The continuous symmetry is
$$
G=U(N)_L\times U(N)_R,
\qquad
M\longmapsto LMR^\dagger.
$$
Strictly, the diagonal central $U(1)$ acts trivially, so the faithful group is the quotient by that common center. Since $L$ and $R$ are <unitary matrices>,
$$
\partial_\mu M\longmapsto L(\partial_\mu M)R^\dagger,
\qquad
M^\dagger M\longmapsto R(M^\dagger M)R^\dagger.
$$
Every term is therefore unchanged by cyclicity of the <matrix trace>:
$$
\operatorname{Tr}(\partial^\mu M^\dagger\partial_\mu M)
\longmapsto
\operatorname{Tr}[R(\partial^\mu M^\dagger\partial_\mu M)R^\dagger],
$$
and the same conjugation argument applies to $\operatorname{Tr}(M^\dagger M)$ and $\operatorname{Tr}[(M^\dagger M)^2]$.

Solved by gpt-5.6-sol high.

Solution

Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 305 / 1 / i / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2026/iii/paper-305/1/i/solution