research37 min studyLesson 714 in Physics

Audit a chiral spectrum for gauge consistency

Convert every fermion to an all-left-handed convention and check independent gauge and mixed gravitational anomaly coefficients.

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What you’ll understand

  • Count charges and multiplicities in an all-left-handed spectrum.
  • Check cubic, mixed, and gauge-gravitational anomaly conditions independently.
  • Distinguish perturbative cancellation from a complete global consistency audit.

A chiral gauge theory may contain different left- and right-handed representations, but gauge redundancy must remain an exact property of its quantum definition. An uncanceled gauge anomaly spoils the identities needed to remove unphysical gauge degrees of freedom. Cancellation is consequently a constraint on the entire matter content, not an optional adjustment to an observable. A spectrum can pass one anomaly test and fail another, so the bookkeeping should separate all independent invariant coefficients.

Put every Weyl field in the same convention

Use the left-handed anomaly-polynomial sign and normalized Abelian charge convention of the preceding lesson. Replace a right-handed Weyl fermion in representation R and charge q by its left-handed charge conjugate in the conjugate representation and charge minus q. Then sum all left-handed contributions. Keeping a right-handed field with its original charge and adding rather than subtracting its cubic contribution would falsely make a vectorlike Dirac pair anomalous. Charge conjugation changes the representation as well as the handedness label.

Multiplicity matters. An Abelian anomaly sum counts every independent component of a multiplet under the other gauge factors, including color and weak-isospin dimension. A mixed non-Abelian-squared–Abelian sum instead contains the quadratic index of the non-Abelian representation and dimensions of any remaining factors. Distinguish this index from representation dimension. For the fundamental of SU(N), the convention tr⁡(TaTb)=δab/2\operatorname{tr}(T^aT^b)=\delta^{ab}/2 gives index one half, not N.

C3=∑leftdiqi3,C1=∑leftdiqi,I6U(1)=C36x3−C124x p1(T)C_3=\sum_{\rm left}d_iq_i^3,\qquad C_1=\sum_{\rm left}d_iq_i,\qquad I_6^{U(1)}=\frac{C_3}{6}x^3-\frac{C_1}{24}x\,p_1(T)
d_i counts degeneracy from other group representations, without counting a field twice. C₃ must vanish for the pure Abelian perturbative gauge anomaly, and C₁ for its mixed gravitational anomaly when gravitational backgrounds are allowed. Charges are in the same normalized U(1) lattice; no extra coupling factor is hidden in these algebraic cancellation tests.↔ Scroll wide equations horizontally.

Test each independent invariant tensor

For a simple non-Abelian gauge factor, the local four-dimensional anomaly is proportional to the symmetrized cubic trace. The sum over all left-handed representations must vanish. Real or pseudoreal representations have no such cubic local anomaly, but that statement does not eliminate possible global anomalies. For product groups, mixed terms impose further conditions. A single non-Abelian generator has zero trace in ordinary irreducible representations of a simple group, explaining why some apparent mixed triangles vanish automatically while those with two non-Abelian vertices need explicit checking.

The Standard Model provides a useful multiplicity diagnostic. In one generation, normalized hypercharge y=6Yy=6Y gives left-handed multiplets with multiplicities and charges (6,1),(3,−4),(3,2),(2,−3),(1,6)(6,1),(3,-4),(3,2),(2,-3),(1,6). The negative charge of the conjugated right-handed up quark is particularly easy to miss. Both their linear and cubic charge sums vanish, as do the independent color-squared–hypercharge and weak-squared–hypercharge coefficients. This cancellation links quark and lepton contributions rather than occurring separately for every individual species.

∑itr⁡Ri ⁣(Ta{Tb,Tc})=0,∑idiother qiT(Ri)=0,tr⁡R(TaTb)=T(R)δab\sum_i\operatorname{tr}_{R_i}\!\left(T^a\{T^b,T^c\}\right)=0,\qquad \sum_i d_i^{\rm other}\,q_iT(R_i)=0,\qquad \operatorname{tr}_{R}(T^aT^b)=T(R)\delta^{ab}
The first condition is for the cubic anomaly of a gauged simple factor, with an all-left-handed sum. The second is the mixed G²U(1) condition; d_i^other counts factors other than G and U(1). Distinct Abelian factors also require all mixed cubic charge sums. These equations test perturbative local anomalies.↔ Scroll wide equations horizontally.

Know what cancellation does and does not establish

A vectorlike pair cancels its own local anomaly because the conjugate representation has the opposite cubic coefficient. A genuinely chiral anomaly-free spectrum can also exist without such pairs, so anomaly cancellation does not require every fermion to admit an invariant mass. Conversely, adding arbitrary scalar fields does not cancel a Weyl triangle contribution. Additional transforming fields and Wess–Zumino or Green–Schwarz terms can participate in cancellation in a larger effective theory, but those structures must be specified rather than silently assumed.

Passing the local tests is necessary for an ordinary stand-alone gauge theory, but not a proof of every global or dynamical property. Large gauge transformations can reveal a further sign obstruction, and the global form of the gauge group and allowed spacetime structures affect the complete audit. Nor does cancellation establish confinement, a mass gap, or phenomenological viability. It establishes the absence of the tested quantum gauge obstruction, leaving those other questions to additional analysis.

An anomaly-free spectrum without opposite-charge pairs

Five left-handed Weyl singlets have U(1) charges 1,5,−7,−8,91,5,-7,-8,9, with no other gauge factor. Check the pure Abelian and mixed gravitational coefficients. Is cancellation simply due to opposite-charge vectorlike pairs?

  1. The linear sum is C1=1+5−7−8+9=0C_1=1+5-7-8+9=0. Thus the mixed U(1)-gravitational polynomial term vanishes in the declared normalization. Each listed field has multiplicity one.

  2. The cubic sum is C3=1+125−343−512+729=0C_3=1+125-343-512+729=0. The positive cubes total 855 and the negative cubes total −855, so the pure Abelian local anomaly also cancels.

  3. No charge occurs together with its opposite, so the cancellation cannot be decomposed into the obvious opposite-charge Dirac pairs within this list. This is a chiral spectrum passing the stated local Abelian tests; no claim about its strong-coupling phase follows.

The result

Both local coefficients vanish despite the absence of opposite-charge partners. Chiral matter can be anomaly free without being vectorlike, provided the entire spectrum satisfies all required conditions.

Explore the model

Interactive lab

Interactive lab · function

See why cubic weights differ from linear weights

Reset and set both controls to coefficient a = 1 and position x = 1; the signed area is 1/3, so three times area is 1. Reset and set both controls to a = 1 and x = 2; area is 8/3, so three times area is 8. This exact parabola integral illustrates cubic scaling of a charge weight. It neither counts fermion multiplets nor tests a full gauge theory's anomaly cancellation.

See why cubic weights differ from linear weightsThis lab uses only y = ax². Playback traces the curve within the fixed integration interval from 0 to x; it does not estimate numerical error or change the chosen point. Current readings: y at chosen x: 0.608; Tangent slope · 2ax: 1.35; Signed area · ax³/3: 0.182. All parameters are adjustable with the labeled controls below.−2.4−1.201.22.4−9−4.504.59xy(x, ax²)
y = ax²TangentSigned integral
y at chosen x
0.608
Tangent slope · 2ax
1.35
Signed area · ax³/3
0.182
t = 0 s

This lab uses only y = ax². Playback traces the curve within the fixed integration interval from 0 to x; it does not estimate numerical error or change the chosen point. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forDoubling a charge multiplies its cubic weight by eight, while its linear gravitational weight only doubles.

Check your understanding

How should a right-handed charge-q Weyl fermion be included in an all-left-handed anomaly sum?

Try it yourself

Another all-left-handed Abelian spectrum has charges 3,4,5,−63,4,5,-6, each with multiplicity one. Check C₃ and C₁ separately. Does vanishing of the pure gauge coefficient suffice if the theory must also couple consistently to arbitrary gravitational backgrounds without additional cancellation fields?

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

Use the cube identity 27 + 64 + 125 = 216, then perform the independent linear sum.

Compare with the solution

The cubic coefficient is C3=27+64+125−216=0C_3=27+64+125-216=0, but C1=3+4+5−6=6C_1=3+4+5-6=6. The pure U(1) local gauge anomaly cancels, while the mixed term is −(6/24)xp1=−xp1/4-(6/24)xp_1=-xp_1/4 and remains nonzero. Thus the stipulated full gauge-gravitational consistency requirement fails. A flat-background triangle test alone would have missed this independent obstruction; no unspecified compensating sector should be inferred.

Take it with you

  • Convert handedness and representation consistently before summing.
  • Cubic, mixed, and gravitational coefficients impose independent tests.
  • Local anomaly cancellation is essential but does not establish global consistency or dynamics by itself.

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