Is there a definition of Heisenberg double for quasi-Hopf algebras? $\newcommand{\dmod}{\text{-}\mathrm{mod}}$Let $A$ be a finite-dimensional quasi-Hopf algebra over a field $k$ and $A\dmod$ be the category of finite dimensional left $A$-modules.
Since $A\dmod$ is a rigid monoidal category, its Drinfel’d center $\mathcal{Z}(A\dmod)$ is a rigid braided monoidal category. So we obtain quasi triangular quasi hopf algebra $D(A)$ from $\mathcal{Z}(A\dmod)$ by using tannaka duality.
Question 1:Is there a (categorical) definition of Heisenberg double $H(A)$ for quasi hopf algebra $A$?
Question 2: If the answer to the above question is yes,is $D(A)$ a subalgebra of $H(A)\otimes H(A)^{\mathrm{op}}$?
 A: $\newcommand{\dmod}{\text{-}\mathrm{mod}}$I've been asking myself that same question for a while, and I'm fairly certain the answer is "no". Let me expand a bit:

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*First of all, in the finite dimensional Hopf case the Heisenberg double is isomorphic, as an algebra, to $\operatorname{End}(A)$ so its category of modules is just equivalent to $\mathrm{Vect}$. Of course this equivalence do not commute with the fiber functor (ie it doesn't preserve dimensions of the underlying vector spaces) but it already suggests there's no way to get something meaningful just from $A\dmod$ as an abstract monoidal category.

*Closely related to that is the fact that there is no good definition of a Hopf module over a quasi-Hopf algebra (while there's one for Yetter-Drinfeld modules) (modules over $H(A)$ is equivalent, this time in a fiber functor preserving way, to the category of Hopf modules).

*I believe the secret reason behind this failure, is that the equation of which the canonical element of $H(A)$ is a universal solution (the pentagon equation) do not quite makes sense categorically (it somehow involves the flip of vector spaces). On the other hand there are I think "braided" version of the Heisenberg double in the literature which might or might not be to your liking.

*That being said, there is a good definition of Hopf-bimodules for quasi-Hopf algebra: even in the quasi-case, $A$ is a coalgebra in the category of $A$-bimodules (with monoidal structure coming from the Hopf structure, not the tensor product over $A$) so you can talk about $A$-comodules in there.

*If $C=A\dmod$, then $C$ is a module category over itself and the category of Hopf bimodules can be identified with the category of right exact $C$-module functors from $C$ to itself. It turns out this is in turn equivalent to just $C$ (this is the quasi-Hopf version of the fundamental theorem of Hopf modules), so abstractly this is again kind of boring but concretely at least you get the correct thing on the nose (in the sense that, again, in the Hopf case the equivalence from $C$-modules endofunctors of $C$, and Hopf bimodules, is compatible with the fiber functor).

*Perhaps more concretely, if $A$ happens to be quasi-triangular/braided, there there is a way to construct (an algebra isomorphic to) $H(A)$ as an algebra in $C$ which thus makes sense for quasi-Hopf algebras as well (but in that case, you'll get something which is truly not an associative algebra in vector spaces). I can expand on that if you like. There is probably a way to remove the braided condition at the cost of making this construction in the category of $A$-bimodules instead but I haven't thought too much about it.

Edit: About your last question: I just realize that although I don't think there is a categorical construction of $H(A)\dmod$, there might be one of $H(A) \otimes H(A^{\mathrm{op}})\dmod$. Namely, again, $A$ is a coalgebra in the category of $A$-bimodules, so you can take $A$-bicomodules in there and this should give you the correct thing. There should then be a formal way to see this has a functor to $Z(A\dmod)$ (which is the same as $C$-bimodule functors from $C$ to itself), answering your question...
