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
13 votes
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What do "pivotal" and "spherical" mean for (unitary) fusion categories on the level of the $F$-symbols?

Any unitary fusion category has a canonical spherical structure. See Example 2.12 of this paper and the references therein. It also follows from a more general result Prop 8.23 in this paper. So you ...
Noah Snyder's user avatar
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12 votes
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When is the endofunctor category of a monoidal category braided? When is it ribbon? Fusion? Modular?

The category $Func(C,C)$ is very rarely braided. It's a bit like asking "when is the endomorphism algebra of a vector space commutative?" For example, if $C=Vect\oplus Vect$, then $Func(C,C)$ is ...
André Henriques's user avatar
11 votes
Accepted

A cohomology theory for fusion categories

There is no such cohomology theory known (in particular, this is not related to Davydov-Yetter cohomology which is about deformations and vanishes for finite groups). In my mind this is a very ...
Noah Snyder's user avatar
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11 votes
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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 ...
Dave Penneys's user avatar
  • 5,192
11 votes
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Existence of a finite group with a given decomposition for a tensor square of one irreducible complex representation

I think that there is indeed no such finite group $G$, whether simple or otherwise. Note first that the representation $5_{1}$ can be assumed to be faithful ( for if $K$ is its kernel, then the group $...
Geoff Robinson's user avatar
11 votes
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Local fusion categories

Let $\mathcal{R}$ be a fusion category and $\beta : \mathcal{R} \to \mathrm{Vec}$ an additive monoidal functor. I first claim that $\beta$ is automatically faithful. (I know why you use "top ...
Theo Johnson-Freyd's user avatar
11 votes
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Is there a strongly noncommutative fusion category?

Consider the symmetric group group $G = S_3$ of order $6$. Then $\mathrm{H}^3_{\mathrm{gp}}(G;\mathrm{U}(1)) \cong \mathbb Z/6\mathbb Z$. Choose a generator $\omega \in \mathrm{H}^3_{\mathrm{gp}}(G;\...
Theo Johnson-Freyd's user avatar
10 votes

What are the necessary conditions for a real number to be a cyclotomic integers?

Since you are looking for restrictions on quantum dimensions of objects in unitary fusion categories, you also want your cyclotomic integers to be totally real, as they are Frobenius-Perron ...
Dave Penneys's user avatar
  • 5,192
10 votes
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Mac Lane strictness theorem and categorifiability of fusion rings

@Todd Trimble is correct; a strict fusion category is just a fusion category that is strict as a monoidal category. (Being strict or being fusion is just a property of a monoidal category that can ...
Dave Penneys's user avatar
  • 5,192
10 votes

What do "pivotal" and "spherical" mean for (unitary) fusion categories on the level of the $F$-symbols?

This is really an answer to the question in the comment of Noah's answer. Let $F$ be a solution to the pentagon equations for some fusion category $\mathcal C$. Pivotal structures on $\mathcal C$ ...
Matthew Titsworth's user avatar
9 votes
Accepted

Are there irreducible multi-fusion categories that are not fusion categories?

Matrix categories, $\mathrm{End}(\mathrm{Vec}^{\oplus n})$. (The identity on each copy of $\mathrm{Vec}$ are summands of the identity.) (You can generalize this example by putting fusion categories ...
Noah Snyder's user avatar
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9 votes
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A property forcing the Frobenius-Schur indicator to be positive

Here is a more general statement, see also Lemma 1.2 in [1]. Lemma: Let $Z$ be a self-dual $kG$-module which admits a non-degenerate $G$-invariant symmetric (alternating) bilinear form $b$. Suppose ...
Mikko Korhonen's user avatar
9 votes
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Are there non-homeomorphic 3-manifolds with the same Turaev-Viro-Barrett-Westbury invariants?

I asked Alexis Virelizier (coauthor of the book mentioned above) by e-mail. Here is his answer (reproduced with his authorisation): The answer is yes. See Theorem 1.1 (page 2291) in the following ...
Sebastien Palcoux's user avatar
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
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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
  • 7,952
8 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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8 votes
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Brauer-Picard for a fusion category coming from a quantum group

As far as I know, no one has written this up, but I think you should be able to find the Brauer-Picard groupoid for quantum groups at roots of unity by the following techniques. Now that I've written ...
Noah Snyder's user avatar
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8 votes
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Open questions on (finite) tensor categories

We made a list of open problems at an AIM conference a few years ago.
Noah Snyder's user avatar
  • 27.6k
8 votes

Why are fusion categories interesting?

(Unitary) fusion categories are interesting in physics because they classify gapped phases on the boundary of 2+1D quantum states of matter. Similarly, unitary modular tensor categories are ...
Xiao-Gang Wen's user avatar
8 votes
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How does the Tannaka duality work for weak Hopf algebras and fusion categories?

The procedure is more or less the standard Tannakian reconstruction argument. The first thing you need is a "forgetful" fiber functor $F:C\to Vect$, then you consider $R=End(F)$ the natural ...
AT0's user avatar
  • 1,417
7 votes
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What are the necessary conditions for a real number to be a cyclotomic integers?

I suspect you might be looking for the following fact: If $x$ is a cyclotomic integer, and $p$ a prime does not divide the discriminant, then the minimal polynomial of $x$ factors modulo $p$ into ...
Scott Morrison's user avatar
7 votes

Is there an integral fusion ring which is not of Frobenius type?

If $3 \not \in S$ then the answer to Question($S$) is yes. There are integral commutative fusion rings which are not of Frobenius type. Examples: Non-simple: rank $4$, FPdim $15$, type $[1,1,2,3]$, ...
Sebastien Palcoux's user avatar
7 votes
Accepted

Is there a fusion category not Grothendieck equivalent to a unitary one?

Yes, according to Andrew Schopieray. He just provided a categorifiable fusion ring, of rank 6 and multiplicity 2, without pseudounitary categorification (so without unitary categorification), in the ...
Sebastien Palcoux's user avatar
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 ...
Victor Ostrik's user avatar
7 votes
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Drinfeld center of a Deligne tensor product

By Cororllary 3.26 of arxiv:1009.2117, any braided tensor functor out of a non-degenerately braided fusion category is automatically fully faithful. Since $Z(\mathcal{C})\boxtimes Z(\mathcal{D})$ is ...
Dave Penneys's user avatar
  • 5,192
6 votes
Accepted

Does every enriched functor preserve tensors?

No. The functor "take a vector space to its double dual" is linear but does not preserve tensors with infinite-dimensional vector spaces. Enriched functors are automatically "lax tensored". An ...
Theo Johnson-Freyd's user avatar
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

What is a true invariant of $G$-crossed braided fusion categories?

Modular tensor categories give (via Reshetikhin-Turaev) a 321 oriented TFT. This gives a huge source of invariants, in particular any closed oriented 3-manifold gives a numerical invariant of MTCs. ...
Noah Snyder's user avatar
  • 27.6k
6 votes
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Conditions on the fusion data of symmetric fusion category

The information in the $\theta$'s is very weak, for example if a FC admits a symmetric structure with $\theta_i=-1$ for at least one $i$, it also admits a symmetric structure with $\theta\equiv 1$. ...
César Galindo's user avatar

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