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Timeline for Hopf dual of the Hopf dual

Current License: CC BY-SA 4.0

18 events
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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