Sums of twisted products of Kloosterman Sums

Question

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 $$ \begin{align} F(a, 1, 1) &= \sum_{b \bmod p} S(a, b; p) S(a, b + 1; p) \tag{2} \\ F(a, 1, \chi) &= \sum_{b \bmod p} \chi(b) S(a, b; p) S(a, b + 1; p). \tag{3} \end{align}$$

Answer 1

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.