Timeline for Action of certain endomorphisms on Pontriyagin dual
Current License: CC BY-SA 3.0
7 events
| when toggle format | what | by | license | comment | |
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| Oct 9, 2016 at 20:03 | comment | added | Uri Bader | Maybe I should mention: a character $\chi$ on a ring $R$ will have the property you seek iff its kernel does not contain any non-trivial ideal. On a field for example, any non-trivial character will do. | |
| Oct 9, 2016 at 19:58 | vote | accept | rohitna | ||
| Oct 11, 2016 at 2:20 | |||||
| Oct 9, 2016 at 19:58 | comment | added | rohitna | Great, it works! The bilinear form I mentioned above is not non-degenerate. I am accepting your answer. | |
| Oct 9, 2016 at 19:46 | comment | added | Uri Bader | @rohitna note that for every $b\in m$, $Rb=kb$. It follows that if $b\in\ker(\chi)$ then $Rb\subset\ker(\chi)$. In particular, $Rb_0\subset\ker(\chi)$, that is $\chi$ does vanish on $Rb_0$. | |
| Oct 9, 2016 at 19:34 | comment | added | Uri Bader | @rohitna for your last comment, with your notation, I believe $\psi_0=\psi_{x-y}$. Thus $a\to \psi_a$ is not an isomorphism. Note that indeed $x-y$ is in the kernel of your bilinear form. | |
| Oct 9, 2016 at 18:23 | comment | added | rohitna | This may not work though as $\chi$ may not vanish on $R b_0$. In fact we have a non-degenerate $\mathbb{F}_p$-valued bilinear form on $R$ given by $(a,b) = f(ab)$ where $f \colon R \to \mathbb{F}_p$ is "sum of coefficient function". Choose a non-trivial character $\chi$ of $\mathbb{F}_p$. Then the map from $R$ to the Pontriyagin dual given by $a \mapsto \psi_a$ is an isomorphism. Here $\psi_a(b) = \chi((a,b))$ . So if all this is right then the result should be true for the ring you mentioned. | |
| Oct 9, 2016 at 11:59 | history | answered | Uri Bader | CC BY-SA 3.0 |