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.  

Let the adjoint map $a \mapsto \overline{a}=\mathcal{F}(\mathcal{F}(a))$ the $2$-clicks rotation (180°), and let $e\_1$ be the trivial biprojection. 


Using diagrams (and irreducibility), we see easily that if $a$ is a projection, then $e\_1 \le a \* \overline{a}$. 


*Question*: Is there a (weak?) Frobenius reciprocity?  

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


By Frobenius reciprocity, you mean that for $a$, $b$, $c$ projections and $\alpha>0$:  

$\alpha c \le a \* b$ $\Rightarrow$ $\alpha b \le \overline{a} \* c$ and $\alpha a \le c \* \overline{b}$  

By *weak* Frobenius reciprocity, you mean the same but on the range support (and so without $\alpha$).



Is there a Frobenius reciprocity for the coproduct?
Yes, but in general, it's an "ultra-weak" Frobenius reciprocity. 


Let $a, b$ be positive operators. The notation $a \preceq b$ means that the support projection of $a$ is a subprojection of the support projection of $b$.


**Lemma:**  

Let $p \in P\_{2,+}$ be a projection, then $e\_1 \preceq p \*\overline{p} $, and $e\_1 \preceq p \*\overline{q} $ (with $q$ a projection) iff $p.q \neq 0$.  

*Proof:*  

 The computation by diagrams of $e\_1 . (p \*\overline{p} )$, using irreducibility and $tr(p) \neq 0$, proves the existence. The same for $e\_1 . (p \*\overline{q} )$, using the assumption and that $p.q \neq 0$ iff $tr(p.q) \neq 0$ (because $tr(pq(pq)\*)=tr(pqq\*p)=tr(pqp)=tr(pq)$), proves the second assertion. $\square$ 


**Ultra-weak Frobenius reciprocity**:  

Let $a,b,c \in P\_{2,+}$ be (minimal) projections, then $a \preceq b\*c$ $\Rightarrow$ $ b' \preceq a\*\overline{c}$ and $ c' \preceq \overline{b}\*a$, with $b',c'$ (minimal) projections and $b.b',c.c' \neq 0$.  

*Proof:*  

First if $ a \preceq b\*c$, then $e\_1 \preceq \overline{a}\*(b\*c)$ , but by associativity $\overline{a}\*(b\*c) = (\overline{a}\*b)\*c $, so $e\_1 \preceq (\overline{a}\*b)\*c $, then by previous lemma, $\overline{c'} \preceq \overline{a}\*b$. So $c'=\overline{\overline{c}} \preceq \overline{\overline{a}\*b}=\overline{b}\*a$.  

Idem, $e\_1 \preceq (b\*c)\*\overline{a} $ and $ b' \preceq a\*\overline{c}$. $\square$ 


**Corollary**: If $a, b, c$ are minimal projections and $\alpha a \le b\*c$ ($\alpha \neq 0$), then:  

if $b$ is central then $\beta b \le a\*\overline{c}$ with $\beta=\alpha \frac{tr( \overline{a})}{tr( \overline{b})}$  

if $c$ is central then $\gamma c \le \overline{b}\*a$ with $\gamma=\alpha \frac{tr(a)}{tr(c)}$.  

*Proof:* About $\alpha, \beta, \gamma$, the formulas come by computing with $\le$ instead of $\preceq$. $\square$ 


**Abelian case**: It follows that we have a kind of Frobenius reciprocity if $P\_{2,+} = \mathbb{C}^n$, i.e. is abelian.


