I'm wondering about the exact reason why anyons escape the spin-statistic theorem (SST), see e.g. http://en.wikipedia.org/wiki/Spin–statistics_theorem.

I've read somewhere (the wikipedia page is sufficient I believe to understand this point, to be honest I don't remember where I read this) that the reason is just that anyons do not belong to the relativistic sector. Despite Lorentz invariance is still the cornerstone of the SST, anyons can have fractional exchange statistic because they belong to the Galilean invariance. See also this SE question about the SST and its demonstration.

On the other hand, it seems to me that this argument might well be wrong. Indeed, I've the feeling that the exact reason is that the permutation group is not enough to understand the exchange of two particles. If one discusses the rotation group instead of the permutation group, then one ends up with the Leinaas and Mirheim [please see the bottom of the question for the reference] argument, saying that in 2D, the homotopy group of SO(2) is $\mathbb{Z}$, no more $\mathbb{Z}_{2}$ as for SO($n$) with $n\geq 3$.

I'm wondering about the quantum field community perspective ? Do they continue to discuss the Lorentz invariance as the key point to obtain SST, or do they generally accept the Leinaas and Mirheim point of view [see below].

The two arguments might be almost equivalent, except that I believe the permutation group does not care about the space dimension. Moreover, I think it has no one-dimensional-non-trivial-projective representation, as required for anyons to exist, am I correct ?

NB: There are some alternative proof of the SST linked from the Wikipedia page about it. I gave the link at the beginning of this question.


2 Answers 2


You are right. In a space-time with one time dimension and $D$ spatial dimensions, finding possible different statistics is equilalent to look at the fundamental group (first homotopy group) of $SO(D)$

For $D=1$, the fundamental group is trivial.

For $D=2$, the fundamental group is $\mathbb{Z}$.

For $D>=3$, the fundamental group is $\mathbb{Z}_{2}$

So, it explains, why, in 3 spatial dimensions, there are only 2 kinds of statistics (fermions and bosons), while the situation is different with 2 spatial dimensions.

For quantum point of view, you will have to find unitary representations of this fundamental group.

  • 1
    $\begingroup$ Thanks a lot. Would you also agree to say that the argument Lorentz vs. Galilean invariance is erroneous (at least misleading) as well ? Thanks for your answer. $\endgroup$
    – FraSchelle
    May 27, 2013 at 8:57
  • 4
    $\begingroup$ I think that this has nothing to do with Lorentz vs. Galilean invariance. In $D$ spatial dimensions and 1 time dimension, The full relativist symmetry group is $ISO(1, D)$ (translations + $SO(1,D)$). So with 2 spatial dimensions, the relativist group is $ISO(1,2)$ ($3$ translations + $SO(1,2)$ group). So anyons in 2 spatial dimensions are as "relativist" as fermions and bosons in 3 dimensions, from my point of view. $\endgroup$
    – Trimok
    May 27, 2013 at 9:51

An alternative view to that of Trimok's answer is (not related to field theory, but I think is worth putting the discussion here): the spin-statistics are essentially representations of braid groups. In (2+1)D, the braid group is nontrivial. Whereas in (3+1)D, the braid groups are "trivial." (i.e. they are equivalent to symmetric groups) Here is a reference to this viewpoint.


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