Conformal Transformations in d Dimensions

Goal

A conformal field theory is not merely a scale-invariant theory with nice power laws. It is a theory whose local observables transform covariantly under the full conformal group. Before studying the conformal algebra, Ward identities, representations, correlation functions, and the AdS/CFT dictionary, we need a precise geometric answer to one question:

What are the spacetime transformations that preserve angles but not necessarily distances?

This page develops that answer in flat $d$-dimensional spacetime. The punchline is simple but extremely powerful:

$$\boxed{ \text{For } d\ge 3,\text{ every local conformal transformation is generated by translations, rotations, dilatations, and special conformal transformations.} }$$

The corresponding finite-dimensional group is locally $SO(d+1,1)$ in Euclidean signature and $SO(d,2)$ in Lorentzian signature. The Lorentzian version is the same group that appears as the isometry group of $AdS_{d+1}$. This is the first reason CFT is the correct boundary language for AdS/CFT.

Conformal transformations: the metric definition

Work first in flat Euclidean space $\mathbb R^d$ with metric

$$ds^2 = \delta_{\mu\nu} dx^\mu dx^\nu.$$

A coordinate transformation

$$x^\mu \mapsto x'^\mu(x)$$

is called conformal if it rescales the metric by a position-dependent positive factor:

$$\frac{\partial x'^\rho}{\partial x^\mu} \frac{\partial x'^\sigma}{\partial x^\nu} \delta_{\rho\sigma} = \Omega(x)^2\delta_{\mu\nu}.$$

Equivalently,

$$ds'^2=\Omega(x)^2 ds^2.$$

The function $\Omega(x)$ is the local scale factor or Weyl factor induced by the transformation. This equation says that the Jacobian matrix of the transformation is, at each point, an orthogonal transformation times a scale:

$$J(x) \in \mathbb R_+\times SO(d).$$

Therefore infinitesimal vectors at the same point have their lengths rescaled by the same factor, while the angle between them is unchanged.

In Lorentzian signature, replace $\delta_{\mu\nu}$ by

$$\eta_{\mu\nu}=\operatorname{diag}(-,+,\ldots,+)$$

or by the opposite convention. The defining condition becomes

$$\frac{\partial x'^\rho}{\partial x^\mu} \frac{\partial x'^\sigma}{\partial x^\nu} \eta_{\rho\sigma} = \Omega(x)^2\eta_{\mu\nu}.$$

The local linear transformation is then a Lorentz transformation times a scale.

A conformal transformation has Jacobian $J$ satisfying $J^T\eta J=\Omega^2\eta$. Thus two tangent vectors $u,v$ are mapped to $Ju,Jv$ with the same angle $\theta$, while all local lengths are multiplied by the same factor $\Omega(x)$. In Lorentzian signature, the analogous statement is that null directions are preserved.

A warning is worth making early: a conformal transformation is a coordinate transformation that induces a Weyl rescaling of the metric. A Weyl transformation is a direct rescaling of the metric at fixed coordinates,

$$g_{\mu\nu}(x)\mapsto e^{2\omega(x)}g_{\mu\nu}(x).$$

These are related but not identical. In flat-space CFT, conformal transformations are special diffeomorphisms that preserve the flat metric up to Weyl rescaling. In curved-space CFT, Weyl transformations become independent background transformations and lead to the Weyl anomaly in even dimensions.

The conformal Killing equation

To find the allowed conformal transformations, start infinitesimally:

$$x'^\mu=x^\mu+\epsilon^\mu(x).$$

The Jacobian is

$$\frac{\partial x'^\rho}{\partial x^\mu} = \delta^\rho_\mu+\partial_\mu\epsilon^\rho.$$

Keeping only first order terms in $\epsilon$, the transformed metric is

$$\delta_{\rho\sigma} \frac{\partial x'^\rho}{\partial x^\mu} \frac{\partial x'^\sigma}{\partial x^\nu} = \delta_{\mu\nu} + \partial_\mu\epsilon_\nu + \partial_\nu\epsilon_\mu.$$

A conformal transformation requires this to equal

$$\Omega(x)^2\delta_{\mu\nu} = \left(1+2\sigma(x)\right)\delta_{\mu\nu}$$

to first order. Hence

$$\partial_\mu\epsilon_\nu+ \partial_\nu\epsilon_\mu = 2\sigma(x)\delta_{\mu\nu}.$$

Taking the trace gives

$$2\partial_\rho\epsilon^\rho=2d\sigma,$$

so

$$\sigma(x)=\frac{1}{d}\partial_\rho\epsilon^\rho.$$

Therefore the infinitesimal conformal transformation is governed by the conformal Killing equation

$$\boxed{ \partial_\mu\epsilon_\nu+ \partial_\nu\epsilon_\mu = \frac{2}{d}\delta_{\mu\nu}\partial_\rho\epsilon^\rho. }$$

In Lorentzian signature the same equation holds with $\delta_{\mu\nu}$ replaced by $\eta_{\mu\nu}$:

$$\partial_\mu\epsilon_\nu+ \partial_\nu\epsilon_\mu = \frac{2}{d}\eta_{\mu\nu}\partial_\rho\epsilon^\rho.$$

This equation is the infinitesimal version of angle preservation. It says that the symmetric part of $\partial_\mu\epsilon_\nu$ is pure trace. The antisymmetric part is a local rotation; the trace is a local scale transformation; no shear is allowed.

Solving the conformal Killing equation for $d\ge 3$

The conformal Killing equation is highly restrictive in dimensions $d\ge 3$. To see this, define

$$\sigma=\frac{1}{d}\partial\cdot \epsilon.$$

Then

$$\partial_\mu\epsilon_\nu+ \partial_\nu\epsilon_\mu=2\sigma\delta_{\mu\nu}.$$

Differentiate and combine cyclic permutations of the indices. One obtains

$$\partial_\mu\partial_\nu\epsilon_\rho = \delta_{\mu\rho}\partial_\nu\sigma + \delta_{\nu\rho}\partial_\mu\sigma - \delta_{\mu\nu}\partial_\rho\sigma.$$

Taking traces then implies, for $d\ge3$,

$$\partial_\mu\partial_\nu\sigma=0.$$

Thus $\sigma$ is at most linear in the coordinates, and $\epsilon^\mu(x)$ is at most quadratic. The most general solution is

$$\boxed{ \epsilon^\mu(x) = a^\mu + \omega^\mu{}_{\nu}x^\nu + \lambda x^\mu + 2(b\cdot x)x^\mu -b^\mu x^2. }$$

Here

$$\omega_{\mu\nu}=-\omega_{\nu\mu}, \qquad x^2=\delta_{\mu\nu}x^\mu x^\nu, \qquad b\cdot x=b_\mu x^\mu.$$

The four terms have distinct geometric meanings:

$$\begin{array}{ccl} a^\mu && \text{translation},\\ \omega^\mu{}_{\nu}x^\nu && \text{rotation},\\ \lambda x^\mu && \text{dilatation},\\ 2(b\cdot x)x^\mu-b^\mu x^2 && \text{special conformal transformation}. \end{array}$$

Counting parameters gives

$$\underbrace{d}_{\text{translations}} + \underbrace{\frac{d(d-1)}{2}}_{\text{rotations}} + \underbrace{1}_{\text{dilatation}} + \underbrace{d}_{\text{special conformal}} = \frac{(d+1)(d+2)}{2}.$$

This is the dimension of $SO(d+1,1)$ in Euclidean signature, or $SO(d,2)$ in Lorentzian signature.

This finite-dimensionality is a major distinction between $d\ge3$ and $d=2$. In two dimensions, the conformal Killing equation becomes the Cauchy-Riemann equation, and every holomorphic map is locally conformal. That infinite-dimensional enhancement is why two-dimensional CFT has Virasoro symmetry. In this course, however, AdS/CFT preparation requires us first to understand the finite-dimensional conformal group in arbitrary $d$.

The finite transformations

The infinitesimal solution exponentiates to four basic finite transformations.

Translations

A translation is

$$x'^\mu=x^\mu+a^\mu.$$

It has

$$\Omega(x)=1.$$

Translations preserve all distances and angles. They are part of the Poincare group.

Rotations and Lorentz transformations

In Euclidean signature,

$$x'^\mu=R^\mu{}_{\nu}x^\nu, \qquad R^T R=1.$$

In Lorentzian signature,

$$x'^\mu=\Lambda^\mu{}_{\nu}x^\nu, \qquad \Lambda^T\eta\Lambda=\eta.$$

Again,

$$\Omega(x)=1.$$

These transformations preserve the metric exactly.

Dilatations

A dilatation is

$$x'^\mu=\lambda x^\mu, \qquad \lambda>0.$$

It rescales the line element by

$$ds'^2=\lambda^2 ds^2,$$

so

$$\Omega(x)=\lambda.$$

Scale invariance alone would mean invariance under these transformations, together with translations and rotations or Lorentz transformations. Conformal invariance further includes the special conformal transformations.

Inversion

The inversion is

$$x'^\mu=\frac{x^\mu}{x^2}.$$

It is not connected continuously to the identity, but it is the simplest way to construct the finite special conformal transformations.

Let

$$x'^\mu=\frac{x^\mu}{x^2}.$$

Then

$$\frac{\partial x'^\mu}{\partial x^\nu} = \frac{1}{x^2}\left(\delta^\mu_\nu- 2\frac{x^\mu x_\nu}{x^2}\right).$$

The matrix in parentheses is a reflection in the direction of $x^\mu$, so it is orthogonal. Therefore

$$ds'^2=\frac{ds^2}{(x^2)^2},$$

and

$$\Omega(x)=\frac{1}{x^2}$$

in Euclidean signature. The inversion maps short distances near the origin to large distances near infinity. Thus global conformal transformations are most naturally defined not on $\mathbb R^d$ alone, but on its conformal compactification.

Special conformal transformations

A special conformal transformation, abbreviated SCT, is

$$\boxed{ x'^\mu = \frac{x^\mu-b^\mu x^2}{1-2b\cdot x+b^2x^2}. }$$

Define

$$D_b(x)=1-2b\cdot x+b^2x^2.$$

Then

$$x'^\mu=\frac{x^\mu-b^\mu x^2}{D_b(x)}.$$

The induced scale factor is

$$\Omega(x)=\frac{1}{D_b(x)}.$$

To first order in $b^\mu$,

$$x'^\mu = \left(x^\mu-b^\mu x^2\right)\left(1+2b\cdot x+O(b^2)\right),$$

so

$$\delta x^\mu = 2(b\cdot x)x^\mu-b^\mu x^2.$$

This matches the quadratic term in the conformal Killing vector.

The most transparent construction of an SCT is

$$\boxed{ \mathrm{SCT}_b=I\circ T_{-b}\circ I, }$$

where

$$I:x^\mu\mapsto \frac{x^\mu}{x^2}, \qquad T_{-b}:y^\mu\mapsto y^\mu-b^\mu.$$

Indeed, after the first inversion,

$$y^\mu=\frac{x^\mu}{x^2}.$$

After translation,

$$y^\mu\mapsto y^\mu-b^\mu.$$

After the second inversion,

$$x'^\mu = \frac{y^\mu-b^\mu}{(y-b)^2} = \frac{x^\mu-b^\mu x^2}{1-2b\cdot x+b^2x^2}.$$

This formula is worth remembering. Many calculations with SCTs become simple if you think of them as inversion-translation-inversion.

How distances transform

The most useful practical formula is the transformation of squared distances.

For an inversion,

$$x_i'^\mu=\frac{x_i^\mu}{x_i^2},$$

one finds

$$(x_i'-x_j')^2 = \frac{(x_i-x_j)^2}{x_i^2 x_j^2}.$$

For a special conformal transformation,

$$x'^\mu = \frac{x^\mu-b^\mu x^2}{D_b(x)}, \qquad D_b(x)=1-2b\cdot x+b^2x^2,$$

one obtains

$$\boxed{ (x_i'-x_j')^2 = \frac{(x_i-x_j)^2}{D_b(x_i)D_b(x_j)}. }$$

This identity is central in the derivation of CFT two- and three-point functions. It tells us that the distance between two points is not invariant, but transforms by a product of local scale factors:

$$(x_i'-x_j')^2 = \Omega(x_i)\Omega(x_j)(x_i-x_j)^2$$

for the convention $\Omega(x)=D_b(x)^{-1}$.

This product structure is the miracle. Because a primary operator transforms with a local power of $\Omega(x)$, two-point and three-point functions can be covariant even though ordinary distances are not invariant.

Active and passive viewpoints

There are two common ways to describe the same transformation.

In the passive viewpoint, we change coordinates but describe the same physical insertion. In the active viewpoint, we move the operator insertion from $x$ to $x'$ in the same coordinate system. Both are useful. Confusion often comes from mixing conventions.

For a scalar primary operator $\mathcal O$ of scaling dimension $\Delta$, the active transformation law is usually written as

$$\boxed{ \mathcal O'(x')= \Omega(x)^{-\Delta}\mathcal O(x). }$$

For a dilatation $x'=\lambda x$, this becomes

$$\mathcal O'(\lambda x)=\lambda^{-\Delta}\mathcal O(x),$$

which is the familiar scaling law.

For an operator with spin, the local orthogonal or Lorentz rotation in the Jacobian also acts on the operator indices. Schematically,

$$\mathcal O'_A(x') = \Omega(x)^{-\Delta} D_A{}^B(R(x))\mathcal O_B(x),$$

where $D(R)$ is the appropriate finite-dimensional spin representation. The local rotation factor is harmless for scalar operators but essential for currents, stress tensors, spinors, and general spinning primaries.

A first consequence: the scalar two-point function

The full derivation of conformal correlation functions comes later, but the two-point function is simple enough to preview now.

Let $\mathcal O_1$ and $\mathcal O_2$ be scalar primary operators with dimensions $\Delta_1$ and $\Delta_2$. Translation and rotation invariance imply

$$\langle \mathcal O_1(x_1)\mathcal O_2(x_2)\rangle =f(x_{12}^2), \qquad x_{12}=x_1-x_2.$$

Scale covariance implies

$$f(\lambda^2 x_{12}^2) = \lambda^{-\Delta_1-\Delta_2}f(x_{12}^2),$$

so

$$\langle \mathcal O_1(x_1)\mathcal O_2(x_2)\rangle = \frac{C_{12}}{|x_{12}|^{\Delta_1+\Delta_2}}.$$

Special conformal covariance then requires

$$\Delta_1=\Delta_2$$

unless the coefficient vanishes. Thus

$$\boxed{ \langle \mathcal O_1(x_1)\mathcal O_2(x_2)\rangle = \begin{cases} \displaystyle \frac{C_{12}}{|x_{12}|^{2\Delta}}, & \Delta_1=\Delta_2=\Delta,\\ 0, & \Delta_1\ne\Delta_2, \end{cases} }$$

up to operator mixing among primaries with the same quantum numbers. In an orthonormal basis, one usually chooses

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

This is the first sign that conformal symmetry is far stronger than scale invariance alone.

Global issues and conformal compactification

The formulas above are local transformations on flat space. Some of them are singular on ordinary $\mathbb R^d$. For example, inversion is singular at $x=0$, and an SCT is singular where

$$D_b(x)=1-2b\cdot x+b^2x^2=0.$$

This is not a pathology of the conformal group. It is telling us that flat $\mathbb R^d$ is not the natural global space on which conformal transformations act. The natural space is the conformal compactification.

In Euclidean signature, the compactification is the sphere

$$S^d.$$

Flat space is obtained from $S^d$ by removing one point, just as the complex plane is obtained from the Riemann sphere by removing the point at infinity. Global conformal transformations act smoothly on $S^d$.

In Lorentzian signature, the conformal compactification is related to

$$S^{d-1}\times \mathbb R$$

with appropriate global identifications. This will become important in radial quantization and in the state-operator correspondence, where dilatations on $\mathbb R^d$ become time translations on the cylinder.

Why $d=2$ is special

For $d\ge3$, the conformal Killing equation forces $\epsilon^\mu(x)$ to be at most quadratic in $x$. That is why the group is finite-dimensional.

For $d=2$, introduce complex coordinates

$$z=x^1+ix^2, \qquad \bar z=x^1-ix^2.$$

The conformal Killing equation becomes the statement that

$$z\mapsto f(z), \qquad \bar z\mapsto \bar f(\bar z)$$

is locally conformal whenever $f$ is holomorphic. Since a holomorphic function has infinitely many Taylor coefficients, the local conformal symmetry becomes infinite-dimensional.

The globally well-defined conformal transformations on the Riemann sphere are only the Mobius transformations

$$z\mapsto \frac{az+b}{cz+d}, \qquad ad-bc\ne0.$$

But local conformal transformations are much larger. Quantum mechanically, this leads to the Virasoro algebra, central charge, minimal models, modular invariance, and all the exact two-dimensional CFT technology. For AdS/CFT in general dimension, however, the finite-dimensional $SO(d,2)$ structure remains the essential starting point.

AdS/CFT checkpoint

The conformal group of a Lorentzian $d$-dimensional CFT is

$$SO(d,2).$$

The isometry group of $AdS_{d+1}$ is also

$$SO(d,2).$$

This equality is not decorative. It is the kinematic backbone of AdS/CFT.

In Poincare coordinates, Euclidean $AdS_{d+1}$ has metric

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

A boundary dilatation

$$x^\mu\mapsto \lambda x^\mu$$

extends into the bulk as

$$z\mapsto \lambda z, \qquad x^\mu\mapsto \lambda x^\mu.$$

The metric is unchanged:

$$\frac{R^2}{(\lambda z)^2} \left(d(\lambda z)^2+d(\lambda x)_\mu d(\lambda x)^\mu\right) = \frac{R^2}{z^2}\left(dz^2+dx_\mu dx^\mu\right).$$

A boundary SCT extends to a bulk isometry as

$$x'^\mu = \frac{x^\mu-b^\mu(x^2+z^2)}{1-2b\cdot x+b^2(x^2+z^2)}, \qquad z' = \frac{z}{1-2b\cdot x+b^2(x^2+z^2)}.$$

At the boundary $z\to0$, this reduces to the CFT special conformal transformation

$$x'^\mu = \frac{x^\mu-b^\mu x^2}{1-2b\cdot x+b^2x^2}.$$

So the bulk isometry acts on the boundary precisely as a conformal transformation. This is why a bulk scalar field of mass $m$ is associated with a boundary primary operator of dimension $\Delta$, and why the relation between $m^2R^2$ and $\Delta$ is fixed by representation theory rather than by a dynamical accident.

The slogan is:

$$\boxed{ \text{AdS isometries become CFT conformal transformations at the boundary.} }$$

Common pitfalls

The first common mistake is to identify conformal transformations with arbitrary local rescalings of coordinates. They are not arbitrary. For $d\ge3$, the conformal Killing equation is so restrictive that only finitely many transformations are allowed.

The second mistake is to confuse conformal transformations with Weyl transformations. A conformal transformation is a diffeomorphism whose pullback rescales the metric. A Weyl transformation rescales the metric directly. A curved-space CFT cares about both, but their roles are conceptually different.

The third mistake is to omit special conformal transformations. Scale invariance plus Poincare invariance is weaker than conformal invariance. In many unitary relativistic QFTs, scale invariance is expected or known under suitable assumptions to enhance to conformal invariance, but the logical distinction matters.

The fourth mistake is to treat inversion as an ordinary small transformation. Inversion is not connected to the identity, but it is extremely useful because it generates SCTs by conjugating translations.

Summary

A conformal transformation rescales the flat metric locally:

$$ds'^2=\Omega(x)^2ds^2.$$

Infinitesimally, it is generated by a conformal Killing vector satisfying

$$\partial_\mu\epsilon_\nu+ \partial_\nu\epsilon_\mu = \frac{2}{d}\eta_{\mu\nu}\partial\cdot\epsilon.$$

For $d\ge3$, the general solution is

$$\epsilon^\mu = a^\mu + \omega^\mu{}_{\nu}x^\nu + \lambda x^\mu + 2(b\cdot x)x^\mu-b^\mu x^2.$$

The finite transformations are translations, rotations or Lorentz transformations, dilatations, and special conformal transformations:

$$x'^\mu = \frac{x^\mu-b^\mu x^2}{1-2b\cdot x+b^2x^2}.$$

In Lorentzian signature, these transformations generate $SO(d,2)$, the same group as the isometry group of $AdS_{d+1}$. This is the first structural bridge from CFT to AdS/CFT.

Exercises

Exercise 1: Derive the conformal Killing equation

Let

$$x'^\mu=x^\mu+\epsilon^\mu(x).$$

Starting from

$$\frac{\partial x'^\rho}{\partial x^\mu} \frac{\partial x'^\sigma}{\partial x^\nu} \delta_{\rho\sigma} = \Omega(x)^2\delta_{\mu\nu},$$

show that, to first order in $\epsilon$,

$$\partial_\mu\epsilon_\nu+ \partial_\nu\epsilon_\mu = \frac{2}{d}\delta_{\mu\nu}\partial_\rho\epsilon^\rho.$$

Solution

The Jacobian is

$$\frac{\partial x'^\rho}{\partial x^\mu} = \delta^\rho_\mu+\partial_\mu\epsilon^\rho.$$

Therefore

$$\delta_{\rho\sigma} \frac{\partial x'^\rho}{\partial x^\mu} \frac{\partial x'^\sigma}{\partial x^\nu} = \delta_{\mu\nu} + \partial_\mu\epsilon_\nu + \partial_\nu\epsilon_\mu +O(\epsilon^2).$$

Write

$$\Omega(x)^2=1+2\sigma(x)+O(\epsilon^2).$$

Then

$$\partial_\mu\epsilon_\nu+ \partial_\nu\epsilon_\mu = 2\sigma\delta_{\mu\nu}.$$

Taking the trace gives

$$2\partial_\rho\epsilon^\rho=2d\sigma,$$

so

$$\sigma=\frac{1}{d}\partial_\rho\epsilon^\rho.$$

Substitution gives the conformal Killing equation.

Exercise 2: Show that inversion is conformal

For

$$x'^\mu=\frac{x^\mu}{x^2},$$

show that

$$ds'^2=\frac{ds^2}{(x^2)^2}.$$

Solution

Compute the Jacobian:

$$\frac{\partial x'^\mu}{\partial x^\nu} = \frac{1}{x^2}\left(\delta^\mu_\nu-2\frac{x^\mu x_\nu}{x^2}\right).$$

Define

$$R^\mu{}_{\nu}(x) = \delta^\mu_\nu-2\frac{x^\mu x_\nu}{x^2}.$$

This is a reflection matrix. It obeys

$$R^T R=1.$$

Thus

$$\delta_{\mu\nu}dx'^\mu dx'^\nu = \frac{1}{(x^2)^2}\delta_{\mu\nu}dx^\mu dx^\nu.$$

Therefore inversion is conformal with

$$\Omega(x)=\frac{1}{x^2}.$$

Exercise 3: Verify the infinitesimal SCT

Start from the finite transformation

$$x'^\mu = \frac{x^\mu-b^\mu x^2}{1-2b\cdot x+b^2x^2}.$$

Expand to first order in $b^\mu$ and show that

$$\delta x^\mu = 2(b\cdot x)x^\mu-b^\mu x^2.$$

Solution

To first order in $b$,

$$\frac{1}{1-2b\cdot x+b^2x^2} = 1+2b\cdot x+O(b^2).$$

Therefore

$$x'^\mu = \left(x^\mu-b^\mu x^2\right) \left(1+2b\cdot x\right)+O(b^2).$$

Keeping first order terms,

$$x'^\mu = x^\mu+2(b\cdot x)x^\mu-b^\mu x^2+O(b^2).$$

Hence

$$\delta x^\mu=x'^\mu-x^\mu = 2(b\cdot x)x^\mu-b^\mu x^2.$$

Exercise 4: Derive the distance transformation under an SCT

Let

$$x'^\mu = \frac{x^\mu-b^\mu x^2}{D_b(x)}, \qquad D_b(x)=1-2b\cdot x+b^2x^2.$$

Show that

$$(x_i'-x_j')^2 = \frac{(x_i-x_j)^2}{D_b(x_i)D_b(x_j)}.$$

Solution

The cleanest proof uses

$$\mathrm{SCT}_b=I\circ T_{-b}\circ I.$$

Translations do not change distances. Inversion transforms distances as

$$(I(x_i)-I(x_j))^2 = \frac{(x_i-x_j)^2}{x_i^2x_j^2}.$$

After the first inversion and translation, define

$$y_i=\frac{x_i}{x_i^2}-b.$$

Then

$$(y_i-y_j)^2 = \left(\frac{x_i}{x_i^2}-\frac{x_j}{x_j^2}\right)^2 = \frac{(x_i-x_j)^2}{x_i^2x_j^2}.$$

The second inversion gives

$$(x_i'-x_j')^2 = \frac{(y_i-y_j)^2}{y_i^2y_j^2}.$$

But

$$y_i^2 = \left(\frac{x_i}{x_i^2}-b\right)^2 = \frac{1-2b\cdot x_i+b^2x_i^2}{x_i^2} = \frac{D_b(x_i)}{x_i^2}.$$

Therefore

$$(x_i'-x_j')^2 = \frac{(x_i-x_j)^2}{D_b(x_i)D_b(x_j)}.$$

Exercise 5: Use SCT covariance to fix scalar two-point functions

Assume that two scalar primary operators have a two-point function

$$\langle\mathcal O_1(x_1)\mathcal O_2(x_2)\rangle = \frac{C_{12}}{|x_{12}|^{\Delta_1+\Delta_2}}.$$

Using special conformal covariance, show that this can be nonzero only if

$$\Delta_1=\Delta_2.$$

Solution

Under an SCT,

$$|x_{12}'|^2 = \frac{|x_{12}|^2}{D_b(x_1)D_b(x_2)}.$$

Thus

$$|x_{12}'|^{\Delta_1+\Delta_2} = \frac{|x_{12}|^{\Delta_1+\Delta_2}}{D_b(x_1)^{(\Delta_1+\Delta_2)/2}D_b(x_2)^{(\Delta_1+\Delta_2)/2}}.$$

The transformed functional form gives

$$\frac{C_{12}}{|x_{12}'|^{\Delta_1+\Delta_2}} = \frac{C_{12}D_b(x_1)^{(\Delta_1+\Delta_2)/2}D_b(x_2)^{(\Delta_1+\Delta_2)/2}}{|x_{12}|^{\Delta_1+\Delta_2}}.$$

On the other hand, primary covariance gives one local factor per operator. Since

$$\Omega(x)=D_b(x)^{-1},$$

we get

$$\langle\mathcal O_1'(x_1')\mathcal O_2'(x_2')\rangle = D_b(x_1)^{\Delta_1}D_b(x_2)^{\Delta_2} \langle\mathcal O_1(x_1)\mathcal O_2(x_2)\rangle.$$

For consistency at arbitrary $x_1,x_2,b$, the powers of $D_b(x_1)$ and $D_b(x_2)$ must match:

$$\Delta_1=\frac{\Delta_1+\Delta_2}{2}, \qquad \Delta_2=\frac{\Delta_1+\Delta_2}{2}.$$

Hence

$$\Delta_1=\Delta_2.$$

Exercise 6: Boundary limit of the bulk SCT

Consider the $AdS_{d+1}$ transformation

$$x'^\mu = \frac{x^\mu-b^\mu(x^2+z^2)}{1-2b\cdot x+b^2(x^2+z^2)}, \qquad z' = \frac{z}{1-2b\cdot x+b^2(x^2+z^2)}.$$

Show that its boundary limit as $z\to0$ is the CFT special conformal transformation.

Solution

As $z\to0$,

$$x^2+z^2\to x^2,$$

and

$$1-2b\cdot x+b^2(x^2+z^2) \to 1-2b\cdot x+b^2x^2.$$

Therefore

$$x'^\mu \to \frac{x^\mu-b^\mu x^2}{1-2b\cdot x+b^2x^2},$$

which is exactly the flat-space SCT on the boundary. Also,

$$z' \to \frac{z}{1-2b\cdot x+b^2x^2},$$

so the boundary $z=0$ maps to itself.

Further reading

For the global conformal group and the basic Ward-identity consequences, read the global conformal invariance chapter of Di Francesco, Mathieu, and Senechal. For this course, the most important next step is not yet Virasoro symmetry, but the higher-dimensional conformal algebra: $P_\mu$, $M_{\mu\nu}$, $D$, and $K_\mu$.