Timeline for A technical problem on the contragredient representation in the context of locally compact totally disconnected groups
Current License: CC BY-SA 3.0
7 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Nov 10, 2011 at 14:53 | vote | accept | Murat Güngör | ||
| Nov 9, 2011 at 20:33 | comment | added | Faisal | As far as this question is concerned, admissibility is important because it gives us the equivalence $\tilde{\tilde{\pi}}=\pi$ (which I used in (2)). This equivalence doesn't hold for general inadmissible $\pi$. | |
| Nov 9, 2011 at 20:21 | comment | added | Matthew Daws | @Faisel: Sure! But I think the original question wanted to know why there is an algebraic vector in the perp of $E_1$. That is, your answer seemed a bit brief, given what the original question asked (so I think maybe it won't be easy to understand). However, on a close reading of the original question, I see that $\pi$ is assumed "admissible". I personally don't understand what implications this has, but it seems to imply lots of powerful things; so maybe one needs to fully understand this definition...?? | |
| Nov 9, 2011 at 20:14 | comment | added | Faisal | In symbols, $\tilde{V} = (V^\ast)^\infty$. | |
| Nov 9, 2011 at 20:04 | comment | added | Faisal | The contragredient is by definition the "smooth dual" representation (i.e. the subrep of the dual rep on the subspace of smooth vectors). | |
| Nov 9, 2011 at 19:57 | comment | added | Matthew Daws | If I understand the question, the worry is about algebraic (=smooth) vectors-- here $x\in E$ is smooth if the stabilizer of $x$ for the $\pi$ action is an open subgroup of $G$. Could you say some words about this?? | |
| Nov 9, 2011 at 18:51 | history | answered | Faisal | CC BY-SA 3.0 |