The chiral multiplet

We now know what a supercharge is and which algebra its transformations must obey. Which fields can carry that algebra? The first useful answer is a chiral multiplet: a scalar, a two-component fermion, and an auxiliary scalar. Their transformation laws explain why a scalar and a fermion belong together, and why an extra bosonic field appears before any equations of motion are imposed.

The prerequisite is The three-dimensional N=2 algebra. Our task is concrete: start with a specified configuration of all three fields, apply a transformation, and check its spin, parity, dimensions, and charge. The next lesson will test closure on the fermion; constructing and varying the action comes after that.

The setting and the three fields

Work in flat three-dimensional Euclidean space with metric $\delta_{\mu\nu}$. We use the full $\mathcal N=2$ transformations, which are the complexification of four real Lorentzian Poincaré supercharges. All background sources vanish:

$$A_\mu=V_\mu=C_\mu=H=0.$$

Take a gauge-neutral, massless chiral multiplet with central-charge eigenvalue $z=0$ and no superpotential interaction. There is no gauge rank to specify. The Euclidean fermion and its tilded partner remain independent, as in Euclidean continuation.

Write the collection of component fields as

$$\Phi=(\phi,\psi_\alpha,F_\Phi).$$

Here $\phi$ and $F_\Phi$ are complex even scalars. The field $\psi_\alpha$ is an odd spinor with two complex components. The subscript on $F_\Phi$ distinguishes the auxiliary field from a free energy or a prepotential. The symbol $\Phi$ presently names the collection; no superspace expansion is needed.

“Chiral” refers to the supersymmetry representation: one set of odd transformations annihilates its lowest scalar. It does not mean that a three-dimensional spinor has been projected onto a four-dimensional Weyl chirality.

We retain the gamma matrices and contraction

$$\gamma^\mu=(\sigma_3,-\sigma_1,-\sigma_2),\qquad u v=u^T C v,\qquad C=\begin{pmatrix}0&-1\\1&0\end{pmatrix}.$$

Both spinors in this matrix formula are lower-index columns. Thus $uv=u_2v_1-u_1v_2$.

A transformation sends each field to a different kind of expression

Let $\zeta$ and $\widetilde\zeta$ be constant commuting spinor labels. As in the preceding lesson, $\delta_\zeta$ and $\delta_{\widetilde\zeta}$ are odd operations. Set $\delta=\delta_\zeta+\delta_{\widetilde\zeta}$. The flat, zero-central-charge specialization of the source convention is

$$\begin{aligned} \delta\phi&=\sqrt2\,\zeta\psi,\\ \delta\psi&=\sqrt2\,\zeta F_\Phi -\sqrt2\,i\gamma^\mu\widetilde\zeta\,\partial_\mu\phi,\\ \delta F_\Phi&=-\sqrt2\,i\widetilde\zeta\gamma^\mu\partial_\mu\psi. \end{aligned}$$

These are equation (6.8) of the three-manifold paper, version 3, with the background and central terms set to zero. The factor $\sqrt2$ fixes the field normalization, while the derivative factors of $i$ follow the algebra and Euclidean conventions. Changing those factors independently would change the representation.

The first equation converts the scalar into a fermion contracted with a spinor label. The second has two distinct contributions. The $\zeta$ operation produces the auxiliary field; the $\widetilde\zeta$ operation produces a derivative of the scalar. The derivative is necessary because composing supersymmetries must produce a translation. The third equation converts the auxiliary field into a derivative of the fermion.

With $\zeta=0$ we have $\delta_{\widetilde\zeta}\phi=0$. This is the chirality of the lowest component. With $\widetilde\zeta=0$, the sequence is scalar to fermion to auxiliary field, with $\delta_\zeta F_\Phi=0$. This describes the organization of a representation, rather than three unrelated fields to which we have assigned similar names.

Because $\delta$ is odd, $\delta\phi$ and $\delta F_\Phi$ are odd and $\delta\psi$ is even. An ordinary infinitesimal field change uses a further constant odd coefficient $\rho$:

$$\Delta=\rho\,\delta,\qquad \phi'=\phi+\rho\,\delta\phi,\qquad \psi'=\psi+\rho\,\delta\psi.$$

The complete variation $\Delta$ is even and preserves field parity. Place $\rho$ consistently on the left; it anticommutes with odd fields. Without it, adding the output of an odd operation directly to a field would mix parities.

Derive the dimensions and R-charges

Brackets denote engineering mass dimension. A dimensionless action in three dimensions has a Lagrangian of dimension three. Each derivative has dimension one, so a scalar kinetic term with two derivatives requires $2+2[\phi]=3$, while a first-order fermion kinetic term requires $1+2[\psi]=3$. Thus

$$[\phi]=\frac12,\qquad [\psi]=1.$$

The transformation labels have dimension $-1/2$, so the first equation passes the check

$$[\zeta]+[\psi]=-\frac12+1=\frac12=[\phi].$$

The auxiliary contribution to $\delta\psi$ then determines

$$[F_\Phi]=[\psi]-[\zeta]=\frac32.$$

The derivative contribution gives the same dimension as $\psi$: $-1/2+1+1/2=1$. Finally, the right side of $\delta F_\Phi$ has dimension $-1/2+1+1=3/2$. These checks constrain the allowed structures, but do not determine their signs or prove closure.

Let $r$ denote the R-charge of $\phi$. An R-symmetry acts on supercharges as well as on fields. The fixed-label operation $\delta_\zeta$ lowers field R-charge by one, so the fermion has charge $r-1$ and the auxiliary has charge $r-2$.

Equivalently, when checking covariance of a parameterized transformation, assign the labels their compensating charges

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

Then both terms in $\delta\psi$ have charge $r-1$:

$$R(\zeta F_\Phi)=1+(r-2)=r-1,\qquad R(\widetilde\zeta\,\partial_\mu\phi)=-1+r=r-1.$$

This formal assignment to a label is different from saying that the fixed odd operation raises a field’s R-charge.

Field	Rotation representation	Parity	Engineering dimension	R-charge
$\phi$	Scalar, spin $0$	Even	$1/2$	$r$
$\psi_\alpha$	Spinor, spin $1/2$	Odd	$1$	$r-1$
$F_\Phi$	Scalar, spin $0$	Even	$3/2$	$r-2$

The transformation rules do not select $r$. In particular, being a scalar does not give $F_\Phi$ the same dimension or charge as $\phi$. Determining a distinguished R-symmetry in a conformal or interacting theory is a further physical question.

Work through a complete component configuration

Choose a profile depending on $x^1$ only:

$$\phi=Ae^{ikx^1},\qquad \psi=\vartheta\begin{pmatrix}u\\v\end{pmatrix}e^{ikx^1}, \qquad F_\Phi=Be^{ikx^1}.$$

The amplitudes $A,B,u,v$ commute, and $\vartheta$ is a constant odd generator. This is a convenient particular fermion profile, not the most general one. Take real $k$ and choose the label directions

$$\zeta=s\begin{pmatrix}1\\0\end{pmatrix}, \qquad \widetilde\zeta=t\begin{pmatrix}0\\1\end{pmatrix}.$$

The constants $s,t$ commute and have dimension $-1/2$. Assign $[A]=1/2$, $[B]=3/2$, $[u]=[v]=1$, and $[\vartheta]=0$. We impose no field equation: “off shell” means precisely that the component profiles are being varied without restricting them to solutions.

First perform the scalar contraction:

$$\zeta\psi =s(1,0)C\,\vartheta(u,v)^T e^{ikx^1} =-s\vartheta v\,e^{ikx^1}.$$

For the fermion derivative term, $\gamma^1\widetilde\zeta=-t(0,1)^T$ and $\partial_1\phi=ikAe^{ikx^1}$. Hence

$$-\sqrt2\,i\gamma^1\widetilde\zeta\,\partial_1\phi =-\sqrt2\,tkA\begin{pmatrix}0\\1\end{pmatrix}e^{ikx^1}.$$

The two factors of $i$ and the lower gamma eigenvalue must all be included. Finally, $\widetilde\zeta\gamma^1\partial_1\psi =itk\vartheta u\,e^{ikx^1}$. The complete result is

$$\begin{aligned} \delta\phi&=-\sqrt2\,s\vartheta v\,e^{ikx^1},\\ \delta\psi&=\sqrt2 \begin{pmatrix}sB\\-tkA\end{pmatrix}e^{ikx^1},\\ \delta F_\Phi&=\sqrt2\,tk\vartheta u\,e^{ikx^1}. \end{aligned}$$

Every output has the dimension of the field on the left and its opposite parity. Multiplication by $\rho$ restores the parity for an actual even variation.

Several limits make the roles visible. A varying scalar with $\psi=F_\Phi=0$ produces a fermion through the $kA$ term. A constant nonzero auxiliary field produces a fermion through the $B$ term. A constant scalar with $\psi=F_\Phi=0$ has zero variation. These are statements about configurations. Calling a configuration a vacuum additionally requires an action and its equations; a nonzero auxiliary profile alone is not yet evidence of spontaneous supersymmetry breaking.

Independent Euclidean partners and the auxiliary field

An anti-chiral multiplet is $\widetilde\Phi=(\widetilde\phi,\widetilde\psi,\widetilde F_\Phi)$. Its flat transformations are

$$\begin{aligned} \delta\widetilde\phi&=-\sqrt2\,\widetilde\zeta\widetilde\psi,\\ \delta\widetilde\psi&=\sqrt2\,\widetilde\zeta\widetilde F_\Phi +\sqrt2\,i\gamma^\mu\zeta\,\partial_\mu\widetilde\phi,\\ \delta\widetilde F_\Phi&=-\sqrt2\,i\zeta\gamma^\mu\partial_\mu\widetilde\psi. \end{aligned}$$

These follow from equation (6.10) in the same flat specialization. Their R-charges are $(-r,1-r,2-r)$; their dimensions and parities match the corresponding untilded fields. In Euclidean signature these are independent variables. Obtain their rules from the declared representation, rather than by imposing an undeclared pointwise dagger operation.

An auxiliary field has an algebraic equation of motion in the standard action, rather than a differential equation describing propagation. The source’s free action contains $-\widetilde F_\Phi F_\Phi$ without derivatives; its two auxiliary equations set both fields to zero. This limited observation follows from equation (6.32); the complete action and its invariance will be derived separately.

Why keep the auxiliary before solving those equations? The transformation of $F_\Phi$ contains $\partial_\mu\psi$. Setting $F_\Phi=0$ while allowing arbitrary fermion profiles need not be preserved by supersymmetry. Exercise 2 exposes this directly. Eliminating an auxiliary consistently and proving closure on the remaining fields is a different operation from simply deleting it from the transformation table.

For the Lorentzian free multiplet, the scalar has two real propagating degrees of freedom and the complex fermion has two on shell. Before imposing equations, $\phi$ and $F_\Phi$ together supply four real bosonic components, matching the four real fermionic components of $\psi$. This count is a useful diagnostic, not a closure proof. The independent Euclidean tilded variables are the complexified partners; they are not additional particle species to be counted again.

One short algebra check is already possible. For a single commuting label,

$$\delta_\zeta^2\phi =\sqrt2\,\zeta(\sqrt2\,\zeta F_\Phi) =2(\zeta\zeta)F_\Phi=0.$$

The last equality uses the antisymmetric scalar contraction of two identical commuting spinors. This tests one component of the same-charge algebra. The mixed transformations on the fermion are the more demanding next calculation.

Exercises

1. Change the two spinor directions

Keep the worked configuration but take $\zeta=s(0,1)^T$ and $\widetilde\zeta=t(1,0)^T$. Compute all three odd variations. Explain which sign in the scalar variation changed.

First hint

Use $\zeta\psi=\zeta_2\psi_1-\zeta_1\psi_2$.

Second hint

Now $\gamma^1\widetilde\zeta=+\widetilde\zeta$. Keep $\partial_1=ik$ on every profile.

Complete solution

The scalar contraction is now $s\vartheta u e^{ikx^1}$. The derivative term of the fermion is $\sqrt2\,tkA(1,0)^T e^{ikx^1}$, while its auxiliary term is $\sqrt2\,sB(0,1)^T e^{ikx^1}$. Also $\widetilde\zeta\gamma^1\partial_1\psi=itk\vartheta v e^{ikx^1}$. Therefore

$$\begin{aligned} \delta\phi&=\sqrt2\,s\vartheta u\,e^{ikx^1},\\ \delta\psi&=\sqrt2\begin{pmatrix}tkA\\sB\end{pmatrix}e^{ikx^1},\\ \delta F_\Phi&=\sqrt2\,tk\vartheta v\,e^{ikx^1}. \end{aligned}$$

The scalar sign changed because the nonzero label component now multiplies the positive term in the antisymmetric contraction. It did not come from exchanging two odd fields: the label commutes. At $k=0$, both derivative contributions vanish, leaving only the auxiliary contribution to $\delta\psi$.

2. Can the auxiliary simply be set to zero?

Take $\phi=F_\Phi=0$ and $\psi=\vartheta h(x^1)(0,1)^T$, where $h$ is a commuting differentiable function. Choose $\zeta=0$ and $\widetilde\zeta=t(1,0)^T$. Calculate $\delta F_\Phi$. Does the restriction $F_\Phi=0$ define an invariant set of arbitrary off-shell configurations?

First hint

Only the derivative of $\psi$ contributes.

Second hint

The matrix $C\gamma^1$ equals $\begin{pmatrix}0&1\\1&0\end{pmatrix}$.

Complete solution

The contraction gives $\widetilde\zeta\gamma^1\partial_1\psi=t\vartheta h'(x^1)$, hence

$$\delta F_\Phi=-\sqrt2\,i\,t\vartheta h'(x^1).$$

For a nonconstant profile this is generally nonzero, even though $F_\Phi$ initially vanished. The corresponding even change is $\Delta F_\Phi=\rho\,\delta F_\Phi$. Thus the unrestricted off-shell set with $F_\Phi=0$ is not invariant. For a constant $h$, the expression vanishes, which checks the derivative dependence. This particular calculation does not establish the full on-shell algebra; it identifies why removing the auxiliary requires more care.

3. Transform the independent anti-chiral scalar

Set $\widetilde\phi=\widetilde A e^{ikx^1}$ and $\widetilde\psi=\widetilde F_\Phi=0$. Choose $\zeta=s(1,0)^T$, $\widetilde\zeta=0$. Find the anti-chiral variations. Check the R-charge of the nonzero term, and explain why it cannot be inferred by first setting all tilded fields equal to daggers while leaving the labels untouched.

First hint

The anti-chiral fermion has a derivative term with a plus sign in front of $i$.

Second hint

Use $R(\zeta)=1$ and $R(\widetilde\phi)=-r$. Conjugating a Lorentzian transformation also acts on its parameters and index conventions.

Complete solution

Only the fermion changes:

$$\delta\widetilde\phi=0,\qquad \delta\widetilde\psi =-\sqrt2\,sk\widetilde A \begin{pmatrix}1\\0\end{pmatrix}e^{ikx^1}, \qquad \delta\widetilde F_\Phi=0.$$

The two factors of $i$ produce the minus sign. The nonzero term has charge $1-r$, which is the charge of $\widetilde\psi$, and dimension $-1/2+1+1/2=1$. At $k=0$ it vanishes.

The Euclidean partners and the two labels are independent. A Lorentzian conjugation relation would involve a simultaneous prescription for fields, parameters, and spinor indices, followed by continuation. Applying a dagger to only one part of the Euclidean calculation changes the stated problem and need not preserve its transformation laws.

Checkpoint and next step

You can now start with explicit component profiles and compute their supersymmetry variations, with the role of each field visible. You should also be able to explain why two scalar fields in the same multiplet have different dimensions and R-charges.

Before moving on, answer this without consulting the table: which term transforms a varying scalar into a fermion, and which term prevents an arbitrary configuration with $F_\Phi=0$ from remaining in that set?

The remaining test is whether composing transformations on $\psi$ gives precisely the expected translation. Follow that calculation with and without the auxiliary field to turn its role from a component count into an explicit algebraic result.