In what sense do Goldstone bosons live in the coset?

Question

Goldstone's theorem says that if a group, $G$, is broken into its subgroup, $H$, then massless particles will appear. The number of massless particles are given by the dimension of the coset, $G/H$. It is then often said that the Goldstone boson's live in the coset. In what sense is this statement true? The Lagrangian is not invariant under transformations of the coset so what does this "living" explicitly mean?

To be explicit we can consider the linear sigma model:
begin{equation} 
{cal L} = frac{1}{2} partial _mu phi ^i partial^mu phi ^i - frac{m ^2 }{2} phi ^i phi ^i  - frac{ lambda }{ 4} ( phi ^i phi ^i ) ^2 
end{equation}

We define,
begin{align} 
& phi _i equiv  pi  _i quad forall i neq N 
& phi _N equiv sigma 
end{align} 
and give $sigma$ a VEV.

The spontaneously broken Lagrangian is,
begin{equation} 
{cal L} = frac{1}{2} partial _mu pi _i partial ^mu pi _i + frac{1}{2} ( partial _mu sigma ) ^2 - frac{1}{2} ( 2 mu ^2 ) sigma ^2 - lambda v sigma ^3 - frac{ lambda }{ 4} sigma ^4 - frac{ lambda }{ 2} pi _i pi _i sigma ^2 - lambda v pi _i pi _i sigma - frac{ lambda }{ 4} ( pi _i pi _i ) ^2 
end{equation} 
The Goldstone bosons, $pi_i$, exibit a $O(N-1)$ symmetry, but this is not the coset group symmetry. So where in the Lagrangian do we see this symmetry?

Frederic Brünner · Accepted Answer

I understand the statement in the following way:

Pions, which are pseudo-goldstone bosons of chiral symmetry breaking, are described by the introduction of a unitary matrix $U(x)$, defined as

$$U(x)=text{exp}left(2ipi^a(x)T^af_pi^{-1}right),$$

where $pi^a$ is the pion field, $f_pi$ is the pion decay constant and $T^a$ are the generators of the broken symmetry, i.e. the coset space. The pion Lagrangian can be written down in terms of $U(x)$

$$mathcal{L}=-frac14 f_pi^2text{Tr}partial^mu U^daggerpartial_mu U,$$

which by expanding the exponential form results in

$$mathcal{L}=-frac12partial^mu pi^apartial_mu pi^a+dots,$$

where dots denote higher order terms. Thus, the statement that goldstone bosons live in the coset space can be related to the fact that the fields themselves are linked to the generators of the coset.

This can be understood in terms of Goldstone's theorem: if the original Lagrangian exhibits a continuous symmetry, the number of goldstone bosons is equal to the number of generators of the broken symmetry. Take for example the linear sigma model: if your original theory is $O(N)$-symmetric, it has $N(N-1)/2$ symmetries. If the symmetry is broken spontaneously, you end up with $O(N-1)$, leaving you with $(N-1)(N-2)/2$ symmetries. The amount of broken symmetries is the difference, i.e. $N-1$. But this is precisely the number of pions you have in your theory. We can conclude that the pions are linked directly to the broken symmetries, i.e. the coset space.

TwoBs · Answer

It is in fact a very simple matter if you use a different parametrization of the fields.  Since we care about the Goldstone bosons only, just send $lambdarightarrow infty$ so that the Higgslike state decouples. Moving to the following parametrization
$$
phi_i(x)=U(x)langle phi_irangle ,,qquad U(x)=e^{i hat{T}^a pi^a(x)},,qquad langlephi_irangle=left(0,0,0ldots,vright)^T
$$
(where $hat{T}^a$ are the broken generators) you immediately see that there is a gauge redundancy in the definitions of the pion fields $pi^a(x)$ since we are allowed rotate them with an $x-$dependent transformation $h(x)$ of the unbroken group $H$, namely
$$
phi_i(x)=U(x)langle phi_irangle=U(x)h(x)langle phi_irangle,.
$$
In other words, the pion field is defined only up to this equivalence $U(x)sim U(x)h(x)$, which is the statement that they live on the coset space $G/H$.

In what sense do Goldstone bosons live in the coset?

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