Timeline for Etale cohomology approach on τ(n)
Current License: CC BY-SA 3.0
16 events
when toggle format | what | by | license | comment | |
---|---|---|---|---|---|
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)qn−1. But your reasoning is fine as k(3k−1)/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 ∑k∈Z(−1)kqk(3k−1)/2 is raised to the 12th power, there are a total of O(x12) terms qr for all r with x≤r≤2x. 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 |