Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 305 / 3

A gauge anomaly makes gauge redundancy inconsistent and must cancel. A chiral anomaly is the quantum nonconservation of a classically conserved axial current and may be a physical effect. A 't Hooft anomaly is an obstruction to gauging a global symmetry; it is preserved by renormalization-group flow and must be reproduced by the infrared theory, through massless degrees of freedom, symmetry breaking, or topological order.

The axial $U(1)$ is anomalous, while vector fermion number survives. In a left-handed convention, put ${\tilde{\psi }}_R=({\psi }_R{)}^c$. The continuous quantum symmetry and charges areup to finite quotients and the surviving discrete axial symmetry.

Normalize a fundamental cubic anomaly to $A({\mathbf{N}}_{\mathbf{f}})=1$ and its Dynkin index to $I({\mathbf{N}}_{\mathbf{f}})=1$. The two gauge colours giveThe left and charge-conjugated right fermions have opposite vector charges and equal multiplicities, soIf the theory confined with unbroken chiral symmetry, gauge-invariant operators contain an even number of $SU(2)$ fundamentals and are bosonic, so no massless composite fermions could match the nonzero chiral anomalies. 't Hooft anomaly matching therefore forces chiral symmetry breaking, with its infrared Goldstone theory carrying the anomaly.

There is no perturbative $SU(2)$ gauge anomaly because both the fundamental and adjoint representations are real or pseudoreal and have vanishing cubic gauge index. The Witten SU(2) anomaly also vanishes: the $N_f$ Dirac fermions give $2N_f$ fundamental Weyl doublets, an even number, while the adjoint Weyl fermion is in an integer-isospin real representation.

In left-handed variables, both ${\psi }_L$ and ${\tilde{\psi }}_R$ have $U(1{)}_X$ charge $-1$. Cancellation of the mixed $SU(2{)}_{\mathrm{gauge}}^{2}U(1{)}_X$ anomaly requireswhere $I(2)=1$ and $I(\mathrm{adj})=4$. Therefore

The adjoint fermion is a flavour singlet, soThere are $4N_f$ fundamental Weyl components of charge $-1$ and three adjoint components of charge $q=N_f/2$. Hence

For $N_f=3$, $q=3/2$. Under $SU(3{)}_L\times SU(3{)}_R\times U(1{)}_X$, the proposed left-handed composites transform asFor example, the charge of $\chi ={\bar{\psi }}_R{\lambda }_L{\psi }_L$ is $-1+3/2-1=-1/2$, and the antisymmetric product of two $SU(3)$ fundamentals is an antifundamental.

For $SU(3{)}_L^3$, the three right-flavour copies of $\chi$ contribute $3$, while $\eta$ contributes $-1$, giving $2$. For $SU(3{)}_L^{2}U(1{)}_X$, these fields contribute $3(-1/2)+(-1/2)=-2$. All fifteen components of $\chi$, $\eta$, and $\tilde{\eta }$ have charge $-1/2$, soThe ultraviolet formulas at $N_f=3$ give $2$, $-2$, $-15/8$, and $-15/2$ respectively. Every 't Hooft anomaly therefore matches.