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Timeline for Hopf dual of the Hopf dual

Current License: CC BY-SA 4.0

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Sep 10, 2021 at 21:03 comment added Konstantinos Kanakoglou oops! of course. So i guess my question was a little naive .... thank you anyway!
Sep 9, 2021 at 12:11 comment added Marco Farinati I'm saying that a counit is a finite dimensional representation. If an algebra has no finite dim reps at all, it can't be Hopf.
Sep 8, 2021 at 22:29 comment added Konstantinos Kanakoglou @Marco Farinati if i understand correctly, your comment essentially says that if it has a counit then it has (non-trivial) fin dim representations. So, algebras not admitting fin dim reps cannot be hopf. Have i understood correctly or am i missing something?
May 9, 2019 at 23:27 comment added Marco Farinati the Weyl algebra (in characteristic zero) can not be a Hopf algebra because it can not have a counit, because it has no finite dimensional representations.
Aug 20, 2018 at 22:16 history edited Konstantinos Kanakoglou CC BY-SA 4.0
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Aug 20, 2018 at 19:37 history edited Konstantinos Kanakoglou CC BY-SA 4.0
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Aug 20, 2018 at 19:37 comment added Konstantinos Kanakoglou ok, this was also in the same spirit as before. i will remove it to avoid confusion.
Aug 20, 2018 at 19:34 comment added darij grinberg Infinite field extensions aren't Hopf algebras (the counit must be an algebra homomorphism!).
Aug 20, 2018 at 19:32 history edited Konstantinos Kanakoglou CC BY-SA 4.0
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Aug 20, 2018 at 18:50 history edited Konstantinos Kanakoglou CC BY-SA 4.0
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Aug 20, 2018 at 18:47 comment added Konstantinos Kanakoglou not as far as i know! you are right. i just mentioned this example as an easy case for which the restricted dual is trivial.
Aug 20, 2018 at 18:36 comment added darij grinberg But is the Weyl algebra Hopf?
Aug 20, 2018 at 18:29 history answered Konstantinos Kanakoglou CC BY-SA 4.0