Linked Questions

21 votes
2 answers
1k views

A new combinatorial property for the character table of a finite group?

Let $G$ be a finite group and $\Lambda = (\lambda_{i,j})$ its character table with $\lambda_{i,1}$ the degree of the ith character. Consider the following combinatorial property of $\Lambda$: for ...
Sebastien Palcoux's user avatar
11 votes
2 answers
392 views

What is known about arbitrary subfactors of integer index?

Let $N\subset M$ be an inclusion of ${\rm II}_1$ factors of finite index, $[M:N]<\infty$. I would be mostly interested in the hyperfinite case, $N\simeq M\simeq R$, but let us just take them ...
Richard's user avatar
  • 1,363
10 votes
1 answer
1k views

Are there workable algebraic geometry approaches for the pentagon equation?

A pentagon equation is a system of polynomial equations of degree $3$ with several variables and integer coefficients, given by a fusion ring. A fusion ring is given by a finite set of integer ...
Sebastien Palcoux's user avatar
16 votes
1 answer
2k views

The cyclic subfactors theory: a quantum arithmetic?

Context: First recall some results: Actions of finite groups on the hyperfinite type $II_{1}$ factor $R$ (Jones 1980). A Galois correspondence for depth 2 irreducible subfactors (Izumi-Longo-Popa ...
4 votes
1 answer
698 views

Abelian subfactors, a relevant concept?

Through the questions below, this post asks whether the concept of abelian subfactor is relevant. Remark : here abelian qualifies an inclusion of II$_1$ factors $(N \subset M)$, $N$ is not an abelian ...
Sebastien Palcoux's user avatar
7 votes
0 answers
473 views

Is there a non-trivial Hopf algebra without left coideal subalgebra?

Let $H$ be a finite dimensional Hopf ${\rm C}^{\star}$-algebra. A $\star$-subalgebra $I$ of $H$ is a left coideal if $\Delta(I) \subset H \otimes I$. $H$ is called maximal if it has no left coideal $\...
Sebastien Palcoux's user avatar
5 votes
0 answers
344 views

Interpolated simple integral fusion categories of Lie type

$\DeclareMathOperator\PSL{PSL} \DeclareMathOperator\Rep{Rep}$The idea motivating this post is that there should exist a global understanding of the unitary fusion categories $\Rep(G(q))$, with $G(q)$ ...
Sebastien Palcoux's user avatar
7 votes
0 answers
329 views

An alternative Cauchy theorem on Hopf algebras

Let $\mathbb{A}$ be a finite dimensional Hopf ${\rm C}^{\star}$-algebra. There already exists a generalization of Cauchy theorem using exponent, see [KSZ06]. We are interesting in an alternative ...
Sebastien Palcoux's user avatar
6 votes
0 answers
259 views

Is there an integral simple fusion ring rank<6, FPdim>60 and Frobenius type?

A fusion ring is a finite dimensional $\mathbb{Z}$-module $\mathbb{Z}\mathcal{B}$ together with a distinguished basis $\mathcal{B} = \{ h_1,...,h_r\}$ and fusion rules $ h_i \cdot h_j = \sum_k n_{ij}^...
Sebastien Palcoux's user avatar
4 votes
0 answers
238 views

Is an integral simple fusion ring, categorifiable?

A fusion ring $\mathcal{F}$ (see here p 28) is integral if the Perron-Frobenius dimension $d(h_i)$ of its basic elements $\{h_1,...,h_r\}$, are integers. Its rank is $r$ and its dimension is $\sum d(...
Sebastien Palcoux's user avatar
6 votes
0 answers
236 views

Existence of a Kac algebra for a given fusion ring in a particular class

A $n$-dimensional Kac algebra (i.e., a Hopf C*-algebra), admits finitely many irreducible representations, whose cardinal $r$ is called its rank, the increasing sequence $(d_{1},d_{2},d_{3}, ..., d_{r}...
Sebastien Palcoux's user avatar
3 votes
0 answers
264 views

Are there workable numerical approaches for the pentagon equation?

Warning: this post is the "numerical" analog of Are there workable algebraic geometry approaches for the pentagon equation? I've replaced "algebraic geometry" by "numerical" in its content, ...
Sebastien Palcoux's user avatar
5 votes
0 answers
198 views

Is there a perfect integral fusion category with PFdim = 2 mod 4?

A finite group $G$ is perfect if $G = G^{(1)} := \langle [G,G] \rangle$. Proposition: A finite group $G$ is perfect iff every $1$-dimensional complex representation of $G$ is trivial. proof: First if $...
Sebastien Palcoux's user avatar