# 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?

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 $$\Delta_{\cal O} = \Delta = \frac{d}{2} + \sqrt{ \frac{d^2}{4} + m^2 L^2 } .$$