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。
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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
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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
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