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 thm 4.1 p18).

Question: Is the coproduct of central operators, also central?
Else, is it (nevertheless) true for the central projections?


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] thm 5.1 p39, relation (3)).

It's false in general:

As observed by Vijay Kodiyalam, 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] 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 with GAP.
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 $\mathcal{P}_{2,+}(\mathcal{R}^{S_4} \subset \mathcal{R} ^{S_2})$


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