## A ternary divisor variation

Here is a sketch of the argument mentioned in the previous post (which arose from the discussions with Étienne Fouvry, Philippe Michel, Paul Nelson, etc, but presentation mistakes are fully mine…).

Theorem. We have
$\frac{1}{Q}\sum_{r\sim R}\sum_{s\sim S}\sum_{a\bmod q,\ P(a)= 0\bmod q}\Bigl|S(a,q)-MT_q\Bigr|=O\Bigl(\frac{X}{Q\log^A X}\Bigr)$
provided
$Q^{3/2}S^{1/2}
where
(1) we put
$S(a,q)=\sum_{m\sim M}\alpha_m\sum_{mn_1n_2n_3\equiv a\bmod{q},\ n_i\sim N_i}1,$
and denote by $MT_q$ the expected main term;
(2) the parameters are $X=MN=MN_1N_2N_3$, $Q=RS$, the modulus is $q=rs$, the moduli $r$ and $s$ are coprime and squarefree, and $P\in \mathbf{Z}[X]$ is the usual polynomial associated to an admissible tuple.

If we take $R=Q^{1/2-\epsilon}$ and $S=Q^{1/2+\epsilon}$, we get a non-trivial result for $Q$ as large as $N^{4/7}X^{-\epsilon}$.

(In fact, the special shape of $P$ will play no role in this argument, and any non-constant polynomial will work just as well.)

More precisely, I will give a proof which is — except for its terseness — essentially complete for $r$ and $s$ of special type, and we anticipate only technical adjustments to cover the general case — we will write this down carefully of course.

Before starting, a natural question may come to mind: given that qui peut le plus, peut le moins, can one give an analogue result for the usual divisor function? Recall that, for the latter, the (individual) exponent of distribution has been known to be at least $2/3$ for a long time (by work of Linnik and Selberg independently, both using the Weil bound for Kloosterman sums.) This exponent has not been improved, even on average over $q$ (although Fouvry succeeded, on average over $q$, in covering the range $X^{2/3-\epsilon}\leq q\leq X^{1-\epsilon}$) despite much effort. However, Fouvry and Iwaniec (with an Appendix by Katz to treat yet another complete exponential sum over finite fields) proved already twenty years ago that one could improve it to $2/3+1/48$ if one averages for a fixed $r \leq X^{3/8}$ over special moduli $q=rs$ with $rs^2\leq X^{1-\epsilon}$ — this gives in particular a nice earlier illustration of the usefulness of factorable moduli for this type of questions.

So, to work. For fixed $r$ and $s$, we begin by applying the Poisson summation formula to the three “smooth” variables $n_i$ (the smoothing is hidden in the notation $n_i\sim N_i$); the simultaneous zero frequencies $(h_1,h_2,h_3)=(0,0,0)$ give the main term, as they should, and the other degenerate cases are easier to handle than the contribution of the non-zero $h_i,$ so the main secondary term for a given $q$ and $a$ is given by
$S_2=\frac{N}{q^2}\sum_m \alpha_m\sum_{1\leq |h_i|\\ H_i} K_3(a\bar{m}h_1h_2h_3;q),$
where the dual lengths are $H_i=Q/N_i$, so that the total number of frequencies is $H_1H_2H_3=Q^3/N$, and where $K_3$ is a normalized hyper-Kloosterman sum modulo $rs$:
$K_3(u;q)=\frac{1}{q}\sum_{xyz=u}\psi_q(x+y+z),\quad\quad \psi_q(x)=e(x/q).$

(Below I will usually not repeat the range of summations when they are unchanged from one line to the next.)

Now we sum over $r$ and $s$, move the sum over $m$ outside, and apply the Cauchy-Schwarz inequality to the $r$ sum, for a fixed $(s,a_s,m)$, where $a_s$ is $a$ modulo $s$. To prepare for this step, we use the Chinese Remainder Theorem to split the condition $P(a)=0\bmod rs$, and to factor the hyper-Kloosterman sum as a sum modulo $r$ times one modulo $s$.

The contribution of a fixed $(s,a_s,m)$ is
$T=\frac{1}{R}\sum_{r}\sum_{P(a_r)=0\bmod{r}}\Bigl|\sum_{h_i}K_3(a\bar{m}h_1h_2h_3;rs)\Bigr|$
and we can bound
$T\ll H^{1/2+\epsilon} U^{1/2},$
where
$U=\frac{1}{R^2}\sum_{r_1,r_2} \lambda(r_1,a_1)\lambda(r_2,a_2) \sum_{1\leq |h|\ll H} \overline{K_3(a_1\bar{s}^3\bar{m}h;r_1)}K_3(a_2\bar{s}^3\bar{m}h;r_2)\overline{K_3(a_s\bar{r}_1^3\bar{m}h;s)}K_3(a_s\bar{r}_2^3\bar{m}h;s)$
for some coefficients $\lambda(\cdot,\cdot)$ which are bounded.

The point of this is that we have smoothed the variable $h=h_1h_2h_3$ by eliminating its multiplicity, and that the range of this variable can be quite long; as long as $H>(R^2S)^{1/2}$, completing the sum in the Polya-Vinogradov (or Poisson) style will be useful.

Now we continue with $U$. It is here that it simplifies matters to have $R and to do as if $r$ and $s$ were primes (this is a technicality which experience shows should give no loss in the final, complete, analysis.)

So we consider the sum over $r_1, r_2$. The diagonal contribution where $r_1=r_2$ is $\ll H^{1+\epsilon}R^{-1+\epsilon}$.

In the non-diagonal terms, we distinguish whether $r_1\equiv r_2\bmod s$ or not. If not, we complete the two hyper-Kloosterman sums. This gives two complete exponential sums modulo $r_1, r_2$ and modulo $s$. The latter is the Friedlander-Iwaniec sum, in its incarnation as "Borel" correlations of hyper-Kloosterman sums (see the remark at the end of our note on these sums; this identification is already in Heath-Brown's paper, and Philippe realized recently that he had also encountered them in a paper on lower bounds for exponential sums.)

Both sums give square root cancellation (using Deligne’s work, of course) except for the $s$ sum if $r_1^3\equiv \pm r_2^3\bmod s$. But we may just push these to the second case. Thus, the contribution $U_0$ of these non-exceptional terms gives
$U_0\ll (R^2S)^{1/2+\epsilon}\Bigl(\frac{H}{R^2S}+1\Bigr)$
(counting the number of complete sums in the $h$-interval).

On the other hand, the exceptional $(r_1,r_2)$ are still controlled by the diagonal terms (because of the condition $R; there is a minor trick involved here if the cubic roots of unity exist modulo $s$, but I'll gloss over that.)

Now we can gather everything, and one checks that we end up with a bound
$\sum_{r,s}\sum_a S_2 \ll X^{\epsilon} QM\Bigl(\frac{1}{R^{1/2}}+\frac{R^{1/4}N^{1/2}}{Q^{5/4}}\Bigr).$

We need this to be $\ll MNQ^{-1} (\log MN)^{-A}$ for any $A$, and we see that we succeed as long as
$Q^2R^{-1/2}
as stated in the theorem (the second condition is implied by the first if $R). And as mentioned just afterwards, if we have $R=Q^{1/2-\epsilon}$, this gives a good distribution up to $X^{-\epsilon}N^{4/7}$ (note that epsilons may change from one inequality to the next.)

Remark. As the reader can see, we do not use either Weyl shifts, or cancellation in Ramanujan sums. The latter might appear in a more precise analysis, however, and give some extra gain.

### 8 Responses to “A ternary divisor variation”

1. Terence Tao wrote:

Very nice!

I think all of the terms with $h_1h_2h_3=0$ can be thrown into the main term because there is no dependence on $a$ in these terms.

The analysis here is in many ways closer to Zhang’s treatment of Type I/II sums (or of FI) based on the dispersion method than the Type III sums, although you take advantage of the smooth coefficients you have here by performing a Fourier expansion before applying the dispersion method. One price to pay for doing so when one works with general moduli rather than prime moduli is that you don’t get to automatically reduce to the r_1,r_2 coprime case in the non-diagonal setting, but I agree with you that this is a minor technical difficulty and one presumably needs to sum over all the different possible values of (r_1,r_2), which presumably are all dominated by the extreme cases of the diagonal r_1=r_2 and the coprime case (r_1,r_2)=1.

Wednesday, June 26, 2013 at 1:04 | Permalink
2. v08ltu wrote:

The elimination of multiplicity is the essentially what H-B uses (saying it dates back a long time), see middle page 42, “We have now arrived at the crux of the whole proof”… then writing $h=rs$ and applying Cauchy to eliminate multiplicity has “The effect of this key step has been to replace a sum over two variables whose ranges are rather short, by a sum over a longer range”.

Wednesday, June 26, 2013 at 4:11 | Permalink
3. Kowalski wrote:

I just highlighted this step to clarify what goes on for the readers. It is certainly very very old to create a longer variable in some way and then have to deal with a new multiplicity (I’m sure one can find it in Vinogradov and Linnik somewhere, and for a concrete example, consider, e.g., the Burgess bound.)

Wednesday, June 26, 2013 at 7:23 | Permalink
4. Terence Tao wrote:

Emmanuel, I am wondering about the completions of sums step when estimating U_0. Does this introduce an additional factor such as e_{r_1 r_2 s}( ah ) to the Kloosterman correlation? If one had not removed the diagonal and was still staring at U then there is enough positivity to complete sums without introducing this Fourier phase, but U_0 does not seem to have enough positivity to avoid this. Presumably the completed sums still have square root cancellation even with the Fourier twist, but it might be more difficult to do the algebraic geometry needed to attain it.

Thursday, June 27, 2013 at 4:19 | Permalink
5. Kowalski wrote:

Yes, it puts in the additive characters, but the sum is still a special case of the Friedlander-Iwaniec sum, as written by Heath-Brown and others, i.e., it has the shape

$\sum_x K_3(ax,s) \overline{K_3(bx,s)} e(cx/s)$

(the identity writing this in F-I terms is at the end of the note on the F-I sum). As I wrote, it is also a special case of the much more general “correlation sums” in the F-K-M papers, and we now can handle these in great generality.