The Disney Junior star will appear in dining, meet and greets and shows at Walt Disney World.
Disney's Hollywood Studios celebrates the premiere of Disney Channel's new multiplatform brand, Disney Junior, with a lively new show. Fans of "Playhouse Disney-Live on Stage!" will thrill to this new production which brings back three audience favorites (Mickey Mouse Clubhouse, Handy Manny and Little Einsteins) and also introduces elements of "Jake and the Never Land Pirates" based on Disney Junior's newest series airing in February 2011. Jake, the main character, leads his young pirate pals (including pirate parrot Skully) in fun adventures. Stories from the show will come to life in the parks with colorful puppet versions of some of the most popular Disney Junior characters, bright visuals, whimsical scenery and interactive moments.
The new show opens on March 4 2011 and replaced "Playhouse Disney - Live on Stage!".


Legal & General has unveiled its first Build to Rent (BTR) scheme in Scotland to help meet the increasing demand for affordable, quality rental accommodation across the UK.
It will provide funds for Drum Property Group to deliver 324 apartments within Glasgow's new Buchanan Wharf waterside regeneration area, which will also house the new Northern European hub of Barclays Bank.
Consisting of two 18 storey towers, current design iterations include dining space, a gym, a residents' lounge and a games room.
Buchanan Wharf, which overlooks the River Clyde, will also offer a 4,250 sq ft communal roof terrace and is walking distance from Glasgow Central Station and the International Financial Services District, linked by the Tradeston Bridge.
Legal & General has identified Glasgow as a key target for regeneration. Over the last 10 years, the city has seen particularly low levels of residential development, compared with other major cities.
Just 517 new homes were completed in 2017, falling substantially short of the 2,500 new homes needed per annum, according to Glasgow City Council.
Dan Batterton, Head of Build To Rent at LGIM Real Assets, said: "Glasgow has seen very low levels of residential development over the last decade. This acute supply and demand imbalance has made it an important target for us, demonstrating our commitment to quality asset selection, with sites cherry-picked by assessing needs-based demand to provide long-term, stabilised cashflows for investors.
Scotland, and Glasgow in particular, has been a key target for Legal & General as part of it Future Cities initiative, according to "Paul Miller, MD principal investing at Legal & General Capital.


text:
                 Converse Skater Backpack

                 Sold: 12

                 $48   $60 tag price

                     Alcott Poplin Short Sleeve Check Shirt

                     Sold: 18

                     $40   $50 tag price

                     Intelligent Nutrients

                     Sold: 23

                     $40   $50 tag price

redpajama_set_name: RedPajamaGithub


Looking for the cheapest car insurance in Big Prairie, Ohio? We can help you compare the best deals. Simply enter your Zip Code in the form above and you will be on your way to finding cheap Big Prairie, OH auto insurance quotes from the best providers in Ohio.
One of the most important types of insurance that you can have if you are an adult is car insurance. It is legally required for all adult drivers. The type of insurance that you get can differ on each vehicle that you own. It depends upon whether or not you are making payments, or if you own the vehicle outright. Older vehicles will only need some form of liability insurance, whereas brand-new vehicles will need liability, collision, and comprehensive coverage. It can be very expensive if you own multiple vehicles, and if your rates have recently increased, it is definitely time to look for better premiums and coverage from a different provider. Here are a few of the best cheaper vehicle insurance coverage online tips that you can use as you are searching the web for the best rates available in Holmes county.
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This guide is useful when you need to replace the wifi card due to it not working properly. In this guide, you will remove the wifi cover to access and then replace the wifi card.
Note: When installing new WiFi card be sure to connect the black wire to the right where there is a small letter "B" imprinted and place white on the opposite side.
Flip Laptop into an upside-down position.
Push latches (indicated in red) outward.
Lift battery by notch (indicated in green) to remove.
Remove the plastic cover closest to the battery by first unscrewing the single 7mm screw (in red) with the phillips #00 screwdriver.
Now, lift the cover by the notch under the screw (in red) using a plastic opening tool.
Remove the black and white wire by carefully pulling them off with the plastic lever (you will hear a light snapping sound).
Unscrew the two 6 mm screws (indicated in red) located at the bottom of the WiFi card and store them in a safe place.
Gently remove the wifi card from the laptop.


text:
Home · Celebrity
Aretha Franklin's Profound Impact On Gospel Music: 'She Gave Us Hope Of Reaching The World'
Ron Galella, Ltd./WireImage
By Paula Rogo	· Updated October 23, 2020
On the second night of the live recording of Aretha Franklin's 1972 gospel double album Amazing Grace in a Los Angeles Church, her father, Rev. C. L. Franklin, made an appearance. Likely one of the most famous pastors in Black America at the time, the reverend took a moment to tell a story about his own famous daughter. He told the congregation about bumping into a woman at a Detroit dry cleaners. The woman had seen his daughter Aretha on television the night before and told the reverend the performance was decent. "But I will be glad when she comes back to the church," the woman said. Franklin had the perfect response that ultimately captures the breadth of Aretha's career in music: "Listen, baby. Let me tell you something. If you want to know the truth, she has never left the church. All you have to do is have something in here, and the ability to hear, and the ability to feel, and you will know that Aretha is still a gospel singer."
It is a fitting story and reminder of Aretha's deep and perennial connection to the church and gospel music. Indeed, for the legend who passed away in her Detroit home on Thursday, it was in the church of her father, Detroit's New Bethel Baptist Church, where she found her voice singing alongside her sisters Erma and Carolyn. It was the church that gave the Queen of Soul her soul. "I didn't cross the line" from gospel to pop, Aretha told the late Gwen Ifill in 2015. "Gospel goes with me wherever I go. Gospel is a constant with me." It is no wonder then that Amazing Grace— backed by the incomparable Southern California Community Choir — was also Aretha's best-selling album, and one of the most commercially successful gospel records of all time. And as Marc Lamont Hill put it, Amazing Grace became the "soundtrack to so many Sundays in Black America."
Aretha Franklin's "Amazing Grace" is still the greatest gospel album ever made. It was the soundtrack to so many Sundays in Black America.
— Marc Lamont Hill (@marclamonthill) August 16, 2018
Gospel artists like Fred Hammond remember the album, and Aretha's voice, as part of the very fabric of their upbringing. "My mom played this album Sunday's b4 Church/Sunday school," Hammond said in an Instagram post following the singer's death. "I was just a kid and I loved it musically. But now I know why a mother raising 3 boys in #Detroit played it so much. The #queensinfluence"
My mom played this album Sunday's b4 Church/sunday school. I was just a kid and I loved it musically. But now I know why a mother raising 3 boys in #Detroit played it so much. The #queensinfluence
A post shared by Fred Hammond (@realfredh) on Aug 16, 2018 at 8:52am PDT
While Aretha was looking to the late great Queen of Gospel Mahalia Jackson as her musical influence (and also being mentored by Clara Ward of the famous Ward gospel singers of Philadelphia), gospel artists like Hammond, Yolanda Adams and Erica Campbell had their eyes on Franklin.
"Aretha Franklin has had a profound impact on the Gospel music genre," Erica Campbell told ESSENCE in a statement. "I loved her. She gave us hope of reaching the world with our voices the way she did." Campbell recounts receiving a call from Franklin to produce for her — asking Campbell, her husband Warryn, and sister Tina to create something similar to Mary Mary's 2008 hit, "God in Me." Though the song never happened, Campbell remembers how big of a compliment it was to even be considered. "She was supportive and reached out personally when she liked your music," she said. "Many of us [in the gospel community] had relationships with her."
LONG LIVE THE QUEEN🙏🏾
A post shared by Kirk Franklin (@kirkfranklin) on Aug 16, 2018 at 1:21pm PDT
All Hail The QUEEN!!! Prayers and comfort to the Franklin Family as we say Good Bye to a TRUE LEGEND. My heart is overwhelmed because Mrs Aretha had such a great influence on My Musical Life. I grew up listening to her almost every day. Business Woman, Philanthropist, Mother..... her Achievements go on and on. I'm so glad I got a chance to tell her how much I love her. 💝💝💝
A post shared by Yolanda Adams (@yolandaadams) on Aug 16, 2018 at 7:54am PDT
https://www.instagram.com/marvinsapp/?hl=en
A post shared by Tasha Cobbs Leonard (@tashacobbsleonard) on Aug 16, 2018 at 7:35am PDT
Campbell added: "Our hearts are broken but she left an amazing roadmap to sing songs for the world (not just the church) with everything in your soul, to be authentic and to handle your business. She had the most magnificent voice and there will never be another." And although we may have lost one of the greatest and most soulful voices of our lifetime — likely ever— the Rev. Jesse Jackson perfectly explained to CNN how Aretha's voice would live on. "The Earth lost a lot of music today," he said, "and heaven must be excited because it has a new lead singer for the gospel choir." Rest in power, Aretha.
TOPICS: Aretha Franklin Clara Ward Erica Campbell Fred Hammond gospel music Kirk Franklin Mahalia Jackson Marvin Sapp Mary Mary Tasha Cobbs Tina Campbell Warryn Campbell Yolanda Adams

redpajama_set_name: RedPajamaCommonCrawl


Sabarimala: The Lord Ayyappa temple here re-opened on Tuesday for five-day monthly pujas under a heavy security blanket as hundreds of police personnel were deployed in and around the hill shrine in view of possible protest from right-wing activists, opposing entry of young women.
Despite security arrangements, a large number of devotees, including those from neighbouring states, turned up to offer prayers when the shrine was opened at 5 pm. Chief priest Vasudevan Nampoothiri opened the sanctum sanctorum in the presence of 'tantri' (head priest) Kandararu Rajeevaru.
There would be no specific pujas on Tuesday and the shrine would be closed at 10 pm with the rendition of 'harivarasanam', the customary lullaby for Lord Ayyappa, temple authorities said.
The temple would remain open for five days till February 17 for the monthly pujas during the Malayalam month of Kumbam, temple authorities said. Several special rituals including 'kalabhabhishekam', 'sahasrakalasam' and 'laksharchana' would be performed during the five days, they said.
There has been a sense of anxiety in and around Sabarimala in view of the reopening as the shrine had witnessed frenzied protests during the recently concluded annual pilgrimage season over the entry of women of menstruating age.
A facebook group named Navothana Kerala Sabarimalayilekku, had announced to send women to Sabarimala once the shrine opens for Kumbha monthly pujas. In reply, Sangh Parivar organisations had said they would stop women if they tried to enter the shrine.
Though no prohibitory orders have been clamped down in and around the shrine, the entire area from the base camp to 'Sannidhanam'(temple complex) is under tight police security. The police had sought to enforce prohibitory orders in Sabarimala but the district administration has decided to take action according as the situation develops.
The BJP and right-wing outfits had spearheaded the agitations against implementation of the September 28 Supreme Court verdict permitting women of all age groups into the shrine.
The areas in and around the temple had witnessed widespread violence after the CPI(M)-led LDF government decided to implement the apex court order. Marking the culmination of the over two-month-long stormy annual pilgrimage season, the shrine was closed on January 20.
Traditionally, women of menstruating age or in the age group of 10 to 50 years were not allowed to enter the temple as the deity, Lord Ayyappa, is believed to be "naishtika brahmachari" or perennial celibate.


text:
§ 1602. Vermont Deaf, Hard of Hearing, and Deafblind Advisory Council
(a) Creation; purpose. There is created a Vermont Deaf, Hard of Hearing, and DeafBlind Advisory Council to promote diversity, equality, awareness, and access among individuals who are Deaf, Hard of Hearing, or DeafBlind.
(b) Membership. The Advisory Council shall consist of the following members:
(1) 16 members of the public, appointed by the Governor in a manner that ensures geographically diverse membership, including:
(A) nine or fewer members who are Deaf, Hard of Hearing, or DeafBlind, provided each population is represented and that if a member represents an organization for persons who are Deaf, Hard of Hearing, or DeafBlind, no other member on the Advisory Council shall also represent that organization;
(B) two members who are each a parent or guardian of a child who is Deaf, Hard of Hearing, or DeafBlind;
(C) two members who serve persons who are Deaf, Hard of Hearing, or DeafBlind in a professional capacity, provided that these members do not represent the same organization;
(D) a professional deaf-education specialist who understands all communication and language modes;
(E) a professional interpreter; and
(F) an audiologist or hard-of-hearing education specialist;
(2) the Senior Counselor for the Deaf and Hard of Hearing in the Department's Division of Vocational Rehabilitation or designee;
(3) the Secretary of Education or designee;
(4) the Secretary of Human Services or designee;
(5) the director of the Department for Children and Families' Children's Integrated Services or designee;
(6) the director of the Vermont Early Detection and Intervention Project;
(7) a representative of the Vermont Association of the Deaf;
(8) a representative of the Vermont chapter of the Hearing Loss Association of America;
(9) a superintendent, selected by the Vermont Superintendents Association; and
(10) a special education administrator, selected by the Vermont Council of Special Education Administrators.
(c) Powers and duties.
(1) The Advisory Council shall assess the services, resources, and opportunities available to children in the State who are Deaf, Hard of Hearing, or DeafBlind. It may consider and make recommendations to the General Assembly and the Governor on the following:
(A) the educational rights of children who are Deaf, Hard of Hearing, or DeafBlind, including full communication and language access in all educational environments and accessibility of qualified teachers, interpreters, and paraprofessionals;
(B) appropriate and ongoing educational opportunities that recognize each child's unique learning needs, including access to a sufficient number of communication or language mode peers and exposure to adult role models who are Deaf, Hard of Hearing, or DeafBlind;
(C) adequate family supports that promote both early development of communication skills and informed participation by parents and guardians in the education of their children;
(D) identification of all losses of or reductions in services arising from the closures of the Austine School for the Deaf and the Vermont Center for the Deaf and Hard of Hearing and evaluation of the adequacy of existing services and resources, as well as identification of those resources not currently available, adequate, or accessible to children;
(E) opportunities to restore and expand educational opportunities to children in the State who are Deaf, Hard of Hearing, or DeafBlind and their families; and
(F) appropriate data collection and reporting requirements concerning students with disabilities.
(2) The Advisory Council shall assess the services, resources, and opportunities available to adults and elders in the State who are Deaf, Hard of Hearing, or DeafBlind. It may consider and make recommendations to the General Assembly and the Governor on the following:
(A) the needs of and opportunities for adults and elders within the State who are Deaf, Hard of Hearing, or DeafBlind and their families;
(B) the adequacy and systemic coordination of existing services and resources for adults and elders throughout the State who are Deaf, Hard of Hearing, or DeafBlind and their families;
(C) proposed legislation and administrative rules pertaining to adults and elders who are Deaf, Hard of Hearing, or DeafBlind; and
(D) delivery models in other states as a point of comparison for the adequacy and systemic coordination of Vermont's existing services and resources for adults and elders who are Deaf, Hard of Hearing, or DeafBlind.
(d) Assistance. The Advisory Council shall have the administrative, technical, and legal assistance of the Agencies of Education and of Human Services. The Advisory Council and Department may consult with national experts on education of persons who are Deaf, Hard of Hearing, or DeafBlind as necessary to fulfill their obligations under this section.
(e) Reports. On or before January 15 of each year, notwithstanding 2 V.S.A. § 20(d), the Advisory Council shall submit a written report to the Senate and House Committees on Education, the Senate Committee on Health and Welfare, the House Committee on Human Services, and the Governor with any findings and recommendations. A reading of each report shall be video recorded using ASL to ensure accessibility.
(f) Appointments; meetings.
(1) The Commissioner of Disabilities, Aging, and Independent Living shall convene the first meeting of the Advisory Council on or before July 1, 2016 and shall select interpreting services, computer assisted captioning in real time (CART), or FM listening systems for the meeting if a member so requests.
(2) At its first meeting, the Advisory Council shall elect a chair and vice chair.
(3) The Chair shall select interpreting services, CART, or FM listening systems for any Advisory Council meeting if a member so requests.
(4) The Advisory Council may meet up to eight times each year to perform its functions under this section. The Secretaries of Education and of Human Services may jointly authorize additional meetings.
(5) The Advisory Council may organize its members into subcommittees to carry out the purposes of this section, including subcommittees designed to address specific age groups within the Deaf, Hard of Hearing, and DeafBlind population.
(g) Reimbursement.
(1) Members of the Advisory Council who are not State employees or otherwise compensated or reimbursed for their attendance shall be entitled to per diem compensation and reimbursement of expenses pursuant to 32 V.S.A. § 1010, payable by the Department.
(2) The Agency of Human Services shall pay for interpreting services, CART, or FM listening systems necessary to conduct all Advisory Council meetings.
(3) The Agency of Education, Department of Health, and Department of Disabilities, Aging, and Independent Living shall share costs for interpreting services, CART, or FM listening systems necessary to conduct all Advisory Council subcommittee meetings. (Added 2015, No. 107 (Adj. Sess.), § 1, eff. May 12, 2016; amended 2019, No. 128 (Adj. Sess.), § 20.)


THERE is a lot more to France than popular tourist attractions such as the Eiffel Tower and the Louvre. There's the royal chateaux in the Loire region, Premier Cru vineyards in Burgundy and the Pope's Palace in Provence.
"One of the most beautiful ways to experience the finest things in France is to take a cruise on one of its many canals and rivers," says James Newcombe, publisher at www.french-waterways.com, an authoritative source of information on France's waterways.
Some of the most famous canals wind through the gorgeous wine-making areas of Sancerre, Chablis and Cotes du Rhone, while the historic 17th century Canal du Midi links the Mediterranean to the Atlantic through a 360-km network of waterways.
Hotel barges are large flat-bottomed boats converted into luxurious private floating hotels. They typically carry only eight to 12 guests, and some as few as four, so they can be chartered privately for an exclusive experience. Hotel barges travel slowly, giving guests an up, close and intimate feel for French life and culture, and travel a short distance each day, docking at a new mooring each evening.
Guest accommodation is luxurious and spacious, some providing dressing rooms, baths and four-poster beds. All are fitted with air-conditioning and modern wi-fi communications and some will even provide private telephones and TVs.
Fleur de Lys, from the fleet run by Belmond, is one of the most luxurious barges on the waters of Europe and is regularly chartered by celebrities and dignitaries. It features its own private pool, grand piano and three extremely spacious suites, each with a separate dressing area and bathroom featuring twin sinks and power showers.
The Golden Odyssey, a private hotel barge operated by an expert husband-and-wife team, is decorated in a sumptuous manner. The wood-panelled salon is furnished with Persian rugs, antique furniture and an open fire. It has an impressive teak sun-deck and a temperature-controlled spa pool.
Hotel barge chefs are highly qualified and experienced and love the opportunity to prepare Cordon Bleu quality food, from tasty breakfasts of fresh croissant pastries and seasonal fruit, to tempting hors d'oeuvres, right through to five course fine dining menus – all accompanied by the best of local wines.
Indeed, the abundance of fresh local produce is what tempted Rick Stein, world famous award-winning chef, to pick a hotel barge cruise as the setting for his memorable TV series, 'French Odyssey'.
While hotel barge cruises offer an unparalleled opportunity to get away from the hustle and bustle of life, there's also no fear of boredom. If guests don't feel like sitting back and basking in the beauty of the slowly passing countryside, there's a range of all-inclusive activities, from wine-tasting to visiting ancient chateaux, from guided city walks to tours of monasteries.
All hotel barges have bicycles on board and most barges have an executive mini-bus travelling alongside devoted to providing fascinating excursions. The crew can also organise additional special activities, like hot air ballooning, horse riding and golfing.
While the luxury and the slow pace of barges are very attractive, what really sets them apart from other types of cruise is the top-class personal service on board. In many cases, there is one crew member for every two guests. Hotel barge teams invariably speak English and are used to catering for many international cultures.
Although every little nook and corner of France has marvellous stories to divulge, there are some waterways that are more loved than others: cruises through Paris and Champagne, Burgundy, Provence, the Loire, Aquitaine and Bordeaux.
The Canal du Midi (a Unesco World Heritage site) is a big favourite for its stunning beauty and magnificent architecture. Belmond's Alouette sails on the Midi from Béziers to Carcassonne in the sun-drenched Languedoc region, France's most productive wine area.


text:
\section{Introduction}
\renewcommand{\ann}[1]{}
General relativity is a Lagrangian theory, i.e., the Einstein equations are derived as the Euler-Lagrange equation of the Einstein-Hilbert functional
\begin{equation}
\int_N(\bar R-2\varLambda),
\end{equation}
where $N=N^{n+1}$, $n\ge 3$, is a globally hyperbolic Lorentzian manifold, $\bar R$ the scalar curvature and $\varLambda$ a cosmological constant. We also omitted the integration density in the integral. In order to apply a Hamiltonian description of general relativity, one usually defines a time function $x^0$ and considers the foliation of $N$ given by the slices
\begin{equation}
M(t)=\{x^0=t\}.
\end{equation}
We may, without loss of generality, assume that the spacetime metric splits
\begin{equation}\lae{1.3}
d\bar s^2=-w^2(dx^0)^2+g_{ij}(x^0,x)dx^idx^j,
\end{equation}
cf.\@\xspace \cite[Theorem 3.2]{cg:qgravity}. Then, the Einstein equations also split into a tangential part
\begin{equation}
G_{ij}+\varLambda g_{ij}=0
\end{equation}
and a normal part
\begin{equation}
G_{\alpha\beta}\nu^\alpha\nu^\beta-\varLambda=0,
\end{equation}
where the naming refers to the given foliation. For the tangential Einstein equations one can define equivalent Hamilton equations due to the groundbreaking paper by Arnowitt, Deser and Misner \cite{adm:old}. The normal Einstein equations can be expressed by the so-called Hamilton condition
\begin{equation}\lae{1.6}
\protect\mathcal H=0,
\end{equation}
where $\protect\mathcal H$ is the Hamiltonian used in defining the Hamilton equations. In the canonical quantization of gravity the Hamiltonian is transformed  to a partial differential operator of hyperbolic type $\hat{\protect\mathcal H}$ and the possible quantum solutions of gravity are supposed to satisfy the so-called Wheeler-DeWitt equation
\begin{equation}\lae{1.7}
\hat{\protect\mathcal H}u=0
\end{equation}
in an appropriate setting, i.e., only the Hamilton condition \re{1.6} has been quantized, or equivalently, the normal Einstein equation, while the tangential Einstein equations have been ignored.

In \cite{cg:qgravity} we solved the equation \re{1.7} in a fiber bundle $E$ with base space ${\protect\mathcal S_0}$,
\begin{equation}
{\protect\mathcal S_0}=\{x^0=0\}\equiv M(0),
\end{equation}
and fibers $F(x)$, $x\in{\protect\mathcal S_0}$,
\begin{equation}
F(x)\su T^{0,2}_x({\protect\mathcal S_0}),
\end{equation}
the elements of which are the positive definite symmetric tensors of order two, the Riemannian metrics in ${\protect\mathcal S_0}$. The hyperbolic operator $\hat{\protect\mathcal H}$ is then expressed in the form
\begin{equation}\lae{1.10}
\hat{\protect\mathcal H}=-\varDelta-(R-2\varLambda)\varphi,
\end{equation}
where $\varDelta$ is the Laplacian of the DeWitt metric given in the fibers, $R$ the scalar curvature of the metrics $g_{ij}(x)\in F(x)$, and $\varphi$ is defined by
\begin{equation}\lae{1.11}
\varphi^2=\frac{\det g_{ij}}{\det\rho_{ij}},
\end{equation}
where $\rho_{ij}$ is a fixed metric in ${\protect\mathcal S_0}$ such that instead of densities we are considering functions. The Wheeler-DeWitt equation could be solved in $E$ but only as an abstract hyperbolic equation. The solutions could not be split in corresponding spatial and temporal eigenfunctions.

The underlying mathematical reason for the difficulty was the presence of the term $R$ in the quantized equation, which prevents the application of separation of variables, since the metrics $g_{ij}$ are the spatial variables. In a recent paper \cite{cg:qgravity3} we overcame this difficulty by quantizing the Hamilton equations instead of the Hamilton condition.

 As a result we obtained the equation
 \begin{equation}\lae{1.12}
-\varDelta u=0
\end{equation}
in $E$, where the Laplacian is the Laplacian in \re{1.10}. The lower order terms of $\hat{\protect\mathcal H}$
\begin{equation}
(R-2\varLambda)\varphi
\end{equation}
were eliminated during the quantization process. However, the equation \re{1.12} is only valid provided $n\not=4$, since the resulting equation actually looks like
\begin{equation}
-(\frac n2-2)\varDelta u=0.
\end{equation}
This restriction seems to be acceptable, since $n$ is the dimension of the base space ${\protect\mathcal S_0}$ which, by general consent, is assumed to be $n=3$. The fibers add additional dimensions to the quantized problem, namely,
\begin{equation}
\dim F=\frac {n(n+1)}2\equiv m+1.
\end{equation}
The fiber metric, the DeWitt metric, which is responsible for the Laplacian in \re{1.12} can be expressed in the form
\begin{equation}
ds^2=-\frac{16(n-1)}n dt^2+\varphi G_{AB}d\xi^Ad\xi^B,
\end{equation}
where the coordinate system is
\begin{equation}\lae{1.17}
(\xi^a)= (\xi^0,\xi^A)\equiv (t,\xi^A).
\end{equation}
The $(\xi^A)$, $1\le A\le m$, are coordinates for the hypersurface
\begin{equation}\lae{1.18}
M\equiv M(x)=\{(g_{ij}):t^4=\det g_{ij}(x)=1,\forall\, x\in{\protect\mathcal S_0}\}.
\end{equation}
We also assumed that ${\protect\mathcal S_0}=\R[n]$ and that  the metric $\rho_{ij}$ in \re{1.11} is the Euclidean metric $\delta _{ij}$. It is well-known that $M$ is a symmetric space
\begin{equation}
M=SL(n,\R[])/SO(n)\equiv G/K.
\end{equation}
It is also easily verified that the induced metric of $M$ in $E$ is isometric to the Riemannian metric of the coset space $G/K$.

Now, we were in a position to use separation of variables, namely, we wrote a solution of \re{1.12} in the form
\begin{equation}
u=w(t)v(\xi^A),
\end{equation}
where $v$ is a spatial eigenfunction of the induced Laplacian of $M$
\begin{equation}\lae{1.21}
-\varDelta_Mv\equiv -\varDelta v=(\abs\lambda^2+\abs\rho^2)v
\end{equation}
and $w$ is a temporal eigenfunction satisfying the ODE
\begin{equation}\lae{1.22}
\Ddot w+m t^{-1}\dot w+\mu_0 t^{-2}w=0
\end{equation}
with
\begin{equation}
\mu_0=\frac{16(n-1)}n(\abs \lambda^2+\abs\rho^2).
\end{equation}

The eigenfunctions of the Laplacian in $G/K$ are well-known and we chose the kernel of the Fourier transform in $G/K$ in order to define the eigenfunctions. This choice also allowed us to use Fourier quantization similar to the Euclidean case such that the eigenfunctions are transformed to Dirac measures and the Laplacian to a multiplication operator in Fourier space.

In the present paper we like to quantize the Einstein-Hilbert functional combined with the functionals of the other fundamental forces of nature, i.e., we look at  the Lagrangian functional
\begin{equation}\lae{1.1}
\begin{aligned}
J&=\alpha_N^{-1}\int_{\tilde\varOmega}(\bar R-2\varLambda)-\int_{\tilde\varOmega}\tfrac14 \gamma_{\bar a\bar b}\bar g^{\mu\rho_2}\bar g^{\lambda\rho_1}F^{\bar a}_{\mu\rho_1}F^{\bar b}_{\rho_2\lambda}\\
&\hphantom =\;\;\,-\int_{\tilde\varOmega}\{\tfrac12 \bar g^{\mu\lambda}\gamma_{\bar a\bar b}\varPhi^{\bar a}_\mu\bar\varPhi^{\bar b}_\lambda+V(\varPhi)\}\\
&\hphantom =\;\;+\int_{\tilde\varOmega}\{\tfrac12[\tilde\psi_{I}E^\mu_a\gamma^a(D_\mu \psi)^{I}+\overline{\tilde\psi_{I}E^\mu_a\gamma^a(D_\mu \psi)^{I}}]+m\tilde\psi_{I}\psi^{I}\},
\end{aligned}
\end{equation}
where $\alpha_N$ is a positive coupling constant,  $\tilde\varOmega\Su N=N^{n+1}$ and $N$ a globally hyperbolic spacetime with metric $\bar g_{\alpha\beta}$, $0\le \alpha,\beta\le  n$, where the metric splits as in \re{1.3}.

The functional $J$ consists of the Einstein-Hilbert functional, the Yang-Mills and Higgs functional and a massive Dirac term.

The Yang-Mills field $(A_\mu)$
\begin{equation}
A_\mu=f_{\bar c}A^{\bar c}_\mu
\end{equation}
corresponds to the adjoint representation of a compact, semi-simple Lie group $\protect\mathcal G$ with Lie algebra $\mf g$. The $f_{\bar c}$,
\begin{equation}
f_{\bar c}=(f^{\bar a}_{\bar c\bar b})
\end{equation}
are the structural constants of $\mf g$.

We assume the Higgs field $\varPhi=(\varPhi^{\bar a})$ to have complex valued components.

The spinor field $\psi=(\psi^I_A)$ has  a spinor index $A$, $1\le A\le n_1$, and a colour index $I$, $1\le I\le n_2$. Here, we suppose that the Lie group has a unitary representation $R$ such that
\begin{equation}
t_{\bar c}=R(f_{\bar c})
\end{equation}
are antihermitian matrices acting on ${\protect\mathbb C}^{n_2}$. The symbol $A_\mu\psi$ is now defined by
\begin{equation}
A_\mu\psi=t_{\bar c}\psi A^{\bar c}_\mu.
\end{equation}

There are some major difficulties in achieving a quantization of the functional in \re{1.1}. First we were unable to quantize the corresponding Hamilton equations, hence, we quantized the Hamilton condition which has the form
\begin{equation}
H_G+H_{YM}+H_D+H_H=0,
\end{equation}
where the subscripts refer to gravity, Yang-Mills, Dirac and Higgs. On the left-hand side are the Hamilton functions of the respective fields. They depend on the  Riemannian metrics $g_{ij}$, the Yang-Mills connections and the spinor and Higgs fields. We were not able to quantize the non-gravitational Hamilton\ann{s} functions for arbitrary metrics $g_{ij}$, but we propose the following model: Choosing the fiber coordinates as in \re{1.17} the fiber metrics $g_{ij}$ can be written in the form
\begin{equation}
g_{ij}(x)=t^{\frac4n}\sigma_{ij}(x),
\end{equation}
where
\begin{equation}
\sigma_{ij}\in M.
\end{equation}
If we were able to express the non-gravitational Hamiltonians as
\begin{equation}\lae{1.32}
H_{YM}=t^{ p} \tilde H_{YM}, \quad H_D=t^p\tilde H_D,\quad H_H=t^p\tilde H_H,
\end{equation}
where the embellished Hamiltonians depend on $\sigma_{ij}$, then, by choosing in addition $n=3$ and $\sigma_{ij}=\delta _{ij}$, these Hamiltonians could be quantized by the known methods of QFT, if the Lie groups would be chosen appropriately. The Wheeler-DeWitt equation would then not be solved for all $(t,\sigma_{ij})$ but only for $(t,\delta _{ij})$. However, the spatial eigendistributions of the Hamilton operator $\hat H_G$, i.e., the  eigendistributions of the Laplacian of $M$, cf.\@\xspace \re{1.21}, would still be used but they would be evaluated at $\sigma_{ij}=\delta _{ij}$.

In \rs{4} we shall prove that the expressions in \re{1.32} are indeed valid with $p=-\frac23$ provided $n=3$ and provided that the mass term\ann{the} in the Dirac Lagrangian and the Higgs Lagrangian are slightly modified. The embellished Hamiltonians are then standard Hamiltonians without any modifications, for details we refer to \rs{4}. The Hamilton constraint then has the form
\begin{equation}
\begin{aligned}
H&=H_G+H_{YM}+H_H+H_D\\
&=H_G+t^{-\frac23}(\tilde H_{YM}+\tilde H_H+\tilde H_D)\\
&\equiv H_G+t^{-\frac23} \tilde H_{SM}=0,
\end{aligned}
\end{equation}
where the subscript $SM$ refers to the fields  of the Standard Model or to a  corresponding subset of fields. The solutions of the Wheeler-DeWitt equation
\begin{equation}
\hat Hu=0
\end{equation}
can then be achieved by using separation of variables. We proved:
\begin{thm}\lat{1.1}
Let $n=3$, $v=e_{\lambda,b_0}$ and let $\psi$ be an  eigendistribution of $\tilde H_{SM}$ when $\sigma_{ij}=\delta _{ij}$ such that
\begin{equation}
- \varDelta_M e_{\lambda,b_0}=(\abs\lambda^2+1)e_{\lambda,b_0},
\end{equation}
\begin{equation}
\tilde H_{SM}\psi=\lambda_1\psi,\qquad\lambda_1\ge 0,
\end{equation}
and let $w$ be a solution of the  ODE
\begin{equation}\lae{4.147.1}
\begin{aligned}
t^{-m}\pde{}{t}(t^m \pde wt)&+\frac{32}3 (\abs\lambda^2+1)t^{-2}w+\frac{32}3\alpha_N^{-1}\lambda_1 t^{-\frac23}w\\
& \quad+\frac{64}3\alpha_N^{-2}\varLambda t^2w=0
\end{aligned}
\end{equation}
then
\begin{equation}
u=w e_{\lambda,b_0}\psi
\end{equation}
is a solution of the Wheeler-DeWitt equation
\begin{equation}
\hat Hu=0,
\end{equation}
where $e_{\lambda,b_0}$ is evaluated at $\sigma_{ij}=\delta _{ij}$ and where we note that $m=5$.
\end{thm}
We shall refer to $e_{\lambda,b_0}$ and $\psi$ as the spatial eigenfunctions and to $w$ as the temporal eigenfunction.

\begin{rem}
We could also apply the respective Fourier transforms to $-\tilde \varDelta e_{\lambda,b_0}$ resp.\@\xspace $\tilde H_{SM}\psi$ and consider
\begin{equation}
w\hat e_{\lambda,b_0}\hat\psi
\end{equation}
as the solution in Fourier space, where $\hat\psi$ would be expressed with the help of the ladder operators.
\end{rem}
The temporal eigenfunctions are analyzed in \rs{5}. They must satisfy an ODE of the form
\begin{equation}\lae{5.2.1}
\Ddot w+5 t^{-1} \dot w+m_1 t^{-2}w+m_2^2 t^{-\frac23}w+m_3 t^2w=0,
\end{equation}
where
\begin{equation}\lae{5.3.1}
m_1\ge\frac{32}3,\quad m_2\ge 0,\quad m_3\in\R[].
\end{equation}
For simplicity we shall only state the result when $m_3=0$ which is tantamount to setting $\varLambda=0$.
\begin{thm}
Assume $m_3=0$ and $m_2>0$, then the solutions of the ODE \re{5.2.1} are generated by
\begin{equation}
J(\tfrac32\sqrt{m_1-4}\,i,\tfrac32 m_2 t^\frac23) t^{-2}
\end{equation}
and
\begin{equation}
J(-\tfrac32\sqrt{m_1-4}\,i,\tfrac32 m_2 t^\frac23) t^{-2},
\end{equation}
where $J(\lambda,t)$ is the Bessel function of the first kind.
\end{thm}
\begin{lem}\lal{5.2.1}
The solutions in the theorem above diverge to complex infinity if $t$ tends to zero and they converge to zero if $t$ tends to infinity.
\end{lem}

\section{Definitions and notations}
Greek indices $\alpha$, $\beta$ range from $0$ to $n$, Latin $i,j,k$ from $1$ to $n$ and we stipulate $0\le a, b\le n$ but $1\le a', b'\le n$. Barred indices $\bar a$ refer to the Lie algebra $\mf g$, $1\le \bar a\le n_0=\dim\mf g$.

$\gamma_{\bar a\bar b}$ is the Cartan-Killing metric.

The Dirac matrices are denoted by $\gamma^a$ and they satisfy
\begin{equation}
\gamma^a\gamma^b+\gamma^b\gamma^a=2\eta^{ab}I,
\end{equation}
where $\eta_{ab}$ is the Minkowski metric with signature $(-,+,\ldots,+)$. $\gamma^0$ is antihermitian and $\gamma^{a'}$ Hermitian.

The indices $a,b$ are always raised or lowered with the help of the Minkowski metric, Greek indices with the help of the spacetime metric $\bar g_{\alpha\beta}$.

The $\gamma^a$ act in
\begin{equation}
{\protect\mathbb C}^{2^{\frac{n+1}2}},
\end{equation}
if $n$ is odd and in
\begin{equation}
{\protect\mathbb C}^{2^\frac n2}\oplus \;{\protect\mathbb C}^{2^\frac n2},
\end{equation}
if $n$ is even. In both cases we simply refer to these spaces as
\begin{equation}
{\protect\mathbb C}^{n_1},
\end{equation}
i.e., the spinor index $A$ has range $1\le A\le n_1$.

The colour index $I$ has range $1\le I\le n_2$ and hence a spinor field $\psi^I_A$ has values in
\begin{equation}
{\protect\mathbb C}^{n_1}\otimes {\protect\mathbb C}^{n_2}.
\end{equation}

Finally, a Hermitian form $\spd\cdot\cdot$ is antihermitian in the first argument.

\section{Spinor fields}\las{3}
The Lagrangian of the spinor field is stated in \re{1.1}. Here, $\psi=(\psi^I_A)$ is a multiplet of spinors with spin $\tfrac12$; $A$ is the spinor index, $1\le A\le n_1$, and $I$, $1\le I\le n_2$, the \textit{colour} index.
We shall also lower or raise the index $I$ with the help of the Euclidean metric $(\delta _{IJ})$.

Let $\varGamma_\mu$ be the spinor connection
\begin{equation}
\varGamma_\mu=\tfrac14 \omega_{\mu\hphantom ba}^{\hphantom\mu b}\gamma_b\gamma^a,
\end{equation}
then the covariant derivative $D_\mu\psi$ is defined by
\begin{equation}
D_\mu\psi=\psi_{,\mu}+\varGamma_\mu\psi+A_\mu\psi.
\end{equation}

Let $(e^b_\lambda)$ be a $n$-bein such that
\begin{equation}
\bar g_{\mu\lambda}=\eta_{ab}e^a_\mu e^b_\lambda,
\end{equation}
where $(\eta_{ab})$ is the Minkowski metric, and let $(E^\mu_a)$ be its inverse
\begin{equation}
E^\mu_a=\eta_{ab}\bar g^{\mu\lambda}e^b_\lambda,
\end{equation}
cf.\@\xspace \cite[p. 246]{eguchi:book}.

The covariant derivative of $E^\alpha_a$ with respect to $(\bar g_{\alpha\beta})$ is then given by
\begin{equation}
E^\alpha_{a;\mu}=E^\alpha_{a,\mu}+\cha \mu\beta\alpha E^\beta_a
\end{equation}
and
\begin{equation}
\omega_{\mu\hphantom ba}^{\hphantom\mu b}=E^\lambda_{a;\mu}e^b_\lambda=-E^\lambda_ae^b_{\lambda;\mu},
\end{equation}
hence the spin connection $\varGamma_\mu$ can be expressed as
\begin{equation}
\varGamma_\mu=\tfrac14 \omega_{\mu\hphantom ba}^{\hphantom\mu b}\gamma_b\gamma^a=\tfrac14 E^\lambda_{a;\mu}e^b_\lambda \gamma_b\gamma^a=-\tfrac14 E^\lambda_ae^b_{\lambda;\mu}\gamma_b\gamma^a.
\end{equation}

We shall first show:
\begin{lem}\lal{3.1}
Let $\bar g_{\alpha\beta}$ be a fixed spacetime metric that is split by the time function $x^0$, then there exists an orthonormal frame $(e^a_\lambda)$ such that
\begin{equation}
e^0_k=0,\qquad 1\le k\le n,
\end{equation}
and
\begin{equation}
e^{a'}_{k;0}=e^{a'}_{,0}-\cha k0\lambda e^{a'}_\lambda=0
\end{equation}
for all $1\le a'\le n$ and $1\le k\le n$.
\end{lem}
\begin{proof}
Assume that
\begin{equation}
\bar g_{00}=-w^2,
\end{equation}
then define the conformal metric
\begin{equation}
\tilde g_{\alpha\beta}=w^{-2}\bar g_{\alpha\beta}.
\end{equation}
The curves
\begin{equation}
(\gamma^\alpha(t,x))=(t,x^i),\qquad x\in{\protect\mathcal S_0},
\end{equation}
are then geodesics with respect to $\tilde g_{\alpha\beta}$. Let $(\hat e^{a'}_\lambda)$, $1\le a'\le n$, be an orthonormal frame in $T^{0,1}({\protect\mathcal S_0})\hra T^{0,1}(N)$ such that
\begin{equation}
\hat e^{a'}_0=0\qquad\forall \, 1\le a'\le n.
\end{equation}
The $\hat e^{a'}$ depend on $x=(x^i)\in {\protect\mathcal S_0}$. Let $(\tilde e^{a'}_\lambda)(t,x)$ be the solutions of the flow equations
\begin{equation}
\begin{aligned}
\frac D{dt}\tilde e^{a'}_\lambda&=0,\\
\tilde e^{a'}_\lambda(0,x)&=\hat e^{a'}_\lambda(x),
\end{aligned}
\end{equation}
i.e., we parallel transport $\hat e^{a'}$ along the geodesics. Setting
\begin{equation}
(\tilde e^0_\lambda)=(1,0,\ldots,0)
\end{equation}
the $(\tilde e^a_\lambda)$ are then an orthonormal frame of $1$-forms in $(N,\tilde g_{\alpha\beta})$ such that the $\tilde e^a$ satisfy
\begin{equation}\lae{3.16}
\tilde e^a_{\lambda:0}=0\qquad\forall\, 0\le a\le n,
\end{equation}
where we indicate covariant differentiation with respect to $\tilde g_{\alpha\beta}$ by a colon.

Define $e^a_\lambda$ by
\begin{equation}
e^a_\lambda=w \tilde e^a_\lambda,
\end{equation}
then the $e^a_\lambda$ are orthonormal frames in $(N,\bar g_{\alpha\beta})$. The Christoffel symbols $\cha \alpha\beta\gamma$ resp.\@\xspace $\tilde \varGamma^\gamma_{\alpha\beta}$ are related by the formula
\begin{equation}\lae{3.18}
\begin{aligned}
\cha \alpha\beta\gamma= \tilde \varGamma^\gamma_{\alpha\beta} - w^{-1} w_\alpha \delta ^\gamma_\beta +w^{-1}w_\beta \delta ^\gamma_\alpha -w^{-1} \check w^\gamma \tilde g_{\alpha\beta},
\end{aligned}
\end{equation}
where
\begin{equation}
\check w^\gamma=\tilde g^{\gamma\lambda}w_\lambda.
\end{equation}

In view of \re{3.16} we then infer
\begin{equation}
0=\tilde e^{a'}_{j:0}=\dot{\tilde e}^{a'}_j-\tilde \varGamma^k_{0j}\tilde e^{a'}_k
\end{equation}
and we deduce further
\begin{equation}
\begin{aligned}
e^{a'}_{j;0}&=\dot w\tilde e^{a'}_j+w\dot{\tilde e}^{a'}_j-\cha 0jk w \tilde e^{a'}_k\\
&=\dot w \tilde e^{a'}_j+\tilde\varGamma ^k_{0j}w\tilde e^{a'}_k -\cha 0jk w\tilde e^{a'}_k\\
&=0
\end{aligned}
\end{equation}
because of \re{3.18}.
\end{proof}
Subsequently we shall always use these particular orthonormal frames.

We are now able to simplify the expressions for the spin connections
\begin{equation}
\varGamma_\mu= -\tfrac14 E^\lambda_a e^b_{\lambda;\mu}\gamma_a\gamma^b.
\end{equation}
We have
\begin{equation}
\begin{aligned}
4\varGamma_0&=-E^\lambda_a e^b_{\lambda;0}\gamma_b\gamma^a\\
&=-E^\lambda_ae^0_{\lambda;0}\gamma_0\gamma^a-E^\lambda_ae^{b'}_{\lambda;0}\gamma_{b'}\gamma^a\\
&=-E^0_0 e^0_{0;0}\gamma_0\gamma^0-E^i_{a'}e^0_{i;0}\gamma_0\gamma^{a'}-E^0_0e^{b'}_{0;0}\gamma_{b'}\gamma^0-E^i_{a'}e^{b'}_{i;0}\gamma_{b'}\gamma^{a'}\\
&=-E^i_{a'}e^0_{i;0}\gamma_0\gamma^{a'}-E^0_0e^{b'}_{0;0}\gamma_{b'}\gamma^0
\end{aligned}
\end{equation}
in view of \rl{3.1} and the fact that
\begin{equation}
e^0_{0;0}=0.
\end{equation}

The matrices $\gamma_0\gamma^{a'}$ and $\gamma_{b'}\gamma^0$ are hermitian, since $\gamma^0$ is antihermitean, $\gamma^{a'}$ hermitean and there holds
\begin{equation}
\gamma_0\gamma^{a'}=-\gamma^{a'}\gamma_0.
\end{equation}

Hence, the quadratic form
\begin{equation}
\tilde\psi E^0_a \gamma^a\varGamma_0\psi=-iE^0_0\bar\psi\varGamma_0\psi
\end{equation}
is imaginary and will be eliminated by adding its complex conjugate. $\varGamma_0$ can therefore be ignored which we shall indicate by writing
\begin{equation}
\varGamma_0\simeq 0.
\end{equation}
A similar notation should apply to other terms that will be cancelled when adding the complex conjugates.

Let us consider $\varGamma_k$:
\begin{equation}
\begin{aligned}
4\varGamma_k&=-E^\lambda_a e^b_{\lambda;k}\gamma_b\gamma^a\\
&=-E^\lambda_ae^0_{\lambda;k}\gamma_0\gamma^a-E^\lambda_ae^{b'}_{\lambda;k}\gamma_{b'}\gamma^a\\
&=-E^0_0 e^0_{0;k}\gamma_0\gamma^0-E^i_{a'}e^0_{i;k}\gamma_0\gamma^{a'}-E^0_0e^{b'}_{0;k}\gamma_{b'}\gamma^0-E^i_{a'}e^{b'}_{i;k}\gamma_{b'}\gamma^{a'}.
\end{aligned}
\end{equation}
The first term on the right-hand side vanishes, since
\begin{equation}
e^0_{0;k}=w_k-\cha0k0w=0.
\end{equation}
Furthermore, there holds
\begin{equation}
e^0_{i;k}=-\cha ik0w=-\tfrac12\dot g_{ik}w^{-1}
\end{equation}
and
\begin{equation}
e^{b'}_{0;k}=-\cha 0kje^{b'}_j=-\tfrac12 g^{lj}\dot g_{kl}e^{b'}_j,
\end{equation}
yielding
\begin{equation}\lae{3.32}
\begin{aligned}
4\varGamma_k&=\tfrac12 \dot g_{ik}w^{-1}E^i_{a'}\gamma_0\gamma^{a'}+\tfrac12 w^{-1}g^{lj}\dot g_{kl} e^{b'}_i\gamma_{b'}\gamma^0-E^i_{a'}e^{b'}_{i;k}\gamma_{b'}\gamma^{a'}\\
&=w^{-1}\dot g_{ik}E^i_{a'}\gamma_0\gamma^{a'}-E^i_{a'}e^{b'}_{i;k}\gamma_{b'}\gamma^{a'},
\end{aligned}
\end{equation}
since
\begin{equation}
\gamma_0\gamma^{a'}=-\gamma^{a'}\gamma_0.
\end{equation}
The first term on the right-hand side of \re{3.32} has to be eliminated because of the presence of $\dot g_{ik}$. To achieve this fix a Riemannian metric $\rho_{ij}=\rho_{ij}(x)\in T^{0,2}({\protect\mathcal S_0})$ and define the function $\varphi$ by
\begin{equation}\lae{3.34.1}
\varphi=\sqrt{\frac{\det g_{ij}}{\det \rho_{ij}}}
\end{equation}
and the spinors $\chi=(\chi^i_A)$ by
\begin{equation}\lae{3.35}
\chi=\sqrt\varphi\psi,
\end{equation}
then
\begin{equation}
\dot \chi=\sqrt\varphi\dot\psi+\tfrac14 g^{ij}\dot g_{ij}\chi
\end{equation}
and
\begin{equation}
\chi_{,k}=\tfrac12 \varphi_k\varphi^{-1/2}\chi +\sqrt\varphi \psi_{,k}.
\end{equation}
Looking at the real part of the quadratic form
\begin{equation}
i\tilde \chi E^k_{a'}\gamma^{a'}\chi_{,k}
\end{equation}
we deduce that
\begin{equation}
\chi_{,k}\simeq \sqrt\varphi \psi_{,k}.
\end{equation}

Moreover, we infer
\begin{equation}
\begin{aligned}
i\tilde\psi E^k_{c'}\gamma^{c'}\varGamma_k\psi&= i\bar \psi E^k_{c'}\gamma^0\gamma^{c'}\varGamma_k\psi\\
&=\tfrac14 i\bar\psi E^k_{c'}E^j_{a'}w^{-1}\dot g_{jk}\gamma^0\gamma^{c'}\gamma_0\gamma^{a'}\psi\\
&\quad -\tfrac14 i \bar\psi E^k_{c'}E^j_{a'}e^{b'}_{j;k}\gamma^0\gamma^{c'}\gamma_{b'}\gamma^{a'}\psi.
\end{aligned}
\end{equation}
We now observe that
\begin{equation}
\gamma^0\gamma^{c'}\gamma_0\gamma^{a'}=-\gamma^0\gamma_0\gamma^{c'}\gamma^{a'}=-\gamma^{c'}\gamma^{a'},
\end{equation}
hence
\begin{equation}
E^k_{c'}E^j_{a'}\gamma^0\gamma^{c'}\gamma_0\gamma^{a'}=-E^k_{c'}E^j_{a'}\gamma^{c'}\gamma^{a'}=-g^{jk}
\end{equation}
and we conclude
\begin{equation}
\begin{aligned}
i\tilde\psi E^\mu_c\gamma^cD_\mu\psi\varphi&\simeq-i\bar\chi \dot\chi w^{-1}\\
&\quad +i\bar\chi E^k_{c'}\gamma^0\gamma^{c'}\{\chi_{,k}-\tfrac14 E^j_{a'}e^{b'}_{j;k}\gamma_{b'}\gamma^{a'}\chi +  A_k\chi\}
\end{aligned}
\end{equation}
\begin{rem}
The term in the braces is the covariant derivative of $\chi$ with respect to the spin connection $\tilde\varGamma_k$
\begin{equation}\lae{3.44}
\tilde\varGamma^{b'}_{ka'}=\tfrac14\tilde\omega^{b'}_{ka'}=-\tfrac14 E^j_{a'}e^{b'}_{j;k}\gamma_{b'}\gamma^{a'}
\end{equation}
and the Yang-Mills connection $(A_\mu)$ satisfying $A_0=0$ such that
\begin{equation}\lae{3.45}
\tilde D_k\chi=\chi_{,k}+\tilde\varGamma_k\chi+ A_k\chi.
\end{equation}
The gauge transformations for both the Yang-Mills connection as well as for the spin connection do not depend on $x^0$ but only on $x\in{\protect\mathcal S_0}$. In case of the Yang-Mills connection this has already been proved in \cite[Lemma 2.6]{cg:uqtheory} while the proof for the spin connection $\tilde\varGamma_k$ follows from \re{3.44} and \re{3.32} if we only consider Lorentzian metrics of the form
\begin{equation}
d\bar s^2=-dt^2+g_{ij}(x)dx^idx^j
\end{equation}
in a product manifold $N=I\times {\protect\mathcal S_0}$, as will be the case after the quantization of the Dirac field.
\end{rem}
Summarizing the preceding results we obtain:
\begin{lem}\lal{3.3}
The Dirac Lagrangian can be expressed in the form
\begin{equation}\lae{3.46}
\begin{aligned}
L_D&=\tfrac{i}2(\bar\chi_I\dot\chi^I-\dot{\bar\chi}^{I}\chi_{I})w^{-1}\varphi^{-1}+m i\bar\chi_I\gamma^0\chi^I\varphi^{-1}\\
&\quad-\tfrac{i}2\{\bar\chi_I\gamma^0E^k_{a'}\gamma^{a'}\tilde D_k\chi^I-\overline{\bar\chi_I\gamma^0E^k_{a'}\gamma^{a'}\tilde D_k\chi^I}\}\varphi^{-1},
\end{aligned}
\end{equation}
where $\chi$ and $\tilde D_k$ are defined in \re{3.35} resp.\@\xspace \re{3.45}.
\end{lem}

\section{Quantization of the Lagrangian}\las{4}
We consider the functional
\begin{equation}\lae{4.1}
\begin{aligned}
J&=\alpha_N^{-1}\int_{\tilde\varOmega}(\bar R-2\varLambda)-\int_{\tilde\varOmega}\tfrac14 \gamma_{\bar a\bar b}\bar g^{\mu\rho_2}\bar g^{\lambda\rho_1}F^{\bar a}_{\mu\rho_1}F^{\bar b}_{\rho_2\lambda}\\
&\hphantom =\;\;\,-\int_{\tilde\varOmega}\{\tfrac12 \bar g^{\mu\lambda}\gamma_{\bar a\bar b}\varPhi^{\bar a}_\mu\bar\varPhi^{\bar b}_\lambda+V(\varPhi)\}\\
&\hphantom =\;\;+\int_{\tilde\varOmega}\{\tfrac12[\tilde\psi_{I}E^\mu_a\gamma^a(D_\mu \psi)^{I}+\overline{\tilde\psi_{I}E^\mu_a\gamma^a(D_\mu \psi)^{I}}]+m\tilde\psi_{I}\psi^{I}\},
\end{aligned}
\end{equation}
where $\alpha_N$ is a positive coupling constant and $\tilde\varOmega\Su N$.

We use the action principle that, for an arbitrary $\tilde\varOmega$ as above, a solution $(A,\varPhi,\psi,\bar g)$ should be a stationary point of the functional with respect to compact variations. This principle requires no additional surface terms for the functional.

As we proved in \cite{cg:qgravity} we may only consider metrics $\bar g_{\alpha\beta}$ that split with respect to some fixed globally defined time function $x^0$ such that
\begin{equation}\lae{4.2}
d\bar s^2=-w^2 (dx^0)^2+g_{ij}dx^idx^j
\end{equation}
where $g(x^0,\cdot)$ are Riemannian metrics in ${\protect\mathcal S_0}$,
\begin{equation}
{\protect\mathcal S_0}=\{x^0=0\}.
\end{equation}
The first functional on the right-hand side of \re{4.1} can be written in the form
\begin{equation}\lae{5.4}
\alpha^{-1}_N\int_a^b\int_\varOmega\{\tfrac14G^{ij,kl}\dot g_{ij}\dot g_{kl}+R-2\varLambda\}w\varphi,
\end{equation}
where
\begin{equation}
G^{ij,kl}=\tfrac12\{g^{ik}g^{jl}+g^{il}g^{jk}\}-g^{ij}g^{kl}
\end{equation}
is the DeWitt metric,
\begin{equation}
(g^{ij})=(g_{ij})^{-1},
\end{equation}
$R$ the scalar curvature of the slices
\begin{equation}
\{x^0=t\}
\end{equation}
with respect to the metric $g_{ij}(t,\cdot)$, and where we also assumed that $\tilde\varOmega$ is a cylinder
\begin{equation}
\tilde\varOmega=(a,b)\times\varOmega,\qquad\varOmega\Su {\protect\mathcal S_0},
\end{equation}
such that $\tilde\varOmega\su U_k$ for some $k\in {\protect\mathbb N}$, where the $U_k$ are  special coordinate patches of $N$ such that there exists a local trivialization in $U_k$ with the properties that there is a fixed  Yang-Mills connection
\begin{equation}
\bar A=(\bar A^{\bar a}_\mu)=f_{\bar a}\bar A^{\bar a}_\mu dx^\mu
\end{equation}
satisfying
\begin{equation}
\bar A^{\bar a}_0=0\qquad \text{in}\; U_k,
\end{equation}
cf.\@\xspace \cite[Lemma 2.5]{cg:uqtheory}. We may then  assume that the Yang-Mills connections $A=(A^{\bar a}_\mu)$ are of the form
\begin{equation}
A^{\bar a}_\mu (t,x)=\bar A^{\bar a}_\mu (0,x)+\tilde A^{\bar a}_\mu (t,x),
\end{equation}
where $(\tilde A^{\bar a}_\mu )$ is a tensor, see \cite[Section 2]{cg:uqtheory}.

The Riemannian metrics $g_{ij}(t,\cdot)$ are elements of the bundle $T^{0,2}({\protect\mathcal S_0})$. Denote by $E$ the fiber bundle with base ${\protect\mathcal S_0}$ where the fibers $F(x)$ consists of the Riemannian metrics $(g_{ij})$. We shall consider each fiber to be a Lorentzian manifold equipped with the DeWitt metric. Each fiber $F$ has dimension
\begin{equation}
\dim F=\frac{n(n+1)}2\equiv m+1.
\end{equation}
Let $(\xi^r)$, $0\le r\le m$, be  coordinates for a local trivialization such that
\begin{equation}
g_{ij}(x,\xi^r)
\end{equation}
is a local embedding. The DeWitt metric is then expressed as
\begin{equation}
G_{rs}=G^{ij,kl}g_{ij,r}g_{kl,s},
\end{equation}
where a comma indicates partial differentiation.
In the new coordinate system the curves
\begin{equation}
t\ra g_{ij}(t,x)
\end{equation}
can be written in the form
\begin{equation}
t\ra \xi^r(t,x)
\end{equation}
and we infer
\begin{equation}
G^{ij,kl}\dot g_{ij}\dot g_{kl}=G_{rs}\dot\xi^r\dot\xi^s.
\end{equation}
Hence, we can express \re{5.4} as
\begin{equation}\lae{3.49}
J=\int_a^b\int_\varOmega \alpha_n^{-1}\{\tfrac14 G_{rs}\dot\xi^r\dot\xi^sw^{-1}\varphi+(R-2\varLambda)w\varphi\},
\end{equation}
where we now refrain from writing down the density $\sqrt\rho$ explicitly, since it does not depend on $(g_{ij})$ and therefore should not be part of the Legendre transformation.  Here we follow Mackey's advice in \cite[p. 94]{mackey:book} to always consider rectangular coordinates when applying canonical quantization, which can be rephrased that the Hamiltonian has to be a coordinate invariant, hence no densities are allowed.

Denoting the Lagrangian \textit{function} in \re{3.49} by $L$, we define
\begin{equation}
\pi_r= \pde L{\dot\xi^r}=\varphi G_{rs}\frac1{2\alpha_N}\dot\xi^sw^{-1}
\end{equation}
and we obtain for the Hamiltonian function $\hat H_G$
\begin{equation}\lae{5.18}
\begin{aligned}
\hat H_G&=\dot\xi^r\pde L{\dot\xi^r}-L\\
&=\varphi G_{rs}\big(\frac1{2\alpha_N}\dot\xi^rw^{-1}\big)\big(\frac1{2\alpha_N}\dot\xi^sw^{-1}\big) w\alpha_N-\alpha_N^{-1}(R-2\varLambda)\varphi w\\
&=\varphi^{-1}G^{rs}\pi_r\pi_s w\alpha_N-\alpha^{-1}_N(R-2\varLambda)\varphi w\\
&\equiv H_G w,
\end{aligned}
\end{equation}
where $G^{rs}$ is the inverse metric. Hence,
\begin{equation}
H_G=\alpha_N\varphi^{-1}G^{rs}\pi_r\pi_s-\alpha_N^{-1}(R-2\varLambda)\varphi
\end{equation}
is the Hamiltonian that will enter the Hamilton constraint, for details see \cite[Chapter 1.4]{cg:qgravity-book}.

Let us recall that the fibers $F$ can be considered to be Lorentzian manifolds, even globally hyperbolic manifolds, equipped with the DeWitt metric $(\varphi G^{ij,kl})$, where $\varphi$ is a time function, cf.\@\xspace \cite[Theorem 1.4.2]{cg:qgravity-book}. In the fibers we can introduce new coordinates, $(\xi^a)=(\xi^0,\xi^A)\equiv (t,\xi^A)$ , $0\le a\le m$, and $1\le A\le m$, such that
\begin{equation}\lae{4.22}
t=\sqrt\varphi
\end{equation}
and $(\xi^A)$ are coordinates for the hypersurface
\begin{equation}
M=\{\varphi=1\}=\{\xi^0=1\}.
\end{equation}
The Lorentzian metric in the fibers can then be expressed in the form
\begin{equation}\lae{4.24}
ds^2=-\frac{16(n-1)}n dt^2+t^2 G_{AB}d\xi^Ad\xi^B,
\end{equation}
where $(G_{AB})$ is a Riemannian metric on $M$ which is independent of $t$. When we work in a local trivialization of the bundle $E$ the coordinates $(\xi^A)$ are independent of $x$. The time coordinate $t$ is also independent of $x$, cf.\@\xspace \cite[Lemma 1.8]{cg:qgravity}. Moreover, the fiber elements $(g_{ij})$ can be expressed in the form
\begin{equation}\lae{4.25}
g_{ij}=t^\frac4n \sigma_{ij},
\end{equation}
where $(\sigma_{ij})$ is an element of $M$, i.e.,
\begin{equation}
t(\sigma_{ij})=1,
\end{equation}
or equivalently,
\begin{equation}
\det{\sigma_{ij}}=\det{\rho_{ij}}.
\end{equation}

Next, let us look at the Yang-Mills Lagrangian which can be expressed as
\begin{equation}\lae{3.39}
L_{YM}=\tfrac12\gamma_{\bar a\bar b}g^{ij}\tilde A^{\bar a}_{i,0}\tilde A^{\bar b}_{j,0}w^{-1}\varphi-\tfrac14 F_{ij}F^{ij}w\varphi.
\end{equation}
Let $E_0$ be the adjoint bundle
\begin{equation}
E_0=(S_0,\mf g,\pi,\Ad(\protect\mathcal G))
\end{equation}
with base space ${\protect\mathcal S_0}$, where the gauge transformations only depend on the spatial variables $x=(x^i)$. Then the mappings $ t\ra\tilde A^{\bar a}_i(t,\cdot)$ can be looked at as curves in $T^{1,0}(E_0)\otimes T^{0,1}({\protect\mathcal S_0})$, where
the fibers of $T^{1,0}(E_0)\otimes T^{0.1}({\protect\mathcal S_0})$ are the tensor products
\begin{equation}
\mf g\otimes T^{0,1}_x({\protect\mathcal S_0}),\qquad x\in {\protect\mathcal S_0},
\end{equation}
which are vector spaces equipped with metric
\begin{equation}
\gamma_{\bar a\bar b}\otimes g^{ij}.
\end{equation}
For our purposes it is more convenient to consider the fibers to be Riemannian manifolds endowed with the above metric. Let $(\zeta^p)$, $1\le p\le n_1n$, where $n_0=\dim\mf g$, be  local coordinates and
\begin{equation}
(\zeta^p)\ra \tilde A^{\bar a}_i(\zeta^p)\equiv \tilde A(\zeta)
\end{equation}
be a local embedding, then the metric has the coefficients
\begin{equation}\lae{4.33}
G_{pq}=\spd{\tilde A_p}{\tilde A_q}=\gamma_{\bar a\bar b}g^{ij}\tilde A^{\bar a}_{i,p}\tilde A^{\bar b}_{j,q}.
\end{equation}
Hence, the Lagrangian $L_{YM}$ in \re{3.39} can be expressed in the form
\begin{equation}
L_{YM}=\tfrac12G_{pq}\dot\zeta^p\dot\zeta^qw^{-1}\varphi-\tfrac14F_{ij}F^{ij}w\varphi
\end{equation}
and we deduce
\begin{equation}
\tilde\pi_p=\pde{L_{YM}}{\dot\zeta^p}=G_{pq}\dot\zeta^qw^{-1}\varphi
\end{equation}
yielding the Hamilton function
\begin{equation}
\begin{aligned}
\hat H_{YM}&=\pi_p\dot\zeta^p-L_{YM}\\
&=\tfrac12 G_{pq}(\dot\zeta^pw^{-1}\varphi)(\dot\zeta^qw^{-1}\varphi)w\varphi^{-1}+\tfrac14F_{ij}F^{ij}w\varphi\\
&=\tfrac 12G^{pq}\tilde\pi_p\tilde\pi_qw\varphi^{-1}+\tfrac14F_{ij}F^{ij}w\varphi\\
&\equiv H_{YM}w.
\end{aligned}
\end{equation}
Thus, after introducing a normal Gaussian coordinate system such that $w=1$,  the Hamiltonian that will enter the Hamilton constraint equation is
\begin{equation}
H_{YM}=\tfrac 12 \varphi^{-1}{G}^{pq}\tilde\pi_p\tilde\pi_q+\tfrac14F_{ij}F^{ij}\varphi.
\end{equation}

Combining, now, \re{4.22}, \re{4.25} and \re{4.33}  we infer that the Yang-Mills Hamiltonian can be expressed as
\begin{equation}\lae{3.51}
H_{YM}=t^{\frac4 n-2}\tilde G^{pq}\tilde\pi_p\tilde\pi_q+\tfrac14F_{ij}F^{ij} t^{2-\frac8n},
\end{equation}
where the indices in the last term are raised with respect to the metric $\sigma_{ij}$, i.e.,
\begin{equation}
F^{ij}=\sigma^{ik}\sigma^{jl} F_{kl}.
\end{equation}
In ase $n=3$ the exponents of $t$ in \re{3.51} are equal
\begin{equation}
\frac43-2=2-\frac83=-\frac23
\end{equation}
and we can write
\begin{equation}
\begin{aligned}
H_{YM}&=t^{-\frac23}\{\tilde G^{pq}\tilde\pi_p\tilde\pi_q+\tfrac14F_{ij}F^{ij}\} \\
&\equiv t^{-\frac23}\tilde H_{YM}.
\end{aligned}
\end{equation}
Moreover, if $(\sigma_{ij})$ as well as $(\rho_{ij})$ would be equal to the Euclidean metric $(\delta _{ij})$, then the quantization of $\tilde H_{YM}$ would be achieved by known methods of QFT.

Hence, we shall try to express the Hamiltonians of the other physical forces like the Dirac and Higgs Hamiltonians, when evaluated for
\begin{equation}
\sigma_{ij}=\rho_{ij}=\delta _{ij}
\end{equation}
and in case $n=3$ in the form
\begin{equation}\lae{4.43}
H_D=t^{-\frac23}\tilde H_D
\end{equation}
resp.\@\xspace
\begin{equation}\lae{4.44}
H_{H}=t^{-\frac23}\tilde H_{H}
\end{equation}
such the quantization of the spatial Hamiltonian
\begin{equation}
\tilde H_{YM}+\tilde H_D+\tilde H_{H}
\end{equation}
would be well known, and in the end, all spatial Hamiltonians of the Standard Model could be incorporated.

Let us first consider the Dirac Hamiltonian. In the Dirac Lagrangian $L_D$, defined in equation \fre{3.46}, the volume density $\sqrt g$ is missing, i.e., in order to define the Hamiltonian we have to multiply the Lagrangian with $\sqrt g$ or, since we would like to work with functions instead of densities, we have to multiply the Lagrangian with $\varphi$.

In addition we shall also consider---at least locally---a normal Gaussian coordinate system such that $w=1$. Then, the final Dirac Lagrangian has the form
\begin{equation}\lae{4.46}
\begin{aligned}
L_D&=\tfrac{i}2(\bar\chi_I\dot\chi^I-\dot{\bar\chi}^{I}\chi_{I})+m i\bar\chi_I\gamma^0\chi^I\\
&\quad-\tfrac{i}2\{\bar\chi_I\gamma^0E^k_{a'}\gamma^{a'}\tilde D_k\chi^I-\overline{\bar\chi_I\gamma^0E^k_{a'}\gamma^{a'}\tilde D_k\chi^I}\},
\end{aligned}
\end{equation}

The spinorial variables  $\chi^I_A$ are anticommuting Grassmann variables. They  are elements of a  Grassmann algebra with involution, where the involution corresponds to the complex conjugation and will be denoted by a bar.

The $\chi^{I}_A$ are complex variables and we define its real resp.\@\xspace imaginary parts as
\begin{equation}
\xi^{I}_A=\tfrac1{\sqrt 2}(\chi^{I}_A+\bar\chi^{I}_A)
\end{equation}
resp.\@\xspace
\begin{equation}
\eta^{I}_A=\tfrac1{\sqrt 2 i}(\chi^{I}_A-\bar\chi^{I}_A).
\end{equation}
Then,
\begin{equation}\lae{4.6.1}
\chi^{I}_A=\tfrac1{\sqrt 2}(\xi^{I}_A+i\eta^{I}_A)
\end{equation}
and
\begin{equation}\lae{4.7.1}
\bar\chi^{I}_A=\tfrac1{\sqrt 2}(\xi^{I}_A-i\eta^{I}_A).
\end{equation}

With these definitions we obtain
\begin{equation}
\frac{i}2(\bar\chi_{I}\dot\chi^{I}-\bar{\dot\chi}^{I}\chi_{I})=\frac{i}2(\xi^A_{I}\dot\xi^{I}_A+\eta^A_{I}\dot\eta^{I}_A).
\end{equation}

Casalbuoni  quantized a Bose-Fermi system in \cite[section 4]{casalbuoni:fermi} the results of which can be applied to spin $\frac12$ fermions. The Lagrangian in \cite{casalbuoni:fermi} is the same as the main part  our Lagrangian in \fre{4.46}, and the left derivative is used in that paper, hence we are using left derivatives as well such that  the conjugate momenta of the odd variables are, e.g.,
\begin{equation}
\pi^A_{I}=\frac{\pa L}{\pa \dot\xi^{I}_A}=-\frac i2\xi^A_{I},
\end{equation}
and thus the conclusions in \cite{casalbuoni:fermi} can be applied.

The Lagrangian has been expressed in real variables---at least the important part of it---and it follows that the odd variables $\xi^{I}_A,\eta^{I}_A$ satisfy, after introducing anticommutative Dirac brackets as in \cite[equ. (4.11)]{casalbuoni:fermi},
\begin{equation}
\{\xi_{I}^A,\xi^{J}_B\}^*_+=-i\delta ^{J}_I\delta _{B}^A,
\end{equation}
\begin{equation}
\{\eta_{I}^A,\eta^{J}_B\}^*_+=-i\delta ^{J}_I\delta _{B}^A,
\end{equation}
and
\begin{equation}
\{\xi_{I}^A,\eta^{J}_B\}^*_+=0,
\end{equation}
cf.\@\xspace \cite[equ. (4.19)]{casalbuoni:fermi}.

In view of \re{4.6.1}, \re{4.7.1} we then derive
\begin{equation}
\{\bar\chi_{I}^A,\chi^{J}_B\}^*_+=-i\delta ^{J}_I\delta _{B}^A,
\end{equation}
where $\bar\chi_{I}^A$ are the conjugate momenta.

Canonical quantization---with $\bar h=1$---then requires that the corresponding operators $\hat\chi^{I}_A, \hat{\bar\chi}_{J}^B$ satisfy the anticommutative rules
\begin{equation}\lae{4.15.1}
[\hat\chi^{I}_A,\hat{\bar\chi}_{J}^B]_+=i\{\chi^{I}_A,\bar\chi_{J}^B\}^*_+=\delta ^{I}_J\delta _{A}^B
\end{equation}
and
\begin{equation}
[\hat{\bar\chi}_{I}^A,\hat{\bar\chi}_{J}^B]_+=[\hat\chi^{I}_A,\hat{\chi}^{J}_B]_+=0,
\end{equation}
cf.\@\xspace \cite[equ. (3.10)]{casalbuoni:odd} and \cite[equ. (5.17)]{casalbuoni:fermi}.

From \re{4.46} we then deduce that the spinorial Hamilton function is equal to
\begin{equation}\lae{4.65}
\begin{aligned}
 H_D&=
\tfrac{i}2\{\bar\chi_I\gamma^0E^k_{a'}\gamma^{a'}\tilde D_k\chi^I-\overline{\bar\chi_I\gamma^0E^k_{a'}\gamma^{a'}\tilde D_k\chi^I}\}\\
&\quad-m i\bar\chi_I\gamma^0\chi^I.
\end{aligned}
\end{equation}
When we try to quantize this Hamilton function then the vielbein $e^{a'}_k$ and its inverse $E^k_{a'}$ will correspond to a given element  $g_{ij}(x)$ in the fiber $F$ which can be expressed as in \re{4.25} and we deduce that the vielbein
\begin{equation}
\tilde e^{a'}_k=t^{-\frac2n}e^{a'}_k
\end{equation}
and its inverse
\begin{equation}
\tilde E^k_{a'}=t^{\frac2n}E^k_{a'}
\end{equation}
correspond to the metric $\sigma_{ij}$. Furthermore, the covariant derivative $\tilde D_k\chi^I$ is independent of $t$, in view of \re{3.44} and \fre{3.45}. Thus, the Hamilton function $H_D$ can expressed as
\begin{equation}\lae{4.65.1}
\begin{aligned}
 H_D&=
t^{-\frac23}\big (\tfrac{i}2\{\bar\chi_I\gamma^0\tilde E^k_{a'}\gamma^{a'}\tilde D_k\chi^I-\overline{\bar\chi_I\gamma^0\tilde E^k_{a'}\gamma^{a'}\tilde D_k\chi^I}\}\big)\\
&\quad-m i\bar\chi_I\gamma^0\chi^I,
\end{aligned}
\end{equation}
i.e., the main  part has already the form that we looked for in \re{4.43}, provided $n=3$, only the mass term spoils the necessary configuration. To overcome this setback we either have to omit the mass term or to modify it by multiplying the mass term  in \fre{1.1} with the factor
\begin{equation}
\varphi^{-\frac1n},
\end{equation}
where $\varphi$ is defined in \fre{3.34.1}. Note that $\varphi=1$ if
\begin{equation}
g_{ij}=\rho_{ij}=\delta _{ij}
\end{equation}
as is the case in QFT. Anyway, either by omitting or by modifying the mass term the Dirac Hamilton function can be expressed in the required form\ann{$t^{-\frac23}$}
\begin{equation}
H_D=t^{-\frac23}\tilde H_D,
\end{equation}
where the underlying Riemannian metric is $\sigma_{ij}$ provided $n=3$.

The remaining Hamiltonian is the Hamiltonian of the Higgs field. The Higgs Lagrangian is defined by
\begin{equation}\lae{4.3.1}
L_H=-\tfrac12\bar g^{\alpha\beta}\gamma_{ab}\varPhi^a_\alpha\varPhi^b_\beta-V(\varPhi),
\end{equation}
where $V$ is a smooth potential. We assume that in a local coordinate system $\varPhi$ has real coefficients. The covariant derivatives of $\varPhi$ are defined by a connection $A=(A^a_\mu)$ in $E_1$
\begin{equation}
\varPhi^a_\mu=\varPhi^a_{,\mu}+f^a_{cb}A^c_\mu\varPhi^b.
\end{equation}
As in the preceding section we work in a local trivialization of $E_1$ using the temporal gauge, i.e.,
\begin{equation}
A^a_0=0,
\end{equation}
hence, we conclude
\begin{equation}
\varPhi^a_0=\varPhi^a_{,0}.
\end{equation}

Expressing the density $g$ as in \re{4.2} we obtain Lagrangian
\begin{equation}
L_H=\tfrac12 \gamma_{ab}\varPhi^a_{,0}\varPhi^b_{,0}\varphi-\tfrac12g^{ij}\gamma_{ab}\varPhi^a_i\varPhi^b_j\varphi-V(\varPhi)\varphi,
\end{equation}
where, again, we used local coordinates such that $w=1$. In order to apply our approach, outlined in \re{4.44}, we have to modify the Lagrangian. Instead of the above Lagrangian we have to consider
\begin{equation}
L_{H mod}=\{\tfrac12 \gamma_{ab}\varPhi^a_{,0}\varPhi^b_{,0}-\tfrac12g^{ij}\gamma_{ab}\varPhi^a_i\varPhi^b_j\}\varphi^{1+\gamma_1}-V(\varPhi)\varphi^{1+\gamma_2}.
\end{equation}
Let us define
\begin{equation}
p_a=\pde{L_H}{\dot\varPhi^a},\qquad\dot\varPhi^a=\varPhi^a_{,0},
\end{equation}
then we obtain the Hamilton function
\begin{equation}
\begin{aligned}
 H_{H mod}&=p_a\dot\varPhi^a-L_H\\
&=\tfrac12\gamma^{ab}p_ap_b\varphi^{-(1+\gamma_1)}+\tfrac12 g^{ij}\gamma_{ab}\varPhi^a_i\varPhi^b_j\varphi^{1+\gamma_1}+V(\varPhi)\varphi^{1+\gamma_2}.
\end{aligned}
\end{equation}
After quantization the $g_{ij}$ are elements of the fiber $F$, i.e.,
\begin{equation}
g_{ij}=t^\frac4n\sigma_{ij}.
\end{equation}
If $n=3$, then $\gamma_1$ has to be chosen such that
\begin{equation}
-2(1+\gamma_1)=-\frac43+2(1+\gamma_1)=-\frac23
\end{equation}
which is the case if
\begin{equation}
\gamma_1=-\frac23.
\end{equation}
For $\gamma_2$ we obtain
\begin{equation}
2(1+\gamma_2)=-\frac23
\end{equation}
yielding
\begin{equation}
\gamma_2=-\frac43.
\end{equation}

Thus, the Hamilton function of the modified Higgs field has the required form
\begin{equation}\lae{4.10}
H_{H mod}=t^{-\frac23}\tilde H_{H mod},
\end{equation}
where
\begin{equation}
\tilde H_{Hmod}=\tfrac12\gamma^{ab}p_ap_b+\tfrac12 \sigma^{ij}\gamma_{ab}\varPhi^a_i\varPhi^b_j+V(\varPhi)
\end{equation}
is a standard Hamiltonian of a Higgs field in QFT by choosing $\sigma_{ij}=\delta _{ij}$ and $\varPhi$,  $V(\varPhi)$ as well as the Yang-Mills connection appropriately.

Combining the four Hamilton functions in \re{5.18}, \re{3.51}, \re{4.10} and \re{4.65.1} the Hamilton constraint has the form
\begin{equation}\lae{4.81}
\begin{aligned}
H&=H_G+H_{YM}+H_H+H_D\\
&=H_G+t^{-\frac23}(\tilde H_{YM}+\tilde H_H+\tilde H_D)\\
&\equiv H_G+t^{-\frac23} \tilde H_{SM}=0,
\end{aligned}
\end{equation}
where we omitted the subscript $mod$ and where $SM$ refers to the fields  of the Standard Model or to a  corresponding subset of fields.

The Hamiltonian
\begin{equation}
H_G=\alpha_N\varphi^{-1}G^{rs}\pi_r\pi_s-\alpha_N^{-1}(R-2\varLambda)\varphi
\end{equation}
we quantize as in our former papers \cite{cg:qgravity} and \cite{cg:qgravity2b} to obtain
\begin{equation}\lae{4.83}
H_G=-\alpha_N \varDelta-\alpha_N^{-1}R t^2+ 2\alpha_N^{-1}\varLambda t^2,
\end{equation}
where the Laplacian is the Laplacian of the metric \re{4.24} acting\ann{action} in the fibers $F$ of $E$. The Laplacian acts on smooth functions $u$ of the form $u=u(g_{ij})$. Choosing the Gaussian  coordinate system $(\xi^a)=(t,\xi^A)$ such that the fiber metric has form as in \re{4.24}, then, the hyperbolic term $-\varDelta u$ can be expressed as
\begin{equation}
-\varDelta u=\frac n{16(n-1)}t^{-m}\pde{}{t}(t^m \pde ut)-t^{-2}\bar\varDelta u,
\end{equation}
where $\bar\varDelta$ is the Laplacian of the hypersurface
\begin{equation}
M=\{t=1\}.
\end{equation}
Using separation of variables by considering functions $u$ which are products
\begin{equation}\lae{3.57}
u(t,\xi^A)=w(t)  v(\xi^A),
\end{equation}
where $v$ is a spatial eigenfunction, or eigendistribution, of the Laplacian $\bar\varDelta$
\begin{equation}
-\bar\varDelta v=\lambda v.
\end{equation}

The hypersurface
\begin{equation}
M=\{\varphi=1\}
\end{equation}
can be considered to be a subbundle of $E$, where each fiber $M(x)$ is a hypersurface in the fiber $F(x)$ of $E$. We shall use the same notation $M$ for the  subbundle as well as for the hypersurface and in general we shall omit the reference to the base point $x\in {\protect\mathcal S_0}$. Furthermore, we specify the metric $\rho_{ij}\in T^{0,2}({\protect\mathcal S_0})$, which we used to define $\varphi$, to be equal to the Euclidean metric such that in Euclidean coordinates
\begin{equation}
\varphi^2=\frac{\det g_{ij}}{\det \delta _{ij}}=\det g_{ij}.
\end{equation}
Then, it is well-known that each $M(x)$ with the induced metric $(G_{AB})$ is a symmetric space, namely, it is isometric to the coset space
\begin{equation}\lae{4.3}
G/K=SL(n,\R[])/SO(n),
\end{equation}
cf.\@\xspace \cite[equ.(5.17), p. 1123]{dewitt:gravity} and \cite[p. 3]{jorgenson:book}. The eigenfunctions in symmetric spaces, and especially of the coset space in \re{4.3}, are well-known, they are the so-called \textit{spherical functions}. One can also define a Fourier transformation for functions in $L^2(G/K)$ and prove a Plancherel formula, similar to the Euclidean case, cf.\@\xspace \cite[Chapter III]{helgason:ga}. Also similar to the Euclidean case we shall use the Fourier kernel to define the eigenfunctions, or eigendistributions, cf.\@\xspace \cite[Section 5]{cg:qgravity3}.

Let
\begin{equation}\lae{4.9}
G=NAK
\end{equation}
be an Iwasawa decomposition of $G$, where $N$ is the subgroup of unit upper triangle matrices, $A$ the abelian subgroup of diagonal matrices with strictly positive diagonal components and $K=SO(n)$. The corresponding Lie algebras are denoted by
\begin{equation}
\mf g, \mf n,\mf a\tup{\, and\, }\mf k.
\end{equation}
Here,
\begin{equation}
\begin{aligned}
\mf g&= \tup{real matrices with zero trace}\\
\mf n&=\tup{subspace of strictly upper triangle matrices with zero diagonal}\\
\mf a&=\tup{subspace of diagonal matrices with zero trace}\\
\mf k&=\tup{subspace of skew-symmetric matrices}.
\end{aligned}
\end{equation}
The Iwasawa decomposition is unique. When
\begin{equation}
g=nak
\end{equation}
we define the maps $n, A, k$ by
\begin{equation}\lae{4.13}
g=n(g)A(g)k(g).
\end{equation}
We also use the expression $\log A(g)$, where $\log$ is the matrix logarithm. In case of diagonal matrices
\begin{equation}
a=\diag(a_1,\dots, a_n)
\end{equation}
with positive entries
\begin{equation}
\log a=\diag(\log a_i),
\end{equation}
hence
\begin{equation}
A(g)=e^{\log A(g)}.
\end{equation}
\begin{rem}\lar{4.1}
(i) The Lie algebra $\mf a$ is a (n-1)-dimensional real algebra, which, as a vector space, is equipped with a natural real, symmetric scalar product, namely, the trace form
\begin{equation}
\spd{H_1}{H_2}=\tr (H_1H_2),\qquad H_i\in\mf a.
\end{equation}

\cvm
(ii) Let $\mf a^*$ be the dual space of $\mf a$. Its elements will be denoted by Greek symbols, some of which have a special meaning in the literature. The linear forms are also called \textit{additive characters}.

\cvm
(iii) Let $\lambda\in \mf a^*$, then there exists a unique matrix $H_\lambda\in\mf a$ such that
\begin{equation}
\lambda(H)=\spd{H_\lambda}H\qquad\forall\, H\in \mf a.
\end{equation}
This definition allows to define a dual trace form in $\mf a^*$ by setting for $\lambda,\mu\in \mf a^*$
\begin{equation}\lae{4.25.1}
\spd \lambda \mu=\spd{H_\lambda}{H_\mu}.
\end{equation}

\cvm
(iv) The Lie algebra $\mf g$ is a direct sum
\begin{equation}
\mf g=\mf n+\mf a+\mf k.
\end{equation}
Let $E_{ij}$, $1\le i<j\le n$, be the matrices with component $1$ in the entry $(i,j)$ and other components zero, then these matrices form a basis of $\mf n$. For $H\in \mf a$, $H=\diag (x_i)$, the Lie bracket in $\mf g$, which is simply the commutator, applied to $H$ and $E_{ij}$ yields
\begin{equation}
[H,E_{ij}]=(x_i-x_j)E_{ij}\quad\forall\, H\in\mf a.
\end{equation}
Hence, the $E_{ij}$ are the eigenvectors for the characters $\alpha_{ij}\in \mf a^*$ defined by
\begin{equation}\lae{4.104}
\alpha_{ij}(H)=x_i-x_j.
\end{equation}
Here, $E_{ij}$ is said to be an eigenvector of $\alpha_{ij}$, if
\begin{equation}
[H,E_{ij}]=\alpha_{ij}(H) E_{ij}\qquad\forall\,H\in \mf a.
\end{equation}
The eigenspace of $\alpha_{ij}$ is one-dimensional. The characters $\alpha_{ij}$ are called the \textit{relevant} characters, or the $(\mf a,\mf n)$ characters. They are also called the positive restricted roots.
\end{rem}

The Fourier theory in $X=G/K$ which we have summarized in  \cite[Section 6]{cg:qgravity3} uses the functions
\begin{equation}\lae{4.102}
e_{\lambda,b}(x)=e^{(i\lambda+\rho)\log A(x,b)},\qquad(\lambda,b)\in \mf a^*\times B,\; x\in X,
\end{equation}
as the Fourier kernel, where
\begin{equation}
B=K/M.
\end{equation}
 Here, $M$ is the centralizer of $A$ in $K$ and $\rho$ a special character with norm
\begin{equation}
\spd\rho\rho=\frac1{12}(n-1)^2n,
\end{equation}
cf.\@\xspace \cite[Lemma 1]{cg:qgravity3}. If $n=3$ then
\begin{equation}\lae{4.109.1}
\abs\rho^2=1.
\end{equation}
For a precise definition of $A(x,b)\in A$ we refer to \cite[p.19]{cg:qgravity3}, where also references to the corresponding mathematical literature are given, especially to Helgason's book \cite[Chapter III]{helgason:ga}.

The Fourier transform for functions $f\in C^\infty_c(X,{\protect\mathbb C})$ is then defined by
\begin{equation}\lae{4.62}
\hat f(\lambda,b)=\int_Xf(x)e^{(-i\lambda +\rho)\log A(x,b)}dx
\end{equation}
for $\lambda\in\mf a^*$ and $b\in B$, or, if we use the definition in \re{4.102}
\begin{equation}
e_{\lambda,b}(x)=e^{(i\lambda+\rho)\log A(x,b)},
\end{equation}
by
\begin{equation}
\hat f(\lambda,b)=\int_Xf(x)\overline e_{\lambda,b}(x)dx.
\end{equation}
The functions $e_{\lambda,b}$ are real analytic in $x$ and are eigenfunctions of the Laplacian, cf.\@\xspace \cite[Prop. 3.14, p. 99]{helgason:ga},
\begin{equation}\lae{4.109}
-\tilde\varDelta e_{\lambda,b}=(\abs \lambda^2+\abs\rho^2)e_{\lambda,b},
\end{equation}
where
\begin{equation}
\abs \lambda^2=\spd \lambda\lam,
\end{equation}
 cf.\@\xspace \re{4.25.1}, and similarly for $\abs\rho^2$. We also denote the Fourier transform by $\protect\mathcal F$ such that
 \begin{equation}
\protect\mathcal F(f)=\hat f.
\end{equation}
Its inverse $\protect\mathcal F^{-1}$ is defined in $R(\protect\mathcal F)$ by
\begin{equation}
f(x)=\frac 1{\abs W}\int_B\int_{\mf a^*}\hat f(\lambda,b)\abs{\mf c(\lambda)}^{-2}d\lambda db,
\end{equation}
where $\mf c(\lambda)$ is Harish-Chandra's  $\mf c$-function, $W$ is the Weyl group and
\begin{equation}
\abs W=\card W,
\end{equation}
the number of elements in $W$, in our case $\abs W=n!$.

In equation \re{4.109} we identified
\begin{equation}
\tilde\varDelta=\varDelta_M=\varDelta_X.
\end{equation}
In \cite{cg:qgravity3} we finally dropped  the embellishment and simply wrote $\varDelta$ when referring to the above Laplacian but at the moment we refrain from doing so in order to avoid confusion.

We shall consider the eigenfunctions $e_{\lambda,b}$ as tempered distributions of the Schwartz space $\protect\mathscr S(X)$ and shall use their Fourier transforms
\begin{equation}
\hat e_{\lambda,b}=\delta _{(\lambda,b)}=\delta _\lambda \otimes \delta _b
\end{equation}
as the spatial eigenfunctions of
\begin{equation}
\protect\mathcal F(-\varDelta)=m(\mu)=(\abs \mu^2+\abs\rho^2),
\end{equation}
which is a multiplication operator, such that
\begin{equation}
\protect\mathcal F(-\varDelta)\hat e_{\lambda,b}=m(\mu)\hat e_{\lambda,b}=(\abs \lambda^2+\abs\rho^2)\hat e_{\lambda,b},
\end{equation}
cf.\@\xspace \cite[Section 6]{cg:qgravity3} for details.

Looking at the Fourier transformed eigenfunctions
\begin{equation}
\hat e _{\lambda,b}=\delta _\lambda\otimes \delta _b
\end{equation}
it is obvious that the dependence on $b$ has to be eliminated, since there is neither a physical nor a mathematical motivation to distinguish between $e_{\lambda,b}$ and $e_{\lambda,b'}$. We discarded the integration over $B$ in \cite[Section 6]{cg:qgravity3} and picked instead a special element $b_0\in B$, namely,
\begin{equation}\lae{6.63}
b_0=eM,\qquad e=\id\in K,
\end{equation}
and only consider the eigenfunctions $e_{\lambda,b_0}$ with corresponding Fourier transforms
\begin{equation}
\delta _{\lambda}\equiv \delta _\lambda\otimes \delta _{b_0}=\hat e_{\lambda,b_0}, \qquad \lambda\in \mf a^*.
\end{equation}
For a justification, see \cite[Lemma 4]{cg:qgravity3} and the arguments preceding the referenced Lemma.

The eigenfunctions $e_{\lambda,b_0}$ depend on the characters $\lambda\in \mf a^*$, but not all characters are physically relevant. For a definition of the physically relevant characters let me cite \cite[Remark 2, p. 18]{cg:qgravity3}:
\begin{rem}\lar{4.3}
The  characters $\alpha_{ij}$, $1\le i<j\le n$,  in \re{4.102} will represent the \textit{elementary gravitons} stemming from the degrees of freedom in choosing the coordinates
\begin{equation}
g_{ij},\qquad 1\le i<j\le n,
\end{equation}
of a metric tensor. The diagonal elements offer in general additional $n$ degrees of freedom, but in our case, where we  consider metrics satisfying
\begin{equation}
\det g_{ij}=1,
\end{equation}
only $(n-1)$  diagonal components can be freely chosen, and we shall choose the first $(n-1)$  entries, namely,
\begin{equation}
g_{ii},\qq1\le i\le n-1.
\end{equation}
The corresponding additive characters are named $\alpha_i, 1\le i\le n-1$, and are defined by
\begin{equation}
\alpha_i(H)=h_i,
\end{equation}
if
\begin{equation}
H=\diag(h_1,\ldots, h_n).
\end{equation}
The characters $\alpha_i$, $1\le i\le n-1$, and $\alpha_{ij}$ $1\le i<j\le n$, will represent the $\frac{(n+2)(n-1)}2$ \textit{elementary gravitons} at the character level. We shall normalize the characters by defining
\begin{equation}\lae{4.46.1}
\tilde\alpha_i=\norm{H_{\alpha_i}}^{-1}\alpha_i
\end{equation}
and
\begin{equation}\lae{4.47}
\tilde\alpha_{ij}=\norm{H_{\alpha_{ij}}}^{-1}\alpha_{ij}
\end{equation}
such that the normalized characters have unit norm, cf.\@\xspace \re{4.25.1}.
\end{rem}
We can now define the corresponding forms in $\mf a^*$ with arbitrary energy levels:
\begin{definition}\lad{5.9}
Let $\lambda\in \R[]_+$ be arbitrary. Then we consider the characters
\begin{equation}
\lambda \tilde{\alpha}_i\quad \wedge \quad \lambda \tilde{\alpha}_{ij},
\end{equation}
where we recall that the terms embellished by a tilde refer to the corresponding unit vectors. Then the eigenfunctions representing the elementary gravitons are $e_{\lambda\tilde \alpha_i,b_0}$ and $e_{\lambda\tilde\alpha_{ij},b_0}$.

The corresponding eigenvalue with respect to $-\tilde\varDelta$ is $\abs\lambda^2+\abs\rho^2$, where by a slight abuse of notation $\abs \lambda^2 = \lambda^2$ and $\abs \rho^2=\spd\rho\rho$. Note that $\abs\rho^2=1$ if $n=3$, cf.\@\xspace \re{4.109.1}.

We would also like to define a \textit{zero-point energy} eigenfunction by choosing $\lambda\in \mf a^*=0$. The corresponding eigenfunction would be $e_{0,b_0}$ satisfying
\begin{equation}
-\tilde\varDelta e_{0,b_0}=\abs \rho^2 e_{0,b_0}=e_{0,b_0}.
\end{equation}
if $n=3$.
\end{definition}

We are now able to quantize the Hamiltonian $H$ in \re{4.81}. For brevity we denote the quantized Hamiltonians, which are operators, by using  the same symbols as for the Hamilton functions. The Hamilton operator $H_G$ we express as in \re{4.83}
\begin{equation}
\begin{aligned}
H_G u&=\ann{-}\alpha_N \frac n{16(n-1)}t^{-m}\pde{}{t}(t^m \pde wt)v-\alpha_N t^{-2}w\bar\varDelta v\\
&\quad -\alpha_N^{-1}t^{2-\frac4n}R(\sigma_{ij})wv + 2\alpha_N^{-1}\varLambda t^2 wv,
\end{aligned}
\end{equation}
where we used the separation of variables in \re{3.57}, the form of the metric in \re{4.25}, namely,
\begin{equation}
g_{ij}=t^{\frac4n} \sigma_{ij}
\end{equation}
and the relation  between the  scalar curvatures of conformal metrics
\begin{equation}
R(g)=t^{-\frac4n} R(\sigma).
\end{equation}.

Let us recall that for the quantization of $\tilde H_{SM}$ we shall specify $\sigma_{ij}=\delta _{ij}$, such that the spatial eigendistributions, or approximate eigendistributions, $\psi$ satisfying
\begin{equation}
\tilde H_{SM}\psi=\lambda_1\psi,\qquad \lambda_1\ge 0
\end{equation}
can be derived by applying standard methods of QFT. We then solve the Wheeler-DeWitt equation
\begin{equation}
Hu=0
\end{equation}
not for all $(t,\sigma_{ij})\in \R[+]\times M$ but only for $(t,\delta _{ij})$, where $t>0$ is arbitrary. Thus, we shall solve
\begin{equation}
-\tilde \varDelta v=(\abs \lambda^2+\abs\rho^2)v
\end{equation}
by using
\begin{equation}
v=e_{\lambda,b_0}
\end{equation}
for arbitrary $\sigma_{ij}\in M$, but we shall evaluate $e_{\lambda, b_0}$ only at $\sigma_{ij}=\delta _{ij}$. Furthermore, we observe that for $x=gK\in X$ and $b=kM\in B$ we have
\begin{equation}
A(x,b)=A(gK,kM)=A(k^{-1}g),
\end{equation}
cf.\@\xspace \cite[equ. (202), p. 18]{cg:qgravity3},  hence, if $b=b_0$, i.e., if $k=e=\id$ then
\begin{equation}
A(x,b_0)=A(g).
\end{equation}
Moreover, let
\begin{equation}
\pi: G/K \ra M
\end{equation}
be the isometry, then
\begin{equation}
\pi(gK)=gg^*,
\end{equation}
where $g^*$ is the adjoint. Thus, if $g=(\delta _{ij})=e$ we infer
\begin{equation}
\sigma_{ij}=\delta _{ij}\in M\im  e_{\lambda,b_0}(\sigma_{ij})=1,
\end{equation}
and we have proved:
\begin{thm}\lat{4.4}
Let $n=3$, $v=e_{\lambda,b_0}$ and let $\psi$ be an  eigendistribution of $\tilde H_{SM}$ when $\sigma_{ij}=\delta _{ij}$ such that
\begin{equation}
-\tilde \varDelta e_{\lambda,b_0}=(\abs\lambda^2+1)e_{\lambda,b_0},
\end{equation}
\begin{equation}
\tilde H_{SM}\psi=\lambda_1\psi,\qquad\lambda_1\ge 0,
\end{equation}
and let $w$ be a solution of the  ODE
\begin{equation}\lae{4.147}
\begin{aligned}
t^{-m}\pde{}{t}(t^m \pde wt)&+\frac{32}3 (\abs\lambda^2+1)t^{-2}w+\frac{32}3\alpha_N^{-1}\lambda_1 t^{-\frac23}w\\
& \quad+\frac{64}3\alpha_N^{-2}\varLambda t^2w=0
\end{aligned}
\end{equation}
then
\begin{equation}
u=w e_{\lambda,b_0}\psi
\end{equation}
is a solution of the Wheeler-DeWitt equation
\begin{equation}
Hu=0,
\end{equation}
where $e_{\lambda,b_0}$ is evaluated at $\sigma_{ij}=\delta _{ij}$ and where we note that $m=5$.
\end{thm}
We shall refer to $e_{\lambda,b_0}$ and $\psi$ as the spatial eigenfunctions and to $w$ as the temporal eigenfunction.

\begin{rem}
We could also apply the respective Fourier transforms to $-\tilde \varDelta e_{\lambda,b_0}$ resp.\@\xspace $\tilde H_{SM}\psi$ and consider
\begin{equation}
w\hat e_{\lambda,b_0}\hat\psi
\end{equation}
as the solution in Fourier space, where $\hat\psi$ would be expressed with the help of the ladder operators.
\end{rem}
In the next section we shall analyze the temporal eigenfunctions.

\section{Temporal eigenfunctions}\las{5}

The temporal eigenfunctions have to satisfy the ODE \re{4.147} or equivalently
\begin{equation}\lae{5.1}
\begin{aligned}
\Ddot w+5 t^{-1} \dot w&+\frac{32}3 (\abs\lambda^2+1)t^{-2}w+\frac{32}3\alpha_N^{-1}\lambda_1 t^{-\frac23}w\\
& \quad+\frac{64}3\alpha_N^{-2}\varLambda t^2w=0,
\end{aligned}
\end{equation}
where we used that $m=5$, since we assume $n=3$. Let us denote the other constants in front of the three lower order terms by $m_i$, $m_2^2$ resp.\@\xspace $m_3$, then the ODE looks like
\begin{equation}\lae{5.2}
\Ddot w+5 t^{-1} \dot w+m_1 t^{-2}w+m_2^2 t^{-\frac23}w+m_3 t^2w=0,
\end{equation}
where
\begin{equation}\lae{5.3}
m_1\ge\frac{32}3,\quad m_2\ge 0,\quad m_3\in\R[].
\end{equation}
The ODE \re{5.2} has two linearly independent solutions which are smooth and defined for all $t>0$. However, if $m_2$ as well as $m_3$ are both different from zero, then the solution cannot be expressed by known functions like variants of the Bessel functions. Only if this is not valid the solutions can be expressed by known functions.
\begin{thm}\lat{5.1}
Assume $m_3=0$ and $m_2>0$, then the solutions of the ODE \re{5.2} are generated by
\begin{equation}
J(\tfrac32\sqrt{m_1-4}\,i,\tfrac32 m_2 t^\frac23) t^{-2}
\end{equation}
and
\begin{equation}
J(-\tfrac32\sqrt{m_1-4}\,i,\tfrac32 m_2 t^\frac23) t^{-2},
\end{equation}
where $J(\lambda,t)$ is the Bessel function of the first kind.
\end{thm}
\begin{proof}
We used Mathematica to obtain these solutions. The verification that these functions are indeed solutions is straightforward.\ann{straight forward}
\end{proof}
\begin{lem}\lal{5.2}
The solutions in the theorem above diverge to complex infinity if $t$ tends to zero and they converge to zero if $t$ tends to infinity.
\end{lem}
\begin{proof}
The results can be derived by looking at a series expansion of the corresponding Bessel functions near the origin resp.\@\xspace near infinity.
\end{proof}
Next, let us consider the solutions when $m_2=0$ and $m_3\not=0$. Then we distinguish two cases $m_3>0$ resp.\@\xspace $m_3<0$. For a better distinction we shall express $m_3$ in the form
\begin{equation}
m_3=m_4^2,\quad m_4>0,
\end{equation}
in the first case and as
\begin{equation}
m_3=-m_4^2, \quad m_4>0,
\end{equation}
in the second case.
\begin{thm}\lat{5.3}
Assume $m_2=0$ and $m_3>0$, then the solutions of the ODE \re{5.2} are generated by the functions
\begin{equation}
J(\tfrac12\sqrt{m_1-4}\,i,\tfrac12 m_4 t^2) t^{-2}
\end{equation}
and
\begin{equation}
J(-\tfrac12\sqrt{m_1-4}\,i,\tfrac12 m_4 t^2) t^{-2},
\end{equation}
where $J(\lambda,t)$ is the Bessel function of the first kind.
\end{thm}
Similarly we obtain in the second case:
\begin{thm}\lat{5.4}
Assume $m_2=0$ and $m_3<0$, then the solutions of the ODE \re{5.2} are generated by the functions
\begin{equation}
I(\tfrac12\sqrt{m_1-4}\,i,\tfrac12 m_4 t^2) t^{-2}
\end{equation}
and
\begin{equation}
I(-\tfrac12\sqrt{m_1-4}\,i,\tfrac12 m_4 t^2) t^{-2},
\end{equation}
where $I(\lambda,t)$ is the modified Bessel function of the first kind. In Mathematica this function is denoted by \tup{BesselI[$\lambda,t]$}.
\end{thm}
The arguments in the proof of \rt{5.1} also apply in case of \rt{5.3} and \rt{5.4}.
\begin{lem}\lal{5.5}
The  solutions in \rt{5.3} resp.\@\xspace \rt{5.4} diverge to complex infinity if $t$ tends to zero as well as if $t$ tends to infinity.
\end{lem}
\begin{proof}
Same arguments as in the proof of \rl{5.2} apply.
\end{proof}

\section{Conclusions}\las{6}

The temporal eigenfunctions in the theorems of the previous section all become unbounded if $t\ra 0$, which can be described as a big bang on a quantum level. Furthermore, if we consider $t<0$, then the functions
\begin{equation}
\tilde w(t)=w(-t),\qquad t<0,
\end{equation}
also satisfy the ODE \re{5.1} for $t<0$, if we replace $t^{-\frac23}$ by $\abs t^{-\frac23}$,  i.e., they are also temporal eigenfunctions  if the light cone in $E$ is flipped.

Thus, we conclude
\begin{thm}
The quantum model we derived for gravity combined with the forces of the Standard Model can be described by products of spatial and temporal eigenfunctions of corresponding self-adjoint operators with a continuous spectrum.

We have a zero-point energy state as a spatial eigendistribution of the gravitational Hamiltonian with smallest eigenvalue $\abs\rho^2=1$ which could be considered to be the source of the dark energy.

Furthermore, we have a big bang singularity in $t=0$. Since the same quantum model is also valid by switching from $t>0$ to $t<0$, with appropriate changes to the temporal eigenfunctions, one could argue that at the big bang two universes with different time orientations could have been created such that, in view of the CPT theorem, one was filled with matter and the other with anti-matter.
\end{thm}

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redpajama_set_name: RedPajamaArXiv


I have now come to the end of the line.
English and with several small jpgs).
through since you tried to set up your studio in Spain.
have to live in N. America for aid.
which would melt me into my shoes.
with success. You write beautifully.
>I have now come to the end of the line.
>English and with several small jpgs).
listed in PMI, CM, CT magazines at the end of their articles.
Too bad you are not in the USA...... you'd have a great lawsuit!


If there was any doubt that camp has begun prior to today, those doubts have all but been erased. "Here we go!" has been the battle cry today, and it has been sung by campers, counselors, and directors alike. Rockmont is a place of new friendships, new adventures, new challenges, and new beginnings. Every morning we encourage our campers and staff to seek out something or someone new and to see where that can lead. This is a place of exploration, inspiration, and exhilaration. This is camp. It has finally begun. Here we go!
Manhood dictates that we seek out adventure and challenge. Those adventures and challenges can be found in stepping outside of our comfort zones and trying something new, something fresh, and something that has not been done before. This is especially true of the current session here on the mountain, as we have started the summer with the debut of our newest program, Base Camp, and continued our traditional "beginner" camp experience with Starter Camp. Both programs give our campers the chance to explore something new, exciting, and unfamiliar.
Today we woke up to the sound of reveille a little later than yesterday, giving our older Base Camp guys the chance to sleep in to 8:30—sleep that was probably a necessity for the huge day that lay ahead. After a traditional breakfast of pancakes, sausage, and cereal, attention was focused on to the day at hand. After Morning Watches filled with songs, chants, and a devotion, the plan of the morning was laid out: pack. Full of energy and motivation from the thrill of something new and unknown, campers from Hemlock, Poplar, and Sycamore darted to their cabins to prepare for the multi-night adventure to Black Mountain Campground. They have embarked to explore the creation that God has given us to take in and examine. They will return with a deeper understanding of this world and undoubtedly with incredible stories to tell. We look forward to sharing those stories of adventure with you in future updates!
While older campers used the later bugle for a little extra sleep, our younger campers embraced the later wake-up with additional games of Thunderball, fishing, disc golf, and creature catching (lizards, skinks, and newts, oh my!). Although Starter Camp is a traditional, reoccurring session at Rockmont, the entire premise of Starter Camp is centered on new beginnings and experiences, as well. The vast majority of campers who participate in Starter Camp have no first-hand camp knowledge and few have been away from their parents for such an extended period. It is remarkable how quickly campers are able to embrace the mixed emotions that occur on opening day as they bravely stare down a week of the unknown. Where there could have been uncertainty, there was understanding. Where there could have been timidity, there was bravery. Where there could have been a feeling of uneasiness around new friends, there was a bond of brotherhood… all while having a whole lot of FUN! This morning the Birches enjoyed a game of Gladiator on the dam, the Buckeyes played Powerball, and the Hickory tribe enjoyed some great fellowship with a morning of cabin activities. After a delicious lunch of tacos and a great rendition of a favorite camp song, "Hey Burrito," Bear and Deer Camp joined forces to play Super Mario Brothers. Good news, parents: Princess Peach has been safely returned to her castle! Following a dinner of roast beef, broccoli, and baked potatoes, Bear Camp combined forces to engage in a game of Battle Ball in the gym while the Hickory tribe fought bravely in a game of Braveheart on the Inspiration Point hill. It was a day full of fun, great weather, and good times—a day your children will relive in the vivid stories they tell of challenge and adventure on the drive home Friday.
Once again, camp has begun. It has finally begun. Here we go!