We consider AdS$\_{d+1}$ in Poincaré coordinates:
$$
ds^2=\frac{1}{z^2}\left(-dt^2+dz^2+dx\_{d-1}^2\right),
$$
where we set the AdS radius to unity. We study a scalar in this background with action
$$
S=-\frac{1}{2}\int d^{d+1}x\,\sqrt{-g}\left(\partial\_\mu\phi\partial^\mu\phi+m^2\phi^2\right).
$$
I want to understand the following statement:

> 
> For $m^2=-\frac{(d-1)(d+1)}{4}$, there is a  **hidden Weyl symmetry** present in the action which simplifies the form of the scalar correlators in the vacuum state.
> 
> 
> 

What I was able to show is the following: If we consider a scalar with a different action
$$
S=-\frac{1}{2}\int d^{d+1}x\,\sqrt{-g}\left(\partial\_\mu \phi \partial^\mu \phi+ \frac{d-1}{4d}R\phi^2\right).
$$
i.e. with coupling to the scalar curvature instead of a mass term, then this action is indeed invariant under Weyl transformations
$$
\begin{align}
g\_{\mu\nu} &\to \Omega^2 g\_{\mu\nu}, \\
 \phi &\to \Omega^{\frac{1-d}{2}}\phi,
\end{align}
$$
with a spacetime dependent $\Omega=\Omega(x)$.

For AdS$\_{d+1}$ the Ricci scalar is $R=-d(d+1)$, so with a non-dynamical AdS background one can rewrite the action as
$$
S=-\frac{1}{2}\int d^{d+1}x\,\sqrt{-g}\left(\partial\_\mu \phi \partial^\mu \phi- \frac{(d-1)(d+1)}{4}\phi^2\right),
$$
so one finds the massive scalar with the $m^2=-\frac{(d-1)(d+1)}{4}$.

1. Now, when gravity is not dynamical and the Ricci scalar is not explicit in the action any more, how does this Weyl symmetry of the gravitational theory affect the scalar theory?
2. Does the hidden Weyl symmetry lead to a conformal symmetry of the scalar theory?