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Mar 27, 2015 at 19:51 comment added Turbo posted on chat too.
Mar 27, 2015 at 9:21 comment added Turbo sorry but could you make 'translation' in "Deligne observed that there was a nice translation of this idea into Grothendieck's cohomology theory" more explicit.
Mar 27, 2015 at 5:51 comment added GH from MO @JeremyRouse: You need to raise to the 24th power to get n=1τ(n)qn1. But your reasoning is fine as k(3k1)/2 is a quadratic sequence.
Mar 27, 2015 at 3:43 vote accept Turbo
Mar 27, 2015 at 3:43 vote accept Turbo
Mar 27, 2015 at 3:43
Mar 27, 2015 at 3:43 vote accept Turbo
Mar 27, 2015 at 3:43
Mar 27, 2015 at 3:42 vote accept Turbo
Mar 27, 2015 at 3:42
Mar 27, 2015 at 3:42 vote accept Turbo
Mar 27, 2015 at 3:42
Mar 26, 2015 at 16:48 comment added Jeremy Rouse Let us continue this discussion in chat.
Mar 26, 2015 at 16:16 comment added Jeremy Rouse I think Qiaochu's heuristic can be refined a little bit. In particular, fix a number x. When kZ(1)kqk(3k1)/2 is raised to the 12th power, there are a total of O(x12) terms qr for all r with xr2x. Heuristically, one would expect about the same number of terms qr for each r[x,2x] and so the coefficient of qr would be a sum O(x11) signs. The variance is then O(x11/2). So it is a case of "cancellations" achieving averaging augmentation.
Mar 26, 2015 at 13:00 comment added Turbo "Qiaochu Yuan says Consider also the following heuristic argument. The pentagonal number theorem in fact lets us write τ(n−1) as a sum of O(n^{12}) signs. Assume that these signs are randomly distributed. Then one expects their sum to have absolute value O(n^6) by a straightforward variance calculation. This is the same sort of argument that correctly suggests that Gauss sums should have absolute value around \sqrt{p}. But in fact Ramanujan's conjecture is better than this by a factor of \sqrt{n}. I don't know where this extra savings comes from even heuristically."
Mar 26, 2015 at 12:54 comment added Turbo So this is not case of mysterious cancellations improving on random averaging argumentation?
Mar 26, 2015 at 12:53 comment added Jeremy Rouse Convergence for s > 1 gives that |\alpha_{p}^{n}| < p and this inequality for all n gives |\alpha_{p}| = 1. On the etale cohomology side, Deligne proves a bound on the Frobenius eigenvalues of a variety X of the strength that is roughly the same as |\tau(p)| \ll p^{6}. Applying this bound to the variety X^{n} for all n yields the Riemann hypothesis. (See page 301 of Deligne's paper.)
Mar 26, 2015 at 12:51 comment added Jeremy Rouse Yes. One way of describing this is saying that Langlands's observation is that the Ramanujan conjecture would follow from knowledge of the poles of L_{n}(s) = L(\Delta \otimes \Delta \otimes \cdots \otimes \Delta, s) (the n-fold Rankin-Selberg convolution of \Delta - which factors as a product of symmetric power L-functions). The reason is that if \tau(p) = p^{11/2} (\alpha_{p} + \alpha_{p}^{-1}) with \alpha_{p} \in \mathbb{C}, then (1 - \alpha_{p}^{n} p^{-s})^{-1} is one of the local factors of L_{n}(s).
Mar 26, 2015 at 7:18 comment added Turbo "Langlands observed that knowledge of the poles of symmetric power L-functions attached to Δ would be sufficient to conclude that |τ(p)|≤2p11/2, and Deligne observed that there was a nice translation of this idea into Grothendieck's cohomology theory" Is there a way to make this more explicit?
Mar 26, 2015 at 1:57 history answered Jeremy Rouse CC BY-SA 3.0