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Anyon theories are required to be associative, i.e. when fusing three anyons with labels $a,b,c$, we have $$(a\times b) \times c = a\times (b \times c)$$ This associativity is extended to the fusion and splitting spaces. Vectors in these spaces are represented by fusion trees as in the image. The image suggests that there are precisely two ways of fusing $a,b,c$ to $d$

enter image description here

My question is simple, what about first fusing $a$ with $c$ and then with $b$? Should this not be another possibility that is not covered by the diagrams. Indeed, in his paper https://arxiv.org/abs/cond-mat/0506438 Kitaev explicitly points out that the anyon theory is established on a line and that the order of anyons on that line matters. Still the theory is claimed to describe particles in 2D. I cannot see how this fits, as in 2D, the anyons need not be arranged on a line but can be literally anywhere.

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It depends on the context whether it's possible to fuse $a$ with $c$ first.

If we just have a fusion category, this process is ambiguous. In this case, our quasiparticles live in one spatial dimension, so their positions are linearly ordered. To fuse $a$ with $c$, we would need to close our space into a circle, and then there would be a choice of boundary condition which could affect the fusion outcome.

On the other hand, if we are in two spatial dimensions, then we have a braided fusion category (which is more algebraic data!), and we can freely move $a$ to the other side of $b \otimes c$ to obtain the fusion $(b \otimes c) \otimes a$, which could be compared to $a \otimes (b \otimes c)$ using the $R$-symbol. It's not enough to just have the associator $F$! The consistency relations between $F$ and $R$ are captured by the hexagon equation.

Note that the 16 Ising categories in Kitaev's paper come with this braiding, so they can describe quasiparticles in 2d.

However, if we are just given a fusion category, we can use it to produce the Levin-Wen model, and so obtain a model with 2d anyons. However, the 2d anyons are not the objects of the fusion category we started with! Instead they form the Drinfeld center, which one can think of as the universally smallest braided envelope of the fusion category (not all fusion categories are braided). Our original fusion category describes quasiparticles constrained to live in a certain universal boundary condition you can think of as a TQFT version of the Dirichlet boundary condition.

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  • $\begingroup$ Thank you for the answer. So your main argument seems to be that braiding introduces commutativity which captures what I think is physical reality for anyons: They are not ordered in any way in 2d and one can fuse them in any order at will. Is that correct? Up to this point, I thought, that the fusion of anyons is described by the mathematics of fusion categories while braiding is an additional structure you put to describe the exotic exchange statistics. But you suggest, that even to successfully describe fusion of anyons in the physical sense, a braiding is necessary? $\endgroup$
    – Marsl
    Commented Apr 21, 2019 at 10:58
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    $\begingroup$ The important point here is that the natural twist map $\tau: a \otimes b \rightarrow b \otimes a$ for braided categories doesn't square to $1$. There isn't a canonical relation between $a \otimes b$ and $b \otimes a$, because you have many variants: $\tau$, $\tau^{-1}$, $\tau^{3}$, etc. This is sometimes (mistakenly, I think) interpreted as the tensor product being noncommutative, which seems to me as nonsense by definition of the tensor product. $\endgroup$
    – Prof. Legolasov
    Commented Apr 21, 2019 at 11:02
  • $\begingroup$ @Marsl Yes I think that is a good summary. $\endgroup$
    – Ryan Thorngren
    Commented Apr 21, 2019 at 11:15
  • $\begingroup$ @SolenodonParadoxus While that's true, a fusion category doesn't even come with such a map! And some cannot even be equipped with one! This is what I call the $R$-symbol btw, following Kitaev and others. $\endgroup$
    – Ryan Thorngren
    Commented Apr 21, 2019 at 11:16
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    $\begingroup$ @Marsl you're proposing to "commute" the tensor factors by changing order (quote what about first fusing $a$ with $c$ and then $b$ unquote). There's no natural map that sends $b \otimes c$ to $c \otimes b$ and squares to 1, so that proposal doesn't make sense. $\endgroup$
    – Prof. Legolasov
    Commented Apr 21, 2019 at 22:34

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