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We first recall the embedding theorem for finite depth subfactor planar algebras:
The planar algebra generated by a (finite depth) subfactor, is embeddable into the planar algebra generated by its principal graph.

In the same spirit, we could imagine to build a planar algebra generated by a fusion ring.
Then, the planar algebra generated by a (finite depth) subfactor could be embeddable into the planar algebra generated by the fusion ring of its principal even part.

Question : How generate a planar algebra from a fusion ring ?

Edit : we show how build a planar algebra from a fusion ring. Previously, we explain where this construction comes from, for showing it could be relevant. We finally ask about an embedding theorem :

In the finite index, depth 2 and irreducible case, a subfactor is of the form $R^{\mathbb{A}} \subset R$ with $\mathbb{A}$ a finite dimensional Kac algebra.

As C$^{*}$-algebra, a finite dimensional Kac algebra $\mathbb{A}$ admits finitely many irreducible representations
$H_{1}$, ..., $H_{r}$, of increasing dimension $n_{1} = 1$, $n_{2}$, ... , $n_{r}$.

The comultiplication $\Delta$ gives an action of $\mathbb{A}$ on $H_{i} \otimes H_{j}$ which decomposes as follows : $$ H_{i}\otimes H_{j} = \bigoplus_{k}{M_{ij}^{k} \otimes H_{k}} $$ with $M_{ij}^{k}$ the multiplicity space of dimension $n_{ij}^{k}$, so that : $\sum n_{i}.n_{j} = \sum n_{ij}^{k} . n_{k}$

The fusion ring coming from $\mathbb{A}$ is generated by the basis $(h_{i})$ and the relations : $$ h_{i}. h_{j} = \sum_{k}{n_{ij}^{k} h_{k}} $$

The inclusion matrix of $[\Delta(\mathbb{A}) \subset \mathbb{A} \otimes \mathbb{A}]$ is $\Lambda = (\Lambda_{(i,j)}^{k})$ with $\Lambda_{(i,j)}^{k} = n_{ij}^{k}$ (see this link).

Now, the inclusion matrix $\Lambda$ gives a bipartite graph $\mathcal{G}(\Lambda)$, and so a graph planar algebra $\mathcal{P}(\Lambda)$.

Question : Does $\mathcal{P}(R^{\mathbb{A}} \subset R)$ embed into $\mathcal{P}(\Lambda)$ ?

Generalization : Let $\mathcal{F}$ be a fusion ring with structure constants $(n_{ij}^{k})$, let $\mathcal{G}(\Lambda)$ be the bipartite graph of matrix $\Lambda = (\Lambda_{(i,j)}^{k})$ with $\Lambda_{(i,j)}^{k} = n_{ij}^{k}$. The fusion ring planar algebra generated by $\mathcal{F}$, is the graph planar algebra $\mathcal{P}(\Lambda)$ of $\mathcal{G}(\Lambda)$.

Question : If $\mathcal{F}$ comes from a subfactor $(N\subset M)$, does $\mathcal{P}(N\subset M)$ embed into $\mathcal{P}(\Lambda)$ ?

Remark : Every finite depth, finite index, irreducible subfactor, is of the form $R^{\mathbb{I}} \subset R$, with $\mathbb{I} \subset \mathbb{A}$,
a left coideal subalgebra of a finite dimensional weak Kac algebra $\mathbb{A}$ (see here).

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Sebastien Palcoux
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  • A fusion ring gives many bipartite graphs, do you choose one specific? – Marcel Bischoff Mar 02 '15 at 06:22
  • @MarcelBischoff: the first question asks about an embedding theorem for fusion ring independently of any bipartite graph. The third question asks about a specific embedding for the specific bipartite graph called $\Lambda$. – Sebastien Palcoux Mar 02 '15 at 06:33
  • The even part of a subfactor gives a fusion category $\mathcal F$. Let us see this as a 2-category with one 0-object $N$. The completion of the subfactor planar algebra is, in general, a 2-category with two 0-objects $N$, $M$, which contains $\mathcal F$ as a sub-2-category. I guess, one can show, that a planar algebra associated with $\mathcal F$ embeds into the graph planar algebra associated with the fusion ring. But why should in general also the subfactor planar algebra (which is bigger) embed? – Marcel Bischoff Mar 02 '15 at 21:24
  • You mentioned that the inclusion matrix $\Lambda$ gives a bipartite graph $\mathcal{G}(\Lambda)$. Could you provide more details? Say how many even/odd vertices. How many edges between two vertices. It appears that $\Lambda$ is a $r\times r^2$ matrix. See my question: http://mathoverflow.net/questions/222171/how-to-calculate-the-principal-graphs-of-a-fusion-ring-with-a-given-simple-objec – Xiao-Gang Wen Oct 30 '15 at 02:32
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    @Xiao-GangWen: yes $r^2$ even vertices, $r$ odd vertices and $n_{i,j}^k$ edges between the vertices $(i,j)$ and $k$. We don't expect this bipartite graph to be a principal graph. – Sebastien Palcoux Oct 30 '15 at 08:59

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