I have been studying the paper "Gapped Boundary Phases of Topological Insulators via

Weak Coupling" https://arxiv.org/abs/1602.04251

On page 13, it talks about the spin/charge relation.

Let $a^{1}$, $a^{2}$, ... ,$a^{n}$ be $U(1)$-gauge fields, and $A$ be classical electromagnetic background field. We consider the following $3d$ Chern-Simons action

$$S=\int_{W}\left(\frac{k_{ij}}{4\pi}a^{i}\wedge da^{j}+\frac{q_{i}}{2\pi}A\wedge da^{i}\right).$$

Suppose the theory satisfies the usual spin/charge relation. If the action is well-defined mod $2\pi\mathbb{Z}$ on any $spin_{\mathbb{C}}$ manifold $W$, with $spin_{\mathbb{C}}$ connection $A$, then the condition for the coefficients is

$$q_{i}\equiv k_{ii}\quad\rm{mod}\ 2\quad\quad\quad\quad\quad(1)$$

Consider the Wilson operator

$$\exp\left(in^{i}k_{ij}\oint a^{j}\right)$$

with integer $n^{i}$. Using standard formula for Abelian Chern-Simons theory, this operator has spin $n^{i}k_{ij}n^{j}/2$. Its coupling to $A$ shows that its charge is $q_{i}n^{i}$. This Wilson operator should satisfy the usual spin/charge relation and it gives the relation (1).

My questions are

1. What is spin/charge relation?

2. How to derive the relation (1)?

3. How to derive the spin and charge of the Wilson operator?

**New Edition**: Solving the equation of motion of the action, one has

$$\left(\frac{k_{ij}}{4\pi}+\frac{k_{ji}}{4\pi}\right)da^{j}=\frac{q_{i}}{2\pi}dA=\frac{k_{ij}}{2\pi}da^{j}$$

or $da^{i}=\frac{q_{i}}{k_{ij}}dA$. Plugging this EOM back to the Wilson operator, one has

$$\exp{\left(in^{i}k_{ij}\oint a^{j}\right)}=\exp{\left(in^{i}g_{i}\oint A\right)}$$

**So this probably gives the value of the charge, is that correct? **

**How to compute its spin then?** In 2+1 dimensions, the spin is the unitary representation of $SO(2)$, which is just $U(1)$. But the Wilson operator is $U(1)$-invariant, so it is a scalar.

Is the EOM,

$$da^{j}=\frac{q_{i}}{k_{ij}}dA$$

also related with the relation (1) for $A$ is a $spin_{\mathbb{C}}$-connection?