= Solution
Both the canonical <kinetic term> and $V(\phi^{\mathsf T}\phi)$ are invariant under the <orthogonal group> $O(n)$. The stated nonzero <vacuum expectation value> has length $|v|$. Its stabilizer is the <orthogonal group> $O(n-1)$ acting on the perpendicular subspace. Hence
$$
\boxed{N_{\mathrm{GB}}=\frac{n(n-1)}2-\frac{(n-1)(n-2)}2=n-1.}
$$
The two nonzero coordinates of this particular <vacuum> do not break more generators than an axis-aligned <vacuum>: the two choices are related by a common <orthogonal transformation>.
Explicitly, with $s=\phi^{\mathsf T}\phi$,
$$
\mathsf M^2=2V'(v^2)\mathsf I+4V''(v^2)vv^{\mathsf T}.
$$
<stationary point> at nonzero $v$ requires $V'(v^2)=0$. The $n-1$ transverse <eigenvalues> vanish, while the radial squared <mass> is $4v^2V''(v^2)$. If this curvature also vanishes, there is an additional accidental massless radial mode; the number of symmetry-protected <Goldstone bosons> remains $n-1$. If $v=0$, the <symmetry> is unbroken and none is required.
Back to article page
