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}+ \...
Francesco Polizzi's user avatar
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 ...
dvitek's user avatar
  • 1,691
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 ...
Noah Snyder's user avatar
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9 votes
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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,...
Fedor Petrov's user avatar
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 ...
Marcel Bischoff's user avatar
8 votes
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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.
Adrien's user avatar
  • 8,234
8 votes
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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 ...
Dave Penneys's user avatar
  • 5,327
7 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 ...
Victor Ostrik's user avatar
7 votes
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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, "...
Eric S.'s user avatar
  • 596
7 votes
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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 ...
S. Carnahan's user avatar
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6 votes
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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 ...
Marcel Bischoff's user avatar
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 ...
eddy ardonne's user avatar
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 ...
Geoff Robinson's user avatar
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: ...
Qiaochu Yuan's user avatar
5 votes
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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, ...
Noah Snyder's user avatar
  • 27.8k
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).
user156832's user avatar
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. ...
Bruce Westbury's user avatar
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 ...
Bugs Bunny's user avatar
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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 ...
Chris Schommer-Pries's user avatar
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 $V_L$ ...
S. Carnahan's user avatar
  • 45k
4 votes
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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{...
Noah Snyder's user avatar
  • 27.8k
4 votes
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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 ...
Alex Degtyarev's user avatar
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\...
Jack L.'s user avatar
  • 1,423
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-...
Milo Moses's user avatar
  • 2,799
3 votes
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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. ...
Adrien's user avatar
  • 8,234
3 votes
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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)...
Sean Sanford's user avatar
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 ...
Milo Moses's user avatar
  • 2,799
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 ...
Sebastien Palcoux's user avatar
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....
Makoto Yamashita's user avatar
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).
Milo Moses's user avatar
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