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A conjugate SU(3)×U(1) chiral pair with zero local polynomial

Physical scope

A worked algebraic check of the flat-space one-loop background model with positive-chirality species (3,+1) and (bar 3,-1); no global-anomaly, dynamical-gauge, empirical, or continuum claim is made.

Example

Use the flat four-dimensional background data with global group $S U (3) \times U (1)_{V}$ and the common consistent regulator fixed above. Take the finite spectrum to consist of positive-chirality Weyl species $E = (3, + 1)$ and $E^{*} = (\overset{ˉ}{3}, - 1)$ . On a formal six-dimensional extension, write the anti-Hermitian $S U (3)$ curvature as $F_{A}$ and the $U (1)_{V}$ curvature as $f = d B$ . The defining-representation curvature is $F_{E} = F_{A} + i f I_{3}$ , with $tr F_{A} = 0$ , so

$I_{6} (E) = \frac{i}{24 π^{2}} (tr F_{A}^{3} + 3 i f \land tr F_{A}^{2} + 3 (i f)^{3}) .$

The dual species has the opposite polynomial. The total polynomial and its common linear consistent descent therefore vanish, so this example has zero perturbative local anomaly class in the selected scheme. The six-dimensional extension is only the formal degree-six polynomial domain; the physical model remains four-dimensional.

Example references

Bilal, §§7.1–7.2, 9.4, and 10.4

Alvarez, §§2.2–2.3

Facts & Assumptions

Given: The flat-space spectrum $E \oplus E^{*}$ , smooth background curvatures, the normalized degree-six polynomial, and the common source-backed large-mass Pauli–Villars one-loop descent input.

[F1]

Flat-space chiral anomaly-matching data and regulator convention: the selected backgrounds are nondynamical, the common regulator uses one linear consistent descent for each positive-chirality species, and the total polynomial is the sum over species.

[F2]

Degree-six cubic trace polynomial for a chiral representation: for $E = (3, + 1)$ , $F_{E} = F_{A} + i f I_{3}$ and $I_{6} (E) = \frac{i}{24 π^{2}} tr (F_{E}^{3})$ .

[F3]

A conjugate representation reverses the cubic anomaly polynomial: the dual connection has $F_{E^{*}} = - F_{E}^{T}$ , so its cubic polynomial and canonical transgression are the negatives of those for $E$ .

[F4]

Local anomaly cancellation for a conjugate chiral pair: under these model and regulator assumptions, the relative local anomaly class of the conjugate pair is zero.

Verification

1.1F1F2algebra

Since $i f I_{3}$ commutes with the even two-form $F_{A}$ , the binomial expansion gives $tr (F_{E}^{3}) = tr (F_{A}^{3}) + 3 i f \land tr (F_{A}^{2}) + 3 (i f)^{2} \land tr (F_{A}) + (i f)^{3} tr (I_{3})$ . The third term vanishes because $tr F_{A} = 0$ , and $tr (I_{3}) = 3$ , yielding the displayed polynomial.

2.1F2F3step 1.1algebra

The dual representation reverses this entire cubic trace, including its pure $S U (3)$ , mixed $U (1)_{V}$ – $S U (3)^{2}$ , and pure $U (1)_{V}^{3}$ terms, by [F3]. Summing the two species therefore gives $I_{6} (E) + I_{6} (E^{*}) = 0$ .

3.1F1F4step 2.1given∎

The common linear descent of the zero polynomial has zero four-form representative, so [F4] applies and gives zero class in $H^{1, 4} (s ∣ d_{H})$ . This verifies the example under the stated regulator input only; it is not a test of global transformations.

A conjugate SU(3)×U(1) chiral pair with zero local polynomial

Example

Example references

Facts & Assumptions

Verification

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