Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 305 / 3

A gauge anomaly makes the gauge symmetry inconsistent and must cancel. An Adler-Bell-Jackiw anomaly breaks a classical global chiral symmetry quantum mechanically but does not by itself invalidate the gauge theory. A 't Hooft anomaly is an obstruction to gauging a global symmetry; it remains invariant along renormalization-group flow and constrains the infrared spectrum or symmetry-breaking pattern.

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In the normalization supplied, a left-handed Weyl fermion in the two-index antisymmetric representation has cubic $SU(N)$ gauge-anomaly coefficient $N-4$, while an antifundamental has coefficient $-1$. Cancellation of the gauge anomaly requiresso

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Classically the $p$ antifundamentals admit $U(p)$ flavour rotations and $\lambda$ admits an independent phase rotation, giving $SU(p)\times U(1{)}_{\psi }\times U(1{)}_{\lambda }$ up to finite quotients. A $U(1)$ with charges $q_{\lambda },q_{\psi }$ is free of the mixed $SU(N{)}^{2}U(1)$ anomaly whenUsing $p=N-4$, chooseThe orthogonal axial phase has a chiral anomaly, while this combination survives. The continuous quantum global symmetry is therefore $SU(p)\times U(1)$, again up to possible finite central quotients.

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Use the charges $q_{\lambda }=N-4$ and $q_{\psi }=-(N-2)$ and the index convention $I(\mathbf{p})=1$. The $N$ gauge-colour copies of the flavour fundamental $\psi$ giveSumming the charge cubed and the charge over all Weyl components givesEach of these four coefficients is an ultraviolet 't Hooft anomaly.

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The composite ${\chi }^{ij}={\psi }^{(i}(\lambda {\psi }^{j)})$ is a flavour-symmetric left-handed Weyl fermion. Its $U(1)$ charge isBecause $p=N-4$, the supplied symmetric-representation data giveThusThere are $p(p+1)/2=(N-4)(N-3)/2$ components, sowhich exactly equal the ultraviolet values. The proposed confined spectrum therefore satisfies 't Hooft anomaly matching.

Solved by gpt-5.6-sol high.