How does the $U(1)$ global symmetry break in the gauged $XY$ model? I'm studying the particle vortex duality, and I'm confused how we're able to say that in the Coulomb phase, the "hidden" $U(1)$ global magnetic symmetry spontaneously breaks.
gauged XY model: $\mathcal{L}  = \frac{1}{2g^2}f_{\mu\nu} f^{\mu\nu} + |\mathcal{D} _\mu \phi|^2 - \mu |\phi|^2 - \lambda |\phi|^4 $
This theory has $U(1)$ gauge symmetry and a less obvious $U(1)$ global symmetry corresponding to the conservation of the magnetic current $j^\mu = \frac{g^2}{\pi} \partial^\mu \sigma$, where $\sigma$ is the dual photon ($d \sigma = \frac{-g^2}{\pi}\star da$). 
In the coulomb phase phase corresponding to $\lambda > 0, \mu > 0$ we have an obvious unbroken $U(1)$ gauge symmetry. However, we say that the $U(1)$ global symmetry is broken, and thus, this phase corresponds to the SSB phase of the global XY model.
I don't see why this is the case, as when I change the variables from $\phi$ to $\sigma$ the Lagrangian still retains the shift symmetry $\sigma \rightarrow \sigma + const.$ modulo 2$\pi$. Thus there's no change in the magnetic current. Where am I going wrong in this reasoning?
 A: Sponteneous symmetry breaking occurs when the ground state has a different (reduced) symmetry from the Hamiltonian/Lagrangian. So $\mathcal{L}$ retains the $U(1)$ or shift symmetry, but you have to look at whether or not the ground state also does.

The "gauged $XY$" model is also called the Abelian-Higgs model.
The first is the usual $U(1)$ gauge symmetry associated with $\phi \rightarrow e^{\mathrm{i}\varphi}\phi$. Making this a local symmetry introduces the coupling to the gauge field $\alpha_\mu$ (in your gauge covariant derivative $\mathcal{D}_\mu$).
Another $U(1)$ symmetry is identified (only because we are in $2+1$ dimensions) that does not depend on the metric $g_{\mu\nu}$ and is hence distinctly topological, of the Chern-Simons type of topological field theories.
Your photon $\alpha_\mu$ has $3$ degrees of freedom, because we are in $2+1$ dimensions. Because it's a gauge field, one d.o.f. is taken out, and we are left with $2$. Depending on the sign of $\mu$, this will remain $2$ (Higgs phase, topological $U(1)$ symmetry unbroken) or will become $1$ (Coulomb phase, topological $U(1)$ symmetry broken).
$\lambda > 0$ is taken everywhere.

*

*Let us first look at the $\mu<0$ case.

The potential $V(\phi) = \mu |\phi|^2 + \lambda |\phi|^4$ now looks like this (with $\lambda = -\mu = 1$):

The potential has a ring of minima at $|\phi|=1$, but the ground state corresponds to a specific choice of a point on this ring. So the Lagrangian (pre-SSB) still has $U(1)$ symmetry, but the ground state does not. You could of course re-write the Lagrangian post-SSB in terms of real fields $\phi$ so that the $U(1)$ would not be there any longer.
The gauge $U(1)$ symmetry is spontaneously broken so that $\phi$ gains an expectation value $\langle \phi \rangle \neq 0$, and you have to re-write your Lagrangian (density) as per the Higgs mechanism $\phi \rightarrow \phi_0 + \tilde{\phi}$. This results in the "would-be" Goldstone boson to be eaten up by the gauge field $\alpha_\mu$ giving mass $m$ to the photon. Because the photon is massive, excitations (creating photons) are now gapped as they cost an energy $\propto m$.
The massive photon $\alpha_\mu$ has $2$ degrees of freedom (two polarisations), end of story.

*

*$\mu >0$:

In this case, the potential $V(\phi)$ has a global minimum at $\phi=0$ meaning the gauge $U(1)$ symmetry is not broken.
Hence, the photon $\alpha_\mu$ remains massless, resulting in an additional redundancy (on top of the gauge condition) which reduces the degrees of freedom from $2$ to $1$.
So your photon $\alpha_\mu$ can be described by a scalar field $\sigma$, known as the dual photon.
Let's treat this as a scalar field like we treated $\phi$ earlier. Its "potential" $V(\sigma) = 0$, and again the ground state of $\sigma$ corresponds to a specific choice on this potential.
The original topological $U(1)$ symmetry has become the freedom to translate $\sigma \rightarrow \sigma + \mathrm{const}$, technically $\mathrm{mod}\,2\pi$ as per the Dirac quantisation condition.
Breaking this topological $U(1)$ symmetry corresponds to choosing a specific value for $\sigma$.
