In spontaneous symmetry breaking, global symmetry broken by gauged subgroup? My question is simple. Given a group $G$ broken to a subgroup $H$, gauging a possibly different subgroup Hg breaks explicitly the global symmetry $G$, generating what is known as pseudo-Goldstone bosons. Why is this?
The usual answer I get is that the gauging determines a specific direction in field space, but I really don't understand this statement,  how having a subgroup $Hg$ gauged can break the global $G$ explicitly?. What I have in mind is that inside the gauge transformations $Hg$ there are included global ones as well, so this is what is confusing me.
 A: How can gauging only a subgroup not break the symmetry? Isn't any group element that doesn't preserve my gauge subrgoup going to change the action. 
Let's I have an $SU(2)$ symmetry, and some scalar field $\rho$ in the fundamental. If I gauge the $U(1)$ subgroup corresponding to rotations by $\sigma_z$. I have a Lagrangian
$$\mathcal{L} = |(\partial + A \sigma_z)\psi|^2 + V(|\psi|^2)$$
Doesn't this manifestly break the $SU(2)$ symmetry? By your point about global symmetries it follows that all of the point interactions must remain invariant, but the Lagrangian as a whole is not invariant. 
This example could physically correspond to two fields which are oppositely charged but otherwise symmetric. The waves in the $z$ axis correspond then to charged waves like a plasmon, whereas the rotations in the $x-y$ plane are local gauge transforms. So they are clearly different.
Added
Everyone seems to be having trouble seeing that this breaks the symmetry, and wants to say that the operation $\psi \rightarrow U\psi$, $A\sigma_z \rightarrow U A\sigma_z U^{\dagger}$, where $U$ is an $SU(2)$ matrix, is a "symmetry" of the Lagrangian. Let me write out the full path integral:
$$\int\mathcal{D}A\mathcal{D}\psi\exp\{i\int d^dx \mathcal{L}\}$$
$$\mathcal{L} = (\partial_\mu A_\nu - \partial_\nu A_\mu)^2 +|(\partial_\mu + A_\mu \sigma_z)\psi|^2 + V(|\psi|^2)$$
The variable $A_\mu$ is an $SU(2)$ scalar. It is just a regular 1-form like in electromagnetism. The integration measure (which also contains whatever unimportant gauge fixing) is just a regular integral over a 1-form field. Now for example to get Noether's theorem I need to have a change of variables that leaves my path integral invariant. The map $A\sigma_z \rightarrow U A\sigma_z U^{\dagger}$ is not a change of variables - I cannot get it by changing $A$, since $A$ doesn't know anything about $SU(2)$. There is no way I can change $x\sigma_z$ to $y\sigma_x$ by making a substitution $y=f(x)$.
You could rewrite the whole thing as 
$$\int\mathcal{D}\hat{A}\mathcal{D}\delta(\hat{A}_{x,y})\psi\exp\{i\int d^dx \mathcal{L}\}$$
$$\mathcal{L} = (\partial_\mu A_\nu - \partial_\nu\hat{A}_\mu)^2 +|(\partial_\mu + \hat{A}_\mu )\psi|^2 + V(|\psi|^2)$$
where $\mathcal{D}\hat{A}$ is the measure for an $SU(2)$ field. But to correspond to what I wrote you need that delta function to in the measure. Otherwise you've gauged the entire $SU(2)$ symmetry which is obviously invariant. Now you can make change of variables $A\rightarrow U\hat{A}U^\dagger$, but this doesn't leave your path integral invariant because of that huge delta function.
A: Main Reference Zee (Quantum Mechanics in a Nutshell).
1) Global symmetry
A global symmetry means that the Lagrangian is invariant by a transformation whose parameters are constant.
For a continuous global symmetry, if the symmetry of the Lagrangian is the group $G$, and if the symmetry of the vacuum is the group $H$, a subgroup of $G$, you have ($dim G-dim H$) Goldstone bosons.
For instance, take a complex scalar field $\Phi$ with a Mexican hat potential , so that the total Lagrangian density is $L = \partial \phi^\dagger \partial \phi + \mu^2 \phi^\dagger \phi - \lambda (\phi^\dagger \phi)^2$.
The group symmetry is here $G=O(2)$
Define $\phi = \rho e^{i\theta}$
Breaking the symmetry means choosing for the vacuum the minima for the potential, and a particular angle, that is : 
$\rho_V = v, \theta_V = \theta_0$
The group $H$ is trivial here.
Define :
$\rho = v + \chi$, where $v = \sqrt{\frac{\mu^2}{2\lambda}}$
Developping the Lagrangian, you get a term $v^2(\partial \theta)^2$, which is the dynamical part for a massless field $\theta$, so  $\theta$ is our Goldstone boson (There is one because $dim G - dim H = 1 - 0 = 1 $).
So, we see, that spontaneous symmetry breaking could arise in a global continuous symmetry.
2) Local symmetry
A local symmetry means that the Lagrangian is invariant by a transformation whose parameters are functions of space-time.
"Gauging" means (continous) local symmetry.
So you don't need "gauging" to have a spontaneous symmetry breaking.
With a local symmetry,  some of the Goldstone Bosons are "eaten" by the Gauge field ($A_\mu$), so that these gauge fields (which are massless) become massive. In a 4d space-time dimension, a massless Gauge field has $2$ degrees of freedom, while a massive gauge field has $3$ degrees of freedom. To do that, the Gauge field has to "eat" one degree of freedom (one Goldstone boson)
3) Global symmetry as a special case of Local Symmetry
In the set of local symmetry, global symmetry is a very special case (a very special subset), where transformation parameters are constant. So, if you want, you can consider that global symmetry are "included" into local symmetry.
