From representations to local fields

A gauge singlet in a tensor product is a candidate ingredient of an interaction. To obtain a local scalar operator, one must also contract Lorentz indices and impose the exchange symmetry of repeated fields. These operations interact: a contraction that is allowed for two distinguishable representation factors can vanish when the factors are copies of the same field.

The local-operator workbench makes this distinction explicit. Specify a spacetime, a global gauge group, field species, and a monomial. The result is the dimension of the corresponding space of complex invariant polynomials, with Bose and Fermi statistics already imposed. This guide develops the field-space construction behind that answer; the method reference gives the exact calculation.

Which Lorentz representation?

In four-dimensional Lorentzian spacetime, the connected spin group is $\operatorname{Spin}^{+}(1,3)\cong SL(2,\mathbb C)$, viewed as a real Lie group. Its complexified Lie algebra is

$$\mathfrak{so}(1,3)\otimes_{\mathbb R}\mathbb C \cong\mathfrak{sl}_2(\mathbb C)\oplus\mathfrak{sl}_2(\mathbb C).$$

Finite-dimensional complex field representations are labeled by $(j_\ell,j_r)$. The two spin labels refer to these algebra factors; neither is the weak-isospin label. Undotted and dotted spinor indices distinguish the two fundamental representations. See Dreiner–Haber–Martin, Section 2 for explicit two-component conventions.

Field kind	Lorentz label	Spin components	Statistics	Canonical dimension in four dimensions
Scalar	$(0,0)$	$1$	Commuting	$1$
Left Weyl	$(\tfrac12,0)$	$2$	Anticommuting	$\tfrac32$
Right Weyl	$(0,\tfrac12)$	$2$	Anticommuting	$\tfrac32$

The canonical dimensions follow by assigning dimension four to the Lagrangian and one to a derivative in the usual scalar and fermion kinetic terms. They are engineering dimensions, not anomalous dimensions at an interacting fixed point.

For finite polynomial invariants, one may replace the complexified Lorentz algebra by its compact real form, represented by $SU(2)_\ell\times SU(2)_r$. A vector annihilated by the compact Lie algebra is also annihilated by its complexification, and conversely. The invariant subspace is therefore the same. This replacement is an algebraic projection device; the physical setup remains Lorentzian. It does not assert an isomorphism between the Lorentzian spin group and $\operatorname{Spin}(4)$.

These field representations also differ from unitary one-particle representations of the Poincaré group. The calculation concerns field components at one spacetime point, without derivatives, momenta, or an on-shell little-group construction. “Scalar” here means invariant under the connected Lorentz spin group; no separate parity or CP condition is imposed.

Build the complete one-field space

For a species $\Phi$ with $f$ flavors, form

$$R_\Phi =V_{j_\ell,j_r}\otimes V_{\mathrm{gauge}} \otimes F_\Phi, \qquad F_\Phi=\mathbb C^f.$$

The gauge and Lorentz groups act trivially on $F_\Phi$. Flavors are independent copies of the same field representation, and all their coefficient tensors are counted. Selecting three flavors does not impose invariance under an additional $U(3)$ flavor group.

For example, a left-handed quark doublet with three flavors has

$$R_Q=\mathbb C^2_\ell\otimes\mathbb C^3_c \otimes\mathbb C^2_w\otimes\mathbb C^3_f, \qquad \dim R_Q=36.$$

The two weak components are part of one species. The three colors are also components of that species. Neither should be entered as separate field rows. Distinct rows instead denote independent fields; repeated copies within a row are controlled by its selected power.

If a monomial contains $m$ copies of $\Phi$, its space is

$$\operatorname{Sym}^m R_\Phi \quad\text{for a scalar}, \qquad \Lambda^m R_\Phi \quad\text{for a Weyl fermion}.$$

The power acts on the entire representation, including flavor. In particular, $\Lambda^m(V_{\mathrm{spin}}\otimes V_{\mathrm{gauge}})$ cannot be replaced by the tensor product of the separate exterior powers. At degree two,

$$\begin{aligned} \operatorname{Sym}^2(A\otimes B) &=(\operatorname{Sym}^2A\otimes\operatorname{Sym}^2B) \oplus(\Lambda^2A\otimes\Lambda^2B),\\ \Lambda^2(A\otimes B) &=(\Lambda^2A\otimes\operatorname{Sym}^2B) \oplus(\operatorname{Sym}^2A\otimes\Lambda^2B). \end{aligned}$$

These decompositions follow by splitting the two-factor exchange on each space into its eigenvalues $+1$ and $-1$. The total exchange eigenvalue is their product. This is the essential reason that gauge symmetry, spin, and statistics must be treated together.

Three bilinears that expose the difference

Let $\epsilon_{12}=+1$ and use undotted indices for left Weyl fields. For anticommuting spinors, the contracted bilinear is symmetric in the two spinors, because exchanging them introduces both a Grassmann sign and an antisymmetric spinor contraction. This is the sign identity in Dreiner–Haber–Martin, equation (2.60).

Gauge-singlet Weyl fields

For sterile fields $N_a^\alpha$, define

$$B_{ab}=\epsilon_{\alpha\beta}N_a^\alpha N_b^\beta.$$

Exchanging $a,b$, moving the fields past one another, and renaming $\alpha,\beta$ gives $B_{ba}=B_{ab}$. Consequently there are

$$\dim\operatorname{Sym}^2\mathbb C^f =\frac{f(f+1)}2$$

independent bilinears: $1,3,6$ for one, two, or three flavors. A single Weyl field can have a Majorana bilinear. Grassmann nilpotence sets the square of an individual component to zero; it does not set the product of its two independent spin components to zero.

Neutral weak-doublet Weyl fields

Now give the Weyl fields a weak index $i$, with zero abelian charge, and consider

$$D_{ab}=\epsilon_{\alpha\beta}\epsilon_{ij} \psi_a^{\alpha i}\psi_b^{\beta j}.$$

There are two antisymmetric invariant tensors, so the spin-and-weak contraction is symmetric under exchanging complete field indices. Grassmann exchange then gives $D_{ba}=-D_{ab}$. The count is

$$\dim\Lambda^2\mathbb C^f=\frac{f(f-1)}2.$$

The answers are $0,1,3$. In particular, one neutral weak doublet has no such bilinear even though both $\mathbf2_\ell^{\otimes2}$ and $\mathbf2_w^{\otimes2}$ contain a singlet. Use a compatible global gauge group for this example, such as the direct product. This illustrative field content is not an anomaly-cancellation claim.

Commuting weak doublets

For one scalar doublet,

$$\epsilon_{ij}H^iH^j=H^1H^2-H^2H^1=0.$$

Two independently declared doublets can instead give $\epsilon_{ij}H_1^iH_2^j\ne0$, provided their other charges allow it. For one Higgs species, a conjugate row permits $H^\dagger H$. With one doublet and its conjugate, the invariants at powers $(k,k)$ are generated by $(H^\dagger H)^k$, giving one invariant for $k=1,2,3$ within the workbench’s dimension bound.

Chirality and conjugation

A left and a right Weyl representation multiply as

$$(\tfrac12,0)\otimes(0,\tfrac12)=(\tfrac12,\tfrac12).$$

This is a Lorentz vector, so two such uncontracted fields alone do not make a scalar, even when their gauge representations are conjugate. A derivative or another vector can change the question, but lies outside the derivative-free bilinear calculation. A necessary selection rule for the supported scalar/Weyl fields is that the total numbers of left and right Weyl factors are each even. It is not sufficient once gauge symmetry and repeated-field statistics are included.

Complex conjugation exchanges the two Weyl chiralities, conjugates the gauge representation, and reverses the abelian charge. A Hermitian-conjugate field is represented by its own row when it occurs in a monomial. In the complex polynomial calculation, its components and the original components are treated as algebraically independent variables. The real or Hermitian condition on a Lagrangian is a subsequent operation.

For this reason a count of six means a six-dimensional complex invariant space. It is not automatically six real parameters, six masses, or six observables after flavor-basis transformations. Those questions require additional information about the action, other couplings, and the allowed field redefinitions.

Start with a valid physical setup

The workbench first checks that every declared field is a representation of the selected global gauge group. For the standard diagonal quotients

$$G_d=\frac{SU(3)\times SU(2)\times U(1)}{\Gamma_d}, \qquad d\in\{1,2,3,6\},$$

the integer abelian normalization and the embedded subgroup matter. A collection of fields can have a neutral tensor product even when an individual field fails to descend to $G_d$. Such a setup is rejected before counting; cancellation in a composite does not repair the field representation. See the Standard Model global group and the descent method.

Once the fields descend, calculate a specified monomial, inspect the count and selection rules, and export the complete setup. The worked physical setups include the contrasting bilinears above, the Weinberg operator, and Yukawa and four-fermion examples. Their purpose is to make each assumption reproducible: the field content, conjugation convention, flavor multiplicities, global group, and selected powers all belong to the question.