Let $\mathcal{P}$ be an irreducible finite index-depth subfactor planar algebra. The $2$-boxes space $\mathcal{P}\_{2,+}$ is equipped with the coproduct $(a,b) \mapsto a\*b = \mathcal{F}(\mathcal{F}^{-1}(a).\mathcal{F}^{-1}(b))$ with $\mathcal{F}: \mathcal{P}\_{2,\pm} \to \mathcal{P}\_{2,\mp}$ the $1$-click rotation. 


It's known that the coproduct of positive operators is also positive (see [here](http://arxiv.org/abs/1308.5656) thm 4.1 p18). 


*Question*: Is the coproduct of central operators, also central?  

Else, is it (nevertheless) true for the central projections?



Is the coproduct of central operators, also central?
It's obviously true if $\mathcal{P}\_{2,+}$ is abelian.  

It's also true for the irreducible depth $2$ case:  

There is a nice direct diagrammatic proof using the splitting ([[KLS]](http://link.springer.com/article/10.1007%2FBF02829677) thm 5.1 p39, relation (3)). 


It's *false* in general:


As observed by [Vijay Kodiyalam](http://www.imsc.res.in/~vijay/), by duality the following sentences are equivalent: 


* (1) $\forall a,b \in \mathcal{P}\_{2,+}$ with $a.c = c.a$ and $b.c = c.b$ $\forall c$, then $(a\*b).c=c.(a\*b)$ $\forall c$
* (2) $\forall x, y \in \mathcal{P}\_{2,-}$ with $x\*z = z\*x$ and $y\*z = z\*y$ $\forall z$, then $(x.y)\*z=z\*(x.y)$ $\forall z$


(1) corresponds to the question, and there is the following counter-example for (2): 


Let $(H \subset G)$ be an inclusion of finite groups.  

Let the subfactor $(\mathcal{R}^ G \subset \mathcal{R} ^ H)$ with $\mathcal{R}$ the hyperfinite ${\rm II}\_1$ factor, and its planar algebra $\mathcal{P}$.  

 Then $\mathcal{P}\_{2,-} = \bigoplus\_{i \in I} \mathbb{C}e\_i$ as an algebra, indexed as the double cosets partition $G = \coprod\_{i \in I} Hg\_iH$ (see [[JS]](http://www.cambridge.org/us/academic/subjects/mathematics/abstract-analysis/introduction-subfactors) p141) and the coproduct $\*$ computes as follows:
 $$e\_i \* e\_j \sim \sum\_{k \in K} e\_k \text{ and } Hg\_iHg\_jH = \coprod\_{k \in K} Hg\_kH$$ **Example:** $(H \subset G) = (S\_2 \subset S\_4)$ then $\mathcal{P}\_{2,-} = \bigoplus\_{i=1}^7 \mathbb{C}e\_i $.  

We obtain the following coproduct table (up to dividing by $\delta = \sqrt{12}$): 


$ \begin{array}{c|c} 
\* & e\_1 & e\_2 & e\_3 & e\_4 & e\_5 & e\_6 & e\_7 \newline \hline 
 e\_1 & e\_1 & e\_2 & e\_3 & e\_4 & e\_5 & e\_6 & e\_7 \newline \hline 
 e\_2 &e\_2 & 2e\_1+ e\_2 & e\_4+ e\_5 & e\_3+ e\_5 & e\_3+ e\_4 & e\_6+ 2e\_7 & e\_6 \newline \hline 
 e\_3 & e\_3 & e\_5+ e\_6 & 2e\_1+ e\_3 & e\_4+ 2e\_7 & e\_2+ e\_6 & e\_2+ e\_5 & e\_4 \newline \hline
 e\_4 & e\_4 & e\_4+ 2e\_7 & e\_2+ e\_5 & e\_5+ e\_6 & e\_2+ e\_6 & 2e\_1+ e\_3 & e\_3 \newline \hline 
 e\_5 & e\_5 & e\_3+ e\_6 & e\_2+ e\_4 & e\_3+ e\_6 & 2e\_1+ 2e\_7 & e\_2+ e\_4 & e\_5 \newline \hline
e\_6 & e\_6 & e\_3+ e\_5 & e\_6+ 2e\_7 & 2e\_1+ e\_2 & e\_3+ e\_4 & e\_4+ e\_5 & e\_2 \newline \hline
 e\_7 & e\_7 & e\_4 & e\_6 & e\_2 & e\_5 & e\_3 & e\_1 
 \end{array}$ 


This table without the coefficients was obtained by a computation on the [double cosets](http://www.math.uiuc.edu/Software/GAP-Manual/DoubleCosets.html) with [GAP](http://www.gap-system.org/).  

Next, for finding the coefficients: in the ith colomn (or the ith lign), each $e\_j$ must appear $\vert H g\_i H \vert / \vert H \vert$ times, with $\vert H g\_i H \vert = \vert H \vert = 2$ for $i \in \{1,7\}$ and $\vert H g\_i H \vert=4$ for $i \in \{2,3,4,5,6\}$ (and next all is divided by $\delta = [G:H]^{1/2}$). 


$A = e\_2+e\_3+e\_7$ and $B=e\_5 + e\_7$ are central for the coproduct, whereas $A.B = e\_7$ is not!  

So this contradicts (2), and so (1) is not true for $\mathcal{P}\_{2,+}(\mathcal{R}^{S\_4} \subset \mathcal{R} ^{S\_2})$ 


