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Let $(H \subset G)$ be an inclusion of finite groups.
Let the subfactor $(\mathcal{R} \rtimes H \subset \mathcal{R} \rtimes G)$ with $\mathcal{R}$ the hyperfinite ${\rm II}_1$ factor, and its planar algebra $\mathcal{P}$.

On the $2$-boxes space of a planar algebra, there is the coproduct $a * b$ (defined for example here p4).
For the group-subgroup subfactor planar algebra $\mathcal{P}$ above, the $2$-boxes space $\mathcal{P}_2 = \bigoplus_{i \in I} \mathbb{C}e_i$ as an algebra, indexed as the double cosets partition $G = \coprod_{i \in I} Hg_iH$ (see Jones-Sunder p141).

Question 1: How compute the coproduct on $\mathcal{P}_2$?

Remark: my guess is that $e_i * e_j \sim \sum_{k \in K} e_k$ with $Hg_iHg_jH = \coprod_{k \in K} Hg_kH$.
Is it true? How prove that? [the relation $\sim$ means same support]

I've written a program computing my guess.
Example: For $(\mathcal{R} \rtimes S_2 \subset \mathcal{R} \rtimes S_4)$, with $\mathcal{P}_2 = \bigoplus_{i=1,\dots 7} \mathbb{C}e_i $,
I've found the following coproduct table (up to $\sim$):

$ \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 & e_1+ e_2 & e_4+ e_5 & e_3+ e_5 & e_3+ e_4 & e_6+ e_7 & e_6 \newline \hline e_3 & e_3 & e_5+ e_6 & e_1+ e_3 & e_4+ e_7 & e_2+ e_6 & e_2+ e_5 & e_4 \newline \hline e_4 & e_4 & e_4+ e_7 & e_2+ e_5 & e_5+ e_6 & e_2+ e_6 & e_1+ e_3 & e_3 \newline \hline e_5 & e_5 & e_3+ e_6 & e_2+ e_4 & e_3+ e_6 & e_1+ e_7 & e_2+ e_4 & e_5 \newline \hline e_6 & e_6 & e_3+ e_5 & e_6+ e_7 & e_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}$

Question 2: What's this coproduct for $(\mathcal{R} ^ G \subset \mathcal{R} ^ H)$?
Example: for $(\mathcal{R} ^{S_4} \subset \mathcal{R} ^{S_2})$ with $\mathcal{P}_2 = \mathbb{C} \oplus \mathbb{C} \oplus \mathbb{C} \oplus M_2(\mathbb{C}) $, what's the coproduct?

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1 Answer 1

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Zhengwei's (email) answer for the question 1:

Your guess is right. Group subgroup subfactors can be considered as a biprojection cut down of group subfactors. We can drive its coproduct from the group case.

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