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Oct 13, 2021 at 4:41 vote accept yohei ohta
Oct 13, 2021 at 4:43
Oct 12, 2021 at 18:02 vote accept yohei ohta
Oct 13, 2021 at 4:40
Oct 10, 2021 at 6:52 comment added yohei ohta Thanks for your reply. I'm still lacking in basic knowledge, so I'll study a little more. May I ask you another question? Could you explain how to construct $H(A)$ when $A$ is quasi-triangular/braided?
Oct 8, 2021 at 13:01 comment added Adrien As I said, $A$ is however always a coalgebra in $A$-bimod, but the functor that send a bimodule to its underlying left module is monoidal iff $A$ is Hopf (since this functor is basically obtained by applying the forgeftful functor to the right module structure) which is another explanation for the point I made above. You can reformulate this in various other ways, but the bottom line is that any definition of Hopf module use explicitly or implicitly the fact that the forgetful functor is monoidal, unlike the definition of Yetter-Drinfeld modules.
Oct 8, 2021 at 12:58 comment added Adrien I claim that it doesn't exist (although of course saying a definition doesn't exists is not really something you can prove). The issue is that $A$ equipped with the $A$-action given by left multiplication, is a coalgebra in $A$-mod if and only if $A$ is actually Hopf (the coporoduct is always a map in this category, but is not coassociative unless the associator is trivial). (continued)
Oct 8, 2021 at 10:23 comment added yohei ohta Thanks for your answer! You write that "there is no good definition of a Hopf module over a quasi-Hopf algebra", does this mean that no such definition has been found? Or does it mean that it doesn't exist? Quasi-Hopf Algebras A Categorical Approach has a description about Quasi-Hopf bimodules, but I wondered that there is no description about “Hopf module for Quasi-Hopf algebra” corresponding to the Yetter Drinfeld module.
Oct 7, 2021 at 14:57 history edited Adrien CC BY-SA 4.0
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Oct 7, 2021 at 14:04 history edited Adrien CC BY-SA 4.0
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Oct 7, 2021 at 13:50 history answered Adrien CC BY-SA 4.0