Check a duality with anomalies and a mass deformation

A duality dictionary makes a strong claim: distinct microscopic fields describe the same infrared physics. Matching one charge table cannot establish that claim. We will ask two different questions. Do the theories respond consistently to background global-symmetry fields? And do they still agree after a deformation changes the low-energy gauge dynamics?

The preparation is the Seiberg-duality dictionary and anomaly matching. Work in flat four-dimensional Lorentzian $\mathcal N=1$ theories, with four real supercharges. Electric color is $SU(3)$, magnetic color is $SU(2)$, and each theory initially has five quark/antiquark pairs. The magnetic theory also has twenty-five singlet chirals.

Count fermions before taking traces

An anomaly trace counts left-handed Weyl fermions, not chiral scalars. The symmetric anomaly-free scalar charges from the dictionary give the following fermion data:

Fermions	Number of components	R-charge $R_{\mathrm f}$	Baryon charge
Electric gaugino	$8$	$1$	$0$
Electric quarks	$15$	$-3/5$	$1$
Electric antiquarks	$15$	$-3/5$	$-1$
Magnetic gaugino	$3$	$1$	$0$
Magnetic quarks	$10$	$-2/5$	$3/2$
Magnetic antiquarks	$10$	$-2/5$	$-3/2$
Magnetic meson fermions	$25$	$-1/5$	$0$

For example, $15=3$ colors times $5$ flavors. There is one adjoint gaugino per theory, with eight or three color components. The R-charge in every matter row is its scalar charge minus one.

The mixed gravitational–R anomaly is proportional to $\operatorname{Tr}R$. Directly,

$$\begin{aligned} \operatorname{Tr}R\big|_{\mathrm e} &=8+30\left(-\frac35\right)=-10,\\ \operatorname{Tr}R\big|_{\mathrm m} &=3+20\left(-\frac25\right)+25\left(-\frac15\right)=-10. \end{aligned} \tag{Q10.1}$$

The cubic R anomaly similarly gives

$$\begin{aligned} \operatorname{Tr}R^3\big|_{\mathrm e} &=8+30\left(-\frac35\right)^3=\frac{38}{25},\\ \operatorname{Tr}R^3\big|_{\mathrm m} &=3+20\left(-\frac25\right)^3+25\left(-\frac15\right)^3 =\frac{38}{25}. \end{aligned} \tag{Q10.2}$$

The singlet fermions matter even though they have no magnetic color. Omitting them fails both tests.

Check non-Abelian signs and the mixed anomalies

Normalize the cubic anomaly of one left-flavor fundamental Weyl fermion to $+1$; its antifundamental contributes $-1$. Electric $Q$ supplies three color copies of a fundamental of $SU(5)_L$. Magnetic $q$ supplies two copies of an antifundamental, while $M$ supplies five fundamentals, one per right-flavor index. Thus

$$k_{SU(5)_L^3}\big|_{\mathrm e}=3, \qquad k_{SU(5)_L^3}\big|_{\mathrm m}=-2+5=3. \tag{Q10.3}$$

In the chosen right-flavor convention all these representations are conjugated, so the right-flavor cubic coefficient is $-3$ on both sides.

Conjugation does not reverse a quadratic Dynkin index. Set $T(\mathbf5)=T(\overline{\mathbf5})=1/2$. The remaining nonzero mixed coefficients follow from the same rows:

Coefficient	Electric sum	Magnetic sum	Common value
$SU(5)_L^2R$	$(3/2)(-3/5)$	$(2/2)(-2/5)+(5/2)(-1/5)$	$-9/10$
$SU(5)_R^2R$	$(3/2)(-3/5)$	$(2/2)(-2/5)+(5/2)(-1/5)$	$-9/10$
$SU(5)_L^2B$	$(3/2)(1)$	$(2/2)(3/2)$	$3/2$
$SU(5)_R^2B$	$(3/2)(-1)$	$(2/2)(-3/2)$	$-3/2$
$B^2R$	$30(1)^2(-3/5)$	$20(3/2)^2(-2/5)$	$-18$

The traces $\operatorname{Tr}B$, $\operatorname{Tr}B^3$ and $\operatorname{Tr}(BR^2)$ vanish: opposite-baryon partners have equal multiplicities and equal R-charge. Terms containing a single generator of either $SU(5)$ factor vanish by tracelessness, including mixed terms involving both non-Abelian factors. This accounts for the perturbative continuous global anomalies of the displayed symmetry algebra. Seiberg, (3.4)

These are background-current anomalies. They need not vanish, unlike an uncanceled anomaly of a dynamical gauge symmetry. Their equality is a necessary infrared test, not a reconstruction of correlation functions, kinetic terms, or every global-form datum. We continue to use an anomaly-free R assignment; no superconformal identification was needed for these sums.

Give one electric flavor a mass

Retain the dictionary’s normalization:

$$\mathcal M=\widetilde Q Q=\mu_*M,\qquad W_{\mathrm m}=hM q\widetilde q =\frac1\mu\mathcal M q\widetilde q,\qquad \mu=\frac{\mu_*}{h}.$$

Here $\mathcal M$ has engineering dimension two, $M$ dimension one, and $h\ne0$ is dimensionless. Define

$$\mathcal L_{\mathrm e}=\Lambda_{\mathrm e}^{\,4},\qquad \mathcal L_{\mathrm m}=\Lambda_{\mathrm m},\qquad \mathcal L_{\mathrm e}\mathcal L_{\mathrm m}=\mu^5 . \tag{Q10.4}$$

Add $m\mathcal M^5{}_5$ with nonzero complex mass $m$. The electric theory loses one flavor, leaving $SU(3)$ with four flavors and scale

$$\mathcal L_4=\Lambda_{(3,4)}^{\,5}=m\mathcal L_{\mathrm e}. \tag{Q10.5}$$

In magnetic variables the same deformation is a linear singlet term:

$$W=\frac1\mu\sum_{i,j,\alpha} \mathcal M^j{}_i q^{i\alpha}\widetilde q_{j\alpha} +m\mathcal M^5{}_5 . \tag{Q10.6}$$

The diagonal singlet equation requires $q^{5\alpha}\widetilde q_{5\alpha}=-\mu m$. A convenient representative is

$$q^5=(v,0),\qquad \widetilde q_5=(\widetilde v,0),\qquad v\widetilde v=-\mu m . \tag{Q10.7}$$

For locally canonical quark kinetic terms, D-flatness requires $|v|=|\widetilde v|$. Their phases can satisfy any nonzero complex product in (Q10.7); no reality restriction on $m$ is needed. Positive wavefunction coefficients modify the equality of coordinate norms, not the holomorphic product condition.

A nonzero fundamental vector has trivial stabilizer in $SU(2)$: a determinant-one unitary matrix fixing $(v,0)^T$ must be the identity. Thus the magnetic gauge group is completely Higgsed, not reduced to a photon. The source states this flow as $SU(N_f-N_c)\to SU(N_f-N_c-1)$; in this example the endpoint has no nontrivial gauge group. Intriligator–Seiberg, (5.11)–(5.13)

Identify the light fields and their cubic interaction

Let $i,j=1,\ldots,4$ now label the retained flavors. The off-diagonal singlet equations set $q^{i1}=\widetilde q_{j1}=0$ at the light-field restriction. Their massive partners are $\mathcal M^5{}_i,\mathcal M^j{}_5$. The fifth diagonal singlet pairs with the fluctuation changing $v\widetilde v$.

At tree level there are eight off-diagonal chiral pairs and one diagonal pair, removing eighteen chiral directions. Three more chiral directions enter the three massive vector multiplets. Starting with $25+10+10=45$ chiral components therefore leaves

$$45-18-3=24 =16+4+4.$$

They are the light meson block $\widehat{\mathcal M}$ and the second-color components $u_i=q^{i2}$, $\widetilde u_j=\widetilde q_{j2}$. Their tree superpotential is

$$W_{\mathrm{tree,light}} =\frac1\mu\sum_{i,j=1}^4 \widehat{\mathcal M}^j{}_i u_i\widetilde u_j. \tag{Q10.8}$$

To express this in electric composites, define the four-flavor baryons by

$$B_i=\frac1{3!}\varepsilon_{iabc}B_{abc},\qquad \widetilde B^{\,j}=\frac1{3!}\varepsilon^{jabc}\widetilde B^{abc}, \qquad \varepsilon_{1234}=1 .$$

The complementary-flavor map of the preceding lesson implies

$$B_i=\kappa v\,u_i,\qquad \widetilde B^{\,j}=\widetilde\kappa\widetilde v\,\widetilde u_j, \qquad \kappa\widetilde\kappa=-\frac{\mathcal L_{\mathrm e}}{\mu^2}. \tag{Q10.9}$$

For the sign, $B_i=-\kappa b^{i5}$, while $b^{i5}=\det(q^i,q^5)=-v u_i$; the two minus signs cancel. The tilde calculation is identical. Consequently

$$\mu\kappa\widetilde\kappa v\widetilde v =m\mathcal L_{\mathrm e}=\mathcal L_4, \qquad W_{\mathrm{tree,light}} =\frac{\widetilde B^T\widehat{\mathcal M}B}{\mathcal L_4}. \tag{Q10.10}$$

This derives the coefficient, including both scale phases, rather than assigning an arbitrary normalization to the surviving magnetic components.

Recover the determinant term

The cubic term is not the whole answer. A completely Higgsed magnetic $SU(2)$ instanton produces an additional interaction. Its nonzero coefficient can be checked using the already normalized one-flavor ADS result, as follows.

Hold a full-rank light block $\widehat{\mathcal M}$ fixed temporarily. Its four holomorphic magnetic masses form the matrix $\widehat{\mathcal M}/\mu$. Integrate those flavors out first. The remaining one-flavor $SU(2)$ scale is

$$\mathcal L_1^{\mathrm m} =\mathcal L_{\mathrm m}\det(\widehat{\mathcal M}/\mu) =\frac{\mathcal L_{\mathrm m}}{\mu^4}\det\widehat{\mathcal M}.$$

The one-flavor ADS superpotential is $\mathcal L_1^{\mathrm m}/N$, where $N=q^5\widetilde q_5$. Including the remaining singlet $T=\mathcal M^5{}_5$, the relevant heavy-field expression is

$$W_{\mathrm{heavy}} =\frac{\mathcal L_1^{\mathrm m}}N+\frac{T N}{\mu}+mT .$$

Its $T$ equation fixes $N=-\mu m$; its $N$ equation fixes $T=\mu\mathcal L_1^{\mathrm m}/N^2$. In particular $T$ is shifted from its tree value. Holding the light block fixed is essential: we are computing its effective superpotential, not imposing its vacuum equations during this elimination.

The two terms proportional to $T$ cancel after substitution, leaving

$$W_{\mathrm{inst}} =-\frac{\mathcal L_{\mathrm m}}{\mu^5m} \det\widehat{\mathcal M} =-\frac{\det\widehat{\mathcal M}}{\mathcal L_4}. \tag{Q10.11}$$

For a controlled starting regime, take the four magnetic thresholds of comparable size $M_H$ and locally weak, canonically normalized couplings. Choose $M_H\gg\rho\gg|\mathcal L_1^{\mathrm m}|^{1/5}$, where $\rho^2=|\mu m|$ is the Higgs scale squared. Such a window exists when $M_H\gg|\mathcal L_{\mathrm m}|$, since $|\mathcal L_1^{\mathrm m}|^{1/5}\sim (|\mathcal L_{\mathrm m}|M_H^4)^{1/5}\ll M_H$. Holomorphy then continues the polynomial interaction beyond this starting regime.

The nonzero ADS coefficient is independent dynamical input, not a consequence of matching charges. In instanton language, four flavor-mass insertions lift the additional matter zero modes and produce the determinant. This is the mechanism and normalization in Intriligator–Seiberg, (5.15)–(5.17), with the original broken-group calculation described in Seiberg, (4.13)–(4.14).

Combining the two contributions gives

$$W_{\mathrm{conf}} =\frac{\widetilde B^T\widehat{\mathcal M}B -\det\widehat{\mathcal M}}{\mathcal L_4}. \tag{Q10.12}$$

This is the $SU(3)$, four-flavor confining description: sixteen mesons and eight baryons, with no remaining magnetic gauge field. Here $[\widehat{\mathcal M}]=2$, $[B]=[\widetilde B]=3$ and $[\mathcal L_4]=5$, so each term has dimension three.

For the massless four-flavor theory the symmetric scalar R assignments are $R(Q)=1/4$, $R(\widehat{\mathcal M})=1/2$, $R(B)=R(\widetilde B)=3/4$; both terms have R-charge two. This is a new fixed-mass symmetry assignment, not the five-flavor table carried through unchanged. Alternatively one can keep the original assignment by treating $m$, of R-charge $6/5$, as a transforming spurion.

We have recovered a dynamical interaction with a sign and normalization that the anomaly sums could not determine. The result does not supply an exact Kähler metric or prove equality of every observable of the original massless pair.

Exercises

1. Diagnose a conjugation error in an anomaly

A calculation assigns the magnetic antifundamental quarks a negative quadratic flavor index and obtains $k_{SU(5)_L^2R}=-1/10$. Locate the error and recompute the coefficient. Then match $\operatorname{Tr}(B^2R)$ and explain why meson fermions contribute to the first check but not the second.

Hint 1

In the conjugate representation $T^a\mapsto-(T^a)^T$. Two insertions and three insertions behave differently under this replacement.

Hint 2

Use two magnetic color copies of the left antifundamental and five right-flavor copies of the meson left fundamental.

Solution

The two minus signs in $\operatorname{tr}(T^aT^b)$ cancel, so $T(\overline{\mathbf5})=+1/2$. The conjugation sign reversal applies to the cubic flavor anomaly, not this quadratic index. The magnetic result is

$$2\left(\frac12\right)\left(-\frac25\right) +5\left(\frac12\right)\left(-\frac15\right) =-\frac9{10},$$

matching $3(1/2)(-3/5)$ electrically.

For the Abelian trace, the electric result is $30(1)^2(-3/5)=-18$; the magnetic result is $20(3/2)^2(-2/5)=-18$. Meson fermions have nonzero R-charge and left-flavor generators, but zero baryon charge. They contribute to $SU(5)_L^2R$ and not to $B^2R$. Gauginos are flavor- and baryon-neutral in both calculations.

2. Is there an unbroken photon?

Use canonical quark and gauge kinetic terms, $t^a=\sigma^a/2$ and $D_\mu q=\partial_\mu q-ig_{\mathrm m}A_\mu^at^aq$. For the Higgs representative (Q10.7), calculate the three-by-three vector mass-squared matrix. Assume $|v|=|\widetilde v|=\rho>0$ and normalize a real vector mass term as $\frac12(M_V^2)_{ab}A_\mu^aA^{b\mu}$.

Hint 1

Both quark kinetic terms contribute. Symmetrize the generator product because $A^aA^b$ is symmetric in color labels.

Hint 2

For Pauli generators, $\{t^a,t^b\}=\delta^{ab}\mathbf1/2$.

Solution

Each scalar contributes $g_{\mathrm m}^2\rho^2\delta_{ab}A^a_\mu A^{b\mu}/4$. The antifundamental contribution has the same sign: the sign of its covariant derivative disappears in this squared term. Their sum is $g_{\mathrm m}^2\rho^2 A_\mu^aA^{a\mu}/2$, hence

$$(M_V^2)_{ab}=g_{\mathrm m}^2\rho^2\delta_{ab}.$$

All three eigenvalues are positive. There is no unbroken photon, and three chiral directions supply the longitudinal modes and scalar partners of the massive vector multiplets. This is a leading locally canonical Higgs calculation; the exact holomorphic relation is $v\widetilde v=-\mu m$, not an assertion that wavefunction corrections leave pole masses unchanged.

3. Detect the missing determinant using a vacuum

In the four-flavor composite theory take $\widehat{\mathcal M}=\operatorname{diag}(x,x,x,0)$ with $x\ne0$, and initially set every baryon to zero. Compare its F equations with and without the determinant in (Q10.12). Restore F-flatness while keeping this meson fixed by turning on suitable baryons. Interpret the result using electric minors.

Hint 1

The determinant vanishes at the displayed matrix, but its derivative with respect to the fourth diagonal entry need not vanish.

Hint 2

The cofactor matrix has just one nonzero entry, the fourth diagonal one. Both $B$ and $\widetilde B$ can lie in the meson’s null direction.

Solution

The cubic term alone has zero F derivatives when both baryons vanish, and incorrectly permits the stated point. In the full superpotential,

$$\frac{\partial W_{\mathrm{conf}}}{\partial\widehat{\mathcal M}^4{}_4} =-\frac{x^3}{\mathcal L_4}\ne0.$$

Keeping only $B_4,\widetilde B^{\,4}$ nonzero and choosing $B_4\widetilde B^{\,4}=x^3$ cancels this derivative. All other meson cofactors vanish, and $\widehat{\mathcal M}B=\widehat{\mathcal M}^T\widetilde B=0$, so the baryon equations also vanish.

The electric minor identity is $B_{123}\widetilde B^{123}=x^3$. Our signed convention gives $B_4=-B_{123}$ and $\widetilde B^{\,4}=-\widetilde B^{123}$, leaving the product positive. The determinant interaction enforces exactly the relation that the cubic term alone misses. Its value happens to vanish at this point; its derivative, and therefore its physical effect on the vacuum equations, does not.

The anomaly sums and the deformation test probe different information. Their agreement supports the proposed infrared equivalence, while the missing-determinant example shows how a plausible partial description can fail. Next, identify the conformal R-symmetry and compute physical operator dimensions. The same three-color, five-flavor example provides a concrete test of the meson unitarity bound.