3
$\begingroup$

Let $(N \subset M)$ be an irreducible finite index depth $n$ subfactor. Let $P = P(N \subset M)$ its planar algebra.
Let $(B_i)$ be the finite sequence of $N$-$N$-bimodules appearing in the principal graph.
Let $2m = n$ if $n$ even, else $2m=n+1$.
Let $p_i \in P_{2m,+}$ be the minimal central projection related to the $N$-$N$-bimodule $B_i$.

Question: Is there a planar tangle $T: P_{2m,+} \otimes P_{2m,+} \to P_{2m,+}$ such that $T(p_i \otimes p_j) = \sum_{k} n_{ij}^k p_k $ with $B_i \boxtimes B_j = \bigoplus_k M_{ij}^k \otimes B_k$ and $dim(M_{ij}^k)= n_{ij}^k$ (the fusion coefficients)?

Else, is there such a $T$ if we only consider the range support? the central support?

Remark: If $n = 2$, such a $T$ exists, it's the coproduct (see here).
Then, a generalization of the coproduct on $P_{2m,+}$ could do the job.

Improve this question
$\endgroup$
3
  • 1
    $\begingroup$ Is there a reason that you want to have all projections on the same number of strings. For example take $A_n$ than the objects are the Jones-Wenzl projections on $0,...,n-1$ strings. In this case I can give you the tangle. See page 97: books.google.com/… $\endgroup$
    – Marcel Bischoff
    Commented Dec 27, 2014 at 18:16
  • $\begingroup$ @MarcelBischoff: I'm agree with this approach without the same number of strings, if it's more natural. $\endgroup$
    – Sebastien Palcoux
    Commented Dec 29, 2014 at 6:58
  • $\begingroup$ I mean you can easily get a projection on $m+2n$ strings (with $n\in\mathbb N$) from a projection on $m$ strings. $\endgroup$
    – Marcel Bischoff
    Commented Dec 30, 2014 at 20:19

1 Answer 1

Reset to default
1
$\begingroup$

I would think about something like this, where the caps and cups stand for $b$ strings the left one for vertical lines for $a$ strings and the right for $c$ strings. It has to be normalized to give again a sum of projections though. And this gives a map $P_{a+c}\otimes P_{c+b} \to P_{a+b}$. And it will not work if there are multiplicities $>1$.

enter image description here

Added Remark:

Note also that for finite depth, there is some $k$ such that $N\subset M_k$ is depth 2, where $M_k$ comes from the iterating Jones' basic construction. Then you can consider the planar algebra of $N\subset M_k$ (I guess the buzzword is cabling) and reduce to the depth 2 case. Then you can use the co-product. But, as far as I understand (see Zhengwei's answer here), this might cause problems, because $N\subset M_k$ is in general not irreducible.

Improve this answer
$\endgroup$
4
  • $\begingroup$ Every fusion category is the tensor category of representations of a weak Hopf algebra (ENO). For the fusion category of the $N$-$N$ bimodules in the principal graph of $(N \subset M)$, it should be the weak Kac algebra coming from the depth $2$ subfactor $(N \subset M_k)$. $\endgroup$
    – Sebastien Palcoux
    Commented Dec 30, 2014 at 21:27
  • $\begingroup$ Is the second sentence a question? $\endgroup$
    – Marcel Bischoff
    Commented Dec 30, 2014 at 21:46
  • $\begingroup$ In the second sentence I write "it should be", because I did not check, but it would be natural to be this. Do you know if it's effectively this? $\endgroup$
    – Sebastien Palcoux
    Commented Dec 31, 2014 at 8:35
  • 1
    $\begingroup$ Yes, if you choose $k$ right and choose the right weak Kac algebra from the two (dual to each other) associated to the subfactor. Note that the weak Hopf algebra is far from unique. A somehow canonical example you get if you take the object $A=\bigoplus_{i,j }M_i\boxtimes M_j$ ($\{M_i\}$ iso-classes of irreducible N-N bimodules in your fusion category), which has naturally the structure of a Q-system and gives you an reducible, self-dual subfactor $N\subset M_A$, which has depth 2, so a weak Kac algebra. $\endgroup$
    – Marcel Bischoff
    Commented Dec 31, 2014 at 17:19

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .