How to get $\mathbb{Z}_2$ gauge symmetry from $\mathrm{U}(1)$ gauge theory?

Question

I'm reading this paragraph from Tong's gauge theory lecture note.

This is the previous phrase of my question:

This is my understanding of getting $\mathbb{Z}_N$ gauge symmetry from $\mathrm{U}(1)$ gauge symmetry: If the charge of the complex scalar is $N$, we could choose the vev of the scalar field to be $t \, e^{i N \alpha}$ where t is a real number, $\alpha=\frac{x}{N}\pi$ and x takes values $x = 0, 1, ... N-1$ such that the Lagrangian will be in the form where Higgs mechanism is transparent.

The picture I had in mind was

1. I have a complex scalar field $\Phi(x)$. I could expand it as \begin{equation} \Phi(x)=(v+\rho(x))e^{i\phi(x)} \end{equation} if it has non-zero expectation value $\langle\Phi(x)\rangle=v$
2. I have a $U(1)$ guage symmetry to utilize \begin{equation} \phi(x)\rightarrow \phi(x)+ N \alpha(x) \end{equation}
3. I choose $\alpha(x)=-\frac{\phi(x)}{N} +\beta\pi$ where $\beta $ is a constant that could take values in $0,\frac{1}{N}, \cdots ,\frac{N-1}{N}$.
4. Therefore, there is an extra $Z_N$ gauge symmetry left because all choices of $\beta$ give the same result.

Later Tong made a comment in his lecture note:

the $\mathrm{U}(1)$ gauge theory for A is actually Higgsed down to $\mathbb{Z}_N$, a fact which is clear in our initial formulation in $L_1$.

Is this just simply saying that the scalar field is of charge $N$ in $\mathrm{U}(1)$ symmetry which is transparent from the kinetic term in $L_1$? ($L_1$ is defined in the quote below)

Here we take the $U(1)$ gauge theory as our starting point. We can focus on the phase, $\phi\in[0,2\pi)$ of the scalar field. We have a gauge symmetry $\phi\to\phi + Nα$ where $α ∼ α + 2π$ is also periodic. In the Higgs phase, the scalar kinetic term is $$\mathcal{L}_1 = t^2 (d\phi − N A) ∧ \star (d\phi − N A)$$ for some $t ∈ \mathbf{R}$ which is set by the expectation value of the scalar. In the low-energy limit, $t^2 → ∞$ and we have $A = \frac{1}{N} d\phi$ which tells us that the connection must be flat. However, something remains because the holonomy around any non-contractible loop can be 2π $\frac{1}{2\pi}\oint A ∈ \frac{1}{N} \mathbf{Z}$.

Another question is why emphasizing the flat connection. Is this because he wants to emphasize the holonomy around any loop can be calculated?

Answer 1

That's not what happens. Instead what happens is, in steps, as follows.

1. You have a complex scalar field, $\Phi(x)$. This you can always write as $$ \Phi(x) = t(x)\ \mathrm{e}^{\mathrm{i}\phi(x)}, $$ where both $t(x)$ and $\phi(x)$ are real scalar fields.
2. The fact that $\Phi(x)$ has charge $N$, means that under a $\mathrm{U}(1)$ gauge transformation it transforms as $$\Phi(x)\mapsto \mathrm{e}^{\mathrm{i}N\alpha(x)}\ \Phi(x).$$
3. The fact that it condenses means that $t(x)=t\in\mathbb{R},$ a constant. $t$ gives the VEV of the field.
4. If you work out the kinetic terms for $\Phi(x)$ $$ L_{0.5} = \overline{\mathrm{d}_A\Phi}\wedge\star\, \mathrm{d}_A\Phi, $$ writing $\Phi(x)$ as in 1., with the covariant derivative, $\mathrm{d}_A$, compensating for the transformation at 2., in the condensed phase described at 3., you get precisely Tong's $L_1$: $$ L_1 = t^2\left(\mathrm{d}\phi-N A\right)\wedge\star\,\left(\mathrm{d}\phi-N A\right).$$
5. The rest of the argument for why this is a $\mathbb{Z}_N$ gauge theory is precisely the last few sentences in Tong's notes, namely the gauge connection is forced to be a flat connection with a non-trivial, $\mathbb{Z}_N$-valued holonomy. I.e. this is a $\mathbb{Z}_N$-gauge field. As for why he's emphasising that the connection is flat, that is because connections for discrete groups need to be flat (cf. this physics.SE answer of mine).
6. You can also see the $\mathbb{Z}_N$ness of the gauge symmetry from the BF version of the action, i.e. Tong's $L_2$. Why such a BF action gives rise to a $\mathbb{Z}_N$ gauge theory is asked and answered in this physics.SE post.