R-symmetry, flavor symmetry, and gauge symmetry

A phase rotation can act on a supersymmetric theory in several different ways. It may rotate every field in a multiplet by the same phase, rotate the supercharges themselves, or express a local redundancy of the field description. These possibilities lead to different charge assignments and different constraints on interactions.

Starting from the free chiral theory, we will assign charges to its components and test a cubic interaction. We then promote a phase angle to a function of position and calculate why an ordinary derivative stops transforming covariantly.

Our main example is the massless, gauge-neutral chiral multiplet in flat three-dimensional Euclidean space, with all background sources and the central charge set to zero. Its full $\mathcal N=2$ algebra is the complexification of four real Lorentzian supercharges. The charge arguments use the classical action and transformation laws. A supersymmetric gauge completion will be a later construction.

The first distinction is how a symmetry acts on supercharges

Let $\mathsf T$ be the generator of an ordinary continuous flavor symmetry. “Flavor” here means an internal global symmetry that commutes with the supercharges:

$$[\mathsf T,Q_\alpha]=[\mathsf T,\widetilde Q_\alpha]=0.$$

An R-symmetry instead rotates supercharges. In our normalization its generator obeys

$$[\mathsf R,Q_\alpha]=-Q_\alpha, \qquad [\mathsf R,\widetilde Q_\alpha]=+\widetilde Q_\alpha.$$

These assignments preserve the supersymmetry algebra: the mixed anticommutator has net R-charge zero, as do translations. This is the convention stated after equation (1.2) of the three-manifold paper. The role of the R-rotation as a symmetry of the supercharge algebra is also described after Aharony and collaborators’ equation (2.1).

For a field $X$ of charge $q_X$, a constant phase rotation is $X\mapsto e^{i\beta q_X}X$. We use the same charge notation for the generator acting on the field, $\mathsf T X=q_X X$. The formulas below concern infinitesimal generators; a compact group’s global period and charge lattice are additional data.

Suppose the scalar has $\mathsf R\phi=r\phi$ and $\mathsf T\phi=q\phi$. Acting with a component of $Q$ lowers its R-charge:

$$\mathsf R(Q\phi) =[\mathsf R,Q]\phi+Q(\mathsf R\phi) =(r-1)Q\phi.$$

The resulting component is a fermion. A second action gives the auxiliary component, with charge $r-2$. By contrast,

$$\mathsf T(Q\phi)=Q(\mathsf T\phi)=q\,Q\phi,$$

so the flavor charge does not change. The chiral multiplet therefore has

Component	R-charge	Flavor charge
$\phi$	$r$	$q$
$\psi_\alpha$	$r-1$	$q$
$F_\Phi$	$r-2$	$q$

The independent anti-chiral components carry the opposite charges: $(-r,1-r,2-r)$ for R-symmetry and $(-q,-q,-q)$ for flavor. This is a representation assignment; no Euclidean dagger condition is required.

The commuting labels in the component rules carry compensating charges when we transform the entire parameterized formula:

$$R(\zeta)=+1,\qquad R(\widetilde\zeta)=-1, \qquad T(\zeta)=T(\widetilde\zeta)=0.$$

For example, $\delta\phi=\sqrt2\,\zeta\psi$ has R-charge $1+(r-1)=r$ and flavor charge $q$. In $\delta\psi$, both $\zeta F_\Phi$ and $\gamma^\mu\widetilde\zeta\,\partial_\mu\phi$ have R-charge $r-1$ and flavor charge $q$. With labels held fixed, $\delta_\zeta$ lowers R-charge by one and $\delta_{\widetilde\zeta}$ raises it by one. Transforming the respective labels makes the complete formula covariant.

Check the free action, then test an interaction

The free Lagrangian is

$$\mathcal L_0= \partial_\mu\widetilde\phi\,\partial_\mu\phi -i\widetilde\psi\gamma^\mu\partial_\mu\psi -\widetilde F_\Phi F_\Phi.$$

A spacetime derivative has zero internal charge. Its three terms have R-charges

$$(-r)+r=0,\qquad (1-r)+(r-1)=0,\qquad (2-r)+(r-2)=0,$$

and flavor charge $-q+q=0$ each. Thus the free action admits the phase assignments above for any $r$ and $q$ at the level of infinitesimal symmetries.

An interaction can restrict this freedom. A superpotential $W(\phi)$ is a holomorphic function of chiral scalar fields: it depends on $\phi$, without independent $\widetilde\phi$ variables. Its supersymmetric interaction uses the highest, auxiliary component of the associated composite chiral multiplet. Since that component’s R-charge is two less than its lowest component’s charge, the selection rules are

$$R(W)=2,\qquad T(W)=0$$

for fixed, uncharged couplings. The F-component construction and its R-charge condition are stated in equation (6.14). The numerical coefficients and complete interaction will be derived in the superpotential lesson; the charge test can already be performed.

For the candidate $W=g\phi^3/3$, the component interaction contains structures of the form $gF_\Phi\phi^2$ and $g\phi(\psi\psi)$. Their relative coefficient is unnecessary for determining their charges:

$$\begin{aligned} R(F_\Phi\phi^2)&=(r-2)+2r=3r-2,\\ R(\phi\psi\psi)&=r+2(r-1)=3r-2,\\ T(F_\Phi\phi^2)&=T(\phi\psi\psi)=3q. \end{aligned}$$

The scalar fermion contraction $\psi\psi$ is nonzero for odd spinors; in our convention it is $-2\psi_1\psi_2$. A fixed neutral $g\ne0$ therefore permits the R-assignment

$$r=\frac23,\qquad R(\psi)=-\frac13,\qquad R(F_\Phi)=-\frac43.$$

But the free multiplet’s flavor rotation with $q=1$ fails the continuous flavor test: the interaction carries charge three. It retains the discrete phases $e^{i\beta}$ satisfying $e^{3i\beta}=1$. With the flavor angle normalized to period $2\pi$, these form $\mathbb Z_3$.

There is a separate dimensional check. With the free engineering dimensions $[\phi]=1/2$, $[\psi]=1$, and $[F_\Phi]=3/2$, both interaction structures have dimension $5/2$. Thus $[g]=1/2$ makes the Lagrangian have dimension three. The allowed R-charge $2/3$ is not the free engineering dimension $1/2$. We have determined a symmetry assignment, without calculating an interacting scaling dimension.

Why the R-generator can mix with a flavor generator

If both $\mathsf R_0$ and $\mathsf T$ are symmetries, then for a real coefficient $t$ the infinitesimal generator

$$\mathsf R_t=\mathsf R_0+t\mathsf T$$

still gives $[\mathsf R_t,Q]=-Q$, because $[\mathsf T,Q]=0$. Its scalar charge is

$$r_t=r_0+tq,$$

and its fermion and auxiliary charges are $r_t-1$ and $r_t-2$. This freedom is called R-symmetry mixing. Whether a particular combination has a chosen compact global period is a separate global question.

For the free chiral with $q=1$, the cubic interaction selects the combination $t=2/3-r_0$. After that interaction is added, the continuous flavor symmetry used for arbitrary further shifts is broken. The symmetry constraints must always be checked for the action actually under discussion.

A superconformal theory has a distinguished R-generator in its superconformal algebra. That additional role is not determined by the statement $[\mathsf R,Q]=-Q$ alone. For a free three-dimensional chiral, its scalar charge is $1/2$, as specified in the text following equation (6.31). Choosing other R-generators of the flat free theory does not change its free scaling dimensions.

The presence of a supersymmetry algebra also does not guarantee that every algebra rotation is a symmetry of every action. An interaction can preserve supersymmetry while breaking a candidate continuous R-symmetry. Exercise 2 makes this distinction concrete.

A local phase needs a connection

Now test a different operation: replace a constant phase by an arbitrary smooth function $\alpha(x)$ on a coordinate patch,

$$\phi'(x)=e^{iq_g\alpha(x)}\phi(x).$$

The subscript on $q_g$ distinguishes a proposed gauge charge from the global flavor charge above. In a compact $U(1)$ example we may normalize a minimally charged field to have $q_g=1$. All components of a matter multiplet have the same ordinary gauge charge.

The ordinary derivative transforms as

$$\partial_\mu\phi' =e^{iq_g\alpha} \left(\partial_\mu\phi+iq_g(\partial_\mu\alpha)\phi\right).$$

The extra term prevents the derivative from transforming like the field. Introduce a connection $a_\mu$, with the gauge coupling absorbed into its normalization, and define

$$D_\mu\phi=(\partial_\mu-iq_ga_\mu)\phi, \qquad a_\mu'=a_\mu+\partial_\mu\alpha.$$

Then the unwanted terms cancel:

$$\begin{aligned} D_\mu'\phi' &=(\partial_\mu-iq_ga_\mu-iq_g\partial_\mu\alpha) (e^{iq_g\alpha}\phi)\\ &=e^{iq_g\alpha}D_\mu\phi. \end{aligned}$$

This is the sign convention of the residual matter transformations and covariant derivative in equations (6.22)–(6.24). The Abelian field strength $f_{\mu\nu}=\partial_\mu a_\nu-\partial_\nu a_\mu$ is unchanged because partial derivatives commute.

In a gauge theory, local gauge transformations supported away from the boundary describe the same physical configuration. They are redundancies of the description; a charged elementary field by itself is not a gauge-invariant local observable. Neutral products such as $\widetilde\phi\phi$ are candidates for such observables. Boundary conditions can give transformations at the boundary a different role, so the statement about redundancy includes the indicated support condition.

Replacing derivatives by $D_\mu$ establishes gauge covariance. It is not a complete construction of a supersymmetric gauge theory: the connection belongs to a vector multiplet, and additional fields and interactions are needed. For this Abelian example the gauge group would have rank one; that rank is unrelated to the label $\mathcal N=2$.

A background connection does not itself make the symmetry dynamical

A global flavor symmetry can also be coupled to a prescribed background connection. Transforming the charged fields and the source together makes their coupling covariant under local changes of description. At a fixed zero source, the constant phase rotation remains the original global flavor symmetry.

The background is an input, not a field integrated over. Making a gauge connection dynamical changes the theory: one supplies its dynamics and integrates over its gauge-equivalence classes, together with its supersymmetric partners. The paper explicitly distinguishes dynamical vector multiplets from sources for global flavor symmetries in the paragraph preceding equation (6.26).

Maintaining rigid supersymmetry also restricts the allowed background multiplet configurations; an arbitrary prescribed $a_\mu(x)$ need not preserve the original supercharges. Background sources for global symmetries are discussed from another perspective in Aharony and collaborators, section 2.3. We will use this distinction when coupling supersymmetric theories to curved backgrounds.

Question	R-symmetry	Flavor symmetry	Ordinary gauge symmetry
Does its generator rotate supercharges?	Yes	No	Supercharges are gauge neutral
Chiral component charges	$(r,r-1,r-2)$	$(q,q,q)$	$(q_g,q_g,q_g)$
Role in this lesson	Global symmetry with a nontrivial action on supersymmetry	Global internal symmetry	Local redundancy once a gauge theory is specified
Condition on a superpotential	R-charge two	Flavor neutral	Gauge invariant

The last row applies with fixed invariant couplings. Treating a coupling as a transforming external quantity has a different meaning, examined below.

Exercises

1. Two multiplets and two candidate terms

Take two gauge-neutral chiral multiplets with scalars $X,Y$, and candidate superpotential

$$W=mXY+\lambda X^3,$$

where both couplings are fixed, nonzero, and uncharged. Find the continuous R-charge assignments and the possible ordinary flavor charges $(q_X,q_Y)$ acting by phases. What changes if $\lambda=0$ while $m\ne0$?

First hint

Each nonzero superpotential monomial must have R-charge two and flavor charge zero.

Second hint

The R-constraints are inhomogeneous, whereas the flavor constraints are homogeneous.

Complete solution

The two R-constraints are $r_X+r_Y=2$ and $3r_X=2$, so $r_X=2/3$ and $r_Y=4/3$. Their fermions have charges $-1/3,1/3$, and their auxiliaries have charges $-4/3,-2/3$.

The flavor constraints are $q_X+q_Y=0$ and $3q_X=0$. Thus both flavor charges vanish for a continuous phase symmetry of this form.

If $\lambda=0$, only the sum constraints remain. The flavor assignment $(q_X,q_Y)=(1,-1)$ generates a continuous symmetry, and the R-family can be written $(r_X,r_Y)=(1+t,1-t)$. This is exactly mixing with that flavor generator. For example, the component term $mF_XY$ then has R-charge $(r_X-2)+r_Y=0$ and flavor charge $1-1=0$, checking the conditions directly.

2. Transforming a coupling is not a symmetry at fixed coupling

For one multiplet, consider

$$W=\frac m2\phi^2+\frac g3\phi^3$$

with fixed nonzero $m,g$. Can both terms preserve a continuous phase R-symmetry when the couplings are neutral? Now formally assign charges to $m,g$ so that the expression transforms covariantly for arbitrary scalar charges $r,q$. Find those assignments and explain what they mean.

First hint

With neutral couplings, the R-condition must hold separately for the quadratic and cubic terms.

Second hint

A transforming coupling must compensate the charge of the field monomial. Holding its numerical value fixed removes that compensation unless its charge vanishes.

Complete solution

Neutral couplings would require $2r=2$ and $3r=2$ simultaneously, which has no solution. This rules out the stated continuous phase R-symmetry; it does not by itself rule out supersymmetry.

Formal covariance can instead be obtained with

$$R(m)=2-2r,\qquad R(g)=2-3r, \qquad T(m)=-2q,\qquad T(g)=-3q.$$

A parameter given such a formal transformation is often called a spurion. These assignments compare actions with transformed couplings. They do not supply a symmetry of one action with fixed nonzero numerical couplings. For example, $r=2/3$ makes $g$ neutral but assigns $R(m)=2/3$; a nonzero fixed mass then breaks that candidate R-symmetry.

For the flavor assignment $q=1$, the quadratic term alone retains a $\mathbb Z_2$ phase subgroup and the cubic term alone a $\mathbb Z_3$ subgroup. With both present, $e^{2i\beta}=e^{3i\beta}=1$ implies $e^{i\beta}=1$, leaving no nontrivial phase subgroup of that flavor rotation. This finite check agrees with the continuous charge test.

3. A constant field in a different gauge

On a coordinate patch, begin with $\phi=v$ constant and $a_\mu=0$. Choose $\alpha(x)=kx^1$ and perform the local transformation with charge $q_g$. Compute $\partial_1\phi'$, $D_1'\phi'$, and $f_{\mu\nu}'$. Does the nonconstant phase create a physical gradient energy? Would prescribing this connection as a source, without integrating over it, already construct a dynamical gauge theory?

First hint

Transform the connection as well as the scalar.

Second hint

The transformed connection is $a_1'=k$, $a_2'=a_3'=0$. Compare the covariant derivative with its initial value.

Complete solution

The transformed scalar is $\phi'=v e^{iq_gkx^1}$, so

$$\partial_1\phi'=iq_gk\phi', \qquad D_1'\phi'=(\partial_1-iq_gk)\phi'=0.$$

All components of $f_{\mu\nu}'$ vanish because the transformed connection is constant. The covariant gradient remains zero, and no physical gradient energy is created in a gauge-covariant kinetic term. The ordinary derivative alone would give a misleading answer.

A prescribed source is still external data. Coupling to it can test a global symmetry and its response; a dynamical gauge theory additionally requires the connection’s dynamics and integration over gauge fields, with the supersymmetric completion when appropriate. The local patch calculation makes no claim about whether a gauge parameter is single-valued on a compact space.

Which symmetry survives the interaction?

You can now identify a charge by more than its name: ask how its generator acts on supercharges, whether its parameter is global or local, and whether a connection is prescribed or dynamical. Then test every term with the couplings held fixed.

For the free chiral, an R-family and a continuous flavor phase are both available. The cubic candidate selects a particular R-charge and breaks that continuous flavor phase.

For the round-sphere route, continue with Frames and spin connections to prepare for curved backgrounds. The component-theory sequence continues with Superpotentials and a first interacting model, which constructs the full interaction and determines its scalar potential; charge balance alone does not establish those dynamics.