# Enumerating lattice points in a product of balls, in limit with dimension

Fix $$(L_n)_n$$ to be a sequence of lattices, each $$L_n\subset \mathbb{R}^n$$, where both the effective-inradius and effective-outradius go to 1 (i.e. the Voronoi region of the lattices approach a ball in the appropriate senses, see [1]). Also, for some $$r_1,\dots, r_m>1$$ take $$K_n = \prod_{m=1}^M (r_m\mathcal{B}^n)$$ where $$\mathcal{B}^n$$ is the unit $$n$$-ball.

Finally, take $$(U_n)_n,(x_n)_n$$ to be arbitrary sequences of unitary transformations, vectors, respectively, in $$\mathbb{R}^{Mn}$$. I am interested in approximating the growth of $$|(L_n)^M\cap (U_nK_n+x_n)|$$ as $$n$$ grows, i.e. determining the following limit (which I believe exists in general) $$$$\lim_n \frac{1}{n} \log \left(|(L_n)^M\cap (U_nK_n+x_n)|\right)$$$$

I believe for such nice lattices and such a simple base set, lattice counting should behave approximately like integration in large dimension, so the limit is eventually like $$\lim_n \frac{1}{n}\log(\operatorname{vol}(K_n)/\operatorname{vol}(\mathcal{B}^n)^M) = \log(\prod_mr_m)$$ but I cannot substantiate this. Does anyone know any easy paths to this sort of result?

• The problem arises in calculation of an entropy for a network information theory problem. – enthdegree Dec 29 '18 at 0:30
• Ok one way you might be able to do it it is if you replace the fixed lattice sequence $(L_n)_n$ with a sequence of random lattices, each one taken from a Minkowski-Hlawa-Siegel ensemble that also has the aforementioned properties. But this is horrifically ugly. – enthdegree Dec 29 '18 at 0:55
• It holds on average over all lattices of a fixed determinant, due to results by Rogers. The proof is a bit involved, I will publish it and replicate it here after peer review! – enthdegree Jan 7 at 4:01