L*********s
发帖数: 3063 | 1 Terry Tao: one can use the Type I estimates to raise \sigma up to 1/6, which
closes off the Type III case completely and allows for a slightly more
elementary proof of Zhang’s theorem in that the full strength of Deligne’s
proof of the Weil conjectures is no longer needed
Ben Green：It means one could in principle teach the whole proof in a
graduate course.
Terry Tao：Just recording the results of a discussion I had with Ben on this
. It does look like Vaughan’s identity with U=V=x^{1/3}, together with a
Zhang-Type I estimate that works for \sigma=1/6, is enough to establish
Zhang
’s theorem; the Vaughan-Type II sums can be handled by Zhang’s Type I/II
analysis, and the Vaughan-Type I sums can be handled by either Zhang Type I/
II or Zhang Type 0, depending on the exact scales involved. So one does not
need either the Heath-Brown identity or Deligne’s theory to prove Zhang’s
theorem, just Weil’s bound on completed exponential sums. |
m*****n
发帖数: 1631 | |
v*******e
发帖数: 11604 | 3 楼主翻译错了，a slightly more elementary proof of Zhang’s theorem只是比他的
初等一点点，不说明它是初等证明。 |
L*********s
发帖数: 3063 | 4 这些解析上的高等技术，比起Etale cohomology on topos来只能算是初等了
【在 v*******e 的大作中提到】
: 楼主翻译错了，a slightly more elementary proof of Zhang’s theorem只是比他的
: 初等一点点，不说明它是初等证明。
|
p*********g
发帖数: 5964 | 5 那你这些所谓的初等，高等要看对谁来说。
这些解析上的高等技术，比起Etale cohomology on topos来只能算是初等了
【在 L*********s 的大作中提到】
: 这些解析上的高等技术，比起Etale cohomology on topos来只能算是初等了
|
m****m
发帖数: 2211 | 6 老张说他的证明其实并不是使用的经典的方法
前面的证明大部分是经典的方法
证明的最后一部分用是很新的东西
是代数几何的东西
Tao这里说的是什么意思？难道是说老张后面的代数几何的东西可以不用？
which
’s
this
【在 L*********s 的大作中提到】
: Terry Tao: one can use the Type I estimates to raise \sigma up to 1/6, which
: closes off the Type III case completely and allows for a slightly more
: elementary proof of Zhang’s theorem in that the full strength of Deligne’s
: proof of the Weil conjectures is no longer needed
: Ben Green：It means one could in principle teach the whole proof in a
: graduate course.
: Terry Tao：Just recording the results of a discussion I had with Ben on this
: . It does look like Vaughan’s identity with U=V=x^{1/3}, together with a
: Zhang-Type I estimate that works for \sigma=1/6, is enough to establish
: Zhang
|
