse and ride thundered down the lane and pulled to stop beside their carriage. He dismounted in haste and approached Lord Nathan.</span></blockquote><blockquote class="tr_bq" style="line-height: normal; mso-pagination: none; text-indent: 0in;"><span style="font-family: inherit;">“Sir, I am looking for Lord Nathan Kerr.” </span></blockquote><blockquote class="tr_bq" style="line-height: normal; mso-pagination: none; text-indent: 0in;"><span style="font-family: inherit;">“I am Lord Nathan Kerr.” Nathan met the man who handed him a sealed letter. “Thank you. See my housekeeper for payment.”</span></blockquote><blockquote class="tr_bq" style="line-height: normal; mso-pagination: none; text-indent: 0in;"><span style="font-family: inherit;">“Not required, my lord. All of the posts were prepaid.”</span></blockquote><blockquote class="tr_bq" style="line-height: normal; mso-pagination: none; text-indent: 0in;"><span style="font-family: inherit;">“All of them? How many do you have for delivery?”</span></blockquote><blockquote class="tr_bq" style="line-height: normal; mso-pagination: none; text-indent: 0in;"><span style="font-family: inherit;">“Three in total. One for you, one for a Mr. Bingley, and I need to attend Pemberley to deliver one to Mr. Fitzwilliam Darcy.”</span></blockquote><blockquote class="tr_bq" style="line-height: normal; mso-pagination: none; text-indent: 0in;"><span style="font-family: inherit;"><span style="mso-spacerun: yes;"> </span>“I can take the letter for Mr. Darcy,” Georgiana interjected, having listened to their conversation. “I am Miss Darcy of Pemberley and am on my way home.”</span></blockquote><blockquote class="tr_bq">The rider handed her a sealed envelope and she heard Lord Nathan say when he glanced down, "Why, it's from George." </blockquote><div class="separator" style="clear: both; text-align: center;"><a href="https://3.bp.blogspot.com/-NgPbIaeLv0c/W4XehbXT76I/AAAAAAAADu8/cPLObZ1Q7mEJ2hrrgp8fuk-UZ5oqbzkzgCPcBGAYYCw/s1600/Always%2BWith%2BLove.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="100" data-original-width="265" src="https://3.bp.blogspot.com/-NgPbIaeLv0c/W4XehbXT76I/AAAAAAAADu8/cPLObZ1Q7mEJ2hrrgp8fuk-UZ5oqbzkzgCPcBGAYYCw/s1600/Always%2BWith%2BLove.png" /></a></div>Sue Barrhttp://www.blogger.com/profile/06347249118943488564noreply@blogger.com0tag:blogger.com,1999:blog-8317810105915057272.post-83543567075987159642019-04-23T02:17:00.000-04:002019-05-13T11:32:41.610-04:00Monday Meditation ~ Meaning of Easter<h2 style="clear: both; text-align: center;"><span style="color: #3d85c6; font-family: Georgia, Times New Roman, serif; font-size: x-large;"><i>April 23, 2019</i></span></h2><div class="separator" style="clear: both; text-align: center;"><a href="https://4.bp.blogspot.com/--dxO6FtIoFc/XNBGbuvD1nI/AAAAAAAAEFY/fc1-vV6LMOIUI7VOBHAOgeSnEumEcCJRQCEwYBhgL/s1600/Monday%2BMeditation.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="200" data-original-width="800" height="160" src="https://4.bp.blogspot.com/--dxO6FtIoFc/XNBGbuvD1nI/AAAAAAAAEFY/fc1-vV6LMOIUI7VOBHAOgeSnEumEcCJRQCEwYBhgL/s640/Monday%2BMeditation.png" width="640" /></a></div>For God so loved the world that He gave His only begotten Son, that whosoever believes in Him shall not perish but have everlasting life.<br /><div style="text-align: right;">John 3:16</div><br />Have a safe and blessed Easter<br /><br /><div class="separator" style="clear: both; text-align: center;"><a href="https://1.bp.blogspot.com/-XBvFW49Raok/XL6tXVwfU0I/AAAAAAAAEDg/4HM2E0bv7Ds3887yuoi1HfjEVpGyuO0VQCLcBGAs/s1600/Easter.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="1600" data-original-width="1200" height="640" src="https://1.bp.blogspot.com/-XBvFW49Raok/XL6tXVwfU0I/AAAAAAAAEDg/4HM2E0bv7Ds3887yuoi1HfjEVpGyuO0VQCLcBGAs/s640/Easter.png" width="480" /></a></div><br />Sue Barrhttp://www.blogger.com/profile/06347249118943488564noreply@blogger.com0tag:blogger.com,1999:blog-8317810105915057272.post-29297348503972139762019-04-12T05:00:00.000-04:002019-04-12T05:00:02.314-04:00First Page Friday - FIANCE FOR HIRE<div class="separator" style="clear: both; text-align: center;"><a href="https://2.bp.blogspot.com/-7CrrO4cKT2o/XKd1-_YRfVI/AAAAAAAAEC0/XiBke9NwH1EySeIcm0b8Kk2VlTQ4YOMQACLcBGAs/s1600/FPF%2B-%2BTransparent.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="200" data-original-width="800" src="https://2.bp.blogspot.com/-7CrrO4cKT2o/XKd1-_YRfVI/AAAAAAAAEC0/XiBke9NwH1EySeIcm0b8Kk2VlTQ4YOMQACLcBGAs/s1600/FPF%2B-%2BTransparent.png" /></a></div><div class="separator" style="clear: both; text-align: center;"></div><div>Hello dear readers. This week I stole the spotlight for my own contemporary romance, <span style="font-family: "georgia" , "times new roman" , serif;"><a href="https://www.books2read.com/u/3yPldv" style="font-style: italic; font-weight: bold;" target="_blank">Fiance for Hire</a>, </span>a quick novella I had a blast writing.</div><div><br /></div><div><b>Book Blurb:</b></div><div><span style="font-family: inherit;"><br /></span></div><div><div class="MsoNoSpacing" style="text-indent: 0in;"><span lang="EN-US"><span style="font-family: inherit;">Never tell a bold faced lie - unless you can follow through with flawless precision.... <o:p></o:p></span></span></div><div class="MsoNoSpacing" style="text-indent: 0in;"><br /></div><div class="MsoNoSpacing" style="text-indent: 0in;"><span lang="EN-US"><span style="font-family: inherit;">Kristen Wainwright was dumped in the most humiliating way possible and to add insult to injury, her ex-boyfriend has taken up with the office mean queen, Janine. Unable to tolerate the snide comments and subtle insults Kristen, in a moment of weakness, tells Janine that 'yes, she's coming to the Christmas gala' and 'yes, she's bringing someone' - her new fiancé. Now all she has to do is find one. <o:p></o:p></span></span></div><div class="MsoNoSpacing" style="text-indent: 0in;"><br /></div><div class="MsoNoSpacing" style="text-indent: 0in;"><span lang="EN-US"><span style="font-family: inherit;">How hard can that be?</span><span style="font-family: "calibri" , sans-serif;"><o:p></o:p></span></span></div></div><div><br /></div><div><b>First Page:</b></div><div><blockquote class="tr_bq" style="line-height: normal; mso-pagination: none; text-align: justify;"><span lang="EN-US" style="font-family: "georgia" , serif;"><i>Dear Diary:</i></span></blockquote><blockquote class="tr_bq" style="line-height: normal; mso-pagination: none; text-align: justify;"><i><span lang="EN-US" style="font-family: "georgia" , serif; font-size: 8pt;"><o:p> </o:p></span><span style="font-family: "georgia" , serif;">Jeremy made reservations at La Crème, the place where our very first date took place four years ago. For months he’s been dropping hints about how he has to ‘look to his future’ and ‘get that promotion’. He’s going to propose, I know it in my bones. It’s only logical. What shows more willingness for responsibility than marriage, a white picket fence, and kids?</span></i></blockquote><blockquote class="tr_bq" style="line-height: normal; mso-pagination: none; text-align: justify;"><i><span lang="EN-US" style="font-family: "georgia" , serif; font-size: 8pt;"><o:p> </o:p></span><span style="font-family: "georgia" , serif;">I can barely stand it.</span></i></blockquote><blockquote class="tr_bq" style="line-height: normal; mso-pagination: none; text-align: justify;"><span lang="EN-US" style="font-family: "georgia" , serif;"><o:p> </o:p></span><span style="font-family: "georgia" , serif;">I closed the diary and hugged it to my chest. Unable to contain my excitement I jumped off the bed and ran to my closet. Only the very best would do as it wasn’t every day a girl became engaged. Clothes were raked back and forth across the rail, my despair mounting as outfit after outfit was discarded. Too frilly, too young, too drab, too homespun.</span></blockquote><blockquote class="tr_bq" style="line-height: normal; text-align: justify;"><span lang="EN-US" style="font-family: "georgia" , serif;">I had nothing to wear.</span></blockquote><blockquote class="tr_bq" style="line-height: normal; text-align: justify;"><span lang="EN-US" style="font-family: "georgia" , serif;">“Good gravy, Kristen. Are you purging your closet?” Grandma Joy paused in the doorway to my room. “I can barely see your bed.”</span></blockquote><blockquote class="tr_bq" style="line-height: normal; text-align: justify;"><span lang="EN-US" style="font-family: "georgia" , serif;">“No Grams. I’m trying to find something to wear Friday night.”</span></blockquote><blockquote class="tr_bq" style="line-height: normal; text-align: justify;"><span lang="EN-US" style="font-family: "georgia" , serif;">“After four years, I think he’s seen everything.” Grams said with a shake of her head and dropped fresh linen at the end of my bed. “What’s wrong with that nice dress you wore to Ruthie’s wedding?”</span></blockquote><blockquote class="tr_bq" style="line-height: normal; text-align: justify;"><span lang="EN-US" style="font-family: "georgia" , serif;">“Grams,” I rolled my eyes and pushed aside a beige top, followed by a brown sweater and shuddered. What was I thinking when I bought these blah, boring clothes? “That was over two years ago and this night is special. I need something with a wow factor of ten.”</span></blockquote><blockquote class="tr_bq" style="line-height: normal; text-align: justify;"><span style="font-family: "georgia" , serif;">“I don’t know why you think you have to change who you are for that boy…” Grams let the sentence hang, her displeasure evident by her tone of voice. “You looked beautiful in that dress.”</span></blockquote><blockquote class="tr_bq" style="line-height: normal; text-align: justify;"><span lang="EN-US" style="font-family: "georgia" , serif;">I cast a glance over my shoulder and noted Grams stood with her arms crossed and her brow furrowed in frustration.</span></blockquote><blockquote class="tr_bq" style="line-height: normal; text-align: justify;"><span lang="EN-US" style="font-family: "georgia" , serif;">“Thanks, Grams, but I think you’re biased.” As always, my heart tugged at her words. Grams raised me from the time I was born. My mom and dad died in a car accident when I was only a few months old.</span></blockquote><blockquote class="tr_bq" style="line-height: normal; text-align: justify;"><span lang="EN-US" style="font-family: "georgia" , serif;">“Well, it’s the truth,” she said as she walked out of my room. “He should be thankful for what he already has in you. Don’t be too long, dinner is in ten minutes.”</span></blockquote><blockquote class="tr_bq" style="line-height: normal; text-align: justify;"><span lang="EN-US" style="font-family: "georgia" , serif;">“I’ll be right down,” I promised and continued shoving clothes back and forth.</span></blockquote><blockquote class="tr_bq" style="line-height: normal; text-align: justify;"><span lang="EN-US" style="font-family: "georgia" , serif;">My imagination reached a fever pitch and as I ate I practiced my new name, Kristen Elaine Wainwright-Danvers. Hmmm…maybe I wouldn’t hyphenate it. Kristen Elaine Danvers had a nice flow.</span></blockquote><blockquote class="tr_bq" style="line-height: normal; text-align: justify;"><i><span lang="EN-US" style="font-family: "georgia" , serif;">Hello, have you met my husband, Jeremy Danvers? </span></i><span lang="EN-US" style="font-family: "georgia" , serif;">And he would say with pride<i>, this is my wife, Kristen.<o:p></o:p></i></span></blockquote><blockquote class="tr_bq" style="line-height: normal; text-align: justify;"><span lang="EN-US" style="font-family: "georgia" , serif;">Visions of intertwined initials on the wedding train filled my head - <b><i>J & K</i></b>. I’d have day lilies in my bouquet, the bridesmaids would wear something in soft pink and Jeremy and his groomsmen would be handsome in deep blue tuxes. We’d dance our first dance while friends blew bubbles around us. Everything would look magical in our wedding videos.</span></blockquote><div class="MsoNormal" style="line-height: normal;"><span style="font-family: "georgia" , serif;"><o:p><br /></o:p></span><span style="font-family: "georgia" , serif;"><o:p>Oh dear... we all know how that's going to turn out. Good thing for me, otherwise there would have been NO story! This all came about because of one line that kept running through my head. "I have a fabulous idea." Funny how one tiny phrase bears fruit in the form of a fun, romantic comedy.</o:p></span></div><div class="MsoNormal" style="line-height: normal;"><span style="font-family: "georgia" , serif;"><o:p><br /></o:p></span></div><div class="MsoNormal" style="line-height: normal;"><span style="font-family: "georgia" , serif;"><o:p>Have a wonderful weekend, until then, </o:p></span></div><div class="separator" style="clear: both; text-align: center;"><a href="https://3.bp.blogspot.com/-NgPbIaeLv0c/W4XehbXT76I/AAAAAAAADu8/cPLObZ1Q7mEJ2hrrgp8fuk-UZ5oqbzkzgCPcBGAYYCw/s1600/Always%2BWith%2BLove.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="100" data-original-width="265" src="https://3.bp.blogspot.com/-NgPbIaeLv0c/W4XehbXT76I/AAAAAAAADu8/cPLObZ1Q7mEJ2hrrgp8fuk-UZ5oqbzkzgCPcBGAYYCw/s1600/Always%2BWith%2BLove.png" /></a></div><div class="MsoNormal" style="line-height: normal;"><span style="font-family: "georgia" , serif;"><o:p><br /></o:p></span></div></div>Sue Barrhttp://www.blogger.com/profile/06347249118943488564noreply@blogger.com0tag:blogger.com,1999:blog-8317810105915057272.post-13749399955913699662019-04-05T05:00:00.000-04:002019-05-05T12:01:58.658-04:00First Page Friday - Guest - Suzan LAUDER<div class="separator" style="clear: both; text-align: center;"><a href="https://2.bp.blogspot.com/-_Pj-PYpvWIk/XKV94BgL29I/AAAAAAAAECc/JZTLhUYG2MkfXHPikhlKu-XFUQNp8kvwwCLcBGAs/s1600/FPF%2B-%2BBlog%2B-%2BLauder.png" imageanchor="1" style="clear: left; float: left; margin-bottom: 1em; margin-right: 1em;"><img border="0" data-original-height="200" data-original-width="800" src="https://2.bp.blogspot.com/-_Pj-PYpvWIk/XKV94BgL29I/AAAAAAAAECc/JZTLhUYG2MkfXHPikhlKu-XFUQNp8kvwwCLcBGAs/s1600/FPF%2B-%2BBlog%2B-%2BLauder.png" /></a></div><br />I am beyond thrilled to have with me this Friday an old, old, old, old... (did I say old?) friend. Suzan and I go WAY back. Most of our joint memories revolve around high school, but we both attended the same church in our small town so I have an inkling we knew each other as wee little girls in Sunday School. Suzan was a year older than me, and when you're at an age where you insist on everyone knowing you are eight and a HALF, one year is a big difference. However, by the time I was in Grade Nine and she in Grade Ten, a solid friendship had begun.<br /><br />Through the magic of Facebook, (which <span style="font-family: "georgia" , "times new roman" , serif;"><i>can </i></span>be used for good), we re-connected. In fact, it was Suzan who introduced me to #JAFF. I had read some fan fiction years before, (Pamela Aiden's Fitzwilliam Darcy - Gentleman trilogy), but never knew there was a whole genre revolving specifically around Jane Austen's characters. I felt like I'd come home when I began to peek around A Happy Assembly.<br /><br />Today, Suzan is sharing from her latest novel, <span style="font-family: "georgia" , "times new roman" , serif;"><i>The Mist of Her Memory</i></span><br /><br /><b>Book Blurb:</b><br /><span style="font-family: inherit; white-space: pre-wrap;"><br /></span><span style="font-family: inherit; white-space: pre-wrap;">Sequestered in her Aunt and Uncle Gardiner’s London home, Elizabeth Bennet is recovering from an incident that stole her memories. She still suffers strange, angry voices in her head and recalls events people tell her never happened—even those who love her don’t believe her. Elizabeth can barely stand it!</span><br /><span style="font-family: inherit; white-space: pre-wrap;"><br /></span><span style="font-family: inherit; white-space: pre-wrap;">Mr. Darcy’s longs for any hint of his beloved’s well-being, yet he lacks the information he so desperately seeks because he has been forbidden contact with Elizabeth. His frustration mounts when he is told of the taunting and torment she endured in Meryton because of her mental impairment.</span><br /><span style="font-family: inherit; white-space: pre-wrap;"><br /></span><span style="font-family: inherit; white-space: pre-wrap;">Which of Elizabeth’s recollections bear the closest resemblance to the truth? What became of Mr. Wickham and Lydia’s elopement? How is Mr. Darcy to restore his romance with Elizabeth when she is so closely shielded by her aunt and uncle? What happened that fateful morning in Lambton? Who attacked her so brutally to have caused grievous, near-fatal injuries? Does she remain in danger? Elizabeth cannot remember!</span><br /><span style="font-family: inherit; white-space: pre-wrap;"><br /></span><span style="font-family: inherit; white-space: pre-wrap;">Get ready to be on the edge of your seat for this romantic suspense and mystery novel by bestselling author Suzan Lauder, based on Jane Austen's </span><i style="font-family: inherit; white-space: pre-wrap;">Pride and Prejudice</i><span style="font-family: inherit; white-space: pre-wrap;">.</span><br /><span style="font-family: inherit; white-space: pre-wrap;"><br /></span><span style="font-family: inherit; white-space: pre-wrap;"><b>First Page:</b></span><br /><div dir="ltr" style="line-height: 1.2; margin-bottom: 0pt; margin-top: 0pt; text-align: center;"><br /></div><div dir="ltr" style="line-height: 1.2; margin-bottom: 0pt; margin-top: 0pt;"><span style="background-color: transparent; color: black; font-family: "arial"; font-size: 12pt; font-style: italic; font-variant: normal; font-weight: 400; text-decoration: none; vertical-align: baseline; white-space: pre;">It was a gloomy prospect, and all that she could do was to throw a mist over it, and hope when the mist cleared away, she should see something else.</span></div><div dir="ltr" style="line-height: 1.2; margin-bottom: 0pt; margin-top: 0pt; text-align: right;"><span style="font-family: "arial"; font-size: 12pt; text-indent: 36pt; vertical-align: baseline; white-space: pre-wrap;">–</span><span style="font-family: "arial"; font-size: 12pt; font-style: italic; text-indent: 36pt; vertical-align: baseline; white-space: pre-wrap;">Mansfield Park</span><span style="font-family: "arial"; font-size: 12pt; text-indent: 36pt; vertical-align: baseline; white-space: pre-wrap;">, by Jane </span><span style="font-family: "arial"; font-size: 12pt; text-indent: 36pt; vertical-align: baseline; white-space: pre-wrap;">Austen</span></div><div dir="ltr" style="line-height: 1.2; margin-bottom: 0pt; margin-top: 0pt;"><span style="font-family: "arial"; font-size: 12pt; font-style: italic; text-indent: 36pt; white-space: pre-wrap;"><br /></span></div><div dir="ltr" style="line-height: 1.2; margin-bottom: 0pt; margin-top: 0pt; text-align: right;"><span style="font-size: 12pt; font-style: italic; text-indent: 36pt; white-space: pre-wrap;"><span style="font-family: "georgia" , "times new roman" , serif;">Early October 1812</span></span></div><div dir="ltr" style="line-height: 1.2; margin-bottom: 0pt; margin-top: 0pt; text-align: right;"><span style="font-size: 12pt; font-style: italic; text-indent: 36pt; white-space: pre-wrap;"><span style="font-family: "georgia" , "times new roman" , serif;">Gracechurch Street, London</span></span></div><br /><blockquote class="tr_bq" style="line-height: 1.2; margin-bottom: 0pt; margin-top: 0pt;"><span style="font-size: 12pt; text-indent: 36pt; white-space: pre-wrap;"><span style="font-family: "georgia" , "times new roman" , serif;">Elizabeth held her breath as she sneaked towards the servants’ entrance of her uncle’s home. She had but a moment before she would be missed, barely enough time to slip from her chamber down the servants’ stairs. Stepping gingerly to avoid the noise made by the loose board on the seventh stair, she made her way to the bottom and opened the door to the rear vestibule only a crack so no one would hear it creak.</span></span></blockquote><br /><blockquote class="tr_bq" style="line-height: 1.2; margin-bottom: 0pt; margin-top: 0pt;"><span style="font-size: 12pt; text-indent: 36pt; white-space: pre-wrap;"><span style="font-family: "georgia" , "times new roman" , serif;">Did she remember correctly? She had seen a cloak of sorts hanging in the back entrance as she wandered up and down stairs whilst her aunt made calls in the mornings. On those days, they did not walk in the park, and her aunt refused to allow Elizabeth to take her walks on her own. Could they not understand how valuable her exercise and time outside of the stale air of the house were to her? She would be perfectly safe—she had never fallen faint on her own.</span></span></blockquote><br /><blockquote class="tr_bq" style="line-height: 1.2; margin-bottom: 0pt; margin-top: 0pt;"><span style="font-size: 12pt; text-indent: 36pt; white-space: pre-wrap;"><span style="font-family: "georgia" , "times new roman" , serif;">But as much as she tried to stay out of the servants’ way on the days when she ambled about the house for exercise, her presence here had been reported at some point. Uncle Gardiner told her he preferred that she did not walk about on the back stairs lest she get in the servants’ way.</span></span></blockquote><br /><blockquote class="tr_bq" style="line-height: 1.2; margin-bottom: 0pt; margin-top: 0pt;"><span style="font-size: 12pt; text-indent: 36pt; white-space: pre-wrap;"><span style="font-family: "georgia" , "times new roman" , serif;">She peeked through the crack in the door. A woman’s hooded cape was suspended on the hook near the doorway, just the thing she would need when the time came to escape. The drab, homespun servant’s cape made her skin itch whenever she imagined wearing it, yet the pathetic garment was a symbol of hope—if she dared fetch it. Its distance from her situation taunted her, a reminder of how she was trapped here against her will.</span></span></blockquote><br /><blockquote class="tr_bq" style="line-height: 1.2; margin-bottom: 0pt; margin-top: 0pt;"><span style="font-size: 12pt; text-indent: 36pt; white-space: pre-wrap;"><span style="font-family: "georgia" , "times new roman" , serif;">A wave of nausea hit her, and her vision blackened and blurred a mite. She shut the door silently and, closing her eyes, leaned her head against the door frame until the moment passed and she was able to make her way back up to the family rooms.</span></span></blockquote><br /><blockquote class="tr_bq" style="line-height: 1.2; margin-bottom: 0pt; margin-top: 0pt;"><span style="font-size: 12pt; text-indent: 36pt; white-space: pre-wrap;"><span style="font-family: "georgia" , "times new roman" , serif;">“Elizabeth?” Her aunt emerged at the very moment when she was able to make it appear as though she was about to enter the family dining room.</span></span></blockquote><br /><blockquote class="tr_bq" style="line-height: 1.2; margin-bottom: 0pt; margin-top: 0pt;"><span style="font-size: 12pt; text-indent: 36pt; white-space: pre-wrap;"><span style="font-family: "georgia" , "times new roman" , serif;">“Forgive my tardiness, Aunt. I had a loose seam that needed mending.”</span></span></blockquote><br /><blockquote class="tr_bq" style="line-height: 1.2; margin-bottom: 0pt; margin-top: 0pt;"><span style="font-size: 12pt; text-indent: 36pt; white-space: pre-wrap;"><span style="font-family: "georgia" , "times new roman" , serif;">It was a lie. But her slow ability with the needle would not be questioned, and her maid had been dismissed a good quarter of an hour ago. Elizabeth pursed her lips. She was so frustrated! Not that she intended to escape right now, but the sensation of freedom had wrapped itself around her and given her hope, if only for a moment.</span></span></blockquote><br />I followed and read this story in pre-production and enjoyed it so very much. Once again, Suzan has given us characters that leap off the page. Don't hesitate to gobble this story up when it releases. Until then,<br /><div class="separator" style="clear: both; text-align: center;"><a href="https://3.bp.blogspot.com/-NgPbIaeLv0c/W4XehbXT76I/AAAAAAAADu8/cPLObZ1Q7mEJ2hrrgp8fuk-UZ5oqbzkzgCPcBGAYYCw/s1600/Always%2BWith%2BLove.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="100" data-original-width="265" src="https://3.bp.blogspot.com/-NgPbIaeLv0c/W4XehbXT76I/AAAAAAAADu8/cPLObZ1Q7mEJ2hrrgp8fuk-UZ5oqbzkzgCPcBGAYYCw/s1600/Always%2BWith%2BLove.png" /></a></div><br /><div><span style="background-color: transparent; color: black; font-family: "arial"; font-size: 12pt; font-style: normal; font-variant: normal; font-weight: 400; text-decoration: none; vertical-align: baseline; white-space: pre;"><br /></span></div>Sue Barrhttp://www.blogger.com/profile/06347249118943488564noreply@blogger.com11tag:blogger.com,1999:blog-8317810105915057272.post-24878578601196569802019-03-30T13:55:00.000-04:002019-05-05T12:00:47.087-04:00Finalist - Hook, Line & Sinker ContestBragging rights are about to happen.<br /><br />Back in October 2018 I entered the first three pages of '<span style="font-family: "georgia" , "times new roman" , serif;"><b><i>Mary: Pride & Prejudice continued... Book Four</i></b></span>' into the RWA Hudson River Hook, Line, and Sinker contest. I'm happy to report that I made it to the finals (with a score of 149/150) and just yesterday I found out that I placed fifth over all. Not the win I'd like to crow about, but holy cow, fifth. AND if I scored almost perfect - what were the other contestants scores??? I think we're going to see a batch of excellent stories out this year. Can't wait.<br /><br />I thought I'd share with you my almost winning entry and hope you enjoy it as much as the judges did.<br /><div class="separator" style="clear: both; text-align: center;"><a href="https://4.bp.blogspot.com/-ij6A0o_55QY/W4XlGWx3EuI/AAAAAAAADvg/kGCx_8iw4mUuVmQiBgX2g7zGKPGzgMggwCPcBGAYYCw/s1600/Mary-title.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="747" data-original-width="1600" height="149" src="https://4.bp.blogspot.com/-ij6A0o_55QY/W4XlGWx3EuI/AAAAAAAADvg/kGCx_8iw4mUuVmQiBgX2g7zGKPGzgMggwCPcBGAYYCw/s320/Mary-title.png" width="320" /></a></div><br /><blockquote class="tr_bq" style="line-height: normal; mso-pagination: none; text-indent: 0in;"><span lang="EN-US"><span style="font-family: "georgia" , "times new roman" , serif;">At one and twenty, Miss Mary Bennet knew she was already considered a spinster by the village of Meryton. Her youngest sister Lydia married at the tender age of fifteen, followed quickly by Jane and Lizzy respectively. Then Kitty followed suit a scant two years later and Mary had watched from the sidelines as her sisters fell in love and married.</span></span></blockquote><blockquote class="tr_bq" style="line-height: normal; mso-pagination: none; text-indent: 0in;"><span lang="EN-US"><span style="font-family: "georgia" , "times new roman" , serif;">She was well aware that she was not traditionally ladylike like her siblings. Too forthright in her speech, she’d rather spend an evening playing the pianoforte to an evening in the company of friends, or heaven forfend, attend a ball. She had a pleasing figure, all her own teeth and, if she had one lick of vanity, it was her thick, beautifully curly, mahogany locks of hair.</span></span></blockquote><blockquote class="tr_bq" style="line-height: normal; mso-pagination: none; text-indent: 0in;"><span lang="EN-US"><span style="font-family: "georgia" , "times new roman" , serif;">Papa had finally settled on her a substantial dowry. With all her siblings out from under the eaves of Longbourn, he had the resources to add to the family coffers and she now had a nice tidy sum of four thousand pounds for any man willing to make an offer. And there was the rub. Someone needed to make an offer.</span></span></blockquote><blockquote class="tr_bq" style="line-height: normal; mso-pagination: none; text-indent: 0in;"><span lang="EN-US"><span style="font-family: "georgia" , "times new roman" , serif;">If anyone were to catalog all her attributes, they’d wonder how she’d gone so long without one single proposal. Without one single kiss. She didn’t count the quick peck the cobbler’s son gave her behind the church when she was fourteen. He’d pressed his lips against hers and tried to push his tongue into her mouth, which she promptly bit.</span></span></blockquote><blockquote class="tr_bq" style="line-height: normal; text-indent: 0in;"><span lang="EN-US"><span style="font-family: "georgia" , "times new roman" , serif;">He barely spoke to her after that and frankly, she hadn’t cared. If kissing involved groping hands and tongues being shoved into one’s mouth, she didn’t wish to be kissed again. However, all her sisters seemed to <i style="mso-bidi-font-style: normal;">like</i> the fact their husbands kissed them, so maybe he’d got it all wrong. </span></span></blockquote><blockquote class="tr_bq" style="line-height: normal; text-indent: 0in;"><span lang="EN-US"><span style="font-family: "georgia" , "times new roman" , serif;">She hastened her pace upon entering the village of Meryton. No one was even mildly bothered that she’d walked the mile and a half from Longbourn alone. She enjoyed the freedom of movement normally attributed to doddering old spinsters and widow’s long past looking for another husband. She was, to put it succinctly, unmarriageable.</span></span></blockquote><blockquote class="tr_bq" style="line-height: normal; text-indent: 0in;"><span lang="EN-US"><span style="font-family: "georgia" , "times new roman" , serif;">“Good afternoon, Miss Bennet. Lovely day, ain’t it?” Mrs. Sheffield greeted her while sweeping the wooden boardwalk outside her shop.</span></span></blockquote><blockquote class="tr_bq" style="line-height: normal; text-indent: 0in;"><span lang="EN-US"><span style="font-family: "georgia" , "times new roman" , serif;">“Good afternoon, Mrs. Sheffield. It is a lovely day.”</span></span></blockquote><blockquote class="tr_bq" style="line-height: normal; text-indent: 0in;"><span lang="EN-US"><span style="font-family: "georgia" , "times new roman" , serif;">“Have you read the newspaper today?”</span></span></blockquote><blockquote class="tr_bq" style="line-height: normal; text-indent: 0in;"><span lang="EN-US"><span style="font-family: "georgia" , "times new roman" , serif;">“Papa keeps the paper to himself. Was there something of importance?”</span></span></blockquote><blockquote class="tr_bq" style="line-height: normal; text-indent: 0in;"><span lang="EN-US"><span style="font-family: "georgia" , "times new roman" , serif;">“Oh, I’d say there were sumpthin’ important. Wait right ‘ere.”</span></span></blockquote><blockquote class="tr_bq" style="line-height: normal; text-indent: 0in;"><span lang="EN-US"><span style="font-family: "georgia" , "times new roman" , serif;">Mrs. Sheffield leaned her broom against the outer wall and disappeared into the store, returning with <i style="mso-bidi-font-style: normal;">The</i> <i style="mso-bidi-font-style: normal;">Gazette</i>, which she handed to Mary and said, “Page Three”.</span></span></blockquote><blockquote class="tr_bq" style="line-height: normal; text-indent: 0in;"><span lang="EN-US"><span style="font-family: "georgia" , "times new roman" , serif;">Mary took the paper and opened it to the suggested page. Emblazoned across the top was a bold headline, proclaiming: <i style="mso-bidi-font-style: normal;"><b>HRH, the Prince Regent Celebrates End of War</b></i>. The article started by saying that to celebrate the end of the Peninsular War, the Prince Regent wished to stage an elaborate garden party for every person to whom titles were bestowed in the past fifteen years. Following a quick blurb of where and when the party was to be held, a list of all invitees was listed, marching down the page in five straight lines.</span></span></blockquote><blockquote class="tr_bq" style="line-height: normal; text-indent: 0in;"><span lang="EN-US"><span style="font-family: "georgia" , "times new roman" , serif;">“I fail to see how a party the Prince Regent is holding affects me, Mrs. Sheffield.”</span></span></blockquote><blockquote class="tr_bq" style="line-height: normal; text-indent: 0in;"><span lang="EN-US"><span style="font-family: "georgia" , "times new roman" , serif;">“I suggest y’take a look at the guest list. The names are in alphabetical order.”</span></span></blockquote><blockquote class="tr_bq" style="line-height: normal; text-indent: 0in;"><span lang="EN-US"><span style="font-family: "georgia" , "times new roman" , serif;">To humor the kind woman, Mary began reading the names, gasping out loud when she came to the letter ‘<i style="mso-bidi-font-style: normal;">B’</i>. Mrs. Sheffield began to cackle at the look on her face.</span></span></blockquote><blockquote class="tr_bq" style="line-height: normal; text-indent: 0in;"><span lang="EN-US"><span style="font-family: "georgia" , "times new roman" , serif;">“I told you it were important.”</span></span></blockquote><blockquote class="tr_bq" style="line-height: normal; text-indent: 0in;"><span lang="EN-US"><span style="font-family: "georgia" , "times new roman" , serif;">“Excuse me.” Mary handed the paper back to Mrs. Sheffield. “I must go home.”</span></span></blockquote><blockquote class="tr_bq" style="line-height: normal; text-indent: 0in;"><span lang="EN-US"><span style="font-family: "georgia" , "times new roman" , serif;">With that she spun on her heel and once out of sight of the village, nearly ran all the way back to Longbourn. When she arrived home, panting, the house was in a state of uproar.</span></span></blockquote><blockquote class="tr_bq" style="line-height: normal; text-indent: 0in;"><span lang="EN-US"><span style="font-family: "georgia" , "times new roman" , serif;">“Mr. Bennet. What are we to do?” Her mother’s voice carried through the window to Papa’s book room. The quiet reply of her father could not be heard and it wasn’t until Mary actually entered the home that more of the conversation filtered out into the front hall.</span></span></blockquote><blockquote class="tr_bq" style="line-height: normal; text-indent: 0in;"><span lang="EN-US"><span style="font-family: "georgia" , "times new roman" , serif;">“But I am forty-two years of age. How can I go through this now?” Mama’s voice had taken on near hysterical proportions.</span></span></blockquote><blockquote class="tr_bq" style="line-height: normal; text-indent: 0in;"><span lang="EN-US"><span style="font-family: "georgia" , "times new roman" , serif;">“I am but fifty-three!”</span></span></blockquote><blockquote class="tr_bq" style="line-height: normal; text-indent: 0in;"><span lang="EN-US"><span style="font-family: "georgia" , "times new roman" , serif;">The door to Papa’s book room opened and upon seeing Mary, he beckoned her inside. With a fair bit of apprehension, she did as he bade and joined them. Mama paced in front of the window while Papa settled at his desk. </span></span></blockquote><blockquote class="tr_bq" style="line-height: normal; text-indent: 0in;"><span lang="EN-US"><span style="font-family: "georgia" , "times new roman" , serif;">“Your mother and I have some news and would appreciate if you kept this knowledge to yourself for a small amount of time.”</span></span></blockquote><blockquote class="tr_bq" style="line-height: normal; text-indent: 0in;"><span lang="EN-US"><span style="font-family: "georgia" , "times new roman" , serif;">“Everyone will know about it soon enough,” Mama cried out and threw herself into the closest chair.</span></span></blockquote><blockquote class="tr_bq" style="line-height: normal; text-indent: 0in;"><span lang="EN-US"><span style="font-family: "georgia" , "times new roman" , serif;">Mary suffered a quick internal debate on the merits of whether to share that the villagers of Meryton were well apprised of their good fortune or not. Deciding to cross that bridge when it arose, she pretended she had no foreknowledge and said, “I shall be the soul of discretion.”</span></span></blockquote><blockquote class="tr_bq" style="line-height: normal; text-indent: 0in;"><span lang="EN-US"><span style="font-family: "georgia" , "times new roman" , serif;">“Thank you, although you won’t have to carry this secret for too long because, as your mother stated, the news will become evident in fairly short order.”</span></span></blockquote><blockquote class="tr_bq" style="line-height: normal; text-indent: 0in;"><span lang="EN-US"><span style="font-family: "georgia" , "times new roman" , serif;">At that her mother wailed again and bolted for the door, flinging it open before rushing up the stairs. Startled, Mary watched, her mouth open. Was her mother not pleased that Papa had been elevated to the rank of Baron and she was now Lady Bennet?</span></span></blockquote><blockquote class="tr_bq" style="line-height: normal; text-indent: 0in;"><span lang="EN-US"><span style="font-family: "georgia" , "times new roman" , serif;">“Don’t worry, Mary,” Papa said. “Your mother is only casting up her accounts because of the babe.”</span></span></blockquote>How fun is that? Mary won't be published until later in the year. Given my current health status (surgery coming up in ten days) I'm struggling to get Georgiana done. I'm so close, only another twenty thousand words and I can begin edits and push my latest baby out into the world. Thanks for sticking with me through all of this.<br /><br />Until then,<br /><div class="separator" style="clear: both; text-align: center;"><a href="https://3.bp.blogspot.com/-NgPbIaeLv0c/W4XehbXT76I/AAAAAAAADu8/cPLObZ1Q7mEJ2hrrgp8fuk-UZ5oqbzkzgCPcBGAYYCw/s1600/Always%2BWith%2BLove.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="100" data-original-width="265" src="https://3.bp.blogspot.com/-NgPbIaeLv0c/W4XehbXT76I/AAAAAAAADu8/cPLObZ1Q7mEJ2hrrgp8fuk-UZ5oqbzkzgCPcBGAYYCw/s1600/Always%2BWith%2BLove.png" /></a></div><br /><br />Sue Barrhttp://www.blogger.com/profile/06347249118943488564noreply@blogger.com5tag:blogger.com,1999:blog-8317810105915057272.post-15592352005121207842019-03-29T08:39:00.000-04:002019-05-05T12:01:30.928-04:00First Page Friday - Guest - Jennifer ALTMAN<div class="separator" style="clear: both; text-align: center;"><a href="https://3.bp.blogspot.com/-jI1rJKb0d_E/XJ4OYam2VHI/AAAAAAAAECE/l5oeUBKYheoamiVByh6tDknTRw5ruGw4ACLcBGAs/s1600/FPF%2B-%2BBlog%2B-%2BAltman.png" imageanchor="1" style="clear: left; float: left; margin-bottom: 1em; margin-right: 1em;"><img border="0" data-original-height="200" data-original-width="800" src="https://3.bp.blogspot.com/-jI1rJKb0d_E/XJ4OYam2VHI/AAAAAAAAECE/l5oeUBKYheoamiVByh6tDknTRw5ruGw4ACLcBGAs/s1600/FPF%2B-%2BBlog%2B-%2BAltman.png" /></a></div><div class="separator" style="clear: both; text-align: left;">Welcome Jennifer Altman. Today we explore her novel <a href="http://getbook.at/ToConquerPride" target="_blank"><b><i><span style="color: #351c75; font-size: large;">To Conquer Pride</span></i></b></a>. I'd read this story a while back and remember it fondly.</div><br /><b>Book Blurb:</b><br /><br /><div class="MsoNormal" style="line-height: normal; margin-bottom: .0001pt; margin-bottom: 0in; mso-pagination: none;"><i><span lang="EN-US" style="background: white; font-size: 12pt;"><span style="font-family: inherit;">The course of true love never did run smooth...<o:p></o:p></span></span></i></div><div class="MsoNormal" style="background: white; line-height: normal; margin-bottom: .0001pt; margin-bottom: 0in; mso-pagination: none;"><br /></div><div class="MsoNormal" style="background: white; line-height: normal; margin-bottom: .0001pt; margin-bottom: 0in; mso-pagination: none;"><span lang="EN-US" style="font-size: 12pt;"><span style="font-family: inherit;">When Fitzwilliam Darcy departs Hunsford after his disastrous proposal to Elizabeth Bennet, he does not expect their paths to cross again. Indeed, knowing the lady's true feelings for him, he makes every effort to see that they do not. But when a chance encounter leaves him stranded in an abandoned cottage with the one woman he can never have, Darcy quickly realizes there is more at risk than just Elizabeth's reputation.</span></span></div><div class="MsoNormal" style="background: white; line-height: normal; margin-bottom: .0001pt; margin-bottom: 0in; mso-pagination: none;"><span lang="EN-US" style="font-size: 12pt;"><span style="font-family: inherit;"><br /></span></span></div><div class="MsoNormal" style="background: white; line-height: normal; margin-bottom: .0001pt; margin-bottom: 0in; mso-pagination: none;"><span lang="EN-US" style="font-size: 12pt;"><span style="font-family: inherit;">Elizabeth Bennet knows Mr. Darcy is the last man in the world whom she could ever be prevailed on to marry. Until the morning he hands her a letter, his countenance as dark and forbidding as the windswept sky. Now, trapped in a snowstorm with the one person she was certain she despised, Elizabeth is startled to discover that her feelings are not at all what she expected.</span></span></div><div class="MsoNormal" style="background: white; line-height: normal; margin-bottom: .0001pt; margin-bottom: 0in; mso-pagination: none;"><span lang="EN-US" style="font-size: 12pt;"><span style="font-family: inherit;"><br />But is one night alone together enough to alter the course of their future?</span></span></div><div class="MsoNormal" style="background: white; line-height: normal; margin-bottom: .0001pt; margin-bottom: 0in; mso-pagination: none;"><span lang="EN-US" style="font-size: 12pt;"><span style="font-family: inherit;"><br />Can any man as proud as Mr. Darcy be expected to offer for the same woman a second time?</span></span></div><div class="MsoNormal" style="background: white; line-height: normal; margin-bottom: .0001pt; margin-bottom: 0in; mso-pagination: none;"><span lang="EN-US" style="font-size: 12pt;"><span style="font-family: inherit;"><br />In this tale of serendipity and second chances, literature's unlikeliest couple must conquer pride, prejudice, and faulty first impressions in the elusive quest for their own happily ever after.</span><o:p></o:p></span></div><br /><b>First Page:</b><br /><blockquote class="tr_bq" style="line-height: normal; margin-bottom: .0001pt; margin-bottom: 0in; mso-pagination: none;"><span style="font-family: "georgia" , "times new roman" , serif;"><span lang="EN-US" style="font-size: 12.0pt; mso-bidi-font-family: Calibri; mso-bidi-theme-font: minor-latin; mso-fareast-font-family: "Times New Roman";"><i>Under different circumstances the rocking of the carriage might have been soothing; however, on this brisk November day, the motion offered little comfort to the compartment’s sole occupant.</i></span></span></blockquote><blockquote class="tr_bq" style="line-height: normal; margin-bottom: .0001pt; margin-bottom: 0in; mso-pagination: none;"><i style="font-family: Georgia, "Times New Roman", serif; font-size: 12pt; text-indent: 0.2in;">Fitzwilliam Darcy stared out the frosted windowpane, but his thoughts were far removed from the passing Kent countryside. From the outset, he had dreaded this journey. Indeed, if it were up to him, he would not have come within fifty miles of Hunsford for as long as he lived. But despite the accusations leveled at him in that very village, Darcy was a man who knew his duty, so when Lady Catherine had requested—nay, demanded—his assistance, he had come. Even so, he kept his visit as brief as possible, arriving yesterday morning and working late into the night. He spoke to no one, save his aunt’s steward, going so far as to take meals in his chambers. And not his usual chambers, either. No, he had made a point of requesting a different apartment. Never again would he step foot in the room where he had dared to dream of a life spent happily with Miss Elizabeth Bennet. The room where he had poured out his soul in that ill-conceived letter…</i></blockquote><blockquote class="tr_bq" style="line-height: normal; margin-bottom: .0001pt; margin-bottom: 0in; mso-pagination: none;"><i style="font-family: Georgia, "Times New Roman", serif;"><span style="font-size: 12pt; text-indent: 0.2in;">Releasing a rough sigh, Darcy pressed his forehead to the cold glass. Seven months. It had been seven months since his previous trip to Rosings. Seven months since he had last seen </span><span style="font-size: 12pt; text-indent: 0.2in;">her</span><span style="font-size: 12pt; text-indent: 0.2in;">.</span></i></blockquote><blockquote class="tr_bq" style="line-height: normal; margin-bottom: .0001pt; margin-bottom: 0in; mso-pagination: none;"><i style="font-family: Georgia, "Times New Roman", serif;"><span style="font-size: 12pt; text-indent: 0.2in;">Darcy’s head fell back against the cushions. Just thinking about Elizabeth was like a spill of salt to an open wound—and yet he could do nothing to </span><span style="font-size: 12pt; text-indent: 0.2in;">stop</span><span style="font-size: 12pt; text-indent: 0.2in;"> thinking. God knows he had tried.</span></i></blockquote><blockquote class="tr_bq" style="line-height: normal; margin-bottom: .0001pt; margin-bottom: 0in; mso-pagination: none;"><i style="font-family: Georgia, "Times New Roman", serif; font-size: 12pt; text-indent: 0.2in;">Oh, it had not been difficult at first. After leaving Rosings, he had gone straight to London to set things right with Bingley, and then there was Wickham to contend with. That endeavor had proven a bit more challenging, but at least it had kept Darcy from dwelling on his disappointed hopes. By mid-summer he was back at Pemberley, and though he made every effort to throw himself into the management of his estate, even the brandy he had taken to consuming on a nightly basis did little to dull his memories. For no matter what he did, Elizabeth Bennet haunted his dreams, both sleeping and awake. Sometimes she would appear to him the way he wished to remember her—her smile radiant, her expression bright with humor. But other times he recalled the way she had looked at him the day of his disastrous proposal—her fine eyes dark with anger.</i></blockquote><blockquote class="tr_bq" style="line-height: normal; margin-bottom: .0001pt; margin-bottom: 0in; mso-pagination: none;"><i style="font-family: Georgia, "Times New Roman", serif; text-indent: 0.2in;"><span lang="EN-US" style="font-size: 12.0pt; mso-bidi-font-family: Calibri; mso-bidi-theme-font: minor-latin; mso-fareast-font-family: "Times New Roman";">Had you behaved in a more gentlemanlike manner…</span></i></blockquote><blockquote class="tr_bq" style="line-height: normal; margin-bottom: .0001pt; margin-bottom: 0in; mso-pagination: none;"><i style="font-family: Georgia, "Times New Roman", serif; font-size: 12pt; text-indent: 0.2in;">An icy wind swept into the carriage and Darcy started, realizing they were no longer moving. Turning to face the open door, he regarded the footman silhouetted against the lead-gray sky. When Darcy spoke, his voice was tight.</i></blockquote><blockquote class="tr_bq" style="line-height: normal; margin-bottom: .0001pt; margin-bottom: 0in; mso-pagination: none;"><i style="font-family: Georgia, "Times New Roman", serif; font-size: 12pt; text-indent: 0.2in;">“Why are we stopping? I thought I made it clear I wished to travel straight through.”</i></blockquote><br />There you have it, an exciting start to another wonderful #JAFF novel. I, myself, have pondered often how Darcy behaved in the months between Hunsford and the fateful meeting with Miss Elizabeth Bennet. Some of my favourite stories revolve around his quest to become a more 'gentlemanlike' person. I enjoyed this story very much and hope you do as well.<br /><br />You can find out more about Jennifer and her book on her <a href="https://www.facebook.com/JAltmanAuthor/" target="_blank">Facebook Page</a>. Until next week,<br /><div class="separator" style="clear: both; text-align: center;"><a href="https://3.bp.blogspot.com/-NgPbIaeLv0c/W4XehbXT76I/AAAAAAAADu8/cPLObZ1Q7mEJ2hrrgp8fuk-UZ5oqbzkzgCPcBGAYYCw/s1600/Always%2BWith%2BLove.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="100" data-original-width="265" src="https://3.bp.blogspot.com/-NgPbIaeLv0c/W4XehbXT76I/AAAAAAAADu8/cPLObZ1Q7mEJ2hrrgp8fuk-UZ5oqbzkzgCPcBGAYYCw/s1600/Always%2BWith%2BLove.png" /></a></div><br /><br />Sue Barrhttp://www.blogger.com/profile/06347249118943488564noreply@blogger.com5tag:blogger.com,1999:blog-8317810105915057272.post-48572699153011467772019-03-22T05:00:00.000-04:002019-05-05T12:02:23.813-04:00First Page Friday - Guest - Riana EVERLY<div class="separator" style="clear: both; text-align: center;"><a href="https://4.bp.blogspot.com/-QUQSfwgQm-g/XJPCOvMfw5I/AAAAAAAAEBk/gmwKuHIU9ikSpLafw47IttnIRoyOSxrXwCLcBGAs/s1600/First%2BPage%2BFriday%2B%25281%2529.png" imageanchor="1" style="clear: left; float: left; margin-bottom: 1em; margin-right: 1em;"><img border="0" data-original-height="200" data-original-width="800" src="https://4.bp.blogspot.com/-QUQSfwgQm-g/XJPCOvMfw5I/AAAAAAAAEBk/gmwKuHIU9ikSpLafw47IttnIRoyOSxrXwCLcBGAs/s1600/First%2BPage%2BFriday%2B%25281%2529.png" /></a></div><br />Welcome Riana Everly to First Page Friday. Today, Rianna is sharing from her Pride and Prejudice variation novel: <a href="https://mybook.to/ThroughaDifferentLens" style="font-family: Georgia, "Times New Roman", serif; font-style: italic; font-weight: bold;" target="_blank">Through a Different Lens</a><span style="font-family: inherit;">. I don't know about you, but I love reading stories of Darcy and Elizabeth. There's something about an enemy to friend trope that takes hold of your imagination. Regardless of how many stories I read, I never tire of them. Without further ado, here is more details about Riana's book.</span><br /><br /><b>Book Blurb:</b><br /><span style="font-family: inherit;"><br /></span><span style="font-family: inherit;"><span style="background: white; color: #333333;">A tale of second glances and second chances</span><span style="color: #333333;"><br /><br /><span style="background: white;">Elizabeth Bennet has disliked the aloof and arrogant Mr. Darcy since he insulted her at a village dance several months before. But an unexpected conversation with a startling turn of phrase suddenly causes her to reassess everything she thought she knew about the infuriating and humourless gentleman. </span><br /><br /><span style="background: white;">Elizabeth knows something of people who think differently. Her young cousin in London has always been different from his siblings and peers, and Lizzy sees something of this boy’s unusual traits in the stern gentleman from Derbyshire whose presence has plagued her for so long. She approaches him in friendship and the two begin a tentative association. But is Lizzy's new understanding of Mr. Darcy accurate? Or was she right the first time? And will the unwelcome appearance of a nemesis from the past destroy any hopes they might have of happiness?</span><br /><br /><span style="background: white;">Warning: This variation of Jane Austen's classic Pride and Prejudice depicts our hero as having a neurological difference. If you need your hero to be perfect, this might not be the book for you. But if you like adorable children, annoying birds, and wonderful dogs, and are open to a character who struggles to make his way in a world he does not quite comprehend, with a heroine who can see the man behind his challenges, and who celebrates his strengths while supporting his weaknesses, then read on! You, too, can learn what wonders can be found when we see the familiar through a different lens.</span></span></span><br /><span style="font-family: inherit;"><span style="color: #333333;"><span style="background: white;"><br /></span></span></span><span style="font-family: inherit;"><span style="color: #333333;"><span style="background: white;"><b>First Page:</b></span></span></span><br /><div style="text-align: center;"><span style="font-family: "georgia" , "times new roman" , serif; font-size: large;"><span style="color: #333333;"><span style="background: white;"><i>Chapter One</i></span></span></span></div><div style="text-align: center;"><span style="font-family: "georgia" , "times new roman" , serif; font-size: large;"><span style="color: #333333;"><span style="background: white;"><i>Ill Qualified to Recommend Himself</i></span></span></span></div><blockquote class="tr_bq"><span style="color: #333333;"><span style="background: white;"><i><span style="font-family: "georgia" , "times new roman" , serif;">It was an evening much like many others over the past few weeks. The small party were gathered in the salon after an uncomfortable dinner, to amuse, delight, and take advice from the doyenne of the house. The meal had seemed endless, with one overly fine dish superseding another, testimony more to the expense of a fine French chef than to the consideration due to the palates of the assembled guests. Likewise, the conversation, more of a series of interrogatory demands by the lady of the house than an exchange of light and pleasant thoughts to lend enjoyment to the meal. Now, the last of the dishes cleared away and the company retired to the salon, Elizabeth sat perched upon the uncomfortable sofa, seeking something amusing to say that would astound those gathered around. Beside her sat her dear Charlotte, whom she was visiting, and nearby, Charlotte's husband, Mr. Collins, who held the living at Hunsford, adjacent to the grand manor house of Rosings. Also in the room were the mistress of the house herself, Lady Catherine de Bourgh, domineering and fierce of temperament, her sickly daughter Anne, who seemed more intimidated than truly ill, Anne's companion Mrs. Jenkins, and Charlotte's timid sister Maria, who hardly spoke a word.</span></i></span></span></blockquote><blockquote class="tr_bq"><span style="color: #333333;"><span style="background: white;"><i><span style="font-family: "georgia" , "times new roman" , serif;">In all of these particulars, the scene had been repeated many times since Elizabeth first arrived at Hunsford for a prolonged visit with her friend; recently, however, two more members had joined their party, one adding to its pleasure, the other to its awkwardness.</span></i></span></span></blockquote><blockquote class="tr_bq"><span style="background: white;"><i><span style="font-family: "georgia" , "times new roman" , serif;"><span style="color: #333333;">The increased pleasure was due entirely to the newly formed </span><span style="color: #333333;">acquaintance</span><span style="color: #333333;"> with Lady Catherine's nephew, Colonel Fitzwilliam, on a short leave from the army to visit his aunt and cousin and help tend to affairs of her estate. The colonel and Elizabeth had quickly formed a comfortable and easy friendship, for the gentleman was intelligent, quick-witted, and extremely good company. Elizabeth had taken an immediate liking to him and was pleased when he sought her companionship, either in the salon or whilst walking through the park.</span></span></i></span></blockquote><blockquote class="tr_bq"><span style="background: white;"><span style="color: #333333; font-family: "georgia" , "times new roman" , serif;"><i>His friend, however, was a far less of a source of pleasure. Silent, stiff and brooding, the colonel's constant shadow was none other than Mr. Darcy, whom Elizabeth had come to know and rather dislike several months ago at her home in Hertfordshire, near the village of Meryton.</i></span></span></blockquote><span style="background: white;"><span style="color: #333333; font-family: inherit;">There you have it, dear reader. Mr. Darcy is still looked upon unfavorable by Miss Elizabeth Bennet. (Sigh) He really was quite the dunderhead, wasn't he? I'm hoping Ms. Everly will take us on a wild ride with enough twists and turns to satisfy our angst ridden desires.</span></span><br /><span style="background: white;"><span style="color: #333333; font-family: inherit;"><br /></span></span><br /><div style="text-align: center;"><span style="background: white;"><b style="font-family: inherit;"><span style="background-attachment: initial; background-clip: initial; background-image: initial; background-origin: initial; background-position: initial; background-repeat: initial; background-size: initial; color: #333333;"><a href="https://rianaeverly.com/" target="_blank">Website</a> | </span><span style="color: #333333;"><span style="background-attachment: initial; background-clip: initial; background-image: initial; background-origin: initial; background-position: initial; background-repeat: initial; background-size: initial;"><a href="https://rianaeverly.com/blog/" target="_blank">Blog</a> | </span></span><span style="color: #333333;"><span style="background-attachment: initial; background-clip: initial; background-image: initial; background-origin: initial; background-position: initial; background-repeat: initial; background-size: initial;"><a href="https://www.facebook.com/RianaEverly/" target="_blank">Facebook</a> | </span></span><span style="color: #333333;"><span style="background-attachment: initial; background-clip: initial; background-image: initial; background-origin: initial; background-position: initial; background-repeat: initial; background-size: initial;"><a href="https://twitter.com/RianaEverly" target="_blank">Twitter</a> | </span></span><span style="color: #333333;"><span style="background-attachment: initial; background-clip: initial; background-image: initial; background-origin: initial; background-position: initial; background-repeat: initial; background-size: initial;"><a href="https://www.amazon.com/Riana-Everly/e/B076C6HY27" target="_blank">Amazon</a></span></span></b></span></div><span style="background: white;"><span style="color: #333333; font-family: inherit;"><br /></span></span><span style="background: white;"><span style="color: #333333; font-family: inherit;">Learn more about Riana by clinking on the links shown above. I'm sure she'd love to hear from you. Until then, </span></span><br /><div class="separator" style="clear: both; text-align: center;"><a href="https://3.bp.blogspot.com/-NgPbIaeLv0c/W4XehbXT76I/AAAAAAAADu8/cPLObZ1Q7mEJ2hrrgp8fuk-UZ5oqbzkzgCPcBGAYYCw/s1600/Always%2BWith%2BLove.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="100" data-original-width="265" src="https://3.bp.blogspot.com/-NgPbIaeLv0c/W4XehbXT76I/AAAAAAAADu8/cPLObZ1Q7mEJ2hrrgp8fuk-UZ5oqbzkzgCPcBGAYYCw/s1600/Always%2BWith%2BLove.png" /></a></div><br /><br />Sue Barrhttp://www.blogger.com/profile/06347249118943488564noreply@blogger.com2tag:blogger.com,1999:blog-8317810105915057272.post-31704644569779679952019-03-08T05:00:00.000-05:002019-05-05T12:02:47.347-04:00First Page Friday ~ Guest ~ Jayne BAMBER<div class="separator" style="clear: both; text-align: center;"><a href="https://3.bp.blogspot.com/-zjJoLBs6JFs/XH2DQM3eiGI/AAAAAAAAEAg/gQKhjoiAI5A_eyQlYWdhT7gxtmQ1zOC5ACLcBGAs/s1600/FPF%2B0304-JayneBamber1.png" imageanchor="1" style="clear: left; float: left; margin-bottom: 1em; margin-right: 1em;"><img border="0" data-original-height="200" data-original-width="800" src="https://3.bp.blogspot.com/-zjJoLBs6JFs/XH2DQM3eiGI/AAAAAAAAEAg/gQKhjoiAI5A_eyQlYWdhT7gxtmQ1zOC5ACLcBGAs/s1600/FPF%2B0304-JayneBamber1.png" /></a></div>Welcome Jayne Bamber to First Page Friday with her debut novel: "<a href="https://www.amazon.com/Happier-her-Friends-than-Relations/dp/1792619928/ref=tmm_pap_swatch_0?_encoding=UTF8&qid=1546417382&sr=8-1" target="_blank"><b><span style="font-family: "georgia" , "times new roman" , serif;"><i>Happier in Her Friends Than Relations</i></span></b></a>". The title speaks for itself. I have to admit - full disclosure here - I read this story as Jayne posted it on A Happy Assembly and looked forward to each and every post. I had a bit of a girl crush on Lady Rebecca Fitzwilliam. We all need a friend like her. I also liked how Jayne wove characters from Sense and Sensibility so seamlessly into our beloved Pride and Prejudice.<br /><br /><b><span style="font-family: inherit;">Book Blurb:</span></b><br /><div class="MsoNormal"><span style="font-family: inherit;"><br /></span></div><div class="MsoNormal"><span style="font-family: inherit;"><o:p></o:p></span></div><div class="MsoNormal"><span style="font-family: inherit;">In this angsty fusion of two of Jane Austen’s most beloved novels, Pride & Prejudice and Sense & Sensibility, the actions of their sisters cause challenging chain reactions for one of literature’s most celebrated couples...<o:p></o:p></span></div><div class="MsoNormal"><span style="font-family: inherit;"><br /></span></div><div class="MsoNormal"><span style="font-family: inherit;">Fitzwilliam Darcy is faced with a family crisis of epic proportions after the fallout of his sister Georgiana’s ill-fated elopement in Ramsgate, while his friend Charles Bingley is persuaded to abandon his scheme of renting Netherfield Park.<o:p></o:p></span></div><div class="MsoNormal"><span style="font-family: inherit;"><br /></span></div><div class="MsoNormal"><span style="font-family: inherit;">Elizabeth Bennet journeys to London to recover her spirits, after Jane’s unexpected marriage changes the sisters’ relationship forever, and the consolation of the Gardiners proves insufficient.<o:p></o:p></span></div><div class="MsoNormal"><span style="font-family: inherit;"><br /></span></div><div class="MsoNormal"><span style="font-family: inherit;">The bonds of friendship offer Elizabeth a lifeline after a series of tragic events causes her to fear for her future. The support she receives from her new neighbor Marianne Brandon, and snarky socialite Lady Rebecca Fitzwilliam prove she is truly happier in her friends than relations.<o:p></o:p></span></div><div class="MsoNormal"><span style="font-family: inherit;"><br /></span></div><div class="MsoNormal"><span style="font-family: inherit;"><b>First Page:</b></span></div><div class="MsoNormal" style="text-align: center;"><span style="font-family: "georgia" , "times new roman" , serif;"><i>Prologue<o:p></o:p></i></span></div><blockquote class="tr_bq"><span style="font-family: "georgia" , "times new roman" , serif;"><i>The several hours Elizabeth Bennet had to wait between learning of her elder sister’s engagement, and finding the opportunity to speak with her privately, felt like an eternity. The house had been in uproar since breakfast, when their odious cousin Mr. Collins, a dubiously welcome guest in their home these two weeks, had requested a private audience with Jane. While Jane had acquiesced to Mr. Collins’s request with forbearance, Elizabeth had no idea that her sister truly meant to accept him, and was shocked when Mrs. Bennet fluttered into the drawing room less than an hour later, rejoicing over their good fortune. Jane and her intended had barely made an appearance, to receive the bewildered congratulations of her younger sisters, before Jane was promptly whisked away by her mother to make the requisite calls around the neighborhood, announcing the news. </i></span></blockquote><blockquote class="tr_bq"><span style="font-family: "georgia" , "times new roman" , serif;"><i>Mr. Collins was unceremoniously left behind by his future mother-in-law, and remained in the drawing room to extol, at length, on the anticipated virtues of his future life to his young cousins. Claiming the excuse of a headache, Elizabeth was the first of them to flee the uncomfortable interlude, and she remained out of doors, alone with her agitated thoughts, until her elder sister had returned. </i></span></blockquote><blockquote class="tr_bq"><span style="font-family: "georgia" , "times new roman" , serif;"><i>Elizabeth entered the room she shared with Jane and found her sister seated at the vanity, staring blankly at her own reflection; the sight of it tore at Elizabeth’s heart. She sat down on the bench next to her, covering Jane’s hand with her own. “Are you well, Jane?” </i></span></blockquote><blockquote class="tr_bq"><span style="font-family: "georgia" , "times new roman" , serif;"><i>Jane nodded absently, offering a thin smile. “Indeed, Lizzy, I am quite content.” </i></span></blockquote><blockquote class="tr_bq"><span style="font-family: "georgia" , "times new roman" , serif;"><i>“Jane, you cannot be serious. You cannot really tie yourself to that awful man!” </i></span></blockquote><blockquote class="tr_bq"><span style="font-family: "georgia" , "times new roman" , serif;"><i>Jane’s expression was one of innocent shock. “I am serious. It has been announced all over the neighborhood. And, truly, he is not so very bad, though I know you dislike him. He has made me an honorable offer.” </i></span></blockquote><blockquote class="tr_bq"><span style="font-family: "georgia" , "times new roman" , serif;"><i>Elizabeth let out a derisive snort. “Honorable indeed! As if he isn’t pompous enough already, he seeks to puff himself up even more by securing a wife who is far too good for him. And Mamma is so eager to see one of us settled that she does not care if it is to a foolish toad! No, Jane, you deserve better than that.” </i></span></blockquote><blockquote class="tr_bq"><span style="font-family: "georgia" , "times new roman" , serif;"><i>Jane recoiled, visibly distraught. “Lizzy, that is unkind. It was very generous of Mr. Collins to select a wife from amongst us.” </i></span></blockquote><blockquote class="tr_bq"><span style="font-family: "georgia" , "times new roman" , serif;"><i>Jane might have said more, but Elizabeth cut her off. “He likely knows that no other woman would have him! At least with us, in his mind, he has some chance of success, as he seems to think us so indebted to him. But you needn’t be, Jane, for you are five times too lovely to be wasted on such a man, when you could do so much better!” </i></span></blockquote><blockquote class="tr_bq"><span style="font-family: "georgia" , "times new roman" , serif;"><i>At this, Jane’s expression hardened, and she launched herself up from the bench beside Elizabeth, to pace the room. “Could I? I am two and twenty years old, I have hardly any dowry to speak of, and no other prospects. The northern gentleman who was rumored to be leasing Netherfield never came, and who knows what other opportunities may ever come our way. I cannot take any more chances, Lizzy. The truth is none of us may ever receive a better offer."</i></span></blockquote>Well... dear Jane-ites. What do you think of our sweet Jane marrying the odious Mr. Collins??? I'd love to behave like Lydia and spill all the beans, but then I'd take away your delight of this book with its unexpected twists and turns. This is a sweet romance even though some of the characters are decidedly NOT sweet and behave in atrocious manners. 'Nuff said.<br /><div class="separator" style="clear: both; text-align: center;"><a href="https://3.bp.blogspot.com/-NgPbIaeLv0c/W4XehbXT76I/AAAAAAAADu8/cPLObZ1Q7mEJ2hrrgp8fuk-UZ5oqbzkzgCPcBGAYYCw/s1600/Always%2BWith%2BLove.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="100" data-original-width="265" src="https://3.bp.blogspot.com/-NgPbIaeLv0c/W4XehbXT76I/AAAAAAAADu8/cPLObZ1Q7mEJ2hrrgp8fuk-UZ5oqbzkzgCPcBGAYYCw/s1600/Always%2BWith%2BLove.png" /></a></div><br />Sue Barrhttp://www.blogger.com/profile/06347249118943488564noreply@blogger.com2tag:blogger.com,1999:blog-8317810105915057272.post-62657773102679778312019-03-01T05:00:00.000-05:002019-03-01T05:00:02.296-05:00First Page Friday ~ MAN OF HER DREAMS<div class="separator" style="clear: both; text-align: center;"><a href="https://2.bp.blogspot.com/-G95jCBPp3nM/XHGR8MFGh9I/AAAAAAAAD_s/vL2ntq6HzM4vWhzOP0-ZnyfKTyxJ0KMoQCLcBGAs/s1600/First%2BPage%2BFriday%2B%25283%2529.png" imageanchor="1" style="clear: left; float: left; margin-bottom: 1em; margin-right: 1em;"><img border="0" data-original-height="200" data-original-width="800" src="https://2.bp.blogspot.com/-G95jCBPp3nM/XHGR8MFGh9I/AAAAAAAAD_s/vL2ntq6HzM4vWhzOP0-ZnyfKTyxJ0KMoQCLcBGAs/s1600/First%2BPage%2BFriday%2B%25283%2529.png" /></a></div>Man of Her Dreams is the second contemporary novel I self-published after I exited the traditional publishing world and launched my own career.<br /><br /><b><span style="font-size: large;">Blurb:</span></b><br /><br /><div class="separator" style="clear: both; text-align: center;"></div>Making it in a mans' world as a home renovation expert is tough, but Lindsay Swanson is up for the job, leaving no time for family and friends, let alone any kind of social life.<br /><br />Jared Kane moved to Ravenwood for some much needed peace and quiet to finish his next book, which is hard to do when he's hired his sexy neighbor to renovate his kitchen.<br /><br />Things start to heat up between the two neighbors until Lindsay overhears Jared talking about her notebook. A notebook wherein she and her BFF - after one too many bad dates and one too many glasses of wine - concocted a checklist of the perfect guy.<br /><br />Is Jared the real deal, or has he been molding himself into the man of her dreams to sneak past her defenses?<br /><br /><b><span style="font-size: large;">First Page:</span></b><br /><blockquote class="tr_bq" style="line-height: normal; text-indent: 0in;"><span style="font-family: "georgia" , "times new roman" , serif;"><i>Lindsay Swanson turned into the Food Mart parking lot and pulled to a stop near the entrance. Waves of heat shimmered from the hood of the truck, making the near empty parking lot look like a Saharan mirage. Sweat trickled down the side of her temple and she swiped it away before twisting her hair into a Ravenwood Hooligan’s ball cap.</i></span></blockquote><blockquote class="tr_bq" style="line-height: normal; text-indent: 0in;"><span style="font-family: "georgia" , "times new roman" , serif;"><i>She climbed out of the truck and a fluttering movement caught her eye. With a sigh she peeled a rogue piece of drywall tape off her tee-shirt. When you owned your own renovation business, things got messy.</i></span></blockquote><blockquote class="tr_bq" style="line-height: normal; text-indent: 0in;"><span style="font-family: "georgia" , "times new roman" , serif;"><i>Through the front window she spotted one of the cashiers working a check-out line and groaned. Just her luck, Carla would be working today. Now everyone in town would know how pathetic her life was. She could hear her now. </i></span><i style="font-family: Georgia, "Times New Roman", serif; text-align: justify; text-indent: 0in;">Poor Lindsay, she only bought a loaf of bread. No one to cook for...</i></blockquote><blockquote class="tr_bq" style="line-height: normal; text-indent: 0in;"><span style="font-family: "georgia" , "times new roman" , serif;"><i>For a brief moment, she considered hopping back into her truck and driving to her parent’s farm for a home cooked meal. Mom wouldn’t mind an extra body at the table, but Lindsay wasn’t up for the subtle guilt that would be sprinkled throughout the meal.</i></span></blockquote><blockquote class="tr_bq" style="line-height: normal; text-indent: 0in;"><span style="font-family: "georgia" , "times new roman" , serif;"><i>Over a savory pot roast, Mom would say, ‘A few grandchildren would be nice while we can enjoy them – without assistance from a walker.’ Or, ‘When are you going to find a nice young man?’ Lindsay would always remind her that she had a business to run. Finding a man and having a family wasn’t a top priority right now.</i></span></blockquote><br /><div class="MsoNormal" style="line-height: normal;"><br /></div><br />Sue Barrhttp://www.blogger.com/profile/06347249118943488564noreply@blogger.com0tag:blogger.com,1999:blog-8317810105915057272.post-19256256635500099252019-02-23T05:00:00.000-05:002019-05-05T12:47:55.702-04:00WEEKEND WRITING WARRIORS #70<div class="separator" style="clear: both; text-align: center;"><a href="https://2.bp.blogspot.com/-vJ4jS8JPBrY/WqVAJzFXvjI/AAAAAAAADaA/q0BqnY24_ZEme-A-zSEgoUBqZGd0WvFVgCPcBGAYYCw/s1600/wewriwa2.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="210" data-original-width="676" height="198" src="https://2.bp.blogspot.com/-vJ4jS8JPBrY/WqVAJzFXvjI/AAAAAAAADaA/q0BqnY24_ZEme-A-zSEgoUBqZGd0WvFVgCPcBGAYYCw/s640/wewriwa2.png" width="640" /></a></div>Welcome back for another week of authors teasing you with eight to ten lines from their current Work In Progress. We continue on with Maxwell (the Duke of Adborough), Mr. Fitzwilliam Darcy and his cousin Richard, better known as Colonel Fitzwilliam. Max has just divulged there is a desperate fortune hunter trying to ferret out information about Mr. Darcy's sister, Georgiana.<br /><blockquote class="tr_bq"><i style="font-family: Georgia, "Times New Roman", serif;">“Is there nobody who can love Georgiana for who she is and not for money?” The colonel flopped back into his chair. “How many fortune hunters need we fight off?”</i></blockquote><blockquote class="tr_bq"><i style="font-family: Georgia, "Times New Roman", serif;">Darcy cut a sharp glance at his cousin and the colonel abruptly quit speaking although Max was quite sure he still fumed silently. He finished his drink and stood.</i></blockquote><div class="MsoNormal" style="mso-pagination: none;"><span style="font-family: "georgia" , "times new roman" , serif;"><i><o:p></o:p></i></span></div><blockquote class="tr_bq" style="mso-pagination: none;"><span style="font-family: "georgia" , "times new roman" , serif;"><i>“I must be off, Mother has opened Kerr House to receive visitors and demanded I make an appearance to lend support to Caroline and Catherine against would be vipers.”</i></span></blockquote><div class="MsoNormal" style="mso-pagination: none;"><span style="font-family: "georgia" , "times new roman" , serif;"><i><o:p></o:p></i></span></div><blockquote class="tr_bq" style="mso-pagination: none;"><span style="font-family: "georgia" , "times new roman" , serif;"><i>“Should I enlist your aid for Elizabeth and Georgiana?” Darcy asked as he rang for a footman.</i></span></blockquote><div class="MsoNormal" style="mso-pagination: none;"><span style="font-family: "georgia" , "times new roman" , serif;"><i><o:p></o:p></i></span></div><blockquote class="tr_bq" style="mso-pagination: none;"><span style="font-family: "georgia" , "times new roman" , serif;"><i>“I defy anyone to slander your wife and sister, Lady Matlock would have them for breakfast and spit out the bones. You forget, I’ve seen the Colonel’s mother at work.”</i></span></blockquote><div class="MsoNormal" style="mso-pagination: none;"><span style="font-family: "georgia" , "times new roman" , serif;"><i><o:p></o:p></i></span></div>There you have it. Poor Georgiana. IS there anyone who will love her for herself and NOT the thirty thousand pounds that comes with her as soon as she says 'I do'?<br /><br />Please use this <a href="http://www.wewriwa.com/" target="_blank"><b><span style="font-family: "georgia" , "times new roman" , serif;"><i>LINK</i></span> </b></a>to visit the Weekend Writing Warriors web site and check out other authors. We are an eclectic group and you never know, you might find your next favourite author in this bunch.<br /><div class="separator" style="clear: both; text-align: center;"><a href="https://3.bp.blogspot.com/-NgPbIaeLv0c/W4XehbXT76I/AAAAAAAADu8/cPLObZ1Q7mEJ2hrrgp8fuk-UZ5oqbzkzgCPcBGAYYCw/s1600/Always%2BWith%2BLove.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="100" data-original-width="265" src="https://3.bp.blogspot.com/-NgPbIaeLv0c/W4XehbXT76I/AAAAAAAADu8/cPLObZ1Q7mEJ2hrrgp8fuk-UZ5oqbzkzgCPcBGAYYCw/s1600/Always%2BWith%2BLove.png" /></a></div><br />P.S.: Hug someone you love today.Sue Barrhttp://www.blogger.com/profile/06347249118943488564noreply@blogger.com18tag:blogger.com,1999:blog-8317810105915057272.post-39041643319764360242019-02-22T09:56:00.000-05:002019-05-05T12:04:01.493-04:00First Page Friday ~ ACCORDING TO PLAN<div class="separator" style="clear: both; text-align: center;"><a href="https://1.bp.blogspot.com/-ZsP-GqLl3XU/XHGQ86-27UI/AAAAAAAAD_g/RDGMFWBNNDkOv3utNPz7mAVgUr3QJeMLgCLcBGAs/s1600/First%2BPage%2BFriday%2B%25282%2529.png" imageanchor="1" style="clear: left; float: left; margin-bottom: 1em; margin-right: 1em;"><img border="0" data-original-height="200" data-original-width="800" src="https://1.bp.blogspot.com/-ZsP-GqLl3XU/XHGQ86-27UI/AAAAAAAAD_g/RDGMFWBNNDkOv3utNPz7mAVgUr3QJeMLgCLcBGAs/s1600/First%2BPage%2BFriday%2B%25282%2529.png" /></a></div><div class="separator" style="clear: both; text-align: center;"></div><div class="separator" style="clear: both; text-align: center;"></div>This was the first manuscript I ever sold in 2010. In 2014 I received my publishing rights back for ATP and seven other publications. In 2015 I re-published According to Plan under my own name (after taking out the naughty bits...). Today, I'm sharing the first page from Tank and Shelby's story.<br /><br /><b><span style="font-size: large;">Blurb</span></b>:<br /><br /><div class="separator" style="clear: both; text-align: center;"></div>Shelby Stewart's been hired for the biggest case of her career. Locate missing socialite Harrison Grant. To complicate matters, her estranged husband lands on her doorstep. He's also on the hunt for Harrison, albeit for a completely different reason. Harry is the prime suspect in the grisly murder of a call girl in L.A.<br /><br />Agent Jake Steele, aka Tank, is deep undercover as a P.I., and sometimes there's collateral damage in his line of work. His marriage to Shelby is one of them. With her life in danger, he lets her think he left for another woman. When this case is over, he means to win back the heart and trust of Shelby, the only woman he's ever loved.<br /><br />However, there's another person who has their eye on Shelby and will stop at nothing to possess her.<br /><br />From an attempted kidnapping to an explosion with deadly consequences, this is not your average missing person's case.<br /><br /><b><span style="font-size: large;">First Page:</span></b><br /><blockquote class="tr_bq" style="line-height: normal; text-indent: 0in;"><span lang="EN-US"><span style="font-family: "georgia" , "times new roman" , serif;"><i>A hint of rain lingered in the air without breaking the stifling heat. Seated in my office, which I’m sure was a broom closet in a former life, sweat threatened to trickle down the inside of my bra. I hated being cooped up on days like this and longed to be outside. At least if I was on the job, I could duck into a coffee shop for some air-conditioned relief, but long overdue paperwork threatened to topple off my desk. Despairing this day would never end, freedom arrived with the ringing of my desk phone. I pounced on it like a cat with a new toy.</i></span></span></blockquote><blockquote class="tr_bq" style="line-height: normal; text-indent: 0in;"><span lang="EN-US"><span style="font-family: "georgia" , "times new roman" , serif;"><i>“Stewart Agencies. This is Shelby.”</i></span></span></blockquote><blockquote class="tr_bq" style="line-height: normal; text-indent: 0in;"><span lang="EN-US"><span style="font-family: "georgia" , "times new roman" , serif;"><i>“Our son is missing,” a deep, modulated voice said with no preamble.</i></span></span></blockquote><blockquote class="tr_bq" style="line-height: normal; text-indent: 0in;"><span lang="EN-US"><span style="font-family: "georgia" , "times new roman" , serif;"><i> I found that odd, but with the angst that rides alongside a missing family member the man obviously did not want to waste precious time. I grabbed a note pad and asked, “How old is your son, Mr...?</i></span></span></blockquote><blockquote class="tr_bq" style="line-height: normal; text-indent: 0in;"><span lang="EN-US"><span style="font-family: "georgia" , "times new roman" , serif;"><i>“Grant. Raymond Grant. Harrison is thirty.”</i></span></span></blockquote><blockquote class="tr_bq" style="line-height: normal; text-indent: 0in;"><span lang="EN-US"><span style="font-family: "georgia" , "times new roman" , serif;"><i>I fell against the back of my chair, admittedly stunned by the caller’s identity. Raymond Thurston Grant was one of our state’s more influential political insider, a behind the scenes kind of guy. It was rumored he even had the ear of the President. Harrison, the spoiled chubby little cherub, was his only son. How could he be missing? My take on Harrison was that he wouldn’t stray far from the family home, or more specifically, the family money. Plus, mommy’s apron strings only stretched so far. Still….</i></span></span></blockquote><blockquote class="tr_bq" style="line-height: normal; text-indent: 0in;"><span lang="EN-US"><span style="font-family: "georgia" , "times new roman" , serif;"><i>“How long has he been missing?”</i></span></span></blockquote><blockquote class="tr_bq" style="line-height: normal; text-indent: 0in;"><span lang="EN-US"><span style="font-family: "georgia" , "times new roman" , serif;"><i>“I don’t have time for small talk, Miss Stewart. Be at our home in forty-five minutes.”</i></span></span></blockquote><blockquote class="tr_bq" style="line-height: normal; text-indent: 0in;"><span lang="EN-US"><span style="font-family: "georgia" , "times new roman" , serif;"><i>Before I could reply, Mr. Grant hung up.</i></span></span></blockquote><br /><div class="MsoNormal" style="line-height: normal;"><o:p> There you go. I hope you enjoyed this vignette.</o:p></div><div class="separator" style="clear: both; text-align: center;"><a href="https://3.bp.blogspot.com/-NgPbIaeLv0c/W4XehbXT76I/AAAAAAAADu8/cPLObZ1Q7mEJ2hrrgp8fuk-UZ5oqbzkzgCPcBGAYYCw/s1600/Always%2BWith%2BLove.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="100" data-original-width="265" src="https://3.bp.blogspot.com/-NgPbIaeLv0c/W4XehbXT76I/AAAAAAAADu8/cPLObZ1Q7mEJ2hrrgp8fuk-UZ5oqbzkzgCPcBGAYYCw/s1600/Always%2BWith%2BLove.png" /></a></div><div class="MsoNormal" style="line-height: normal;"><o:p><br /></o:p></div><br />Sue Barrhttp://www.blogger.com/profile/06347249118943488564noreply@blogger.com0tag:blogger.com,1999:blog-8317810105915057272.post-70789912944881644692019-02-15T09:12:00.000-05:002019-02-15T09:12:55.520-05:00Happy Valentine's DayHubby and I don't "celebrate" Valentine's Day. It's just not what we do. I think it's because we say 'I love you' every day. The only exception being when he's away on a trip, and even then he usually FaceTime's me, or sends a text.<br /><br />With hubby having served in the military and being away (anywhere from three weeks to months at at time, AND losing many, many friends during that time) we make sure to always say I love you. You truly do not know from one minute to the next if you will still walk this earth.<br /><br />The biggest tragedy we faced was in 1985 when two Hercules crashed mid-air. We lost ten friends in a split second. The part that sent shivers running up and down my spine was the fact hubby was invited to join them. It was a three plane formation (rare for Herc's), but he'd just returned from a search and rescue mission and was bagged. All he wanted to do was go home and catch some zzzzz's. Other friends and family were not so lucky.<br /><br />This is why every day is Valentine's day in our house and why I urge you to always hug someone you love. You just don't know...<br /><br /><div class="separator" style="clear: both; text-align: center;"><a href="https://2.bp.blogspot.com/-DIupFlFVFEk/XGbHZU7jfII/AAAAAAAAD9c/-JymmZ1RF4Q79z0nIT53pAifdI0M2WkcQCLcBGAs/s1600/Valentine.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="800" data-original-width="800" height="640" src="https://2.bp.blogspot.com/-DIupFlFVFEk/XGbHZU7jfII/AAAAAAAAD9c/-JymmZ1RF4Q79z0nIT53pAifdI0M2WkcQCLcBGAs/s640/Valentine.png" width="640" /></a></div><br /><div class="separator" style="clear: both; text-align: center;"><a href="https://3.bp.blogspot.com/-NgPbIaeLv0c/W4XehbXT76I/AAAAAAAADu8/cPLObZ1Q7mEJ2hrrgp8fuk-UZ5oqbzkzgCPcBGAYYCw/s1600/Always%2BWith%2BLove.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="100" data-original-width="265" src="https://3.bp.blogspot.com/-NgPbIaeLv0c/W4XehbXT76I/AAAAAAAADu8/cPLObZ1Q7mEJ2hrrgp8fuk-UZ5oqbzkzgCPcBGAYYCw/s1600/Always%2BWith%2BLove.png" /></a></div>P.S.: Hug someone you love today....Sue Barrhttp://www.blogger.com/profile/06347249118943488564noreply@blogger.com0tag:blogger.com,1999:blog-8317810105915057272.post-32864563267530376432019-02-09T05:00:00.000-05:002019-05-05T12:47:30.645-04:00WEEKEND WRITING WARRIORS #69<div class="separator" style="clear: both; text-align: center;"><a href="https://2.bp.blogspot.com/-vJ4jS8JPBrY/WqVAJzFXvjI/AAAAAAAADaA/q0BqnY24_ZEme-A-zSEgoUBqZGd0WvFVgCPcBGAYYCw/s1600/wewriwa2.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="210" data-original-width="676" height="196" src="https://2.bp.blogspot.com/-vJ4jS8JPBrY/WqVAJzFXvjI/AAAAAAAADaA/q0BqnY24_ZEme-A-zSEgoUBqZGd0WvFVgCPcBGAYYCw/s640/wewriwa2.png" width="640" /></a></div>What a week! I won an award for my book CAROLINE (<a href="https://suebarrauthor.blogspot.com/2019/02/i-won-award.html" target="_blank">Read More</a>), and am still waiting to hear back from the doctor. Normally I wouldn't care because I retired early from my job, but we're trying to plan a family vacation and Murphy Law states that the minute you pick a date and are non-refundable... then they'll call with a day sometime in the middle of your fabulous vacay.<br /><br />At the time of this writing we are in the midst of a freezing rain warning. Blech. Top Gun gets home from Santiago, Chili tonight where the weather is 37C, which for you non-Celsius users is 98.6F. This translates in any formula as HOT, HOT, HOT.<br /><br />Go figure. I've got a skating rink down my driveway and he had to stay inside so he wouldn't burn. You know where I'd rather be.<br /><table cellpadding="0" cellspacing="0" class="tr-caption-container" style="margin-left: auto; margin-right: auto; text-align: center;"><tbody><tr><td style="text-align: center;"><a href="https://4.bp.blogspot.com/-fZJVvQz9Xlc/XFraaIZEf4I/AAAAAAAAD80/_u0dUkJp9dYv98vkdE1peJnFUMFG79ZEACLcBGAs/s1600/freezing%2Brain1.jpg" imageanchor="1" style="clear: left; margin-bottom: 1em; margin-left: auto; margin-right: auto;"><img border="0" data-original-height="240" data-original-width="341" height="448" src="https://4.bp.blogspot.com/-fZJVvQz9Xlc/XFraaIZEf4I/AAAAAAAAD80/_u0dUkJp9dYv98vkdE1peJnFUMFG79ZEACLcBGAs/s640/freezing%2Brain1.jpg" width="640" /></a></td></tr><tr><td class="tr-caption" style="text-align: left;"><br /></td></tr></tbody></table>Onto this weeks snippet for Weekend Writing Warriors, which is what we are all here for. I'm sharing from Georgiana. (As God as my witness, I WILL finish her this year - I'm channeling Scarlet O'Hara here, in case you wondered) The Duke of Adborough is with Darcy and his cousin the Colonel, about to tell them how Sir Reginald Slade may attempt to court Darcy's sister Georgiana solely for her substantial dowry.<br /><div class="separator" style="clear: both; text-align: center;"></div><div><blockquote class="tr_bq" style="mso-pagination: none;"><span style="font-family: "georgia" , "times new roman" , serif;"><i>“That brings me to the purpose of my visit.” He set down his drink and braced his hands on his thighs and said, “George heard rumors of Sir Reginald Slade making inquiries about Miss Darcy.”</i></span></blockquote><blockquote class="tr_bq" style="mso-pagination: none;"><span style="font-family: "georgia" , "times new roman" , serif;"><i>“What type of inquiries?”</i></span></blockquote><div class="MsoNormal" style="mso-pagination: none;"><span style="font-family: "georgia" , "times new roman" , serif;"><i><o:p></o:p></i></span></div><blockquote class="tr_bq" style="mso-pagination: none;"><span style="font-family: "georgia" , "times new roman" , serif;"><i>“Of her dowry and value of any inheritance.”</i></span></blockquote><blockquote class="tr_bq" style="mso-pagination: none;"><span style="font-family: "georgia" , "times new roman" , serif;"><i>“He must know she gets none of Pemberley, not with Elizabeth giving Darcy two fine sons,” the Colonel blurted out before he stood and began to pace.</i></span></blockquote><blockquote class="tr_bq" style="mso-pagination: none;"><span style="font-family: "georgia" , "times new roman" , serif;"><i>“This by itself is not earth shattering, Richard. Adborough must have good reason to bring this to our attention.”</i></span></blockquote><div class="MsoNormal" style="mso-pagination: none;"><span style="font-family: "georgia" , "times new roman" , serif;"><i><o:p></o:p></i></span></div><blockquote class="tr_bq" style="mso-pagination: none;"><span style="font-family: "georgia" , "times new roman" , serif;"><i>“George told me Slade is in over his head, with his estate mortgaged to the rafters and all his holdings are nothing but vowels in the hands of those he has wagered against and lost. You’d do well to direct your sister to avoid him at all costs.”</i></span></blockquote>To test drive other authors and read posts from their Works In Progress, please click <a href="http://www.wewriwa.com/" target="_blank">HERE </a><br /><blockquote class="tr_bq" style="mso-pagination: none;"><span style="font-family: "georgia" , "times new roman" , serif;"><i></i></span><br /><div class="separator" style="clear: both; text-align: center;"><span style="font-family: "georgia" , "times new roman" , serif;"><i><a href="https://3.bp.blogspot.com/-NgPbIaeLv0c/W4XehbXT76I/AAAAAAAADu8/cPLObZ1Q7mEJ2hrrgp8fuk-UZ5oqbzkzgCPcBGAYYCw/s1600/Always%2BWith%2BLove.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="100" data-original-width="265" src="https://3.bp.blogspot.com/-NgPbIaeLv0c/W4XehbXT76I/AAAAAAAADu8/cPLObZ1Q7mEJ2hrrgp8fuk-UZ5oqbzkzgCPcBGAYYCw/s1600/Always%2BWith%2BLove.png" /></a></i></span></div><span style="font-family: "georgia" , "times new roman" , serif;"><i></i></span></blockquote><div class="MsoNormal" style="mso-pagination: none;"><o:p></o:p></div></div>Sue Barrhttp://www.blogger.com/profile/06347249118943488564noreply@blogger.com12tag:blogger.com,1999:blog-8317810105915057272.post-73147955990423868222019-02-06T09:12:00.000-05:002019-02-06T09:12:18.583-05:00Do We Deserve Grace?First, let's define what 'grace' means.<br /><br />Grace means to get something you do not deserve. Hand in hand with grace walks mercy. Mercy means you do not get the punishment you deserve. God is abundant in grace and mercy and we can read all about it in the Bible. I'm going to share one verse and also a video of Matt Redman singing his song, "Your Grace Finds Me".<br /><blockquote class="tr_bq"><span style="font-family: Georgia, Times New Roman, serif;"><i>Therefore, since we have been made righteous by faith, we have peace with God through Lord Y'Shua Messiah also through Whom we have had access by faith into this <b>grace</b> in which we stand and boast in hope of the glory of God.</i></span></blockquote><div style="text-align: right;">Romans 5: 1-2<br /><br /></div><div style="text-align: center;"><iframe allow="accelerometer; autoplay; encrypted-media; gyroscope; picture-in-picture" allowfullscreen="" frameborder="0" height="315" src="https://www.youtube.com/embed/Mqd8MoiCbcI" width="560"></iframe><br /><br /><div class="separator" style="clear: both; text-align: center;"><a href="https://3.bp.blogspot.com/-NgPbIaeLv0c/W4XehbXT76I/AAAAAAAADu8/cPLObZ1Q7mEJ2hrrgp8fuk-UZ5oqbzkzgCPcBGAYYCw/s1600/Always%2BWith%2BLove.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="100" data-original-width="265" src="https://3.bp.blogspot.com/-NgPbIaeLv0c/W4XehbXT76I/AAAAAAAADu8/cPLObZ1Q7mEJ2hrrgp8fuk-UZ5oqbzkzgCPcBGAYYCw/s1600/Always%2BWith%2BLove.png" /></a></div><br /></div>Sue Barrhttp://www.blogger.com/profile/06347249118943488564noreply@blogger.com0tag:blogger.com,1999:blog-8317810105915057272.post-17444059867051876842019-02-04T11:36:00.000-05:002019-05-05T12:05:01.038-04:00I Won An Award!I received the loveliest surprise this morning in my e-mail. I won the Jane Austen Reader's Bi-Monthly award for my book <span style="font-family: "georgia" , "times new roman" , serif;"><i>CAROLINE</i></span>.<br /><br />I was quite worried about my entry because most Janeites absolutely HATE Caroline Bingley and have a hard time reconciling their minds around her having a change of character and heart. Thank you to Charlotte Bream and her team for choosing my book.<br /><br />Here is the review, link to the blog site added on the bottom if you wish to verify and/or check out this blog for other books and recommendations.<br /><br /><div style="text-align: center;"><span style="font-family: "georgia" , "times new roman" , serif; font-size: large;"><b>Editorial Review</b></span></div><blockquote class="tr_bq" style="text-align: center;"><i style="font-family: Georgia, "Times New Roman", serif;">A poignant tale of Caroline Bingley's struggle to overcome Mr. Darcy's rebuttal of her affections, and his eventual wedlock to Miss Elizabeth Bennet unfolds. With a hard lesson learned for the young lady with a scissor sharp tongue, Caroline is forced to consider her future. A bleak picture emerges, and while she must at all costs be present at Darcy's wedding, she did not foresee Pemberley as the prime venue for the wedding breakfast. With stoicism Caroline strives to impress on the Misses Elizabeth and Jane Bennet, she no longer bears hard feelings toward either of them. All the while she remains embittered in belief her life is in tatters and her heart shredded by Darcy. Quite oblivious to another gentleman who views her as Darcy never has, Caroline is soon faced with a dilemma. Totally unaware the thrill of the chase excites her admirer she finds him exhilarating company. He is what he is at heart while somewhat reformed in character. To her chagrin he becomes her salvation and mentor in may respects. Soon a Caroline she never new existed emerges from the dark shadow of all she had despised about herself. Hence envy, heartache, gratitude and love, is blended with the skill of excellent prose and solid pot. And within the pictorial splendour of the Pemberley Estate is where a happy ever after for Caroline is realized. Therefore, the Jane Austen Award is hereby bestowed to Sue Barr for Caroline: a P&P Continuation.</i></blockquote><div class="separator" style="clear: both; text-align: center;"><a href="https://1.bp.blogspot.com/-4TEewqRMveM/XFho-Ic_utI/AAAAAAAAD8o/d0C-MRx0J0A0z_IK0NzpM6BV8m2fT2CLwCLcBGAs/s1600/JaneAustenAward.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="342" data-original-width="277" height="320" src="https://1.bp.blogspot.com/-4TEewqRMveM/XFho-Ic_utI/AAAAAAAAD8o/d0C-MRx0J0A0z_IK0NzpM6BV8m2fT2CLwCLcBGAs/s320/JaneAustenAward.png" width="259" /></a></div><div style="text-align: center;"><a href="https://janeaustenreadersaward.blogspot.com/2019/02/latest-jane-austen-awardregency-award.html" target="_blank">Jane Austen Reader's Award Blog</a></div><div style="text-align: center;"><br /></div><div class="separator" style="clear: both; text-align: center;"><a href="https://3.bp.blogspot.com/-NgPbIaeLv0c/W4XehbXT76I/AAAAAAAADu8/cPLObZ1Q7mEJ2hrrgp8fuk-UZ5oqbzkzgCPcBGAYYCw/s1600/Always%2BWith%2BLove.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="100" data-original-width="265" src="https://3.bp.blogspot.com/-NgPbIaeLv0c/W4XehbXT76I/AAAAAAAADu8/cPLObZ1Q7mEJ2hrrgp8fuk-UZ5oqbzkzgCPcBGAYYCw/s1600/Always%2BWith%2BLove.png" /></a></div><div style="text-align: center;"><br /></div>Sue Barrhttp://www.blogger.com/profile/06347249118943488564noreply@blogger.com0tag:blogger.com,1999:blog-8317810105915057272.post-83803379729643192482019-02-02T05:00:00.000-05:002019-05-05T12:47:03.043-04:00WEEKEND WRITING WARRIORS #68<div class="separator" style="clear: both; text-align: center;"><a href="https://2.bp.blogspot.com/-vJ4jS8JPBrY/WqVAJzFXvjI/AAAAAAAADaA/q0BqnY24_ZEme-A-zSEgoUBqZGd0WvFVgCPcBGAYYCw/s1600/wewriwa2.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="210" data-original-width="676" height="196" src="https://2.bp.blogspot.com/-vJ4jS8JPBrY/WqVAJzFXvjI/AAAAAAAADaA/q0BqnY24_ZEme-A-zSEgoUBqZGd0WvFVgCPcBGAYYCw/s640/wewriwa2.png" width="640" /></a></div><br />Hello fellow Warriors. I'm back. My health is a bit dodgy, but I'm still kickin' and causing dear hubby tons of grief. Surgery looms in my future, date yet to be determined. Other than that, this past week saw me refresh my memory banks about where I was in my two stories (Georgiana and Mary). This led to the inevitable editing I do whenever I read my prose. I've never figured out how to shut the internal editor off, and I'm coming to the conclusion that a HUGE portion of Georgiana is going to drift onto the cutting floor. I can't seem to get past the last scene and that means - Cut the little darlings. *Ouch*<br /><br />As I struggled with Georgiana, this lovely scene popped into my head and it doesn't suit sweet Miss Darcy. This snippet, from my future <span style="font-family: "georgia" , "times new roman" , serif;"><i><b>ONLY A ROGUE WILL DO</b></i></span>, is here for your perusal.<br /><br />It's so fresh, I haven't transferred it from the notebook I keep by my bed.<br /><blockquote class="tr_bq"><span style="font-family: "georgia" , "times new roman" , serif;"><i>He plucked the dance card from her fingers and before she could protest, opened the tiny booklet and began to read the names she'd scribbled down before entering the ball room.</i></span></blockquote><blockquote class="tr_bq"><span style="font-family: "georgia" , "times new roman" , serif;"><i>"Hmmm... Baron Bootstrap... Lord Runamuk..." he murmured, his lips tilting into a grin. "I see Sir Trapdoor didn't make his dance, poor soul, do you believe he... fell through something?"</i></span></blockquote><blockquote class="tr_bq"><span style="font-family: "georgia" , "times new roman" , serif;"><i>"Sometimes I think I absolutely hate you." </i></span><i style="font-family: georgia, "times new roman", serif;">She snatched the card out of his hand and tucked it into her reticule. Before she could turn and escape his all knowing gaze and self-important gloating, he touched her arm and stayed her movement.</i></blockquote><blockquote class="tr_bq"><span style="font-family: "georgia" , "times new roman" , serif;"><i>"Forgive me, I have no right to make fun of how you deal with this. May I please see the card again."</i></span></blockquote><blockquote class="tr_bq"><span style="font-family: "georgia" , "times new roman" , serif;"><i>She almost shook off his hand, but something in his eyes - compassion, perhaps? - had her reach into her bag and hand him the card. He promptly borrowed a pen from Miss ______, standing nearby, and wrote his name next to the supper set.</i></span></blockquote>I like these two already.<br /><br />To view the work of other authors participating in this weekly writing exercise, please click <a href="http://www.wewriwa.com/" target="_blank">HERE</a>. And, before you go to bed tonight, hug someone you love.<br /><div class="separator" style="clear: both; text-align: center;"><a href="https://3.bp.blogspot.com/-NgPbIaeLv0c/W4XehbXT76I/AAAAAAAADu8/cPLObZ1Q7mEJ2hrrgp8fuk-UZ5oqbzkzgCPcBGAYYCw/s1600/Always%2BWith%2BLove.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="100" data-original-width="265" src="https://3.bp.blogspot.com/-NgPbIaeLv0c/W4XehbXT76I/AAAAAAAADu8/cPLObZ1Q7mEJ2hrrgp8fuk-UZ5oqbzkzgCPcBGAYYCw/s1600/Always%2BWith%2BLove.png" /></a></div><br /><div class="separator" style="clear: both; text-align: center;"></div><br />Sue Barrhttp://www.blogger.com/profile/06347249118943488564noreply@blogger.com22
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"birth": {
"place": {
"name": "B\u00fcren an der Aare, Schweiz",
"placeName": "B\u00fcren an der Aare",
"placeType": "inhabited_place"
},
"time": {
"startYear": 1941
}
},
"birthYear": 1941,
"date": "born 1941",
"fc": "Markus Raetz",
"gender": "Male",
"id": 1812,
"mda": "Raetz, Markus",
"movements": [],
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"url": "http://www.tate.org.uk/art/artists/markus-raetz-1812"
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A newly published study in Cell Reports suggests people can improve their skin's innate UV protection simply by eating in sync with their circadian rhythms, i.e., at "normal" times. While further study is needed, habits such as late night snacking could diminish humans' vulnerability to sunburn, skin aging and even skin cancer.
In an animal model, researchers found that mice, which are nocturnal but were fed during the day, experienced disruptions in their skin's natural circadian rhythms, which are the inverse of humans. This negatively impacted the ability of an innate enzyme, xeroderma pigmentosum group A (XPA), to protect them against UV radiation.
To perform the test, different feeding times were set for various animal groups, and liver processes and genetic expressions were then measured. Researchers identified shifts in circadian skin patterns distinct from normal patterns in the liver, as well as alterations in the expression of several diurnally expressed genes in the skin, including that of the key DNA repair factor XPA.
Furthermore, changes to sensitivities to UVB exposure were observed. Mice that ate mainly or only at night formed more cyclobutane pyrimidine dimers (CPDs) when exposed to UVB during the night than during the day. In contrast, mice fed during the day exhibited a reverse pattern. Together, these results indicate the timing of the food intake had a pronounced influence on skin biology—representing a modifiable regulator of skin health.
Joseph S. Takahashi, Ph.D., chairman of Neuroscience at the University of Texas (UT) Southwestern Medical Center and co-author on the study, commented on these observations.
“This finding is surprising. I did not think the skin was paying attention to when we are eating,” said Takahashi. He added, "It is likely that if you have a normal eating schedule, then you will be better protected from UV during the daytime."
Score another point for holistically focused beauty and personal care. It's the future.
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SERVICES
Foundation Background
TETAF was created after the Texas Legislature passed laws prohibiting the state from providing services offered by other entities, even national entities. In the case of the trauma designation process, the state turned to the American College of Surgeons to provide verification surveys for all levels of trauma centers. In 2006, concerned with the high cost of ACS survey services, the Texas trauma community – with strong support from the 22 trauma regional advisory councils (RACs) – launched the Texas EMS, Trauma and Acute Care Foundation, a 501(c)(3) nonprofit organization. TETAF provides operational support and helps fill the gap when federal, state and local governments and private entities cannot provide services on a statewide basis. TETAF was granted authority by the Texas Department of State Health Services in 2008 to provide verification surveys for Level III and Level IV designated trauma facilities, and in 2009 the first TETAF staff was hired to begin fully implementing TETAF’s mission.
TETAF has recently released a new document providing information and answering questions about the TETAF survey process to help hospitals and Trauma Program Managers. Click here to access this helpful FAQ document.
Partnering with the Texas Department of State Health Services, TETAF staff provides technical assistance resources in program development as well as coordination of survey services to ensure a successful Stroke Support Facility (Level III) designation process.
Partnering with the Texas Department of State Health Services, TETAF staff provides technical assistance resources in trauma program development as well as coordination of survey services to ensure a successful designation process.
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MIAMI, February 26 – The Miami HEAT announced today that they have signed forward Michael Beasley to a 10-day contract. Per club policy, terms of the deal were not disclosed.
Beasley most recently played in the Chinese Basketball Association, signing a one-year deal with the Shanghai Sharks on Oct. 9, 2014. He averaged 27.1 points, 10.4 rebounds, 5.2 assists and 1.92 steals in 37 games while shooting 51.3 percent from the floor. His time with the Sharks was highlighted by capturing MVP honors at the 2015 CBA All-Star Game where he came off the bench to score a CBA All-Star Game record 59 points.
The 6’9 1/2”, 235-pound forward begins his third stint with the HEAT. He was originally drafted by Miami with the second pick of the 2008 NBA Draft and played two seasons with the HEAT before being traded to Minnesota on July 12, 2010. He re-joined the HEAT last season, signing as a free agent on Sept. 11, 2013. Beasley appeared in 55 games (two starts) with the HEAT last season and averaged 7.9 points, 3.1 rebounds and 15.1 minutes while shooting 49.9 percent from the floor, 38.9 percent from three-point range and 77.2 percent from the line. In his three seasons as a member of the HEAT he has appeared in 214 games (99 starts) and averaged 12.7 points, 5.2 rebounds, 1.1 assists and 24.1 minutes while shooting 46.6 percent from the floor, 34.6 percent from three-point range and 78.34 percent from the foul line.
In addition to his time with the HEAT, Beasley has also played with the Minnesota Timberwolves and Phoenix Suns during his six-year NBA career. He has appeared in 409 NBA regular season games (199 starts) and has averaged 13.2 points, 4.9 rebounds, 1.3 assists and 24.9 minutes while shooting 45 percent from the floor, 34.8 percent from three-point range and 75.8 percent from the foul line. In 12 postseason games (five starts), all with Miami, he has averaged 11.4 points, 6.7 rebounds and 26.1 minutes while shooting 40.9 percent from the field, 33.3 percent from three-point range and 76.9 percent from the foul line.
Beasley will wear number 30.
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On Tuesday night, Axios reporter Alexi McCammond tweeted out that former NBA star Charles Barkley told her that “I don’t hit women but if I did I would hit you.”
McCammond said Barkley’s comments came after she asked about clarification for which candidate he was supporting in the 2020 Democratic primary.
She says Barkley spoke glowingly about Deval Patrick, but then a member of Pete Buttigieg’s campaign approached the group. When Barkley then said he was a fan of Buttigieg, McCammond pointed out that he had just voiced his support for Patrick. He then made the comment to her.
McCammond also said the comments had been made off the record, an agreement she would normally respect were it not for the nature of those comments.
UPDATE: Barkley has apologized for the comment.
As many online pundits pointed out, Barkley has a troubling history of comments made about violence toward women. Most were couched as jokes, but it’s still a worrying thing that he hasn’t seemed to learn a lesson about this.
McCammond concluded by saying she didn’t like being a part of the story.
“It’s not about me or my feelings,” she tweeted. “But it’s about refusing to allow this culture to perpetuate because of silence on these issues. It’s easier and less awkward to be silent, but that helps NO ONE but the perpetrator.”
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Q:
Reduce 0-1 knapsack prob. to a SAT prob
Is there any way to reduce the 0-1 knapsack problem to a SAT problem in Conjunctive Norm Form?
A:
You could always work out the digital circuits necessary to implement adders and comparators and then turn the result of that into conjunctive normal form. You can get circuits into CNF form without expanding them exponentially by making up intermediate variables which represent the outputs of small sections of circuit.
Each node of a circuit amounts to a=f(b, c) where a is the output, b and c the input, and f is some simple function like & or |. You can create a CNF function that is true only when a really is the result of f(b, c) and it can't be too unwieldy, because it is a function on only three variables.
You can rewrite any circuit into a large number of terms of the form a=f(b, c) and all you have to do with the CNF versions of these is to AND them all together. Assuming you want to solve for the output being true, you then just stick on the output variable as the final component of that big AND.
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Reaction kinetics and transformation products of 1-naphthol by Mn oxide-mediated oxidative-coupling reaction.
In this study, the transformation of 1-naphthol via oxidative-coupling reaction was investigated using Mn oxides. 1-Naphthol was transformed completely by birnessite, which is one of the natural Mn oxides present in soil. The surface area-normalized specific rate constant, k(surf), for 1-naphthol was determined to be 9.66 x 10(-4)L/m(2)min using observed pseudo-first-order rate constants with respect to birnessite loading. The transformation of 1-naphthol was dependent on the solution pH, and the pseudo-first-order rate constants increased from 0.028 at pH 11 to 0.075 at pH 2 at a birnessite loading of 0.625 g/L. GC and LC mass spectroscopic analysis of the supernatants were performed after separating the reaction solution into hydrophobic and hydrophilic fractions by solvent extraction. The major transformation products were found to be 1,4-naphthoquinone(1,4-NPQ) and naphthol polymerized products with a molecular weight (m/z) ranging from 400 to 2000. Transformation of 1,4-NPQ, to the polymerized products by an additional birnessite loading was also verified. The DOC concentrations of the supernatants before and after the reaction were analyzed and the rate of oligomeric precipitate formation was measured.
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Regular
I play a lot of first person shooters but never really got into the Battlefied series. With that being said, I picked this one up to play with friends and I really enjoy it. For some reason I feel like I'm terrible at it compared to other shooters but hopefully I'll get better with some practice. Also, I prefer the smaller gametypes as conquest just feels like a cluster sometimes. My favorite game mode I've played so far is Rush (I think that's what it was called). It's still a larger lobby but I feel like it's less getting shot in the back all the time.
Also, there was an update yesterday that fixed/corrected the class levels. They were mistakenly giving out Level 10's and their weapons. My buddy earned that Martini-Henry sniper and had it taken away. Kind of a bummer.
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The production of ice as previously practiced requires apparatus involving a variety of moving parts and devices for establishing a solid product, to harvest, and its transport to storage. Mechanical refrigeration is usually involved, by which gas is compressed, liquified and then expanded; in which the principles of evaporation are utilized to absorb heat from water and thereby transforming it into ice. The water to be frozen must be supplied in measured quantity and contained within vessels from which it is extracted as ice; a time schedule must be established for the heat absorption period, extraction period, and the vessel refill period, and all of which requires time and temperature control of the water supply and coordinated operation of the refrigeration equipment. In practice, unforeseeable variables enter into ice production in the form of atmospheric conditions, inlet water temperatures, and variations in plant efficiencies; all of which are unpredictable and require complex sensing and control means. Therefore, it is a general object of this invention to provide for minimal complexity in the process and/or operation of apparatus producing ice, and a system therefor which is inherently operable continuously at optimum efficiencies dependent upon prevailing conditions.
The present invention employs mechanical refrigeration in its basic form and combined therewith the fewest number of elements, and excepting the refrigeration means per se this invention virtually eliminates moving parts and relies upon the thermo control of refrigerant flow and product water supply. Accordingly, it is an object of this invention to provide an apparatus that operates in reliance upon the principles of freezing, thawing and ejection of the product. With the present invention, the product cycle involves applied heat absorption to a stall filled with water and applied until iced, followed by applied heating of the stall filled with ice until the ice-to-stall interfaces are thawed and accompanied by application of water under pressure to refill said stall by floating the ice as a product therefrom, and the product cycle repeated by reinstating the applied heat absorption in place of said heating. Floatation water is recovered and reapplied as refill water to this "freeze-thaw-ejection cycle".
Quantity production of ice, regardless of size, on an economical basis is an object of this invention, and to this end the aforementioned freeze-thaw ejection cycle is progressively applied to a series of stalls for the efficient and continuous production of ice. With the preferred form of the present invention, the product cycle involves progressively applied heat absorption through a series of stalls filled with water and applied until all stalls are iced, followed by progressively applied heating through said series of stalls filled with ice until the ice-to-stall interfaces are thawed and accompanied by application of water pressure to refill said stalls by progressively floating the ice as a product therefrom, and the product cycle repeated by reinstating the progressively applied heat absorption in place of said progressive heating. The progressive application of the aforementioned freeze-thaw-ejection cycle involves the heat conductive capability of a series of ice stalls cooperatively related to heat absorption and heating applied sequentially thereto.
It is an object of this invention to provide apparatus for carrying out the method referred to, and to this end there is a stall filled with water from a pressure supply thereof and means alternately absorbing heat from and applying heat thereto, the water supply and said means being responsive to a stall temperature sensor and a control therefrom responsive to thawing and operating said means to absorb heat until the stall is iced and responsive to freezing and operating said pressure supply to apply water and simultaneously operating said means to apply heat until the stall is thawed.
It is also an object of this invention to provide apparatus for continuously carrying out the method referred to, and to this end there is a series of stalls filled with water from a common pressure supply thereof and means alternately absorbing heat progressively from and progressively applying heat thereto, the water supply and said means being responsive to a selectively positioned stall temperature sensor and a control therefrom responsive to thawing and operating said means to absorb heat until the stalls are iced and responsive to freezing and operating said pressure supply to apply water and simultaneously operating said means to apply heat until the stalls are thawed.
It is still another object of this invention to provide an ice stall and water filling manifold therefor combined so as to discharge product ice by means of floatation. With the present invention the ice stall is displaced from vertical and the product ice discharged therefrom by hydraulic floatation. In addition to floatation in the strict sense, the hydraulic action involves displacement whereby the ice is ejected as a piston.
It is a further object of this invention to provide apparatus of the character thus far referred to that is self-sufficient and entirely automatic, with means to conserve floatation water, and to conserve operating energy when a full harvest is obtained. The harvest bin is a cold-sink for storage, while the floatation water is maintained at a constant level for accurate pressure application by means of a pump. Control is entirely by thermal response, there being a single thermostatic switch to automatically control refrigerant flow and supply water to the stalls. The harvest switch operates independently in response to the temperature of ice reaching the brim of the storage bin.
|
{
"pile_set_name": "USPTO Backgrounds"
}
|
Undo the accidental dismissal of notification cards.
Get more information about watch battery performance including usage by apps and features.
Info about storage space including how much memory is available on the watch.
Though the design of your Moto 360 may be timeless, the software for the watch doesn’t have to be. Instead, it continues to get better and better. Last month we gave you more choice for your Moto 360 including new bands, watch faces, and a new feature called Moto Body . Today, we’re announcing a new update to Android Wear that brings you even more ways to customize your Moto 360 while improving the performance.With this upgrade you can now choose from a variety of beautiful, new third-party watch faces through the Google Play Store, created by designers like Rebecca Minkoff and Xogram. You can even choose a fun watch face from the makers of one of the most popular Android games Plants vs. Zombies. This is in addition to the existing ability to design your own watch face through Moto Connect. Once you receive the update, there will be a link to the watch faces collection in the Android Wear companion app The update also introduces new settings. Swiping down from the top of the screen brings new features like Theatre mode, which turns the display off even when receiving notifications, and Sunlight mode, which temporarily maximizes brightness. You can also manage interruptions and access deeper settings quickly. And if you have a phone running Android 5.0, Lollipop , the choices you make on your Moto 360 automatically change the Interruptions setting on your phone.Here are some of the other highlights you can expect after installing the update:For more information check out the release notes . Or for more on Motorola updates and repairs, visit motorola.com/mymoto360 We are enhancing this most recent update to Android Wear with a new, optimized build that squashes a few bugs. Some users reported that some watch faces appear off center on their Moto 360. This affects users who are using different language settings on their watch and phone. It is now fixed. We are rolling out this update to users in phases, and all users should have it by the end of the week.
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{
"pile_set_name": "OpenWebText2"
}
|
See related research by Switzer et al., <http://breast-cancer-research.com/content/14/5/R125>
Unlike the constitutive nitric oxide (NO) synthase isoforms, the inducible isoform of NO synthase - nitric oxide synthase 2 (NOS2) - is capable of producing sustained intracellular levels of NO, and it is increasingly appreciated that protein S-nitrosylation, the covalent modification of cysteine thiol by NO, is important for NOS2-dependent signal transduction.
In the previous issue of *Breast Cancer Research*, Switzer and coworkers delineate a novel pathway for S-nitrosylation in regulation of estrogen receptor (ER)-negative breast cancer invasion \[[@B1]\]. This work adds to the growing appreciation that S-nitrosylation can regulate myriad pathways important for tumorigenesis, including gene transcription, apoptosis and DNA repair \[[@B2]-[@B4]\], and it complements another recent study by these coworkers showing that S-nitrosylation and activation of epidermal growth factor receptor is associated with the induction of epithelial-to-mesenchymal transition as well as chemoresistance in ER-negative breast cancer \[[@B5]\]. More generally, these data begin to explain why expression of NOS2 correlates with aggressive tumor phenotypes and poor clinical outcomes in a variety of malignancies, including breast cancer, lung cancer, colon cancer and prostate cancer \[[@B6]-[@B9]\].
The small GTPase Ras was one of the earliest described regulatory targets of S-nitrosylation. Modification of a single cysteine residue that is located in the nucleotide-binding region of wild-type Ras and is conserved among all Ras isoforms (Cys118 in human H-Ras) stimulates guanine nucleotide exchange and downstream pathways, including activation of mitogen-activated protein kinase signaling \[[@B10]\]. S-nitrosylation of wild-type Ras by endothelial NO synthase has been shown to promote pancreatic tumor growth \[[@B11]\]. Switzer and colleagues identify a role for S-nitrosylation of wild-type Ras in ER-negative breast cancer \[[@B1]\]. They find that the only elements common to genes upregulated in high NOS2-expressing breast cancer are binding sites for the Ets-1 transcription factor. Using a model ER-negative breast cancer cell line, they further show that NO induces S-nitrosylation of wild-type Ras, leading to phosphorylation and activation of Ets-1 through the Ras/MEK/ERK pathway. Knockdown of Ets-1 inhibits the NO-dependent expression of basal-like breast cancer markers and attenuates NO-dependent cancer cell invasion. These findings delineate a mechanism by which NOS2 promotes an aggressive tumor phenotype in ER-negative breast cancer.
While NOS2 expression is high in ER-negative tumors \[[@B9]\], it is not constitutively expressed in the triple-negative breast cancer cell line MDA-MB-468 (lacking ER, progesterone receptor and epidermal growth factor receptor 2) that was utilized in this and related studies \[[@B1],[@B5]\]. Indeed, the loss of NOS2 expression in cultured tumor cells *ex vivo*is not uncommon and could reflect a deficiency in cytokine signaling and/or an inability to replicate the hypoxic tumor environment. Regardless of mechanism, this deficit in NOS2 expression can make challenging the study of the (patho)physiological functions of the enzyme in cell culture systems. To model NOS2-based signaling as it occurs in the solid tumor, Switzer and colleagues employ a long-lived NO donor as well as adenoviral expression of NOS2 \[[@B1]\]; in earlier work, they also co-cultured these cells with macrophages stimulated to produce high levels of NOS2-derived NO \[[@B5]\]. While these conditions may adequately recapitulate pathophysiological NO levels within the tumor, further work is needed to determine the extent to which these manipulations fully reconstitute NOS2-dependent signal transduction. These questions may in part be addressed by defining the precise mechanisms underlying enhanced expression of NOS2 in breast cancer and other malignancies. In addition to the environmental factors of cytokine secretion and hypoxia, aberrant regulation of NOS2 transcription, mRNA stability, and proteasomal degradation in the tumor cells present potential mechanisms.
There is increasing recognition that denitrosylases - in particular, S-nitrosoglutathione reductase (GSNOR) and thioredoxin (Trx) - are critical modulators of S-nitrosothiol (SNO) homeostasis \[[@B12]\]. As the name implies, GSNOR metabolizes S-nitrosoglutathione, a small molecule intermediate in S-nitrosylation and denitrosylation. This enzyme may have general housekeeping functions, one of which is to protect against indiscriminate protein S-nitrosylation (nitrosative stress) that may occur as a result of excessive NOS2 activity. In this regard, GSNOR deficiency has been tied to inhibition of DNA repair and to hepatocellular carcinoma \[[@B3]\], and the enzyme is also deficient in some lung cancers \[[@B13]\]. Trx reduces protein SNOs directly, and inhibition of thioredoxin reductase (TrxR) can increase protein SNOs by blocking Trx turnover. Trx has another major role as a modulator of oxidative stress, and it is in this capacity that chemotherapeutics targeting TrxR are believed to affect cancer cell killing. Given the potential for protein S-nitrosylation to stimulate tumor growth and invasion, TrxR inhibitors may have unexpected deleterious effects. More generally, the balance of SNO formation/degradation is clearly an important factor in malignant cell transformation and thus warrants further investigation.
Given the potential role of NOS2 in tumor chemoresistance and aggressiveness, it is important to note that NOS2 inhibitors have had some success in inhibiting tumor growth \[[@B14],[@B15]\]. At the same time, interventions that increase SNO levels, either through application of NO donors and low-mass SNOs or through inhibition of SNO-metabolizing enzymes, are being proposed as therapeutics in numerous diseases \[[@B16]\]. Given recent findings in the cancer field, it may be necessary to weigh the risk-benefit of therapies designed to increase protein S-nitrosylation for those individuals at high risk of developing cancer.
Abbreviations
=============
ER: estrogen receptor; GSNO: S-nitrosoglutathione reductase; NO: nitric oxide; NOS2: nitric oxide synthase 2; SNO: S-nitrosothiol; Trx: thioredoxin; TrxR: thioredoxin reductase.
Competing interests
===================
The authors declare that they have no competing interests.
Acknowledgements
================
This work was supported in part by National Institutes of Health grants HL106121 (to MWF and HEM) and HL092994 (to HEM).
|
{
"pile_set_name": "PubMed Central"
}
|
From the historic Wensleydale Creamery in England, this latest creation quickly made waves at the New York Fancy Food Show last summer. Creamy, buttery cheddar is perfectly blended
with tangy, caramelized onions to produce a new English favorite. As versatile as it is addictive, this new cheddar, also known as Abbot's Gold, has livened up everything from simple sandwiches to hot dogs and
grilled portabellas in the igourmet kitchen. A must try for fans of Cotswold, Harlech and Tintern.
A rich, creamy cheese...can’t eat too much at once, but every bite is delicious!
By Jannahfrom Olympia WAon Feb 04, 2018
Not what I expected
Well, it wasn't what we expected. Not cheddar-y really. Interesting consistency, was on the fence about it. Flavor was good but the carmelized onions altered the taste of the cheese. I like it better right out of the fridge than after sitting out for a bit. It made an excellent cheese sauce for mushroom ravioli!
By Krisfrom Floridaon Dec 28, 2017
WOW !!!
unbelievable. This is my first review of any product and let me tell you this. JUST BUY THIS CHEESE !! you will not be disappointed.
By Sandifrom PAon Dec 25, 2017
great cheese
love this cheese. in fact this was my Christmas gift to my sister and my two friends at work who keep asking me when I will place another order. If you haven't tried this I would highly recommend it
By allsashcafrom St. Petersburg, FLon Nov 04, 2017
One of my favorites
I hide this one from my adult children. I could, and do, eat this one at breakfast, lunch and dinner. I love it plain as well as mixed in with scrambled eggs, on a warm sandwich, or crumbled on a salad. Yum! Love it with a cold glass of cider as well!
By Mountaingirlfrom COon Oct 09, 2017
Sublime
The ultimate comfort food for cheese lovers.
By Michelle T witterfrom NCon Sep 05, 2017
Tastious cheese ever
Had a party and it was a hit.
By Michelle T Witterfrom NCon Sep 05, 2017
Delicious
The best cheese in the world!!!
By RGFfrom Southern Illinoison Aug 31, 2017
A winner !
I've served this twice now at literary salons that I host. Extremely popular cheese.
By Elizabeth from Auburn, WAon Aug 27, 2017
Love this
This cheese is wonderful. It is a rich and creamy, sort of mellow cheddar that is spiked with carmelized onions. The onions are sweet and they come in st the end of the bite, so first you get the rich creamy cheddar then the sweet onion taste. This cheese is great by itself, on crackers, even on burgers. My favorite way to eat this is on a sweet onion cracker with a small piece of sweetheart ham and a spread of German sweet or spicy mustard. It also goes well with fruit.
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{
"pile_set_name": "Pile-CC"
}
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Honeybadger Overkill
Tyrann Mathieu has been dubbed the Honeybadger of LSU and as you may know The Honeybadger "Takes What He Wants" and "Don't Give a S&*T". Take a listen from last nights Alabama stomping of LSU and let us know if you think Brent Musburger overused "HONEYBADGER".
We've also added the original (NSFW)Honeybadger video in case you have been living under a rock for the past year.
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{
"pile_set_name": "Pile-CC"
}
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Interaction of amiodarone and desethylamiodarone with solubilized nuclear thyroid hormone receptors.
The mechanisms of action of the potent antiarrhythmic drug amiodarone are unknown. However, amiodarone and its abundant metabolite, desethylamiodarone, bear a striking structural resemblance to thyroid hormones. In addition, certain cardiac electrophysiologic effects of amiodarone treatment are similar to those of hypothyroidism. These facts suggest that amiodarone or desethylamiodarone could be acting, in part, by blocking thyroid hormone action. Because thyroid hormones are known to act through nuclear receptor proteins, the binding of amiodarone and desethylamiodarone was measured to nuclear extracts derived from human lymphocytes, bovine atrium and ventricle and rat liver. The capacity of increasing concentrations of amiodarone and desethylamiodarone nuclear extracts to block receptor binding of radiolabeled triiodothyronine (T3) in a standard in vitro competition assay was tested. Nuclear extracts demonstrated only minimal binding to amiodarone. However, all receptor preparations had substantial affinities (KD) for the desethyl analog: lymphocyte, 8.6 microM; atrium, 35.0 microM; ventricle, 26.9 microM and liver, 8.6 microM. Desethylamiodarone accumulates in very large quantities in parenchymatous organs during long-term amiodarone treatment. Taking its usual therapeutic serum level (about 4 microM or 2.7 micrograms/ml) as an estimate of intranuclear concentration, desethylamiodarone would partially saturate nuclear thyroid hormone receptors in several different tissues, including the heart. Thus, amiodarone treatment may exert some of its electrophysiologic effects by metabolic conversion to desethylamiodarone. This metabolite may then exclude thyroid hormone from nuclear receptor sites within the myocardium.
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{
"pile_set_name": "PubMed Abstracts"
}
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Thaduka
Thaduka is a genus of butterflies in the family Lycaenidae, the blues. It is monotypic, containing only the species Thaduka multicaudata, the many-tailed oak-blue, which is found in India, Burma and Indochina.
Description
References
Category:Arhopalini
Category:Monotypic butterfly genera
Category:Taxa named by Frederic Moore
Category:Lycaenidae genera
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{
"pile_set_name": "Wikipedia (en)"
}
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/*
* AAC defines
*
* This file is part of FFmpeg.
*
* FFmpeg is free software; you can redistribute it and/or
* modify it under the terms of the GNU Lesser General Public
* License as published by the Free Software Foundation; either
* version 2.1 of the License, or (at your option) any later version.
*
* FFmpeg is distributed in the hope that it will be useful,
* but WITHOUT ANY WARRANTY; without even the implied warranty of
* MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE. See the GNU
* Lesser General Public License for more details.
*
* You should have received a copy of the GNU Lesser General Public
* License along with FFmpeg; if not, write to the Free Software
* Foundation, Inc., 51 Franklin Street, Fifth Floor, Boston, MA 02110-1301 USA
*/
#ifndef AVCODEC_AAC_DEFINES_H
#define AVCODEC_AAC_DEFINES_H
#ifndef USE_FIXED
#define USE_FIXED 0
#endif
#if USE_FIXED
#include "libavutil/softfloat.h"
#define FFT_FLOAT 0
#define FFT_FIXED_32 1
#define AAC_RENAME(x) x ## _fixed
#define AAC_RENAME_32(x) x ## _fixed_32
typedef int INTFLOAT;
typedef unsigned UINTFLOAT; ///< Equivalent to INTFLOAT, Used as temporal cast to avoid undefined sign overflow operations.
typedef int64_t INT64FLOAT;
typedef int16_t SHORTFLOAT;
typedef SoftFloat AAC_FLOAT;
typedef int AAC_SIGNE;
#define FIXR(a) ((int)((a) * 1 + 0.5))
#define FIXR10(a) ((int)((a) * 1024.0 + 0.5))
#define Q23(a) (int)((a) * 8388608.0 + 0.5)
#define Q30(x) (int)((x)*1073741824.0 + 0.5)
#define Q31(x) (int)((x)*2147483648.0 + 0.5)
#define RANGE15(x) x
#define GET_GAIN(x, y) (-(y) * (1 << (x))) + 1024
#define AAC_MUL16(x, y) (int)(((int64_t)(x) * (y) + 0x8000) >> 16)
#define AAC_MUL26(x, y) (int)(((int64_t)(x) * (y) + 0x2000000) >> 26)
#define AAC_MUL30(x, y) (int)(((int64_t)(x) * (y) + 0x20000000) >> 30)
#define AAC_MUL31(x, y) (int)(((int64_t)(x) * (y) + 0x40000000) >> 31)
#define AAC_MADD28(x, y, a, b) (int)((((int64_t)(x) * (y)) + \
((int64_t)(a) * (b)) + \
0x8000000) >> 28)
#define AAC_MADD30(x, y, a, b) (int)((((int64_t)(x) * (y)) + \
((int64_t)(a) * (b)) + \
0x20000000) >> 30)
#define AAC_MADD30_V8(x, y, a, b, c, d, e, f) (int)((((int64_t)(x) * (y)) + \
((int64_t)(a) * (b)) + \
((int64_t)(c) * (d)) + \
((int64_t)(e) * (f)) + \
0x20000000) >> 30)
#define AAC_MSUB30(x, y, a, b) (int)((((int64_t)(x) * (y)) - \
((int64_t)(a) * (b)) + \
0x20000000) >> 30)
#define AAC_MSUB30_V8(x, y, a, b, c, d, e, f) (int)((((int64_t)(x) * (y)) + \
((int64_t)(a) * (b)) - \
((int64_t)(c) * (d)) - \
((int64_t)(e) * (f)) + \
0x20000000) >> 30)
#define AAC_MSUB31_V3(x, y, z) (int)((((int64_t)(x) * (z)) - \
((int64_t)(y) * (z)) + \
0x40000000) >> 31)
#define AAC_HALF_SUM(x, y) (((x) >> 1) + ((y) >> 1))
#define AAC_SRA_R(x, y) (int)(((x) + (1 << ((y) - 1))) >> (y))
#else
#define FFT_FLOAT 1
#define FFT_FIXED_32 0
#define AAC_RENAME(x) x
#define AAC_RENAME_32(x) x
typedef float INTFLOAT;
typedef float UINTFLOAT;
typedef float INT64FLOAT;
typedef float SHORTFLOAT;
typedef float AAC_FLOAT;
typedef unsigned AAC_SIGNE;
#define FIXR(x) ((float)(x))
#define FIXR10(x) ((float)(x))
#define Q23(x) ((float)(x))
#define Q30(x) ((float)(x))
#define Q31(x) ((float)(x))
#define RANGE15(x) (32768.0 * (x))
#define GET_GAIN(x, y) powf((x), -(y))
#define AAC_MUL16(x, y) ((x) * (y))
#define AAC_MUL26(x, y) ((x) * (y))
#define AAC_MUL30(x, y) ((x) * (y))
#define AAC_MUL31(x, y) ((x) * (y))
#define AAC_MADD28(x, y, a, b) ((x) * (y) + (a) * (b))
#define AAC_MADD30(x, y, a, b) ((x) * (y) + (a) * (b))
#define AAC_MADD30_V8(x, y, a, b, c, d, e, f) ((x) * (y) + (a) * (b) + \
(c) * (d) + (e) * (f))
#define AAC_MSUB30(x, y, a, b) ((x) * (y) - (a) * (b))
#define AAC_MSUB30_V8(x, y, a, b, c, d, e, f) ((x) * (y) + (a) * (b) - \
(c) * (d) - (e) * (f))
#define AAC_MSUB31_V3(x, y, z) ((x) - (y)) * (z)
#define AAC_HALF_SUM(x, y) ((x) + (y)) * 0.5f
#define AAC_SRA_R(x, y) (x)
#endif /* USE_FIXED */
#endif /* AVCODEC_AAC_DEFINES_H */
|
{
"pile_set_name": "Github"
}
|
Reducing misinformation effects in older adults with cognitive interview mnemonics.
We examined the effect of a prior Modified Cognitive Interview on young and older adults' recall of a short film of a staged crime and subsequent reporting of misinformation. Participants viewed the film followed the next day by misinformation presented in a postevent summary. They were then interviewed with either a Modified Cognitive Interview or a control interview followed by a recognition memory test. A Modified Cognitive Interview elicited more correct details and improved overall accuracy compared to a control interview in both age groups, although the young adults recollected three times more correct information in a Modified Cognitive Interview than the older adults. In both age groups, correct recollections of person and action details were higher in a Modified Cognitive Interview than a control interview. Importantly, older adults who were interviewed with a Modified Cognitive Interview were not susceptible to misinformation effects.
|
{
"pile_set_name": "PubMed Abstracts"
}
|
5" Obsidian Crystal Skull 1131
An excellent grounding stone. Produces very blunt answers to one's vision. A protective stone. Stabilise energies. Shields against negativity. A good healing stone. Ancients used it for scrying and mirror magic
|
{
"pile_set_name": "Pile-CC"
}
|
Episode #1 Episode #2 Episode #3 Have you ever wanted to test your Hearthstone knowledge? Do you think you know everything about the cards, interactions and maybe even flavor texts? Then welcome to the first episode of our own Hearthstone Quiz! Each episode is going to feature 9 questions divided into 3 categories – Easy, Medium […]
List of Episodes
Introduction
Have you ever wanted to test your Hearthstone knowledge? Do you think you know everything about the cards, interactions and maybe even flavor texts? Then welcome to the first episode of our own Hearthstone Quiz!
Each episode is going to feature 9 questions divided into 3 categories – Easy, Medium and Hard. It’s not perfect, because question difficulty might be subjective, but we hope that everyone finds some challenges when doing the Quiz.
When it comes to the scoring, you can calculate your own points if you want. It’s not necessary, but if you want to see your score and compare it to other people, here is the deal. Easy questions are worth 2 points, Medium questions 3 points and Hard questions 4 points each! It means that in total, you can score up to 27 points. If the question has more than one part, you need to answer everything correctly to get points.
We’ve also asked some of our pro players to answer the questions, so you can you can compare your answers with theirs. This episode features Hoarth and Kabi.
Let’s start with Hearthstone Quiz #1!
Difficulty: Easy
Question 1. What happens when you use crazed-alchemist crazed-alchemist doomsayer doomsayer
[spoiler title=”Answer to Question 1″]
Hoarth: It dies.
Kabi: Dies unless there is a passive buff aura that would give him an extra life as it gets swapped, like the old blood-imp.
Answer: Since you change the minion’s health to 0, it instantly dies.[/spoiler]
Question 2. When you Silence the ancient-of-war ancient-of-war druid-of-the-claw druid-of-the-claw
[spoiler]
Hoarth: It becomes a 4/6, non taunted.
Kabi: Stays the same minus Taunt, as it is not a buff, but a card replacement.
Answer: When it comes to the Ancient of War, the +5 Attack or Health is considered a buff, thus it goes away when Silenced. With Druid of the Claw, both forms are different minions, not buffs. The Bear Form is a 4/6 minion with Taunt, so it stays a 4/6 and only the Taunt gets removed.[/spoiler]
Question 3. If a minion that already attacked this turn is given Windfury (for example by windspeaker windspeaker enhance-o-mechano enhance-o-mechano
[spoiler]
Hoarth: Yes.
Kabi: Yes.
Answer: Yes, it can. [/spoiler]
Difficulty: Medium
Question 4. Is there a minion that can attack more than 2 times per turn?
[spoiler title=”Answer to Question 4″]
Hoarth: Yes – V-07-TR-0N.
Kabi: Voltron.
Answer: When it comes to collectible minions, there is none. But if you meet the conditions on mimirons-head, at the start of your next turn you spawn v-07-tr-0n – a minion with Mega-Windfury. He is able to attack 4 times per turn. [/spoiler]
Question 5. Does Spell Power (on Paladin’s side) affect damage dealt by the eye-for-an-eye eye-for-an-eye
[spoiler title=”Answer to Question 5″]
Hoarth: No.
Kabi: Yes, damage Secrets are affected by Spell Power. For example, explosive-trap does 3 damage if you have a bloodmage-thalnos in play.
Answer: Since the Secrets are considered Spells, their damage is affected by the Spell Power. What’s worth noting is that the Spell Power bonuses are counted when the Secret is procced, not when it’s used.[/spoiler]
Question 6. Under what conditions a minion with Stealth gets unstealthed – when it receives damage, deals damage or both?
[spoiler title=”Answer to Question 6″]
Hoarth: Deals damage.
Kabi: Minions do not get unstealthed when they take damage, they get unstealthed when they do damage.
Answer: Minions only lose Stealth once they deal damage to something: by attacking it, by being attacked and hitting it back (e.g. by ogre-brute missing the original target and attacking minion in Stealth) or by their effects. Popping a Divine Shield is also considered as dealing damage. So, for example, if you holy-nova enemy stranglethorn-tiger, it stays in Stealth. But if you Stealth your knife-juggler and play another minion, activating the card’s ability, Knife Juggler gets out of Stealth because he has dealt damage with its effect.[/spoiler]
Difficulty: Hard
Question 7. Both you and your enemy are at 1 health with an empty board. You’re out of cards in the deck and will take fatigue damage with next card draw. You drop knife-juggler knife-juggler novice-engineer novice-engineer
[spoiler]
Hoarth: It is a tie.
Kabi: It’s a draw, the juggle will resolve, but so will the Battlecry.
Answer: Both of the actions are triggered when you play a minion – Novice Engineer. It means that minion completely resolves only after both the juggle (dealing 1 damage to enemy) and card draw (taking fatigue damage). By the time the minion is resolved, you and your enemy are dead, so the game ends in a draw, meaning that no one wins it.[/spoiler]
Question 8. When Warrior attacks a minion with a 1/1 gorehowl gorehowl
[spoiler title=”Answer to Question 8″]
Hoarth: It becomes a 0/1, and the Warrior can attack with it again only if he uses something like captain-greenskin.
Kabi: It becomes a 0/1, can’t attack.
Answer: The weapon becomes 0/1, just as the Gorehowl effect states (attack gets reduced by 1). You can’t attack with characters that have 0 attack (minions, Heroes), so the Warrior is unable to attack with it. Only if the weapon gets replaced or the attack gets buffed, the Warrior can attack again. It still, however, counts for the sake of effects that require you to have an equipped weapon – like southsea-deckhand‘s Charge. [/spoiler]
Question 9. Priest steals enemy Warlock’s minion with shadow-madness shadow-madness power-overwhelming power-overwhelming
[spoiler]
Hoarth: It dies.
Kabi: Dies, it says “at the end of the turn” not “your turn”.
Answer: Blizzard never confirmed whether it’s a bug or intended behavior, but the minion actually doesn’t die. “End of turn” in card’s text means “end of owner’s turn” in this context. Priest used Shadow Madness on a minion, then Power Overwhelming. “End of the turn” effects are resolved in the order which they were used. It means that Shadow Madness is resolved first, returning the minion to Warlock. Then, Power Overwhelming doesn’t kill the minion, because the current owner is Warlock, and it’s not the end of his turn yet. Warlock can still attack with the minion one time and only at the end of his turn the minion dies.
EDIT
As it has been found out recently, the outcome depends on whether you are the dominant player or not. I’ve never heard of this before, even though I’ve been reading about the in-game interactions a lot, thus the confusion around the question. You can find out more by watching this video.[/spoiler]
Scoring
Our pro team did pretty well in this episode. The only question that was tough for them was Question 9. But since the interaction might be bugged, making a mistake is understandable. The final score looks as follows:
Closing
That’s it for the first episode of Hearthstone Quiz. I encourage you to share your scores, questions and ideas in the comment section below 🙂
If you have any ideas for questions we can use in future, hit me up at stonekeephs@gmail.com and if they aren’t on my list already, I’ll be sure to add them in future, giving you a credit for coming up with a question.
|
{
"pile_set_name": "OpenWebText2"
}
|
Scopula usticinctaria
Scopula usticinctaria is a moth of the family Geometridae. It is found on Borneo and Peninsular Malaysia.
Adults have very dark red margins on both wings.
References
Category:Moths described in 1861
usticinctaria
Category:Moths of Asia
|
{
"pile_set_name": "Wikipedia (en)"
}
|
Pages
Tuesday, July 28, 2009
Comic-Con: Women in Manga
On Saturday, I attended this female-oriented panel at San Diego Comic-Con International, moderated by Eva Volin, chair, Great Graphic Novels for Teens. The panel included a wide variety of women from across the manga scene, from publishers and artists to reporters and librarians. While this is a nearly blow-by-blow account, there are some minor omissions, especially since I was taking notes by hand.
The panel included Leyla Aker, an editor with Viz; Lillian Diaz-Przybyl, an editor with Tokyopop; Deb Aoki, manga editor for About.com; Becky Cloonan, artist and creator of Tokyopop's East Coast Rising; Robin Brenner, librarian, and JuYuon Lee, an editor at Yen Press. The room, albeit small, was filled to capacity with young girls, women and men of all ages. Of course, everyone was interested in hearing about what the comprised group of experts thought about the role of women in the manga publishing industry.
I think much of what was said during this panel can be summed up in a quote that Volin used to kick off the panel, "In manga, women sit in the 'teacup' next to the 'kettle' of men."
The role of women in publishing
Volin started by asking Aker about her transition from traditional publishing to her role at Viz, where she oversees the seinen imprint IKKI. For Aker, her biggest frustration was that while women often served in editorial roles, the decision-makers continued to be men, noting that the disparity wasn't nearly as bad as it is in Tokyo publishing houses, where there are still limited opportunities for women. In referencing the disparity in the U.S., Aker said, "You will be frustrated by it if you let it."
Next up was Diaz-Przybyl, who described her experience at Tokyopop. When she joined the staff a few years ago, she was the first Japanese speaker in editorial, but, at the same time, half of the editorial department was comprised of women.
Lee, who started working in publishing in her native South Korea, mentioned that the disparity was less of an issue there because of the smaller size of publishing houses. In her experience, editorial departments are staffed primarily by women. But, she noted that the major difference between the U.S. and Korean markets was the cultural connotations used in the States. While everything is simply known as "comics" in Korea, there are several different terms used here, including manga, manhwa, graphic novels and comics.
Finding comics for women
Volin then asked Cloonan how she discovered manga. Cloonan started reading traditional superhero comics and, growing up, thought that working in comics was an unattainable job. However, she discovered Rumiko Takahashi's Ranma 1/2 in floppy format at her local comic book store and that changed her life, especially since it was written by a woman.
From here the conversation changed with Aoki asking, "Who is marketing well to older teens and women?"
As evidenced by the dearth of good josei titles in the "Best and Worst Manga" panel on Thursday, Aoki pointed out that most of the titles aimed at an older female demographic are, for the most part, trashy, Harlequin-style romances. From what she's seen, older teens are reading seinen, but there's still a growing market.
And while the focus of the conversation was on sales, Brenner noted that for her, a librarian, the issue was less about sales, but reader interest.
"I don't care how much they sell, but how much do they circulate?" Brenner said. "More and more women are coming up to me - they are diverse readers and they're vocal; men pick up books and leave, while women pick up books and talk to you."
While the anecdotal information provided by the panelists showed that women are looking for "smart" manga, the question still remained: is the market growing to accommodate an older, more mature, female audience?
The U.S. market
According to Aker, the U.S. manga market can sustain older readers, but it's still developing. She noted a recent study comparing subscribers of Shonen Jump and the now-defunct Shojo Beat. While a whopping 40 percent of Shonen Jump's subscribers are female, only 5 percent of Shojo Beat's subscribers were male.
"Girls will read shonen, but boys won't read shojo," Aker said. As other panelists noted, this is a trend seen across the publishing industry and isn't confined only to manga.
Once again, the panelists looked at the gender composition of the manga industry, specifically artists. Aker and Diaz-Pryzbyl noted that there were mangaka who specifically chose to remain gender ambiguous with their pen names and not allow their likenesses to be seen. By doing so, the artists insured that their gender wouldn't work to their disadvantage, but it certainly shows the gender discrimination still afoot in male-dominated Japan.
Will women read manga?
The conversation turned to the success of josei titles, with Diaz-Przybyl noting that there was an inherent need for cultural understanding mature titles. Diaz-Przybyl mentioned that in Tramps Like Us the main character's constant reference to her boyfriend as her sempai despite the fact they were in a serious relationship. Basically, there is a barrier of cultural understanding inherent in titles for older readers.
But, perhaps Aker's reference to some marketing research conducted on Viz's behalf truly told the story. Evidently, a focus group showed that "women will pick up manga if 'the pictures are taken out.'" So, it wasn't the content itself, but possibly the presentation of the story in a graphic novel format that turns women off.
Cloonan chimed in with an anecdote of her own, noting that she has bought manga for her mother, a voracious reader. But, her mother has never understood the appeal of manga, nor read any of the books she's been given.
Not 'real' books
The conversation then turned to a well-worn topic -- that comics are not considered "real" books. Both Volin and Brenner mentioned the flak they receive from fellow librarians who don't consider manga literature. There were also mentions of "reading manga in public," which I've done myself as mentioned in my "Goodbye, Shojo Beat" post. By reading in public, the panelists hoped to dispel some of the stereotypes surrounding graphic novels as a genre.
As time was running out, the conversation began to wind down, with Lee mentioning that there simply aren't enough titles for older women readers, as they're looking for content they can empathize and connect with. But, there was hope in the younger girls market eventually maturing into readers who are still determined to enjoy their manga.
Geeking out
In possibly the funniest moment during the entire panel, Aoki mentioned her determination to read manga. "I've been reading manga since I was 8," Aoki said, "I would sit there with my Japanese-English dictionary and write in the margins."
To which Aker replied with a sigh, "Oh, Deb!" which was met by a laugh from the audience. Volin also chimed in with a well-timed narrator's remark, "Deb just raised the bar of geekdom."
At this point, the panel took questions from the audience regarding a variety of topics and I specifically asked what the implications of the Shojo Beat's shut down had on the larger women's manga market.
The tough questions
Aker took the question in stride, noting that when the magazine shut down, her colleagues and she knew that this question would inevitably arise. For the most part, Aker said that the magazine was unsustainable in the current economy, noting the recent demise of other magazines.
She also took the time to once again highlight the gender disparity in readership between Shojo Beat and Shonen Jump. Unfortunately, time had run out by the time Aker finished answering my question, so the room had to empty in order to accommodate the next panel.
My reflections and thoughts
Much of what the panelists said was all too true -- that women simply aren't reading comics in the numbers that men are right now, so, of course, the market is catering to that. But, there's hope in the young girls and women who are dedicated to reading manga.
Perhaps Shojo Beat was simply ahead of its time, but for now, the market is tightening because of the recession. However, I think there is light at the end of the tunnel, especially with some of the licenses discussed and announced at San Diego Comic-Con International.
Specifically, Viz's license of Ooku, by Fumi Yoshinaga of Antique Bakery fame, and Yen Press's announcement of Bunny Drop, both josei titles, give me hope that publishers aren't dismissing the market entirely and are simply in a 'wait and see' mode. If nothing else, this panel was a refreshing, female-dominated look at how women are affecting -- and being affected by -- the U.S. manga industry.
6 comments:
@mbeasi Thanks! I'm glad you enjoyed my (nearly) blow-by-blow account. Overall, I think this panel showed that there is still optimism in the industry and that everyone is just waiting to see what happens once the economy recovers.
"Girls will read shonen, but boys won't read shojo," Aker said. As other panelists noted, this is a trend seen across the publishing industry and isn't confined only to manga.
Ugh yes, it rather annoys me how it's taken for granted that fiction geared towards guys are the default that everyone can and should read, whereas if it was fictional work geared towards women, suddenly the guys can't read it anymore. *headdesks x1000*
the general female population just don't have any experience reading manga/comic in this country. i've tried to get my friend(30 something ) to read manga b/c she has a little girl and suggested some josei for her and shojo for her girl. she doesn't like the manga but likes the k-drama adaptation of the same manga(hana yori dango) Not being a comic book reader , for her, comic is hard to sustain her attention span. she belongs to a generation who prefers watching tv and reading text than reading pictures.. older ppl are set in their ways
but her daughter does like the comic and manga i suggested and took to it very well.
i also grew up partly in taiwan this is completely not the case there. everyone grew up with manga there and a 30 and 40 year old have no problem reading in the format of manga..
unfortunately women(20 something and above) in US have been turned off by superhero comics for 30/40 years and comic is just not a reading format they're used to.
even in tv shows there's less geared toward women and more to men.. whereas in asia tv drama is really dominated by what's favored by women/girls.. (idol drama for example).. it's very culturally diff.
@Softpill I totally agree that it's mostly a culturally dominant attitude that "women simply do not read comics." For me, I grew up reading non-superhero comics (think Disney and Archie), but I also read a wide variety of novels and books. By the time I hit junior and senior high, I was no longer reading comics and was instead drawn to other visual media and literature. Of course, I was brought back to comics with my discovery of manga. It had everything I wanted -- short formats, meaningful stories, interesting characters and entertainment at its best.
Unfortunately, I know few women in RL who appreciate manga and/or comics, but thankfully young girls are interested in it as a viable alternative (or in addition) to other books. I've shared some shojo books with my co-worker's daughter who is a reluctant reader and she's loved them.
And, as a whole, you're right -- American media is still dominated by content created by men for men with a few exceptions. But, I think that will eventually change as more women become vocal and persistent about the types of media they want as part of a larger neo-feminist movement. Of course, I could be wrong...
Review Rankings
Required reading: these are titles that are not only something anyone can enjoy regardless of genre preference, but they're also classics (or titles that will soon enough become classics) that should be on anyone's bookshelf.
Highly recommended: these are titles that are highly recommended, but for various reasons are not for everyone.
If only...: as you might guess, these are books that would be SO much better "if only..." *something* was changed. Not highly recommended, but not necessarily "meh" material just yet.
Meh: I'll use this whenever a title doesn't do it for me, and is something you could read out of boredom or borrow from the library.
Just say no!: this will hopefully not be used often, but is basically the most negative I can get regarding a book without getting nasty!
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---
author:
- 'Steven A. Balbus'
date: 'Received ; accepted '
title: Nonlinear Scale Invariance in Local Disk Flows
---
Introduction
============
Disks with Keplerian rotation profiles are linearly stable by the Rayleigh criterion of outwardly increasing specific angular momentum, but are extremely sensitive to the presence of magnetic fields. A weakly magnetized disk is linearly unstable if its angular velocity decreases outward, a condition met by Keplerian and almost all other astrophysical rotation profiles (Balbus & Hawley 1991). The underlying physics behind this magnetorotational instability (MRI) is well-understood, and the breakdown of the flow into fully developed turbulence has been convincingly demonstrated in a large series of numerical simulations (Balbus 2003 for a review).
Not all astrophysical disks need have the requisite minimum ionization level to sustain magnetic coupling, however. Protostellar disks, for example, may have an extended “dead zone” near the midplane on radial scales from $\sim 0.1$ to $\sim 10$ AU (Gammie 1996; Fromang, Terquem, & Balbus 2001). This, along with other similar cases (e.g. CV disks, cf. Gammie & Menou 1998), has led to speculation that there are also hydrodynamical mechanisms by which Keplerian flow is destabilized (Gammie 1996).
Before the advent of the MRI, such reasoning was orthodox. The pioneering work of Shakura & Sunyaev (1973), for example, invoked nonlinear, high Reynolds number shear instabilities as a likely destabilizing mechanism that would lead to turbulence (see also Crawford & Kraft 1956). Since, for nonaxisymmetric disturbances, there is still no proof either of linear or nonlinear stability, this mechanism continues to attract adherents (Dubrulle 1993, Richard & Zahn 1999, Richard 2003).
Theoretical analysis may have hit an impasse, but the intervening years have in fact seen a stunning rise in the capabilities of numerical simulation. These have shown no indication of local nonlinear rotational instabilities (Hawley, Balbus, & Winters 1999), in Keplerian disks. They do, however, reveal nonlinear shear instabilities when Corilois forces are absent, or when the disk is marginally stable (constant specific angular momentum). Indeed, even linear instability is possible in some Rayleigh-stable disks, provided that global physics is introduced (Goldreich, Goodman, & Narayan 1986; Blaes 1987), a result that has been numerically confirmed (Hawley 1991).
The numerical stability findings have been criticized on the grounds that the effective Reynolds number of the codes is too low, and that this damps the nonlinear instabilities: the latter require yet-to-be resolved spatial scales in order to reveal themselves (Richard & Zahn 1999). In this paper, we show that the local disk equations possess a scale invariance that implies any solution to the governing equations must be present on all scales. In other words, for every small scale velocity flow, there is an exact large scale counterpart with the same long term stability behavior. The absence of instability at large scale therefore implies the absence at small scales as well. Conversely, any true small scale instabilities (those present in a shear layer, for example), must also have large scale counterparts, and therefore instability should be found even at crude numerical resolutions. This is indeed the case. Our findings suggest that if nonlinear hydrodynamical instabilities were present in Keplerian disks, such unstable disturbances must involve dynamics beyond the local approximation, and are not an inevitable nonlinear outcome of differential rotation.
The Local Approximation
=======================
In cylindrical coordinates $(R, \phi, z)$, the fundamental equations of motion for a flow in which viscous effects are negligible are mass conservation \[fun0\] [t]{} + [[ ]{}]{}[[ ]{}]{}([ ]{})= 0, and the dynamical equations, \[fun1\] [v\_Rt]{} + [ ]{}[[ ]{}]{}[[ ]{}]{}v\_R - [v\_\^2R]{} = - [1]{}[PR]{} - [R]{} \[fun2\] [v\_t]{} + [ ]{}[[ ]{}]{}[[ ]{}]{}v\_+ [v\_v\_RR]{} = - [1R]{}[P]{} \[fun3\] [v\_zt]{} + [ ]{}[[ ]{}]{}[[ ]{}]{}v\_z = - [1]{}[Pz]{} -[z]{} Our notation is standard: ${ \mbox{\boldmath{$v$}} }$ is the velocity field, $\rho$ the mass density, $P$ the gas pressure, and $\Phi$ is the Newtonian point mass potential for central mass $M$: = - [GM(R\^2 + z\^2)\^[1/2]{}]{}. $G$ is the gravitational constant.
The local limit consists of the following series of approximations. First, we assume that $R$ is large and $z \ll R$, so that \[phis\] - [GMR]{} (1 - [z\^22R\^2]{}) and , Choose a fiducial value of $R$, say $R_0$. Denote the angular velocity as $\Omega(R)$ (we assume a dependence only upon $R$), and let $\Omega_0 = \Omega(R_0)$. We next erect local Cartesian coordinates x = R - R\_0, y = R\_0(-\_0 t) which corotate with the disk at $R=R_0$. Let \[w\] [ ]{} [ ]{} - R\_0 t [ ]{} be the velocity relative to uniform rotation at $\Omega= \Omega_0$. In the local approximation, the magnitude of ${ \mbox{\boldmath{$w$}} }$ is assumed to be small compared with $R\Omega_0$.
The undisturbed angular velocity is Keplerian, \[kep\] \^2 = [GMR\^3]{} Substitution of equations (\[phis\]-\[kep\]) into equations (\[fun1\]–\[fun2\]) and retaining leading order, yields the so-called [*local*]{} or [*Hill*]{} equations (e.g., Balbus & Hawley 1998): \[hill0\] [t]{} + [[ ]{}]{}[[ ]{}]{}([ ]{})= 0, \[hill1\] ( [t]{} + [ ]{}[[ ]{}]{})[w\_R]{} - 2w\_= - x[d\^2d R]{} - [1]{} [Px]{} \[hill2\] ( [t]{} + [ ]{}[[ ]{}]{})[w\_]{} + 2w\_R = - [1]{} [Py]{} \[hill3\] ( [t]{} + [ ]{}[[ ]{}]{})[w\_z]{} = - z\^2 - [1]{} [Pz]{} The “0” subscript has been dropped in the $2\Omega$ terms in equations (\[hill1\]) and (\[hill2\]), and in the derivative term on the right of equation (\[hill1\]). The time derivative is taken in the corotating frame, viz.: = [ t]{} + \_0 Equations (\[hill0\]–\[hill3\]) are well known, and have been used extensively in both numerical and analytical studies. The fundamental approach dates from nineteenth century treatments of the Earth-moon system (Hill 1878).
Scale symmetry in the Hill Equations
====================================
The local equations of motion incorporate an important symmetry in their structure. Let [ ]{}([ ]{}, t), ([ ]{}, t), P([ ]{}, t), where ${ \mbox{\boldmath{$r$}} }=(x, y, z)$, be an exact solution to the Hill equations (\[hill0\]–\[hill3\]). Then, if $\alpha$ is an arbitrary constant, (1/)[ ]{}([ ]{}, t), ([ ]{}, t), (1/\^2) P([ ]{}, t) is also an exact solution to the same equations. The proof is a simple matter of direct substitution.
An equivalent formulation of the scaling symmetry is [ ]{}([ ]{}/, t)(1/) [ ]{}([ ]{}, t) ([ ]{}/, t) ([ ]{}, t) P ([ ]{}/, t)(1/\^2) P ([ ]{}, t). In this form, with $\epsilon \ll 1$, we see that any solution of the Hill equations that involves very small length scales has a rescaled counterpart solution with exactly the same time dependence. In particular, any solution corresponding to a breakdown into turbulence must be present on both large and small scales.
The implications of this scaling symmetry are of particular importance for understanding and testing the possible existence of local nonlinear instabilities in Keplerian disks. The key point is that any such instability would have to exist not just at small scales, but at all scales. Finite difference numerical codes would find such instabilities, if they existed. Indeed, a constant specific angular momentum profile is nonlinearly unstable, and is found to be so even at resolutions as crude as $32^3$. By way of contrast, local Keplerian profiles show no evidence of nonlinear instability at resolutions up to $256^3$, instead converge to the same stable solution in codes with completely different numerical diffusion properties (Hawley, Balbus, & Winters 1999). The argument that small scale flow structure is somehow being repressed is simply untenable.
To see how the Reynolds number changes with scale, assume that a flow is characterized by an effective kinematic viscosity $\nu$. The scaling argument we have just given applies to inviscid equations, so we should not expect it to hold in the presence of viscosity. The Reynolds number associated with the small scale solution is Re\_[s]{} = [w l ]{} The Reynolds number associated with the large scale solution is Re\_[l]{} = [wl]{} where $w$ here means $w(l, t)$, the value of the velocity function evaluated at a fiducial value length $l$ and time $t$. $Re_l =Re_s/\epsilon^2 \ll Re_{s}$ because at larger scales both the velocity and the length scales increase by a factor of $1/\epsilon$. In a numerical simulation, strict scaling invariance is not obeyed. Instead, the large scale solutions approach the inviscid limit, while their sufficiently small scale counterparts are damped. But by behaving nearly inviscidly, the large scale solutions capture the behavior of the Hill system at all scales.
What this Result Does Not Show
==============================
Obviously, scale invariance does not constitute a proof of nonlinear stability in any Keplerian flow. There are several points we have not covered.
First, the local approximation ignores boundary conditions. In laboratory flows, the fluid is always bounded by hard walls, and boundary layers form. A recent laboratory confirmation of the MRI also finds finite amplitude velocity fluctuations in a magnetically stable flow, for example. But the source of such disturbances are boundary layers (Sisan et al. 2004).
The Hill equations emerge in the limit $R\rightarrow\infty$, and therefore curvature terms drop out of the analysis. Instabilities that depend, for example, upon inflection points or vorticity maxima in the background rotation profile would not appear in this limit. Nothing precludes them from forming in the $w$ velocity profile, however, and if such instabilities were present they should manifest on large scales as well as small. In any case, the criticism of the numerical simulations is that extremely small structure is being lost, and that high Reynolds number differential rotation is supposedly intrinsically unstable. It is very difficult to see how large scale curvature could play an essential destabilizing role here. In these equations, the curvature terms are nonsingular perturbations. Planar Couette and Poiseuille flows break down into turbulence without assistance from geometrical curvature.
Our Hill analysis together with numerical simulations would also suggest that a non-Keplerian disk with, say, $\Omega \propto R^{-1.8}$ is nonlinearly stable. But an annulus supporting such a profile is in fact [*linearly*]{} unstable (Goldreich, Goodman, & Narayan 1986), transporting angular momentum outward even in its linear phase. The point is that the annulus supports edge modes that become unstable, and these global modes do not exist in the local approximation. The existence of a similar instabilities in disks found in nature cannot be ruled out, though to date none afflicting Keplerian disks have been found.
The disk thermal structure could also be unstable, at least in principle. Nothing presented in this work bears on these types of instabilities.
Finally, there are technical loopholes to the argument presented in this paper. What if the unstable solution required not just some small scales to be resolved, but very disparate scales? Why this should be so is far from clear, but this possibility cannot be ruled out [*a priori.*]{} Indeed, one could imagine that a fractal structure is required down to infinitesimal scales. Rescaling would not bring such a solution to larger characteristic length scales, by definition. This solution is obviously not characterized by a critical Reynolds number, above which it is necessary to be seen. The critical Reynolds number would be infinity! This is not the argument made by proponents of nonlinear high Reynolds number instability. Such a solution may remain a mathematical possibility, but not one that can be realized in nature.
Conclusion
==========
The local dynamics of Keplerian or other astrophysical disk profiles can be captured by a an established formalism known as the local, or Hill, approximation. The resulting system of equations has an exact scale invariance, so that any flow characterized by very small scales has an exact large scale counterpart with same stability properties. This feature of the Hill equations implies that finite difference codes at available resolutions are sufficient to explore the possibility of [*local*]{} nonlinear shear instabilities in astrophysical disks. If simulations accurately describe the large scale behavior of the Hill system, there is nothing more to uncover at small scales; it is simply renormalized large scale behavior. The absence of any observed instabilities in Keplerian numerical studies, coupled with the ready manifestation of such instabilities in local shear layers and constant specific angular momentum systems, suggests that any putative nonmagnetic disk instability would have to incorporate physics beyond simple differential rotation.
I thank C. Gammie, J. Hawley, K. Menou, and C. Terquem for useful comments. This work is supported by NASA grants NAG5-13288 and NNG04GK77G.
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