In the theory of non-abelian anyons, essential information is stored in the fusion multiplicities or Verlinde coefficients $N_{ab}^c$.

Having the Pants Decomposition in mind, it is possible to use these coefficients to calculate the dimension of the topological/fusion Hilbert space for $p$ anyons 3, which is

$${\rm \ dim} H= \sum_{\{a_i\}}N_{b_1 b_2}^{a_1} N_{a_1 b_3}^{a_2}...N_{a_{p-3} b_{p-1}}^{b_p}$$.

My question is: What is the effect of braiding of , for example, anyon 1 with anyon 2 on the dimension? What is the effect of a twist?

Knowing that $N_{ab}^c=Z(S^2\times S^1; R_a, R_b, R_c)$ (Witten `89) braiding of anyons $a,b$ should give rise to a phase, when considering subsections 4.4 and 4.6 of 4 or chapter 18 of 5. Can this affect the overall dimension $dim H$?

  • $\begingroup$ Braiding and Dehn twists are both unitary operations on the fusion space. So I don't think they could change the dimension. $\endgroup$ Jun 23, 2014 at 15:11
  • $\begingroup$ Well, according to the prescription given in source 4 of my post one also has: $Z(S^2\times S^1; R_i)=\delta_{i 0}$ and $Z(S^2\times S^1; R_i, R_j)=\delta_{i j}$. If we braid charge i and j this could be unraveled giving a twist, so $\delta$ times a phase. The same holds for the fusion multiplicities and should have an impact in the product of $N$s above. The $S^1$ in the partition function arises from compactifying the time interval. Source 4 says that $Z(S^2\times S^1; no braid)\neq Z(...; braid)$ if I don't get it wrong. Maybe I confuse things... $\endgroup$
    – Hamurabi
    Jun 23, 2014 at 16:00
  • $\begingroup$ Mark is correct that Braiding and Dehn twists do not change the dimension. I am putting together the explanation for this that I know and feel makes this most clear, but fair warning that doing so requires changing our language a bit to one more suited to talking about anyons. $\endgroup$ Jun 23, 2014 at 17:23
  • $\begingroup$ Cool. I am eager to read your explanation. Maybe you could adress the point about the partition function with and without braiding then because from this point of view I am getting most of the trouble. $\endgroup$
    – Hamurabi
    Jun 23, 2014 at 17:46

1 Answer 1


My question is: What is the effect of braiding of , for example, anyon 1 with anyon 2 on the dimension? What is the effect of a twist?

Short answer: None. Consider that for anyons $N_{ab}^c=N_{ba}^c$ and that twisting is really just a braiding with some special stuff.

Longer answer: In order for this to make sense, we have to dig a little deeper and clear out some of the debris involved in going through the TQFT details and get to a more concise description of anyons and how to deal with them. Algebraically, anyons are described by things called modular tensor categories, or sometimes modular categories for short. Going through the details of how we get from one to the other is complicated (I can recommend this as a starting point for exploration - specifically the section relating CS to the Reshetikin-Turaev construction, as well as this for getting to the heart of the matter. For a more complete description of the algebraic description of anyons, see appendix E of Kitaev's paper here. Kitaev's paper is a personal favourite and much imitated.)

For those who can take it on faith that have said above is correct but would rather not slog through things to find the relevant definitions, we need a few things:

A category $\mathcal C$ consists of two things: A collection of objects $\mathcal C_0$. For a modular category, this collection is a finite set. The second thing is that for each pair of objects $x,y\in \mathcal C_0$, we have a collection $C_1(x,y)$ of maps (called hom-spaces) from $x$ to $y$, i.e. functions $f:x\rightarrow y$.

For the case of a modular category, these hom spaces are finite dimensional vector spaces, and in particular modular categories of physical interest are called unitary and these include a choice of inner product making them Hilbert spaces. The simplest example of a modular category is actually the category of finite dimensional complex vector spaces. Objects are FDC vector spaces and maps are linear maps between them (which we know also form vector spaces).

A monoidal category is a category that also has a way of "multiplying" two objects together. Modular categories have the really nice property that doing this gives us back a "sum" of so called "simple objects." Since we've already mentioned $Vec_\mathbb C$, what we mean by these two things is pretty simple: multiplication is the tensor product of vector spaces and sum is the direct sum of vector spaces. What our statement above then translates to is that the tensor product of finite dimensional vector spaces decomposes as a direct sum of one dimensional complex vector spaces (which we knew already). In this case then, we have only one simple object: $\mathbb C$. In general, we can think of our simple objects as types of anyons.

Now let's think about the hom-spaces. The decomposition of our product $a\otimes b$ into a sum of some $c$'s gives us maps from one to the other, and so for each $\{a,b,c\}$ as above we have a Hilbert space $\mathcal C_1(a\otimes b,c)$. These are precisely our fusion spaces and their dimensions are $N_{ab}^c$. These coefficients satisfy the Verlinde formula, though how we get it from all of this would take us away from the topic of this answer.

If we think about multiple fusions, we can extend our ideas about vector spaces in the same way: If I tensor three vector spaces together, two things must happen:

  1. The results of tensoring the first two then the second must be isomorphic to tensoring the second and then the first.
  2. Because of this, our semi-simplicity (that tensors decompose into sums), and the fact that our hom-spaces are vector spaces, we must have that $$\mathcal C_1(a\otimes b\otimes c, d)=$$ $$\bigoplus \mathcal C_1(a\otimes b,e)\bigotimes \mathcal C_1(e\otimes c,d)\cong \bigoplus \mathcal C_1(b\otimes c, f)\bigotimes \mathcal C_1(a\otimes f, d).$$ Thus $$\sum_e N_{ab}^e N_{ec}^d = \sum_f N_{bc}^f N_{af}^d.$$ The isomorphism between these two decompositions (which constitute two different choices of basis) is unitary and where we get our F-matrices from.

Now that we have our fusion spaces, we need to know what a braiding is. Since I've mentioned that a monoidal category has a notion of product, our braiding can be thought of as a notion of that product being commutative, that is $a\otimes b \cong b\otimes a$ in some way. The details of this are going to require that our hom spaces be isomorphic though, i.e. $\mathcal C_1(a\otimes b, c)\cong \mathcal C_1(b\otimes a,c)$. So, we must have that $N_{ab}^c=N_{ba}^c$. Since we are interested in physical situations, our braiding is then just a unitary matrix $R_{ab}^c$ with sides of length $N_{ab}^c$ taking $\mathcal C_1(b\otimes a, c)\rightarrow \mathcal C_1(a\otimes b,c)$.

From this, we get that our exchange matrix for two anyons a and b is just the direct sum of our matrices $R_{ab}^c$. But our R-matrices are isomorphisms, so our direct sum is an isomorphism, and so the dimensions can't change.

Taking this, let's think about a twist. One thing we've not had to deal with so far is where anyons come from and how that's modelled in our modular category, but that's pretty simple. For a given simple object (anyon type) $a$, the simple object (anyon type) $a^{*}$ such that $N_{a a^{*}}^0\neq 0$ is referred to as the dual. From a physical standpoint, we'd call this the antiparticle of $a$. Because our category is unitary, we get for free that $(a^{*})^{*}=a$. So we have a choice of creation operators: $\mathbb C \rightarrow a^{*}\otimes a$ and $\mathbb C \rightarrow a \otimes a^{*}$. Likewise we have two choices of annihilation operator.

Putting things together then - and omitting some important, but potentially occluding details - we can think of a twist on an anyon $a$ as having $a$, creating an $a^{*}\otimes a$ pair on the left of it, braiding the two copies of $a$ and then annihilating. The dimensionality of our system at the beginning and end of the process is then the same again.

  • $\begingroup$ I had a first read. Thank you! If I was to rephrase the answer in layman's words could I say that the effect of twisting and braiding doesn't change the dimension since the fusion H. space is the H. space of the anyonic system and as such already knows about these transformations? A second question would still direct to the partition function on $S^2\times I$ with sources which could be braided. If one braided source a with b and then compactified the time interval to the circle $S^1$ the partition function would not be the same as in the unbraided case. How can I connect this to your answer? $\endgroup$
    – Hamurabi
    Jun 23, 2014 at 21:58

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