Timeline for Etale cohomology approach on $\tau(n)$

Current License: CC BY-SA 3.0

16 events

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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 $24$ th power to get $\sum_{n=1}^\infty\tau(n)q^{n-1}$ . 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 $\sum_{k \in \mathbb{Z}} (-1)^{k} q^{k(3k-1)/2}$ is raised to the $12$ th power, there are a total of $O(x^{12})$ terms $q^{r}$ for all $r$ with $x \leq r \leq 2x$ . Heuristically, one would expect about the same number of terms $q^{r}$ for each $r \in [x,2x]$ and so the coefficient of $q^{r}$ would be a sum $O(x^{11})$ signs. The variance is then $O(x^{11/2})$ . 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

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Timeline for Etale cohomology approach on τ(n)\tau(n)

Current License: CC BY-SA 3.0

Timeline for Etale cohomology approach on $\tau(n)$