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## Hot answers tagged planar-algebras

16

Overview of an explanation : Jones-Wassermann subfactors for the loop algebra : Let $\mathfrak{g} = \mathfrak{sl}_{2}$ be the Lie algebra, $L\mathfrak{g}$ its loop algebra and $\mathcal{L}\mathfrak{g} = L\mathfrak{g} \oplus \mathbb{C}\mathcal{L}$ the central extension : $$[X^{a}_{n},X^{b}_{m}] = [X^{a},X^{b}]_{m+n} + m\delta_{ab}\delta_{m+n}\mathcal{L}$$ ...

6

(Unitary) fusion categories are interesting in physics because they classify gapped phases on the boundary of 2+1D quantum states of matter. Similarly, unitary modular tensor categories are interesting in physics because they classify gapped phases of 2+1D quantum states of matter. I have two papers to explain the connections arXiv:1405.5858 and arXiv:...

5

In the formula, $tr(b)$ should be replaced by $tr(|b|)$, i.e., $\|b\|_1$. $$b_\alpha*b_\beta=\sum_\gamma\frac{\|b_\alpha\|_1\|b_\beta\|_1\bar{c}_{\alpha\beta}^\gamma}{\sqrt{n}\|b_\gamma\|_1}b_\gamma.$$ 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 ...

4

You could define the composition of planar tangles in a way that is obviously well-defined. Something like: Scale the input planar tangle to half the size of the hole it goes in. Rotate the input disk so the two stars are on the same radial line. Finally, connect endpoints using the unique arcs that are linear functions in polar coordinates and do not cross ...

4

Your answer will work, except if the $b_i$'s have trace zero as Theo points out. If the $b_i\in \mathcal{P}_{n_i}$ has trace zero, just use $1_{n_i}+b_i$ instead of $b_i$, where $1_{n_i}$ is $n_i$ parallel strands. Then you can cap this off to get a scalar as before. By what you remarked above, you'll be able to recover $1_{n_i}+b_i$, from which you can ...

4

The answer is positive. For ease of notation, let me write $a=\sum_g \alpha_g u_g$ and $b = \sum_g \beta_g u_g$. Let's equip $\mathbb{C}G$ with the positive inner product induced by the standard basis. In terms of the usual normalized trace on a group algebra, this inner product is given by $(x,y)\mapsto\mathrm{tr}(x^*y)$. So when you decompose $\mathbb{C}G$...

3

There are various ways to generalize the equation. I give two different ones here: (1) In terms of bimodules: Take a finite index irreducible subfactor $N \subset M$. Suppose $P_1$ and $P_2$ are two intermediate subfactors and $P=P_1 \cap P_2$. Then we have the following equation for $N-N$ bimodules: \dim(P_1\mathbin{\mathop{\otimes}\limits_{P}} P_2)\dim(...

3

According to this paper of Nikshych-Vainerman, a finite-dimensional weak Kac algebra is precisely a finite-dimensional weak Hopf ${\rm C}^{\star}$-algebra with an involutive antipode ($S^2 = id$), and they give (p306) an explicit example of finite-dimensional weak Hopf ${\rm C}^{\star}$-algebra which is not a weak Kac algebra. More generally, any finite ...

3

Yes finitely many disks. Anything that forgets to say that is in error (my apologies if it’s my papers). One way to understand why is by thinking about the 1-dimensional case. The 2-box space of a “linear algebra” (algebra for the operad of 1-dimensional tangles) is just an ordinary associative algebra, and your question is whether one should allow ...

3

Can you clarify whether there exists a notion of direct product in this setting with the desired properties? If so, the asymptotics you are predicting only seem consistent with the hypothesis that there are no such objects besides groups. (I guess they do actually exist or you wouldn't ask this question.) The first version of this post suggested taking the ...

3

The answer is no. In fact, the group planar algebra has no non-trivial planar ideals. If $P_\bullet$ is a spherical planar algebra with non-zero modulus, and if $P_{0,\pm}$ are one dimensional, then every planar ideal is contained in the planar ideal $N_\bullet$ of negligible elements, i.e., those $x\in P_{n,\pm}$ for which tr$(xy)=0$ for all $y\in P_{n,\pm}... 3 You should first read the new Wikipedia page: Planar algebra. Next, a neat course by Vijay Kodiyalam is available on YouTube, see here (HD). 2 This depends on your definition of maximal. There's an intermediate von Neumann algebra which is not a factor, but no intermediate subfactors. I'll translate this question into a question about tensor categories. The even part of the subfactor is the category of graded vector spaces$\mathrm{Vec}(\mathbb{Z})$where I'll denote the 1-dimensional vector ... 2 This is an answer to the question in the OPs answer. I just took a brief look into Rehren - Weak C Hopf symmetry* in Quantum Groups Symposium at "Group21", eds. H.-D. Doebner et al., Goslar 1996 Proceedings, Heron Press, Sofia (1997), pp. 62-69. https://arxiv.org/abs/q-alg/9611007 which does the type III case. The point is that$S^{-1}(q)=S^\ast(q^...

2

The answer is yes. Moreover, $K_1\cap K_2 \subset M$ is a dihedral group subfactor, so the lattice of intermediate subfactors between $K_1 \cap K_2$ and $M$ is clear. Proof: Let us look at the dual lattice. Suppose $\hat{K_1},\hat{K_2}$ are index two intermediate subfactors of $M\subset \hat{N}$. Let the $e+p_1, e+p_2$ be the corresponding biprojections. ...

1

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 ...

1

Zhengwei's (email) answer for the question 1: Your guess is right. Group subgroup subfactors can be considered as a biprojection cut down of group subfactors. We can drive its coproduct from the group case.

1

Yes, any finite lattice with at most six elements can be realized as an intermediate subfactor lattice. The lattice $L_{19}$ and $L_{20}$ are realized as group-subgroup subfactors. This is proved in the proposition 1 of the paper On intervals in subgroup lattices of finite groups (2008) of Michael Aschbacher. Thanks to Mikko Korhonen who indicates me ...

1

Yes, there is a proof for $B_n$ with $n \le 4$ in this paper (Corollary 6.12 p29).

1

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 $>1$. Added ...

1

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) $\forall a,b \in \mathcal{P}_{2,+}$ ...

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