Certainly, the Voronoi cell of any lattice tiles the span of the lattice. But, the other polyhedron you mentioned, it would be shocking to me if that also tiles for an arbitrary lattice.
Wendy, I'm not sure what language you were having trouble with in the original post. Maybe it's just
Voronoi Cell.
A lattice is a vector space with integer coefficients. A lattice is said to be generated by a set of vectors S if every element of the lattice is a linear combination (with integer coeffiecients) of members of S. Put another way, a lattice is said to be generated by a set of vectors S if every element of the lattice is a sum of members of the closure (under additive inverses) of S.
For the matrix given, the following points (amongst infinitely many others) are also in the lattice: (phi,1,0), (-phi,-1,0), (phi,1+phi,1), (phi,1-phi,-1), (35*phi-3,35+2*phi,2-3*phi), etc.
A lattice generated by a set of vectors S is invariant under translation by any member of S.
For interesting lattices, there is a minimum distance between points in the lattice. A lattice could be degenerate. For example, the lattice formed by the vectors [ 1 ] and [ sqrt(2) ] is degenerate. There is no minimum distance between lattice points. We'll ignore degenerate lattices from here on.
Let's consider the lattice as a subset of R<sup>n</sup> for large enough n. The Voronoi cell of the origin is the set of points in R<sup>n</sup> which are as close or closer to the origin than any other point in the lattice. Because our lattice is not degenerate, this Voronoi cell is more than just the origin. It is, in fact, easy to show that it's a polytope. (For each point in the lattice other than the origin, make a line segment between the point and the origin. The perpendicular bisector of that segment is a hyperplane. Hang onto the (bounded) halfspace containing the origin. Now, take the intersection of all such halfspaces.)
Because of the invariance under translation by members of S, the Voronoi cell of every point is identical to the one for the origin. Also, by the way the Voronoi cells are constructed, there can be no gaps between adjacent Voronoi cells. So... they must tile.
Now, if you draw the Voronoi cell for every lattice point and highlight the lattice points, the final question is what does the polyhedron look like if you take for vertices all of the lattice points whose Voronoi cells share faces with the Voronoi cell of the origin. I'd have to think a bit more, but it seems to me that for an arbitrary lattice, this set of points might not be convex.