# n-dimensional Delaunay Triangulation of Lattices

I have several questions concerning the Delaunay triangulation of a high dimensional lattice. Given an $n$-dimensional lattice $L$ and its Delaunay triangulation (partition of $R^n$ into simplices such that the circumsphere of any simplex does not contain any lattice point).

1. Does the center of a circumsphere of a Delaunay simplex have to be contained inside the simplex?
2. Does a set of any $n$ independent vectors such that each vector coincides with a Delaunay edge form a basis of $L$?
3. Is there a basis $B = \{b_1,b_2,\ldots , b_n\}$ of $L$ such that the $n$ dimensional parallelepiped $B[0,1)^n$ can be decomposed into Delaunay simplices?

Any partial answer for any question or pointing to future reading would be greatly appreciated.

For (2), observe that a positive answer here would imply all Delaunay simplices to be unimodular (i.e., have volume equal to $1/n!$ times the volume of a fundamental parallelepiped). This holds for $n\le 4$ but starts to fail for $n\ge 5$. See, for example, my paper "Lattice Delone simplices with super-exponential volume" (arXiv, journal) and the references therein.
For (1) I can give an explicit counter-example in dimension three. Consider the lattice generated by $(1,a,a)$, $(a,1,a)$ and $(a,a,1)$, for a very small positive number $a$. When $a=0$ the Delaunay "triangulation" is not a triangulation, but the decomposition of $R^3$ into unit cubes. The effect of making $a\ne 0$ (but small) is that these cubes are refined into simplices, all of which have (by continuity) their center very close to $(1/2,1/2,1/2)$. But it turns out that one of these simplices has the origin and the three basis vectors as vertices, so the center of the circumsphere is outside of it.