# Question about the existence of non-Abelian anyon

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?
• My feeling is they're implicitly assuming that the state space of a collection of particles at specified positions is one-dimensional -- which is violated by non-Abelian anyons. – Dominic Else Nov 22 '17 at 0:40
• Here's a relatively friendly introduction, covering why 2-d space allows more than just fermions/bosons and how this is related to the braid group, plus how to construct (non-abelian) representations of the braid group: "The braid group, representations and non-abelian anyons" (people.kth.se/~dogge/files/KEX2015-weinberger.pdf). – Chiral Anomaly Dec 17 '18 at 0:42

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 .

• Can one have free anyons? Free non abelian anyons? Or do they have to be engineered that way? – SuperCiocia Dec 18 '18 at 22:00
• @SuperCiocia what do you mean by free? Nothing is for free :P – Ryan Thorngren Dec 19 '18 at 12:05
• No interactions – SuperCiocia Dec 19 '18 at 12:13
• @SuperCiocia Even then it's not clear what you mean. A "free" nonabelian finite gauge theory has nonabelian quasiparticles. – Ryan Thorngren Dec 19 '18 at 13:56
• Okay. But I thought that experimentally they only found them in interacting systems. Also, I seem to recall reading somewhere that free anyons (even abelian ones) are not comprabile with Lorentz symmetry? – SuperCiocia Dec 19 '18 at 14:39

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.

• wait so can we make non-abelian fermions and bosons in, say, $d=3$? – SuperCiocia Dec 16 '18 at 23:58
• @SuperCiocia That would be called parastatistics. However, they aren't truly non-abelian anyons, which are forbidden anyway by the spin-statistics theorem. One can show that by field redefinitions, you simply recover a bunch of fermions and bosons, possibly with additional degeneracies. – knzhou Dec 17 '18 at 11:50
• @SuperCiocia For example, people once tried to describe quarks as parafermions. But it is equivalent to say they are regular fermions with a local, color degree of freedom. True non-abelian anyons (in $d = 2$) can't be described this way. – knzhou Dec 17 '18 at 11:53
• Coul you point me to a reference where these field redefinitions are detailed please? – SuperCiocia Dec 18 '18 at 22:02