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

*

*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 $\langle\Phi(x)\rangle=v$

*I have a $U(1)$ guage symmetry to utilize
\begin{equation}
\phi(x)\rightarrow \phi(x)+ N \alpha(x)
\end{equation}

*I choose $\alpha(x)=-\frac{\phi(x)}{N} +\beta\pi$ where $\beta $ is a constant that could take values in $0,\frac{1}{N}, \cdots ,\frac{N-1}{N}$.

*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?
 A: That's not what happens. Instead what happens is, in steps, as follows.

*

*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.

*The fact that $\Phi(x)$ has charge $N$, means that under a $\mathrm{U}(1)$ gauge transformation it transforms as
$$\Phi(x)\mapsto \mathrm{e}^{\mathrm{i}N\alpha(x)}\ \Phi(x).$$

*The fact that it condenses means that
$t(x)=t\in\mathbb{R},$ a constant. $t$ gives the VEV of the field.

*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 $L_1$:
$$ L_1 = t^2\left(\mathrm{d}\phi-N A\right)\wedge\star\,\left(\mathrm{d}\phi-N A\right).$$

*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).

*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.

