Conformal Transformations in d Dimensions

At a critical point the RG flow has stopped, the correlation length is infinite, and correlation functions become scale covariant. Scale covariance is already a major constraint, but it is usually not the full spacetime symmetry of a local critical theory. The larger symmetry is conformal symmetry: transformations that may stretch lengths by a position-dependent factor, but do not shear angles.

This page is the kinematic bridge from RG scaling to conformal field theory. We derive the conformal Killing equation, explain why two dimensions are exceptional, classify the finite conformal transformations in $d>2$, study inversion and special conformal transformations, and then use inversion to constrain two-point functions of primary fields. The next page will build on this by deriving the conformal forms of three- and four-point functions and introducing the operator product expansion.

Required background. Lesson 13 derives fixed-point scaling and explains the stress-tensor criterion that promotes scale symmetry to conformal symmetry.

Helpful background. Lesson 12 supplies the scaling-dimension and Ising-operator conventions used in the correlator examples.

Conformal maps as local similarities

Conformal factor convention

We work in Euclidean $d$ dimensions with metric $\delta_{\mu\nu}$. A finite conformal transformation is written

$$x'^\mu=f^\mu(x),$$

and its local scale factor $\Omega(x)$ is defined by

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

Thus $\Omega(x)$ multiplies infinitesimal lengths, while $\Omega(x)^2$ multiplies the metric.

A differentiable map is conformal when its Jacobian is locally a scale times an orthogonal matrix:

$$J(x)=\Omega(x)R(x), \qquad R(x)^T R(x)=\mathbf 1.$$

Equivalently,

$$J(x)^T J(x)=\Omega(x)^2\mathbf 1,$$

or in components,

$${\partial f^\rho\over\partial x^\mu} {\partial f^\rho\over\partial x^\nu} =\Omega(x)^2\delta_{\mu\nu}.$$

A small circle is therefore mapped to a small circle, not to an ellipse. A right angle remains a right angle, while the radius is multiplied by $\Omega(x)$.

A conformal transformation is locally a scale factor $\Omega(x)$ times an orthogonal transformation $R(x)$. Infinitesimally this condition becomes the conformal Killing equation.

An isometry is the special case $\Omega=1$. A global dilation $x'=\lambda x$ has $\Omega=\lambda$. Inversion has a nonconstant scale factor, $\Omega(x)=1/x^2$, and is the simplest transformation that exposes the difference between scale invariance and full conformal invariance.

Infinitesimal form: the conformal Killing equation

Let

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

Then

$$dx'^\mu=dx^\mu+\partial_\nu\xi^\mu\,dx^\nu,$$

so, to first order in $\xi$,

$$ds'^2 =\left(\delta_{\mu\nu}+\partial_\mu\xi_\nu+ \partial_\nu\xi_\mu\right)dx^\mu dx^\nu.$$

A conformal transformation may only change the metric by a scalar factor. Hence

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

Taking the trace gives

$$\sigma(x)={1\over d}\partial_\rho\xi^\rho,$$

and therefore

$$\boxed{ \partial_\mu\xi_\nu+ \partial_\nu\xi_\mu ={2\over d}(\partial\cdot\xi)\delta_{\mu\nu}. }$$

This is the conformal Killing equation. Ordinary Killing vectors obey the same equation with the right-hand side set to zero. Conformal Killing vectors are allowed to have a trace part; that trace is the infinitesimal local scale transformation.

The elementary solutions are

$$\begin{aligned} \text{translations:}\qquad &\xi^\mu=a^\mu,\\ \text{rotations:}\qquad &\xi^\mu=\omega^\mu{}_{\nu}x^\nu, \qquad \omega_{\mu\nu}=-\omega_{\nu\mu},\\ \text{dilations:}\qquad &\xi^\mu=\alpha x^\mu,\\ \text{special conformal:}\qquad &\xi^\mu=2(b\cdot x)x^\mu-b^\mu x^2. \end{aligned}$$

The special conformal vector is worth checking once. Differentiating gives

$$\partial_\mu\xi_\nu+\partial_\nu\xi_\mu=4(b\cdot x)\delta_{\mu\nu}, \qquad \partial\cdot\xi=2d(b\cdot x),$$

so it indeed satisfies the conformal Killing equation.

Two dimensions: holomorphic freedom

In two Euclidean dimensions write

$$z=x^1+i x^2, \qquad \xi(z,\bar z)=\xi^1(x^1,x^2)+i\xi^2(x^1,x^2).$$

For $d=2$, the conformal Killing equation becomes

$$\partial_1\xi^1=\partial_2\xi^2, \qquad \partial_1\xi^2=-\partial_2\xi^1.$$

These are precisely the Cauchy–Riemann equations, so

$$\partial_{\bar z}\xi(z)=0.$$

The antiholomorphic component satisfies

$$\partial_z\bar\xi(\bar z)=0.$$

Thus the local conformal transformations are

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

with holomorphic and antiholomorphic functions, at least locally where the derivatives do not vanish. For a real orientation-preserving map of the Euclidean plane, the two functions are complex conjugates. In the complexified conformal algebra they are treated as independent left- and right-moving sectors.

In two dimensions the conformal Killing equation splits into holomorphic and antiholomorphic equations. Reality ties the two functions together for a real Euclidean map; the complexified conformal algebra treats the two sectors independently.

The word “local” matters. On the Riemann sphere, the globally nonsingular one-to-one orientation-preserving conformal maps are only the Möbius transformations

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

But local conformal transformations are generated by infinitely many vector fields,

$$l_n=-z^{n+1}\partial_z, \qquad \bar l_n=-\bar z^{n+1}\partial_{\bar z}, \qquad n\in\mathbb Z,$$

which obey

$$[l_m,l_n]=(m-n)l_{m+n}, \qquad [\bar l_m,\bar l_n]=(m-n)\bar l_{m+n}, \qquad [l_m,\bar l_n]=0.$$

The quantum version of this algebra, after central extension, will become the Virasoro algebra.

The finite group for d greater than two

For $d>2$, the conformal Killing equation is much more rigid. Differentiating and permuting indices gives

$$\partial_\mu\partial_\nu\xi_\rho =\delta_{\mu\rho}\partial_\nu\sigma +\delta_{\nu\rho}\partial_\mu\sigma -\delta_{\mu\nu}\partial_\rho\sigma, \qquad \sigma={1\over d}\partial\cdot\xi.$$

A further compatibility condition implies that $\sigma$ is at most linear in $x$, so $\xi^\mu(x)$ is at most quadratic. The general solution is

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

Counting parameters gives

$$N=d+{d(d-1)\over2}+1+d ={(d+1)(d+2)\over2}.$$

For $d>2$, the Lie algebra of Euclidean conformal transformations is

$$\mathfrak{so}(d+1,1).$$

Equivalently, the identity component of the conformal group is locally isomorphic to $SO_0(d+1,1)$. Global statements require the conformal compactification $S^d$ and a discrete quotient; inversion lies outside the identity component.

For $d>2$, the conformal Killing equation has a finite-dimensional solution space. Its Lie algebra is generated by translations $P_\mu$, rotations $M_{\mu\nu}$, dilations $D$, and special conformal transformations $K_\mu$.

A useful differential-operator basis is

$$P_\mu=\partial_\mu, \qquad M_{\mu\nu}=x_\mu\partial_\nu-x_\nu\partial_\mu,$$ $$D=x\cdot\partial, \qquad K_\mu=2x_\mu x\cdot\partial-x^2\partial_\mu.$$

With these sign conventions,

$$[D,P_\mu]=-P_\mu, \qquad [D,K_\mu]=K_\mu, \qquad [P_\mu,K_\nu]=2\delta_{\mu\nu}D-2M_{\mu\nu}.$$

The finite-dimensional algebra for $d>2$ should be compared with the infinite-dimensional local algebra in $d=2$. That contrast is one of the central structural facts of conformal field theory.

Inversion and special conformal transformations

Inversion is

$$I:\qquad x'^\mu={x^\mu\over x^2}.$$

Its Jacobian is

$${\partial x'^\rho\over\partial x^\mu} ={1\over x^2}\left(\delta^\rho{}_{\mu}-{2x^\rho x_\mu\over x^2}\right).$$

The matrix in parentheses is an orthogonal reflection, so

$${\partial x'^\rho\over\partial x^\mu} {\partial x'^\rho\over\partial x^\nu} ={1\over (x^2)^2}\delta_{\mu\nu}.$$

Thus inversion has local scale factor

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

For two points,

$$x'_1={x_1\over x_1^2}, \qquad x'_2={x_2\over x_2^2},$$

we find

$$\begin{aligned} |x'_1-x'_2|^2 &=\left({x_1\over x_1^2}-{x_2\over x_2^2}\right)^2\\ &={1\over x_1^2}+{1\over x_2^2}-{2x_1\cdot x_2\over x_1^2x_2^2}\\ &={|x_1-x_2|^2\over x_1^2x_2^2}. \end{aligned}$$

Equivalently,

$$|x'_{12}|^2=\Omega(x_1)\Omega(x_2)|x_{12}|^2.$$

Under inversion, $x^\mu\mapsto x^\mu/x^2$, distances transform with one local scale factor from each endpoint: $|x'_{12}|^2=\Omega(x_1)\Omega(x_2)|x_{12}|^2$.

A special conformal transformation is inversion, followed by translation, followed by inversion. With the convention

$$K_b=I\circ T_{-b}\circ I,$$

one obtains

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

Expanding to first order in $b$ gives

$${x'}^\mu=x^\mu+2(b\cdot x)x^\mu-b^\mu x^2+O(b^2),$$

which is exactly the infinitesimal special conformal vector.

Primary scalar fields and two-point functions

A scalar primary operator $O(x)$ of dimension $\Delta$ transforms under a conformal map as

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

For inversion, this reads

$$O'(x')=(x^2)^\Delta O(x).$$

The corresponding covariance law for scalar-primary correlators is

$$\boxed{ \left\langle O_1(x'_1)\cdots O_n(x'_n)\right\rangle =\prod_{i=1}^n\Omega(x_i)^{-\Delta_i} \left\langle O_1(x_1)\cdots O_n(x_n)\right\rangle. }$$

For a scalar two-point function of identical primaries,

$$\langle O(x_1)O(x_2)\rangle={C_O\over |x_1-x_2|^{2\Delta}},$$

inversion covariance is immediate:

$$\begin{aligned} \langle O(x'_1)O(x'_2)\rangle &={C_O\over |x'_1-x'_2|^{2\Delta}}\\ &={C_O\over [\Omega(x_1)\Omega(x_2)|x_1-x_2|^2]^\Delta}\\ &=\Omega(x_1)^{-\Delta}\Omega(x_2)^{-\Delta} {C_O\over |x_1-x_2|^{2\Delta}}. \end{aligned}$$

For spinning primaries one must also rotate the indices by the local orthogonal matrix

$$R^\mu{}_{\nu}(x)=\Omega(x)^{-1}{\partial x'^\mu\over\partial x^\nu}.$$

This extra rotation is essential for currents and the stress tensor, but scalar primaries already contain the main idea.

Now consider two scalar primaries of dimensions $\Delta_i$ and $\Delta_j$. Translation, rotation, and scale invariance allow

$$\langle O_i(x_1)O_j(x_2)\rangle ={C_{ij}\over |x_1-x_2|^{\Delta_i+\Delta_j}}.$$

Conformal invariance is stronger. Under inversion, the transformed functional form gives

$${C_{ij}\over |x'_1-x'_2|^{\Delta_i+\Delta_j}} ={C_{ij}(x_1^2x_2^2)^{(\Delta_i+\Delta_j)/2}\over |x_1-x_2|^{\Delta_i+\Delta_j}}.$$

The primary covariance law gives instead

$${C_{ij}(x_1^2)^{\Delta_i}(x_2^2)^{\Delta_j} \over |x_1-x_2|^{\Delta_i+\Delta_j}}.$$

These agree for arbitrary $x_1$ and $x_2$ only if $\Delta_i=\Delta_j$, or else $C_{ij}=0$. Hence

$$\boxed{ \langle O_i(x_1)O_j(x_2)\rangle=0 \qquad \text{unless } \Delta_i=\Delta_j. }$$

If several operators have the same dimension and the same spin and internal quantum numbers, their two-point functions form a constant matrix. In a unitary theory one usually chooses a basis that diagonalizes this matrix.

Relation to the Ising fields

In the two-dimensional critical Ising theory, the identity $1$, spin field $\sigma$, disorder field $\mu$, and energy field $\varepsilon$ can all be represented as scalar conformal fields. Their dimensions are

$$\Delta_1=0, \qquad \Delta_\sigma=\Delta_\mu={1\over8}, \qquad \Delta_\varepsilon=1.$$

Conformal covariance therefore predicts

$$\langle \sigma(x)\sigma(0)\rangle\propto {1\over |x|^{1/4}}, \qquad \langle \mu(x)\mu(0)\rangle\propto {1\over |x|^{1/4}}, \qquad \langle \varepsilon(x)\varepsilon(0)\rangle\propto {1\over |x|^2}.$$

It also predicts, in a basis adapted to conformal symmetry, that

$$\langle \sigma(x)\varepsilon(0)\rangle=0,$$

because $\Delta_\sigma\ne\Delta_\varepsilon$. The equality $\Delta_\sigma=\Delta_\mu$ is consistent with Kramers–Wannier duality, but it does not mean that $\sigma$ and $\mu$ are the same local operator. They are mutually nonlocal: taking an order field around a disorder field detects a branch cut. In the diagonal local Ising CFT, one chooses a mutually local operator algebra rather than treating both lattice realizations as independent local primaries. Conformal invariance fixes powers and tensor structures; the operator algebra still remembers the order–disorder construction.

A primary scalar field carries a conformal weight at its insertion point. Inversion then forces a scalar two-point function to vanish unless the two operators have the same scaling dimension.

Summary

Conformal symmetry is the symmetry of local angle preservation. Its infinitesimal form is the conformal Killing equation

$$\partial_\mu\xi_\nu+ \partial_\nu\xi_\mu ={2\over d}(\partial\cdot\xi)\delta_{\mu\nu}.$$

In two dimensions this equation becomes the Cauchy–Riemann equation, producing infinitely many local conformal maps. In $d>2$ it has only the finite-dimensional solution space generated by translations, rotations, dilations, and special conformal transformations.

Inversion is the key finite transformation. Its distance identity

$$|x'_{12}|^2={|x_{12}|^2\over x_1^2x_2^2}$$

makes the primary transformation law transparent and shows why scalar two-point functions vanish between primaries of unequal dimension.

Common pitfalls

The factor $\Omega(x)$ multiplies lengths, not squared lengths. The metric transforms with $\Omega(x)^2$.

The statement that two-dimensional conformal maps are arbitrary holomorphic functions is local. Globally regular maps on the sphere are Möbius transformations.

For a real Euclidean conformal map, the antiholomorphic function is the complex conjugate of the holomorphic one. Treating the two sectors as independent refers to the complexified algebra.

Inversion is singular at the origin and exchanges the origin with infinity. Algebraic identities involving inversion are usually cleanest on the conformal compactification.

The conformal group is locally related to $SO(d+1,1)$, but the exact global group depends on connected components and a discrete quotient. The Lie-algebra statement $\mathfrak{conf}(\mathbb R^d)\cong\mathfrak{so}(d+1,1)$ is unambiguous.

Scale invariance alone permits a two-point function proportional to $|x|^{-\Delta_i-\Delta_j}$. The condition $\Delta_i=\Delta_j$ comes from the larger conformal group.

Exercises

Exercise 1: Special conformal Killing vector

Verify that

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

satisfies the conformal Killing equation.

Solution

Differentiate:

$$\partial_\nu\xi_\mu =2b_\nu x_\mu+2(b\cdot x)\delta_{\mu\nu}-2b_\mu x_\nu.$$

Therefore

$$\partial_\nu\xi_\mu+\partial_\mu\xi_\nu =4(b\cdot x)\delta_{\mu\nu}.$$

The divergence is

$$\partial_\mu\xi^\mu =2(d+1)(b\cdot x)-2(b\cdot x) =2d(b\cdot x).$$

Thus

$${2\over d}(\partial\cdot\xi)\delta_{\mu\nu} =4(b\cdot x)\delta_{\mu\nu},$$

which matches the symmetrized derivative.

Exercise 2: Cauchy–Riemann equations

Show that the two-dimensional conformal Killing equation is equivalent to the Cauchy–Riemann equations for $\xi=\xi^1+i\xi^2$.

Solution

For $d=2$,

$$\partial_i\xi_j+\partial_j\xi_i=(\partial_k\xi_k)\delta_{ij}.$$

The $11$ component gives

$$2\partial_1\xi^1=\partial_1\xi^1+\partial_2\xi^2,$$

so

$$\partial_1\xi^1=\partial_2\xi^2.$$

The $12$ component gives

$$\partial_1\xi^2+\partial_2\xi^1=0.$$

These are exactly the Cauchy–Riemann equations, so $\partial_{\bar z}\xi=0$.

Exercise 3: Inversion of a separation

Derive the inversion distance formula

$$|x'_1-x'_2|^2={|x_1-x_2|^2\over x_1^2x_2^2}, \qquad x'_i={x_i\over x_i^2}.$$

Solution

Compute directly:

$$\begin{aligned} |x'_1-x'_2|^2 &=\left({x_1\over x_1^2}-{x_2\over x_2^2}\right)^2\\ &={1\over x_1^2}+{1\over x_2^2}-{2x_1\cdot x_2\over x_1^2x_2^2}\\ &={x_1^2+x_2^2-2x_1\cdot x_2\over x_1^2x_2^2}\\ &={|x_1-x_2|^2\over x_1^2x_2^2}. \end{aligned}$$

Exercise 4: Equal dimensions from inversion covariance

Use inversion covariance to show that a nonzero scalar-primary two-point function requires equal scaling dimensions.

Solution

Start from the scale-invariant form

$$\langle O_i(x_1)O_j(x_2)\rangle ={C_{ij}\over |x_1-x_2|^{\Delta_i+\Delta_j}}.$$

At inverted points this becomes

$${C_{ij}(x_1^2x_2^2)^{(\Delta_i+\Delta_j)/2} \over |x_1-x_2|^{\Delta_i+\Delta_j}}.$$

Primary covariance gives instead

$${C_{ij}(x_1^2)^{\Delta_i}(x_2^2)^{\Delta_j} \over |x_1-x_2|^{\Delta_i+\Delta_j}}.$$

If $C_{ij}\ne0$, the powers of $x_1^2$ and $x_2^2$ must match independently, so

$${\Delta_i+\Delta_j\over2}=\Delta_i \qquad\text{and}\qquad {\Delta_i+\Delta_j\over2}=\Delta_j.$$

Thus $\Delta_i=\Delta_j$. If the dimensions differ, $C_{ij}=0$.

Exercise 5: Infinitesimal special conformal transformation

Show that

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

has infinitesimal form

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

Solution

Expand the denominator:

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

Then

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

so

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

Subtracting $x^\mu$ gives the result.

References

• P. Di Francesco, P. Mathieu, and D. Sénéchal, Conformal Field Theory, for the standard two-dimensional treatment.
• S. Rychkov, EPFL Lectures on Conformal Field Theory in D ≥ 3 Dimensions, for higher-dimensional conformal kinematics and correlators.

Further reading

• J. Cardy, Scaling and Renormalization in Statistical Physics, for the critical-phenomena route to conformal invariance.
• A. M. Polyakov, Gauge Fields and Strings, especially the discussion of conformal field theory, stress tensors, and random surfaces.
• J. Zinn-Justin, Quantum Field Theory and Critical Phenomena, for the RG and critical-exponent perspective.