For an irreducible finite depth finite index subfactor $(N \subset M)$, there is a structure of fusion category given by the even part of its principal graph. The simple objects $(X_i)_{i \in I}$ of depth $0$ or $2$ correspond to the projections $p_i = I_{H_i}$ on the $2$-boxes space $\mathcal{P}_2(N \subset M) = \bigoplus_{i \in I} End(H_i) $ as an algebra.
Warning: $I$ is the index set for the simple objects of depth $0$ or $2$, but there can have others.

On a $2$-boxes space, there is a structure of coproduct $a∗b$ (defined for example [here][1] p4).

Question: How compute the coproduct $p_j * p_j$ from the fusion $X_i \boxtimes X_j$?

My guess is that $p_i * p_j \sim \sum_{k \in K} p_k$ with $K=\{ k \in I \ \vert \ X_k \le X_i \boxtimes X_j \}$ (truncated fusion).
Is it true? What's the exact formula? How prove it?

Else, if there is a minimal projection $u \preceq p_i∗p_j$ such that its central support $Z(u)=p_k$, but $p_k \not \preceq p_i∗p_j$, is it still true that $X_k$ appears as a summand of $X_i ⊠ X_j$?
In other words, is it true that $Z(p_i * p_j) \le \sum_{k \in K} p_k$? Is it an equality? (weak truncated fusion)


In the formula, $tr(b)$ should be replaced by $tr(|b|)$, i.e., $\|b\|_1$.


Unfortunately the proof needs the irreducibility. I do not know how to generalize it to weak Kac algebras.

As Dave mentioned, one direction is clear. It is hard to determine when $(p_i*p_j)p_k=0$.

If $P_{2,\pm}$ are abelian, then I have a criterion for the extremal case: $(p_i*p_j)p_k \neq 0$ for all permitted fusion rule if and only if $P$ has a Yang-Baxter relation (see this paper).

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This is a great question. I think about it every once in a while, but I don't have a complete answer for you. Your guess is a completely natural one, and morally I think it should be true.

First, when you write $p_i * p_j \sim \sum_k p_k$, I assume what you mean is that we have $p_i*q_j = \sum_{k=1}^n \lambda_k p_k$ where each $\lambda_k>0$.

One side of the argument is easy when the planar algebra is irreducible. First, $p_i * p_j$ is positive (it's of the form $a^*a$ for a certain $a$ with 2 strings on one side and 4 strings on the other) and non-zero (just cap $p_j$). Now if $p_i*q_j = \sum_{k=1}^n \lambda_k p_k$ where all $\lambda_k>0$, we must have that each $X_k$ appears as a summand of $X_i \boxtimes X_j$. This follows since $(p_i*p_j)p_k = \lambda_kp_k \neq 0$ corresponds to a nonzero morphism $X_i \boxtimes X_j \to X_k$.

The tricky part is the other direction, where you want to show that such a morphism results in a minimal projection appearing in the sum in the expression for the coproduct. I think Liu has the best results on this to date in [Section 4, arXiv:1308.5656]. There he shows that a bound is possible. The hard part is determining when $(p_i * p_j)p_k = 0$ in the planar algebra, and Liu has nice results on this as well.

In more detail, he defines the rank of a positive element $x\in P_{2,+}$ to be the number of minimal projections $n$ needed to write $x=\sum_{i=1}^n \lambda_i p_i$ with all $\lambda_i>0$. For $x$ a projection, this is exactly the rank of the projection.

Theorem (Liu). If $\overline{p_i}$ is the dual of $p_i$ (rotate by 180), $$ \operatorname{rank}(\overline{p_i}* p_j)\leq \operatorname{dim}(p_j \mathcal{P}_{3,+} p_i), $$ where the right hand side is the number of length 2 paths between $p_i$ and $p_j$.

This result has higher analogs in the $2n$-box space, but there you need paths of length $2n$.

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  • $\begingroup$ You've written (at the end) about higher analogs. Is there an extension of the coproduct to every $n$-box space (or at least every $2n$-box space)? In fact there are several natural possible planar tangle generalizing the planar tangle of the coproduct on $2$-box space, which one is the good one? Finally, is it reasonable to generalize the relation between fusion and coproduct to such a generalized coproduct? $\endgroup$ – Sebastien Palcoux Sep 5 '14 at 12:30
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    $\begingroup$ Take the usual coproduct tangle on the 2-box space, and cable it by $n$ strands. This is the coproduct on the $2n$-box space. $\endgroup$ – Dave Penneys Sep 9 '14 at 18:26
  • $\begingroup$ Let $\mathcal{F}$ be the $1$-click rotation (or Fourier transform). Is the coproduct the same that the convolution $a * b = \mathcal{F}(\mathcal{F}^{-1}(a).\mathcal{F}^{-1}(b))$ ? If so, it exists for all the $n$-boxes space (with $n$ odd or even) and the tangle seems different to what you wrote in your last comment (see the diagrammatic computation here). What do you think? $\endgroup$ – Sebastien Palcoux Sep 22 '14 at 21:22
  • $\begingroup$ Yes, that is another associative multiplication, which could be considered a reasonable extension of the coproduct on the $2$-box space to the $2n$-box space. However, the coproduct used for the higher analog of Liu's result is the cabled one. There are many multiplications which deserve to be investigated further! $\endgroup$ – Dave Penneys Sep 22 '14 at 22:45
  • $\begingroup$ Dave, the coproduct is not a "truncated fusion" (see my answer). Nevertheless, if there is a minimal projection $u < p_i * p_j$ such that its central support $Z(u) = p_k$, but $p_k \not \le p_i * p_j$, is it still true that $X_k$ appears as a summand of $X_i \boxtimes X_j$? $\endgroup$ – Sebastien Palcoux Dec 27 '14 at 17:52

Depth $2$ case:

Let $(N \subset M)$ be a depth $2$ irreducible subfactor of finite index $[M:N]=: \delta^{2}$.
Let $\mathbb{A}:=P_{2,+}(N \subset M)$ the finite dimensional Kac algebra. Let the inner product $(a,b) = tr(a^*b)$ on $\mathbb{A}$ with $tr()$ the normalized trace. Let $(b_i)_i$ be an orthonormal basis of $\mathbb{A}$, i.e. $(b_i, b_j) = \delta_{i,j}$.

Consider the comultiplication $\Delta(a)$, the coproduct multiplication $a * b$, and their constant structure:

$$\Delta(b_i) = \sum_{j,k} c_{jk}^i \cdot(b_j \otimes b_k)$$ $$ b_j * b_k = \sum_{i} d_{jk}^i \cdot b_i$$ Proposition: $d_{jk}^i = \delta \cdot \overline{c_{jk}^i}$
Proof: by computing the inner product $(b_i,b_j∗b_k)$ in two different manners: the first using the constant structure of $b_j∗b_k$ and the second by computing $tr(b_i^*.(b_j∗b_k))$ diagrammatically. We use the irreducibility, the planar algebraic formulation of the comultiplucation by splitting, and the orthonormality of the basis $(b_i)_i$. $\square$

Corollary: $b_j * b_k = \delta \sum_{i} \overline{c_{jk}^i} \cdot b_i $

$\mathbb{A} = \bigoplus_{i \in I} End(H_i)$ with $(H_i)_i$ the sequence of irreducible complex representations of $\mathbb{A}$.
If $H_i \otimes H_j = \bigoplus_{ij} M_{ij}^k \otimes H_k$ (exactly the fusion rules for the depth $2$ case) and $n_{ij}^k = \dim(M_{ij}^k)$ then it's proved here that the inclusion matrix of $(\Delta(\mathbb{A}) \subset \mathbb{A} \otimes \mathbb{A})$ is $\Lambda = (\Lambda_{(ij)}^{k})$ with $\Lambda_{(ij)}^{k} = n_{ij}^{k}$.

So $p_i * p_j \sim \sum_k n_{ij}^{k}.p_k$, and my guess for the depth $2$ case follows ( $\sim$ means "same support").

General case:

Remark: An irreducible depth $d$ finite index subfactor $(N \subset M)$ is an intermediate of the depth 2 subfactor $(N \subset M_{d-2})$, which is non nec. irreducible (see here prop. 9.1.1 p 37). So the result could follow by generalizing the argument on Kac algebras above to an argument on weak Kac algebras.

The coproduct is not a (central) "truncated fusion", because the coproduct does not keep the centralness. The dual $(S_2 \subset S_4)$ subfactor planar algebra is a counter-example (see here).

Anyway, the coproduct could be a non-central truncation of the fusion.

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