Creating n-dimensional lattices from lower dimensional parts Suppose I have n orthogonal unit vectors (in Euclidean space). These unit vectors may be used to describe an n-dimensional unit hypercube. A subset of these unit vectors may be combined to make lower dimensional faces on the n-dimensional unit hypercube. What I mean is that one or more of these unit vectors may be used to define $b_1$ 1-D edges, $b_2$ 2-D faces, $b_3$ 3-D cells, etc. (up to $b_{n-1}$ $(n-1)$-dimensional faces). For example, in 3D, you may take the x,y,z unit vectors to form three 1-D edges (along the three basis vectors) and three 2-D faces (in the xy, xz, and yz plane).
Is there a rule whereby I may create an n-dimensional lattice (for now say cubic lattice for simplicity) using these lower dimensional faces (the $b_1$ 1-faces, the $b_2$ 2-faces, etc)? 
For ex: By inspection, I may draw a 6x6 lattice in two dimensions as a collection of 36 letter L's (one 0-D corner + two 1-D edges) and one big ($6\times 6$) rotated letter 'L' to close the figure. Along those lines I may make a $k^3$ cubic lattice as a combination of $k^3\times$ (one 0-D corner $+$ three 1-D edges $+$ three 2-D faces) plus one large figure similar to the others (3 faces) rotated to close the lattice. 
I think the solution is that I can build a $k^n$ (n-dimensional) lattice from the $b_1$ 1-D edges (the n orthogonal unit vectors), the $b_2$ combinations of 2-D faces that these unit vectors may describe, the $b_3$ combinations of 3-D cells these unit vectors may describe, etc., and than have one identical (but larger) figure with the same number of 1-faces, 2-faces, etc. rotated to close the lattice. This is a solution I came up with, but I'd like to hear people who are more trained in mathematics and lattices than I (I haven't found a proof one way or the other....this is just a solution that seems reasonable).
If this method is correct, than I may describe a $k \times k \times k \times k$ (4-D) lattice as a collection of $k^4 \times$ (one 0-D face $+$ four 1-D faces $+$ six 2-D faces $+$ four 3-D faces) and one large (one 0-D face $+$ four 1-D faces $+$ six 2-D faces $+$ four 3-D faces) rotated to close the figure. (In this example, I'm not looking to completely cover $R^4$ with lower dimensional pieces, I just want to be able to build a lattice from lower dimensional pieces that approximates $R^4$ when the lattice becomes a sufficiently large.)
(Technically, I am re-asking this question. I'd like to attach a bounty, but I don't have enough rep to attach a bounty yet, and have been very busy since the first ask. If mods really don't like this, i'll take this question down.)
 A: What you wrote somehow reminds me of the way Plato proved that there's only the known "Platonic Bodies". In a similar vein, it's possible to prove that there's only so many substantially different 3-d crystal lattices.
Technically, you can restrict the question to integer lattices (there might be rational or even irrational offsets, however, you can always count [fractions of] unit cells). Then, the question boils down to the one if you can reach the corners of edges (2d basis vectors) via the (rotate) 1d basis vector, i.e. via a unimodular 2d transformation of a 1d projection of this basis vector to 2d, so that you get a basis of a plane of faces. Then, you could ask the question if the 2d basis vectors can be rotated so that you get a basis of the 3d lattice of the "crystal" and so on. That way, you'll arrive at the same statement about the number of different crystals.
If memory serves me right, there's exactly one possibility for a 4d crystal - the analogon to a "simple cubic" crystal, that is: the edges at every integer point. The google search for this concept is getting clogged by some physicists, however.
