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Consider the usual lattice $M:=\mathbb{Z}^2\subseteq\mathbb{R}^2$, and let $v_1,v_2\in\mathbb{Z}^2\subseteq\mathbb{R}^2$ be two non-zero lattice points which are $\mathbb{Z}$-linearly independent. Consider the closed cone lattice $$ C=\{sv_1+tv_2:s,t\geq 0\}\subseteq\mathbb{R}^2. $$ For each radius $r$ we let $B_r$ stant for the closed ball of radius $r$ centered at the origin. Let $\alpha_1,\alpha_2\in\mathbb{R}\backslash\mathbb{Z}$ be non-integral real numbers. Is it known if the difference $$ \sum_{(n_1,n_2)\in M\cap C\cap B_r} e^{2\pi i (\alpha_1 n_1+\alpha_2 n_2)}-\int_{C\cap B_r}e^{2\pi i (\alpha_1 x_1+\alpha_2 x_2)} dx_1 dx_2 $$ is bounded independently of the radius r ?

Note that the 1-dimensional analogue of this is true.

added: Assuming that the difference above is uniformly bounded in $r$ for balls in the usual euclidean metric I would be tempted to think that the boundedness is independent of the chosen metric (i.e of the shape of the ball). Moreover, if it holds for two dimensional lattice cones I would expect it to hold as well for lattice cones in arbitrary dimensions.

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  • $\begingroup$ What do you mean by 1 dimensional analogue? $\endgroup$
    – Asaf
    Commented Jan 27, 2023 at 23:00
  • $\begingroup$ You can use for example the Euler-Maclaurin formula (up to order 2) to show that the difference between the exponential sum (so here over a consecutive set of integers) and the corresponding integral is bounded. The bound only depends on $0<\alpha=\alpha_1<1$ and looking at my calculation it looks like $1+\pi \alpha/6+\pi/3\cdot\alpha^2/(1-\alpha)$ which is independent of the length of the sum. $\endgroup$
    – Hugo Chapdelaine
    Commented Jan 27, 2023 at 23:38
  • $\begingroup$ should probably replaced the denominator $(1-\alpha)$ by $\min\{\alpha,1-\alpha\}$. $\endgroup$
    – Hugo Chapdelaine
    Commented Jan 28, 2023 at 3:08
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    $\begingroup$ Perhaps, each of the summands is bounded by itself? This holds if you replace the region of summation/integration $C\cap B_r$ with, for instance, the triangle with vertices $0$, $Mv_1$ and $Mv_2$ ($M\in\mathbb N$). Then the two summands can be found using the discrete and continuous version of Brion's theorem. In fact, both summands will have the form $P+a^MQ+b^MR$ for constants $P,Q,R,a,b\in\mathbb C$ with $|a|=|b|=1$. Of course, quite a bit more care (or some trick I'm not seeing) is needed to translate this argument from a simplex to a sector of the ball. $\endgroup$
    – Igor Makhlin
    Commented Jan 28, 2023 at 12:39
  • $\begingroup$ Very good point @imakhlin, may be each term is bounded independently of $r$. I also agree that the shape of the ball with which we intersect is cumbersome and kind of artificial. Could you point out to me this paper from Brion that you mentioned. I would be happy to learn more about the technique he is using. $\endgroup$
    – Hugo Chapdelaine
    Commented Jan 28, 2023 at 14:19

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