quantum field theory - In what sense do Goldstone bosons live in the coset?

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?

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

$$U(x)=\text{exp}\left(2i\pi^a(x)T^af_\pi^{-1}\right),$$

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 $U(x)$

$$\mathcal{L}=-\frac14 f_\pi^2\text{Tr}\partial^\mu U^\dagger\partial_\mu U,$$

which by expanding the exponential form results in

$$\mathcal{L}=-\frac12\partial^\mu \pi^a\partial_\mu \pi^a+\dots,$$

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.