Repost from math.SE since no answer after two months, but feel free to close if not appropriate:

Everything is finite-dimensional over a field $k$.

Let $B$ be a bialgebra with $B\text{-mod}$ its category of modules. Suppose now that we fix a rigid structure on $B\text{-mod}$ such that the dual of a module is given by the the dual vector space, with some action. In other words, taking duals in $B\text{-mod}$ commutes on the nose with the canonical forgetful functor to $Vect$.

My question: Is $B$ necessarily a Hopf algebra? $^\text{see Edit below}$

The reason for this question is that from the above data I can easily define a quasi-Hopf algebra structure on $B$, whose coassociator is trivial. A quasi-Hopf algebra for me is like a Hopf algebra, except that its not quite coassociative and that the antipode has been replaced by a triple $(S,\alpha, \beta)$, where $S$ is still the antipode, but $\alpha, \beta \in B$ are now some elements implementing the evaluation and coevaluation in $B\text{-mod}$. This triple has to satisfy some axioms, namely the zig-zag equations of the rigid structure. ${}^\star$


My question formulated differently: Is it necessarily the case that $\alpha = \beta = 1$ in the quasi-Hopf algebra I obtained above?

${}^\star$ A source for this construction is eg Section 3.5 in "Quasi-Hopf algebras - a categorical approach" by Bulacu, Caenepeel, Panaite, and van Oystaeyen.

Edit: Sorry, there is a "mistake" in the above. My two questions are not equivalent. As Adrien pointed out, antipode triples are not unique, but instead two distinct ones are related by some invertible element (this is easily seen by noting that a choice of rigid structure is structure, but different choices are related by a unique natural isomorphism). If the coassociator is trivial, one can show (as in Drinfeld's paper) that if $B$ admits some antipode triple $(S,\alpha,\beta)$ then it admits an antipode triple $(S,1,1)$. In particular, with the latter it is a Hopf algebra.

However, my second question was: For a fixed rigid structure on $B\text{-mod}$, are $\alpha$ and $\beta$ from the reconstruction automatically trivial? And I suppose the answer to this is "no", since for any invertible element $u$ in a Hopf algebra $H$ with antipode $S$, we get a quasi-Hopf structure (i.e. antipode triple) $(S(h) = uhu^{-1}, \alpha = u, \beta = u^{-1})$, such that we are in the situation of my question.


2 Answers 2


Yes, it is true that if a quasi-Hopf algebra has a trivial coassociator, then it's equivalent to an actual Hopf algebra (with $\alpha=\beta=1$). In other words, if you know the category is rigid (i.e. if you know duals exists) then there is a particular choice for this duality (canonically isomorphic to any other one) for which the corresponding antipode is Hopf on the nose.

This was proven in Drinfeld's original "Quasi-Hopf algebras" paper (see his second remark after equation 1.20) and is also a well-known feature of Tanakian reconstruction (the proof is pretty much the same). See e.g. http://schauenburg.perso.math.cnrs.fr/papers/tdaha.ps. Note, as it is done in this paper, that the general formalism works well only if you think about finite dimensional comodules over a bialgebra, in which case it is Hopf iff the category is rigid. But if $B$ is finite-dimensional then your category is the same as f.d. $B^*$-comodules so everything's fine.

  • $\begingroup$ I am a bit confused. What do you mean by equivalent? My (second) question was whether $\alpha$ and $\beta$ are equal to 1 in the situation described above. In my situation, I wanted the rigid structure to be fixed and immutable. $\endgroup$
    – Jo Mo
    Mar 22, 2021 at 17:35
  • $\begingroup$ I have added a relevant edit to the question. $\endgroup$
    – Jo Mo
    Mar 22, 2021 at 17:46
  • $\begingroup$ I'm not sure I understand your edit. Being rigid for a monoidal category is a property: either it is or it isn't, and then any two choices for a particular duality are canonically, naturally equivalent. So as you say even if you start with an Hopf algebra you can always cook up some choice of duality that is not equal to the standard one defined by the antipode, but those two choices are canonically isomorphic. (continued) $\endgroup$
    – Adrien
    Mar 22, 2021 at 17:58
  • $\begingroup$ In other words: for the fixed choice you made for a dual $V^*$ of some module $V$, there is a unique actual antipode on $B$ and a unique, canonical and natural isomorphism of $B$-module from your $V^*$ to the one induced by this antipode. It doesn't get better than that. $\endgroup$
    – Adrien
    Mar 22, 2021 at 17:59
  • $\begingroup$ (I mean a unique one compatible with the duality of course). $\endgroup$
    – Adrien
    Mar 22, 2021 at 18:06

If the category of modules has duals the forgetful functor preserves them because a strong monoidal functor preserves duals.


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