The composite $\chi^{ij}=\psi^{(i}(\lambda\psi^{j)})$ is a flavour-symmetric left-handed Weyl fermion. Its $U(1)$ charge is
$$
q_\chi=q_\lambda+2q_\psi=(N-4)-2(N-2)=-N.
$$
Because $p=N-4$, the supplied symmetric-representation data give
$$
A(\operatorname{Sym}^2\mathbf p)=p+4=N,
\qquad
I(\operatorname{Sym}^2\mathbf p)=p+2=N-2.
$$
Thus
$$
\mathcal A_{SU(p)^3}=N,
\qquad
\mathcal A_{SU(p)^2U(1)}=(N-2)(-N)=-N(N-2).
$$
There are $p(p+1)/2=(N-4)(N-3)/2$ components, so
$$
\mathcal A_{U(1)^3}=\frac{p(p+1)}2(-N)^3,
\qquad
\mathcal A_{U(1)\text{-grav}^2}=\frac{p(p+1)}2(-N),
$$
which exactly equal the ultraviolet values. The proposed confined spectrum therefore satisfies <'t Hooft anomaly matching>.

Solved by gpt-5.6-sol high.

Solution

Paper 305 / 3 / iv / Solution

The three Weyl components form one fundamental representation of the global $SU(3)$, so coupling $G$ to background gauge fields reveals a nonzero $SU(3)^3$ <'t Hooft anomaly>. This does not destroy the global symmetry in flat space or make the theory inconsistent; it constrains any infrared description through <'t Hooft anomaly matching>. If $G$ were instead intended as a dynamical gauge symmetry, the same uncancelled <gauge anomaly> would make the model inconsistent and additional chiral matter would be required. Ordinary quantum corrections cannot generate the forbidden superpotential, and the <non-renormalization theorem> protects its absence.

Solved by gpt-5.6-sol high.

Paper 307 / 1 / vii / Solution

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't Hooft anomaly matching requires the massless infrared degrees of freedom, topological sector, or symmetry-breaking pattern to reproduce every anomaly of an unbroken global symmetry measured in the ultraviolet theory.


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