bio | website | plusepsilon.de |
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location | Uni Hamburg | |
age | 28 | |
visits | member for | 3 years, 5 months |
seen | 4 hours ago | |
stats | profile views | 7,968 |
Mar 16 |
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How can I find the spectrum of this operator?
Pls use one/two dollar signs for math |
Mar 16 |
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How can I find the spectrum of this operator?
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Mar 16 |
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What can the theory of automorphic forms for $SL(n,\mathbb{Z})$ say about $SL(n,\mathbb{Z})$?
Otherwise, I'd suggest to allow general congruence subgroups, and the obvious thing is that they contain the representation theory of $SL_N(Z/M)$, but I think there are easy ways to study this rep.theory. |
Mar 16 |
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What can the theory of automorphic forms for $SL(n,\mathbb{Z})$ say about $SL(n,\mathbb{Z})$?
I think your question has no obvious/easy answer because automorphic forms fix a specific discrete subgroup $\Gamma$ up front, so you can simply give get information for subgroups containing $\Gamma$, and $SL(n,Z)$ is known to be maximal. |
Mar 15 |
answered | Generic topology on a field |
Mar 10 |
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Adelization of modular forms: What measure on $L^2(G_\mathbb{Q}Z_{\mathbb{R}}\backslash G_{\mathbb{A}})$?
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 |
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Adelization of modular forms: What measure on $L^2(G_\mathbb{Q}Z_{\mathbb{R}}\backslash G_{\mathbb{A}})$?
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 |
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Adelization of modular forms: What measure on $L^2(G_\mathbb{Q}Z_{\mathbb{R}}\backslash G_{\mathbb{A}})$?
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 |
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Adelization of modular forms: What measure on $L^2(G_\mathbb{Q}Z_{\mathbb{R}}\backslash G_{\mathbb{A}})$?
You have a finite volume Radon measure. It is automatically regular. That's why I am confused. |
Mar 7 |
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Adelization of modular forms: What measure on $L^2(G_\mathbb{Q}Z_{\mathbb{R}}\backslash G_{\mathbb{A}})$?
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 |
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Adelization of modular forms: What measure on $L^2(G_\mathbb{Q}Z_{\mathbb{R}}\backslash G_{\mathbb{A}})$?
Möbius transformations, where $z$ is equivalent to $-z$. Classically one uses upper plus lower halfplane. What is regularity? You mean normalization? |
Mar 7 |
revised |
Adelization of modular forms: What measure on $L^2(G_\mathbb{Q}Z_{\mathbb{R}}\backslash G_{\mathbb{A}})$?
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Mar 7 |
answered | Adelization of modular forms: What measure on $L^2(G_\mathbb{Q}Z_{\mathbb{R}}\backslash G_{\mathbb{A}})$? |
Mar 6 |
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Can we expect $\left\||f|^{2}f-|g|^{2}g\right\|\leq C ||f-g||$ in the Banach algebra $A(\mathbb T)$ ?
Ah okay, I didn't carefully read, where $f$ and $g$ are living! |
Mar 6 |
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Can we expect $\left\||f|^{2}f-|g|^{2}g\right\|\leq C ||f-g||$ in the Banach algebra $A(\mathbb T)$ ?
Your constant depends on $a$ and $b$. The point is a uniform constant. But maybe your estimates can be made sharp disproving the conjecture of the OP. |
Feb 26 |
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Objections to and arguments for the simplicity of all Riemann zeros
I can't see why two zeros are close, then the maximal distance on the curve is close. That seems not to be true. Can you elaborate what you mean "by the max princip". |
Feb 26 |
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Objections to and arguments for the simplicity of all Riemann zeros
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Feb 26 |
accepted | Supercuspidal with Iwahori fixed vector |
Feb 25 |
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Joint representation of the semi-direct product of the metaplectic group and Heisenberg group
The semidirect product representation are classified by the Mackey machine. |
Feb 24 |
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Objections to and arguments for the simplicity of all Riemann zeros
Okay, I see the problem. One could take a sequence $a_n$ with $a_n = 10^{\log \lceil n \rceil}$ and $b_n$ growing like $n \log(n)$ but being at least one apart from $\{ 10^m: m \in \mathbb{Z} \}$. Thank you. |