Timeline for Modularity of $E_2$ on congruence subgroups
Current License: CC BY-SA 3.0
11 events
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| Mar 27, 2012 at 22:12 | vote | accept | Eugene | ||
| Mar 27, 2012 at 21:03 | comment | added | David Loeffler | Sorry for the confusion, guys -- my original answer is obvious nonsense (I was somehow muddling the holomorphic and non-holomorphic $E_2$'s). I've replaced the bogus remark with an expansion of wood's last comment. | |
| Mar 27, 2012 at 21:02 | history | edited | David Loeffler | CC BY-SA 3.0 |
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| Mar 27, 2012 at 20:56 | history | edited | David Loeffler | CC BY-SA 3.0 |
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| Mar 27, 2012 at 18:14 | comment | added | wood | Right, if you do the Hecke trick, the non-holomorphic part will occur naturally. I am only saying that the difference of two holomorphic functions is holomorphic. In fact the correct transformation behavior $E_2(-1/z) - z^2 E_2(z)=-2 \pi i z$ also shows that $E_2$ cannot be modular for any level. | |
| Mar 27, 2012 at 17:48 | history | edited | Kevin Ventullo | CC BY-SA 3.0 |
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| Mar 27, 2012 at 17:07 | comment | added | paul garrett | Weight-two holomorphic Eisenstein series aren't defined directly by the usual series, which doesn't converge. Rather, consider a larger family of Eisenstein series with additional complex parameter $s$, show meromorphic continuation in $s$, and evaluate at $s_o$ most likely to produce something holomorphic in $z$. However, the obstruction is non-trivial in general. For Hilbert modular forms over fields other than $\mathbb Q$ the obstruction vanishes. This game is sometimes called "Hecke summation". The obstruction tends to be in the constant term, so the trick mentioned above succeeds. | |
| Mar 27, 2012 at 10:58 | comment | added | wood | In fact the formulas is $E_2(-1/z) - z^2 E_2(z)=-2 \pi i z$. But correcting with a non-holomorphic summand gives really modular transformation behavior. | |
| Mar 27, 2012 at 10:48 | comment | added | wood | How is this possible? $E_2(z)$ is holomorphic in $z$, isn't it. So also $z^2E_2(z)$ is holomorphic and also $E_2(-1/z)$ is also holomorphic (except at $z=0$). So then the difference has to be holomorphic as well. | |
| Mar 27, 2012 at 7:26 | history | edited | David Loeffler | CC BY-SA 3.0 |
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| Mar 27, 2012 at 7:04 | history | answered | David Loeffler | CC BY-SA 3.0 |