8
$\begingroup$

I have a question about the argumentation at the beginning of section 15 in this paper. The goal is to estimate the sum $$V(\beta) = 2 \sum_{(z_1,z_2)=1} \beta_{z_1} \overline{\beta_{z_2}} \sum_{\substack{de > X \\ r_1s_2 \equiv r_2s_1 \text{ } (4d)}} \frac{\varphi(de)}{de} f \Big( \frac{|\Delta|}{de} \Big) \left(\dfrac{s_1s_2}{e}\right) \left(\dfrac{r_1r_2}{d}\right) \log \Big(2 \big| \frac{z_1z_2}{\Delta} \big| \Big),$$ where the summation is restricted to those $z_1,z_2 \in \mathbb{Z}[i]$ fulfilling $$z_1 \equiv z_2 \equiv z_0 \text{ (mod } 8 \text{)},$$ where $z_0$ is assumed to be primary, i.e., $z_0 \equiv 1$ (mod $2(i+1)$), and $$(z_1, \prod_{p \leq P} p) = (z_2, \prod_{p \leq P} p) = 1,$$ where $P$ is a constant to be specified later. Here, we write $z_j = e r_j + i s_j$ for $j=1,2$.

Further, $\Delta = \Delta(z_1,z_2) = \mathrm{Im}(z_1 \overline{z_2}) = \frac{1}{2i}(\overline{z_1}z_2-z_1 \overline{z_2})$, $|\Delta| \leq N$ and $f$ is a function that can be bounded (in absolute value) by $1$. Moreover, we define $$\beta_z = q(\mathrm{arg}(z))p(z \overline{z}) \mu(z \overline{z}) \sum_{c \mid z \overline{z}, c \leq C} \mu(c),$$ where $q$ is a $2\pi$-periodic $C^2$-function supported on $\phi < \mathrm{arg}(z) \leq \phi + 2 \pi \theta$ such that $q^{(j)} \ll \theta^{-j}$ for $j=0,1,2$, and $p$ is a twice differentiable (or even smooth) function supported on $N' \leq n \leq N'(1+\theta)$ such that $p^{(j)} \ll (\theta N)^{-j}$ for $j=0,1,2$, where $N < N' < 2N$, and $\theta^{-1}$ will be chosen to be a large power of $\log(N)$.

In particular, $\beta_z$ is supported on the polar box $$\{ z \mid N' < |z|^2 \leq (1+\theta)N', \phi < \mathrm{arg}(z) \leq \phi + 2 \pi \theta \}$$ of volume $\pi \theta^2 N' \ll \theta^2 N$.

Now they claim that the condition $(z_1,z_2)=1$ can be dropped at a cost of $\mathcal{O}(N^2P^{-1})$. Moreover: "To get it at this cost, apply Lemma 2.2 with respect to $k=4$ to reduce $d$ before estimating trivially."

(The relevant part of) Lemma 2.2 says that, for any fixed $k \geq 2$, any $n \geq 1$ has a divisor $d \leq n^{1/k}$ such that $$\tau(n) \leq (2 \tau(d))^{\frac{k \log(k)}{\log(2)}}.$$

I guess that they want to consider the sum $2 \sum_{(z_1,z_2) \neq 1} ...$, a similar argumentation also appears between the equations (5.9) and (5.10) in their paper. Further, it is clear that the terms involving $\phi$, $f$ or the Legendre symbols can be bounded by $1$, and the $\log$-term can be bounded by $\log(N)$ (which can be compensated later, since $\theta^{-1}$ is a large power of $\log(N)$).

However, this is the point where I get stuck. I do not understand what they mean by "reducing $d$", I am not sure how to estimate the $\beta_z$, and I do not know how the volume of the polar box and the fact that
$$(z_1, \prod_{p \leq P} p) = (z_2, \prod_{p \leq P} p) = 1$$ come into play to finally get their desired bound. Could anyone please help me?

$\endgroup$
  • $\begingroup$ The amount of different notations per volume unit is amazing! I must admit you spent a lot of time explaining most of them, but some are still unclear. What is $X$? Wat is $C$? What is $N$? Did I miss any others? $\endgroup$ – Vincent Jan 8 '19 at 10:42
  • 1
    $\begingroup$ @Vincent: They are all constants. $N$ first appears in equation (4.3) in the definition of a certain bilinear form $B(x;N)$. $P$ and $N$ seem to depend on $x$ (not to be confused with $X$), where $x$ is the major variable (and is explained in equation (4.2)). Further, $C$ is specified in Prop. 4.1, and $X \leq N^{1/9}$ by Prop. 15.1. However, the problem is that sometimes variables change their meaning during the text. Here, I am refering to this version of the paper: arxiv.org/pdf/math/9811185v1.pdf $\endgroup$ – Algebrus Jan 8 '19 at 10:55
  • $\begingroup$ I agree with Vincent, unless one already knows that paper's structure, it's hard to parse this surfeit of notation in the context of a self-contained MO question. $\endgroup$ – literature-searcher Jan 8 '19 at 12:26
4
$\begingroup$

Note first the $\beta_z$ are bounded by $\tau(|z|^2)$.

What you then have as a bound with removing the gcd in 15.2 is a sum over $g=\gcd(z_1,z_2)$ as $$\sum_{|g|\ge P}\tau(g)^2\sum_{|z_i|\le N/|g|\atop \text{$z_i$ in box}}\tau(|z_1|^2)\tau(|z_2|^2)\sum_{e|\gcd({\rm Re}(z_1),{\rm Re}(z_2))}\sum_{d\le N\atop g^2\Delta(z_1,z_2)\equiv 0 (4d)}(\log N)$$

The innermost log-term comes from (7.3), everything else being bounded by 1. Note that we've saved essentially $P^2$ from the sums over both $z_1$ and $z_2$ being smaller by that factor, though we lose back one $P$ from the $g$-sum (see below). See 4.4 and perhaps 5.24 and 4.11 for the size of $P$ relative to everything else.

The remaining $d$-sum can be taken over the divisors of $g^2\Delta$, and similarly with the $e$-sum over the relevant divisors with the real parts of $z_i$.

Now we can argue as with (10.8) and (10.9) and its usage of Lemma 2.2 (though perhaps with a different letter-name for $d$). This introduces some large amount of log factors.

As they say, the box factor (with $\theta$) saves more logarithms than we lose from the arithmetics, and we are reduced to something like $$\theta^2N^2(\log N)^{\text{big}}\sum_{|g|\ge P} \frac{\tau(g)^4}{g^2}$$ which indeed saves us the $P$ we desire.

| cite | improve this answer | |
$\endgroup$

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.