Past exam of the mathematics course of the University of Cambridge / 2003 / iii / Paper 51 (Conformal Field Theory)

The pullback of the flat Euclidean metric under $x^{\prime }{}^{\mu }=x^{\mu }+ϵ{\omega }^{\mu }(x)$ isFor a conformal transformation it must be a scalar multiple of ${\delta }_{\mu \nu }$. Taking the trace determines that multiple. With $\rho ={\partial }_{\alpha }{\omega }^{\alpha }$, the necessary and sufficient infinitesimal condition is the Conformal Killing equationDifferentiate its $(\alpha ,\beta )$ instance with respect to $x^{\gamma }$, its $(\alpha ,\gamma )$ instance with respect to $x^{\beta }$, and its $(\beta ,\gamma )$ instance with respect to $x^{\alpha }$. Adding the first two and subtracting the last, commuting partial derivatives, leaves $2{\partial }_{\beta }{\partial }_{\gamma }{\omega }_{\alpha }$. Hence flat conformal Killing integrability givesContracting $\beta$ with $\gamma$ gives $\Delta {\omega }_{\alpha }=(2-d){\partial }_{\alpha }\rho /d$. Its divergence is $\Delta \rho =(2-d)\Delta \rho /d$, so $2(d-1)\Delta \rho =0$: thus $\boxed{\Delta \rho =0\text{ for }d>1}$. Taking instead the $\alpha$ divergence of the displayed second-derivative identity givesConsequently $\boxed{{\partial }_{\beta }{\partial }_{\gamma }\rho =0\text{ for }d>2}$.

On a connected flat domain, write $\rho =d\lambda +2db\cdot x$, with constant $\lambda$ and $b$. Integrating the second-derivative identity, and imposing the Conformal Killing equation on its remaining affine part, gives the complete Conformal Killing vector fieldIndeed the quadratic part has precisely the required Hessian matrix, while the symmetric part of the remaining linear term must be $\lambda {\delta }_{\mu \nu }$. The constants generate translations, rotations, dilations and special conformal transformations, respectively. There are $d$, $d(d-1)/2$, $1$ and $d$ independent parameters, so the full conformal group has dimensionHere “full” means local transformations, or global transformations of the conformal compactification of Euclidean space; its connected group is locally $SO(d+1,1)$. Requiring every transformation to be nonsingular on all of uncompactified ${\mathbb{R}}^d$ would exclude the special conformal transformations with finite poles.

The stress-energy tensor generates translations. One may improve it by adding a divergence without changing the translation charges under suitable boundary conditions, and choose it symmetric in a relativistic theory. If ${\Theta }^{\mu \nu }$ is symmetric and conserved, a spacetime vector field $\omega$ defines the current $j^{\mu }={\Theta }^{\mu \nu }{\omega }_{\nu }$. For a Conformal Killing vector field,Thus a traceless conserved stress-energy tensor produces all the conformal charges. The dilation current is $x_{\nu }{\Theta }^{\mu \nu }$, and the special conformal currents are $(2x_{\lambda }{x}_{\nu }-x^2{\eta }_{\lambda \nu }){\Theta }^{\mu \nu }$. A trace that is merely the divergence of a virial current gives scale invariance; conformal invariance requires the additional improvement that removes the trace. These statements distinguish an actual traceless tensor from a trace discarded without justification.

For a free massless scalar field, the canonical stress-energy tensor has trace $(1-d/2)(\partial \varphi {)}^2$. For $d>1$, the improved stress-energy tensor of a free massless scalar isIts improvement is identically conserved. On $\square \varphi =0$, the identity $\square {\varphi }^2=2(\partial \varphi {)}^2$ shows that its trace vanishes. In $d=2$, no improvement is needed: the canonical stress-energy tensor is already traceless. In Lorentzian null coordinates, conservation and tracelessness imply that its two surviving components depend separately on $x^{+}$ and $x^{-}$. After Euclidean continuation, these become the holomorphic stress-energy tensor $T(z)$ and its antiholomorphic counterpart $\bar{T}(\bar{z})$, withLocal conformal changes are arbitrary holomorphic and antiholomorphic coordinate changes, not just the three global Möbius transformation generators in either sector.

To make the free-field quantum normalization explicit, rescale the field to $\phi =\sqrt{4\pi }\varphi$, so that its plane contraction is $\phi (z,\bar{z})\phi (w,\bar{w})\sim -\log |z-w{|}^2$. Define $J=i\partial \phi$, so $J(z)J(w)\sim (z-w{)}^{-2}$. Normal ordering givesA single Wick contraction shows that $J$ is a weight-one primary operator. The two double contractions in $T(z)T(w)$ produce $\frac{1}{2}(z-w{)}^{-4}$; single contractions, followed by Taylor expansion at $w$, produce $2T(w)/(z-w{)}^2+\partial T(w)/(z-w)$. Thus the operator product expansion isThe same holds independently for $\bar{T}$, with $\bar{c}=1$. The undifferentiated noncompact scalar has a logarithmic contraction and a zero-mode subtlety; its derivative current has the ordinary primary transformation law.

The holomorphic stress-energy tensor implements infinitesimal conformal transformations through contour charges $Q_v=\frac{1}{2\pi i}\oint dzv(z)T(z)$. A primary field $\mathcal{O}$ of weight $h$ obeysFor example the free-field vertex operator $:e^{i\alpha \phi }:$ has holomorphic weight $h={\alpha }^2/2$, as follows by contracting $\partial \phi$ with the exponential twice. In correlators the contour may encircle each insertion, giving the conformal Ward identity $⟨T(z){\prod }_i{\mathcal{O}}_i(z_i)⟩={\sum }_i[h_i/(z-z_i{)}^2+{\partial }_{z_i}/(z-z_i)]⟨{\prod }_i{\mathcal{O}}_i(z_i)⟩$. This also yields the global conformal Ward identities for chiral correlators.

Writing $T(z)={\sum }_n{L}_n{z}^{-n-2}$ and integrating the stress-tensor operator product expansion gives the Virasoro algebraIts central charge is the quantum correction to the classical conformal algebra; the correction vanishes on $L_{-1},L_0,L_1$. Equivalently, in the pullback convention a finite holomorphic change $z\mapsto f(z)$ sends $T$ to $(f^{\prime }{)}^{2}T(f)+\frac{c}{12}\{f,z\}$, where $\{f,z\}=f^{\prime \prime \prime }/f^{\prime }-\frac{3}{2}(f^{\prime \prime }/f^{\prime }{)}^2$ is the Schwarzian derivative. This anomalous term vanishes for Möbius transformations. On a curved background the related trace anomaly is proportional to its curvature; it does not destroy the flat-space conformal symmetry of this massless theory.

Let $G_n(z_1,\dots ,z_n)$ denote the ordered vacuum correlator. Inserting the vacuum-preserving Möbius transformation operator and its inverse givesFor $\gamma (z)=z+ϵv(z)$, the three vector fields $v=1,z,z^2$ yield the global conformal Ward identities for chiral correlatorsAll formulas below concern separated insertion points and a consistent local choice of branches for fractional powers.

For one insertion, the translation identity makes $G_1$ constant and the dilation identity gives $h_1{G}_1=0$. ThusFor two insertions, translation invariance gives $G_2=f(z_{12})$, where $z_{ij}=z_i-z_j$. Dilations imply $z_{12}{f}^{\prime }+(h_1+h_2)f=0$, so $f=C_{12}{z}_{12}^{-h_1-h_2}$. Substituting into the special conformal identity leaves $(h_1-h_2)z_{12}f=0$. Thereforewith $G_2=0$ when the weights differ.

For three insertions write $G_3=C_{123}{z}_{12}^{-a}{z}_{13}^{-b}{z}_{23}^{-c}$. Under a Möbius transformation, $\gamma (z_i)-\gamma (z_j)=z_{ij}\sqrt{{\gamma }^{\prime }(z_i){\gamma }^{\prime }(z_j)}$ with compatible local branches. The covariance condition therefore fixesSolving givesThis is the complete solution because Möbius transformations act transitively on ordered triples of distinct points: the ratio of any two covariant nonzero solutions has no remaining invariant variable and is constant. Three-point covariance imposes no equality or triangle inequality among the weights. Additional fusion rules or other symmetries may force $C_{123}=0$, but none follow from the stated covariance and vacuum assumptions alone.

For more than three generic points, three positions may be fixed by a Möbius transformation, leaving $n-3$ independent complex conformal cross-ratios. Correlators consequently have a covariant prefactor multiplied by an arbitrary function of these invariants. For example one convenient four-point form isThe Möbius invariance of the cross-ratio guarantees covariance for any $F$. Thus global conformal symmetry fixes one-, two- and three-point functions up to constants, but does not fix higher-point functions. Their remaining functions encode dynamics.

For $d\ge 2$, let $\omega$ be a Conformal Killing vector field, put $\sigma =\partial \cdot \omega /d$, and assign the massless scalar field scaling dimension $\Delta =(d-2)/2$. In a finite conformal transformation with metric scale ${\Omega }^2$, the transformation law is ${\varphi }^{\prime }(x^{\prime })=\Omega (x{)}^{-\Delta }\varphi (x)$. Its infinitesimal fixed-coordinate form isFor $\mathcal{L}=(\partial \varphi {)}^2/2$, contract the Conformal Killing equation with ${\partial }_{\mu }\varphi {\partial }_{\nu }\varphi$. Direct differentiation then givesHere $2(1+\Delta )=d$. Flat conformal Killing integrability gives $\square \sigma =0$, with the Lorentzian version obtained by replacing the flat metric consistently. Thus the action variation is a boundary integral, vanishing for suitable boundary conditions. This proves conformal invariance with field dimension $(d-2)/2$, without using the field equation. The connected finite symmetry follows by integrating the infinitesimal generators. Equivalently its conformal currents follow from the improved stress-energy tensor of a free massless scalar.

In one dimension the metric conformality condition is vacuous. The same action-variation calculation contains $-{\varphi }^2{\omega }^{\prime \prime \prime }/4$, so the free action is invariant under the projective subgroup ${\omega }^{\prime \prime \prime }=0$, rather than arbitrary reparametrizations. The usual field-theory conformal-invariance claim above concerns $d\ge 2$.

In $d=2$, $\Delta =0$, so the field transforms simply as a scalar. There is no improvement term and no derivative of the conformal scale in the variation. All local independent reparametrizations of the two null coordinates preserve the action; after Euclidean continuation they are holomorphic and antiholomorphic maps. The conformal algebra is therefore infinite-dimensional, in contrast to $d>2$.

Now use Lorentzian coordinates with spatial period $\ell$. The canonical momentum is $\pi =\dot{\varphi }$, and canonical quantization imposes $[\varphi (t,x),\pi (t,y)]=i{\delta }_{\ell }(x-y)$, with the periodic delta distribution ${\delta }_{\ell }(s)={\ell }^{-1}{\sum }_{n\in \mathbb{Z}}{e}^{2\pi ins/\ell }$. Solving the wave equation and imposing reality gives a convenient oscillator normalization:Strict spatial periodicity excludes a winding term. The canonical commutation relation is equivalent towith ${\alpha }_n^{†}={\alpha }_{-n}$ and the corresponding tilded relation. Both oscillator families commute with $q,p$. To verify the normalization, the zero-mode commutator supplies $i/\ell$, while each oscillator family supplies $i(2\ell {)}^{-1}{\sum }_{n\ne 0}{e}^{2\pi in(x-y)/\ell }$. Their sum is exactly $i{\delta }_{\ell }(x-y)$. The positive-index modes are annihilation operators and the negative-index modes are creation operators, giving two independent bosonic Fock spaces together with the free-particle zero mode.

Define ${\alpha }_0={\tilde{\alpha }}_0=p/\sqrt{4\pi }$ and the normal ordering quadratic generatorsThese are the Fourier modes of the two chiral components of the stress-energy tensor. The oscillator operator commutators immediately give $[L_m,{\alpha }_n]=-n{\alpha }_{m+n}$. Applying this to both factors of $L_n$ gives $(m-n)L_{m+n}$, except for a scalar term when $m+n=0$, produced by reordering creators and annihilators. Its value is independent of the momentum sector. Evaluate it in the formal zero-momentum oscillator vacuum. For $m>0$,The two contractions of each pair explain the factor $1/2$. Hence the free-boson Virasoro central term yieldsThe tilded modes give a second commuting Virasoro algebra with $\bar{c}=1$. The adjoint condition $L_m^{†}=L_{-m}$ and the positive oscillator norms make this a unitary highest-weight Virasoro representation in each momentum sector. In a fixed momentum sector, $L_0=p^2/(8\pi )+{\sum }_{n>0}{\alpha }_{-n}{\alpha }_n$, so the oscillator ground state is a highest-weight state of weight $p^2/(8\pi )$ in either sector. For a noncompact field the zero mode is a free particle: definite momentum ground states are generalized states, rather than a normalizable zero-mode vacuum. The oscillator representation and central term remain well defined. With the standard cylinder vacuum-energy subtraction, the Hamiltonian isThe constant is the Casimir energy of a two-dimensional conformal field theory; omitting it amounts to a different additive Hamiltonian convention, not a change of the plane Virasoro algebra.

Use $[L_m,L_n]=(m-n)L_{m+n}+\frac{c}{12}(m^3-m){\delta }_{m+n,0}$ and $L_n^{†}=L_{-n}$. A normalized highest-weight vector obeys $L_0|h⟩=h|h⟩$ and $L_n|h⟩=0$ for $n>0$. Its Virasoro descendants are products of negative modes. Their inner products are determined entirely by $c,h$ and the Virasoro algebra; for example $\Vert L_{-1}|h⟩{\Vert }^2=2h$, so unitarity requires $h\ge 0$. An irreducible unitary highest-weight Virasoro module is the Virasoro Verma module quotient by its null submodules, and its surviving inner product must be positive definite.

The classification for $c<1$ is the Unitary Virasoro discrete series:where $m$ is an integer at least $2$, $1\le r\le m-1$, and $1\le s\le m$. The identification $(r,s)\sim (m-r,m+1-s)$ leaves $m(m-1)/2$ distinct modules. The two generating singular-vector levels are $rs$ and $(m-r)(m+1-s)$; all their Virasoro descendants are removed in the irreducible quotient. For $m=2$ the only module is the trivial $c=h=0$ module; for $m\ge 3$ these are the nontrivial unitary discrete-series modules. There are no other irreducible unitary highest-weight representations with $c<1$. This is the classification theorem, rather than a condition inferred only from finitely many descendant norms.

At $c=1/2$, $m=3$ and the possible weights areConstruct them with a single free chiral Majorana fermion:The free chiral Majorana fermion conformal field theory has $T=-:\psi \partial \psi :/2$. In the Neveu–Schwarz sector, $r\in \mathbb{Z}+1/2$, while in the Ramond sector, $r\in \mathbb{Z}$. With positive modes moved to the right and the zero-mode product antisymmetrized, the corresponding generators areThe oscillator relation gives $[L_n,b_r]=-(r+n/2)b_{n+r}$. The stress-tensor operator product expansion, or the reordering of the quadratic generators, then gives the Virasoro algebra with $c=1/2$. One can check both constants directly in the ground states. In the NS sector,In the R sector, $b_0^2=1/2$ and $L_{-1}|R⟩=\frac{1}{2}{b}_{-1}{b}_0|R⟩$, whose norm squared is $1/8$. The relation $[L_1,L_{-1}]=2L_0$ requires $2h_R=1/8$, so $h_R=a=1/16$. Thus the Ramond sector zero-mode shift is essential.

The three constructions in Ising Virasoro modules from a Majorana fermion are explicit. The NS vacuum is a highest-weight vector of weight $0$. The state $b_{-1/2}|0⟩$ is a highest-weight vector of weight $1/2$: applying any positive $L_n$ gives $-(n/2-1/2)b_{n-1/2}|0⟩=0$, including the zero coefficient for $n=1$. Since $L_n$ is quadratic in fermions, the two cyclic modules lie in the even and odd fermion parity sectors, respectively. For R, choose a normalized ground state annihilated by all $b_n$ with $n>0$ and with $b_0|R⟩=\pm |R⟩/\sqrt{2}$. All positive $L_n$ annihilate it and its weight is $1/16$. The two zero-mode sign choices give isomorphic Virasoro modules; one choice suffices to construct the third module.

The oscillator inner products are positive, and each cyclic highest-weight representation just constructed is irreducible. Indeed any nonzero proper invariant graded submodule would have a lowest-energy vector $w$ of weight greater than the cyclic ground weight. Every positive $L_n$ annihilates $w$; thus $w$ is orthogonal to every negative-mode descendant of the original ground vector, by moving the modes across the inner product. It is also orthogonal to that ground vector by its different $L_0$ eigenvalue. Since those Virasoro descendants span the cyclic module, $w$ would have zero norm, a contradiction. This proves irreducibility directly inside the positive fermionic Fock space.

In fact the cyclic modules fill the entire even NS, odd NS and chosen R Fock spaces. Otherwise an orthogonal remaining lowest vector would be the ground vector of a further unitary highest-weight Virasoro module; its weight would lie in ${\mathbb{Z}}_{>0}$, $1/2+{\mathbb{Z}}_{>0}$ or $1/16+{\mathbb{Z}}_{>0}$, respectively. None belongs to the classified list $\{0,1/2,1/16\}$. As a check, their Virasoro characters, defined by ${\chi }_h(q)=\mathrm{Tr}{q}^{L_0-c/24}$, areThe NS signs project onto even and odd occupation number; the R product uses the single chosen ground state. Null vectors of the abstract Virasoro Verma modules are automatically zero in this construction. In particular $L_{-1}|0⟩=0$, and the weight-$1/2$ and weight-$1/16$ ground states obey, respectively, $(L_{-2}-\frac{3}{4}{L}_{-1}^2)|1/2⟩=0$ and $(L_{-2}-\frac{4}{3}{L}_{-1}^2)|1/16⟩=0$, as is also verified by their level-two Gram matrices.

Write the affine modes as $J_n^a=T_n^a$, and put $Q=Q^{\mathfrak{g}}$. The Sugawara construction requires a restricted module, meaning that sufficiently large positive current modes annihilate each fixed vector, and a noncritical level $2k+Q\ne 0$. A positive-energy highest-weight representation has the required restriction. DefineNormal ordering puts positive modes to the right, making each sum finite on a fixed finite-energy vector. The invariant sum over the current index $a$ is the quadratic Casimir operator construction for the affine current algebra.

Here is the normal ordering calculation responsible for its denominator. In commuting $S_m$ with $J_n^b$, the two affine central terms supply $-2kn{J}_{m+n}^b$. The remaining terms contain $i{f}^{abc}$ times the two ways to replace a current by $J_{p+n}^c$. Relabeling the summation index and using antisymmetry of $f^{abc}$ cancels the un-reordered terms. Reordering the modes that cross the creation/annihilation boundary leaves $n$ signed terms, whose two structure constants contract by $f^{abc}{f}^{abd}=Q{\delta }^{cd}$. Their sum is $-Qn{J}_{m+n}^b$. ThusThe adjoint-Casimir contribution is genuinely quantum. For an explicit check of its coefficient, let $|0⟩$ be the affine vacuum, with all $J_n^a|0⟩=0$ for $n\ge 0$. ThenThe first term acts by the adjoint Casimir operator $Q$, and only $p=1$ contributes $2k$ to the second. This also verifies that a current has conformal weight one under the displayed normalization.

Now commute the quadratic generators with one another. Acting on either current factor produces $(m-n)L_{m+n}^{\mathfrak{g}}$; the residual full contractions are scalar and vanish unless $m+n=0$. The Jacobi identity fixes their cubic dependence on $m$, and the plane vacuum convention $L_{-1}|0⟩=L_0|0⟩=L_1|0⟩=0$ fixes the combination $m^3-m$. To determine its coefficient explicitly, noteCommuting a single $J_1^a$ through the numerator gives $(2k+Q)J_{-1}^a|0⟩$: the two central contractions give $2k{J}_{-1}^a$, and the zero-mode commutator in the structure-constant term gives $Q{J}_{-1}^a$. The remaining $J_1^a$ contraction gives $k$ for each adjoint index. ThereforeFor an algebraic module without a positive inner product, the same calculation is its vacuum matrix element and fixes the universal central term. ConsequentlyThis is a representation of the Virasoro algebra, not merely its centerless part.

For a complex semisimple Lie algebra, a Cartan subalgebra $\mathfrak{h}$ is a maximal abelian subalgebra of semisimple elements; equivalently it is the complexification of a maximal torus algebra in the compact real form. Its dimension is the rank of a semisimple Lie algebra, $r$. Choose an orthonormal Cartan basis with currents $H_n^i$. They have no structure-constant term:For $k\ne 0$, their abelian Sugawara construction isEach Cartan current has weight one under both $L^{\mathfrak{g}}$ and $L^{\mathfrak{h}}$, so $K_m=L_m^{\mathfrak{g}}-L_m^{\mathfrak{h}}$ commutes with all $H_n^i$, hence also with $L_n^{\mathfrak{h}}$. Subtracting the two commuting stress tensors yields the Coset constructionIn the intended nontrivial compact unitary theory at positive admissible level, this coset acts on a positive inner-product space, with $K_m^{†}=K_{-m}$. Its vacuum satisfies $K_n|0⟩=0$ for $n\ge -1$, soThus $c^{\mathfrak{g}}\ge r$. Also $Q>0$ for a compact simple Lie algebra and $k>0$, so $2k/(2k+Q)<1$, which proves the upper bound directly. Hence, under these physical representation hypotheses,The upper inequality is strict at finite positive level; $c^{\mathfrak{g}}$ tends to $\dim \mathfrak{g}$ as the level grows.

The positivity hypothesis is necessary and is not implied by the bare affine commutation relation. For example, at $k=0$ its trivial representation has all currents zero and gives $c^{\mathfrak{g}}=0<\mathrm{rank}\mathfrak{g}$. Even positive arbitrary levels need not satisfy the bound: with $\mathfrak{g}=\mathfrak{su}(2)$, $f^{abc}={ϵ}^{abc}$, $Q=2$ and $k=1/4$, the formula gives $c^{\mathfrak{g}}=3/5<1$. Such a level does not give the intended compact unitary positive-energy theory. Thus the last bound is a statement about nontrivial unitary admissible-level representations; it cannot hold for every representation in the unrestricted algebraic formulation. At $2k+Q=0$ the displayed Sugawara construction itself is undefined.