Timeline for Adelization of modular forms: What measure on $L^2(G_\mathbb{Q}Z_{\mathbb{R}}\backslash G_{\mathbb{A}})$?
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12 events
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| Mar 10, 2014 at 23:27 | comment | added | Marc Palm | Okay, trivial is a strong word. Sorry, for that. I meant that it follows immediately from the definitions for weight zero plus a coordinate change. Compact subgroups are never an issue in harmonic analysis. The point of my answer was simply that the volume of the adelic quotient is usually considered as a rational number and $\Gamma \backslash H$ to be something else. You have to give $GL_2(Z_p)$ and $O(2)$ the right volume for both measures to coincide. The usual choice to give a compact subgroup having unit volume fails here. | |
| Mar 10, 2014 at 17:56 | comment | added | Fabian Werner | Well, if you think this process is trivial (see my answer) then you are much more clever than I am :) | |
| Mar 10, 2014 at 11:14 | comment | added | Marc Palm | ask if they are equal instead, so one can give you a definite answer. It depends on the normalization of the measures of the compact subgroups for example. | |
| Mar 10, 2014 at 11:11 | comment | added | Marc Palm | Ok. I misread. I think the invariance is trivial though. Working with GL(2) instead of SL(2) works only well for weight zero Maass wave forms. What is tricky is the normalization? There are differences between SL(2) or GL(2), which you seem to ignore.A discrete series of GL(2,R) splits up into two discrete series, when restricted to SL(2,R), holomorphic and antiholomorphic part. The normalization is the only issue in your question. And there is no natural quotient measure on quotient, but a one-parameter family. Your question is pretty long. I suggest you define the measures on both space, | |
| Mar 8, 2014 at 14:44 | comment | added | Fabian Werner | That is why I said that regularity is NOT the problem here. | |
| Mar 8, 2014 at 13:32 | comment | added | Marc Palm | You have a finite volume Radon measure. It is automatically regular. That's why I am confused. | |
| Mar 7, 2014 at 18:36 | comment | added | Fabian Werner | No, I meant regularity = outer/inner/... regularity of the measure. How to act an $\Gamma\backslash H$ by Moebius transformations? The action of $GL_2^+(R)$ is a left action and we have divided out something on the left that does not commute with $GL_2^+(R)$! | |
| Mar 7, 2014 at 18:18 | comment | added | Marc Palm | The point is that the normal(=Tamagawa) measure does not coincide with the normal measure. The Tamagawa measure is a rational number, the measure on $X$ is a multiple of $1/\pi$. So no if you mean normal=Tamagawa. | |
| Mar 7, 2014 at 18:12 | comment | added | Marc Palm | Möbius transformations, where $z$ is equivalent to $-z$. Classically one uses upper plus lower halfplane. What is regularity? You mean normalization? | |
| Mar 7, 2014 at 16:43 | comment | added | Fabian Werner | But this is precisely what the question is about: why is this pushforward measure [that originates in the long diagram described above] from the one on $H$ right invariant under the whole group $G_A$? Usually, the algebraic invariance is easy and the regularity is tricky, but here it is the other way around! What is the action of GL$_2(R)$ supposed to do on $\Gamma\backslash H$ ? | |
| Mar 7, 2014 at 15:38 | history | edited | Marc Palm | CC BY-SA 3.0 |
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| Mar 7, 2014 at 15:30 | history | answered | Marc Palm | CC BY-SA 3.0 |