## LATEX

Noncommutative geometry

We consider Euclidean ${\bf AdS}^2$ (a pseudosphere in Minkowski spacetime  with metric $\eta=(-1,+1,+1)$) given by the embedding relation
\begin{eqnarray}
-X_1^2+X_2^2+X_{3}^2=-R^2.\label{emb}
\end{eqnarray}
The Poincare coordinates $(t,z)$ and  the canonical coordinates $(x,y)$ are defined by the metric
\begin{eqnarray}
ds^2&=&\frac{R^2}{z^2}(dz^2+dt^2)\nonumber\\
&=&R^2dx^2+(dy-ydx)^2.\label{metric}
\end{eqnarray}
The relation between $(t,z)$ and $(x,y)$ is given explicitly by
\begin{eqnarray}
z=R\exp(-x)~,~t=\exp(-x)y.
\end{eqnarray}
$z$ is the radial coordinate of ${\bf AdS}^2$ whereas $u=1/z$ is the energy scale of the conformal field theory   ${\bf CFT}_1$ living on its boundary at $z\longrightarrow 0$ or equivalently $u\longrightarrow\infty$ which is a timelike curve parameterized by $t$. In fact, in the case of ${\bf AdS}^2$we have actually two disconnected one-dimensional boundaries located at $u\longrightarrow\pm \infty$.

The explicit relation between the global  embedding coordinates $X_a$ and the local coordinates $(x,y)$ is given by
\begin{eqnarray}
X_3&=&\frac{1}{2R}e^{-x}y^2+\frac{R}{2}e^{-x}-\frac{R}{2}e^x\nonumber\\
X_2&=&y\nonumber\\
X_1&=&\frac{1}{2R}e^{-x}y^2+\frac{R}{2}e^{-x}+\frac{R}{2}e^x.\label{co}
\end{eqnarray}
The radius $z=1/u$ is given in terms of the  global  embedding coordinates $X_a$ by the relation
\begin{eqnarray}
&&u=\frac{1}{z}=\frac{X_1-X_3}{R^2}.
\end{eqnarray}
The generators of the $SO(1,2)$ isometry group of Euclidean ${\bf AdS}^{2}$ are given by

\begin{eqnarray}
iK^1&=&-\frac{1}{R}e^{-x}y\partial_x-X_3\partial_y\nonumber\\
iK^2&=&\partial_x\nonumber\\
iK^3&=&\frac{1}{R}e^{-x}y\partial_x+X_1\partial_y.\label{Ks}
\end{eqnarray}
These are the generators of the conformal group on the boundary of Euclidean ${\bf AdS}^2$ which satisfy the algebra (with $f^{ab}~_c=\epsilon^{ab}~_c$)

\begin{eqnarray}
[K^a,K^b]=if^{ab}~_cK^c.\label{lie}
\end{eqnarray}
${\bf AdS}^2$ is the co-adjoint orbit $SO(1,2)/SO(2)$ which is a symplectic manifold and thus the canonical quantization of the corresponding Poisson structure, which is given by the inverse of the symplectic form on ${\bf AdS}^2$, produces the noncommutative ${\bf AdS}^2_{\theta}$.

See \cite{Ho:2000fy,Ho:2000br,Jurman:2013ota,Pinzul:2017}.

The noncommutative ${\bf AdS}^2_{\theta}$ is given by the embedding relation
\begin{eqnarray}
-\hat{X}_1^2+\hat{X}_2^2+\hat{X}_{3}^2=-R^2.\label{well1}
\end{eqnarray}
This is a noncommutative space in which the coordinate operators $\hat{X}^a$ are found to satisfy the commutation relations
\begin{eqnarray}
[\hat{X}^a,\hat{X}^b]=i\kappa f^{ab}~_c\hat{X}^c.\label{well2}
\end{eqnarray}
More explicitly, the coordinate operators $\hat{X}^a$ are given by
\begin{eqnarray}
\hat{X}^a=\kappa J^a.
\end{eqnarray}
The $J^a$ are the generators of the Lie group $SO(1,2)=SU(1,1)$ in the irreducible representations of the Lie algebra (\ref{lie}) given by the discrete series $D_j^{\pm}$ with $j=\{1/2,1,2/3,2,3/2,...\}$ (we will only consider integer values of $j$) \cite{barg}. These  are infinite dimensional unitary irreducible representations corresponding to the lowest  and highest weight states given respectively by the Hilbert spaces

\begin{eqnarray}
{\cal H}_j^+=\{|jm\rangle; m=j,j+1,j+2,...\}.
\end{eqnarray}
\begin{eqnarray}
{\cal H}_j^-=\{|jm\rangle; m=-j,-j-1,-j-2,...\}.
\end{eqnarray}
The Casimir is given by $C=-J_1^2+J_2^2+J_3^2=-j(j-1).{\bf 1}$ whereas the action of the generators $J^1$ and  $J^{\pm}=J^{2}\pm iJ^3$ is given by

\begin{eqnarray}
&&J^1|jm\rangle =m|jm\rangle\nonumber\\
&&J^+|jm\rangle=\sqrt{m(m+1)-j(j-1)}|jm+1\rangle\nonumber\\
&&J^-|km\rangle=\sqrt{m(m-1)-j(j-1)}|km-1\rangle.
\end{eqnarray}
The requirement that the Casimir operator must be negative removes the integer value $j=1$ and thus  $j=\{2,3,...\}$. Indeed, the   integer $j$ is such that \cite{Pinzul:2017wch}
\begin{eqnarray}
\frac{R^2}{\kappa^2}=j(j-1).
\end{eqnarray}
The commutative limit $\kappa\longrightarrow 0$ corresponds therefore to $j\longrightarrow\infty$.

The radius operator is obviously defined by
\begin{eqnarray}
&&\hat{u}=\frac{1}{\hat{z}}=\frac{\hat{X}_1-\hat{X}_3}{R^2}.
\end{eqnarray}
We compute immediately the expectation value $\langle jm|\hat{u}|jm\rangle=-\kappa m/R^2$ which approaches $\mp \infty$ (corresponding to the two boundaries of noncommutative ${\bf AdS}^2_{\theta}$) as $m\longrightarrow \pm\infty$ (corresponding to the two representations $D_j^{\pm}$). Indeed, in the representation $D_j^+$ and $D_j^-$ we find as eigenvalues of $\hat{u}$ the commutative values $u\leq 0$ and $u\geq 0$ respectively. Hence, near the boundaries we have very large values of $m$ \cite{Pinzul:2017wch}.

The operators $\hat{x}$ and $\hat{y}$ associated with the canonical coordinates $x$ and $y$ are found to satisfy the canonical commutation relation
\begin{eqnarray}
[\hat{x},\hat{y}]=i\kappa.\label{MW}
\end{eqnarray}
This structure allows us to introduce a Weyl map \cite{weyl} and a Moyal-Weyl star product \cite{Moyal:1949skv2,Groenewold:1946kpv2} in the usual way.

The weyl map $\pi$ allows us to map operators $\hat{F}(\hat{x},\hat{y})$ in the Hilbert spaces ${\cal H}_j^{\pm}$ back to functions $\pi(\hat{F})(x,y)$ on the commutative ${\bf AdS}^2$ such that the operator product $\hat{F}.\hat{G}$ of two operators $\hat{F}$ and $\hat{G}$ is mapped to the star product $\pi(\hat{F})*\hat{\pi}(\hat{G})$ of the two corresponding functions $\pi(\hat{F})$ and $\hat{\pi}(\hat{G})$, namely
\begin{eqnarray}
\pi(\hat{F}.\hat{G}(\hat{x},\hat{y}))=\pi(\hat{F})*\hat{\pi}(\hat{G})(x,y).
\end{eqnarray}
By construction the map of the canonical operators $\hat{x}$ and $\hat{y}$ is precisely the canonical coordinates $x$ and $y$, viz
\begin{eqnarray}
\pi(\hat{x})=x~,~\pi(\hat{y})=y.
\end{eqnarray}
The star product is given explicitly by  (with $(\bar{x}^1,\bar{x}^2)=(x,y)$ and $\bar{\theta}^{ab}=\kappa \epsilon^{ab}$)

\begin{eqnarray}
f\bar{*}g(\bar{x})&=&\exp(\frac{i}{2}\bar{\theta}^{ab}\frac{\partial}{\partial\xi^a}\frac{\partial}{\partial\eta^b})f(\bar{x}+\xi)g(\bar{x}+\eta)|_{\xi=\eta=0}.
\end{eqnarray}
Or equivalently

\begin{eqnarray}
f\bar{*}g(\bar{x})&=&f({x},y)\exp\big(\frac{i\kappa}{2}(\overleftarrow{{\partial}}_x\overrightarrow{{\partial}}_y-\overleftarrow{{\partial}}_y\overrightarrow{{\partial}}_x)\big)g({x},y).
\end{eqnarray}
In particular, we have for $f=f(x)$ the results

\begin{eqnarray}
f\bar{*}g(\bar{x})=f(x+\frac{i\kappa}{2}\overrightarrow{{\partial}}_y)g(x,y)~,~ g\bar{*}f(\bar{x})=g(x,y)f(x-\frac{i\kappa}{2}\overleftarrow{{\partial}}_y).\label{star1}
\end{eqnarray}
Similarly, we have for $g=g(y)$ the results

\begin{eqnarray}
g\bar{*}f(\bar{x})=g(y-\frac{i\kappa}{2}\overrightarrow{{\partial}}_x)f(x,y)~,~ f\bar{*}g(\bar{x})=f(x,y)g(y+\frac{i\kappa}{2}\overleftarrow{{\partial}}_x).\label{star2}
\end{eqnarray}

The relation between the embedding coordinate operators $\hat{X}_a$ and the canonical coordinate operators $\hat{x}$ and $\hat{y}$ is given from (\ref{co}) by the equations
\begin{eqnarray}
\hat{X}_3&=&\frac{1}{2R}\hat{y}e^{-\hat{x}}\hat{y}+\frac{R}{2}e^{-\hat{x}}-\frac{R}{2}e^{\hat{x}}\nonumber\\
\hat{X}_2&=&\hat{y}\nonumber\\
\hat{X}_1&=&\frac{1}{2R}\hat{y}e^{-\hat{x}}\hat{y}+\frac{R}{2}e^{-\hat{x}}+\frac{R}{2}e^{\hat{x}}.\label{ord}
\end{eqnarray}
It is then straightforward to check  (using $[\exp(\pm \hat{x}),\hat{y}]=\pm i\kappa\exp(\pm \hat{x})$) that
\begin{eqnarray}
[\hat{X}^3,\hat{X}^2]=i\kappa\hat{X}^1~,~[\hat{X}^1,\hat{X}^2]=i\kappa\hat{X}^3~,~[\hat{X}^3,\hat{X}^1]=i\kappa\hat{X}^2.
\end{eqnarray}
These are precisely equations (\ref{well2}). In the same manner we can verify equation (\ref{well1}). This shows explicitly that the operator ordering chosen in (\ref{ord}) is the correct one.

The derivation operators $\hat{\partial}_x$ and $\hat{\partial}_y$ associated with the canonical operators $\hat{x}$ and $\hat{y}$ are given immediately (since (\ref{MW}) defines a Moyal-Weyl plane) by
\begin{eqnarray}
\hat{\partial}_x=-\frac{1}{i\kappa}[\hat{y},]~,~\hat{\partial}_y=\frac{1}{i\kappa}[\hat{x},].
\end{eqnarray}
The derivation operator $\hat{K}^2$ associated with the coordinate operator $\hat{X}_2=\hat{y}$ is then given by
\begin{eqnarray}
i\hat{K}^2=\frac{i}{\kappa}[\hat{X}_2,]=\frac{i}{\kappa}[\hat{y},]=\hat{\partial}_x.
\end{eqnarray}
By construction this has the correct commutative limit $i\hat{K}^2\longrightarrow iK^2=\partial_x$. Similarly, the derivation operators $\hat{K}^1$ and $\hat{K}^3$ should be defined by the relations
\begin{eqnarray}
i\hat{K}^1=\frac{i}{\kappa}[\hat{X}_1,]~,~i\hat{K}^3=\frac{i}{\kappa}[\hat{X}_3,].
\end{eqnarray}
We have explicitly (where the arrow indicates the Weyl map)
\begin{eqnarray}
i\hat{K}^{1,3}(\hat{f})&=&\frac{i}{\kappa}[\hat{X}_{1,3},\hat{f}]\nonumber\\
&=& \frac{i}{\kappa}\bigg(\frac{1}{2R}[\hat{y}e^{-\hat{x}}\hat{y},\hat{f}]+\frac{R}{2}[e^{-\hat{x}},\hat{f}]\pm \frac{R}{2}[e^{\hat{x}},\hat{f}]\bigg)\nonumber\\
&\longrightarrow &\frac{i}{\kappa}\bigg(\frac{1}{2R}[{y}*e^{-{x}}*{y},{f}]_*+\frac{R}{2}[e^{-{x}},{f}]_*\pm \frac{R}{2}[e^{{x}},{f}]_*\bigg).
\end{eqnarray}
It is not difficult to check explicitly (by using (\ref{star1}) and (\ref{star2})) that this definition of $\hat{K}^1$ and $\hat{K}^3$ will tend in the commutative limit $\kappa\longrightarrow 0$ to the generators $K^1$ and $K^3$ given explicitly in equation (\ref{Ks}). In other words, the generators $\hat{K}^a$ are the noncommutative  ${\bf AdS}^2_{\theta}$ Killing vectors  in the same way that the generators $K^a$ are the commutative ${\bf AdS}^2$ Killing vectors.

Action and commutative limit

We are now in a position to write down an action principle for scalar fields $\hat{\Phi}$ on  noncommutative  ${\bf AdS}^2_{\theta}$ using the Killing vectors $\hat{K}^a$ as outer derivations. See in particular \cite{Pinzul:2017wch}. First we note the dimension of the various objects: $[t],[z]\sim R$, $[x]\sim 1$, $[y]\sim R$, $[\kappa]\sim R$ and $[X^a]\sim R$. In particualr note that $x$ is dimensionless and thus the noncommutativity parameter $\bar{\theta}^{ab}\equiv \kappa\epsilon^{ab}$ is of dimension length and not of dimension area. The scalar field $\hat{\Phi}$ in two dimension is canonically of dimension $0$. We can then write down immediately the free action
\begin{eqnarray}
S=\frac{2\pi R \kappa}{2}Tr\bigg(-\frac{1}{R^2\kappa^2}[\hat{X}^a,\hat{\Phi}][\hat{X}_a,\hat{\Phi}]+m^2\hat{\Phi}^2\bigg).\label{action}
\end{eqnarray}
The Weyl map takes the trace $Tr$ to the integral on ${\bf AdS}^2$ as follows (the metric $g$ in the canonical coordinates is read from the second line of (\ref{metric}) and it is clear that $\sqrt{{\rm det}g}=1$)
\begin{eqnarray}
2\pi Tr\longrightarrow \int \frac{dx dy}{\sqrt{{\rm det}\bar{\theta}}}=\int \frac{dx dy}{\kappa}.
\end{eqnarray}
Recall that the integral over a symplectic manifold ${\cal M}^{d}$, such as    ${\bf AdS}^2$, is given in terms of its symplectic structure $\omega$, which is here $\omega=-dx\wedge dy/\kappa$, by the formula (where $d=2n$)
\begin{eqnarray}
(2\pi)^nTr \longrightarrow \int \frac{\omega^n}{n!}=\int \frac{d^{2n}x}{\sqrt{{\rm det}\bar{\theta}}}.
\end{eqnarray}
In terms of the star product the above action reads then (where ${\cal X}^a=\pi(\hat{X}^a)$)
\begin{eqnarray}
S&=&\frac{R}{2}\int dx dy\bigg(-\frac{1}{R^2\kappa^2}[{\cal X}^a,\Phi]_**[{\cal X}_a,\Phi]_*+m^2\Phi*\Phi\bigg).\label{actionstar}
\end{eqnarray}
Recall that ${\bf AdS}^2$ is a manifold with two boundaries in which boundary terms play an important role. However, for simplicity we start by neglecting boundary terms. Thus, the integral over the star product of two functions $f$ and $g$ is equal (up to boundary terms)  to the integral over the ordinary product of these  two functions, viz
\begin{eqnarray}
\int dx dy f*g(x,y)=\int dx dy f(x,y).g(x,y)+{\rm boundary~ terms}.
\end{eqnarray}
By concentrating on the bulk fields for the moment we can use this equation to obtain the action
\begin{eqnarray}
S
&=&\frac{1}{2R\kappa^2}\int dx dy\bigg([{\cal X}^1,\Phi]_*^2-[{\cal X}^2,\Phi]_*^2-[{\cal X}^3,\Phi]_*^2+R^2\kappa^2m^2\Phi^2\bigg)\nonumber\\
&=& \frac{1}{2R\kappa^2}\int dx dy\bigg([y*e^{-x}*y,\Phi]_*[e^x,\Phi]_*+R^2[e^{-x},\Phi]_*[e^{x},\Phi]_*-[y,\Phi]_*^2+R^2\kappa^2m^2\Phi^2\bigg).
\end{eqnarray}

A relatively short calculation using (\ref{star1}) and (\ref{star2}) gives the results
\begin{eqnarray}
[y,\Phi]_*=-i\kappa\partial_x \Phi.
\end{eqnarray}
\begin{eqnarray}
[e^{\pm x},\Phi]_*=\pm i\kappa e^{\pm x}\Delta_y \Phi.
\end{eqnarray}
\begin{eqnarray}
[y*e^{-x}*y,\Phi]_*=-i\kappa e^{-x}\bigg[2yS_y\partial_x\Phi -\frac{\kappa^2}{4}\partial_x^2\Delta_y\Phi+y^2\Delta_y\Phi+\frac{\kappa^2}{4}\Delta_y\Phi\bigg].
\end{eqnarray}
The operators $\Delta_y$ and $S_y$ are given explicitly by
\begin{eqnarray}
\Delta_y=\frac{2}{\kappa}\sin \frac{\kappa}{2}\partial_y~,~S_y=\cos \frac{\kappa}{2}\partial_y.
\end{eqnarray}
We get then the action
\begin{eqnarray}
S
&=& \frac{1}{2R}\int dx dy\bigg((\partial_x\Phi)^2+(R^2+\frac{\kappa^2}{4}+y^2)(\Delta_y\Phi)^2+2y\Delta_y\Phi S_y\partial_x\phi -\frac{\kappa^2}{4}\Delta_y\Phi\partial_x^2\Delta_y\Phi+R^2m^2\Phi^2\bigg).
\end{eqnarray}
We use now the identity $1=S_y²+\kappa^2 \Delta_y^2/4$ or equivalently
\begin{eqnarray}
\int dxdy (\partial_x\Phi)^2&=&\int dx dy\partial_x\Phi S_y^2\partial_x\Phi+\frac{\kappa^2}{4}\int dx dy \partial_x\Phi\Delta_y^2\partial_x\Phi\nonumber\\
&=& \int dx dy (S_y\partial_x\Phi)^2-\frac{\kappa^2}{4}\int dx dy (\Delta_y\partial_x\Phi)^2.
\end{eqnarray}
In the last line we have neglected boundary terms and used the results
\begin{eqnarray}
&&\int dxdy f(x,y).\Delta_yg(x,y)=-\int dxdy \Delta_yf(x,y).g(x,y)+{\rm boundary~terms}\nonumber\\
&&\int dxdy f(x,y).S_yg(x,y)=\int dxdy S_yf(x,y).g(x,y)+{\rm boundary~terms}.\end{eqnarray}
The action then becomes

\begin{eqnarray}
S
&=& \frac{1}{2R}\int dx dy\bigg((S_y\partial_x\Phi+y\Delta_y\Phi)^2+(R^2+\frac{\kappa^2}{4})(\Delta_y\Phi)^2+R^2m^2\Phi^2\bigg).
\end{eqnarray}
The classical equation of motion follows directly from this action. We get (the operators $\partial_x$ and $S_y$ commute but the order of the operators $\Delta_y$ and $y$ matters)

\begin{eqnarray}
(\partial_xS_y+\Delta_yy)(S_y\partial_x+y\Delta_y)\Phi+(R^2+\frac{\kappa^2}{4})\Delta_y^2\Phi-R^2m^2\Phi=0.\label{eom}
\end{eqnarray}
In the commutative limit we have $\kappa\longrightarrow 0$ and thus $\Delta_y=\partial_y+O(\kappa^2)$, $S_y=1+O(\kappa^2)$. As a consequence the above action reduces to
\begin{eqnarray}
S
&=& \frac{1}{2R}\int dx dy\bigg((\partial_x\Phi+y\partial_y\Phi)^2+R^2(\partial_y\Phi)^2+R^2m^2\Phi^2\bigg).
\end{eqnarray}
This read in the Poincare coordinates

\begin{eqnarray}
S
&=& \frac{1}{2R}\int \frac{1}{R}\sqrt{g}dt dz\bigg(z^2(\partial_z\Phi)^2+z^2(\partial_t\Phi)^2+R^2m^2\Phi^2\bigg)\nonumber\\
&=& \frac{1}{2}\int \sqrt{g}d^2x\bigg((\partial_a\Phi)(\partial^a\Phi)+m^2\Phi^2\bigg).
\end{eqnarray}
This is the correct commutative action.

In the near-boundary limit $z\longrightarrow 0$ we have  $\Delta_y=\frac{z}{R}\partial_t+O(z^3)$ and $S_y=1+O(z^2)$ since $\partial_y=\frac{z}{R}\partial_t$ (recall that $z=R\exp(-x)$ and $t=zy/R$). Thus, $S_y\partial_x+y\Delta_y=\partial_x+t\partial_t+O(z^2)=-z\partial_z+O(z^2)$. We obtain then the near-boundary action
\begin{eqnarray}
S
&=& \frac{1}{2R}\int \frac{1}{R}\sqrt{g}dt dz\bigg(z^2(\partial_z\Phi)^2+(1+\frac{\kappa^2}{4R^2})z^2(\partial_t\Phi)^2+R^2m^2\Phi^2\bigg)\nonumber\\
&=& \frac{1}{2}\int \sqrt{g^{\prime}}d^2x^{\prime}\bigg((\partial_a^{\prime}\Phi)(\partial^{\prime a}\Phi)+m^2\Phi^2\bigg).\label{nb}
\end{eqnarray}
The primed coordinates are given by $x^1=t^{\prime}$ and $x^2=z$ where the rescaled time parameter  $t^{\prime}$ is related to the original Poincare time $t$ by the relation
\begin{eqnarray}
t^{\prime}=\frac{t}{\sqrt{1+\frac{\kappa^2}{4R^2}}}.\label{time}
\end{eqnarray}
In order to take into account the effect of boundary terms in the classical equation of motion we return to the original action (\ref{action}) and compute the variation
\begin{eqnarray}
\delta S&=&2\pi R \kappa Tr\delta\hat{\Phi}\bigg(\frac{1}{R^2\kappa^2}[\hat{X}^a,[\hat{X}_a,\hat{\Phi}]]+m^2\hat{\Phi}\bigg)\nonumber\\
&+& \pi R \kappa Tr\bigg(-\frac{2}{R^2\kappa^2}[\hat{X}^a,\delta\hat{\Phi}[\hat{X}_a,\hat{\Phi}]]-\frac{1}{R^2\kappa^2}[[\hat{X}^a,\hat{\Phi}],[\hat{X}_a,\delta\hat{\Phi}]]\nonumber\\
&+&m^2[\hat{\Phi},\delta\hat{\Phi}]\bigg).\label{var}
\end{eqnarray}
The first line produces the equation of motion of bulk fields given by  (\ref{eom}) which can then also be put in the form
\begin{eqnarray}
\frac{1}{R^2\kappa^2}[\hat{X}^a,[\hat{X}_a,\hat{\Phi}]]+m^2\hat{\Phi}=0.
\end{eqnarray}
The second line of  (\ref{var}) represents boundary terms. These terms are all given by the trace of a commutator which should vanish identically if the trace were finite dimensional. It is straightforward to check that these boundary terms (at least up to order $\kappa^2$) yields the commutative result \cite{Pinzul:2017wch}

\begin{eqnarray}
-\int dt (\delta \Phi\partial_z\Phi)|_{z=0}.
\end{eqnarray}

Conformal field theory and the ${\bf AdS}_2/{\bf CFT}_1$ correspondence

The near-boundary action (\ref{nb}) can be equivalently interpreted as an action over a commutative ${\bf AdS}^2$ with Poincare coordinates $x^1=t$ and $x^2=z$ and a rescaled field $\Phi^{\prime}$ given by
\begin{eqnarray}
\Phi^{\prime}=(1+\frac{\kappa^2}{4R^2})^{1/4}\Phi.\label{nc2c}
\end{eqnarray}
\begin{eqnarray}
S&=& \frac{1}{2}\int \sqrt{g}d^2x\bigg((\partial_a\Phi^{\prime})(\partial^a\Phi^{\prime})+m^2\Phi^{\prime2}\bigg).\label{nb1}
\end{eqnarray}
The near-boundary action is therefore an ordinary free scalar field theory on commutative ${\bf AdS}^2$ (with rescaled fields). This means in particular that noncommutative ${\bf AdS}^2_{\theta}$ is asymptotically a commutative ${\bf AdS}^2$ (since the bounadry limit $z\longrightarrow 0$ corresponds to $x,y\longrightarrow \infty$ keeping the time $t=\exp(-x)y$ fixed).

Furthermore, it should be noted that the near-boundary action  (\ref{nb1}) is in fact exact in $\kappa$ and as such we will take it as our  first approximation of the original free scalar field theory on noncommutative ${\bf AdS}^2_{\theta}$ given by the action (\ref{action}).

The equation of motion which follows from action (\ref{nb1}) is the commutative limit of the equation of motion  (\ref{eom}). This is given explicitly by
\begin{eqnarray}
z^2(\partial_z^2+\partial_t^2)\Phi^{\prime}-m^2R^2\Phi^{\prime}=0.\label{nb1eom}
\end{eqnarray}
We will assume that the mass $m^2$ satisfies the so-called Breitenlohner-Freedman  bound $m^2>-1/4R^2$. The solution near the boundary $z=\epsilon$  is then given by \cite{Ramallo:2013bua}

\begin{eqnarray}
\Phi^{\prime}(t,z=\epsilon)= A(t)\epsilon^{1-\Delta}.\label{div}
\end{eqnarray}
The exponent $\Delta$ is the so-called scaling dimension of the field and it is given by
\begin{eqnarray}
\Delta=\frac{1}{2}+\sqrt{\frac{1}{4}+m^2R^2}.
\end{eqnarray}
For $m^2>0$ the exponent $1-\Delta$ is negative and hence the field (\ref{div}) is divergent. The quantum field theory source $\varphi^{\prime}(t)$, i.e. the scalar field living on the boundary, is then identified with $A(t)$, viz
\begin{eqnarray}
\varphi^{\prime}(t)={\rm lim}_{\epsilon\longrightarrow 0}\epsilon^{\Delta-1}\Phi^{\prime}(t,\epsilon).\label{bva}
\end{eqnarray}
$\varphi^{'}(t)$ is the scalar field representing  the  anti-de Sitter scalar field $\Phi^{\prime}(t,z)$ at the boundary $z=0$, i.e. the boundary field $\varphi^{\prime}$ is the holographic dual of the bulk field $\Phi^{\prime}$. The scaling dimension of the source $\varphi^{\prime}$ is given by $1-\Delta$.

Let ${\cal O}^{\prime}(t,z)$ be the dual operator of the scalar field $\Phi^{\prime}(t,z)$. Their coupling is a boundary term of the form (by using also equation (\ref{nc2c}))
\begin{eqnarray}
S_{\rm bound}&=&\int dt\sqrt{\gamma}\Phi^{\prime}(t,\epsilon){\cal O}^{\prime}(t,\epsilon)\nonumber\\
&=&(1+\frac{\kappa^2}{4R^2})^{1/2}\int dt\sqrt{\gamma}\Phi^{}(t,\epsilon){\cal O}^{}(t,\epsilon).\label{bt0}
\end{eqnarray}

Obviously, we must have
\begin{eqnarray}
{\cal O}^{\prime}(t,z)=(1+\frac{\kappa^2}{4R^2})^{1/4}{\cal O}(t,z). \label{nc2c1}
\end{eqnarray}
It is clear that by rescaling the time coordinate in  (\ref{nb1}) we obtain the bulk action (\ref{nb}) whereas rescaling time in (\ref{bt0}) yields a canonically normalized boundary term. The time rescaling is naturally given by (\ref{time}).

The boundary term (\ref{bt0}) can also be rewritten  in terms of the dual operator ${\cal O}^{\prime}(t)$ of the scalar field $\varphi^{\prime}(t)$ as
\begin{eqnarray}
S_{\rm bound}
&=&\int dt \frac{R}{\epsilon} \epsilon^{1-\Delta}\varphi^{\prime}(t){\cal O}^{\prime}(t,\epsilon)\nonumber\\
&=&R \int dt \varphi^{\prime}(t){\cal O}^{\prime}(t).\label{bt}
\end{eqnarray}
Where
\begin{eqnarray}
{\cal O}^{\prime}(t,\epsilon)=\epsilon^{\Delta}{\cal O}^{\prime}(t).
\end{eqnarray}
This is the wave function renormalization of the operator ${\cal O}^{\prime}$ as we move into the bulk. This also shows explicitly that $\Delta$ is the scaling dimension of the dual operator ${\cal O}^{\prime}$ since going from $z=0$ to $z=\epsilon$ is a dilatation operation in the quantum field theory.

The boundary term (\ref{bt}) can be written in terms of the noncommutative scalar field $\varphi(t)$ and its dual operator ${\cal O}(t)$ as
\begin{eqnarray}
S_{\rm bound}
&=&R \int dt \varphi^{}(t){\cal O}^{}(t).\label{bt1}
\end{eqnarray}
The noncommutative boundary scalar field $\varphi(t)$ and its dual operator ${\cal O}(t)$ are related to the boundary scalar field $\varphi^{\prime}(t)$ and its dual operator ${\cal O}^{\prime}(t)$ by equations (\ref{nc2c}) and (\ref{nc2c1}) respectively.

The CFT living on the boundary is completely determined by the correlation functions
\begin{eqnarray}
\langle{\cal O}^{\prime}(t_1)...{\cal O}^{\prime}(t_n)\rangle.
\end{eqnarray}
On the boundary with lagrangian ${\cal L}$ the calculation of these correlation functions  proceeds as usual by introducing the generating functional
\begin{eqnarray}
Z_{\rm CFT}[J]=\int ~\exp({\cal L}+\int dt J(t){\cal O}^{\prime}(t))=\langle\exp(\int dt J(t){\cal O}^{\prime}(t))\rangle .
\end{eqnarray}
Then we have immediately
\begin{eqnarray}
\langle{\cal O}^{\prime}(t_1)...{\cal O}^{\prime}(t_n)\rangle=\frac{\delta^n\log Z_{CFT}[J]}{\delta J(t_1)...\delta J(t_n)}|_{J=0}.
\end{eqnarray}
The operator ${\cal O}^{\prime}(t)$ is sourced by the scalar field $\varphi^{\prime} (t)$ living on the boundary which is related to the boundary value of the bulk scalar field $\Phi^{\prime}(t,z)$ by the relation (\ref{bva}). The boundary scalar field $\Phi_0^{\prime}(t)$ is actually divergent and it is simply defined by
\begin{eqnarray}
\Phi_0^{\prime}(t)=\Phi^{\prime}(t,0).
\end{eqnarray}
The AdS/CFT correspondence \cite{Gubser:1998bc,Witten:1998qj} states that the CFT generating functional with source $J=\Phi_0^{\prime}$ is equal to the path integral on the gravity side evaluated over a bulk field which has the value $\Phi_0^{\prime}$ at the boundary of AdS. We write
\begin{eqnarray}
Z_{\rm CFT}[\Phi_0^{\prime}]\equiv Z_{\rm grav}[\Phi^{\prime}\longrightarrow\Phi_0^{\prime}] =\int_{\Phi^{\prime}\longrightarrow \Phi_0^{\prime}} {\cal D}\Phi^{\prime}\exp(S_{\rm grav}[\Phi^{\prime}]).
\end{eqnarray}
In the limit in which classical gravity is a good approximation the gravity path integral can be replaced by the classical amplitude given by the classical on-shell gravity action, i.e.

\begin{eqnarray}
Z_{\rm CFT}[\Phi_0^{\prime}]=\exp(S_{\rm grav}^{\rm on-shell}[\Phi^{\prime}\longrightarrow\Phi_0^{\prime}]).
\end{eqnarray}
Typically the on-shell gravity action is divergent requiring holographic renormalization  \cite{Henningson:1998gx,deHaro:2000vlm,Skenderis:2002wp}. The on-shell action gets renormalized and the above prescription becomes
\begin{eqnarray}
Z_{\rm CFT}[\Phi_0^{\prime}]=\exp(S_{\rm grav}^{\rm renor}[\Phi^{\prime}\longrightarrow\Phi_0^{\prime}]).
\end{eqnarray}
The correlation functions are then renormalized as

\begin{eqnarray}
\langle{\cal O}^{\prime}(t_1)...{\cal O}^{\prime}(t_n)\rangle=\frac{\delta^n S_{\rm grav}^{\rm renor}[\Phi^{\prime}\longrightarrow\Phi_0^{\prime}]}{\delta\varphi^{\prime}(t_1)...\delta\varphi^{\prime}(t_n)}|_{\varphi^{\prime}=0}.
\end{eqnarray}
In our case the bulk action $S_{\rm grav}$ is proportional (with a proportionality constant denoted by $-\eta$) to the near-boundary action  (\ref{nb1})  which is an  approximation of the action (\ref{action}) of free scalar fields on noncommutative ${\bf AdS}^2_{\theta}$.

The on-shell action $S_{\rm grav}^{\rm on-shell}$ is a boundary term obtained from $S_{\rm grav}$ by substituting the solution of the classical equation of motion (\ref{nb1eom}) (which is also required to be a regular solution in the IR limit $z\longrightarrow\infty$).

The on-shell action is found to be divergent requiring the addition of a counter term in the form of a quadratic local term living on the  boundary of AdS space given explicitly by \cite{Ramallo:2013bua}
\begin{eqnarray}
S^{}_{\rm ct}
&=&\frac{\eta}{2}\eta_1 \int_{} \sqrt{\gamma}dt \phi^2.
\end{eqnarray}
After some calculation we find that $\eta_1=-(1-\Delta)/R$ and as a consequence the  renormalized action reduces to
\begin{eqnarray}
S^{\rm renor}_{\rm grav}
&=&-\frac{\eta}{2}(2\Delta -1)\frac{\Gamma(1-\nu)}{\Gamma(1+\nu)}\int_{} \frac{d\omega}{2\pi} \varphi^{\prime}(\omega)(\frac{\omega}{2})^{2\nu}\varphi^{\prime}(-\omega).
\end{eqnarray}
The exponent $\nu$ is given by
\begin{eqnarray}
\nu^2=\frac{1}{4}+m^2R^2=(\Delta-\frac{1}{2})^2.
\end{eqnarray}
The two-point function is then given immediately by

\begin{eqnarray}
\langle{\cal O}^{\prime}(t){\cal O}^{\prime}(0)\rangle=\frac{2\nu \eta}{\sqrt{\pi}}\frac{\Gamma(\frac{1}{2}+\nu)}{\Gamma(-\nu)}\frac{1}{|t|^{2\Delta}}.
\end{eqnarray}
This is the correct behavior of a conformal field of scaling dimension $\Delta$, i.e. the exponent $\Delta$ is indeed the scaling dimension of the boundary operator ${\cal O}^{\prime}(x)$.

The final step is to substitute the operator rescaling (\ref{nc2c1}) to obtain the noncommutative two-point function

\begin{eqnarray}
\langle{\cal O}^{}(t){\cal O}^{}(0)\rangle=(1+\frac{\kappa^2}{4R^2})^{-1/2}\frac{2\nu \eta}{\sqrt{\pi}}\frac{\Gamma(\frac{1}{2}+\nu)}{\Gamma(-\nu)}\frac{1}{|t|^{2\Delta}}.
\end{eqnarray}
By exapnding in powers of $\kappa^2$ and setting $m^2=0$ we obtain the result of \cite{Pinzul:2017wch} which was directly computed from the noncommutative action (\ref{actionstar}) $\kappa-$expanded up to the order of $\kappa^2$.

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