Background
Homotopy classes in the path integral
Following the answer to this question about the role of homotopy classes in path integrals, it seems reasonable to me that, when calculating the propagator using the path integral formulation, we should first do the integral separately for paths within the same homotopy class of the configuration space, and then add together these contributions with some weight factors. Mathematically this would look something like
$$ K \, \sim \sum_{\alpha \in \pi_{1}(X)} \chi(\alpha) K^{\alpha}, $$
where $\pi_{1}(X)$ is the fundamental group of the configuration space $X$, $K^{\alpha}$ is the partial amplitude associated with the contributions from all paths within the homotopy class $\alpha$, and the $\chi(\alpha)$ are some weights to be determined.
Now, in their 1970 paper, Laidlaw and DeWitt purport to show that the weights $\chi$ must form a scalar unitary representation of the fundamental group $\pi_{1}(X)$. The proof is not too long, but I won't include it here for the sake of brevity.
Fundamental group of configuration space
For $n$ indistinguishable particles with hardcore interactions in $d$ dimensions, the configuration space is
$$ X = Y(n,d)/S_{n}, $$
where $S_{n}$ is the permutation group, quotiented out because the particles are indistinguishable, and $Y(n,d)$ is the set of all $n$-tuples of vectors in $\mathbb{R}^{d}$ such that no two vectors coincide, i.e.
$$ Y(n,d) = \{ y = (\mathbf{x}_{1},\dots,\mathbf{x}_{n}) : \mathbf{x}_{i} \in \mathbb{R}^{d} \hspace{0.5em} \text{and} \hspace{0.5em} \mathbf{x}_{i} \neq \mathbf{x}_{j} \}. $$
It is well known that the fundamental group of this configuration space is different in $d=2$ compared with higher dimensions, namely
$$ \pi_{1}(X) = \begin{cases} B_{n}, \quad d=2 \\ S_{n}, \quad d>2 \end{cases} $$
where $B_{n}$ is the braid group.
Anyonic statistics
Following Laidlaw and DeWitt's claim that the weights must form a scalar unitary representation of $\pi_{1}(X)$, we note that $S_{n}$ has only two one-dimensional (unitary) representations: the trivial representation $\chi(\alpha) = 1$, and the sign representation $\chi(\alpha) = \pm 1$ depending on the sign of the permutation $\alpha$. The former case corresponds to bosons, while the latter corresponds to fermions. Hence for $d>2$ these are the only possibilities.
However, for $d=2$, we have $\pi_{1}(X) = B_{n}$, which has a whole family of one-dimensional unitary representations parametrized by a single angle $\theta$ as
$$ \chi(\alpha) = e^{i \theta W(\alpha)}, $$
where $W(\alpha)$ is the winding number of the braid $\alpha$. This shows that in 2 dimensions we can get Abelian anyons, namely particles which acquire a phase $\theta \in [0,2\pi]$ as they are moved round each other.
However, it is also well known that in $d=2$ we can also get non-Abelian anyons, which in some sense corresponds to taking the weights to be elements of a non-commutative representation of $\pi_{1}(X) = B_{n}$.
I am aware of how to get non-Abelian statistics for a system of quasiparticles with some energy degeneracy by looking at the non-Abelian Berry phase. However, it seems to me that Laidlaw and DeWitt's result that the weights should be from a one-dimensional rep of the fundamental group limits us to Abelian statistics in the context of the path integral.
Question
How do non-Abelian exchange statistics appear in the path integral formulation, and is this consistent with Laidlaw and DeWitt's result?