Timeline for Non-vanishing of L-series of modular forms (easy case?)
Current License: CC BY-SA 2.5
10 events
when toggle format | what | by | license | comment | |
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Apr 5, 2011 at 14:52 | vote | accept | Aaaa | ||
Apr 2, 2011 at 3:12 | comment | added | GH from MO | Siegel zeros are very close to the edge of the critical strip, i.e. very far from the center. In some sense they are the worst (potential) violations of GRH. | |
Apr 2, 2011 at 2:00 | answer | added | Rob Harron | timeline score: 9 | |
Apr 2, 2011 at 1:36 | comment | added | user14084 | Why are Siegel zeros relevant here? Aren't these zeroes close to the center of critical strip? Even for k=3, s=1 is pretty from the center. | |
Apr 2, 2011 at 1:01 | comment | added | GH from MO | @Peter: Actually Hoffstein and Ramakrishnan proved that for GL(2) cusp forms $L(s,f)$ has no Siegel zeros. See imrn.oxfordjournals.org/content/1995/6/279.extract | |
Apr 2, 2011 at 0:31 | comment | added | Peter Humphries | Well then this resolves your question, because the Grand Riemann Hypothesis for automorphic $L$-functions says that all the zeroes must occur on the critical line. However, there is no proof of this fact: even zero-free regions for automorphic forms do not eliminate the case of a Siegel zero (if $f$ is self-dual). It is much more common to write the normalisation as $L(s,f) = \sum^{\infty}_{n = 1}{a_f(n) n^{-s}}$, where $f(z) = \sum^{\infty}_{n = 1}{a_f(n) n^{(k - 1)/2} e(nz)}$, as this then ensures that $L(s,f)$ has critical line $\Re(s) = 1/2$. | |
Apr 2, 2011 at 0:21 | comment | added | user14084 | My normalization of the $L$-series is $L(f,s) = \sum_n a_n n^{-s}$ where $f=\sum a_n q^n$, and so $1$ is not on the critical line if $k>2$. | |
Apr 1, 2011 at 23:32 | answer | added | GH from MO | timeline score: 1 | |
Apr 1, 2011 at 22:16 | comment | added | Stopple | How are you normalizing the functional equation? Is $s=1$ meant to be on the critical line? | |
Apr 1, 2011 at 22:12 | history | asked | Aaaa | CC BY-SA 2.5 |