Why there is an isomorphism between the coset group and the Goldstone bosons?

Question

Suppose we have $n$ Goldstone bosons which is obtained from the fact that the ground state $\eta$ is invariant under a subgroup $H$ of $G$. Each of these Goldstone bosons will be described by an independent field $\phi_{i}$ which is a smooth real function on Minkowski space $M^{4}$. These fields are collected in an $n$ -component vector $\Psi=\left(\phi_{1}, \ldots, \phi_{n}\right)$, defining the real vector space $$ V \equiv\left\{\Psi: M^{4} \rightarrow \mathbb{R}^{n} \mid \phi_{i}: M^{4} \rightarrow \mathbb{R} \text { smooth }\right\} $$
Now we define the map $F:G\times V \longrightarrow V$ by

\begin{gathered} F(e, \Psi)=\Psi \quad \forall \quad \Psi \in V, \quad e \text { identity of } G \\ F\left(g_{1}, F\left(g_{2}, \Psi\right)\right)=F\left(g_{1} g_{2}, \Psi\right) \quad \forall \quad g_{1}, g_{2} \in G, \quad \forall \quad \Psi \in V \end{gathered}

For the subgroup $H$ of $G$ the set $g H=\{g h \mid h \in H\}$ defines the left coset of $g$ and we define $G / H=\{g H \mid g \in G\}$ to be the set of all left cosets.

Now let $F':G / H \times V \longrightarrow V$ and $0\in V$. We have that $0$ is mapped onto the same vector in $\mathbb{R}^{n}$ under all elements of a given coset $g H$ since $$ F'(g h, 0)=F(g, F(h, 0))=F(g, 0) \quad \forall \quad g \in G \quad \text { and } \quad h \in H $$ The map is injective since for two elements $g$ and $g^{\prime}$ of $G$ where $g^{\prime} \notin g H$. Let us assume $F'(g, 0)=F'\left(g^{\prime}, 0\right)$ : $$ 0=F'(e, 0)=F(e, 0)=F\left(g^{-1} g, 0\right)=F\left(g^{-1}, F(g, 0)\right)=F\left(g^{-1}, F\left(g^{\prime}, 0\right)\right)=F\left(g^{-1} g^{\prime}, 0\right) $$ However, this implies $g^{-1} g^{\prime} \in H$ or $g^{\prime} \in g H$ in contradiction to the assumption $g^{\prime} \notin g H$ and therefore $F'(g, 0)=F'\left(g^{\prime}, 0\right)$ cannot be true. Let $n_G$ be the dimension of the lie group G and $n_h$ the dimension of the lie subgroup $H$. Goldstone theorem says that $n=n_G-n_h$. But it is a fact that for lie groups we have the dimension of $G / H$ is $n_{G / H}=n_G-n_h$

From the considerations above in this notes A Chiral Perturbation Theory Primer they claim that there exists an isomorphic mapping between the quotient $G / H$ and the Goldstone-boson fields.

But I am not seeing the isomorphism since $G / H$ has $n$ elements and the $V$ has infinite elements.

Am I missing something here?

Answer 1

How many Goldstone bosons?

A system with a spontaneously broken symmetry is one where the vacuum is not invariant under the full continuous, semisimple symmetry group $G$ of the theory, but only a subgroup $H\subset G$. We say that the symmetry $G$ is spontaneously broken to $H$. In general, the equation defining the mass matrix of the field content in this system is $$ \underbrace{\left(\frac{\partial^2 V}{\partial\phi_a\partial\phi_b}\right)\bigg|_{\phi=v}}_{M_{ab}}(T^A)_{ac}v_c=0 $$

where $v_c=\langle\phi_c\rangle$ is the vacuum expectation value (VEV) and $V(\phi)$ is the potential of the system. Here the generators which do annihilate the vacuum - $(T^A)_{ac}v_c=0$ - span the Lie algebra $\mathfrak h\equiv\mathfrak{Lie}(H)\subset\mathfrak g\equiv\mathfrak{Lie}(G)$.

Diagonalising the mass matrix $M_{ab}$, there are as many massless Goldstone modes as there are null eigenvalues of $M_{ab}$. This is in turn equal to the number of broken generators, i.e. generators that do not annihilate the VEV $v_c$, which is $\dim G-\dim H=\dim(G/H)$.

This number is the dimension of $G/H$ as a manifold - for instance, the quotient $\dim(\mathrm{SU}(2)/\mathrm{U}(1))=\dim(\mathrm{SU}(2))-\dim(\mathrm{U}(1))=3-1=2$. This means that the quotient space is a 2-dimensional manifold (which, indeed is isomorphic to $S^2$).

The isomorphism between $G/H$ and the Goldstone fields

Each Goldstone boson field is denoted by $\varphi_i : M_4\to\mathbb R$. There are a finite number of these fields ($i\in\{1...\dim G-\dim H\}$), which are packaged into the $(\dim G-\dim H)$-component vector $\vec{\boldsymbol\varphi}\equiv(\varphi^1, \varphi^2 ... \varphi^{\dim G-\dim H})$. The claimed mapping is not between $G/H$ and the vector space formed by all possible Goldstone fields, but rather between $G/H$ and the specific Goldstone fields associated to the symmetry breaking of the model under consideration. These Goldstone bosons are then seen to "live in the coset space", since their low-energy dynamics are described by a non-linear sigma model with $G/H$ as the target manifold; also see In what sense do Goldstone bosons live in the coset?.

Proof: We have a left G-action on the vector space of all Goldstone multiplets $M_\phi$, denoted $\kappa_g : M_\phi\to M_\phi, g\in G$. Call the VEV $v_a$, as above. For $h$ in the unbroken subgroup $H\subset G$, $\kappa_h = \mathrm{id}$ (by definition) so $\kappa_{gh}(v^a)=\kappa_g(v^a)$. Finally, since $\kappa_g$ is bijective when restricted to each left coset, there is an exact isomorphism between the overall field configuration of all the Goldstone modes $\vec{\boldsymbol\varphi}$, and the left cosets $gH, g\in G$.

Example:

This is easier than it sounds: here is an explicit example to illustrate this. Focus on only the Goldstone sector for now, and coonsider an $O(N)\to O(N-1)$ symmetry breaking pattern induced by the following VEV (or any of its $O(N)$-equivalent friends) $$ \langle\phi\rangle\equiv[0\ 0\ ...\ v]^\top $$ This particular choice of VEV is clearly invariant under the remnant $O(N-1)$ matrices of the form $$ h_{ab}=\begin{bmatrix}\mathbf{\tilde h}&0\\0&1\end{bmatrix} $$

Thus, when the Goldstone fields are described by the costless rotations "along the gutter" of the potential: $$\varphi_a=U_{ab}(x)[0\ 0\ ... \ v\!+\!\rho(x)]^\top$$

However, $U_{ab}$ and $U_{ab}h_{bc}$ lead to physically equivalent field configurations for $h\in O(N-1)$, so we should mod out by the invariant subgroup to obtain the true vacuum manifold . Hence the dynamics of the $N-1$ Goldstone bosons here are described by excitations along $O(N)/O(N-1)\cong\mathbb S^{N-1}$, as promised.

As an aside, the corresponding low-energy effective action (classically exact) for these Goldstone fields is given by the action $$ S=\frac{v^2}{8}\int\mathrm d^4x\ G_{ab}(\varphi)\partial_\mu\varphi^a\partial_\nu\varphi^b \\ $$ where $G_{ab}(\varphi)$ is the Riemannian metric on $G/H$, and the Goldstone fields are the coordinates on this manifold - $\varphi^a(x) : \mathbb R^4\to G/H$.

Answer 2

The dimension of $G/H$ is $n_G-n_H$, with a minus, not a plus. This is a quotient, not a product.

The space $G/H$ is naturally a manifold. The claim is that its tangent space is ismomorphic to $V$. Both the tangent space and $V$ are vector spaces, they both have infinitely-many elements but a finite basis. The claim is that the bases for both have the same number of elements (and there is a natural pairing).

It is not true that $G/H$ has $n$ elements. It is a continuous manifold, whose dimension is $n$. (Unless $G$ and $H$ have the same rank, in which case $G/H$ is indeed a finite group and $n\equiv0$. But in that case there are no Goldstone bosons either, so the isomorphism still holds). In the same way the dimension of $\mathbb R^n$ is $n$; clearly, $\mathbb R^n$ has infinitely many elements (but a finite basis, with precisely $n$ elements).