# Does lattice mod preserve direction?

For high enough dimension $$n$$ there are lattices $$L_n$$ in $$\mathbb{R}^n$$ whose Voronoi partition's base regions encompass all but a negligible proportion of a $$(1-\varepsilon)$$-ball, and also nearly contain a $$(1+\varepsilon)$$-ball to all but a negligible proportion.[1] How much can such base regions behave like a ball in other ways?

One can define a "ball mod" as $$mod_\ast(v)=(v/\|v\|)\cdot [\|v\| - \ \operatorname{round}(\|v\|/2)]$$, which just reduces $$v$$ by lengths of 2 until it is in the unit ball at the origin.

One can also define "round" and "mod" operations over space up to a lattice $$L$$ and its Voronoi partition as: $$R_L(v)=\arg\min_{\ell\in L}\|v-\ell\|$$ and $$mod_L(v)=v-R_L(v).$$

Fixing radius $$R>0$$, for high enough dimension $$n$$, then does a lattice $$L$$ of the sort described above have that: $$\|mod_\ast(v)-\operatorname{mod}_L(v)\|<\varepsilon$$ for most $$v\in B(0,R)$$ ?

• I guess no, since I suspect there are not enough Voronoi regions adjacent to the base region for a lattice-mod-reduction of a vector just slightly outside of the base region to commonly approximately correspond to that vector's length-2 reduction. – enthdegree Nov 29 '18 at 23:58