text:
Chandigarh Elections
Youth paraded in underwear, stripped naked, thrashed
This story is from August 31, 2018
Yudhvir Rana | TNN | Updated: Aug 31, 2018, 6:06 IST
Picture used for representational purpose
AMRITSAR: A youth was thrashed and paraded in an underwear on the streets in Majithia Road area of Amritsar over his alleged ties with his former fiancée and then stripped naked by a group of 12 people before being assaulted again. Police swung into action after a video, presumably made by the alleged accused, went viral on social media on Thursday. However, no arrests have been made in the case.
Amristar additional deputy commissioner of police Lakhbir Singh said the victim had gone to Sandhu Colony on Majitha Road to meet his former fiancée when he was accosted by her cousins and his friends. He said police had identified three alleged accused but not made any arrests so far.
A case has been registered under various sections of the IPC and the Information Technology Act.
In the video, the youth, with a blood-stained face, is seen begging his assaulters with folded hands to let him go. In another clip, he is seen running barefoot in an underwear and being chased by some youths on motorcycles. In yet another clip, he is seen naked and being thrashed.
Sources said the youth was engaged to a woman from Sandhu Colony but the engagement was later called off. However, the two remained in touch much against the wishes of her family members. "When the youth went to meet her on Wednesday, he was stopped by the woman's cousin and his friends. They blamed him for imposing himself on the woman and thrashed him," said sources.
Navjot Singh Sidhu quits Punjab Cabinet, posts June 10 letter addressed to Rahul Gandhi on Twitter
2 dead, 12 injured in Kurukshetra after private bus gutted in fire


MANCHESTER City stars David Silva, Nolito and Sergio Aguero have scored a slap-up meal at one of Mallorca's most famous restaurants.
The City trio dined out at Palma's Forn De Sant Joan, which has attracted other top sporting stars like Andy Murray, Rafa Nadal and Arsene Wenger.
Ciudadanos leader Albert Rivera is another of the famous faces who has recently eaten at the stylish restaurant in the heart of trendy La Llotja.


Muhammadu Buhari's administration has not fulfilled agreements it made with academic staff in Nigeria's public universities.
All 37,504 academic staff of Nigeria's public universities have been on an indefinite strike since early November. This followed a meeting of the union's national executive council which ruled that no substantial progress had been made on a range of past agreements between the Academic Staff Union of Universities, which represents academic staff, and the federal government.
Nigeria has 43 federal universities and 47 state universities which all fall under public universities. All academic activities at public universities – teaching, supervision, examination – are on hold, affecting about 1.3 million students. The admission of new students is also being affected.
The vast majority of Nigerians attend a public university. This year, of the 1.6 million students who wrote the final high school exams, 94% picked public universities. This suggests that it would be wise for the Nigerian government to invest in this critical sector, where the interest of the country's future lies.
Students in most public universities pay less than N50,000 a year. This means that the institutions get very little fee income. As a result, government remains the major funder of universities. But it hasn't met its obligations even though many institutions face serious infrastructure decay. Most of the public universities generate funds from business ventures. But this isn't enough to cover costs.
It's not the first time academic staff have gone on strike. In 2013 they held a six-month strike and in 2017 they had a five-week strike. After the 2017 strike a memorandum of action was signed with Muhammadu Buhari's administration. The agreement set out what steps would be taken to implement outstanding agreements.
The union called the latest strike because the government still hasn't met the commitments it made. These include shortfalls in salary payments as well as academic allowances. For its part, the government argues that it can't meet the demands of striking lecturers to the detriment of the country's infrastructure, health, and security.
Other outstanding issues are government's promise to provide funding to revitalise public universities and changes to the University Pension Fund that would enable staff to manage the pension funds themselves.
So what's behind the recurrent strikes in Nigeria's educational system? The answer is the seeming lack of the political will by the federal government to honour agreements. What's also irking academics is that the government isn't ensuring the environment at universities is conducive to teaching and learning. And that conditions aren't good enough to attract foreign scholars.
The main factor is that the federal government can't be trusted to deliver on its promises. For example, the Academic Staff Union of Universities suspended its strike in 2017 on the promise of federal government action within six months. Almost a year later, Buhari's government continues to drag its feet on fully implementing the agreements.
The issue of insufficient funding of Nigeria's public universities goes back a long way. In 2012, the federal government constituted a panel to undertake a needs assessment of public universities. The report indicated that universities need N1.3 trillion for a modest revitalisation. The fund was to be paid in tranches over the next five years. But only the Goodluck Jonathan government released N200 billion in 2013.
This underfunding has serious consequences. Infrastructure is decaying. State universities struggle to attract foreign scholars. And Nigerian universities perform poorly in global rankings.
When it comes to salaries, the union has acknowledged that some progress has been made in payment of shortfalls. But in a letter to the Minister of Education a month ago, the union asked the government to pay the outstanding balances.
Other issues are the failure to release an operational license to the Nigerian University Employees Pension Company. The union has only been given approval in principle but cannot hit the ground running without an operational license. It feels the federal government is frustrating it because operating pension fund administrators are friends of government.
Another reason for the strike is the government's planned re-introduction of the Education Bank which was scrapped in 2001. It was designed to help indigent students access loans to attend universities. The union believes that re-introducing the bank is an attempt to re-introduce tuition fees to public universities.
The government denies this. It argues the Education Bank would only provide loans to students.
The strike is happening at a strategic time. Nigerians will be going to the polls in February 2019; the incumbent Buhari is seeking re-election.
Nigeria's public officials tend to pay attention to issues affecting the general populace only during election periods. The hope is that government officials will see education as a vital investment in the future – and commit funds to institutions of higher learning.


text:
Following collections have been recently added:
- Athens, GA, Georgia Museum of Art, University of Georgia: 42 sculptures (8 March 2018)
- Bristol, RI, Colt State Park and Linden Place: 9 sculptures (8 March 2018)
- Cornish, NH, Saint-Gaudens National Historic Site: 1 sculpture (8 March 2018)
- Montreal, QC, Canada, outdoor sculpture: 9 sculptures (8 March 2018)
- Montreal, QC, Canada, Congrégation de Notre-Dame, Musée Marguerite-Bourgeoys: 1 sculpture (8 March 2018)
- Québec, QC, Canada, outdoor sculpture: 1 sculpture (8 March 2018)
- Saint Paul, MN, James J. Hill Business Reference Library: 1 sculpture (8 March 2018)
- Washington, D.C., The Carnegie Library: 1 sculpture (8 March 2018)
- New Haven, CT, Yale University Art Gallery: 1123 additional sculptures, mainly medals (16 June 2018)
- Saratoga Springs, NY, The Frances Young Tang Teaching Museum and Art Gallery at Skidmore College: 3 sculptures (26 June 2018)
- Asheville, NC, Biltmore Estate: 53 sculptures (13 August 2018)
- Edwardsville, IL, The University Museum, Southern Illinois University: 1 sculpture (13 August 2018)
- Glens Falls, NY, The Hyde Collection: 9 sculptures (13 August 2018)
- Hagerstown, MD, Washington County Museum of Fine Arts: 20 sculptures (not complete) (13 August 2018)
- San Francisco, CA, former Grand Lodge: 1 sculpture (13 August 2018)
- Shreveport, LA, The R.W. Norton Art Gallery: 97 sculptures (not complete) (13 August 2018)
- Washington, D.C., Frederick Douglass National Historic Site (U.S. National Park Service): 1 sculpture (13 August 2018)
- Washington, D.C., The Society of the Cincinnati, Anderson House: 4 sculptures (13 August 2018)
- Woodstock, VT, Marsh – Billings – Rockefeller National Historic Park: 1 sculpture (13 August 2018)
- Boston, MA, McMullen Museum of Art, Boston College: 16 sculptures (24 August 2018)
- Richmond, VA, Virginia Museum of Fine Arts: 181 sculptures (24 August 2018)
- Ocala, FL, Appleton Museum of Art: 28 sculptures (5 October 2018)
- Baltimore, MD, The Walters Art Museum: 57 additional sculptures (16 May 2017)
- Lenox, MA, Frelinghuysen Morris House and Studio: 4 sculptures (16 May 2017)
- New Harmony, IN, Roofless Church: 2 sculptures (16 May 2017)
- Portland, ME, Portland Museum of Art: 47 sculptures (16 May 2017)
- Amherst, MA, Mead Art Museum, Amherst College: 41 sculptures (2 July 2017)
- Deerfield, MA, Historic Deerfield: 3 sculptures (2 July 2017)
- Northampton, MA, Smith College Museum of Art: 107 sculptures (2 July 2017)
- South Hadley, MA, Mount Holyoke College Art Museum: 8 sculptures (2 July 2017)
- Goldendale, WA, Maryhill Museum of Art: 162 sculptures (13 September 2017)
- Middlebury, VT, Middlebury College Museum of Art: 22 sculptures (13 September 2017)
- Newport, RI, Rough Point: 7 sculptures (13 September 2017)
- Stockbridge, MA, Chesterwood: 6 sculptures (13 September 2017)
- Newport, RI, Redwood Library: 4 sculptures (13 November 2017)
- Worcester, MA, Worcester Art Museum: 18 sculptures (13 November 2017)
- outdoor sculptures by Léon Hermant: 57 sculptures (7 January 2016)
- Charlottesville, VA, Monticello: 11 sculptures (8 January 2016)
- New Brunswick, NJ, Zimmerli Art Museum: 201 sculptures (8 January 2016)
- Washington, D.C., United States Capitol: 15 sculptures (16 February 2016)
- Baltimore, MD, Mt. Vernon Place: 9 sculptures (22 February 2016)
- Memphis, TN, Memphis Brooks Museum of Art: 14 sculptures (22 February 2016)
- Tarrytown, NY, Lyndhurst: 2 sculptures (22 February 2016) (more to come)
- Williamstown, MA, Williams College Museum of Art: 26 sculptures (22 February 2016)
- Brunswick, ME, Bowdoin College Museum of Art: 196 sculptures (29 February 2016)
- Minneapolis, MN, The Minneapolis Institute of Art: 133 sculptures (4 March 2016)
- Washington, D.C., George Washington University: 5 sculptures (4 March 2016)
- Washington, D.C., The Kreeger Museum: 15 sculptures (4 March 2016)
- Washington, D.C., Library of Congress: 16 sculptures (4 March 2016)
- Philadelphia, PA, University of Pennsylvania Art Collection, Dr. Thomas W. Evans Collection: 16 sculptures (14 March 2016)
- Rochester, NY, The Memorial Art Gallery at the University of Rochester: 37 sculptures (25 March 2016)
- Easton, PA, Lafayette College: 13 sculptures (7 April 2016)
- Milwaukee, WI, Milwaukee Art Museum: 40 sculptures (7 April 2016)
- Raleigh, NC, North Carolina Museum of Art: 40 sculptures (10 May 2016)
- Eugene, OR, Jordan Schnitzer Museum of Art, University of Oregon: 2 sculptures (18 October 2016)
- works by Léonard Crunelle (1872-1944): 43 sculptures (October-December 2016)
- Portland, OR, Portland Art Museum: 26 sculptures (18 October 2016)
- Manchester, NH, Currier Museum of Art: 18 sculptures (18 October 2016)
- Nashville, TN, Vanderbilt University Fine Arts Gallery: 8 sculptures (11 November 2016)
- Madison, WI, Chazen Museum of Art: 104 sculptures (20 December 2016)
- St. Louis, MO, Mildred Lane Kemper Art Museum, Washington University: 23 sculptures (22 December 2016)
- New York, NY, Brooklyn Museum: 183 sculptures (30 December 2016)
- Washington, D.C., The National Gallery of Art, Corcoran Collection
- New Washoe City, NV, Bowers Mansion: 1 sculpture (13 January 2015)
- Notre Dame, IN, University of Notre Dame, Snite Museum of Art: 79 sculptures (27 February 2015)
- Providence, RI, Columbus Square: 1 sculpture (27 February 2015)
- Wilmington, DE, Nemours Mansion and Gardens: 16 sculptures (3 March 2015)
- South Hadley, MA, Mount Holyoke College Museum of Art: 7 sculptures (3 March 2015)
- Phoenix, AZ, Arizona Capitol Museum: 3 sculptures (10 March 2015)
- Philadelphia, PA, Philadelphia Museum of Art: 243 sculptures (25 March 2015)
- Princeton, NJ, Princeton University Art Museum: 116 medals (25 March 2015)
- Philadelphia, PA, Philadelphia Museum of Art: 45 sculptures (10 April 2015)
- San Antonio, TX, The Marion Koogler McNay Art Museum: 67 sculptures (15 May 2015)
- Washington, D.C., The Phillips Collection: 21 sculptures (10 June 2015)
- Stanford, CA, Iris & B. Gerald Cantor Center for Visual Arts at Stanford University: 259 sculptures (2 July 2015)
- Charlottesville, VA, Monticello: 11 sculptures (21 August 2015)
- Indianapolis, IN, Indiana State Library: 29 sculptures (21 August 2015)
- New Brunswick, NJ, Zimmerli Art Museum: 197 sculptures (21 August 2015)
- New Haven, CT, Yale Center for British Art: 7 sculptures (21 August 2015)
- New York, NY, Whitney Museum of American Art: 32 sculptures (21 August 2015)
- Seattle, WA, Seattle Art Museum: 11 sculptures (21 August 2015)
- Sleepy Hollow, NY, Kykuit: 13 sculptures (21 August 2015)
- miscellaneous outdoor sculpture: 30 sculptures (21 August 2015)
- Phoenix, AZ, Phoenix Art Museum: 69 sculptures (15 September 2015)
- Providence, RI, Rhode Island School of Design Museum of Art: 60 sculptures (15 September 2015)
- Tampa, FL, Henry B. Plant Museum: 20 sculptures (15 September 2015)
- Norfolk, VA, Chrysler Museum of Art: 119 sculptures (1 October 2015)
- Norman, OK, Fred Jones Jr. Museum of Art: 7 sculptures (30 October 2015)
- several Virgin of Lourdes (30 October 2015)
- San Antonio, TX, Our Lady of the Lake University: 3 sculptures (30 October 2015)
- St. Louis, MO, Saint Louis Art Museum: 71 sculptures (5 November 2015)
- Princeton, NJ, Princeton University Art Museum: 12 medals (many more to come!) (19 December 2014)

redpajama_set_name: RedPajamaCommonCrawl


Someone is pestering you so much you can't go over all the necessary details. They've just about driven you to distraction. Before you let them drive you completely crazy, understand that the last thing they want is for you to be well informed. They have their reasons.
You're on the receiving end of some workplace admiration on Monday, performance related or otherwise. On Tuesday or Wednesday you're in a frugal state of mind, but don't pinch budget pennies where they'll cut into morale – or productivity. Toward the end of the workweek, on the other hand, you're in a pioneering mode – and the cutting edge is no place to worry about cutting costs (plenty of time for that later). This weekend, back up your files and keep badges, cell phones, and key cards directly on your person – or plan to spend time backtracking.
Have you realized that you don't want something as much as you thought? It really is time for an honest career assessment. You have to know what you need in order to make plans to get it. If you want something and pull out all stops, you increase the odds of getting it. Heads up because the April 5 new Aries moon whets your appetite and makes you want to take strategic risks with available opportunities in the 30 days following the phase.
A cosmically busy time starts on the tenth when the sun squares Saturn. The planets symbolize a laser focus on important things. Also that day, Jupiter's Sagittarius retrograde lengthens your attention span while extending the time to take advantage of opportunities.
On April 16, inquiring Mercury enters your risk-taking creativity sign, with a lucky Venus transit soon (stay focused).
Sunday is a lucky day. Search and prepare resumes if job hunting. Send them on Monday, another lucky day that's perfect for interviews and meetings.
Capricorn Pluto turns retrograde in your money sign on the twenty-fourth, followed by Saturn's reverse turn the next week. Retrogrades bring back things you've given up on - or thought you had. You didn't really. Take action on them next month.


text:
\section{Introduction}

In 1977 Saito and Kurokawa \cite{Kurokawa78} conjectured that cusp forms
in $S_{2k-2}(\SL_2(\mathbf{Z}))$ can be lifted to Siegel modular forms of
weight
$k$
and that this lifting is Hecke-equivariant in an appropriate sense. This
conjecture was proved in a series of papers by Maass \cite{Maass79},
Andrianov \cite{Andrianov79}, and Zagier \cite{Zagier81}, and later
reformulated and proved using the language of representation theory by
Piatetski-Shapiro \cite{Piatetski-Shapiro83}.
In the 1980s Kojima \cite{Kojima} and Gritsenko \cite{Gritsenko90} proved
the
existence of a similar lifting from the space of modular forms of level 4
and non-trivial character to the space of hermitian modular forms.
The group $\U(2,2)$ for which the hermitian modular forms were defined was
associated with
the extension $\mathbf{Q}(i)/\mathbf{Q}$. Following \cite{Kojima} we will refer to the
image
of this lifting as the \textit{Maass space} and to the lifting itself as
the \textit{Maass lifting}. In 1991 Krieg
\cite{Krieg91} extended the results of \cite{Kojima} and
\cite{Gritsenko90} to imaginary quadratic fields and
showed Hecke-equivariance of the Maass lifting for a certain family of
Hecke
operators living in the local Hecke algebra at $p$, for $p$ any
inert prime. As \cite{Kojima}, \cite{Gritsenko90} and \cite{Krieg91}
all treated hermitian modular forms as classical objects (i.e., as
functions defined on a higher-dimensional analogue of the complex upper
half-plane), their methods did not yield Hecke-equivariance of the lift
under
the action of local Hecke algebras at split primes, except when the class
number of $F$ was one (\cite{Gritsenko90}).

In this article we define an adelic version of the Maass space by
imposing a condition on the Fourier coefficients of hermitian modular
forms regarded as functions on $\U(2,2)(\mathbf{A})$ (Definition \ref{Maass
form}).
In fact our definition reduces to that of Krieg's when the
class number of $F$ is one. We also show that the adelic Maass space,
which we
denote by $\mathcal{M}_k$, is invariant
under the action of local Hecke algebras at all primes $p$ not ramified
in $F/\mathbf{Q}$ (Theorem \ref{thmmain}).

To prove Hecke-equivariance of the Maass space we assume that the class
number $h_F$ of $F$ is odd, and
in that
case we show that $\mathcal{M}_k$ is isomorphic to $h_F$
copies of
Krieg's Maass space (Proposition \ref{oddclass}), hence we obtain a
lifting from
$M_{k-1}(D_F,
\chi_F)^{h_F}$ to $\mathcal{M}_k$ together with a homomorphism from the Hecke
algebra of $\mathcal{M}_k$ into the space of endomorphisms of $M_{k-1}(D_F,
\chi_F)^{h_F}$ generated by the classical Hecke operators $T_p$ (for $p$
split), $T^2_{p}$ (for $p$ inert) and the group of permutations of the
factors in $M_{k-1}(D_F,
\chi_F)^{h_F}$ (Theorems \ref{heckedesc} and \ref{heckedesc2}).

In \cite{Klosin06preprint} the author used Hecke-equivariance of the Maass
lifting
for the
extension $\mathbf{Q}(i)/\mathbf{Q}$ proved in \cite{Gritsenko90} to construct
congruences between Maass forms and non-Maass forms and as a consequence
give evidence for the Bloch-Kato conjecture (Theorem 7.11 and Corollary
9.9 in
\cite{Klosin06preprint}). We hope to be able to use Theorem \ref{thmmain}
to
extend the results of \cite{Klosin06preprint} to all imaginary quadratic
fields of
odd class number.

We believe that our assumption on the class number of $F$ is unnecessary
and we make it here only to make the proofs simpler. We work with the
Hecke algebra for the algebraic group $\U(2,2)$ instead of $\GU(2,2)$
which is the case in \cite{Gritsenko90} and \cite{Krieg91}, but our
results easily imply analogous ones for $\GU(2,2)$. Finally we want to
point out that
the Maass lifting we obtain agrees with a lifting that has recently
been defined by Ikeda
\cite{Ikedapreprint2005} using a different approach. Our result provides
an explicit description of the image of that lifting as well as explicit
formulas for the descent of Hecke algebras.

\section{Definitions}

Let $F$ be an imaginary quadratic field with ring of integers $\mathcal{O}_F$.
We denote by $\Cl_F$ the class group of $F$ and set $h_F:= \# \Cl_F$. For
any affine group scheme $X$ over $\mathcal{O}_F$ and any $\mathbf{Z}$-algebra $A$, we
denote by $x \mapsto \overline{x}$
the automorphism of $(\Res_{\mathcal{O}_F/\mathbf{Z}}X)(A)$ induced by the non-trivial
automorphism of
$F/\mathbf{Q}$. Note that $(\Res_{\mathcal{O}_F/\mathbf{Z}}X)(A)$ can be identified with a
subgroup of $\GL_n(A \otimes \mathcal{O}_F)$ for some $n$. In what follows we
always specify such an identification. Then for $x \in
(\Res_{\mathcal{O}_F/\mathbf{Z}}X)(A)$ we write $x^t$ for the transpose of $x$, and set
$x^*:= \overline{x}^t$ and $\hat{x}:= (\overline{x}^t)^{-1}$. Moreover, we write
$\diag(a_1, a_2, \dots, a_n)$ for the $n\times n$-matrix with $a_1, a_2,
\dots a_n$ on the diagonal and all the off-diagonal entries equal to zero.

To the extension $F/\mathbf{Q}$ we associate the quasi-split unitary group
$$\U(n,n):= \{ g \in \Res_{\mathcal{O}_F/\mathbf{Z}} \GL_{2n/\mathcal{O}_F} \mid g J_n g^* =
J_n\},$$
where $J_n=\left[ \begin{smallmatrix} & -I_n \\
I_n\end{smallmatrix} \right]$ and $I_n$ stands for the $n \times n$ identity matrix. For $q
\in \Res_{\mathcal{O}_F/\mathbf{Z}}\GL_{n/\mathcal{O}_F}$, we set $p_q:= \left[ \begin{smallmatrix} q \\
& \hat{q} \end{smallmatrix} \right] \in \U(n,n)$. We will also write $G$ for $\U(2,2)$ and
$J$ for $J_2$.

For a prime ideal $\mathfrak{p}$ of $F$, let $F_{\mathfrak{p}}$ denote the completion of $F$
with respect to the valuation induced by $\mathfrak{p}$. Set $F_p:= F
\otimes_{\mathbf{Q}} \mathbf{Q}_p$ and $\mathcal{O}_{F,p}= \mathcal{O}_F \otimes_{\mathbf{Z}} \mathbf{Z}_p$. Note
that if
$p$ is
inert or ramified in $F$, then $F_p/\mathbf{Q}_p$ is a degree two extension of
local fields and $a \mapsto \overline{a}$ induces the non-trivial automorphism
in $\Gal(F_p/\mathbf{Q}_p)$, while if $p$ splits in $F$,
then $F_p \cong \mathbf{Q}_p \times \mathbf{Q}_p$, and $a \mapsto \overline{a}$ corresponds
on the right-hand side to the automorphism defined by $(a,b) \mapsto
(b,a)$. We
denote the isomorphism $\mathbf{Q}_p \times
\mathbf{Q}_p \xrightarrow{\sim} F_p$ by $\iota_p$. For a matrix $g=(g_{ij})$
with entries
in $\mathbf{Q}_p \times \mathbf{Q}_p$ we also
set $\iota_p(g)= ((\iota_p(g_{ij}))$. For a split prime $p$ the map
$\iota_p^{-1}$ identifies $G(\mathbf{Q}_p)$ with $$G_p =
\{(g_1,g_2) \in \GL_{4}(\mathbf{Q}_p)
\times \GL_{4}(\mathbf{Q}_p) \mid g_1 J g_2^t = J \}.$$ Note that the map
$(g_1,g_2) \mapsto g_1$ gives a (non-canonical) isomorphism $G(\mathbf{Q}_p)
\cong \GL_4(\mathbf{Q}_p)$.

Denote by $\mathbf{A}$ (resp. $\mathbf{A}_{\textup{f}}$, $\mathbf{A}_F$, $\mathbf{A}_{F,\textup{f}}$) the ring of adeles of
$\mathbf{Q}$ (resp. finite adeles of $\mathbf{Q}$, adeles of $F$, finite adeles of
$F$). For an adele $a$, we write $a_{\textup{f}}$ for its finite part.
Set $\hat{\mathbf{Z}}:=
\prod_p \mathbf{Z}_p$, $\hat{\mathcal{O}}_F:= \mathcal{O}_F
\otimes_{\mathbf{Z}} \hat{\mathbf{Z}}$, $K'= \GL_2(\hat{\mathcal{O}}_F)$ and
$K=G(\hat{\mathbf{Z}})$. Note that $K$ is a maximal compact subgroup of
$G(\mathbf{A}_{\textup{f}})$. For any rational prime $p$, we denote by $j_p$ the canonical
embedding $G(\mathbf{Q}_p) \hookrightarrow G(\mathbf{A}).$

Let $M_n$ denote the additive group of $n \times n$ matrices.
Set $$S=\{ h \in \Res_{\mathcal{O}_F/\mathbf{Z}} M_{2/\mathcal{O}_F} \mid h^* = h\}, $$ $$T=\{h \in
S(\mathbf{Q})
\mid \textup{tr}\hspace{2pt} (S(\mathbf{Z})h) \subset \mathbf{Z} \},$$
$$T_p=\{h \in
S(\mathbf{Q}_p)
\mid \textup{tr}\hspace{2pt} (S(\mathbf{Z}_p)h) \subset \mathbf{Z}_p \},$$ and $$T_{\mathbf{A}} = \prod_p
T_p\subset M_2(\mathbf{A}_{\textup{f}}).$$
For a matrix $h=(h_p)_p \in T_{\mathbf{A}}$,
and a prime $p$, set $$\epsilon_p(h) = \max \{n \in \mathbf{Z} \mid
\frac{1}{p^n} h_p \in T_p \}$$ and $$\epsilon(h) = \prod_p
p^{\epsilon_p(h)}.$$
Note that since $h \in T_{\mathbf{A}}$, we have $\epsilon_p(h) \geq
0$ for every $p$.

\section{Some coset decompositions}

To shorten notation in this section we put $U_n= U(n,n)$. It is well-known
(see e.g., \cite{Bump97}, Theorem 3.3.1) that for any
finite subset $\mathcal{B}$ of $\GL_n(\mathbf{A}_{F,\textup{f}})$ of cardinality $h_F$ with the
property that the canonical homomorphism $c_F$ defined as the
composite $\mathbf{A}_F^{\times} \twoheadrightarrow
\mathbf{A}_F^{\times} / F^{\times} \mathbf{C}^{\times} \hat{\mathcal{O}}_F^{\times} \cong
\Cl_F$ restricted to $\det \mathcal{B}$ is a bijection, the following
decomposition
holds $$\GL_n(\mathbf{A}_F) = \bigsqcup_{b \in \mathcal{B}} \GL_n(F) \GL_n(\mathbf{C}) b
\GL_n(\hat{\mathcal{O}}_F).$$
We will call any such $\mathcal{B}$ a \textit{base}. We always assume that a base
comes with a fixed ordering, so in particular if we consider a tuple
$(f_b)_{b \in \mathcal{B}}$
indexed by elements of $\mathcal{B}$, and apply a permutation $\sigma$ to the
elements
$f_b$, we do not consider the tuples $(f_b)_{b \in \mathcal{B}}$ and
$(f_{\sigma^{-1}(b)})_{b \in
\mathcal{B}}$ to be the same.

Put $$H:= \{ a
\in \Res_{\mathcal{O}_F/\mathbf{Z}} \mathbf{G}_{m/\mathcal{O}_F} \mid a \overline{a} = 1\},$$ and let $\varphi:
\Res_{\mathcal{O}_F/\mathbf{Z}} \mathbf{G}_{m/\mathcal{O}_F}
\rightarrow H$ be defined by $\varphi(a)=a/\overline{a}$. Here $\mathbf{G}_m$ denotes
the multiplicative group.

\begin{lemma} \label{Hlemma} For every rational prime $p$, we have $\det
U_n(\mathbf{Z}_p) =
\varphi(\mathcal{O}_{F,p}^{\times})$. \end{lemma}

\begin{proof} We have $\det U_n(\mathbf{Z}_p)  \subset
\varphi(\mathcal{O}_{F,p}^{\times})$ by
\cite{Shimura97}, Lemma 5.11(4). We
will show the other containment. Every element of
$\varphi(\mathcal{O}_{F,p}^{\times})$
can be
written as $a/\overline{a}$
for some $a \in \mathcal{O}_{F,p}^{\times}$. For such an $a$,
set $x_a:=
\diag(a,1,1, \dots, \overline{a}^{-1},1,1,\dots)$, where $\overline{a}^{-1}$ is on the
$(n+1)$st position. We certainly have $x_a \in U_n(\mathbf{Z}_p)$ and
$\det
x_a =
a/\overline{a}$. \end{proof}

\begin{prop} \label{decomp657} Assume $2 \nmid h_F$. For any base
$\mathcal{B}$ the following
decomposition holds
\begin{equation} \label{Undecomp} U_n(\mathbf{A}) = \bigsqcup_{b
\in \mathcal{B}} U_n(\mathbf{Q}) U_n(\mathbf{R}) p_b U_n(\hat{\mathbf{Z}}).\end{equation} \end{prop}

\begin{proof} First note that $\{ c_F(\det p_b)\}_{b
\in \mathcal{B}}= \Cl_F$ as long as $2 \nmid h_F$. [Indeed, $\det p_q =
\det q/
\overline{\det q}$. Since $N_{F / \mathbf{Q}} \det q \in
\mathbf{A}^{\times} \subset F^{\times} \mathbf{C}^{\times} \hat{\mathcal{O}}_F^{\times}$, i.e.,
$\det q \hspace{2pt} \overline{\det q}$ represents a principal fractional ideal of $F$,
we
have
$c_F(\overline{\det q}) = c_F(\det q^{-1})$ and hence $c_F(\det
p_b) =
c_F(\det b)^2$. Since $2 \nmid h_F$, $\alpha
\mapsto
\alpha^2$ is an
automorphism of $\Cl_F$. Hence $c_F(\det \mathcal{B}) = \Cl_F$ implies $\{
c_F(\det
p_b)
\}_{b \in \mathcal{B}} = \Cl_F$.] In the notation of Lemma
8.14 of \cite{Shimura97}, we have $D= U_n(\mathbf{R})
U_n(\hat{\mathbf{Z}})$. By
part (3) of that lemma, the map $x \mapsto \det x$ defines a bijection
between $U_n(\mathbf{Q})\setminus U_n(\mathbf{A})/D$ and $H(\mathbf{A})/H(\mathbf{Q}) \det D$.
Set
$U_0$ (in the notation of part (4) of that lemma) to be
$\mathbf{C}^{\times} \hat{\mathcal{O}}_F^{\times}$.
Then part (4) of that
lemma asserts that $\varphi$ gives an isomorphism of
$\mathbf{A}_F^{\times}/\mathbf{A}^{\times}
F^{\times} U_0$ with $H(\mathbf{A})/H(\mathbf{Q}) \varphi(U_0)$. By Lemma \ref{Hlemma}
we
have $\varphi(U_0)= \det D$. Thus the composite of $x
\mapsto \det x$ with $\varphi$ gives a bijection of $U_n(\mathbf{Q})\setminus
U_n(\mathbf{A})/D$ onto $\mathbf{A}_F^{\times}/\mathbf{A}^{\times}
F^{\times} U_0$, which in turn can be identified with $\Cl_F$ since
$\mathbf{A}^{\times} =
\mathbf{Q}^{\times} \times \mathbf{R}_+ \times \hat{\mathbf{Z}}^{\times} \subset F^{\times}
U_0$.
\end{proof}

\begin{cor} \label{scalarcor} If $(h_F, 2n)=1$ a base $\mathcal{B}$ can be
chosen
so that for all $b \in \mathcal{B}$ the matrices $b$ and $p_b$ are
scalar matrices and $b b^* = b^* b = I_n$. \end{cor}

\begin{proof} It follows from the Tchebotarev density theorem, that
elements of $\Cl_F$ can be represented by prime ideals. Since all
the inert
ideals are principal, $\Cl_F$ can be represented by prime ideals
lying over split primes of the form $(p)=\mathfrak{p} \overline{\mathfrak{p}}$. Let $\Sigma$ be a
representing
set consisting of such ideals $\mathfrak{p}$. As $(2n,h_F)=1$, the set
$\Sigma^{2n}$
consisting of elements of $\Sigma$
raised to the power $2n$ is also a representing set for
$\Cl_F$. Moreover, as
$\mathfrak{p} \overline{\mathfrak{p}}$ is a principal ideal, $\overline{\mathfrak{p}} = \mathfrak{p}^{-1}$ as
elements of $\Cl_F$, hence $\Sigma':=\{ \mathfrak{p}^n \overline{\mathfrak{p}}^{-n}\}_{\mathfrak{p}\in
\Sigma}$
also represents all the elements $\Cl_F$. Elements
of $\Sigma'$ can be written adelically as $\alpha_{\mathfrak{p}}^n$,
with $\alpha_{\mathfrak{p}}=(1,1, \dots,
1,
p, p^{-1}, 1,
\dots) \in \mathbf{A}_{F,\textup{f}}$, where $p$ appears on the $\mathfrak{p}$-th place and $p^{-1}$
appears at the $\overline{\mathfrak{p}}$-th place. Set $b_{\mathfrak{p}}= \alpha_{\mathfrak{p}} I_n$.
Then we can take $\mathcal{B}=\{ b_{\mathfrak{p}}\}_{\mathfrak{p}^n \overline{\mathfrak{p}}^{-n} \in \Sigma'}$
and we have $p_{b_{\mathfrak{p}}}=\alpha_{\mathfrak{p}}
I_{2n}$. It is also clear that $b b^*=b^* b =I_n$.
\end{proof}

\section{Hermitian modular forms}

From now on let $k$ be a positive integer divisible by $\#
\mathcal{O}_F^{\times}$. Let $\mathbf{i} = \left[ \begin{smallmatrix} i \\ & i \end{smallmatrix} \right] $ and set $$\mathcal{H}:=
\{Z \in M_2(\mathbf{C}) \mid -\mathbf{i} (Z-Z^*) >0\}.$$ The group
$G(\mathbf{R})$ acts on $\mathcal{H}$ - an element $g=\left[ \begin{smallmatrix} A&B \\ C&D \end{smallmatrix} \right] \in
G(\mathbf{R})$
(with $A, B, C, D \in M_2(\mathbf{C})$) sends $Z
\in \mathcal{H}$ to $gZ:=(AZ+B)(CZ+D)^{-1}$. Set $j(g,Z):= \det(CZ+D)$. For a
congruence
subgroup $\Gamma \subset
G(\mathbf{Q})$, define $\mathcal{M}'_k(\Gamma)$ to be the $\mathbf{C}$-space consisting of
functions $\phi: \mathcal{H} \rightarrow \mathbf{C}$ satisfying
$$(\phi|_k \gamma)(Z):= j(\gamma,Z)^{-k}\phi(\gamma Z) = \phi(Z)$$ for all $\gamma \in
\Gamma.$
Denote by
$\mathcal{M}_k(\Gamma)$ the subspace of
$\mathcal{M}'_k(\Gamma)$ consisting of holomorphic functions. We call elements of
$\mathcal{M}_k(\Gamma)$ the \textit{$\Gamma$-hermitian modular forms (of weight
$k$)}.

Let $\mathcal{M}'_k$ denote the $\mathbf{C}$-space consisting of functions $f
: G(\mathbf{A})
\rightarrow \mathbf{C}$ satisfying the following conditions: \begin{itemize}
\item $f (\gamma g) = f(g)$ for all $\gamma \in G(\mathbf{Q})$, $g \in G(\mathbf{A})$,
\item $f (g \kappa) = f(g)$ for all $\kappa \in K$, $g \in
G(\mathbf{A})$,
\item $f(hg) = j(h_{\infty}, g_{\infty} \mathbf{i})f(g)$ for all $g = \gamma g_{\infty}
\kappa
\in G(\mathbf{Q}) G(\mathbf{R}) K$, $h=(h_{\infty},1) \in G(\mathbf{R}) G(\mathbf{A}_{\textup{f}})$. \end{itemize}

Let $\mathcal{C} \subset G(\mathbf{A}_{\textup{f}})$ be a finite subset such that $$G(\mathbf{A}) =
\bigsqcup_{c
\in \mathcal{C}} G(\mathbf{Q}) G(\mathbf{R}) c K.$$ In particular if $h_F$ is odd we can
take $\mathcal{C} = \{p_b\}_{b \in \mathcal{B}}$ for any base $\mathcal{B}$. Let $f \in \mathcal{M}'_k$.
For $g\in
G(\mathbf{A})$, write $g = \gamma g_{\infty} c \kappa \in G(\mathbf{Q}) G(\mathbf{R}) c K$ for a
unique $c\in \mathcal{C}$ and set $Z:= g_{\infty} \mathbf{i}$. Set $f_c(Z) =
j(g_{\infty}, \mathbf{i})^{k}f(cg)$. The
map
$f
\mapsto (f_c)_{c \in \mathcal{C}}$ defines a $\mathbf{C}$-linear isomorphism
$\Phi_{\mathcal{C}}: \mathcal{M}'_k \xrightarrow{\sim} \prod_{c \in \mathcal{C}} \mathcal{M}'_k(\Gamma_c),$
where
$\Gamma_c:= G(\mathbf{Q}) \cap
(G(\mathbf{R}) cK c^{-1})$ (cf. \cite{Shimura97}, p. 80). If $h_F$ is odd, $\mathcal{B}$
is a base
and $\mathcal{C} = \{p_b\}_{b \in \mathcal{B}}$, we write $\Gamma_b$ instead of $\Gamma_{p_b}$ and
$f_b$ instead of $f_{p_b}$ for $b \in \mathcal{B}$, and $\Phi_{\mathcal{B}}$ instead of
$\Phi_{\mathcal{C}}$.

\begin{definition} A function $f \in \mathcal{M}'_k$ whose image under the
isomorphism $\Phi$ lands in $\prod_{c \in \mathcal{C}} \mathcal{M}_k(\Gamma_c)$ will be called
a
\textit{hermitian modular form of weight $k$}. The
space of
hermitian modular forms of weight $k$ will be denoted by $\mathcal{M}_k$.
\end{definition} We clearly have \begin{equation} \label{prod1} \mathcal{M}_k \cong \prod_{c
\in \mathcal{C}}
\mathcal{M}_k(\Gamma_c).\end{equation}

For every $q \in G(\mathbf{A}_{\textup{f}})$, $f_q \in \mathcal{M}_k(\Gamma_q)$ possesses a Fourier
expansion $$f_q(Z) =
\sum_{h
\in S(\mathbf{Q})}
c_q(h) e^{2 \pi i \textup{tr}\hspace{2pt} (hZ)}.$$ Similarly, every $f \in \mathcal{M}_k$ possesses a
Fourier expansion, i.e., for every $q \in
\GL_2(\mathbf{A}_F)$, and every $h \in S(\mathbf{Q})$ there exists a complex number
$c_f(h,q)$ such that one has $$f\left( \left[ \begin{matrix} I_2 &
\sigma \\ & I_2 \end{matrix} \right] \left[ \begin{matrix} q \\ & \hat{q} \end{matrix} \right]
\right) = \sum_{h \in S(\mathbf{Q})} c_f(h,q) e_{\mathbf{A}} (\textup{tr}\hspace{2pt} h \sigma)$$ for every
$\sigma \in S(\mathbf{A})$. Here $e_{\mathbf{A}}$ is defined in the following way.
Let $a = (a_v) \in \mathbf{A}$, where $v$ runs over all the places of $\mathbf{Q}$. If
$v=\infty$, set $e_v(a_v) = e^{2 \pi i a_v}$. If $v =p$, set $e_v(a_v) =
e^{-2 \pi i y}$, where $y$ is a rational number such that $a_v-y \in
\mathbf{Z}_p$. Then we set
$e_{\mathbf{A}}(a) = \prod_v e_v(a_v)$.

\begin{definition} Let $\mathcal{B}$ be a base. We will say that $q \in
\GL_2(\mathbf{A}_F)$ belongs to a class $b \in \mathcal{B}$ if there exist $\gamma \in
\GL_2(F)$, $q_{\infty} \in \GL_2(\mathbf{C})$ and $\kappa\in K'$ such that $q = \gamma
b q_{\infty} \kappa$. \end{definition}

\begin{rem} It is clear that the class of $q$ depends only on $q_{\mathbf{f}}$.
\end{rem}

\begin{lemma} \label{sameclass} Suppose $r \in \GL_2(\mathbf{A}_F)$ and $q \in
\GL_2(\mathbf{A}_F)$ belong to the same class and $r_{\mathbf{f}} = \gamma q_{\mathbf{f}}
\kappa \in \GL_2(F) q_{\mathbf{f}} \GL_2(\hat{\mathcal{O}}_F)$. Then
\begin{equation} \label{fourierrel} c_f(h,r) = \left( \overline{\det r_{\infty}} /
\overline{\det q_{\infty}}\right)^{k }e^{-2 \pi \textup{tr}\hspace{2pt} (r_{\infty}^* h r_{\infty} -
q_{\infty}^*
\gamma^* h \gamma q_{\infty})} (\det \gamma^*)^{-k} c_f(\gamma^* h \gamma, q). \end{equation} \end{lemma}

\begin{proof} It follows from the proof of part (4) of
Proposition 18.3 of \cite{Shimura97}, that \begin{equation} \label{f12} c_f(h,r) =
\left(
\overline{\det
r_{\infty}}\right)^k e^{-2 \pi \textup{tr}\hspace{2pt} (r_{\infty}^* h r_{\infty})}
c_{p_{r_{\textup{f}}}}(h),\end{equation}
where
$$f_{p_{r_{\textup{f}}}}(Z) = \sum_{h \in S} c_{p_{r_{\textup{f}}}}(h) e^{2 \pi i \textup{tr}\hspace{2pt}
hZ}.$$ As is easy
to see (cf. for example the Proof of Lemma 10.8 in \cite{Shimura97}),
$f_{p_{r_{\textup{f}}}} = f_{p_{q_{\textup{f}}}} |_k \left[ \begin{smallmatrix} \gamma^{-1} \\ & \gamma^* \end{smallmatrix} \right]$.
Hence
\begin{equation} \label{f13} c_{p_{r_{\textup{f}}}}(h) = (\det \gamma^*)^{-k}
c_{p_{q_{\textup{f}}}}(\gamma^* h
\gamma).\end{equation} The Lemma follows from combining (\ref{f12}) with (\ref{f13}).
\end{proof}

\section{The Maass space} \label{The Maass space}

\begin{definition} \label{Maass form} Let $\mathcal{B}$ be a base. We say that $f\in
\mathcal{M}_k$ is a $\mathcal{B}$-\textit{Maass form} if there exist functions $c_{b,
f}:
\mathbf{Z}_{\geq 0} \rightarrow \mathbf{C}$, $b \in \mathcal{B}$, such that for every $q \in
\GL_2(\mathbf{A}_F)$ and every $h \in S(\mathbf{Q})$ the Fourier coefficient $c_f(h,q)$
satisfies
\begin{multline} \label{Maass condition} c_f(h,q) =
\left(\overline{\det q_{\infty}}\right)^k
e^{-2\pi\textup{tr}\hspace{2pt}(q_{\infty}^* h q_{\infty})} \left( \det \gamma_{b, q}^*\right)^{-k}
\times
\\
\times \sum_{\substack{d \in \mathbf{Z}_{+} \\ d \mid \epsilon(q_{\mathbf{f}}^* h
q_{\mathbf{f}})}}
d^{k-1} c_{b, f}\left( D_F d^{-2} \det h \hspace{2pt} \prod_{p} p^{\val_p(\det
q_{\mathbf{f}}^*q_{\mathbf{f}})}\right),\end{multline} where $q_{\mathbf{f}} = \gamma_{b,
q} b
\kappa_q \in \GL_2(F) b K'$ for a unique $b
\in \mathcal{B}$. \end{definition}

\begin{rem} Note that by \cite{Shimura97}, Proposition 18.3(2), $c_f(h,q)
\neq 0$ only if $(q^* h q)_p \in T_p$, so $\epsilon_p(q_{\mathbf{f}}^* h
q_{\mathbf{f}})\geq 0$. Also, note that Definition \ref{Maass form} is
independent of the decomposition $q_{\mathbf{f}} = \gamma_{b,
q} b
\kappa_q\in \GL_2(F) b K'$. Indeed,
if $q_{\mathbf{f}} = \gamma'_{b,
q} b
\kappa'_q \in \GL_2(F) b K'$ is
another decomposition of $q_{\mathbf{f}}$, then
$$\det \gamma'_{b,
q} \det \gamma_{b,
q}^{-1} = \det (\kappa_q (\kappa'_q)^{-1})\in \hat{\mathcal{O}}_F^{\times} \cap
F^{\times} =
\mathcal{O}_F^{\times},$$
so $\det (\gamma'_{b,
q})^k =  \det \gamma_{b,
q}^k$. \end{rem}

\begin{definition} \label{Maass space} The $\mathbf{C}$-subspace of $\mathcal{M}_k$
consisting of $\mathcal{B}$-Maass forms
will be called the $\mathcal{B}$-\textit{Maass space}. \end{definition}

Let $T'_{\mathbf{A}}$ (resp. $T'_p$) be defined in the same way as $T_{\mathbf{A}}$
(resp. $T_p$) except we do not require that $h \in T'_{\mathbf{A}}$ (resp.
$T'_p$) be hermitian. Define $\epsilon'$, $\epsilon'_p$ in the same way
as $\epsilon$, $\epsilon_p$ except replace $T$ with $T'$. Then
$\epsilon = \epsilon'|_{T}$. \begin{lemma}
\label{epsilon1} If $h=(h_p)_p \in T'_{\mathbf{A}}$ and $\kappa=(\kappa_p)_p \in
K'$, then $$\epsilon'(h \kappa) = \epsilon'(\kappa h) = \epsilon'(h).$$
\end{lemma}

\begin{proof} If $h_p = p^n h'$, $h' \in T'_p$, then $\kappa_p h_p = p^n
\kappa_p
h' \in p^n T'_p$. On the other hand, if $\kappa_p h_p \in p^n T'_p$, then
$h_p =
\kappa_p^{-1} (\kappa_p h_p) \in p^n T'_p$ by the above argument. So, $h_p
\in
p^n
T'_p$ if and only if $\kappa_p h_p \in p^n T'_p$, so $\epsilon_p(h \kappa)
=
\epsilon_p(h)$. The other equality is proved in the same way. \end{proof}

\begin{cor} \label{epsilon} If $q \in \GL_2(\mathbf{A}_F)$ and $r \in
\GL_2(\mathbf{A}_F)$ are in the same class, and $r_{\mathbf{f}} =\gamma q_{\mathbf{f}} \kappa
\in \GL_2(F)
q_{\mathbf{f}}K'$, then $\epsilon(r_{\mathbf{f}}^* h
r_{\mathbf{f}}) = \epsilon( q_{\mathbf{f}}^* \gamma^* h
\gamma q_{\mathbf{f}})$. \end{cor}

\begin{prop} \label{to check} Choose a base $\mathcal{B}$ and let $f \in \mathcal{M}_k$.
If
there exist functions $c^*_{b,f}: \mathbf{Z}_{\geq 0} \rightarrow \mathbf{C}$, $b \in
\mathcal{B}$, such
that for every $b \in \mathcal{B}$ and every $h \in S(\mathbf{Q})$, the Fourier coefficient
$c_f(h,b)$ satisfies
condition (\ref{Maass condition}) with $c_{b,f}^*$ in place of $c_{b,f}$,
then $f$ is a $\mathcal{B}$-Maass form and one has $c_{b,f}=c^*_{b,f}$ for every
$b \in \mathcal{B}$. \end{prop}

\begin{proof} Fix $\mathcal{B}$ and $f \in \mathcal{M}_k$. Suppose there exist
$c_{b,f}^*$
such that (\ref{Maass condition}) is satisfied for all pairs $(h,b)$. Let
$q = \gamma b x \kappa = (\gamma x, \gamma b \kappa)\in \GL_2(\mathbf{C}) \times
\GL_2(\mathbf{A}_{F,\textup{f}})$, where $\gamma \in \GL_2(F)$, $x \in \GL_2(\mathbf{C})$ and $\kappa
\in K'$. Then by Lemma \ref{sameclass}, $$c_f(h,q) = \left( \overline{\det
q_{\infty}}\right)^k e^{-2 \pi \textup{tr}\hspace{2pt} (q_{\infty}^* h q_{\infty} -
\gamma^* h \gamma )} (\det \gamma^*)^{-k} c_f(\gamma^* h \gamma, b).$$ Since condition
(\ref{Maass
condition}) is satisfied for $(h,b)$, we know that
$$c_f(h,b)= e^{-2 \pi \textup{tr}\hspace{2pt} h}
\sum_{\substack{d \in \mathbf{Z}_{+} \\ d \mid \epsilon(b^* hb)}}
d^{k-1} c^*_{b, f}\left( D_F d^{-2} \det h \hspace{2pt} \prod_{p} p^{\val_p(\det
b^*b)}\right).$$ Thus $$c_f(\gamma^*  h \gamma,b) = e^{-2 \pi \textup{tr}\hspace{2pt}(\gamma^* h \gamma)}
\sum_{\substack{d \in \mathbf{Z}_{+} \\ d \mid \epsilon(b^*\gamma^* h\gamma b)}}
d^{k-1} c^*_{b, f}\left( D_F d^{-2} \det (\gamma^* h\gamma) \hspace{2pt} \prod_{p}
p^{\val_p(\det
b^*b)}\right).$$ So, \begin{multline} c_f(h,q) = \left(\overline{\det
q_{\infty}}\right)^k
e^{-2\pi\textup{tr}\hspace{2pt}(q_{\infty}^* h q_{\infty})} \left( \det \gamma^*\right)^{-k}
\times\\
\times \sum_{\substack{d \in \mathbf{Z}_{+} \\ d \mid \epsilon(b^* \gamma^* h
\gamma b)}}
d^{k-1} c^*_{b, f}\left( D_F d^{-2} \det h \det(\gamma^* \gamma) \hspace{2pt} \prod_{p}
p^{\val_p(\det
b^* b)}\right).\end{multline}
The claim now follows since $\epsilon(b^* \gamma^* h
\gamma b) = \epsilon(q_{\mathbf{f}}^* h q_{\mathbf{f}})$ by Corollary \ref{epsilon} and $\det
(\gamma^* \gamma) \in \mathbf{Q}_+$, so $\det (\gamma^* \gamma) = \prod_p p^{\val_p(\det \gamma^*
\gamma)}$.
\end{proof}

\begin{prop} \label{independence} If $\mathcal{B}$ and $\mathcal{B}'$ are two bases, then
the $\mathcal{B}$-Maass space and the $\mathcal{B}'$-Maass space coincide, i.e., the
notion of a Maass form is independent of the choice of the base.
\end{prop}

\begin{proof} Let $\mathcal{B}$ and $\mathcal{B}'$ be
two bases. Write $q_{\mathbf{f}} =
\gamma_{b,q} b
\kappa_{\mathcal{B}}= \gamma_{b',q} b' \kappa_{\mathcal{B}'}$ with $b \in \mathcal{B}$,
$b'\in \mathcal{B}'$, $\gamma_{b,q},
\gamma_{b',q} \in \GL_2(F)$ and $\kappa_{\mathcal{B}},  \kappa_{\mathcal{B}'} \in K'$.
Suppose $f$ is a $\mathcal{B}$-Maass form, i.e.,
there
exist functions $c_{b,f}$ for $b \in \mathcal{B}$, such that for every $q$ and
$h$,
$$c_f(h,q) = t \hspace{2pt} \det(\gamma_{b,q}^*)^{-k}
\sum_{\substack{d \in \mathbf{Z}_{+} \\ d \mid \epsilon(q_{\mathbf{f}}^* h
q_{\mathbf{f}})}}
d^{k-1} c_{b, f}(s),$$ where $t=\left(\overline{\det q_{\infty}}\right)^k
e^{-2\pi\textup{tr}\hspace{2pt}(q_{\infty}^* h q_{\infty})} $ and
$s= D_F d^{-2} \det h \hspace{2pt}
\prod_{p} p^{\val_p(\det
q_{\mathbf{f}}^*q_{\mathbf{f}})}$. Our goal is to show that there exist
functions $c_{b',f}$ for $b' \in \mathcal{B}'$,
such that for every $q$ and $h$, \begin{equation} \label{want1} c_f(h,q) = t \hspace{2pt}
\det(\gamma_{b',q}^*)^{-k}
\sum_{\substack{d \in \mathbf{Z}_{+} \\ d \mid \epsilon(q_{\mathbf{f}}^* h
q_{\mathbf{f}})}}
d^{k-1} c_{b', f}(s).\end{equation} We have \begin{equation} \label{dets1} \det \gamma_{b,q} =
\det
\gamma_{b',q}
\det (b' b^{-1}) \det (\kappa_{\mathcal{B}}^{-1} \kappa_{\mathcal{B}'}).\end{equation}
Since $\det(b'b^{-1})$ corresponds to a principal fractional ideal, say
$(\alpha_{b,b'})$, under the map $((\alpha_{b,b'})_{\mathfrak{p}}) \mapsto
\prod_{\mathfrak{p}}
\mathfrak{p}^{\val_{\mathfrak{p}}((\alpha_{b,b'})_{\mathfrak{p}})}$, using \cite{Bump97}, Theorem
3.3.1, we
can write $\det(b' b^{-1}) = \alpha_{b,b'} \kappa_{b,b'}\in
\mathbf{A}_{F,\textup{f}}^{\times}$ with
$\kappa_{b,b'} \in \hat{\mathcal{O}}_F^{\times}$. Then it follows from (\ref{dets1})
that
$$\beta:= \kappa_{b,b'} \det(\kappa_{\mathcal{B}}^{-1} \kappa_{\mathcal{B}'}) \in
\hat{\mathcal{O}}_F^{\times} \cap F^{\times} = \mathcal{O}_F^{\times}.$$
Hence $\beta^k=1$. Thus $\left(\det \gamma_{b,q}^*\right)^{-k} = \left(\det
\gamma_{b',q}^*\right)^{-k} \overline{\alpha_{b,b'}}^{-k}$. Note that
$\alpha_{b,b'}^{-k}$ is
well defined and only depends on $b$ and $b'$ (i.e., it is independent
of $q$ and $h$). Set $c_{b', f}(n) =
\overline{\alpha_{b,b'}}^{-k}c_{b, f}(n)$. Then it is clear that $c_{b', f}$
satisfies (\ref{want1}).
\end{proof}

\begin{definition} \label{maass2} From now on we will refer to $\mathcal{B}$-Maass
forms simply
as
\textit{Maass forms}. Similarly we will talk about the \textit{Maass space}
instead of $\mathcal{B}$-Maass spaces. This is justified by Proposition
\ref{independence}. The Maass space will be denoted by
$\mathcal{M}^{\textup{M}}_k$. \end{definition}

We now recall the definition of Maass space introduced in \cite{Krieg91}.
We will refer to it as the $G(\mathbf{Z})$-Maass space. Consider the space
$\mathcal{M}_k(G(\mathbf{Z}))$ of $G(\mathbf{Z})$-hermitian modular forms of weight $k$. We
say that $\phi
(Z)= \sum_{h \in T} c_{\phi}(h) e^{2 \pi i \textup{tr}\hspace{2pt} (hZ)} \in \mathcal{M}_k(G(\mathbf{Z}))$
is a
\textit{$G(\mathbf{Z})$-Maass form} if there exists a function $\alpha_{\phi}:
\mathbf{Z}_{\geq 0}
\rightarrow \mathbf{C}$ such that for every $h \in T$, one has \begin{equation}
\label{kriegcond} c_{\phi}(h) = \sum_{\substack{ d \in \mathbf{Z}_+\\ d \mid
\epsilon(h)}} d^{k-1} \alpha_{\phi}(D_F d^{-2} \det h).\end{equation} The subspace of
$\mathcal{M}_k(G(\mathbf{Z}))$ consisting of $G(\mathbf{Z})$-Maass forms will be denoted by
$\mathcal{M}_k^{\textup{M}}(G(\mathbf{Z}))$.

\begin{prop} \label{oddclass} If $2 \nmid h_F$, then the Maass space
$\mathcal{M}_k^{\textup{M}}$ is isomorphic
(as a
$\mathbf{C}$-linear space) to $\# \mathcal{B}$ copies of the $G(\mathbf{Z})$-Maass space
$\mathcal{M}^{\textup{M}}_k(G(\mathbf{Z}))$.
\end{prop}

\begin{proof} Since the Maass space is independent of the choice of
a base $\mathcal{B}$ by Proposition \ref{independence}, we may choose $\mathcal{B}$
as in Corollary \ref{scalarcor}. The map $\Phi_{\mathcal{B}}: \mathcal{M}_k
\rightarrow \prod_{b \in
\mathcal{B}} \mathcal{M}_k(G(\mathbf{Z}))$ is an isomorphism. Let $f \in \mathcal{M}_k^{\textup{M}}$
and set $(f_b)_{b \in \mathcal{B}}= \Phi_{\mathcal{B}}(f)$. Set $\alpha_{f_b}:= c_{b,f}$.
Then using
(\ref{f12}), and the fact that the matrices $b$
commute with
$h$ and
$b^*b=1$, we see that condition (\ref{Maass
condition}) for $c_f(h,b)$ translates
into condition (\ref{kriegcond}) for $c_{f_b}(h)$.
Hence $\Phi_{\mathcal{B}}(\mathcal{M}^{\textup{M}}_k) \subset
\prod_{b \in \mathcal{B}} \mathcal{M}^{\textup{M}}_k(G(\mathbf{Z}))$. On the other hand if $(f_b)_{b
\in \mathcal{B}} \in \prod_{b \in \mathcal{B}} \mathcal{M}^{\textup{M}}_k(G(\mathbf{Z}))$, set $c_{b,f}:=
\alpha_{f_b}$. Then conditions
(\ref{kriegcond}) for $c_{f_b}(h)$ translate into conditions
(\ref{Maass condition}) for $c_f(b,h)$. By Proposition
\ref{to check} this implies
that $f$ is a Maass form.
\end{proof}

\section{Invariance under Hecke action}

It was proved in \cite{Krieg91} that the $G(\mathbf{Z})$-Maass space is
invariant under the action of a certain Hecke operator $T_p$ associated
with a prime $p$
which is inert in $F$. On the other hand Gritsenko in \cite{Gritsenko90}
proved the invariance of the $G(\mathbf{Z})$-Maass space under all the Hecke
operators when the class number of $F$ is equal to 1. In this section we
show that if the class number of $F$ is odd, then the Maass space
$\mathcal{M}_k^{\textup{M}}$ is
in
fact invariant under all the local Hecke algebras (for primes $p \nmid
D_F$).

\subsection{The Hecke algebra}

From now on assume that $h_F$ is odd. Let $p$ be a rational prime and
write $K_p$ for $G(\mathbf{Z}_p)$. Let $\mathcal{H}_p$
be the $\mathbf{C}$-algebra
generated by
double cosets $K_p g K_p$, $g \in G(\mathbf{Q}_p)$ with the usual law of
multiplication (cf. \cite{Shimura97}, section 11).
If $K_pgK_p \in \mathcal{H}_p$, there exists a finite set $A_g \subset
G(\mathbf{Q}_p)$ such that $K_p g K_p = \bigsqcup_{\alpha \in A_g} K_p \alpha$.

For $f \in \mathcal{M}_k$, $g \in G(\mathbf{Q}_p)$, $h \in G(\mathbf{A})$, set
$$([K_p g K_p]f)(h) = \sum_{\alpha
\in A_g } f(h j_p(\alpha)^{-1}).$$ It is clear that $[K_p g K_p]f \in \mathcal{M}_k$.

\begin{thm} \label{thmmain} Let $p \nmid D_F$ be a rational prime. The
Maass
space is invariant under the action of $\mathcal{H}_p$, i.e., if $f \in
\mathcal{M}^{\textup{M}}_k$, and $g \in G(\mathbf{Q}_p)$, then $[K_pgK_p]f \in \mathcal{M}^{\textup{M}}_k$.
\end{thm}

\begin{proof} We will only present the proof in the case when $p$
splits in $F/\mathbf{Q}$. For such a prime $p$ the invariance of
$\mathcal{M}_k^{\textup{M}}$ under the action of $\mathcal{H}_p$ follows from Lemma
\ref{generation} and Propositions
\ref{invariance}, \ref{invariance2} and \ref{invariance3} below. If $p$ is
inert one
can proceed along the same lines,
however, it is the case when $p$ splits that is essentially new.
Indeed, if $p$ is inert, the elements of
$\mathcal{H}_p$ respect the decomposition
(\ref{prod1}), hence the statement of the theorem reduces to an assertion
about the action of $\mathcal{H}_p$ on $\mathcal{M}_k(G(\mathbf{Z}))$. Then
the method used in \cite{Gritsenko90} can be adapted to prove the theorem.
See
also Theorem 7 in \cite{Krieg91} which proves the invariance of the
$G(\mathbf{Z})$-Maass space for a certain family of Hecke operators in $\mathcal{H}_p$.
\end{proof}

Let $p$ be a prime which splits in $F$. Write $(p)=\mathfrak{p} \overline{\mathfrak{p}}$.
Recall that $G(\mathbf{Q}_p) \cong \GL_4(\mathbf{Q}_p)$, and
an element
$A$
of $G(\mathbf{Q}_p)$ can be written as $A=(A_1, A_2) \in \GL_4(\mathbf{Q}_p) \times
\GL_4(\mathbf{Q}_p)$ with $A_2 = -J (A_1^t)^{-1} J.$ Note that if
$a,b,c,d \in M_2(\mathbf{C})$ then $$-J \left[ \begin{matrix} a &
b \\ c &d \end{matrix} \right] J = \left[ \begin{matrix} d & -c \\ -b
&a \end{matrix} \right].$$ Set \begin{itemize} \item $T_{\mathfrak{p}} := K_p \iota_p((\diag(
p^{-1}, 1,1,1), \diag(1,1,p,1) ))K_p$
\item $U_{\mathfrak{p}}:= K_p \iota_p((\diag (p^{-1}, p^{-1}, 1,1) ,
\diag(1,1,p,p)))K_p,$
\item $\Delta_{\mathfrak{p}} := K_p \iota_p((pI_4,p^{-1}I_4)) K_p.$ \end{itemize}

\begin{lemma} \label{generation} The $\mathbf{C}$-algebra $\mathcal{H}_p$ is generated
by the operators
$T_{\mathfrak{p}}$, $T_{\overline{\mathfrak{p}}}$, $U_{\mathfrak{p}}$, $\Delta_{\mathfrak{p}}$ and their
inverses. \end{lemma}

\begin{proof} This follows from the theory of Hecke algebras on
$GL_4(\mathbf{Q}_p)$. \end{proof}

\subsection{Diagonalizing hermitian matrices mod $p^n$}

We begin by proving an analogue of Proposition 7 of \cite{Krieg91} for a
split prime $p$.

\begin{prop} \label{diagonal} Let $n$ be a positive integer, $p$ a
prime number split in $F$ and
write $(p) = \mathfrak{p} \overline{\mathfrak{p}}$. Let $h \in T$, $h \neq 0$. Then there exist
$a$, $d \in \mathbf{Z}_+$ with $p \nmid a$ and $u \in \SL_2(\mathcal{O}_F)$ such that
$$u^* h u \equiv \epsilon(h) \left[ \begin{matrix} a \\ &d \end{matrix} \right] \pmod{p^nT}.$$
\end{prop}

In fact it is enough to prove the following lemma.

\begin{lemma} \label{diagonal2} Proposition \ref{diagonal} holds if $T$ is
replaced by the subgroup of hermitian matrices inside $M_2(\mathcal{O}_F)$.
\end{lemma}

\begin{proof} Without loss of generality we may assume that
$\epsilon(h)=1$. Let $(M, M^t)$ be the image of $h$ under the composite
$$M_2(\mathcal{O}_F) \twoheadrightarrow M_2(\mathcal{O}_F/p^n) \xrightarrow{\sim}
M_2(\mathcal{O}_F/\mathfrak{p}^n) \oplus M_2(\mathcal{O}_F/\overline{\mathfrak{p}}^n) \cong M_2(\mathbf{Z}/p^n) \oplus
M_2(\mathbf{Z}/p^n).$$ Since the canonical map $\SL_2(\mathcal{O}_F) \rightarrow
\SL_2(\mathcal{O}_F/p^n) \cong \SL_2(\mathbf{Z}/p^n) \oplus \SL_2(\mathbf{Z}/p^n )$ is surjective
(\cite{Serre70}, p. 490), it is enough to find $A_1, A_2 \in
\SL_2(\mathbf{Z}/ p^n \mathbf{Z})$ such that \begin{equation} \label{conj1} A_2^t M A_1 = \left[ \begin{smallmatrix}
\alpha \\
& \delta \end{smallmatrix} \right]\end{equation}
with $\alpha \neq 0$ mod $p$.
The existence of such $A_1$ and $A_2$ is clear.
\end{proof}

\subsection{Invariance under $T_{\mathfrak{p}}$}

\begin{lemma} \label{decomp1} We have the following decomposition
\begin{equation} \begin{split} T_{\mathfrak{p}}  = & \bigsqcup_{a,b,c \in \mathbf{Z}/p \mathbf{Z}} K_p
\left(
\left[ \begin{smallmatrix} p^{-1} &
ap^{-1} & bp^{-1} & cp^{-1} \\ & 1\\ && 1\\ &&&1 \end{smallmatrix} \right], \left[ \begin{smallmatrix} 1&&b\\ &1
& c \\ && p \\ && -a & 1 \end{smallmatrix} \right] \right)  \sqcup \\
& \bigsqcup_{d,e \in \mathbf{Z}/p \mathbf{Z}}  K_p \left( \left[ \begin{smallmatrix} 1 \\ & p^{-1} &
dp^{-1}
& ep^{-1} \\ && 1 \\ &&& 1 \end{smallmatrix} \right] , \left[ \begin{smallmatrix} 1 &&& d \\ & 1 && e \\ && 1 \\
&&& p \end{smallmatrix} \right] \right)\sqcup \\
& \bigsqcup_{f \in \mathbf{Z}/p \mathbf{Z}} K_p \left( \left[ \begin{smallmatrix} 1\\ & 1 \\ && p^{-1} &
p^{-1} f \\ &&& 1 \end{smallmatrix} \right] , \left[ \begin{smallmatrix} p \\ -f & 1 \\ && 1 \\ &&& 1 \end{smallmatrix} \right]
\right) \sqcup \\
& K_p \left( \left[ \begin{smallmatrix} 1 \\ & 1 \\ && 1 \\ &&& p^{-1} \end{smallmatrix} \right], \left[ \begin{smallmatrix} 1 \\ & p
\\ && 1 \\ &&& 1 \end{smallmatrix} \right] \right).
\end{split} \end{equation} \end{lemma}

\begin{proof} This follows from an analogous decomposition for
$\GL_4(\mathbf{Q}_p)$.
\end{proof}

Let $g = T_{\mathfrak{p}} f$. Then for $q \in \GL_2(\mathbf{A}_F)$ and $\sigma \in S(\mathbf{A})$,
we
can write $$g \left( \left[ \begin{matrix} q & \sigma \hat{q} \\ & \hat{q} \end{matrix} \right] \right)
= \sum_{h \in S} c_g(h,q) e_{\mathbf{A}}(\textup{tr}\hspace{2pt} (h \sigma)).$$ Define the following
matrices $$\alpha'_a = \left(\left[ \begin{matrix} p & -a \\ & 1 \end{matrix} \right], I_2 \right) \in
\GL_2(\mathbf{Q}_p) \times \GL_2(\mathbf{Q}_p)
\quad
a=0,1, \dots,
p-1$$ and $$\alpha'_p = \left( \left[ \begin{matrix} 1 \\ & p \end{matrix} \right], I_2 \right),\in
\GL_2(\mathbf{Q}_p) \times \GL_2(\mathbf{Q}_p).$$
For $a=0,1,
\dots, p$, set $\alpha_a= \iota_p(\alpha'_a) \in \GL_2(F_p)$.
\begin{lemma} \label{Fourier1} One has the following formula
$$c_g(h,q) = p^2 \sum_{a=0}^{p}
 c_f(h, q \alpha_a) + \sum_{a=0}^{p} c_f(h, q \hat{\alpha}_a).
$$\end{lemma}

\begin{proof} This is a straightforward calculation using Lemma
\ref{decomp1}. \end{proof}

\begin{prop} \label{invariance} The Maass space is invariant under the
action of $T_{\mathfrak{p}}$, i.e., if $f\in \mathcal{M}_k^{\textup{M}}$, then $g\in
\mathcal{M}^{\textup{M}}_k$.
\end{prop}

\begin{proof} Choose a base $\mathcal{B}$ as in Corollary \ref{scalarcor}. In
particular we have $\epsilon(b^* h
b) = \epsilon(h)$ and $\val_p (\det b^* b) = 0$. By Propositions \ref{to
check} and \ref{independence}
it is enough to show that there exist functions $c_{b, g}: \mathbf{Z}_+
\rightarrow \mathbf{C}$ ($b \in \mathcal{B}$) such that \begin{equation} \label{condrep} c_g(h,b) =
e^{-2\pi\textup{tr}\hspace{2pt}  h}
\sum_{\substack{ d \in \mathbf{Z}_+ \\ d \mid \epsilon(h)}} d^{k-1} c_{b,g}
\left( D_F d^{-2} \det h \right).\end{equation} For $b \in \mathcal{B}$, set $b' = b
\alpha_p$. Note that all of the matrices: $b \alpha_a$,
$b \hat{\alpha}_a$, ($a=0,1, \dots, p$) belong to the same
class $b'$. Denote any of those matrices by $q$. Then $q =
\gamma_{b', q} b' \kappa_q \in \GL_2(F) b' K'$ and it is easy to see that
$$\det \gamma_{b', q}^k = \begin{cases} 1 & q = b \alpha_a, \hspace{5pt} a = 0, 1,
\dots, p \\ p^{-k} & q = b \hat{\alpha}_a \hspace{5pt} a = 0, 1,
\dots, p ,
\end{cases}$$ and $$\val_p(\det q^* q ) =\begin{cases} 1 & q = b
\alpha_a \hspace{5pt} a = 0, 1,
\dots, p \\ -1 & q = b \hat{\alpha}_a \hspace{5pt} a = 0, 1,
\dots, p .
\end{cases}$$ Write $h = \epsilon(h) h'$. One has $\epsilon(h') = 1$.
Set $D= D_F \det h$ and $D' = D_F \det h'$.
Using Lemma
\ref{Fourier1} and the fact that $f$ is a Maass form, we obtain
\begin{multline} \label{Fourier2} c_g(h,b) = e^{-2 \pi \textup{tr}\hspace{2pt} h} \times
\left( p^2
\sum_{a=0}^p \sum_{\substack {d \in \mathbf{Z}_+ \\ d \mid \epsilon(\alpha_a^* h
\alpha_a)}} d^{k-1} c_{b',f}(D d^{-2} p) + \right. \\
+ \left. p^k \sum_{a=0}^p \sum_{\substack {d \in \mathbf{Z}_+ \\ d \mid
\epsilon(\hat{\alpha}_a^*
h
\hat{\alpha}_a)}} d^{k-1} c_{b',f}(D d^{-2} p^{-1})\right). \end{multline}
Using Proposition \ref{diagonal}, one can relate $\epsilon(\alpha_a^* h \alpha_a)$
and $\epsilon(\hat{\alpha}_a^* h \hat{\alpha}_a)$ to $\epsilon(h)$ for $a=0,1,
\dots, p$, and then
(\ref{Fourier2}) becomes \begin{multline} \label{mess} c_g(h,b)  =
e^{-2 \pi
\textup{tr}\hspace{2pt} h} p^2 \sum_{0} A^{(1)}_d +
e^{-2 \pi \textup{tr}\hspace{2pt} h} \times \\
\times \begin{cases} p^3 \sum_0
A^{(1)}_d
& p \nmid D', p \nmid \epsilon(h)\\
p^k(p+1) \sum_{-1} A^{(-1)}_d + p^3 \sum_0 A_d^{(1)} & p \nmid D', p \mid
\epsilon(h)
\\
p^2(p-1) \sum_0 A_d^{(1)} + p^2 \sum_1 A_d^{(1)} + p^k \sum_0  A^{(-1)}_d&
p \mid
D', p
\nmid
\epsilon(h)\\
p^2(p-1) \sum_0 A_d^{(1)} + p^2 \sum_1 A_d^{(1)} + p^k \sum_0  A^{(-1)}_d
+
p^{k+1}
\sum_{-1}
 A^{(-1)}_d
& p \mid
D', p \mid \epsilon(h), \end{cases} \end{multline}
where $\sum_n A_d^{(m)}= \sum_{d \mid
p^n \epsilon(h)}A_d^{(m)}$, $A^{(m)}_d = d^{k-1} c_{b',f}(D d^{-2} p^m)$.

For $D$ in the image of the map $h
\mapsto D_F \epsilon(h)^{-2} \det h$ and $b \in \mathcal{B}$ we make the following
definition
\begin{equation} \label{maass1} c_{b,g}(D) =  p^2(p+1 ) c_{b', f} (Dp) + p^k (p+1)
c_{b',f} (Dp^{-1}) ,\end{equation} where we assume that
$c_{b',f}(n)=0$ when $n \not\in \mathbf{Z}_+$. If $D$ is not in the image of
that map, we set $c_{b,g}(D)=0$. Note that we clearly have
$$c_g(h,b) = e^{-2 \pi \textup{tr}\hspace{2pt} h} c_{b,g} (D_F \det h)$$ for every $h$ with
$\epsilon(h)=1$. Thus to check if $g$ lies in the Maass space we just need
to check that (\ref{condrep}) holds with $c_{b,g}$ defined by
(\ref{maass1}).
This is an easy calculation using (\ref{mess}). \end{proof}

\subsection{Invariance under $U_{\mathfrak{p}}$}

This is completely analogous to the proof for $T_{\mathfrak{p}}$, hence we only
include the relevant formulas for the reader's convenience.
\begin{lemma} We have the following decomposition:
\begin{equation} \begin{split} U_{\mathfrak{p}} = & \bigsqcup_{b,c,d,e \in \mathbf{Z}/ p \mathbf{Z}} K_p
\left( \left[ \begin{smallmatrix} p^{-1} & & bp^{-1} & dp^{-1} \\ & p^{-1} & cp^{-1} & ep^{-1}
\\ && 1 \\ &&&1 \end{smallmatrix} \right] , \left[ \begin{smallmatrix} 1 && b & c \\ & 1 & d & e \\ &&p \\ &&& p
\end{smallmatrix} \right] \right) \sqcup \\
& \bigsqcup_{a,c,f \in \mathbf{Z}/p \mathbf{Z}} K_p \left( \left[ \begin{smallmatrix} 1\\
-fp^{-1}
& p^{-1} &  cp^{-1} \\ && 1 \\ && -ap^{-1} & p^{-1} \end{smallmatrix} \right], \left[ \begin{smallmatrix} 1 & a
&& c \\ &p
\\ &&1&f \\ &&& p \end{smallmatrix} \right] \right) \sqcup\\
& \bigsqcup_{e,f \in \mathbf{Z}/p\mathbf{Z}} K_p\left( \left[ \begin{smallmatrix} 1 \\ -fp^{-1} & p^{-1}
&&
ep^{-1} \\ && p^{-1} \\ &&&1 \end{smallmatrix} \right], \left[ \begin{smallmatrix} p \\ & 1 && e \\ && 1& f \\
&&& p \end{smallmatrix} \right]\right) \sqcup \\
& \bigsqcup_{a,b \in \mathbf{Z}/ p \mathbf{Z}} K_p\left( \left[ \begin{smallmatrix} p^{-1} && bp^{-1} \\ &
1
\\ && 1 \\ && -ap^{-1} & p^{-1}
\end{smallmatrix} \right] , \left[ \begin{smallmatrix} 1 & a& b \\ & p \\ && p \\ &&&1 \end{smallmatrix} \right] \right) \sqcup \\
& \bigsqcup_{d\in \mathbf{Z}/ p \mathbf{Z}} K_p\left( \left[ \begin{smallmatrix} p^{-1} &&& dp^{-1} \\ &1
\\
&& p^{-1} \\ &&& 1 \end{smallmatrix} \right] , \left[ \begin{smallmatrix} p\\ & 1& d \\ &&p \\ &&&1 \end{smallmatrix} \right]
\right) \sqcup \\
& K_p \left( \left[ \begin{smallmatrix} 1 \\ & 1 \\ && p^{-1} \\ &&& p^{-1} \end{smallmatrix} \right], \left[ \begin{smallmatrix} p
\\
&
p \\ && 1 \\ &&&1 \end{smallmatrix} \right] \right).\end{split} \end{equation}

\end{lemma}

As before, let $g = T_{\mathfrak{p}} f$. Define matrices:
$$\beta'_p = \left( \left[ \begin{matrix} p \\ & p \end{matrix} \right], I_2\right) \in
\GL_2(\mathbf{Q}_p)\times
\GL_2(\mathbf{Q}_p),$$
$$\gamma'_a = \left( \left[ \begin{matrix} 1 \\ a & p \end{matrix} \right], I_2\right) \in \GL_2(\mathbf{Q}_p)
\times \GL_2(\mathbf{Q}_p),\quad a=0,1,\dots, p-1,$$
$$\gamma'_p = \left( \left[ \begin{matrix} p \\ & 1 \end{matrix} \right], I_2 \right) \in \GL_2(\mathbf{Q}_p)
\times \GL_2(\mathbf{Q}_p),$$ and set $\beta_p = \iota_p(\beta'_p)\in
\GL_2(F_p)$, $\gamma_a= \iota_p(\gamma'_a)\in \GL_2(F_p)$ ($a=0,1, \dots,
p$).

\begin{lemma} One has the following formula
$$c_g(h,q) = p^4 c_f(h,q \beta_p) + c_f(h, q \hat{\beta}_p) + p
\sum_{a=0}^p \sum_{b=0}^p c_f(h, q \gamma_a
\hat{\gamma}_b).$$ \end{lemma}

\begin{prop} \label{invariance2} The Maass space is invariant under the
action of $U_{\mathfrak{p}}$. \end{prop}

\begin{proof} This is similar to the proof of Proposition
\ref{invariance}. Let $b$, $D$, $D'$, $h$, $h'$ be as in that proof. Then
for all
$a$, $c$, we
see that $b\beta_p$, $b \hat{\beta}_p$, $b \gamma_a \hat{\gamma}_c$ all
lie in the same class $b'= b \beta_p$.
One has $$\det
\gamma_{b', q}^k = \begin{cases} 1 & q = b \beta_p \\ p^{-k} & q = b
\gamma_a \hat{\gamma}_c, \hspace{5pt} a,c \in \{ 0,1, \dots, p \} \\ p^{-2k} & q =
b \hat{\beta}_p, \end{cases}$$ and $$\val_p (\det q^* q) = \begin{cases} 2
& q = b \beta_p \\ 0 & q = b
\gamma_a \hat{\gamma}_c, \hspace{5pt} a,c \in \{ 0,1, \dots, p \} \\ -2 & q =
b \hat{\beta}_p. \end{cases}$$

Using Proposition \ref{diagonal} as in the proof of Proposition
\ref{invariance} we obtain
\begin{multline} c_g(h,b) = e^{-2 \pi \textup{tr}\hspace{2pt} h} p^4 \sum_1A^{(2)}_d + p^{2k}
\sum_{-1} A^{(-2)}_d + p^{k+1}(p+1) \sum_0 A^{(0)}_d + p^{k+3} \sum_{-1}
A^{(0)}_d+\\
+ e^{-2 \pi \textup{tr}\hspace{2pt} h} p^{k+1} \begin{cases} p \sum_{-1} A^{(0)}_d & p
\nmid D'\\
p \sum_0 A^{(0)}_d & p \mid D', \hspace{2pt} p^2 \nmid D'\\
\sum_1 A^{(0)} + (p-1) \sum_0 A^{(0)}& p^2 \mid D',\end{cases}
\end{multline}
where if there is no $d$ dividing $p^n \epsilon(h)$, we set $\sum_n = 0$.
For $D$ in the image of the map $h \mapsto D_F \epsilon(h)^{-2} \det h$,
we make the following definition: \begin{multline} \label{for1}
c_{b,g}(D) =
p^4c_{b',f}(Dp^2) + (p^{k+3}+p^{k+2} + p^{k+1}) c_{b',f}(D) + \\
+ \begin{cases} 0 & p \nmid D \\
p^{k+2} c_{b'f}(D) & p \mid D, \hspace{2pt} p^2 \nmid D \\
p^{k+2} c_{b',f}(D) + p^{2k} c_{b',f}(Dp^{-2}) & p^2 \mid D. \end{cases}
\end{multline} We now check as in the proof of Proposition
\ref{invariance} that $g$ is a Maass form.
\end{proof}

\subsection{Invariance under $\Delta_{\mathfrak{p}}$}

Let $g = \Delta_{\mathfrak{p}}f $ and set $\delta_p = \iota_p((p^{-1}I_4, p I_4))$.
Then we have $$c_g(h,q) = c_f(h, q \delta_p).$$
\begin{prop} \label{invariance3} The Maass space is invariant under the
action of $\Delta_p$. \end{prop}

\begin{proof}
Let $\mathcal{B}$ be as before. For $b \in \mathcal{B}$ set $b' = b
\delta_p$. One clearly has $\val_p( \delta_p^* \delta_p) = 0$ and
$\epsilon(\delta_p^* h \delta_p) = \epsilon(h)$. Hence one can define
\begin{equation} \label{for2} c_{b,g}(D) =
c_{b',f}(D).\end{equation} The claim is now clear. \end{proof}

\section{Descent} \label{Descent}

Assume that $h_F$ is odd. Let $\chi_F$ be the quadratic Dirichlet
character attached to the extension $F/\mathbf{Q}$. For a positive integer $n$,
set $$a_F(n) =
\# \{ \alpha \in (i D_F^{-1/2} \mathcal{O}_F)/\mathcal{O}_F \mid D_F N_{F/\mathbf{Q}}(\alpha)
\equiv -n \pmod{D_F}\}.$$ Let $\mathcal{B}$ be a base as in Corollary
\ref{scalarcor}.

\begin{thm} \label{desc4} There exists a $\mathbf{C}$-linear injection of vector
spaces
$$\Desc_{\mathcal{B}}: \mathcal{M}_k^{\textup{M}} \hookrightarrow \prod_{b \in \mathcal{B}} M_{k-1}
(D_F,
\chi_F),$$
such that $\Desc(f) = (F_b)_{b \in \mathcal{B}}$ with $$a_{F_b}(n) = i
\frac{a_{F}(n)}{\sqrt{D_F}} c_{b,f}(n),$$ where $a_{F_b}(n)$ is the
$n$-th Fourier coefficient of $F_b$. The map $\Desc_{\mathcal{B}}$
depends on the choice of $\mathcal{B}$. \end{thm}

\begin{proof} This follows immediately from \cite{Krieg91}, Theorem 6 and
formula (4) using our assumption on $\mathcal{B}$ and (\ref{f12}). \end{proof}

\begin{rem} Krieg in \cite{Krieg91} explicitly describes the image of the
descent map he defines and denotes it by $G_{k-1}(D_F, \chi_F)^*$. The
image of $\Desc$ is exactly $\prod_{b \in \mathcal{B}} G_{k-1}(D_F, \chi_F)^*
\subset \prod_{b \in \mathcal{B}} M_{k-1}(D_F, \chi_F)$.
\end{rem}

Let $S_{\mathcal{B}}$ denote the group of permutations of $\mathcal{B}$. For $b \in \mathcal{B}$,
let $\sigma_{\mathfrak{p}, n} \in S_{\mathcal{B}}$ be the unique permutation such that $b
\alpha_p^n$ is
in the class of $\sigma_{\mathfrak{p},n}(b)$. If $A=(a_b)_{b \in \mathcal{B}}$ is an
ordered tuple
indexed by elements of $\mathcal{B}$, we define $\sigma_{\mathfrak{p},n}A =
(a_{\sigma_{\mathfrak{p},n}^{-1}(b)})_{b \in \mathcal{B}}$. Moreover, for $b \in \mathcal{B}$
and $n \in \mathbf{Z}$, write $\gamma_{b,\mathfrak{p},n}$ for an element of $G(\mathbf{Q})$ such
that
$b
\alpha_p^n = \gamma_{b,\mathfrak{p},n} \sigma_{\mathfrak{p},n}(b) \kappa$ for $\kappa \in K$, and
denote
by $\gamma_{\mathcal{B},\mathfrak{p},n}$ the $\mathcal{B}$-tuple $(\det\gamma_{b,\mathfrak{p},n}^*)_{b \in \mathcal{B}}$.

For
a
rational prime $p \nmid D_F$ denote by $T_p$ the
operator acting on $\prod_{b \in \mathcal{B}} M_{k-1}(D_F, \chi_F)$ which sends
$(F_b)_{b \in \mathcal{B}}$ to $(F'_b)_{b \in \mathcal{B}}$, where
$F_b(z) =
\sum_{n=1}^{\infty} a(n) e^{2 \pi i nz}$, $F'_b(z) =
\sum_{n=1}^{\infty} a'(n)
e^{2 \pi i nz}$ with $a'(n) = a(np) + \chi_F(p)p^{k-2}
a(n/p)$. Here $a(m)=0$ if $m \not\in \mathbf{Z}_{\geq 0}$. Denote by
$\mathbf{T}_p$
the
$\mathbf{C}$-subalgebra of endomorphisms of $\prod_{b \in \mathcal{B}} M_{k-1}(D_F,
\chi_F)$ generated by $T_p$, the group $S_{\mathcal{B}}$ and $\mathcal{B}$-tuples of
complex numbers acting on $\prod_{b \in \mathcal{B}} M_{k-1}(D_F,
\chi_F)$ in an obvious way.

\begin{thm} \label{heckedesc} Let $p$ be a rational prime which splits in
$F/\mathbf{Q}$. There
exists a $\mathbf{C}$-algebra map $$\Desc_{\mathcal{B},p}: \mathcal{H}_p \rightarrow
\mathbf{T}_p,$$ such
that for
every $H \in \mathcal{H}_p$ the
following diagram
$$\xymatrix@C7em{\mathcal{M}_k \ar[r]^{H} \ar[d]^{\Desc_{\mathcal{B}}} & \mathcal{M}_k
\ar[d]_{\Desc_{\mathcal{B}}} \\ \prod_{b \in \mathcal{B}} M_{k-1} (D_F, \chi_F)
\ar[r]^{\Desc_{\mathcal{B},p}(H)} & \prod_{b \in \mathcal{B}} M_{k-1} (D_F,
\chi_F)}$$ commutes. Moreover, one has \begin{equation} \begin{split}
\Desc_{\mathcal{B},p}(T_{\mathfrak{p}}) & = (\gamma_{\mathcal{B},\mathfrak{p},1})^{-k} p^2 (p+1) T_p \circ
\sigma_{\mathfrak{p},1}, \\
\Desc_{\mathcal{B},p}(U_{\mathfrak{p}})& =(\gamma_{\mathcal{B},\mathfrak{p},2})^{-k} p^4( T^2_p
+p^{k-1}+p^{k-3})\circ
\sigma_{\mathfrak{p},2} \\
\Desc_{\mathcal{B},p}(\Delta_{\mathfrak{p}}) &=
(\gamma_{\mathcal{B},\mathfrak{p},4})^{-k}\sigma_{\mathfrak{p},4}.\end{split} \end{equation}

\end{thm}

\begin{proof} This follows from Theorem \ref{desc4},
and formulas (\ref{maass1}), (\ref{for1}), and (\ref{for2}).
Just for illustration, we include the argument in the case of $T_{\mathfrak{p}}$.
Let $f \in \mathcal{M}_k^{\textup{M}}$ and set $g=T_{\mathfrak{p}}f$. Fix $b \in \mathcal{B}$ and write
$b_1$ for $\sigma_{\mathfrak{p},1}(b)$ and $b'$ for $b
\alpha_{p}$. There exist diagonal matrices $\gamma \in G(\mathbf{Q})$ and $\kappa \in K$
such that $b'=\gamma b_1 \kappa$ (hence also $b_1 = \gamma^{-1} b' \kappa^{-1}$).
Identify $\mathcal{M}^{\textup{M}}_k$ with $\prod_{b \in \mathcal{B}} \mathcal{M}_k^{\textup{M}}(G(\mathbf{Z}))$ via
$f' \mapsto (f'_b)_{b \in \mathcal{B}}$. For $\phi \in \mathcal{M}_k^{\textup{M}}(G(\mathbf{Z}))$, $h
\in T$, denote
by $c_{\phi}(h)$ the $h$-Fourier coefficient of $\phi$. We will study the
action of $T_{\mathfrak{p}}$ on $c_{f_{b_1}}(h)$. Since $f$ is a Maass form it is
enough to consider $h$ of the form $\left[ \begin{smallmatrix} 1&* \\ *&*\end{smallmatrix} \right]$. Fix such an
$h$. Set $D=D_F \det h$. Then by (\ref{f12}) and (\ref{maass1}),
$$c_{g_b}(h) = e^{2 \pi \textup{tr}\hspace{2pt} h} c_g(h,b) = p^2 (p+1) (c_{b',f}(Dp) +
p^{k-2} c_{b',f}(Dp^{-1})).$$ By (\ref{f12}) and (\ref{Maass condition}),
we have \begin{equation} \begin{split} c_{f_{b_1}}\left(h \left[ \begin{smallmatrix} 1 \\ & p \end{smallmatrix} \right]
\right) &
= e^{2 \pi \textup{tr}\hspace{2pt}
h \left[ \begin{smallmatrix} 1 \\ & p \end{smallmatrix} \right]} c_f \left( h \left[ \begin{smallmatrix} 1 \\ & p \end{smallmatrix} \right], b_1 \right)
=
c_{b_1,f}(Dp) =
(\det \gamma^*)^k c_{b',f}(Dp)\\
 c_{f_{b_1}}\left(h \left[ \begin{smallmatrix} 1 \\ & 1/p \end{smallmatrix} \right] \right) & = e^{2 \pi \textup{tr}\hspace{2pt}
h \left[ \begin{smallmatrix} 1 \\ & 1/p \end{smallmatrix} \right]} c_f \left( h \left[ \begin{smallmatrix} 1 \\ & 1/p \end{smallmatrix} \right],
b_1
\right) = c_{b_1,f}(Dp^{-1}) =\\
&= (\det \gamma^*)^k c_{b',f}(Dp^{-1}),\end{split} \end{equation} where we have used the
fact that $\epsilon(h) = \epsilon\left(h \left[ \begin{smallmatrix} 1 \\ & p \end{smallmatrix} \right]\right) =
\epsilon\left(h \left[ \begin{smallmatrix} 1 \\ & 1/p \end{smallmatrix} \right]\right)=1$ for $h$ as above
(the last equality holding for $h$ such that $h \left[ \begin{smallmatrix} 1 \\ & 1/p
\end{smallmatrix} \right] \in T$). This gives us $$c_{g_b}(h) = p^2 (p+1) (\det \gamma^*)^{-k}
(c_{b_1,f}(Dp) + p^{k-2} c_{b_1,f}(Dp^{-1})).$$ The claim now follows from
Theorem \ref{desc4}.
\end{proof}

For completeness we also include the analogue of Theorem \ref{heckedesc}
for an inert
$p$. It can be proved in the same way or can be
deduced
from the results of section 3 of \cite{Gritsenko90}.

Let $p$ be an inert prime. The Hecke algebra $\mathcal{H}_p$ is generated by the
double cosets $T_{\mathfrak{p}}:=K_p \diag(p^{-1}, 1, p, 1 ) K_p$ and $U_{\mathfrak{p}}:=K_p
 \diag (p^{-1}, p^{-1}, p,p ) K_p$.

\begin{thm} \label{heckedesc2} Let $p$ be a rational prime which is inert  in
$F/\mathbf{Q}$. There
exists a $\mathbf{C}$-algebra map $$\Desc_{\mathcal{B},p}: \mathcal{H}_p \rightarrow \mathbf{T}_p,$$
such
that for
every $H \in \mathcal{H}_p$ the
following diagram
$$\xymatrix@C7em{\mathcal{M}_k \ar[r]^{H} \ar[d]^{\Desc_{\mathcal{B}}} & \mathcal{M}_k
\ar[d]_{\Desc_{\mathcal{B}}} \\ \prod_{b \in \mathcal{B}} M_{k-1} (D_F, \chi_F)
\ar[r]^{\Desc_{\mathcal{B},p}(H)} & \prod_{b \in \mathcal{B}} M_{k-1} (D_F,
\chi_F)}$$ commutes.  Moreover, one has \begin{equation} \begin{split}
\Desc_{\mathcal{B},p}(T_{\mathfrak{p}}) & = p^{-k+4}(p^2+1) T_p^2+ p^4 + p^3+ p
- 1,\\
\Desc_{\mathcal{B},p}(U_{\mathfrak{p}}) & = p^8(T_p^4 + (p+3)p^{k-2} T_p^2 +
p^{2k-4}(p^2+p+1)).\end{split} \end{equation} \end{thm}

redpajama_set_name: RedPajamaArXiv


Matthew Hochstrasser has accumulated over 20 years of diverse experience as an entrepreneur and business executive. He has skillfully completed numerous transactions in the world of finance and real estate. Throughout the last ten years he has assisted multiple companies in achieving their business funding goals. Matthew has acted as a spokesperson for various Fortune 500 companies. He also has owned and operated several successful businesses.
Matthew was educated at La Salle Academy where he learned to operate in a highly efficient manner. In college he studied marketing and won numerous awards for his academic mastery. Matthew studied at The University of Erasmus and in Italy. He has a keen understanding of diverse cultures. Matthew has had a Real Estate License since 1997. Many companies have hired Matthew to assist in developing and implementing marketing strategies. He has been heavily involved with many charitable organizations, including the Art Institute for Children.
Matthew's experiences in sales, deal structures, business development, and negotiations bring forth a powerful wealth of experience. The dynamic experiences achieved throughout his career have proven to be an asset to both buyers and sellers of businesses, as they navigate through this complex process.


For full nutrition info, click here!
Is there a vegetarian or vegan protein powder you recommend for the smoothie… to replace the whey?
Hi Ashley! You can use hemp protein powder for a vegan option.
How many carbs are in one serving of this Smoothie?
Hi Juliet! You can find the full nutrition facts here!


Hair Extensions or Hair Integrators are basically adding synthetic or human hair to one's existing hair.
At Present, It isn't just an extension of your head, but has become the extension of your personality. It is the hottest addition to the world of hair care and hair style. With High Quality of Berkowits' Hair Extensions in Delhi, You can get Instant Longer, Thicker & Fuller Hair.
Although its primary use continues to be to add volume and length to one's existing hair, it is also being used by many to 'wear' various styles. Women have started using human Hair Extensions in different colors to substitute hair coloring as it protects their hair from harmful chemicals.
At Berkowits, we believe all Hair Extensions should be DIY (Do It Yourself). You should have the option of putting it on and removing it at your own will. Therefore, we emphasize clip-On Hair Extensions online over any other method.
It consists of 4 to 7 pieces of Hair Extensions that are meant to add overall volume. If you are looking for frontal volume, this might be your answer. Below is a picture of a model wearing a 4 Piece Hair Extension.
Hair Volumizers are a single piece of many hair extensions that are meant to add volume and length from the back. If you are looking for additional length, this is just what you need. Pictures of a model wearing a Volumizer either sides.
Highlighters are single pieces of Hair Extensions that are available in various colors such as Amber Red, Lightening Blue, Copper Blonde, Silver white, copper blonde etc, It is a great option at best price that matches the style of Hair Coloring while avoiding exposure to harmful chemicals.
PRP stands for Platelet Rich Plasma. the fundamental idea at work behind this treatment is rather simple – the platelet present in your blood.
Micro Transplantation of Autologous Skin Cells or Autologous Stem Cell Treatment is a breakthrough treatment pioneered in North India.


In the name of God Amen. I John Ellis at present in St. John's in the Island of Newfoundland Gentleman but residing in Guysborough Nova Scotia do make this my last will and testament- I will and bequeath unto my executors hereafter named and the survivor of them all the land and property both real and personal of what nature and description soever and wheresoever of which I shall die possessed or entitled to upon trust for the sole use and behoof of my dear wife Jane for and during the term of her natural life and after her decease upon trust for the use and behoof of my children living at the time of my decease, share and share alike- I appoint William H. Mare of St. John's aforesaid, Merchant and William Vallance Whiteway of St. John's Barrister at law, executors of this my last will and testament and I hereby revoke all former wills by me made.
(Signed) John Ellis (LS) Signed sealed and declared by me the said John Ellis as and for my last will and testament at St. John's aforesaid this fifth day of June Anno Domini eighteen hundred and sixty one- in presence of Signed T. H. Mabin, Signed, Richard Rewell.


https://www.globe-flight.de/DJI-S800-Spreading-Wings-Arm-Spare_3 0556 DJI S800 Arm (rotation left, green LED) https://www.globe-flight.de/media/image/product/598/lg/dji-s800-spreading-wings-arm-spare_3.jpg 99.00 289.00 EUR InStock /Special sales and used items /DJI Drones & Multicopters/Multicopter parts DIY/Multicopter Frames/DJI/DJI S800 / S800 EVO Spares This Product include all parts, that you only have to change the whole copter arm and start again.
This Article is the Copter-Arm with the "O" marking.
This Product include all parts, that you only have to change the whole copter arm and start again.
The Arm includes the Motor, Brushlesscontroller and the Prop.


text:
It's a Pivotal Week in Harris' 'Fraught, Fascinating' Tenure
All political eyes on the vice president as she heads to Guatemala and Mexico
By John Johnson, Newser Staff
Posted Jun 7, 2021 9:34 AM CDT
Vice President Kamala Harris waves goodbye prior boarding Air Force Two at Andrews Air Force Base, Md., Sunday, June 6, 2021, en route to Guatemala City. (AP Photo/Jacquelyn Martin)
Kamala Harris has embarked on her first foreign trip as vice president, though things got off to a bumpy start. Her plane out of Joint Base Andrews had to turn back after 30 minutes because of an unspecified "technical issue" on Sunday, reports the AP. Harris, though, boarded a second plane and has since landed in Guatemala City. Now comes the hard part of trying to put a dent in the number of migrants heading north to the US—and, in the bigger picture, navigating an especially complicated political terrain. Coverage:
The trip: Harris visits Guatemala on Monday and Mexico on Tuesday, all part of her mandate as VP to ease the immigration crisis. She'll be focusing on what officials have called "root issues" behind the surge, including corruption. The latter "really does sap the the wealth of any country, and in Central America is at a scale where it is a large percentage of GDP across the region," says special envoy Ricardo Zuniga.
The map: A look at a map shows why Guatemala is key, notes USA Today. Migrants from El Salvador and Honduras to the south must pass through Guatemala on their way to Mexico and the US border. Plus, the US has better relations with Guatemala than with the other two nations in what's known as the Northern Triangle.
The risk: An analysis by Lisa Lerer in the New York Times wonders whether Harris is "drawing the shortest straws in the White House." Because she's not only leading the White House mission on immigration but on voting-rights reform. Both "are politically fraught problems with no easy solutions," and Lerer digs into the details. The rewards are great if Harris succeeds, but failure puts her political future—including a possible 2024 run—in jeopardy because her critics love to pile on. Is "Biden setting up Kamala Harris for failure in 2024," asks a headline on an op-ed in USA Today by Peter Funt.
The criticism: For a sample of how Harris' critics on the right already are pouncing, see this column by Charles Cooke in the National Review. He calls Harris a "phony" under a headline of "The Democrats Have a Kamala Harris Problem." Cooke predicts Democratic regret in the near future. "By 2024, Republicans may well be politically ascendant—and, if they are, the Democrats are going to wish that they hadn't allowed a dud such as Kamala Harris to become the most vibrant face of their brand."
The dilemma: Harris "can't win," writes Frank Bruni in the New York Times. "She confronts the confines in which a vice president has to operate on top of the similar confines in which Black people and women in positions of power are often expected to operate," he writes. "It's a Goldilocks double or even triple whammy. Too strong a voice and you're stepping outside of your place. Too soft a voice and you're timidly failing to rise to the occasion." All in all, he adds, "this is one fraught, fascinating vice presidency."
Summing up: In an analysis at CNN, Stephen Collinson goes over much of the above. "Neither the immigration nor the voting rights assignments offer much promise for success," he writes. "But they are an opportunity for Harris to create her own White House legacy—and will test her skills and qualifications for the one executive office higher than the one she already holds."
(Read more Kamala Harris stories.)


text:
Borderlands 2 Launches, Delivers Deeper Story, Same Satisfying, Fun Gameplay
by Joel Hruska — Tuesday, September 18, 2012, 02:03 PM EDT
The original Borderlands was a surprise smash hit, selling millions of copies more than its developer, Gearbox, ever thought it would. The first game's randomized weapon generation, unique cell-shaded graphics, steep difficulty, and excellent co-op was a potent combination -- more than enough to overcome complaints about its thin story. Borderlands 2 launched this morning, and early reviews paint it as a further improvement on its predecessor.
The dark humor, crazy NPCs, and frenetic combat of the original game are faithfully replicated and the story is better; Kotaku's Tina Amini writes "The story in Borderlands 2 occasionally gets serious, dark even, and it was something I felt and believed rather than just listened to. I have never been so invested in Pandora and its citizens until this sequel." PC Gamer praises the titles' tweaks to the weapon randomizations, claiming that the end result is an array of weapons that are better tuned to specific instances.
This is a game designed for co-op, quests can be long, brutal, and it's easy to run short of ammo -- but it's also a world that's meant to be explored, observed, and enjoyed. This time around, the main antagonist is a man named Handsome Jack. Jack took credit for the discovery of the Eridian Vault at the end of the original Borderlands and he has a plan to bring the lawless planet of Pandora under control. Needless to say, it's not a very <b>nice</b> plan, and involves killing most of the population and turning the world into an industrial mining center. When the game opens, you're fighting to death in a gladiatorial combat ring for Jack's amusement, then left for dead when your popularity is perceived as a threat. The Guardian Angel from Borderlands revives you, which is where the game begins.
Concept art for the upcoming Mechromancer character
As before, you've got the choice of four unique character classes (Gunzerker, Assassin, Commando, Siren), and the four characters from Borderlands all make NPC appearances at various points in the game. A fifth class, the Mechromancer, drew fire earlier this year for offering what designer John Hemingway referred to as the "girlfriend skill tree," and isn't ready for release yet. She's expected to be added by a DLC content patch at some point in the future.
Ellie. The developers specifically designed her to portray a different view of femininity than games typically show
Borderlands 2 is expected to be one of the major games of Fall 2012, and Nvidia decided to bundle the title with its GeForce 660 Ti that launched last month. Thus far, reactions from players have been just as strong as reviewer impressions; there are some quirks around the game's interface and quest tracking, but nothing as off-putting as, say, Skyrim's default interface. If you've already picked up the game, or are planning to, what's your impression thus far? Does it live up to what Borderlands delivered?
Tags: Gaming, Borderlands, Gearbox, Borderlands 2, Mechromancer, girlfriend mode, co-op


Airbus Aerial is a comprehensive service provider that combines various layers of aerospace data capture infrastructure to help customers solve the problems they have.
It uses commercial drones to offer solutions to sectors that include asset management, catastrophe response, corridor monitoring and site inspection.
Its parent company, Dutch-based Airbus, will be keen to continue to push its subsidiary towards expansion next year in order to keep the business in a prominent position as part the global airspace market.


Daffodil is a common name that refers to the Narcissus genus of flowers. These hardy bulbs are native to Europe, North Africa and Asia. Most species of Narcissus bloom in the spring, but there are several varieties that bloom during the autumn.
Daffodils are easeir to grow than other varieties of bulbous plants. Also, they are typically avoided by garden pests such as rabbits and deer, making them much easier to protect and maintain.
Select daffodil bulbs in the color and size you prefer. Larger daffodil bulbs will grow up to be large flowers, and little bulbs will be on the smaller side when they are full grown flowers. The bulbs you select should be not be hollow; the best bulbs will feel firm to the touch and be heavy for their size.
Plant the daffodil bulbs in a sunny, well-lit area. The soil they are planted in should be well-drained. It is advisable to plant daffodil bulbs on a slight slope to ensure better drainage. The soil should be relatively dry. If it is damp, be sure to add dry sand before planting.
Ideally, daffodil bulbs should be planted with about six inches between each bulb. They should be planted at a depth about three times the height of the bulb, or at least six inches deep.
Before placing the daffodil bulb in the hole you've dug, sprinkle a small amount of daffodil booster on the floor of the hole.
Set the bulb in the bottom of the hole, with the pointed side facing up.
Cover the daffodil bulb with dry soil and pat the surface when finished.
Daffodils do not require watering.
When foliage begins to turn yellow, use pruning shears to remove dead leaves and flowers.


Class and event information for Longfield Suite Prestwich.
The Longfield Suite is a civic suite run by Bury council. Based in the heart of Prestwich village it has a large fully sprung dance floor and fully licensed bar.
You can contact Longfield Suite Prestwich direct by completing the form below and clicking the 'Send' button.
The organiser last reviewed this information on 04/08/2013 18:27:30. This page has been viewed 4834 time(s).


text:
A 'Grease' TV Show Is in the Works
"Grease is the word!"
By Southern Living
Credit: Fotos International/Getty Images
If you can't get enough of the 1978 musical romp known affectionately as Grease, this news is for you. Get excited, because there's a Grease-inspired spinoff coming your way: It's a musical television series called Grease: Rydell High, and it's headed to the new streaming platform HBO Max, which will launch in 2020. The production is being spearheaded by Paramount TV, the organization responsible for bringing the live televised musical version of Grease to FOX in 2016.
While the series won't be a complete remake of the now-classic film starring John Travolta and Olivia Newton-John, it will incorporate many of the elements we remember fondly about the original. According to The Hollywood Reporter, "The series is still set in the 1950s and will feature big musical numbers from the era combined with new original songs." It promises to be a rollicking good time with poodle skirts and dance numbers aplenty, though there are still many details to come, as the production is in its beginning stages.
Paramount is also currently developing a Grease prequel called Summer Lovin'. John August, the writer behind Big Fish, Charlie's Angels, and Charlie and the Chocolate Factory, is set to write the movie, which will take place before the events of the 1978 film. A new television show and a prequel, too? With all these Grease-inspired productions underway, it's making us want to re-watch the original—and sing along, of course. Anyone want to do the hand jive?
WATCH: Science Says Rewatching Your Favorite TV Series Is Good for Your Soul
Do you have a favorite movie musical? Is Grease one of your all-time favorites? What are you watching these days, and what new series are you looking forward to tuning into next year?


Smartphone users record front camera videos for very similar purposes as for still images. As with stills, there is an abundance of mobile apps for editing videos and applying filters and effects. Social media and messaging apps such as Facebook, Instagram, Snapchat, and WhatsApp all allow for sharing video clips as they do for sharing photos. However, in addition to sharing pre-recorded clips, many apps now also allow video live-streaming. Furthermore, people now frequently use front cameras for video calls with applications such as Skype or FaceTime.
The use cases, methods, and equipment we use for our video image quality testing and evaluation are very similar to those for stills, too, so be sure to also read our article on "How we test smartphone front camera still image quality," as it will provide you with useful background knowledge for this article.
Despite the similarities, there are image quality aspects that are unique to video. For example, video stabilization is an important attribute that has a big impact on the overall quality of a video recording. The same is true for continuous focus, as an unstable autofocus can easily ruin an otherwise good clip. The most crucial difference between still image and video testing, however, is the addition of a temporal (time) dimension to most other image quality attributes. The tests for most static attributes that look only at a single frame of a video clip—for example, target exposure, dynamic range, or white balance—are pretty much identical to the equivalent tests for still images. However, a video clip is not a single static image, but is rather many frames that are recorded and played back in quick succession. We therefore also have to look at image quality from a temporal point of view—for example, how stable are video exposure and white balance in uniform light conditions? How fast and smooth are exposure or white balance transitions when light levels or sources change during recording?
Temporal testing attributes aside, other testing conditions and procedures are almost identical to those for still images. Our Selfie video testing protocol requires that we shoot several hours of video of test charts and scenes in the DxOMark image quality laboratory, as well as a variety of indoor and outdoor "real-life" scenes. Light conditions include bright outdoor light and many types of artificial light sources in indoor conditions. Our studio test light levels range from a bright 1000 lux down to a very low 1 lux.
As with front camera stills, Selfie videos should be optimized for portraiture at close shooting distances. For video testing, we shoot at subject distances of 30cm and 50 cm, using single and multiple subjects of varying skin tones so as to reliably cover as many use cases as possible. Like the Photo Score, we calculate the DxOMark Selfie Video score using both objective measurements and perceptual scores from a panel of image quality experts. We then condense objective and perceptual measurements into Video sub-scores, to which we apply our proprietary formulas and weighting systems to finally arrive at the Overall score.
This section describes in more detail the video attributes we test and analyze in order to compute our Selfie Video sub-scores.
Our Exposure score looks at target exposure, contrast, and dynamic range from both a static and temporal point of view. As with still images, we weight subject and background exposure differently depending on shooting distance. For closeups, the face is the central image element. With the camera further away from the subject, the background becomes more important. We undertake exposure tests in the lab at light levels from 1 to 1000 lux, using a range of test charts and equipment. In addition, we shoot outdoors on bright sunny days and indoors in our offices.
Static exposure testing is pretty much identical to still image testing and is undertaken under several light sources.
The sample graph below shows how the camera's video exposure system reacts to changes in light conditions, ranging from fairly small steps (5 to 30 lux) to extreme changes in brightness (30 to 1000 lux).
In addition to static tests, our temporal tests look at the effects of changing light levels.
Our experts undertake perceptual evaluation of selfies and group selfies taken at different subject distances and at different light levels.
We test color in the lab at light levels from 1 to 1000 lux, using Gretag ColorChecker charts, our own proprietary DxOMark chart, and a gray chart. We also shoot outside on sunny days and inside the DxOMark offices. As for stills, we use our ellipsoid scoring system for white balance and color rendering.
Our ellipsoid scoring system for white balance: color coordinates within the green circle achieve full score.
Points within the red circle are still acceptable but score lower.
We test color at different light levels and under different light sources.
In the first section of this article, you saw a sample graph that illustrates the problems that can occur during a sudden change of light during video recording. The graphs below shows what happens during slower changes in lighting conditions. For this test, we slowly ramp the light level up from 20 lux to 300 lux, and in the process also slightly change the color temperature. We then do the same test in reverse.
In the left graph you can see a smooth white balance transition, without any sudden changes or other irregularities. In the right graph, the transition is pretty smooth as well, but there is a noticeable overshoot between 15 and 16 seconds. The white balance system overcompensates slightly and then readjusts.
For our perceptual analysis of Color, we check white balance accuracy and repeatability for 2 distances and several light sources, and use group portraits to check the rendering of different types of skin tones. In addition, we check our sample clips for color shading under several tungsten and fluorescent illuminants.
We use group video selfies to perceptually evaluate skin tones.
We evaluate depth of field by checking the detail in the background of portrait clips and the sharpness of subjects in group portraits which are not in the focal plane.
Focus Stability is the only temporal attribute we use in focus testing. At this point in time, not many smartphones come with an autofocus system in the front camera. AF systems can help achieve better sharpness, but can also cause problems in terms of stability while recording.
We use group video selfies to evaluate depth of field.
We test Texture and Noise in the lab at light levels from 1 to 1000 lux and under different light sources, using Dead Leaves charts and the same life-like mannequin we use in our still image testing. We also test outside on sunny days and inside the DxOMark offices using human subjects. We do all our tests at subject distances of 30 and 50cm.
We use a series of life-like mannequins for texture and noise evaluation.
We test texture and edge acutance in static scenes and in scenes with both camera and subject motion. It is important to test both scenarios, as frame-stacking algorithms can easily reduce noise in static scenes, but can often struggle to achieve the same results when some scene elements are in motion. To test the impact of camera shake on video texture and detail, we perceptually analyze footage shot during tripod movement as well as footage recorded handheld.
We analyze both texture and edge detail in smartphone video footage.
To perceptually analyze texture, we concentrate on the areas of the face that should always show good detail—for example, eyes, eyebrows, eyelashes, lips, and beard. To perceptually analyze noise, we look closely at the skin tones of our video subjects.
We look at the same set of artifacts for video testing as for still images, including the ones in the list below, but also keep our eyes open for any other type of video-specific artifacts we might come across.
Stabilization is just as important for front camera video as it is for main camera video. We test video stabilization by hand-holding the camera without motion and by shooting while walking at a subject distance of 30cm and at arms-length. For video calls–one of the most important use cases for front camera video—users typically hand-hold the device close to their faces (30 to 40 cm). For this kind of static video, the stabilization algorithms should counteract all hand-motion, but not react to any movements of the subject's head.
For group selfie videos, we have to check that stabilization works well for all the faces in the frame. In the sample below, you can see that only the main subject's head is well-stabilized, and that there is noticeable deformation on the other subjects in the video.
Walking with the camera is the most challenging use case for stabilization systems, as walking movement has a strong amplitude which requires much more heavy-handed stabilization than a static scene. Inefficient video stabilization in such cases can often result in intrusive deformation effects.