19
votes
Sum of squares and divisibility
This is not a complete answer, but just a way to transform the problem into one that can be attacked by brute force in some known way.
Write $d_i^2=N/n_i$. Then your relation becomes $$\frac{1}{n_1}+ \...
12
votes
Accepted
Mapping class group of torus, why is $(ST)^3=S^2$?
Flip the direction of rotation for $S$, or choose the other meridian for $T$.
We can see this at the level of matrices. Define
$$S_1 = \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}, \qquad ...
10
votes
When modular tensor categories are equivalent?
A tensor category includes the information of a tensor product, which is something that takes objects and returns objects. This means that a tensor functor can't just "preserve tensor product" it ...
9
votes
Accepted
Sum of squares and divisibility
I hope so. But please double check (or, better, simplify) the argument below.
Denote $N=qs^2$ for $q$ squarefree. Then each $d_i$ divides $s$, say $d_i=s/m_i$ and we get $$q=1/s^2+\sum_{i=1}^r 1/m_i^2,...
8
votes
Is the central charge of a Drinfeld center always 0?
The Drinfeld center of a spherical fusion category has topological central charge $0\pmod 8$ see Remark 5.19 in
Müger, Michael:
From subfactors to categories and topology. II. The quantum double ...
8
votes
Accepted
On the existence of a square root for a modular tensor category
A characterization of Drinfeld centers of fusion categories is given in this paper as braided fusion categories containing a so-called Lagrangian algebra.
8
votes
Accepted
Are there (non Lagrangian) algebras of Turaev-Viro TQFTs which cannot be completed to Lagrangian algebras?
This is never possible. Indeed, if $A\in Z(\mathcal{C})$ is a condensable (connected, separable, commutative) algebra, then the condensed theory is $Z(\mathcal{C})_A^{\operatorname{loc}}$, which is ...
7
votes
Accepted
Finite groups G with Rep(G) Grothendieck equivalent to a modular category
Here is a necessary condition for a group $G$ such that Rep($G$) is Grothendieck equivalent
to a modular category:
there is a bijection between irreducible complex characters of $G$ and conjugacy ...
7
votes
Accepted
How to make a premodular category a modular tensor category?
A premodular category is always spherical, so you can take the Drinfel'd center:
http://arxiv.org/pdf/math/0111205v1.pdf
EDIT: I probably should have pointed out, as Marcel does in the comments, "...
7
votes
Accepted
Does unitarity and modularity constrain fusion multiplicities to be 0,1?
This is false for $D(G)$, when $G$ is sufficiently complicated. For a finite group $G$, the representation category of $D(G)$ has irreducible objects parametrized by pairs $(g, V)$ where $g$ is a ...
6
votes
Accepted
Module categories for Fibonacci anyons
There is only one equivalence class of indecomposable module categories, namely the trivial one.
Let us look into the possible algebras. They are $1$ and $1\oplus \tau$, and both have a unique ...
6
votes
How do I calculate the modular fusion category from a given Lie algebra and level in Chern-Simons theory?
I stumbled over this older question. I actually wrote a program, that takes the type of algebra (A,B,...,G), the rank, level, and appropriate root of unity as an input. It uses the associated quantum ...
6
votes
Sum of squares and divisibility
The answer of Francesco Polizzi recasts the problem into a form in which known results prove at least that there are (at most) a finite number of exceptions.
For any positive integer $s$, E. Landau ...
6
votes
Is there a non-degenerate quadratic form on every finite abelian group?
Yes. It's necessary and sufficient to show that every finite abelian group admits a nondegenerate quadratic form valued in a finite cyclic group. The following slightly stronger statement is true: ...
5
votes
Accepted
Automorphisms of a modular tensor category
For most quantum group categories all braided autoequivalences are classified by Cain Edie-Michell in this paper. The kind you're interested in, which is called "gauge auto-equivalences" there, ...
5
votes
Automorphisms of a modular tensor category
Not much known in general, but in this paper https://arxiv.org/abs/1312.7466, Davydov gives a description of them for Drinfeld Centers of Vec(G).
5
votes
How do I calculate the modular fusion category from a given Lie algebra and level in Chern-Simons theory?
There are two constructions of the modular fusion category. The conformal field theory approach is to take representations of the affine Kac-Moody algebra of given level and define a tensor product. ...
5
votes
Tensor functor between rigid tensor categories preserves $\text{Hom}$-objects
Your $F$ preserves dual objects because it preserves the duality situations: quadruples $(X,Y,\alpha : X\otimes Y \rightarrow I, \beta: I \rightarrow Y \otimes X)$.
It remains to recall the formula ...
4
votes
Internal Hom of Deligne' tensor product
That equation is not correct. You should be suspicious because the definition of the $\mathcal{C}$-module category structure on $\mathcal{M} \boxtimes \mathcal{N}$ doesn't use the $\mathcal{C}$-module ...
4
votes
Accepted
Modular tensor category associated to an even integral lattice and the lattice automorphism
Edit: I've thought about this question again, and I think the answer is more positive than what I said in an earlier version.
I will assume $L$ is positive-definite, since we need that to make $V_L$ ...
4
votes
Accepted
How nontrivial can "central extensions of ribbon fusion categories" be?
Deequivariantization (of which modularization is a special case) is inverse to equivariantization. That is, you can recover $\mathcal{C}$ as the category of G-equivariant objects in $\tilde{\mathcal{...
4
votes
Accepted
Do all non-degenerate quadratic forms come from positive even lattices?
Edited: I have missed your "positive". The signature of the lattice modulo 8 depends on the form only (some people call this Brown invariant and van der Blij theorem; Nikulin below calls this just the ...
4
votes
Accepted
Is there a non-degenerate quadratic form on every finite abelian group?
Thanks to the Fundamental Theorem of Abelian Groups, let
$$G:=\prod_{k=1}^{n}\{z:z^{m_k}=1\,,z\in\mathbb{S}\}\,,$$
and let $\chi(m)=2$ if $m$ is odd and $\chi(m)=1$ if $m$ is even. Then define
$$q\...
3
votes
Non-cyclotomic modular fusion categories
A detailed discussion of your question can be found in Davidovitch et. al's paper "On arithmetic modular tensor categories". They say that it is still an open problem whether there are non-...
3
votes
Accepted
Bialgebras with rigid representation theory
Yes, it is true that if a quasi-Hopf algebra has a trivial coassociator, then it's equivalent to an actual Hopf algebra (with $\alpha=\beta=1$). In other words, if you know the category is rigid (i.e. ...
3
votes
Accepted
Relationship between fusion category and its Drinfel'd center
The short answer is no.
Suppose you have a fully faithful monoidal functor $(F,J):\mathcal C\to\mathcal B$, where $\mathcal C$ and $\mathcal B$ are fusion and $J_{X,Y}:F(X)\otimes F(Y)\to F(X\otimes Y)...
3
votes
Simple modular tensor category and zero entries in its S-matrix
I may be misunderstanding the definition of simple (I am using your post Strongly simple fusion categories: the known examples? as reference), but I believe that the $(A_1,7)_{1/2}$ quantum group MTC ...
3
votes
Simple modular tensor category and zero entries in its S-matrix
Here is an anwser to Question 2 given by Andrew Schopieray by email:
<< Having a zero in your column (or row) is preserved under Galois conjugacy of characters. All columns having a zero but ...
2
votes
How to make a premodular category a modular tensor category?
Whether there is a "minimal" way to do this is still open, as far as I understand. One concrete formulation is in Mueger's "On the structure of modular categories" (Conjecture 5.2), see http://arxiv....
2
votes
Unitary structures on fusion categories
Reutter's recent paper "uniqueness of unitary structure for unitarizable fusion categories" answers your question in the affirmative (link: https://arxiv.org/pdf/1906.09710.pdf).
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