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Dec 27, 2018 at 15:40 answer added student timeline score: 4
Apr 28, 2017 at 15:38 history edited Wolfgang
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Jan 2, 2011 at 10:34 answer added GH from MO timeline score: 5
May 28, 2010 at 18:22 history bounty ended David Hansen
May 27, 2010 at 17:04 comment added David Hansen ...was I wrong for thinking that a new, global, proof of the Ramanujan conjecture would be interesting?!
May 21, 2010 at 17:33 history bounty started David Hansen
May 11, 2010 at 13:06 comment added David Hansen Junkie, yes, that is the normalization I am using.
May 11, 2010 at 2:48 comment added Junkie Actually, I am now confused by your normalization of the half-line. Are the coefficients bounded by 1, so that the symmetry line is $Re(s)=1/2$, and the edge is $Re(s)=1$. This is normalization that Ogg uses.
May 11, 2010 at 2:42 comment added Junkie "...another approach to the Ramanujan conjecture, essentially through Langlands functoriality. ... if we knew for all $n$ that the $L$-functions $L(sym^n f)$ were holomorphic and nonvanishing in the halfplane $Re(s)\ge 1$, the Ramanujan conjecture [follows]..." I don't know about functorality - a variant of the weaker claim is in a paper of Ogg springerlink.com/content/k24u6q718u3v770w Ogg, A. P. A remark on the Sato-Tate conjecture. Invent. Math. 9 1969/1970 198--200. Hs shows a holomorphic continuation past 1/2-line (follows from functorality?) implies no zeros on 1-line.
May 10, 2010 at 16:05 history edited David Hansen CC BY-SA 2.5
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May 10, 2010 at 15:48 comment added Emerton I think that Langlands claims the deduction of Ramanujan from functoriality for symmetric powers to be his own. One could also note that the same argument appears in Deligne's proof of the Riemann hypothesis.
May 10, 2010 at 15:32 history asked David Hansen CC BY-SA 2.5