R-mixing and F-maximization

The first SQED matrix integral was simplest at matter R-charge $r=1/2$. Convenience did not make that the interacting theory’s superconformal charge. We now use the dependence on $r$ to identify the physical current.

The preparation is R-symmetry and flavor symmetry and conformal fixed points, together with that matrix integral. Only the later SQED–XYZ comparison also requires the duality dictionary.

Throughout, the flat parent is three-dimensional $\mathcal N=2$, with four real supercharges. Its Euclidean round-sphere construction preserves the corresponding rigid subalgebra, with the two independent spinor families used in the determinant calculation. Superconformal theories have additional conformal generators; extremizing a sphere partition function is not a change in the parent supersymmetry count.

A first extremum: one free chiral

For a gauge-neutral free chiral multiplet of trial R-charge $r$ and zero real mass, the regulated determinant is $Z_\chi(r)=\exp[\mathcal L(1-r)]$. Recall that its logarithm is fixed by

$$\mathcal L'(u)=-\pi u\cot(\pi u),\qquad \mathcal L(0)=0.$$

The branch is continued smoothly from $u=0$ to the real interval $0<u<1$. Define the real sphere free energy $f(r)=-\log|Z_\chi(r)|=-\mathcal L(1-r)$. For the first calculation, stay in $0<r<1$, which includes the physical answer and has no determinant singularity. Differentiate, keeping the minus sign in $1-r$:

$$\begin{aligned} f'(r)&=\mathcal L'(1-r) =\pi(1-r)\cot(\pi r),\\ f''(r)&=-\pi\cot(\pi r) -\pi^2(1-r)\csc^2(\pi r). \end{aligned} \tag{O05.1}$$

The first derivative is positive below $r=1/2$ and negative above it. At the stationary point,

$$r_*=\frac12,\qquad f(r_*)=\frac{\log 2}{2},\qquad f''(r_*)=-\frac{\pi^2}{2}. \tag{O05.2}$$

The value follows from $\mathcal L(1/2)=-\log 2/2$, already fixed by the regulated determinant. Locally, $f(1/2+\epsilon)=\log 2/2-\pi^2\epsilon^2/4+O(\epsilon^3)$. The negative curvature checks that we have maximized $-\log|Z|$, or equivalently minimized $|Z|$.

The flat free field has scaling dimension $1/2$ throughout this calculation. Varying a curvature coupling did not give it a continuously variable flat-space dimension. Instead, the extremum selected the R-current that belongs to its superconformal algebra.

What is allowed to mix?

An R-current rotates the supercharges. A flavor current commutes with them. Adding a conserved Abelian flavor current to an R-current therefore gives another R-current with the same action on the supercharges. Write the corresponding charges as

$$R(\mathbf t)=R_0+\sum_a t_a F_a,\qquad r_i(\mathbf t)=r_i^{(0)}+\sum_a t_a q_i^a. \tag{O05.3}$$

Here $F_a$ are independent flavor charges, $q_i^a$ are the charges of a chiral field or operator, and $t_a$ are real dimensionless mixing parameters. The current $R_0$ is a reference choice, not necessarily the superconformal one.

Only symmetries of the theory with its couplings held fixed belong in this family. For a nonzero superpotential monomial, its total R-charge must remain two, and each mixing flavor must give it charge zero. Assigning a compensating charge to a fixed coupling is a useful spurion convention; it does not create an additional conserved flavor current. A dynamical gauge transformation is also not an independent global current to maximize over.

These statements concern current algebra and local background couplings. A generic real combination need not generate a compact circle with the same period as each original integral charge generator. The compact flavor-charge lattice and permitted background counterterms remain fixed while one varies the local trial current.

On the sphere, changing $\mathbf t$ changes the R-dependent couplings to the supersymmetric background. It is related by analytic continuation to imaginary flavor masses. It does not mean turning on a physical real mass and moving away from the conformal vacuum. The dimensionless combinations are $\ell m_a$, where $\ell$ is the sphere radius and $m_a$ is a physical mass. At a conformal point, with all physical masses and FI parameters zero, the finite quantity

$$\mathcal F(\mathbf t)=-\log|Z_{S^3}(\mathbf t)| =\operatorname{Re}[-\log Z_{S^3}(\mathbf t)] \tag{O05.4}$$

is dimensionless. Use the same supersymmetric subtraction, source normalization and background contact-term prescription throughout. The complex logarithm itself need not be real.

Why the physical stationary point is a local maximum

At a unitary superconformal fixed point, use all independent nontrivial Abelian flavor currents that can mix with the R-current. In the supersymmetric conformal prescription, the result is

$$\left.\partial_a\mathcal F\right|_*=0,\qquad \left.\partial_a\partial_b\mathcal F\right|_* =-\frac{\pi^2}{2}\tau_{ab}. \tag{O05.5}$$

The normalization of $\tau_{ab}$ is fixed by the separated-point correlation function

$$\langle j^a_\mu(x)j^b_\nu(0)\rangle =\frac{\tau_{ab}}{16\pi^2} \bigl(\delta_{\mu\nu}\partial^2-\partial_\mu\partial_\nu\bigr) \frac1{|x|^2},\qquad x\ne0. \tag{O05.6}$$

Each current has dimension two, and the right-hand side scales as $|x|^{-4}$, as required. Unitarity makes $\tau$ positive definite for the stated independent currents. Thus every nonzero permitted direction has negative second variation of $\mathcal F$. These are the local maximum and normalization proved in Closset–Dumitrescu–Festuccia–Komargodski–Seiberg, (4.9)–(4.11). The earlier Jafferis argument, (5.1), established the stationarity prescription and left the general second-derivative analysis open.

A short way to understand the sign is to use the supersymmetric mass source. In source conventions where a mixing displacement is the imaginary part of $w=\ell m+i t$, the real-mass response at the conformal point is $+\pi^2\tau/2$. Two derivatives along the imaginary direction supply $i^2=-1$. Computing the real-mass coefficient requires the regulated current Ward identity, including contact terms; it does not follow from simply integrating a singular separated-point function. The same source, (4.6)–(4.8), gives this coefficient. Reversing the sign used to name a mass source reverses the imaginary continuation but leaves this second-derivative sign unchanged.

For the unit-charge free chiral, (O05.2) gives $\tau=1$, agreeing with that source’s independent Gaussian calculation, (5.7). This checks both the factor $\pi^2/2$ and our use of $\mathcal F$.

The theorem is local. It does not instruct us to optimize over singular determinants, over directions broken by fixed couplings, or over every formal extremum of a meromorphic continuation. A null direction from a redundant or purely contact current must not be mistaken for a negative-norm state. Nor does a successful extremization alone establish that a proposed theory flows to the assumed unitary fixed point.

Compare SQED with XYZ

This section uses the SQED–XYZ dictionary. The fixed gauge theory is compact $U(1)$ with one charge pair $Q,\widetilde Q$, level $k=0$, $W=0$, and zero physical masses and FI parameter. Its infrared description has three gauge-neutral chirals with $W=XYZ$ and the chosen holomorphic dictionary $X=M=Q\widetilde Q$, $Y=V_+$, $Z=V_-$.

There are two useful mixing directions. The axial flavor gives $Q,\widetilde Q$ equal charge $+1$; the topological flavor gives $V_\pm$ charges $\pm1$. Let $r$ be the common matter trial R-charge and let $t$ be the coefficient of the unit-normalized topological charge. The induced charges are

Chiral operator	Axial charge	Topological charge	Trial R-charge
$X=M$	$2$	$0$	$2r$
$Y=V_+$	$-1$	$+1$	$1-r+t$
$Z=V_-$	$-1$	$-1$	$1-r-t$

Their R-charges sum to two for every $r,t$, so this is precisely a two-parameter family compatible with the cubic superpotential. Charge conjugation exchanges $Y,Z$ and reverses $t$; the physical current must respect this symmetry, but we can retain $t$ to check its curvature.

In the preceding lesson’s FI convention, the factor was $e^{2\pi i\eta s}$. Topological mixing is its continuation $\eta\to-it$, so

$$\begin{aligned} Z(r,t)&=\int_{\mathbb R}ds\, \exp\!\left[2\pi t s+\mathcal L(1-r-is)+\mathcal L(1-r+is)\right],\\ &\hspace{12mm}0<r<1,\qquad |t|<1-r. \end{aligned} \tag{O05.7}$$

The variable $s=\ell\sigma$ is the dimensionless constant gauge scalar. The displayed domain makes the integral absolutely convergent: its two tails behave as $\exp[-2\pi(1-r)|s|+2\pi t s]$. It also keeps all three XYZ trial charges between zero and two. No change in compact gauge group or level accompanied this analytic continuation.

The weight in (O05.7) is positive for real $r,t$. Define an expectation by dividing its integral by $Z$. Differentiating a normalized expectation subtracts its disconnected part, giving

$$\partial_t\mathcal F=-2\pi\langle s\rangle,\qquad \partial_t^2\mathcal F =-4\pi^2\bigl(\langle s^2\rangle-\langle s\rangle^2\bigr)<0. \tag{O05.8}$$

Thus reflection symmetry makes $t=0$ stationary, and the positive variance makes it a maximum along this direction. This sign follows directly from the gauge integral. The physical real FI response had the opposite sign because its insertion was imaginary.

To find the axial extremum analytically, use the one-flavor specialization of the partition-function identity in Willett–Yaakov, section 3.1, pp. 8–10:

$$Z(r,t)= \exp\!\left[\mathcal L(1-2r)+\mathcal L(r-t)+\mathcal L(r+t)\right]. \tag{O05.9}$$

The same determinant normalization and matched background contact terms are understood. This is an exact integral identity, used here as an input rather than derived from symmetry. It tests this observable of the duality; equality of one partition function is not by itself a proof of every duality claim. In the ordinary convergence domain above no further contour continuation is needed.

Taking the logarithm expresses the interacting answer in terms of the already defined single-chiral function:

$$\mathcal F(r,t)=f(2r)+f(1-r+t)+f(1-r-t). \tag{O05.10}$$

The three contributions arise after localization; the cubic constraint remains essential even though its coupling does not appear explicitly. One cannot independently set all three charges to the free-field value.

Here $f(R)=-\mathcal L(1-R)$ uses the same smooth determinant continuation throughout $0<R<2$, including negative $1-R$. The first free-chiral calculation needed only $0<R<1$; it did not restrict the domain of this continued function.

At $r=1/3,t=0$, every argument of $f$ is $2/3$. The first derivatives cancel because the axial coefficient vector is $(2,-1,-1)$ and the topological vector is $(0,1,-1)$. Set $c=f''(2/3)=\pi/\sqrt3-4\pi^2/9<0$. Differentiating once more gives

$$\begin{gathered} (r_*,t_*)=\left(\frac13,0\right),\qquad \left.\begin{pmatrix} \mathcal F_{rr}&\mathcal F_{rt}\\ \mathcal F_{tr}&\mathcal F_{tt} \end{pmatrix}\right|_* =\begin{pmatrix}6c&0\\0&2c\end{pmatrix},\\ R(X)=R(Y)=R(Z)=\frac23,\qquad \mathcal F_*=-3\mathcal L(1/3)\simeq0.872372. \end{gathered} \tag{O05.11}$$

Both eigenvalues are negative. The gauge-invariant scalar chiral operators therefore have scaling dimensions $2/3$. The value $r_*=1/3$ assigned to gauge-charged $Q$ is not a violation of the scalar unitarity bound: that bound constrains local gauge-invariant operators in the physical Hilbert space.

There is a useful check that does not differentiate the XYZ answer. In the gauge integral put $g(r,s)=2\operatorname{Re}\mathcal L(1-r+is)$. Direct differentiation gives $\mathcal F_r=-\langle g_r\rangle$ and $\mathcal F_{rr}=-\langle g_{rr}\rangle-\operatorname{Var}(g_r)$. Numerically integrating these insertions and $s^2$ reproduces (O05.11) and $\langle s^2\rangle_*=2/9-1/(2\pi\sqrt3)$. The regulated function is evaluated before integrating over $s$, and exponentially bounded Cartan tails are retained in the error estimate. This probes the stationary point and both independent curvatures, not just the value of $Z$.

The superconformal statement concerns the vacuum at the branch intersection. It neither identifies the two descriptions’ classical ultraviolet kinetic metrics nor sets trial charges equal to dimensions away from that fixed point.

When a symmetry appears only in the infrared

An accidental symmetry is a conserved symmetry of the infrared theory that was not a symmetry of the ultraviolet theory with its interactions fixed. If its current can mix with the infrared R-current, the ultraviolet family (O05.3) may not contain the correct answer.

The local theorem still holds for the actual infrared SCFT. What fails is applying it to an incomplete family computed along a chosen ultraviolet flow; this distinction is explicit at the end of FM section 4. Establish which sector becomes free, or otherwise identify the additional infrared current, before extending the mixing space. Simply forcing a trial charge up to a unitarity bound is not a derivation of a corrected partition function.

Exercise 3 gives a controlled elementary example: an irrelevant superpotential disappears at long distances, so a constraint that was valid at nonzero coupling must be removed. The prerequisite fixed-point discussion matters here. A trial extremum cannot replace an analysis of which interactions and symmetries survive the flow.

Exercises

1. Change the flavor basis for two free chirals

Two independent free chirals have trial charges $r_1=1/2+t+u$ and $r_2=1/2+t-u$. Find the Hessian of $\mathcal F=f(r_1)+f(r_2)$ at the physical point and infer $\tau$ in this flavor basis. If the second flavor generator is replaced by three times itself, does the location or physical meaning of the maximum change?

Hint 1

Each second derivative contains two factors of the corresponding flavor charge.

Hint 2

Write the two charge vectors as columns of a matrix and use (O05.2). A new generator normalization requires a corresponding change in its mixing coordinate.

Solution

The charge matrix is $A=\left(\begin{smallmatrix}1&1\\1&-1\end{smallmatrix}\right)$. The free-field first derivatives vanish at $t=u=0$, and

$$H=-\frac{\pi^2}{2}A^{\mathsf T}A =-\pi^2 I_2,\qquad \tau=2I_2.$$

The off-diagonal cancellation is a check: the two fields have opposite products of the two flavor charges. Replacing the second generator by three times itself replaces $u$ by $3u'$ in the charge formula. The Hessian in coordinates $(t,u')$ is $-\pi^2\operatorname{diag}(1,9)$, and $\tau'=\operatorname{diag}(2,18)$. The physical current and maximum remain $r_1=r_2=1/2$. Hessian entries depend on charge normalization; negative definiteness and the selected current do not.

2. Extract a fluctuation from the interacting extremum

In the SQED example, use the topological direction to derive the exact variance of $s$ at the conformal point. Predict the quadratic response to a small physical real FI parameter $\eta$ at fixed $r=1/3$. Explain why this variance differs from $1/4$ in the preceding lesson’s $r=1/2$ example.

Hint 1

Compare the $tt$ entry in (O05.11) with the normalized integral identity (O05.8).

Hint 2

For real FI, the exponent is $2\pi i\eta s$. Its second derivative inserts $(2\pi i s)^2$, not $(2\pi s)^2$.

Solution

Reflection symmetry gives $\langle s\rangle=0$. Equating $-4\pi^2\langle s^2\rangle=2c$ yields

$$\langle s^2\rangle =-\frac{c}{2\pi^2} =\frac29-\frac1{2\pi\sqrt3}>0.$$

For the physical source, $Z(\eta)/Z(0)=1-2\pi^2\eta^2\langle s^2\rangle+O(\eta^4)$. Consequently

$$\mathcal F(\eta)=\mathcal F_*+ \left(\frac{4\pi^2}{9}-\frac{\pi}{\sqrt3}\right)\eta^2 +O(\eta^4).$$

The coefficient is $-c>0$, the opposite curvature to imaginary FI mixing. Both $s$ and $\eta$ are dimensionless; the physical FI-source mass is proportional to $\eta/\ell$ in the preceding convention.

At $r=1/2$ the localized weight was proportional to $\operatorname{sech}(\pi s)$, giving variance $1/4$. At $r=1/3$ it is a different weight, constructed with the superconformal R-current. This is a change of the prescribed observable, not an inconsistency of the two integrals. A Cartan moment should not be relabeled as an ordinary local-field two-point function without the source and Ward-identity dictionary.

3. Diagnose an incomplete mixing family

Consider a weakly coupled three-dimensional effective theory of two gauge-neutral chirals $X,Y$, near its zero-field vacuum, with $W=\lambda X^2Y^3$. Tune away relevant deformations and assume the effective description remains in the basin of the free fixed point below its cutoff. Use canonical free-field dimensions to determine whether $\lambda$ survives in the infrared. For nonzero fixed $\lambda$, what R-charge constraint and continuous flavor generator remain? Can that trial family contain the free infrared superconformal current?

Hint 1

The superpotential has mass dimension two in three dimensions; a canonically normalized free chiral has dimension $1/2$.

Hint 2

Test both the proposed free values and the scalar unitarity inequalities against the nonzero-coupling R-charge constraint. Then reconsider that constraint at the endpoint of the flow.

Solution

The monomial has dimension $5/2$, so $[\lambda]=-1/2$. The dimensionless coupling $\widehat\lambda(E)=\lambda E^{1/2}$ tends to zero as the energy $E$ tends to zero in the stated weak-coupling regime. This is an effective-theory argument, not a proposed interacting ultraviolet completion.

At nonzero fixed coupling the constraint is $2r_X+3r_Y=2$. A surviving flavor generator gives charges $(3,-2)$, since $2(3)+3(-2)=0$. The free infrared values $(1/2,1/2)$ instead give $2r_X+3r_Y=5/2$. They are absent from that restricted family.

Indeed, both $X,Y$ are gauge-invariant chirals. Imposing $r_X,r_Y\ge1/2$ at a proposed unitary fixed point would already make the left-hand side at least $5/2$. Thus the assumption of a fixed point with this nonzero superpotential and both fields present as such chirals is incompatible with those bounds.

At the free endpoint the interaction vanishes, and the independent phase rotations of $X$ and $Y$ are restored. The enlarged mixing space has no superpotential constraint. Its free energy is $f(r_X)+f(r_Y)$, maximized at $(1/2,1/2)$ with value $\log2$ and Hessian $-\pi^2 I_2/2$. We changed the actual infrared symmetry and interaction data; we did not repair an incorrect ultraviolet extremum by clamping a charge by hand.

The practical exit calculation is to list the conserved mixing directions, impose only the interactions that survive at the fixed point, and test the resulting stationary point in every independent direction. Source derivatives and contact terms explains how to interpret the complex response while keeping its allowed local ambiguity explicit.