Timeline for Hopf dual of the Hopf dual
Current License: CC BY-SA 4.0
18 events
when toggle format | what | by | license | comment | |
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S Aug 25, 2018 at 17:00 | history | bounty ended | CommunityBot | ||
S Aug 25, 2018 at 17:00 | history | notice removed | CommunityBot | ||
Aug 23, 2018 at 15:51 | vote | accept | Nadia SUSY | ||
Aug 20, 2018 at 18:29 | answer | added | Konstantinos Kanakoglou | timeline score: 2 | |
Aug 18, 2018 at 20:44 | comment | added | darij grinberg | @Adrien: Thank you! This is a beautiful counterexample. I've worked it into my answer now. | |
Aug 18, 2018 at 11:24 | answer | added | Adrien | timeline score: 3 | |
Aug 18, 2018 at 9:00 | comment | added | Adrien | @darijgrinberg for $k=\mathbb{C}$ and letting $G$ be the group $(\mathbb{C},+)$, I think the Hopf dual of $k[x]$ is $k[x^*]\otimes k[G]$ where $x^*$ is defined by $x^*(x^n)=\delta_{1,n}$. | |
Aug 18, 2018 at 4:25 | comment | added | zibadawa timmy | @darijgrinberg For $A=k[x]$ you can use Example 9.1.7 in Susan's book. The Hopf dual of $A$ is the linearly recursive functions on $k[x]$, and she provides another characterization and isomorphisms. | |
Aug 18, 2018 at 4:08 | comment | added | zibadawa timmy | Chapter 9 in Susan Montgomery's book is entirely dedicated to exploring properties of the Hopf dual, including several isomorphisms and density results. I'm not sure right now if it contains any results that are immediately applicable to the question at hand, but if you haven't already done so it may be worth looking through. | |
Aug 17, 2018 at 20:38 | comment | added | darij grinberg | I'm wondering if the result is true for $A$ being graded of finite type (i.e., each degree is finite-dimensional). Even for $A = k\left[x\right]$ I don't quite understand the Hopf dual $A^o$ -- it might be something like the ring of rational power series. | |
Aug 17, 2018 at 16:55 | comment | added | Nadia SUSY | No, I do not assume finite dimensionality. | |
Aug 17, 2018 at 16:11 | answer | added | darij grinberg | timeline score: 13 | |
Aug 17, 2018 at 15:53 | comment | added | Sam Hopkins | Ok, I see. It seems likely the answer is then "no" | |
Aug 17, 2018 at 15:51 | comment | added | darij grinberg | @SamHopkins: That would make the question trivial. | |
Aug 17, 2018 at 15:50 | comment | added | Sam Hopkins | Do you assume that $A$ is finite-dimensional? | |
S Aug 17, 2018 at 15:38 | history | bounty started | Nadia SUSY | ||
S Aug 17, 2018 at 15:38 | history | notice added | Nadia SUSY | Draw attention | |
Aug 15, 2018 at 15:12 | history | asked | Nadia SUSY | CC BY-SA 4.0 |