Holographic Correlators and the Breitenlohner–Freedman Bound

A holographic correlation function is a response to a boundary source. To calculate it, one solves a bulk field equation, imposes an interior condition, and differentiates a renormalized action. The normalization of that action matters as much as the two near-boundary powers. This page carries those steps through for a scalar field and then explains how changing its boundary condition changes the dual operator.

The central results are the scalar relation $m^2R^2=\Delta(\Delta-d)$, the normalized two-point function, and the Breitenlohner–Freedman (BF) bound $m^2R^2\geq-d^2/4$. The last statement concerns linear scalar dynamics on fixed AdS with suitable boundary conditions. Within a limited negative-mass range, two ordinary boundary quantizations are possible; at the endpoints, logarithms require separate treatment. A cubic interaction will finally produce a three-point function.

The calculations use the shared Fourier and source definitions and build on the boundary-source prescription on page 31. We work in boundary dimension $d\geq3$, with flat boundary metric and a free bulk scalar except when explicitly adding the leading cubic interaction.

Scalar fields and their boundary powers

Consider the positive Euclidean kinetic action

$$S_{\rm E}[\psi] =\frac{\mathcal N}{2}\int d^{d+1}X\sqrt g\, \left(g^{MN}\partial_M\psi\partial_N\psi+m^2\psi^2\right), \qquad \mathcal N>0,$$

in the Poincaré metric

$$ds^2=\frac{R^2}{z^2}\left(dz^2+dx^i dx^i\right), \qquad z>0.$$

Here $z$ and $x^i$ have dimensions of length. We take $\psi$ dimensionless, so $\mathcal N$ has dimensions $\text{length}^{1-d}$. A negative $m^2$ does not change the sign of the kinetic term; whether it causes an instability is a question about the radial gradient energy and the boundary condition.

Since $\sqrt g=R^{d+1}z^{-d-1}$, the equation $(\nabla^2-m^2)\psi=0$ becomes

$$z^2\partial_z^2\psi-(d-1)z\partial_z\psi +z^2\partial_i^2\psi-m^2R^2\psi=0.$$

Near $z=0$, a power $z^\alpha$ obeys the indicial equation

$$\alpha(\alpha-d)=m^2R^2.$$

Define its two roots by

$$\Delta_\pm=\frac d2\pm\nu, \qquad \nu=\sqrt{\frac{d^2}{4}+m^2R^2}.$$

For now take real $\nu>0$ and avoid integer $\nu$. The expansion has the form

$$\psi(z,x) =z^{\Delta_-}\left[J(x)+z^2a_{(2)}(x)+\cdots\right] +z^{\Delta_+}\left[A(x)+\cdots\right].$$

In standard quantization, $J$ is the source for an operator of dimension $\Delta_+$. Its dimensions are $[J]=\text{length}^{-\Delta_-}$, making $\int d^d x\,J\mathcal O$ dimensionless. The coefficient $A$ has dimensions $\text{length}^{-\Delta_+}$ and will determine the expectation value. The intermediate coefficients are local derivatives of the source; they are not independent responses.

For a Euclidean Fourier mode of magnitude $q=|\mathbf q|>0$, substitution gives

$$\begin{aligned} a_{(2j)}(q) &=\frac{q^2a_{(2j-2)}(q)}{4j(j-\nu)}, \qquad a_{(0)}=J,\\ a_{(2)}(q)&=\frac{q^2J(q)}{4(1-\nu)}. \end{aligned}$$

This recurrence explains both an important subtraction and an exception. If $\nu>1$, the $z^{\Delta_-+2}$ source term occurs before the response. If $\nu$ is a positive integer, the recurrence becomes singular at $j=\nu$, signaling a logarithm rather than a failure of the bulk solution.

Interior regularity fixes the response

Writing $\psi=z^{d/2}f(qz)$ turns the radial equation into the modified Bessel equation. The two independent solutions are $I_\nu$ and $K_\nu$. At positive Euclidean momentum, $I_\nu$ grows exponentially toward $z\to\infty$, while $K_\nu$ decays. Interior regularity therefore selects

$$\psi(z,q) =\frac{q^\nu}{2^{\nu-1}\Gamma(\nu)}\, z^{d/2}K_\nu(qz)J(q).$$

The prefactor makes the coefficient of $z^{\Delta_-}$ exactly $J$. For noninteger $\nu$, the small-argument expansion is

$$\begin{aligned} K_\nu(y)={}&2^{\nu-1}\Gamma(\nu)y^{-\nu} \left[1+\frac{y^2}{4(1-\nu)}+\cdots\right]\\ &+2^{-\nu-1}\Gamma(-\nu)y^\nu[1+O(y^2)]. \end{aligned}$$

It follows that

$$A(q)=b_\nu(q)J(q), \qquad b_\nu(q)=\frac{\Gamma(-\nu)}{2^{2\nu}\Gamma(\nu)}(q^2)^\nu.$$

The two boundary coefficients are thus independent local data for the differential equation, but they are related once the interior condition is imposed. Different states or interior geometries generally change that relation.

Extracting a response after the source terms

An exact example makes the subtraction visible. In $AdS_5$ with $\nu=3/2$, equivalently $m^2R^2=-7/4$,

$$\psi(z,q)=J(q)z^{1/2}F(qz), \qquad F(y)=(1+y)e^{-y}.$$

The expansion $F(y)=1-y^2/2+y^3/3+\cdots$ contains an analytic $q^2$ source descendant before the nonanalytic $q^3$ response. Thus

$$\frac{A}{J}=\frac{q^3}{3}, \qquad \lim_{y\to0}\frac{F(y)-1+y^2/2}{y^3}=\frac13.$$

For the regular $\nu=3/2$ mode, the plotted residual isolates the coefficient of the faster boundary power. Both axes are dimensionless, with $y=qz$ and $q>0$. The boundary lies at $y=0$; the limit $1/3$ gives $A/J=q^3/3$. The curve is calculated from the exact Bessel solution, not from a truncated power series.

Prescribing a source at a finite cutoff

If the boundary value is imposed at $z=\epsilon$, the exact solution instead reads

$$\psi(z,q) =\left(\frac z\epsilon\right)^{d/2} \frac{K_\nu(qz)}{K_\nu(q\epsilon)}\,\psi_\epsilon(q).$$

The denominator ensures $\psi(\epsilon,q)=\psi_\epsilon(q)$. It must be retained while varying the cutoff action. Its subleading terms can contribute when the radial derivative is evaluated at the same cutoff. Taking the boundary limit prematurely gives an incorrect two-point normalization, as the appendix of Freedman, Mathur, Matusis, and Rastelli explains.

The rescaled cutoff source $J_\epsilon=\epsilon^{-\Delta_-}\psi_\epsilon$ approaches $J$. One may use either as the independent source, provided the relation between them is kept through the calculation and the limit is taken consistently.

Renormalized action and the two-point function

On a solution, integration by parts reduces the action to a surface term. At the lower boundary of the region $z\geq\epsilon$, the outward normal points toward decreasing $z$:

$$n^z=-\frac\epsilon R, \qquad \sqrt\gamma=\left(\frac R\epsilon\right)^d.$$

Consequently,

$$S_{\rm reg} =-\frac{\mathcal N R^{d-1}}2 \int_{z=\epsilon}d^d x\, \epsilon^{1-d}\psi\partial_z\psi.$$

It is useful to denote the dimensionless coefficient produced by this reduction as

$$\mathcal C_\psi\equiv\mathcal N R^{d-1}.$$

The factor of $R^{d-1}$ comes from the induced measure and the unit normal. It is part of the physical normalization of every response below.

Subtract divergences before differentiating

For $0<\nu<1$, the leading local counterterm is sufficient:

$$S_{\rm ct}^{(0)} =\frac{\mathcal N}{2}\int_{z=\epsilon}d^d x\sqrt\gamma\, \frac{\Delta_-}{R}\psi^2.$$

To see its finite effect, keep the cross term between $Jz^{\Delta_-}$ and $Az^{\Delta_+}$. The boundary action contributes $-\mathcal C_\psi d\int JA/2$, while the counterterm contributes $\mathcal C_\psi\Delta_-\int JA$. Their sum is

$$S_{\rm E,ren}^{\rm nonlocal} =-\nu\mathcal C_\psi\int d^d x\,J(x)A(x).$$

For $\nu>1$, divergent source descendants require derivative counterterms as well. Away from $\nu=1$, the next term is

$$S_{\rm ct}^{(2)} =\frac{\mathcal N}{2}\int d^d x\sqrt\gamma\, \frac{R}{2(\nu-1)}\psi\Box_\gamma\psi.$$

Here $\Box_\gamma$ is the Laplacian of the induced metric. Higher descendants require further local terms; at integer $\nu$ one of the subtractions is logarithmic. The nonlocal cross term above survives this procedure, while the finite local terms depend on the subtraction scheme. This is the scalar holographic-renormalization construction developed in de Haro, Solodukhin, and Skenderis, section 5.1.

With $W_{\rm E}=\log Z=-S_{\rm E,ren}$, variation gives

$$\langle\mathcal O(x)\rangle_J =\frac{\delta W_{\rm E}}{\delta J(x)} =2\nu\mathcal C_\psi A(x)+\text{local terms}.$$

The factor of two is essential: $A$ is itself a linear functional of $J$. Equivalently, write the quadratic functional as $W_{\rm E}=\tfrac12\int_q J(q)G_{\rm E}(q)J(-q)$, with the shared Fourier measure. Then

$$G_{\rm E}(q) =2\nu\mathcal C_\psi \frac{\Gamma(-\nu)}{2^{2\nu}\Gamma(\nu)}(q^2)^\nu +\text{local terms}.$$

This expression applies to noninteger $\nu>0$. A finite local counterterm changes a polynomial in $q^2$, hence a delta function or its derivatives in position space. It does not change the answer at distinct insertion points.

Position-space normalization

The Fourier transform must be understood as a distribution with local subtractions. Analytic continuation of the power-law transform gives, away from $x=0$,

$$\int\frac{d^d q}{(2\pi)^d}e^{iq\cdot x}(q^2)^\nu =\frac{4^\nu\Gamma(d/2+\nu)}{\pi^{d/2}\Gamma(-\nu)} \frac1{|x|^{d+2\nu}}.$$

Combining the two coefficients yields

$$\langle\mathcal O(x)\mathcal O(0)\rangle =\frac{2\nu\mathcal N R^{d-1}\Gamma(\Delta_+)} {\pi^{d/2}\Gamma(\nu)}\, \frac1{|x|^{2\Delta_+}}, \qquad x\ne0.$$

The coefficient is positive for $\mathcal N>0$ and $\nu>0$. A negative coefficient of a nonlocal momentum-space expression, such as the $\nu=1/2$ result on page 31, does not imply a negative separated-point norm: the distributional Fourier transform also has a sign.

Integer indices and the massless AdS5 scalar

When $\nu=n$ is a positive integer, the pole of $\Gamma(-\nu)$ multiplies a local polynomial. Subtracting that pole leaves a logarithm. Its nonlocal coefficient is

$$G_{\rm E}(q) =\frac{(-1)^{n+1}\mathcal C_\psi} {2^{2n-1}[(n-1)!]^2} (q^2)^n\log\frac{q^2}{\mu^2} +\text{local terms}.$$

For example, put $\nu=n+\delta$ and expand $(q^2)^\nu=(q^2)^n[1+\delta\log q^2+\cdots]$. The product of this logarithm with the Gamma-function residue gives the stated coefficient. The scale $\mu$ makes the logarithm dimensionless; changing it adds a local term.

A massless scalar in $AdS_5$ has $\nu=2$, $\Delta_+=4$, and regular profile $F(qz)=(q^2z^2/2)K_2(qz)$. Thus

$$\begin{aligned} G_{\rm E}(q)&=-\frac{\mathcal C_\psi}{8} q^4\log\frac{q^2}{\mu^2}+\text{local terms},\\ \langle\mathcal O(x)\mathcal O(0)\rangle &=\frac{24\mathcal C_\psi}{\pi^2|x|^8},\qquad x\ne0. \end{aligned}$$

Exercise 2 restores the ten-dimensional dilaton normalization and connects this result to the absorption spectral density on page 30. The coincident-root case $\nu=0$ requires a different source expansion and is treated below.

The bulk-to-boundary kernel

The same linear solution can be written directly in position space:

$$\psi(z,x)=\int d^d y\,K_\Delta(z,x;y)J(y), \qquad \Delta=\Delta_+>\frac d2,$$

where

$$K_\Delta(z,x;y) =c_\Delta\left(\frac z{z^2+|x-y|^2}\right)^\Delta, \qquad c_\Delta=\frac{\Gamma(\Delta)} {\pi^{d/2}\Gamma(\Delta-d/2)}.$$

Substitution verifies the scalar equation with $m^2R^2=\Delta(\Delta-d)$. The coefficient is fixed by the source, not by the bulk kinetic coefficient. To check it, set $y=x+zu$ in the convolution:

$$z^{\Delta-d}\psi(z,x) =c_\Delta\int d^d u\, \frac{J(x+zu)}{(1+u^2)^\Delta}.$$

For a smooth bounded source, the integrable weight at $\Delta>d/2$ permits the limit inside the integral. Using radial coordinates and a beta integral gives

$$\int d^d u\,(1+u^2)^{-\Delta} =\pi^{d/2}\frac{\Gamma(\Delta-d/2)}{\Gamma(\Delta)},$$

so $z^{\Delta-d}\psi\to J$. In this sense, $z^{\Delta-d}K_\Delta$ approaches a boundary delta function. At a fixed separated point $x\ne y$, however,

$$K_\Delta(z,x;y) \sim c_\Delta\frac{z^\Delta}{|x-y|^{2\Delta}}.$$

This tail determines the response to a localized source. Multiplying it by $2\nu\mathcal C_\psi$ reproduces the separated-point correlator. The delta-function limit and the separated-point tail are two different statements about the same kernel, not competing boundary conditions.

The BF bound and the allowed boundary quantizations

The roots become complex if $m^2R^2<-d^2/4$. A radial-energy calculation explains why this threshold is the stability bound, and why its use requires a boundary condition.

In Lorentzian AdS, separate a mode as

$$\psi(z,t,\mathbf x) =z^{(d-1)/2}u(z)e^{-i\omega t+i\mathbf k\cdot\mathbf x}.$$

The Klein–Gordon equation becomes a one-dimensional spectral problem:

$$H_\nu u=(\omega^2-\mathbf k^2)u, \qquad H_\nu=-\frac{d^2}{dz^2} +\frac{\nu^2-1/4}{z^2}.$$

The radial part of the Klein–Gordon norm is proportional to

$$\int_0^\infty dz\,z^{1-d}|\psi|^2 =\int_0^\infty dz\,|u|^2.$$

This is the norm relevant to Lorentzian scalar modes. It is not the Euclidean spacetime integral $\int\sqrt g\,|\psi|^2$.

Gradient energy can support a negative mass squared

For a smooth function $u$ of compact support in $z>0$, integration by parts gives the Hardy identity

$$\int_0^\infty dz\left(|u'|^2-\frac{|u|^2}{4z^2}\right) =\int_0^\infty dz\,z \left|\frac{d}{dz}\left(\frac u{\sqrt z}\right)\right|^2 \geq0.$$

It follows that the quadratic form of $H_\nu$ is nonnegative when $\nu^2\geq0$. Closing this form gives the standard positive, or Friedrichs, realization of the radial operator. Below the bound, the extra negative inverse-square potential makes the form unbounded below; Exercise 3 constructs explicit trial functions.

The conclusion is that

$$m^2R^2\geq-\frac{d^2}{4}$$

allows stable free scalar dynamics with an appropriate boundary condition. It does not assert that every boundary condition is positive, nor does it prove nonlinear stability of the coupled gravitational theory. The relation between AdS boundary conditions and positive self-adjoint extensions is analyzed systematically by Ishibashi and Wald.

When both powers have finite norm

Near the boundary, the two radial behaviors are

$$u_\pm(z)\sim z^{1/2\pm\nu}.$$

The faster behavior has finite boundary norm for $\nu\geq0$. The slower one requires

$$\int_0 dz\,z^{1-2\nu}<\infty, \qquad \nu<1.$$

Thus two distinct, ordinary power-law quantizations are available in the strict window

$$-\frac{d^2}{4}<m^2R^2<-\frac{d^2}{4}+1, \qquad 0<\nu<1.$$

In standard quantization, the slower coefficient is the source and the operator has dimension $\Delta_+$. In alternate quantization, the faster coefficient is the source and the operator has dimension $\Delta_-$. This is a change in the boundary condition defining the theory. Both choices have to be distinguished from turning on a source within one fixed theory. Their relation is developed by Klebanov and Witten, section 2.

For $AdS_5$, $\Delta_\pm=2\pm\sqrt{4+m^2R^2}$. The ordinary alternate branch is available only for $-4<m^2R^2<-3$. At $m^2R^2=-4$ the roots coincide and a logarithmic mode appears. At $m^2R^2=-3$, the slower mode reaches $\Delta_-=1$ and its ordinary Klein–Gordon norm diverges logarithmically. The diagram shows the mass and dimension conditions; positivity also depends on the chosen boundary condition.

For $d\geq3$, the scalar unitarity bound is $\Delta\geq(d-2)/2$, apart from the identity operator. The upper endpoint $\nu=1$ saturates this bound for $\Delta_-$. It is not part of the ordinary finite-norm alternate window: obtaining a free-field or singleton limit there requires an additional limiting construction. At the lower endpoint $\nu=0$, both logarithmic and nonlogarithmic radial behaviors have finite boundary norm, but the two power-law dimensions have coalesced. That case also needs separate treatment.

These statements describe allowed free bulk boundary quantizations. Embedding an arbitrary chosen scalar mass in a complete interacting AdS/CFT pair is an additional physical requirement.

Alternate quantization and mixed boundary conditions

The change of quantization has a useful functional description. In this calculation, choose the subtraction scheme with finite local quadratic source terms removed, so the standard response is exactly $2\nu\mathcal C_\psi A$. Such terms would otherwise also change the conjugate variable in the Legendre transform. For $0<\nu<1$, let

$$\chi=2\nu\mathcal C_\psi.$$

Choose the alternate source to be the faster coefficient $A$ itself. The appropriate Legendre transformation is

$$W_-[A] =\left[W_+[J]-\chi\int d^d x\,J A\right]_{\rm stationary\ in\ J}.$$

Stationarity says $\delta W_+/\delta J=\chi A$, precisely the standard response relation. Differentiating the transformed functional with respect to its independent source gives

$$\langle\mathcal O_-\rangle_A=-\chi J.$$

For a quadratic functional, $G_+=\chi b_\nu$ and direct elimination of $J$ gives

$$G_-(q)=-\frac{\chi^2}{G_+(q)} =-\frac{\chi}{b_\nu(q)}.$$

The square of $\chi$ records the chosen source normalization. If the independent source were instead $\chi A$, the corresponding kernel would be $-1/G_+$. A Legendre transform without a specified source variable leaves precisely this ambiguity.

As a check, at $\nu=1/2$ the regular solution has $b_{1/2}=-q$. Hence

$$G_+(q)=-\mathcal C_\psi q, \qquad G_-(q)=\frac{\mathcal C_\psi}{q}.$$

In four boundary dimensions, the latter transforms to $\mathcal C_\psi/(4\pi^2|x|^3)$, the positive two-point function of an operator with $\Delta_-=3/2$.

A double-trace deformation changes the boundary condition

Add $+\tfrac f2\int\mathcal O_-^2$ to the Euclidean action and couple an external source $\eta$ through $-\int\eta\mathcal O_-$. At the Gaussian level, or at leading large $N$ when the relevant correlators factorize, the expectation value $v$ responds to the effective source $\eta-fv$:

$$v=G_-\,(\eta-fv), \qquad G_f(q)=\frac{G_-(q)}{1+fG_-(q)}.$$

Since $v=-\chi J$, the bulk source relation is

$$\eta=A-\chi fJ.$$

This is a mixed boundary condition relating the two asymptotic coefficients. The coupling has mass dimension $2\nu$, so it is relevant in the alternate theory. This relation between multi-trace interactions and boundary conditions was formulated by Witten.

The sign of $f$ matters. For $\nu=1/2$,

$$G_f(q)=\frac{\mathcal C_\psi}{q+\mathcal C_\psi f}.$$

A negative $f$ produces a pole at positive Euclidean $q$. Exercise 4 identifies the corresponding negative radial eigenvalue and growing Lorentzian mode. The mass is above the BF bound, but this boundary condition is unstable. For positive $f$, the same formula illustrates the flow toward standard-quantization scaling after its local term is removed.

At the BF bound: a logarithmic source

At $\nu=0$, the two powers merge. Use the independent asymptotic data

$$\psi(z,x)=z^{d/2} \left[J(x)\log(z\mu)+A(x)+\cdots\right].$$

Here the source $J$ is explicitly the coefficient of the logarithm. The scale $\mu$ only specifies how the nonlogarithmic coefficient is separated from it. Interior regularity at $q>0$ gives

$$\psi(z,q)=-J(q)z^{d/2}K_0(qz), \qquad \frac{A(q)}{J(q)}=\log\frac{q}{2\mu}+\gamma_{\rm E}.$$

This follows from $K_0(y)=-\log(y/2)-\gamma_{\rm E}+O(y^2\log y)$. The sign in the solution is fixed by the chosen positive coefficient of $\log(z\mu)$.

The required leading counterterm is now

$$S_{\rm ct}^{\rm BF} =\frac{\mathcal N}{2}\int_{z=\epsilon}d^d x\sqrt\gamma\, \left[\frac{d}{2R} +\frac1{R\log(\epsilon\mu)}\right]\psi^2.$$

Combining it with the oriented boundary action gives

$$S_{\rm E,ren}^{\rm BF} =\frac{\mathcal C_\psi}{2}\int d^d x\,J A +\text{local terms}.$$

Consequently the logarithmic-source normalization yields

$$\begin{aligned} G_{\rm E}^{\rm BF}(q) &=-\mathcal C_\psi\log\frac q\mu+\text{local terms},\\ \langle\mathcal O(x)\mathcal O(0)\rangle_{\rm BF} &=\frac{\mathcal C_\psi\Gamma(d/2)}{2\pi^{d/2}|x|^d}, \qquad x\ne0. \end{aligned}$$

Exercise 5 derives the finite term explicitly. Simply putting $\nu=0$ into the standard prefactor $2\nu$ would miss this result, because the basis of boundary modes and the normalization of the source have changed.

Under $\mu\to e^t\mu$, the same bulk field has $A\to A-tJ$. This changes the response by a local multiple of the source. In particular, the source-free condition $J=0$, which excludes the logarithmic branch, is scale invariant. A condition $A=0$ with unrestricted $J$ is not invariant under this change. There is therefore one scale-invariant source-free choice at this coalescing-root point, rather than two ordinary quantizations with different dimensions; more general mixed conditions introduce scale dependence.

A cubic interaction gives a three-point function

Now consider three distinct scalar fields with standard dimensions $\Delta_i>d/2$ and the interaction

$$S_{\rm int}=g_{123}\int d^{d+1}X\sqrt g\, \psi_1\psi_2\psi_3.$$

For dimensionless fields, $g_{123}$ has dimensions $\text{length}^{-d-1}$. There is no $1/3!$ for three distinct fields with this definition of the coupling. For an identical field, writing $g\psi^3/3!$ instead is convenient because three source derivatives then supply $3!$.

At first order in $g_{123}$, insert the three free solutions into $S_{\rm int}$. Corrections to the solutions do not contribute through the quadratic action at this order: its first variation vanishes on the free solution with the renormalized sources held fixed. Using $W=-S_{\rm E,ren}$, three source derivatives therefore give the connected tree-level result

$$\langle\mathcal O_1(x_1)\mathcal O_2(x_2)\mathcal O_3(x_3)\rangle =-g_{123}\int d^{d+1}X\sqrt g\, \prod_{i=1}^3K_{\Delta_i}(X;x_i).$$

The leading contact Witten diagram for three distinct scalar fields integrates one cubic vertex over AdS. Each line represents a source-normalized kernel $K_{\Delta_i}(X;x_i)$. The overall sign follows from $W_{\rm E}=-S_{\rm E,ren}$. The drawing is schematic: the lines are kernel connections, not particle trajectories, and the vertex can lie anywhere in the integration region.

Conformal powers and the contact integral

Translation and rotation invariance restrict the answer to functions of the three separations. Under inversion, $|x_{ij}|^2$ transforms to $|x_{ij}|^2/(|x_i|^2|x_j|^2)$. Scalar-primary covariance then requires that the powers attached to point $i$ add to $2\Delta_i$. Define

$$\Sigma=\Delta_1+\Delta_2+\Delta_3, \qquad a_{ij}=\frac{\Delta_i+\Delta_j-\Delta_k}{2},$$

where $k$ is the remaining index. The answer has the form

$$\langle\mathcal O_1\mathcal O_2\mathcal O_3\rangle =\frac{C_{123}} {|x_{12}|^{2a_{12}}|x_{23}|^{2a_{23}}|x_{31}|^{2a_{31}}}.$$

The integral determines $C_{123}$, which conformal symmetry alone does not fix. One efficient calculation uses an inversion to send $x_3$ to infinity and rescales the correlator by $|x_3|^{2\Delta_3}$. The remaining integral, before the kernel prefactors, is

$$\int_0^\infty dz\int d^d x\, \frac{z^{\Sigma-d-1}} {[z^2+|x-x_1|^2]^{\Delta_1} [z^2+|x-x_2|^2]^{\Delta_2}}.$$

Combine the two denominators with a Feynman parameter $t$. Completing the square in $x$ introduces $t(1-t)|x_{12}|^2$. The $x$ integral and the radial integral use

$$\begin{aligned} \int d^d x\,(x^2+B)^{-p} &=\pi^{d/2}\frac{\Gamma(p-d/2)}{\Gamma(p)}B^{d/2-p},\\ \int_0^\infty dz\,\frac{z^{s-1}}{(z^2+B)^p} &=\frac{B^{s/2-p}}2 \frac{\Gamma(s/2)\Gamma(p-s/2)}{\Gamma(p)}. \end{aligned}$$

The final parameter integral is $\int_0^1dt\,t^{a_{13}-1}(1-t)^{a_{23}-1}$. Combining these beta functions gives

$$\begin{aligned} C_{123}={}&-g_{123}R^{d+1} \left(\prod_{i=1}^3c_{\Delta_i}\right)\frac{\pi^{d/2}}2\\ &\times \frac{\Gamma((\Sigma-d)/2) \Gamma(a_{12})\Gamma(a_{23})\Gamma(a_{31})} {\Gamma(\Delta_1)\Gamma(\Delta_2)\Gamma(\Delta_3)}. \end{aligned}$$

The factor $R^{d+1}$ is the volume factor of the dimensionful AdS metric. This calculation, including derivative vertices, is developed in Freedman, Mathur, Matusis, and Rastelli.

The displayed integral is directly convergent for separated insertions when $\Sigma>d$ and every $a_{ij}>0$, in addition to the standard kernel conditions $\Delta_i>d/2$. At an extremal relation such as $\Delta_3=\Delta_1+\Delta_2$, a Gamma function has a pole. The unregulated formula then cannot be used as a finite answer. Couplings, boundary terms, and the limiting prescription must be examined in the particular bulk theory.

Common pitfalls

Reading the response before subtracting descendants. When $\nu>1$, local source terms can precede the faster independent power. The response coefficient is what remains after those determined terms are removed.

Dropping the bulk normalization or the sign of the boundary. The product $\mathcal N R^{d-1}$ and the outward normal both enter the action. Once the source and $W_{\rm E}=-S_{\rm E,ren}$ are fixed, the nonlocal sign is fixed too.

Using the BF inequality without a boundary condition. The inequality permits a positive realization of the free radial operator. A mixed condition can still introduce a negative eigenvalue, as Exercise 4 shows.

Treating an endpoint as an interior point of the mass window. At $\nu=0$ the source is logarithmic; at $\nu=1$ the ordinary slow-mode norm diverges. Neither endpoint is obtained by copying the two-power discussion unchanged.

Exercises

1. A response hidden behind a derivative counterterm

For $d=4$, $\nu=3/2$, verify the exact solution

$$\psi(z,q)=J z^{1/2}(1+qz)e^{-qz}.$$

Compute the regulated action and both displayed counterterms. Obtain the nonlocal two-point coefficient and its position-space normalization. Show what divergence remains if the derivative counterterm is omitted.

Solution

Here $m^2R^2=-7/4$. Direct differentiation gives

$$z^2\psi''-3z\psi'-(q^2z^2-7/4)\psi=0.$$

For one real Fourier amplitude, with the product $J(q)J(-q)$ denoted by $J^2$, the three action terms are

$$\begin{aligned} S_{\rm reg}&=-\frac{\mathcal C_\psi}{2}\epsilon^{-3}\psi\psi',\\ S_{\rm ct}^{(0)}&=\frac{\mathcal C_\psi}{4}\epsilon^{-4}\psi^2,\\ S_{\rm ct}^{(2)}&=-\frac{\mathcal C_\psi}{2}q^2\epsilon^{-2}\psi^2. \end{aligned}$$

All fields and derivatives on the right are evaluated at $z=\epsilon$. The negative sign of the last term follows from $\Box_\gamma\psi=-\epsilon^2q^2\psi/R^2$.

Writing $F=(1+q\epsilon)e^{-q\epsilon}$, the first two terms combine exactly to

$$S_{\rm reg}+S_{\rm ct}^{(0)} =-\frac{\mathcal C_\psi J^2}{2\epsilon^2}FF' =\frac{\mathcal C_\psi J^2q^2}{2\epsilon} (1+q\epsilon)e^{-2q\epsilon},$$

where $F'$ is the derivative with respect to $\epsilon$. Adding the derivative counterterm gives

$$S_{\rm reg}+S_{\rm ct}^{(0)}+S_{\rm ct}^{(2)} =-\frac{\mathcal C_\psi J^2q^3}{2} (1+q\epsilon)e^{-2q\epsilon}.$$

Therefore

$$S_{\rm E,ren}=-\frac{\mathcal C_\psi}{2}J^2q^3, \qquad G_{\rm E}(q)=\mathcal C_\psi q^3.$$

The general position-space coefficient with $\Delta_+=7/2$ becomes

$$\frac{3\mathcal C_\psi\Gamma(7/2)}{\pi^2\Gamma(3/2)} =\frac{45\mathcal C_\psi}{4\pi^2}.$$

Thus the separated-point correlator is $45\mathcal C_\psi/(4\pi^2|x|^7)$. Omitting $S_{\rm ct}^{(2)}$ leaves $\mathcal C_\psi J^2q^2/(2\epsilon)$: identifying the finite response in the field does not by itself renormalize its action.

2. Dilaton normalization and a local scale dependence

Use the dimensionless ten-dimensional dilaton action

$$S_{\rm E,\phi}^{(2)} =\frac1{4\kappa_{10}^2}\int d^{10}X\sqrt g\,(\partial\phi)^2$$

on $AdS_5\times S^5$ with common radius $R$. For an $S^5$-independent fluctuation, determine $\mathcal C_\phi$, the position-space correlator, and the positive-frequency spectral density. Show that changing the subtraction scale changes only a local term. Finally, track the correlator under a constant source change $J'=aJ$, $a\ne0$.

Solution

The sphere volume is $\pi^3R^5$. Comparing the reduced action with $\mathcal N_\phi\int\sqrt{g_5}(\partial\phi)^2/2$ gives

$$\mathcal N_\phi=\frac{\pi^3R^5}{2\kappa_{10}^2}, \qquad \mathcal C_\phi=\mathcal N_\phi R^3 =\frac{\pi^3R^8}{2\kappa_{10}^2} =\frac{N^2}{8\pi^2}.$$

The last step uses the shared D3 charge and gravitational normalization, equivalently $R^8=\kappa_{10}^2N^2/(4\pi^5)$. The sphere integral supplies length to the fifth power; $\kappa_{10}^2$ has length to the eighth power, so $\mathcal N_\phi$ has length dimension $-3$ and $\mathcal C_\phi$ is dimensionless.

The massless result becomes

$$\begin{aligned} G_{\rm E}(q)&=-\frac{N^2}{64\pi^2}q^4 \log\frac{q^2}{\mu^2}+\text{local terms},\\ \langle\mathcal O_\phi(x)\mathcal O_\phi(0)\rangle &=\frac{3N^2}{\pi^4|x|^8},\qquad x\ne0. \end{aligned}$$

Using the shared continuation $G_R=-G_{\rm E}$ at $q^2=\mathbf k^2-(\omega+i0)^2$, put $s=\omega^2-\mathbf k^2>0$. For $\omega>0$, the logarithm becomes $\log(s/\mu^2)-i\pi$, so

$$\operatorname{Im}G_R=-\frac{N^2}{64\pi}s^2, \qquad \rho=-2\operatorname{Im}G_R=\frac{N^2}{32\pi}s^2.$$

This is the normalization measured by absorption on page 30. Classical supergravity gives the leading large-$N$ coefficient; it does not distinguish the exact finite-$N$ interacting-sector factor from the free center-of-mass contribution.

The scale derivative is

$$\mu\frac{\partial G_{\rm E}}{\partial\mu} =\frac{N^2}{32\pi^2}q^4.$$

Its inverse Fourier transform is proportional to $\Box^2\delta^{(4)}(x)$. Hence a scale change leaves both separated-point correlators and the positive timelike spectral density unchanged.

Finally, preserving $J\mathcal O=J'\mathcal O'$ requires $\mathcal O'=\mathcal O/a$. Therefore $G'=G/a^2$. A source rescaling changes the normalization explicitly; it cannot be used to change the sign of a real operator’s two-point norm.

3. Constructing a negative-energy trial mode below BF

Let $\nu^2=-\sigma^2<0$. For the radial operator $H_\nu$, use $u(z)=\sqrt z\,f(\log(z/z_*))$, where $z_*>0$ is a fixed reference length, to construct a compactly supported trial function with negative quadratic form. Explain how a dilation makes the form unbounded below at fixed norm, and why the argument extends from a piecewise smooth trial function to smooth compactly supported functions.

Solution

For real $u$, the pointwise identity underlying Hardy’s inequality is

$$|u'|^2-\frac{u^2}{4z^2} =z\left|\left(\frac u{\sqrt z}\right)'\right|^2 +\frac{d}{dz}\left(\frac{u^2}{2z}\right).$$

The last term integrates to zero for compact support away from the endpoints. Let $t=\log(z/z_*)$. Then $dz/z=dt$, and the quadratic form is

$$Q[u]=\int_{-\infty}^\infty dt\, \left[(f')^2-\sigma^2f^2\right].$$

Choose $f(t)=\sin(\pi t/L)$ on $0<t<L$ and zero elsewhere. This continuous function has a square-integrable weak derivative. Its support corresponds to $z_*\leq z\leq z_*e^L$, so its radial norm is finite. Direct integration gives

$$Q[u]=\frac L2\left(\frac{\pi^2}{L^2}-\sigma^2\right)<0 \quad\text{if}\quad L>\frac\pi\sigma.$$

For instance, $L=2\pi/\sigma$ gives $Q=-3\pi\sigma/4$. Although the derivative jumps at the endpoints, the trial function can be approximated in the first-derivative norm by smooth functions supported in a slightly enlarged finite interval. On this interval $z$ stays away from zero, so the potential is bounded and the quadratic form converges. A sufficiently close smooth approximation still has negative $Q$.

Normalize such a smooth $u$ to unit radial norm and define

$$u_\ell(z)=\ell^{-1/2}u(z/\ell),\qquad \ell>0.$$

A change of variable gives $\|u_\ell\|_2=1$ and $Q[u_\ell]=\ell^{-2}Q[u]$. Taking $\ell\to0$ makes $Q$ arbitrarily negative. Each individual function still has compact support strictly inside $z>0$. Thus the radial form is unbounded below already on the common compact-support domain; choosing a boundary extension cannot repair this below-BF defect.

This is a statement about the free linear scalar operator. It does not analyze nonlinear scalar potentials or gravitational backreaction.

4. An instability produced by a mixed boundary condition

Take $d=4$, $\nu=1/2$ and alternate quantization. Starting with $G_-(q)=\mathcal C_\psi/q$, add $+f\mathcal O_-^2/2$ to the Euclidean action. Derive the response kernel and the source-free bulk boundary condition. Find a negative radial eigenvalue when $f<0$. For $f>0$, identify the nonlocal infrared scaling after removing the leading contact term.

Solution

The linear response equation $v=G_-(\eta-fv)$ gives

$$G_f(q)=\frac{\mathcal C_\psi}{q+\mathcal C_\psi f}.$$

At $\nu=1/2$, the field expansion is $\psi=z^{3/2}(J+zA+\cdots)$, so $u(0)=J$ and $u'(0)=A$. Since $\chi=\mathcal C_\psi$, the source-free condition $\eta=A-\mathcal C_\psi fJ=0$ becomes

$$u'(0)=\mathcal C_\psi f\,u(0).$$

The radial potential vanishes for $\nu=1/2$, hence $H=-\partial_z^2$. A square-integrable bound mode has

$$u(z)=e^{-qz},\qquad H u=-q^2u,\qquad q>0.$$

Its boundary condition requires $-q=\mathcal C_\psi f$. Such a mode exists precisely when $f<0$, and its eigenvalue is $-(\mathcal C_\psi f)^2$. At zero spatial momentum, $\omega^2$ equals that negative eigenvalue, giving an exponentially growing solution. The same $q$ is the positive Euclidean pole of $G_f$.

For $f>0$ the denominator has no pole at positive $q$. At $q\ll\mathcal C_\psi f$,

$$G_f(q)=\frac1f-\frac{q}{\mathcal C_\psi f^2}+O(q^2).$$

The first term is local. The leading nonlocal term is proportional to $-q$, the standard-quantization kernel for $\Delta_+=5/2$, with a source normalization depending on $f$. This is the leading-large-$N$ double-trace flow. The sign and the pole calculation rely on the stated deformation of the Euclidean action; they would change if the sign of that deformation were changed.

5. Renormalizing the logarithmic BF source

At $\nu=0$, hold the logarithmic coefficient $J$ fixed as the cutoff is removed. Derive the finite action from the stated BF counterterm, calculate its two-point kernel, and show how the scale change $\mu\to e^t\mu$ acts on $A$. Explain why the $\nu>0$ two-point formula cannot simply be evaluated at zero.

Solution

Set $L=\log(\epsilon\mu)$ and $F=JL+A$. Up to terms vanishing with the cutoff, $\psi(\epsilon)=\epsilon^{d/2}F$ and

$$\epsilon\psi'(\epsilon) =\epsilon^{d/2}\left(\frac d2F+J\right).$$

The two surface contributions are therefore

$$\begin{aligned} S_{\rm reg}&=-\frac{\mathcal C_\psi}{2} \left(\frac d2F^2+JF\right),\\ S_{\rm ct}^{\rm BF}&=\frac{\mathcal C_\psi}{2} \left(\frac d2+\frac1L\right)F^2. \end{aligned}$$

Their sum simplifies before the limit:

$$S_{\rm reg}+S_{\rm ct}^{\rm BF} =\frac{\mathcal C_\psi}{2}\left(JA+\frac{A^2}{L}\right) \longrightarrow\frac{\mathcal C_\psi}{2}JA.$$

For a general source, restore the boundary integral. Interior regularity gives $A(q)=[\log(q/(2\mu))+\gamma_{\rm E}]J(q)$. Taking source derivatives of $W=-S_{\rm ren}$ with $J$ as the independent source yields

$$G_{\rm E}(q) =-\mathcal C_\psi\log\frac q\mu+\text{constant}.$$

The constant is a contact term. Differentiate the analytically continued Fourier power at $\nu=0$ to obtain, away from the origin,

$$\int\frac{d^d q}{(2\pi)^d}e^{iq\cdot x}\log q =-\frac{\Gamma(d/2)}{2\pi^{d/2}|x|^d}.$$

The product gives the positive coefficient quoted in the text. Keeping the field fixed under $\mu\to e^t\mu$ requires

$$J\log(z\mu)+A =J\log(ze^t\mu)+(A-tJ).$$

Hence $A$ changes by $-tJ$, and $G_{\rm E}$ changes by the local constant $\mathcal C_\psi t$.

The regular standard solution at small positive $\nu$ has normalization $q^\nu/[2^{\nu-1}\Gamma(\nu)]\sim2\nu$. To keep a finite logarithmic source, its source must be rescaled according to $J_{\rm BF}=-2\nu J_{\rm standard}$. The correlator must then be divided by $4\nu^2$ before taking the limit, and a local pole must be subtracted. That procedure recovers $-\mathcal C_\psi\log q$; setting $\nu=0$ with the old source held fixed instead gives a degenerate normalization.

6. A normalized cubic contact diagram

For three distinct scalars with $\Delta_1=\Delta_2=\Delta_3=3$ in $AdS_5$, compute the complete coefficient of the contact three-point function for the coupling $g_{123}$ used above. Derive its distance powers from inversion covariance. Explain what changes for one identical scalar with interaction $g\psi^3/3!$, and why an extremal relation cannot be substituted into the convergent integral without further analysis.

Solution

Let the distance powers be $a$, $b$, $c$ on $|x_{12}|$, $|x_{23}|$, $|x_{31}|$. Inversion covariance at the three insertions requires

$$a+c=6,\qquad a+b=6,\qquad b+c=6.$$

Thus $a=b=c=3$. These equations come from the factors of $|x_i|$ acquired under inversion, not from independently rescaling the neighborhood of only one point.

To evaluate the coefficient, send $x_3$ to infinity. The Feynman-parameter representation combines the two remaining denominators of power three. Integrating over the four boundary coordinates supplies

$$\frac{\pi^2\Gamma(4)}{\Gamma(3)^2} \int_0^1dt\,t^2(1-t)^2 \int_0^\infty dz\, \frac{z^4}{[z^2+t(1-t)|x_{12}|^2]^4}.$$

With $B=t(1-t)|x_{12}|^2$, the $z$ integral is

$$\frac{B^{-3/2}}2 \frac{\Gamma(5/2)\Gamma(3/2)}{\Gamma(4)}.$$

The remaining $t$ integral is $B(3/2,3/2)=\Gamma(3/2)^2/\Gamma(3)$. Therefore the scalar integral before the three kernel constants is

$$\frac{\pi^2}{2} \frac{\Gamma(5/2)\Gamma(3/2)^3}{\Gamma(3)^3} \frac1{|x_{12}|^3} =\frac{3\pi^4}{512|x_{12}|^3}.$$

Each source-normalized kernel has $c_3=2/\pi^2$. Restoring the other separations by covariance, the AdS measure $R^5$, and the interaction sign gives

$$\langle\mathcal O_1(x_1)\mathcal O_2(x_2)\mathcal O_3(x_3)\rangle =-\frac{3g_{123}R^5}{64\pi^2} \frac1{|x_{12}|^3|x_{23}|^3|x_{31}|^3}.$$

The dimensionless product $g_{123}R^5$ multiplies length to the power $-9$, as required for three operators of dimension three.

For an identical field, the three source derivatives of $-g\psi^3/3!$ give six identical terms. The factorial cancels, leaving the same integral with $g$ in place of $g_{123}$. Inserting an uncanceled $1/3!$ for three distinct fields would instead change the definition of the coupling and the answer by a factor of six.

At an extremal relation $\Delta_3=\Delta_1+\Delta_2$, one has $a_{12}=0$. The radial beta function develops a $\Gamma(0)$ pole, expressing a genuine failure of the stated convergence domain. A finite extremal correlator in a specific theory may involve a vanishing cubic coupling, boundary contributions, or a regulated limit; none follows by assigning an arbitrary finite value to that Gamma function.

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