(Sorry for the length of the question, I'm trying to communicate what is bothering me as thoroughly as possible)

In the construction of modular tensor categories (MTC) from ground zero, we put structures one by one:

  • Tensor Product Structure $\to$ Monoidal Categories
  • Dual Objects $\to$ Rigid Monoidal Categories
  • Representation of Braid Group $\to$ Rigid Braided Monoidal Categories
  • Abelian and $k$-Linear Category
  • Semisimplicty
  • Finiteness conditions $\to$ Locally Finiteness and Finite Number of Isomorphism Class of Simple Objects
  • $\mathrm{End}(\mathbf{1})=k$: The unit object is simple.
  • $X\simeq X^{**}$ $\to$ Pivotal Categories

Demanding the above we get a Ribbon Fusion Category. If we look back at these structures there is a very good and mathematically natural reason why we demand them. The most non-trivial reasons are probably related to $\mathrm{End}(\mathbf{1})=k$ and finiteness conditions. But even then there are good reasons, for example $\mathrm{End}(\mathbf{1})=k$ guarantees not only that unit is simple, but also that quantum traces are a $k$-number. The finiteness conditions on the other hand gives us access to Jordan-Holder and Krull-Schmidt theorems, and isomorphisms like $(X\otimes Y)^*\simeq Y^*\otimes X^*$. So every last demand is quite reasonable and natural for a mathematician to study!

Then comes the last structure of an MTC:

The following structures on a ribbon fusion category are equivalent and this last structure defines a Modular Tensor Category

  • (Turaev) The matrix $\tilde{s}$ with components $\tilde{s}_{ij}=\mathrm{tr}(\sigma_{{X_i}X_j}\circ \sigma_{{X_j}X_i})$ is invertible (trace is quantum trace, $\sigma$ is braiding isomorphism and $X_i$ are simple)
  • (Bruguières, Müger) Multiples of unit $\mathbf{1}$ are the only transparent objects of the category. (An object $X$ is called transparent if $\sigma_{XY}\circ\sigma_{YX}=\mathrm{id}_{Y\otimes X}$ for all objects $Y$.)
  • Our premodular category $\mathscr{C}$ is factorizable, i.e. the functor $\mathscr{C}\boxtimes \mathscr{C}^\text{rev}\to \mathcal{Z}(\mathscr{C})$ is an equivalence of categories.

I've been desperately searching for a good reason on why this last demand is also mathematically natural to ask. If I may be so bold, seems a bit random to me! The best I could came up with is that

An MTC is the polar opposite to a symmetric ribbon fusion category. In other words in a symmetric category every object is transparent, so it is, in a way, maximally transparent. While an MTC is minimally transparent (maximally opaque maybe!).

But this still doesn't satisfy my irk! So what? What about the categories in between? Then there are indirect reasons why one should demand this structure:

By demanding this last structure a lot of `nice' things happen.

  • We have a projective representation of $\mathrm{SL}(2,\mathbb{Z})$ in our MTC: Namely matrices $s=\tilde{s}/\sqrt{D}$ with $D$ being the global dimension, $t$ with $t_{ij}=\theta_i\delta_{ij}$ and $c$ (charge conjugation) $c_{ij}=\delta_{ij^*}$ such that $$(st)^3=\zeta^3s^2, \quad s^2=c, \quad ct=tc, \quad c^2=1$$
  • Modular Functors and their $\mathscr{C}$-extensions and they are very closely related to MTCs.
  • Verlinde Formula: The data $s,t,\zeta$ determines an MTC uniquely.
  • Vafa Theorem and it's generalizations: The scalars $\theta_i$ and $\zeta$ are roots of unity.

and even more nice things...

But still... Unless there is some good reason why we demand this last conditions naturally, all of this seems like a happy coincidence. So basically my question is

Is there a deep fundamental mathematical (or physical, physical is also fine) reason behind this last structure, which makes this condition a priori?

  • 2
    $\begingroup$ If you consider rational chiral conformal field theory you naturally get MTCs. If you look into Moore and Seiberg axioms you can easily check their data indeed gives a MTC. Moreocer there is a precise statement for vertex operator algebras and conformal nets saying that under certain technical assumptions the category of representations is a MTC. Then MTCs are equivalent to 321 topological field theories $\endgroup$ Mar 5, 2016 at 4:36
  • $\begingroup$ @MarcelBischoff Yes but why? :D $\endgroup$ Mar 5, 2016 at 6:02
  • 3
    $\begingroup$ Note that $\mathcal{C}^\mathrm{rev}$ is the dual to $\mathcal{C}$ in the bicategory of algebras over $Z(\mathcal{C})$ and bimodules between them. So the factorizability condition says that the dual to $\mathcal{C}$ is in fact inverse to $\mathcal{C}$ under $\boxtimes$. Equivalently it says that $\mathcal{C}$ is $\boxtimes$-invertible. For if $\mathcal{C}$ is invertible, its dual must be its inverse. The appearance of $Z(\mathcal{C})$ is inevitable because the factorizability condition can't hold with respect to anything else. This all categorifies the story of Azumaya algebras. $\endgroup$
    – Tim Campion
    Aug 14, 2017 at 22:17
  • 1
    $\begingroup$ There is a slight mistake in your definition of ribbon fusion categories. You also have to assume that the categorical dimensions arising from the pivotal structure are all real. This is called the spherical axiom. $\endgroup$ Aug 21, 2018 at 11:35

2 Answers 2


Interesting question! As far as I know, there are at least two secretly equivalent answers.

You somehow already gave the first one: a modular tensor category is the same as a modular functor (though the precise statement is quite subtle, see the beautiful introduction to this paper: https://arxiv.org/abs/1509.06811). A modular functor is, roughly, a collection of compatible (projective) representations of mapping class groups, so in particular you need a representation of $SL_2(\mathbb{Z})$. It turns out this is also sufficient, in the sense that a premodular tensor category gives representations of mapping class groups in genus 0, and modularity is exactly what you need to extend this to a higher genus.

A somehow more conceptual reason is related to the fact that those theories have an anomaly, i.e. you get only projective representations of MCG's, and invariant of 3-fold with some extra structure (I learned this point of view from Walker and Freed-Teleman). The origin of the story is the 3d Chern-Simons TFT introduced by Witten using Feynman integrals. It turns out that there is a 4d TFT around, a very simple topological version of Yang-Mills theory. This theory is simple because it involves integrating 4-forms on 4-folds, and those forms are actually exact. So if Feynman integrals really were integrals, by Stokes theorem this would be trivial on a closed manifold, and for a 4-fold W with boundary we could define $$Z_{CS}(\partial W):=Z_{YM}(W).$$ Since any oriented 3-fold bound a 4-fold this would indeed be enough to define your theory.

It turns out $Z_{YM}$ is not trivial, but it is close to be: it is an invertible theory. It means in particular that it attaches 1-dimensional vector spaces to 3-fold, and that every 4-fold $W$ gives an isomorphism $$Z_{YM}(\partial W) \longrightarrow Z_{YM}(\emptyset)=\mathbb{C}$$ which depends on the choice of $W$ only up to bordism. Hence you get a number defined up to this choice, and the bordism class of $W$ is precisely the extra structure occurring in this anomaly I was talking about.

How does it relate to your question? Well, it is expected that every premodular category gives rise to a 4-dimensional TFT. Roughly speaking, to know whether this theory is invertible, it's enough to check what you get for the 2-sphere, which is a category, and it turns out this category is equivalent to the Müger center, i.e. the category of transparent objects. At the categorical level, invertibility means being equivalent to the category of vector spaces.

Therefore, the modularity condition is exactly what you need to go from a 4d TFT to a 3d TFT with anomaly.


Adrien's answer is fantastic, and that is certainly one of my major intuitions. However, there is also a much more elementary explanation that does not use any powerful mathematics. It only requires you know a bit about quantum error correcting codes/stabilizer codes.

We use the fact that MTCs are anyon models. That is, its simple objects should correspond to quasiparticles in a 2+1 topological quantum phase of matter. Braiding on the category level should correspond to just moving your particles around each other.

Bruguières' condition for non-degeneracy says the following very simple physical condition: for every non-trivial particle type, there should be another particle which braids non-trivially with it.

Why is that? Well, the answer comes from stabilizer codes. As shown in Kitaev's original paper on topological quantum computation, anyons come naturally from stabilizers on lattices. That is, you have some big Hilbert space, and you get your code subspace by restricting to the common +1 eigenspace of all your stabilizers. Failures of these stabilizers (i.e. stabilizers which act non-trivially on a given state) signal the presence of a quasiparticle.

Now, suppose we have a particle type. This will have some stabilizers associated to it, and it will move by applying some other matrices. The classic example is the toric code, where your stabilizers might be some tensors of the Pauli matrix $\sigma_X$, and the corresponding movement matrix is $\sigma_Z$. Here, by "movement" we mean that $\sigma_X\sigma_Z=-\sigma_Z\sigma_X$, so it flips a -1 eigenvalue (having a quasiparticle present) to a +1 eigenvalue (not having a quasiparticle present) and vice-versa. Thus, when it is applied it changes eigenvalues such that the anyon moves from one site to another.

The key observation is as follows:

If you move an anyon around a null-homotopic loop, it should act by the identity. This "loop=identity" condition is a new stabilizer, and its failures have associated anyon types. This new anyon type will necessarily braid non-trivially with your starting anyon type, and hence every non-trivial anyon type braids non-trivially with somebody else.

Why do these quasiparticle types braid non-trivially with each other? Well, one particle type moves by the stabilizer of the other. Stabilizers and movers only commute up to a non-trivial action (this is what makes the movers actually move things), and hence the conclusion follows.

It shouldn't be surprising that the non-degeneracy condition has an explanation in terms of stabilizers and anyons: everything about MTCs does. Sometimes you just have to go looking for it.

This also explains how the quantum double models work a bit better. You have one copy of a finite group $G$ to start with, and then you want all of their null-homotopic loops to act by the identity. This creates a whole new copy of quasiparticles corresponding to the dual of $G$. This gives you your two nice families of quasiparticles coming from your two "twisted" copies of $G$.


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