2
$\begingroup$

I have a doubt about the number of solutions for the system $$x_1^j+\cdots+x_s^j=y_1^j+\cdots+y_s^j,\quad(1\leq j\leq k)$$ with $1\leq x_i,y_i\leq X$. It is a big breakthrough of Bourgain-Demeter-Guth, Wooley that the number of solutions has the upper bound $$X^{\epsilon}(X^s+X^{2s-k(k+1)/2}).$$ I think is known that these are optimal i.e. we can prove the same lower bound, but is this more difficult or we can prove this first? So is it trivial? Not for the diagonal solutions but the $$X^{2s-k(k+1)/2}$$ term. Thanks!!!

$\endgroup$
4
  • 1
    $\begingroup$ please edit to make your question self contained and understandable. What is $X$ for example? $\endgroup$ – kodlu Oct 23 '19 at 0:22
  • 2
    $\begingroup$ Yeah it’s trivial once you know why it’s trivial :). Write down the integral $\int_{[0,1]^k} |\sum_{n << X} e(a_1 n + a_2 n^2 + \cdots + a_k n^k)|^{2s} da_1 da_2 \cdots da_k$ and then only integrate over $|a_i| << X^{-i}$. $\endgroup$ – alpoge Oct 23 '19 at 3:05
  • $\begingroup$ What is meant by solutions of a theorem? $\endgroup$ – Gerry Myerson Oct 23 '19 at 5:22
  • $\begingroup$ I messed up my account but want to thank alpoge, PeteM and Thomas Bloom for your answers, and kodlu for fixing the question to adress comments. I would like to accept the answer of Thomas Bloom but I cant now. $\endgroup$ – krish Oct 25 '19 at 13:01
4
$\begingroup$

PeteM and alpoge have already given one explanation, but I thought I'd spell that out explicitly and give a second way to see it as well.

Let $J_s(X)$ count the number of solutions to $x_1^j+\cdots+x_s^j = y_1^j+\cdots y_s^j$ for $1\leq j\leq k$ with $1\leq x_i,y_i \leq X$. The lower bound you mention is that $J_s(X) \gg X^{2s-k(k+1)/2}$.

Let $f(\alpha_1,\ldots,\alpha_k) = \sum_{1\leq x\leq X}e(\alpha_1x+\cdots+\alpha_kx^k)$ where $e(t)=e^{2\pi it}$. Orthogonality implies that $$ J_s(X) = \int_{[0,1]^k} \lvert f(\alpha_1,\ldots,\alpha_k)\rvert^{2s} \mathrm{d}\alpha_1\cdots \mathrm{d}\alpha_k.$$

Method 1: Note that since $\lvert 1-e(t)\rvert \ll t$ if $\lvert \alpha_1\rvert \leq 1/100X$ and so on up to $\lvert \alpha_k\rvert \leq 1/100X^k$ then $\lvert f(\alpha_1,\ldots,\alpha_k)\rvert \gg X$. Since the integrand $\lvert f\rvert^{2s}$ is non-negative, we can give a lower bound for $J_s(X)$ just from integrating over this range of $\alpha_i$. This gives

$$ J_s(X) \gg X^{-1}\cdot X^{-2}\cdots X^{-k} \cdot X^{2s} = X^{2s-k(k+1)/2}.$$

Method 2: Consider the slight generalisation $J_s(X;n_1,\ldots,n_k)$ which counts solutions to $x_1^j+\cdots+x_s^j-y_1^j-\cdots-y_s^j = n_j$ for $1\leq j\leq k$ and $x_i,y_i\leq X$. In particular, note that $J_s(X)=J_s(X;0,\ldots,0)$. By orthogonality we have $$ J_s(X;n_1,\ldots,n_k) = \int_{[0,1]^k} \lvert f(\alpha_1,\ldots,\alpha_k)\rvert^{2s} e(-\alpha_1n_1-\cdots -\alpha_kn_k)\mathrm{d}\alpha_1\cdots \mathrm{d}\alpha_k.$$

By the triangle inequality it follows that $J_s(X;n_1,\ldots,n_k)\leq J_s(X)$ (this fact can also be proved in 'physical space' without using orthogonality). Now note that $$ \sum_{n_1,\ldots,n_k} J_s(X;n_1,\ldots,n_k) = X^{2s},$$ since every $2s$-tuple gives a solution to some $n_1,\ldots,n_k$. Finally, observe that $J_s(X;n_1,\ldots,n_k)$ is only non-zero if $1\leq n_1\leq sX$ and so on up to $1\leq n_k\leq sX^k$. This implies that $$ X^{2s} = \sum_{n_1,\ldots,n_k} J_s(X;n_1,\ldots,n_k) \ll J_s(X) X\cdot X^2\cdots X^k,$$ and the lower bound follows.

$\endgroup$
2
$\begingroup$

I'm sure I saw an argument for this once. It goes something like: only look at the part of your integral really close to 0, where your exponential sum is almost exactly 1, and this alraedy contributes at least the right amount of solutions for the 2s-k(k+1)/2 exponent. If anyone else knows how this goes, please let me know.

(The X^s I think you know comes from diagonal solutions).

$\endgroup$

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.