Why auxiliary fields make the algebra close

The auxiliary scalar in the chiral multiplet has no independent propagating particle. Why keep it? We can answer by composing supersymmetry transformations on the fermion. One composition passes through a scalar derivative; the other passes through the auxiliary field. Both contributions are needed to produce a translation for an arbitrary fermion profile.

Off-shell closure means that the algebra holds without imposing equations of motion. On-shell closure means that additional terms vanish when the field equations hold. These are statements about how transformations compose, distinct from whether an action is invariant. We will calculate the difference explicitly.

The two routes through a fermion transformation

Work in flat three-dimensional Euclidean space, with all background sources zero and a gauge-neutral, free chiral multiplet of central charge $z=0$. We retain the full complexified $\mathcal N=2$ algebra, whose Lorentzian parent has four real Poincaré supercharges. There is no gauge sector in this example.

The even scalars are $\phi,F_\Phi$ and the odd spinor is $\psi$. Use constant commuting spinor labels $\zeta,\widetilde\zeta$, so the operations $\delta_\zeta,\delta_{\widetilde\zeta}$ are odd. The transformations needed here are

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

These are the flat specialization of the three-manifold paper’s equation (6.8), version 3. Our contractions are $u v=u^TCv$ for lower-index columns, with

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

The mixed algebra should give a translation along the constant vector $K^\mu=\zeta\gamma^\mu\widetilde\zeta$. To test it on $\psi$, compose the operations in both orders. They commute with spacetime derivatives, and their labels commute with everything:

$$\begin{aligned} \delta_\zeta\delta_{\widetilde\zeta}\psi &=-\sqrt2\,i\gamma^\mu\widetilde\zeta\, \partial_\mu(\sqrt2\,\zeta\psi)\\ &=-2i(\gamma^\mu\widetilde\zeta) (\zeta\,\partial_\mu\psi),\\ \delta_{\widetilde\zeta}\delta_\zeta\psi &=\sqrt2\,\zeta \left(-\sqrt2\,i\widetilde\zeta\gamma^\mu \partial_\mu\psi\right)\\ &=-2i\zeta (\widetilde\zeta\gamma^\mu\partial_\mu\psi). \end{aligned}$$

The first route varies $\phi$ inside the derivative term of $\delta_{\widetilde\zeta}\psi$. The second varies $F_\Phi$ inside $\delta_\zeta\psi$. Their sum is not yet written as a translation: two spinor outer products must be combined.

The spinor identity that completes the calculation

An outer product of a column and a row is a matrix. Define $\mathsf M=\zeta\widetilde\zeta^TC$ and $\mathsf N=\widetilde\zeta\zeta^TC$, and write $s=\zeta\widetilde\zeta$. The matrix completeness identity from the spinor lesson gives

$$\mathsf M=-\frac{s}{2}\mathbf1+\frac{K^\nu}{2}\gamma_\nu, \qquad \mathsf N=\frac{s}{2}\mathbf1+\frac{K^\nu}{2}\gamma_\nu.$$

The scalar contraction changes sign when the commuting labels are exchanged, whereas the vector contraction does not. This accounts for the different scalar coefficients.

Now multiply the second matrix by $\gamma^\mu$ on the left and the first on the right:

$$\begin{aligned} \gamma^\mu\mathsf N+\mathsf M\gamma^\mu &=\frac{s}{2}\gamma^\mu-\frac{s}{2}\gamma^\mu +\frac{K^\nu}{2} \left(\gamma^\mu\gamma_\nu+\gamma_\nu\gamma^\mu\right)\\ &=K^\mu\mathbf1. \end{aligned}$$

Only the Clifford relation was used in the last step. Acting on any spinor column $\chi$ gives the identity in the form needed for closure:

$$(\gamma^\mu\widetilde\zeta)(\zeta\chi) +\zeta(\widetilde\zeta\gamma^\mu\chi)=K^\mu\chi.$$

Here $\chi$ may be odd. The parameter spinors commute, and no two odd fields have been interchanged. Taking $\chi=\partial_\mu\psi$ and summing over $\mu$ therefore yields

$$\boxed{\{\delta_\zeta,\delta_{\widetilde\zeta}\}\psi =-2iK^\mu\partial_\mu\psi.}$$

This is precisely the flat $z=0$ algebra in equation (6.1). No Dirac equation has entered. In particular, the contribution through $F_\Phi$ did not disappear merely because the field is auxiliary.

Eliminate the auxiliary and keep track of what is lost

For the free theory, the auxiliary equations set $F_\Phi=\widetilde F_\Phi=0$. This follows from the term $-\widetilde F_\Phi F_\Phi$ in equation (6.32). Algebraic elimination of auxiliary fields is also explained in Bilal’s section 4.3, equations (4.31)–(4.32), in four-dimensional conventions. Here we keep our three-dimensional transformations.

Denote the reduced transformations on $(\phi,\psi)$ by a hat. They have the same scalar rule and derivative term as before, but

$$\widehat\delta_\zeta\psi=0.$$

The route through $F_\Phi$ is now absent. Reusing the identity just proved,

$$\begin{aligned} \{\widehat\delta_\zeta,\widehat\delta_{\widetilde\zeta}\}\psi &=-2i(\gamma^\mu\widetilde\zeta)(\zeta\,\partial_\mu\psi)\\ &=-2iK^\mu\partial_\mu\psi +2i\zeta(\widetilde\zeta\,\mathcal E_\psi), \end{aligned}$$

where we have defined the spinor

$$\mathcal E_\psi=\gamma^\mu\partial_\mu\psi.$$

The free fermion equation is $\mathcal E_\psi=0$. To identify it, only the kinetic term is needed: varying $-i\widetilde\psi^TC\gamma^\mu\partial_\mu\psi$ with respect to the independent tilded spinor gives $C\mathcal E_\psi=0$, equivalent to $\mathcal E_\psi=0$ because $C$ is invertible. The complete action and its invariance will be treated in the next lesson.

The extra term is thus an equation-of-motion remainder. It vanishes on solutions, but generally survives on arbitrary field configurations. Its coefficient is exactly the negative of the missing auxiliary contribution:

$$\delta_{\widetilde\zeta}\delta_\zeta\psi =-2i\zeta(\widetilde\zeta\,\mathcal E_\psi).$$

Eliminating the auxiliary is a legitimate description of the free theory. It changes how its supersymmetry algebra is represented: the reduced transformations close on shell.

A profile for which the missing term is visible

Choose a positive reference mass $M$ and the commuting labels

$$\zeta=M^{-1/2}\begin{pmatrix}1\\0\end{pmatrix}, \qquad \widetilde\zeta=M^{-1/2}\begin{pmatrix}0\\1\end{pmatrix}, \qquad K^\mu=M^{-1}(1,0,0).$$

Let $\vartheta$ be a constant odd generator of dimension zero, and take

$$\psi=\vartheta \begin{pmatrix}u(x^1)\\v(x^1)\end{pmatrix},$$

with commuting functions $u,v$ of mass dimension one. A prime denotes an $x^1$ derivative. Direct matrix multiplication gives

$$\begin{aligned} \delta_\zeta\delta_{\widetilde\zeta}\psi &=-\frac{2i\vartheta}{M} \begin{pmatrix}0\\v'\end{pmatrix},\\ \delta_{\widetilde\zeta}\delta_\zeta\psi &=-\frac{2i\vartheta}{M} \begin{pmatrix}u'\\0\end{pmatrix}. \end{aligned}$$

The two terms supply the two components of $-2i\partial_1\psi/M$. The reduced algebra retains only the first term. Since

$$\mathcal E_\psi=\vartheta \begin{pmatrix}u'\\-v'\end{pmatrix},$$

its remainder is $2i\vartheta(u',0)^T/M$, as predicted.

For a particularly sharp test, choose $u=a x^1$, $v=0$, with constant $a$ of mass dimension two. The reduced mixed anticommutator is zero, while the required translation is $-2i\vartheta(a,0)^T/M$. The equation-of-motion remainder cancels that translation, explaining the failure. With the auxiliary present, its route supplies the missing result.

This remains true if the original configuration has $F_\Phi=0$: its variation is generally nonzero. Evaluating a completed two-step transformation at $F_\Phi=0$ is different from deleting $F_\Phi$ and its transformation before composing. The constant-profile limit $u'=v'=0$ makes both routes vanish and is a useful check.

Check the remaining components and the bracket convention

The scalar calculation already gave $\{\delta_\zeta,\delta_{\widetilde\zeta}\}\phi =-2iK^\mu\partial_\mu\phi$. On the auxiliary field the only nonzero mixed composition is

$$\delta_\zeta\delta_{\widetilde\zeta}F_\Phi =-2i(\widetilde\zeta\gamma^\mu\zeta)\partial_\mu F_\Phi =-2iK^\mu\partial_\mu F_\Phi.$$

The same-family operations square to zero. The only less immediate case is

$$\delta_{\widetilde\zeta}^{\,2}F_\Phi =-2(\widetilde\zeta\gamma^\mu\gamma^\nu\widetilde\zeta) \partial_\mu\partial_\nu\phi=0.$$

Indeed, the gamma product is $\delta^{\mu\nu}+i\varepsilon^{\mu\nu\rho}\gamma_\rho$. Its scalar part multiplies $\widetilde\zeta\widetilde\zeta=0$; its antisymmetric part contracts a symmetric pair of derivatives. Applying the same square to the sum of any two labels also shows that their same-family anticommutator vanishes. Thus the full chiral multiplet realizes the flat algebra off shell.

For ordinary even variations, attach independent odd constants: $\Delta_1=\lambda_1\delta_\zeta$ and $\Delta_2=\lambda_2\delta_{\widetilde\zeta}$. As derived in the algebra lesson, the odd operation crossing the other odd constant gives

$$[\Delta_1,\Delta_2]=-\lambda_1\lambda_2 \{\delta_\zeta,\delta_{\widetilde\zeta}\}.$$

Consequently the full even commutator translates $\psi$ by $2i\lambda_1\lambda_2K^\mu\partial_\mu\psi$. For the reduced transformations it contains the additional term $-2i\lambda_1\lambda_2\zeta(\widetilde\zeta\mathcal E_\psi)$. The off-shell versus on-shell distinction is the same in either bracket convention.

The dimensions also agree: $\mathcal E_\psi$ has dimension two, and the two spinor labels together have dimension minus one. The remainder therefore has the fermion’s dimension one. With compensating label R-charges $(+1,-1)$, it has the same charge $r-1$ as $\psi$.

Exercises

1. Can one pair of parameters miss a failure?

Use the labels in the worked example but take $\psi=\vartheta f(x^1)(0,1)^T$ with nonconstant $f$. Does the reduced mixed anticommutator give the expected translation for this pair? Does that prove off-shell closure? Repeat the test after exchanging the two label directions.

First hint

The remainder tests a particular projection $\widetilde\zeta\mathcal E_\psi$, not every component of $\mathcal E_\psi$.

Second hint

For the original labels, $\widetilde\zeta^TC=M^{-1/2}(1,0)$. After exchanging directions, the tilde row is $M^{-1/2}(0,-1)$.

Complete solution

Here $\mathcal E_\psi=-\vartheta f'(0,1)^T$. The original tilde row annihilates this spinor, so the remainder vanishes and the reduced algebra gives the expected translation for this pair, despite $\mathcal E_\psi\ne0$.

After exchanging directions, $K^1$ is still $1/M$, while

$$\widetilde\zeta\mathcal E_\psi =\frac{\vartheta f'}{\sqrt M}, \qquad 2i\zeta(\widetilde\zeta\mathcal E_\psi) =\frac{2i\vartheta f'}M\begin{pmatrix}0\\1\end{pmatrix}.$$

This cancels the required translation, and the reduced mixed anticommutator is zero. The new pair reveals the failure. Closure must hold for every allowed parameter and profile; one successful projection is insufficient. If the remainder vanishes for all nonzero $\zeta$ and both independent tilde directions, then both components of $\mathcal E_\psi$ must vanish.

2. Fix the auxiliary transformation coefficient

Keep every transformation unchanged except

$$\delta_{\widetilde\zeta}F_\Phi =-c\sqrt2\,i\widetilde\zeta\gamma^\mu\partial_\mu\psi,$$

where $c$ is a dimensionless constant. Find the mixed fermion anticommutator. Which value gives off-shell closure, and why could a test performed only on solutions fail to determine it?

First hint

Only the route through $F_\Phi$ is multiplied by $c$.

Second hint

Write the derivative route as the desired translation plus the remainder, then add the modified auxiliary route.

Complete solution

The result is

$$\{\delta_\zeta,\delta_{\widetilde\zeta}\}\psi =-2iK^\mu\partial_\mu\psi +2i(1-c)\zeta(\widetilde\zeta\mathcal E_\psi).$$

For arbitrary profiles and labels the second term vanishes only for $c=1$. The choice $u=a x^1,v=0$ in the worked example supplies a direct nonzero test of its coefficient. If every test first sets $\mathcal E_\psi=0$, the remainder vanishes for any $c$, so the mixed fermion calculation alone would not determine the coefficient. Off-shell closure is a stronger requirement than agreement on solutions.

3. Do the reduced transformations preserve the free equations?

For the reduced multiplet, assume both $\mathcal E_\psi=0$ and $\partial^2\phi=0$, where $\partial^2=\delta^{\mu\nu}\partial_\mu\partial_\nu$. Show that the hatted transformations preserve this pair of equations. Why is the scalar equation needed as well as the fermion equation?

First hint

Apply a transformation to $\gamma^\mu\partial_\mu\psi$. Constant labels commute with the derivatives.

Second hint

The Clifford relation and commuting derivatives imply $(\gamma^\mu\partial_\mu)^2=\partial^2\mathbf1$.

Complete solution

Since $\widehat\delta_\zeta\psi=0$, $\widehat\delta_\zeta\mathcal E_\psi=0$. The other operation gives

$$\widehat\delta_{\widetilde\zeta}\mathcal E_\psi =-\sqrt2\,i\gamma^\nu\gamma^\mu\widetilde\zeta\, \partial_\nu\partial_\mu\phi =-\sqrt2\,i\widetilde\zeta\,\partial^2\phi=0.$$

Thus preserving the fermion equation also uses the scalar equation. Conversely,

$$\widehat\delta_\zeta(\partial^2\phi) =\sqrt2\,\zeta\,\partial^2\psi =\sqrt2\,\zeta\gamma^\mu\partial_\mu\mathcal E_\psi=0, \qquad \widehat\delta_{\widetilde\zeta}(\partial^2\phi)=0.$$

The complete free solution set is preserved. This calculation explains why “on shell” refers to the field equations of the theory together, rather than permission to discard a convenient term in a single transformation. The identical argument applies to smooth local solutions; no boundary conditions or normalizability claim has been used.

From the algebra to the action

The auxiliary field supplies an explicit term in the fermion algebra. Keeping it gives closure on arbitrary profiles; eliminating it leaves the Dirac equation as the condition for reduced closure. A component count suggested this possibility, but the two composed transformations have now established it.

Before continuing, identify which route disappears when $F_\Phi$ is removed and explain why a configuration with $F_\Phi=0$ can still require that route. The next calculation, building the free chiral theory, asks a separate question: how do the scalar, fermion, and auxiliary terms in an action vary so that their sum is invariant?

Use the chiral-closure laboratory to reproduce this algebra with ordered odd symbols, remove the auxiliary before composition, and test an exchange-sign error that a scalar-only check misses.