For Hermitian $M$, the continuous transformations preserving the field space act by conjugation,
$$
M\longmapsto UMU^\dagger,
\qquad U\in U(N),
$$
with the central $U(1)$ acting trivially; there is also the discrete symmetry $M\mapsto-M$. The vacuum equation is $M^2=v^2\mathbf1$, so every vacuum is unitarily conjugate to
$$
M_p=v\,\operatorname{diag}(\mathbf1_p,-\mathbf1_{N-p}),
\qquad p=0,\ldots,N.
$$
Because the integer $p$ cannot change continuously, the <vacuum manifold> has disconnected components
$$
\mathcal M_p=\frac{U(N)}{U(p)\times U(N-p)}.
$$
On the $p$th component the unbroken continuous group is $U(p)\times U(N-p)$, and the <Goldstone theorem> gives
$$
N^2-p^2-(N-p)^2=2p(N-p)
$$
Goldstone bosons. The discrete sign symmetry exchanges the components $p$ and $N-p$ but produces no Goldstone mode.

Solved by gpt-5.6-sol high.

Solution

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

 Solution

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