7
$\begingroup$

For $m,n,c \in \mathbb{N}$, let $S(m,n;c)$ denote the Kloosterman sum $$ S(m,n;c) := \sum_{\substack{1 \leq a < c \\ \gcd(a,c) = 1}} e \left( \frac{ma + n\overline{a}}{c} \right) $$ where $e(n) = e^{2 \pi i n}$ and $\overline{a}$ denotes the multiplicative inverse of $a \bmod c$.

In my research, involving producing a subconvexity bound for automorphic L-functions, I've recently come across a twisted shifted sum of Kloosterman sums. Let $\chi(\cdot)$ denote a Dirichlet character mod a prime $p$. Then I'm looking at $$ F(a, h, \chi) = \sum_{b \bmod p} \chi(b) S(a, b; p) S(a, b + h; p) \tag{1}.$$

I've never seen sums like this appear, but it looks pretty complicated. A first thing to consider might be an upper bound. We can produce a trivial upper bound using the Weil bound for Kloosterman sums, which indicates that $(1)$ is bounded above by $p^{2 + \epsilon}$, independently of $a,h, \chi$. But I think we should expect much smaller, at the most $p^{3/2 + \epsilon}$.

So I am wondering if someone has considered sums similar to $(1)$. I would also be interested in considerations of the similar but simpler sums $$ \begin{align} F(a, 1, 1) &= \sum_{b \bmod p} S(a, b; p) S(a, b + 1; p) \tag{2} \\ F(a, 1, \chi) &= \sum_{b \bmod p} \chi(b) S(a, b; p) S(a, b + 1; p). \tag{3} \end{align}$$

$\endgroup$
  • $\begingroup$ Sum (2) is linear over $b$ and can be easily calculated. $\endgroup$ – Alexey Ustinov Dec 16 '15 at 0:34
  • 6
    $\begingroup$ These estimates are likely covered by the general trace weight machinery of Fouvry, Kowalski, and Michel, though one may need some algebraic geometry to actually deploy this machinery for your specific sums: people.math.ethz.ch/~kowalski/trace-functions-pisa.pdf $\endgroup$ – Terry Tao Dec 16 '15 at 3:44
5
$\begingroup$

Your sum is $$\sum_{i=0}^2 (-1)^i \operatorname{tr}(\operatorname{Frob}_p, H^i_c( \mathbb A^1_{\overline{\mathbb F}_p}, \mathcal{K}\ell_2 (ab) \otimes \mathcal{K}\ell_2 (a(b+h)) \otimes \mathcal L_\chi)$$

The $H^1$ term corresponds to a square-root canncelation bound, the $H^0_c$ term vanishes by definition, and so it is sufficient to show the $H^2_c$ term vanishes.

By Poincare duality, the $H^2_c$ term is equal to the monodromy coinvariants of $\mathcal{K}\ell_2 (ab) \otimes \mathcal{K}\ell_2 (a(b+h)) \otimes \mathcal L_\chi$. We can show this vanishes by the following arguments:

($\chi$ nontrivial) There are no local monodromy invariants at $0$, since $\mathcal{K}\ell_2(ab)$ has unipotent local monodromy at $0$, whereas $\mathcal L_\chi$ has local monodromy a nontrivial tame character.

($h$ nontrivial) One can do something similar using the irreducibility of $\mathcal{K}\ell_2$. However this is superfluous because, as noted by Alexey Ustinov in the comments, the sum simplifies in the $\chi$ trivial case.

$\endgroup$

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.