Question about the existence of non-Abelian anyon

Question

My question from reading this paper:

Michael G. G. Laidlaw and Cécile Morette DeWitt, Feynman Functional Integrals for Systems of Indistinguishable Particles. Phys. Rev. D 3, 1375 (971).

Definition: the configuration space of $n$ indistinguishable particles in $d-$dim Eucildean space $\mathbb{R}^d$ is $M_n=(\mathbb{R}^{dn}-D)/S_n$ with $D=\{(\mathbf{r}_1,\cdots,\mathbb{r}_n) | \mathbf{r}_i=\mathbf{r}_j \text{ for some}\ i\neq j\}$, $S_n$ the permutation group.

In this paper they proved the following theorem:

In $d-$dimensional space, the statistics of $n$ indistinguishable particles is classified by different $1$-dimensional irreducible representation of $\pi_1(M_n)$, the fundamental group of $M_n$.

For $d\ge3$, $\pi_1(M_n)=S_n$, there are only two types of $1$-dim rep. of $S_n$, i.e. Boson and Fermion. This theorem excludes the possibility of Parastatistics, i.e. higher dimensional rep. of permutation group.

For $d=2$, $\pi_1(M_n)=B_n$ the Braid group. So $1$-dim rep. of $B_n$ is Abelian anyon, see Yong-Shi Wu, General Theory for Quantum Statistics in Two Dimensions.

My questions:

1. From this theorem, it seems that only Abelian anyon can occur in $2$-dim, because it requires $1$-dim rep. of fundamental group. So why can there exist non-Abelian anyon, i.e. the higher dim. rep. of $B_n$ that violates this theorem?
2. Or is there some loophole or assumption in this theorem that allows the non-Abelian anyon to break?

Answer 1

Dominic's comment is a precise answer to this question.

Consider a sphere with several marked points where we have made quasiparticle insertions. By definition, the space of ground states of this system is 1-dimensional iff all of these quasiparticles are abelian. In this case, the braid group indeed acts on the ground states by adiabatic braiding, yielding a 1-dimensional representation which is just a bunch of phases.

However, in the case of nonabelian anyons, we have degenerate ground states and the braid group still acts but now irreducible representations of dimension >1 can appear.

Many people want to implement quantum logic gates with these braiding operations. For instance if you have $2n$ Majorana anyons on the sphere the ground state is $2^{n-1}$-fold degenerate, and braiding these Majoranas gives various Clifford gates on this space, which cannot be simultaneously diagonalized (and are also not computationally universal, by the way). See this paper for instance https://www.nature.com/articles/npjqi20151 .

Answer 2

From the bounty text:

How can non-abelian anyons exist, when the group they are based on (SO(2)) is Abelian?

Indeed, when you have two particles in 2D, the configuration space is homotopic to $SO(2)$, and $\pi_1(SO(2)) = \mathbb{Z}$, an abelian group which has no irreducible representations with dimension greater than $1$. So you can't see non-abelian anyon effects with only two particles, you just pick up phases.

Now you might think that with $n$ particles, the configuration space is homotopic to $SO(2)^{n(n-1)/2}$, for the relative angles of all pairs of anyons. Then the fundamental group would be $\mathbb{Z}^{n(n-1)/2}$, which remains abelian. However, it's really more subtle than this, because the anyons are identical particles: a process by which the particles switch places is a valid loop in configuration space. This tells us the fundamental group contains $S_n$, at the very least, which is already non-abelian.

This logic also holds for $d > 2$. But for $d = 2$ the fundamental group is even more complicated, because different braids corresponding to the same permutation of the particles may not be deformable to each other. (This is exactly what allows abelian anyons for $n = 2$.) In fact, the fundamental group $B_n$ is an infinite group, which contains $S_n$ as a quotient. It is also non-abelian, allowing for non-abelian anyons.