Timeline for Hopf dual of the Hopf dual
Current License: CC BY-SA 4.0
13 events
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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 |