Follow up question on "Wilson Loops as Raising Operators" This is a follow-up question on the topic that I opened a few days ago,
Wilson Loops as raising operators.
The paper 

Topological Degeneracy of Quantum Hall Fluids. X.G. Wen, A. Zee. Phys. Rev. B 58 no. 23 (1998), pp. 15717-15728. arXiv:cond-mat/9711223.

gives a nice derivation of the explicit ground states of the $U(1)$ Chern-Simons Theory on a torus in Section 2 on Abelian Quantum Hall States. 
In particular Eq. (12) gives the generic form of a ground state 
$\psi(y) = \sum_{n=-\infty}^{\infty} c_{n} \ e^{i\ 2\pi ny}$. 
Due to the fact that the theory lives on a torus the ground state manifold 
is found to be $k$-fold degenerate. 
My question: 
Is it possible (by direct calculation) to obtain the relations 
\begin{align}
W(b)|n \rangle &= |n + 1 \text{ mod } |k| \rangle,
\nonumber \\
W(a) |n \rangle &= e^{2\pi i n /k} |n \rangle.
\end{align}
from the previous question? 
I don't have a particularly strong background in field theory so I am feeling somewhat uneasy when it comes to the explicit evaluation of the Wilson Loop (with its exponentiated gauge field and the path ordering) acting on the constructed state.
I am looking forward to your responses.
 A: The correspondence is : $  W(a)= e^{2i \pi y}, \quad W(b) = e^{-2i \pi x}$. 
From $[x,y] = \frac{i}{2\pi k}$ $(9)$, we get : $W(a)W(b) = e^{ \large \frac{2i\pi}{k}} W(b) W(a)$ (see precedent answer). 
As it is explained in the text between formulae $(19)$ and $(20)$, $x$ and $y$ may only take discrete values $\frac{1}{k},\frac{2}{k},....., \frac{k-1}{k},1$. 
A direct consequence of that is : $W^k(a) = W^k(b) = \mathbb{Id}$, as wished, because of the identifications $x= x+1$ and $y = y+1$. 
[EDIT]
The Wilson loops
Here is a non rigorous argument, but this is my feeling. If we look at equation $(4)$ of your paper, we see that :
$$a_0(x_1,x_2,t)=0, \quad a_1(x_1,x_2,t)= 2 \pi \frac{x(t)}{L_1},  \quad a_2(x_1,x_2,t)= 2 \pi \frac{y(t)}{L_2}\tag{4}$$
Now, choose for the path $(a)$ : $x_2$ increasing , $0 \le x_2 \le L_2$, and $x_1,t$ constant (so $dx_1=0$)  we would have : 
$$\oint_a (a_i dx^i) = \oint_a (a_2 dx^2) = \int_0^{L_2}  2 \pi \frac{y(t)}{L_2} = 2 \pi y(t) \tag{5}$$
So, we would have : 
$$W(a) = e^{ \large i\oint_a (a_i dx^i)} =  e^{2 i\pi y}  \tag{6}$$
In the same spirit choose the path $(b)$ : $x_1$ decreasing , $0 \le x_1 \le L_1$, and $x_2,t$ constant (so $dx_2=0$)  we would have : 
$$\oint_b (a_i dx^i) = \oint_a (a_1 dx^1) = \int_{L_1}^0  2 \pi \frac{x(t)}{L_1} = -2 \pi x(t) \tag{7}$$
So, we would have : 
$$W(b) = e^{\large  i\oint_b (a_i dx^i)} =  e^{-2 i\pi x}  \tag{8}$$
The basis
Because the momentum $p$ is $p=2\pi kx = -i\frac{\partial}{\partial y}$, we may rewrite $W(b)$ as $W(b) = e^{\large -\frac{1}{k}\frac{\partial}{\partial y}}$.
So, we have : $W(b) \psi(y) = \psi(y-\frac{1}{k})$
It is natural, then to postulate the following basis : 
$$\psi_n(y) = \langle n|y\rangle = \delta (ky - n) \tag{9}$$
We see that : 
$$W(a) \psi_n(y) = e^{2i \pi y} \delta (ky - n)= e^{2i \pi \large \frac{n}{k}}\delta (ky - n) = e^{2i \pi \large \frac{n}{k}} \psi_n(y) \tag{10}$$
$$W(b) \psi_n(y) = \psi_n(y - \frac{1}{k}) = \delta(k (y -\frac{1}{k}) - n) =   \delta (ky - (n+1)) =  \psi_{n+1}(y)\tag{11}$$
