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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