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Problem Sets and Projects

This appendix is the working laboratory for the course. The lectures explain the logic of conformal field theory; the problems make that logic usable. The goal is not to accumulate tricks, but to develop the habits that matter for AdS/CFT:

symmetryWard identitiesoperator datacrossinglarge-Nbulk interpretation.\text{symmetry} \longrightarrow \text{Ward identities} \longrightarrow \text{operator data} \longrightarrow \text{crossing} \longrightarrow \text{large-}N \longrightarrow \text{bulk interpretation}.

A good solution should usually include three layers: the physical meaning, the algebra, and the holographic checkpoint. A correct formula with no explanation is not yet a good solution; a beautiful explanation with no formula is not yet a CFT solution. The sweet spot is both.

The problems below are organized as cumulative problem sets. The first half builds the core CFT machinery; the second half turns it into AdS/CFT readiness. Many problems include complete solutions. Some mini-projects are deliberately open-ended, because graduate-level mastery requires producing a coherent derivation, not only checking boxes.

A reasonable one-semester rhythm is:

WeeksSuggested assignmentMain skill
1—2Problem Sets 1—2QFT data, RG, critical exponents
3—4Problem Sets 3—4conformal group, correlators, Ward identities
5—6Problem Sets 5—6radial quantization, OPE, conformal blocks
7—8Problem Sets 7—8Virasoro, minimal models, modular invariance
9—10Problem Sets 9—10thermal CFT, entanglement, supersymmetry
11—12Problem Sets 11—12large-NN, N=4\mathcal N=4 SYM, pre-dictionary
13—14one mini-projectsynthesis and presentation

Throughout this page, local operators are normalized in CFT conventions, not Lagrangian conventions. For example, for scalar primaries we often write

Oi(x)Oj(0)=δijx2Δi,\langle \mathcal O_i(x)\mathcal O_j(0)\rangle = \frac{\delta_{ij}}{|x|^{2\Delta_i}},

after choosing an orthonormal basis of scalar primaries with equal spin and global-symmetry quantum numbers.

Problem 1.1: Connected correlators from W[J]W[J]

Section titled “Problem 1.1: Connected correlators from W[J]W[J]W[J]”

Let

Z[J]=exp(ddxJ(x)O(x)),W[J]=logZ[J].Z[J] = \left\langle \exp\left(\int d^d x\,J(x)\mathcal O(x)\right)\right\rangle, \qquad W[J]=\log Z[J].

Show that functional derivatives of W[J]W[J] generate connected correlation functions. In particular, show that

δWδJ(x)=O(x)J,\frac{\delta W}{\delta J(x)}=\langle \mathcal O(x)\rangle_J,

and

δ2WδJ(x)δJ(y)=O(x)O(y)JO(x)JO(y)J.\frac{\delta^2 W}{\delta J(x)\delta J(y)} = \langle \mathcal O(x)\mathcal O(y)\rangle_J - \langle \mathcal O(x)\rangle_J\langle \mathcal O(y)\rangle_J.

Explain why this is the field-theory ancestor of the AdS/CFT statement

ZCFT[J]=Zbulk[ϕ=J].Z_{\mathrm{CFT}}[J]=Z_{\mathrm{bulk}}[\phi_{\partial}=J].
Solution

By definition,

δZ[J]δJ(x)=O(x)exp(ddyJ(y)O(y)).\frac{\delta Z[J]}{\delta J(x)} = \left\langle \mathcal O(x) \exp\left(\int d^d y\,J(y)\mathcal O(y)\right)\right\rangle.

Dividing by Z[J]Z[J] gives the expectation value in the source-deformed ensemble,

δWδJ(x)=1Z[J]δZ[J]δJ(x)=O(x)J.\frac{\delta W}{\delta J(x)} = \frac{1}{Z[J]}\frac{\delta Z[J]}{\delta J(x)} = \langle \mathcal O(x)\rangle_J.

For the second derivative,

δ2WδJ(x)δJ(y)=1Zδ2ZδJ(x)δJ(y)1Z2δZδJ(x)δZδJ(y).\frac{\delta^2 W}{\delta J(x)\delta J(y)} = \frac{1}{Z}\frac{\delta^2 Z}{\delta J(x)\delta J(y)} - \frac{1}{Z^2}\frac{\delta Z}{\delta J(x)}\frac{\delta Z}{\delta J(y)}.

The first term is O(x)O(y)J\langle \mathcal O(x)\mathcal O(y)\rangle_J and the second is the product of one-point functions. Therefore

δ2WδJ(x)δJ(y)=O(x)O(y)J,c.\frac{\delta^2 W}{\delta J(x)\delta J(y)} = \langle \mathcal O(x)\mathcal O(y)\rangle_{J,c}.

Higher derivatives of WW similarly generate connected correlators.

In AdS/CFT, J(x)J(x) is not just a bookkeeping device. It is the boundary value of a bulk field ϕ\phi dual to O\mathcal O. Therefore the same operation that differentiates WCFT[J]W_{\mathrm{CFT}}[J] with respect to JJ becomes, on the bulk side, a variation of the renormalized on-shell gravitational action with respect to the boundary condition of ϕ\phi.

Problem 1.2: Contact terms and source redefinitions

Section titled “Problem 1.2: Contact terms and source redefinitions”

Suppose one changes the source-dependent generating functional by a local counterterm

W[J]W[J]+a2ddxJ(x)2.W[J]\mapsto W[J]+\frac{a}{2}\int d^d x\,J(x)^2.

What happens to the two-point function? Why is this a contact-term ambiguity rather than a change of separated-point CFT data?

Solution

Taking two derivatives gives

δ2δJ(x)δJ(y)[a2ddzJ(z)2]=aδ(d)(xy).\frac{\delta^2}{\delta J(x)\delta J(y)} \left[\frac{a}{2}\int d^d z\,J(z)^2\right] = a\delta^{(d)}(x-y).

Therefore

O(x)O(y)cO(x)O(y)c+aδ(d)(xy).\langle \mathcal O(x)\mathcal O(y)\rangle_c \mapsto \langle \mathcal O(x)\mathcal O(y)\rangle_c+a\delta^{(d)}(x-y).

For xyx\ne y, the correlator is unchanged. The local counterterm only changes coincident-point information. This is why separated-point CFT data such as scaling dimensions and OPE coefficients are scheme-independent, while contact terms require conventions.

In holographic renormalization, precisely such local source counterterms are added to make the on-shell bulk action finite. They change contact terms but not the nonlocal correlator.

Deform a CFT by scalar primaries,

S=SCFT+igiddxOi(x),Δi=Δ(Oi).S=S_{\mathrm{CFT}}+\sum_i g_i\int d^d x\,\mathcal O_i(x), \qquad \Delta_i=\Delta(\mathcal O_i).

Define dimensionless couplings

λi(μ)=giμΔid.\lambda_i(\mu)=g_i\mu^{\Delta_i-d}.

Show that, to first order near the fixed point,

βi(λ)=μdλidμ=(Δid)λi+O(λ2).\beta_i(\lambda)=\mu\frac{d\lambda_i}{d\mu} =(\Delta_i-d)\lambda_i+O(\lambda^2).

Classify relevant, marginal, and irrelevant operators.

Solution

Since gig_i has engineering dimension

[gi]=dΔi,[g_i]=d-\Delta_i,

the dimensionless coupling is λi=giμΔid\lambda_i=g_i\mu^{\Delta_i-d}. Holding the bare deformation fixed and differentiating with respect to μ\mu gives

μdλidμ=(Δid)giμΔid+O(λ2)=(Δid)λi+O(λ2).\mu\frac{d\lambda_i}{d\mu} =(\Delta_i-d)g_i\mu^{\Delta_i-d}+O(\lambda^2) =(\Delta_i-d)\lambda_i+O(\lambda^2).

Thus:

OperatorConditionLinearized behavior
relevantΔi<d\Delta_i<dgrows toward the IR
marginalΔi=d\Delta_i=ddecided by O(λ2)O(\lambda^2) terms
irrelevantΔi>d\Delta_i>ddies toward the IR

The phrase “toward the IR” means lowering the physical energy scale. Since the beta function above is written with respect to μ\mu, the sign should be interpreted carefully: for Δi<d\Delta_i<d, the dimensionful effect of gig_i becomes larger at long distances.

Problem 2.2: Critical exponents from operator dimensions

Section titled “Problem 2.2: Critical exponents from operator dimensions”

Let a statistical critical point be described by a CFT. Suppose the thermal deformation is controlled by a scalar operator ϵ\epsilon of dimension Δϵ\Delta_\epsilon, and the magnetic field couples to a scalar order parameter σ\sigma of dimension Δσ\Delta_\sigma:

S=SCFT+tddxϵ(x)hddxσ(x).S=S_{\mathrm{CFT}}+t\int d^d x\,\epsilon(x)-h\int d^d x\,\sigma(x).

Show that

ν=1dΔϵ,η=2Δσd+2,\nu=\frac{1}{d-\Delta_\epsilon}, \qquad \eta=2\Delta_\sigma-d+2,

and

γ=d2ΔσdΔϵ.\gamma=\frac{d-2\Delta_\sigma}{d-\Delta_\epsilon}.
Solution

The thermal coupling tt has RG eigenvalue

yt=dΔϵ.y_t=d-\Delta_\epsilon.

The correlation length scales as the inverse RG scale at which the dimensionless thermal perturbation becomes order one:

ξt1/yt.\xi\sim |t|^{-1/y_t}.

Thus

ν=1yt=1dΔϵ.\nu=\frac{1}{y_t}=\frac{1}{d-\Delta_\epsilon}.

At criticality,

σ(x)σ(0)1x2Δσ.\langle \sigma(x)\sigma(0)\rangle\sim \frac{1}{|x|^{2\Delta_\sigma}}.

In statistical-mechanics notation,

σ(x)σ(0)1xd2+η.\langle \sigma(x)\sigma(0)\rangle\sim \frac{1}{|x|^{d-2+\eta}}.

Equating powers gives

2Δσ=d2+η,2\Delta_\sigma=d-2+\eta,

so

η=2Δσd+2.\eta=2\Delta_\sigma-d+2.

The susceptibility is the integrated connected two-point function cut off at the correlation length:

χξddx1x2Δσξd2Δσ.\chi\sim \int^{\xi} d^d x\,\frac{1}{|x|^{2\Delta_\sigma}} \sim \xi^{d-2\Delta_\sigma}.

Using ξtν\xi\sim |t|^{-\nu} gives

χtν(d2Δσ).\chi\sim |t|^{-\nu(d-2\Delta_\sigma)}.

Therefore

γ=ν(d2Δσ)=d2ΔσdΔϵ.\gamma=\nu(d-2\Delta_\sigma) =\frac{d-2\Delta_\sigma}{d-\Delta_\epsilon}.

Problem Set 3: Conformal geometry and algebra

Section titled “Problem Set 3: Conformal geometry and algebra”

Problem 3.1: Finite special conformal transformations

Section titled “Problem 3.1: Finite special conformal transformations”

A special conformal transformation can be written as inversion, translation, inversion:

xμxμx2,xμx2xμx2bμ,xμx2bμxμ.x^\mu\mapsto \frac{x^\mu}{x^2}, \qquad \frac{x^\mu}{x^2}\mapsto \frac{x^\mu}{x^2}-b^\mu, \qquad \frac{x^\mu}{x^2}-b^\mu\mapsto x'^\mu.

Show that

xμ=xμbμx212bx+b2x2.x'^\mu=\frac{x^\mu-b^\mu x^2}{1-2b\cdot x+b^2x^2}.

Then expand to first order in bμb^\mu and identify the infinitesimal vector field.

Solution

Let

yμ=xμx2bμ.y^\mu=\frac{x^\mu}{x^2}-b^\mu.

The final inversion gives

xμ=yμy2.x'^\mu=\frac{y^\mu}{y^2}.

Compute

y2=(xx2b)2=1x22bxx2+b2=12bx+b2x2x2.y^2=\left(\frac{x}{x^2}-b\right)^2 =\frac{1}{x^2}-\frac{2b\cdot x}{x^2}+b^2 =\frac{1-2b\cdot x+b^2x^2}{x^2}.

Therefore

xμ=xμ/x2bμ(12bx+b2x2)/x2=xμbμx212bx+b2x2.x'^\mu = \frac{x^\mu/x^2-b^\mu}{(1-2b\cdot x+b^2x^2)/x^2} = \frac{x^\mu-b^\mu x^2}{1-2b\cdot x+b^2x^2}.

To first order in bb,

xμ=(xμbμx2)(1+2bx)+O(b2),x'^\mu=(x^\mu-b^\mu x^2)(1+2b\cdot x)+O(b^2),

so

δxμ=xμxμ=2(bx)xμbμx2.\delta x^\mu=x'^\mu-x^\mu =2(b\cdot x)x^\mu-b^\mu x^2.

The corresponding infinitesimal vector field is

ξb=(2(bx)xμbμx2)μ.\xi_b=\left(2(b\cdot x)x^\mu-b^\mu x^2\right)\partial_\mu.

Up to the conventional factor of i-i, this is the special conformal generator bμKμb^\mu K_\mu.

Problem 3.2: The commutator [Kμ,Pν][K_\mu,P_\nu]

Section titled “Problem 3.2: The commutator [Kμ,Pν][K_\mu,P_\nu][Kμ​,Pν​]”

Using the differential-operator representation

Pμ=iμ,D=ixρρ,Mμν=i(xμνxνμ),P_\mu=-i\partial_\mu, \qquad D=-ix^\rho\partial_\rho, \qquad M_{\mu\nu}=i(x_\mu\partial_\nu-x_\nu\partial_\mu),

and

Kμ=i(2xμxρρx2μ),K_\mu=-i(2x_\mu x^\rho\partial_\rho-x^2\partial_\mu),

show that

[Kμ,Pν]=2i(δμνDMμν)[K_\mu,P_\nu]=2i(\delta_{\mu\nu}D-M_{\mu\nu})

in Euclidean signature.

Solution

It is enough to act on a test function f(x)f(x). Write

Kμ=iVμ,Pν=iν,K_\mu=-iV_\mu, \qquad P_\nu=-i\partial_\nu,

where

Vμ=2xμxρρx2μ.V_\mu=2x_\mu x^\rho\partial_\rho-x^2\partial_\mu.

Then

[Kμ,Pν]=[Vμ,ν].[K_\mu,P_\nu]=-[V_\mu,\partial_\nu].

For a vector field V=VρρV=V^\rho\partial_\rho,

[V,ν]=(νVρ)ρ.[V,\partial_\nu]=-(\partial_\nu V^\rho)\partial_\rho.

Here

Vμρ=2xμxρx2δμρ,V_\mu{}^\rho=2x_\mu x^\rho-x^2\delta_\mu{}^\rho,

so

νVμρ=2δμνxρ+2xμδνρ2xνδμρ.\partial_\nu V_\mu{}^\rho =2\delta_{\mu\nu}x^\rho+2x_\mu\delta_\nu{}^\rho-2x_\nu\delta_\mu{}^\rho.

Thus

[Kμ,Pν]=(2δμνxρρ+2xμν2xνμ).[K_\mu,P_\nu] =\left(2\delta_{\mu\nu}x^\rho\partial_\rho+2x_\mu\partial_\nu-2x_\nu\partial_\mu\right).

Using

xρρ=iD,xμνxνμ=iMμν,x^\rho\partial_\rho=iD, \qquad x_\mu\partial_\nu-x_\nu\partial_\mu=-iM_{\mu\nu},

we obtain

[Kμ,Pν]=2iδμνD2iMμν=2i(δμνDMμν).[K_\mu,P_\nu] =2i\delta_{\mu\nu}D-2iM_{\mu\nu} =2i(\delta_{\mu\nu}D-M_{\mu\nu}).

Different sign conventions for MμνM_{\mu\nu} shift the sign in the second term; the physics is unchanged once conventions are used consistently.

Problem Set 4: Correlation functions and Ward identities

Section titled “Problem Set 4: Correlation functions and Ward identities”

Use conformal invariance to derive the form of the scalar three-point function

O1(x1)O2(x2)O3(x3).\langle \mathcal O_1(x_1)\mathcal O_2(x_2)\mathcal O_3(x_3)\rangle.

Assume Oi\mathcal O_i are scalar primaries of dimensions Δi\Delta_i.

Solution

Translation and rotation invariance imply that the correlator depends only on distances xij=xixjx_{ij}=|x_i-x_j|. Scale covariance suggests

O1O2O3=C123x12ax23bx13c.\langle \mathcal O_1\mathcal O_2\mathcal O_3\rangle = \frac{C_{123}}{x_{12}^{a}x_{23}^{b}x_{13}^{c}}.

Under xiλxix_i\mapsto \lambda x_i, the correlator must scale as

λ(Δ1+Δ2+Δ3).\lambda^{-(\Delta_1+\Delta_2+\Delta_3)}.

Thus

a+b+c=Δ1+Δ2+Δ3.a+b+c=\Delta_1+\Delta_2+\Delta_3.

Special conformal covariance, or equivalently inversion covariance, fixes how powers attach to each point. Under inversion xμxμ/x2x^\mu\mapsto x^\mu/x^2,

xijxijxixj,x_{ij}\mapsto \frac{x_{ij}}{|x_i||x_j|},

and a scalar primary transforms with a factor xi2Δi|x_i|^{2\Delta_i}. Matching the powers of each xi|x_i| gives

a+c=2Δ1,a+b=2Δ2,b+c=2Δ3.a+c=2\Delta_1, \qquad a+b=2\Delta_2, \qquad b+c=2\Delta_3.

Solving,

a=Δ1+Δ2Δ3,a=\Delta_1+\Delta_2-\Delta_3, b=Δ2+Δ3Δ1,b=\Delta_2+\Delta_3-\Delta_1, c=Δ1+Δ3Δ2.c=\Delta_1+\Delta_3-\Delta_2.

Therefore

O1(x1)O2(x2)O3(x3)=C123x12Δ1+Δ2Δ3x23Δ2+Δ3Δ1x13Δ1+Δ3Δ2.\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_{23}^{\Delta_2+\Delta_3-\Delta_1} x_{13}^{\Delta_1+\Delta_3-\Delta_2}} }.

The coefficient C123C_{123} is dynamical CFT data.

For a conserved spin-one current in a dd-dimensional CFT, the two-point function is

Jμ(x)Jν(0)=CJIμν(x)x2(d1),\langle J_\mu(x)J_\nu(0)\rangle = \frac{C_J I_{\mu\nu}(x)}{x^{2(d-1)}},

where

Iμν(x)=δμν2xμxνx2.I_{\mu\nu}(x)=\delta_{\mu\nu}-2\frac{x_\mu x_\nu}{x^2}.

Show that it is conserved for x0x\ne 0:

μJμ(x)Jν(0)=0.\partial^\mu\langle J_\mu(x)J_\nu(0)\rangle=0.
Solution

Let r2=x2r^2=x^2. We need

μ(Iμνr2(d1))=0.\partial^\mu\left(\frac{I_{\mu\nu}}{r^{2(d-1)}}\right)=0.

Write

Iμνr2(d1)=δμνr2d22xμxνr2d.\frac{I_{\mu\nu}}{r^{2(d-1)}} = \frac{\delta_{\mu\nu}}{r^{2d-2}} -2\frac{x_\mu x_\nu}{r^{2d}}.

The derivative of the first term is

μ(δμνr2d2)=(2d2)xνr2d.\partial^\mu\left(\frac{\delta_{\mu\nu}}{r^{2d-2}}\right) =-(2d-2)\frac{x_\nu}{r^{2d}}.

For the second term,

μ(xμxνr2d)=(d+1)xνr2d2dx2xνr2d+2=(d+1)xνr2d2dxνr2d=(1d)xνr2d.\partial^\mu\left(\frac{x_\mu x_\nu}{r^{2d}}\right) = \frac{(d+1)x_\nu}{r^{2d}}-2d\frac{x^2x_\nu}{r^{2d+2}} = \frac{(d+1)x_\nu}{r^{2d}}-2d\frac{x_\nu}{r^{2d}} = (1-d)\frac{x_\nu}{r^{2d}}.

Multiplying by 2-2 gives

2μ(xμxνr2d)=2(d1)xνr2d.-2\partial^\mu\left(\frac{x_\mu x_\nu}{r^{2d}}\right) =2(d-1)\frac{x_\nu}{r^{2d}}.

This cancels the derivative of the first term. Therefore the correlator is conserved away from contact terms at x=0x=0.

Problem Set 5: Radial quantization and unitarity

Section titled “Problem Set 5: Radial quantization and unitarity”

Problem 5.1: Cylinder energies from scaling dimensions

Section titled “Problem 5.1: Cylinder energies from scaling dimensions”

Let a scalar primary O\mathcal O of dimension Δ\Delta create a radial-quantization state

O=O(0)0.|\mathcal O\rangle=\mathcal O(0)|0\rangle.

Use the plane-cylinder map r=eτr=e^\tau to show that the cylinder Hamiltonian is the dilatation operator and that

HcylO=ΔO.H_{\mathrm{cyl}}|\mathcal O\rangle=\Delta |\mathcal O\rangle.
Solution

On Rd{0}\mathbb R^d\setminus\{0\}, write the flat metric as

ds2=dr2+r2dΩd12.ds^2=dr^2+r^2d\Omega_{d-1}^2.

With r=eτr=e^\tau,

ds2=e2τ(dτ2+dΩd12).ds^2=e^{2\tau}(d\tau^2+d\Omega_{d-1}^2).

After a Weyl transformation, this is the cylinder metric on Rτ×Sd1\mathbb R_\tau\times S^{d-1}. Translation in cylinder time τ\tau is radial scaling on the plane:

ττ+arear.\tau\mapsto \tau+a \quad\Longleftrightarrow\quad r\mapsto e^a r.

Thus the Hamiltonian generating cylinder time translations is the dilatation generator DD.

A primary obeys

DO(0)0=ΔO(0)0,D\mathcal O(0)|0\rangle=\Delta\mathcal O(0)|0\rangle,

so

HcylO=ΔO.H_{\mathrm{cyl}}|\mathcal O\rangle=\Delta|\mathcal O\rangle.

This is the seed of the AdS interpretation: CFT operator dimensions are energies of states on the cylinder, matching global AdS energies.

For a scalar primary Δ|\Delta\rangle, positivity of descendants implies the unitarity bound

Δd22,\Delta\ge \frac{d-2}{2},

except for the identity operator. Show how this bound arises from the norm of the level-two scalar descendant P2ΔP^2|\Delta\rangle.

Solution

In radial quantization,

Pμ=Kμ.P_\mu^\dagger=K_\mu.

The norm of the level-two scalar descendant is

P2Δ2=ΔK2P2Δ.\|P^2|\Delta\rangle\|^2 =\langle \Delta|K^2P^2|\Delta\rangle.

Using the conformal algebra and the primary conditions

KμΔ=0,MμνΔ=0,DΔ=ΔΔ,K_\mu|\Delta\rangle=0, \qquad M_{\mu\nu}|\Delta\rangle=0, \qquad D|\Delta\rangle=\Delta|\Delta\rangle,

one finds, up to a positive normalization convention,

ΔK2P2ΔΔ(Δd22)ΔΔ.\langle \Delta|K^2P^2|\Delta\rangle \propto \Delta\left(\Delta-\frac{d-2}{2}\right)\langle \Delta|\Delta\rangle.

The level-one descendants already require Δ0\Delta\ge 0. For a non-identity scalar, positivity of the level-two norm then requires

Δd22.\Delta\ge \frac{d-2}{2}.

At saturation, P2ΔP^2|\Delta\rangle is null. In operator language,

2O=0,\partial^2\mathcal O=0,

so the operator behaves like a free scalar field. This is why saturation of a unitarity bound usually means a shortening condition.

Problem 6.1: Crossing equation for identical scalars

Section titled “Problem 6.1: Crossing equation for identical scalars”

For identical scalar primaries ϕ\phi of dimension Δϕ\Delta_\phi, write

ϕ(x1)ϕ(x2)ϕ(x3)ϕ(x4)=1x122Δϕx342ΔϕG(u,v),\langle \phi(x_1)\phi(x_2)\phi(x_3)\phi(x_4)\rangle = \frac{1}{x_{12}^{2\Delta_\phi}x_{34}^{2\Delta_\phi}}G(u,v),

with cross-ratios

u=x122x342x132x242,v=x142x232x132x242. u=\frac{x_{12}^2x_{34}^2}{x_{13}^2x_{24}^2}, \qquad v=\frac{x_{14}^2x_{23}^2}{x_{13}^2x_{24}^2}.

Derive the crossing equation relating G(u,v)G(u,v) and G(v,u)G(v,u) under x1x3x_1\leftrightarrow x_3.

Solution

Under the exchange x1x3x_1\leftrightarrow x_3, the cross-ratios transform as

uv. u\leftrightarrow v.

The same four-point function can be written in the exchanged channel as

ϕ1ϕ2ϕ3ϕ4=1x232Δϕx142ΔϕG(v,u).\langle \phi_1\phi_2\phi_3\phi_4\rangle = \frac{1}{x_{23}^{2\Delta_\phi}x_{14}^{2\Delta_\phi}}G(v,u).

Equating this with the original representation gives

1x122Δϕx342ΔϕG(u,v)=1x232Δϕx142ΔϕG(v,u).\frac{1}{x_{12}^{2\Delta_\phi}x_{34}^{2\Delta_\phi}}G(u,v) = \frac{1}{x_{23}^{2\Delta_\phi}x_{14}^{2\Delta_\phi}}G(v,u).

Using

x122x342x142x232=uv,\frac{x_{12}^2x_{34}^2}{x_{14}^2x_{23}^2}=\frac{u}{v},

we get

G(u,v)=(uv)ΔϕG(v,u).\boxed{ G(u,v)=\left(\frac{u}{v}\right)^{\Delta_\phi}G(v,u) }.

This is the simplest bootstrap equation. Expanded in conformal blocks, it becomes a constraint on the spectrum and OPE coefficients appearing in the ϕ×ϕ\phi\times\phi OPE.

Problem 6.2: Block expansion and positivity

Section titled “Problem 6.2: Block expansion and positivity”

Assume the identical-scalar four-point function has the conformal block expansion

G(u,v)=Oϕ×ϕλϕϕO2GΔ,(u,v),G(u,v)=\sum_{\mathcal O\in \phi\times\phi}\lambda_{\phi\phi\mathcal O}^2G_{\Delta,\ell}(u,v),

where the exchanged operators are primaries of dimension Δ\Delta and spin \ell. Explain why the coefficients are nonnegative in a unitary CFT, after choosing real normalized operators.

Solution

The OPE has the schematic form

ϕ(x)ϕ(0)OλϕϕOCΔ,(x,)O(0),\phi(x)\phi(0)\sim \sum_{\mathcal O}\lambda_{\phi\phi\mathcal O}\,C_{\Delta,\ell}(x,\partial)\mathcal O(0),

where CΔ,C_{\Delta,\ell} is fixed by conformal symmetry. In a basis where two-point functions of primaries are diagonal and positive,

Oa(x)Ob(0)δab,\langle \mathcal O_a(x)\mathcal O_b(0)\rangle\propto \delta_{ab},

the four-point function contribution of a given conformal family is proportional to the product of the OPE coefficient in the left OPE and the OPE coefficient in the right OPE. For identical external operators these two coefficients are the same:

λϕϕOλϕϕO=λϕϕO2.\lambda_{\phi\phi\mathcal O}\lambda_{\phi\phi\mathcal O} =\lambda_{\phi\phi\mathcal O}^2.

Reflection positivity allows the coefficients to be chosen real. Hence

λϕϕO20.\lambda_{\phi\phi\mathcal O}^2\ge 0.

This positivity is the key input that turns crossing symmetry from a formal associativity equation into a powerful numerical bootstrap constraint.

Problem 6.3: Generalized free field spectrum

Section titled “Problem 6.3: Generalized free field spectrum”

Let O\mathcal O be a generalized free scalar of dimension Δ\Delta. Its four-point function is

G(u,v)=1+uΔ+(uv)Δ.G(u,v)=1+u^\Delta+\left(\frac{u}{v}\right)^\Delta.

Explain why the O×O\mathcal O\times\mathcal O OPE contains double-trace primaries with schematic form

[OO]n,O2nμ1μOtraces,[\mathcal O\mathcal O]_{n,\ell} \sim \mathcal O\,\partial^{2n}\partial_{\mu_1}\cdots\partial_{\mu_\ell}\mathcal O-\text{traces},

and dimensions

Δn,=2Δ+2n+.\Delta_{n,\ell}=2\Delta+2n+\ell.
Solution

The generalized free four-point function is obtained by Wick-like pairings, even though O\mathcal O need not be a fundamental free field. In the 123412\to34 channel, the identity comes from the disconnected pairing

O1O2O3O4.\langle \mathcal O_1\mathcal O_2\rangle\langle \mathcal O_3\mathcal O_4\rangle.

The remaining pairings must be reproduced by an infinite tower of nontrivial conformal families in the O×O\mathcal O\times\mathcal O OPE. Since the theory is generalized free at leading order, the two-particle operators behave like normal-ordered products of two independent one-particle operators.

Each O\mathcal O contributes dimension Δ\Delta. Derivatives contribute one unit of dimension. A spin-\ell symmetric traceless tensor needs \ell uncontracted derivatives. Radial excitations are represented by 2n2n extra contracted derivatives. Therefore

Δn,=Δ+Δ++2n=2Δ+2n+.\Delta_{n,\ell}=\Delta+\Delta+\ell+2n =2\Delta+2n+\ell.

In AdS, these are precisely two-particle states made from two identical bulk quanta in global AdS. The integer nn is a radial excitation number and \ell is angular momentum.

Problem Set 7: Essential two-dimensional CFT

Section titled “Problem Set 7: Essential two-dimensional CFT”

Problem 7.1: Virasoro algebra from the TTT T OPE

Section titled “Problem 7.1: Virasoro algebra from the TTT TTT OPE”

Assume the holomorphic stress tensor has OPE

T(z)T(w)c/2(zw)4+2T(w)(zw)2+T(w)zw.T(z)T(w) \sim \frac{c/2}{(z-w)^4} +\frac{2T(w)}{(z-w)^2} +\frac{\partial T(w)}{z-w}.

With modes

Ln=12πidzzn+1T(z),L_n=\frac{1}{2\pi i}\oint dz\,z^{n+1}T(z),

derive

[Lm,Ln]=(mn)Lm+n+c12m(m21)δm+n,0.[L_m,L_n]=(m-n)L_{m+n}+\frac{c}{12}m(m^2-1)\delta_{m+n,0}.
Solution

The commutator is computed by nested contour integrals:

[Lm,Ln]=1(2πi)2wdwwn+1z=wdzzm+1T(z)T(w).[L_m,L_n] = \frac{1}{(2\pi i)^2}\oint_w dw\,w^{n+1}\oint_{z=w}dz\,z^{m+1}T(z)T(w).

Insert the singular part of the OPE. The central term is

c212πidwwn+112πiz=wdzzm+1(zw)4.\frac{c}{2}\frac{1}{2\pi i}\oint dw\,w^{n+1} \frac{1}{2\pi i}\oint_{z=w}dz\,\frac{z^{m+1}}{(z-w)^4}.

Using

12πiz=wdzzm+1(zw)4=13!w3wm+1=(m+1)m(m1)6wm2,\frac{1}{2\pi i}\oint_{z=w}dz\,\frac{z^{m+1}}{(z-w)^4} =\frac{1}{3!}\partial_w^3 w^{m+1} =\frac{(m+1)m(m-1)}{6}w^{m-2},

the central contribution becomes

c12m(m21)12πidwwm+n1=c12m(m21)δm+n,0.\frac{c}{12}m(m^2-1)\frac{1}{2\pi i}\oint dw\,w^{m+n-1} = \frac{c}{12}m(m^2-1)\delta_{m+n,0}.

The 2T(w)/(zw)22T(w)/(z-w)^2 term gives

2(m+1)12πidwwm+n+1T(w)=2(m+1)Lm+n.2(m+1)\frac{1}{2\pi i}\oint dw\,w^{m+n+1}T(w) =2(m+1)L_{m+n}.

The T(w)/(zw)\partial T(w)/(z-w) term gives

12πidwwm+n+2T(w)=(m+n+2)Lm+n,\frac{1}{2\pi i}\oint dw\,w^{m+n+2}\partial T(w) =-(m+n+2)L_{m+n},

after integration by parts. Combining these two noncentral terms gives

2(m+1)(m+n+2)=mn.2(m+1)-(m+n+2)=m-n.

Hence

[Lm,Ln]=(mn)Lm+n+c12m(m21)δm+n,0.[L_m,L_n]=(m-n)L_{m+n}+\frac{c}{12}m(m^2-1)\delta_{m+n,0}.

Consider a Virasoro highest-weight state h|h\rangle. A level-two descendant has form

χ=(L2+aL12)h.|\chi\rangle=(L_{-2}+aL_{-1}^2)|h\rangle.

Find the conditions for χ|\chi\rangle to be null, meaning

L1χ=0,L2χ=0.L_1|\chi\rangle=0, \qquad L_2|\chi\rangle=0.
Solution

Use

[L1,L2]=3L1,[L1,L1]=2L0.[L_1,L_{-2}]=3L_{-1}, \qquad [L_1,L_{-1}]=2L_0.

Since L1h=0L_1|h\rangle=0,

L1L2h=3L1h.L_1L_{-2}|h\rangle=3L_{-1}|h\rangle.

Also

L1L12h=[L1,L1]L1h+L1[L1,L1]h.L_1L_{-1}^2|h\rangle =[L_1,L_{-1}]L_{-1}|h\rangle+L_{-1}[L_1,L_{-1}]|h\rangle.

Using L0L1h=(h+1)L1hL_0L_{-1}|h\rangle=(h+1)L_{-1}|h\rangle and L0h=hhL_0|h\rangle=h|h\rangle, this gives

L1L12h=2(h+1)L1h+2hL1h=(4h+2)L1h.L_1L_{-1}^2|h\rangle =2(h+1)L_{-1}|h\rangle+2hL_{-1}|h\rangle =(4h+2)L_{-1}|h\rangle.

Therefore L1χ=0L_1|\chi\rangle=0 implies

3+a(4h+2)=0,3+a(4h+2)=0,

so

a=32(2h+1).a=-\frac{3}{2(2h+1)}.

Next use

[L2,L2]=4L0+c2,[L2,L1]=3L1.[L_2,L_{-2}]=4L_0+\frac{c}{2}, \qquad [L_2,L_{-1}]=3L_1.

Thus

L2L2h=(4h+c2)h.L_2L_{-2}|h\rangle=\left(4h+\frac{c}{2}\right)|h\rangle.

For the other term,

L2L12h=[L2,L1]L1h+L1[L2,L1]h.L_2L_{-1}^2|h\rangle =[L_2,L_{-1}]L_{-1}|h\rangle+L_{-1}[L_2,L_{-1}]|h\rangle.

The second term vanishes because L1h=0L_1|h\rangle=0. The first gives

3L1L1h=3(2h)h=6hh.3L_1L_{-1}|h\rangle=3(2h)|h\rangle=6h|h\rangle.

So L2χ=0L_2|\chi\rangle=0 implies

4h+c2+6ah=0.4h+\frac{c}{2}+6ah=0.

Substituting a=3/[2(2h+1)]a=-3/[2(2h+1)] gives the level-two degeneracy condition

4h+c29h2h+1=0.4h+\frac{c}{2}-\frac{9h}{2h+1}=0.

Equivalently,

c=2h(58h)2h+1.c=\frac{2h(5-8h)}{2h+1}.

This is the algebraic origin of second-order BPZ equations.

Problem Set 8: Modular invariance and Cardy growth

Section titled “Problem Set 8: Modular invariance and Cardy growth”

Problem 8.1: Cardy formula from modular invariance

Section titled “Problem 8.1: Cardy formula from modular invariance”

For a unitary compact 2D CFT with central charge cc and discrete spectrum, the torus partition function is

Z(τ,τˉ)=TrqL0c/24qˉLˉ0c/24,q=e2πiτ.Z(\tau,\bar\tau)=\mathrm{Tr}\,q^{L_0-c/24}\bar q^{\bar L_0-c/24}, \qquad q=e^{2\pi i\tau}.

Take a rectangular torus with τ=iβ/(2π)\tau=i\beta/(2\pi). Use modular invariance to derive the asymptotic density of high-energy states,

S(E)2πceffE3,S(E)\sim 2\pi\sqrt{\frac{c_{\mathrm{eff}}E}{3}},

where E=L0+Lˉ0c/12E=L_0+\bar L_0-c/12 and ceffc_{\mathrm{eff}} is the effective central charge. State the simplification for a unitary CFT with vacuum dimension zero.

Solution

For a rectangular torus,

Z(β)=TreβE,E=L0+Lˉ0c12.Z(\beta)=\mathrm{Tr}\,e^{-\beta E}, \qquad E=L_0+\bar L_0-\frac{c}{12}.

Modular SS invariance relates the low-temperature and high-temperature limits:

Z(β)=Z(4π2β).Z(\beta)=Z\left(\frac{4\pi^2}{\beta}\right).

At low temperature, the partition function is dominated by the state of lowest shifted energy. Let

E0=ceff12.E_0=-\frac{c_{\mathrm{eff}}}{12}.

Then for large 4π2/β4\pi^2/\beta,

Z(4π2β)exp(4π2βceff12)=exp(π2ceff3β).Z\left(\frac{4\pi^2}{\beta}\right) \sim \exp\left(\frac{4\pi^2}{\beta}\frac{c_{\mathrm{eff}}}{12}\right) = \exp\left(\frac{\pi^2c_{\mathrm{eff}}}{3\beta}\right).

Thus the high-temperature canonical free energy is controlled by

logZ(β)π2ceff3β.\log Z(\beta)\sim \frac{\pi^2c_{\mathrm{eff}}}{3\beta}.

The microcanonical density ρ(E)\rho(E) follows by inverse Laplace transform:

ρ(E)dβexp(βE+π2ceff3β).\rho(E)\sim \int d\beta\,\exp\left(\beta E+\frac{\pi^2c_{\mathrm{eff}}}{3\beta}\right).

The saddle satisfies

E=π2ceff3β2,β=πceff3E.E=\frac{\pi^2c_{\mathrm{eff}}}{3\beta^2}, \qquad \beta_*=\pi\sqrt{\frac{c_{\mathrm{eff}}}{3E}}.

The saddle exponent is

S(E)=βE+π2ceff3β=2πceffE3.S(E)=\beta_*E+\frac{\pi^2c_{\mathrm{eff}}}{3\beta_*} =2\pi\sqrt{\frac{c_{\mathrm{eff}}E}{3}}.

For a unitary compact CFT with vacuum dimension zero,

ceff=c.c_{\mathrm{eff}}=c.

For independent left- and right-moving energies, the more refined formula is

S2πc6(L0c24)+2πcˉ6(Lˉ0cˉ24).S\sim 2\pi\sqrt{\frac{c}{6}\left(L_0-\frac{c}{24}\right)} +2\pi\sqrt{\frac{\bar c}{6}\left(\bar L_0-\frac{\bar c}{24}\right)}.

This is the CFT side of the BTZ black-hole entropy match.

Problem 8.2: Why modular invariance is not crossing symmetry

Section titled “Problem 8.2: Why modular invariance is not crossing symmetry”

Both crossing symmetry and modular invariance are consistency conditions. Explain the difference between them.

Solution

Crossing symmetry is associativity of the local operator algebra. It compares different OPE decompositions of the same correlation function, for example the ss- and tt-channel decompositions of a four-point function on the sphere:

(12)(34)versus(14)(23).(12)(34)\quad \text{versus}\quad (14)(23).

Modular invariance is consistency under large diffeomorphisms of a higher-genus Euclidean spacetime, most famously the torus:

τaτ+bcτ+d,(abcd)SL(2,Z).\tau\mapsto \frac{a\tau+b}{c\tau+d}, \qquad \begin{pmatrix}a&b\\c&d\end{pmatrix}\in SL(2,\mathbb Z).

It constrains the spectrum and Hilbert-space trace

Z(τ,τˉ)=TrHqL0c/24qˉLˉ0cˉ/24.Z(\tau,\bar\tau)=\mathrm{Tr}_{\mathcal H}\,q^{L_0-c/24}\bar q^{\bar L_0-\bar c/24}.

So:

ConstraintObjectPhysical meaning
crossinglocal correlatorsOPE associativity
modular invariancepartition functions and higher-genus amplitudesconsistency of quantization on nontrivial manifolds

In rational 2D CFT the two are deeply related through chiral algebras, fusion, and modular tensor categories, but they are not the same statement.

Problem Set 9: Thermal CFT and entanglement

Section titled “Problem Set 9: Thermal CFT and entanglement”

For a homogeneous thermal CFT in flat dd-dimensional spacetime, show that

ϵ=(d1)p,\epsilon=(d-1)p,

where ϵ\epsilon is the energy density and pp is the pressure. Then show that

pTd,sTd1.p\propto T^d, \qquad s\propto T^{d-1}.
Solution

For a homogeneous isotropic thermal state,

Tμν=diag(ϵ,p,p,,p)\langle T^\mu{}_{\nu}\rangle =\mathrm{diag}(-\epsilon,p,p,\ldots,p)

in Lorentzian signature. Conformal invariance implies the stress tensor is traceless, up to anomalies. In flat space at finite temperature there is no curvature anomaly, so

Tμμ=ϵ+(d1)p=0.\langle T^\mu{}_{\mu}\rangle=-\epsilon+(d-1)p=0.

Hence

ϵ=(d1)p.\epsilon=(d-1)p.

Dimensional analysis gives the free energy density

f(T)=adTd,f(T)=-a_d T^d,

where ada_d is theory-dependent. Since p=fp=-f,

p=adTd.p=a_dT^d.

The entropy density is

s=pT=dadTd1.s=\frac{\partial p}{\partial T}=d a_d T^{d-1}.

This scaling is the CFT side of the planar AdS black-brane thermodynamics.

Problem 9.2: First law of entanglement for a ball

Section titled “Problem 9.2: First law of entanglement for a ball”

For the vacuum state of a CFT, the modular Hamiltonian of a ball BB of radius RR centered at the origin is

KB=2πx<Rdd1xR2x22RT00(0,x).K_B=2\pi\int_{|\vec x|<R} d^{d-1}x\,\frac{R^2-|\vec x|^2}{2R}T_{00}(0,\vec x).

Use the entanglement first law

δSB=δKB\delta S_B=\delta\langle K_B\rangle

to compute the first-order change in entanglement entropy for a homogeneous small energy-density perturbation δT00=δϵ\delta\langle T_{00}\rangle=\delta\epsilon.

Solution

Substitute the constant perturbation into δKB\delta\langle K_B\rangle:

δSB=2πδϵx<Rdd1xR2r22R.\delta S_B =2\pi\delta\epsilon \int_{|\vec x|<R} d^{d-1}x\,\frac{R^2-r^2}{2R}.

Let n=d1n=d-1 be the number of spatial dimensions. The volume element is

dnx=Sn1rn1dr,d^n x=S_{n-1}r^{n-1}dr,

so

r<Rdnx(R2r2)=Sn10Rdrrn1(R2r2).\int_{r<R} d^n x\,(R^2-r^2) =S_{n-1}\int_0^R dr\,r^{n-1}(R^2-r^2).

Compute

0Rdrrn1(R2r2)=Rn+2nRn+2n+2=2Rn+2n(n+2).\int_0^R dr\,r^{n-1}(R^2-r^2) =\frac{R^{n+2}}{n}-\frac{R^{n+2}}{n+2} =\frac{2R^{n+2}}{n(n+2)}.

Thus

δSB=2πδϵ12RSn12Rn+2n(n+2).\delta S_B =2\pi\delta\epsilon\frac{1}{2R} S_{n-1}\frac{2R^{n+2}}{n(n+2)}.

Using n=d1n=d-1,

δSB=2πSd2Rd(d1)(d+1)δϵ.\boxed{ \delta S_B =\frac{2\pi S_{d-2}R^d}{(d-1)(d+1)}\,\delta\epsilon }.

This result is a key CFT input in the derivation of linearized Einstein equations from entanglement equilibrium.

Problem Set 10: Supersymmetry and protected data

Section titled “Problem Set 10: Supersymmetry and protected data”

Problem 10.1: BPS shortening as a norm condition

Section titled “Problem 10.1: BPS shortening as a norm condition”

Suppose a superconformal primary O|\mathcal O\rangle obeys a schematic anticommutator

{S,Q}=ΔR,\{S,Q\}=\Delta-R,

on the relevant component of the multiplet. Show that unitarity implies

ΔR,\Delta\ge R,

and that saturation implies shortening.

Solution

In radial quantization, SS is the adjoint of QQ:

Q=S.Q^\dagger=S.

Therefore

QO2=OSQO.\|Q|\mathcal O\rangle\|^2 =\langle \mathcal O|S Q|\mathcal O\rangle.

If O|\mathcal O\rangle is annihilated by SS, then on this primary state

OSQO=O{S,Q}O.\langle \mathcal O|S Q|\mathcal O\rangle = \langle \mathcal O|\{S,Q\}|\mathcal O\rangle.

Using the schematic algebra,

QO2=(ΔR)OO.\|Q|\mathcal O\rangle\|^2 =(\Delta-R)\langle \mathcal O|\mathcal O\rangle.

Unitarity requires the norm to be nonnegative, so

ΔR.\Delta\ge R.

If Δ=R\Delta=R, then

QO2=0.\|Q|\mathcal O\rangle\|^2=0.

The descendant QOQ|\mathcal O\rangle is null and is removed from the physical Hilbert space. The multiplet is shorter than a generic long multiplet. This is the algebraic meaning of BPS protection.

Problem 10.2: Half-BPS chiral primaries in N=4\mathcal N=4 SYM

Section titled “Problem 10.2: Half-BPS chiral primaries in N=4\mathcal N=4N=4 SYM”

The six real scalars of N=4\mathcal N=4 SYM transform in the vector representation of SO(6)RSO(6)_R. Define a null auxiliary vector YIY^I satisfying YY=0Y\cdot Y=0, and write

Op(x,Y)=YI1YIpTr(ΦI1ΦIp)+multi-trace/subtraction terms.\mathcal O_p(x,Y)=Y^{I_1}\cdots Y^{I_p}\mathrm{Tr}\left(\Phi^{I_1}\cdots\Phi^{I_p}\right)+\text{multi-trace/subtraction terms}.

Explain why the null condition packages the symmetric traceless representation [0,p,0][0,p,0] of SU(4)RSU(4)_R, and state the protected dimension.

Solution

The product YI1YIpY^{I_1}\cdots Y^{I_p} is manifestly symmetric in the SO(6)SO(6) indices. Traces are encoded by contractions with δIJ\delta_{IJ}. Since

YY=δIJYIYJ=0,Y\cdot Y=\delta_{IJ}Y^IY^J=0,

any trace part vanishes when contracted with the null polarization vector. Thus Op(x,Y)\mathcal O_p(x,Y) packages the symmetric traceless rank-pp representation of SO(6)RSO(6)_R.

Using

SO(6)RSU(4)R,SO(6)_R\simeq SU(4)_R,

this representation has Dynkin label

[0,p,0].[0,p,0].

The operator is half-BPS, so its dimension is protected:

Δ=p.\Delta=p.

In AdS/CFT, these operators are dual to Kaluza—Klein modes on S5S^5 with harmonic degree pp.

Problem 11.1: Connected correlator scaling

Section titled “Problem 11.1: Connected correlator scaling”

Let Oi\mathcal O_i be normalized single-trace operators in a large-NN gauge theory, with two-point functions of order one:

Oi(x)Oj(0)O(N0).\langle \mathcal O_i(x)\mathcal O_j(0)\rangle\sim O(N^0).

Use planar counting to argue that connected kk-point functions scale as

O1OkcN2k.\langle \mathcal O_1\cdots \mathcal O_k\rangle_c\sim N^{2-k}.

Explain the AdS interpretation.

Solution

For adjoint matrix fields, vacuum ribbon diagrams scale as

N22g,N^{2-2g},

where gg is the genus. A single-trace insertion creates one boundary on the ribbon surface. A connected planar diagram with kk single-trace insertions therefore scales as

N2kN^{2-k}

before accounting for normalization. Choosing operators normalized to have order-one two-point functions gives the stated scaling:

O1OkcN2k.\langle \mathcal O_1\cdots \mathcal O_k\rangle_c\sim N^{2-k}.

In particular,

OOcN0,\langle \mathcal O\mathcal O\rangle_c\sim N^0, OOOc1N,\langle \mathcal O\mathcal O\mathcal O\rangle_c\sim \frac{1}{N}, OOOOc1N2.\langle \mathcal O\mathcal O\mathcal O\mathcal O\rangle_c\sim \frac{1}{N^2}.

The AdS interpretation is that 1/N1/N is a bulk interaction strength. More precisely, in standard large-NN holographic theories,

GNbulk1N2.G_N^{\mathrm{bulk}}\sim \frac{1}{N^2}.

Thus large-NN factorization corresponds to the classical bulk limit. Connected correlators are suppressed because bulk quantum fluctuations are suppressed.

Problem 11.2: Large-NN anomalous dimensions from logarithms

Section titled “Problem 11.2: Large-NNN anomalous dimensions from logarithms”

Suppose a four-point function has a conformal block expansion containing double-trace operators with dimensions

Δn,=Δn,(0)+1N2γn,+O(N4).\Delta_{n,\ell}=\Delta_{n,\ell}^{(0)}+\frac{1}{N^2}\gamma_{n,\ell}+O(N^{-4}).

Show why expanding the blocks produces terms proportional to

1N2γn,logu.\frac{1}{N^2}\gamma_{n,\ell}\log u.
Solution

In the small-uu limit, a conformal block behaves schematically as

GΔ,(u,v)uΔ/2×function of v.G_{\Delta,\ell}(u,v)\sim u^{\Delta/2}\times \text{function of }v.

Substitute

Δ=Δ(0)+1N2γ+O(N4).\Delta=\Delta^{(0)}+\frac{1}{N^2}\gamma+O(N^{-4}).

Then

uΔ/2=uΔ(0)/2uγ/(2N2)=uΔ(0)/2(1+γ2N2logu+O(N4)). u^{\Delta/2} =u^{\Delta^{(0)}/2} u^{\gamma/(2N^2)} =u^{\Delta^{(0)}/2} \left(1+\frac{\gamma}{2N^2}\log u+O(N^{-4})\right).

Therefore anomalous dimensions appear in the four-point function as logarithmic terms:

1N2γn,logu.\frac{1}{N^2}\gamma_{n,\ell}\log u.

In AdS, these logarithms encode shifts in two-particle energies caused by weak bulk interactions.

Problem Set 12: The pre-AdS/CFT dictionary

Section titled “Problem Set 12: The pre-AdS/CFT dictionary”

Problem 12.1: Scalar mass-dimension relation

Section titled “Problem 12.1: Scalar mass-dimension relation”

A scalar field in Euclidean AdSd+1\mathrm{AdS}_{d+1} with radius LL obeys

(2m2)ϕ=0.(\nabla^2-m^2)\phi=0.

Near the boundary in Poincare coordinates,

ds2=L2dz2+dxμdxμz2,z0,ds^2=L^2\frac{dz^2+dx^\mu dx_\mu}{z^2}, \qquad z\to 0,

assume

ϕ(z,x)zαϕ0(x).\phi(z,x)\sim z^\alpha \phi_0(x).

Show that

m2L2=α(αd).m^2L^2=\alpha(\alpha-d).

Therefore the two possible exponents are

α=Δ,α=dΔ,\alpha=\Delta, \qquad \alpha=d-\Delta,

where

m2L2=Δ(Δd).m^2L^2=\Delta(\Delta-d).
Solution

Near the boundary, neglect xx-derivatives relative to radial scaling. The scalar Laplacian in AdSd+1\mathrm{AdS}_{d+1} gives

2ϕ1L2zd+1z(z1dzϕ).\nabla^2\phi \sim \frac{1}{L^2}z^{d+1}\partial_z\left(z^{1-d}\partial_z\phi\right).

For ϕ=zαϕ0(x)\phi=z^\alpha\phi_0(x),

zϕ=αzα1ϕ0,\partial_z\phi=\alpha z^{\alpha-1}\phi_0,

and

z1dzϕ=αzαdϕ0.z^{1-d}\partial_z\phi =\alpha z^{\alpha-d}\phi_0.

Then

z(z1dzϕ)=α(αd)zαd1ϕ0.\partial_z\left(z^{1-d}\partial_z\phi\right) =\alpha(\alpha-d)z^{\alpha-d-1}\phi_0.

Multiplying by zd+1/L2z^{d+1}/L^2 gives

2ϕα(αd)L2zαϕ0.\nabla^2\phi \sim \frac{\alpha(\alpha-d)}{L^2}z^\alpha\phi_0.

The wave equation becomes

α(αd)L2zαϕ0m2zαϕ0=0,\frac{\alpha(\alpha-d)}{L^2}z^\alpha\phi_0-m^2z^\alpha\phi_0=0,

so

m2L2=α(αd).m^2L^2=\alpha(\alpha-d).

Writing one root as Δ\Delta, the other is dΔd-\Delta, and

m2L2=Δ(Δd).m^2L^2=\Delta(\Delta-d).

The boundary source and response are associated with the two independent asymptotic coefficients.

Explain how the following CFT data appear in a weakly coupled AdS dual:

CFT datumBulk interpretation
scalar primary dimension Δ\Delta?
spin-\ell primary?
conserved current JμJ_\mu?
stress tensor TμνT_{\mu\nu}?
OPE coefficient CijkC_{ijk}?
large central charge CTC_T?
Solution

The dictionary is:

CFT datumBulk interpretation
scalar primary dimension Δ\Deltamass of a bulk scalar, m2L2=Δ(Δd)m^2L^2=\Delta(\Delta-d)
spin-\ell primarybulk spin-\ell field or composite multi-particle state
conserved current JμJ_\mubulk gauge field AMA_M
stress tensor TμνT_{\mu\nu}bulk metric fluctuation gMNg_{MN}, i.e. the graviton
OPE coefficient CijkC_{ijk}cubic bulk coupling, after normalization
large central charge CTC_Tsmall Newton constant, CTLd1/GNC_T\sim L^{d-1}/G_N

The point is that CFT data are not merely boundary observables. They are the nonperturbative definition of the bulk theory. Weakly coupled Einstein gravity is a special regime of CFT data: large CTC_T, sparse low-spin single-trace spectrum, large-NN factorization, and controlled anomalous dimensions.

Each mini-project should result in a short note of 5—10 pages, or an equivalent webpage. A good project has a clear question, a derivation, one explicit example, and an AdS/CFT checkpoint.

Build a simple numerical bootstrap for identical scalar four-point functions in one dimension or for a truncated higher-dimensional crossing equation.

Deliverables:

  1. Derive the crossing equation.
  2. Choose a finite derivative basis at the crossing-symmetric point.
  3. Implement a linear-functional search.
  4. Plot one exclusion curve or demonstrate one allowed/disallowed spectrum.
  5. Explain which ingredients survive in the full modern bootstrap.

Write a note explaining how the critical Ising model is encoded by CFT data.

Minimum content:

σ×σ=1+ϵ+,\sigma\times\sigma=1+\epsilon+\cdots,

critical exponents from Δσ\Delta_\sigma and Δϵ\Delta_\epsilon, and the difference between the exact 2D solution and the numerical 3D bootstrap logic.

Project C: Virasoro minimal models from null vectors

Section titled “Project C: Virasoro minimal models from null vectors”

Starting from a level-two null state, derive the BPZ differential equation for a four-point function containing a degenerate primary. Then apply the result to one Ising-model correlator.

A strong project also explains why the existence of null states converts symmetry into solvability.

Project D: Modular invariance and black-hole entropy

Section titled “Project D: Modular invariance and black-hole entropy”

Derive the Cardy formula carefully and compare it with the BTZ entropy. The final section should identify which assumptions enter the Cardy derivation and which assumptions enter the gravity derivation.

Project E: Generalized free fields and two-particle AdS states

Section titled “Project E: Generalized free fields and two-particle AdS states”

Use generalized free four-point functions to identify double-trace towers. Explain the interpretation of nn and \ell as radial and angular quantum numbers in global AdS.

A strong project computes at least one explicit low-lying OPE coefficient from mean-field theory.

Project F: Large-spin perturbation theory and bulk locality

Section titled “Project F: Large-spin perturbation theory and bulk locality”

Explain why the crossed-channel identity contribution implies large-spin double-twist operators. Then discuss how anomalous dimensions at large spin encode long-distance bulk forces.

Project G: The ball modular Hamiltonian and linearized gravity

Section titled “Project G: The ball modular Hamiltonian and linearized gravity”

Starting from

KB=2πBdd1xR2r22RT00,K_B=2\pi\int_B d^{d-1}x\,\frac{R^2-r^2}{2R}T_{00},

explain how the entanglement first law maps to the first law of hyperbolic black holes or to the linearized gravitational equations in AdS.

Project H: N=4\mathcal N=4 chiral primaries and S5S^5 harmonics

Section titled “Project H: N=4\mathcal N=4N=4 chiral primaries and S5S^5S5 harmonics”

Explain the map

[0,p,0],Δ=pS5 scalar harmonic of degree p.[0,p,0],\quad \Delta=p \quad\longleftrightarrow\quad S^5\text{ scalar harmonic of degree }p.

Include the mass-dimension relation

m2L2=p(p4)m^2L^2=p(p-4)

for the corresponding scalar in AdS5\mathrm{AdS}_5.

Study the half-BPS Wilson line in N=4\mathcal N=4 SYM as a one-dimensional defect CFT. Identify the preserved symmetry, define defect operators, and explain the displacement operator.

Project J: From this course to AdS/CFT proper

Section titled “Project J: From this course to AdS/CFT proper”

Write a “pre-dictionary” memo: choose ten equations from this CFT course and explain exactly where each enters the AdS/CFT correspondence.

The memo should include at least:

ZCFT[J]=Zbulk[ϕ=J],Z_{\mathrm{CFT}}[J]=Z_{\mathrm{bulk}}[\phi_\partial=J], m2L2=Δ(Δd),m^2L^2=\Delta(\Delta-d), CTLd1GN,C_T\sim \frac{L^{d-1}}{G_N},

and large-NN factorization.

A student is ready to begin a serious AdS/CFT course when the following tasks feel routine:

  1. Derive scalar two- and three-point functions from conformal symmetry.
  2. Explain the difference between source, vev, and operator.
  3. Use radial quantization to translate dimensions into energies.
  4. State and use unitarity bounds.
  5. Write an OPE and a conformal block decomposition.
  6. Derive the identical-scalar crossing equation.
  7. Explain why large-NN factorization means weak bulk coupling.
  8. Identify single-trace and multi-trace operators.
  9. Explain the role of TμνT_{\mu\nu}, JμJ_\mu, and scalar primaries in the bulk dictionary.
  10. Recognize which parts of 2D CFT are special to Virasoro symmetry and which parts generalize to higher dimensions.

The highest-level synthesis is this:

AdS/CFT is not a map from a Lagrangian to a spacetime; it is a map from consistent CFT data to quantum gravity in AdS.\boxed{ \text{AdS/CFT is not a map from a Lagrangian to a spacetime; it is a map from consistent CFT data to quantum gravity in AdS.} }

The problem sets above are designed to make that sentence operational.