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Let $q \in \mathbb{N}$. I am interested in getting an upper bound for the sum $$ \sum_{(a_1, a_2, a_3, q) = 1} \sum_{\mathbf{h} \in (\mathbb{Z}/q\mathbb{Z})^n }e( \frac{a_1}{q}\ell_1(h_1, \ldots, h_n) + \frac{a_2}{q}\ell_2(h_1, \ldots, h_n) + \frac{a_3}{q}\ell_3(h_1, \ldots, h_n) ) $$ where $\ell_i$ is a linear form with integral coefficients, and the only thing we know is that the matrix formed by them have rank $3$ over $\mathbb{Q}$. And $e(z) = e^{2\pi i z}$.

I am hoping to obtain $\ll q^{n-3 - \delta}$ (the implicit constant may depend on the coefficients of the linear forms) for some $\delta>0$ but I have not been able to find a way to do this so far...

I would appreciate any suggestions on how I can approach this problem. Thank you.

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    $\begingroup$ I think there needs to be some additional hypotheses to say something nontrivial. What if all the coefficients of the $\ell_i$ are $0$ modulo $q$? This wouldn't contradict any linear independence hypothesis over $\mathbb{Q}$. In that case, the sum would be of order $q^{n+3}$. A second comment is that the sum over $h_1, \dots, h_n$ is either $0$ or $q^n$, by orthogonality of additive characters, so the only way to get a bound that is $o(q^n)$ is to show it vanishes for all choices of $a_i$. $\endgroup$
    – Matt Young
    Commented Apr 30, 2019 at 17:39
  • $\begingroup$ @MattYoung For your first comment I was thinking maybe that would be a problem for a small $q$ but it wouldn't be an issue once $q$ is sufficiently large (hence the upper bound with $\ll$). But thanks to your second comment I think I have a better idea of what I should do. Thank you! $\endgroup$
    – Johnny T.
    Commented May 1, 2019 at 13:49
  • $\begingroup$ I suppose if you are fixing your linear forms and estimating the exponential sum as $q \rightarrow \infty$, then my objection doesn't apply. It wasn't clear if you wanted your implied constant in the $\ll$ sign to be uniform in the linear forms, or not. $\endgroup$
    – Matt Young
    Commented May 1, 2019 at 15:12
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    $\begingroup$ Since they’re linearly independent there’s a 3x3 minor in the corresponding rank 3 matrix that is a nonzero integer. For the \ell_i to be linearly dependent (with coeffs in (\Z/q)^\times) mod q, then, it follows that q must divide said minor. (Equivalently: (a_1 \ell_1) \wedge (a_2 \ell_2) \wedge (a_3 \ell_3) = (a_1 a_2 a_3) (\ell_1 \wedge \ell_2 \wedge \ell_3), hence is zero if and only if \ell_1\wedge \ell_2\wedge \ell_3\in \bigwedge^3 \Z^n is zero mod q.) Thus for q large all linear forms \sum a_i \ell_i appearing in the exponential sum are nonzero, hence the inner sum is zero for all a_i. $\endgroup$
    – alpoge
    Commented May 1, 2019 at 16:07
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    $\begingroup$ Maybe it’d be better to just say: the three linear forms give a map \Z^n\to \Z^3. Put that map into Smith normal form (this means premultiply (aka change coordinates) by an element of GL_n \Z and postmultiply (aka take linear combinations of the \ell_i) by an element of GL_3 \Z). You’re left with wlog having to deal with the case \ell_i = d_i x_i, where d_1 \vert d_2 \vert d_3, which is evident once q doesn’t divide I believe (this may depend on the wlog) d_1 [—- certainly d_3, which by the way isn’t 0 by the rank hypothesis]. Let me know if I’ve made a mistake in the above! $\endgroup$
    – alpoge
    Commented May 1, 2019 at 16:17

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