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\begin{document}

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\begin{titlepage}

\versionDeformation of the cone  over $Y^{p,q}$

\hbox to \hsize{\hbox{\tt hep-th/0504155}\hss
	\hbox{\small{\tt MCTP-05-73}}}

\vspace{1.7cm}

\centerline{\bf \Large Towards Supergravity Duals of Chiral Symmetry
Breaking}

\vspace{.6cm}
\centerline{\bf \Large in Sasaki-Einstein Cascading Quiver Theories}
\vspace{2cm}

\centerline{ \large Benjamin  A. Burrington, James T. Liu}

\vspace{0.3cm}

\centerline{\large Manavendra
Mahato and Leopoldo A. Pando Zayas}

\vspace{1cm}

\centerline{\it Michigan Center for Theoretical Physics}
\centerline{\it Randall Laboratory of Physics, The University of Michigan}
\centerline{\it Ann Arbor, MI 48109--1040}

\vspace{1cm}

\begin{abstract}
We construct a first order deformation of the complex structure of the cone over
Sasaki-Einstein spaces $Y^{p,q}$ and check supersymmetry
explicitly. This space is a central element in the holographic dual of
chiral symmetry breaking for a large class of cascading quiver
theories. We discuss a solution describing a stack of $N$ D3 branes
and $M$ fractional D3 branes at the tip of the deformed spaces.
\end{abstract}

\end{titlepage}

\section{Introduction and summary}

One of the most promising elements of the AdS/CFT correspondence is
the possibility of providing an alternative method of studying aspects
of confining theories. Unfortunately, there are but a few smooth supergravity
backgrounds potentially dual to confining ${\cal N}=1$ SYM.

Recently, a new class of Sasaki-Einstein manifolds $Y^{p,q}$ has been constructed
\cite{se,GMSW}. Given a Sasaki-Einstein manifold $X^5$ one can
consider a stack of $N$ D3 branes at the tip of the cone over
$X^5$. Taking the Maldacena limit leads to a duality between string
theory on $AdS_5\times X^5$ and a superconformal gauge theory living
in the world volume of the D3 branes. In this context the infinite family of
spaces $Y^{p,q}$ was shown to be dual to superconformal quiver gauge
theories \cite{MS,sequiver}. Remarkably, the correspondence has
provided a better treatment of some of the gauge theory questions.
In particular, the irrational nature of some R-charges has
been elucidated in the field theory using gravity input \cite{MS,friends}.

Since $Y^{p,q}$ spaces are generalizations of $T^{1,1}$ one naturally
wonders about the possibility of further generalizing these spaces
in a way that parallels the program carried by Klebanov and
collaborators for the conifold \cite{igor,KS}. In fact, the first step
in this direction
has already
been taken in a recent collaboration including Klebanov \cite{HEK}. The pinnacle of
Klebanov's program is the Klebanov-Strassler (KS) background \cite{KS}, which is a
smooth supergravity solution constructed as a warped deformed
conifold.

In this paper we take a step in the construction of
generalizations of the KS background. Namely, we construct a first
order deformation of the complex structure of the cone over $Y^{p,q}$.
This deformation should lead, upon
warping, to a supergravity background describing the IR of the
recently constructed cascading solution \cite{HEK} on the cone over $Y^{p,q}$.

One important motivation for this work is that the existence of a
supergravity background allows for explicit computation of
some dynamical quantities, like the spectrum of various states
including mesons, quantum corrections to Regge trajectories and possibly Goldstone
bosons.

In section \ref{review} we review some of the key aspects of the
$Y^{p,q}$ spaces and their dual quiver gauge theories. Our aim is to
set up the notation and technical motivation for later sections. In
particular, we address various technical aspects as the supergravity
realization of chiral symmetry breaking and the algebra of vielbeins.
In section \ref{deformation}, after recalling the general theory of deformation
of CY spaces from the differential geometric point of view, we
present the first order deformation for the cone over $Y^{p,q}$. This
is the main technical result of the paper. Section \ref{deformation} also
contains the result of explicitly checking that the deformation is
supersymmetric. This guarantees that the deformation has $SU(3)$
holonomy and that indeed corresponds to a complex K\"ahler manifold. Section
\ref{kah} reviews the complex coordinates introduced for the cone over
$Y^{p,q}$ in \cite{MS} with the aim of classifying the deformation. We
show that it corresponds to a deformation of the complex
structure. Moreover, in this section we find the K\"ahler potential
for the cone over $Y^{p,q}$, knowing the K\"ahler potential opens a
new venue for understanding deformations of these CY spaces. Section
\ref{warp} discusses placing a stack of $N$ D3 and $M$ fractional D3 branes at the tip of the
deformed space. Since we only know the deformation of the space to
first order, our analysis is approximate. The main technical result in
this section is the explicit construction of the imaginary self dual
3-form in the deformed space. This implies that to first order there
is a supergravity background that should correspond to the chiral
symmetry broken phase of the cascading Sasaki-Einstein quivers
discussed in \cite{HEK}.

Let us end this section with some comments on open questions that we
were not able to answer in this work. The most glaring question is the
existence of a solution beyond first order.
Our paper follows a strictly differential geometric approach
to the deformation. Another unexplored venue is the algebraic
approach. Namely, the complex deformation for the conifold is fairly
simple from the algebraic standpoint. Given the
defining equation $\sum\limits_{a=1}^4 w_a^2=0$, we simply need to
replace it by $\sum\limits_{a=1}^4 w_a^2=\epsilon^2$. There are
various questions that the algebraic approach answers immediately. For
example, chiral symmetry breaking is nothing but the breaking of the
$U(1)$ symmetry that rotates the coordinates $w_a$ in the undeformed
space. In this paper we followed exclusively the differential
geometric approach but we hope to discuss the algebraic one in the
future. Finally, chiral symmetry breaking can also be explored along
the lines of \cite{chiral,chiral2}.

\section*{Note Added}
After the first version of this work appeared in the archive, a series of
papers addressing the complex deformation from the field theory side
have given arguments in favor of supersymmetry breaking
\cite{susybreaking}.
We also became aware of work by Altmann \cite{altmann}
who shows the existence of an obtruction for finding the complex
deformations beyond first order. It is worth mentioning that while
these works make a case for obstruction to a solution built around a
complex deformation of the corresponding cone over $Y^{p,q}$ it is
plausible  that the full solution  is supersymmetric albeit with
$SU(3)$-structure rather than $SU(3)$ holonomy as is the case for the
baryonic branch of the KS solutions as discussed in \cite{minasian}.

\section{Review of superconformal quiver theories and their dual
$AdS_5\times Y^{p,q}$ backgrounds}\label{review}

The gauge theory dual to IIB on $AdS_5\times Y^{p,q}$ has been the
subject of much recent investigation. Here we begin with a summary
of some of the key aspects of the quiver gauge theories with $Y^{p,q}$
duals, as explained in Ref.~\cite{sequiver}.

The quivers for $Y^{p,q}$ can be constructed starting with the quiver
of $Y^{p,p}$ which is naturally related to the quiver theory obtained
from $\mathbb C^3/\mathbb Z_{2p}$. The gauge group is $SU(N)^{2p}$ and the
superpotential is constructed out of cubic and quartic terms in the
four types of fields present: $U,V,Y$ and $Z$, where $U$ and $V$ are
doublets of $SU(2)$:
\begin{equation}
\epsilon_{\alpha\beta} U^\alpha_a V^\beta Y, \qquad \epsilon_{\alpha\beta} U^\alpha_b V^\beta Y,
\qquad \epsilon_{\alpha\beta}ZU_a^\alpha Y U_b^\beta.
\end{equation}
Greek indices $\alpha,\beta=1,2$ are in $SU(2)$, and Latin subindices $a,b$
refer to the gauge group where the corresponding arrow originates.
Equivalently, as explained in \cite{HEK}, the quiver theory for $Y^{p,q}$
can be constructed from two basic cells denoted by $\sigma$ and $\tau$.
Some concrete examples can be found in \cite{sequiver,HEK}.

One key ingredient is that the geometric realization of the
${\bf R}$-symmetry is given by a Killing vector field in the
Sasaki-Einstein geometry of the form
\begin{equation}
\eqlabel{u1inse}
K=\frac13 \frac{\partial}{\partial \psi},
\end{equation}
where the geometrical meaning of the coordinates $\psi$ and $\alpha$
will become apparent momentarily. Breaking the ${\bf R}$-symmetry is
our main goal, and it translates into constructing a supergravity
background without (\ref{u1inse}) as a Killing vector.

\subsection{The cone over $Y^{p,q}$}

Before turning on any deformations, we highlight the structure of the
cone over $Y^{p,q}$, emphasizing its similarities with the conifold,
which is the cone over $T^{1,1}$.  This connection motivates our later
Ansatz for the deformation.

Following \cite{GMSW,MS}, the Sasaki-Einstein metric on $Y^{p,q}$
can be written in the form
\begin{eqnarray}
ds^2&=&\frac{1-cy}{6}(d{\theta}^2+{\sin}^2 \theta d{\phi}^2)
+\frac{1}{w(y)v(y)}dy^2
+\frac{w(y)v(y)}{36}(d\beta-c\cos\theta d\phi )^2\nonumber\\
&&+\frac{1}{9}[d\psi +\cos\theta d \phi+y(d\beta-c \cos\theta d\phi )]^2\\
&=&\frac{1}{6}[(e^{\theta})^2+(e^{\phi})^2+(e^y)^2+(e^{\beta})^2]
+\frac{1}{9}(e^{\psi})^2,
\end{eqnarray}
where
\begin{equation}
w(y)=\frac{2(a-y^2)}{1-cy}, \qquad v(y)=\frac{a-3y^2+2cy^3}{a-y^2}.
\end{equation}
The natural one forms are given by \cite{HEK,SSP}
\begin{eqnarray}
e^{\theta}&=&\sqrt{1-cy}\;d\theta, \nonumber\\
e^{\phi}&=&-\sqrt{1-cy}\;\sin\theta d\phi, \nonumber\\
e^y&=&-\frac{1}{H(y)}dy, \nonumber\\
e^{\beta}&=& H(y)(d\beta-c \cos\theta d\phi), \nonumber\\
e^{\psi}&=& d\psi+\cos\theta d\phi+y(d\beta-c\cos\theta d\phi),
\end{eqnarray}
with $H(y)=\sqrt{w(y)v(y)/6}$.

Before proceeding, it would be convenient to develop the algebra of
exterior derivatives. For this purpose, let
$L(y,\theta)=\frac{\cot\theta}{\sqrt{1-cy}}$
and $K(y)=\frac{cH(y)}{2(1-cy)}$. Then $\frac{dH(y)}{dy}=K(y)-\frac{y}{H(y)}$.
The exterior derivatives of the above forms then satisfy
\begin{eqnarray}
de^{\theta}&=& K(y)e^{y}\wedge e^{\theta}, \nonumber\\
de^{\phi}&=&K(y)e^y\wedge e^{\phi}+L(y,\theta)e^{\theta}\wedge
e^{\phi}, \nonumber\\
de^{y}&=&0, \nonumber\\
de^{\beta}&=&\left (\frac{y}{H(y)}-K(y)\right )
e^y\wedge e^{\beta}-2K(y)e^{\theta}\wedge e^{\phi}, \nonumber\\
\label{de}
de^{\psi}&=&e^{\theta}\wedge e^{\phi}-e^y\wedge e^{\beta}.
\end{eqnarray}

At various stages in the construction, we will refer to the model of
the deformed conifold. In particular, it is worth noting that the
above definitions reduce to those in \cite{KS} describing $T^{1,1}$ in
the limit \cite{GMSW}
\begin{equation}
\eqlabel{limit}
c\rightarrow 0,\qquad
y\rightarrow \cos\theta _2,\qquad
a \rightarrow 3 \quad\hbox{and}\quad
\beta \rightarrow \phi_2.
\end{equation}
Following the discussion of \cite{KS} (see also \cite{mt}), one can further define
the rotated and shifted vielbeins $e^1,e^2,\ldots,e^5$ and
$g^1,g^2,\ldots,g^5$ as
\begin{eqnarray}
e^1=-e^{\beta},&&
e^3=-\cos\psi e^{\phi}-\sin\psi e^{\theta},\qquad e^5=e^{\psi},\nonumber\\
e^2=e^{y},\hphantom{-}&&
e^4=-\sin\psi e^{\phi}+\cos\psi e^{\theta},
\label{eq:ei}
\end{eqnarray}
and
\begin{eqnarray}
g^1=\frac{1}{\sqrt{2}}(e^1-e^3),&&
g^3=\frac{1}{\sqrt{2}}(e^1+e^3),\qquad g^5=e^{5}.\nonumber\\
g^2=\frac{1}{\sqrt{2}}(e^2-e^4),&&
g^4=\frac{1}{\sqrt{2}}(e^2+e^4),
\label{gs}
\end{eqnarray}

Note that, if desired, one could instead have defined
$\tilde{e}^1,\tilde{e}^2,\ldots,\tilde{e}^5$ as
\begin{eqnarray}
\tilde{e}^1=e^{\phi},&&
\tilde{e}^3=\cos\psi e^{\beta}-\sin\psi e^y,\qquad \tilde{e}^5=e^{\psi}.
\nonumber\\
\tilde{e}^2=e^{\theta},&&
\tilde{e}^4=\sin\psi e^{\beta}+\cos\psi e^y,
\label{eq:other}
\end{eqnarray}
The corresponding $\tilde{g}^1,\tilde{g}^2,...,\tilde{g}^5$ would then be
defined in a similar fashion as in (\ref{gs}) with $e^i$ replaced by
$\tilde{e}^i$ for $i=1,2,\ldots,5$.  One should note, however, that
the $\tilde{e}^i$ are physically indistinct from the ${e}^i$ in
the $T^{1,1}$ limit (\ref{limit}).  They are distinct when $c\neq0$, and so
we may expect that they could play a r\^ole in deforming the cone over
$Y^{p,q}$.  We will nevertheless show that the first order deformation
given in terms of the $\tilde e^i$ is no different than that given in
terms of the $e^i$.

Although the angular coordinate $\psi$ appears explicitly in the the
above one forms, the metric remains $\psi$ independent.  As we will see
in the next section, the perturbed solution will break this symmetry
direction in an analogous way to the solution of \cite{KS}.

\section{The first order complex deformation}\label{deformation}

We now briefly review the deformation theory of CY spaces following
the presentation of \cite{moduli}. Let the parameter space
of Calabi-Yau manifolds be the parameter space of Ricci-flat
K\"ahler metrics, and let $g_{mn}$ and $g_{mn}+ \delta g_{mn}$
satisfy
\begin{equation}
R_{mn}(g)=0, \qquad R_{mn}(g+\delta g)=0.
\end{equation}
Then, with the coordinate condition $\nabla^n\delta g_{mn}=0$, one
obtains that $\delta g_{mn}$ satisfies the Lichnerowicz equation
\begin{equation}
\eqlabel{lich}
\nabla^k\nabla_k \delta g_{mn}+ 2R_m{}^p{}_n{}^q \delta g_{pq}=0.
\end{equation}

The connection between the geometro-differential approach and the
algebraic approach to deforming a CY space arises due to an
isomorphism between the solutions of (\ref{lich}) and harmonic forms on
CY. Namely, a solution with mixed indices is associated with a
$(1,1)$-form
\begin{equation}
i\delta g_{\mu\bar{\nu}}dx^\mu \wedge dx^{\bar{\nu}},
\end{equation}
which is harmonic if and only if the variation of the metric satisfies
the Lichnerowicz equation, (\ref{lich}). Similarly, for a variation of
pure type, one can associate a $(2,1)$-form using the holomorphic 3-form
$\Omega$:
\begin{equation}
\Omega_{\kappa \lambda}{}^{\bar{\nu}} \delta g_{\bar{\mu}\bar{\nu}}
dx^{\kappa}\wedge dx^{\lambda}\wedge dx^{\bar{\mu}}.
\end{equation}
This form is harmonic if and only if $\delta g_{\bar{\mu}\bar{\nu}}$
satisfies (\ref{lich}).

With these isomorphisms in place we can classify a deformation as either
K\"ahler or complex.  In particular, variations of pure type (there are
$b_{2,1}$ of them) correspond to variations of the complex structure. Note
that $g+\delta g$ is a K\"ahler metric on a manifold close to the original
one.  There must therefore exist a coordinate system in which the pure parts
of the metric vanish. Under a change of coordinates $x^m \to x^m +
f^m(x)$, the metric variation transforms according to the familiar
\begin{equation}
\delta g_{mn}\to \delta g_{mn} - \frac{\partial f^r}{\partial x^m }
g_{rn} - \frac{\partial f^r}{\partial x^n } g_{mr}.
\end{equation}
If $f^\mu$ is holomorphic then $\delta g_{\mu\nu}$ is invariant. Thus,
the pure part of the variation could be removed by a transformation of
coordinates but it cannot be removed by a holomorphic coordinate
transformation. Thus the pure part of the metric variation corresponds
precisely to changes of the complex structure.

\subsection{First order perturbation}

We are now in a position to construct the first order deformation of the
cone over $Y^{p,q}$.  Using the one-forms of section \ref{review}, the
undeformed metric is given by
\begin{equation}
ds^2 = dr^2+r^2\left(\ft16\left[(e^1)^2+(e^2)^2+(e^3)^2+(e^4)^2\right]
+\ft19(e^5)^2\right).
\label{eq:cypq}
\end{equation}
This has the same form as the conifold; however important nontrivial
dependences on the coordinate $y$ have been introduced through the
definitions of the vielbeins $e^i$.

As mentioned above, most of our intuition for constructing a deformation
of (\ref{eq:cypq}) comes from the analogous case of the deformed conifold.
Here we recall that the deformed conifold metric can be written as
follows \cite{conifold}:
\begin{eqnarray}
ds_{DC}^2= \epsilon^{\fft43}\frac{K(\tau)}{3}\bigg(\frac{1}{3K^3}
\left[4\,d\tau^2+(g^5)^2\right] \!\!&+&\!\!
\cosh(\tau)^2\left[(g^3)^2+(g^4)^2\right]\nonumber \\
\!\!&+&\!\! \sinh(\tau)^2 \left[(g^1)^2+(g^2)^2\right]\bigg),
\end{eqnarray}
where
\begin{equation}
K(\tau) = \frac{ 2^{2/3}(\sinh(4\tau)-4\tau)^{1/3}}{2 \sinh(2\tau)}.
\end{equation}

While the above is given in terms of a radial coordinate $\tau$,
a more natural coordinate for our analysis is given by $r$ where
$r^3=\epsilon^2 \cosh(2 \tau)$.  Furthermore, to understand the first order
deformation, is it also convenient to expand the metric for large $r$;
more precisely, we expand in the dimensionless quantity $\epsilon^2/r^3$.
We find, up to second order in the deformation, that
\begin{eqnarray}
ds_{DC(2)}^2 &=&
dr^2+ r^2\left(\fft{1}{6}\left[(g^1)^2+(g^2)^2+(g^3)^2+(g^4)^2\right]
+\fft{1}{9} (g^5)^2\right) \nonumber \\
&&+\fft16r^2\left[-(g^1)^2-(g^2)^2+(g^3)^2+(g^4)^2\right]
\frac{\epsilon^2}{r^3}\nonumber \\
&&+ \frac{1}{3}\bigg(-\frac{1}{6} r^2\bigl[(g^3)^2+(g^4)^2+(g^1)^2+(g^2)^2
\bigr] \nonumber \\
&& \qquad\qquad+2\left[dr^2+\frac{1}{9}(g^5)^2r^2\right]\bigg)
\left(\ln\left(\frac{2r^3}{\epsilon^2}\right)-1\right)
\frac{\epsilon^4}{r^6}+dr^2\frac{\epsilon^4}{r^6}+\cdots. \nonumber\\
\label{eq:dc2}
\end{eqnarray}
The zeroth order term is of course the undeformed conifold metric itself.
The first order (in $\epsilon^2$) term should be a solution to the
linearized Einstein's equations, and so should be a zero mode of the
Lichnerowicz operator (\ref{lich}).  We make the simple observation that
the first order contribution is transverse and traceless.  The second
order perturbation is ``sourced" by the first order equations, and we
have verified explicitly that the above system indeed satisfies the
Ricci-flatness condition to second order.

We now turn our attention back to the cone over $Y^{p,q}$.  We note that
the combination $-(g^1)^2-(g^2)^2+(g^3)^2+(g^4)^2$ showing up at first
order in the above expansion is the same as $2(e^1e^3+e^2e^4)$.  This
therefore suggests an Ansatz for a first order perturbation for the cone
over $Y^{p,q}$ of the form
\begin{equation}
ds_{(1)}^2= h_{m n} dx^m dx^n = f(y)r^2\frac{\left(e^1e^3+e^2e^4\right)}{3}
\frac{\epsilon^2}{r^3},
\label{eq:firsto}
\end{equation}
where we have included a function of $y$ because of the non trivial
dependence of the $Y^{p,q}$ metric on this coordinate.  Inserting this
Ansatz into the first order deformation equations, we find that $f(y)$
must satisfy a {\it first order ODE} for it to be a zero mode (even though the
Lichnerowicz operator is {\it second} order).  The solution is simply
\begin{equation}
f(y)=\frac{1}{(1-cy)^2}.
\end{equation}
We have dropped an arbitrary multiplicative factor (it only becomes
important in the case of a fully non-perturbative solution).  Note
that the first order perturbation is again transverse and traceless.

Despite some effort, we have unfortunately not been able to satisfy the second
order deformation equations using the first order contributions as a source.
It is likely that, at second order, the deformation of the metric will
non-trivially depend on both $r$ and $y$ coordinates in an inherently
non-separable manner.  Although we have no direct proof, this belief
is supported by various unsuccessful attempts at separating the functional
dependence on $r$ and $y$ in the spirit of the deformed conifold metric
(\ref{eq:dc2}).

When constructing the one-forms $e^i$, we have introduced angular
$\psi$ dependence by ``rotating" $e^{\phi}$ and $e^{\theta}$ together
using $\psi$.  As mentioned above, one could instead have mixed
$e^{y}$ and $e^{\beta}$.  This, however, does not alter the first order
perturbation (up to irrelevant minus signs).  Hence the other mixing
(\ref{eq:other}), while it is perhaps distinct non-perturbatively, is
indistinct to lowest order in $\epsilon^2$. This might signal another
possible type of deformation, but we will not explore it here.

As mentioned above, the rotation between $e^{\phi}$ and $e^{\theta}$
introduces dependence on the coordinate $\psi$.  This dependence on
$\psi$ disappears in the zeroth order metric because $e^{\phi}$ and
$e^{\theta}$ always appear in the $SU(2)$ invariant combination.  However
this is no longer true for the first order perturbation.  This dependence
on $\psi$ is dual to chiral symmetry breaking in the gauge theory, and
will be discussed more explicitly in section \ref{pured3} below.

\subsection{Supersymmetry of the perturbed solution}\label{secSUSY}

Although the first order deformation was simply obtained by demanding
Ricci-flatness to order $\epsilon^2$, it turns out that it is in fact a
complex deformation, at least to this order.  To show this explicitly,
we turn to the Killing spinor equation.  In the absence of fluxes, the
supersymmetry condition takes the form $\nabla \lambda=0$, with the
resulting $\lambda$ a parallel spinor.

However, instead of working with the Killing spinor equation directly,
we focus on the integrability condition
\begin{equation}
\eqlabel{riemProj}
e_a^\mu e_b^\nu\left[\nabla_\mu, \nabla_\nu \right]\lambda
=\ft14 R_{abcd} \Gamma^{cd} \lambda=0.
\end{equation}
Before obtaining the Riemann tensor in the tangent basis, we first make
a convenient choice of vielbein.  To do this, we note that we can
absorb the first order term (\ref{eq:firsto}) into a shift of $e^1$
and $e^2$ according to
\begin{equation}
\eqlabel{1order}
ds^2 = dr^2+r^2\left(\ft16 \left[(e^1+\delta e^3)^2+(e^2+\delta e^4)^2
+ (e^3)^2 + (e^4)^2\right]+\ft19 (e^5)^2\right),
\end{equation}
where we have defined the quantity
\begin{equation}
\eqlabel{delta}
\delta = \frac{\epsilon^2}{r^3(1-cy)^2}.
\end{equation}
This introduces a shift to second order in $\epsilon^2$, which however
is unimportant as we will always work only to first order in $\epsilon^2$.
We now make a natural choice of shifted vielbeins
\begin{eqnarray}
\hat{e}^6_{(s)}= dr,~&&~
\hat{e}^1_{(s)}=\frac{r}{\sqrt{6}}\left(e^1+\delta e^3\right),
\qquad \hat{e}^2_{(s)}=\frac{r}{\sqrt{6}}\left(e^2+\delta e^4\right),
\nonumber\\
\hat{e}^3_{(s)}=\frac{r}{\sqrt{6}}e^3,~&&~
\hat{e}^4_{(s)}=\frac{r}{\sqrt{6}}e^4,
\kern5.8em \hat{e}^5_{(s)}=\frac{r}{3}e^5.
\label{eq:ncsv}
\end{eqnarray}

Using this shifted vielbein basis, we find that the integrability
condition (\ref{riemProj}) is satisfied for spinors $\lambda$ satisfying
the simultaneous projections
\begin{eqnarray}
\left(1+\Gamma^{1256}\right)\lambda &=& 0, \nonumber \\
\left(1-\Gamma^{1234}\right)\lambda &=& 0.
\end{eqnarray}
Writing out the $SO(6)$ generators in the spinor representation as
$T^1=\fft{i}2\Gamma^{12}$, $T^2=\fft{i}2\Gamma^{34}$ and
$T^3=\fft{i}2\Gamma^{56}$, we see that parallel spinors have $SO(6)$
weights $(\fft12,-\fft12,\fft12)$ or $(-\fft12,\fft12,-\fft12)$.
Changing to an $SU(4)$ basis with Cartan generators
\begin{equation}
H^1=\ft12(T^1+T^2),\qquad H^2=\ft1{4\sqrt{3}}(-T^1+T^2+2T^3),\qquad
H^3=\ft1{2\sqrt{6}}(T^1-T^2+T^3),
\end{equation}
we see that the parallel spinors are singlets of the $SU(3)$ corresponding
to $H^1$ and $H^2$.  This is just the standard Calabi-Yau decomposition
$\mathbf4\to\mathbf3+\mathbf1$ under $SU(4)\supset SU(3)$.

More explicitly, we have verified that the integrability condition
(\ref{riemProj}) is satisfied at both zeroth and first order in $\epsilon^2$
for the above projections.  For example%
\footnote{All components have been computed in Maple.  Details of the
calculations are available upon request to the authors.}
\begin{eqnarray}
R_{53 ab}\Gamma^{ab} &=& \frac{6\epsilon^2}{r^5(1-cy)^2}
\Gamma^{51}\left(1+\Gamma^{1256}\right),\nonumber\\
R_{24 ab}\Gamma^{ab} &=& \frac{2(ac^2-1)}{r^2(1-cy)^3}\Gamma^{31}
\left(1-\Gamma^{1234}\right)
+\frac{3\sqrt{6}H(y)\epsilon^2 c}{r^5(1-cy)^3}\Gamma^{51}
\left(1+\Gamma^{1256}\right),
\end{eqnarray}
which are clearly zero when applied to $\lambda$.

\section{Complex coordinates and K\"ahler potential}\label{kah}

Having given the first order deformation and shown that it is
supersymmetric, we now proceed to its classification. As a first step,
we review the complex coordinates introduced in \cite{MS} with some minor
adjustments to fit the present conventions%
\footnote{Note that we have taken $\phi \to -\phi$ with respect
to \cite{MS}.}.
The main new result of this section is an expression for the K\"ahler
potential of the cone over $Y^{p,q}$.

We recall that Ref.~\cite{MS} obtained the complex coordinates for
the cone over Sasaki-Einstein space. Let
\begin{eqnarray}
{\eta}^1&=&\frac{1}{\sin\theta}d\theta-id\phi, \nonumber\\
\tilde{\eta}^2&=&-\frac{dy}{H(y)^2}-i(d\beta-c \cos\theta d\phi), \nonumber\\
\tilde{\eta}^3&=&\frac{3dr}{r}+i\big[d\psi+\cos\theta d\phi+y(d\beta-c
\cos\theta d\phi)\big].
\end{eqnarray}
In terms of these coordinates, one can write the metric as
\begin{equation}
ds^2=r^2\frac{(1-cy)}{6}{\sin}^2\theta {\eta}^1\overline{\eta^1}
+r^2\frac{H(y)^2}{6}{\tilde{\eta}^2}\overline{\tilde{\eta}^2}
+\frac{r^2}{9}{\tilde{\eta}^3}\overline{\tilde{\eta}^3}.
\end{equation}
Unfortunately, ${\tilde{\eta}^2}$ and ${\tilde{\eta}^3}$ are not
integrable. However, integrable one-forms can be obtained by taking linear
combinations of them:
\begin{equation}
{\eta}^2={\tilde{\eta}^2}+c \cos \theta {\eta}^1, \qquad
{\eta}^3={\tilde{\eta}^3}+\cos\theta {\eta}^1+y{\tilde{\eta}^2}.
\end{equation}
In this case, ${\eta}^i=dz_i/z_i$ for $i=1,2,3$, where
\begin{eqnarray}
\label{z1}
z_1&=&\tan\frac{\theta}{2}e^{-i\phi},\\
\label{z2}
z_2&=&\frac{(\sin\theta)^{2c}}{f_1(y)}e^{-i\beta},\\
\label{z3}
z_3&=&r^3\frac{\sin\theta}{f_2(y)}e^{i\psi},
\end{eqnarray}
with
\begin{equation}
f_1(y)=\exp\left(\int\frac{1}{H(y)^2}dy\right),\qquad
f_2(y)=\exp\left(\int\frac{y}{H(y)^2}dy\right).
\end{equation}
As a check, we may take the limit to the conifold (\ref{limit}), in
which case the $z_i$'s reduce to
\begin{equation}
z_1\rightarrow \tan\frac{{\theta}_1}{2}e^{-i{\phi}},\qquad
z_2\rightarrow \tan\frac{{\theta}_2}{2}e^{-i{\beta}},\qquad
z_3\rightarrow r^3\sin{\theta}_1\sin{\theta}_2e^{i\psi},
\end{equation}
along with $f_1=\cot(\theta_2/2), \quad f_2=\csc\theta_2$.

Returning to the Sasaki-Einstein case, note that
(\ref{z1}) can be written as
\begin{equation}
\label{t}
\sin\theta=\frac{2\sqrt{z_1\bar{z}_1}}{1+z_1\bar{z}_1},
\end{equation}
where bars denote complex conjugation.  Therefore
\begin{equation}
\label{dt}
\frac{\partial\sin\theta}{\partial
z_1}=\frac{1}{2z_1}\sin\theta\cos\theta,\qquad
\frac{\partial\sin\theta}{\partial z_2}=0,\qquad
\frac{\partial\sin\theta}{\partial z_3}=0.
\end{equation}
{}From equation (\ref{z2}), an explicit expression for $f_1(y)$ can be
obtained:
\begin{equation}
f_1(y)^2=\frac{({\sin\theta})^{2c}}{z_2\bar{z}_2}=\sigma.
\end{equation}
We are unable, however, to invert this equation to obtain an explicit
expression of $y$ in terms of $\sigma$.  Nevertheless, the above equation
suggests that if an explicit expression for $y$ is possible, then the
complex coordinates will enter into it only in the above combination
denoted by $\sigma$. Moreover, $y$ is independent of $z_3$ and its conjugate,
{\it i.e.} $\partial y/\partial z_3=\partial y/\partial \bar{z}_3=0$.

For suitable values of $y$, the total derivative of $y$ can be evaluated:
\begin{equation}
\frac{dy}{d\sigma}=(\frac{d\sigma}{dy})^{-1}
=\frac{wv}{12f_1(y)^2}.
\end{equation}
This is possible because $y$ can be viewed as a function of $\sigma$ only.
In turn, $\sigma$ is a function of $z_1,z_2$ and their conjugates. One
obtains
\begin{equation}
\frac{\partial \sigma}{\partial z_1}=\frac{c\sigma}{z_1}\cos\theta.
\end{equation}
Then
\begin{equation}
\frac{\partial y}{\partial z_1}=\frac{dy}{d\sigma}
\frac{\partial\sigma}{\partial z_1}=\frac{1}{z_1}\frac{wv}{12}c\cos\theta.
\end{equation}
Similarly, other partial derivatives can be evaluated. One can write
all of them more succinctly using coordinates $u_i=\ln z_i$ as
\begin{equation}
\begin{tabular}{llll}
$\displaystyle\frac{\partial \theta}{\partial u_1}=\frac{1}{2}\sin\theta,$
&$\displaystyle\frac{\partial \theta}{\partial u_2}=0,$
&$\displaystyle\frac{\partial \theta}{\partial u_3}=0,$
&$\displaystyle\frac{\partial \theta}{\partial \bar{u}_i}=
\frac{\partial \theta}{\partial u_i},$\\[2ex]
$\displaystyle\frac{\partial y}{\partial u_1}=\frac{wv}{12}c\cos\theta,$
&$\displaystyle\frac{\partial y}{\partial u_2}=-\frac{wv}{12},$\kern1em
&$\displaystyle\frac{\partial y}{\partial u_3}=0,$\kern1em
&$\displaystyle\frac{\partial y}{\partial \bar{u}_i}=
\frac{\partial y}{\partial u_i},$\\[2ex]
$\displaystyle\frac{\partial r}{\partial u_1}=-\frac{r}{6}(1-cy)\cos\theta,$
\kern1em
&$\displaystyle\frac{\partial r}{\partial u_2}=-\frac{ry}{6},$
&$\displaystyle\frac{\partial r}{\partial u_3}=\frac{r}{6},$
&$\displaystyle\frac{\partial r}{\partial \bar{u}_i}=
\frac{\partial r}{\partial u_i}.$\\
\end{tabular}
\end{equation}

The metric can now be written in terms of the $u_i$ as
\begin{eqnarray}
ds^2&=&\left [r^2\frac{(1-cy)}{6}{\sin}^2\theta
+r^2\frac{wv}{36}c^2{\cos}^2{\theta}
+\frac{r^2}{9}(1-cy)^2{\cos}^2\theta
\right ]du_1 d\bar{u}_1\nonumber\\
&&+\left [ r^2\frac{wv}{36}+\frac{r^2}{9}y^2\right ]du_2 d\bar{u}_2
+\frac{r^2}{9}du_3 d\bar{u}_3\nonumber\\
&&+\left [-r^2\frac{wv}{36}c\cos\theta+\frac{r^2}{9}y(1-cy)\cos\theta\right]
(du_1 d\bar{u}_2+d\bar{u}_1 du_2)\nonumber\\
\label{met}
&&-\frac{r^2}{9}y(du_2 d\bar{u}_3+ d\bar{u}_2 du_3)
-\frac{r^2}{9}(1-cy)\cos\theta (du_1 d\bar{u}_3+d\bar{u}_1 d{u_3}).
\end{eqnarray}
The coordinate $r$ in the limit (\ref{limit}) reduces to the coordinate
$\rho$ as used by \cite{conifold} for the conifold. Recall that the K\"ahler
potential in the conifold case is ${\rho}^2$. Here, one can show that the
K\"ahler potential for the Sasaki-Einstein cone is $r^2$.  For example,
\begin{eqnarray}
{\partial}_{u_1}{\partial}_{\bar{u}_1}r^2&=&
{\partial}_{u_1}\left(-2r\frac{r}{6}(1-cy)\cos\theta \right )\nonumber\\
&=&-\left(\frac{1}{3}\frac{{\partial}r^2}{\partial{u_1}}(1-cy)\cos\theta
+\frac{r^2}{3}\frac{{\partial}(-cy)}{\partial{u_1}}\cos\theta
+\frac{r^2}{3}(1-cy)\frac{{\partial}\cos\theta}{\partial{u_1}}\right)
\nonumber\\
&=&\frac{r^2}{9}(1-cy)^2{\cos}^2\theta
+r^2\frac{wv}{36}c^2{\cos}^2\theta
+\frac{r^2}{6}(1-cy){\sin}^2{\theta},
\end{eqnarray}
which matches with the $(1,\bar{1})$ component of the metric written in
(\ref{met}).  The other components of the metric can be similarly
obtained. Thus the relation $g_{{\mu}\bar{\nu}}
={\partial}_{\mu}{\partial}_{\bar{\nu}}{\cal K}$ holds for the K\"ahler
potential ${\cal K}=r^2$. This is an interesting result, and due to its
simplicity it could provide alternative ways of understanding
deformations and resolutions
of the cone over $Y^{p,q}$. Similarly, the K\"ahler potential
for the four-dimensional K\"ahler-Einstein base turns out to be
$\frac{2}{3}\ln[(1+\frac{1}{z_1\bar{z}_1}) f_2(y)]$.

\subsection{First order perturbation in complex form}

We now arrive at our goal of showing that the deformation obtained in
the previous section is a complex deformation.  Using the complex one-forms
defined earlier, one obtains
\begin{eqnarray}
-\sin\theta{\eta}^1\tilde{\eta}^2&=&
\frac{1}{H(y)^2}dyd\theta+\sin\theta d\phi (d\beta- c\cos\theta d\phi )
\nonumber\\&&
+i\big[d\theta (d\beta-c\cos\theta d\phi )
-\frac{1}{H(y)^2}dy\sin\theta d\phi \big].
\end{eqnarray}
In terms of these complex coordinates, the metric perturbation can be written as
\begin{eqnarray}
\label{gcomplex}
2(e^1e^3+e^2e^4)
&=&2[\cos{\psi}(e^{\beta}e^{\phi}+e^{y}e^{\theta})
+\sin{\psi}(e^{\theta}e^{\beta}-e^ye^{\phi})]\nonumber\\
&=&-2H(y)\sqrt{1-cy}\;[
\cos\psi \,{\rm Re}(-\sin\theta {\eta}^1\tilde{\eta}^2 )
+\sin\psi \,{\rm Im}(-\sin\theta {\eta}^1\tilde{\eta}^2)]\nonumber\\
\label{f-o}
&=&H(y)\sqrt{1-cy}\;\sin\theta [e^{i\psi}{\eta}^1\tilde{\eta}^2
+e^{-i\psi}\bar{\eta}^1\bar{\tilde{\eta}}^2].
\end{eqnarray}
The first order perturbation is of pure type.  We note here that
the $\eta^1 \tilde{\eta}^2$ part is {\it by itself} a zero mode of the
Lichnerowicz operator.  Since the holomorphic (3,0)-form for the
Sasaki-Einstein cone is known \cite{MS}, a complex closed $h_{2,1}$
form can be constructed,
\begin{eqnarray}
h_{(2,1)}&=&-\frac{1}{18}\frac{{\epsilon}^2}{(1-cy)^2}
[H^2du^2\wedge du^3\wedge d\bar{u}^2+{\sin}^2\theta (1-cy)
du^3\wedge du^1\wedge d\bar{u}^1\nonumber\\&&
+cH^2{\cos}\theta du^3\wedge du^1\wedge d\bar{u}^2
+y(1-cy){\sin}^2\theta du^1\wedge du^2\wedge d\bar{u}^1\nonumber\\&&
+H^2\cos\theta du^1\wedge du^2\wedge d\bar{u}^2].
\end{eqnarray}

\section{Warping the deformation}\label{warp}

The general form of the solution we are seeking in this section has
been explained in \cite{HEK}. The main difference is that instead
of warping the cone over $Y^{p,q}$ we need to warp the deformation of the
cone. The solution contains a nontrivial $F_5$ representing the flux
left by the D3 branes after the transition and $G_3$ which is the
implication of considering fractional D3 branes, that is, D5 branes
wrapping a 2-cycle inside the deformed cone. Thus the full IIB solution
is given in terms of the fields
\begin{eqnarray}
\label{ansatz}
ds^2&=& h^{-1/2} \eta_{\mu\nu} dx^\mu dx^\nu + h^{1/2}ds_6^2,\nonumber\\
ds_6^2&=& dr^2 +r^2\left[(e^1_{(s)})^2+(e^2_{(s)})^2+(e^3_{(s)})^2+
(e^4_{(s)})^2+(e^5_{(s)})^2\right],\nonumber\\
F_5&=&(1+*){\cal F}_5=(1+*)dh^{-1}\wedge dx^0\wedge dx^1\wedge dx^2\wedge dx^3,
\nonumber\\
G_3&=&-F_3+\frac{i}{g_s}H_3= i M\Omega_{3}.
\end{eqnarray}
The shifted vielbein basis is given above in section \ref{secSUSY}.
Given this general form of the solution, our goal is to find the
explicit form of $h(r,y)$ and $ \Omega_{3}$. Furthermore, in the case
when the deformation is zero, we expect to recover the solution of \cite{HEK}.

\subsection{Turning on a $G_3$ flux in the deformed cone over $Y^{p,q}$}

In this section we search for an appropriate complex $G_3$ which may be
turned on in the deformed Sasaki-Einstein cone. To have a supersymmetric
solution we would need $G_3$ to be a $(2,1)$ form \cite{gubser}. However,
we will not address the Dolbeault decomposition of $G_3$ directly. Rather,
we will simply concentrate on finding a solution to the equations of motion
with constant dilaton field. The constancy of the dilaton is readily
satisfied by an imaginary self dual complex 3-form $G_3$.  We thus
consider $G_3$ to be proportional to an imaginary self dual 3-form
$\Omega_3$, namely
\begin{equation}
\eqlabel{iself}
*_6\Omega_3=i\Omega_{3}.
\end{equation}
To assist in taking the Hodge dual while at the same time keeping the
radial coordinate dependence explicit, we use a set of shifted vielbeins
\begin{eqnarray}
e^6_{(s)}= dr, ~&&~ e^1_{(s)}=\left(e^1+\delta e^3\right),\qquad
e^2_{(s)}=\left(e^2+\delta e^4\right),\nonumber\\
e^3_{(s)}=e^3, ~&&~ e^4_{(s)}=e^4,
\kern5.7em e^5_{(s)}=e^5,
\end{eqnarray}
which resemble the choice of (\ref{eq:ncsv}), but with $r$-dependence
(and some factors) removed.

In this shifted vielbein basis, the most general Ansatz for an imaginary self
dual $\Omega_{3}$ may be written as
\begin{eqnarray}
\Omega_{(3)}&=&\left(\frac{dr}{r}+ i\frac{e^5}{3}\right)\wedge
\bigg(\alpha_1\left(e^1_{(s)}\wedge e^2_{(s)}+e^3_{(s)}\wedge e^4_{(s)}\right)
+\alpha_2\left(e^1_{(s)}\wedge e^4_{(s)}+e^2_{(s)}\wedge e^3_{(s)}\right)
\nonumber \\
&&\kern8em+i\alpha_3\left(e^1_{(s)}\wedge e^3_{(s)}
-e^2_{(s)}\wedge e^4_{(s)}\right)\bigg) \nonumber \\
&&+\left(\frac{e^5}{3}+i\frac{dr}{r}\right)\wedge \bigg(
i\beta_1\left(e^1_{(s)}\wedge e^2_{(s)} - e^3_{(s)}\wedge e^4_{(s)}\right)
+i\beta_2\left(e^1_{(s)}\wedge e^4_{(s)} - e^2_{(s)}\wedge e^3_{(s)}\right)
\nonumber \\
&&\hphantom{+}\kern8em+\beta_3\left(e^1_{(s)}\wedge e^3_{(s)}
+e^2_{(s)}\wedge e^4_{(s)}\right)\bigg)\nonumber\\
&&+i\lambda_1\bigg(\frac{dr}{r} \wedge \frac{e_{(s)}^5}{3}\wedge e_{(s)}^1
+\frac{i}{6}e^2_{(s)}\wedge e^3_{(s)}\wedge e^4_{(s)}\bigg) \nonumber\\
&&+\lambda_2\bigg(\frac{dr}{r} \wedge \frac{e_{(s)}^5}{3}\wedge e_{(s)}^2
-\frac{i}{6}e^1_{(s)}\wedge e^3_{(s)}\wedge e^4_{(s)}\bigg) \nonumber\\
&&+i\lambda_3\bigg(\frac{dr}{r} \wedge \frac{e_{(s)}^5}{3}\wedge e_{(s)}^3
+\frac{i}{6}e^1_{(s)}\wedge e^2_{(s)}\wedge e^4_{(s)}\bigg) \nonumber\\
&&+\lambda_4\bigg(\frac{dr}{r} \wedge \frac{e_{(s)}^5}{3}\wedge e_{(s)}^4
-\frac{i}{6}e^1_{(s)}\wedge e^2_{(s)}\wedge e^3_{(s)}\bigg),
\label{eq:omeg3}
\end{eqnarray}
where the ten functions $\alpha_i,\beta_i,\lambda_i$ are general (possibly
complex) functions.  The factors of $i$ in front of some of these functions
are chosen for later convenience.  We, however, restrict to the case when
these functions depend only on $r$ and $y$.

Our goal now is to expand the above equations in $\delta$, that is in powers
of $\epsilon^2/r^3$, and then to require $\Omega_{(3)}$ to be closed.  For
this purpose we note that the algebra of the exterior derivative $d$ acting
on the basis $e^i$ is
\begin{eqnarray}
de^1&=&-\left(\frac{y}{H}-K\right)e^1\wedge e^2 + 2K e^3\wedge e^4,\nonumber\\
de^2&=&0, \nonumber\\
de^3&=&-e^5\wedge e^4 + Ke^2\wedge e^3 - \frac{y}{H} e^1 \wedge e^4,
\nonumber\\
de^4&=& e^5 \wedge e^3 + K e^2 \wedge e^4 + \frac{y}{H} e^1 \wedge e^3, \nonumber\\
de^5 &=& -e^1\wedge e^2 +e^3\wedge e^4,
\end{eqnarray}
This follows directly from (\ref{de}) and (\ref{eq:ei}).
We find that taking the exterior derivative of $\Omega_3$ produces 14 out
of 15 possible terms.  To simplify matters we assume that we are looking
for a solution that becomes that of \cite{HEK} in the $\epsilon\rightarrow0$
limit, and so we may take that all functions except $\alpha_1$ are of order
$\epsilon^2$ already.  This allows us to drop certain $\epsilon^4$ terms.
The actual computation is straightforward but quite involved, and we present
some of the partial results in appendix \ref{3form}. The most important
result is that such an imaginary self dual 3-form can be constructed
explicitly, resulting in
\begin{eqnarray}
{\Omega}_3&=&\left ( \frac{dr}{r}+i\frac{e_{(s)}^5}{3}\right )
\wedge \biggl[
\frac{3}{2}\frac{1}{(1-cy)^2}
(e_{(s)}^1\wedge e_{(s)}^2+e_{(s)}^3\wedge e_{(s)}^4)\nonumber\\
&&\kern4em+{\alpha}_2(r,y)(e_{(s)}^1\wedge e_{(s)}^4+e_{(s)}^2\wedge e_{(s)}^3)
+i{\alpha}_3(r,y)(e_{(s)}^1\wedge e_{(s)}^3-e_{(s)}^2\wedge
e_{(s)}^4)\biggr]\nonumber\\
&&\kern-2em+b_1(r, y)\left (\frac{e_{(s)}^5}{3}+i\frac{dr}{r}\right)\wedge
\bigl[(e_{(s)}^1\wedge e_{(s)}^3+e_{(s)}^2\wedge e_{(s)}^4)
+i(e_{(s)}^1\wedge e_{(s)}^4-e_{(s)}^2\wedge e_{(s)}^3)\bigr]\nonumber\\
&&\kern-2em
+il_2(r,y)(e_{(s)}^3+ie_{(s)}^4)\wedge
\left(\!\frac{dr}{r}\wedge \frac{e_{(s)}^5}{3}+\frac{1}{6}e_{(s)}^1\wedge e_{(s)}^2\!\right),\nonumber\\
\label{eq:isd3f}
\end{eqnarray}
where
\begin{eqnarray}
{\alpha}_2(r,y)&=&\frac{3}{4}\frac{{\epsilon}^2}{r^3}\left (1-\frac{1}{(1-cy)^4}\right )
+\frac{3}{4}\frac{{\epsilon}^2}{r^3}\frac{1}{c^2{\cal Q}(y)}
\left [(1-ac^2)\left (\frac{1}{(1-cy)^4}-1 \right )
\right. \nonumber \\&&
\left. -\frac{9}{(1-cy)^2}+\frac{11}{1-cy}
+4-6(1-cy)-3(1-cy)^2+3(1-cy)^3
\right],\quad\\
{\alpha}_3(r,y)&=&-\frac{3}{4}\frac{{\epsilon}^2}{r^3}\left (1-\frac{1}{(1-cy)^4}\right )
+\frac{3}{4}\frac{{\epsilon}^2}{r^3}\frac{1}{c^2{\cal Q}(y)}
\left[ (1-ac^2)\left (\frac{1}{(1-cy)^4}-1 \right )
\right. \nonumber \\&&
\left.-\frac{3}{(1-cy)^2} +\frac{5}{1-cy}
-8+6(1-cy)+3(1-cy)^2-3(1-cy)^3
\right],\\
b_1(r,y)&=&\frac{3}{2}\frac{{\epsilon}^2}{r^3}\left [\ln \frac{2r^3}{{\epsilon}^2}-4
+\frac{3}{(1-ac^2)(1-cy)}\right. \nonumber\\&&
\left. +\frac{c}{2(1-ac^2)}\sum_{i=1}^{3}
\frac{(a+2acy_i+(1-ac^2)y_i^2)}{y_i(1-cy_i)}\ln(|y-y_i|)
\right],\nonumber\\
l_2(r,y)&=&\frac{9}{2}\frac{{\epsilon}^2}{r^3}\frac{1}{cH(1-cy)}
\left[(1-cy)-\frac{1}{(1-cy)}\right]^2.
\label{eq:b1ry}
\end{eqnarray}
Here ${\cal Q}(y)=a-3y^2 +2cy^3$ and $y_i$ are the three roots
resulting from ${\cal Q}(y)=0$.

In the absence of a deformation ($\delta=\epsilon=0$) the above reduces
to the simple result for $\Omega_{(3)}$ given by \cite{HEK}
\begin{equation}
\alpha_1(y)= \frac{3}{2(1-cy)^2}.
\end{equation}
Moreover, in the limit of \cite{HEK} $\Omega_{(3)}$ is a $(2,1)$ form, a fact
that can be established using the complex coordinates of \cite{MS} as
reviewed in section \ref{kah}. Another interesting case, $c=0$ (when $K=0$)
is the KS \cite{KS} solution with
\begin{equation}
\alpha_1\to\fft32, \quad \beta_2=\beta_3\to\fft32 \frac{\epsilon^2}{r^3}
\left(\ln\left(\frac{2r^3}{\epsilon^2}\right)-1\right),
\end{equation}
and all other functions set to zero.  For the KS solution the corresponding
$\Omega_3$ is also (2,1).
In both of the above limiting cases, the fact that they are $(2,1)$
guarantees that the solution is supersymmetric. We have not addressed
the question of the precise Dolbeault decomposition of our $\Omega_3$;
however we hope to return to this question in the future.

\subsection{Other forms}

By comparing the imaginary self dual 3-form obtained above with the
Klebanov-Strassler solution \cite{KS} and the Herzog, Ejaz and Klebanov
solution \cite{HEK}, we may extract the NS-NS 3-form flux $H_3$ and the
R-R form $F_3$ from the real and imaginary parts of $\Omega_3$.  We
find
\begin{eqnarray}
\frac{F_3}{M}&=& {\alpha}_1(y)\biggl[ e^1_{(s)}\wedge e^2_{(s)}
\wedge\frac{e^5_{(s)}}{3}+ e^3_{(s)}\wedge e^4_{(s)}\wedge\frac{e^5_{(s)}}{3}\biggr]\nonumber\\&&
+[{\alpha}_2(r,y)+b_1(r,y)] e^1_{(s)}\wedge e^4_{(s)}\wedge\frac{e^5_{(s)}}{3}
+[{\alpha}_2(r,y)-b_1(r,y)] e^2_{(s)}\wedge e^3_{(s)}\wedge\frac{e^5_{(s)}}{3}
\nonumber\\&&
+[{\alpha}_3(r,y)+b_1(r,y)] e^1_{(s)}\wedge e^3_{(s)}\wedge\frac{dr}{r}
+[-{\alpha}_3(r,y)+b_1(r,y)] e^2_{(s)}\wedge e^4_{(s)}\wedge\frac{dr}{r}
\nonumber\\&&
-l_2(r,y)e^3_{(s)}\wedge \frac{e^5_{(s)}}{3}\wedge \frac{dr}{r}
+\frac{1}{6}l_2(r,y) e^1_{(s)}\wedge e^2_{(s)}\wedge e^3_{(s)},\\
\frac{H_3}{g_sM}&=& {\alpha}_1(y)\left[ e^1_{(s)}\wedge e^2_{(s)}\wedge\frac{dr}{r}+ e^3_{(s)}\wedge e^4_{(s)}\wedge\frac{dr}{r}
\right]\nonumber\\&&
+[{\alpha}_2(r,y)-b_1(r,y)] e^1_{(s)}\wedge e^4_{(s)}\wedge\frac{dr}{r}
+[{\alpha}_2(r,y)+b_1(r,y)] e^2_{(s)}\wedge e^3_{(s)}\wedge\frac{dr}{r}
\nonumber\\&&
+[-{\alpha}_3(r,y)+b_1(r,y)] e^1_{(s)}\wedge e^3_{(s)}\wedge\frac{e^5_{(s)}}{3}
+[{\alpha}_3(r,y)+b_1(r,y)] e^2_{(s)}\wedge e^4_{(s)}\wedge\frac{e^5_{(s)}}{3}
\nonumber\\&&
+l_2(r,y)e^4_{(s)}\wedge \frac{e^5_{(s)}}{3}\wedge \frac{dr}{r}
-\frac{1}{6}l_2(r,y) e^1_{(s)}\wedge e^2_{(s)}\wedge e^4_{(s)}.
\end{eqnarray}
The NS-NS 3-form flux can be derived from a potential $H_3=dB_2$ where
\begin{eqnarray}
\frac{B_2}{g_sM}&=& f(r,y)[ e^1_{(s)}\wedge e^2_{(s)}+
e^3_{(s)}\wedge e^4_{(s)}]\nonumber\\
&&+\left[-\ft{1}{3}{\alpha}_3(r,y)+\ft{1}{3}b_1(r,y)-{\delta}f(r,y)
\right ]e^1_{(s)}\wedge e^4_{(s)}\nonumber\\
&&+\left[-\ft{1}{3}{\alpha}_3(r,y)-\ft{1}{3}b_1(r,y)+{\delta}f(r,y)
\right]e^2_{(s)}\wedge e^3_{(s)}\nonumber\\
&&+\ft{1}{3}l_2(r,y)e^3_{(s)}\wedge \frac{dr}{r},
\end{eqnarray}
with
\begin{equation}
f(r,y)=\frac{1}{2}\frac{1}{(1-cy)^2}\left [\ln \left(
\frac{2r^3}{{\epsilon}^2}\right)-1\right].
\end{equation}
If we take the limit (\ref{limit}), then only the term proportional to
$f(r,y)$ survives. This is in agreement with the Klebanov-Strassler expression
for $B_2$ expanded to first order in $\epsilon^2$, which is
\begin{equation}
\frac{B_2}{g_sM}=\frac{1}{2}\left[\ln\left(\frac{2r^3}{{\epsilon}^2}\right)-1
\right]
[e^1_{(s)}\wedge e^2_{(s)}+ e^3_{(s)}\wedge e^4_{(s)}]+{\cal O}({\epsilon}^4).
\end{equation}
Due to self duality of the complex 3-form, the dilaton field is constant,
$\phi=0$. Since $F_{3\;\mu\nu\rho}{H_3}^{\mu\nu\rho}=0$, the axion vanishes
as well.  The five-form flux is
\begin{eqnarray}
F_5&=&{\cal F}_5+*{\cal F}_5,\nonumber\\
{\cal F}_5 &=& B_2\wedge F_3,\nonumber\\
\frac{1}{g_sM^2}{\cal F}_5&=&
\frac{2}{3}f(r,y){\alpha}_1(y)
e^1_{(s)}\wedge e^2_{(s)}\wedge e^3_{(s)}\wedge e^4_{(s)}\wedge e^5_{(s)}
\nonumber\\&&
-\frac{l_2(r,y)}{6(1-cy)^2}\ln \left (\frac{2r^3}{{\epsilon}^2}\right )
e^1_{(s)}\wedge e^2_{(s)}\wedge e^3_{(s)}\wedge e^5_{(s)}\wedge
\frac{dr}{r}+{\cal O}({\epsilon}^4).
\end{eqnarray}
The effect of the inhomogenous metric is reflected in the five-form at
first order in ${\epsilon}^2$.

\subsection{The warp factor}

The equation for the warp factor can be extracted from the $F_5$ equation
of motion
\begin{equation}
dF_5=H_3\wedge F_3.
\end{equation}
The form of $F_5$ from (\ref{ansatz}) implies that
\begin{equation}
d*dh^{-1}\wedge dx^0\wedge dx^1\wedge dx^2\wedge dx^3= H_3\wedge F_3,
\end{equation}
where
\begin{eqnarray}
*{\cal F}_5&=&*dh^{-1} \wedge dx^0\wedge dx^1\wedge dx^2\wedge dx^3\nonumber\\
&=&-r^5 h'\,e_{(s)}^1\wedge e_{(s)}^2\wedge e_{(s)}^3 \wedge e_{(s)}^4
\wedge e_{(s)}^5\nonumber\\
&&+r^3 \dot{h} H(y) \, e^r \wedge e_{(s)}^1\wedge e_{(s)}^3 \wedge e_{(s)}^4
\wedge e_{(s)}^5
-r^3 \dot{h} H(y)\,\delta\, e^r \wedge e_{(s)}^1\wedge e_{(s)}^2
\wedge e_{(s)}^3 \wedge e_{(s)}^5,\nonumber\\
\end{eqnarray}
where in the last equation we used the fact that $dy=-H(y)(e_{(s)}^2-\delta\,
e_{(s)}^4)$.

\subsubsection{D3 branes on the deformation}\label{pured3}

We consider first the simple case where there are no fractional D3 branes,
that is, the case of zero three-form flux.  The above equations simplify
schematically to the familiar
\begin{equation}
\nabla^2 h=\frac{1}{\sqrt{g}}\partial_{m}\left(\sqrt{g}g^{m
n}\partial_{n} h\right)=\delta(\vec{x}\,).
\end{equation}
The first order perturbation is traceless, implying that the first order
correction to $\det g$ is zero.  Further, if we take that the D3 branes
are at the tip of the cone, the zeroth order harmonic function is simply
$A+{B}/{r^4}$.  The metric perturbation has no $r$ indices, and so the
resulting expansion in $\epsilon^2$ of the above equation trivially yields
\begin{equation}
\nabla^2 h_1=0,
\end{equation}
where $h_1$ is the $\epsilon^2$ correction of $h$.  The first order
correction to the harmonic function can therefore be set to $0$
consistently.

Given this fact, taking the usual limit $N\rightarrow \infty$, with $g_sN$
fixed gives the following geometry
\begin{equation}
ds^2=R^2\left[ds^2_{AdS_5} + ds^2_{Y^{p,q}} +
\frac{\epsilon^2}{r^3}\left(\frac{\left(e^1e^3+e^2e^4\right)}{3(1-cy)^2}\right)\right].
\end{equation}

As we mentioned before, this deformation of the solution corresponds
to chiral symmetry breaking. To see this, note that the vector
$\partial/\partial \psi$ is no longer a Killing vector for the
perturbed metric.  The new metric therefore also breaks the $U(1)_R$
symmetry (of the gauge theory) associated with (\ref{u1inse})
\cite{sequiver}
\begin{equation}
3\frac{\partial}{\partial \psi} - \frac{1}{2}\frac{\partial}{\partial \alpha},
\end{equation}
and so chiral symmetry has been broken. Let us see this argument using
standard AdS/CFT arguments. To a supergravity deformation of the form
\begin{equation}
ar^{\Delta-4}+b r^{-\Delta},
\end{equation}
corresponds a field theory deformation by an operator ${\cal O}$ such
that
\begin{equation}
{\cal H}\to {\cal H}+a {\cal O}, \qquad \langle{\cal O}\rangle=b.
\end{equation}
Applying this to our situation, we have a deformation that goes as
\begin{equation}
\epsilon^2 \; r^{-3},
\end{equation}
which implies an expectation value for a dimension-3 operator which should be
of the form
\begin{equation}
\langle\bar{\Psi}\Psi\rangle=\epsilon^2.
\end{equation}
Thus the gauge theory living in the worldvolume of a stack of D3
branes at the tip of the complex deformation of the cone over
$Y^{p,q}$ corresponds to placing the superconformal quiver gauge
theory in a different vacuum where the operator $\bar{\Psi}\Psi$
has a nonzero vacuum expectation value.

\subsubsection{D3 branes and fractional D3 branes}

The problem of solving for the warp factor in the presence of
fractional D3 branes on $B^{p,q}$ was addressed in \cite{HEK}.
We will take their solution as the zeroth order in epsilon term
(as our solution collapses to theirs in the
$\epsilon \rightarrow 0$ limit).  Curiously, their solution
remains unperturbed, as will be shown here.  First, consider the
resulting equation for $h$,
\begin{equation}
-\nabla^2_{(6)} h = \frac{1}{6}\left|H_3\right|^2.
\end{equation}
Next, note that in the first order deformed solution none of the indices
(in vielbein basis) on the $\mathcal O(\epsilon^2)$ terms agree with those
of the $\mathcal O(\epsilon^0)$ term.  The basis is diagonal, and so the
zeroth order term with indices contracted with the first order term vanishes.
Therefore, there are no $\mathcal O(\epsilon^2)$ source terms on the right
hand side of the above equation.  As before, the zeroth order solution takes
care of the right hand side, leaving us with only the first order equation
\begin{equation}
\frac{1}{\sqrt{g}}\partial_{m}\left[\sqrt{g}g^{m n}
\partial_{n} h_1(r,y)\right]=
\frac{1}{\sqrt{g}}\partial_{m}\left[\sqrt{g}h^{m n}
\partial_{n} h_0(r,y)\right],
\end{equation}
where we read the right hand side of the equation as being a
source term.  Without the fractional D3 branes, the source had
vanished because $h_0$ only depended on $r$, and $h^{m n}$
has no $r$ indices.  However, now $h_0$ depends both on $r$
and $y$.  Interestingly, however, the right hand side still vanishes
because
\begin{equation}
\sqrt{g}h^{m y}\partial_y h_0(r,y)=
\fft{\epsilon^2\sqrt{3}}{54\mathcal Q(y)^{1/2}} \left[\begin{matrix}
0\\
-\cos(\psi)\dot{h}_0(r,y)\sin(\theta)\\
-\sin(\psi)\dot{h}_0(r,y)\\
0\\
-c \cos(\theta) \sin(\psi) \dot{h}_0(r,y)\\
\cos(\theta)\sin(\psi) \dot{h}_0(r,y)
\end{matrix} \right],
\end{equation}
where this has been written in $[r,\theta,\phi,y,\beta,\psi]$ order.  The
divergence of this quantity obviously vanishes, with the first non-zero and
last non-zero terms canceling after the appropriate derivatives are taken.
Thus, again, we may consistently set the first order perturbation to the warp
factor to be zero because the source term is absent.

\section*{Acknowledgments}

We are especially grateful to S.~Franco for various comments and
discussions on topics related to this work. We have enjoyed comments by
C.~N\'u\~nez, J.~Sonnenschein and D.~Vaman. This work is supported by
the US Department of Energy under grant DE-FG02-95ER40899.


text:
Liverpool news: Barcelona chief issues major Ousmane Dembele statement
OUSMANE DEMBELE will not be leaving Barcelona despite talk of a summer move to Liverpool.
Alex HarrisAssistant Sports News Editor
That is according to Barcelona president Josep Maria Bartomeu, who has declared that an exit for the France international is "not on the table".
Liverpool boss Jurgen Klopp has declared himself "interested" in Dembele, who has struggled for Barcelona since joining from Borussia Dortmund last summer for a fee that could reach £142m.
Speaking on Catalan radio station RAC1, Bartomeu said: "Dembele leaving? It's not on the table.
"We're coming to the end of the season and after that, there will be meetings.
"He's an investment for the future.
"He's very young and we expect a lot from him.
"When a player comes to Barca, it's because [head coach Ernesto] Valverde wants them."
The team Cristiano Ronaldo thinks will start for Real Madrid in Champions League final
Dembele has struggled to make an impact for Barcelona during an injury-hit campaign.
The 20-year-old suffered a hamstring injury on his fourth appearance, which sidelined him for around four months.
He has five assists in his last nine league appearances but speculation continues to rage over his future.
Reports in France last weekend suggested that Liverpool are interested in landing Dembele.
Liverpool boss Klopp was asked about a potential swoop in the wake of Sunday's 1-0 defeat at Chelsea.
The German said: "He's on the market? Now I'm interested."
Dembele rejected a move to Liverpool in 2016 when he left Rennes for Dortmund.
Explaining the snub, he said: "I saw Klopp in Paris but I told him I'd made up my mind to sign for BVB."

redpajama_set_name: RedPajamaCommonCrawl


text:
Q: What are the minimum system requirements for Ubuntu Snappy Core? What are the minimum system requirements for Ubuntu Snappy Core?

A: According to this article (not offical docs, I can't find any :/ ), the minimum  requirements are:
Processor – 600 MHz processor (ARMv7 or greater, or x86)
System Memory – 128 MB RAM or greater (The system itself uses 40 MB RAM)
Storage – 4GB flash / storage for factory reset and system rollback

I am  not sure if any other specs need to be atleast something, or whether the above is accurate.
Note that ARMv6 and earlier (used on the old Raspberry Pi models etc) is not supported.
x86 is apparently supported, though a image for 15.04 does not seem to be available, though you can get one for 15.10

A: In addition to Wilf (I do not have 50 reputation points yet to comment :s), the official docs that back this claim are found here:
http://www.ubuntu.com/internet-of-things (scroll to "Widest range of supported devices")

redpajama_set_name: RedPajamaStackExchange


text:
Thousands of Kurds protest Turkish strikes on Kurdish groups in Syria
AFP , Sunday 27 Nov 2022
Thousands of Kurds protested on Sunday in the Syrian city of Qamishli against days of deadly Turkish cross-border strikes targeting Kurdish groups in the country's northeast.
Syrian-Kurdish demonstrators in the Syrian city of Qamishli raise pictures of people killed in Turkish strikes. AFP
Turkey says hits nearly 500 Kurdish targets in Iraq, Syria
'Fear reigns' among Syria's Kurds following Turkish strikes
Turkey strikes in Syria, Iraq a week after Istanbul bombing
Turkey announced last Sunday it had carried out air strikes against semi-autonomous Kurdish zones in north and northeastern Syria, and across the border in Iraq. It has also threatened a ground offensive in those areas of Syria.
Demonstrators in Kurdish-controlled Qamishli, in Hasakeh province, brandished photos of people killed during recent strikes in the semi-autonomous region, an AFP correspondent said.
"Only the will of the Kurdish people remains," said protester Siham Sleiman, 49. "It will not be broken and we remain ready. We will not leave our historic land."
Protesters in Qamishli also chanted in favour of the resistance in "Rojava" -- the name Kurds in Syria give to the area they administer.
"The message that we want to convey to the world is that we are victims of eradication," said Salah el-Dine Hamou, 55.
"How long will we continue to die while other countries watch?"
The PKK has waged an insurgency against the Turkish state since 1984. Turkey alleges that Syrian Kurdish fighters are the PKK's allies.
Some protesters on Sunday carried Kurdish flags alongside photos of PKK leader Abdullah Ocalan -- jailed in Turkey since 1999 -- and shouted slogans against Turkish President Recep Tayyip Erdogan.
Since 2016, Turkey's military has conducted three offensives mostly targeting Kurdish fighters, and captured territory in northern Syria, which is now held by Ankara-backed proxies.
The US-backed Syrian Democratic Forces (SDF), the Kurds' de facto army in the area, led the battle that dislodged Islamic State group jihadist fighters from the last scraps of their Syrian territory in 2019.
Qamishli
cross-border strikes
Also In Region
Pope condemns increasing 'death spiral' in Middle East
Russia's Lavrov urges 'maximum responsibility' in calls with Israeli, Palestinian counterparts
Earthquake kills three, injures more than 800 in northwest Iran
Israeli guards kill allegedly armed Palestinian in West Bank


text:
Kentucky Whiskey-Making In The Wathens Over 100 Years
March 20, 2021 June 18, 2016 by Gary
From 1790-1905, a History of Technology and Innovation
An article in 1905 in The Wine and Spirit Bulletin outlines the history of a famous distilling family, the Wathens, from c. 1790 until the year of writing. The article is, "The Manufacture Of Whisky In Kentucky" and starts at pg. 22 in the June 1 issue. Its focus is technological and on output. I won't summarise too much because the article is clear and easy enough to read, but some salient points follow.
To start a distilling run, founder Henry Wathen mashed in small tubs with water and relied on "spontaneous" fermentation.
This is interesting as a natural ferment was achieved without "backset" (say, as lambic beer is fermented today). However, as soon as backset, or spent residue in the still, was available from the first boils, it was carried to each successive tub in four- gallon buckets, and used in mashing.
For fermentation, as soon as they had top yeast from the fermenter, they used that – this is yeasting-back, an alternate way to sour-mash.
Henry Wathen used all-corn mashing according to this account. To ensure conversion, part was malted on the property. He used two copper stills, a wash and spirit still, just as in Scotland to this day for malt whisky.
With each generation, improvements and changes occurred. The wash still was replaced with a hollowed log still, then a wood two-chamber still, then a wood three-chamber still. The last is a kind of transition between a pot still and a continuous still that is divided by plates through which steam surges and re-distills the mash.
As I glean it, a three-chambered still is really a batch system, like three pot stills side-by-side except except each chamber is placed atop each other. There is no external condensation and re-charging. In effect though, a triple distillation occurs. These were used into the 1900s but became obsolete after Prohibition.
Other successive changes: the copper spirit still was replaced by a horizontal copper doubler, similar to the ones used today.
And finally, a metal, multi-tiered column still, similar to those today, replaced the wood three-chamber still. The account implies the spirit improved with each such change. My only experience with pot stills and bourbon is Woodford Reserve's bourbon. The pot still element of its bourbon is quite strong-tasting, at least at four or five years old, vs. a column distillate of the same age; I prefer the latter and I think the market probably did in the 1800s, too.
Other changes: barley malt replaced corn malt about 1850, for two reasons: it was made by professional maltsters and more reliable. Second, the whiskey tasted better. This ties in to some other accounts, e.g., distiller Charlie Thomason from Willett of Bardstown, KY said c. 1960 that modern bourbon has reduced the amount of barley malt formerly used, to the detriment of flavour. (I will show later that in general a greater amount of barley malt was used in bourbon production c. 1900 than today).
One of the Wathen descendants ensured steam-heating of his warehouse to accelerate maturation and produce again a better product, this is later in the 1800s.
Finally, about 1870, a purpose-made yeast replaces yeasting-back.
This would have been a hop yeast, cultured up before labs isolated a pure-culture yeast. In a word, jug yeast. This account from the same era as the article mentioned describes its preparation.
Just as an index of its modernity, as late as 1905 you find ads for whiskey proudly advertising "no jug yeast".
It appears – this from another account – that the hops neutralized the "bad" organisms to allow the selected yeast to do its work. One can infer the Kentucky distillers' practice of yeast-making probably is not age-old and developed from the mid-century, but further substantiation would be needed.
The best part of the article is, detailed illustrations show how each item mentioned worked. E.g., the famous log still, an early steam-operated affair made from two hollowed poplar logs, is shown. They were arrayed horizontally on each other, not end to end, which suggests this was a batch-type system, more primitive than the wood two-chamber still but similar in principle.
Allowing for some commercial puffery as any article of this type would have, it is still a remarkable account of technological evolution in bourbon production as practiced by one family over time, the sons working with fathers and then improving in their turn as warranted.
One interesting thing, finally. I don't think the article mentions the word bourbon once. The operative word is whiskey and this characterized George Washburne's Wine and Spirit Bulletin in general. The word bourbon often appears in the brand names listed, which is many times, but not too often in the textual narrative. I think this suggests bourbon initially was a retail and wholesale trade term, applied from the beginning from the outside so to speak and literally probably outside Kentucky (in New Orleans, Cincinnati, Chicago, etc.). The Kentucky pros called it whiskey – and lots of them still do.
American Distilling's Julius Freiberg Leaves A Special Mark
Kentucky Burgoo


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The App Store has changed the software business and many struggle to find a way to make a living in it. It's not only a crowded market with 10 apps solving the same problem, you're competing for money that people get more pleasure spending on Clash of Clans. What does it take to find success?
Last year Curtis got serious about turning his indie app, Slopes, into a larger part of his business. Since then Slopes has turned it into a reliable source of recurring revenue which continues to grow. By rethinking many of his past assumptions about app development and sales he's seen a dramatic increase in app revenue since last year. He'd like to share what's worked along the way, and challenge you to rethink what's holding you back.


text:
An Apology for W.P. Esq
An Apology for W.P. Esq in which the conduct of L--- G--- B---h is vindicated from all the cavils thrown out against him.
London : Printed for J. Pridden, 1759.
Bligh, Thomas, > 1685-1775 > Military leadership.
Pitt, William, > Earl of Chatham, > 1708-1778.
Anglo-French War, 1755-1763 > Campaigns.
Great Britain > Politics and government > 1727-1760.
A letter from the Honourable L--t G--l B--gh, to the Rt. Hon. W--m P--t, Esq; se--y of s--te together with His M--y's instructions for the late expeditions on the coast of France.
by: Bligh, Thomas, 1685-1775.
A vindication of Mr. Pitt wherein all the aspersions thrown out against that gentleman, relative to the affair of Rochfort, are unanswerably confuted /
by: Member of Parliament.
A review of Mr. Pitt's administration
by: Almon, John, 1737-1805.
A second letter to the Right Honourable the Earl of B***
by: Author of the first letter to the Right Honourable the Earl of B***.
A Letter to the Right Honourable the Earl of B*** on a late important resignation, and its probable consequences.
A Letter to His E--y L--t G--l B--h
An impartial narrative of the last expedition to the coast of France
by: Eye witness.
The political conduct of the Earl of Chatham
The Answer at large to Mr. P--tt's speech
An Enquiry into the conduct of a late Right Honourable commoner
Lord Ch----m's prophecy, an ode addressed to Lieutenant General G-ge /
Short considerations upon some late extraordinary grants and other particulars of a late patriot's conduct.
by: Detestor of jobbs under all administrations.
The Conduct of a Rt. Hon. gentleman [William Pitt] in resigning the seals of his office justified, by facts, and upon the principles of the British constitution
A short view of the political life and transactions of a late Right Honourable Commoner to which is added, a full refutation of an invidious pamphlet, supposed to be published under the sanction of a very popular nobleman, entitled, "An enquiry into the conduct of a late Right Honourable Commoner."
A Letter to the Right Honourable author of A letter to a citizen with animadversions on the answer thereto, and on the behaviour of the corporation of the city of London : in which his reasons for resigning, the conduct, success and advantages of his administration, his fidelity to his country, capacity for directing the transactions of war, commerce and pacification, are fairly stated and freely considered.
Plain reasons for removing a certain great man from his M--y's presence and councils for ever addressed to the people of England /
by: Carlyle, Alexander, 1722-1805.
An examination of the principles and boasted disinterestedness of a late right honourable gentleman in a letter from an old man of business, to a noble lord.
The history of the life of William Pitt, Earl of Chatham
by: Goodwin, William, 1756-1836.
Anecdotes of the life of the Right Hon. William Pitt, earl of Chatham, and of the principal events of his time with his speeches in Parliament, from the year 1736 to the year 1778 ...
An Authentic account of the part taken by the late Earl of Chatham in a transaction, which passed in the beginning of the year 1778 [in which Sir J. Wright tried to effect a coalition between the Earl of Chatham and the Earl of Bute]
A Letter to His Grace the Duke of N********, on the present crisis in the affairs of Great Britain containing, reflections on a late resignation.
An Earnest address to the people of Great-Britain and Ireland occasioned by the dismission of William Pitt, Esq., from the office of Secretary of State.
A Letter to the Right Honourable the Earl of B***, on a late important resignation, and its probable consequences
An appendix to the Review of Mr. Pitt's administration
Anecdotes of the life of the Right Hon. William Pitt, Earl of Chatham and of the principal events of his time, with his speeches in Parliament from the year 1736 to year 1778.
The Constitution with an address to a great man.
An original camera obscura, or, The court, city, and country magic-lanthorn in which every one may take a peep, laugh, and shake their noddles at each other, go away well pleased ... being an account of the most curious and uncommon collection of manuscripts (warranted originals) ever yet offered to the public : with as curious and uncommon a dedication to the Right Honourable the Earl of Cheatum ... /
by: Smirk, Mr.
An enqviry into the condvct of a late Right Honovrable commoner
by: Cotes, Humphrey.
The Speech of the statue, of the Right Hon. William Pitt, Earl of Chatham. To the virtueous [sic] and patriotic citizens of New-York.
The speeches of the Earl of Chatham
by: Pitt, William, Earl of Chatham, 1708-1778.
A letter from the anonymous author of the Letters versified to the anonymous writer of the Monitor
by: Francis, Philip, 1708?-1773.
An Address to the great man with advice to the public.
Father Abraham's almanac, (on an entire new plan.) For the year of our Lord, 1760 ... To which is added, beautifully engrav'd on a large copper-plate, a fine picture of the Right Honorable William Pitt ... /
Mr. Weatherwise's pocket-almanac, (on an entire new plan.) For 1760. ...
The father of the city of Eutopia, or, The surest road to riches being a narrative of the remarkable life and adventures of an elevated bear : delivered under the similitude of a dream : dedicated to the Rt. Hon. Wm. P---, Esq.
by: C. C., 18th cent.
Correspondence of William Pitt, Earl of Chatham
A speech of the Rt. Hon. the Earl of Chatham. He who regards the honor and safety of his country, let him attend.
Dr. Williamson's narrative of the extraordinary case of a late great commoner in a letter to the president of the College of Physicians.
by: Williamson, Dr.


text:
Q: Submitting a task from inside celery worker task body I have a celery worker, where processing function (alpha_task) is defined. I'd like to submit another job to celery workers from INSIDE this function body (alpa_task). The new job will be responsible for processing just computed result.
So simply saying, i want the result to processed by another pool of celery workers.
One pool is responsible performing the job and another for processing results.
Celery Worker -> Submit job for result processing -> Another Celery Workers
I was thinking to use different routing key and queue, so the worker will NOT have the result processing job  submitted to themselfs, but it doens't appear to work.

*

*I have a following code for celery worker.

app = Celery('celery_worker_A', backend='backendstring', broker='brokerstring')
@app.task
   def process_alpha_task(json_input):
     processor = TaskProcessor()
     r = processor.process(json_input)
     app.send_task('process_result', args=[r], kwargs={},queue='queue_result_processing', routing_key='key_result_processing')
The problem is that the newly submitted task is being sent to same worker which just procesed "alpha" task, and i'm getting error about undefined task called "process_result" which is of course true.
I even tried to run this celery worker with forced queue, so it doesn't poll for any other tasks than alpha_tasks.
celery -A proj worker -Q alpha_tasks.
But it still seems to submit task to themselfs and poll for it, even though i defined other queue in send_task() and routing keys.
How can i achieve my goal, to process results using different celery workers?
//Edit:
I have one more example, should be more clear.
Explicitly running worker to process tasks ONLY from task_processor queue:
celery worker -A task_processor -Q task_processor
task_processor.py

app = Celery('task_processor', backend=config.get('CELERY', 'backend'), broker=config.get('CELERY', 'broker'))

#submit new task and route to resultprocessor queue, Shouldn't be processed by this worker.
def test():
    app.send_task('process_result', args=["ABC"], kwargs={}, queue="resultprocessor", routing_key="resultprocessorkey")

#TASK ROUTED CORRECTLY on worker start, NOT SUBMITTED TO THEMSELFS
test()

@app.task
def process_task(json_input):
    print("Processing task")
    #TASK ROUTING FAILED WHEN CALLED INSIDE BODY OF TASK, SUBMITTING TO THEMSELFS.
    test()
    print("Submitted request to  process result")


text:
The 10 Biggest Startups in India to Watch
It's been said that India may not have potable water or reliable electricity, but it has entrepreneurs. More than 1.3 billion of them at last count. It shouldn't come as too big of a surprise, then, that India is home to a growing number of technology startups—about 19,000 at last count. Not only that, but it is third on the planet in the number of unicorns—private startups valued at $1 billion or more—after the United States and China.
Yes, the land that gave us Delhi Belly and Bollywood isn't just another Third World country in need of a bailout. In 2015 alone, more than 880 deals worth $7.3 billion went down, according to the Economic Times of India. Still, exits remain elusive, with one expert quoted in the story saying serious IPOs are about five years out. We wonder if any of those will involve India's 10 biggest startups. Here's a beautiful breakdown of the world's unicorn population by those bright minds over at CB Insights:
A snapshot of the global world of unicorns. Note that Mu Sigma does not appear on this graphic. Credit: CB Insights
We'll take a quick snapshot of all 10 below, but a few things to note. First, the herd is particularly heavy in the e-commerce space, most of them locked in a Battle Royale with each other and colonial power tech giant Amazon. We wonder when the skies will fill with the buzz of drones delivering plastic isht made in China. Second, the venture capitalists behind these startups are diverse but a couple of familiar names recur with quite a bit of frequency. That includes SoftBank, which is attempting to fulfill CEO Mayaoshi Son's vision of making machines smarter than humans by throwing money at just about anything remotely tech related. Finally, we couldn't help but note that the Indians have more unicorns than former colonial power Britain (at eight). Just saying, mate.
Unicorn Valuation Date Joined Industry Select Investors
Flipkart $10 8/6/2012 eCommerce/Marketplace Accel Partners, Ebay, Iconiq Capital
Snapdeal $7 5/21/2014 eCommerce/Marketplace Intel Capital, Kalaari Capital, Nexus Venture Partners
One97 Communications (operates Paytm) $4.2 5/12/2015 Fintech Intel Capital, Sapphire Ventures, Alibaba Group
Olacabs $3 10/27/2014 On-Demand Accel Partners, SoftBank Group, Sequoia Capital
ReNew Power Ventures $2 2/14/2017 Energy & Utilities Goldman Sachs, JERA, Asian Development Bank
Mu Sigma $1.5 2/7/2013 Big Data Sequoia Capital, General Atlantic
Hike $1.4 8/16/2016 Social Foxconn, Tiger Global management, Tencent
Shopclues $1.1 1/12/2016 eCommerce/Marketplace Nexus Venture Partners, GIC Special Investments, Tiger Global Management
Zomato Media $1 4/10/2015 Social Sequoia Capital, VY Capital
InMobi $1 12/2/2014 Adtech Kleiner Perkins Caufield & Byers, Softbank Corp., Sherpalo Ventures
Flipkart – e-Commerce
Flipkart Group just raised $1.4 billion from Tencent, eBay and Microsoft in April. The huge infusion of capital brings total investments from 13 rounds into India's leading e-commerce company to $4.55 billion, bumping its valuation from $10 billion to $11.6 billion. A sidebar to the mega round included a $500 million purchase by Flipkart of Ebay India. Flipkart is expected to use its war chest in its ongoing battle with Amazon, which has reportedly amassed billions of its own to grab a piece of an e-commerce industry in India valued at nearly $120 billion by 2020.
Another shark circling the waters includes China e-commerce giant Alibaba, which has struggled to gain a foothold in India. Why is everyone hungry for India? It's a matter of mass: The country is expected to have about 320 million online shoppers by the end of the decade, the Motley Fool reported. That's roughly equal to every man, woman and child in the United States. Alibaba had reportedly tried to acquire Snapdeal, another e-commerce unicorn in India. Now there's quite a bit of speculation that Flipkart itself will try to buy Snapdeal with backing from SoftBank (also an investor in Snapdeal and Alibaba). It's more confusing than a Bollywood romance but with less singing and dancing. If you find all of these machinations that interesting, check out this longish article on VentureBeat for a recap on the corporate intrigue.
Oh, by the way, Flipkart offers about 20 million products to 26 million registered users that clock in about eight million visits each day. It employs about 20,000 people.
Snapdeal – e-Commerce
As of this writing, Snapdeal is still an independent startup. But, as we've already discussed, that may not last much longer. Launched in 2010, Snapdeal has raised about $1.76 billion, including $221 million in two separate rounds last year. Rumors of a buyout have left employees, who stand to see their stock options turn to dust, understandably jittery. Reuters has reported that the company's management has acknowledged that a deal may be in the works. That has sent some employees for the exit door, following earlier layoffs. All that implies that Snapdeal might come at a deal when a transaction finally goes down. Snapdeal is rated as the No. 3 e-commerce company in India after Amazon and Flipkart.
One97 Communications – e-Commerce/Fintech
Synonymous with its flagship consumer brand Paytm, One97 Communications carries a valuation of $4.2 billion, according to startup data firm CB Insights. However, we've seen others value the New Delhi-based startup as high as $5 billion. The company has raised a little more than $1 billion, with Alibaba a prime investor. Paytm is reportedly in talks with (you guessed it) SoftBank for a cool $1 billion in further investments. That would boost the company's valuation to $7 billion.
The PayPal of India, with a particular emphasis on mobile technology, Paytm recently launched its services in Canada, mostly revolving around payment of utility bills for now.
Ola Cabs – On-Demand
The Uber of India (with even some of the controversy thrown in), Ola Cabs has raised $1.56 billion since its founding in 2010. The most recent round was led (surprise!) by SoftBank, a $330 million Series E+ round in February. It was one of the few mega deals of the first quarter in 2017 in the on-demand sector, which has been on the decline, according to CB Insights. Ola is in more than 100 cities and counts 450,000 vehicles in its "fleet," from luxury automobiles to auto rickshaws. They are locked in a Mad Max-style battle with Uber for supremacy of India's traffic-choked roads.
Update 09/20/2018: Ola has raised an additional $50 million from Sailing Capital and the China-Eurasian Economic Cooperation Fund (CEECF) at an estimated valuation of $4.3 billion. This brings the company's total funding to$3.3 billion to date.
ReNew Power Ventures – Energy and Utilities
A solar and wind energy company in a nation that plans to build nearly 400 new coal plants, ReNew Power Ventures has pulled in $1.13 billion in funding. Most recently, in February, an energy firm from Japan picked up a 10 percent stake ReNew Power Ventures for $200 million, Times of India reported. Last year, the startup reportedly became the first company in India to achieve 1 GW of commissioned wind and solar energy capacity. It has also won contracts with state-owned Indian Railways to install 5 MW of renewable energy capacity and with Solar Energy Corporation of India.
Mu Sigma – Big Data
This is our kind of company, and the kind of tech startup that we would have expected to dominate the top rankings in tech-loving India. Mu Sigma is a big data analytics firm that has counted Microsoft and WalMart among its clients. Total funding is relatively modest at $211 million. It's a particularly rare breed of unicorn, having actually turned a profit! The company reported a 22 percent growth in net profits at $71.25 million on a revenue of $124. 5 million for 2015-16.
Hike – Social Media
In yet another example of imitation is the best form of flattery (if not creativity), Hike is a social media messaging app that is a cross between WeChat and SnapChat. It has raised $261 million, including a $175 million Series D last year led by Tencent, which developed WeChat (yet more VC incest), and included SoftBank. Hike reportedly has more than 100 million downloads. The Gangnam Style video has been viewed more than 2.8 billion times. Obviously we've reached the pinnacle of civilization with social media.
Shopclues – e-Commerce
You get the feeling that the e-commerce sector in India will experience some serious consolidation in the future? Another Amazon-like online merchant of merchants, Shopclues rocketed to unicorn status in just five rounds of funding and $231 million in equity. A report in the Economic Times suggests the company has even loftier ambitions—a listing on the NASDAQ as early as this year. A company co-founder said Shopclues is on pace to break even, meaning not lose money, by March 2018.
Zomato – Social Media
Ever wonder what happened to U.S.-based Urbanspoon, a restaurant information and rating website? Neither do we. But now that it's on your mind: Zomato scooped it up in 2015 for $55 million. The Indian startup has collected nearly $224 million and counts Sequoia Capital among its chief investors. More than just a restaurant review site, Zomato also provides a platform for table reservations, home delivery and online ordering. It's used by more than 1.2 million restaurants in 24 countries.
Update 10/16/2018: Zomato has taken in an additional $210 million from Alibaba's Ant Financial, as the company grows to compete against rival Swiggy for market leadership. This brings the company's total funding to $653.8 million so far.
InMobi – Adtech
Despite running afoul of the Federal Trade Commission last year for software "glitches" (big air quotes here) for illegally tracking 100 million phones, InMobi enjoyed a good year in 2016. Funded to the tune of $320 million, the Bangalore-based startup reportedly became profitable last year. That's not an easy thing to do in the mobile advertising game, especially when you're competing against companies like Google and Facebook. InMobi's CEO has promised an IPO in the next couple of years.
Will any of these companies make an exit on the scale of Google or Facebook? And will U.S. investors be able to get a piece of the action? We know at least two are considering the jump. We'll continue to monitor their progress, as we also watch the startup sector in India continue to mature.
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