Timeline for Unitary Representations of $GL_2({\mathbb Q}_p)$
Current License: CC BY-SA 3.0
16 events
| when toggle format | what | by | license | comment | |
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| Jul 9, 2012 at 10:52 | history | undeleted | Marc Palm | ||
| Jul 9, 2012 at 8:48 | history | deleted | Marc Palm | ||
| Jul 8, 2012 at 22:31 | comment | added | Marc Palm | Smooth vectors of representation on locally convex spaces are always dense. I am not taking inverse of $\phi$, since $\phi$ is not surjective. We take a closure. Smooth vectors are a dense proper subset. I am only claiming that for two unitarization, you get a densely defined (on smooth vectors), closed operator between them, and that this is all you need. But I don't know the proof, but that is how it appears plausible. The facts remain true and are not my observations. Sorry that I do not remember a reference for this. I come back when I find the time. Best, Marc | |
| Jul 8, 2012 at 19:43 | comment | added | Bugs Bunny | I meant $\phi^{-1}(V)\cap V$... as we have $\phi^{-1}(V)\cap V\rightarrow V$... but from what I can see we can just have $\phi^{-1}(V)\cap V=0$, why not?? | |
| Jul 8, 2012 at 19:36 | comment | added | Bugs Bunny | And you cannot really work with $\overline{V}_{<,>}$ because to extend $(,)$ to this completion you really need to show that the norms are equivalent... | |
| Jul 8, 2012 at 19:33 | comment | added | Bugs Bunny | Still not sure. Let me get my head around $S_1$ and $S_2$ first. The crucial intertwiner $\phi$ is defined $V\rightarrow \overline{V}_{<,>}$, the completion with respect to $<,>$. Are you saying that $\phi^{-1} (V)$ is dense in $V$? O'why could that be true? | |
| Jul 8, 2012 at 17:18 | comment | added | Marc Palm | equal should mean bijective on objects and morphisms | |
| Jul 8, 2012 at 17:17 | comment | added | Marc Palm | Sorry $S_1$ is equal to $S_2$. | |
| Jul 8, 2012 at 17:16 | comment | added | Marc Palm | For $1$: Schur's lemma holds for densely-defined, closed intertwiner. Use the graph norm for extending Schur's lemma. For $2$: There is no nice correspondance between the categories $S_1$ and $S_3$. If you change the equivalence on the Hilbert-reps. to Naimark equivalence, then $S_1$ is contained in $S_3$ and $S_1$ is equal to $S_3$. I am not sure what the correct category terms are here, but I guess you understand what I want to say. Note that there exists unbounded intertwiner between non-unitarizable principal series representations. | |
| Jul 8, 2012 at 15:50 | comment | added | Bugs Bunny | I am less certain what you are saying about $C$. All admissible irreducible reps are smooth and some are not unitarizable, as sure as devine carrots. How does it give a rep in $S_3$, not in $C(S_2)$? I am kind of thinking that $C$ is not surjective, but this should come from a representation on a Hilbert space without any nonzero smooth vectors!! I think $C$ is injective but again one needs to argue why two different reps cannot have isomorphic completions. This should correspond to "too many" smooth vectors forming a reducible representation... | |
| Jul 8, 2012 at 15:36 | comment | added | Bugs Bunny | Thanks! Unfortunately, you have neither explained, nor given any references. Say, the standard proof of uniqueness of the unitary structure in the finite dimensional case goes via a morphism $\phi$ defined by $<\phi (x),y> = (x,y)$ and the endomorphism property $End (V)={\mathbb C}$. Here one has the endomorphism property, indeed, but I do not understand why $\phi$ is well-defined!! Essentially, so defined $\phi (x)$ is an element of the complection, and not necessarily of the original representation. How do you go around it? | |
| Jul 8, 2012 at 9:31 | history | edited | Marc Palm | CC BY-SA 3.0 |
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| Jul 8, 2012 at 8:51 | comment | added | Marc Palm | Btw, either unitarizable smooth or unitary. Unitary is the completed stuff, which does not preserve smoothness, and the smooth vectors are almost never a Hilbert space, at most a pre-Hilbert space. | |
| Jul 8, 2012 at 8:46 | comment | added | Marc Palm | Steinberg=special. | |
| Jul 8, 2012 at 8:43 | history | edited | Marc Palm | CC BY-SA 3.0 |
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| Jul 8, 2012 at 8:35 | history | answered | Marc Palm | CC BY-SA 3.0 |