text:
\section{Introduction}
$U(N)_k$ Chern-Simons theories coupled to fundamental matter provide an interesting class of interacting three-dimensional conformal field theories that are exactly solvable in the 't Hooft limit: $N \rightarrow \infty$, $k \rightarrow \infty$ with the 't Hooft coupling $\lambda \equiv \frac{N}{k}$ held fixed. These theories, which have been intensively studied in the past few years \cite{Giombi:2011kc,  Aharony:2011jz, Aharony:2012nh,  GurAri:2012is, Jain:2013py-2, Jain:2014nza, Jain:2013py-1,Aharony:2012ns,Jain:2013gza,Jain:2012ss,Minwalla:2011ma,Yokoyama:2012fa,Minwalla:2015new}, are particularly interesting because they provide examples of non-supersymmetric dualities. For instance, they are widely believed to be dual to Vasiliev higher-spin gauge theories (see \cite{ Giombi:2012ms} for a review.) They also exhibit a spectacular bosonization duality relating Chern-Simons theory coupled to fundamental fermions to critical Chern-Simons theory coupled to fundamental bosons, which can be thought of as a non-supersymmetric generalisation of the ABJ and Giveon-Kutasov dualities. \cite{Giombi:2011kc,Aharony:2011jz, Aharony:2012nh,  GurAri:2012is, Jain:2013py-2, Jain:2014nza, Jain:2013py-1,Aharony:2012ns,Jain:2012ss, Chang:2012kt, Giveon:2008zn,Benini:1}

The bosonization duality has been tested via three point functions and also in thermal free energy computations, leaving little doubt to its correctness. However, it is still of independent interest to directly test the duality at the level of four-point functions; which are not determined by purely kinematic considerations.

In this paper, we calculate four-point correlation functions of  the primary scalar operator $J^{(0)}$ in the critical and non-critical $U(N)_k$ Chern-Simons theory coupled to fundamental fermions. For a particular choice of external momenta, we are able to obtain a closed form (but highly non-trivial) expression for the four point function of the scalar primary as a function of $\lambda$ -- which we then compare to the free and critical bosonic theories to obtain another independent check of the bosonization duality.

\label{sec:intro}

\subsection{Review of the Bosonization Duality}

The bosonization duality \cite{Aharony:2012nh}, which can be thought of as a non-supersymmetric generalization of the Giveon-Kutasov duality \cite{Giveon:2008zn}, states that a $U(N_f)_{k_f}$ Chern-Simons theory coupled to fermions in the fundamental representation is dual to a $U(N_b)_{k_b}$ Chern-Simons theory coupled to critical bosons in the fundamental representation. The critical theory is obtained by deforming the usual (non-critical) theory by a double trace operator $\lambda_4 \phi^2 \phi^2$ and taking the coupling to infinity. (The coupling $\lambda_4$ should not be confused with $\lambda_b=N_b/k_b$.)

The conjectured duality claims that the two theories are equivalent, with the following relation between parameters:
\begin{align}
k_f  =  -k_b \\
N_f  =  |k_b| - N_b
\end{align}
Though we present the duality in terms of $k$ and $N$, the duality has only been tested in the large $N$, 't Hooft limit; at finite $N$ there will be some shifts of $\pm 1/2$ in the Chern-Simons level for the fermionic theory as discussed in \cite{Aharony:2012nh, GurAri:2012is}. All parameters are defined in a dimensional reduction regularization scheme, used in \cite{Aharony:2012nh}. In terms of $\lambda = \frac{N}{k}$, the duality can be written as:
\begin{align}
\lambda_f  =   \lambda_b-\sign(\lambda_b), \\
N_f  =  N_b\frac{1-|\lambda|_b}{|\lambda|_b},
\end{align}
or
\begin{align}
\lambda_b  =  \lambda_f -\sign(\lambda_f),  \\
 N_b  =  N_f  \frac{1-|\lambda|_f}{|\lambda|_f}.
 \label{duality}
\end{align}
From these results, we have the simple relation $|\lambda_b|  =  1-|\lambda_f|$ and $\sign(\lambda_b)=-\sign(\lambda_f)$.

As both sides of the theory are exactly solvable, the simplest way to verify the duality is to calculate correlation functions on both sides, which we illustrate below.

The two-point function of the scalar primary,  which is defined as $J_f^{0} \equiv \bar{\psi}\psi$, in the fermionic theory is:
\begin{align}
\langle J_f^{0} (-q) J_f^{0} \rangle  =  -N_f \frac{\tan (\pi \lambda_f/2)}{4\pi \lambda_f} |q|.
\end{align}
In the critical bosonic theory, the two-point function is:
\begin{align}
\langle J_b^{0} (-q) J_b^{0} \rangle  =  -N_b \frac{4\pi \lambda_b}{\tan (\pi \lambda_b/2)} |q|.
\end{align}
These two-point functions determine the relative normalisation of the scalar operator in the two descriptions. We see that, taking $J_b^{0}=4 \pi \lambda_b J_f^{0}$\footnote{This relation makes sense because $\lambda$ and $J_f^0$ are odd under parity, while $J_b^0$ is even.}, the two point functions are identical:
\begin{align*}
 \langle J_b^{0} (-q) J_b^{0} \rangle =  -N_b \frac{4\pi \lambda_b}{\tan (\pi \lambda_b/2)} |q| \\
\langle J_f^{0} (-q) J_f^{0} \rangle  =  -N_b \frac{ \cot (\pi \lambda_b/2) }{4\pi \lambda_b} |q| \\
                                      =  N_f \sign(\lambda_b) \frac{ \tan (\pi \lambda_f/2) }{4\pi(1-|\lambda_b|)} |q| \\
                                      =  -N_f \frac{\tan (\pi \lambda_f/2)}{4 \pi \lambda_f} |q|
\end{align*}

The duality now implies that, for the three-point functions:
\begin{align}
\langle J_f^{0} (q_1) J_f^{0}(q_2) J_f^{0}(q_3) \rangle = \left(\frac{1}{4 \pi \lambda_b} \right)^3  \langle J_b^{0} (q_1) J_b^{0}(q_2) J_b^{0}(q_3) \rangle
\end{align}

Using the results in \cite{Aharony:2012nh, GurAri:2012is}, we can explicitly compute that:
\begin{align*}
\langle J_f^{0} (q_1) J_f^{0}(q_2) J_f^{0}(q_3) \rangle  =  0 \\
\langle J_b^{0} (q_1) J_b^{0}(q_2) J_b^{0}(q_3) \rangle  =  0
\end{align*}
which agrees with the duality,.

There are additional predictions for three-point functions; for instance for the three-point function of vector and scalar operators, which are non-zero, and which can be tested on similar lines.

Applying the duality to four-point functions, we obtain:
\begin{eqnarray}
\hspace{-3mm} \left(4 \pi \left(\lambda_f -\sign(\lambda_f) \right) \right)^4 \left \langle J_f^{0} (q_1) J_f^{0}(q_2) J_f^{0}(q_3) J_f^{0}(q_4) \right \rangle &=&  \left \langle J_b^{0} (q_1) J_b^{0}(q_2) J_b^{0}(q_3)  J_b^{0}(q_4) \right \rangle \label{duality-prediction}
\end{eqnarray}
In what follows, we directly calculate the LHS of \eqref{duality-prediction} (for a restricted class of external momenta) and obtain a finite answer in the limit $\lambda_f \rightarrow 1$ (when expressed in terms of $N_b$). The result can then be compared to a calculation the critical bosonic theory at $\lambda_b=0$ on the RHS and we find perfect agreement.

As described below, and in \cite{Aharony:2012nh, GurAri:2012is}, the non-critical bosonic theory is dual to a critical fermionic theory. We also compare the critical fermionic theory to the non-critical bosonic theory and find agreement.

\label{sec:normalisation}
\section{The Exact Ladder Diagram}
\label{sec:ladder}
\begin{figure}
\centering
\includegraphics[scale=0.25]{ladder.png}
\caption{\label{fig:i:1} The diagrammatic definition of the exact ladder diagram, which is the shaded box. All propagators in this diagram are exact propagators. In light cone gauge, this diagram is sufficient to calculate all planar correlation functions of single trace operators.}
\end{figure}

In this paper, we will exclusively use light-cone gauge\footnote{The conventions used in the following sections are those of \cite{Giombi:2011kc, Aharony:2012nh, GurAri:2012is}.   In particular, our light-cone gauge is defined in Euclidean space and described in detail in \cite{Giombi:2011kc}.  We also use the notation that $\gamma^{A}$ may be the $2\times2$ identity matrix $\mathbf 1$ or any of the three $\gamma^\mu$ and $p_s^2=p_1^2+p_2^2=2p_+p_-$.}.
Two crucial features of light-cone gauge are that ghosts decouple and cubic vertices are absent. Therefore any planar correlation function can be evaluated to all orders in $\lambda$ given the exact propagator, first evaluated in \cite{Giombi:2011kc}, and the "exact ladder diagram" defined diagrammatically in Figure \ref{fig:i:1}, which we denote  by $\Gamma_{AB}(p,q,r) \gamma^A \otimes \gamma^B$.

More precisely, we define $\Gamma_{AB}(p,q,r) \gamma^A \otimes \gamma^B$ as the following four-fermion interaction term in the quantum effective action for fermions, obtained after integrating out the gauge field in light cone gauge.
\begin{equation}
S_{eff} = -\frac{1}{2} \int \frac{d^3p d^3q d^3r}{(2\pi)^9} \Gamma_{AB}(p,q,r) \bar{\psi}^i(-p) \gamma^A \psi_j(r) \bar{\psi}^j(-r-q) \gamma^B \psi_i(p+q).
\end{equation}

$S(p)$, the exact large $N$ fermion propagator, valid to all orders in $\lambda$, is defined via $\langle \psi_i(p) \bar{\psi}^j(q) \rangle = \delta_i^j (2\pi)^3 \delta^3(p-q) S(p)$ and is given by  \cite{Giombi:2011kc}:
\begin{equation}
S(p)= \left({-i \gamma^\mu p_\mu + i \lambda^2 \gamma^+ p^- + \lambda p_s \over p^2}\right).
\end{equation}
The gauge propagator is defined via $\langle A_\mu(p) A_\nu(-q) \rangle = \delta^{ab} (2\pi)^3 \delta^3(p-q)  G_{\nu\mu}(p) $ and is given by
\begin{equation}
G_{+3}=\frac{4\pi i}{k} \frac{1}{p^+}.
\end{equation}
The self-consistent Schwinger-Dyson equation, given diagrammatically in Figure \ref{fig:i:1}, where all propagators are taken to be exact propagators $S(p)$, is:
\begin{eqnarray}
 && \Gamma_{AB}(p,q,r) \gamma^A \otimes \gamma^B  =
\left(\frac{-2\pi i}{k} \right) \frac{1}{(p-r)^+} \left (\gamma^+ \otimes \gamma^3 - \gamma^3 \otimes \gamma^+ \right) \nonumber \\ &&
+ \left(\frac{-2N\pi i}{k} \right) \int \frac{d^3l}{(2\pi)^3} \frac{1}{l^+} \Gamma_{CD}(l,q,r) \nonumber \\ && \times  \left( \gamma^+ S(l) \gamma^C \otimes \gamma^D S(l+q) \gamma^3 - \gamma^3 S(l) \gamma^C \otimes \gamma^D S(l+q) \gamma^+ \right).
\end{eqnarray}

In terms of $S(p) = S_A(p) \gamma^A$, we have:
\begin{eqnarray}
 && \Gamma_{AB}(p,q,r) \gamma^A \otimes \gamma^B  =
\left(\frac{-2\pi i}{k} \right) \frac{1}{(p-r)^+} \left (\gamma^+ \otimes \gamma^3 - \gamma^3 \otimes \gamma^+ \right) \nonumber \\
&&
- \left(2\lambda\pi i \right) \int \frac{d^3l}{(2\pi)^3} \frac{1}{l^+} \Gamma_{CD}(l,q,r)S_E(l) S_F(l+q) \nonumber \\
&& \times  \left(
\gamma^+  \gamma^F \gamma^C
\otimes
\gamma^D \gamma^E \gamma^3
-
\gamma^3 \gamma^F \gamma^C
\otimes
\gamma^D  \gamma^E \gamma^+
\right) \label{ladder}
\end{eqnarray}

\subsection{Rewriting the Schwinger-Dyson equation}
\label{rewrite}
There are 16 components of $\Gamma_{AB}$, which appear to be coupled. We now obtain an alternate expression for $\Gamma_{AB}$ that ``diagonalizes'' the Schwinger-Dyson equation \eqref{ladder} and shows that most of the 16 components are not independent.

For this purpose it is convenient to define:
\begin{equation}
4\pi i A_P{}^Q(p,q,r)\gamma^P = \Gamma_{AB}(p,q,r)\gamma^A \gamma^Q \gamma^B .
\label{Adef}
\end{equation}
It is easy to see that the inverse relation is
\begin{equation}
4\pi i \text{ Tr} \left(\gamma_{A} \gamma^P \gamma_{B} \gamma_Q\right) A_P{}^Q(p,q,r) = \Gamma_{AB}(p,q,r)
\label{Adef3}
\end{equation}
(where `Tr' denotes a trace over the gamma matrices.)

Let us also define, following \cite{Giombi:2011kc}
\begin{equation}
\label{contraction}
 H_+(Y) =  \gamma^3 Y \gamma^+ -\gamma^+ Y \gamma^3= 2( Y_I \gamma^+ - Y_- I).
\end{equation}
The Schwinger-Dyson equation \eqref{ladder} can be re-written as
\begin{eqnarray}
4\pi i A_P{}^Q(p,q,r) \gamma^P
 & = & \Gamma_{AB}(p,q,r) \gamma^A \gamma^Q \gamma^B \\
& = &
\left(\frac{-2\pi i}{k} \right) \frac{1}{(p-r)^+} \left (\gamma^+ \gamma^Q \gamma^3 - \gamma^3 \gamma^Q \gamma^+ \right) \nonumber
\\
&&
+ \left(\frac{-2N\pi i}{k} \right) \int \frac{d^3l}{(2\pi)^3} \frac{l}{l^+} \Gamma_{CD}(l,q,r)S_E(l) S_F(l+q) \nonumber
\\
&& \times  \left(
\gamma^+  \gamma^F \gamma^C
\gamma^Q
\gamma^D \gamma^E \gamma^3
-
\gamma^3 \gamma^F \gamma^C
\gamma^Q
\gamma^D  \gamma^E \gamma^+
\right)  \\
& = &
\left(\frac{2\pi i}{k} \right) \frac{1}{(p-r)^+} H_+(\gamma^Q) \nonumber
\\
&&
+ \left(\frac{2N\pi i}{k} \right) \int \frac{d^3l}{(2\pi)^3} \frac{1}{l^+} \Gamma_{CD}(l,q,r)S_E(l) S_F(l+q) \nonumber \times  \nonumber \\ &&
H_+(  \gamma^F \gamma^C
\gamma^Q
\gamma^D \gamma^E )
 \label{boot3}
\\
 A_P{}^Q(p,q,r) \gamma^P
& = &
\left(\frac{1}{2k} \right) \frac{1}{(p-r)^+} H_+(\gamma^Q) \nonumber
\\
&&
+  2\pi i \lambda   \int \frac{d^3l}{(2\pi)^3} \frac{1}{l^+} A_P{}^Q(l,q,r)S_E(l) S_F(l+q) \nonumber
 \times  \nonumber \\ &&
H_+(  \gamma^F
\gamma^P
 \gamma^E )
 \label{boot4}
\end{eqnarray}
Because $H_+$ contains only the identity and $\gamma^+$ components,this means that
\begin{equation}
 A_3{}^Q(p,q,r) = A_-{}^Q(p,q,r) = 0
\end{equation}
for all $Q$. This translates into $8$ linear equations relating various of the 16 components of $\Gamma_{AB}$.   Moreover, the 8 $A_+{}^Q$ and $A_I{}^Q$ are the only non-vanishing components of $A$, and they are independent for different values of $Q$. It is also consistent to set $A_P{}^3=0$ and $A_P{}^+=0$ so we have only 4 equations, which are 2 pairs of 2 coupled integral equations.

Evaluating \eqref{boot4} explicitly, we obtain:
\begin{eqnarray}
 A_+{}^Q(p,q,r) \gamma^+ + A_I{}^Q(p,q,r)
&=&
\left(\frac{1}{2k} \right) \frac{1}{(p-r)^+} H_+(\gamma^Q)
\nonumber
\\
&& \hspace{-2cm}
+ 2\pi i \lambda  \int \frac{d^3l}{(2\pi)^3} \frac{1}{l^+} A_+{}^Q(l,q,r)S_E(l)S_F(l+q) \nonumber
 \times
H_+(  \gamma^F
\gamma^+
 \gamma^E )
\nonumber
\\
&& \hspace{-2cm}
+ 2\pi i \lambda  \int \frac{d^3l}{(2\pi)^3} \frac{1}{l^+} A_I{}^Q(l,q,r)S_E(l)S_F(l+q)
 \times
H_+( \gamma^F
 \gamma^E ) \label{ieqn}
\end{eqnarray}

\subsection{Evaluating the exact ladder diagram when $q_\pm=0$}

We have not yet been able to solve this integral equation for arbitrary $q$. However, if we restrict ourselves to $q_\pm=0$ it is possible to obtain a solution, which will enable us to calculate the four-point function of scalar primaries for a restricted class external momenta.

To motivate our ansatz for the solution, we note that the results of section \ref{rewrite} can also be thought of diagrammatically as follows: Let $f^{(0)}(p,q,r)$ be any $2\times2$ matrix (with spinor indices) that is a function of $p$, $q$ and $r$. We think of $f^{(0)} \delta^i_j$ as representing an arbitrary "contraction" of the ladder diagram on the right, so that the tree level ladder diagram acting on $f^0 \delta^l_m$ is given by
\begin{equation}
\frac{N}{2} \gamma^\nu f^{(0)}(p,q,r) \gamma^\mu G_{\mu\nu}(p-r)\delta^i_j =\frac{N}{2}\delta^i_j G_{+3}(p-r)H_+\left(f^{(0)}(p,q,r)\right) \label{contr}
\end{equation}
as pictured in Figure \ref{fig:i:2}.
\begin{figure}[h]
\centering
\includegraphics[scale=0.15]{f0}
\caption{\label{fig:i:2} A contraction of the tree-level ladder diagram corresponding to   Equation \eqref{contr}. Note that there is no integration over momenta.}
\end{figure}

We then define $f^{(n)}(p,q,r)$ (which can be thought of as the ladder diagram with $n$ "rungs", contracted with $f^{(0)}$ on the right) recursively in terms of $f^{(n-1)}$:
\begin{equation}
\gamma^\nu f^{(n)}(p,q,r) \gamma^\mu G_{\mu\nu}(p-r)=  \frac{N}{2} \gamma^\nu \left(\int \frac{d^3l}{(2\pi)^3}  S(l)  \gamma^\sigma f^{(n-1)}(l,q,r)\gamma^\rho G_{\rho \sigma}(l-r) S(l+q) G_{\mu\nu}(p-l) \right) \gamma^\mu  \label{pre-ladder}
\end{equation}
as pictured in Figure \ref{fig:i:3}.
 \begin{figure}[h]
\centering
\includegraphics[scale=0.15]{fn}
\caption{\label{fig:i:3} The diagrammatic relation for $f^{(n)}$ in terms of $f^{(n-1)}$.}
\end{figure}

Because $G_{+3}=-G_{3+}$ are the only nonzero components of $G_{\mu\nu}$, only two components of $f^{(n-1)}$ contribute to $f^{(n)}$, which are
\begin{equation}
 H_+\left(f^{(n)} \right)=2f_I^{(n)} \gamma^+ - 2f_-^{(n)} \mathbb{I}.
\end{equation}
In terms of these variables, the equation \eqref{pre-ladder} is:
\begin{equation}
{H_+(f^{(n)}(p,q,r)) \over (p-r)^+ }
=  2\pi i \lambda\int \frac{d^3l}{(2\pi)^3}   H_+ \left( S(l)  H_+\left(f^{(n-1)}(l,q,r)\right)S(l+q) \right)\frac{1}{(l-r)^+ (p-l)^+}     \label{pre-ladder2}
\end{equation}

The infinite sum $\sum f^{(n)}$ is related to $A_P^Q$ defined in \eqref{Adef} of the previous subsection and $f_I^{(0)}$  and $f_-^{(0)}$ as follows:
\begin{equation}
\frac{\sum f^{(n)}_I}{k(p-r)^+}=A_+^If_I^{(0)}+A_+^-f_-^{(0)}, ~~~
 -\frac{\sum f^{(n)}_-}{k(p-r)^+}=A_I^I  f_I^{(0)}+A_I^-f_-^{(0)}.
 \label{eq:19}
 \end{equation}.

Let us choose $f^{(0)}$ (which is an arbitrary matrix) such that $f_-^{(0)}= - c f_I^{(0)} r^+$.  The parameter $c$ is arbitrary, and introduced for convenience: when we set $c=0$, equation \eqref{eq:19} determines $A_+^I$ and $A_I^I$ and when $c \rightarrow +\infty$, equation \eqref{eq:19} determines $A_+^-$ and $A_I^-$.

We note that it is consistent to assume $f^{(n)}(p,q,r)$ is independent of $p_3$. Evaluating equation \eqref{pre-ladder}, including the $l_3$ integral, we have:
 \begin{equation}
 \frac{f_I^n(p)}{(p-r)^+}=-2i \lambda  \int \frac{d^2l}{(2\pi )^2}\frac{f_I^{n-1}(l)\left(q_3+2i \lambda  l_s\right) l^+ - f_-^{n-1}(l)\left(2\left(1-\lambda ^2\right)l_s^2\right)}{\left(q_3^2+4l_s^2\right)l_s}\frac{1}{(p-l)^+(l-r)^+}
 \end{equation}
 \begin{equation}
 \hspace{-8 mm}
 \frac{f_-^n(p)}{(p-r)^+}=-2i \lambda  \int \frac{d^2l}{(2\pi )^2}\frac{f_I^{n-1}(l)2l^{+2}-f_-^{n-1}(l)\left(2i \lambda  l_s-q_3\right)l^+}{\left(q_3^2+4l_s^2\right)l_s}\frac{1}{(p-l)^+(l-r)^+}
 \end{equation}
After integrating to obtain the first few terms, we find $f_I^{(n)}(p)$ and $f_-^{(n)}(p)$ to be of the form:
 \begin{align}
 f_I^{(n)}(p)=a_1^{(n)}+a_2^{(n)} \frac{r^+}{p^+}  \label{eq:24a}    \\
 f_-^{(n)}(p)=-b_1^{(n)}p^+-b_2^{(n)} r^+
 \label{eq:24}
 \end{align}

We can sum the series to obtain:
\begin{equation}
\sum f_I^{(n)}(p)- f_I^{(0)} =
-2i \lambda  \int \frac{d^2l}{(2\pi )^2}\frac{\left(\sum f_I^n(l)\left(q_3+2i \lambda  l_s\right)l^+ -\sum f_-^n(l)\left(2\left(1-\lambda ^2\right)l_s^2\right)\right)}{\left(q_3^2+4l_s^2\right)l_s}\frac{(p-r)^+}{(p-l)^+(l-r)^+}
\label{eq:25}
\end{equation}
\begin{equation}
\sum f_-^{(n)}(p)- f_-^{(0)}=-2i \lambda  \int \frac{d^2l}{(2\pi )^2}\frac{\left(\sum f_I^{(n)}(l)2l^{+2}-\sum f_-^{(n)}(l)\left(2i \lambda  l_s-q_3\right)l^+\right)}{\left(q_3^2+4l_s^2\right)l_s}\frac{(p-r)^+}{(p-l)^+(l-r)^+}
\label{eq:26}
 \end{equation}
  From \eqref{eq:24a} and \eqref{eq:24}  we make the  ansatz
\begin{equation}
  \sum f_I^{(n)}(p)=\left(a_1+a_2\frac{r^+}{p^+}\right)f_I^{(0)}  \hspace{10 mm}       \sum f_-^{(n)}(p)=-\left(b_1p^++b_2r^+\right)f_I^{(0)}.
  \end{equation}
We now use the identity
   \begin{equation}
 {\frac{1}{(p-l)^+(l-r)^+}} ={\frac{1}{(p-r)^+}\left(\frac{1}{(p-l)^+}+ \frac{1}{(l-r)^+}\right)}
  \end{equation}
and, following \cite{Aharony:2012nh}, introduce the dimensionless variables:
\begin{align}
 y=\frac{2p_s}{|q_3|} \hspace{5 mm}
 x=\frac{2r_s}{|q_3|} \hspace{5 mm}
 t=\frac{2l_s}{|q_3|} \hspace{5 mm}
 \hat{\Lambda}=\frac{2\Lambda_s}{|q_3|} \hspace{5 mm}
 \hat{\lambda}=\lambda \text{ sign}(q_3)
 \label{eq:27}
\end{align}
 to carry out the angular integrals and rewrite \eqref{eq:25} and \eqref{eq:26}  as
 \begin{align}
  \left(\left(a_1+a_2\frac{r^+}{p^+}\right)-1\right) &=& \int_{x}^{y} \left(a_1\left(1+i \hat{\lambda } t \right)+\frac{b_1}{2}\left(1-\hat{\lambda}^2\right)t^2\right) \, dt+ \int_{x}^0
\left(a_2(1+i \lambda  t)+b_2\right) \, dt\nonumber \\ && +
  \frac{r^+}{p^+}\int_0^{y} \left(a_2\left(1+i \hat{\lambda  }t\right)+\frac{b_2}{2}\left(1-\hat{\lambda }^2\right)t^2\right) \, dt
  \label{eq:30}
 \end{align}
 \begin{align}
 \left(b_1p^++b_2r^+\right)-c r^+ =&&
 r^+\left(\int _{x}^{\Lambda }\left(2a_1+b_1\left(i \hat{\lambda } t-1\right)\right)dt \right.
  \left.+\int_{x}^{y} \left(2a_2+b_2\left(i \hat{\lambda } t-1\right)\right) \, dt \right)\nonumber \\ &&
  + p^+\int_{\Lambda }^{y} \left(2 a_1+b_1\left(i \hat{\lambda } t -1\right)\right) \, dt.
  \label{eq:31}
 \end{align}

To solve these coupled four-variable equations (which could in principle have been obtained directly from equation \eqref{ieqn} with the approporiate ansatz for $A_P^Q$) we differentiate and the resulting differential equation decouples into two sets of two variable coupled equations.

Equating coefficients of $\frac{r^+}{p^+}, p^+$ and $r^+$, we obtain:
\begin{align}
 && \frac{\partial a_1}{\partial y}=-\left(a_1\left(1+i \hat{\lambda } y\right)+\frac{b_1}{2}\left(1-\hat{\lambda }^2\right)y^2\right)
\nonumber \\ &&
 \frac{1}{2}\frac{\partial b_1}{\partial y}=\left(a_1+\frac{b_1}{2}\left(i \hat{\lambda } y-1\right)\right)
 \nonumber \\ &&
 \frac{\partial a_2}{\partial y}=-\left(a_2\left(1+i \hat{\lambda } y\right)+b_2\left(1-\hat{\lambda }^2\right)y^2\right)
 \nonumber \\ &&
 \frac{1}{2}\frac{\partial b_2}{\partial y}=\left(a_2+b_2\left(i \hat{\lambda } y-1\right)\right)
 \end{align}
The general solution of the set is
\begin{eqnarray}
a_i &=& \alpha _i\left(\beta _i(1-i \hat{\lambda}  y )-(1+i \hat{\lambda}  y)e^{-2i \hat{\lambda} \arctan(y)}\right)
\nonumber \\
 b_i&=&2\alpha_i\left(\beta _i+ e^{-2i \hat{\lambda} \arctan(y)}\right)
 \end{eqnarray}
 where the subscript $i=1,2$.
 Requiring that the solutions satisfy the integral equations \eqref{eq:30} and \eqref{eq:31} fixes the integration constants:
\begin{eqnarray}
 \beta_1 &=&-e^{-2i \hat{\lambda}\arctan[\hat{\Lambda}]} \nonumber \\  \beta_2 & = & 1
\nonumber \\
 \alpha _1 & = & \frac{\frac{c}{2}\left((1-i \hat{\lambda} x)-(1+i \hat{\lambda} x)e^{-2i \hat{\lambda}   \arctan[x]}\right)-\left(1+e^{-2i \hat{\lambda} \arctan[x]}\right)}{2e^{-2i \hat{\lambda}   \arctan[x]}\left(1+e^{-2i \hat{\lambda}   \arctan[\hat{\Lambda}]}\right)}
\nonumber \\
\alpha_2 &=&
\frac{\frac{c}{2}\left((1-i \hat{\lambda}  x)e^{-2i \hat{\lambda} \arctan[\hat{\Lambda}]}+(1+i \hat{\lambda}  x)e^{-2i \hat{\lambda} \arctan [x]}\right)
-\left(e^{-2i \hat{\lambda} \arctan[\hat{\Lambda}]}-e^{-2i \hat{\lambda} \arctan[x]}\right)}{2e^{-2i \hat{\lambda} \arctan[x]}\left(1+e^{-2i\hat{\lambda} \arctan(\hat{\Lambda})}\right)}.
\label{coeffc}
\end{eqnarray}

To determine $A_+^I$ and $A_I^I$ we set $c=0$ and to determine $A_+^-$ and $A_I^-$ we take the limit $c\rightarrow \infty$ (i.e., equate coefficients of $c$ on both sides of Equation \eqref{eq:19}). Writing the answers in the form:
$$\tilde{A}_C^B =2k e^{-2i \hat{\lambda} \arctan[x]}\left(1+e^{-2i\hat{\lambda} \arctan(\hat{\Lambda})}\right)(p-r)^+ A_C^B  $$
we have:
  \begin{eqnarray}
  \tilde{A}_I^I &=&\left(e^{-2i \hat{\lambda}   \arctan[\hat{\Lambda}]}-e^{-2i \hat{\lambda}   \arctan[y]}\right)\left(1+e^{-2i \hat{\lambda}   \arctan[x]}\right) \left(\frac{2p^+}{q_3}\right)
 \nonumber \\ && +
 \left(e^{-2i \hat{\lambda}   \arctan[x]}-e^{-2i \hat{\lambda}   \arctan[\hat{\Lambda}]}\right)\left(1+e^{-2i \hat{\lambda}   \arctan[y]}\right)\left(\frac{2r^+}{q_3}\right)
 \label{ladder1} \\
\tilde{A}_I^- &=& -\left(e^{-2i \hat{\lambda}   \arctan[\hat{\Lambda}]}-e^{-2i \hat{\lambda}   \arctan[y]}\right)\left(( i\hat{\lambda} x-1)+(1+ i\hat{\lambda} x)e^{-2i \hat{\lambda}   \arctan[x]}
\right)
\left(\frac{p^+}{r^+}\right)
\nonumber \\ && -
 \left((1+ i\hat{\lambda} x )e^{-2i \hat{\lambda}   \arctan[x]}-( i\hat{\lambda} x -1)e^{-2i \hat{\lambda}   \arctan[\hat{\Lambda}]}\right)\left(1+e^{-2i \hat{\lambda}  \arctan[y]}\right)
 \label{ladder2} \\
\tilde{A}_+^I &= & \left((1+ i\hat{\lambda} y )e^{-2i \hat{\lambda}   \arctan[y]}-( i\hat{\lambda} y-1)e^{-2i \hat{\lambda}  \arctan[\hat{\Lambda}]}\right)\left(1+e^{-2i
\hat{\lambda}   \arctan[x]}\right)
\nonumber \\ && +
\left(e^{-2i \hat{\lambda}   \arctan[\hat{\Lambda}]}-e^{-2i \hat{\lambda}   \arctan[x]}\right)\left(( i\hat{\lambda} y-1)+(1+ i\hat{\lambda} y)e^{-2i \hat{\lambda}   \arctan[y]}\right)\left(\frac{r^+}{p^+}\right)
\label{ladder3}\\
 \hspace{-3 mm}
\tilde{A}_+^- &=& -\left((1+ i\hat{\lambda} y )e^{-2i \hat{\lambda}   \arctan[y]}-( i\hat{\lambda} y-1)e^{-2i \hat{\lambda}   \arctan[\hat{\Lambda}] ]}\right)
 \nonumber \\ &&
 \times \left(\left((1+ i\hat{\lambda} x )e^{-2i \hat{\lambda}   \arctan[x]}+( i\hat{\lambda} x-1)\right)
\left(\frac{q_3}{2r^+}\right)\right)
 \nonumber \\ &&
 -
\left(( i\hat{\lambda} x-1)e^{-2i \hat{\lambda} \arctan[\hat{\Lambda}]}-(1+ i\hat{\lambda} x )e^{-2i \hat{\lambda}   \arctan[x]}\right)
  \nonumber \\ &&
\times \left(( i\hat{\lambda} y-1)+(1+ i\hat{\lambda} y )e^{-2i \hat{\lambda}   \arctan[y]}\right)\left(\frac{q_3}{2p^+}\right)
\label{ladder4}
\end{eqnarray}

In Appendix A, we evaluate the exact vertex for the scalar primary using this result.

\section{Four-Point Correlation Functions}

We now proceed to calculate the (gauge-invariant and parity-invariant) four-point function of the scalar primary. In this section, we evaluate the four-point function in the free fermionic theory, the interacting fermionic theory and the critical interacting fermionic theory. In section \ref{bosonic_theory_section}, we then evaluate the four-point function in the free and critical bosonic theories to test the duality. The four-point functions depend on external momenta $q^{(i)}$; as discussed in the previous section, our calculations are only valid in the special case of $q_\pm=0$ (i.e., only $q_3 \neq 0$). Hence, in what follows, we drop all spacetime-indices and label the four external momenta as $q_1$, $q_2$, $q_3$ and $q_4$.

\subsection{Free Fermionic Theory}

The free ($\lambda_f=0$) four-point function in the $U(N_f)$ fermionic theory is given by
\begin{eqnarray}
 F(q_1,q_2,q_3,q_4) & = & \langle J^{(0)}(-q_1)J^{(0)}(-q_2)J^{(0)}(-q_3)J^{(0)}(-q_4) \rangle \nonumber \\ &=&-N_f \int \frac{d^3k}{(2\pi )^3}\text{ Tr}
\left( S_0(k)S_0(k+q_1)S_0(k+q_1+q_2) S_0(k+q_1+q_2+q_3)\right)  \nonumber \\ && + \text{ permutations}
\end{eqnarray}
where $S_0(q)=(i\slashed q)^{-1}$ is the free fermion propagator and all external momenta are restricted in the 3 direction.

Evaluating this, we obtain
\begin{equation}
F(q_1,q_2,q_3,q_4) =N_f\frac{1}{8(q_1+q_2)(q_1+q_4)}(\mid q_1\mid-\mid q_2 \mid+\mid q_3 \mid-\mid q_4 \mid )+\text{5 permutations}
\label{eq:28}
\end{equation}
which can be simplified to:
\begin{equation}
F(q_1,q_2,q_3,q_4) =N_f\frac{q_1|q_1|+q_2|q_2|+q_3|q_3|+q_4|q_4|}{2(q_1+q_2)(q_1+q_3)(q_2+q_3)} \delta(q_1+q_2+q_3+q_4)
\label{eq:41}
\end{equation}

\subsection{Interacting Fermionic Theory}
\label{Interacting}
We now proceed to calculate the four-point function in the interacting theory.

In the interacting theory, our basic ingredients are the exact propagator, the exact $J^{(0)}$ vertex $V$ \cite{GurAri:2012is}, and the ladder diagram in section \ref{sec:ladder}. The correlator can be written as as a sum of Diagrams $A$, $B$ and $C$ in Figure \ref{fig:i:5} (where it is understood that all propagators are exact),
\begin{equation}
\langle J^{(0)}(p-q)J^{(0)}(-t)J^{(0)}(t-p)J^{(0)}(q) \rangle = [\text{A}]+[\text{B}] + [\text{C}].
\end{equation}
The diagrams $B$ and $C$ involve the ladder diagram.

Diagram $A$ is given by
\begin{align}
(\text{A}) = -N \int \frac{d^3k}{(2\pi)^3} \text{Tr}\left[V(p-q)S(k)V(-t)S(k+t)V(t-p)S(k+p)V(q)S(k+p-q)\right] \nonumber \\ + (5 \text{ permutations})
\label{eq:46}
\end{align}
and diagrams $B$ and $C$ are given by
\begin{align}
(\text{B}) & = & N \int \frac{d^3k}{(2\pi)^3} \text{Tr}[S(r+p)V(q) S(r-p+q)V(p-q) S(r)\gamma^\mu S(k)V(-t)S(k+t)V(t-p) \nonumber \\ && S(k+p)\gamma^\nu \Gamma_{\mu\nu}(k,p,r)] + (5 \text{ permutations}) \\
(\text{C}) & = & N \int \frac{d^3k}{(2\pi)^3} \text{Tr} [(S(r+p-q-t)V(p-q)S(r-t)V(-t)S(r)\gamma^\mu S(k)V(t-p)  S(k+p-t) \nonumber \\ && V(q) S(k+p-q-t)\gamma^\nu \Gamma_{\mu\nu}(k,p-q-t,r)] + (5 \text{ permutations})
\label{eq:47}
\end{align}
It is difficult to solve this integral in closed from for arbitrary $p$, $q$, and $t$. To obtain humanly readable answers that can be easily compared to the bosonic theory, we observe that the limit where two momenta ($ q\rightarrow  0^+ $ and $t\rightarrow  0^+ $ ) vanish is relatively tractable.

Let us first consider the diagrams where the two non-vanishing external momenta are ``diagonal" (as depicted in Figure \ref{fig:i:5} when $ q\rightarrow  0^+ $ and $t\rightarrow  0^+ $). In this limit, the integral is solvable. We find the 2 ``diagonal" permutations of diagram $A$ are given by
\begin{align}
 && -N \int \frac{d^3k}{(2\pi)^3} \text{Tr} [V(p-q)S(k)V(-t)S(k+t) V(t-p)S(k+p)V(q)S(k+q)] \nonumber
 \\&& =  -N \frac{\sec(\frac{\pi \lambda}{2})^2 (\pi \lambda + \sin(\pi \lambda))}{4( 2\pi \lambda |p|)}
 \label{eq:29}
 \end{align}
For diagrams $B$ and $C$, we use \eqref{eq:19}and \eqref{ladder1}-\eqref{ladder2} for the ladder expressions, and obtain
\begin{equation}
\text{2 permutations of B} =N \sec^3\left(\frac{\pi\lambda}{2}\right)\frac{(2 \pi\lambda \cos\left(\frac{\pi\lambda}{2}\right)-7 \sin \left(\frac{\pi\lambda}{2}\right) + \sin \left(\frac{3\pi\lambda}{2}\right))}
{32p \pi \lambda}
\label{eq:59}
\end{equation}
Adding \eqref{eq:29} and \eqref{eq:59} for B as well as C gives
\begin{equation}
-N\frac{2 \sec^2\left(\frac{\pi \lambda}{2}\right) \tan\left(\frac{\pi \lambda}{2}\right) }{4|p| \pi \lambda}
\label{eq:44}
\end{equation}

\begin{figure}
\centering
\includegraphics[scale=0.7]{fourcorrelator2}
\caption{\label{fig:i:5} Diagrams in the interacting theory.}
\end{figure}
The next step is to evaluate the remaining permutations of diagrams $A$, $B$ and $C$ with two adjacent non-vanishing external momenta (i.e., permutations of the external momenta not depicted in Figure $4$). These integrals are more nontrivial, but can be obtained by using the substitution $q \rightarrow p-q$ in \eqref{eq:46} and \eqref{eq:47} which evaluates to
\begin{equation}
-N \frac{1}{|p|}\left(\frac{\sec^2\left(\frac{\pi \lambda}{2}\right) \tan\left(\frac{\pi \lambda}{2}\right) }{4 \pi \lambda}+h(\lambda) \right)
\end{equation}
where $h(\lambda)$ is
\begin{align}
h(\lambda) = \frac{i}{4\left(1+e^{i \pi \lambda}\right)^2}
     \Big(1 + 2 e^{i \pi \lambda} \pi \lambda \cot \left(\pi \lambda \right) + \lambda \psi \left(\frac{1-\lambda}{2}\right)-\lambda \psi \left(\frac{-\lambda}{2}\right) + \nonumber \\
   e^{2i \pi \lambda} \left( 1 + \lambda \psi \left(\frac{\lambda}{2}\right) - \lambda \psi \left(\frac{1+\lambda}{2}\right)\right) \Big)
  \label{hresult}
\end{align}
and $\psi(x)=\frac{\Gamma'(x)}{\Gamma(x)}$ is the Digamma function.
It can be seen that the above equation goes to 0 for $\lambda \rightarrow 0$ and decreases approximately as $-\tan^2(\frac{\pi \lambda}{2})$. This property will be later be of use in the critical theory.

In summary, after adding all the diagrams we get the following result
\begin{equation}
\langle J^{(0)}(p-q)J^{(0)}(-t)J^{(0)}(t-p)J^{(0)}(q) \rangle=-N\frac{2 \sec^2\left(\frac{\pi \lambda}{2}\right) \tan\left(\frac{\pi \lambda}{2}\right) }{|p| \pi \lambda}-\frac{4 N h(\lambda)}{|p|}
\label{eq:45}
\end{equation}
\begin{figure}
\centering

\includegraphics[scale=0.8]{Hfunctiongraph.png}

\label{fig:fermion graphs}
\caption{ The plot of modulus of $H(p,\lambda)$ function and $\tan^3[\frac{\pi \lambda}{2}]$ vs $\lambda$ for momenta $p=1$. It is clear that the magnitude of $H(\lambda)$ rises slower than $ \tan^3[\frac{\pi \lambda}{2}] $.}
\end{figure}

It is worthwhile to notice that both terms in the R.H.S of \eqref{eq:45} are parity invariant.
We can draw a parallel between \eqref{eq:28} ,\eqref{eq:41} and  \eqref{eq:44} ,\eqref{eq:45} respectively.

It is natural to conjecture that this expression generalises to the following expression for general momenta:
\begin{align}
\langle J^{(0)}(-q_1)J^{(0)}(-q_2)J^{(0)}(-q_3)J^{(0)}(-q_4) \rangle=\frac{2\sec^2\left(\frac{\pi \lambda}{2}\right) \tan\left(\frac{\pi \lambda}{2}\right) }{ \pi \lambda}F(q_1,q_2,q_3,q_4)\nonumber \\ + H(q_1,q_2,q_3,q_4,\lambda)
\label{main_result}
\end{align}
where $F(q_1,q_2,q_3,q_4)$ is the four-point function of the scalar operator in the free fermionic theory and $H(q_1,q_2,q_3,q_4,\lambda)$ is an additional structure, which goes to zero as $\lambda \rightarrow 0$.

\subsection{Critical Fermionic Theory}
We now consider the four point function of the scalar primary in the critical fermionic theory described in \cite{Aharony:2012nh}, \cite{GurAri:2012is}, which is conjectured to be dual to the (non-critical) bosonic theory.

Let us briefly review the definition of the critical fermonic theory, which at zero-coupling is essentially the Gross-Neveu model (in three-dimensions): We introduce a field $\sigma$ (without a kinetic term) that couples to the scalar primary as $S_\sigma = \int d^3x \sigma \bar{\psi}\psi$ and perform a path integral over $\sigma$. The equation of motion for $\sigma$ is $\bar{\psi}\psi = 0$; therefore, instead of $\bar{\psi}\psi$, the single trace scalar primary operator in the critical theory is $\sigma$. Notice that, $\bar{\psi}\psi$ has scaling dimension 2, so $\sigma$  has scaling dimension $1$, which matches the scaling dimension of the scalar primary $J^{(0)}_b=\bar{\phi}\phi$ in the non-critical bosonic theory, as required for a duality. Because $\sigma$ has mass dimension $1$, there is the possibility of an additional marginal coupling of the form $\int d^3x N\frac{\lambda^F_6}{3!} \sigma^3$, which is related to the marginal $\phi^6$ coupling in the bosonic theory.

In the large $N$ limit, the exact two point function of $\sigma$ is clearly related the inverse of the two point function of $\bar{\psi}\psi$, via
\begin{equation}
G(q)=\langle \sigma(-q)\sigma \rangle = \left(-\langle J^{(0)}_f(-q) J^{(0)}_f \rangle_{\text{non-critical}} \right)^{-1}, \label{a}
\end{equation}
where two-point function of the scalar primary $J_f^{(0)}$ in the non-critical theory is
\begin{equation}
\langle J_f^{(0)}(-q)J_f^{(0)}\rangle=-\frac{N_f\tan \left(\frac{\pi \lambda_f}{2} \right)}{4\pi \lambda_f} |q|.
\end{equation}

\begin{figure}[h]\centering
\begin{tabular}{ccc}
\includegraphics[scale=0.4]{critical_fermionic_diagram_1.png} &
\includegraphics[scale=0.4]{critical_fermionic_diagram_2.png} &
\includegraphics[scale=0.4]{critical_fermionic_diagram_3.png}

\end{tabular}

\caption{ The diagrams that contribute to the four-point function of $\sigma$  in the critical fermionic theory. The shaded square is the exact scalar four-point function in the non-critical theory, the shaded triangle is the exact three-point function in the non-critical theory and the dashed line with a shaded circle is the $\langle \sigma\sigma \rangle$ propagator. The last diagram includes the contribution from the marginal $\lambda_6$ coupling, represented by black dot. (There is also another diagram with only one $\lambda_6$ vertex and one exact three point function that is not pictured.) \label{fig:critical_fermions}
}
\end{figure}

Directly using the result \eqref{main_result}, it is not hard to calculate the four-point function of the scalar primary operator $\sigma$ in the critical fermionic theory.
The diagrams that contribute to the four-point function of $\sigma$ are  shown in Figure \ref{fig:critical_fermions}.
\begin{align}
&&\langle \hat{\sigma}(-q_1)\hat{\sigma}(-q_2)\hat{\sigma}(-q_3)\hat{\sigma}(-q_4)\rangle=N^4 \tilde{\lambda}_{q.b}^4G(-q_1)G(-q_2)G(-q_3)G(-q_4)\nonumber \\ && (\langle  J^{(0)}(-q_1)J^{(0)}(-q_2)J^{(0)}(-q_3)J^{(0)}(-q_4) \rangle + \nonumber \\ && G(q_1+q_2)(\langle  J^{(0)}(-q_1)J^{(0)}(-q_2)J^{(0)}(q_1+q_2) \rangle \langle  J^{(0)}(-q_3)J^{(0)}(-q_4)J^{(0)}(q_3+q_4) \rangle + \nonumber \\ &&\lambda_6(\langle  J^{(0)}(-q_1)J^{(0)}(-q_2)J^{(0)}(q_1+q_2) \rangle +\langle  J^{(0)}(-q_3)J^{(0)}(-q_4)J^{(0)}(q_3+q_4) \rangle)+\lambda_{6}^2)) + \text{ permutations}
\label{eq:42}
\end{align}

The critical fermionic theory is dual to the non-critical bosonic theory in the limit $\lambda_f \rightarrow 1$ and $\lambda_6=0$. In this limit, there are two surviving terms in \eqref{eq:42}, the first two diagrams of Figure \ref{fig:critical_fermions}. Using \eqref{main_result} and the fermionic three point function \cite{GurAri:2012is} we find
\begin{align}
&\langle \hat{\sigma}(-q_1)\hat{\sigma}(-q_2)\hat{\sigma}(-q_3)\hat{\sigma}(-q_4)\rangle=-\frac{256 \pi^4 \lambda_{f}^4\cot^8 \left(\frac{\pi \lambda_f}{2} \right)}{|q_1||q_2||q_3||q_4|} \times \nonumber \\ &  \Big(\frac{2\sec^2\left(\frac{\pi \lambda_f}{2}\right) \tan \left(\frac{\pi \lambda_f}{2}\right) }{\pi \lambda}F(q_1,q_2,q_3,q_4) +H(q_1,q_2,q_3,q_4) \nonumber \\ & +\frac{N_f\tan^3 \left(\frac{\pi \lambda_f}{2}\right)}{\pi \lambda_{f}} \left(\frac{1}{|q_1+q_2|}+\frac{1}{|q_1+q_3|}+\frac{1}{|q_3+q_3|} \right) \Big)
\\ &
=-\frac{256 \pi^4 \lambda_{f}^4}{|q_1||q_2||q_3||q_4|} \cot^4 \left(\frac{\pi \lambda_{f}}{2} \right)\frac{|\lambda_b|}{|\lambda_f|}\left(F(q_1,q_2,q_3,q_4)+N_f\left(\frac{1}{2|q_1+q_2|}+\frac{1}{2|q_1+q_3|}+\frac{1}{2|q_3+q_3|}\right)\right)
\label{eq:43}
\end{align}
In the last step we have applied the limit $\lambda_f \rightarrow 1$ and used \eqref{duality} to express the answer in terms of $|\lambda_b|$, anticipating the comparison in the next section. We have also used the fact that $H(q_1,q_2,q_3,q_4) \sim \tan^2\left(\frac{\pi \lambda_f}{2}\right)$ to eliminate it from \eqref{eq:43}. Using equation \eqref{norm}, this expression can be compared to the four-point correlator of $\hat{J}^{(0)}_b$ in the bosonic theory which we calculate in the next section.

\section{Comparison to the Bosonic Theory}
\label{bosonic_theory_section}
\subsection{Non-Critical Bosonic Theory}
In the free theory, we have
\begin{align}
\langle J^{(0)}(-q_1)J^{(0)}(-q_2)J^{(0)}(-q_3)J^{(0)}(-q_4) \rangle &&
 \\&& \hspace{-30mm}= N\int \frac{d^3k}{(2\pi )^3}\frac{1}{k^2(k+q_1)^2(k+q_1+q_2)^2(k+q_1+q_2+q_3)^2}
 \label{eq:32}
 \end{align}
 We solve \eqref{eq:32} in the limit $q_\pm=0$. The integral evaluates to
 \begin{align}
&&\frac{N}{2}\Big(\frac{(q_2^2+q_3^2+q_1q_2+q_1q_3+q_2q_3)}{|q_1|q_2q_3(q_1+q_2+q_3)(q_1+q_2)(q_1+q_3)(q_2+q_3)}+  (q_1\rightarrow q_2,q_3) \nonumber \\ &&+ \frac{q_1q_2+q_1q_3+q_2q_3}{q_1q_2q_3|q_1+q_2+q_3|(q_1+q_2)(q_1+q_3)(q_2+q_3)}\nonumber \\ &&- \frac{(q_1+q_3)(q_2+q_3)}{(q_1q_2q_3(q_1+q_2+q_3)|(q_1+q_2)|(q_1+q_3)(q_2+q_3))}+(q_1 \leftrightarrow q_2 \leftrightarrow q_3))\Big)
\label{eq:33}
\end{align}
where the arrows in the numerator imply symmetric terms on replacing $q_1$ with $q_2$ and $q_3$.
\begin{figure}
\centering
\includegraphics[scale=0.7]{bosoncorrection}
\caption{\label{fig:i:7} Correction to the critical scalar four correlator.}
\end{figure}

Using the normalisation relation
\begin{equation}
\hat{\sigma}=-4 \pi \lambda_f \cot \left(\frac{\pi \lambda_f}{2} \right) J^{(0)}_b \label{norm}
\end{equation}
which follows from comparing the two point functions of the scalar primaries in both theories, it is easy to see that this result matches the R.H.S of \eqref{eq:43} -- i.e., that $$ \langle \sigma(-q_1) \sigma(-q_2) \sigma(-q_3) \sigma (-q_4)\rangle = \left(-4\pi \lambda_f \cot \left(\pi \lambda_f /2 \right) \right)^4 \langle J^0_b(-q_1) J^0_b(-q_2)  J^0_b(-q_3)  J^0_b(-q_4)  \rangle,$$ thereby verifying the duality between the critical fermionic theory and the non-critical bosonic theory (for our restricted choice of external momenta).

\subsection{Critical Bosonic Theory}
Next we turn to the four point correlator at the critical fixed point of the theory.  This is accomplished by adding a double trace term to the scalar action. The vertex is given by $-\frac{\lambda_4}{N}(\phi^\dagger \phi)^2$. We next flow to the IR limit with the IR scalar mass zero by tuning $\lambda_4 $ to infinity. The scalar propagator does not get a finite correction from this deformation. The divergent terms can be subtracted by a mass counterterm. Two and three point correlators in the critical theory were discussed in \cite{Aharony:2012nh, GurAri:2012is}.

The four point correlator receives a correction from the $\lambda_4$ deformation in two diagrams, one with a four point diagram Figure~\ref{fig:i:7}(A) and one with a double three-point diagram Figure ~\ref{fig:i:7}(B).
 Diagram ~\ref{fig:i:7}(B) can be evaluated as
 \begin{eqnarray}
&&
\langle J^{(0)}(-q_1)J^{(0)}(-q_2)J^{(0)}(-q_3)J^{(0)}(-q_4) \rangle_{\lambda_4}^B= \nonumber
\\ &&
\langle J^{(0)}(-q_1)J^{(0)}(-q_2)J^{(0)}\rangle_{\lambda_4=0}\langle J^{(0)}(-q_3)J^{(0)}(-q_4)J^{(0)}\rangle_{\lambda_4=0} \nonumber
\\ &\times &
  \prod \limits_{q=q_1,q_2,q_3,q_4,q_1+q_2} \sum \limits_{n=0}^{\infty} \left(-\frac{\lambda_4}{N_b}\langle J^{(0)}(-q)J^{(0)}\rangle_{\lambda_4=0}\right)^n \nonumber \\
  && + \text{ 3 permutations: } (q_1+q_2 )\rightarrow (q_1+q_3),(q_1+q_4)
\label{eq:34}
 \end{eqnarray}
 The values of $\langle J^{(0)}(-q_1)J^{(0)}\rangle_{\lambda_4=0}$ and $\langle J^{(0)}(-q_1)J^{(0)}(-q_2)J^{(0)}\rangle_{\lambda_4=0}$ are given in  \cite{Aharony:2012nh, GurAri:2012is}. We are interested in the free scalar theory hence we take the momentum dependence as

 \begin{gather}
 \langle J^{(0)}(-q_1)J^{(0)}\rangle_{\lambda_4=0}=N_b\frac{1}{8|q|}
 \label{eq:35}
 \\\langle J^{(0)}(-q_1)J^{(0)}(-q_2)J^{(0)}(q_1+q_2)\rangle_{\lambda_4=0}=N_b\frac{1}{4|q_1||q_2||q_1+q_2|}
 \label{eq:36}
 \end{gather}

 \eqref{eq:35} and \eqref{eq:36}can be substituted in \eqref{eq:34} to give
 \begin{align}
 \hspace{-5mm}\frac{-N_b\lambda_4}{8|q_1||q_2||q_3|q_4||q_1+q_2|^2} \prod \limits_{q_1,q_2,q_3,q_4,q_1+q_2}\frac{8|q_1|}{\lambda_4 \left(1+\frac{16|q_1|}{\lambda_4}\right)}+\left((q_1+q_2 )\rightarrow (q_1+q_3),(q_1+q_4)\right)
 \label{eq:37}
 \end{align}
 We define $\tilde{J}^{(0)}=\lambda_4J^{(0)}$ as the scalar operator at the critical fixed point. In the IR limit, taking $ \lambda_4\rightarrow \infty$ expanding the denominator and keeping the leading term we get
 \begin{equation}
 \langle\tilde{J}^{(0)}(-q_1)\tilde{J}^{(0)}(-q_2)\tilde{J}^{(0)}(-q_3)\tilde{J}^{(0)}(q_1+q_2+q_3)\rangle_{\lambda_4}^B=-\frac{8^4N_b}{2}\left(\frac{1}{|q_1+q_2|}+\frac{1}{|q_1+q_3|}+\frac{1}{|q_2+q_3|}\right)
 \label{eq:38}
 \end{equation}
Figure  ~\ref{fig:i:7}(A) turns out to be
  \begin{eqnarray}
&& \hspace{-5mm}\langle J^{(0)}(-q_1)J^{(0)}(-q_2)J^{(0)}(-q_3)J^{(0)}(-q_4)\rangle_{\lambda_4}^A  \nonumber \\ && \hspace{-5mm}
= \langle J^{(0)}(-q_1)J^{(0)}(-q_2)J^{(0)}(-q_3)J^{(0)}(-q_4)\rangle_{\lambda_{4=0}} \hspace{-.5cm} \prod \limits_{q=q_1,q_2,q_3,q_4}\sum \limits_{n=0}^{\infty} (-\frac{\lambda_4}{N_b}\langle J^{(0)}(-q)J^{(0)}\rangle)^n
 \label{eq:39}
  \end{eqnarray}
 We can use \eqref{eq:33} and use the same methods to get
 \begin{eqnarray}
 &&\langle\tilde{J}^{(0)}(-q_1)\tilde{J}^{(0)}(-q_2)\tilde{J}^{(0)}(-q_3)\tilde{J}^{(0)}(q_1+q_2+q_3)\rangle_{\lambda_4}^A=8^4N_b \nonumber \\ &&
 (\text{sign}[q_1](q_2^2+q_3^2+q_1q_2+q_1q_3+q_2q_3)+  (q_1\rightarrow q_2,q_3))+(\text{sign}[q_1+q_2+q_3](q_1q_2+q_1q_3+q_2q_3))\nonumber \\ && \frac{-(\text{sign}[q_1+q_2](q_1+q_3)(q_2+q_3)+(q_1 \leftrightarrow q_2 \leftrightarrow q_3))}{(q_1+q_2)(q_1+q_3)(q_2+q_3)}\text{sign}[(q_1q_2q_3(q_1+q_2)]
 \label{eq:40}
 \end{eqnarray}

 Adding \eqref{eq:38} and \eqref{eq:40} to obtain $\langle \tilde J^{(0)}(-q_1) J^{(0)}(-q_2)J^{(0)}(-q_3)J^{(0)}(-q_4) \rangle $, it is easy to see that the momentum dependence equates to the non-critical free fermion correlator \eqref{eq:41}.
Employing normalisation $\tilde{J}_b^{(0)} = 4\pi \lambda_b J^{(0)}_f$ obtained in Section \ref{sec:normalisation} and applying the limit $\lambda_f \rightarrow 1$ to \eqref{main_result} we find
 \begin{equation}
\langle \tilde{J}_b^{(0)}(-q_1) \tilde{J}_b^{(0)}(-q_2)\tilde{J}_b^{(0)}(-q_3)\tilde{J}_b^{(0)}(-q_4) \rangle = (4\pi\lambda_b)^4 \langle J_f^{(0)}(-q_1) J_f^{(0)}(-q_2)J_f^{(0)}(-q_3)J_f^{(0)}(-q_4) \rangle  \end{equation}
thereby verifying the duality between the critical bosonic theory and the non-critical fermionic theory for our restricted choice of external momenta.

\section{Discussion}

The main result of the paper is \eqref{main_result}, an explicit expression for the four-point function of the scalar primary in a particular limit of external momenta for both the non-critical fermionic theory. We also calculated the four-point function in the critical fermionic theory, and compared to critical and non-critical free bosons, providing an independent confirmation of the bosonization duality introduced in section \ref{sec:intro} at the level of four-point functions.

Our calculations crucially relied on the off-shell exact ladder diagram \eqref{ladder1}-\eqref{ladder4} together with \eqref{eq:19}. It is relatively straightforward to solve the resulting integral equations for the case when $q_\pm=0$. However, if we could generalise the calculation above to the case $q_\pm \neq 0$, we would be able to calculate four-point functions with arbitrary momenta. More importantly, the off-shell ladder diagram is also required for calculating $1/N$ corrections (and $M/N$  corrections in a bifundamental theory, see \cite{Guru:2014ad}) to all orders in $\lambda$. We hope to return to this off-shell ladder diagram in the future. We note that the \textit{on-shell} four point function is calculated to all orders in \cite{Jain:2014nza}, and its supersymmetric generalization \cite{Inbasekar:2015tsa}.

The theories we study possess a slightly-broken higher spin symmetry. In \cite{Maldacena:2011jn, Maldacena:2012sf}, two different classes of large $N$ field theories with a slightly broken higher spin symmetry were found to exist -- "quasi-boson" and "quasi-fermion" theories. The quasi-fermion theory depends on two parameters, $\tilde{N}$ and
$\tilde{\lambda}$.  The quasi-boson theories depend on three parameters, $\tilde{N}$ and
$\tilde{\lambda}$ and $a_3$. The first two parameters essentially correspond to the rank of the gauge group and the 't Hooft coupling $\lambda=\frac{N}{k}$ (in a microscopic description) and the third parameter corresponds to the $\phi^6$ triple-trace coupling which is exactly marginal in the large $N$ limit of the bosonic theory.

While three point functions in conformal field theories are severely constrained by purely kinematic considerations, four-point functions are determined only up-to an undetermined function of two conformal cross-ratios. In particular, conformal invariance restricts the four point function of scalars $J^0$, with scaling dimension $\Delta$ to the   form:
\begin{equation}
\langle J^0(x_1) J^0(x_2) J^0(x_3) J^0(x_4) \rangle = \frac{1}{x_{12}^{2\Delta} x_{23}^{2\Delta}}f(u,v)
\end{equation}
where $f$ is any function of the conformally invariant cross-ratios $u= \frac{x_{12}^2 x_{34}^2}{x_{13}^2 x_{24}^2}$ and $v = \frac{x_{14}^2 x_{32}^2}{x_{13}^2 x_{42}^2}$, satisfying $f(u,v)=\left(u/v\right)^\Delta f(v,u)$. However, dynamically, all correlation functions in a conformal field theory are (in principle) uniquely determined by the three-point functions (i.e., the operator-algebra) and scaling dimensions of the primary operators in the theory via bootstrap arguments.\footnote{We note that the planar four-point function calculated here is expected to receive contributions from the three-point functions of multi-trace as well as single-trace operators. Previous works have checked the duality at the level of planar three-point functions of single-trace operators but not multi-trace operators, so the calculation of this paper provides an independent check of the duality. We thank an anonymous referee for clarifying this point.} It would be interesting to study the four-point functions in theories with a slightly broken higher spin symmetry via bootstrap arguments. We hope the explicit results for Chern-Simons vector models here derived here would be useful in such a program.

\textbf{Acknowledgements:}
The authors thank Shiraz Minwalla and V. Umesh for reading a draft for the paper. SP acknowledges support of a DST Inspire Faculty Award.
SP would like to thank the Centre for High Energy Physics, IISc Bangalore and IISER Pune for hospitality when part of this work was completed and CMS, Durham University and the Rudolf Peierls Centre for Theoretical Physics, Oxford University for hospitality during the final stages of this work.

redpajama_set_name: RedPajamaArXiv


text:
This Is Anfield • Liverpool FC Features •
Top 10 Liverpool goals against Man United
There have been some magical strikes by Liverpool players against their most bitter rivals over the years, from long-range efforts to cleverly worked tap-ins. Henry Jackson (@OnFootballTweet) selects his top 10 Liverpool goals against Man United.
Dirk Kuyt celebrates scoring his hat-trick after scoring the third goal against Manchester United. (Photo by David Rawcliffe/Propaganda)
10. JIMMY CASE (1976/77 FA Cup Final)
Case's terrific turn and finish brought the Reds level in the 1977 FA Cup Final. While the day was to end in disappointment for them, the former midfielder's goal remains one of Liverpool's best in Wembley history.
9. FABIO AURELIO (2008/09 Premier League)
Some would say Andrea Dossena's goal was more memorable because it put the icing on the cake in Liverpool's incredible 4-1 win at Old Trafford back in 2009, but it was Aurelio's goal that was the most pivotal.
Nemanja Vidic had just been sent-off for hauling down Gerrard, and from the resulting free-kick the Brazilian bent a wonderful effort past Edwin van der Sar. It was his finest moment in a Reds shirt.
8. NIGEL CLOUGH (1993/94 Premier League)
With Liverpool 3-0 down to their bitter rivals, Clough's double inspired his side to an unlikely and unforgettable 3-3 draw at Anfield.
It was the midfielder's first goal that was both crucial and high in quality. He fired the ball home from 30 yards in front of the Kop to make it 3-1, bringing his side right back into the game in the process.
7. DANNY MURPHY (2001/02 Premier League)
There may have been more spectacular goals scored by Liverpool players at Old Trafford- Patrik Berger springs to mind- but for sheer importance Murphy's late winner back in 2002 has to make the list.
With the game heading towards a goalless draw, Steven Gerrard's wonderful pass found the midfielder, who caressed a clever finish over Fabien Barthez. It revitalised Liverpool's title challenge that season, and they eventually finished second; above United.
6. IAN RUSH (1992/93 Premier League)
Although the Reds eventually lost this game back in 1993, Rush's superb strike lives long in the memory.
Liverpool's all-time leading goalscorer was struggling to get into the side under Graeme Souness, but this terrific half-volley, past a motionless Schmeichel, showed him just what he was missing.
5. DIRK KUYT (2010/11 Premier League)
Kuyt may take the credit for scoring the goal, but the reason for its brilliance was the genius of Luis Suarez.
El Pistolero had only joined Liverpool a month or so previously, and he introduced himself in a big game with a mesmerising dribble past several United defenders before putting the ball on a plate for Kuyt to open the scoring.
4. ROBBIE FOWLER (1995/96 Premier League)
http://youtu.be/l8v08SP0Lnc
This was the day that Fowler overshadowed the much-talked about return of Eric Cantona from suspension.
'God' scored twice in an enthralling game that ended 2-2, but it was his first that was truly brilliant. With Ian Rush waiting in the middle for a cross, Fowler fired a magnificent effort past Peter Schmeichel to stun the hosts.
3. JAN MOLBY (1984/95 Milk Cup)
This forgotten gem from a Milk Cup clash at Anfield in 1984 was an unstoppable effort from the big Dane. Although not known for his mobility, Molby's long run and stunning shot remains one of the best goals ever scored against the Red Devils.
2. STEVEN GERRARD (2000/01 Premier League)
http://youtu.be/j0O-uBYEOLE
Gerrard was just 20 when he hit this thunderbolt in 2000, and it was further proof of the huge talent the now legendary midfielder possessed. It was the first of many goals against United for the skipper, and it remains comfortably his best.
1. JOHN ARNE RIISE (2001/02 Premier League)
It has to be number one, doesn't it? The Norwegian's unbelievable strike back in 2001 put the Reds 2-0 up, almost breaking the Anfield Road crossbar in the process. They went on to beat Sir Alex Ferguson's side 3-1 at Anfield, in what was a crucial win for Gerard Houllier's side.
What would be your number one? What goals did we miss?
QUIZ: Name the 18 Liverpool and Southampton double agents in Premier League era
Southampton dealt COVID-19 blow as goalkeeper ruled out vs. Liverpool
Tags Dirk Kuyt, Fabio Aurelio, Goals, Ian Rush, Jan Molby, John Arne Riise, Luis Suarez, Man United
"Enjoy the golden sky" – Liverpool legends pay tribute to the late Gerry Marsden
Birthday boy Milner & Reds aim for a record-breaking win at St. Mary's
Jurgen Klopp says Liverpool will make signings "if we can" this winter
Thiago 'in contention' for first Liverpool start since October vs. Southampton
Titles, Thiago & normality – Liverpool fans' top 2020 moments and New Year hopes
A quick guide to Sven Botman – The Dutch centre-back linked with Liverpool
How are Liverpool's 11 loanees performing halfway into 2020/21?
Centre-backs and leftfield signings – Debating Liverpool's January transfer window plans
Reds must arrest away record with vital win – Southampton vs. Liverpool Preview
Gini to return & Thiago time? – Predicting Liverpool's lineup vs. Southampton

redpajama_set_name: RedPajamaCommonCrawl


Pope Benedict XVI announced recently that he will be resigning his position as the leader of the Catholic Church. While a Pope's resignation is not common, it is common for leaders of businesses to resign their positions, either because of age, a sale of the business, a forced resignation by others with an interest in the business, or any number of other reasons. When these situations arise, it is imperative that the business follow the legal and corporate requirements for the changes to be effective and enforceable. If you need help in these or other corporate matters, you will be well-served by seeking the assistance of an experienced business law attorney. Call us at (480) 833-1113 and one of our experienced business lawyers will help you with whatever transition you are facing.


As the world is getting smaller, people travel across the world more frequently. Travel is no longer limited to one holiday per year or summer days at the beach. Fascinating, tropical destinations are more accessible than ever. People who love adventures often prefer spending their holiday at some exotic destinations.
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This is why resort wear options often include fun prints, accentuated accessories. You will also see more natural, breathable fabrics that allow you to relax and be comfortable in hot, humid climates. You can pop over here to get more information related to resort wear and what they are used for.


text:
Spaceflight Now +
Premium video content for our Spaceflight Now Plus subscribers.
Cassini preview
The Cassini spacecraft's arrival at Saturn is previewed in this detailed news conference from NASA Headquarters on June 3. (50min 01sec file)
Relive Cassini's launch
An Air Force Titan 4B rocket launches NASA's Cassini spacecraft at 4:43 a.m. October 15, 1997 from Cape Canaveral, Florida. (5min 15sec file)
Exploring the hills
"A brand new mission" is beginning for the Mars Exploration Rover Spirit as it nears the Columbia Hills as described in this presentation by science team member James Rice. (5min 57sec file)
Exploring Endurance
New pictures from the Mars rover Opportunity as it drives around the rim of Endurance Crater are presented with narration by science team member Wendy Calvin. (5min 25sec file)
Mars rover update
Mission officials and scientists discuss the condition and progress of Mars rovers Spirit and Opportunity plus the latest science news in this briefing from June 2. (40min 55sec file)
Options to save Hubble
NASA Administrator Sean O'Keefe announces plans to examine a robotic servicing mission to the Hubble Space Telescope. (33min 51sec file)
Station supply ship
Ride along with the Progress 14P resupply ship as it makes the final approach and docking to the International Space Station on May 27 as seen by a camera mounted on the craft's nose. (9min 02sec file)
Results from Spitzer
Scientists present new discoveries from the Spitzer Space Telescope, including their findings of raw ingredients for life detected around young stars. (53min 03sec file)
Spacewalk previewed
The Expedition 9 crew describes their upcoming spacewalk in Russian spacesuits, life aboard the space station and the view of Earth in this interview with Bill Harwood of CBS News. (20min 19sec file)
Progress undocking
The Progress 13P cargo ship departs the International Space Station on May 24 carrying trash and unneeded items to burn up in the atmosphere. (2min 56sec file)
AP interviews the crew
The Associated Press interviews the two-man Expedition 9 crew living aboard the International Space Station on May 24. (9min 36sec file)
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Pinwheel Galaxy's hidden wonders revealed
HARVARD-SMITHSONIAN CENTER FOR ASTROPHYSICS NEWS RELEASE
Like nosy neighbors, astronomers are spying on one of the nearest galaxies to our Milky Way. In studying the Pinwheel Galaxy, also known as Messier 33 (M33), they seek not malicious gossip but new knowledge as they search for clues to how galaxies like our own are born, live, and die.
At the 204th meeting of the American Astronomical Society in Denver, Colorado, astronomers from the University of Minnesota, the Harvard-Smithsonian Center for Astrophysics (CfA), and the University of Arizona unveiled new infrared images of M33 taken by NASA's Spitzer Space Telescope. The photos reveal features of the galaxy never before visible.
This Spitzer Space Telescope image of the Pinwheel Galaxy (M33) reveals a multitude of dusty red spiral arms studded with bright knots marking star formation regions. By studying this nearby galaxy in detail, astronomers can see "the Universe in a nutshell." Credit: E. Polomski et al. (University of Minnesota), G. Fazio et al. (SAO)
About 50,000 light-years across, the spiral galaxy M33 is half the diameter of the Milky Way. It lies 3 million light-years from the Milky Way, which places it among the Local Group of galaxies. Its nearness and viewing angle give astronomers an excellent opportunity to study M33's physical and chemical processes.
"With the Andromeda Galaxy, it's one of the two nearest large spiral galaxies comparable to the Milky Way. Since it's so close, we can get a nice panoramic view. It's a great object for detailed study," said Smithsonian astronomer Steven Willner (CfA).
"M33 is a gigantic laboratory where you can watch dust being created in novae and supernovae, being distributed in the winds of giant stars, and being reborn in new stars," said University of Minnesota researcher and lead author Elisha Polomski. By studying M33, "you can see the Universe in a nutshell."
Because it operates at infrared wavelengths, the Spitzer Space Telescope detects details hidden to the human eye and to telescopes that operate in visible light. Spitzer collects light at wavelengths measured in microns-millionths of a meter. The new pictures were taken in light at wavelengths ranging from 3.5 to 24 microns.
"At 3.5 microns, we see stars," said University of Minnesota astronomy professor Robert Gehrz, a member of the M33 observation team. "At eight microns, we see warm dust that's about 130 degrees Fahrenheit. At 24 microns, we're picking up cool dust that's between minus 100 and minus 190 degrees Fahrenheit." Spitzer's cameras also operate at 70 and 160 microns.
Observations of M33's cool components are expected to reveal much about the "metabolism" of galaxies. A galaxy is akin to a living body, in which food substances are broken down to build the body, and the waste and decomposition products of a body are recycled to feed new life. For example, the iron in Earth's core was forged in the bellies of large, luminous stars, and the heavier elements-all the way to uranium, the heaviest naturally occurring element-were created in supernova explosions. The deaths of those stars sprayed interstellar space with dust and gas, some of which clumped together in a disk that coalesced to form the sun and its planets.
The Spitzer team will examine the Pinwheel Galaxy in detail for the next two and a half years, studying the processes that circulate energy and chemical elements through the galaxy to build up, destroy, and recycle the building blocks of stars and planets. The researchers expect to identify new star-forming regions, red giant stars, novae and supernovae, thereby mapping out the evolutionary process of stars in M33 and comparing it to the process in our own Galaxy.
The NASA Jet Propulsion Laboratory (JPL) manages the Spitzer Space Telescope mission for NASA's Office of Space Science, Washington. Science operations are conducted at the Spitzer Science Center at the California Institute of Technology in Pasadena. JPL is a division of Caltech.
Headquartered in Cambridge, Mass., the Harvard-Smithsonian Center for Astrophysics (CfA) is a joint collaboration between the Smithsonian Astrophysical Observatory and the Harvard College Observatory. CfA scientists, organized into six research divisions, study the origin, evolution and ultimate fate of the universe.


text:
aside Why Democrats Are Viewing President's Proposal To End Shutdown As A Non Starter
SPEAKER NANCY PELOSI/ MAR LDR MITCH MCCONNELL
Thanks to Politico, Full text: Trump's shutdown offer to Dems
Here's a good analysis as to why Dems are viewing the republican President Donald Trump's 1/19/19 proposals to end the government shutdown as a non-starter…
On January 19, 2019, Dara Lind and Li Zhoa of VOX penned the following report, "Here's Trump's latest offer to end the shutdown-and why Democrats aren't interested" ("The "deal" Trump is offering on immigration and DACA, explained.")
"President Donald Trump just blinked on the government shutdown. But his standoff with congressional Democrats doesn't appear to be headed to a resolution anytime soon."
"On Saturday, in remarks billed as a "major announcement" on the border and the shutdown, Trump proposed a deal to Democrats. He continues to insist that any bill to reopen the government include billions of dollars for a physical barrier on the US-Mexico border — a "wall" — but is now open to such a bill including other immigration provisions as well."
"Most notably, he's open to extending existing protections for the 700,000 or so immigrants currently protected by the Deferred Action for Childhood Arrivals (DACA) program and the hundreds of thousands of immigrants who currently have legal status under Temporary Protected Status. The Trump administration has moved to sunset DACA, and to end protections for most of the immigrants covered under TPS. Both of those plans are currently held up in litigation."
SEN DURBIN WROTE DEMS' RESPONSE
"Democrats aren't particularly interested in what Trump's proposing. "Democrats were not consulted on this and have rejected similar overtures previously," a Democratic aide told Vox. "It's clearly a non-serious product of negotiations amongst White House staff to try to clean up messes the president created in the first place. POTUS is holding more people hostage for his wall."
"After weeks of all-or-nothing intransigence, Trump's announcement Saturday indicates that the White House realizes they're losing the shutdown in the eyes of most Americans, and are willing to compromise to reopen the government. But Democrats also know the White House is losing the shutdown, and the compromise now on offer is something they are unlikely to take."
DREAMERS/ REUTERS
What Trump's offering: $5.7 billion for the wall in exchange for extensions of existing protections for some immigrants
"Trump's pitching this as a compromise: He wants the wall, Democrats want to help DACA and TPS recipients. But the deal isn't the result of conversations with Democrats. It's reportedly the result of discussions that Vice President Mike Pence and presidential son-in-law Jared Kushner have had with congressional Republicans (most notably Sen. Lindsey Graham (R-SC))"
"And it shows. What Trump's offering — temporary extensions of existing protections for both groups of immigrants — isn't something that Democrats have been wildly enthusiastic about in the past. Furthermore, with Trump's efforts to strip existing protections held up in court, it's essentially a short extension of the status quo."
PENCE/ KUSHNER
"DACA recipients are currently being allowed to extend their protections for two years, just as they could under the Obama administration, while the administration fights in court to end the program. (People who don't already have protections are no longer allowed to apply.) Without knowing when the Supreme Court will rule — or how the Trump administration will proceed if the Supreme Court agrees they can end DACA, since their original plan (issuing no renewals for expirations after March 2018) is obviously moot — it's hard to say for sure that a three-year one-time extension will protect DACA recipients for longer than waiting for the Supreme Court."
Here's what he offered Saturday:
"$5.7 billion in funding for a physical barrier on the US-Mexico border. Trump's not budging on this. The White House has already "conceded" that the barrier will be made of steel poles — which is what experts and border agents wanted anyway — rather than solid concrete. Per a letter sent earlier this month, the administration could build 243 miles of barriers with the $5.7 billion it's requesting, most of which would be built in the Rio Grande Valley.'
"Three years of temporary protections for DACA recipients. On DACA, Trump is embracing a version of Graham's BRIDGE Act, which would extend DACA recipients' existing deportation protections and work permits for three more years. (The original BRIDGE Act applied also to immigrants who were eligible for DACA but not currently protected.) In theory, Congress would use that time to work out a permanent solution for DREAMers; but the last time the White House tried that, by giving Congress six months to address DACA before sunsetting it entirely, the gambit did not succeed. During that debate in late 2017 and early 2018, many Republicans gravitated toward bills that would offer DREAMers access to permanent legal status and ultimately to citizenship — a more moderate approach than what Trump is offering now."
A three-year extension of protections for TPS holders. Trump is also offering to extend ( for three years as well) the legal protections that hundreds of thousands of immigrants have under the Temporary Protected Status program — which is supposed to allow people to stay in the US while their countries recover from war or natural disasters, but which, over the years, has allowed many people to stay and put down roots in the US. TPS, unlike DACA, grants official legal status, but it doesn't offer any way to apply for a green card or citizenship. Trump's efforts to end TPS for most countries are held up in a different court fight — so this proposal, like the DACA proposal, would essentially be a legislative extension of the current judicially-imposed status quo.
$800 million to improve care for children and families at the border — with millions more for enforcement. The rest of Trump's proposal is a modified version of what the White House originally floated to Democrats in negotiations two weeks ago, codified in a letter sent by the Office of Management and Budget. Those demands include $800 million to deal with the actually-urgent humanitarian crisis at the US/Mexico border — the fact that unprecedented numbers of children and families are coming to the US (often to seek asylum) and border agents aren't equipped to deal with them. Trump's also demanding 2,750 more border agents and other law enforcement officials; millions of dollars in screening technology to detect drugs at ports of entry; and the hiring of 75 new immigration judges to address the immigration-court backlog, which is currently the biggest barrier to deporting people quickly (and which the current shutdown has exacerbated).
Modest changes to asylum for Central American children and teenagers. The Trump administration is floating allowing Central American children and teenagers to apply for asylum in their home countries — a modification of an Obama-administration program Trump ended in 2017. In return, they want to change current law to eliminate automatic court hearings for children and teens who come to the US from Central America and other countries — making it much easier to summarily deport them.
SENATPRS GRAHAM/ DURBIN
Trump is in a weakening position on the shutdown — and on immigration
"Trump could have proposed this deal at any time since before the government shut down; Graham has been pushing it for weeks. But as recently as Wednesday, Trump was telling reporters that he was waiting for Democrats to come back to the table to negotiate. And as recently as last week, Vice President Mike Pence told reporters that the president was firmly opposed, in particular, to any wall deal that addressed the DACA issue".
Link to entire report: Here's Trump's latest offer to end the shutdown — and why Democrats …
Bussiness & Consumer News
2019 US government shutdown
Dara Lind
Li Zhoa
US SW border wall
Previous As Per Axios, Democratic Leaders Have Already Nixed President's 1/19/19 Proposal To End Shutdown
Next Deja Vu/ What Happened With The October 2013 Gov Shutdown That lasted 16 Days
Scottie says:
Hello Gronda. Simply put the republicans do not want any immigration , want the ability to deport all immigrants regardless of age or status, and want to trade scraps for a whole meal. I agree this is not acceptable. The republicans seem to not understand we are part of an international community, and like it or not we can not stay shut up in our home alone. Hugs
Dear Scottie,
You've got the picture.
The president's base are now comprised of anti-immigration / racist leaning folks, Evangelicals who harbor antipathy towards LGBTQ AND transgender groups. Both these factions support his wall and anti-immigration rhetoric.
Then there are the rich corporate business folks who care nothing about any of this as long as they get their tax cuts, deregulation orders, etc. Many of these folks could give a hoot about the suffering of hardworking federal personnel going without pay/ or having their paychecks delayed.
They'll start caring when the US economy starts to take a serious hit because of it.
But in the business community, there is a split as most executives recognize the benefits of increasing legal immigration and many support fixing the plight of the DACA DREAMERS.
However a lot of the presidents's staff like Stephen Miller are opposed to even legal immigration.
The other split in the business community has to do with the president resorting to tariffs which sparks trade wars.
But there are some small government hardliners and one of them is the president's point man on the current government shutdown, the chief-of staff Mike Mulvaney. These guys look at the shutdown as a plus because it could lead to smaller government.
Hello Gronda. I agree. But on the small government idea, they love the idea of these people working without pay. I do not see how this is legal. They keep calling more people back in claiming they are essential, but all they really want is for no one to experience any hurt from the shut down, other than those not getting paid. Sounds like slavery to me, all being done to keep the white house poll numbers from tanking. Hugs
Lindi Roze says:
We are still waiting for his tax returns. This administration has also walked back on Russian sanctions… why should anyone believe the shut-down will end until he shows good faith negotiating by providing a bridge to end shutdown while negotiating. Bottomline DJT thinks this is like TV and doesn't care about the average American trying to keep his or her family with a roof over their head and food on the table.
Dear Lindi Roze,
It's extremely difficult to negotiate with a president whose idea of a deal is, I WIN/ YOU LOSE.
Sen McConnell wants to shift blame for this government shutdown on Dems by bringing up the President's Bridge Act which ends the shutdown to be voted on the Senate floor, this upcoming week of January 22, 2019.
He can't pass it without 7 democratic senators joining their GOP colleagues. I'm not sure he can get all GOP senators to back this bill. The chances of the president's bill being passed in the house is just about nil because it won't even have the support of the 'Freedom Caucus' GOP members/ the Tea Party crowd.
We'll see what happens.
Steve Naegele says:
trump on TV making a campaign speech to his followers is not a negotiation offer but an attempt to use public opinion to manipulate the country and appeal to his supporters. In order to negotiate you have to directly interect with the people with whom you are negotiating. Interact is a key word, not telling, not tweeting, not standing up onTV with a TV audience but face to face with people who will make the decisions.
if trump were serious to negotiate in good faith, he would meet in private with Democrats, have a comprehensive position which he can state verbally, have discusssion and find places where there can be sincere compromise.
But as the history of trump has shown, he has no negotiation skills, and his skills of lynig, bullying, threatening, making vile personal attacks on people with achievements he will never have, and making deals on which he reneges, are not the skills of successful good faith negotiators.
Democrats gave trump the opportunity twice, both times from the beginning trump did not have a sincere intention to negotiate in good faith but he used it as he used his TV show, a PR opportunity to rant and have a tantrum like a three year old, i.e. to appeal his supporters
Congress needs to pass a clean bill with no funding for the wall, tell trump if he vetos it, then Congress will override his veto, And Congress needs to very clearly tell trump he is never again to use a terrorist tactic such as attacking Americans and holding them as financial hostages.
Congress should be ready to do this because trump used a terrorist tactic to harm the people of the United States, and if not specifically told to not to do this again, he will continue to do it.
Still to trump supporters this is what makes a strong leader.
OG says:
That is true that congress should pass a bill and over ride the veto, but the two parties will not even speak up each other. They are a long way from coming up with a plan that two thirds of each chamber will agree one a long way.
Trump asked Nsncy how much dhe would agree to budget for the wall if he agree to do what dhe asked for. She said not anything. If she and Chuck will not give up an inch then that leaves it up to congress and the veto.
I urge everyone to vonact their congressmen and urge that they compromise with the other party and act.
Dear OG,
Any good negotiator would have taken whatever Speaker Pelosi said with a grain of salt but would have continued to talk to get a better feel for what might work. No good salesperson worth his/ her salt is going to be stopped by the word "NO" or they'd lose their jobs faster than a speeding bullet.
The president is supposed to be an expert negotiator but he's not living up to this reputation. What his response told me was that he's the one who did not want to deal. He was wanting to do this shutdown. He set her up for a no, she complied and then he was gone. He was never serious.
This does not mean that Speaker Pelosi doesn't present a counter offer to the president's proposal that he presented on 1/19/19.
She rejected the offer he made Saturday in advance snd has made no counter offer. Just crickets from her and Chuck. Reasonable congressmen will have to step forward and reach a compromise.
We are not dealing with what is supposed to be, we are dealing with what is.
There are reasons she outright rejected President Trump's proposals. One was due to fact that she and other Democratic Party leaders were not included in any discussions regarding his plans that he announced on 1/19/19.. This tells her that GOP leadership in the US Congress were trying to do an end run around her.
My thinking is that If his proposals didn't include the immediate ending to this 2019 government shutdown, this is not a deal being done in good faith. If she gives in to his demands, the president can hold American federal workers as hostage, over and over again every time he has a hissy fit about not getting his way. Everyone knows that you don't pay off a blackmailer or a hostage taker just one time. That doesn't mean that negotiations won't happen.
The chances of that Bridge Act passing both Houses of the US Congress is like me winning the lottery. I'm of the school to let the Senate Majority Leader McConnell bring this bill to the Senate Floor to be voted on this week of 1/22/19 as he has been instructed to do by his boss, President Trump.
When this plan blows up, as I strongly suspect will happen, then maybe President Trump will get serious about negotiating a deal. And yes, I'd love to be wrong about all of this, but both houses have not been able to pass any serious legislation pertaining to immigration in over a decade, and so, I'm not holding my breath.
Eventually, Speaker Pelosi will have to present a counter offer but it would be wise of her not to base it on the Bridge Act being passed and signed by the president.. Whoever drafted this proposal was very naive about how this plan would be executed.
Dems in US Congress are acting in unison. But that doesn't mean that negotiations will not happen. The priority has to be end this shutdown asap that's for real and serious. But the president did act and in doing so, he opened the door to possibilities. Still, he has to be made to get that he cannot work around the leader Speaker Pelosi to undercut her and the other Democratic leaders in the US Congress, and that the chances for the Bridge Act actually being passed into law, is a long shot. It's a start. I'm praying that his overture will start the process to end the shutdown asap.
Its beyond me why anyone thinks trump is a "successful negotiator," his whole history is one bad deal after the other and then trump sending in his lawyers to to deal with the mess. His tv show presented him just like we see him now, his basic people skills are lying, threatening, bullying and tweeting personal insults at people. By 2000 no US banks would loan trump money because banks decided trusted could not be trusted—this at a time when almost anyone who walked in the door without dying could get a loan.
Secondly there is no offer from trump, a tv show with trump making a campaign speech is not an offer. trump met twice with Nancy Pelosi , I guess he thought if he acted like a three year old it might arouse her "mother instinct. "
trump is scared of Pelosi, he is afraid to negotiate with her, and now he hides behind McConnell's skirts. She is not at all obligated to make a "counter offer" until trump sits across the table and makes an offer to her, which he has not yet done.
Instead he runs to McConnell crying like a baby and McConnell says "you do not have be an adult, hide behind me and we will lie and bully our way forward." Maybe McConnell thinks trump is going to like him better,
but how can McConnell compete with Kim Jong un, who last week sent trump a letter in which he addressed trump four time as "Your Excellency" and trump is proudly carrying the letter around. Kim is willing to kill everyone in the USA , how would McConnell ever compete with that.?
Steve Naegele,
Dealing with President Trump can be an exercise designed to frustrate the most skilled negotiators, but unfortunately, he's the only game in town as the US Senate Majority Leader Mitch McConnell refuses to let the senators vote on any bill to end this 2019 government shutdown. Where the senators to vote, the shutdown would end immediately with probably enough votes to override a presidential veto.
Nine republican senators have crafted a law to end the possibility of future shutdowns which is a signal that there are republican as well as democratic senators who want to end this shutdown.
The history of the president wanting to force a government shutdown over inadequate funding for his wall is telling.
In 2017, 2018 there were Republican Party majorities in both houses of the US Congress where 3 out of 4 Republican Party members discounted efficacy of wall. This is why they had never voted for a budget more than $1.6 billion dollars for the president's wall.
In Feb 2018, President Trump's administration's own budget was published for the fiscal yr. 2019 where on pg 57, the amount of $1.6 billion is allotted for the president's border wall. On Dec. 18. 2019, the US senators with the president's prior approval, passed a funding bill which included this $1.6 billion before he decided to throw a hissy fit to gum up the works. Based on some negative feedback from right wing media pundits, he decided to demand $5.7 billion dollars before he would agree to sign onto this budget. He wanted to cater to the anti-immigration hard-liners' faction of his base. While they are a minority of the US population, they are a significant portion of the Republican Party's base of voters.
President Trump had wanted his shutdown in 2017. In 2018 he got a 3 day shutdown where the Democratic Party lawmakers were pushing for a way to bar the deportation of about 700,000 young adults who had come to the USA with undocumented parents to where they knew no other country. The former Democratic President Barack Obama had issued an order termed DACA (Deferred Action on Childhood Arrivals), to block their deportation. President Trump had backed out on working with democrats on a law called the Dream Act, to permanently protect these young peoples from being deported. The Democratic Party lawmakers agreed to end the 3-day 2018 weekend shutdown with the promise by republicans that serious negotiations would continue regarding these young adults which went nowhere. Fortunately, the US courts stepped in to bar President Trump from being able to deport them.
But in 2019 President Trump finally got his record breaking government shutdown, surpassing the US past record of 21 days with no end in eight. That he's been wanting a shutdown over his wall being inadequately funded by US Congress in 2017, 2018 , when Republican Party legislators were in the majority position in both houses of the US Congress, is NO SECRET. But he had been talked out of it, as he was told that it would be republicans who would be blamed by the American voters. Even then, the Republican Party Congressional members had never budgeted for more than $1.6 billion dollars for his wall.
But in 2019, President Trump had miscalculated that he could shift the blame for his government shutdown to the democrats who were now in the majority position in the the lower house (US House of Representatives) DEMS, to where he could demand $5.7 billion for his wall that even his own party members would not agree to in 2017, 2018, when they were in the majority beyond the amount of $1.6 billion dollars.
To add salt to injury, President Trump's point man on this government shutdown, his Chief-of-Staff Mick Mulvaney is one of those small government zealots who's not opposed to this shutdown lasting as long as possible because that could lead to the development of a smaller government. In addition, he had been one of the lawmakers who had pushed for the 16 day shutdown in 2013.
All of this brouhaha over the wall is based on the president falsely alleging that there is a US national security crisis at the US SW border where criminals, jihadists are crossing the US Southern border in huge numbers which justifies his demand for additional funds which would build about 200 miles of a wall which falls short of the 700 plus miles that he's been describing to his supporters..
Now, it is true that there is a humanitarian crisis at the border because of the president's own doing.
This is the history of what the Democratic Party leaders have to face while trying to act / negotiate to end this government shutdown asap, as it is harming so many Americans. Giving into the president's unreasonable demands would make the US federal government employees vulnerable to government shutdowns, over and over again, every time President Trump threw another hissy fit.
But try, they must as the goal should be to end the hurt being felt by so many public servants who are being deprived of their paychecks or having them delayed versus the the need to give President Trump a black eye, no matter how much he deserves it.
totally agree with your assessment. Love the clean bill without the wall and take it from there.


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The new small displacement adventure bike from KTM is reaching the end of the development program as the 390 Adventure is caught testing in India.
KTM doesn't hide that one of their new models for 2020 is the 390 Adventure, and a new spy shot shows that this new adventure model is almost ready to start production and reach KTM dealerships by the end of 2019 or early 2020.
According to the photo published by Italian website Insella.it, the bike featured in the photo is the final test mule that KTM is using to fine-tune the last details in India, but almost everything seen on the photo will be used on the final production version.
This 100% adventure bike features higher screen for extra wind protection (maybe adjustable in height), the seat is almost flat and high just like a true enduro bike, and according to the same Italian website the instrument panel, besides the usual information, includes an altimeter and a compass so that even when riding off-road the rider knows the right way to the destination.
With a design that mimics the big adventure bikes from KTM, the 390 Duke will use the same single-cylinder engine with 373,2cc displacement, the exact same engine that we already know from the 390 Duke naked bike. That means the 390 Adventure offers 43 hp and 37 Nm of torque.
The suspension is sourced from WP specialists, but the wheel travel will be much bigger than on the 390 Duke to allow the 390 Adventure to be ridden off-road without any problem. With that in mind, KTM also decided to go for mixed tires fitted to cast wheels, while the brakes get ABS as standard with a special off-road mode.
According to the European press, the new KTM 390 Adventure should get a 5700 euros price tag, which translates to something around RM26.687 at current exchange rates.
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text:
Q: print double with variable number of decimals Sometimes I need double to be printed with 1 decimal (when it has the same value as integer), sometimes I need double to be printed with 6 decimals.
For example:
0.000 should be printed as 0.0
1.000 should be printed as 1.0
1.200 should be printed as 1.200000
1.123456789 should be printed as 1.234568

How can I do this?
Commentary on possible duplicate
I am not sure my question is a duplicate of the other post. I am interested in a VARIABLE number of decimals. That is not touched upon there, as far as I can tell.

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