The R-multiplet through its physical currents

A supersymmetry multiplet need not be a collection of elementary particles. The stress tensor, the supersymmetry current and an R-current are composite operators. Supersymmetry transforms these operators into one another, so their external probes must also fit together. This is the physical content of the R-multiplet.

Begin with superspace, source variations and stress-tensor improvements. We work in flat three-dimensional Euclidean space, as the continuation of an N=2 theory with four real Lorentzian supercharges and a continuous conserved R-symmetry. Superspace makes the complexified algebra available; it does not double the physical supersymmetry. The general construction does not specify a gauge group. Our concrete matter example will be a free gauge-neutral chiral.

We will first identify the currents, then multiply two explicit superfields to recover their source couplings. The superspace constraints will explain how the resulting operators are related. We are not yet choosing a curved supersymmetric background.

Currents before superfields

Write the infinitesimal metric as $g_{\mu\nu}=\delta_{\mu\nu}+2h_{\mu\nu}$. The Euclidean stress convention is $\delta S=-\frac12\int T^{\mu\nu}\delta g_{\mu\nu}$, so its linear coupling is $-T^{\mu\nu}h_{\mu\nu}$. A conserved R-current $j_R^\mu$ couples to an R-connection. The odd supersymmetry currents $\mathcal S^\mu,\widetilde{\mathcal S}^{\mu}$ couple to odd vector-spinor sources, the gravitini $\Psi_\mu,\widetilde\Psi_\mu$.

Two further operators are needed: a vector $j_Z^\mu$, whose integrated Lorentzian charge can contribute to the central charge $Z$, and a scalar $J_Z$. The latter is not itself a conserved vector current. Its gradient makes the conserved antisymmetric current $\mathcal J_{\rm string}^{\mu\nu}=i\epsilon^{\mu\nu\rho}\partial_\rho J_Z$. Commuting ordinary derivatives proves $\partial_\mu\mathcal J_{\rm string}^{\mu\nu}=0$ identically. The name refers to the kind of extended charge it can describe; it does not assert that a particular string state exists.

Here is the operator basis we will derive, with engineering dimensions appropriate to the source normalization:

Operator	Dimension	Source appearing in the linear coupling
$T^{\mu\nu}$	$3$	$-h_{\mu\nu}$
$\mathcal S^\mu,\widetilde{\mathcal S}^{\mu}$	$5/2$	$-\Psi_\mu/2,\ +\widetilde\Psi_\mu/2$, in displayed contraction order
$j_R^\mu$	$2$	$a_\mu^R=A_\mu-\frac32V_\mu$
$j_Z^\mu$	$3$	$C_\mu$
$J_Z$	$2$	$H$

The source dimensions are respectively $0,1/2,1,0,1$, making each density dimension three. The extra dimension of $j_Z$ reflects that $Z$ has mass dimension one, unlike a dimensionless flavor charge. These are operator dimensions fixed by the current/source assignments; the table is not a particle-state count.

The potentials $C_\mu$ and $B_{\mu\nu}=-B_{\nu\mu}$ define

$$V^\mu=-i\epsilon^{\mu\nu\rho}\partial_\nu C_\rho,\qquad H=\frac i2\epsilon^{\mu\nu\rho}\partial_\mu B_{\nu\rho}, \qquad \epsilon^{123}=1. \tag{R02.1}$$

Thus $\partial_\mu V^\mu=0$ is a field-strength identity. $V$ is not an independently adjustable gauge connection. The table uses $(a^R,C,B,h)$ as independent bosonic source coordinates; holding $A$ fixed instead when varying $C$ will add a curl improvement, calculated in Exercise 1.

For compactly supported variations, $A\mapsto A+\partial\lambda_R$ changes the source action by $\int j_R^\mu\partial_\mu\lambda_R=-\int\lambda_R\partial_\mu j_R^\mu$. Its invariance is the Ward identity for R-current conservation. Similarly, $C\mapsto C+\partial\lambda_C$ tests $\partial_\mu j_Z^\mu=0$, since its curl $V$ does not change. These conservation equations hold on the matter equations of motion, or in quantum correlators away from the contact terms at charged insertions. They are not identities for arbitrary off-shell matter configurations.

Package the operators without losing their meaning

Use the independent odd coordinates $\theta_\alpha,\widetilde\theta_\alpha$ and the spinor contractions of the superspace lesson. Define $v^\mu=\theta\gamma^\mu\widetilde\theta$ and $s=\theta\widetilde\theta$. The vector symbol $v^\mu$ in this calculation is a Grassmann bilinear, not a gauge field. In the measure with $\int d^4\theta\,\theta^2\widetilde\theta^2=1$,

$$v^\mu v^\nu=\frac12\delta^{\mu\nu}\theta^2\widetilde\theta^2,\qquad s^2=-\frac12\theta^2\widetilde\theta^2,\qquad s v^\mu=0. \tag{R02.2}$$

For example, $s=\theta_2\widetilde\theta_1-\theta_1\widetilde\theta_2$ and $s^2=-2\theta_1\theta_2\widetilde\theta_1\widetilde\theta_2$. This checks the minus sign rather than treating the odd coordinates as commuting variables.

The needed part of the R-multiplet expansion is the paper’s (2.2):

$$\begin{aligned} \mathcal R_\mu={}&j_{R\mu} -i\theta\mathcal S_\mu-i\widetilde\theta\widetilde{\mathcal S}_\mu\\ &-v^\nu\left(2T_{\mu\nu} +i\epsilon_{\mu\nu\rho}\partial^\rho J_Z\right)\\ &-is\left(2j_{Z\mu} +i\epsilon_{\mu\nu\rho}\partial^\nu j_R^\rho\right) +\cdots . \end{aligned} \tag{R02.3}$$

The omitted cubic and quartic coefficients are fixed derivatives of the lower ones. For the coupling below they cannot contribute, because the source superfield starts at degree two. Notice that the stress tensor occupies a different quadratic component from the central-charge current. An arbitrary vector superfield with a bottom component named “R-current” would not yet enforce these relations.

Let $\mathcal H_\mu$ denote the metric superfield, distinguished from the scalar source $H$. In Wess–Zumino gauge, where removable lower source components have been set to zero, its expansion is

$$\begin{aligned} \mathcal H_\mu={}& \frac12v^\nu(h_{\mu\nu}+B_{\mu\nu})-\frac i2s C_\mu\\ &-\frac i2\theta^2\widetilde\theta\widetilde\Psi_\mu +\frac i2\widetilde\theta^2\theta\Psi_\mu +\frac12\theta^2\widetilde\theta^2(A_\mu-V_\mu). \end{aligned} \tag{R02.4}$$

This source package and the supersymmetric coupling $\delta\mathcal L=2\int d^4\theta\,\mathcal R^\mu\mathcal H_\mu$ are the linearized-supergravity input from (2.4) and (2.7). We will derive its components; we are not deriving the complete nonlinear gravity theory.

Multiply the packages and identify the source

There are three bosonic products. The bottom of $\mathcal R$ multiplies the top of $\mathcal H$. The two quadratic vector bilinears multiply each other, as do the two scalar bilinears. Using (R02.2), symmetry of $T,h$ and antisymmetry of $B$ gives

$$\begin{aligned} 2\int d^4\theta\,\mathcal R^\mu\mathcal H_\mu \big|_{\rm bosonic} ={}&-T^{\mu\nu}h_{\mu\nu}+j_R^\mu(A_\mu-V_\mu)+j_Z^\mu C_\mu\\ &-\frac i2B^{\mu\nu}\epsilon_{\mu\nu\rho}\partial^\rho J_Z +\frac i2C^\mu\epsilon_{\mu\nu\rho}\partial^\nu j_R^\rho . \end{aligned} \tag{R02.5}$$

The last two terms explain the less familiar sources. Under the spacetime integral, integrate the $B$ term by parts to obtain $\frac i2J_Z\epsilon^{\rho\mu\nu}\partial_\rho B_{\mu\nu}=J_ZH$. For the last term, the curl is formally self-adjoint: $\int C\cdot(\nabla\times j_R)=\int j_R\cdot(\nabla\times C)$. Since $\nabla\times C=iV$, it contributes $-j_R\cdot V/2$.

The odd current/source products must retain their order. Multiplication of the linear current terms by the cubic source terms gives $-\mathcal S^\mu\Psi_\mu/2+\widetilde{\mathcal S}^{\mu}\widetilde\Psi_\mu/2$. The signs include moving the odd current coefficients through the superspace coordinates. The result is

$$\boxed{\displaystyle \delta\mathcal L= -T^{\mu\nu}h_{\mu\nu} -\frac12\mathcal S^\mu\Psi_\mu +\frac12\widetilde{\mathcal S}^{\mu}\widetilde\Psi_\mu +j_R^\mu\left(A_\mu-\frac32V_\mu\right) +j_Z^\mu C_\mu+J_ZH .} \tag{R02.6}$$

This reproduces (2.10), up to the displayed integrations by parts. We require decaying fields, periodic fields or boundary conditions eliminating the surface terms. With a physical boundary, additional boundary contributions must be retained.

In particular, the coefficient $3/2$ came from two different products. It cannot be inferred just by inserting an R-connection into a matter derivative. The full background dictionary uses $A-V/2$ in that derivative. What each background field does explains its separate role; it is not another definition of the source in (R02.6).

What the superspace constraints say

With the two-lower-index gamma matrix, define the symmetric bispinor $\mathcal R_{\alpha\beta}=-2(\gamma^\mu)_{\alpha\beta}\mathcal R_\mu$.

The R-multiplet conditions are

$$\widetilde D^\beta\mathcal R_{\alpha\beta} =-4i\widetilde D_\alpha\mathcal J_Z,\qquad D^\beta\mathcal R_{\alpha\beta}=4iD_\alpha\mathcal J_Z,\qquad D^2\mathcal J_Z=\widetilde D^2\mathcal J_Z=0. \tag{R02.7}$$

Here $D_\alpha,\widetilde D_\alpha$ are the superspace derivatives, not a matter covariant derivative or an auxiliary field. The scalar superfield $\mathcal J_Z$ has bottom component $J_Z$ and begins as

$$\mathcal J_Z= J_Z-\frac12\theta\gamma^\mu\mathcal S_\mu +\frac12\widetilde\theta\gamma^\mu\widetilde{\mathcal S}_\mu +is\,T^\mu{}_\mu-v^\mu j_{Z\mu}+\cdots . \tag{R02.8}$$

Thus its coefficients contain the gamma traces of the supercurrents, the stress-tensor trace and the central-charge current. The equations are the source’s (2.1)–(2.2). “Real linear” describes the Lorentzian reality and the linear differential constraints; Euclidean tilded spinors remain independent.

There is a direct component check. Set the fermionic coefficients to zero temporarily, and complete the bosonic linear polynomial by $+\theta^2\widetilde\theta^2\partial^2J_Z/4$. The left derivatives and gamma matrices of the superspace lesson give

$$D_1D_2\mathcal J_Z =i\widetilde\theta^2\,\partial_\mu j_Z^\mu,\qquad \widetilde D_1\widetilde D_2\mathcal J_Z =-i\theta^2\,\partial_\mu j_Z^\mu. \tag{R02.9}$$

The derivatives of $J_Z$ cancel against its quartic term, while those of $T^\mu{}_\mu$ cancel by the spinor identities. The remaining coefficient is precisely the conservation law. This illustrates what a constraint does, rather than interpreting it as an additional field equation for the sources.

In a superconformal theory one can choose the distinguished R-multiplet with $\mathcal J_Z$ redundant, vanishing as an operator modulo the relevant equations and contact terms. Then the stress trace and supercurrent gamma traces vanish at separated points. Vanishing of the single bottom component $J_Z$ on one field configuration is a much weaker statement.

A free chiral exposes the improvement

Take a massless free chiral with scalar R-charge $r$, no superpotential and no dynamical gauge field. Put $\rho=\widetilde\phi\phi$. The conserved flavor current and the R-current are

$$\begin{aligned} j_\mu&=i(\widetilde\phi\,\partial_\mu\phi -\partial_\mu\widetilde\phi\,\phi) -\widetilde\psi\gamma_\mu\psi,\\ j_{R\mu}&=ir(\widetilde\phi\,\partial_\mu\phi -\partial_\mu\widetilde\phi\,\phi) -(r-1)\widetilde\psi\gamma_\mu\psi. \end{aligned} \tag{R02.10}$$

The fermion’s R-charge is $r-1$, explaining its different coefficient. For the scalar part, $\partial^\mu j_\mu=i(\widetilde\phi\,\partial^2\phi-\partial^2\widetilde\phi\,\phi)$. The two first-derivative products cancel. The scalar equations set this to zero; the two Dirac equations likewise cancel the divergence of the fermion bilinear.

The remaining components of (7.12)–(7.13), at zero mass, are

$$\begin{aligned} T_{\mu\nu}^{(r)} &=T_{\mu\nu}^{(0)} +\frac r2(\delta_{\mu\nu}\partial^2-\partial_\mu\partial_\nu)\rho,\\ J_Z^{(r)}&=-i\left(r-\frac12\right)\widetilde\psi\psi,\qquad j_{Z\mu}^{(r)} =-i\left(r-\frac12\right)\epsilon_{\mu\nu\rho}\partial^\nu j^\rho. \end{aligned} \tag{R02.11}$$

Here $T^{(0)}$ denotes the minimally coupled flat stress tensor, including its fermion contribution. The curl current is conserved identically, even before using flavor-current conservation. Changing $r$ to $r+t$ shifts $j_R$ by $t j$ and changes all three operators in (R02.11). This is the component form of the flavor improvement in (2.3). It ties the stress improvement of the earlier lesson to the choice of R-symmetry.

For the free massless theory, $r=1/2$ makes the complete $\mathcal J_Z$ vanish on shell. This is its superconformal R-charge. It is not a prescription to assign every interacting chiral the free value.

For a concrete local test set $\psi=\widetilde\psi=0$ and $\phi=a(x^1+ix^2)$, $\widetilde\phi=a(x^1-ix^2)$, with real constant $a$ of dimension $3/2$. Both scalars are harmonic. Direct differentiation gives

$$j=(2a^2x^2,-2a^2x^1,0),\qquad j_Z=\left(0,0,4ia^2\left(r-\frac12\right)\right),\qquad T^\mu{}_\mu=2a^2(2r-1). \tag{R02.12}$$

The trace follows from the minimal scalar trace $-2a^2$ and $\partial^2\rho=4a^2$. Meanwhile $J_Z=0$ because the fermions vanish. This explicitly demonstrates that a zero bottom component does not mean a zero multiplet. All displayed currents are conserved on this profile, and both the trace and $j_Z$ vanish at $r=1/2$. The growing profile is a local diagnostic, not a finite-energy state or a calculation of total charges.

Exercises

1. Specify what is held fixed in a source derivative

Using the bosonic terms of (R02.6), determine the operator obtained by varying $C_\mu$ while holding $A_\mu$ fixed. Compare with varying $C_\mu$ while holding $a_\mu^R=A_\mu-3V_\mu/2$ fixed. Does the difference change the integrated central charge when the improvement surface term vanishes?

Hint 1

$V=-i\nabla\times C$. The dependence on $C$ occurs both explicitly and through $V$ when $A$ is fixed.

Hint 2

Move the curl from $\delta C$ onto $j_R$ by integration by parts. For the second variation, vary $A$ as well to keep $a^R$ fixed.

Solution

At fixed $A$,

$$\delta_C S=\int d^3x\,\delta C_\mu \left[j_Z^\mu+\frac{3i}{2}\epsilon^{\mu\nu\rho} \partial_\nu j_{R\rho}\right].$$

At fixed $a^R$, $\delta A=3\delta V/2$ cancels that additional term, leaving $\delta_C S=\int j_Z^\mu\delta C_\mu$. Both statements refer to the same functional written in different independent source coordinates.

The difference is a conserved curl. After continuation to a time slice, its contribution to the charge is a spatial boundary term; it vanishes for the stated decay or periodic conditions. With a boundary or nontrivial patching, that conclusion requires an additional boundary analysis. The operator dictionary has not changed the physical central charge by an arbitrary local redefinition.

As a sign check, on a periodic interval take $C_3=\cos x^1$ and $j_{R2}=\sin x^1$, with unit transverse volume and other components zero. Then $V_2=-i\sin x^1$. The integral of $-3j_R\cdot V/2$ is $3i\pi/2$, equal to that of $C\cdot(3i\,\nabla\times j_R/2)$. A minus sign in the curl improvement would fail this nonzero test.

2. Recover the curvature coupling selected by the improvement

Shift the free chiral charge by $r\mapsto r+t$. Show that the corresponding stress improvement is identically conserved and determine its linear metric-source interaction. Use the paper’s curvature sign, $R_P^{(1)}=2(\partial^2 h^\mu{}_\mu-\partial_\mu\partial_\nu h^{\mu\nu})$.

Hint 1

The shift is $\Delta T_{\mu\nu}=\frac t2(\delta_{\mu\nu}\partial^2-\partial_\mu\partial_\nu)\rho$. Differentiate before using any matter equation.

Hint 2

In $-\int h^{\mu\nu}\Delta T_{\mu\nu}$, integrate both derivatives onto $h$. Use the factor two in the definition of $h$.

Solution

Commuting derivatives gives $\partial^\mu\Delta T_{\mu\nu}=\frac t2(\partial_\nu\partial^2-\partial^2\partial_\nu)\rho=0$. Under the same boundary assumptions as the main calculation,

$$-\int h^{\mu\nu}\Delta T_{\mu\nu} =-\frac t2\int\rho (\partial^2h^\mu{}_\mu-\partial_\mu\partial_\nu h^{\mu\nu}) =-\frac t4\int R_P^{(1)}\rho.$$

The curved coupling therefore shifts by $-tR_P\rho/4$ at this order. This is precisely the linearization of the curvature coefficient established in the stress-tensor lesson. It is part of the same R-improvement that shifted $j_R$, $J_Z$ and $j_Z$; it is not an independent choice of metric coupling once that current multiplet is fixed. This calculation establishes a first-order match, not an entire finite-background supersymmetric action.

3. A mass cannot be removed by choosing an R-charge

For the same free chiral, now supply a nonzero real mass $m_{\rm real}$. In the paper’s conventions its central-charge eigenvalue is $z=-m_{\rm real}$. The relevant components from (7.13) become

$$\begin{aligned} J_Z&=-m_{\rm real}\rho+ \left(r-\frac12\right) (2m_{\rm real}\rho-i\widetilde\psi\psi),\\ j_Z^\mu&=-m_{\rm real}j^\mu -i\left(r-\frac12\right)\epsilon^{\mu\nu\rho}\partial_\nu j_\rho. \end{aligned}$$

Can a constant choice of $r$ remove $\mathcal J_Z$? What central charge follows from the displayed current if its curl surface term vanishes?

Hint 1

To remove even the bottom operator $J_Z$ for arbitrary fields, the independent scalar and fermion bilinear coefficients must both vanish.

Hint 2

Eliminating the fermion coefficient fixes $r=1/2$. Then inspect the scalar coefficient. Integrating the vector current removes only the curl term under the stated boundary assumption.

Solution

The fermion bilinear coefficient vanishes only at $r=1/2$, where $J_Z=-m_{\rm real}\rho$ remains. Alternatively $r=1$ cancels the scalar bilinear but leaves $-i\widetilde\psi\psi/2$. No constant $r$ removes both for nonzero mass, so it cannot remove the whole multiplet. A relevant mass deformation has not become conformal by changing an improvement.

If $Q_{\rm fl}$ denotes the flavor charge with the displayed current normalization, the integrated relation is $Z=-m_{\rm real}Q_{\rm fl}$. Its dimension is one, and the unit flavor-charge field has $z=-m_{\rm real}$, agreeing with the given dictionary. Boundary terms or additional topological sectors require their own treatment; the local curl does not license dropping them without checking.

The R-multiplet now has a concrete meaning: its components are physical responses, its constraints relate their conservation and traces, and its improvements change a whole supersymmetric set of operators. Supergravity as background sources explains how to freeze those probes. After the background-field dictionary, the Killing-equation reconstruction connects the linear source multiplet to the rigid spinor equations. The separate flat-space consistency lesson will use these operators to test supersymmetric deformations; that later test is stronger than merely recognizing a source coupling.

Read Section 2 alongside the paper guide to locate these component products in the original source and follow them into the gravitino variation.

Paper guides

Connect this lesson to the original papers and their notation.

• Reading Supersymmetric Field Theories on Three-Manifolds
• Paper guide: Section 2, currents and rigid backgrounds