Determine the R-symmetry by a-maximization

An anomaly-free R-current need not be the current in the superconformal algebra. Adding a conserved flavor current leaves the supercharges’ charges unchanged, but can change the charges of chiral operators. Which combination determines their dimensions?

The answer in four-dimensional $\mathcal N=1$ theories is a-maximization. It turns the Weyl-fermion anomalies of the anomaly lesson into a local extremum problem. We will recover the five-flavor SQCD charges from the conformal-window lesson, then test what the method can and cannot determine. Four real Poincaré supercharges are present; a superconformal fixed point also has four conformal partners. No curved supersymmetric background is needed for the calculation.

Start with one free chiral field

The number $a$ is the coefficient associated with the Euler-density term in the four-dimensional trace anomaly. It is dimensionless. We use the normalization in which a free chiral multiplet contributes $1/48$ and a free vector multiplet contributes $3/16$.

Supersymmetric conformal Ward identities relate this coefficient to anomalies of the superconformal R-current:

$$a=\frac3{32}\left(3\operatorname{Tr}R^3-\operatorname{Tr}R\right). \tag{Q12.1}$$

Every trace runs over independent left-handed Weyl fermions, including their color and flavor multiplicities. It does not run over scalar fields or the two spinor components separately. For a chiral scalar of charge $r$, the fermion has charge $r-1$; a gaugino has charge one. Equation (Q12.1) is a quantum Ward-identity result, not a classical component-counting identity. Intriligator–Wecht, (1.2)

A free chiral $X$ has a phase symmetry, so a trial R-assignment can give its scalar any charge $r$. The associated polynomial is

$$a_\chi(r)=\frac3{32}\left[3(r-1)^3-(r-1)\right], \qquad a_\chi'(r)=\frac3{32}\left[9(r-1)^2-1\right].$$

There are two stationary charges:

$$r=\frac23,\ \frac43,\qquad a_\chi''(r)=\frac{27}{16}(r-1).$$

At $r=2/3$ the second derivative is $-9/16$, so this is a local maximum and $a_\chi=1/48$. At $r=4/3$ the curvature is positive: it is the wrong stationary point. The scalar dimension at the physical free point is $\Delta=3r/2=1$.

The polynomial tends to $+\infty$ as $r\to+\infty$. We are selecting a local maximum, not searching for its largest value over all real charges. Nor is this polynomial an effective potential for $X$: the variable being varied is a current assignment.

Decide which directions may vary

Let $R_0$ be one conserved R-generator with $R_0(Q_\alpha)=-1$. Let $F_I$ be independent Abelian flavor generators that act trivially on the supercharges and commute with the unbroken non-Abelian flavor group. Then

$$R_t=R_0+\sum_I s_I F_I,\qquad s_I\in\mathbb R. \tag{Q12.2}$$

The subscript $t$ means trial. The coefficients are dimensionless. Every included $F_I$ must be a genuine symmetry of the quantum theory: it must be free of anomalies with two dynamical gauge currents and must preserve all nonzero superpotential couplings. A nonzero ‘t Hooft anomaly involving only background sources does not disqualify it.

For each simple dynamical gauge group, with $T(\mathbf{fund})=1/2$, impose

$$T(\mathrm{adj})+\sum_i T(\mathcal R_i)(r_i-1)=0. \tag{Q12.3}$$

For every nonzero superpotential monomial, impose total scalar R-charge two, with its coupling held neutral. If a coupling vanishes at the fixed point, imposing that absent interaction’s charge constraint would remove a possible infrared current incorrectly. An anomalously broken axial rotation is not an extra direction to maximize over.

The current space must also include any accidental infrared symmetries. Identifying that space is physical input. Optimizing a polynomial cannot certify that no current was omitted, or establish that the proposed interacting fixed point exists.

Why the stationary point is a maximum

Define

$$a_t(s)=\frac3{32} \left(3\operatorname{Tr}R_t^3-\operatorname{Tr}R_t\right).$$

The needed superconformal inputs are

$$9\operatorname{Tr}(R_{\rm SC}^2F_I)=\operatorname{Tr}F_I, \qquad \operatorname{Tr}(R_{\rm SC}F_IF_J)=-\frac13\tau_{IJ}. \tag{Q12.4}$$

Here $\tau_{IJ}$ normalizes the separated-point flavor-current correlator. In the source’s Euclidean convention,

$$\langle j^I_\mu(x)j^J_\nu(0)\rangle =\frac{\tau_{IJ}}{16\pi^4} \left(\delta_{\mu\nu}\partial^2-\partial_\mu\partial_\nu\right) \frac1{|x|^4}, \qquad x\ne0. \tag{Q12.5}$$

Reflection positivity makes $\tau$ positive definite on independent nonzero currents. A redundant generator would give a null direction and should first be removed.

Why do anomalies know about this two-point function? Supersymmetry relates the current multiplet’s trace and chiral anomalies. The first identity in (Q12.4) follows because the superconformal three-point function containing two stress-tensor multiplets and one flavor multiplet has a single relevant coefficient: its $RRF_I$ and gravitational–$F_I$ anomalies have a fixed ratio. A free chiral fermion has $R=-1/3$, which fixes that ratio to nine. The second identity follows by turning on flavor background fields in the same Ward identities; its coefficient is the same $\tau$ as in (Q12.5). This is the sourced Ward-identity input of AM, (2.19)–(2.21), rather than a claim that elementary anomaly arithmetic alone proves the theorem.

Now the maximization argument is explicit:

$$\begin{aligned} \frac{\partial a_t}{\partial s_I} &=\frac3{32}\left[9\operatorname{Tr}(R_t^2F_I)-\operatorname{Tr}F_I\right],\\ \frac{\partial^2 a_t}{\partial s_I\partial s_J} &=\frac{27}{16}\operatorname{Tr}(R_tF_IF_J). \end{aligned}$$

At $R_t=R_{\rm SC}$, the gradient vanishes and

$$\boxed{\left.\partial_I\partial_J a_t\right|_{\rm SC} =-\frac9{16}\tau_{IJ}.} \tag{Q12.6}$$

For any nonzero allowed displacement $v_I$, the quadratic change is $-\tfrac9{32}v_I\tau_{IJ}v_J<0$. That gives a strict local maximum. It also explains why solving only the stationary equations is insufficient.

As a normalization check, a free unit-charge chiral’s scalar and Weyl currents contribute respectively $1/(4\pi^4)$ and $1/(2\pi^4)$ to the coefficient of $I_{\mu\nu}(x)/|x|^6$, where $I_{\mu\nu}=\delta_{\mu\nu}-2x_\mu x_\nu/|x|^2$. Since the differential tensor in (Q12.5) equals $12I_{\mu\nu}/|x|^6$, this gives $\tau=1$ and the previously found Hessian $-9/16$.

Work through five-flavor SQCD

Take massless $SU(3)$ SQCD with five fundamental–antifundamental pairs, no superpotential, and the flavor-symmetric vacuum at the origin. The standard interacting conformal-window description is the dynamical input. Its non-Abelian flavor group is $SU(5)_L\times SU(5)_R$.

The common anomaly-free scalar charge is $r_0=1-3/5=2/5$. Baryon number $B$ gives $Q,\widetilde Q$ charges $+1,-1$. Retain its mixing explicitly, even though charge conjugation already suggests the answer:

$$R_t(Q)=\frac25+s,\qquad R_t(\widetilde Q)=\frac25-s.$$

The gauge anomaly vanishes for every $s$: $3+\tfrac52(-3/5+s)+\tfrac52(-3/5-s)=0$. The fermion data are

Weyl fermion	Multiplicity	Trial R-charge	Baryon charge
Gaugino	$8$	$1$	$0$
Quark	$15$	$-3/5+s$	$1$
Antiquark	$15$	$-3/5-s$	$-1$

Consequently,

$$\begin{aligned} \operatorname{Tr}R_t &=8+15\left(-\frac35+s\right)+15\left(-\frac35-s\right)=-10,\\ \operatorname{Tr}R_t^3 &=8+15\left[\left(-\frac35+s\right)^3+ \left(-\frac35-s\right)^3\right] =\frac{38}{25}-54s^2. \end{aligned}$$

The odd powers of $s$ cancel between conjugate flavors. We obtain

$$\boxed{a_t(s)=\frac{273}{200}-\frac{243}{16}s^2.} \tag{Q12.7}$$

The stationary point is $s=0$, its second derivative is $-243/8$, and the fixed-point anomaly is $a=273/200$. Independently reading the mixed anomaly from the table gives $\operatorname{Tr}(R_{\rm SC}B^2)=30(-3/5)=-18$. Thus $\tau_{BB}=54$, reproducing the same Hessian through (Q12.6).

The physical meson has $R=4/5$ and $\Delta=6/5$; a three-quark baryon has $R=6/5$ and $\Delta=9/5$. Both pass scalar-primary unitarity. The elementary quark is gauge variant, so imposing a physical local-operator bound $R(Q)\ge2/3$ would be an error.

As a flow check, the asymptotically free ultraviolet limit contains eight vector and thirty chiral multiplets:

$$a_{\rm UV}=8\frac3{16}+30\frac1{48}=\frac{17}{8}, \qquad a_{\rm UV}-a_{\rm IR}=\frac{19}{25}>0.$$

This is a consistency check of the assumed flow, not a proof of it from positivity. Notice also that the ultraviolet superconformal quark charge is $2/3$, whereas the conserved current used to compute infrared anomalies assigns $2/5$. Anomaly matching preserves the anomalies of a fixed current; it does not identify two different superconformal currents along the flow.

This example tests fermion shifts, multiplicities and the curvature of the trial polynomial. It does not test an asymmetric nonzero mixing solution: charge conjugation already forces $s=0$. Nor does it diagnose every accidental symmetry. The adjoint example in Exercise 2 supplies a genuine remaining parameter; Exercise 3 exhibits a failed restricted maximization.

When a chiral operator becomes free

For a nonidentity gauge-invariant scalar chiral primary, $\Delta=3R_{\rm SC}/2\ge1$. At equality it is a free field. A putative interacting assignment below $R=2/3$ is a warning that the assumed current space or fixed-point description is wrong.

Suppose independent dynamics identifies a decoupled free chiral operator $X$, whose charge under the inherited trial current is $r_X(s)$. At the infrared factorization, the old anomaly polynomial contains the contribution $a_\chi(r_X(s))$ of that free sector. Its own phase current permits an independent charge. Maximizing in this newly included direction replaces that contribution by $a_\chi(2/3)$:

$$a_{\rm corrected}(s) =a_t(s)+\sum_X d_X \left[\frac1{48}-a_\chi(r_X(s))\right]. \tag{Q12.8}$$

Here $d_X$ counts the independent free chiral fields of that charge, not the number of elementary constituents in a composite. This formula follows by subtracting the old free-sector anomaly before adding its superconformal value. It is the prescription derived in Kutasov–Parnachev–Sahakyan, (2.5)–(2.8), whose $\widetilde a$ is $32a/3$.

If $r_X$ depends on $s$, the subtraction changes the derivative used to find the interacting sector’s charge. Merely changing the reported charge of $X$ after extremizing the old polynomial can therefore leave an incorrect answer. Recompute and check that the assumed free fields really decouple and that no other independent primary violates the bound.

Passing those checks is necessary, not a universal certificate. An accidental current can matter even when some affected operators never approach the bound; SQCD’s free magnetic phase provides a concrete example. KPS, discussion of (1.8)–(1.9) A-maximization is exact when applied to the actual superconformal current space. A guessed ultraviolet subset need not contain the answer.

Comparison with F-maximization

This final comparison assumes the three-dimensional sphere construction developed in R-mixing and F-maximization. It is a separate entrance from the four-dimensional calculation above.

Both methods vary a trial R-generator by conserved flavor currents and require control of accidental symmetries. Their functions and dimension formulas differ:

Question	Four-dimensional $\mathcal N=1$	Three-dimensional $\mathcal N=2$
Function at the physical current	Euler trace-anomaly coefficient $a$	Sphere free energy $F=-\log\lvert Z_{S^3}\rvert$
Trial data	Cubic and linear Weyl-fermion anomalies	A supersymmetric sphere partition function with its source prescription
Scalar chiral-primary dimension	$\Delta=3R/2$	$\Delta=R$
Free chiral R-charge	$2/3$	$1/2$

In four dimensions the anomaly polynomial makes the extremization algebraic once the symmetry data are known. There is no corresponding ordinary local Euler trace anomaly in three dimensions whose fermion cubic trace could replace the sphere calculation. Jafferis’s introduction and section 5 establish the sphere-extremization argument and its accidental-current assumption. One extremizes the magnitude, or equivalently the real free energy, with a fixed supersymmetric source and contact-term prescription; a complex partition-function phase is not an independent trial R-current. The sphere lesson separately derives the negative Hessian from current positivity using the later local-maximality theorem. It does not attribute that theorem to Jafferis’s stationarity argument alone.

Neither method means “maximize the number of elementary fields,” and neither proves that a proposed renormalization-group flow ends at an interacting SCFT. Their shared achievement is to select the current in the superconformal algebra once the relevant fixed point, symmetries and observable have been specified.

Practice

1. Repeat the polynomial in magnetic variables

The five-flavor dual has gauge group $SU(2)$, five pairs $q,\widetilde q$, twenty-five singlets $M$, and nonzero $W=hM q\widetilde q$. Use scalar charges $R(q)=R(\widetilde q)=3/5$, $R(M)=4/5$, and baryon charges $B(q)=3/2$, $B(\widetilde q)=-3/2$, $B(M)=0$. Compute the whole trial polynomial for $R_t=R_0+sB$ and its Hessian. Explain why assigning dual quarks baryon charge one would test a different generator.

Hint 1

There are three gauginos, ten quark fermions in each sector, and twenty-five singlet fermions. Subtract one from each scalar R-charge.

Hint 2

Use $(b+d)^3+(b-d)^3=2b^3+6bd^2$. A magnetic baryon contains two quarks, while the electric baryon contains three.

Solution

The trace of $R_t$ is $3+10(-2/5+3s/2)+10(-2/5-3s/2)+25(-1/5)=-10$. The cubic trace is

$$\begin{aligned} \operatorname{Tr}R_t^3 &=3+10\left[\left(-\frac25+\frac32s\right)^3+ \left(-\frac25-\frac32s\right)^3\right] +25\left(-\frac15\right)^3\\ &=\frac{38}{25}-54s^2. \end{aligned}$$

Thus $a_t=273/200-243s^2/16$, exactly the electric polynomial, and $a_t''(0)=-243/8$. The superpotential is neutral under baryon number and has R-charge two for all $s$, so this is an allowed mixing.

An electric baryon has $B=3$; a magnetic two-quark baryon has $B=2(3/2)=3$. Replacing the magnetic elementary charge by one without changing the generator dictionary gives a different current normalization and a different Hessian. Under a consistent rescaling $B'=cB$, the same physical assignment has $s'=s/c$ and $\tau_{B'B'}=c^2\tau_{BB}$; maximality is invariant, its numerical coordinate curvature is not. The operator and charge dictionary is Seiberg, (3.1)–(3.4).

2. Leave a genuine mixing parameter

Now change the theory: add one adjoint chiral $X$ to electric $SU(3)$ with five flavor pairs, still with $W=0$. There is no $\widetilde QXQ$ interaction; this is a four-dimensional $\mathcal N=1$ model, not the $\mathcal N=2$ theory.

Assume its usual interacting fixed-point description without extra accidental currents. Charge conjugation gives $R(Q)=R(\widetilde Q)=y$, but does not fix $R(X)=u$. Use $T(\mathrm{adj})=3$, $\dim(\mathrm{adj})=8$ to derive the constraint, find both stationary values of $u$, and select the local maximum. Test the basic scalar chiral invariants for an immediate free-field obstruction. Explain what the result assumes.

Hint 1

The gauge anomaly gives $3+3(u-1)+5(y-1)=0$. In $a_t$, there are eight adjoint fermions of charge $u-1$, not three.

Hint 2

Eliminate $y$ first. The derivative reduces to a quadratic proportional to $171u^2-1800u+1025$.

Solution

The constraint is $3u+5y=5$, hence $y=1-3u/5$. This direction is a genuine flavor mixing: increasing $u$ by $t$ changes $y$ by $-3t/5$, and its gauge anomaly is $3-5(3/5)=0$.

The complete polynomial is

$$\begin{aligned} a_t(u) &=\frac3{32}\left\{ 16+8\left[3(u-1)^3-(u-1)\right] +30\left[3\left(-\frac35u\right)^3+\frac35u\right]\right\}\\ &=\frac{3u}{400}\left(57u^2-900u+1025\right). \end{aligned}$$

Its stationary values are $u_\pm=(300\pm5\sqrt{2821})/57$. At these points $a_t''(u_\pm)=\pm9\sqrt{2821}/40$. The negative branch is therefore the local maximum:

$$u_*\simeq0.6041151842,\qquad y_*=1-\frac35u_*\simeq0.6375308895,\qquad a_*\simeq2.2749412186.$$

The adjoint is traceless, so there is no gauge-invariant $\operatorname{Tr}X$. The lowest trace operator $\operatorname{Tr}X^2$ has R-charge $2u_*\simeq1.20823$; the meson has $2y_*\simeq1.27506$; an undressed baryon has $3y_*\simeq1.91259$. Higher powers of $X$ add positive charge. Thus the trace, dressed-meson and dressed-baryon generators have no immediate $R<2/3$ obstruction. We have not imposed this bound on the gauge-variant elementary fields.

The baryon-mixing direction omitted by charge conjugation also has negative curvature: its trace is $\operatorname{Tr}(RB^2)=30(y_*-1)<0$. Its mixed Hessian with $u$ vanishes at zero baryon mixing. Hence the stationary point is a local maximum in both Abelian directions.

This finite-rank example uses the general adjoint theory of AM, section 4.1. A stationary point and these operator checks do not by themselves prove the assumed fixed point exists or exclude all accidental currents. They do show a determination that charge conjugation alone could not make.

3. Show why repairing only the meson can fail

Consider ordinary $SU(5)$ SQCD with seven flavors. The conformal-window lesson places it in the free magnetic phase with gauge group $SU(2)$. An incorrect continuation gives electric scalar charges $2/7$ and meson charge $4/7$.

Compute the inherited trial $a$ at zero baryon mixing. Apply (Q12.8) only to the forty-nine mesons. Compare with the actual Gaussian magnetic limit, which has three vector multiplets, twenty-eight quark chiral components and forty-nine singlet chiral fields. Does the discrepancy invalidate a-maximization?

Hint 1

The inherited electric count contains twenty-four gauginos and seventy matter fermions of charge $-5/7$.

Hint 2

At the Gaussian magnetic limit all seventy-seven chiral components have scalar charge $2/3$. Its gauge and Yukawa interactions vanish in the infrared, allowing currents not captured by the interacting electric assignment.

Solution

The inherited traces are $\operatorname{Tr}R=24-50=-26$ and $\operatorname{Tr}R^3=24-1250/49=-74/49$. Thus

$$a_{\rm inherited}=\frac{789}{392}.$$

Since $a_\chi(4/7)=99/5488$, the meson-only correction gives

$$a_{\rm meson\ only} =\frac{789}{392} +49\left(\frac1{48}-\frac{99}{5488}\right) =\frac{316}{147}.$$

The actual free magnetic field content instead gives

$$a_{\rm free\ magnetic} =3\frac3{16}+77\frac1{48} =\frac{13}{6}, \qquad a_{\rm free\ magnetic}-a_{\rm meson\ only}=\frac5{294}.$$

The discrepancy is nonzero. The two computations extremize different current spaces. Free mesons are not the only change: the magnetic gauge and superpotential couplings tend to zero, so additional infrared symmetries affect the identification of the current. In particular, the physical magnetic two-quark baryon has R-charge $4/3$, whereas the inherited electric assignment gives $5(2/7)=10/7$. That baryon never violated the scalar unitarity bound, yet its inherited charge was still wrong.

This is precisely the distinction emphasized in KPS, (1.8)–(1.9). The failure is not of a-maximization applied to the true infrared theory. It is the assumption that one free-sector correction supplies every missing current.

To use the method in a new theory, specify the proposed fixed point, enumerate its allowed conserved-current mixing, compute the fermion traces, select the negative-Hessian stationary point, and test the physical chiral operators and any free sectors. The polynomial calculation is short once those physical choices are correct.