Classically the $p$ identical fields $\psi^i$ have $U(p)=SU(p)\times U(1)_\psi$, and $\lambda$ has an independent $U(1)_\lambda$. One linear combination has an $SU(N)^2U(1)$ anomaly. For the orthogonal combination choose
$$
q_\lambda=p=N-4,
\qquad
q_\psi=-(N-2).
$$
Indeed,
$$
I(\text{antisym})q_\lambda+pI(\overline\square)q_\psi
=(N-2)p-p(N-2)=0.
$$
The continuous quantum global symmetry is therefore
$$
\boxed{SU(p)\times U(1)},
$$
up to discrete identifications. The anomalous orthogonal axial rotation does not survive as a continuous quantum symmetry.


Solution

Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 307 / 3 / ii / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2022/iii/paper-307/3/ii/solution