Holographic Anomalies and the Central Charge of $\mathcal N=4$ SYM

In the scalar example, the boundary value of a bulk field acted as a source for a CFT operator. The most important version of the same idea is obtained by making the source the metric. The boundary metric $g_{(0)ij}$ couples to the CFT stress tensor,

$$\delta S_{\rm CFT}={1\over2}\int d^dx\sqrt{g_{(0)}}\,T^{ij}\delta g_{(0)ij}.$$

Therefore a bulk graviton with boundary value $g_{(0)ij}$ computes stress-tensor correlation functions. In the classical supergravity limit,

$$W_{\rm CFT}[g_{(0)}] = -\log Z_{\rm CFT}[g_{(0)}] \simeq S_{\rm grav}^{\rm ren}[G_{\rm cl};g_{(0)}],$$

where $G_{\rm cl}$ is the asymptotically AdS metric solving the bulk equations with conformal boundary metric $g_{(0)}$. Functional derivatives of $W[g_{(0)}]$ give

$$\langle T^{ij}(x)\rangle = {2\over\sqrt{g_{(0)}}}{\delta W\over \delta g_{(0)ij}(x)},$$

and higher derivatives give connected correlators of $T_{ij}$.

This page explains a particularly clean piece of stress-tensor physics: the Weyl anomaly. In four dimensions it is governed by the central charges $a$ and $c$. For the canonical $AdS_5\times S^5$ duality, the leading large-$N$ result is

$$a=c={N^2\over4}.$$

This equality and its normalization are not just dimensional estimates. They follow from the logarithmic divergence of the on-shell gravitational action and from the reduction of type IIB supergravity on $S^5$.

Stress tensors from gravitons

For a scalar field, the source was a coefficient in the near-boundary expansion of the scalar. For the stress tensor, the source is the boundary value of the metric. In Fefferman—Graham coordinates an asymptotically $AdS_{d+1}$ metric may be written as

$$ds^2=R^2{dz^2\over z^2}+{R^2\over z^2}g_{ij}(z,x)dx^idx^j,$$

with

$$g_{ij}(z,x)\to g_{(0)ij}(x)$$

as $z\to0$. The CFT lives on the conformal class of $g_{(0)ij}$. More precisely, the radial diffeomorphism

$$z\to e^{-\sigma(x)}z$$

acts near the boundary as a Weyl transformation

$$g_{(0)ij}\to e^{2\sigma(x)}g_{(0)ij}.$$

This simple observation is the geometric origin of the holographic trace anomaly: a change of scale in the field theory is a change of radial cutoff in the bulk.

The connected two-point function of the stress tensor is fixed by conformal invariance up to one overall number. In four dimensions one common normalization is

$$\langle T_{ij}(x)T_{kl}(0)\rangle = {C_T\over x^8}\mathcal I_{ij,kl}(x),$$

where $\mathcal I_{ij,kl}$ is a known dimensionless tensor determined by symmetry. With the anomaly convention used below,

$$C_T={40\over\pi^4}c.$$

Thus the coefficient of the graviton kinetic term in $AdS_5$ is the same information as the four-dimensional central charge $c$. Similarly, the stress-tensor three-point function is computed by the cubic graviton vertex in the Einstein action. In a general four-dimensional CFT, $\langle TTT\rangle$ contains more than one independent tensor structure. In a holographic theory governed by pure two-derivative Einstein gravity, the structures occur in the special combination corresponding to

$$a=c.$$

Higher-derivative terms in the bulk action, such as curvature-squared terms, generically shift $a$ and $c$ differently.

Weyl anomalies in even-dimensional CFTs

Classically, a CFT on a curved background has a traceless stress tensor,

$$T^i{}_i=0.$$

Quantum mechanically this statement can fail in even spacetime dimension. The failure is local in the background metric and is called the Weyl anomaly or trace anomaly.

In two dimensions the standard normalization is

$$\langle T^i{}_i\rangle=-{c\over24\pi}\mathcal R,$$

where $c$ is the Virasoro central charge. This is the two-dimensional prototype of a more general phenomenon: the coefficient of the anomaly counts degrees of freedom and is visible in stress-tensor correlators.

In four dimensions the universal part of the anomaly is

$$\langle T^i{}_i\rangle = {c\over16\pi^2}C_{ijkl}C^{ijkl} - {a\over16\pi^2}E_4 + {b\over16\pi^2}\Box\mathcal R.$$

Here $C_{ijkl}$ is the Weyl tensor and

$$E_4 = \mathcal R_{ijkl}\mathcal R^{ijkl} -4\mathcal R_{ij}\mathcal R^{ij} +\mathcal R^2$$

is the Euler density. The $\Box\mathcal R$ coefficient is scheme-dependent: it can be shifted by adding a local $\mathcal R^2$ counterterm to the generating functional. By contrast, $a$ and $c$ are universal CFT data.

A useful identity in four dimensions is

$$C_{ijkl}C^{ijkl}-E_4 = 2\left(\mathcal R_{ij}\mathcal R^{ij}-{1\over3}\mathcal R^2\right).$$

Therefore if $a=c$, the anomaly becomes

$$\langle T^i{}_i\rangle = {a\over8\pi^2}\left(\mathcal R_{ij}\mathcal R^{ij}-{1\over3}\mathcal R^2\right) + \hbox{scheme-dependent terms}.$$

This form is exactly what emerges from two-derivative Einstein gravity in $AdS_5$.

The radial cutoff in AdS is the UV cutoff of the boundary theory. When the cutoff is changed locally, the boundary metric is Weyl-rescaled. Power-law divergences are removable by local counterterms, while the logarithmic divergence produces the Weyl anomaly.

The logarithmic divergence of the AdS action

Take five-dimensional Euclidean Einstein gravity with negative cosmological constant,

$$S_E = -{1\over16\pi G_5}\int_{\mathcal M}d^5x\sqrt G\left(\mathcal R-2\Lambda\right) -{1\over8\pi G_5}\int_{\partial\mathcal M}d^4x\sqrt\gamma\,K +S_{\rm ct}.$$

The Gibbons—Hawking term makes the variational problem well-defined with Dirichlet boundary conditions for the induced metric $\gamma_{ij}$. The counterterm action $S_{\rm ct}$ subtracts divergences as the cutoff surface approaches the boundary.

For $AdS_5$ of radius $R$,

$$\Lambda=-{6\over R^2},$$

and the Einstein equation gives

$$\mathcal R_{MN}=-{4\over R^2}G_{MN},$$

so

$$\mathcal R=-{20\over R^2}.$$

The on-shell bulk Lagrangian is therefore constant:

$$\mathcal R-2\Lambda=-{8\over R^2}.$$

This is why the leading part of the on-shell action is proportional to the regulated hyperbolic volume. In Fefferman—Graham coordinates, after solving the Einstein equations near the boundary and cutting off the spacetime at $z=\epsilon$, the regulated action has the schematic form

$$S_{\rm reg} = {A_4\over \epsilon^4} +{A_2\over \epsilon^2} +A_{\log}\log\epsilon +S_{\rm finite} +O(\epsilon).$$

The coefficients $A_4$ and $A_2$ are local functionals of the boundary metric and can be removed by local counterterms. The coefficient $A_{\log}$ is also local, but it cannot be removed without introducing a scale. It is the anomaly.

For pure Einstein gravity in $AdS_5$, holographic renormalization gives

$$\langle T^i{}_i\rangle = {R^3\over8\pi G_5}\left({1\over8}\mathcal R_{ij}\mathcal R^{ij}-{1\over24}\mathcal R^2\right),$$

up to the scheme-dependent $\Box\mathcal R$ term. Equivalently,

$$\langle T^i{}_i\rangle = {R^3\over64\pi G_5}\left(\mathcal R_{ij}\mathcal R^{ij}-{1\over3}\mathcal R^2\right).$$

Comparing with the four-dimensional anomaly in the special case $a=c$ gives the central result

$$\boxed{a=c={\pi R^3\over8G_5}}.$$

This equation is the holographic analogue of $c=3R/(2G_3)$ in $AdS_3/CFT_2$. In both cases, the central charge is essentially the AdS radius measured in Planck units.

A quick derivation using an $S^4$ boundary

The local anomaly formula is powerful, but the normalization of $a$ can also be read from a very simple background: Euclidean $AdS_5$ with an $S^4$ conformal boundary. Write the metric as

$$ds^2=R^2\left(d\rho^2+\sinh^2\rho\,d\Omega_4^2\right).$$

At large $\rho$, the boundary sphere has radius proportional to $e^\rho$. Cutting off the space at $\rho=\rho_0$ is therefore equivalent to placing a UV cutoff

$$\Lambda_{\rm UV}\sim e^{\rho_0}.$$

The regulated volume contains a term linear in $\rho_0$ because

$$\int^{\rho_0}d\rho\,\sinh^4\rho = \hbox{power divergences}+{3\over8}\rho_0+O(e^{-2\rho_0}).$$

After including the standard boundary terms and counterterms, the coefficient of the logarithmic scale dependence is

$$\Lambda_{\rm UV}{\partial W\over\partial\Lambda_{\rm UV}} = {\pi R^3\over2G_5}.$$

On the CFT side, $S^4$ is conformally flat, so $C_{ijkl}=0$. The integrated Euler density satisfies

$$\int_{S^4}d^4x\sqrt g\,E_4=64\pi^2,$$

because $\chi(S^4)=2$ and $\int E_4=32\pi^2\chi$. Therefore

$$\int_{S^4}d^4x\sqrt g\,\langle T^i{}_i\rangle=-4a$$

with the standard convention

$$\langle T^i{}_i\rangle={c\over16\pi^2}C^2-{a\over16\pi^2}E_4.$$

Depending on whether one differentiates $W=-\log Z$ with respect to the radius or with respect to the UV cutoff, a minus sign may move between the two sides. The invariant comparison of magnitudes gives

$$4a={\pi R^3\over2G_5},$$

hence again

$$a={\pi R^3\over8G_5}.$$

This spherical calculation is a useful sanity check because it reduces the anomaly matching problem to the coefficient of a single logarithm.

Reducing type IIB supergravity on $S^5$

The result $a=c=\pi R^3/(8G_5)$ is a five-dimensional statement. To compare with the D3-brane gauge theory, we need express $G_5$ and $R$ in string variables.

The ten-dimensional Newton constant is

$$G_{10}=8\pi^6g_s^2\alpha'^4.$$

The compactification space is a round five-sphere of radius $R$, with volume

$$\operatorname{Vol}(S^5)=\pi^3R^5.$$

For the graviton zero mode, dimensional reduction gives

$${1\over G_5}={\operatorname{Vol}(S^5)\over G_{10}}.$$

Thus

$$G_5={G_{10}\over \pi^3R^5} ={8\pi^3g_s^2\alpha'^4\over R^5}.$$

The D3-brane five-form flux fixes the radius by

$$R^4=4\pi g_sN\alpha'^2.$$

Combining these relations,

$${R^3\over G_5} ={R^8\over8\pi^3g_s^2\alpha'^4} ={2N^2\over\pi}.$$

Substitution into the holographic central charge formula gives

$$a=c={\pi R^3\over8G_5}={N^2\over4}.$$

The leading stress-tensor normalization is fixed by the five-dimensional Newton constant. Reducing type IIB supergravity on $S^5$ and using the D3-brane flux relation gives $a=c=N^2/4$.

The exact central charges of four-dimensional $SU(N)$ $\mathcal N=4$ super-Yang—Mills are

$$a=c={N^2-1\over4}.$$

The classical supergravity answer sees the leading $N^2/4$ term. The difference $-1/4$ is subleading at large $N$ and is associated with quantum effects and the removal of the decoupled overall $U(1)$ multiplet. The equality $a=c$ is exact in $\mathcal N=4$ SYM; it is already visible at leading order because the bulk theory is ordinary Einstein gravity to lowest order.

The $S^5$ volume converts $G_{10}$ into $G_5$, while five-form flux quantization relates the common radius $R$ to the D3-brane number $N$. These two ingredients are the whole normalization behind the $N^2$ central charge.

Why $N^2$ appears

The factor $N^2$ has a simple gauge-theory meaning. The elementary fields of $\mathcal N=4$ SYM are in the adjoint representation. At large $N$, the number of color degrees of freedom is

$$\dim SU(N)=N^2-1\sim N^2.$$

The gravity computation translates the same statement into Planck units. The combination controlling classical gravitational correlators in $AdS_5$ is

$${R^3\over G_5}.$$

Large $N$ means $R^3/G_5\gg1$, so the five-dimensional gravitational action is large and the saddle-point approximation is reliable. In the string description this is part of the familiar regime

$$N\gg1, \qquad \lambda={R^4\over\alpha'^2}\gg1,$$

with $g_s=\lambda/(4\pi N)$ small enough to suppress string loops. The central charge is therefore a precise measure of the number of degrees of freedom, and the condition $a\sim c\sim N^2\gg1$ is one reason a classical bulk dual can exist.

Why pure Einstein gravity implies $a=c$

Four-dimensional CFTs can have $a\ne c$. In the bulk, this difference is controlled by higher-derivative gravitational interactions. Schematically,

$$S_{5d}= {1\over16\pi G_5}\int d^5x\sqrt G\left(\mathcal R+{12\over R^2}+\alpha R^2\mathcal R_{MNPQ}\mathcal R^{MNPQ}+\cdots\right).$$

Curvature-squared terms modify the tensor structures in the stress-tensor three-point function and generally shift $a$ and $c$ differently. Pure Einstein gravity has only one effective normalization for the graviton interactions, so it cannot generate two independent central charges. That is why the leading holographic prediction for any two-derivative Einstein dual in $AdS_5$ is

$$a=c.$$

This statement is not an accident of $S^5$. It is true for a broad class of leading large-$N$ holographic CFTs whose bulk dual is dominated by a two-derivative Einstein action. When higher-derivative corrections are important, the ratio $a/c$ becomes a sensitive diagnostic of stringy physics in the bulk.

Central charge, entropy, and the same normalization

The same combination $R^3/G_5$ appeared earlier in the entropy density of the near-extremal D3-brane. The black-brane result is

$$s={\pi^2\over2}N^2T^3$$

at strong coupling. The central charge result is

$$a=c={N^2\over4}.$$

Both are consequences of the same fact: classical type IIB supergravity on $AdS_5\times S^5$ is normalized by $1/G_5\sim N^2/R^3$. Thermodynamics, two-point functions, three-point functions, and anomalies all know about the same large number of degrees of freedom.

Exercises

Exercise 1. The $a=c$ anomaly in Ricci form

Show that in four dimensions

$$C_{ijkl}C^{ijkl}-E_4 = 2\left(\mathcal R_{ij}\mathcal R^{ij}-{1\over3}\mathcal R^2\right).$$

Then derive the anomaly for $a=c$ in the convention

$$\langle T^i{}_i\rangle={c\over16\pi^2}C^2-{a\over16\pi^2}E_4.$$

Solution

In four dimensions,

$$C_{ijkl}C^{ijkl} =\mathcal R_{ijkl}\mathcal R^{ijkl} -2\mathcal R_{ij}\mathcal R^{ij} +{1\over3}\mathcal R^2.$$

Also,

$$E_4 =\mathcal R_{ijkl}\mathcal R^{ijkl} -4\mathcal R_{ij}\mathcal R^{ij} +\mathcal R^2.$$

Subtracting gives

$$C^2-E_4 =2\mathcal R_{ij}\mathcal R^{ij}-{2\over3}\mathcal R^2 =2\left(\mathcal R_{ij}\mathcal R^{ij}-{1\over3}\mathcal R^2\right).$$

If $a=c$, then

$$\langle T^i{}_i\rangle ={a\over16\pi^2}(C^2-E_4) ={a\over8\pi^2}\left(\mathcal R_{ij}\mathcal R^{ij}-{1\over3}\mathcal R^2\right).$$

Exercise 2. Integrated anomaly on $S^4$

For a round four-sphere of radius $r$, verify that

$$\int_{S^4}d^4x\sqrt g\,\langle T^i{}_i\rangle=-4a$$

in the standard convention above.

Solution

For a round $S^4$,

$$C_{ijkl}=0.$$

The Euler density integrates to

$$\int_{S^4}d^4x\sqrt g\,E_4=32\pi^2\chi(S^4)=64\pi^2,$$

because $\chi(S^4)=2$. Hence

$$\int_{S^4}d^4x\sqrt g\,\langle T^i{}_i\rangle =-{a\over16\pi^2}64\pi^2 =-4a.$$

Equivalently, using $\mathcal R_{ij}=3g_{ij}/r^2$, one finds $\mathcal R_{ij}\mathcal R^{ij}=36/r^4$ and $\mathcal R^2/3=48/r^4$. The volume is $8\pi^2r^4/3$, so the integral of the Ricci-form anomaly is also $-4a$.

Exercise 3. On-shell Einstein action in $AdS_D$

Let $AdS_D$ have radius $R$. Show that

$$\Lambda=-{(D-1)(D-2)\over2R^2}$$

and

$$\mathcal R-2\Lambda=-{2(D-1)\over R^2}.$$

Solution

The $D$-dimensional Einstein equation with cosmological constant is

$$\mathcal R_{MN}-{1\over2}G_{MN}\mathcal R=-\Lambda G_{MN}.$$

For $AdS_D$,

$$\mathcal R_{MN}=-{D-1\over R^2}G_{MN},$$

so

$$\mathcal R=-{D(D-1)\over R^2}.$$

Substituting into the Einstein equation gives

$$-{D-1\over R^2}+{D(D-1)\over2R^2}=-\Lambda,$$

hence

$$\Lambda=-{(D-1)(D-2)\over2R^2}.$$

Then

$$\mathcal R-2\Lambda =-{D(D-1)\over R^2}+{(D-1)(D-2)\over R^2} =-{2(D-1)\over R^2}.$$

Exercise 4. The five-dimensional Newton constant

Starting from

$$G_{10}=8\pi^6g_s^2\alpha'^4$$

and

$$\operatorname{Vol}(S^5)=\pi^3R^5,$$

derive

$$G_5={8\pi^3g_s^2\alpha'^4\over R^5}.$$

Solution

Dimensional reduction of the Einstein—Hilbert term gives

$${1\over G_5}={\operatorname{Vol}(S^5)\over G_{10}}.$$

Therefore

$$G_5={G_{10}\over\operatorname{Vol}(S^5)} ={8\pi^6g_s^2\alpha'^4\over\pi^3R^5} ={8\pi^3g_s^2\alpha'^4\over R^5}.$$

Exercise 5. The central charge of $\mathcal N=4$ SYM from supergravity

Use

$$R^4=4\pi g_sN\alpha'^2$$

and the result of Exercise 4 to show that

$$a=c={N^2\over4}.$$

Solution

The holographic formula is

$$a=c={\pi R^3\over8G_5}.$$

Using

$$G_5={8\pi^3g_s^2\alpha'^4\over R^5},$$

we obtain

$$a=c={\pi R^3\over8}{R^5\over8\pi^3g_s^2\alpha'^4} ={R^8\over64\pi^2g_s^2\alpha'^4}.$$

Now square the radius relation:

$$R^8=16\pi^2g_s^2N^2\alpha'^4.$$

Hence

$$a=c={16\pi^2g_s^2N^2\alpha'^4\over64\pi^2g_s^2\alpha'^4} ={N^2\over4}.$$

Exercise 6. The stress-tensor two-point coefficient

In the four-dimensional convention

$$C_T={40\over\pi^4}c,$$

find $C_T$ for strongly coupled $SU(N)$ $\mathcal N=4$ SYM at leading large $N$.

Solution

At leading large $N$,

$$c={N^2\over4}.$$

Therefore

$$C_T={40\over\pi^4}{N^2\over4} ={10N^2\over\pi^4}.$$

The exact $SU(N)$ result would replace $N^2$ by $N^2-1$.

Exercise 7. Why higher derivatives can make $a\ne c$

Explain qualitatively why pure Einstein gravity gives $a=c$, while higher-derivative terms in the bulk can produce $a\ne c$.

Solution

Pure Einstein gravity has one overall normalization for the graviton kinetic term and its cubic self-interaction, namely $R^3/G_5$. This fixes the stress-tensor two-point function and the Einstein-gravity stress-tensor three-point structure with a single coefficient. In a four-dimensional CFT language, this corresponds to the special relation $a=c$.

Higher-derivative terms introduce new independent graviton interactions. For example, curvature-squared terms can change the coefficient of the stress-tensor two-point function and the coefficient of the Euler-anomaly term in different ways. Therefore the two independent CFT anomaly coefficients $a$ and $c$ no longer have to be equal.