(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 Functorsand their $\mathscr{C}$-extensions and they are very closely related to MTCs.Verlinde Formula: The data $s,t,\zeta$ determines an MTC uniquely.Vafa Theoremand 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?

sphericalaxiom. $\endgroup$ – Manuel Bärenz Aug 21 '18 at 11:35