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Let $\mathbb{A}$ be a finite dim. weak Hopf $C^*$-algebra (or semisimple finite quantum groupoid) and $\hat{\mathbb{A}}$ its dual.
Let the Fourier transform $\mathcal{F}: \mathbb{A} \to \hat{\mathbb{A}}$ and the convolution $a * b = \mathcal{F^{-1}}(\mathcal{F}(a).\mathcal{F}(b))$.
As an algebra, $\mathbb{A}$ is isomorphic to a direct sum of matrix algebras. Let $B = \{b_1, \dots , b_n \}$ a matrix basis.

Let the structure constants $(c^{\gamma}_{\alpha \beta})$ and $(d^{\gamma}_{\alpha \beta})$ of the comultiplication $\Delta$ and the convolution product $*$ given by $\Delta(b_{\gamma}) = \sum_{\alpha \beta} c^{\gamma}_{\alpha \beta} . b_{\alpha} \otimes b_{\beta}$ and $b_\alpha*b_\beta= \sum_\gamma {d}_{\alpha\beta}^\gamma b_\gamma$.

Question: Is $d^{\gamma}_{\alpha \beta} = \delta \overline{c^{\gamma}_{\alpha \beta}}$?
[with $\delta = dim(\mathbb{A})^{1/2}$]

Remark: See proposition 8.16 p40 of this paper for a proof in the non-weak case.


There is the following definition of the Fourier transform in the planar algebra framework, but I guess there is an equivalent definition in the Hopf algebra framework.

A finite dimensional weak Hopf $C^*$algebra is by definition a finite dimensional weak Kac algebra.
If we consider the finite index depth $2$ subfactor planar algebra $\mathcal{P} = \mathcal{P}(R \subset R\rtimes \mathbb{A})$, with $R$ the hyperfinite ${\rm II}_1$ factor, then $\mathbb{A}$ and $\hat{\mathbb{A}}$ are the $2$-boxes algebras $\mathcal{P}_{2,+}$ and $\mathcal{P}_{2,-}$.
The Fourier transform $\mathcal{F} : \mathcal{P}_{2,\pm} \to \mathcal{P}_{2,\mp}$ is the $1$-click rotation (see for example here p9).

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    $\begingroup$ Could I ask how do you define the Fourier Transform and if yes how do you define the Fourier Transform? $\endgroup$ Sep 19, 2014 at 11:25
  • $\begingroup$ @JpMcCarthy: see the edit. $\endgroup$ Sep 19, 2014 at 12:16
  • $\begingroup$ @JpMcCarthy: your comment seems said that there is no definition for the Fourier transform in the general weak Hopf C*-algebra framework, isn't it? $\endgroup$ Sep 20, 2014 at 19:28
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    $\begingroup$ Excuse me but (at least for me) extracting the needed expression from that text (I guess you mean (5.33) and (5.34) there), or also from the text linked to in the question, purely in terms of $\mathbb A$ is quite a nontrivial task, so it would make your question much more accessible (at least for me) if you could just give a formula for $\cal F$ explicitly - say, in terms of the $b_i$s. $\endgroup$ Mar 24, 2016 at 9:51
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    $\begingroup$ There is a nice series of smooth lectures introducing to the planar algebras, see this link. $\endgroup$ Mar 25, 2016 at 17:22

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