Relationship between scaling dimension and mass in AdS/CFT I've been reading Horatiu Nastase's notes on AdS/CFT, but I was confused about a certain relationship he claimed. If we compactify supergravity on $AdS_5\times S^5$, we may expand the fields in Kaluza-Klein modes
$$\phi(x,y)=\sum_n\sum_{I_n}\phi^{I_n}_{(n)}(x)Y^{I_n}_{(n)}(y),$$
for $x\in AdS_5$, $y\in S^5$. $I_n$ is an index in a representation of the symmetry group and $Y^{I_n}_{(n)}$ are spherical harmonics. Then the field $\phi^{I_n}_{(n)}$ living in $AdS_5$, of mass $m$, corresponds to an operator $\mathcal{O}^{I_n}_{(n)}$ in 4-dimensional $\mathcal{N}=4$ Super Yang-Mills, of dimension $\Delta$. This makes sense, but immeiately after (in equation (8.13)), he claims that
$$\Delta=\frac{d}{2}+\sqrt{\frac{d^2}{4}+m^2R^2},$$
where $R$ is the curvature of the background. Does anyone know how this relationship is derived?
 A: This is a bit long, but I'm going to be super precise.
Let's work in Poincare coordinates in AdS,
$$
ds^2 = \frac{L^2}{z^2} ( dz^2 + dx^\mu dx_\mu ) . 
$$
$\Delta$ is the eigenvalue of scale transformations which acts as $z \to \lambda z$, $x^\mu \to \lambda x^\mu$.
A scalar field of mass $m$ in the bulk satisfies an equation of the form
$$
(\Box - m^2) \Phi(z,x) =  J(z,x)
$$
where $J(z,x)$ is a source which describes how $\Phi$ couples to other fields in the theory. Near the boundary $z=0$, the scalar field has a fall-off $\Phi(z,x) = z^a \phi(x) + \cdots$. To determine $a$, we substitute this expansion into the equation of motion and expand in small $z$. The leading order equation in $z$ fixes $a$ as
$$
a = \Delta ~\text{or}~ d-\Delta, \qquad \Delta = \frac{d}{2} + \sqrt{ \frac{d^2}{4} + m^2 L^2 } . 
$$
In other words, the expansion is
$$
\Phi(z,x) = z^{d-\Delta} [ \phi_0(x) + \cdots ] + z^\Delta [ A(x) + \cdots ] 
$$
$A(x)$ is fixed in terms of $\phi_0(x)$ using regularity in the interior of AdS.
Under the AdS/CFT dictionary, the bulk path integral with a boundary condition $\phi_0$ on the field $\Phi$ is equal to a boundary path integral with a source $
\phi_0$ for the dual operator, i.e.
$$
Z_{bulk}[\phi_0] = Z_{CFT}[\phi_0] = \langle e^{ - \int d^d x \phi_0(x) {\cal O}(x) } \rangle_{CFT} .
$$
Here, ${\cal O}(x)$ is the operator in the CFT which is dual to $\Phi(z,x)$.
Now, in the bulk $z \to \lambda z$, $x^\mu \to \lambda x^\mu$ is simply a diffeomorphism so the bulk path integral is invariant. On the boundary, the same transformation is a scale transformation so the operators transform as
$$
{\cal O}(x) \to {\cal O}'(x)  = \lambda^{-\Delta_{\cal O}} {\cal O}(\lambda x)
$$
where $\Delta_{\cal O}$ is the scaling dimension of ${\cal O}$.
We also need to find out how $\phi_0(x)$ transforms under the same transformation. Since $\Phi(z,x)$ is a scalar field, we have
$$
\Phi'(z',x') = \Phi(z,x) \quad \implies \quad \phi_0'(x') = \lambda^{-(d-\Delta)} \phi_0(x) . 
$$
Finally, since the bulk path integral is invariant under diffeomorphisms, so is the boundary path integral. We must therefore have
\begin{align}
\langle e^{ - \int d^d x \phi_0(x) {\cal O}(x) } \rangle_{CFT} &= \langle e^{ - \int d^d x \phi'_0(x) {\cal O}'(x) } \rangle_{CFT} \\
&= \langle e^{ - \int d^d x' \phi'_0(x') {\cal O}'(x') } \rangle_{CFT}\\
&= \langle e^{ - \int d^d x \lambda^d \lambda^{-(d-\Delta)} \lambda^{-\Delta_{\cal O}}  \phi_0(x) {\cal O}(x) } \rangle_{CFT}\\
&= \langle e^{ - \int d^d x \lambda^{\Delta-\Delta_{\cal O}}  \phi_0(x) {\cal O}(x) } \rangle_{CFT}
\end{align}
Thus, we have
$$
\Delta_{\cal O} = \Delta = \frac{d}{2} + \sqrt{ \frac{d^2}{4} + m^2 L^2 } .
$$
