Anomalies as checks on quantum dynamics

Strong interactions can hide the particles that appear in a microscopic Lagrangian. Is there information we can still calculate and demand that any proposed low-energy description reproduce? An anomaly supplies such information, but only after we identify which symmetry is involved.

This lesson uses four-dimensional supersymmetric actions and sources and currents. Both examples are flat four-dimensional $\mathcal N=1$ theories with four real supercharges, all manifest in that superspace. We first use a dynamical $U(1)$ gauge theory, then a free matter theory with a background $SU(3)$ flavor connection. The dimension is four throughout; these are not three-dimensional parity anomalies.

Count fermions before counting anomalies

A four-dimensional chiral anomaly depends on Weyl fermions. Scalars do not cancel it against their superpartners. Write every independent fermion as left-handed; a right-handed fermion can instead be represented by its left-handed charge conjugate, with conjugate representation and opposite Abelian charges. Do not also count the antiparticle of the same Weyl field.

For Abelian generators $A,B,C$, define the dimensionless coefficients

$$k_{ABC}=\operatorname{Tr}_{\rm Weyl}(ABC) =\sum_i q_{A,i}q_{B,i}q_{C,i}, \qquad k_A=\operatorname{Tr}_{\rm Weyl}A=\sum_iq_{A,i}.$$

Include every gauge and flavor component in these sums, but no factor of two for the two spinor components. The cubic coefficient comes from the three current insertions on a fermion triangle. The linear coefficient measures the mixed anomaly with one $A$ current and two stress tensors. It is often called $A$–gravity–gravity; it is not a purely gravitational anomaly.

For non-Abelian generators the corresponding triangle weight is the symmetrized trace

$$D^{abc}_{\mathcal R} =\frac12\operatorname{tr}_{\mathcal R} \!\left[T^a\{T^b,T^c\}\right].$$

The two orders of the current insertions produce the anticommutator. The regulated chiral-fermion calculation establishes that this group factor multiplies the anomaly; calculating a finite matrix trace alone does not establish that quantum result. The result and the chirality signs are derived in Bilal, §7.1, (7.1)–(7.9).

In a chiral superfield of scalar R-charge $r$, the expansion

$$\Phi=\phi+\sqrt2\theta\psi+\theta^2F,\qquad R(\theta)=1$$

gives $R(\psi)=r-1$ and $R(F)=r-2$. The gaugino $\lambda$ has R-charge $+1$ and lies in the adjoint of the dynamical gauge group. Thus

$$\operatorname{Tr}R=\dim G+\sum_i d_i(r_i-1),\qquad \operatorname{Tr}R^3=\dim G+\sum_i d_i(r_i-1)^3,$$

where $d_i$ includes the matter representation multiplicities and the first term is present when a vector multiplet is present. The adjoint’s vanishing gauge cubic anomaly does not eliminate its R anomaly. Intriligator–Seiberg, the paragraph following (4.2) makes precisely this component-charge distinction.

A first table with a dynamical photon

Take two chiral superfields $Q_+,Q_-$ of gauge charges $+1,-1$, with canonical kinetic terms and $W=0$. Work about their zero expectation values. Choose scalar R-charges one for both. The classical axial rotation $A$ gives both superfields charge $+1$ and gives the vector multiplet charge zero.

These assignments produce the following fermion table:

Left-handed field	Gauge charge $q$	Axial charge $A$	R-charge	Multiplicity
$\psi_+$	$+1$	$1$	$0$	$1$
$\psi_-$	$-1$	$1$	$0$	$1$
$\lambda$	$0$	$0$	$1$	$1$

The $U(1)$ adjoint is neutral and one-dimensional. The gaugino is therefore the third row even though it does not couple to the photon through a non-Abelian adjoint charge.

Now calculate, rather than label the whole theory “anomaly-free”:

$$\begin{aligned} k_{qqq}&=1^3+(-1)^3=0,& k_q&=1-1=0,\\ k_{Aqq}&=1(1)^2+1(-1)^2=2,& k_{Rqq}&=0(1)^2+0(-1)^2+1(0)^2=0,\\ k_R&=0+0+1=1,& k_{RRR}&=0^3+0^3+1^3=1. \end{aligned}$$

The first row tests the local gauge anomaly and its mixed gravitational counterpart. The second row distinguishes the axial and R currents in the presence of the dynamical gauge field. The final row concerns R backgrounds. These numbers describe three different questions, despite all being fermion traces.

A useful independent arithmetic check introduces formal commuting variables $x_q,x_A,x_R$:

$$\sum_{\rm Weyl}(x_q q+x_A A+x_R R)^3 =(x_q+x_A)^3+(-x_q+x_A)^3+x_R^3 =6x_q^2x_A+2x_A^3+x_R^3.$$

The coefficient of $x_q^2x_A$ is $3k_{Aqq}$, not $k_{Aqq}$. The factor three counts its possible placements. This polynomial packages trace data; it is not yet a spacetime anomaly density.

Three meanings of the same obstruction

An anomalous Ward identity can arise even when the classical action is invariant, because its regulated fermion measure cannot preserve all the proposed transformations.

Which transformations are involved?	What a nonzero coefficient means
Dynamical gauge redundancy	The stated ordinary gauge theory is inconsistent unless its full anomaly is canceled.
A proposed global current and two dynamical gauge currents	That ordinary continuous global current is anomalously broken, with gauge Ward identities kept exact.
Exact global symmetries coupled only to background sources	A possible ‘t Hooft anomaly: the quantum theory exists, but the backgrounds cannot all be gauged consistently without additional data.

For our pair, $k_{qqq}=k_q=0$ supplies the displayed gauge checks. The axial value $k_{Aqq}=2$ means that the classical axial current has a quantum divergence involving the dynamical gauge field strength. It is not an exact ordinary continuous symmetry to be matched in the infrared. This statement does not require an instanton calculation or a claim about a residual discrete group.

By contrast, $k_{Rqq}=0$ removes that obstruction for R. The R symmetry exists in the flat quantum theory, although $k_R=k_{RRR}=1$ gives nonzero responses to R and gravitational backgrounds. In this context “nonanomalous R-symmetry” must mean nonanomalous with respect to the dynamical gauge group, not that every R trace vanishes. Tong, §§5.2.1–5.2.3 separates these three uses.

The source language makes the last distinction precise. Let $\mathcal W[a]=-\log Z[a]$ for an Abelian background $a_\mu$. Retaining the prerequisite’s source-current convention, suppose its infinitesimal background transformation gives

$$\delta_\alpha\mathcal W =\int d^4x\,\sqrt g\,\alpha\,\mathcal B[a,g].$$

Since $\delta a_\mu=\partial_\mu\alpha$, integration by parts gives

$$\nabla_\mu\langle j^\mu\rangle=-\mathcal B[a,g].$$

The Euclidean quantity $\mathcal B$ may carry the phase associated with the anomalous partition function. Its spacetime prefactor and its allocation among mixed Ward identities require a definite current/counterterm convention. Our $k$‘s specify the normalized fermion traces, not a convention-independent numerical formula for $\mathcal B$.

Adding a local function of the sources can move terms between Ward identities. A genuine anomaly is the obstruction that no allowed local counterterm removes while preserving all the symmetries in question. With dynamical gauge fields, their Ward identities take priority. Bilal, §6.2, (6.7) explains this distinction. Zero sources can still support an exactly conserved global current even when the sourced theory has such an obstruction.

A flavor trace is not a count of components

For a non-Abelian flavor example, take three free chiral fields $X^i$, $i=1,2,3$, with $K=\sum_i\bar X_iX^i$ and $W=0$. There is no dynamical gauge group. View $X$ as one fundamental of a global $SU(3)_F$, give it Abelian flavor charge $B=1$, and choose scalar R-charge $2/3$. These are actual symmetries of the free action. We need no identification of R with a superconformal generator.

Thus the Weyl fermions form one $\mathbf3$, with $B=1$ and $R=-1/3$. We discuss the displayed continuous Lie-algebra backgrounds; this is not a specification of every finite quotient or discrete anomaly of the full symmetry group.

Fix Hermitian generators by

$$\operatorname{tr}_{\mathbf3}(T^aT^b)=\frac12\delta^{ab}, \qquad \operatorname{tr}_{\mathcal R}(T^aT^b)=T(\mathcal R)\delta^{ab}.$$

The quadratic index is $T(\mathbf3)=1/2$. Define the cubic index $\mathcal A(\mathcal R)$ by comparing the symmetrized triple trace with that of $\mathbf3$, so $\mathcal A(\mathbf3)=1$. A spectator multiplicity counts copies of this representation; it must not count its three components again.

For mixed coefficients,

$$k_{F^2B}=\sum_{\mathcal R}B_{\mathcal R}T(\mathcal R), \qquad k_{F^2R}=\sum_{\mathcal R}R_{\mathcal R}T(\mathcal R).$$

Tong uses $I(\mathcal R)=2T(\mathcal R)$ in his equation (5.8). His fundamental quadratic index is one. Translating this factor is necessary when comparing mixed-anomaly tables; it does not change the physics.

Our complete nonzero cubic and linear coefficients for these generators are

Coefficient	Calculation	Value
$SU(3)_F^3$	One fundamental: $\mathcal A(\mathbf3)$	$1$
$SU(3)_F^2B$	$1\times T(\mathbf3)$	$1/2$
$SU(3)_F^2R$	$(-1/3)T(\mathbf3)$	$-1/6$
$\operatorname{Tr}B$ and $\operatorname{Tr}B^3$	Three components of charge one	$3,\ 3$
$\operatorname{Tr}R$ and $\operatorname{Tr}R^3$	$3(-1/3)$ and $3(-1/3)^3$	$-1,\ -1/9$
$\operatorname{Tr}B^2R$	$3(1)^2(-1/3)$	$-1$
$\operatorname{Tr}BR^2$	$3(1)(-1/3)^2$	$1/3$

Terms with only one $SU(3)$ generator and Abelian insertions vanish because $\operatorname{tr}T^a=0$. There is no gaugino in this example.

Check the distinction with explicit matrices:

$$T^3=\frac12\operatorname{diag}(1,-1,0),\qquad T^8=\frac1{2\sqrt3}\operatorname{diag}(1,1,-2).$$

Direct multiplication gives

$$\operatorname{tr}(T^3)^2=\frac12,\qquad \operatorname{tr}\!\left[(T^3)^2T^8\right]=\frac1{4\sqrt3}.$$

The first trace checks $k_{F^2B}=1/2$. The second is a nonzero cubic flavor component. In the conjugate representation $\bar T^a=-(T^a)^T$: the quadratic trace keeps its sign, while the symmetrized cubic trace reverses it. Hence

$$T(\bar{\mathbf3})=\frac12,\qquad \mathcal A(\bar{\mathbf3})=-1.$$

These nonzero flavor coefficients do not make the free theory inconsistent. They forbid promoting all those backgrounds to ordinary dynamical gauge fields without an anomaly-canceling completion.

What must survive a change of scale?

For an exact global symmetry preserved along the flow, the ultraviolet and infrared theories must reproduce the same background anomaly, up to local counterterms:

$$k^{\rm UV}=k^{\rm IR}$$

in a common normalization. This is ‘t Hooft anomaly matching. A useful argument is to add decoupled spectator fermions canceling the anomaly, weakly gauge the symmetry, and take that new gauge coupling small enough not to affect the original flow. Gauge consistency must hold at both ends, while the spectator contribution is unchanged. The original sectors must therefore match. This argument concerns the full infrared anomaly response, not an assumption that every infrared theory has weakly coupled fermions.

If a description does use weakly coupled massless fermions, their traces perform the test. If the symmetry breaks spontaneously, Goldstone-field interactions can carry its anomaly; one cannot discard them and demand that a fermion-only sum remain equal. A featureless gapped phase with a unique vacuum, unbroken symmetry and no anomaly-carrying infrared sector cannot reproduce a nonzero anomaly.

For a directly controlled check, return to the gauge pair and add $W=mQ_+Q_-$ with fixed nonzero $m$. It has R-charge two and preserves the selected R symmetry. Around the origin, the two matter multiplets become massive; the $U(1)$ vector remains massless. At energies well below the matter mass, in a weakly coupled regime, the R traces are

$$\begin{array}{c|cc} &\operatorname{Tr}R&\operatorname{Tr}R^3\\ \hline \text{above the mass}&0+0+1&0+0+1\\ \text{below the mass}&1&1 \end{array}$$

The gaugino matches both. There is no demand to match the classical axial symmetry, which was already anomalous and is also broken by the fixed mass.

In an interacting proposed duality, matching is a necessary check, not a proof: inequivalent theories can share these sums. Seiberg’s original duality paper, §3, (3.4) uses them as a consistency condition alongside operator assignments and deformations. The full dual gauge-theory comparison requires those additional ingredients.

Exercises

1. Try to mix an anomalous current into R

In the massless $U(1)$ pair, define $R_t=R+tA$ for real $t$. Find the matter scalar and fermion charges, $k_{R_tqq}$, $\operatorname{Tr}R_t$, and $\operatorname{Tr}R_t^3$. Which values give an exact ordinary continuous R current with the dynamical gauge Ward identities preserved?

Hint 1

Mix the scalar charges first, then subtract the R-charge of $\theta$. The gaugino is axial-neutral.

Hint 2

The relevant obstruction is the coefficient with two dynamical gauge insertions. Vanishing purely global cubic and linear traces is not required.

Solution

The matter scalars have R-charge $1+t$ and their Weyl fermions have charge $t$. The gaugino remains of charge one. Therefore

$$k_{R_tqq}=2t,\qquad \operatorname{Tr}R_t=1+2t,\qquad \operatorname{Tr}R_t^3=1+2t^3.$$

Only $t=0$ cancels the mixed dynamical-gauge obstruction. At that value the other two traces are one, consistently with an exact R symmetry and nonzero background anomalies.

For example $t=-1/2$ makes the linear trace vanish but leaves $k_{R_tqq}=-1$. It does not repair the current. Mixing is a statement about candidate generators; it cannot convert an anomalously broken current into an exact one without satisfying the relevant constraints. We have tested the continuous Lie algebra, not asserted a compact global period for every real $t$.

2. Give the flavor multiplet a symmetry-preserving mass

Add three chirals $\widetilde X_i$ in $\bar{\mathbf3}$ of the same flavor group, with $B=-1$ and scalar R-charge $4/3$. Add $W=M\widetilde X_iX^i$, with fixed nonzero $M$ and canonical kinetic terms. Check the preserved symmetries and all anomaly coefficients in the flavor table. What remains at energies much smaller than $|M|$?

Hint 1

The new Weyl fermion has R-charge $+1/3$. Distinguish the sign of the cubic representation index from the sign of its quadratic index.

Hint 2

The mass is an invariant contraction of $\mathbf3$ with $\bar{\mathbf3}$. For mixed Abelian traces, reverse all charges of the original Weyl fermions.

Solution

The contraction is an $SU(3)$ singlet, has $B=1-1=0$, and has scalar R-charge $2/3+4/3=2$. Thus the superpotential and its chiral measure are invariant with $M$ fixed. Its potential is $|M|^2(|X|^2+|\widetilde X|^2)$, so the origin is the unique supersymmetric vacuum and all six chiral multiplets are massive.

The new fermions contribute

$$\begin{aligned} SU(3)^3 &: -1,& SU(3)^2B &: -\frac12,& SU(3)^2R &: +\frac16,\\ \operatorname{Tr}B &: -3,& \operatorname{Tr}B^3 &: -3,& \operatorname{Tr}R &: +1,\\ \operatorname{Tr}R^3 &: +\frac19,& \operatorname{Tr}B^2R &: +1,& \operatorname{Tr}BR^2 &: -\frac13. \end{aligned}$$

Each cancels the original contribution. The quadratic index is positive for both representations; the opposite $B$ and R-charges cancel the mixed coefficients. At $E\ll|M|$, no propagating field remains in this free model. The vanishing anomaly is compatible with that trivially gapped infrared theory. Possible local background terms from integrating out the pair do not constitute a nonzero anomaly obstruction.

A single triplet could not have acquired this invariant quadratic mass by itself. The partner representation both permits the mass and cancels the anomaly. This explains the decoupling check rather than merely asserting that massive particles disappear from a table.

3. Diagnose a proposed infrared spectrum

Suppose a symmetry-preserving flow starts from the single flavor triplet example, and a proposed infrared description contains only three free chiral fields that are $SU(3)_F$ singlets, each with $B=1$ and scalar R-charge $2/3$. Compare all the Abelian anomalies and the flavor anomalies. Does either agreement or disagreement prove the proposed dynamics?

Hint 1

The Abelian fermion charge lists are identical in the two descriptions. Their flavor representations are not.

Hint 2

State explicitly whether the proposal includes Goldstone fields, a further interacting sector or any other contribution to the background anomaly.

Solution

The three singlet Weyl fermions have exactly the original Abelian charges $B=1$, $R=-1/3$, so all six Abelian entries in the table match. Their $SU(3)$ generators vanish, however. The infrared values of $SU(3)^3$, $SU(3)^2B$ and $SU(3)^2R$ are all zero, disagreeing with $1,1/2,-1/6$.

Consequently this cannot be the complete infrared description of such a flow with unbroken $SU(3)_F$ and no additional anomaly-carrying sector. Matching only selected Abelian sums was insufficient. Allowing spontaneous symmetry breaking or an omitted sector changes the proposal and requires computing that sector’s response; it is not an automatic remedy.

Conversely, even agreement with every displayed coefficient would be only a necessary test. It would not fix correlation functions, operator products, vacua or deformation responses. No interacting flow or duality was established by this hypothetical spectrum.

Before comparing quantum descriptions, identify the exact global currents, translate every scalar R-charge to a fermion charge, and fix trace normalization. Then preserve the full anomaly response across the proposed flow. The sums are small; the physical distinction between them does most of the work.