18 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 ...
  • 1,671
11 votes
Accepted

Twists, balances, and ribbons in pivotal braided tensor categories

Question 2: Given a pivotal braided category $\mathcal{C}$, there are 2 ways to endow $\mathcal{C}$ with twists under which $\mathcal{C}$ is a rigid balanced category. Conversely, given a rigid ...
  • 5,097
10 votes

What's the right way to think about "anomalies" in 3d TQFTs?

Here is my understanding from physics point of view. A quantum field theory is anomalous if it lacks of a UV completion. In other words there is no lattice theory in the same dimension, whose ...
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

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,...
  • 90.4k
7 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 ...
  • 26.7k
6 votes
Accepted

On the existence of a square root for a unitary 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.
  • 7,517
6 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, "...
  • 556
6 votes
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Is there a quotient or exact sequence of symmetric, premodular (ribbon fusion) and modular categories?

Short answer: Look for papers on "deequivariantization". (I think the original references are by Müger and Brugières, but I am not sure whether they used the term "deequivariantization".) Longer ...
6 votes
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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 ...
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 ...
5 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

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 ...
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
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Distinct 2D RCFTs with the same underlying MTC

The moonshine module VOA has trivial representation theory, i.e. the MTC is $Vec$ (and so the Frobenius algebra inside will be trivial). So this should give an example (i.e. take a trivial RCFT).
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5 votes
Accepted

Is the modularisation of a unitary fusion category always unitary?

The answer is yes to both questions (see Müger's paper). Müger's version of modularization is done in the unitary setting. The only new information that you need for full generality is that the ...
5 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 ...
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{...
  • 26.7k
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

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 $V_L$ ...
  • 43.3k
4 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, ...
  • 26.7k
4 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).
3 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\...
  • 1,395
3 votes

Gauss-Milgram formula for fermionic topological order?

We just posted a paper http://arxiv.org/abs/1507.04673 addressing this issue. For fermion topological orders, the fermionic version of this formula is $\Theta=\sum_a d_a^2 \theta_a=0$. See eq. 14 of ...
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. ...
  • 7,517
2 votes

Gauss-Milgram formula for fermionic topological order?

There is no fermionic analogue of the Gauss-Milgram formula. It applies to modular topological quantum field theories, while phases built out of fermions are not modular. A simple example, showing ...
2 votes

Distinct 2D RCFTs with the same underlying MTC

This is just a small extension of Eric's answer. But the point is that for every MTC (arising from a VOA) there are infinitely many examples. So many, that it is even hopeless to classify them (the ...
2 votes
Accepted

Do $G$-invariant non-degenerate quadratic forms come from $G$-invariant even lattices?

The group of automorphisms of $A$ which preserve the quadratic form $q$ is known as the orthogonal group $O(A,q)$. Likewise, if $L$ is a free $\mathbb{Z}$-module of finite rank with an even $\mathbb{...

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