Holographic Correlators and the Breitenlohner--Freedman Bound

The preceding page gave the dictionary-level slogan

$$m^2R^2=\Delta(\Delta-d),$$

for a scalar field in $AdS_{d+1}$ dual to a scalar primary operator of dimension $\Delta$ in a $d$-dimensional CFT. This page derives that relation and then uses it to compute the simplest holographic correlation functions.

The key lesson is that an AdS field has two independent near-boundary behaviors. One is interpreted as the source for the dual CFT operator; the other is interpreted as the response, or expectation value. This is the seed of the whole GKPW prescription.

Scalar fields in Euclidean AdS

Work in Euclidean Poincare coordinates

$$ds^2=R^2{dz^2+d\vec x^{\,2}\over z^2}, \qquad z>0, \qquad \vec x\in \mathbb R^d.$$

The conformal boundary is at $z=0$. The deep interior is $z\to\infty$. A free scalar with action

$$S_E[\Phi] = {\mathcal N\over2} \int d^{d+1}X\sqrt g\, \left(g^{MN}\partial_M\Phi\partial_N\Phi+m^2\Phi^2\right)$$

obeys

$${1\over\sqrt g}\partial_M\left(\sqrt g\,g^{MN}\partial_N\Phi\right)-m^2\Phi=0.$$

For the Poincare metric this becomes

$$\left[ z^2\partial_z^2-(d-1)z\partial_z+z^2\partial_i\partial_i-m^2R^2 \right]\Phi=0.$$

Fourier transforming along the boundary,

$$\Phi(z,x)=\int {d^dq\over(2\pi)^d}\,e^{iq\cdot x}\Phi_q(z), \qquad q^2\ge0,$$

gives

$$\left[ z^2{d^2\over dz^2}-(d-1)z{d\over dz}-q^2z^2-m^2R^2 \right]\Phi_q(z)=0.$$

This is a Bessel equation after writing $\Phi_q=z^{d/2}f_q$. The solution regular in the Euclidean interior is

$$\Phi_q(z)=b_q\,z^{d/2}K_\nu(qz), \qquad \nu=\sqrt{{d^2\over4}+m^2R^2}.$$

The other independent solution is $I_\nu(qz)$, which grows exponentially as $z\to\infty$ and is therefore not allowed for the Euclidean vacuum. In Lorentzian signature the analogous choice is not regularity but an appropriate real-time boundary condition, such as infalling behavior at a horizon.

The near-boundary falloffs

Near $z=0$, the $q^2z^2$ term in the wave equation is subleading. Try a power law $\Phi\sim z^\lambda$. Then

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

Thus the two possible exponents are

$$\lambda=\Delta_\pm, \qquad \Delta_\pm={d\over2}\pm\nu, \qquad \nu=\sqrt{{d^2\over4}+m^2R^2}.$$

It is conventional to call

$$\Delta\equiv\Delta_+={d\over2}+\nu.$$

Then the two falloffs are

$$\Phi(z,x) = z^{d-\Delta}\left(\phi_0(x)+\cdots\right) + z^{\Delta}\left(A(x)+\cdots\right).$$

The coefficient $\phi_0(x)$ is the boundary source. The coefficient $A(x)$ is the response. In the CFT, the source couples as

$$\delta S_{\rm CFT}=-\int d^dx\,\phi_0(x)\mathcal O(x),$$

up to a sign convention. The operator $\mathcal O$ has dimension $\Delta$ because the source has dimension $d-\Delta$.

The near-boundary expansion separates the source coefficient $\phi_0$ from the response coefficient $A$. In ordinary quantization, $\phi_0$ sources the dual operator $\mathcal O$, while $A$ determines $\langle\mathcal O\rangle$ after holographic renormalization.

A useful way to remember the powers is to ask how the bulk field behaves under the AdS scaling isometry

$$x^i\to \Lambda x^i, \qquad z\to\Lambda z.$$

The leading source term transforms as

$$z^{d-\Delta}\phi_0(x)\to \Lambda^{d-\Delta}z^{d-\Delta}\phi_0(\Lambda x).$$

Since the source coupling $\int d^dx\,\phi_0\mathcal O$ must be scale invariant, $\phi_0$ has dimension $d-\Delta$ and $\mathcal O$ has dimension $\Delta$.

Source, response, and one-point functions

The statement that $A(x)$ is the response can be made precise. In a cutoff description, specify the boundary value at $z=\epsilon$ and solve the bulk field equation. The on-shell action reduces to a boundary term,

$$S_E^{\rm os} = {\mathcal N\over2} \int_{z=\epsilon} d^dx\sqrt\gamma\,\Phi\,n^M\partial_M\Phi,$$

where $\gamma$ is the induced metric at the cutoff and $n^M$ is the outward-pointing unit normal. This expression diverges as $\epsilon\to0$. The divergences are local in the source and are removed by adding local counterterms at the cutoff surface.

After holographic renormalization, the renormalized on-shell action $S_E^{\rm ren}[\phi_0]$ is a functional of the source. In the classical supergravity limit,

$$Z_{\rm CFT}[\phi_0] = \left\langle \exp\left(\int d^dx\,\phi_0\mathcal O\right)\right\rangle \simeq \exp\left(-S_E^{\rm ren}[\phi_{\rm cl}] \right).$$

Therefore

$$W_{\rm CFT}[\phi_0]\equiv \log Z_{\rm CFT}[\phi_0] \simeq -S_E^{\rm ren}[\phi_{\rm cl}].$$

For ordinary quantization, the one-point function takes the form

$$\langle\mathcal O(x)\rangle_{\phi_0} = (2\Delta-d)\mathcal N\,A(x)+\text{local terms}.$$

The local terms depend on the counterterm scheme. They affect contact terms in correlation functions, but not the nonlocal power-law tails at separated points.

Boundary-to-bulk propagator

The regular solution with prescribed source can be written using the boundary-to-bulk propagator

$$\Phi(z,x)=\int d^dx'\,K_\Delta(z,x;x')\phi_0(x'),$$

where

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

This object is fixed almost entirely by symmetry. It solves the massive scalar equation, is regular in the Euclidean interior, and has the near-boundary distributional behavior

$$K_\Delta(z,x;x')\to z^{d-\Delta}\delta^{(d)}(x-x') \qquad (z\to0).$$

The same solution in momentum space is proportional to $z^{d/2}K_\nu(qz)$. For noninteger $\nu$, the small-$z$ expansion has the schematic form

$$z^{d/2}K_\nu(qz) = c_1(q)z^{d/2-\nu} + c_2(q)z^{d/2+\nu} + \cdots, \qquad {c_2(q)\over c_1(q)}\propto q^{2\nu}.$$

Thus the response coefficient is nonlocal in momentum space:

$$A(q)\propto q^{2\nu}\phi_0(q).$$

When $\nu$ is an integer, the nonlocal term is instead of the form

$$A(q)\propto q^{2\nu}\log q^2\,\phi_0(q),$$

with analytic polynomial terms mixed in. The polynomial terms are contact terms; the logarithm is the physically important nonlocal part.

Two-point functions

Differentiating the renormalized on-shell action twice gives the connected two-point function. With the normalization used above, the separated-point result is

$$\boxed{ \langle\mathcal O(x)\mathcal O(y)\rangle = \mathcal N(2\Delta-d) {\Gamma(\Delta)\over\pi^{d/2}\Gamma(\Delta-d/2)} {1\over |x-y|^{2\Delta}} }$$

up to an overall sign convention tied to the Euclidean source term. The power $|x-y|^{-2\Delta}$ is forced by conformal invariance; holography computes the coefficient in terms of the bulk kinetic normalization.

The example most directly relevant for the D3-brane absorption calculation is a massless scalar in $AdS_5$. Then $d=4$ and

$$m^2=0 \quad\Rightarrow\quad \Delta(\Delta-4)=0.$$

The ordinary choice is $\Delta=4$. In momentum space $\nu=2$, so the nonlocal part of the quadratic on-shell action is proportional to

$$q^4\log q^2.$$

Fourier transforming gives

$$\langle\mathcal O(x)\mathcal O(0)\rangle \propto {1\over |x|^8},$$

which is exactly what a dimension-four operator in a four-dimensional CFT requires. In $AdS_5\times S^5$, the dilaton couples to the SYM Lagrangian density, schematically

$$\mathcal O_\varphi\sim {1\over g_{\rm YM}^2}\operatorname{Tr}F^2+\cdots,$$

which has protected dimension $4$.

Negative mass squared is not automatically an instability

In flat spacetime, a scalar with $m^2<0$ is tachyonic. In AdS, the curvature changes the story. The near-boundary wave equation admits real powers as long as

$$\nu^2={d^2\over4}+m^2R^2\ge0.$$

Therefore AdS is perturbatively stable for

$$\boxed{ m^2R^2\ge -{d^2\over4} }$$

This is the Breitenlohner—Freedman bound. The intuitive reason is that AdS acts like a gravitational box. A moderately negative mass is allowed because the gradient energy associated with the boundary behavior can compensate for it. Instability begins only when the falloff exponents become complex, producing oscillatory behavior near the boundary and destroying the positive-energy theorem.

For $AdS_5$, the bound is

$$m^2R^2\ge -4.$$

The dimension formula becomes

$$\Delta=2+\sqrt{4+m^2R^2}.$$

At the bound, $m^2R^2=-4$, one has $\Delta=2$ and the two roots coincide. Coincident roots are accompanied by logarithmic branches in the near-boundary expansion, much as in an ordinary differential equation with repeated indicial roots.

The BF bound is the left edge of the stable region. For $AdS_5$, the alternate quantization window is $-4<m^2R^2<-3$, where both near-boundary falloffs are normalizable.

Alternate quantization

For most scalar masses, only one of the two falloffs is normalizable. Then the slower falloff must be treated as a source, and the faster falloff is the response. But in the special window

$$0<\nu<1,$$

both falloffs are normalizable. Equivalently,

$$-{d^2\over4}<m^2R^2<-{d^2\over4}+1.$$

In this range there are two consistent CFTs associated with the same bulk scalar. In ordinary quantization, the operator dimension is

$$\Delta_+={d\over2}+\nu.$$

In alternate quantization, the roles of source and response are exchanged, and the operator dimension is

$$\Delta_-={d\over2}-\nu.$$

This is consistent with the scalar unitarity bound in $d$ dimensions,

$$\Delta\ge {d-2\over2},$$

because $\Delta_-$ lies above that bound precisely when $\nu\le1$. For $AdS_5$, the alternate window is

$$-4<m^2R^2<-3,$$

and $1<\Delta_-<2$. At the upper endpoint $m^2R^2=-3$, the alternate dimension reaches the four-dimensional scalar unitarity bound $\Delta=1$.

This is not merely a curiosity. In holographic RG flows, changing the boundary condition of a scalar in the alternate window is dual to adding a double-trace deformation such as

$${f\over2}\int d^dx\,\mathcal O^2.$$

The radial evolution of the boundary condition then encodes the beta function of the double-trace coupling.

Three-point functions and Witten diagrams

The same logic extends beyond two-point functions. Suppose the bulk action contains a cubic interaction

$$S_{\rm int} = {g_{123}\over3!} \int d^{d+1}X\sqrt g\,\Phi_1\Phi_2\Phi_3.$$

At tree level, the connected three-point function is obtained by inserting three boundary-to-bulk propagators and integrating over the interaction point in AdS:

$$\langle\mathcal O_1(x_1)\mathcal O_2(x_2)\mathcal O_3(x_3)\rangle \propto -g_{123}\int_{AdS} d^{d+1}X\sqrt g\, K_{\Delta_1}(X;x_1)K_{\Delta_2}(X;x_2)K_{\Delta_3}(X;x_3).$$

The integral has the form required by conformal invariance,

$$\boxed{ \langle\mathcal O_1(x_1)\mathcal O_2(x_2)\mathcal O_3(x_3)\rangle = {C_{123}\over |x_{12}|^{\Delta_1+\Delta_2-\Delta_3} |x_{13}|^{\Delta_1+\Delta_3-\Delta_2} |x_{23}|^{\Delta_2+\Delta_3-\Delta_1}} }$$

where $x_{ij}=x_i-x_j$. The powers are kinematic; they are dictated by scale invariance and special conformal invariance. The coefficient $C_{123}$ is dynamical and is determined by the bulk cubic coupling and by the normalization of the fields.

A cubic bulk vertex joined to three boundary points by boundary-to-bulk propagators gives the leading large-$N$, strong-coupling contribution to a CFT three-point function.

For canonically normalized scalar propagators, the AdS integral contains the standard factor

$${\pi^{d/2}\over2} {\Gamma\left({\Sigma-d\over2}\right) \Gamma\left({\Delta_1+\Delta_2-\Delta_3\over2}\right) \Gamma\left({\Delta_1+\Delta_3-\Delta_2\over2}\right) \Gamma\left({\Delta_2+\Delta_3-\Delta_1\over2}\right) \over \Gamma(\Delta_1)\Gamma(\Delta_2)\Gamma(\Delta_3)}, \qquad \Sigma=\Delta_1+\Delta_2+\Delta_3,$$

multiplied by the three normalization constants $C_{\Delta_i}$ and by the coupling $g_{123}$. Divergences in this expression signal contact terms, extremal correlators, or the need to treat boundary terms and field redefinitions with more care.

What is universal and what is not

Several parts of the calculation are universal:

• the mass-dimension relation $m^2R^2=\Delta(\Delta-d)$;
• the power laws $|x|^{-2\Delta}$ and the three-point conformal structure;
• the BF bound $m^2R^2\ge -d^2/4$;
• the existence of alternate quantization when $0<\nu<1$.

The numerical coefficients are more delicate. They depend on the normalization of the bulk fields, the normalization of CFT operators, and the holographic counterterm scheme. Scheme dependence only changes contact terms at separated points, but it matters when comparing precise Ward identities, anomalies, and integrated correlators. The next page uses this machinery to compute anomaly coefficients and the central charge of $\mathcal N=4$ SYM from the five-dimensional gravitational action.

Exercises

Exercise 1. The indicial equation

Starting from

$$\left[ z^2\partial_z^2-(d-1)z\partial_z+z^2\partial_i\partial_i-m^2R^2 \right]\Phi=0,$$

ignore the boundary-derivative term near $z=0$ and set $\Phi\sim z^\lambda$. Derive the relation between $\lambda$ and $m^2$.

Solution

With $\Phi=z^\lambda$,

$$\partial_z\Phi=\lambda z^{\lambda-1}, \qquad \partial_z^2\Phi=\lambda(\lambda-1)z^{\lambda-2}.$$

Thus

$$z^2\partial_z^2\Phi-(d-1)z\partial_z\Phi = \left[\lambda(\lambda-1)-(d-1)\lambda\right]z^\lambda = \lambda(\lambda-d)z^\lambda.$$

The leading near-boundary equation is therefore

$$\left[\lambda(\lambda-d)-m^2R^2\right]z^\lambda=0,$$

so

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

The two roots are $\lambda=d/2\pm\sqrt{d^2/4+m^2R^2}$.

Exercise 2. Dimensions of a massless scalar in $AdS_5$

For $d=4$ and $m^2=0$, solve

$$\Delta(\Delta-4)=0.$$

Which root is used for the ordinary quantization of the dilaton in $AdS_5\times S^5$?

Solution

The roots are

$$\Delta=0, \qquad \Delta=4.$$

The ordinary quantization uses

$$\Delta=4.$$

The corresponding source has dimension $d-\Delta=0$, as expected for a coupling. In the D3-brane system, the dilaton source changes the Yang—Mills coupling and couples to a dimension-four operator, schematically $\operatorname{Tr}F^2$ plus its supersymmetric completion.

Exercise 3. The BF bound in $AdS_5$

Use

$$\nu=\sqrt{4+m^2R^2}$$

for $AdS_5$ to find the stability bound. What happens to the two dimensions $\Delta_\pm$ at the bound?

Solution

Stability requires $\nu$ to be real:

$$4+m^2R^2\ge0.$$

Therefore

$$m^2R^2\ge -4.$$

At the bound, $\nu=0$, so

$$\Delta_+=\Delta_-={d\over2}=2.$$

The two power-law solutions coincide. The second independent near-boundary solution then contains a logarithm.

Exercise 4. The alternate quantization window

Show that the alternate quantization window in $AdS_{d+1}$ is

$$-{d^2\over4}<m^2R^2<-{d^2\over4}+1.$$

Then specialize to $AdS_5$.

Solution

Alternate quantization is possible when both falloffs are normalizable and the lower dimension does not violate the scalar unitarity bound. This is the window

$$0<\nu<1, \qquad \nu^2={d^2\over4}+m^2R^2.$$

Squaring gives

$$0<{d^2\over4}+m^2R^2<1.$$

Therefore

$$-{d^2\over4}<m^2R^2<-{d^2\over4}+1.$$

For $d=4$ this becomes

$$-4<m^2R^2<-3.$$

In this range the ordinary dimension is $\Delta_+=2+\nu$ and the alternate dimension is $\Delta_-=2-\nu$.

Exercise 5. Scaling of the boundary-to-bulk propagator

Verify that

$$K_\Delta(z,x;x')=C_\Delta \left({z\over z^2+|x-x'|^2}\right)^\Delta$$

has the correct scaling under $z\to\Lambda z$, $x\to\Lambda x$, and $x'\to\Lambda x'$.

Solution

The numerator scales as

$$z\to\Lambda z.$$

The denominator scales as

$$z^2+|x-x'|^2 \to \Lambda^2\left(z^2+|x-x'|^2\right).$$

Therefore the ratio scales as

$${z\over z^2+|x-x'|^2} \to \Lambda^{-1}{z\over z^2+|x-x'|^2}.$$

Hence

$$K_\Delta\to\Lambda^{-\Delta}K_\Delta.$$

This is appropriate for an object that propagates a boundary insertion of dimension $\Delta$ into the bulk.

Exercise 6. Momentum-space nonlocality

For a scalar in $AdS_{d+1}$ with noninteger $\nu$, the regular solution behaves near the boundary as

$$\Phi_q(z)=z^{d/2-\nu}\phi_0(q)+z^{d/2+\nu}A(q)+\cdots.$$

Use dimensional analysis in momentum space to show that the nonlocal part of $A(q)$ must be proportional to $q^{2\nu}\phi_0(q)$.

Solution

The two coefficients multiply powers whose exponents differ by

$$\left({d\over2}+\nu\right)-\left({d\over2}-\nu\right)=2\nu.$$

Momentum $q$ has inverse-length dimension $1$. To convert the source coefficient into the response coefficient, the regular solution can only use the scale $q$, so the nonlocal relation must have the form

$$A(q)\propto q^{2\nu}\phi_0(q).$$

When $\nu$ is an integer, the two Bessel-series branches mix and the nonlocal term becomes $q^{2\nu}\log q^2$ times the source, up to analytic contact terms.

Exercise 7. Three-point conformal powers

Assume three scalar primaries have dimensions $\Delta_1$, $\Delta_2$, and $\Delta_3$. Show that the three-point form

$$\langle\mathcal O_1(x_1)\mathcal O_2(x_2)\mathcal O_3(x_3)\rangle = {C_{123}\over |x_{12}|^{a}|x_{13}|^{b}|x_{23}|^{c}}$$

has the correct scaling at each point if

$$a=\Delta_1+\Delta_2-\Delta_3, \qquad b=\Delta_1+\Delta_3-\Delta_2, \qquad c=\Delta_2+\Delta_3-\Delta_1.$$

Solution

Near $x_1$, the distances involving $x_1$ are $|x_{12}|$ and $|x_{13}|$. The total power associated with $x_1$ is therefore

$$a+b = (\Delta_1+\Delta_2-\Delta_3)+(\Delta_1+\Delta_3-\Delta_2) = 2\Delta_1.$$

This is the correct local scaling for a scalar operator of dimension $\Delta_1$. Similarly,

$$a+c=2\Delta_2, \qquad b+c=2\Delta_3.$$

Under a common scaling $x_i\to\Lambda x_i$, the denominator scales as

$$\Lambda^{a+b+c} = \Lambda^{\Delta_1+\Delta_2+\Delta_3},$$

which is exactly the expected scaling of a three-point function of scalar primaries.