Timeline for Etale cohomology approach on $\tau(n)$
Current License: CC BY-SA 3.0
16 events
| when toggle format | what | by | license | comment | |
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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 | ||
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| Mar 27, 2015 at 3:43 | vote | accept | Turbo | ||
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| Mar 27, 2015 at 3:42 | vote | accept | Turbo | ||
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| Mar 27, 2015 at 3:42 | vote | accept | Turbo | ||
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| 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 |