1
$\begingroup$

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.

It's known that the coproduct of positive operators is also positive (see here thm 4.1 p18).

Question: Is the coproduct of central operators, also central?
Else, is it (nevertheless) true for the central projections?

$\endgroup$
1
$\begingroup$

It's obviously true if $\mathcal{P}_{2,+}$ is abelian.
It's also true for the irreducible depth $2$ case:
There is a nice direct diagrammatic proof using the splitting ([KLS] thm 5.1 p39, relation (3)).

It's false in general:

As observed by Vijay Kodiyalam, by duality the following sentences are equivalent:

  • (1) $\forall a,b \in \mathcal{P}_{2,+}$ with $a.c = c.a$ and $b.c = c.b$ $\forall c$, then $(a*b).c=c.(a*b)$ $\forall c$
  • (2) $\forall x, y \in \mathcal{P}_{2,-}$ with $x*z = z*x$ and $y*z = z*y$ $\forall z$, then $(x.y)*z=z*(x.y)$ $\forall z$

(1) corresponds to the question, and there is the following counter-example for (2):

Let $(H \subset G)$ be an inclusion of finite groups.
Let the subfactor $(\mathcal{R}^ G \subset \mathcal{R} ^ H)$ with $\mathcal{R}$ the hyperfinite ${\rm II}_1$ factor, and its planar algebra $\mathcal{P}$.
Then $\mathcal{P}_{2,-} = \bigoplus_{i \in I} \mathbb{C}e_i$ as an algebra, indexed as the double cosets partition $G = \coprod_{i \in I} Hg_iH$ (see [JS] p141) and the coproduct $*$ computes as follows: $$e_i * e_j \sim \sum_{k \in K} e_k \text{ and } Hg_iHg_jH = \coprod_{k \in K} Hg_kH$$ Example: $(H \subset G) = (S_2 \subset S_4)$ then $\mathcal{P}_{2,-} = \bigoplus_{i=1}^7 \mathbb{C}e_i $.
We obtain the following coproduct table (up to dividing by $\delta = \sqrt{12}$):

$ \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 & 2e_1+ e_2 & e_4+ e_5 & e_3+ e_5 & e_3+ e_4 & e_6+ 2e_7 & e_6 \newline \hline e_3 & e_3 & e_5+ e_6 & 2e_1+ e_3 & e_4+ 2e_7 & e_2+ e_6 & e_2+ e_5 & e_4 \newline \hline e_4 & e_4 & e_4+ 2e_7 & e_2+ e_5 & e_5+ e_6 & e_2+ e_6 & 2e_1+ e_3 & e_3 \newline \hline e_5 & e_5 & e_3+ e_6 & e_2+ e_4 & e_3+ e_6 & 2e_1+ 2e_7 & e_2+ e_4 & e_5 \newline \hline e_6 & e_6 & e_3+ e_5 & e_6+ 2e_7 & 2e_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}$

This table without the coefficients was obtained by a computation on the double cosets with GAP.
Next, for finding the coefficients: in the ith colomn (or the ith lign), each $e_j$ must appear $\vert H g_i H \vert / \vert H \vert$ times, with $\vert H g_i H \vert = \vert H \vert = 2$ for $i \in \{1,7\}$ and $\vert H g_i H \vert=4$ for $i \in \{2,3,4,5,6\}$ (and next all is divided by $\delta = [G:H]^{1/2}$).

$A = e_2+e_3+e_7$ and $B=e_5 + e_7$ are central for the coproduct, whereas $A.B = e_7$ is not!
So this contradicts (2), and so (1) is not true for $\mathcal{P}_{2,+}(\mathcal{R}^{S_4} \subset \mathcal{R} ^{S_2})$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.