Chern-Simons theory on a plane/sphere with a single charge insertion Consider the pure Chern-Simons theory on the plane $\mathbb{R}^2$ with a single charge insertion in some representation $\rho$ of the group $G$. What does the Hilbert space look like? Is it null or non-null for non-integrable representations $\rho$?
Below is my attempt at tackling this problem and an outline of the difficulties that I ran into.
The same question has a very nice answer for the case of a sphere $S^2$. Here there's no non-contractible loops, thus the loop around the charge insertion must have holonomy of $1$ (the identity element of $G$). Thus it is restricted to the trivial orbit (which is actually a single point), which means that the phase space is a point if $\rho$ is trivial and it doesn't exist if it isn't. Hence, the Hilbert space is 1-dimensional for $\rho$ a trivial representation and 0-dimensional otherwise.
Now for $\mathbb{R}^2$ we don't have any restriction on the holonomy around the charge insertion, besides the fact that it must lie on an orbit which belongs to the discrete series (for consistent quantization of the orbit using Kirillov's method).
But there's also the gauge invariance – we have to factor out by the gauge group $\mathcal{G}$. It is equivalent to saying that the group $G$ acts on the holonomy by conjugation: $h \rightarrow g h g^{-1}$. This constraint is saying that all points on the orbit are gauge-equivalent, so essentially the entire orbit reduces to just one point on the moduli space. Hence, the Hilbert space must be 1-dimensional for any $\rho$.
I'm dissatisfied with the answer that I got. I think I'm missing something important. One problem with it is that it doesn't use the value of the Chern-Simons level $k$ anywhere. What happened to the integrability requirement? I expected the non-integrable representations $\rho$ to lead to the null Hilbert space, because they don't correspond to the representations of the quantum group.
 A: In order to get a nontrivial result, we should consider the problem on a disc (as large as one wishes) with free boundary conditions. Otherwise, if for example, we fix the boundary values, we will have no (phase) space of classical solutions and the Hilbert space becomes trivial.
In addition, since we are eventually quantizing the oscillations of the gauge field on the boundary which has the topology of a circle, we should expect an infinite number of modes, i.e., an infinite dimensional Hilbert space. (In contrast to the cases where the base manifold is boundaryless)
I am mainly following here Elitzur, Moore, Schwimmer and Seiberg who quantize the Chern-Simons theory with a single insertion on the disc (in section 3, pages 7-9.)
Before describing the solution, let me consider first the case where no insertion is present. Here the solution of the Gauss Law constraint is completely determined on the boundary by a group element:
$$ A|_{S^1} = u^{-1} d_{\phi}u $$
($\phi$ is the coordinate along the circle). The boundary action becomes a chiral WZW action. This problems has many treatments in the literature, please see for example Caneschi and Lysiansky. The corresponding Hilbert space is the tensor product of the integrable representations of the loop group $LG$ corresponding to the level $k$.
$$\mathcal{H} = \otimes_{\lambda | \lambda.\delta \le k} \mathcal{H}_{\lambda}$$
($\delta$ is half the sum of the positive roots). $\mathcal{H}_{\lambda}$ are the integrable loop group representations of highest weight $\lambda$.
As explained by Elitzur, Moore, Schwimmer and Seiberg, the addition of a Wilson loop in the path integral is equivalent to adding a source term in the action consisting of the Kirillov-Kostant-Souriau action on $G/T$ at the insertion point. (This result is sometimes attributed to Diakonov and Petrov, please see an elaboration in Alekseev, Chekeres and Mnev (equation 3.6 ).
The gauge freedom in the Kirillov-Kostant Souriau action doesn't remove the holonomy completely but restricts it to a conjugacy class determined completely by the representation $\lambda$ at the insertion point, please see Murayama equation (6.10) where the diagonalization is performed explicitly. In the general case the residual holonomy is given by:
$$e^{\frac{2\pi i}{k} \lambda . H}$$
The solution is still of the form given above, with:
$$u = U e^{\frac{2\pi i}{k} \lambda . H}$$
In order to recognize the Hilbert space Elitzur, Moore, Schwimmer and Seiberg computed the symplectic structure from the effective action:
$$\Omega = \frac{k}{2\pi} \int_{S1} \mathrm{Tr}(U^{-1}\delta U \frac{\partial}{\partial \phi} U^{-1}\delta U) + \frac{1}{2\pi} \int_{S1} \mathrm{Tr}( \lambda . H (U^{-1}\delta U )^2)$$
The additional term restricts the Hilbert space of the disc quantization to a single representation of the Loop group based on the group representation $\lambda$ of the insertion operator.
$$\mathcal{H} = \mathcal{H}_{\lambda}$$
Since, it is the only representation giving the correct holonomy.
