Timeline for Functoriality of the Hopf dual
Current License: CC BY-SA 4.0
19 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| S Nov 21, 2018 at 20:01 | history | bounty ended | CommunityBot | ||
| S Nov 21, 2018 at 20:01 | history | notice removed | CommunityBot | ||
| Nov 21, 2018 at 14:52 | vote | accept | Nadia SUSY | ||
| Nov 16, 2018 at 10:52 | vote | accept | Nadia SUSY | ||
| Nov 21, 2018 at 14:52 | |||||
| Nov 15, 2018 at 11:32 | comment | added | Christoph Mark | Well, you already proved in your post that $j^*$ is functorial if $j$ is an algebra map (regardless wether $j$ vanishes on an ideal of finite codimension in $G$). But you can't make this an "iff"-statement because $j^*$ will always be functorial in the finite-dimensional case. If $j$ is not an algbera map, it may or may not happen that $j^*$ is functorial (see my example above in the comments). | |
| Nov 14, 2018 at 23:53 | answer | added | Konstantinos Kanakoglou | timeline score: 3 | |
| Nov 14, 2018 at 22:27 | comment | added | Nadia SUSY | @KonstantinosKanakoglou If I am correct in seeing that this is an IFF statement, then yes this is enough! Please put it as an answer. | |
| Nov 14, 2018 at 22:04 | comment | added | Konstantinos Kanakoglou | Well, it is functorial -in the sense that $j^*(H^o)\subseteq G^o$- if $j$ vanishes on an ideal of finite codimension in $G$. But does this satisfy you as an answer? Or are you looking for something different? | |
| Nov 14, 2018 at 19:35 | comment | added | Nadia SUSY | I was hoping for a description of when the map is functorial. | |
| Nov 14, 2018 at 16:12 | comment | added | Christoph Mark | Nadia, tell us what you exactly expect from your bounty which you started later. | |
| Nov 14, 2018 at 14:02 | comment | added | Konstantinos Kanakoglou | Generaly, the situation described in the OP, happens for those elements $f\in H^\circ$ for which there exists an ideal of finite codimension in $G$, whose image under $j$ is contained in the kernel of $f$. For example for those linear maps $j$ which vanish on an ideal of finite codimension in $G$ this will always be the case thus $j^*(H^\circ)$ will be inside $G^\circ$. I am not sure however if this is too restrictive for an answer. | |
| Nov 14, 2018 at 13:49 | comment | added | Konstantinos Kanakoglou | Ok it is clear now. Thanks. I am not sure if the OP is looking for a counterexample but In my understanding, your comment would probably deserve to be an answer. | |
| Nov 14, 2018 at 11:31 | comment | added | Christoph Mark | $k[\mathbb{Z}]$ has a basis $e_n$ where $n\in\mathbb{Z}$ which multiplies like $e_ne_m=e_{n+m}$. $e_0$ is the unit with respect to this multiplication. @KonstantinosKanakoglou | |
| Nov 14, 2018 at 1:16 | comment | added | Konstantinos Kanakoglou | @user66288, what do you mean with $e_0$ ? | |
| S Nov 13, 2018 at 18:40 | history | bounty started | Nadia SUSY | ||
| S Nov 13, 2018 at 18:40 | history | notice added | Nadia SUSY | Draw attention | |
| Nov 11, 2018 at 17:06 | comment | added | Christoph Mark | If $j$ is not an algebra map, I do not see a reason why $j^*(H^o)\subseteq G^o$. Indeed, let, for example, be $j\colon k[\mathbb{Z}]\to k[\mathbb{Z}]$ given by the projection map onto $ke_0$. Let $\epsilon\in k[\mathbb{Z}]^o$ - the counit. Then, $\epsilon\circ j\notin k[\mathbb{Z}]^o$ since the only ideal contained in $\operatorname{ker}\epsilon\circ j$ is $0$ which has not finite codimension in $k[\mathbb{Z}]$. | |
| Nov 11, 2018 at 14:57 | comment | added | lambda | If $H = \mathbb C$ then $j$ is an element of $G^*$, and $j^*$ sends the identity to $j$ itself. So in this case the maps with that property are exactly those in $G^o$. | |
| Nov 11, 2018 at 13:29 | history | asked | Nadia SUSY | CC BY-SA 4.0 |