Three-Point Functions and Cubic Couplings

The main idea

The previous two pages developed the two ingredients needed for the first interacting calculation in AdS/CFT:

1. A boundary source $\phi_{(0)}(x)$ determines a classical bulk solution through a bulk-to-boundary propagator.
2. The renormalized on-shell action generates CFT correlation functions.

Two-point functions are mostly kinematics plus normalization. For a scalar primary, conformal symmetry fixes

$$\langle \mathcal O(x_1)\mathcal O(x_2)\rangle = \frac{C_{\mathcal O}}{|x_{12}|^{2\Delta}}, \qquad x_{ij}=x_i-x_j.$$

The first genuinely dynamical CFT data appear in three-point functions. For scalar primaries,

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

at separated points. Conformal symmetry fixes the entire position dependence. The number $C_{123}$ is dynamical. Equivalently, after normalizing two-point functions, it is the OPE coefficient.

In holography, the leading large-$N$ contribution to $C_{123}$ is computed by a cubic interaction in the bulk:

$$S_{\rm int} = \lambda_{123} \int_{\rm AdS}d^{d+1}X\sqrt g\,\phi_1\phi_2\phi_3.$$

The corresponding diagram is the simplest contact Witten diagram: three bulk-to-boundary propagators meet at one integrated bulk point.

A cubic bulk vertex computes a scalar CFT three-point function. Each boundary source $J_i(x_i)$ creates a classical bulk field through $K_{\Delta_i}(X;x_i)$; the interaction point $X$ is integrated over AdS. The result has the unique scalar conformal structure, while the overall coefficient is determined by the renormalized cubic coupling.

This page is conceptually important because it explains how the dynamical bulk Lagrangian becomes the CFT operator algebra. Masses determine dimensions. Kinetic terms determine two-point normalizations. Cubic couplings determine three-point coefficients.

What the CFT side is asking for

A CFT is not defined only by its list of scaling dimensions. For local scalar primaries, the basic data include

$$\Delta_i, \qquad C_i, \qquad C_{ijk}, \qquad \ldots$$

where

$$\langle \mathcal O_i(x)\mathcal O_j(0)\rangle = \frac{C_i\delta_{ij}}{|x|^{2\Delta_i}}$$

in a diagonal basis. The operator product expansion then has the schematic form

$$\mathcal O_i(x)\mathcal O_j(0) \sim \sum_k \frac{C_{ijk}}{C_k} \frac{\mathcal O_k(0)}{|x|^{\Delta_i+\Delta_j-\Delta_k}} + \text{descendants}.$$

If we define normalized operators

$$\widehat{\mathcal O}_i = \frac{\mathcal O_i}{\sqrt{C_i}},$$

then

$$\langle \widehat{\mathcal O}_i(x)\widehat{\mathcal O}_j(0)\rangle = \frac{\delta_{ij}}{|x|^{2\Delta_i}},$$

and the normalized OPE coefficient is

$$\boxed{ \lambda_{ijk}^{\rm CFT} = \frac{C_{ijk}}{\sqrt{C_iC_jC_k}}. }$$

This distinction is not pedantic. Holographic computations naturally produce numbers in a chosen bulk-field normalization. To compare with bootstrap conventions, one must divide by the square roots of the two-point coefficients.

For scalar primaries, the three-point position dependence is fixed by translations, rotations, dilatations, and special conformal transformations. It is useful to define

$$\delta_{12}=\frac{\Delta_1+\Delta_2-\Delta_3}{2}, \qquad \delta_{13}=\frac{\Delta_1+\Delta_3-\Delta_2}{2}, \qquad \delta_{23}=\frac{\Delta_2+\Delta_3-\Delta_1}{2}.$$

Then the scalar three-point function is

$$\langle \mathcal O_1(x_1)\mathcal O_2(x_2)\mathcal O_3(x_3)\rangle = \frac{C_{123}}{|x_{12}|^{2\delta_{12}}|x_{13}|^{2\delta_{13}}|x_{23}|^{2\delta_{23}}}.$$

The goal of the bulk calculation is to compute $C_{123}$.

The bulk setup

Work in Euclidean Poincare AdS$_{d+1}$,

$$ds^2 = \frac{L^2}{z^2}(dz^2+d x^\mu d x_\mu), \qquad z>0.$$

Consider three scalar fields $\phi_i$ with masses

$$m_i^2L^2 = \Delta_i(\Delta_i-d), \qquad i=1,2,3.$$

A convenient schematic Euclidean action is

$$S_E = \sum_{i=1}^3 \frac{\mathcal N_i}{2} \int d^{d+1}X\sqrt g \left( g^{ab}\partial_a\phi_i\partial_b\phi_i+m_i^2\phi_i^2 \right) + \lambda_{123} \int d^{d+1}X\sqrt g\,\phi_1\phi_2\phi_3 + S_{\rm ct}.$$

Here $\mathcal N_i$ fixes the two-point normalization of $\mathcal O_i$. The coupling $\lambda_{123}$ is written in the same field normalization. In a top-down compactification, $\mathcal N_i$ and $\lambda_{123}$ are not arbitrary: they come from reducing the ten- or eleven-dimensional supergravity or string action. In a bottom-up model, they are phenomenological parameters specifying the CFT data one wants to model.

For three distinct fields the cubic term has no symmetry factor. If the interaction is instead

$$S_{\rm int}=\frac{\lambda}{3!}\int\sqrt g\,\phi^3,$$

the three functional derivatives with respect to the same source produce a factor $3!$, so the final answer again contains $\lambda$ rather than $\lambda/3!$.

At leading order in $\lambda_{123}$, the classical solution sourced by $J_i(x)$ is just the free solution:

$$\phi_i(X) = \int d^d x\,K_{\Delta_i}(X;x)J_i(x)+O(J^2).$$

The normalized Euclidean bulk-to-boundary propagator is

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

The normalization is chosen so that

$$\lim_{z\to0}z^{\Delta-d}K_\Delta(z,x;y) = \delta^{(d)}(x-y).$$

Substituting the free solutions into the cubic term gives the cubic part of the renormalized generating functional at separated points:

$$S_{\rm ren}^{(3)}[J_1,J_2,J_3] = \lambda_{123} \int_{\rm AdS}d^{d+1}X\sqrt g \prod_{i=1}^3 \left[ \int d^d x_i\,K_{\Delta_i}(X;x_i)J_i(x_i) \right] + \text{local contact terms}.$$

Using the convention of the previous pages, where functional derivatives of $S_{\rm ren}$ generate correlators, the separated-point three-point function is

$$\boxed{ \langle \mathcal O_1(x_1)\mathcal O_2(x_2)\mathcal O_3(x_3) \rangle_{\rm conn} = \lambda_{123} \int_{\rm AdS}d^{d+1}X\sqrt g\, K_{\Delta_1}(X;x_1)K_{\Delta_2}(X;x_2)K_{\Delta_3}(X;x_3) + \text{contact terms}. }$$

Different Euclidean sign conventions move an overall sign between the action and the generating functional. That sign is not the interesting physics here. The physical data are the separated-point coefficient, the normalization of the operators, and the tensor structure.

The scalar contact integral

Set $L=1$ for the moment and strip off the normalization constants $C_{\Delta_i}$. Define

$$D_{\Delta_1\Delta_2\Delta_3}(x_1,x_2,x_3) = \int_0^\infty\frac{dz}{z^{d+1}} \int d^d x \prod_{i=1}^3 \left( \frac{z}{z^2+|x-x_i|^2} \right)^{\Delta_i}.$$

Let

$$\Sigma=\Delta_1+\Delta_2+\Delta_3.$$

For non-extremal dimensions in the domain where the integral converges, and elsewhere by analytic continuation plus holographic renormalization, the result is

$$\boxed{ D_{\Delta_1\Delta_2\Delta_3} = \frac{\pi^{d/2}}{2} \frac{ \Gamma\left(\frac{\Sigma-d}{2}\right) \Gamma(\delta_{12}) \Gamma(\delta_{13}) \Gamma(\delta_{23}) }{ \Gamma(\Delta_1)\Gamma(\Delta_2)\Gamma(\Delta_3) } \frac{1}{ |x_{12}|^{2\delta_{12}} |x_{13}|^{2\delta_{13}} |x_{23}|^{2\delta_{23}} }. }$$

Restoring the propagator normalizations and the AdS radius gives

$$\langle \mathcal O_1(x_1)\mathcal O_2(x_2)\mathcal O_3(x_3)\rangle = \frac{\mathfrak C_{123}}{ |x_{12}|^{2\delta_{12}} |x_{13}|^{2\delta_{13}} |x_{23}|^{2\delta_{23}} },$$

with

$$\boxed{ \mathfrak C_{123} = \lambda_{123}L^{d+1} \prod_{i=1}^3 C_{\Delta_i} \frac{\pi^{d/2}}{2} \frac{ \Gamma\left(\frac{\Sigma-d}{2}\right) \Gamma(\delta_{12}) \Gamma(\delta_{13}) \Gamma(\delta_{23}) }{ \Gamma(\Delta_1)\Gamma(\Delta_2)\Gamma(\Delta_3) } }$$

up to the field-normalization and sign conventions already stated.

This formula is one of the cleanest examples of the AdS/CFT dictionary. The bulk integral knows about the boundary conformal group because AdS isometries act as boundary conformal transformations. The bulk-to-boundary propagator transforms like a primary of dimension $\Delta$, and the invariant AdS measure does the rest.

Why the answer had to have this form

There is a quick conceptual derivation of the position dependence. Under an AdS isometry corresponding to a boundary conformal transformation $x\mapsto x'$, the bulk point transforms as $X\mapsto X'$, the measure is invariant,

$$d^{d+1}X\sqrt g=d^{d+1}X'\sqrt g,$$

and the bulk-to-boundary propagator transforms as

$$K_\Delta(X;x) = \left|\frac{\partial x'}{\partial x}\right|^{\Delta/d} K_\Delta(X';x').$$

Therefore the integral transforms as a product of three scalar primaries. There is no conformal cross-ratio for three points, so the answer must be the unique scalar three-point structure. The only nontrivial information left is the coefficient.

This is a good diagnostic principle:

If a tree-level scalar contact diagram in pure AdS gives a separated-point answer that is not the scalar CFT three-point structure, the calculation has violated the boundary condition, the normalization, or conformal covariance.

A worked special case: three equal dimensions

For $\Delta_1=\Delta_2=\Delta_3=\Delta$, one has

$$\delta_{12}=\delta_{13}=\delta_{23}=\frac{\Delta}{2}, \qquad \Sigma=3\Delta.$$

The stripped contact integral becomes

$$D_{\Delta\Delta\Delta} = \frac{\pi^{d/2}}{2} \frac{ \Gamma\left(\frac{3\Delta-d}{2}\right) \Gamma\left(\frac{\Delta}{2}\right)^3 }{ \Gamma(\Delta)^3 } \frac{1}{|x_{12}|^{\Delta}|x_{13}|^{\Delta}|x_{23}|^{\Delta}}.$$

Including the propagator normalization gives

$$\mathfrak C_{\Delta\Delta\Delta} = \lambda L^{d+1} \left[ \frac{\Gamma(\Delta)}{\pi^{d/2}\Gamma(\Delta-d/2)} \right]^3 \frac{\pi^{d/2}}{2} \frac{ \Gamma\left(\frac{3\Delta-d}{2}\right) \Gamma\left(\frac{\Delta}{2}\right)^3 }{ \Gamma(\Delta)^3 }.$$

For example, for three marginal scalar operators in $d=4$, so $\Delta=4$, the normalized integral part is proportional to

$$\frac{1}{|x_{12}|^4|x_{13}|^4|x_{23}|^4},$$

as conformal invariance demands. The number multiplying this structure depends on the normalization of the bulk fields and on the cubic coupling.

Derivative cubic couplings

Real supergravity actions rarely contain only $\phi_1\phi_2\phi_3$. They contain derivative interactions, mixing terms, gauge couplings, and terms produced by dimensional reduction. A simple example is

$$S_{\rm int}^{\nabla} = a_{123} \int d^{d+1}X\sqrt g\, \phi_3\nabla_a\phi_1\nabla^a\phi_2.$$

For separated scalar three-point functions, many derivative interactions reduce to the same conformal structure with a modified coefficient. Using the free equations of motion

$$\nabla^2\phi_i=m_i^2\phi_i,$$

and integrating by parts, one obtains

$$\int\sqrt g\, \phi_3\nabla_a\phi_1\nabla^a\phi_2 = \frac{1}{2} \left(m_3^2-m_1^2-m_2^2\right) \int\sqrt g\,\phi_1\phi_2\phi_3 + \text{boundary terms}.$$

Thus this derivative coupling shifts the effective scalar cubic coefficient by

$$\lambda_{123}^{\rm eff} = \lambda_{123} + \frac{a_{123}}{2} \left(m_3^2-m_1^2-m_2^2\right) + \cdots,$$

where the dots denote other derivative structures and possible boundary contributions.

The phrase “boundary terms” should not be skipped over. At separated, non-extremal points, many such terms are contact terms or vanish after renormalization. In extremal and near-extremal correlators, however, boundary terms can carry finite physical information. This is one reason why top-down three-point computations require the properly reduced and renormalized action, not only the bulk equations of motion.

For spinning operators the situation is richer. Derivative couplings are no longer merely coefficient shifts. They produce different conformal tensor structures. A cubic vertex such as

$$A^a\phi_1\nabla_a\phi_2$$

computes a current-scalar-scalar three-point function, while graviton couplings compute stress-tensor correlators. Ward identities then relate some three-point coefficients to two-point normalizations.

From cubic couplings to OPE coefficients

Suppose the holographic two-point functions are

$$\langle \mathcal O_i(x)\mathcal O_i(0)\rangle = \frac{C_i}{|x|^{2\Delta_i}},$$

and the cubic computation gives

$$\langle \mathcal O_1(x_1)\mathcal O_2(x_2)\mathcal O_3(x_3)\rangle = \frac{\mathfrak C_{123}}{|x_{12}|^{2\delta_{12}}|x_{13}|^{2\delta_{13}}|x_{23}|^{2\delta_{23}}}.$$

Then the normalized OPE coefficient is

$$\boxed{ \lambda_{123}^{\rm CFT} = \frac{\mathfrak C_{123}}{\sqrt{C_1C_2C_3}}. }$$

This is often the number that should be compared with conformal-bootstrap conventions.

In a top-down compactification, the cubic coupling itself often has the schematic form

$$\lambda_{123}^{(d+1)} \sim \frac{1}{16\pi G_D} \int_Y d^q y\sqrt{g_Y}\, \mathcal Y_1(y)\mathcal Y_2(y)\mathcal Y_3(y) \times \left(\text{derivatives, flux factors, metric factors}\right),$$

where $D=d+1+q$ and $\mathcal Y_i$ are internal harmonics or wavefunctions on the compact space $Y$. This immediately explains many selection rules. For $\mathrm{AdS}_5\times S^5$, the internal harmonics transform under $SO(6)$, so a three-point function is allowed only when the tensor product of the corresponding $SO(6)$ representations contains a singlet.

The internal-space overlap is the bulk version of a CFT statement: global symmetry representations constrain which OPE coefficients can be nonzero.

Large-$N$ scaling

The overall size of cubic Witten diagrams is controlled by the gravitational coupling. For a classical Einstein-like bulk dual, the effective action has the schematic form

$$S_{\rm bulk} \sim C_T \int d^{d+1}X\sqrt g \left[ \frac{1}{2}(\nabla\phi)^2 + \frac{1}{2}m^2\phi^2 +g_3\phi^3 +g_4\phi^4 +\cdots \right],$$

where

$$C_T \sim \frac{L^{d-1}}{G_{d+1}}$$

is proportional to the stress-tensor two-point coefficient of the CFT. In the canonical AdS$_5$/CFT$_4$ example, $C_T\sim N^2$.

With the source-normalized fields above, connected $n$-point functions of unnormalized single-trace operators scale as

$$\langle \mathcal O_1\cdots\mathcal O_n\rangle_{\rm conn} \sim C_T$$

at tree level, because the whole classical action is multiplied by $C_T$. After normalizing each operator to have a unit two-point function,

$$\widehat{\mathcal O}_i=\frac{\mathcal O_i}{\sqrt{C_T}},$$

the connected tree-level scaling becomes

$$\boxed{ \langle \widehat{\mathcal O}_1\cdots\widehat{\mathcal O}_n \rangle_{\rm conn}^{\rm tree} \sim C_T^{1-n/2}. }$$

Thus normalized scalar three-point coefficients scale as

$$\lambda_{123}^{\rm CFT} \sim C_T^{-1/2} \sim \frac{1}{N}$$

for matrix large-$N$ theories with $C_T\sim N^2$. This is the same large-$N$ factorization seen from the boundary. In canonically normalized bulk fields,

$$\phi_{\rm can} \sim \sqrt{C_T}\,\phi,$$

the cubic coupling is explicitly suppressed by $1/\sqrt{C_T}$. Bulk perturbation theory is therefore the same expansion as large-$N$ factorization.

Extremal and near-extremal subtleties

The contact integral contains gamma functions

$$\Gamma(\delta_{12}), \qquad \Gamma(\delta_{13}), \qquad \Gamma(\delta_{23}).$$

If one dimension equals the sum of the other two, for example

$$\Delta_3=\Delta_1+\Delta_2,$$

then

$$\delta_{12}=0,$$

and the naive bulk integral diverges. This is the extremal case. Geometrically, the divergence comes from a near-boundary region of the bulk integral. In many protected supergravity examples, the corresponding cubic coupling vanishes at the same rate as the integral diverges, so the product is finite after analytic continuation. In other cases, boundary terms are essential.

The lesson is simple but important:

Extremal three-point functions cannot be computed by blindly inserting dimensions into the non-extremal contact integral.

One must use the renormalized action, including boundary terms and the correct limiting procedure. These subtleties are especially common for protected chiral-primary correlators in supersymmetric examples.

There are also logarithmic cases. When dimensions make powers in the near-boundary expansion collide, holographic renormalization produces logarithmic counterterms. These affect contact terms and anomaly-related terms. At separated points the nonlocal conformal structure remains fixed, but the precise local terms are scheme-dependent.

Contact terms and scheme dependence

The formula above is a separated-point result. Local counterterms can add terms such as

$$\delta^{(d)}(x_1-x_2)\frac{1}{|x_{13}|^{2\Delta_3}}, \qquad \Box_{x_1}\delta^{(d)}(x_1-x_2)\delta^{(d)}(x_1-x_3),$$

and similar descendants. These are contact terms. They matter for Ward identities, anomalies, integrated correlators, and source-dependent one-point functions. They do not change the separated-point OPE coefficient $\lambda_{123}^{\rm CFT}$.

Bulk field redefinitions can also move terms between different cubic vertices and boundary terms. For example, redefining

$$\phi_3\to \phi_3+a\phi_1\phi_2$$

changes the apparent cubic Lagrangian. A physical separated-point CFT three-point coefficient cannot depend on such a choice. The invariant object is the properly renormalized on-shell action as a functional of the boundary sources.

Practical recipe

To compute a scalar three-point coefficient holographically:

1. Identify the operators and fields. Determine $m_i^2L^2=\Delta_i(\Delta_i-d)$ and the correct standard or alternate quantization.
2. Normalize the quadratic action. Compute the two-point coefficients $C_i$ in the same conventions.
3. Find the cubic action. Include derivative terms, mixing terms, boundary terms, and compactification overlap factors.
4. Evaluate the contact integral. Reduce derivative vertices when possible, or compute the appropriate tensor integral.
5. Renormalize. Separate the physical separated-point coefficient from contact terms and scheme-dependent local terms.
6. Normalize the OPE coefficient. Divide by $\sqrt{C_1C_2C_3}$ if comparing to unit-normalized CFT operators.
7. Check symmetries. Global charges, spins, parity, supersymmetry, and Ward identities should match the CFT expectations.

This recipe generalizes. Four-point contact diagrams use the same logic but are no longer fixed by conformal symmetry because four points have cross-ratios. Exchange diagrams require bulk-to-bulk propagators. Loops require the quantum bulk theory. Those are the subject of the next Witten-diagram page.

Common mistakes

Mistake	Why it is wrong	Safer statement
“The cubic coupling is the OPE coefficient.”	It is only true after propagator normalizations, AdS integrals, two-point normalizations, and renormalization are accounted for.	Cubic couplings determine OPE coefficients.
Forgetting two-point normalization	Bootstrap OPE coefficients usually use unit-normalized operators.	Always compute $\lambda_{123}^{\rm CFT}=\mathfrak C_{123}/\sqrt{C_1C_2C_3}$.
Ignoring derivative terms	Supergravity cubic actions often contain derivatives and field mixing.	Reduce derivative vertices on shell only after checking boundary terms.
Treating extremal limits as ordinary integrals	Gamma functions can diverge and cubic couplings may vanish simultaneously.	Use analytic continuation and the full renormalized action.
Confusing contact terms with separated OPE data	Counterterms change local distributions but not separated-point scalar structures.	Specify whether the result is at separated points or as a distribution.
Assuming large $N$ means classical Einstein gravity	Large $N$ suppresses loops, but locality also requires a large gap to stringy or higher-spin states.	Classical supergravity requires both large $C_T$ and a sparse low-dimension single-trace spectrum.

Exercises

Exercise 1: Fix the scalar three-point exponents

Assume the scalar three-point function has the form

$$G(x_1,x_2,x_3) = \frac{C}{|x_{12}|^{a}|x_{13}|^{b}|x_{23}|^{c}}.$$

Use scaling at each insertion to show that

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

Solution

Under a global scale transformation $x_i\to \Lambda x_i$, the whole correlator must scale as

$$G(\Lambda x_1,\Lambda x_2,\Lambda x_3) = \Lambda^{-(\Delta_1+\Delta_2+\Delta_3)}G(x_1,x_2,x_3),$$

so

$$a+b+c=\Delta_1+\Delta_2+\Delta_3.$$

A sharper way is to apply inversion or, equivalently, demand the correct conformal weight at each point. The factors containing $x_1$ are $|x_{12}|^{-a}$ and $|x_{13}|^{-b}$, so insertion $1$ has weight

$$a+b=2\Delta_1.$$

Similarly,

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

Solving these three equations gives

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

Exercise 2: Evaluate the contact integral by Schwinger parameters

Use

$$\frac{1}{A^\Delta} = \frac{1}{\Gamma(\Delta)} \int_0^\infty d\alpha\,\alpha^{\Delta-1}e^{-\alpha A}$$

to derive the position dependence of

$$D_{\Delta_1\Delta_2\Delta_3} = \int_0^\infty\frac{dz}{z^{d+1}} \int d^d x \prod_{i=1}^3 \left( \frac{z}{z^2+|x-x_i|^2} \right)^{\Delta_i}.$$

You do not need to reproduce every gamma-function coefficient, but you should obtain the powers of $|x_{ij}|$.

Solution

Write each denominator with a Schwinger parameter:

$$(z^2+|x-x_i|^2)^{-\Delta_i} = \frac{1}{\Gamma(\Delta_i)} \int_0^\infty d\alpha_i\, \alpha_i^{\Delta_i-1} e^{-\alpha_i(z^2+|x-x_i|^2)}.$$

The exponent involving $x$ is

$$\sum_i\alpha_i|x-x_i|^2 = A|x-\bar x|^2 + \frac{1}{A}\sum_{i<j}\alpha_i\alpha_j|x_{ij}|^2,$$

where

$$A=\alpha_1+\alpha_2+\alpha_3, \qquad \bar x=\frac{1}{A}\sum_i\alpha_i x_i.$$

The $x$ integral gives a Gaussian factor $A^{-d/2}$. The $z$ integral is also elementary after combining $e^{-Az^2}$ with the numerator $z^{\Sigma-d-1}$:

$$\int_0^\infty dz\,z^{\Sigma-d-1}e^{-Az^2} = \frac{1}{2}A^{-(\Sigma-d)/2} \Gamma\left(\frac{\Sigma-d}{2}\right).$$

Now change variables to $\alpha_i=A u_i$ with $u_1+u_2+u_3=1$. The remaining integral over the simplex has exponential

$$\exp\left[-A\sum_{i<j}u_iu_j|x_{ij}|^2\right].$$

The final $A$ integral and simplex integral produce the unique homogeneous conformal structure. Dimensional covariance and symmetry among the three points then fix the powers to be

$$D_{\Delta_1\Delta_2\Delta_3} \propto \frac{1}{|x_{12}|^{2\delta_{12}}|x_{13}|^{2\delta_{13}}|x_{23}|^{2\delta_{23}}},$$

with

$$\delta_{12}=\frac{\Delta_1+\Delta_2-\Delta_3}{2}, \qquad \delta_{13}=\frac{\Delta_1+\Delta_3-\Delta_2}{2}, \qquad \delta_{23}=\frac{\Delta_2+\Delta_3-\Delta_1}{2}.$$

Keeping track of the beta-function integrals gives the gamma-function coefficient displayed in the main text.

Exercise 3: Reduce a derivative vertex

Show that, for free external scalar solutions satisfying $\nabla^2\phi_i=m_i^2\phi_i$,

$$\int\sqrt g\,\phi_3\nabla_a\phi_1\nabla^a\phi_2 = \frac{1}{2}(m_3^2-m_1^2-m_2^2) \int\sqrt g\,\phi_1\phi_2\phi_3 + \text{boundary terms}.$$

Solution

Use the identity

$$\nabla_a\phi_1\nabla^a\phi_2 = \frac{1}{2} \left[ \nabla^2(\phi_1\phi_2)-\phi_1\nabla^2\phi_2-\phi_2\nabla^2\phi_1 \right].$$

Multiplying by $\phi_3$ and integrating gives

$$I = \frac{1}{2} \int\sqrt g\,\phi_3\nabla^2(\phi_1\phi_2) - \frac{1}{2}(m_1^2+m_2^2) \int\sqrt g\,\phi_1\phi_2\phi_3.$$

Integrate the first term by parts twice:

$$\int\sqrt g\,\phi_3\nabla^2(\phi_1\phi_2) = \int\sqrt g\,(\nabla^2\phi_3)\phi_1\phi_2 + \text{boundary terms}.$$

Using $\nabla^2\phi_3=m_3^2\phi_3$ yields

$$I = \frac{1}{2}(m_3^2-m_1^2-m_2^2) \int\sqrt g\,\phi_1\phi_2\phi_3 + \text{boundary terms}.$$

Exercise 4: Large-$N$ normalization of a three-point coefficient

Assume a matrix large-$N$ CFT has $C_T\sim N^2$. A single-trace operator $\mathcal O$ is normalized so that

$$\langle \mathcal O(x)\mathcal O(0)\rangle\sim \frac{N^2}{|x|^{2\Delta}},$$

and the connected three-point function scales as

$$\langle \mathcal O\mathcal O\mathcal O\rangle_{\rm conn}\sim N^2.$$

Define $\widehat{\mathcal O}=\mathcal O/N$. How does

$$\langle \widehat{\mathcal O}\widehat{\mathcal O}\widehat{\mathcal O}\rangle_{\rm conn}$$

scale with $N$?

Solution

Each normalized operator contributes a factor $1/N$. Therefore

$$\langle \widehat{\mathcal O}\widehat{\mathcal O}\widehat{\mathcal O}\rangle_{\rm conn} = \frac{1}{N^3} \langle \mathcal O\mathcal O\mathcal O\rangle_{\rm conn} \sim \frac{N^2}{N^3} = \frac{1}{N}.$$

This agrees with the bulk statement that canonically normalized cubic couplings are suppressed by $1/N$ in theories with $C_T\sim N^2$.

Exercise 5: Detect an extremal divergence

Let $\Delta_3=\Delta_1+\Delta_2$. Which gamma function in the contact integral diverges? What is the expected position dependence of the separated three-point function after a proper limiting procedure?

Solution

If

$$\Delta_3=\Delta_1+\Delta_2,$$

then

$$\delta_{12} = \frac{\Delta_1+\Delta_2-\Delta_3}{2} =0.$$

The contact integral contains $\Gamma(\delta_{12})=\Gamma(0)$, so the naive non-extremal formula diverges. This is the extremal divergence.

The separated-point conformal structure is still fixed by conformal symmetry. Substituting $\Delta_3=\Delta_1+\Delta_2$ gives

$$\langle \mathcal O_1(x_1)\mathcal O_2(x_2)\mathcal O_3(x_3)\rangle \propto \frac{1}{ |x_{13}|^{2\Delta_1} |x_{23}|^{2\Delta_2} },$$

with no power of $|x_{12}|$. The coefficient must be computed by the correct renormalized limiting procedure, including possible vanishing cubic couplings and boundary terms.

Further reading

• D. Z. Freedman, S. D. Mathur, A. Matusis, and L. Rastelli, Correlation functions in the CFT$_d$/AdS$_{d+1}$ correspondence. The classic early computation of scalar and current three-point functions from AdS integrals.
• E. D’Hoker and D. Z. Freedman, Supersymmetric gauge theories and the AdS/CFT correspondence. A detailed lecture-note treatment of Witten diagrams, supergravity fields, and correlators.
• K. Skenderis, Lecture notes on holographic renormalization. The systematic framework for counterterms, one-point functions, Ward identities, and contact terms.
• J. Penedones, TASI lectures on AdS/CFT. A modern CFT-first approach emphasizing boundary operators, AdS perturbation theory, and correlators.
• D. Z. Freedman, K. Pilch, S. S. Pufu, and N. P. Warner, Boundary terms and three-point functions. A useful warning about boundary terms and extremal three-point subtleties.