text:
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@property (nonatomic, strong) UISearchController *searchController;
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    self.searchController.delegate = self;
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    self.searchController.searchBar.delegate = self;
    self.searchController.searchBar.scopeButtonTitles = @[@"All", KeywordCategoryShopping, KeywordCategoryRestaurant, KeywordCategoryMovie];

    self.searchController.hidesNavigationBarDuringPresentation = YES;
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}

#pragma mark - UISearchResultsUpdating
-(void)updateSearchResultsForSearchController:(UISearchController *)searchController {
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    JWSearchResultsController *resultsController = (JWSearchResultsController *)searchController.searchResultsController;
    resultsController.filterString = searchController.searchBar.text;
}

#pragma mark - UISearchControllerDelegate
- (void)willPresentSearchController:(UISearchController *)searchController
{
    self.searchController.searchBar.selectedScopeButtonIndex = 0;
}

#pragma mark - UISearchBarDelegate
- (void)searchBar:(UISearchBar *)searchBar selectedScopeButtonIndexDidChange:(NSInteger)selectedScope
{
    if (!self.searchController.active) {
        return;
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    JWSearchResultsController *resultsController = (JWSearchResultsController *)self.searchController.searchResultsController;
    resultsController.searchBarScopeIndex = selectedScope;
    resultsController.filterString = self.searchController.searchBar.text;
}

@end


Background: Chin augmentation is the most common facial reshaping procedure that employs an implant to create its effect. Most people seek it to enhance the projection of the lower face as a stand alone procedure or as a complement to other facial procedures. (e.g., rhinoplasty) The outcome success is fairly high for chin implants, citing a 74% satisfaction rate on the patient website Real Self.
But not all patients have the desired outcome they seek for a variety of reasons. One of the more common is that of an inadequate aesthetic result (not enough chin augmentation or not the correct type of dimensional change) or the realization that they desire a compressive overall jawline enhancement.
The most powerful jawline augmentation effect comes from a custom jawline implant. To no surprise this is so because the total and connected jawline coverage of the implant which has a greater aesthetic impact due to the bony surface area covered. Designing a custom jawline implant in a patient who has an existing chin implant is actually an advantage and not a hindrance. In the required 3D CT scan used in custom implant designing the existing chin implant is both seen as well is digitally removed to create the new implant design. Also the known dimensions of what is both 'right and wrong' with the indwelling chin implant helps guide how a new implant design would offer aesthetic improvement.
Case Study: This young male has a history of chin implant augmentation, which while providing improvement, still lacked the desired horizontal projection and was crooked. The existing chin implant had 9mms of central horizontal projection with a rounded shape. A 3D CT scan shows the asymmetric position of the chin implant.
A custom jawline implant design was done that increased the horizontal projection of the chin to 15mms with a square shape. To prevent the jawline from become too wide, a thin linear connection was done back to the jaw angles which had a width of 7mms and a more defined jaw angle shape.
Under general anesthesia and through his exiting submental incision, the chin implant was removed as well as all capsular tissue on the bone. Through intraoral posterior vestibular incisions, the subperiosteal jaw angle pockets were created and then connected to that of the chin beneath the mental nerves. The jawline implant was then placed as a single piece in a front to back placement technique. Double screw fixation was done at the chin and jaw angles.
The placement of a chin implant, albeit initially successful, may ultimately create the desire for an even greater lower facial augmentation result. This unexpected aesthetic sequelae is actually not rare. While some patient jump in and go right fo a total jawline augmentation effect initially, others approach that end result either in a deliberate or unintended manner.
1) A significant percent of patients who present for custom jawline implants have existing implants in place.
2) Indwelling implants help guide the new custom jawline implant design.
3) How much extra chin augmentation can be obtained in any dimension is not limitless due to the tightness of the soft tissue chin pad and must be carefully considered in the new implant design.


text: Harry Dean Stanton, man! I don't think I've ever seen him be bad in a movie, though he's been in some bad movies, like everybody. But this is the best I've ever seen him. He's playing against type, because usually there is at least a hint of menace in his craggy face (though he played an angel in One Magic Christmas, which just goes to show you that casting against type often yields the best performances, casting directors please take note). He wanders into the picture from the desert, a lost soul in a suit and tie and baseball cap. He's just this side of catatonic. But he has a brother who comes to get him after a 4-year absence. He gradually returns to the human race, and it is a long, slow trip, done amazingly well. He has a son, and a wife, and of course we're dying to find out more about them and whatever cataclysmic event tore them apart. Wim Wenders makes you wait, and it's worth it. Fraulein Nastassja Kinski (who I learn is fluent in English, German, French, Italian and Russian) turns in a great performance as a small-town Texas girl; could have fooled me, and I grew up in Texas. Nine-year-old Hunter Carson is very good as the son. And some scenes were filmed in a peep show that never could have existed (not out in the open, anyway) in Port Arthur, Texas, back when I was going to high school 5 miles from there. What more could I ask for?

redpajama_set_name: RedPajamaCommonCrawl


text:
Dark Days () — третий студийный альбом американской ню-метал-группы Coal Chamber, выпущенный 7 мая 2002 года на лейбле Roadrunner Records. Это первый альбом, в записи которой участвовала новая бас-гитаристка группы Надя Пойлен — она заменила Райану Фосс-Роуз, которая покинула группу после записи альбома. Так же Dark Days стал последним релизом группы, после которого Coal Chamber распались по причине музыкальных разногласий. Вскоре, в 2015 году группа вновь объединилась и выпустила альбом Rivals.

С альбома был выпущен один единственный сингл — «Fiend» в 2002 году, на который был снят видеоклип, показывающий группу то репетирующей в гараже, то выступающей перед фанатами на концерте.

Альбом получил преимущественно смешанные отзывы, но при этом он является самым популярным среди фанатов группы. Dark Days считается самым агрессивным и тяжёлым альбомом группы, в то время как предыдущие релизы группы были более мягче по звучанию. Альбом в основном сочетает в себе стили первого и второго альбомов, скрещивая элементы ню- и готик-метала.

Об альбоме
Dark Days был записан в 2002 году. Басистка Райана Фосс-Роуз покинула группу после записи этого альбома в связи с рождением дочери от барабанщика группы Sevendust Моргана Роуза. Ей нашли замену в лице Нади Пойлен, которая заменяла Райану во время беременности между двумя альбомами группы. Райана поссорилась с фронтменом Дез Фафарой, сказав, что она и её муж «нашли Христа» и она навсегда покидает Coal Chamber.

В мае 2002 года было объявлено о распаде Coal Chamber после ссоры между двумя участниками группы — вокалистом Дез Фафарой и гитаристом Мигелем Расконом во время шоу в Лаббоке, штат Техас. После словесной перепалки Мигель ударил Деза гитарой по голове, после чего фронтмен заявил, что «это последнее выступление Coal Chamber» и покинул сцену. Группа продолжила выступать без участия вокалиста, но вскоре прекратила шоу. Перед тем как покинуть сцену барабанщик Майк Кокс разбил свою ударную установку. Спустя долгое время Coal Chamber всё же сыграли летом на фестивале Last Call with Carson Daly, где по-мимо них так же выступали группы American Head Charge, Lollipop Lust Kill и Medication. Это было последнее живое выступление группы.

Позже Дез выпустил пластинку под названием Giving the Devil His Due — сборник, состоящий из би-сайдов, ремиксов и демоверсий песен Coal Chamber. После он приступил к работе в своей группе DevilDriver.

Песня «Glow» вошла в саундтрек кинофильма Царь скорпионов. Композиция под названием «Something Told Me» входит в саундтрек к фильму Обитель зла.

Список композиций
Автор текстов песен Дез Фафара. Музыка — Coal Chamber, за исключением песни «Rowboat» — Саймон Дэниелс.

Участники записи

Coal Chamber
 Дез Фафара — вокал
 Мигель Раскон — бэк-вокал, гитара
 Райана Фосс-Роуз — бас-гитара
 Майк Кокс — барабаны

Производственный персонал
 Росс Хогарт — продюсер, микширование, звукорежиссёр
 Тед Дженсен — мастеринг
 Джереми Блейр — ассистент звукорежиссёра
 Пол Р. Браун — графический дизайнер, фотограф

Примечания

Ссылки
 Обзор альбома на AllMusic
 Информация об альбоме на Discogs.com
 Интервью группы для журнала Uranium на YouTube

Альбомы Coal Chamber
Альбомы Roadrunner Records
Альбомы на английском языке
Альбомы США
Альбомы ню-метала
Альбомы готик-метала

redpajama_set_name: RedPajamaWikipedia


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text:
Mountain Mischief
Sherri Cole works at Achieving Balance in her life and other's
W. Kenneth Medley II • May 6, 2019 at 12:00 AM
Society today can sometimes discourage taking credit when credit is due with so many people seeking self-indulgence for mediocre performance. To say Sherri Cole is modest would be an understatement.
She is a mother, business owner, athlete and empowering representative of a demographic that is sometimes downplayed in the "mainstream." Cole is the founder of Achieving Balance, a business with a mission statement the reads, "to educate, inspire and embolden beginner and intermediate off road cyclists through positive and professional training."
More importantly she is a mountain biker that represents a demographic that is ultimately driving a multi-billion dollar industry.
Fast Five:
Dogs or Cats: Dogs
Current Book: I just finished, "Un**** Yourself: Get out of your head and into your life" by Gary John Bishop.
Favorite Movie: "Tombstone"
Dinner w/any historical figure: Abraham Lincoln
Favorite Season: Spring/Summer
Will you tell me a little about your business Achieving Balance?
Achieving Balance was created to give a new way for new, beginner or intermediate riders to safely ride mountain bikes. We do group rides to kind of give a sense of community and to open doors to those who are not used to riding bikes or those who are. We have all levels. It provides a neutral or safe place for people to ride in groups.
Why cycling?
I don't know it just makes you feel good. It makes you feel clean after you ride, you feel happy. It is like being a kid.
Can you tell me a little about yourself?
I grew up in Blountville. Nathan (husband) and I live in Johnson City now. We have been here for nine or 10 years. I have not been into mountain biking forever. I started riding a bike for physical therapy after knee surgery. I got on a bike and felt like I was 10, so we rode and it would be for a mile then we would stop and die. Then that turned into 10 miles, 20 and 30 miles and so forth and so on. Cycling has become a life-changing event for us. Cycling has changed our health, our mental outlook and it has changed our community. It has given us a new passion that we didn't have before that is important to all of us in the family.
You have children, do they cycle?
We have one cycling little child (laughs)…Ryan he is a NICA kid and he also works at a local bike shop. He likes to do it leisurely. He is not really a racer or group rider. He is more like, 'I want to go ride my bike,' kind of guy.
How did you become a face for female mountain biking in the Tri-Cities, and what motivates you to bring people into the mountain biking community, especially woman?
I don't see myself has that, (laughs), at all. When I started riding I was very fortunate. I had Dwayne Letterman and Jeff, and they kind of took me under their wings and taught me how to bike. Not everyone has that. You go out on the trails and you see women in small groups, trying to get through and they look uncomfortable. Then when you put them in a group and they are smiling and they are happy and learning from each other. It is this overwhelming sense of amazement to see what you can do with a group of women.
What type of empowerment do you get from mountain biking?
Empowerment no, it's community. It is more about community. Men have groups, you know, they have the road club, SORBA and they have things like that, but you do not see woman coming together to be a bigger body. I think that is important, for woman to have voice, a social circle and a safe place.
What type of challenges have you faced?
Starting a business was only that I needed insurance. (Achieving Balance) wasn't originally intended to be a business. Obviously, there are the financial issues that all small-business owners have to deal with. Really it is just getting the point across that it is not a club; it is not a race team, it is just a community. That has been the biggest struggle.
What is has been your motivation to build that community?
It is about building a bridge. Mountain biking has brought so much happiness and health to our family. I really just want to pass that along; that and growing a sense of community, bonding with other women and teaching them how to be safe and how to have fun, and that it is not a man's sport. It is very much a woman's sport and that it is exhilarating and exciting. It is dangerous but it is just incredible.
What are the challenges that you face on the trails?
(Laughs)…Balance and that is why we are all about balance right. We are trying to achieve balance. Something that is important to me is being able to balance life and being able to balance personal goals, and community goals. Balance is such an important thing in our lives and if you think about it, when you are mountain biking you are achieving that balance. It kind of reflects on so many aspects of your life
For more stories follow @kmedleyjcpress on Facebook and Twitter.


text:
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		/// Required method for Designer support - do not modify
		/// the contents of this method with the code editor.
		/// </summary>
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		{
			components = new System.ComponentModel.Container();
		}

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}

redpajama_set_name: RedPajamaGithub


text:
\section{Introduction}

After Weil's local rigidity theorem of uniform lattices in semisimple Lie groups, there have been many generalizations in different contexts.
Due to Margulis' superrigidity and Corelette's theorem, lattices in higher rank semisimple Lie groups and in quaternionic, octonionic hyperbolic groups are very rigid. Hence it is only meaningful to study the embedding of uniform lattices
in real and complex hyperbolic spaces into bigger Lie groups.

Several studies have been done for complex hyperbolic lattices in different Lie groups.
In terms of maximal representations, Burger and Iozzi studied the representations of a lattice in $SU(1,p)$ with values in a Hermitian Lie group $G$ \cite{BI1, BI2}. Koziarz and Maubon \cite{KM}  studied the similar representations
in rank 2 Hermitian Lie groups. Pozzetti \cite{Po} deals with maximal representations of complex hyperbolic lattices
in $SU(m,n)$.
Recently Oscar-Garcia and Toledo \cite{GT} proved a global rigidity of complex hyperbolic lattices in quaternionic hyperbolic spaces.  More precisely, they defined the Toledo invariant $c(\rho)$ of a complex hyperbolic lattice $\Gamma$ under the representation
$\rho:\Gamma\rightarrow PSp(n,1)$ by
$$\int_{M} f^*_\rho \omega \wedge \omega_0^{n-2}$$ where $f_\rho$ is a descended map to $M=\Gamma\backslash PSU(n,1)/S(U(n)\times U(1))$ from a  $\rho$-equivariant
map from $H^n_{\mathbb C}$ to $H^n_{\mathbb H}$, $\omega$ is the quaternionic K\"ahler form on $H^n_{\mathbb H}$ and $\omega_0$ is the complex K\"ahler form
on $M$. They showed that this invariant $c(\rho)$ satisfies Milnor-Wood inequality and the maximality holds if and only if the representation
stabilizes a copy of $H^n_{\mathbb C}$ inside $H^n_{\mathbb H}$. Such a use of Toledo invariant goes back to Toledo \cite{Tol89} where he proves that a maximal representation from a surface group into $SU(1,q)$ fixes a complex geodesic.
Hernandez \cite{Her91} also studied maximal representations from a surface group into $SU(2,q)$ and showed that the image must stabilize a symmetric space associated to the group $SU(2,2)$.

In this paper we attempt to generalize their result to different quaternionic K\"ahler manifolds. The first goal would be to prove a similar result in
$$\Gamma\subset SU(n,1) \subset SU(2n,2)$$  using  Toledo invariant
$$c(\rho)=\int_M f_\rho^*\omega^{\frac{n}{2}}$$ for $n$ even where $\omega$ is the quaternionic K\"ahler 4-form on the associated symmetric space of $SU(2n,2)$.
This Toledo invariant is constant on each connected component of the character variety
$\chi(\Gamma, SU(2n,2))$. Hence it can be used to distinguish different components of the character variety.

As a starting point, we consider the simplest case
$$\Gamma\subset SU(2,1) \subset SU(4,2).$$
 This case is interesting since the symmetric space
of  $SU(4, 2)$ has both Hermitian and quanternionic
structures and it is worth to study the interplay between them.
We will consider several different embeddings coming from the natural
  holomorphic, totally real and symmetric square representations, and obtain
\begin{thm}There are at least 4 distinct connected components in
$\chi(\Gamma,SU(4,2))$
 where $\Gamma\subset SU(2,1)$ is a uniform lattice.
\end{thm}
Here the group $SU(4,2)$ acts on $\text{Hom}(\Gamma, SU(4,2))$ via conjugation on the target group and the character
variety is defined by
$$\chi(\Gamma, SU(4,2))=\text{Hom}(\Gamma, SU(4,2))//SU(4,2)$$ in the sense of geometric invariant
theory.

This is one of the first examples known in higher dimensional complex hyperbolic lattices. For different examples of character variety $\chi(\Gamma, SU(2,1))$, see \cite{Domingo}.
It is known in surface group case that there are $6(g-1)+1$
distinct components in $\chi(\pi_1(S), PSU(2,1))$ \cite{GKL,Xia}.
 Indeed, in \cite{GKL}, a discrete faithful representation into $SU(2,1)$ is constructed that on each component of $S\setminus \Sigma_0$, where $\Sigma_0$ is a set of disjoint simple closed geodesics, the representation  stabilizes either a complex line or a totally real plane. Then the Toledo invariants are maximal
on pieces contained in complex line, are zero on pieces contained in totally real plane. Hence one can realize any even integer
between $\chi(S)$ and $-\chi(S)$. This implies that there are $6(g-1)+1$ distinct components in $\chi(\pi_1(S), PSU(2,1))$.

To prove the global rigidity, the common technique known so far is to consider a holomorphic horizontal lifting of a $\rho$-equivariant map to a proper period domain (or twistor space) where one can do  complex geometry. It was successful in the case that Oscar-Garcia and Toledo considered in \cite{GT}. But in general, for higher rank case, there even does not exist a horizontal holomorphic lifting. At the last section of this paper, we give two cases where there exists or does not exist a   holomorphic horizontal lifting of a symmetric square representation.
\begin{thm}Let $\iota:B\rightarrow \cal X$ be a totally geodesic map inducing the symmetric square representation where $B=SU(2,1)/S(U(2)\times U(1))$ and $\cal X=SU(4,2)/S(U(4)\times U(2))$. Then it lifts to a  holomorphic horizontal map to the period domain
$D=SU(4,2)/S(U(3)\times U(1)\times U(2))$.
\end{thm}
See Section \ref{symmetric} for the definition of the symmetric square representation.

We thank D. Toledo for numerous discussions and suggestions for various period domains for liftability problem.
We also thank B. Klingler for a suggestion for possible different
period domains. Lastly, we thank  Mathematics department at Stanford
University where the first author spent a sabbatical year and the
second
author visited in June 2014 while part of this paper was  written.

\section{Quaternionic structure of $SU(2n,2)$}\label{Quaternionic}
\subsection{Quaternionic K\"ahler manifold in general}
A Riemannian manifold $M$ of real dimension $4n$ is quaternionic K\"ahler if its holonomy group is contained in $Sp(n)Sp(1)$. We denote by $\cal P_M$ the canonical $Sp(n)Sp(1)$-reduction of the principal bundle of orthogonal frames of $M$, and by $\cal E_M$ the canonical three-dimensional parallel subbundle $\cal P_M\times_{Sp(n)Sp(1)}{\mathbb R}^3$ of $End(TM)$. Since $Sp(n)Sp(1)$-module $\wedge^4 ({\mathbb R}^{4n})^*$ admits a unique trivial submodule of rank 1, any quaternionic K\"ahler manifold $M$ admits a nonzero closed 4-form $\omega$, canonical up to homothety. In \cite{Sal}, it is proved that the form $\omega$ (properly normalized) is the Chern-Weil form of the first Pontryagin class
$p_1(\cal E_M)\in H^4(M,{\mathbb Z})$.

Let $N$ be a smooth closed manifold and $\rho:\pi_1(N)\rightarrow G$ a representation into a quaternionic K\"ahler group $G$, i.e., the associated symmetric space is a quaternionic K\"ahler noncompact irreducible symmetric space $X$. Choose any $\rho$-equivariant smooth map $\phi:\tilde N\rightarrow  X$. The pullback $\phi^*\cal E_X$ descends to a bundle over $N$, still denoted $\phi^*\cal E_X$. By the functoriality of characteristic classes, the 4-form $\phi^*\omega$ represents the Pontryagin class $p_1(\phi^*\cal E_X)\in H^4(N,{\mathbb Z})$. As $X$ is contractible, any two $\rho$-equivariant maps give rise to the same class depending only on $\rho$. Then by the integrality of the Pontryagin class, the quaternionic Toledo invariant $c(\rho)=\int_N \phi^*\omega^{\frac{n}{2}}$ is constant on each connected component of the character variety.

\subsection{K\"a{}hler and Quaternionic structures of $SU(2n,2)/S(U(2n)\times U(2))$}
Let $G=SU(p,q)$, $p\ge q$, be  in its standard realization as linear
transformations on $\mathbb C^{p+q}=
\mathbb C^{p}\oplus\mathbb C^{q}$
preserving the indefinite Hermitian form of signature $(p,q)$.
We shall later specify $G$ to the case $SU(2n, 2)$ or $SU(n, 1)$.
Let $\cal X$ be the Hermtian symmetric space $\cal X=G/K$,
$K=S(U(p)\times U(q))$.
We recall briefly \cite{Sa}
the Harish-Chandra realization of the symmetric
space $\cal X$ into $M_{p\times q}$  which might
be useful in understanding various totally geodesic
embeddings in our present paper.
Fix $V^+_0=\mathbb C^{p},V^-_0=\mathbb C^q$
mutually orthogonal subspaces of ${\mathbb C}^{p+q}$ which are positive and negative definite respectively with respect to the  Hermitian form $h^{\mathbb C}$.
Fix orthonormal basis $\{e_1,\cdots,e_{p}\},\{e_{p+1},e_{p+q}\}$
 of $V^+_0,V^-_0$ respectively. Then $G$ acts on the set $\cal X$ of $q$-dimensional negative definite subspaces. Any other $q$-dimensional negative definite subspace $V^-$ is a graph of a unique linear map $A_{p\times q}=(z_{ij})$ from $V^-_0$ so that
$$\sum_{i=1}^{p}e_i z_{ij}+e_{p+j},\ j=1,\dots, q$$ form a basis of
$V^-$.
Hence $\cal X$ is identified with
$$\{Z\in M_{p\times q}: I_q - Z^t\bar Z>0\}.$$

The center of maximal compact subgroup $K$ of
is parameterized by the center of $U(p)$ and
it defines a complex K\"ahler structure. To be more precise
let $\mathfrak g$ be the Lie
algebra of  $G$, and
$\mathfrak g=\mathfrak t \oplus \mathfrak p$ its
 Cartan decomposition, where $\mathfrak k$ is the Lie algebra of $K$, with
$\mathfrak p$ consisting of matrices of the form
$$
\begin{pmatrix}
  0 & A\\
  A^* & 0 \end{pmatrix},
A\in M_{p\times q}.
$$
The real tangent space at $0=eK$ of $\cal X=G/K$
 is identified with $\mathfrak p$ with a complex structure $J$ on $T_o\cal X$ is
  $$J \begin{pmatrix}
  0 & A\\
  A^* & 0 \end{pmatrix}=\begin{pmatrix}
       0 & iA\\
       -iA^* & 0 \end{pmatrix}.$$ The K\"ahler metric on $T_o\cal X$ is
 $$g_0(X,Y)=2\text{Tr}(YX)=4 \text{Re} \text{Tr}(B^*A),\ \text{for}\       X=\begin{pmatrix}
  0 & A\\
  A^* & 0 \end{pmatrix}, Y=\begin{pmatrix}
  0 & B\\
  B^* & 0 \end{pmatrix}.$$
 The corresponding complex K\"ahler form is
  $$\Omega_0(X,Y)=g_0(JX, Y).$$

Let now $G=SU(2n, 2)$. The second factor $U(2)$ of $K$ defines a quaternionic structure as follows.
The holomorphic tangent
space of $\cal X$ at $0$ is identified with
$\begin{pmatrix}
0 &  M_{2n\times 2}\\
0  & 0
\end{pmatrix}$  where $M_{2n\times 2}$ denotes $2n$ by 2 complex
matrix.
The real tangent space will be parametrized and identified
by the holomorphic tangent space.
The three elements of $SU(2)$
$$\begin{pmatrix}
i &  0\\
0  & -i
\end{pmatrix} ,             \begin{pmatrix}
                            0 &  1 \\
                            -1 & 0    \end{pmatrix},   \begin{pmatrix}
                                                      0 &  i \\
                                                      i &
                                                      0\end{pmatrix}$$
acts on the tangent space as the quaternionic multiplication
by $i,j,k$  as follows. The adjoint action of $\begin{pmatrix}
i &  0\\
0  & -i
\end{pmatrix}$ is
$$\begin{pmatrix}
I_{2n} &   0 \\
0 & \begin{pmatrix}
     -i & 0 \\
     0  & i \end{pmatrix} \end{pmatrix}      \begin{pmatrix}
                                             0 & \begin{pmatrix}
                                                   x_1 & y_1 \\
                                                   x_2 & y_2 \\
                                                   \cdots \\
                                                   x_{2n}& y_{2n} \end{pmatrix}\\
                                               0   & 0  \end{pmatrix}         \begin{pmatrix}
                                                                               I_{2n} & 0 \\
                                                                               0 & \begin{pmatrix}
                                                                                i & 0 \\
                                                                                0 & -i \end{pmatrix}\end{pmatrix}$$$$=\begin{pmatrix}
                                                                                0 & \begin{pmatrix}
                                                   x_1i & y_1(-i) \\
                                                   x_2i & y_2(-i) \\
                                                   \cdots \\
                                                   x_{2n}i & y_{2n}(-i) \end{pmatrix}\\
                                               0   & 0  \end{pmatrix} . $$
                                               Hence if we identify
a matrix   $(x, y)\in M_{2n\times 2}=\mathbb C^{2n}\times
\mathbb C^{2n}$ with a quaternionic vector
                                               $q\in {\mathbb H}^{2n}$,
with $\mathbb H=\mathbb C +
\mathbb C j$ being the quaternionic number, by
                                               $$
X=(x, y)\leftrightarrow q_X=(x_1+ y_1 j, x_2+y_2 j, \cdots, x_{2n}+ y_{2n}j),$$ the multiplication by $i$ on the right becomes
                                  \begin{equation*}
                                              \begin{split}
                                            &\quad\,
       (x_1i+y_1(-i)j, x_2i+y_2(-i)j,\cdots,x_{2n}i+y_{2n}(-i)j)\\
&=      (x_1+y_1j, x_2+y_2j,\cdots,x_{2n}+y_{2n}j)i =q_Xi,
                                                 \end{split}
                                               \end{equation*}
i.e.,  the adjoint action of $\begin{pmatrix}
                                                   i & 0\\
                                                   0 &
                                                   -i \end{pmatrix}$
is just the multiplication by $i$ on the right.
It is easy to check that the adjoint action of the other two elements correspond to the multiplication by $j$ and $k$ on the right.
When no confusion would arise we shall just write the identification
$Z\to q_Z$ as $q_Z =Z$.

The quaternionic parallel closed nondegenerate K\"ahler 4-form, at the origin is given by
\begin{eqnarray}\label{kahler}\omega= \omega_i\wedge \omega_i +
  \omega_j
\wedge \omega_j +
\omega_k \wedge \omega_k\end{eqnarray}  where
$$\omega_i(X,Y)=\text{Re}( q_X\cdot {\bar q_Y} i),
\omega_j(X,Y)=\text{Re}(  q_X\cdot {\bar q_Y j}),\
\omega_k(X,Y)=\text{Re}(  q_X\cdot {\bar q_Yk}),$$ and
$p\cdot \bar q=\sum_{m=1}^{2n} p_m \bar q_m$ is the standard quaternionic Hermitian form on ${\mathbb H}^{2n}$.

Then it is easy to check that this $\omega$ and $\Omega_0^2$, where
$\Omega_0$
 is the complex K\"ahler form on $\cal X$
defined above, are linearly independent on $H^4(M,{\mathbb R})$ where $M=\Gamma\backslash \cal X$.

\subsection{Totally geodesic embeddings
of the unit ball $B=SU(n, 1)/U(n)$ in
$X$}
The complex hyperbolic space $H^n_{\mathbb C}$, i.e.
the symmetric space $SU(n, 1)/U(n)$,
will be realized as the unit ball $B$ in $\mathbb C^n$
as in \S2.2.
A natural holomorphic embedding of $H^n_{\mathbb C}=B=
\{(z_1,\cdots,z_n)\in {\mathbb C}^n:\sum |z_i|^2 <1\}$ into $\cal X$ is given by
$$\rho:(z_1,\cdots,z_n)\hookrightarrow  Z=\begin{pmatrix}
                                    z_1 I_2\\
                                    z_2 I_2 \\
                                    \cdots\\
                                    z_n I_2\end{pmatrix}$$ which seems to give rise to the maximal Toledo invariant of $\omega$.
On this holomorphic embedding, the tangent vectors to the image
are $$X=(x_1, x_1j,\cdots, x_n, x_nj)\in {\mathbb H}^{2n},\
x_k=a_k+ib_k\in{\mathbb C}, k=1, \dots, n$$
on which the form $\omega_j$ and $\omega_k$
and
$$\omega(X,Y,Z,W)= \omega_i\wedge \omega_i(X,Y,Z,W)$$$$=\text{Re}(iX\cdot \bar Y)\text{Re}(iZ\cdot \bar W)-   \text{Re}(iX\cdot \bar Z)\text{Re}(iY\cdot \bar W)+\text{Re}(iX\cdot \bar W)\text{Re}(iY\cdot \bar Z)$$ $$=\text{Re}(2i\sum_{i=1}^n x_i\bar y_i)\text{Re}(2i\sum_{i=1}^n z_i\bar w_i)- \text{Re}(2i\sum_{i=1}^n x_i\bar z_i)\text{Re}(2i\sum_{i=1}^n y_i\bar w_i)$$$$+\text{Re}(2i\sum_{i=1}^n x_i\bar w_i)\text{Re}(2i\sum_{i=1}^n y_i\bar z_i)    .$$But when we write $X=\begin{pmatrix}
                                                                0 & A\\
                                                                A^* & 0 \end{pmatrix}, Y=\begin{pmatrix}
                                                                0 & B\\
                                                                B^* & 0 \end{pmatrix}$
$$\Omega_0(X,Y)=g_0(JX,Y)=4 \text{Re Tr}(i B^*A)=4\text{Re} (2i\sum_{i=1}^n x_i\bar y_i).$$ Hence
      $$        \Omega_0^2(X,Y,Z,W)=16\omega(X,Y,Z,W)$$ for tangent vectors $X,Y,Z,W$ at the image of the natural holomorphic embedding of $H^n_{\mathbb C}$. In other words,
      \begin{eqnarray}
\rho^*\Omega_0^2=16 \rho^*\omega, \quad
\rho^*\omega=\frac{1}{16}\rho^*\Omega_0^2
      \end{eqnarray}
 for the natural holomorphic embedding $\rho$ of $H^n_{\mathbb C}$ into $\cal X$.

On the other hand, another  natural embedding
$$SU(n,1)\subset Sp(n,1)\subset SU(2n,2)$$ gives rise to a totally real embedding
\begin{equation}
  \label{eq:tot-real}
                                (z_1,\cdots,z_n)\hookrightarrow  Z=\begin{pmatrix}
                                    \begin{pmatrix}
                                     z_1 & 0\\
                                     0 & \bar z_1 \end{pmatrix}\\
                                    \cdots\\
                                   \begin{pmatrix}
                                     z_n & 0\\
                                     0 & \bar
                                     z_n \end{pmatrix} \end{pmatrix}
\end{equation}
whose Toledo invariant of $\Omega_0$ is zero. Contrary to $SU(1,1)$ case, this totally real embedding is locally rigid for $n>1$, see \cite{KKP}.
On this totally real embedding, the tangent vectors are $X=(x_1, \bar x_1j,\cdots, x_n,\bar x_nj)\in {\mathbb H}^{2n},\ x_i\in{\mathbb C}$, and
$$\omega(X,Y,Z,W)= \omega_i\wedge \omega_i(X,Y,Z,W)=0.$$ Hence the Toledo invariant of $\omega$ also vanishes.    These two special embeddings suggest that the Toledo invariant of $\omega$ is maximal on holomorphic embedding and zero on totally real embedding.  \\

{\bf Warning}: If we identify the holomorphic tangent space of $\cal X$ with ${\mathbb H}^{2n}$ by $(x_1+ j y_1,\cdots,x_{2n}+j y_{2n})$,
$\omega$ vanishes on holomorphic embedding and $16\omega=\Omega_0^2$ on totally real embedding. Hence the convention determines which one has a maximal Toledo invariant. In \cite{GT}, it seems that they use a different convention from ours. Nevertheless we stick to our convention in this paper.

       \section{Symmetric square representation
of SU(2, 1) and related
4-forms}\label{symmetric}

Denote $V=\mathbb C^{2+1}
=\mathbb C^{2}
+\mathbb Ce_3
$ the space $\mathbb C^3$
equipped with the Hermitian metric with signature
$(2, 1)$ and  $B=SU(2, 1)/U(2)$ as in \S2.2.
Recall that it is also identified  as the open domain in $\mathbb P^2$
of lines $\mathbb C(z+e_3)$
with negative metric, i.e. $|z|=|(z_1,z_2)|<1$.

Let $W=V^2$ be the symmetric square of $V$. Then
$W$ is equipped with the square of
the Hermitian metric of $V$ and $W=\mathbb C^4 +\mathbb C^2
=((\mathbb C^{2} )^2 +\mathbb C e_3^2)
\oplus (\mathbb C^{2} \odot e_3)
$
is of signature $(4, 2)$. Here $e_i\odot e_j=\frac{1}{2}(e_i\otimes e_j + e_j\otimes e_i)$.
We fix an orthonormal basis
$\{E_1, E_2, E_3, E_4, E_5, E_6\}$
of $W$ with
$$
E_j=e_j^2, E_4=\frac 1{\sqrt 2}(e_1\otimes e_2 +e_2\otimes e_1)=\sqrt 2 e_1\odot e_2,$$$$
 E_{4+i}=\frac 1{\sqrt 2}(e_3\otimes e_i +e_i\otimes e_3)=\sqrt 2 e_3\odot e_i,
j=1, 2, 3, i=1, 2.
$$

The square of the defining
representation of $H=SU(2, 1)
$ defines
a representation
$$
\iota: H\to G=SU(4, 2),\
g\mapsto \otimes^2 g
$$

As in \S2.2 the symmetric space $\mathcal X$ of
$SU(4, 2)$ will be realized as the open domain
of Grassmannian manifold $Gr(2, W)$
$2$-dimensional complex subspaces in $W$ with negative
metric, and is further identified
with the space of  $4\times 2$ matrices $Z$
with matrix norm $\Vert Z\Vert <1$ under the identification
$$
\{Zx\oplus x; x\in \mathbb C^2\} \mapsto Z.
$$
Recall also the normalization
of the K\"ahler metric on $B$ and on $\mathcal X$
$$
g_B(u, v)=4\text{Re}(u_1 \bar v_1 +u_2 \bar v_2), \,
g_\mathcal X(u, v)=4\text{Re}\text {Tr} v^\ast u
$$
where the real tangent space of $B$ and
$\mathcal X$ at $z=0$ and $Z=0$ are identified with $\mathbb C^2
$ and $M_{4\times 2}$; the respective K\"ahler forms
are  $\Omega_B(u, v)=g_B(iu, v)
$ and $\Omega_\mathcal X =g_\mathcal X(iu, v)$.

The representation $\iota: H\to G$
induces a totally geodesic mapping (with the same notation)
$\iota$: $B \to \mathcal X$. In terms of the above
identification of $B$ and $X$ as submanifolds
of projective and Grassmannian manifolds
the map $\iota$ is
$$
\iota (l)= l\odot l^{\perp}
$$
where $l^{\perp}$ is the orthogonal complement of
$l$ in $V$ and $l\odot l^{\perp}$
is the subspace of vectors
$u\otimes v + v\otimes u$, $u\in l, v\in l^{\perp}$.
We find now the map $\iota_\ast $ at $z=0\in B$.

Fixing the reference line $\mathbb Ce_3
\in \mathbb P^2$
and the plane $\mathbb C^2 \odot e_3
\in  Gr(2,W)$
corresponding to the point $0\in B$
and $0\in \mathcal X$,
the map $\iota$ is
$$
\iota: \exp(tX) \cdot (\mathbb Ce_3)
\mapsto (\exp(tX) \cdot (\mathbb Ce_3))
\odot (\exp(tX) \cdot (\mathbb C^2)),
$$  where
$$
X=\begin{pmatrix}
0 & 0 & a_1\\
0 & 0 & a_2 \\
\bar a_1 & \bar a_2 & 0 \end{pmatrix}
\in \mathfrak p
$$ and $\mathfrak{su}(2,1)=\mathfrak k + \mathfrak p$.
Thus $\iota_\ast (X)
$ is the linear transformation
$$
\iota_\ast (X): \mathbb C^2 \to \mathbb C^4,$$$$
\mathbb C\{E_5, E_6
\}
\mapsto
\mathbb C\{(Xe_3)\odot e_1 +
e_3\odot (Xe_1),
(Xe_3)\odot e_2 +
e_3\odot (Xe_2)
\}.
$$

Note that $X e_1=\bar a_1 e_3,\ Xe_2=\bar a_2 e_3,\ Xe_3=a_1 e_1 + a_2 e_2$ and
$$(Xe_3)\odot e_1 +
e_3\odot (Xe_1)= (a_1 e_1 + a_2 e_2)\odot e_1 + e_3 \odot \bar a_1 e_3$$
$$=a_1 e_1\odot e_1 + a_2 e_2\odot e_1 +\bar a_1 e_3\odot e_3=a_1E_1+\bar a_1 E_3+\frac{a_2}{\sqrt 2} E_4.$$

A similar calculation for the second factor shows that, under the  basis
$\{E_j\}$,
$\iota_\ast(X)$ corresponds to the $4\times 2$ matrix
$$
\begin{bmatrix} a_1
&0\\
0& a_2 \\
 \bar a_1
& \bar a_2
 \\
 \frac{a_2}{\sqrt 2}
&  \frac{a_1}{\sqrt 2}
\end{bmatrix}
= T_{a},
$$
Taking the basis vectors $X=\begin{pmatrix}
                                  0 & 0 & 1\\
                                  0 & 0 & 0\\
                                  1 & 0 & 0
                                  \end{pmatrix}, Y=\begin{pmatrix}
                                  0 & 0 & i\\
                                  0 & 0 & 0\\
                                  -i & 0 & 0 \end{pmatrix},
Z=\begin{pmatrix}
                                  0 & 0 & 0\\
                                  0 & 0 & 1\\
                                  0 & 1 & 0 \end{pmatrix}, W=\begin{pmatrix}
                                  0 & 0 & 0\\
                                  0 & 0 & i\\
                                  0 & -i & 0 \end{pmatrix}$ we find the
corresponding images in ${\mathbb H}^4$ under $\iota_\ast$
$$\iota_\ast(X)=(1,0,1,\frac{1}{\sqrt 2}j),\ \iota_\ast(Y)=(i,0,-i,\frac{k}{\sqrt 2}),$$$$\iota_\ast(Z)=(0,j,j,\frac{1}{\sqrt 2}),\ \iota_\ast(W)=(0,k,-k,\frac{i}{\sqrt 2})$$
and that
\begin{eqnarray}\label{value}
\omega(\iota_\ast(X), \iota_\ast(Y), \iota_\ast(Z), \iota_\ast(W))=\frac{11}{4}.
\end{eqnarray}

Namely
$$
\iota^\ast \omega=\frac{11}{64} \Omega_B^2
$$
where $\Omega_B$
is the K\"ahler form on $B$.

\section{Character variety $\chi(\Gamma, SU(4,2))$}
\begin{thm}There are at least 4 distinct connected components in $\chi(\Gamma,SU(4,2))$ where $\Gamma\subset SU(2,1)$ is a uniform lattice.
\end{thm}
\begin{proof}Let $X=\begin{pmatrix}
                                  0 & 0 & 1\\
                                  0 & 0 & 0\\
                                  1 & 0 & 0
                                  \end{pmatrix}, Y=\begin{pmatrix}
                                  0 & 0 & i\\
                                  0 & 0 & 0\\
                                  -i & 0 & 0 \end{pmatrix},
Z=\begin{pmatrix}
                                  0 & 0 & 0\\
                                  0 & 0 & 1\\
                                  0 & 1 & 0 \end{pmatrix}, W=\begin{pmatrix}
                                  0 & 0 & 0\\
                                  0 & 0 & i\\
                                  0 & -i & 0 \end{pmatrix}$ be the standard basis of $T_oB=\mathfrak p$ such that
      $$\Omega^2_B(X,Y,Z,W)=\Omega_B(X,Y)\Omega_B(Z,W)=4\text{Tr}(YJX)\text{Tr}(WJZ)=4\cdot 4=16.$$
Consider first the holomorphic
embedding  $\rho$ in \S2.
The images of the above vectors under $\rho_*$,
written as block $3\times 3$-matrix with each entry being $2\times 2$
matrix, are
$$\rho_*(X)=\begin{pmatrix}
                                 0& 0& I_2\\
                                   0&  0 & 0 \\
                                  I_2& 0 & 0          \end{pmatrix},\
\rho_*(Y)=\begin{pmatrix}
                                 0 & 0 & i I_2\\
                                   0& 0 & 0 \\
                                 -iI_2 & 0 &0
                                  \end{pmatrix},$$$$
\rho_*(Z)=\begin{pmatrix}
                                  0&0 &0  \\
                                  0&0  &  I_2 \\
                                  0 & I_2 & 0 \\
                                  \end{pmatrix},\
\rho_*(W)=\begin{pmatrix}
                                                    0& 0 & 0
\\
                                  0 & 0 & iI_2 \\
                                  0 &  -iI_2 & 0 \\
 \end{pmatrix},$$ which correspond to
$$\rho_*(X)=(1,j,0,0),\ \rho_*(Y)=(i,ij,0,0),$$$$\rho_*(Z)=(0,0,1,j),\ \rho_*(W)=(0,0,i,ij)$$ in ${\mathbb H}^2$ coordinates, see Section \ref{Quaternionic}.
Then by Equation (\ref{kahler}) $$\rho^*\omega(X,Y,Z,W)=4, \ \text{i.e.}\ \rho^* \omega=\frac{1}{4} \Omega_B^2$$ whereas for the square representation $\iota$, by Equation (\ref{value})
$$\iota^*\omega(X,Y,Z,W)=\frac{11}{4},\ \text{i.e.}\ \iota^*
\omega=\frac{11}{64} \Omega_B^2.$$ For the totally real embedding
  (\ref{eq:tot-real}), the pull-back form vanishes. This implies that
the quaternionic Toledo invaraints are
$$\int_{\Gamma\backslash H^2_{\mathbb C}} \rho^*\omega=\frac{1}{4}\int_{\Gamma\backslash H^2_{\mathbb C}} \Omega_B^2=\frac{1}{4}\operatorname{vol}(\Gamma\backslash H^2_{\mathbb C}),$$$$\ \int_{\Gamma\backslash H^2_{\mathbb C}} \iota^*\omega=\frac{11}{64}\int_{\Gamma\backslash H^2_{\mathbb C}} \Omega_B^2=\frac{11}{64}\operatorname{vol}(\Gamma\backslash H^2_{\mathbb C}),\ 0$$ respectively.

The last representation with a different Toledo invariant is given by the embedding $\phi:(z_1,\cdots,z_n)\rightarrow ((z_1,0),\cdots,(z_n,0))$ which produces that
\begin{equation}
  \label{eq:phi-p-b}
\phi^* \omega=\frac{1}{16} \Omega_B^2.
\end{equation}
Since the quaternionic Toledo invariant is constant on each connected component, we are done.
\end{proof}
Note that for a lattice $\Gamma\subset SU(2,1)$, the holomorphic
embedding
$\rho$ corresponds to the diagonal embedding
$\gamma\rightarrow (\gamma,\gamma)\in SU(2,1)\times SU(2,1)\subset SU(4,2)$, and the totally real embedding to
$\gamma\rightarrow(\gamma,\overline\gamma)$ whereas the last example in
the previous theorem corresponds to the embedding $\gamma\rightarrow (\gamma,id)\in SU(2,1)\times SU(2,1)\subset SU(4,2)$.

In this direction, Toledo constructed the following examples \cite{Domingo}.
There are examples of two complex hyperbolic surfaces $X=\Gamma\backslash H^2_{\mathbb C}$ and $Y=\Gamma'\backslash H^2_{\mathbb C}$ with a surjective holomorphic map
$f:X\rightarrow Y$ with $0<\operatorname{deg}(f)< \frac{\operatorname{vol}(X)}{\operatorname{vol}(Y)}$, which  induces a group homomorphism $f_*: \Gamma\to \Gamma'$.
See also \cite{DM, Mos} for the constructions of various
subgroups $\Gamma'\subset \Gamma$ of finite index.
(The volumes $\operatorname{vol}(X)$ and $\operatorname{vol}(Y)$ can  be further computed
by using the Chern-Gauss-Bonnet theorem for orbifolds.)
Consider the following representation
$$
 \Gamma\stackrel{f_*}{\rightarrow} \Gamma'\stackrel{\phi}{\rightarrow}
SU(4,2),
$$
 where $\phi$ is the restriction of the holomorphic embedding (5)
above.
Then the quaternionic Toledo invariant of this representation is
$$\int_X f^*(\phi^*\omega)
=\int_X
f^*(\frac{1}{16}\Omega^2_B)
=\frac{1}{16}
{\operatorname{deg}(f)\operatorname{vol}(Y)} < \frac{1}{16}
{\operatorname{vol}(X)},
$$
with $\frac{1}{16}
{\operatorname{vol}(X)}$ being the smallest  among the Toledo invariants
in Theorem 4.1 except zero case.
We obtain thus
an improvement of Theorem 4.1 in this case, viz

\begin{Prop}
Let $\Gamma\subset \Gamma'$ be as above. There exist at least
$5$  distinct components in $\chi(\Gamma,SU(4,2))$.
\end{Prop}

Some  versions of local rigidity for the representations
in some of the components above have been studied in
\cite{KKP, Klingler-inv}.

\section{Holomorphic Lifting to Various Period Domains}
In \cite{GT} the authors
study  some holomorphic liftings of mappings
from the complex hyperbolic ball
to quaternionic hyperbolic ball to holomorphic
mapping to the (pseudo-Hermitian) twister space, which enable
them to apply a variant of Schwarz lemma
and to prove local rigidity theorems. Following a
suggestion of Toledo we shall study holomorphic
liftings in our context.

\subsection{Non-lifting property}

Let $D=SU(4, 2)/S(U(4)\times U(1)\times U(1))$
be a twistor space.
We shall realize it as an open
subset in a homogeneous flag manifold.
Let $W^\ast$ be the dual space equipped with the
$G$-invariant metric of signature $(4, 2)$,
and
let $\{\epsilon_j\}$ in $W^\ast$
be the dual basis of $\{E_j\}$.
Let $D_1$ be the set
of orthogonal pairs $(l, \lambda)$
in
 $\mathbb P(W)\times \mathbb  P(W^\ast)
$,
 i.e.,
satisfying  $\epsilon(e)=0$ for all $(e, \epsilon)
\in l\times \lambda$. As a homogeneous
manifold of $SU(4,2)$ $D=SU(4, 2)/S(U(4)\times U(1)\times U(1))$
can be realized as the subset
 $(l, \lambda)$ such that
$l$ and $\lambda$ are negative
definite.
Indeed, first it is elementary
to see that $SU(4,2)$
acts transitively on the subset of lines.
Second we need to check that a stabilizer of $(l,\lambda)$ is
$S(U(4)\times U(1)\times U(1))$.
 A stabilizer of the negative 2-plane
$l+(\ker \lambda)
^{\perp}$ in $W$
is $S(U(4)\times U(2))$
 and a stabilizer
in  $S(U(4)\times U(2))$ of the
pair $(l, \ker \lambda)$
 of subspaces in $W$, equivalently
the pair  $(l,  \lambda)$ in
 $\mathbb P(W)\times \mathbb  P(W^\ast)
$, is
exactly $U(1)\times U(1)$. Hence as a differentiable manifold $D$ has such a realization.

Then as
an open set in $D_1\subset \mathbb P(W)\times \mathbb  P(W^\ast)$,
$D$ is equipped with the corresponding complex structure.

In general if a homogeneous manifold $G/(L\times U(1))$ has $U(1)$ factor in the stabilizer, it inherits a complex structure as follows.
Let $\mathfrak u(1)={\mathbb R} i H$ and consider the root space decomposition of $\mathfrak g^{\mathbb C}$ under the action of $H=\begin{pmatrix}
0  & 0 \\
0 & \begin{pmatrix}
 1 & 0 \\
 0 & -1 \end{pmatrix} \end{pmatrix}$. Set $\mathfrak b$ to be the Borel subalgebra consisting of zero and negative eigenspaces. The positive eigenspace constitutes the holomorphic tangent space as an
open set $$G/(L\times U(1)) \subset G^{\mathbb C}/B.$$

We find the holomorphic tangent space of $D$ in this context.
To find a realization of the complex tangent
space we fix the pair
$(\mathbb C E_6,
\mathbb C \epsilon_5)$ as a base point of
$D\subset D_1$. The space
$D_1$ is a complex homogeneous
space of $SL(6, \mathbb C)$, $D_1=SL(6, \mathbb C)/B$, where $B$ is
the Borel subgroup whose Lie algebra consists
of elements in $\mathfrak{sl}(6, \mathbb C)$
of the special form.

To justify this, note that $B$ is equal to the stabilizer of $(\mathbb C E_6,
\mathbb C \epsilon_5)$.
Hence $B$ should have the block matrix of form,
$$\begin{pmatrix}
* & * & 0 \\
0 & * &  0 \\
* & * & *   \end{pmatrix},
$$
the size of the  matrix being $(4+1+1)\times (4+1+1)$.
Alternatively $\mathfrak b$ consists of
non-positive roots  spaces
of $H$, i.e. eigenspaces of $ad(H)$  of
of non-positive eigenvalues, where
$$H=\begin{pmatrix}
0  & 0 \\
0 & \begin{pmatrix}
 1 & 0 \\
 0 & -1 \end{pmatrix} \end{pmatrix}.
$$
Thus holomorphic
tangent space $\mathfrak n$ consists of elements of $\mathfrak g^{\mathbb C}$
of the form, the size of the matrix being
the same as above,
$$\begin{pmatrix}
0 & 0  & * \\
* &  0 & * \\
0 & 0 &  0 \end{pmatrix}.
$$

\begin{lemma}\label{holo} The above quadratic map $\iota: B\to \mathcal X$
does not lift to a holomorphic horizontal mapping into $D=SU(4, 2)/S(U(4)\times U(1)\times U(1))$.
       \end{lemma}

\begin{proof}

Suppose $F$ is a holomorphic horizontal lifting.
The  complexification of $F_\ast$,
still denoted by $F_\ast$, maps $\mathfrak b^+$,
the holomorphic tangent space of $B$
to holomorphic tangent space $\mathfrak n$ (up
to changing of base point under SU(2)-action).
In particular the image of $\iota_\ast$
of $\mathfrak b^+$ is contained in
$\pi_\ast (\frak n)$ where $\pi:D\rightarrow \cal X$ is the natural projection. In particular
$\iota_\ast(\mathfrak b^+)
$ is a subspace of $\pi_\ast (\mathfrak n)$.
Using the above formula for $\mathfrak n$
we find that elements in $\pi_\ast (\mathfrak n)
\supset \iota_\ast(\mathfrak b^+)$
are of the form
$$\begin{pmatrix}
0 & 0  & * \\
* &  0 & 0 \\
0 & 0 & 0 \end{pmatrix}.
$$

However our computations above show that
for
\begin{eqnarray}\label{horiz}
S=\begin{pmatrix}
0 & 0 & a_1\\
0 & 0 & a_2 \\
0 & 0 & 0 \end{pmatrix}
=\frac 12(
\begin{pmatrix}
0 & 0 &  a_1\\
0 & 0 &  a_2 \\
 \bar a_1 &  \bar a_2 & 0 \end{pmatrix}\\\nonumber
-\sqrt{-1}
\begin{pmatrix}
0 & 0 & i a_1\\
0 & 0 & i a_2 \\
-i \bar a_1 & -i \bar a_2 & 0 \end{pmatrix}
)
\in \mathfrak b^+
\end{eqnarray}
its image $\iota_\ast(S)$ is
$$
\iota_\ast(S)=
\begin{bmatrix}
0 & U\\
V& 0 \end{bmatrix}
$$
where
$$
U=
\begin{pmatrix}
a_1 & 0\\
0 & a_2 \\
0 & 0 \\
\frac{a_2}{\sqrt 2}
&\frac{a_1}{\sqrt 2}
 \end{pmatrix},\,
V=
\begin{pmatrix}
0 & 0 &a_1 &0\\
0 & 0& a_2 &0 \end{pmatrix}.
$$
This is a contradiction to the form
of $\pi_\ast(\mathfrak n)$.
\end{proof}

We may construct similarly the twister
cover $SU(2m,2)/S(U(2m)\times U(1)\times U(1))$
of $\mathcal X=SU(2m,2)/S(U(2m)\times U(2))$ as above and
consider the question of holomorphic
lifting of maps from $B$ to $\mathcal X$. The
above proof leads to  a simple
necessary condition for the existence.
\begin{co}Given a representation $\rho:\Gamma\subset SU(n,1)\rightarrow SU(2m,2)$, with a $\rho$-equivariant map $f$
on the associated symmetric spaces $B=SU(n,1)/S(U(n)\times U(1)),\
\mathcal X=SU(2m,2)/S(U(2m)\times U(2))$
 and a fixed base point $o=[K]\in SU(n,1)/S(U(n)\times U(1))$,
let
$$Df_o\begin{pmatrix}
  0 & X \\
  X^* & 0\end{pmatrix}= \begin{pmatrix}
                        0 & U \\
                        U^* & 0\end{pmatrix}$$ be a differential map at the base point, where $X\in {\mathbb C}^n, U=(U_1,U_2)\in M_{2m\times 2}$.
                        For $f$ to have a holomorphic lift to the
                        twistor space, every component of  $U_1$  is
                        an conjugate $\mathbb C$-linear  in $X$,
and every component of  $U_2$  is a
                        $\mathbb C$-linear in $X$.
Here we regard $Df_o$ as a map from ${\mathbb C}^n$ to $M_{2m\times 2}={\mathbb C}^{4m}$.
\end{co}
\begin{proof}Note that $Df_o$ is a real linear map between real tangent spaces $T_o B$ and $T_{f(o)} \mathcal X$.
For $X=(z_1,\cdots, z_n)$ and $z_i=x_i+i y_i$, let $X=(x_1,\cdots, x_n, y_1,\cdots,y_n)=(x,y)$, with the same notation, be the corresponding coordinates in ${\mathbb R}^{2n}$. Then $i X$ corresponds to $$iX=(-y_1,\cdots,-y_n,x_1,\cdots,x_n)=(-y,x)$$ as usual.
For $f$ to lift to the holomorphic map to the twistor space, the equation (\ref{horiz}) should read
 $$Df_o(X-\sqrt{-1} iX)= (U_1',U_2')=(0, U_2').$$ Hence from $U_1'=0$, we get
 $$    \begin{pmatrix}
     A & B \\
     C & D \end{pmatrix} \begin{bmatrix}
                          x \\
                          y \end{bmatrix} -\sqrt{-1} \begin{pmatrix}
     A & B \\
     C & D \end{pmatrix} \begin{bmatrix}
                         -y \\
                          x \end{bmatrix}=0.$$ It is
$$         \begin{pmatrix}
           Ax+By \\
           Cx+Dy \end{pmatrix}+ \begin{pmatrix}
                                 -Cy+Dx \\
                                 -(-Ay+Bx)\end{pmatrix}=0.$$
From this we get
$$A=-D,\ B=C.$$  This exactly implies that every component function of
$U_1$ is
conjugate $\mathbb C$-linear in $X=(z_1,\cdots,z_n)$ variables.
Using the equation for $U^*$, a similar calculation shows that every
component function
 of $U_2$ is $\mathbb C$-linear in $z_i$ variables for $f$ to have a holomorphic lift to the twistor space.
\end{proof}

When $\rho$ is reductive, we can choose $f$ to be harmonic. Hence if we write $Df=(f_1,\cdots,f_{2m}; g_1,\cdots, g_{2m})$, $f_i$ are anti-holomorphic and $g_i$ are holomorphic for $f$ to have a holomorphic lift to the twistor space.

\subsection{Lifting to a different period domain}
We consider a different map from $B$ to  $D=SU(4,2)/S( U(3)\times U(1)\times U(2))$ and show that there exists a holomorphic horizontal lifting of $\iota$.

Let $f$ associate $  S^2 L^{\perp}, L^2, L\odot L^{\perp}$ to a negative line $L$ in $V$.
Then
 $S^2 L^{\perp}$ is a positive 3-dimensional space in $W$, $L^2$ is a positive line in $W$, and $L\odot L^{\perp}$ is a negative plane in $W$. Hence their stabilizers are
$U(3), U(1)$ and $U(2)$ respectively. Therefore
$$f: B\rightarrow D=SU(4,2)/S(U(3)\times U(1)\times U(2)).$$
Since $$\iota(L)=(L\odot L^{\perp},(L\odot L^{\perp})^\perp),$$ $f(L)=(  ( S^2 L^{\perp}, L^2),\iota(L))$ is a lifting of $\iota$ to $D$.

We claim that $f$ is holomorphic with respect to a complex structure on $D$ induced by $S(U(3)\times U(1))$.
In the  explicit coordinates, if $L={\mathbb C} e_3$, then $L^\perp=\langle e_1, e_2\rangle$ and
$$S^2L^\perp
=\langle e_1^2, e_2^2, e_1\odot e_2\rangle=\langle E_1,E_2,
E_4\rangle,\
L^2=\langle e_3^2\rangle=\langle E_3\rangle,$$
$$L\odot L^\perp=\langle e_1\odot e_3, e_2\odot e_3\rangle=\langle E_5,E_6\rangle.$$

The center of the stabilizer $U(3)$ of $S^2L^\perp$ is generated
by the diagonal matrix, under the above decomposition of
$W=
S^2L^\perp +L^2 +L\odot L^\perp =\mathbb C^3+\mathbb C +\mathbb C^2$.
$$
\text{diag}(1, 1, 0, 1, 0, 0)
$$
This means that under the adjoint action of
 an element in the center of $U(3)\times U(1)\times U(2)$
$$
\text{diag}(1, 1, -3, 1, 0, 0),
$$ the positive eigenspace constitutes
the holomorphic tangent space of $D$:
$$\begin{pmatrix}
0 & 0 & 0 & 0 & * & * \\
0 & 0 & 0 & 0 & * & * \\
0 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & * & * \\
0 & 0 & * & 0 & 0 & 0 \\
0 & 0 & * & 0 & 0 & 0 \end{pmatrix}.
$$

Hence the claim follows from the fact that the holomorphic tangent vector in $B$
$$\label{hori}
S=\begin{pmatrix}
0 & 0 & a_1\\
0 & 0 & a_2 \\
0 & 0 & 0 \end{pmatrix}
=\frac 12(
\begin{pmatrix}
0 & 0 &  a_1\\
0 & 0 &  a_2 \\
 \bar a_1 &  \bar a_2 & 0 \end{pmatrix}\\
-\sqrt{-1}
\begin{pmatrix}
0 & 0 & i a_1\\
0 & 0 & i a_2 \\
-i \bar a_1 & -i \bar a_2 & 0 \end{pmatrix}
)
\in \mathfrak b^+
$$ is mapped to $\iota_\ast(S)$
$$
\iota_\ast(S)=
\begin{bmatrix}
0 & U\\
V& 0 \end{bmatrix}
$$ as in the proof of Lemma \ref{holo}
where
$$
U=
\begin{pmatrix}
a_1 & 0\\
0 & a_2 \\
0 & 0 \\
\frac{a_2}{\sqrt 2}
&\frac{a_1}{\sqrt 2}
 \end{pmatrix},\,
V=
\begin{pmatrix}
0 & 0 &a_1 &0\\
0 & 0& a_2 &0 \end{pmatrix}.
$$

Now we show the horizontality, i.e., the image lies in the form $L\odot L^\perp$. For any smooth curve in $B$,
denote it by $L(t)=\langle v_0+ w(t) \rangle$ where $w(t)\subset v_0^\perp$, a differentiable family of lines, such that $w(0)=0, w'(0)\in v_0^\perp$. Then we can write $L(t)^\perp= \langle v(t)\rangle^\perp$ where $v(0)=v_0,\ v'(0)=w'(0)\in v_0^\perp$.

  Since $L(t)\odot L(t)^\perp$ is already horizontal,
it suffices to show the horizontality of $L(t)^2$ and $S^2(L(t)^\perp)$.
But $$L(t)^2=\langle (v_0 + w(t))\odot (v_0 + w(t))\rangle=\langle v_0^2+ v_0\odot w(t)+w(t)^2\rangle.$$ Hence
$$\frac{d}{dt}|_{t=0} L(t)^2=v_0\odot w'(0)\in L(0)\odot L(0)^\perp.$$
Similar calculation shows that
$$\frac{d}{dt}|_{t=0} S^2(L(t)^\perp)=\frac{d}{dt}|_{t=0} \langle
v(t)^\perp \odot v(t)^\perp \rangle$$
$$=\langle v'(0)^\perp \odot
v_0^\perp\rangle\subset \langle v_0\odot v_0^\perp\rangle\subset
L(0)\odot L(0)^\perp,$$
completing the proof.

redpajama_set_name: RedPajamaArXiv


For Princeton University's Sherrerd Hall, AWS designed, engineered and manufactured the factory-glazed, unitized curtainwall system, unitized storefront system, balustrade glass hand railing system, zero sightline doors, and exterior aluminum doors. The curtainwall glass features a custom frit pattern on the #2 and #3 glass surfaces.
We also supplied the skylight system, perforated soffit panels, interior stairwell and atrium railing system. The interior atrium railing was custom-designed specifically for this project. The atrium also boasts 8-foot and 12-foot tall glass with a frost pattern covering half of the glass lite.


What is the best way to implement above?
There are some squeeze page type plugins/themes out there. I know one of them (Optimize Press) some of our plugins have issue with due to how it works. You can see from search results.
You could of course go the multisite route and have your main domain as a the squeeze page and then subsites as you need them.
Thank you for your reply. Optimize Press is out for me as it's theme-based which I'll be using another theme for the main WordPress. Your multisite route suggestion sounds great and feasible (though Optimize Press specifically stated it won't work for multisite installation).


Letters for Term 3 clubs have been sent home.
As a reminder, please ensure you have the correct payment for all Forever Sport Clubs as the office is unable to issue change.
Payment must be made before the first session.
Please note - Smoothie Making Club is now FULL.


I've been working on a front-end application that uses ESLint for linting and Mocha as our testing framework. We've tried a few solutions to integrate linting into our workflow but it can easily be ignored. To make sure linting errors were given the attention they deserved we decided to turn linting errors into test failures.
To get started, create a file called called eslint-test.js in your test directory.
The first thing we need to do is collect an array of file paths we want to lint and get the results from ESLint after telling it our environment information and to use our .eslintrc.
Now that we have the results we can define our describe block and generate tests in that block. Let's add that to the end of our eslint-test.js file.
For each result we want to extract the filePath and messages. filePath will be used to generate a good test name. We'll use messages to check if we have any messages and if we do, we fail the test with a well formatted reason built from messages.
Finally we can loop over each of the messages and generate a failure message telling us the line number, column number, failure message, and the ESLint rule name.
Now that we have our ESLint test generator we can run our test suite and see it in action.
It works! Our failure tells us the failure's file, line number, column number, and linting error!
This combined with GitHub's required checks can make it much harder to merge code with poor style in your application. You can also take this same idea and apply it to other tools or linters like Hound.


text:
Brookdale student murder trial: Did she have any role in dumping body?
The lead detective in the case against a Brookdale student accused of murder was cross-examined during trial Thursday.
Brookdale student murder trial: Did she have any role in dumping body? The lead detective in the case against a Brookdale student accused of murder was cross-examined during trial Thursday. Check out this story on app.com: https://on.app.com/2MxGdZ3
Kala Kachmar, Asbury Park Press Published 5:00 a.m. ET Aug. 17, 2018
Brookdale Community College student Raquel Garajau listens as Brandon Wobser testifies during her trial for involvement in the murder of Trupal Patel. Doug Hood, Asbury Park Press
Brookdale honors student Raquel Garajau appears before Superior Court Judge Joseph W. Oxley during the opening of the trial for the murder of pot dealer Trupal Patel at the Monmouth County Courthouse in Freehold, NJ Tuesday July 24, 2018. (Photo: Tanya Breen)Buy Photo
FREEHOLD - A lead detective in the murder of Trupal Patel testified Thursday that it wasn't until seven months after Raquel Garajau was arrested for the murder that he realized she may not have been involved in a plan to dump the victim's body.
Garajau and her boyfriend, Joseph Villani, are accused of robbing and killing Patel, 29, of Brick, who was a known marijuana dealer. Authorities also charged the couple with conspiracy related to the death.
The prosecution alleges Villani shot Patel on Feb. 6, 2017, and that he and Garajau dumped the victim's body in Shark River Park in Wall. Villani admitted to shooting Patel in self-defense, while Garajou maintains she had nothing to do with the murder or dumping. Villani will be tried separately.
READ: Murder trial: Body dumped for $300 and an ounce of weed
A key witness, Brandon Wobster, testified on Aug. 1 that he helped move the body and chose where it would be dumped after determining that Villani's ideas were too risky.
Police didn't become aware of Wobster's direct involvement until October 2017, at least seven months after the couple was arrested, according to testimony Ocean Township Police Detective Steven Walker.
Brookdale Community College student Raquel Garajau listens as Brandon Wobser testifies during her trial for involvement in the murder of Trupal Patel. Freehold, NJ Wednesday, August 1, 2018 @dhoodhood (Photo: Doug Hood , )
Wobster, who was given immunity by the Monmouth County Prosecutor's Office, also testified that Garajau had nothing to do with the body dumping.
Niko Kacandes, a lifelong friend of Garajau, testified on July 15 that Garajau spoke openly about Villani's plan to rob a marijuana dealer. Kacandes testified he never heard anyone mention that the drug dealer had been killed or was going to be killed.
Most of Thursday's testimony was cross-examination of Walker, the state's last witness in the case. Defense attorney Robert Honecker used evidence presented by the prosecution to try to show that while Garajau knew about Patel's death, there's no evidence that she played a role in planning or executing it.
Honecker, while cross-examining Walker, asked if when putting the case together, Walker already had in the back of his mind that Garajau was known to use the Gulley Road entrance — the road off which Patel's body was dumped — when she goes to Shark River Park.
MORE: Murder trial: Cellphones show suspect near victim
Honecker also asked if another piece of evidence was in his mind: a text message from Garajau's phone to Villani's phone sent around the time Patel's body was allegedly dumped asking if Villani was in Wall.
In response, Walker said those statements were "correct."
Honecker pointed out that during Wobster's testimony, which Walker as a witness was sequestered from hearing, he said he noticed Villani and Garajau turned on location sharing on their phones so she could check his whereabouts whenever she wanted. Honecker also said several individuals "involved in the investigation" had shared locations.
"Once you became aware they shared locations, in regards to dumping the body in Shark River Park by Garajau and Villani, it was no longer was a pre-planned event by the two of them," Honecker said to Walker on the stand.
Walker said Wobster's testimony only showed that she didn't plan the dumping of the body.
Text messages between Garajau's and Villani's phones that were discussed during Walker's testimony on Wednesday show that Villani may have purchased a rifle on Jan. 19, 2017, and then cleaned it.
A message from Garajau's phone to Villani's said "What would make you get up and clean a gun?" to which there was no text response, according to testimony.
MORE: Brookdale student's murder trial: 'I can't believe I was dating a murderer'
There were also messages between the couple from Feb. 6 that indicated Garajau told Villani to take his time "bleaching everything," including the "entire door and inside," according to testimony.
Wobster also testified that Patel's body, along with a lot of blood, was on the floor of Villani's garage the night he helped move it on Feb. 6.
A message from Garajau's phone to Villani's phone also said not to leave the keys and to take the bag of stuff out to be thrown in the ocean because she "touched his phone," according to the messages.
The last cell phone "ping" to Patel's phone was in Long Branch at about 11:30 p.m. on Feb. 7, 2017 — a day after Patel was allegedly killed in Villani's garage in Ocean Township, said Stephanie Dugan, a prosecutor in the case, during the redirect examination of Walker.
Walker also testified that all texts to Villani from Garajau's phone were deleted before she voluntarily turned her phone over to the Monmouth County Prosecutor's Office on Feb. 24, 2017. None of the text messages between Garajau's phone and the phones of the individuals allegedly involved with the couple in the events surrounding Patel's murder were deleted, he also said.
The trial will continue Tuesday, Aug. 21 with the defense's witnesses before Judge Joseph W. Oxley in Monmouth County Superior Court.
Kala Kachmar: @NewsQuip; 732-643-4061; kkachmar@gannettnj.com
Read or Share this story: https://on.app.com/2MxGdZ3
Vietnam vet's flagpole request shot down by homeowners association
Toms River school board member charged after chasing other driver, police say
Judge who cited alleged rapist's 'good family' in ruling agrees to step down
Court begins removal proceedings against 'close your legs' judge
Cops: Teen lured victim before Jackson beating
Deer Tick stuck on man's eyeball; popping sound heard when removed


FILE PHOTO: The complainant, 42-year-old Dr Archana Nagar, a resident of Anoop Nagar, had purchased a fully automatic washing machine from Bharti Retail Limited at Malhar Mall for Rs 14,690.
INDORE: District Consumer Dispute Redressal Forum directed Whirlpool India Ltd to either pay a compensation of Rs 17,690 or provide a new washing machine to a consumer for reportedly selling a defective washing machine and not fixing it even during the warranty period.
The complainant, 42-year-old Dr Archana Nagar, a resident of Anoop Nagar, had purchased a fully automatic washing machine from Bharti Retail Limited at Malhar Mall for Rs 14,690. However, within a few days the machine stopped working properly. She lodged a complaint about it to the retailer, but the retailer could not fix the machine. Subsequently, Nagar wrote a mail to the washing machine company, but it didn't respond to her complaint. Later, the company did send a technician to repair the machine, but failed to resolve the problem despite her several attempts. She then moved the consumer forum against the company and dealer, seeking a compensation for poor services.
On the other hand, the retailer claimed that they only sell products of the company and are not manufacturers, therefore they should not be held liable for any defect in the machine. The company officials claimed that the all their machines are thouroughly tested before being launched in the market for sale and therefore they are also not liable to pay any compensation to the consumer.
After scrutinizing all the evidences, the forum noted that the machine was in warranty period and could have been repaired or replaced. Therefore, the forum upheld the order in favour of consumer and ordered the company to either fix the machine within two months for free or replace it with the new washing machine of same model else pay the purchase price of machine Rs 14,690. Besides, it has been asked to pay the complainant Rs 2000 for mental harassment and Rs 1000 for litigation expenses.


Anti wrinkle cream for decolletage. It' s a big business.
As bad as décolletage wrinkles are, they do not have to be accepted as inevitable. by Arden Fanning Andrews.
The Best Anti- Aging Creams for Your Neck, Décolleté, and Chest Getty Images Boost elasticity, tighten crepey skin, and smooth wrinkles overnight with the help of these powerful anti- aging creams. Clarins recognizes that the neck is a fragile area.
This article shares the 7 best treatment options for repairing and maintaining the delicate skin of the décolletage. Anti wrinkle cream for decolletage.
Give this anti- wrinkle cream by Olay a spin for line- free skin. Photographed by David Sims, Vogue, February.
Treat your sensitive skin with. Apply a wrinkle cream to your chest each day.
Use at night after washing with Refresh Face Wash and exfoliating with Scrubalicious, then apply Ultra Glow to your entire face, neck and decollete. Ultra Glow Retiol Moisturizering Night Cream is and anti- aging and anti- wrinkle cream that also contains powerfull antioxidants to help repair the skin from sun damaged and environment toxins.
Clarins Extra- Firming Neck Anti- Wrinkle Rejuvenating Cream. The 15 Best Neck- Firming Creams and Serums to Fight Wrinkles.
Clarins Extra- Firming Neck Anti- Wrinkle Rejuvenating Cream. July 28, 1: 35 PM.
Though creams won' t necessarily remove wrinkles from your decolletage decollete area, look for products made with the active ingredients decolletage of alpha hydroxy, retinol, vitamin C, kinetin, copper peptides or coenzyme Q10 to reduce the appearance of lines and wrinkles. In Lab tests, Olay' s.
How We Chose the Best Wrinkle Cream Anti- aging serums made up a $ 130 billion market in, a number that the Infinium Global Research Group projects will continue to rise. The New Wrinkle You Didn' t See Coming— decolletage And How to Make It Disappear.
Olay Total Effects Anti- Aging Night Firming Cream $ 25. Not only does this jar look glitzy in your medicine cabinet, but the BareMinerals Ageless Genius Firming & Wrinkle Smoothing Neck Cream also infused with literal gold particles, which act as anti.
The fast- action formula decolletage starts reducing wrinkles instantly with a blend of hyaluronic acid, vitamin B, and amino- peptides, so if you' re in a rush to reduce age marks on a budget, Olay has found the way.


text:
ashley asks....
ASHLEY ASKS…… Reece Gold
By ladybug388 on January 21, 2023 • ( Leave a comment )
Juncos Hollinger Racing will be returning to Indy NXT competition for the 2023 campaign, with a familiar face behind the wheel as Reece Gold makes the jump up after placing second in the Indy Pro 2000 Championship Presented by Cooper Tires standings. The American recently shared his thoughts with NEWS FROM THE PITS.
What are your thoughts going into the 2023 season?
I'm super excited to continue my time with Juncos-Hollinger Racing – my first season in Indy NXT, and their first season back in Indy NXT. The field looks really competitive and I'm super excited to continue with my team. Everyone I've been working with the past few years, I'm really looking forward to it.
What are your goals for this year knowing it's your rookie campaign?
As a racecar driver, you always want to win. But I think it's my rookie campaign so I definitely set realistic goals but I'd love to just be fighting for podiums and wins, and be on pace, and do my best, and give the team what they deserve and what they give me in return.
What track are you most looking forward to on the schedule?
Probably Laguna Seca. I love just the track, the history, the whole city is awesome.
What is one thing you feel you could've done better last year to make your Indy Pro 2000 Championship season better?
I think overall it was pretty successful. I think certain races I just lacked, missed the pace and set-up, and I think those two weekends cost us. But overall, it was a pretty good season. I think there were some small mistakes on my end that I could've fixed.
How did you get started in racing?
I started when I was five-years-old. My dad bought me a go-kart. I hopped in it and drove around the neighborhood. From there, I moved up the ranks.
What's been the most memorable moment in racing so far for you?
That's hard. There are a lot of memorable moments. I'd say one of them has to be my first win on the Road to Indy at Indianapolis (Motor Speedway). (It's) just an awesome place to get your first win at the speedway. That's definitely one of them.
What would it mean to you one day if you were able to race in the NTT IndyCar Series?
That would be – I've been racing since I was five and it's always been a dream of mine. So that'd be absolutely amazing to drive an IndyCar.
Now is there anything out there besides open-wheel that you'd love to get behind the wheel of?
I've always liked SportsCars and I've been able to drive a few, but I'd love to be able to do LeMans or Daytona. That's definitely a bucket list race right there.
Lastly, what would be one piece of advice for the next racer out there hoping to get to where you are?
That's hard. There's a lot of factors that go into racing. But I think just stay focused, give it your all, and if you really give it your hardest, you can make it professionally.
Categories: ashley asks...., Feature, Interview, Road to Indy
Tagged as: ashley asks...., ashley mccubbin, featured, indy nxt, reece gold, Road to Indy
ASHLEY ASKS…… Benjamin Pedersen
Braun Pushes Meyer Shank Acura to Fastest GTP Lap Yet at Roar


text:
Photo: biletskiy/Shutterstock
The best place to go in Italy for every type of traveler
Travel Trip Planning Italy
Alex Bresler
Italy's not just a place. It's a lifestyle.
It's reaching for olive oil, rather than butter, then drizzling on a little extra. Taking regular espresso breaks but swearing off milky coffee after breakfast. Staying up late without staying out all night, and honoring superstitions passed down by your nonna.
Business hours are a loose concept, but you can set your watch by riposo, Italy's answer to Spain's siesta. The whole country's a style icon, but it's hardly lacking character or natural beauty, from fishing villages radiating Mediterranean vibes to low-key luxe mountain towns oozing Alpine cool. Then there's the bigger, historic, pleasantly walkable cities.
As destinations go, Italy's a safe recommendation for almost anyone, yet there's always the question of where to spend those precious vacation days. Rather than trying to bounce around the entire country like you're on a cruise, opt to settle into one region, really get to know it, and embrace the local lifestyle.
To help you put together your perfect itinerary, here's our guide to imitable Italy for every type of traveler.
For the food: Naples or Emilia-Romagna
Photo: Vladimir Staykov/Shutterstock
Grittier than canal-cut Venice and Florence with its Renaissance roots, yet every bit as rich, Naples can be hit or miss with travelers. Until you ask about the food, that is. The city's often credited with inventing pizza, and the classic Neapolitan style is protected both by UNESCO and a government outfit tasked with certifying only the most authentic pizzerias. Margherita pies are just the beginning in Naples, though. Save room for deep-fried street foods; seafood pasta like spaghetti alle vongole, made with fresh clams; and sweets like sfogliatella, a croissant-like pastry filled with ricotta and candied fruit.
Photo: javarman/Shutterstock
Travel north from Naples and you'll eventually hit Bologna, the capital of the Emilia-Romagna region and a powerhouse of flavor. You'll do well there if your metabolism can withstand meaty pasta like green lasagne, tagliatelle al ragù, and tortellini stuffed with cheese and ham. If you prefer to graze rather than gorge, take it back to the basics in Parma with salty, nutty Parmigiano Reggiano and prosciutto di Parma tastings, or swing by Modena to sample the best balsamic vinegar you'll ever taste (and, if you can swing it, a meal at world-famous Osteria Francescana).
For the wine: Tuscany
Photo: theskaman306/Shutterstock
Postcards were invented to capture places like Tuscany, where vineyards, olive and cypress groves, and swaths of sunflowers fleck the hills outside landmark cities like Florence and Pisa. Anyone can appreciate the panoramas Tuscany's landscape paints, but none more than oenophiles, who quite literally relish drinking in the scenery. Vino aficionados can whet their palates with classic Chiantis at acclaimed estates like Antinori, Castellare di Castellina, and Fattoria la Loggia before sampling vintages not made from Sangiovese grapes, especially the region's widely grown and increasingly popular Cabernet Sauvignons.
Really, though, Tuscany's been perfecting its viticulture since the eighth century, so you'd have to try pretty hard to be disappointed here.
For finer things and late nights: Milan
Photo: MarinaD_37/Shutterstock
Milan may be Italy's fashion capital, if not the world's, but la dolce vita dictates more than just the dress code in the modern, glamorous metropolis. Manage not to max out your credit card at the Galleria Vittorio Emanuele II, the oldest shopping arcade in Italy, and you'll realize just how many other luxuries await. Think Michelin-caliber meals at restaurants like two-star Seta and one-star Alice Ristorante, bookended by pre-dinner drinks at chic aperitivo bars and nightcaps at trendy cocktail joints like Nottingham Forest.
The fun doesn't have to end there, however. Milan is a young, vibrant city with clubs that stay open late, some as late as 6:00 AM. Just be sure to get some sleep, as you won't want to miss on the city's famed fine arts, including Leonardo da Vinci's "The Last Supper," housed in the Santa Maria delle Grazie church.
For hiking, cycling, and snow sports: the Dolomites
Photo: Gaspar Janos7/Shutterstock
The crown jewel of the Italian Alps, strung across northeast Italy near the Austrian border, the Dolomite range is an 18-peak, UNESCO-protected wonder that reaches an altitude of nearly 10,000 feet. Come in summer for via ferrata hiking up cliff faces, perhaps in Parco Naturale Adamello Brenta near Trentino, or mountain biking around the Val Gardena valley. Alternately, visit ritzy, sporty Cortina d'Ampezzo during ski season to sail down the same slopes as Bond in For Your Eyes Only and 1956's Winter Olympians. Consider a day trip to the Dolomites next time you're in Venice, only a couple of hour's south by car; however you get there, don't leave without seeing the beautiful 15th-century Reifenstein castle in South Tyrol.
For art and architecture: Florence
Photo: Catarina Belova/Shutterstock
Florence is, on the whole, an architectural masterpiece: a sea of whitewashed, terracotta-topped structures with the city's centerpiece, the bulbous Duomo, rising a few stories above the rest. The epicenter of commerce and culture from the 14th to 16th centuries, aptly called the birthplace of the Renaissance, Florence continues to draw visitors for its exquisite art. Tour the Uffizi Gallery to see precious works like Botticelli's "Birth of Venus" and the Galleria dell'Accademia for Michelangelo's "David." Double down with a visit to the Palazzo Vecchio to see the "David" copy, famous in its own right, and nearby Loggia dei Lanzi statue garden.
Later, climb to the top of the Duomo, also known as the Florence Cathedral, for an unbeatable lookout over the city. Youthful but not excessively young, historic but not stodgy, Florence is a treat for art and architecture lovers of all ages; even just standing on the Ponte Vecchio bridge, watching the sun sink into the horizon and dip below the Arno river, is pure magic.
For lovebirds: Verona
Photo: kavalenkava/Shutterstock
Romeo and Juliet is more of a cautionary tale starring teenage hormones than it is the #relationshipgoals love story it's sometimes cited as, but that doesn't make Verona, where Shakespeare's iconic play was set, any less romantic. While Casa di Giulietta and Juliet's balcony are obvious tourist gimmicks, the city's wide piazzas, Roman and Renaissance architecture, museums, galleries, and cozy restaurants specializing in creamy risotto make it ideal for couples. Steal away to nearby Lake Garda for a day if you're craving seclusion.
For summer sunbathers: the Amalfi Coast
Photo: Oleg Voronische/Shutterstock
Most people think Positano and its Marina Grande beach when they hear Amalfi, but the area's 30 miles of coastline house seaside towns both rightfully popular and nearly undiscovered. We recommend Ravello for a quieter yet equally stunning getaway, located midway between Positano and Salerno, not far from the town of Amalfi itself.
Ravello shares the coast's iconic cliffside architecture but does one better with exquisite gardens overlooking the Tyrrhenian Sea at the 13th-century Villa Rufolo. Visit during summer to tan and catch the city's lively music and arts festival, sometimes called the Wagner Festival after the German composer who visited the city in 1880 and was presumably as smitten as you'll be. The best beach near Ravello is Castiglione; elsewhere, sun-seekers may enjoy sandy Vietri sul Mare beach, secret Gavitello beach in Praiano, or family-friendly Maiori and Minori beaches.
For ruins and relics: Pompeii or Matera
Photo: Romas_Photo/Shutterstock
Pompeii, the lost Roman city toppled by Mount Vesuvius in 79 AD, has been a magnet for tourists and archaeologists alike since it was rediscovered in the mid-18th century, remarkably well preserved under a thick layer of ash. Now is a particularly exciting time to visit as a 2,000-year-old shrine was newly uncovered late last year. Full-price tickets are a reasonable 15 euros. Dedicate at least three hours to exploring Pompeii (it's a city-sized attraction, after all) and tack on Herculaneum, a neighboring city also destroyed in the eruption, if you have time.
Photo: Svetla Momchilova/Shutterstock
Matera doesn't have Pompeii's name recognition, but that's part of the appeal. Unlike the ancient, ash-buried city, Matera showcases two impressive networks of prehistoric dwellings, called sassi, cut from southern Italy's limestone cliffs. It's one of the oldest continuously inhabited settlements on Earth, albeit one plagued with poverty, disease, and neglect until more recently. All that's changing, though. Since its UNESCO inscription in 1993 and more recent nomination as one of 2019's European Capitals of Culture, Matera has more tourist appeal than ever before with a growing number of accommodations and dining options.
For family vacation: Rome
Photo: Boris Stroujko/Shutterstock
Italy's dynamic capital is at once an obvious choice for a family vacation and a curious one. It's not the easiest city to navigate, and the near-constant crush of tourists certainly doesn't help the oppressive traffic, confusing public transport, lengthy wait times, and the like.
But the good far outweighs the bad and the ugly in Rome, and it's perfect for groups of all ages, budgets, and interests. Alongside world-class, educational attractions like the Colosseum and Roman Forum, the city's filled with fun to-dos that kids will eat up, like puppet theater in the park, tossing coins into the Trevi Fountain, posing with a giant marble foot, and sticking a hand in the Mouth of Truth, a stone frieze that's said to chomp down on any arms belonging to liars. Then there's Vatican City, technically an independent city-state within Rome's bounds, where you can admire Michelangelo's works of art inside the Sistine Chapel and St. Peter's Basilica.
Walking tours are readily available both across the city and underground through the catacombs below. Don't forget to refuel with a few scoops of gelato after a full day of exploring.
For a crowd-free retreat: Lake Iseo
Photo: Swen Stroop/Shutterstock
Lake Iseo is what Lake Como was before Clooney put it on the map. Tucked away between Bergamo and Brescia, it's the best-kept secret of the Lombard lakes, though chances are the glassy, snaking, mountain-ringed body of water won't stay that way for long.
Hug the lake on a scenic drive, take a refreshing dip right from shore, or get out on the water whether by kayak, canoe, or speedboat. Then venture to nearby Valle Camonica to see Stone Age carvings, followed by a jaunt to the neighboring Franciacorta wine country, which is quickly carving space for itself as one of Italy's premier sparkling-wine regions, when you want to channel your inner A-lister without the superstar price tag.
For off-season travelers: Venice
Photo: Rostislav Glinsky/Shutterstock
With its meandering gondolas, glinting marble architecture, and turquoise lagoon, Venice really is as stunning as they say. It's also as crowded, a problem the city has taken measures to combat. To do your part to preserve what some consider the world's most beautiful city, visit Venice in winter when the crowds clear out and locals come out of hiding. It'll be misty, and cold, but pack appropriately and the only chill you'll feel is at Saint Mark's Basilica, where you'll have a shot at an actual moment of silence. If you'd rather make some noise, go in February for the pre-Lent Venice Carnival festival, famous for its masks, costumes, and masquerade balls.
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Home / Home / MRC
MRC Public Hearing
Melvin Price Locks and Dam, Alton, Ill.
Friday, August 12, 2016 at 9 a.m.
Additional MRC Information
Select...MRC HistoryCurrent MRC MembersMRC - Low Water InspectionMotor Vessel MISSISSIPPI HistoryMRC Website
MRC History
Current MRC Members
MRC - Low Water Inspection
Motor Vessel MISSISSIPPI History
MRC Website
2016 MRC Public Hearing
The Mississippi River Commission will host a public hearing Friday August 12 at Melvin Price Locks and Dam in Alton, Ill., aboard the Motor Vessel MISSISSIPPI. The hearing is open to the public and scheduled to begin at 9 a.m.
Anyone with an interest is invited to present their views and suggestions on matters affecting the water resources infrastructure needs in the valley, including flood control and the Mississippi River and Tributaries project, environmental issues, recreation, navigation and others.
The agenda for the August 12 public hearing is as follows:
Summary report by Maj. Gen. Michael Wehr, president of the Commission, on national and regional issues affecting the U.S. Army Corps of Engineers and Commission programs and projects on the Mississippi River and its tributaries.
St. Louis District overview on current project issues in the St. Louis District by District Commander Col. Anthony Mitchell.
Presentations to the Commission by local organizations and members of the public giving views or comments on any issue affecting the programs or projects of the Commission and the Corps of Engineers.
The MRC, established in 1879, is composed of seven members, each nominated by the President of the United States and vetted by the Senate. Three of the organization's members are officers of the Corps of Engineers; one member is from the National Oceanic and Atmospheric Administration; and three members are civilians, two of whom are civil engineers.
General duties of the commission include recommendation of policy and work programs, the study of and reporting upon the necessity for modifications or additions to the flood control and navigation project, recommendation upon any matters authorized by law, and making semi-annual inspection trips. The duties of the commission include the entire length of the Mississippi River from its headwaters at Lake Itasca, Minn., to Head of Passes, La., where the Mississippi River empties into the Gulf of Mexico.
The purpose of the public meetings is to maintain a dialogue, an exchange of viewpoints and ideas flowing between the watershed interests, the public and the Corps. Presentations by the public are made orally, but a copy of the remarks should be presented to the commission for the official record and written response.
The public hearing process is unique to the Mississippi River Commission and the U.S. Army Corps of Engineers. The benefits of hearing the issues and concerns first hand through the public hearing process are invaluable to the commission and the Corps. Also, the interaction with congressional, federal and state interests, local boards and non-government organizations and the public is crucial to the decision-making process for the nation's water resources.
Since 1879, the seven-member Presidentially appointed Mississippi River Commission has developed and matured plans for the general improvement of the Mississippi River from the Head of Passes to the Headwaters. The Mississippi River Commission brings critical engineering representation to the drainage basin, which impacts 41% of the United States and includes 1.25 million square miles, over 250 tributaries, 31 states and two Canadian provinces.
Additional MRC Public Hearing
August 8 8:00 a.m. St. Paul, Minn.
August 11 2:00 p.m. Hannibal, Mo.
August 12 9:00 a.m. Alton, Ill.
August 15 9:00 a.m. Caruthersville, Mo.
August 16 9:00 a.m. Helena, Ark.
August 17 2:00 p.m. Natchez, Miss.
August 19 9:00 a.m. Morgan City, La.


text:
est un film muet américain réalisé par Carl M. Leviness et sorti en 1916.

Synopsis

Fiche technique
 Titre :
 Réalisation : Carl M. Leviness
 Scénario : Frank Borzage
 Société de production : American Film Manufacturing Company
 Pays d'origine :
 Format : Noir et blanc — 35 mm — 1,33:1 — Muet
 Dates de sortie :
  :

Distribution
 Vivian Rich
 Gayne Whitman

Liens externes

Film américain sorti en 1916
Film dramatique américain
Court métrage américain
Film muet américain
Film américain en noir et blanc


text:
FTC approves Nielsen acquisition of Arbitron
11:27 PM 2013
by Marc Castro 
The US Federal Trade Commission recently approved the USD1.26 billion acquisition of Arbitron Inc by Nielsen Holdings NV. Nielsen is the biggest research firm tracking US television ratings.
The deal was cleared after Nielsen was required to divest and/or license its assets related to Arbitron's audience measurement services for different platforms. The process was formalized through a settlement agreement, as confirmed in a statement made today.
According to FTC Chairwoman Edith Ramirez, "The evidence provided us with a strong reason to believe that absent a remedy, the deal was likely to harm emerging competion in the ara of cross-platform audience measurement."
With the acquisition, Nielson would be able to offer advertisers a unified system in the measurement of audiences across multiple media platforms. This would allow advertisers an easier time in making decisions on advertising time be in on television, radio or online. The ratings firms sets the advertising costs for the USD70 billion US television advertising market. The company announced the acquisition of Arbitron last December, as the latter measures radio audience share.
Federal Trade Commission, Nielsen, Acquisition


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Speak with a care advisor: (866)-986-0215
Living in Community – Part One
In 1845 Henry David Thoreau was told by a fellow writer to "Go…build yourself a hut," and live a solitary life in the woods. Walden; or, Life in the Woods, the chronicle of that two year period, has inspired countless others to live the introspective life—at least for a time. Others work out their beliefs in the midst of communal living. Aristotle observed that "Man is by nature a social animal," and John Donne agreed with him when he wrote that "No man is an island, entire of itself; every man is a piece of the continent, a part of the main."
Human beings need other human beings. Our language, cognitive abilities, emotions and sense of self all develop from birth in relationship with those around us. When we live in isolation from others, creative ability declines and loneliness inevitably results. According to neuroscientist John Cacioppo, author of Loneliness: Human Nature and the Need for Social Connection, loneliness can negatively affect our health, particularly immune and cardiovascular function—even contributing to premature aging.
Wendell Berry is a contemporary writer who touches on many of the same ideas that once occupied Thoreau. Of community Berry writes, "I believe that the community—in the fullest sense: a place and all its creatures—is the smallest unit of health and that to speak of the health of an isolated individual is a contradiction in terms."
Even though Thoreau decided to live for a while all by himself in a hut next to Walden pond, he was supported with food from family and friends—especially Ralph Waldo Emerson—and he kept chairs ready for visitors, which he claimed to have more of while alone in the woods than when he lived in the city. Thoreau needed others to sustain even his temporary solitude.
We all need relationships to sustain and support wellbeing. But as we age, the desire to remain in place—long after we become isolated in that space—is often stronger than the desire for community. Unfortunately, the one time in our modern day when we can be most isolated from relationships—in the latter stages of life, when companions die or we become infirm and less able to travel—is the time in our lives when it can be most dangerous to live alone.
Part Two of Living in Community explores some of those dangers—and looks at some of the surprising benefits for those who move from isolation back to communal living.
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