Timeline for Question on the proper sub-representation of induced representation
Current License: CC BY-SA 4.0
15 events
| when toggle format | what | by | license | comment | |
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| May 24, 2020 at 11:58 | comment | added | Monty | @Subhajit Jana, Oh I am sorry for having not given enough explanation. I mean small supports of $e$ in $U^-$ modulo $P$. | |
| May 24, 2020 at 11:45 | comment | added | Subhajit Jana | What do you mean by ''$f$ has small support near $e$''? By definition, $f(n)=f(e)$, for $n\in N$, which means that $f$ can not have compact support in $G$. If you meant to say ''small support'' in $N\backslash G$ that also seems unlikely. Because $f(nm)=\sigma(m)v$ for $m\in M$. This is nonzero (for $v\neq 0$), for instance, if $M$ is a torus. | |
| May 23, 2020 at 4:16 | comment | added | Monty | @LSpice, how about to restrict $\pi$ to the image of intertwining operator appearing in the proof of analytic continuation and functional equation of Eisenstein series? It looks so difficult to analyze the image of intertwining operator. | |
| May 23, 2020 at 4:11 | history | edited | Monty | CC BY-SA 4.0 |
I added the PS.
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| May 22, 2020 at 16:57 | comment | added | Monty | @LSpice, Thank you very much for your comments. I revised my question! | |
| May 22, 2020 at 16:55 | history | edited | Monty | CC BY-SA 4.0 |
added 432 characters in body
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| May 22, 2020 at 16:44 | comment | added | LSpice | Oh, good point. I was thinking that $f$ is identically $f(e)$ on $U^-$, but actually it's identically $f(e)$ on $U$. Now that I don't mix up the parabolic and its opposite, I also doubt that this one holds for all subrepresentations, though again without an explicit counterexample. (I think that these revised questions are more interesting, and, if you move them into the question, I would remove my close vote.) | |
| May 22, 2020 at 16:42 | comment | added | Monty | @LSpice, Thank you very much for sharing your wisdom. You think that it might be impossible to choose $f$ which has small support. On your answer to my second question, I am wondering why $(\int_{U^- \cap K} \pi(u^-) f du^-)(e)=f(e)$. Why this does holds? | |
| May 22, 2020 at 16:26 | comment | added | LSpice | I don't have a specific counterexample, but I suspect that there's too much flexibility with subrepresentations to insist that they all must contain functions of small support. As for $U^-$-invariance, choose a compact open subgroup $K$ such that $G = P K$, and replace $f$ by $\int_{U^- \cap K} \pi(u^-)f\,\mathrm du^-$. | |
| May 22, 2020 at 16:07 | review | Close votes | |||
| May 22, 2020 at 17:16 | |||||
| May 22, 2020 at 16:04 | comment | added | Monty | @LSpice, can we choose such $f$ which is either $U^{-}$-invariant or sufficiently small support near $e$? (here, $U^{-}$ is the unipotent radical of the opposite parabolic subgroup $P^{-}$ of $P$) These are possible if $\pi$ is the full induced representation. But I am wondering if it holds for a proper subrepresentation $\pi$. How do u think of it? | |
| May 22, 2020 at 15:52 | comment | added | Monty | @LSpice, Thank you very much! It helps me a lot! | |
| May 22, 2020 at 15:40 | history | edited | LSpice | CC BY-SA 4.0 |
TeX and minor proofreading
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| May 22, 2020 at 15:35 | comment | added | LSpice | The image of $\operatorname{Res}^G_M \pi \to \sigma$, $f \mapsto f(e)$, is an $M$-submodule, since $\pi(m)f \mapsto f(m) = \sigma(m)f(e)$. So it is all of $V$ unless every vector in $\pi$ vanishes at $e$, which implies that $\pi$ is the $0$ representation, since $\pi(g)f \mapsto f(g)$ for all $f$ in the space of $\pi$ and all $g \in G$. (Also, I encourage you never to use $\nu$ for a vector; I guarantee it will lead to confusion at some point.) | |
| May 22, 2020 at 13:53 | history | asked | Monty | CC BY-SA 4.0 |