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The product formula on finite groups states that for $H_1, H_2$ subgroups of $G$, then $$ |H_1H_2| \cdot |H_1 \cap H_2|=|H_1| \cdot |H_2| $$ This statement could be generalized to any finite index irreducible subfactor planar algebra $\mathcal{P}$ by:
For any biprojection $b_1, b_2 \in \mathcal{P}_{2,+}$, then $$ tr(S(b_1 * b_2)) \cdot tr(b_1 \wedge b_2)=tr(b_1) \cdot tr(b_2) $$ with $(\_ * \_)$ the coproduct, $S(\_)$ the range support, and $tr(\_)$ the normalized trace.

Question: Is this generalization true?


Note that $tr(b_1 * b_2) = \delta tr(b_1)tr(b_2)$, with $\delta^{-2} = tr(e_1)$ and $e_1$ the trivial biprojection.

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  • $\begingroup$ A result of Zeph Landau states that $b_1∗b_2=\delta tr(b_1b_2)S(b_1∗b_2)$ (see Theorem 3.10 here). It follows that $tr(S(b_1∗b_2))tr(b_1b_2)=tr(b_1)tr(b_2)$. $\endgroup$ Feb 25, 2017 at 12:58
  • $\begingroup$ Before that, see Sano-Watatani paper, Remark p235. $\endgroup$ Apr 7, 2017 at 17:32

1 Answer 1

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There are various ways to generalize the equation. I give two different ones here:

(1) In terms of bimodules: Take a finite index irreducible subfactor $N \subset M$. Suppose $P_1$ and $P_2$ are two intermediate subfactors and $P=P_1 \cap P_2$. Then we have the following equation for $N-N$ bimodules: $$\dim(P_1\mathbin{\mathop{\otimes}\limits_{P}} P_2)\dim(P)=\dim(P_1)\dim(P_2).$$

Proof: \begin{align*} \dim(P_1)\dim(P_2)&=\dim(P_1\mathbin{\mathop{\otimes}\limits_{N}}P_2)\\ &=\dim(P_1\mathbin{\mathop{\otimes}\limits_{P}}P\mathbin{\mathop{\otimes}\limits_{N}}P\mathbin{\mathop{\otimes}\limits_{P}}P_2)\\ &=\dim(P_1\mathbin{\mathop{\otimes}\limits_{P}} P_2)\dim(P\mathbin{\mathop{\otimes}\limits_{P}}P)\\ &=\dim(P_1\mathbin{\mathop{\otimes}\limits_{P}} P_2)\dim(P). \end{align*}

When $M=N\rtimes G$, $P_i=N\rtimes H_i$, for $i=1,2$, one obtains the group case.

(2) In terms of two boxes in planar algebras: We have an inequality in your formulation.

Proof: Let $b,b_1,b_2$ be biprojections corresponding to $P,P_1,P_2$ in case (1). Then \begin{align*} tr(S(b_1*b_2))&=\dim(P_1P_2)\leq\dim(P_1\mathbin{\mathop{\otimes}\limits_{N}}P_2);\\ tr(b)&=\dim(P);\\ tr(b_1)&=\dim(P_1);\\ tr(b_2)&=\dim(P_2). \end{align*} By the equation in case (1), we have that $$tr(S(b_1*b_2))tr(b)\leq tr(b_1)tr(b_2).$$ The equality holds iff $b_1b_2=b$. This condition always holds for commutative algebras, but not for non-commutative algebras.

Instead we can prove the following equality using the exchange relation of biprojections. $$tr(S(b_1*b_2))tr(b_1b_2)=tr(b_1)tr(b_2).$$ (I used this formula to show that the lower bound of the angle between minimal biprojections is $60^\circ$.)

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  • $\begingroup$ An interesting corollary is that if two biprojections $b_1$ and $b_2$ bicommute (i.e. $b_1b_2 = b_2b_1$ and $b_1 * b_2 = b_2 * b_1$) then $|b_1 \vee b_2 : b_1| = |b_2 : b_1 \wedge b_2|$. $\endgroup$ Sep 8, 2016 at 3:13

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