Lower bound on exponential sums

Question

Let $k\geq 2$. Consider the following norm of exponenetial sum: $$ I(N,p,k)=\int_0^1\int_0^1 \left|\sum_{n=0}^N e^{2\pi i (n x+n^k y)}\right|^p dxdy. $$ Bourgain mentioned on Page 118 of

https://math.mit.edu/classes/18.158/bourgain-restriction.pdf

that $I(N,6,2)\gtrsim N^3\log N$, where he referenced the following article: https://www.researchgate.net/publication/259308546_The_method_of_trigonometric_sums_in_number_theory.

But I did not find an explicit result in the above article that leads directly to the lower bound above.

So my questions are:

1. What is the idea to prove the above lower bound? The famous Vinogradov's mean value theorem deals with upper bounds of $I(N,p,2)$, but not lower bounds.

2. What is a reasonably sharp lower bound for $I(N,p,3)$, or particularly, $I(N,6,3)$? Note that this may not be in the direct form of Vinogradov's mean value theorem, as the $n^2$ term is missing here.

Answer 1

There are a few things to clear up.

The first is that, on the page in the Bourgain paper you mention, he actually proves the lower bound $I(N,6,2)\gg N^3\log N$ from the fact that

$$ \left\lvert\sum_{n=0}^N e(nx+n^2y)\right\rvert \gg N/q^{1/2}$$

whenever $\lvert x-b/q\rvert \ll 1/N$ and $\lvert y-a/q\rvert \ll 1/N^2$ for some fixed $1\leq a< q\leq N^{1/2}$ with $(a,q)=1$ and $1\leq b\leq q$. (The proof is simply summing the contribution from all such $a$ and $b$). It is this estimate which he invokes a reference for, rather than the lower bound for $I(N,6,2)$.

Secondly, the reference he gives is not to the paper you link to (which is a 1986 paper by Karatsuba-Vinogradov) but instead to Vinogradov's 1954 book with a similar title, usually translated to 'The Method of Trigonometric Sums in the Theory of Numbers'. I don't have a copy of this to hand to check the reference, but a quick search turned up a short note by Tamahiro Oh (https://www.maths.ed.ac.uk/~toh/Files/WeylSum.pdf) proving exactly this Weyl sum lower bound.

Finally, for $I(N,6,3)$, the situation is quite different, and here in fact an asymptotic formula is known: $$ I(N,6,3) = 6N^3 + O(N^2(\log N)^5).$$ This is a result of Vaughan and Wooley (On a certain nonary cubic form and related equations. Duke Mathematical Journal, 80(3), 669–735, 1995).

Answer 2

The result for $I(N,6,2)$ was proved by Rogovskaya N. N. in the article An asymptotic formula for the number of solutions of a certain system of equations. The proof is elementary. Main idea is to replace the system $$x_ 1+x_ 2+x_ 3=y_ 1+y_ 2+y_ 3,\quad x^ 2_ 1+x^ 2_ 2+x^ 2_ 3=y^ 2_ 1+y^ 2_ 2+y^ 2_ 3 $$ by $$a_1+a_2+a_3=0,\quad a_1b_1+a_2b_2+a_3b_3=0,$$ where $a_i=x_i-y_i$ and $b_i=x_i+y_i$. Then you can count solutions of the last equation and sum the result over $a_i.$ The answer is nice $${\mathcal N}(P)=18\pi^{-2}P^ 3\log P+{\mathcal O}(P^ 3).$$ Probably this is the only case when trigonometric integral was calculated explicitely.