Residual symmetry group of a scalar field theory Given a Lagrangian
$$\frac{1}{2} (\partial_\mu \phi)^2 - \frac{\lambda}{4!}(\phi^2 - v^2)^2$$
for a real scalar field theory with $\vec{\phi} = (\phi_1,\phi_2,...,\phi_n)^T$ and $O(n)$ symmetry. Why is the residual symmetry group (or little group) given by $O(n\!-\!1)$ when spontaneous symmetry is broken?
 A: O(n) means you may rotate any n-vector to any other of the same length, or a suitably normalized combination of others. So you make a choice to rotate your reference vector to say, $\phi_1=v(1,0,0,..,0)^T$.
Its little group rotating the n-1 components indexed by 2,3,...,n among themselves is thus O(n-1), and it has the obvious $(n-1)(n-2)/2$ generators of that group acting linearly on your fields. The ones you "lost" (not really, the symmetry generators are still there, transforming $\phi_1$ to the other components, in a nonlinear manner) are the
$n-1$ ones realized nonlinearly, corresponding to massless Goldstone bosons (show this!). Your Goldstone bosons are $\phi_a$ with a=2,3,...,n-1, while $\phi_1$ is massive, the σ or Higgs.
Specifically,
$$
  \Delta_{ij}\phi_k= -\Delta_{ji}\phi_k= \theta_{ij}(\delta_{ik}\phi_i - \delta_{jk}\phi_j),
$$
So
$$
  \Delta_{ij}\phi_1= 0  
$$
For the O(n-1) Δs involving only indices 2,3,...n.
Further,
$$
  \Delta_{1j}\phi_1=  \theta_{1j}\phi_j,
$$
for only one index, j in that set: these do not leave your reference vector invariant. There are n-1 of them and shift the $\phi_j$ s by a constant when you redefine $\phi'_1$ to have a vanishing  vacuum value.
