3
$\begingroup$

In http://131.220.77.52/files/preprints/diophantine/bruedern/wpminicubefinal.pdf, Bruedern and Wooley mention the following fact on the bottom of page 6: Let $$\Phi(\alpha) = \sum_{h\le 6H}\sum_{P<x\le 2P}e(\alpha h(3x^2 + 3xh + h^2))$$ They then assert that $$\int_{0}^1\left|\Phi(\alpha)\right|^4d\alpha\ll H^3P^{2 + \epsilon}$$ for all $\epsilon>0$

However, I don't have access to the paper mentioned in which this fact is proven. Can someone show me a proof of this? It is supposed to be along the lines of the proof of Hua's Lemma, but I can't see how this is the case.

$\endgroup$
  • 4
    $\begingroup$ Google Scholar gives this publicly available pdf for the cited paper. $\endgroup$ – Dougal Aug 27 '14 at 6:27
  • 3
    $\begingroup$ Dougal's link is valuable of course, but it's amazing to me that a reference to the paper gets more upvotes here than a response by the author himself. $\endgroup$ – Lucia Aug 28 '14 at 3:05
7
$\begingroup$

First, the paper is Bruedern and Wooley!

Next ... by Cauchy's inequality, one has $$|\Phi(\alpha)|^2\le 6H\Psi(\alpha),$$ say, where $$\Psi(\alpha)=\sum_{h\le 6H}\left| \sum_{P<x\le 2P}e(\alpha h(3x^2+3xh+h^2))\right|^2.$$ Thus the mean value in question is bounded above by $6H$ times $$\int_0^1\Psi(\alpha)|\Phi(\alpha)|^2\, d\alpha ,$$ and by orthogonality that counts the number of solutions of the Diophantine equation $$3h(x_1-x_2)(x_1+x_2+h)=g_1(3y_1^2+3y_1g_1+g_1^2)-g_2(3y_2^2+3y_2g_2+g_2^2),$$ with $h,g_1,g_2\le 6H$ and $P<x_1,x_2,y_1,y_2\le 2P$. When $x_1=x_2$ and $h$ is free, then given $g_2$ and $y_2$, one finds that $g_1$ and $y_1$ are determined by a divisor estimate. So the number of diagonal solutions is $\ll HP\cdot (HP)^{1+\epsilon}=(HP)^{2+\epsilon}$. Meanwhile, when $x_1\ne x_2$, the right hand side here is also non-zero. Then given a fixed choice for $g_1,g_2,y_1,y_2$, one finds that $x_1-x_2$ and $h$ are determined by a divisor function estimate. There are consequently $\ll (HP)^2\cdot (HP)^\epsilon$ such solutions. Combine these estimates, throw back in the factor $6H$ that arose from applying Cauchy's inequality, and one gets the estimate claimed.

$\endgroup$
3
$\begingroup$

Inventiones papers up to 1996 are publicly available here.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.