Timeline for Characterizing the real analytic Eisenstein series
Current License: CC BY-SA 3.0
21 events
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| Nov 18, 2014 at 20:36 | history | edited | Hugo Chapdelaine | CC BY-SA 3.0 |
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| Oct 26, 2014 at 21:15 | comment | added | GH from MO | I am glad we sorted this out! | |
| Oct 26, 2014 at 15:51 | comment | added | Hugo Chapdelaine | Dear GH, yes you are right. Sorry for not having understood the point of your comment. So as you wrote, one may write down the appropriate second order linear differential equation and then find its general solution. Another way to see it is to notice that at $s=1/2$ there is a cancellation of the poles of $\zeta(2s)$ and $\zeta(2-2s)$ so that the next leading term turns out to be precisely $\log(y)\sqrt{y}$. Thanks for being persistant! | |
| Oct 26, 2014 at 15:39 | history | edited | Hugo Chapdelaine | CC BY-SA 3.0 |
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| Oct 26, 2014 at 8:04 | comment | added | GH from MO | I meant (6.12) in Section 1.6 of Bump's book. | |
| Oct 26, 2014 at 7:54 | comment | added | GH from MO | What you wrote is not correct, and what I wrote was not correct either. The constant term of your $E(z,s)$ is a function $a_0(y,s)$ such that $y^2a_0''(y,s)+s(1-s)a_0(y,s)=0$, where derivation is with respect to $y$. For $s\neq 1/2$ this yields that $a_0(y,s)=c_1y^s+c_2y^{1-s}$, while for $s=1/2$ this yields that $a_0(y,1/2)=c_1\sqrt{y}\log y+c_2\sqrt{y}$. The constants $c_{1,2}$ can be extracted from (6.12) in Bump: Automorphic forms and representations. In particular, for $s=1/2$ we get that the constant term of your $E(z,1/2)$ equals $a_0(y,1/2)=\sqrt{y}\log y+(\gamma-\log(4\pi))\sqrt{y}$. | |
| Oct 26, 2014 at 1:51 | vote | accept | Hugo Chapdelaine | ||
| Oct 23, 2014 at 19:57 | answer | added | Luis Garcia | timeline score: 4 | |
| Oct 23, 2014 at 16:15 | answer | added | Matt Young | timeline score: 4 | |
| Oct 23, 2014 at 14:55 | history | edited | Hugo Chapdelaine | CC BY-SA 3.0 |
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| Oct 23, 2014 at 14:52 | comment | added | Hugo Chapdelaine | Dear Gunter Harder, I know what is wrong. In the formula appearing in the display if it is $\frac{\partial}{\partial s}E(z,s)$ and therefore this is why you pick up a $\log(y)$. So I think that what I wrote is correct. | |
| Oct 23, 2014 at 3:29 | comment | added | GH from MO | $E(z,1/2)$ is rather special, and its constant term is proportional to $\sqrt{y}\log y$. See the last display in Chapter 3 of Iwaniec: Spectral methods of automorphic forms. | |
| Oct 23, 2014 at 1:33 | history | edited | Hugo Chapdelaine | CC BY-SA 3.0 |
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| Oct 23, 2014 at 1:24 | history | edited | Hugo Chapdelaine | CC BY-SA 3.0 |
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| Oct 23, 2014 at 1:22 | comment | added | Hugo Chapdelaine | Dear Kunnysan, thanks for pointing out the inconsistancy with the functional equation and the shift with the constant! | |
| Oct 23, 2014 at 1:18 | comment | added | Hugo Chapdelaine | Dear GH, thanks for the comment you are perfectly. I'll add the Euler factor with the factor $1/2$ so that I at least get the right residues! | |
| Oct 23, 2014 at 0:43 | comment | added | GH from MO | @Kunnysan: You have to be careful. Eisenstein series do not lie in the $L^2$-space, not even those which contribute to the spectral decomposition: $E(z,s)$ with $\Re(s)=1/2$. | |
| Oct 22, 2014 at 23:56 | comment | added | Subhajit Jana | From the spectral decomposition $$L^2(\Gamma\backslash G)=L^2_{cusp}\oplus \mathbb{C}\oplus L_{cont}^2,$$ Any function satisfying (2),(3) and (5) should be in the continuous spectrum. Therefore it can be described by the given Eisenstein series (as it has only one cusp at $\infty$. (2), (3) and (5) imply (1) and (4) with @GHfromMO's correction. | |
| Oct 22, 2014 at 23:41 | comment | added | Subhajit Jana | But, $\Delta(E(z,s)+c)=s(1-s)E(z,s)\neq s(1-s)(E(z,s)+c)$ | |
| Oct 22, 2014 at 22:59 | comment | added | GH from MO | Your (4) and (5) are not correct. To fix them, you should modify your definition of $E(z,s)$ by including the factor $\pi^{-s}\Gamma(s)$ in front of the $(m,n)$-sum. See for example Theorem 1.6.1 in Bump: Automorphic forms and representations, and note that $E(z,s)$ there denotes the Eisenstein series with the extra factors included (Bump also has a factor of $1/2$). | |
| Oct 22, 2014 at 20:41 | history | asked | Hugo Chapdelaine | CC BY-SA 3.0 |