Build the free chiral theory

An algebra acting on fields does not yet tell us which dynamics respects it. We now build an action for the chiral multiplet and check its variation term by term. The important cancellation joins the scalar and fermion kinetic terms; another fixes the sign of the auxiliary term. Neither cancellation will use a field equation.

The prerequisite is Why auxiliary fields make the algebra close. Keep the component rules from The chiral multiplet available. Our calculation concerns a gauge-neutral, massless multiplet in flat three-dimensional Euclidean space, with the full complexified $\mathcal N=2$ algebra of four real Lorentzian supercharges. There is no gauge sector or superpotential. Set

$$A_\mu=V_\mu=C_\mu=H=0,\qquad z=0.$$

The scalar R-charge $r$ remains arbitrary. All tilded component fields are independent variables during the algebraic calculation.

Construct a candidate from familiar kinetic terms

Normalize the scalar kinetic term to one and consider

$$\mathcal L(b,c)= \partial_\mu\widetilde\phi\,\partial_\mu\phi +b\,\widetilde\psi\gamma^\mu\partial_\mu\psi +c\,\widetilde F_\Phi F_\Phi.$$

This is the massless quadratic kinetic ansatz, with no additional interactions. Every term is even, has engineering dimension three, and has R-charge zero: the two factors in each term carry opposite charges. The unknown constants $b,c$ set relative normalizations. Dimensions and charges alone cannot fix them.

We use lower spinor columns, the contraction $uv=u^TCv$, and

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

Two useful identities are

$$(\gamma^\mu)^TC=-C\gamma^\mu,\qquad \gamma^\nu\gamma^\mu =\delta^{\nu\mu}\mathbf1+i\epsilon^{\nu\mu\rho}\gamma^\rho, \qquad \epsilon^{123}=1.$$

Let $\delta_\zeta$ and $\delta_{\widetilde\zeta}$ be odd operations with constant commuting spinor labels. The graded product rule is therefore

$$\delta(XY)=(\delta X)Y+(-1)^{|X|}X(\delta Y).$$

In particular, the fermion term varies as

$$\delta(\widetilde\psi\gamma^\mu\partial_\mu\psi) =(\delta\widetilde\psi)\gamma^\mu\partial_\mu\psi -\widetilde\psi\gamma^\mu\partial_\mu(\delta\psi).$$

The minus sign is essential. A spacetime derivative is even and introduces no graded sign.

For the ansatz, the $\zeta$ variation has a scalar-gradient contribution proportional to $1-ib$. The $\widetilde\zeta$ variation has an auxiliary contribution proportional to $b-ic$. More explicitly, the structures whose coefficients must cancel are

$$\begin{aligned} \delta_\zeta\mathcal L(b,c)&\supset \sqrt2(1-ib)(\partial_\mu\widetilde\phi)\zeta\partial_\mu\psi,\\ \delta_{\widetilde\zeta}\mathcal L(b,c)&\supset \sqrt2(b-ic)\widetilde F_\Phi \widetilde\zeta\gamma^\mu\partial_\mu\psi. \end{aligned}$$

Setting $b=-i$ and $c=-1$ cancels both. The calculation below shows that the remaining terms are precisely divergences. We obtain

$$S_E=\int d^3x\,\mathcal L,\qquad \mathcal L= \underbrace{\partial_\mu\widetilde\phi\,\partial_\mu\phi}_{\mathcal L_s} +\underbrace{-i\widetilde\psi\gamma^\mu\partial_\mu\psi}_{\mathcal L_\psi} +\underbrace{-\widetilde F_\Phi F_\Phi}_{\mathcal L_F}.$$

This is the flat, neutral, zero-central-charge specialization of equation (6.32) in the three-manifold paper, version 3. Its source-dependent covariant derivatives in (6.31) both reduce to ordinary derivatives here. All curvature and background terms vanish, so the flat action does not select a particular $r$.

The first variation: two cancellations and a current

Set $\widetilde\zeta=0$. The needed transformations are

$$\begin{aligned} \delta_\zeta\phi&=\sqrt2\zeta\psi, &\delta_\zeta\psi&=\sqrt2\zeta F_\Phi, &\delta_\zeta F_\Phi&=0,\\ \delta_\zeta\widetilde\phi&=0, &\delta_\zeta\widetilde\psi&=\sqrt2i\gamma^\nu\zeta\,\partial_\nu\widetilde\phi, &\delta_\zeta\widetilde F_\Phi&=-\sqrt2i\zeta\gamma^\mu\partial_\mu\widetilde\psi. \end{aligned}$$

They are the same flat specialization of equations (6.8) and (6.10). Separate the three variations:

$$\begin{aligned} \delta_\zeta\mathcal L_s &=\sqrt2(\partial_\mu\widetilde\phi)\zeta\partial_\mu\psi,\\ \delta_\zeta\mathcal L_\psi &=\sqrt2(\partial_\nu\widetilde\phi) (\gamma^\nu\zeta)\gamma^\mu\partial_\mu\psi +\sqrt2i\widetilde\psi\gamma^\mu\zeta\,\partial_\mu F_\Phi,\\ \delta_\zeta\mathcal L_F &=\sqrt2iF_\Phi\zeta\gamma^\mu\partial_\mu\widetilde\psi. \end{aligned}$$

There is no field equation in these expressions. To compare the first two lines, move the gamma matrix off the first spinor using its transpose identity:

$$(\gamma^\nu\zeta)\gamma^\mu\partial_\mu\psi =-\zeta\gamma^\nu\gamma^\mu\partial_\mu\psi.$$

The identity part of $\gamma^\nu\gamma^\mu$ cancels $\delta_\zeta\mathcal L_s$. The remainder is

$$-\sqrt2i\epsilon^{\nu\mu\rho} (\partial_\nu\widetilde\phi)\zeta\gamma^\rho\partial_\mu\psi.$$

It is a divergence, because the putative extra term contains a symmetric second derivative contracted with an antisymmetric tensor:

$$\begin{aligned} &\partial_\mu\!\left[ -\sqrt2i\epsilon^{\nu\mu\rho} (\partial_\nu\widetilde\phi)\zeta\gamma^\rho\psi \right]\\ &\quad=-\sqrt2i\epsilon^{\nu\mu\rho} (\partial_\nu\widetilde\phi)\zeta\gamma^\rho\partial_\mu\psi, \qquad \epsilon^{\nu\mu\rho}\partial_\mu\partial_\nu\widetilde\phi=0. \end{aligned}$$

For the auxiliary terms, the symmetric matrix $C\gamma^\mu$ gives $\widetilde\psi\gamma^\mu\zeta=\zeta\gamma^\mu\widetilde\psi$: one spinor is even, so their exchange supplies no Grassmann sign. The last two terms combine as

$$\sqrt2i\left[ (\partial_\mu F_\Phi)\zeta\gamma^\mu\widetilde\psi +F_\Phi\zeta\gamma^\mu\partial_\mu\widetilde\psi \right] =\partial_\mu(\sqrt2iF_\Phi\zeta\gamma^\mu\widetilde\psi).$$

We have therefore exhibited the complete result

$$\delta_\zeta\mathcal L=\partial_\mu B_\zeta^\mu,\qquad B_\zeta^\mu= -\sqrt2i\epsilon^{\nu\mu\rho} (\partial_\nu\widetilde\phi)\zeta\gamma^\rho\psi +\sqrt2iF_\Phi\zeta\gamma^\mu\widetilde\psi.$$

Here $B_\zeta^\mu$ means the boundary current appearing in the variation. It is not by itself a definition of the conserved Noether supercurrent.

The second variation: the other half of the proof

Now set $\zeta=0$. The nonzero transformations are

$$\begin{aligned} \delta_{\widetilde\zeta}\psi&=-\sqrt2i\gamma^\nu\widetilde\zeta\,\partial_\nu\phi, &\delta_{\widetilde\zeta}F_\Phi&=-\sqrt2i\widetilde\zeta\gamma^\mu\partial_\mu\psi,\\ \delta_{\widetilde\zeta}\widetilde\phi&=-\sqrt2\widetilde\zeta\widetilde\psi, &\delta_{\widetilde\zeta}\widetilde\psi&=\sqrt2\widetilde\zeta\widetilde F_\Phi. \end{aligned}$$

The three action terms give

$$\begin{aligned} \delta_{\widetilde\zeta}\mathcal L_s &=-\sqrt2(\widetilde\zeta\partial_\mu\widetilde\psi)\partial_\mu\phi,\\ \delta_{\widetilde\zeta}\mathcal L_\psi &=-\sqrt2i\widetilde F_\Phi\widetilde\zeta\gamma^\mu\partial_\mu\psi +\sqrt2\widetilde\psi\gamma^\mu\gamma^\nu\widetilde\zeta\, \partial_\mu\partial_\nu\phi,\\ \delta_{\widetilde\zeta}\mathcal L_F &=+\sqrt2i\widetilde F_\Phi\widetilde\zeta\gamma^\mu\partial_\mu\psi. \end{aligned}$$

The auxiliary contributions cancel directly. The antisymmetric part of $\gamma^\mu\gamma^\nu$ annihilates the symmetric second derivative. For the remaining scalar contraction, $\widetilde\psi\widetilde\zeta=-\widetilde\zeta\widetilde\psi$. Thus

$$\begin{aligned} \delta_{\widetilde\zeta}\mathcal L &=-\sqrt2\left[ (\widetilde\zeta\partial_\mu\widetilde\psi)\partial_\mu\phi +(\widetilde\zeta\widetilde\psi)\partial_\mu\partial_\mu\phi \right]\\ &=\partial_\mu B_{\widetilde\zeta}^\mu,\qquad B_{\widetilde\zeta}^\mu=-\sqrt2(\widetilde\zeta\widetilde\psi)\partial_\mu\phi. \end{aligned}$$

The two currents look different because the chosen fermion kinetic term puts its derivative on $\psi$. Moving that derivative by integration by parts changes the displayed boundary currents, while leaving the bulk action on a space without boundary unchanged.

For the full operation, $\delta=\delta_\zeta+\delta_{\widetilde\zeta}$,

$$\delta S_E=\int_{\partial M}d\Sigma_\mu\, (B_\zeta^\mu+B_{\widetilde\zeta}^\mu).$$

This vanishes on a flat torus with periodic fields and spinors, or on $\mathbb R^3$ when the fields fall off sufficiently fast. On an actual boundary, one must specify conditions or boundary terms that cancel the flux. “A total derivative” is the intermediate result; vanishing boundary flux is what proves invariance of the action. Multiplying $\delta$ by an odd constant $\rho$ gives the usual even variation $\Delta=\rho\delta$ and multiplies this identity by $\rho$.

A local variation that integrates to zero

Take a periodic coordinate $x=x^1$ of length $2\pi/k$, with $k>0$. Work per unit transverse area and choose

$$\phi=A\sin(kx),\qquad \widetilde\psi=\vartheta\begin{pmatrix}0\\1\end{pmatrix}\cos(kx), \qquad \widetilde\zeta=t\begin{pmatrix}1\\0\end{pmatrix}, \qquad \zeta=0,$$

with the remaining fields zero. The constants $A,t$ commute and $\vartheta$ is an odd amplitude, with $[A]=1/2$, $[t]=-1/2$, and $[\vartheta]=1$. This is a configuration in the independent complexified variables; no conjugation relation is imposed.

Since $\widetilde\zeta\widetilde\psi=-t\vartheta\cos(kx)$,

$$B_{\widetilde\zeta}^1 =\sqrt2tA k\vartheta\cos^2(kx).$$

Directly varying the scalar and fermion terms gives equal contributions,

$$\delta\mathcal L_s=\delta\mathcal L_\psi =-\sqrt2tA k^2\vartheta\sin(kx)\cos(kx), \qquad \delta\mathcal L_F=0.$$

Their sum is $\partial_1B_{\widetilde\zeta}^1$, and its integral over the period is zero. The Lagrangian density need not be invariant at each point. This example checks both the sign and coefficient of the boundary identity with nonconstant fields.

Equations of motion, contours, and a fermion phase

Varying $\widetilde F_\Phi$ and $F_\Phi$ independently gives $F_\Phi=\widetilde F_\Phi=0$. These are algebraic equations. In contrast, varying $\widetilde\phi$ gives $\partial^2\phi=0$, and varying the independent fermion row gives $\gamma^\mu\partial_\mu\psi=0$. These differential equations describe propagating fields. They were not inputs to the supersymmetry proof.

The negative auxiliary term does not mean that a propagating field has a wrong-sign kinetic energy. It has no derivative. It does mean that the Euclidean integral needs a stated contour. If one set $\widetilde F_\Phi=F_\Phi^*$, its factor in $e^{-S_E}$ would grow as $e^{+|F_\Phi|^2}$. A convergent contour for the free auxiliary Gaussian is instead

$$F_\Phi=u+iv,\qquad \widetilde F_\Phi=-u+iv=-F_\Phi^*, \qquad u,v\in\mathbb R.$$

On this contour $-\widetilde F_\Phi F_\Phi=u^2+v^2$. Independently, the scalar contour $\widetilde\phi=\phi^*$ makes its gradient term nonnegative. These are integration prescriptions on complexified bosonic fields, not substitutions into the transformation algebra. Fermions use Berezin integration over independent odd variables. The local identity above does not alone establish a quantum Ward identity: the measure, integration prescription, and boundary behavior must also respect the transformation.

There is a useful normalization comparison with Euclidean continuation. That lesson’s massless fermion action was $\chi_E\gamma^\mu\partial_\mu\psi_E$. To match the present convention while keeping $\psi_E=\psi$, set

$$\chi_E=-i\,\widetilde\psi_P^{\,T}C.$$

Then $\chi_E\gamma^\mu\partial_\mu\psi_E=-i\widetilde\psi_P\gamma^\mu\partial_\mu\psi$, as required. If the independent column there is defined by $\chi_E=\widetilde\psi_E^{\,T}C$, it satisfies $\widetilde\psi_E=-i\widetilde\psi_P$. The subscript $P$ identifies the paper normalization used throughout this lesson. This phase relates two complete kinetic conventions; the contraction dictionary alone does not supply it. A field redefinition also changes that field’s transformation rule.

Exercises

1. Change the auxiliary sign

Replace $-\widetilde F_\Phi F_\Phi$ by $+\widetilde F_\Phi F_\Phi$ while keeping every transformation fixed. Find the uncancelled $\widetilde\zeta$ variation. Show that it need not integrate to zero by using periodic profiles $\widetilde F_\Phi=B\cos(kx)$, $\psi=\vartheta(0,1)^T\sin(kx)$, $\widetilde\zeta=t(1,0)^T$, with the other fields and $\zeta$ zero.

First hint

The change in the Lagrangian is $2\widetilde F_\Phi F_\Phi$. Vary this difference instead of repeating the whole proof.

Second hint

Use $C\gamma^1=\begin{pmatrix}0&1\\1&0\end{pmatrix}$ and integrate $\cos^2(kx)$ over one period.

Complete solution

The extra term gives

$$\delta_{\widetilde\zeta}\mathcal L_{\rm wrong} -\partial_\mu B_{\widetilde\zeta}^\mu =-2\sqrt2i\,\widetilde F_\Phi \widetilde\zeta\gamma^\mu\partial_\mu\psi.$$

For the assigned profiles the original boundary current is zero, while the right side is $-2\sqrt2iBt k\vartheta\cos^2(kx)$. Per unit transverse area,

$$\delta_{\widetilde\zeta}S_{\rm wrong} =-2\sqrt2i\pi Bt\vartheta.$$

This is generally nonzero despite periodic boundary conditions. The auxiliary sign is fixed by invariance for the chosen transformations. One cannot change that sign alone and describe the result as the same convention.

2. Keep the boundary

On $0\le x=x^1\le L$, take $\phi=Akx$, $\widetilde\psi=\vartheta kx(0,1)^T$, $\widetilde\zeta=t(1,0)^T$, $\zeta=0$, and all other fields zero. Compute the action variation per unit transverse area. Does the bulk cancellation prove invariance on this interval without further boundary information?

First hint

Compute $B_{\widetilde\zeta}^1(L)-B_{\widetilde\zeta}^1(0)$.

Second hint

Here $\partial_1\phi=Ak$ and $\widetilde\zeta\widetilde\psi=-t\vartheta kx$. Check the answer by integrating the local scalar-term variation.

Complete solution

The current is $B_{\widetilde\zeta}^1=\sqrt2tAk^2\vartheta x$, so

$$\delta S_E=\sqrt2tAk^2\vartheta L.$$

Directly, $\delta\mathcal L_s=\sqrt2tAk^2\vartheta$ and the other two variations vanish because $\partial_1^2\phi=F_\Phi=\widetilde F_\Phi=0$. Integrating gives the same result. The boundary flux is nonzero; these unconstrained boundary data do not define a supersymmetric action by themselves. Appropriate boundary conditions or boundary contributions are additional physical choices.

3. Auxiliary equations do not remove Gaussian fluctuations

For one rescaled auxiliary mode, use the convergent contour above and the weight $e^{-a(u^2+v^2)}$, with $a>0$ and measure $du\,dv$. Calculate its integral and the normalized expectations $\langle F_\Phi\rangle$ and $\langle\widetilde F_\Phi F_\Phi\rangle$. Explain why a nonzero second expectation is consistent with the classical auxiliary equation.

First hint

Use polar coordinates, or multiply two one-dimensional Gaussian integrals.

Second hint

Differentiating the integral with respect to $a$ inserts $-(u^2+v^2)$. On this contour $\widetilde F_\Phi F_\Phi=-(u^2+v^2)$.

Complete solution

The integral is

$$\int_{\mathbb R^2}du\,dv\,e^{-a(u^2+v^2)}=\frac{\pi}{a}.$$

Reflection symmetry gives $\langle F_\Phi\rangle=0$. Differentiation yields

$$\langle u^2+v^2\rangle =-\frac{\partial_a(\pi/a)}{\pi/a}=\frac1a, \qquad \langle\widetilde F_\Phi F_\Phi\rangle=-\frac1a.$$

The classical equation identifies the stationary point $F_\Phi=\widetilde F_\Phi=0$. A Gaussian integral still samples fluctuations around that point. Since this mode has no derivative term, these fluctuations do not describe a propagating auxiliary particle. Increasing $a$ suppresses them, as the result checks.

Checkpoint

We have a complete off-shell action and its two explicit variation currents. The proof depends on the graded product rule, the Clifford identity, the auxiliary coefficient, and a boundary prescription.

Before proceeding, explain which cancellation fails if the auxiliary sign is reversed, and why $\delta\mathcal L\ne0$ can coexist with $\delta S_E=0$. The next question is which internal symmetries act on this action and how an R-symmetry differs from an ordinary flavor symmetry.