The vector multiplet

A three-dimensional supersymmetric Yang–Mills theory contains a gauge field, a gaugino, a real scalar, and an auxiliary field. We will identify their roles, calculate how two supersymmetry operations act on the gauge potential, and see why the same field list can describe quite different dynamics.

The preceding interaction lesson established a matter theory. Adding gauge symmetry is a further choice: the interactions must respect the symmetry being gauged. In particular, its generic quadratic-plus-cubic superpotential for one chiral field cannot simply be assigned a nonzero gauge charge. Here we first construct the gauge sector and test its coupling to a charged matter field. A complete interacting gauge theory must pass both tests.

Our physical parent is three-dimensional Lorentzian $\mathcal N=2$ supersymmetry with four real Poincaré supercharges. The first count concerns a free, massless Yang–Mills multiplet. We then use flat Euclidean space and the spinor conventions of the previous lessons for component calculations. The smallest gauge group is $U(1)$; it has one generator. There is no curved background in these calculations.

First calculation: count the propagating fields

Write the multiplet as

$$(a_\mu,\ \sigma,\ \lambda_\alpha,\ \widetilde\lambda_\alpha,\ D).$$

Here $a_\mu$ is the dynamical gauge potential, $\sigma$ is a real scalar in the Lorentzian theory, $\lambda$ is a complex two-component fermion called the gaugino, and $D$ is a real Lorentzian auxiliary scalar. The tilded gaugino is its Lorentzian partner, not another independent complex Lorentzian fermion. After complexification for Euclidean calculations, $\lambda$ and $\widetilde\lambda$ are independent.

Consider a nonzero null momentum in Lorentzian signature $(+--)$:

$$p^\mu=(E,E,0),\qquad E>0.$$

A Maxwell plane wave has polarization $\varepsilon^\mu$ satisfying $p_\mu\varepsilon^\mu=0$, with the gauge identification $\varepsilon^\mu\sim\varepsilon^\mu+c\,p^\mu$. The constraint leaves two components; the gauge direction lies inside that constrained space because $p^2=0$, and removes one more. Thus a three-dimensional massless photon has one physical polarization.

The real scalar supplies one additional bosonic mode. The massless two-by-two Dirac operator at nonzero null momentum has rank one, leaving one complex polarization for the gaugino, or two real fermionic modes. The auxiliary $D$ has no kinetic term and contributes no propagating mode:

Field	On-shell contribution per gauge generator
Maxwell field $a_\mu$	One real bosonic mode
Real scalar $\sigma$	One real bosonic mode
Complex gaugino, with its Lorentzian partner	Two real fermionic modes
Auxiliary $D$	None

The result is $2$ bosonic and $2$ fermionic modes. The scalar also has a simple reduction interpretation: suppressing dependence on one spatial coordinate of a four-dimensional vector leaves its extra gauge-field component as a three-dimensional scalar. This is a classical component reduction, not a calculation of finite-radius quantum effects. Aharony et al., section 2.2.

Off shell, the bookkeeping is different. Locally, before equations of motion, $a_\mu$ has $3-1=2$ components modulo gauge transformations; adding $\sigma,D$ gives $4$ real bosonic components. The Lorentzian complex two-component gaugino has $4$ real components. This $4/4$ count is about field functions modulo gauge freedom, not physical particle polarizations. For a group $G$, the perturbative massless count is per generator and multiplies by $\dim G$, not by its rank.

Fix the fields, charges, and normalization

Use lowercase $a_\mu$ for the dynamical gauge potential. The uppercase $A_\mu$ used in the three-manifold paper is a fixed background connection for R-symmetry. They are different fields. In our flat calculation all those fixed background sources vanish.

For the Abelian gauge transformation, with local parameter $\Lambda(x)$,

$$a_\mu\longmapsto a_\mu+\partial_\mu\Lambda,\qquad f_{\mu\nu}=\partial_\mu a_\nu-\partial_\nu a_\mu.$$

The other Abelian vector components are gauge neutral. The component fields above are what remain after a conventional partial gauge choice called Wess–Zumino gauge. Supersymmetry must be accompanied by a gauge adjustment to stay in that choice; this will explain a gauge transformation in the closure calculation. No superspace expansion is needed here. P, (6.18)–(6.20).

We absorb the gauge coupling into the connection appearing in a covariant derivative, and put the coupling in front of the kinetic action. With $g_{\rm YM}>0$ and $g_{\rm YM}=e$ in P’s notation,

$$\mathcal L_{\rm YM} =\frac1{g_{\rm YM}^2} \left[ \frac14f_{\mu\nu}f_{\mu\nu} +\frac12\partial_\mu\sigma\,\partial_\mu\sigma -i\widetilde\lambda\gamma^\mu\partial_\mu\lambda -\frac12D^2 \right].$$

This is the flat Abelian specialization of P, (6.30). The engineering dimensions and R-charges are

Object	Mass dimension	R-charge
$a_\mu,\sigma$	$1$	$0$
$\lambda,\widetilde\lambda$	$3/2$	$+1,-1$
$D$	$2$	$0$
$g_{\rm YM}^2$	$1$	$0$

Every term in $\mathcal L_{\rm YM}$ has dimension three. For example, $[f^2/g_{\rm YM}^2]=4-1=3$. Canonically normalized free fields are $a/g_{\rm YM}$, $\sigma/g_{\rm YM}$, and $\lambda/g_{\rm YM}$; their dimensions are respectively $1/2,1/2,1$. The two sets of dimensions describe different field normalizations.

The negative Euclidean $D^2$ coefficient requires care, just as the chiral auxiliary term did. The algebra uses complexified fields. For the free auxiliary Gaussian a decaying contour is $D=ih$ with real $h$, giving $+h^2/(2g_{\rm YM}^2)$. This contour choice does not redefine the Lorentzian auxiliary as imaginary, and it is not a conjugation condition on the Euclidean gauginos.

Transform the vector multiplet

The constant lower-column spinors $\zeta,\widetilde\zeta$ are commuting labels. They multiply odd operations $\delta_\zeta,\delta_{\widetilde\zeta}$, so products of fields obey the graded Leibniz rule. A physical even variation would include an additional odd parameter, as in the algebra lesson.

Define the dual field strength

$$b^\rho=\frac12\epsilon^{\rho\mu\nu}f_{\mu\nu},\qquad \epsilon^{123}=1.$$

The flat specialization of P’s transformations is

$$\begin{aligned} \delta a_\mu&=-i(\zeta\gamma_\mu\widetilde\lambda +\widetilde\zeta\gamma_\mu\lambda),\\ \delta\sigma&=-\zeta\widetilde\lambda+\widetilde\zeta\lambda,\\ \delta\lambda&=i\zeta D-i\gamma^\mu\zeta\,b_\mu -i\gamma^\mu\zeta\,\partial_\mu\sigma,\\ \delta\widetilde\lambda&=-i\widetilde\zeta D -i\gamma^\mu\widetilde\zeta\,b_\mu +i\gamma^\mu\widetilde\zeta\,\partial_\mu\sigma,\\ \delta D&=\zeta\gamma^\mu\partial_\mu\widetilde\lambda -\widetilde\zeta\gamma^\mu\partial_\mu\lambda . \end{aligned}$$

Here $\delta=\delta_\zeta+\delta_{\widetilde\zeta}$; retain only one label to obtain one family. Spinor products mean $uv=u^TCv$ with $C=\left(\begin{smallmatrix}0&-1\\1&0\end{smallmatrix}\right)$. These rules have no central-charge parameter: the gauge multiplet itself has zero central charge. P, (6.20).

The three terms in $\delta\lambda$ distinguish three bosonic data: the auxiliary, the gauge curvature, and the scalar gradient. Their coefficients can be checked by closure; replacing $\sigma$ by a constant mass parameter would discard one of those required variations.

Closure displays a gauge transformation

Define the commuting-label bilinears $s=\zeta\widetilde\zeta$ and $K^\mu=\zeta\gamma^\mu\widetilde\zeta$. The scalar contraction is antisymmetric, while the vector contraction is symmetric:

$$\widetilde\zeta\zeta=-s,\qquad \widetilde\zeta\gamma^\mu\zeta=K^\mu.$$

Apply both operations to $\sigma$:

$$\begin{aligned} \{\delta_\zeta,\delta_{\widetilde\zeta}\}\sigma &=-\zeta\,\delta_{\widetilde\zeta}\widetilde\lambda +\widetilde\zeta\,\delta_\zeta\lambda\\ &=i(s-s)D+i(K^\mu-K^\mu)b_\mu -2iK^\mu\partial_\mu\sigma\\ &=-2iK^\mu\partial_\mu\sigma. \end{aligned}$$

Both cancellations occur before using a field equation. For the gauge field, insert the gaugino transformations into

$$\{\delta_\zeta,\delta_{\widetilde\zeta}\}a_\mu =-i\zeta\gamma_\mu\delta_{\widetilde\zeta}\widetilde\lambda -i\widetilde\zeta\gamma_\mu\delta_\zeta\lambda.$$

Using $\gamma_\mu\gamma_\nu=\delta_{\mu\nu} +i\epsilon_{\mu\nu\rho}\gamma^\rho$, the sum and difference of the required contractions are

$$\begin{aligned} \zeta\gamma_\mu\gamma_\nu\widetilde\zeta +\widetilde\zeta\gamma_\mu\gamma_\nu\zeta &=2i\epsilon_{\mu\nu\rho}K^\rho,\\ \zeta\gamma_\mu\gamma_\nu\widetilde\zeta -\widetilde\zeta\gamma_\mu\gamma_\nu\zeta &=2s\delta_{\mu\nu}. \end{aligned}$$

The $D$ terms cancel again. The field-strength and scalar-gradient terms give

$$\begin{aligned} \{\delta_\zeta,\delta_{\widetilde\zeta}\}a_\mu &=-2iK^\nu f_{\nu\mu}+2s\,\partial_\mu\sigma\\ &=-2iK^\nu\partial_\nu a_\mu+\partial_\mu\Lambda_{\rm cl},\\ \Lambda_{\rm cl}&=2iK^\nu a_\nu+2s\sigma. \end{aligned}$$

The first term is a translation. The second is precisely a gauge transformation, now with a field-dependent parameter $\Lambda_{\rm cl}$. Closure on a gauge potential is allowed to have this form. Gauge-invariant field strengths only see the translation because the curl of $\partial_\mu\Lambda_{\rm cl}$ vanishes.

This calculation verifies the bosonic part of the off-shell algebra. The full free Abelian rules also close on $\lambda,\widetilde\lambda,D$, and each same-family operation squares to zero; the role of $D$ is to permit these statements without imposing the Dirac equation. A component count alone would not have checked their signs.

A worked configuration

Choose Cartesian coordinates $x^1,x^2,x^3$ and

$$a_2=B x^1,\quad a_1=a_3=0,\quad \sigma=c x^1,\quad D=d,\quad \lambda=\widetilde\lambda=0,$$

where $B,c,d$ are constants. Take $\zeta=(1,0)^T$ and $\widetilde\zeta=(0,1)^T$. Then $b_3=B$, $s=-1$, and $K=(1,0,0)$ in our gamma basis. Direct substitution gives

$$\delta_\zeta\lambda= \begin{pmatrix}i(d-c)\\-B\end{pmatrix}, \qquad \delta_{\widetilde\zeta}\widetilde\lambda= \begin{pmatrix}B\\-i(d+c)\end{pmatrix}.$$

The initial bosonic variations vanish because the initial gauginos vanish. Their second variations need not vanish: the first operation has produced gauginos. Indeed,

$$\{\delta_\zeta,\delta_{\widetilde\zeta}\}\sigma=-2ic,\qquad \{\delta_\zeta,\delta_{\widetilde\zeta}\}(a_1,a_2,a_3) =(-2c,-2iB,0).$$

Translation accounts for the $a_2$ entry. Since $\Lambda_{\rm cl}=-2cx^1$, its gradient accounts for $a_1$. The result is independent of $d$, as the off-shell cancellation predicted. This local polynomial profile checks the algebra; it is not assumed to solve the equations of motion or to have finite total action.

Couple a charged matter field covariantly

Let $\phi$ have dimensionless Abelian charge $q$, so $\phi'=e^{iq\Lambda}\phi$. Its derivative is

$$\mathscr D_\mu\phi=(\partial_\mu-iqa_\mu)\phi.$$

Check it before using it:

$$\begin{aligned} \mathscr D'_\mu\phi' &=\big[\partial_\mu-iq(a_\mu+\partial_\mu\Lambda)\big] (e^{iq\Lambda}\phi)\\ &=e^{iq\Lambda} \big[\partial_\mu\phi+iq(\partial_\mu\Lambda)\phi -iqa_\mu\phi-iq(\partial_\mu\Lambda)\phi\big]\\ &=e^{iq\Lambda}\mathscr D_\mu\phi . \end{aligned}$$

The inhomogeneous terms cancel. The opposite-charge field uses $\mathscr D_\mu\widetilde\phi=(\partial_\mu+iqa_\mu)\widetilde\phi$, so $\mathscr D_\mu\widetilde\phi\,\mathscr D_\mu\phi$ is gauge invariant.

Gauge covariance of a derivative is necessary but does not construct the whole supersymmetric matter coupling. For example, P’s charged fermion transformation contains

$$\delta\psi=\sqrt2\zeta F_\Phi -\sqrt2i(z-q\sigma)\widetilde\zeta\,\phi -\sqrt2i\gamma^\mu\widetilde\zeta\,\mathscr D_\mu\phi .$$

Here $z$ is a fixed eigenvalue of the central-charge generator; $\sigma(x)$ is a dynamical field. We may set $z=0$ and still have a nonzero $\sigma$ coupling. Applying the two operations to $\phi$ gives the useful consistency check

$$\begin{aligned} \{\delta_\zeta,\delta_{\widetilde\zeta}\}\phi &=-2i\big[K^\mu\mathscr D_\mu\phi+s(z-q\sigma)\phi\big]\\ &=-2iK^\mu\partial_\mu\phi +iq\Lambda_{\rm cl}\phi-2isz\,\phi . \end{aligned}$$

The gauge parameter is exactly the one found from $a_\mu$; the remaining central-charge action is separate. A constant scalar in a fixed flavor-source multiplet can encode a real mass. That restricted background use does not identify a general dynamical $\sigma(x)$ with the central operator. P, (6.23)–(6.24) and the discussion following (6.32).

For matrix-valued fields, the compatible Hermitian convention is $a_\mu=a_\mu^AT^A$, $\operatorname{Tr}_N(T^AT^B)=\delta^{AB}/2$ for the ordinary fundamental trace of $SU(N)$, with

$$f_{\mu\nu}=\partial_\mu a_\nu-\partial_\nu a_\mu-i[a_\mu,a_\nu], \qquad \mathscr D_\mu\sigma=\partial_\mu\sigma-i[a_\mu,\sigma].$$

The bosonic Yang–Mills normalization is then $g_{\rm YM}^{-2}\operatorname{Tr}_N [\tfrac12 f_{\mu\nu}f_{\mu\nu} +(\mathscr D_\mu\sigma)^2-D^2]$, which reproduces the displayed coefficients per generator. In a fundamental representation $\mathscr D=\partial-ia$; on an adjoint it acts by commutator. Non-Abelian supersymmetry also needs commutator interactions. The Abelian component rules above are not a complete non-Abelian action.

The action determines what propagates

The same Abelian fields admit a supersymmetric Chern–Simons term:

$$\mathcal L_{\rm CS} =\frac{k}{4\pi} \left(i\epsilon^{\mu\nu\rho}a_\mu\partial_\nu a_\rho -2D\sigma+2i\lambda\widetilde\lambda\right).$$

This is P, (6.28). The parameter $k$ is dimensionless. Its globally allowed values require the gauge-group and spin data; the following conclusion uses only the local classical equations at $k\ne0$.

For pure Chern–Simons theory, without a Yang–Mills term or matter, varying $D$ sets $\sigma=0$; varying $\sigma$ sets $D=0$; the algebraic gaugino equations set both gauginos to zero. Varying $a_\mu$ gives

$$\epsilon^{\mu\nu\rho}\partial_\nu a_\rho=0, \qquad\text{equivalently}\qquad f_{\mu\nu}=0.$$

On a contractible patch the connection is pure gauge. There are no local propagating bulk oscillators in this pure theory. Global flat connections and boundary degrees of freedom are not excluded by that local statement.

The $2/2$ massless count at the start therefore belongs to the Yang–Mills action, not to every use of the vector-multiplet field list. Adding both actions gives another regime: the Yang–Mills derivatives remain and the excitations can become massive. Supersymmetry, gauge rank, and propagation count are three different pieces of information.

Exercises

1. Count generators rather than rank

Linearize a pure $SU(2)$ Yang–Mills theory about its zero-field vacuum, with no Chern–Simons term. Count its local massless bosonic and fermionic modes. Why does replacing $\dim SU(2)$ by its rank give the wrong answer? How does this differ from a vacuum with a nonzero adjoint scalar that leaves only a $U(1)$ gauge subgroup massless?

First hint

The adjoint has one component for each of the three generators. Linearization about zero does not discard the two non-Cartan gauge fields.

Second hint

Each massless generator has one photon polarization, one scalar mode, and two real gaugino modes. A different scalar vacuum can change which components remain massless.

Complete solution

Since $\dim SU(2)=3$, the massless perturbative count is

$$n_B=3(1+1)=6,\qquad n_F=3(2)=6.$$

The rank is one, the number of independent Cartan generators, not the number of gauge-field components around the unbroken vacuum. Multiplying by the rank would omit two entire massless vector multiplets.

For a generic nonzero adjoint scalar, commutators with its expectation value distinguish the two non-Cartan directions from the commuting Cartan direction. The former become massive; the latter supplies one massless $U(1)$ vector multiplet with $2/2$ modes. This statement counts the massless sector about that different classical vacuum; it does not erase the massive excitations from the theory. Quantum Coulomb-branch dynamics is a separate calculation.

2. Can the preceding superpotential be gauged?

Give a single chiral field a nonzero $U(1)$ charge $q$. For fixed neutral coefficients $m,g$, test $W=m\phi^2/2+g\phi^3/3$ for gauge invariance. Does replacing ordinary derivatives by covariant derivatives repair any failure? Compare a pair $\phi_+,\phi_-$ of charges $+1,-1$ with $W=m\phi_+\phi_-$.

First hint

Under a local phase transformation, the quadratic and cubic monomials acquire phases with charges $2q$ and $3q$.

Second hint

For a fixed gauge redundancy, each nonzero term with a fixed neutral coefficient must be gauge invariant. A cancellation between different powers of an arbitrary field cannot do this.

Complete solution

Infinitesimally,

$$\delta_\Lambda W =iq\Lambda\left(m\phi^2+g\phi^3\right).$$

For nonzero $m,g$ this vanishes for all fields and local $\Lambda$ only if $q=0$. The failure is in a nondifferentiated interaction, so changing derivatives cannot repair it. Even retaining just one of these two monomials with a fixed neutral nonzero coefficient forbids a nonzero continuous $U(1)$ charge for that single field.

For the pair, $\phi_+\phi_-\mapsto e^{i\Lambda}e^{-i\Lambda}\phi_+\phi_-=\phi_+\phi_-$. The mass superpotential passes the gauge-invariance test. Its components still require the supersymmetric gauge couplings; this charge calculation alone does not prove the completed action invariant. Treating a charged coupling as a transforming background would define additional data, not rescue the original fixed-coupling model.

3. Add Yang–Mills and Chern–Simons terms

Keep both displayed Abelian actions, with no matter. Eliminate $D$ algebraically and determine the mass squared of the canonically normalized scalar $\sigma_c=\sigma/g_{\rm YM}$. Explain why the result does not have the pure Chern–Simons propagation count.

First hint

The terms containing $D$ are $-D^2/(2g_{\rm YM}^2)-kD\sigma/(2\pi)$. Vary before substituting, and keep the sign of the Euclidean auxiliary quadratic term.

Second hint

After substitution, compare the scalar potential with $\tfrac12m_\sigma^2\sigma_c^2$. The kinetic term for $\sigma_c$ is still present.

Complete solution

The auxiliary equation gives

$$-\frac{D}{g_{\rm YM}^2}-\frac{k\sigma}{2\pi}=0, \qquad D_*=-\frac{k g_{\rm YM}^2}{2\pi}\sigma.$$

Completing the square provides a sign check:

$$-\frac{D^2}{2g_{\rm YM}^2}-\frac{kD\sigma}{2\pi} =-\frac1{2g_{\rm YM}^2} \left(D+\frac{k g_{\rm YM}^2\sigma}{2\pi}\right)^2 +\frac{k^2g_{\rm YM}^2}{8\pi^2}\sigma^2.$$

Thus

$$\mathcal L_\sigma^{\rm reduced} =\frac12(\partial_\mu\sigma_c)^2 +\frac12\left(\frac{k g_{\rm YM}^2}{2\pi}\right)^2\sigma_c^2, \qquad m_\sigma=\frac{|k|g_{\rm YM}^2}{2\pi}.$$

The dimensions are correct because $[k]=0$ and $[g_{\rm YM}^2]=1$. At $k=0$ the scalar returns to the massless Yang–Mills result. At nonzero $k$ it is a massive propagating field, unlike the algebraically constrained scalar of pure Chern–Simons theory.

On the free contour $D=ih$, the linear term is oscillatory and the quadratic part decays. The real saddle above generally lies off that contour; completing the Gaussian by an allowed contour shift gives the same positive scalar potential. Algebraic elimination alone is not a claim that the Euclidean field was integrated over real $D$.

Checkpoint

You can now distinguish the vector multiplet’s field content from its dynamics, verify its scalar and gauge-potential closure, and recognize the same gauge parameter acting on matter. In P, equations (6.18)–(6.24) specify that component structure; equations (6.28) and (6.30) choose different actions for it.

Before packaging these fields into superspace, reproduce the worked configuration’s $a_1$ variation and explain why it is a gauge transformation rather than a new central charge. Then check why a gauge-invariant matter derivative is only the first step toward a supersymmetric gauge theory.

Superspace as a way to organize known fields performs that packaging explicitly, recovering the chiral action and introducing the gauge-invariant linear field-strength superfield.

For the curved-space continuation, The curved vector multiplet combines these fields with the Killing spinor equations. The derivative of a label changes the auxiliary variation and the compensating gauge transformation.