# The planar algebra generated by the biprojections

Let $(N_1 \subset M_1)$ and $(N_2 \subset M_2)$ be two irreducible finite index subfactors.
Let $\mathcal{B}_i$ be the set of all the biprojections of $\mathcal{P}_{2+}(N_i \subset M_i)$.
Let $\mathcal{P}(\mathcal{B}_i)$ be the planar algebra generated by $\mathcal{B}_i$.

Suppose it exists a bijection $\phi: \mathcal{B}_1 \to \mathcal{B}_2$ such that:
- $b \le b' \Leftrightarrow \phi(b) \le \phi(b')$ [same lattice]
- $tr(\phi(b)) = tr(b)$ [same indices]

Main question: Are $\mathcal{P}(\mathcal{B}_1)$ and $\mathcal{P}(\mathcal{B}_2)$ isomorphic as planar algebras?

Remark: It's true if the lattice of biprojections is a single chain, because the planar algebra generated by the biprojections is the (generalized) Fuss-Catalan planar algebra, which is completely determined by the trace of the biprojections (see [BJ] and [L]).

Optional part: the planar algebra freely generated by a finite weighted lattice

Let $\mathcal{L}$ be a finite lattice and let $\mathcal{I} = \{ 4cos^2(\pi/n) \ \vert \ n=3,4,5, \dots \} \cup [4,\infty]$
Let $\tau: \mathcal{L} \to \mathcal{I}$ such that: $b \le b' \Rightarrow \frac{\tau(b')}{\tau(b)}\in \mathcal{I}$. Let's call $(\mathcal{L},\tau)$ a weighted lattice.

A chain of vertices $b_0 < b_1 < \dots < b_n$ is called complete if:
- $b\le b_0 \Rightarrow b=b_0$
- $b_n \le b \Rightarrow b=b_n$
- $b_i \le b \le b_{i+1} \Rightarrow b=b_i$ or $b_{i+1}$

Let $\mathcal{C}$ be the (finite) set of all the complete chains of $\mathcal{L}$. At each chain $\mathcal{c} \in \mathcal{C}$ we associate the Fuss-Catalan planar algebra $\mathcal{P}(\mathcal{c},\tau):=TL_{\delta_1} * TL_{\delta_2} * \dots * TL_{\delta_n}$ with $\mathcal{c} = (b_0^\mathcal{c} , b_1^\mathcal{c}, \dots , b_n^\mathcal{c})$ and $\delta_i = (\frac{\tau(b_{i+1}^\mathcal{c})}{\tau(b_{i}^\mathcal{c})})^{1/2}$.
Let $\mathcal{B}_{\mathcal{c}}=\{e_0^\mathcal{c}, e_1^\mathcal{c}, \dots, e_n^\mathcal{c} \}$ be the set of the corresponding biprojections in $\mathcal{P}_{2+}(\mathcal{c},\tau)$.

Let $\mathcal{P}(\mathcal{C},\tau)$ be the "free" planar algebra satisfying : $\forall \mathcal{c} \in \mathcal{C}$, there is an injection $i_{\mathcal{c}}: \mathcal{P}(\mathcal{c},\tau) \hookrightarrow \mathcal{P}(\mathcal{C},\tau)$.
[The definition of $\mathcal{P}(\mathcal{C},\tau)$ needs to be detailed. In general it's not a subfactor planar algebra]

Let $\mathcal{J}_m$ be the ideal generated by $\{ e_i^\mathcal{c}- e_j^\mathcal{c'} \ \vert \ \forall \mathcal{c}, \mathcal{c'}, \forall i,j \text{ with } b_i^\mathcal{c}= b_j^\mathcal{c'} \in \mathcal{L} \}$
Let $\mathcal{P}(\mathcal{L},\tau)$ be the quotiented planar algebra $\mathcal{P}(\mathcal{C},\tau) / \mathcal{J}_m$

The weight $\tau$ is called admissible if $\mathcal{P}(\mathcal{L},\tau)$ can be quotiented into an irreducible subfactor planar subalgebra $\mathcal{P}(N \subset M)$ with $(\mathcal{L}(N\subset M),tr)$ equivalent to $(\mathcal{L},\tau)$.

Optional questions:

Let $\mathcal{L}$ be a finite lattice:

1) Are all the weights $\tau$ admissible?
No, the subfactors of index $4$ do not realize all the lattices of height $2$.
But the question is still open if we restrict to the distributive lattices.

2) Is there an admissible weight $\tau$ ?
If so, every finite lattice can be realized as an intermediate subfactors lattice (see [W] about that).

3) What happen if we quotient $\mathcal{P}(\mathcal{L},\tau)$ by the ideal generated by all the non-evaluable closed diagrams?
Can such a process (or something close) running for some specific weights $\tau$?

Let $(N \subset M)$ be an irreducible finite index subfactor and $\mathcal{B}$ the set of all its biprojections:

4) Does the planar algebra $\mathcal{P}(\mathcal{B})$ embedd into a quotient of $\mathcal{P}(\mathcal{L}(N\subset M),tr)$ ? (in the same spirit than [JP])

5) Are all the irreducible subfactor planar algebras quotient of $\mathcal{P}(\mathcal{L}(N\subset M),tr)$ and having a weighted lattice equivalent to $(\mathcal{L}(N\subset M),tr)$, isomorphic to the planar algebra $\mathcal{P}(\mathcal{B})$?
If so, the answer of the main question above is yes.