Timeline for Is the preimage of a Hopf subalgebra a Hopf subalgebra?
Current License: CC BY-SA 4.0
10 events
| when toggle format | what | by | license | comment | |
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| Aug 31, 2023 at 12:52 | answer | added | Bugs Bunny | timeline score: 2 | |
| Aug 26, 2023 at 6:46 | comment | added | Sergei Akbarov | @VictorOstrik I think you should post a detailed answer. | |
| Aug 25, 2023 at 22:06 | comment | added | Ali Taghavi | So is it a trivial question if we ask the preimage of every hopf subalgebra is a coideal or cosubalgebra(we forget the antipod)? | |
| Aug 25, 2023 at 21:34 | comment | added | Victor Ostrik | For example assume that first group is of order 4 and the second group is of order 2. Then dimension of $\varphi^{-1}(H)$ is 3. This can't be Hopf subalgebra -- e.g. by Nichols-Zoeller theorem dimension of any Hopf subalgebra divides the dimension of Hopf algebra. | |
| Aug 25, 2023 at 18:20 | comment | added | Sergei Akbarov | @VictorOstrik ah, yes, so this means that $\varphi^{-1}(H)$ is bigger than the group algebra of the kernel... But why isn't $\varphi^{-1}(H)$ necessarily a Hopf subalgebra? | |
| Aug 25, 2023 at 18:02 | comment | added | Victor Ostrik | Assume that two elements $g$ and $h$ of the first group map to the same non-unit element of the second group. Then $g-h$ is in the pre-image. | |
| Aug 25, 2023 at 17:41 | comment | added | Sergei Akbarov | @VictorOstrik I would think that in this case $\varphi^{-1}(H)$ is the group algebra of the kernel of this homomorphism of groups. That is not true? | |
| Aug 25, 2023 at 17:12 | comment | added | Victor Ostrik | I think you can get a counterexample by looking at a special case when $F$ and $G$ are group algebras, $\varphi$ is induced by a surjective homomorphism of groups and $H$ is subalgebra spanned by the unit element. | |
| Aug 25, 2023 at 15:15 | history | edited | YCor |
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| Aug 25, 2023 at 13:25 | history | asked | Sergei Akbarov | CC BY-SA 4.0 |