Given two linearly independent vector $v_1 , v_2 \in \mathbb{R}^2$ which span the lattice $\Gamma := \langle v_1 , v_2 \rangle_{\mathbb{Z }}$ I want to show, that there is a universal constant $M$, independent of the given lattice, such that the number of lattice points in $B_r(0)$, the ball with radius $r>0$ and center $0$, is $$\leq M \frac{r^2}{A} .$$Here, $A$ is the covolume of the lattice, meaning $A = \det \big(v_1 , v_2\big)$.
It is easy to see that bound when $v_1$ and $v_2$ are orthogonal, because then you can count the lattice points of $\mathbb{Z}^2$ in an ellipse. But I am having trouble doing the same thing, when $v_1 , v_2$ aren't orthogonal. Because then the ellipse you get when transforming the whole thing including the circle to the $\mathbb{Z}^2$-situation is somehow rotated and I don't know how to do the estimates then.
Best Answer
In fact that statement, as stated, is false. It is true, when $r$ is big enough, so that $B_r(0)$ contains a basis of $\Gamma$. A proof can look as follows: take a fundamenal domain $$\mathcal{F} := \{\lambda_1 v_1 + \lambda_2 v_2 : \lambda_1 , \lambda_2 \in [0,1)\}$$(associated to the basis $v_1 , v_2 \in \Gamma \cap B_r(0)$) and take the union $\mathcal{G}$ of all translates of $\mathcal{F}$, that have nonempty intersection with $B_r(0)$. This is a covering of $B_r(0)$. Because both basis vectors lie in $B_r(0)$ we have $\vert \vert v_1 + v_2 \vert \vert \leq 2 r$ and therefore we have $\mathcal{G} \subset B_{3r}(0)$.
Now that helps, because the number of lattice points in $B_r(0)$, let's call that $N(r)$, are less or equal to the minimal number of fundamental domains needed to cover $B_r(0)$. But this number is equal to $$\frac{\operatorname{vol}\mathcal{G}}{\operatorname{vol}\mathcal{F}}.$$ In total, we get $$N(r) \leq \frac{\operatorname{vol}\mathcal{G}}{\operatorname{vol}\mathcal{F}} \leq \frac{\operatorname{vol}B_{3r}(0)}{\operatorname{vol}\mathcal{F}} = 9 \pi \frac{r^2}{\operatorname{vol}\mathcal{F}}.$$