text:
Le Conseil des ministres (Amharique: የሚኒስትሮች ምክር ቤት, Ye ministeroch meker bét) est une institution politique éthiopienne établie par le Constitution de 1994. Il détient, avec le Premier ministre, les pouvoirs exécutifs suprêmes du Gouvernement fédéral (article 72-1), ses membres sont collectivement responsables devant la Chambre des représentants des peuples (article 72-2). Il est constitué du Premier ministre, du Vice-Premier ministre, les ministres et éventuellement d'autres membres désignés dans le cadre de la loi (article 76-1). Tous les nominations doivent être approuvées par la Chambre des représentants des peuples (article55-13); par conséquent, lorsque la coalition du Conseil des ministres perd la majorité à la Chambre des représentants des peuples, il sera dissous (article 60-2). Il est également responsable devant le Premier ministre (article 76-2). Ce dernier est le président et le représentant du Conseil des ministres dont il coordonne les activités (article 74-4). Par ailleurs, les lois, politiques, règlementations, directives et autres décisions adoptées par le Conseil des ministres sont toutes supervisées par le Premier ministre (article 74-5).

Composition du gouvernement

Pouvoirs et fonctions du Conseil des ministres

Les pouvoirs et les fonctions du Conseil des ministres sont décrits à l'article 77 de la Constitution :
 il assure la mise en œuvre des lois et décisions adoptées par la Chambre des représentants des peuples (article 77-1);
 il décide de l'organisation structurelle des ministères et d'autres organes gouvernementaux qui sont responsables devant lui, il doit coordonner leurs activités et assurer un leadership (article 77-2);
 il propose un budget fédéral annuel, lorsqu'il est approuvé par la Chambre des représentants des peuples il doit le mettre en œuvre (article 77-3);
 il doit assurer la bonne exécution des politiques et stratégies financières et monétaires au niveau national; il doit administrer la Banque nationale, décider de l'émission de monnaie, emprunter de l'argent au niveau national et international, et réguler les opérations concernant les devises étrangères (article 77-4);
 il doit protéger les brevets et les droits d'auteur (article 77-5);
 il doit élaborer et mettre en œuvre les politiques et les stratégies économiques, sociales et de développement du pays (article 77-6);
 il doit présenter des standards de mesures et un calendrier uniformes (article 77-7);
 il doit élaborer la politique étrangère et superviser sa mise en œuvre (article 77-8);
 il doit assurer l'observance de la loi et de l'ordre (article 77-9);
 il a le pouvoir de déclarer l'état d'urgence. S'il le déclare, il devra soumettre, dans la limite de temps prévue par la Constitution, une proclamation déclarant l'état d'urgence au vote de la Chambre des représentants des peuples (article 77-10);
 il soumet des propositions de loi à la Chambre des représentants des peuples dans le domaine de ses compétences, y compris des propositions de loi sur les déclarations de guerre (article 77-11) ou les proclamations d'état de guerre (article 55-9);
 il doit s'acquitter d'autres responsabilités que la Chambre des représentants des peuples ou le Premier ministre pourraient lui confier (article 77-12);
 il doit mettre en œuvre les règlementations en conformité avec les pouvoirs qui lui sont attribués par la Chambre des représentants des peuples (article 77-13).

Déclaration de l'état d'urgence

En ce qui concerne la déclaration de l'état d'urgence, le rôle du Conseil des Ministres est décrit à l'article 93 de la Constitution :
 le Conseil des ministres dispose du pouvoir de décréter l'état d'urgence lorsque le territoire national est envahi ou lorsqu'apparaît un trouble à l'ordre public, mettant en danger l'ordre constitutionnel et ne pouvant être contrôlé par les organes et les personnes chargés en temps normal de faire respecter la loi ; il peut également proclamer l'état d'urgence à la suite d'une catastrophe naturelle ou lorsqu'une épidémie surgit (article 93-1-a). Cet état d'urgence peut rester en vigueur pendant six mois et pourra être renouveler (article 93-3);
 lorsuqu'il le déclare, il devra soumettre, dans une limite de temps prévue par la Constitution, une proclamation au vote du Chambre des représentants des peuples (article 77-10);
 après l'entrée en vigueur de l'état d'urgence, le Conseil des ministres dispose, en conformité avec les règlementations qu'il a adoptées, des pouvoirs nécessaires afin de protéger la paix interne, la souveraineté nationale, la sécurité publique, afin également d'assurer le respect de la loi et de l'ordre (article 93-4-a);
 le Conseil des ministres peut suspendre les droits démocratiques et politiques mentionnés dans la Constitution dans la mesure où cette suspension peut permettre d'écarter les circonstances ayant amené à la proclamation de l'état d'urgence (article 93-4-b);
 au cours de l'exercice de ses pouvoirs durant l'état d'urgence, le Conseil des ministres ne peut pas suspendre ou limiter les droits prévus aux articles 1, 18, 25, 39-1 et 39-2 de la Constitution (article 93-4-b);
 le Conseil des ministres reçoit, avec le Premier ministre, des propositions de mesures correctives élaborées par la Commission d'Enquête sur l'état d'urgence (article 94-6-c).

Voir aussi
 Politique de l'Éthiopie

Source
Tous les articles cités font référence (sauf mention contraire) à la Constitution.

Politique en Éthiopie

redpajama_set_name: RedPajamaWikipedia


To get better at baseball, Orioles second baseman Jonathan Schoop went back to playing soccer.
After a rookie season full of ups and downs, Orioles second baseman Jonathan Schoop wanted to spend this offseason focused on self-improvement.
He found that opportunity back home in Curacao, where he spent the offseason training with a local soccer team to improve his agility. Instead of fielding grounders, Schoop dribbled a ball around cones.
Soccer is big in Curacao, and was Caribbean island's most popular sport until a wave of players made it to the majors, a path blazed by slugger Andruw Jones and, more recently, a talented group of young middle infielders that includes Schoop, Atlanta Braves shortstop Andrelton Simmons, Texas Rangers second baseman Jurickson Profar and New York Yankees shortstop Didi Gregorius. Now baseball is just as popular there.
But most kids, like the 23-year-old Schoop, grew up playing both sports. And from his days playing soccer as a kid, he knew it could help him get in better shape. He dropped 10 pounds from his tall frame in the offseason, going from 235 to 225.
"The field was just a five-minute bike ride from my house," Schoop said. "You know you have to be in shape before you have to run for 90 minutes. The drills I did were to make my agility better. It's running with quick feet. You stop and go there, run, dribble, pass. I didn't do it because I wanted to lose weight, because last year I wasn't feeling bad or anything. But I want to do something to get better.
Orioles left-hander Brian Matusz got his first chance to start an exhibition game this spring and made the most of it, giving up just two hits over three scoreless innings in Saturday's 3-2 loss to the Tampa Bay Rays at Ed Smith Stadium.
"I want to be better than last year. That's my goal. No matter how I do it, I want to get better every year. Everybody wants to improve here, and I want to be a better player than last year. I feel like if I do those things, I'll get better."
At last year's spring training, Schoop was expected to start the season at Triple-A Norfolk, but a strong spring — he hit .385 in Grapefruit League play — earned Schoop a roster spot. He filled in at third base for injured teammate Manny Machado for the first month of the season before assuming the starting role at second base, where he made 102 starts.
Orioles catcher Matt Wieters won't throw from the catching position in today's minor league exhibition game against the Puerto Rican developmental team today, but before the game, Wieters' throws were timed during a pregame workout.
Orioles manager Buck Showalter liked the results.
Schoop's strong defense allowed the Orioles to stick with him throughout some difficult times at the plate. Schoop hit just .209, struck out 122 times and drew 13 walks in 481 plate appearances.
But he also hit 16 homers, showing a power stroke rare for middle infielders. Six of those homers gave the Orioles the lead, including two off Yankees right-hander Masahiro Tanaka that broke scoreless ties at Yankee Stadium.
Last season included its share of adversity for Schoop, but his immense talent, eagerness to learn and desire to get better have the Orioles eager to see what the 23-year-old can do in his second full big league season.
"You notice the little things," Orioles manager Buck Showalter said. "But with Jon … he never assumes anything. He's always got a great grip on it. Every day he's in the big leagues, he's going to approach it like it's a gift, an honor. I love his makeup. We went through some growing pains offensively last year. We'll see what return we get for it. He's been pretty locked in here offensively" in spring training.
Schoop enters Sunday's B game against the Pittsburgh Pirates second on the club in total bases this spring (nine), with two doubles and a homer in six games. Schoop said he also feels lighter on his feet in the field.
"I think I feel better," Schoop said. "I feel better moving around. You're not carrying as much weight. You have more range, I think. You can also put more speed on base. That's how I think. I feel it. I feel stronger from losing weight. Less weight, I feel like I can run better."
The Orioles also have seen the difference his offseason work has made on defense. He has shown more agility, covered more ground and had a quicker first step. Showalter recently said that Schoop seemingly makes one great defensive play a game in Grapefruit League play.
"One of the things he didn't do well for us last year, and he didn't do well in the minor leagues when I had him, is go back on pop-ups, going back on balls over his head," said Orioles third base and infield coach Bobby Dickerson, who worked with Schoop in the minors as the team's infield coordinator. "He's already gone back on two. He went back on one the other day. He came out of nowhere to get to it. He didn't get close to those balls last year.
"Those are outs that turn the lineup over. And I've been harping on him since he was in rookie ball, those extra outs let [good hitters] hit again. … And he knows if he handles the ball in his range fine, if he can take a few more steps, add a little more range and get a few more extra outs that he wasn't getting before, he can be much better. And he's shown it."
Those little things can make Schoop, already a good defender, a great one. The Orioles rave about what he can do around the second base bag, especially for someone who's 6 foot 3. Dickerson contends that the way Schoop turns a double play, with how he pivots and shows his arm strength, is the best in baseball.
"You can tell he looks a little different," Showalter said. "He was pretty good last year. I thought he was the best second baseman in the league defensively. Who's better? I think this year, if he's with us playing second base every day, he'll contend for a Gold Glove. He's still growing, little by little."
Having J.J. Hardy, who has won three straight Gold Gloves at shortstop, next to Schoop has eased his transition to the big leagues. Dickerson uses Hardy to point out the right way to play defensively, and Hardy is always willing to give Schoop some veteran advice.
"It's fun working with him because he has such a good time," Hardy said. "He keeps me feeling a little bit younger … or maybe it makes me feel older. One of the two. But it's fun watching him work because he does it with purpose. Buck talks about doing everything with purpose to get better, and he definitely does that."
At the plate, Schoop most needs to improve his discipline, which will help raise his batting average and lower his strikeout total.
"Of course, I want to show that I can hit more for average," Schoop said. "I think I have to swing at more strikes. I'm not going to go in there and trying to walk, because I don't think I'm that type of hitter. I like to swing the bat, but if I can calm that a little bit and only swing at strikes, I'll be better. I'll cut down on my strikeouts a little bit more. I think strikeouts are going to come regardless, but I've just got to swing at strikes. Of course, you're going to swing at some balls, but try to minimize that and good things will happen. … But if you stay focused on every pitch, every play, good things will happen."
Schoop said he hopes having a full year of experience in the majors will help. Now that he has seen major league pitching, he's more familiar with the pitchers he'll face and what they'll try to do to get him to chase balls outside the strike zone.
"You've got to think about the game," Schoop said. "Don't just go in there open-minded and just swing. You can be aggressive, but you've got to recognize situations, what they're trying to do with you, what they're trying to accomplish. If they throw you the opposite [of what you're expecting], you just have to tip your hat and go in. But don't get yourself out."
Hardy said Schoop's offensive improvement is already evident this spring.
"Just his first couple of at-bats this year in spring training, you see him laying off sliders that he swung at all year last year," Hardy said. "And he's aware of that type of stuff. He comes up to me and he'll let me know. He'll say, 'You see me lay off those sliders? I spit on those.' He's aware of what he's doing, and I think he understands, too, that the sky's the limit. He can be really, really good. He expects it right now. He thinks he should be … I think he has the chance to be very, very good. He's young, and it takes some time."


text:
\section{Introduction}

The holographic computation of the correlation functions has been an important subject in AdS/CFT correspondence.
Soon after the proposal of the correspondence\cite{Maldacena:1997re,Gubser:1998bc,Witten:1998qj}, the two-point and three-point functions of the operators in the protected sectors have been computed on both sides\cite{Freedman:1998tz, Chalmers:1998xr,Lee:1998bxa,Arutyunov:1999en}. On the field theory side, the correlation functions were
discussed in the free field limit, while on the string side, they were computed in the supergravity approximation. The perfect match has been found due to the non-renormalization theorem, strongly supporting the correspondence. However, for the operators beyond the protected sectors, the holographic computation usually becomes difficult.

Very recently, there has been renewed interest in the holographic computation of the correlation functions of semiclassical strings. The semiclassical string states correspond to the composite operators with large quantum numbers in the field theory\cite{Gubser:2002tv}. And from the dictionary of the AdS/CFT correspondence, at leading order their semiclassical energies, whose quantum corrections are suppressed by the large quantum number, give the quantum dimensions of the corresponding operators, which are the eigenvalues of the dilatation operator and are calculable with the help of integrability. This relation provides not only a nontrivial check of AdS/CFT correspondence beyond the BPS protected sector, but also leads to a better understanding of integrable structures on both sides of the correspondence. The study of the correlation functions of semi-classical string states
is expected to shed light on the string interaction and integrability in super-Yang-Mills theory at non-planar level.

There is a clear physical picture in holographic computation of the
correlation functions of semiclassical string states. One needs to
insert the string state vertex operators in the string path
integral. At strong coupling, the path integral is dominated by a
saddle point. Therefore one has to solve the equations of motions
with the vertex operators as sources, and find the string
configurations whose ends shrink and approach to the boundary of
AdS$_5$ at the insertion points of the composite operators
corresponding to the string states.  The holographic computation of
the two point functions is relatively easy and has been illustrated
in \cite{Buchbinder:2010gg, Janik,Tsuji:2006zn} . The computation
for the higher point functions becomes much more difficult, as the
string configurations satisfying the appropriate boundary conditions
are hard if not impossible to obtain. Nevertheless, using a strategy
in the holographic study of operator product expansion(OPE) of Wilson line and
Wilson surface
operators\cite{Berenstein:1998ij,Corrado99,Giombi:2006de,Chen:2007zzr},
the three-point and four-point functions of two very massive string
states and one or two light string states have been
investigated\cite{Zarembo:2010rr,
Costa:2010rz,Roiban:2010fe,Buchbinder:2010ek}. For other related
studies, see
\cite{Buchbinder:2010vw,Grossardt:2010xq,Hernandez:2010tg,Ryang:2010bn,Arnaudov:2010kk,Georgiou:2010an,Escobedo:2010xs,Russo:2010bt,Park:2010vs}.

In this article, we would like to study the holographic computation of two-point and three-point functions of the giant gravitons with open strings. The giant gravitons are the D3-banes wrapping trivial cycles without collapsing due to their coupling to the background flux\cite{McGreevy00,Grisaru00,{Hashimoto00}}. As the open string may end on the D-brane, one can consider the system of the giant gravitons with open strings, which correspond to determinant-like operators mapping to a class of open spin chain\cite{Berenstein:2005vf,Berenstein:2005fa}. As D-brane is non-perturbative object, it is not clear what kind of string state vertex operators should be inserted in the string path integral. One
possibility is to consider the boundary conformal field theory, which may count the presence of D-brane. However, for the case at hand, the string world-sheet picture for the giant graviton wrapping a sphere of finite size seems to make no sense. Nevertheless, inspired by the study of the usual semi-classical closed
string state, we expect the similar picture could be carried over, at least for the maximal giant graviton wrapping an $S^3$ in $S^5$. As to the leading order of string coupling $g_s$ and $\alpha'$, the D-brane and open string decouple, we may compute their contributions separately. Namely  we expect that D-brane contribution is from a ``fat'' D3-brane configuration whose ends shrink and approach to the insertion points at the AdS boundary. For the open string sector, the contribution is from the open string configuration connecting the insertion points, ending on the brane and
satisfying appropriate boundary conditions. For the two-point functions, the computation is relatively simple. For the three-point function, the holographic computation is quite difficult. As in other cases, we consider the cases with two massive states corresponding to the giant gravitons with open strings and one gravity state corresponding to the half-BPS chiral primary operator. Due to the presence of the giant gravitons, we have to consider the contribution of D3-branes to the three-point structure constants, besides the one from open string. It is remarkable that the D-brane contribution is vanishing, possibly due to the BPS property of the giant graviton in our case.

There is another novel feature in our treatment. We propose to use the Routhian to handle the string state contribution. For all the cases studied in the literature, our prescription gives the same answer. But for the cases discussed in this paper, we have to use the Routhian. This seems to suggest that our prescription is more useful.

In the next section, we give a brief review of open string configurations ending on the maximal giant graviton branes in $S^5$, which correspond to the open spin chains. In Sec. 3, we discuss the two-point functions of the giant gravitons with open strings, including the cases corresponding to the ground state of $Z=0$ open spin chain, the ground state of $Y=0$ open spin chain and the open string in AdS$_5$ and ending on $Z=0$ brane. In Sec.~4, we
investigate the three-point functions. In the small coupling limit, we compute the three-point functions in the free field limit, and in the strong coupling limit, we holographically compute it in the limit of operator product expansion. We find the perfect match for the structure constant $c^Y_{123}$ involving the ground state operator of the $Y=0$ open spin chain.
Last section is devoted to some conclusions and discussions.

\section{Open strings ending on  giant graviton branes\label{open}}
In this section we shall be describing giant graviton branes and
open strings ending on the maximal giant graviton branes. We shall
also comment on the ${\cal N}=4$ SYM operators corresponding to the
D-branes plus the open string states. In this note we shall be
mainly concerned about giant graviton branes moving in the  $S^5$
part of the 10d spacetime. The $S^5$
 can be described by three complex embedding coordinates $(X,Y,Z)$ satisfying
 the constraint,
 \be
 |X|^2+|Y|^2+|Z|^2=1
 \ee
 where we set the AdS radius to be unity.
 Using the parametrization,
 \be
 X=\sin\th\,\cos\alpha \, e^{i\phi_1},\hs{3ex}
Y=\sin\th\,\sin\alpha \, e^{i\phi_2}, \hs{3ex} Z=\cos\th\, e^{i\phi}\,,
\ee
one finds that
 the  $S^5$ metric becomes
 \be ds^2_{S^5}=d\th^2+\cos^2\th\,
d\phi^2+\sin^2\th\, d\Omega_3^2\,,
\ee
with
\be
d\Omega_3^2=\cos^2\alpha\, d\phi_1^2+d\alpha^2+\sin^2\alpha\, d\phi_2^2\,.
\ee
Defining $r=\sin \th$, one can alternatively write the $S^5$ metric as
\be
ds^2_{S^5}=\frac{1}{1-r^2}\,dr^2+(1-r^2)\,d\phi^2+r^2\,d\Omega^2_3\,,
\ee
 where the coordinate $r$ is ranged over the interval $[0,\,1]$. The AdS$_5$ part can be described
 by the embedding coordinates $Y_I$ in $R^{2,4}$ satisfying the constraint
 \be
 -Y_0^2-Y_5^2 + Y_1^2 + Y_2^2 + Y_3^2 + Y_4^2=-1\,.
 \ee
The global AdS  coordinates  are related to the embedding coordinates by
the relations
\bea
 Y_5+iY_0= \cosh\rho\, e^{i t_{ads}}\,,\ \ \ \  Y_k =\sinh \rho\,\, \hat{n}_k\,,
\eea where $k=1,2,3,4$ and $\hat{n}_k$ with $\hat{n}_k \hat{n}_k=1$
describes a unit $S^3$.

The above three complex coordinates $(X, Y, Z)$ are dual to the
following three complex linear combinations \be X=\Phi^1+i\Phi^2,
\hs{3ex}Y=\Phi^3+i\Phi^4,\hs{3ex} Z=\Phi^5+i\Phi^6, \ee of the six
real scalar fields $\Phi^i$ in the ${\cal N}=4$ super Yang-Mills
side.

The D3 giant graviton brane\cite{McGreevy00,Grisaru00,Hashimoto00} of our main interest below wraps an $S^3$ part of the $S^5$
and rotates around the remaining directions of the $S^5$ while taking a pointlike
trajectory given by $\rho=0$ in the AdS$_5$ part of the geometry. These D3 branes preserve a half of the supersymmetries. We first consider a maximal sized giant graviton
brane. One example of such brane is given by the trajectory
$Z=0$ and $\rho=0$, which is called as $Z=0$ brane.   This $Z=0$ brane
corresponds to the gauge invariant SYM operator ${\cal O}_{D3}= {\rm det}\, Z$\,, which is  a
half
 BPS
state.
The dimension of this operator is simply given by its engineering dimension $N$, which is protected against
any quantum corrections. Below we shall also consider the $Y=0$ brane which is related to the $Z=0$ brane by
an appropriate $SO(6)$ rotation and dual to the operator ${\cal O}'_{D3}= {\rm det}\, Y$\,.

The correspondence between open strings ending on the giant graviton
brane and the open spin chain operators in the ${\cal N}=4$ SYM
theory is first considered in Ref.~\cite{HM} and further developed in
Refs.~\cite{correa,ABR}\, (for a recent review, see \cite{Zoubos:2010kh}).
There are two
classes of   open string states of our interest below: one is the
open string ending on the $Z=0$ brane and the other is the open
string ending on   the $Y=0$ brane. We shall choose the open string
vacuum oriented along $Z$ direction. Then the open string ending on
the $Y=0$ brane takes  Neumann  boundary condition on the $Z$ plane.
The open string can also end on the $Z=0$ brane with Dirichlet
boundary condition on the $Z$ plane; an additional localized
boundary degree of freedom is necessary at each boundary of $Z=0$.

In the ${\cal N}=4$ SYM theory side, the $Y=0$ brane open spin chain is represented
by composite operators\cite{HM}
\be
{\cal O}_Y=\epsilon^{j_1\ldots j_{N-1}A}_{i_1\ldots i_{N-1}B}\,
Y^{i_1}_{j_1} \cdots Y^{i_{N-1}}_{j_{N-1}}
(Z \ldots Z \chi_1 Z\ldots Z\chi_2 Z\ldots Z)^B_A,
\label{ybrane}
\ee
where $\chi_1,\chi_2,\ldots$ represent other SYM fields of bulk excitations.
An elementary excitation of a single impurity
is organized by the $SU(1|2)^2$ symmetry. It is clear that the ground state
of the $Y=0$ open spin chain is described by a unique vacuum configuration which
is a ${1\over 4}$ BPS state preserving the  $SU(1|2)^2$ symmetry.
The other is the $Z=0$ brane open spin chain
represented by composite SYM operators\cite{HM},
\bea
{\cal O}_Z = \epsilon^{j_1 \cdots j_{N-1} A}_{i_1 \cdots i_{N-1} B}\,
Z_{j_1}^{i_1} \cdots Z_{j_{N-1}}^{i_{N-1}}
(\chi_{L} Z \cdots Z \chi_1 Z \cdots Z\chi_2 Z \cdots \chi_R)^B_A
\,\,.\label{zbrane}
\eea
An important difference of the $Z=0$ brane case  from that of the $Y=0$ brane
is that the $Z=0$ open
spin chain is connected to
the giant graviton
through boundary
impurities $\chi_L$ and $\chi_R$.
Each  boundary state is organized by the representation of $SU(2|2)^2$ symmetry.
This elementary magnon involves 16 degenerate states with the energy
spectrum\cite{HM}
\be
E_B = \sqrt{1+ {\lambda\over 4\pi^2}}\,.
\label{eb}
\ee
Since the presence of the boundary impurities is essential, the space of corresponding ground states
is organized by the $256$ degenerate states of $SU(2|2)_L \otimes SU(2|2)_R$. Thus
there are no remaining supersymmetries for the ground state of the $Z=0$ open spin chain.

We now turn to the open string description\cite{Bak,BOR} of the open spin chain dynamics.
For the AdS$_5$ part, we use the ansatz
\bea
&& Y_5+ i Y_0 =\cosh \rho(\sigma)\, e^{i\kappa_s \tau}\nn\\
&& Y_1+i Y_2 = \sinh \rho(\sigma)\, \, e^{i\omega
\,\tau}\label{ansatz1} \eea with $Y_3+ i Y_4=0$. On the other hand,
for the $S^5$ part, we use the ansatz \be \theta=\theta(\sigma)\,, \
\ \ \  \phi= \nu\tau \ee with the $S^3$ coordinates $\alpha$,
$\phi_1$ and $\phi_2$ being taken as constants. The string equations
of motion tell us that \bea\label{eqrt}
&& \rho'' = (\kappa_s^2 -\omega^2)\sinh\rho \, \cosh\rho\,,\nn\\
&& \theta'' = \nu^2 \sin \theta\, \cos\theta\,.
\eea
The conformal gauge constraints lead to the condition
\be
{\rho'}^2 + {\theta'}^2= \kappa_s^2 \cosh^2\rho -\omega^2 \sinh^2\rho -\nu^2 \cos^2\theta\,.
\ee
The first equation of (\ref{eqrt}) may be integrated and leads to
\be
{\rho'}^2 = \kappa_s^2 \cosh^2\rho -\omega^2 \sinh^2\rho -\nu^2 k^2\,
\label{rhoeq}
\ee
and the second equation of (\ref{eqrt}) can be integrated to give
\be
{\theta'}^2= -\nu^2 \cos^2\theta +\nu^2 k^2\,
\label{thetaeq}
\ee
where the integration constant is fixed by the conformal gauge constraint.
Now for the $Z=0$ brane, one has the Dirichlet boundary condition $\rho=0$ and $\theta=\pi/2$ at
$\sigma =0,\,\pi$. The solution to the equations (\ref{rhoeq}) and (\ref{thetaeq}) satisfying
the boundary condition at $\sigma=0$ is given by the Jacobi elliptic function as\cite{BOR}
\be
\cosh\rho = {1\over {\rm dn}(\tilde{\omega}\sigma, {\tilde{k}}^2)}\,,\ \ \ \
\sin\theta ={\rm dn} (\nu\sigma, k^2)
\ee
where we have introduced
\be
{\tilde{\omega}}^2 \equiv \omega^2 -\nu^2 k^2\,, \hs{3ex}
{\tilde{\kappa}_s}^2 \equiv \kappa_s^2 -\nu^2 k^2\,, \hs{3ex} {\tilde{k}}^2={{\tilde{\kappa}_s}^2\over{\tilde{\omega}}^2}\,.
\ee
Note that the periodicity of the Jacobi elliptic function ${\rm dn}(x,k^2)$ is given by
$2 K(k^2)$ where we have introduced the complete elliptic integral of the first kind
$K(k^2)$
defined by
\bea
K(k^2)=\int_0^1 dx {1\over \sqrt{1-x^2}\sqrt{1-k^2 x^2}}\,
\eea
Since we are interested in the open string for $0\ \le\ \sigma \ \le\ \pi$ starting
from and ending on the $Z=0$ brane at $\rho=0$, we have the relations
\be
\tilde{\omega}\pi = 2 K({\tilde{k}}^2)\,, \ \ \ \
\nu  \pi = 2 K({{k}}^2) \,.
\ee
The corresponding solution describes a string ending on the $Z=0$ brane spinning
around the $Z$ plane of $S^5$ and $Y_1+iY_2$ of $AdS_5$ at the same time.
The ground state of the $Z=0$ open spin chain
that does not involve any derivative type letters, e.g. $D^S_+ Z$,
is described by $\rho(\sigma)=0$ which leads to the condition
$\kappa_s= \nu k$ with $\tilde{\kappa}_s=0$. Then this ground state configuration carries
the energy $E$ and the $S^5$ R-charge $J$\cite{Bak},
\bea
E &=& \frac{\sqrt{\lambda}}{2\pi} \kappa_s \int_0^\pi d\s \cosh^2 \rho
 = {\sqrt{\lambda}\over \pi} \,\,{k\, K(k^2) } \label{eopen} \\
J &=& \frac{\sqrt{\lambda}}{2\pi} \nu \int_0^\pi d\s \cos^2 \theta
={\sqrt{\lambda}\over \pi} \left[\, K(k^2)-E(k^2)\, \right]\,, \eea
where \bea
 E(k^2)=\int_0^1 dx {\sqrt{1-k^2 x^2}\over \sqrt{1-x^2}}
\eea
is the complete elliptic integral of the second kind.
It is then straightforward to show that
\be
2 E_B= E-J
= {\sqrt{\lambda}\over \pi} \left(1-{4\over e^2} e^{-{2\pi J\over\sqrt{\lambda} }}+ \cdots\right)\,
\ee
in the long string limit of large $J$.
This agrees with $E_B$ in (\ref{eb})  in the large $\lambda$ and the large $J$ limit.
The $J$ dependent correction in this spectrum stems from the
finite size effect of boundary degrees of freedom of the $Z=0$ brane.

Next we turn to the open string description of the $Y=0$ open spin
chains. At $\sigma=0,\,\,\pi$, one has the Dirichlet boundary
condition $\rho=0$ in the AdS$_5$ side and $\theta'=0$ for the $Z$
plane which is parallel to the worldvolume directions of the $Y=0$
brane. Again using the string equations in (\ref{rhoeq}) and
(\ref{thetaeq}), one may construct various solutions but we shall
here simply present a rather trivial solution, \be \theta=0\,, \ \ \
\ \rho=0\, \ee with $\kappa_s=\nu$ with $k=1$. This describes a
pointlike string  rotating in the equator $r=1$, which corresponds
to the ground state of the $Y=0$ open string carrying\cite{Bak} \be E= J=
{1\over 2}{\sqrt{\lambda}\,\nu}\,. \ee One may see that there is no
finite size correction. Since this vacuum state preserves a quarter
of the sixteen supersymmetries, one has the non-renormalization of
the ground state energy $E_g=E-J=0$ using the argument
Ref.~\cite{Basu:2004nt}. The operator dual to this ground state
${\cal O}_Y^J$ will be given in subsection~\ref{field}.

\section{Two-point functions}

\subsection{The crucial role played by Routhian\label{routhian}}

A prescription on the computations of the two-point functions from
the semi-classical string solution was given in \cite{Janik}. One of
the points stressed there is that we cannot simply evaluate the
on-shell actions of the semiclassical solutions. Instead there are
important corrections from convolution with the string-state
wave-functions. For point particles and strings only rotating in
$S^5$ part of the background, these corrections just change the
action into energy. But for the strings rotating in $AdS$ part as
well, the story is more involved. We need to treat certain zero
modes carefully. And the proposed corrections in this case reads
 \bea \Delta S
=-\int d\tau d\sigma (\vec{\Pi}-\vec{\Pi}_0)\cdot (\dot{\vec{Y}}-
\dot{\vec{Y}_0}) \eea where $\vec{Y}_0$ and $\vec{\Pi}_0$ are
respectively so called zero-modes of the coordinate $\vec{Y}$ and
the momentum $\vec{\Pi}$. It is also stated there that we should use
the embedding coordinates $Y$'s which respect the $SO(2, 4)$
symmetry.

Though the above prescription works well for all cases considered there,
we found that this
turns out to be problematic for more
general cases. As an example, for the open string solution in
subsection~\ref{2p open}, we do not obtain any sensible result
following the prescription. And the problem is already there for
the closed string counterpart of the open string solution. Thus
we need some generalization of the prescription.

Through trial and error, we now propose to use the Routhian to
handle the string state contribution. Our suggestion is as follows.
Whenever there is a conserved quantity $Q_a$ for the solution
(except the energy) with the corresponding cyclic coordinate $y^a$,
we perform the Legendre transform
\bea L_R= L -\sum_a Q_{a}
{\dot{y}}^a\,, \eea to yield the Routhian. For the further variation
of the Routhian, the conserved charge $Q_{a}$ should be held fixed.

For the strings only spinning inside $S^5$, the conserved charges
are three spins $J_1, J_2, J_3$ in $S^5$ and the Routhian coincides
with the energy. Our new prescription goes back to the old one in
\cite{Janik} for these strings. It is not hard to see that it is
also the case for the string rotating in both $AdS_5$ and $S^5$
studied there. However as we have already mentioned, the
generalization of the prescription in \cite{Janik} is needed for
more complicated cases.

We also find that, interestingly, the Routhian plays a similar role when
we compute the two-point function from a giant graviton (a D3 brane
inside $S^5$) in the next subsection.

\subsection{Maximal Giant Gravitons\label{2p brane}}

In this subsection, we will compute the two-point function of local
operators dual to maximal giant graviton which is a D3 moving in
$S^5$. More precisely, the operator we focus on is ${\cal
O}_{D3}={\rm det}\,Z$. As we have reviewed, the dual  object in the gravity side is
the $Z=0$ giant graviton brane.

For this computation, we  have to use the Poincare coordinate
for $AdS_5$. The corresponding metric is given by \be
ds_{AdS_5}^2=\frac{1}{z^2}\Big[-dt^2+\sum_{i=1}^3 dx_i^2+dz^2
\Big]\,. \label{metric1} \ee The two-point function we want to
compute in this subsection is \be \langle {\cal
O}^\dagger_{D3}(t=t_f, {\bf x}={\bf 0})\, {\cal O}_{D3}(t=0, {\bf
x}={\bf 0}))\rangle\,,\ee and, as in \cite{Janik}, we need to first
find the suitable classical D3 brane solution.
 The worldvolume coordinates of this D3 brane is
$(\tau, \theta_1, \theta_2, \theta_3)$, where $\theta_i$ are three
directions $(\alpha,\phi_1,\phi_2)$ of the $S^3$ inside $S^5$. The part of
nontrivial embedding involves
$t(\tau),
z(\tau), \phi(\tau)$ directions
with $r$  being a constant\footnote{We will narrow down to
the case with $r=1$ shortly.}. The boundary conditions read \be
\big(\,t(-s/2), z(-s/2)\,\big)=(0, \epsilon),\hs{3ex} \big(\,t(s/2), z(s/2)\,\big)=(t_f,
\epsilon). \ee
The resulting induced metric  becomes
\be ds_{ind}^2=\Big[\,\frac{-\dot{t}^2+\dot{z}^2}{z^2}+(1-r^2)\dot{\phi}^2\,\Big] d\tau^2+r^2d\Omega_3^2
\ee
where dot denotes a derivative with respect to $\tau$

The action of D3-brane is given by \be
I_{D3}=I_{DBI}+I_{CS}=-T_{D3}\int\sqrt{
\gamma}+T_{D3}\int P[C_4],\ee
where $\gamma_{ab}$ is the induced metric of D3 brane, $P[C_4]$ is the pullback of
Ramond-Ramond four-form potential and $T_{D3}$ is the tension of
D3-brane, \be
T_{D3}=\frac{1}{(2\pi)^3g_s\alpha^{\prime2}}=\frac{N}{2\pi^2}.\ee

For the solution at hand, after integration over $S^3$, the DBI part
of the Lagrangian becomes \be {\cal
L}_{DBI}=-T_{D3}\Omega_3r^3\sqrt{\frac{\dot{t}^2-\dot{z}^2}{z^2}-(1-r^2)\dot{\phi}^2}
,\ee
where $\Omega_3=2\pi^2$ is the volume of unit $S^3$. The
Chern-Simons part is the same as the one in \cite{McGreevy00}: \be
{\cal L}_{CS}=\dot{\phi}Nr^4,\ee
with
\be {\cal L}={\cal L}_{DBI}+{\cal
L}_{CS}.\ee

The angular momentum
carried by the giant graviton brane is given by
 \be J=\frac{\p {\cal L}}{\p
\dot{\phi}}=\frac{Nr^3(1-r^2)\dot{\phi}}{\sqrt{\frac{\dot{t}^2-\dot{z}^2}{z^2}-(1-r^2)\dot{\phi}^2}}+Nr^4.
\label{angular1} \ee And
one has
$J=N$ for the maximal case of $r=1$.

It can be checked that
 \be  t= R\tan\kappa\tau+t_0, \hs{3ex}z=\frac{R}{\cos\kappa\tau},
\label{sol} \ee is the solution of the equations of motion.
The boundary condition
$z(\pm s/2)
\epsilon$ implies
\be \frac{ R}{\cos\frac{\kappa s}2}=\epsilon.\ee Since
$\epsilon$ is very small, we can see that $s$ cannot be real. We can
write $s=-i\tilde s$ with $\tilde s$ being real. Then the worldsheet
coordinate $\tau$ and the spacetime coordinate $t$ have to be purely
imaginary. From $t(-s/2)=0, t(s/2)=t_f$, we get \be \kappa\sim
\frac2{\tilde s}\log\frac{|t_f|}{\epsilon}.\ee

As we discussed in the previous subsection in detail, we proposed
that it is the Routhian which gives the correlation function. Now
the only conserved charge besides the energy is the angular momentum
$J$. It is easy to see that for maximal giant graviton, the Routhian
is nothing but the DBI part of the action: \be {\cal
L}_{DBI}=-N\kappa=-\frac{2N}{\tilde s}\log\frac{|t_f|}{\epsilon} \ee
and \be I_{DBI}=\int_{+i{\tilde s}/2}^{-i{\tilde s}/2}{\cal
L}_{DBI}d\tau=-2iN\log\frac{|t_f|}{\epsilon}.\ee We are thus led to
\be\exp(iI_{DBI})=\left(\frac{|t_f|}{\epsilon}\right)^{-2N}\ee which
is the expected result of the two-point function $\langle {\cal
O}_{D3}^\dagger (t_f)\, {\cal O}_{D3}(0) \rangle$.

\subsection{Nonmaximal Giant Gravitons in $S^5$}

We  begin with the D3 brane action \be {\cal L} = -{T_{D3} \over 2
} \sqrt{-\gamma}\left( \gamma^{ab}\partial_a X^m \partial_b
X^n G_{mn}(X) -2\right) +{\cal L}_{CS} \ee where $\gamma_{ab}$
is the world volume metric which is dynamical now. Since ${\cal
L}_{CS}$ is metric independent, the equations for the worldvolume
metric $\gamma_{ab}$ become
\be \gamma_{ab} \left(-1+{1\over
2} \gamma^{cd} \partial_c X^m \partial_d X^n
G_{mn}(X) \right) = \partial_a X^m \partial_b X^n G_{mn}(X)
\,. \ee Taking trace of this equation, one finds \be \gamma^{cd
} \partial_c X^m \partial_d X^n G_{mn}(X) =4\,. \ee
We partially solve the problem by the choice $\gamma_{ij}d\sigma^i
d\sigma^j= d\Omega_3^2$ and integrate the Lagrangian density over
the  $S^3$. The resulting Lagrangian then becomes \be L =-{N\over 2}
r^3 \,{b} \,\left( -\left(  {-{\dot{t}}^2 +{\dot{z}}^2 \over z^2} +
(1-r^2){\dot{\phi}}^2 \right){1\over {b}^2} +1 \right)  + N r^4
\dot{\phi} \ee
where
b=\sqrt{-\gamma_{00}}$. We have set $\dot{r}=0$
consistently since we are interested in the solution with fixed
radius. The angular momentum \be J=Nr^3(1-r^2){\dot{\phi}
\over
}+Nr^4, \label{angular2} \ee
is a conserved quantity as a result of
the equation of motion. By the Legendre transform, one now introduce
the Routhian \be L_R =L-\dot{\phi} J ={N\over 2} r^3  \,\left(
{-{\dot{t}}^2 +{\dot{z}}^2 \over z^2\, b
} - b
\left( {(J/N
-r^4)^2\over (1-r^2) r^6} +1 \right) \right) \ee One may check that
all the resulting equations of motion are equivalent to the ones got from
the Lagrangian before the transformation.  The equation for $b
$ reads
\be
 -{-{\dot{t}}^2 +{\dot{z}}^2 \over z^2\, b^2
 } -  \left(
{(J/N -r^4)^2\over (1-r^2) r^6} +1 \right)=0\,. \ee
Using this,  the
 equation for $r$ becomes \be 6 r^2 \left( {(J/N -r^4)^2\over (1-r^2)
r^6} +1 \right) + r^3 {d\over dr}\left({(J/N -r^4)^2\over (1-r^2)
r^6}\right)=0\,, \ee whose stable solution is $r^2=J/N$. Inserting
this solution to $L_R$, one gets the effective Lagrangian for the
AdS part: \be L_R ={N\over 2} (J/N)^{3\over 2}  \,\left(
{-{\dot{t}}^2 +{\dot{z}}^2 \over z^2\, b
} - b
N/J \right)\,. \ee
The desired solution of this system is given by  (\ref{sol})
with $
=\kappa \sqrt{J/
N}$,
which leads to
\be L_R = -J\kappa\,. \ee
The
treatment afterward is the same as the maximal giant case.

\subsection{The two-point functions from giant gravitons with open
strings \label{2p open}}

Let us consider the
two-point function  $\langle {\cal O}^{\dagger J}_{Z}(x)\, {\cal O}^J_{Z}(0)\rangle$ with the operator ${\cal
O}_Z$ corresponding to the ground state of $Z=0$ open spin
chain without involving any derivative type operators $D^M$. The  open string
solution dual to the single operator is give in section \ref{open}.
We can use the idea in \cite{Janik} to find the open string solution
for two-point function: \bea && x(\tau)=R \tanh\kappa\tau+x_0,\
z(\tau)=\frac{R}{\cosh\kappa\tau}, \nn\\
&& \sin\theta(\sigma)={\rm dn}(\nu\sigma, k^2)\,, \ \ \ \phi=\nu\tau\,,
\label{zsolution}
\eea
with $i\kappa\equiv\kappa_s=\nu k$.
The boundary
conditions $x(-s/2)=0,\  x(s/2)=x, \ z(\pm s/2)=\epsilon$ give the
following constraints: \be x_0=x/2,\ x=2R\tanh\frac{\kappa s}{2},\
\frac{R}{\epsilon}=\cosh \frac{\kappa s}{2}.  \ee

There are two parts of contributions to this two-point function.
The first part is from the Routhian of the D3-brane which is
discussed in subsection \ref{2p brane}. The second part is from the
Routhian of the open string which is just the same as the energy of
the open string. The computations are the same as the ones for the
closed string case in \cite{Janik}. And finally we obtain \be\langle
{\cal O}^{\dagger J}_{Z}(x) {\cal
O}^J_{Z}(0)\rangle=|x|^{-2N-2E_{open}},\ee with $E_{open}$ given in
(\ref{eopen}). The computation of the two-point function $\langle
{\cal O}^{\dagger J}_{Y}(x)\, {\cal O}^J_{Y}(0)\rangle$ from the
point-like open string solution in section \ref{open} can be treated in a  similar manner.
The string solution is
 \bea && x(\tau)=R \tanh\kappa\tau+x_0,\
z(\tau)=\frac{R}{\cosh\kappa\tau}, \nn\\
&&\theta=0\,, \ \ \phi=\nu\tau\,,\ \ i\kappa=\nu,
\label{ysolution}
\eea together with similar constraints from the boundary
conditions.

Now we turn to the open string inside $AdS_5$ \cite{BOR}.
We try the same ansatz as in (\ref{ansatz1}) \bea
&& Y_5+ i Y_0 =\cosh \rho(\sigma)\, e^{i\kappa_s \tau}\nn\\
&& Y_1+i Y_2 = \sinh \rho(\sigma)\, \, e^{i\omega
\,\tau}, \label{ansatz} \eea
with $Y_3= Y_4 =0$ to describe the spinning open-string solution
ending on the giant graviton branes. We choose the range of string
coordinate $\sigma$ to be $0\le \sigma \le \pi$. For the $S^5$ part,
we assume that the string stays at a point in $S^5$ that satisfies
the required open-string boundary conditions. The contribution from
this part vanishes in this case.

The equation of motion for the string becomes \bea \rho''
=(\kappa_s^2-\omega^2) \sinh \rho \cosh \rho\,. \eea Integration of
this equation leads to \bea {\rho'}^2 =\kappa_s^2 \cosh^2
\rho-\omega^2 \sinh^2 \rho + a\,, \label{maineq} \eea where $a$ is
the integration constant. We also have the conformal gauge
constraint \bea
&& \dot{\vec{Y}} \cdot \dot{\vec{Y}}+ {\vec{Y}'} \cdot {\vec{Y}'}=0\\
&& \dot{\vec{Y}} \cdot \vec{Y}'=0. \eea The latter condition is
satisfied automatically for any ansatz of the form in (\ref{ansatz})
and the former one is solved by setting $a=0$. Below for the later
purpose, we shall relax this condition $a=0$ and solve the equation
of motion for general $a$.

The equation (\ref{maineq}) is solved by \bea \cosh \rho(\sigma)=
{1\over {\rm dn}(\bar{\omega}\sigma, k^2)}\,, \label{spinsol} \eea
where we have defined \bea \bar{k}^2\equiv {{\bar{\kappa}}_s^2 \over
{\bar{\omega}}^2}\,, \ \ {\bar{\omega}}^2 \equiv \omega^2 +a\,,
\ \ {\bar{\kappa}}_s^2 \equiv \kappa_s^2 +a \,.\eea
Since we are considering the open string starting
from and ending on the giant graviton at $\rho=0$, we find the
relation \bea \bar{\omega} \pi = 2 {K}(\bar{k}^2)\,. \label{omega}
\eea

\begin{figure}[ht!]
\centering
\includegraphics[width=4cm]{fig1.eps}
\caption{\small The open string starting from and ending on the
giant graviton at $\rho=0$ is rotating in the $(Y_1, Y_2)$ plane. }
\end{figure}

The energy and the angular momentum are given by \bea
&&E={\sqrt{\lambda}\over 2\pi} \kappa_s \int_0^{\pi} d\sigma \cosh^2 \rho,\nn\\
&&S={\sqrt{\lambda}\over 2\pi} \omega \int_0^{\pi} d\sigma \sinh^2
\rho \label{ang} \eea which are related to each other by \bea E=
{\kappa_s\over 2}\sqrt{ \lambda}+ {\kappa_s\over \omega} S\,. \eea
Using the solution in (\ref{spinsol}), one finds \bea
&&E={\sqrt{\lambda}\over \pi} {\kappa_s \over \bar{\omega}} {{ E}(\bar{k}^2)\over 1-\bar{k}^2}\,,\nn\\
&& S={\sqrt{\lambda}\over \pi} {\omega \over \bar{\omega}} \Bigl(
{{ E}(\bar{k}^2)\over 1-\bar{k}^2} -{K}(\bar{k}^2) \Bigr)\,. \label{angular}
\eea

As in \cite{Janik}, we do the transformation $Y_0\to iY_4, Y_4\to
iY_0, \kappa_s\to -i\kappa$ and get \bea
&& Y_4=\cosh \rho (\sigma) \sinh\kappa\tau,\  Y_5=\cosh\rho(\sigma)\cosh\kappa\tau,\nn\\
&& Y_1=\sinh \rho (\sigma) \cos \omega\tau,\  Y_2=\sinh\rho
(\sigma)\sin \omega\tau, \ Y_0=Y_3=0.\label{eqm}\eea
In Poincare
coordinate, this solution is
\bea
x_1&=& e^{\kappa\tau} \tanh\rho (\sigma)\cos \omega\tau  , \nn\\
x_2&=& e^{\kappa\tau} \tanh \rho(\sigma)\sin \omega\tau , \nn\\
z\,\,&=&\frac{e^{\kappa\tau}}{\cosh\rho(\sigma)}\,, \eea
with $t=x_3=0$.

Now we use the same conformal transformation as the one used in
\cite{Janik} and get
\bea
x_1&=&\frac{\tanh\rho(\sigma)\cos \omega\tau e^{\kappa\tau}+\frac1R
e^{2\kappa\tau}}
{1+\frac2R \tanh\rho(\sigma)\cos \omega\tau e^{\kappa\tau}+\frac1{R^2}e^{2\kappa\tau}},\nn\\
x_2&=&\frac{\tanh\rho(\sigma)\sin\omega\tau e^{\kappa\tau}}{1+\frac2R \tanh\rho(\sigma)\cos\omega\tau e^{\kappa\tau}+\frac1{R^2}e^{2\kappa\tau}},\nn\\
z\,\,&=&\frac{\frac1{\cosh\rho(\sigma)}e^{\kappa\tau}}{1+\frac2R
\tanh\rho(\sigma)\cos\omega\tau
e^{\kappa\tau}+\frac1{R^2}e^{2\kappa\tau}},\eea
again with $t=x_3=0$.
  Notice that for
fixed $\tau$, the value of $z$ depends on $\sigma$ for the current
case. So we need a slight modification of the boundary conditions
used in \cite{Janik}. We choose the following boundary conditions:
\be x_1(\tau_i, \sigma)=0, \hs{3ex} x_1(\tau_f, \sigma)=x,  \hs{3ex}x_2(\tau_i,
\sigma)=x_2(\tau_f, \sigma)=0,\ee \be
{\rm min}_{\,\,0\le
\sigma \le \pi}\,
z(\tau_i, \sigma)={\rm min}_{\,\,0\le
\sigma \le \pi}z(\tau_f, \sigma)=\epsilon. \ee From these, we get
\be x=R,  \hs{3ex}\kappa=\frac1{s}(2\log R-2\log\epsilon+\log(1-k^2)),\ee
where $s=\tau_f-\tau_i$ and we have used the fact that the minimum
value of $\rm{dn} (\bar\omega\sigma, k^2)$ is given by
$\sqrt{1-k^2}$ for our spinning string. Now we rescale $\epsilon$,
such that
 \be x=R, \hs{3ex}
\kappa=\frac1{s}(2\log R-2\log\epsilon).\ee

We remark here that one can also use other boundary condition by
changing the value of $z$ at $t=t_i,\, t_f$ into the maximal value of
$z$ or the mean value of $z$ or the value of $z$ at any $\sigma$.
These boundary conditions give the same results if we always use
the rescaled $\epsilon$ like above.

As we discussed in subsection~\ref{routhian}, we need to use the
Routhian to compute the two-point functions. For the present
problem, the conserved quantity is the angular momentum $S$ and the
corresponding coordinate $\phi$ is defined by ${Y_1+iY_2}=
 |Y_1+iY_2|e^{i\phi}$ with $\dot\phi=\omega$.
Evaluation of the action can be expressed as \be I={\sqrt{\lambda}\over 4\pi}\int
d\tau d \sigma \left(\kappa^2 \cosh^2\rho +\omega^2 \sinh^2\rho
-{\rho'}^2 \right). \ee This leads to the corrected action of the
form, \bea I_R =I-\int d\tau S \omega={\sqrt{\lambda}\over 4\pi}\int d\tau
d\sigma \left(\kappa^2 \cosh^2\rho -\omega^2 \sinh^2\rho -{\rho'}^2
\right) \eea where we use the expression for the angular momentum in
(\ref{ang}). Using the equation of motion in (\ref{maineq}), the
above expression can be reduced to \bea I_R={\sqrt{\lambda}\over 2\pi} \int d\tau
d\sigma (\kappa^2 \cosh^2\rho -a)= 2
\left( {\kappa\over
2} \sqrt{\lambda}+{\kappa\over \omega}S -{a\over 4\kappa}\right) \log {R/\epsilon}.
\eea We would then like to minimize the above expression with respect to
the modular parameter $s$ that is proportional to $\kappa={i\kappa_s}$.
One further note that $\bar{\omega}$ will be completely determined
as a function of $\bar{k}$ by (\ref{omega}). Using the definition of $\bar{k}$,
$\bar{\kappa}_s$ is also function of only $\bar{k}$. Using the expression
in (\ref{angular}), we find also that $\omega$ is completely fixed
as a function of $\bar{k}$ when we fix $S$ as a constant. Noting
$a={\bar{\omega}}^2-\omega^2$, one finds that $a$ is also
completely determined as a function of $\bar{k}$. Thus
the extremization
condition
 of $I_R$ with respect to modular parameter $s$ is  equivalent to
that of  extremization of $I_R$ with respect to $k$. As summarized
in the appendix, $d I_R/d\bar{k}^2=0$ gives $a=0$, which is the desired
solution of the Virasoro constraint. One then has \bea e^{i (I^{D3}_R+I_R)}
\propto {1\over (x/\epsilon)^{2\Delta}} \eea with $\Delta=N+
{\k_s\over 2}\sqrt{\lambda}+ {\k_s\over \omega}S\,$ including the contributions from the D3-brane.

\section{Three-point functions}

\subsection{Field theory side \label{field}}
In this subsection, we discuss the computation of three-point
function in  ${\cal N}=4$ SYM theory,
  which will be used for the later comparison
with the semiclassical computation.

We begin our discussion with the correlation functions involving
the vacuum-state operator of the $Y=0$ open spin chain,
\be
{\cal O}^J_Y ={d_N^J}\,\,{1 \over (N-1)!}\,\,\epsilon^{j_1\ldots j_{N-1}A}_{i_1\ldots i_{N-1}B}\,
Y^{i_1}_{j_1} \cdots Y^{i_{N-1}}_{j_{N-1}}
(Z^J)^B_A\,.
\ee
In our convention,
the normalization factor $d_N^J$  is defined by
 the two-point function
\be
\langle{\cal O}^{\dagger J }_{\,\,Y} (x) \, {\cal O}^{ J}_Y (0) \rangle = {1\over (x^2)^{N+J-1}}\,
\ee
with a unit normalization.
Note that the position space propagator of the fields
$X,Y$ and $Z$ is given by
\be
I(x)= {g^2_{\mbox{\tiny YM}}\over 4\pi^2}\, {1\over x^2}\equiv {s_2 \over x^2}\,,
\ee
in the standard convention of the ${\cal N}=4$ SYM theory.

\begin{figure}[ht!]
\centering
\includegraphics[width=7cm]{fig2.eps}
\caption{\small The relevant Feynman diagram of free field evaluation for the two-point function
 of the $Y=0$ ground state
 is depicted. We illustrate here
 the case of
 $N=16$ and $J=11$.
  }
  \label{fig2}
\end{figure}

The relevant Feynman diagram of the free field contractions is depicted in Fig.~\ref{fig2}:
The part involving  separated parallel lines represents the
contractions of elementary fields inside the subdeterminant parts of the operators whereas the remaining
square represents the contractions in the $(Z^J)^A_B$ parts.
With the operator of the normalization
${\cal O}^{J}_Y/d^J_N$, the subdeterminant part has $(N-1)!$ independent ways of contractions
for given indices $A$ and $B$ and the $(Z^J)^A_B$ part involves $N^{J-1}$ factor from the
closed loops of indices. Further including the $N^2$ possibilities of the indices $(A,B)$, one has
 \be
 \langle{\cal O}^{\dagger J }_{\,\,Y} (x) \, {\cal O}^{ J}_Y (0) \rangle\,\, (d^J_N)^{-2}=\big[\,I(x)\,\big]^{N+J-1} N^2\, N^{J-1}\,
 (N-1)!\,.
\ee
Therefore, the normalization
factor is determined as
\be
(d^J_N)^{-2}= (s_2)^{N+J-1}\, N^J\, N!\,.
\ee
This computation is valid for all range of $\lambda$, which can be justified using the
non-renormalization argument of $1/2,1/4$ and $1/8$ BPS operators for the two-point functions in Ref.~\cite{Basu:2004nt}.

Next we turn to the evaluation of the 1/4 BPS three-point correlation function
\be
\langle{\cal O}^{\dagger J+\ell}_Y (x_1) \, {\cal O}^{ J}_Y (x_2) {\cal O}_{Z^{ \ell}} (x_3)\rangle={c^Y_{123}}\,\,{1\over
(x^2_{12})^{N+J-1}\,(x^2_{13})^{\ell}}
\ee
where  ${\cal O}_{Z^{ \ell}}$  denotes  the half BPS chiral primary operator oriented
in the $Z$ direction given by
$
{\cal O}_{Z^{ \ell}} = c_\ell \, {\rm tr}\, Z^{\ell}
$
with $
c_\ell ={(2\pi)^\ell/(\sqrt{\ell} \lambda^{\ell/2}})\,.
$

\begin{figure}[ht!]
\centering
\includegraphics[width=7cm]{fig3.eps}
\caption{\small The relevant Feynman diagram of free field evaluation for $c^Y_{123}$ is depicted here.
We illustrate
the case of
 $N=16$, $J=7$ and $\ell=4$.
  }
 \label{fig3}
\end{figure}

The corresponding Feynman diagram for the free field evaluation is illustrated in Fig.~\ref{fig3}.
We note that there are $J-1$ independent ways to put the operator ${\cal O}_{Z^\ell}$ inside the
box and $\ell$ independent ways to pick a particular $Z$ in ${\rm tr} \, Z^\ell$.
The remaining counting of the combinatoric factors is similar to that of the two-point function.
Therefore, one is led to
\be
c^Y_{123}= d^{J+\ell}_N d^{J}_N c_\ell \,N!\, N^{J+\ell-1}\, (s_2)^{N+J+\ell-1}\, (J\!-\!1)\,
\ell={1\over N} \sqrt{\ell}\,(J\!-\!1)\,,
\ee
which is again  valid for all range of the `t Hooft coupling $\lambda$ due to the
nonrenormalization of the 1/4-BPS three-point correlation function\cite{Basu:2004nt}.

This kind of non-renormalization argument does not apply for
the three-point function involving  the operator
${\cal O}_Z^{J}$ which represents a particular ground state of
the $Z=0$ open spin chain carrying  the ($Z$ plane) R-charge $J+N-1$.

 This is because
the three-point
function of our interest below
\be
\langle{\cal O}^{\dagger J+\ell}_Z (x_1) \, {\cal O}^{ J}_Z (x_2) {\cal O}_{Z^{ \ell}} (x_3)\rangle={c^Z_{123}}\,\,{1\over
(x^2_{12})^{N+J-1+2E_B}\,(x^2_{13})^{\ell}}\,,
\ee
does not preserve  any of the sixteen supersymmetries.
We expect generic perturbative renormalization of the three-point function  and do not have
any field theoretic means to compute
the corresponding structure constant
in the strongly coupled regime.

\subsection{D brane contribution to the structure constants}

In the following two subsections, we shall concern about the
holographic computations of the structure constants $c^Y_{123}$ and
$c^Z_{123}$. This subsection will be devoted to the D-brane
contributions $c_{D3}$. And  we will focus on the open string
contributions $c_{string}$ in the next subsection.
We shall compute the three-point functions in the limit of the
operator product expansion (OPE)
\be R_{12}=|x_1-x_2|\ \ll\
L_{12}\equiv \frac{1}{2}({x_1+x_2})\,,\ee
with the choice of $x_3=0$.
Our prescription of
holographic computations of these three-point functions is along the
way in \cite{Zarembo:2010rr,Costa:2010rz}. The light chiral primary
operator is dual to specific fluctuations of background fields in
string theory side. This fluctuation will lead to the fluctuation
of the classical D3-brane and open string solutions used in the
calculation of the two-point functions. The variant of the on-shell
action of the coupled system of D-brane and open string will give
the three-point functions in the large $N$ and large $\lambda$
limit.
 This can be understood as that we treat the
brane and string as source for the supergravity fields and read off
the OPE coefficients from the coupling to the brane and string of
the bulk supergravity modes dual to the light operators
\cite{Berenstein:1998ij,Giombi:2006de}.

To the leading order of $g_s$ and $\alpha^\prime$, the classical
onshell action of the coupled system of the brane and the string is the sum
of the brane action $I_{D3}$ and the string action $I_{string}$. So
the variation of the on-shell action of the coupled system is equal
to $\delta I_{D3}+\delta I_{string}$. This leads to the fact that
the OPE coefficient $c_{123}$ is the sum of $c_{D3}$ and
$c_{string}$.

The fluctuations of the background $AdS_5\times S^5$  metric
$g_{\m\n}, g_{\a\b}$ corresponds to the half-BPS  chiral primary operators
are
\cite{Lee:1998bxa} \bea h_{\m\n}&=&(-\frac65\, \ell\,
g_{\m\n}s_I+\frac{4}{\ell+1}\nabla_{(\mu}\nabla_{ \nu)} s_I)\cY^I,\\
h_{\a\b}&=&2\,\ell\, g_{\a\b}\, s_I\,\cY^I, \eea where we use $\mu, \nu, \cdots$
for
the coordinate indices of the $AdS_5$ part and $\a, \b,
\cdots$  for
the coordinate indices of the $S^5$ part, and the
round parentheses of $(\mu\nu)$ denotes the symmetric traceless
part. The fluctuation of the background four-form potential is
\cite{Lee:1998bxa}
\bea a_{\m_1\cdots \m_4}&=&-4\epsilon_{\m_1\cdots
\m_5}\nabla^{\m_5}s_I \cY^I,\\
a_{\a_1\cdots\a_4}&=&4\epsilon_{\a_1\cdots \a_5}s_I
\nabla^{\a_5}\cY^I. \eea The spherical harmonic function $\cY^I$ in
the $S^5$ coordinate space is entirely fixed by the choice of the
chiral primary operators: for ${\cal O}_{Z^\ell}=c_\ell{\rm
tr}(Z^\ell)$ and ${\cal O}_{Y^\ell}=c_\ell{\rm tr}(Y^\ell)$,
the corresponding harmonic functions are given by
$2^{-\ell/2}\, Z^\ell$ and $2^{-\ell/2}\, Y^\ell$ respectively.

The linear operation $s^I$ is related to a source
$s_0(\vec{x}^\prime)$ at the boundary of $AdS_5$ through the
following relation: $s^I(\vec{x}, y)=\int d^4\vec{x}^\prime\,
G_\ell(\vec{x}^\prime; \vec{x}, y)s^I_0(\vec{x}^\prime)$ where the
bulk-to-boundary propagator $G_\ell$ is given by \be
G_\ell(\vec{x}^\prime; \vec{x}, y)=g_\ell \left(y\over
y^2+|\vec{x}-\vec{x}^\prime|^2\right)^\ell, \hs{3ex}
g_\ell=\frac{\ell+1}{2^{2-\ell/2}N\sqrt{\ell}}\,. \ee In the OPE
limit $R_{12}\ll L_{12}$, one has \be G_\ell\simeq
g_\ell\,\frac{z^\ell}{L_{12}^{2\ell}}, \ee and also \be
h_{\m\n}\simeq(-2\ell
g_{\m\n}s_I+\frac{4\ell}{z^2}\delta^z_\mu\delta^z_\nu s_I)\cY^I.\ee
The three-point function we want to compute is proportional to
$\frac{\d I_{D3}}{\d s_0(\vec{x}^\prime=0)}$.

We begin with the DBI  action of the D3-brane
\be
I_{DBI}=-T_{D3}\int\sqrt{\gamma}\,.
\ee
Its variation due to the above fluctuation of
the background fields is given by
\bea \d I_{DBI}&=&-\frac12 T_{D3}\int
d^4\sigma \sqrt{ \gamma}\,\gamma^{ab}(h_{\mu\nu}\p_aX^\m
\p_bX^\n+h_{\a\b}\p_aX^\a \p_bX^\b)\nn\\
&=&-\frac12 T_{D3}\int d^4\sigma \sqrt{ \g}\left(-2\ell
s_I\cY^I+\frac{4\ell s_I\cY^I}{\kappa^2z^2}\left(\frac{\p z}{\p
\tilde\tau}\right)^2+6\ell s_I\cY^I\right)\nn \\
&=&-T_{D3}\int d^4\sigma \sqrt{\g}\, 2\,\ell\,
s_I\,\cY^I\left(2-\frac1{\cosh^2\kappa\tilde\tau}\right),\eea where
$\tilde\tau=i\tau$, and we  know that $\tilde\tau$ is real from
subsection~\ref{2p brane}.

Following the prescription just reviewed, replacing $s_I$ by the
bulk-to-boundary propagator, \be G_\ell\simeq
g_\ell\frac{z^\ell}{L_{12}^{2\ell}}=\frac{g_\ell R_{12}^\ell}{2^\ell
L_{12}^{2\ell}\cosh^\ell \kappa\tilde\tau}, \ee one finds that the
contribution of the DBI part of the action to the structure constant
takes the form
\be c_{DBI}=\frac{g_\ell \,\ell
}{2^{\ell-1}} \int d\tilde\tau\, \kappa\,
\frac{\cY^I}{\cosh^{\ell}\kappa\tilde\tau}\left(-2+\frac1{\cosh^2\kappa\tilde\tau}\right).
\ee

The variation of the Chern-Simons term is given by \be\d
I_{WZ}=\frac{N}{2\pi^2}\int
4\sin^3\theta\cos\theta\sin\alpha\cos\alpha  s^I(-\partial_\theta
\cY^I)\dot{\phi}\,d\tau\wedge d\phi_1\w d\alpha \w d\phi_2\,,\ee and
the contribution from this part to the structure constant is
proportional to
 \be c_{WZ}=
 g_\ell \, N\over 2^{\ell+1}\pi^2 } \,
\int
\frac{4\sin^3\theta\cos\theta\sin\alpha\cos\alpha(-\partial_\theta
\cY^I)\dot{\phi}}{\cosh^{\ell}\kappa\tilde\tau}\, d\tilde\tau\wedge
d\phi_1\w d\alpha \w d\phi_2. \ee

Now we use the above formula to show that the contribution from the
$Z=0$ ($Y=0$) brane part to the structure constants $c^Z_{D3}$
($c^Y_{D3}$) is vanishing. We first consider the case of $c^Z_{D3}$ where
the corresponding spherical harmonic function is
$2^{-\ell/2}\, Z^\ell=2^{-\ell/2}\cos^\ell\th e^{i\ell\phi}$. This  is identically
zero on the $Z=0$ brane which is at $\th=\pi/2$. So the contribution
from the DBI part
is vanishing. As for the Chern-Simons part, we
notice that $\cos\theta\,\p_\theta (Z^\ell)$ vanishes on the brane for $\ell \ge 1$, and
this leads to the vanishing of the contributions since $\ell \ge 2$ for the chiral primary operators.

The $Y=0$ brane contribution $c^Y_{D3}$ can be related to the the
$Z=0$ brane contribution with the 1/2 BPS operator ${\cal
O}_{Y^\ell}=c_\ell {\rm tr}(Y^\ell)$ by an appropriate $SO(6)$
rotation. The corresponding spherical harmonic function is given by
$2^{-\ell/2} Y^\ell=2^{-\ell/2}\sin^\ell \th \sin^\ell \alpha
e^{i\ell\phi_2}$. The contribution from the DBI part is proportional
to the integral \be c_{DBI} \ \propto\  \int_0^{2\pi}\!\! \! d\phi_2
\,\, e^{i\ell\phi_2}\, \ee which is zero for $\ell \ne 0$. As for
the contribution from the Chern-Simons part, we have $\p_\theta
(Y^\ell)=0$ on the $Z=0$ brane and the corresponding contribution
again vanishes. This shows that $c^Y_{D3}=c^Z_{D3}=0$.

\subsection{Open string contributions}

In this subsection we compute the open string contribution to
the structure constants $c^Y_{123}$ and $c^Z_{123}$. Since we are
dealing with the three-point functions of one light chiral primary
operator and two heavy operators represented by semiclassical string
trajectory, the method developed in
Ref.~\cite{Zarembo:2010rr,Costa:2010rz} is appropriate. The
derivation of the corresponding formula is originally for the closed
string state. The derivation for our open string states is then
essentially the same as the closed string case in
Ref.~\cite{Zarembo:2010rr} but simply replacing the  closed string
trajectory by
  the open string  one. One finds that
\be
c_{string}= {2^\ell (\ell+1)\sqrt{\ell\lambda}\over 8\pi N R^\ell_{12}} \int d^2\sigma{\cal Y}^I z^\ell
\left[
{(\partial \vec{x})^2 -(\partial z)^2\over z^2}- h_{\alpha\beta}\partial X^\alpha \partial X^\beta
\right]
\ee
where one evaluates the above integral with the on-shell open string trajectory in AdS$_5\times S^5$ and
$\ell$ is the dimension of the light chiral primary operator  as before.
The rest is basically straightforward: we use the semiclassical  solution of the open string state
constructed in the previous section to evaluate
the contribution to the structure constant.

Let us now start with the open string contribution to the structure constant $c^Y_{123}$.
Using the open string trajectory in (\ref{ysolution}), one has
\bea
&& {1\over z^2}\Big[
(\partial \vec{x})^2 -(\partial z)^2
\Big]= {2\kappa^2\over \cosh^2 \kappa \tau} -\kappa^2\nn\\
&& h_{\alpha\beta}\,\partial X^\alpha \partial X^\beta =-\kappa^2
\eea
with ${\cal Y}^I=2^{-\ell/2}\cos^\ell\theta e^{i\ell\phi} =2^{-\ell/2} e^{-\ell\kappa\tau}$.
A straightforward computation then leads to the expression,
\be
c^Y_{string}={J\sqrt{\ell}\,(\ell+1)\over  N \, 2^{\ell+1}} \kappa \int^{\infty}_{-\infty}\! d\tau
{e^{-\ell \kappa \tau}\over \cosh^{\ell+2}\kappa\tau} ={1\over N} J\sqrt{\ell}\,.
\ee
Since the $Y=0$ brane contribution is vanishing as computed in the previous section,
our final semiclassical result is given by
\be
c^Y_{123}= c^Y_{D3}+ c^Y_{string}= {1\over N} J\sqrt{\ell}\,,
\ee
which agrees with the field theory result in the large $J$ limit.

Let us now turn to the case of the $Z=0$ open string contribution
to $c^Z_{123}$: By a straightforward evaluation using the solution (\ref{zsolution}), one has
\bea
&& {1\over z^2}\Big[
(\partial \vec{x})^2 -(\partial z)^2
\Big]= {2\kappa^2\over \cosh^2 \kappa \tau} -\kappa^2\nn\\
&& h_{\alpha\beta}\,\partial X^\alpha \partial X^\beta =\kappa^2 -2\kappa^2\, {\rm sn}^2(\nu\sigma, k^2)
\eea
with ${\cal Y}^I=2^{-\ell/2}\, k^\ell\, {\rm sn}^\ell(\nu\sigma, k^2)  e^{-\ell\kappa\tau/k}$.
This leads to  the expression
\bea
&&c^Z_{string}={J(\ell+1)\sqrt{\ell}\over  N \, 2^{\ell+2}} {k^\ell\over K(k^2)\!-\!E(k^2)} \int^{\infty}_{-\infty} dx
{e^{-\ell x/k}\over \cosh^{\ell}x} \nn\\
&&\ \ \ \ \ \ \ \ \  \times \int_0^{2K(k^2)} ds \,\,{\rm sn}^\ell
(s,k^2)\Big( {\rm sn}^2 (s,k^2)-\tanh^2 x \Big) \,, \eea for the
$Z=0$ open string trajectory. Since $k < 1$, the $x$ integral
diverges which implies a divergent $c^Z_{string}=\infty$.
Therefore we have a trouble in this case since our semiclassical
value of $c^Z_{123}$ diverges whereas the field theoretic one should
be finite for a given $\lambda$. We shall comment on this issue in the last section.

\section{Discussions}

In this paper, we studied the holographic computation of two- and
three-point functions of giant graviton with open strings. We
discussed the string configurations corresponding to the ground
states of $Z=0$ and $Y=0$ open spin chain, and the spinning string
in AdS$_5$ corresponding to the derivative
 type impurities in $Z=0$
open spin chain as well. For the two-point functions, we found both
the D3-brane configuration  and the open string configurations
connecting two insertions of the composite operators at the
boundary. There is only one subtle point: We had to use the Routhian
rather than energy directly in the computation. For the three-point
functions, we consider two giant gravitons with open strings coupled
to a half-BPS chiral primary operator. In the field theory side, we
computed them in the free field limit. In the limit of OPE and with
the configurations for the computation of two-point functions, we
computed the three-point functions holographically. In the
holographic computation, we carefully took the D-brane contribution
into account, which turned out to be vanishing. For the point-like
string dual to the ground state of $Y=0$ open spin chain, we found
perfect agreement. The reason of this agreement may be the
nonrenormalization of this three-point correlation function
\cite{Basu:2004nt} which we mentioned in the field theory
computations. But for the string dual to the $Z=0$ open spin chain,
we obtained a divergent structure constant from holographic
computation. This divergence may be canceled by the $1/N$
corrections to the contributions of D-brane because of the attached
open strings.
These $1/N$ contributions also include quantum corrections of D branes
and there is currently no available method to compute these quantum
corrections for generic, non-BPS states of D branes.
We also wish that an improved understanding of the prescription for
three-point functions will shed light on this subtle issue.

The computation in this paper is based on the picture that both the D-brane and open string contributes to the computation. This is reasonable as the dual operators include both the part corresponding to the giant graviton and the part corresponding to the open spin chain. However, this picture should be changed for other kinds of open spin chain\cite{Chen:2004mu,DeWolfe:2004zt}. For example, in the case studied in \cite{Chen:2004mu}, there are two kinds of integrable spin chain, closed one and open one, which decouple to the leading order. In other words, the presence of D7-brane just provides boundary conditions for the open semiclassical string. To compute the correlation functions of the open string state, the contribution from D-brane could be neglected safely.

In our work, we discussed the giant gravitons in $S^5$. It would be
nice to consider the case with the giant graviton in AdS$_5$.
It is also interesting to compute holographically the correlation
functions of these giant gravitons and non-local operators, such as
Wilson loops.

\section*{Acknowledgement}
DB would like to thank Romuald Janik for helpful discussions. DB was
supported in part by NRF SRC-CQUeST-2005-0049409 and  NRF Mid-career Researcher Program 2011-0013228.
BC was in
part supported by NSFC Grant No. 10775002, 10975005. JW would like
to thank Center for High Energy Physics, Peking University and Kavli
Institute for Theoretical Physics China (KITPC) for hospitality at
various stages of this work.

redpajama_set_name: RedPajamaArXiv


Pumpkins are everywhere this time of year. They decorate doorsteps and kitchen tables. They're for sale outside supermarkets. They overflow from farm stands on the side of the road, and at Northampton's Tuesday Market this week, along with gourds and roots, they were a seasonal centerpiece at most booths.
As shoppers browsed for groceries, a few farmers paused to share some of their seasonal favorites.
Along with pot roasts and pumpkin pie, Claire Brodhead of Red Fire Farm says her favorite dish to make around this time is stuffed honeynut squash.
Honeynut squash is sweeter and richer than butternut squash, Brodhead says, and a little smaller — perfect for one person without much cleanup if the squash is used as a bowl.
"They're a good size," said Brodhead, who most recently roasted a squash with beans, cilantro, onions, and Swish cheese on top.
At Conway-based Hart Farm's booth, Anna Meyer said that besides stuffed vegetables, she enjoys cooking with rutabaga, a root vegetable similar to a potato or turnip.
At the Old Friends Farm booth, Emily Coble said that she adds a teaspoon of fresh turmeric to rice and a teaspoon ginger to her smoothies for a little bit of the seasonal flavor.
Beside her at the cash register, Meghan Ecclesine suggested "roasting sweet potatoes with a little bit of maple syrup," or roasting butternut squash with a scoop of peanut butter.
"Carrots are good with maple syrup, too," Ecclesine said.
With his wife, Rachel Tevis-Robertson, who was helping to manage the booth Tuesday, he often makes a pot roast with a meat and potatoes, onions, garlic, and various seasonal vegetables like cauliflower, artichokes, brussels sprouts, or squash. Because he piles the roast on top of the vegetables, the meat cooks a little bit slower than it would without them, and it's tender on the bottom and slightly glazed on top.
"Me and my wife have slightly different tastes on how we like our meat cooked," Tevis-Robertson said.


text:
Bazil George Assan (1 August 1860 – 16 June 1918) was a Romanian engineer, explorer and economist. Belonging to a wealthy family in Bucharest, Assan was an important figure in the industrialization of the Kingdom of Romania. He studied engineering, commerce and economics, which impulsed him to discover the globe. In 1896, he became the first Romanian to travel to the Arctic, and between 1897 and 1898, he became the first Romanian to travel around the world. His travels were later presented to King Carol I of Romania. Assan died on 16 June 1918 in Montreux, Switzerland.

Biography

Assan was born on 1 August 1860 in Bucharest to a rich family. His father, George Assan, a Romanian from Moldavia, settled in the city and built the first windmill in Romanian history (known as the Assan Windmill) on the shore of the Colentina Lake in 1853. His mother was Alexandrina Assan, who took charge of the company of the family since her spouse's death in 1866 until her son Assan replaced her in 1884. Gheorghe G. Assan, his brother who was born in 1862, studied commerce in Antwerp, Belgium, and held various positions in commercial and other institutions.

Assan studied engineering in Liège, Belgium, and economics in Montreux, Switzerland. In 1884, he returned to Bucharest to work at his father's company with his brother. These two modernized the company and built two factories, one for soap and the other for paint and varnish. Assan was one of the persons who helped the Kingdom of Romania industrialize at more or less the same rate as other European countries. He also designed a canal between Cernavodă and Constanța (later the Danube–Black Sea Canal), publishing an article in 1899 about it.

Assan, together with the aviation and automobile enthusiast George Valentin Bibescu and the baron Barbu Bellu, was the first person to bring an automobile to Romania. The first two requested one from the brand FN Herstal. Since Bibescu did not have time to register it, Assan became in 1900 the first person to receive a plate, 1B. However, Bibescu wanted to be the first one, so in 1901, the Capital Prefecture gave him the plate 0B, which would technically make him the owner of the first plate in Romania.

Not much is known about Assan's family. He married a Romani woman and had three daughters. Assan named them according to where he was when they were born. Thus, the first was called Athenaida after Athens, Consuella after Spain and Pacifica after the Pacific Ocean.

In 1904, Assan built a  high grain silo, which became the tallest building in Bucharest at the time. He also built the first edible corn oil factory in the country. Between 1906 and 1914, he built a neoclassical house in Bucharest known as the Assan House. He died on 16 June 1918 in Montreux.

Expeditions
Assan, stimulated by his knowledge of economics, commerce and industry, was greatly interested in discovering the world. Therefore, in the summer of 1896, he embarked on the Norwegian ship Erling Jart and, along with other scientists of various nationalities, traveled to the Arctic. Assan and the scientists studied the geological structure and natural resources of the archipelago of Svalbard. Thus, he became the first Romanian to travel to the Arctic.

Afterwards, between 1897 and 1898, departing from Constanța, he passed through Constantinople, Alexandria, Ceylon, Singapore, Hong Kong, Shanghai, Nagasaki, Tokyo, Yokohama, San Francisco, New York and London, establishing several trade agreements and becoming the first Romanian to travel around the world. His expeditions were published in the Royal Society of Geography of Romania, of which he was a member. After returning, he presented his conference Călătorie în regiunile polare nordice ("Journey in the northern Polar regions") to King Carol I and Prince Ferdinand I, and days later to the Queen Elisabeth of Wied. In 1899, he presented to the public Călătoria împrejurul Pământului ("Journey around the Earth"), his second conference. Between December 1897 and the middle of 1898, Assan made another trip, this time with more economic and cultural intentions.

See also
 List of Romanian explorers

References

Assan
Assan
Assan
Assan
Assan
Assan
Assan


text:
Pseudopanurgus boylei is een vliesvleugelig insect uit de familie Andrenidae. De wetenschappelijke naam van de soort is voor het eerst geldig gepubliceerd in 1896 door Cockerell.

Andrenidae


Published January 17, 2018 By Anna T. This post may contain affiliate links. Regardless, I only recommend sites I've researched and/or used and trust.
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This site has a slightly broader range of employment opportunities than most of the other businesses on this list. Here, your job might consist of answering simple written questions, and providing phone consultations, or it could be much more involved. For instance, you could get hired in a semi-permanent capacity as part of a larger project for some big company. It just depends on the level of skill you bring to the table, and the availability of work in your field of expertise.
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This is another company that focuses on phone consultations, but they have a very different emphasis. Most calls to 6ya are apparently related to technical issues with various products, and they particularly need experts who are able to provide either telephone-based tech support, or mechanical expertise.
According to their site, most calls last about 6 minutes, and you'll be paid for every single call you handle. You do the work through a smartphone app which lets you know when a call request is available. You can take whichever calls you want, and set your own schedule. There are many categories available, including subjects such as audio equipment, plumbing, sewing, and computers.
Go here to learn more about 6ya.
This is another site where customers browse a list of experts and set up a scheduled phone call. Ranking high in their search results appears to be based on a rating system.
To get a higher ranking (which means more business), they suggest you take special care to fill out your profile with as many details as you can come up with. Customers want experts with the best qualifications, so modesty apparently doesn't pay off. It's also important to provide timely service to customers since their search algorithm uses response time as one of the factors when determining rank.
Go here to check out OnFrontiers.
This is a site where you can work as a tutor through a chat interface. Unlike the homework site mentioned above, Yup tutors aren't allowed to offer answers directly. Instead, they work with students to help them come up with answers on their own. The company's rules in this regard are pretty strict, and it's clear that they're focusing this site on parents who want to offer their children a way of getting homework help, but don't want to worry about the kids using the service to cheat.
Yup only offers tutoring in three subjects—math, chemistry, and physics. To apply, you'll need to have suitable qualifications in your subject of choice, and pass a difficult test.
Go here to visit Yup Tutoring and sign up.
This site has a slightly looser approach to everything. Here you can provide any sort of consultation you might want. They have the philosophy that everyone is an expert on some subject or other, so why not sell that expertise?
Are you good with Photoshop? Great. Sign up on Ether, and you can get paid to give people photo editing help.
Do you know a lot about business strategy? If so, you could set yourself up as a business strategy phone consultant.
Do you have a lot of relationship experience? Become a telephone counseler, helping people out with their romantic problems. Basically, on Ether the sky's the limit.
To start working, you set your rate, and perform a procedure to have calls forwarded to your real phone number. You'll only get calls from customers who've already pre-paid your per-minute rate, and you can set things up so you only get calls when you're interested in getting them.
Go here to learn more about Ether and sign up.
This is another site where you can make money tutoring people. To sign up you'll need considerable training in your chosen field, and you'll have to pass a difficult exam.
They offer tutoring in a much wider variety of subjects than Yup.com. You communicate with students via an advanced virtual classroom with a white board, and built in chat functions. You can also use voice chat during your sessions, which should make it much easier to get your point across.
You can check out our full Tutor.com review to learn more, or visit the Tutor.com site here.
This site focuses on providing advice. If you look carefully, you'll find quite a few different categories available, but there's clearly a focus on things like family and relationship coaching, as well as various forms of personal counseling.
You can set your own rate, and handle your consultations through either phone calls or the built in chat interface.
You can check out Ingenio here.
This is a site where you can become a teacher and offer classes on just about any topic imaginable. They apparently have a very simple format for creating classes, along with online tools that make the process easier.
One cool thing about this site is that there's not a whole lot of vetting. You don't have to bring any flashy qualifications to teach a class. If you've got enough practical experience to put together a decent course, you can offer it, and they'll let consumers decide whether or not what your offering is worth any money.
You can sign up to be a teacher on Skillshare here.
What makes this approach to money-making interesting is that there are so many variations on it, and quite a few of them are pretty easy to get started with. Some of these jobs can also be very, very lucrative. There are people in certain high demand industries making hundreds of dollars an hour for their consulting services, and in many cases all they're doing is sitting at home on the phone, answering questions that are easy for them because of their prior expertise.
If you know of any other sites where you can sell your expertise that weren't mentioned here, or if you have personal experience with any of the sites listed above, let us know in the comments section, and tell us how they work.
My e-book, How To Find REAL Work From Home Jobs, contains 150 legit companies, specific instructions on finding work at home jobs yourself, and also many, many resources you can trust for searching/following.
Thanks a lot for the invaluable info. I have tried Studypool and Yup. The studypool test is pretty simple. It only requires you to prove that you can write a good grammar essay. It has lots of jobs but bidding is very competitive so as a new tutor you have to set really low bid prices to win a bid. Due to this, I found their site somewhat overrated in terms of the ease by which you get jobs. Yup on the other hand has a very tough test, extremely tough. I got a 52% when I needed to get a 70% to pass. You can, however, retake the test after 3 months. Eagerly waiting for 3 months. LOL. Thanks anyway for this article. Really helpful. You might also consider adding Chegg and Course Hero. They're also nice tutoring sites.


text:
How Should Republicans Deal With Donald Trump?
July 20, 2015 • Donald Trump is rising in the polls and is getting all the attention when he delivers controversial speeches. A look at how the other candidates, and the Republican establishment, are responding.
Politics In The News: Iran Nuclear Talks
July 6, 2015 • Iran is clearly a matter of huge concern to the Obama administration. But are these nuclear talks making much of a splash on the presidential campaign trail or even foreign affairs issues in general?
Politics In The News: Obama's Successful Week In Washington
June 29, 2015 • With Republican help, President Obama received fast-track authority to negotiate trade deals. Then, the Supreme Court upheld subsidies to Obamacare and legalized same-sex marriage.
Politics In The News: Republican Presidential Field Gets Crowded
May 18, 2015 • GOP presidential hopefuls converged in Iowa for the Lincoln Day Dinner. More candidates are entering the presidential race, in part, because there is no obvious next-in-line for the GOP nomination.
Politics In The News: Patriot Act
May 11, 2015 • Lawmakers in Congress will be negotiating whether to extend the Patriot Act. That's the law written following the September 11th attacks, designed to thwart future acts of terrorism in the U.S.
More Republicans To Jump Into GOP Presidential Mix
May 4, 2015 • Pediatric neurosurgeon Ben Carson is set to formally announce his candidacy for president in Detroit on Monday. Also getting into the mix on Monday is former Hewlett-Packard CEO Cary Fiorina.
Same-Sex Supreme Court Case Raises Political Issues
April 27, 2015 • Public opinion on the subject of gay marriage has changed dramatically, and it will be interesting to see how the Republican presidential contenders come down on the issue.
Clinton Returns To New Hampshire, Site Of 2008 Comeback
April 20, 2015 • After talking to voters in Iowa last week, Hillary Clinton will be in New Hampshire, where she already has 19 paid staffers. She wants to tell voters the election is about them — it's not about her.
Politics In The News: Religious Freedom Act
March 30, 2015 • Indiana Gov. Mike Pence on Sunday said he supports an effort to "clarify the intent" of a new law that would allow businesses in the state to cite religious objections to refuse to serve gay people.
2016 Presidential Campaign Has Its First Official Candidate
March 23, 2015 • Republican Ted Cruz jumped into the race for the presidency, announcing his intentions in a tweet shortly after midnight Monday. Cruz is the first major candidate to outright declare he is running.
Politics Roundup: A Balancing Act For 2016
March 16, 2015 • The Republican field for 2016 is more crowded than the Democratic side. With Hillary Clinton unchallenged, the GOP can focus its attacks on a single candidate. That has some Democrats worried.
Netanyahu In D.C., DHS Funding Still Not Settled
March 2, 2015 • Israeli Prime Minister Benjamin Netanyahu addresses Congress on Tuesday. Funding for the Department of Homeland Security is being negotiated by the House and Senate, they disagree over what to do.
Homeland Security Warns Of Terrorist Threats Against U.S. Malls
February 23, 2015 • The secretary of Homeland Security has issued a warning after a video purportedly from the al-Qaida-linked militant group al-Shabab called for Muslims to attack shopping malls.
Politics In The News: Fiscal Budget For 2016
February 2, 2015 • It's not uncommon for presidential budgets of either party to be declared dead on arrival when they hit Capitol Hill. Is this exercise really about budgets or about politics?
A Closer Look At Saturday's Iowa Freedom Summit
January 26, 2015 • Some potential Republican candidates, such as Mitt Romney and Jeb Bush, stayed away from the event in Des Moines. Was that a smart move?

redpajama_set_name: RedPajamaCommonCrawl


A promising self-expanding stent, demonstrated in a study - presented at the annual American College of Cardiology conference - to reduce cardiac complications, is aiming for the US market.
A self-expanding stent, manufactured by the French firm Stentys, has been proven to reduce the incidence of cardiac complications versus standard stents in circulation currently. The company now plans to expand into many new territories around the world, including the $3 billion United States market, according to a Reuters report.
The American College of Cardiology presented positive findings on the Stentys self-expanding stent at the ACC's annual conference this weekend.
According to the study, which featured a universe of 1000 heart patients, the Stentys self-expanding mesh design produced negative cardiac events at a 9.3 percent rate compared to 11.1 percent from standard stents currently in use worldwide. Additional studies have found the self-expanding stent to yield a death rate of 2 percent versus 3.9 percent from traditional stents in market.
Citing a late 2011 analysis from Global Industry Analysts, Inc., the global stent market is projected to reach $9.8 billion by 2017.
Stentys hopes to receive approval from the U.S. Food and Drug Administration by 2015.


text:
Q: c# flatten nested object I'm having a nested class :
    class Item
    {
        public Int32 Id { get; set; }
        public Int32 Pid { get; set; }
        public String Name { get; set; }
        public IEnumerable<Item> Children { get; set; }

    }

Now I want to flatten this so I can get the name of all Items and their childeren.
Problem here is that I don't know how many levels deep this goes.
I had a look at :
How to flatten nested objects with linq expression
This is great if you know how many levels you have, which I don't.
So :
        var r = from b in items
                from c in b.Children
                from d in c.Children
                ...
                select new { b.Name, c = c.Name, d = d.Name ... };

does pretty much what I need, but I don't know how many levels deep I need to go, also if one item does not have a child it doesn't return anything.
I would need some recursive routine on this I guess, but I can't seem to find it.
I looked at IEnumerable but I don't really understand this yet :)
So any help would be very much appriciated.

A: You are right, you need recursion:
public IEnumerable<Item> GetAllChildren(Item item)
{
    return item.Children.Concat(item.Children.SelectMany(GetAllChildren));
}

To get all names you can project the result:
var allDescendantNames = GetAllChildren(item).Select(child => child.Name).ToList();

redpajama_set_name: RedPajamaStackExchange