The [product formula](http://groupprops.subwiki.org/wiki/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?




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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.



Generalization of the product formula on subfactors
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 $60^\circ$.)


