Chiral multiplet closure

Before this lesson: Why auxiliary fields make the algebra close.

Can a computer show exactly what disappears from supersymmetry when an auxiliary field is removed? In this laboratory you will compose the transformations before imposing a field equation, repeat the calculation after removing the auxiliary, and inspect the remaining polynomial. A second experiment introduces a wrong parameter-exchange sign. The scalar test still passes, so checking the whole multiplet matters.

Prepare with why auxiliary fields make the algebra close. That lesson owns the analytic proof. Here we turn its rules into editable code, test products containing two independent odd fields, and compare with a calculation that can be done by hand.

Specify the calculation

Work locally in flat three-dimensional Euclidean space. All background sources vanish; the chiral is free and gauge neutral, with central charge $z=0$ . Its complexified $\mathcal N=2$ algebra has a Lorentzian parent with four real Poincaré supercharges. There is no gauge rank, boundary condition or curved-background spinor equation in this test.

The component fields are even $\phi,F$ and odd $\psi=(\psi_1,\psi_2)^T$ . Use constant commuting columns $\zeta,\widetilde\zeta$ , with

C=\begin{pmatrix}0&-1\\1&0\end{pmatrix},\qquad \gamma^1=\begin{pmatrix}1&0\\0&-1\end{pmatrix},\quad \gamma^2=\begin{pmatrix}0&-1\\-1&0\end{pmatrix},\quad \gamma^3=\begin{pmatrix}0&i\\-i&0\end{pmatrix}. \tag{LAB02.1}

The contractions are $uv=u^TCv$ and $u\gamma^\mu v=u^TC\gamma^\mu v$ ; transpose does not take a complex conjugate. These are P’s Appendix A.1 conventions.

The source’s operations with commuting labels are odd. To implement even variations with the ordinary product rule, introduce independent odd constants $\rho_1,\rho_2$ and set $\Delta_1=\rho_1\delta_\zeta$ , $\Delta_2=\rho_2\delta_{\widetilde\zeta}$ . Every odd parameter anticommutes with every odd field and its derivatives. The required flat specialization of P (6.8) is

\begin{aligned} \Delta_1\phi&=\sqrt2\rho_1\zeta^TC\psi,& \Delta_2\phi&=0,\\ \Delta_1\psi&=\sqrt2\rho_1\zeta F,& \Delta_2\psi&=-\sqrt2i\rho_2\gamma^\mu\widetilde\zeta\,\partial_\mu\phi,\\ \Delta_1F&=0,& \Delta_2F&=-\sqrt2i\rho_2\widetilde\zeta^TC\gamma^\mu\partial_\mu\psi. \end{aligned} \tag{LAB02.2}

The parameters themselves do not vary. Because $\Delta_i$ is even, $\Delta_i(XY)=(\Delta_iX)Y+X(\Delta_iY)$ , including when both $X,Y$ are odd. Signs arise when the resulting factors are put back into their chosen order. Adding an odd-derivation sign to this product rule would count a sign twice.

For example, the two paths on the fermion give

\begin{aligned} \Delta_1\Delta_2\psi &=-2i\rho_2\rho_1(\gamma^\mu\widetilde\zeta) (\zeta^TC\partial_\mu\psi),\\ \Delta_2\Delta_1\psi &=-2i\rho_1\rho_2\zeta (\widetilde\zeta^TC\gamma^\mu\partial_\mu\psi). \end{aligned} \tag{LAB02.3}

The first line must be reordered before subtraction. With $K^\mu=\zeta^TC\gamma^\mu\widetilde\zeta$ , the expected result on every component $X$ is

[\Delta_1,\Delta_2]X =2i\rho_1\rho_2K^\mu\partial_\mu X. \tag{LAB02.4}

This is the even-variation version of P (6.1). The program substitutes the rules in both orders; it does not insert (LAB02.4) as its calculated answer.

Represent fields and their derivatives

A jet here means a field value together with finitely many of its derivatives at one point. For instance, psi1, d1_psi1 and d12_psi1 represent $\psi_1$ , $\partial_1\psi_1$ and $\partial_1\partial_2\psi_1$ . They are independent odd generators for the initial algebraic calculation. A symmetric derivative multi-index identifies $\partial_1\partial_2$ with $\partial_2\partial_1$ , appropriate to this flat, ungauged example.

The program represents an ordered odd monomial by an increasing tuple of generator numbers. A polynomial is a dictionary from those tuples to commuting SymPy coefficients. The coefficients can contain arbitrary scalar and auxiliary jets. Multiplication checks for a repeated odd generator, which makes that monomial zero, and then counts exchanges needed to sort the concatenated word. It never uses commuting symbols as stand-ins for fermions.

The order is fixed:

\rho_1<\rho_2<\text{all jets of }\psi_1<\text{all jets of }\psi_2.

The direction of the order is conventional; using it consistently is essential. The two parameters are kept independent. Setting them equal would make $\rho_1\rho_2=0$ and conceal the mixed algebra.

Mathematical operation	Downloaded implementation
Ordered odd multiplication	`Algebra.mul`
Even scalar and auxiliary jets	`Algebra.even`
Odd parameter and fermion jets	`Algebra.odd`
A spacetime derivative	`Algebra.derivative`
Ordinary even Leibniz rule and chain rule	`Algebra.derive`
One actual component transformation	`Algebra.transform`
Both compositions followed by subtraction	`comm` inside `run`

Two derivative orders suffice for these transformations, including the same-family bracket on $F$ . This is not a derivative expansion: second derivatives are retained independently, and the program raises an error if an operation requires a higher jet. It does not silently set higher derivatives to zero.

The formal tests cover all four components, both same-family brackets, and the products $\phi\psi_1$ , $\psi_1\psi_2$ , and $F\psi_2$ . The labels have arbitrary commuting symbolic components. Checking $\psi_1\psi_2$ forces the calculation to handle parameter–field and field–field exchanges, which a linear component test alone might miss.

Remove the auxiliary at the correct stage

The free auxiliary equation from the $-\widetilde F F$ term in P (6.32) sets $F=0$ . If we remove it before composing, the rule $\Delta_1\psi$ disappears. Write a hat on these reduced transformations. The diagnostic should be

\begin{aligned} \mathcal R_\psi &=[\widehat\Delta_1,\widehat\Delta_2]\psi -2i\rho_1\rho_2K^\mu\partial_\mu\psi\\ &=-2i\rho_1\rho_2\zeta (\widetilde\zeta^TC\mathcal E_\psi), \qquad \mathcal E_\psi=\gamma^\mu\partial_\mu\psi. \end{aligned} \tag{LAB02.5}

The program computes this remainder and compares it with the displayed expression. No equation of motion is imposed in obtaining it. On solutions of the free Dirac equation it vanishes, while it generally survives on arbitrary jets. The scalar still closes. This is the distinction between on-shell and off-shell closure, not a failure of the free theory.

Evaluating a completed two-step transformation at a configuration with $F=0$ is different: $\Delta_2F$ need not vanish there. The complete algebra still contains that route. The computation therefore deletes the auxiliary rule explicitly, rather than merely substituting a zero value for $F$ at the end.

Download and reproduce

Download the complete lab (ZIP) with the program, pinned dependencies, reference output and run instructions.

Save the editable Python calculation, pinned requirements, and complete expected output in one folder. The tested environment is Python 3.13.7, SymPy 1.13.3 and mpmath 1.3.0. Only SymPy is used directly; mpmath is its pinned supporting dependency.

In Windows PowerShell, from that folder:

python -m venv .venv
.venv/Scripts/python.exe -m pip install -r requirements.txt
.venv/Scripts/python.exe chiral_closure.py

On macOS or Linux use python3 -m venv .venv, then .venv/bin/python in the remaining commands. Package installation needs registry access; running the downloaded calculation needs no network, external data or notebook service.

The default output includes

{
  "total": 29,
  "off_shell_closure": true,
  "reduced_fermion_closes_for_arbitrary_jets": false
}

These are 29 grouped exact assertions, not 29 distinct physical results. Four check complete component identities, three check products, eight check same-family composition, four check the reduced remainder, four check the displayed profile below, and six check elementary ordering. Each component identity compares every coefficient of its resulting polynomial.

There is no sampling error, tolerance, numerical cutoff or spacetime lattice. Exact arithmetic does not prevent coding the wrong transformation. The source translation, explicit profile and changed-sign test address that separate risk. The downloadable script is editable and defines its rules visibly.

A hand benchmark with two independent odd fields

Introduce a positive reference mass $M$ and dimensionless coordinates $u^\mu=Mx^\mu$ . Scale the fields by $\phi/M^{1/2}$ , $\psi/M$ , $F/M^{3/2}$ and the spinor labels by $M^{1/2}$ . The program uses these dimensionless quantities, equivalent to setting $M=1$ . The odd constants have dimension zero. Restoring these powers gives $[\phi]=1/2$ , $[\psi]=1$ , $[F]=3/2$ and $[\zeta]=[\widetilde\zeta]=-1/2$ .

Take the dimensionless labels $\zeta=(1,0)^T$ , $\widetilde\zeta=(0,1)^T$ and two independent odd constants $\theta_1,\theta_2$ , ordered after the two $\rho$ ‘s. Set

\psi=\begin{pmatrix}\theta_1(u^1)^2\\\theta_2u^1u^2\end{pmatrix}, \quad K=(1,0,0),\qquad \mathcal E_\psi= \begin{pmatrix}2u^1\theta_1-u^1\theta_2\\-u^2\theta_2\end{pmatrix}. \tag{LAB02.6}

The second term in the upper Dirac component comes from $\gamma^2\partial_2\psi$ . Thus the expected ordered answers are

\begin{aligned} [\Delta_1,\Delta_2]\psi &=\rho_1\rho_2 \begin{pmatrix}4iu^1\theta_1\\2iu^2\theta_2\end{pmatrix},\\ [\widehat\Delta_1,\widehat\Delta_2]\psi &=\rho_1\rho_2 \begin{pmatrix}2iu^1\theta_2\\2iu^2\theta_2\end{pmatrix}. \end{aligned} \tag{LAB02.7}

Their difference is exactly (LAB02.5). This example changes in two spatial directions and uses two distinct odd generators; it is not a single commuting spinor substituted into a fermion formula. The program evaluates this profile after composing the arbitrary-jet transformations. In the profile output only, the word labels psi1 and psi2 denote $\theta_1,\theta_2$ , as its printed dictionary explains.

A separate check represents $\rho_1,\rho_2,\theta_1,\theta_2$ by tensor-product fermionic creation matrices. Their matrix products enforce the same signs without sorting tuples. Direct polynomial-profile differentiation in that representation reproduces both component answers and the nonlinear product calculation below. These are auxiliary matrices for an algebra test, not a truncation of the quantum field theory’s Hilbert space.

Change one input

1. Introduce a wrong parameter exchange

Run

.venv/Scripts/python.exe chiral_closure.py --wrong-parameter-sign

This changes only the exchange of $\rho_2$ with $\rho_1$ to a plus. Before reading the result, determine whether the scalar, fermion and auxiliary tests must all detect it. Would passing only the scalar test establish closure?

Hint 1

For the scalar, only $\Delta_2\Delta_1\phi$ is nonzero. Its parameter product is already $\rho_1\rho_2$ .

Hint 2

The first line of (LAB02.3) instead contains $\rho_2\rho_1$ . The auxiliary’s nonzero mixed composition has that order too.

Solution

The diagnostic reports

"closure_matches_translation": {
  "phi": true,
  "psi1": false,
  "psi2": false,
  "F": false
}

The scalar cannot see the altered exchange. The other tests require it, so their residual polynomials are nonzero for generic labels and jets. The program exits successfully because it detected the deliberately introduced error; it does not certify a second supersymmetry convention. In particular, nilpotency of each parameter by itself would not fix their relative exchange sign. Restore the ordinary run before making physical inferences.

2. Test a product rather than one component

For (LAB02.6), compute the full and reduced commutators on $\psi_1\psi_2$ . Put every answer in the order $\rho_1\rho_2\theta_1\theta_2$ . Why can replacing $\theta_1,\theta_2$ by one common odd constant conceal this test?

Hint 1

The commutator of two even derivations is even, so its product rule has a plus sign. Use both components of (LAB02.7).

Hint 2

Moving the pair $\rho_1\rho_2$ through one odd field gives two minus signs. In the reduced upper-component contribution, $\theta_2^2=0$ .

Solution

The original product is $\psi_1\psi_2=\theta_1\theta_2(u^1)^3u^2$ . Its full translation is

[\Delta_1,\Delta_2](\psi_1\psi_2) =6i(u^1)^2u^2\rho_1\rho_2\theta_1\theta_2.

The upper reduced contribution is proportional to $\theta_2^2$ and vanishes. The lower one remains, giving

[\widehat\Delta_1,\widehat\Delta_2](\psi_1\psi_2) =2i(u^1)^2u^2\rho_1\rho_2\theta_1\theta_2.

The difference is $-4i(u^1)^2u^2\rho_1\rho_2\theta_1\theta_2$ , also obtained by multiplying the upper-component remainder in (LAB02.5) by $\psi_2$ . A common generator would make the original product and both answers zero. That would be a poor test of the product’s exchange signs, even though a one-generator profile can still reveal a linear component failure.

3. Change a coordinate dependence

Replace the profile by $\psi=(\theta_1(u^2)^2,0)^T$ , keeping the same labels. Is the reduced remainder zero? Is the fermion on shell? Find a tilde-label direction that detects any missed remainder.

Hint 1

Here $K\cdot\partial=\partial_1$ annihilates the profile, while $\mathcal E_\psi=\gamma^2\partial_2\psi$ need not vanish.

Hint 2

The original tilde row is $(1,0)$ . The choice $\widetilde\zeta=(1,0)^T$ instead has row $(0,-1)$ .

Solution

Directly, $\mathcal E_\psi=(0,-2u^2\theta_1)^T$ . The original tilde row projects this to zero, so the reduced remainder vanishes for that pair even though the profile is not a Dirac solution. The full and reduced mixed commutators both vanish there.

Keep $\zeta=(1,0)^T$ but choose $\widetilde\zeta=(1,0)^T$ . The new vector is $K=(0,1,i)$ , so the required translation is $4iu^2\rho_1\rho_2\theta_1(1,0)^T$ . The remainder is its negative, $-4iu^2\rho_1\rho_2\theta_1(1,0)^T$ , and the reduced commutator again vanishes. This time it fails the required translation. Both real spatial derivatives and complex parameter bilinears are allowed in the complexified Euclidean algebra. Closure for arbitrary allowed labels is stronger than success for a single projection.

The laboratory establishes local classical transformation identities in the stated free model. It does not test action invariance, integration contours, quantum anomalies or closure with gauge, central-charge or curved-background terms. Those require their own transformations and assumptions; the exactness of the present arithmetic does not supply them.