Is there a Frobenius reciprocity for the coproduct?

Question

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 $\alpha$).

Answer 1

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.