We will start with an atom at <0, 0, 0>. The points <1, 1, 1>, <1, −1, −1>, <−1, 1, −1>, <−1, −1, 1> are the corners of a regular tetrahedron whose center is <0, 0, 0>. We assume that there are atoms at those four points and that they are the four neighbors of <0, 0, 0>. The bond length is √(3) in these convenient length units.

The atom at <1, 1, 1> also has four neighbors including <0, 0, 0>. I claim that the other three are at <2, 2, 0>, <2, 0, 2> and <0, 2, 2>.

All atoms have integer coordinates and there are two populations:

C0
atoms whose coordinates are all even and whose sum is 0 mod 4.
C1
atoms whose coordinates are all odd and whose sum is 3 mod 4.
Together these two populations constitute all of the (carbon) atoms in the diamond lattice. An atom of one population has as neighbors, four atoms which are of the other population.

### Symmetries

It is clear that (x, y, z) → (−x, −y, z) is a symmetry as well as (x, y, z) → (y, z, x). (x, y, z) → (x, z, y) is a reflection which is a symmetry. Indeed in this note “etc.” always means permute the preceding elements according to (x, y, z) → (y, z, x) two more times. These above symmetries generate the group of the tetrahedron. A translation of any atom to one of the same population is also a symmetry.
Cubic ice has a similar structure.

### Where are the cavities in diamond?

The form of the descriptions of the two populations above leads one’s attention to changing “0 mod 4” to “2 mod 4” and “3 mod 4” to “1 mod 4”, since the sum is already constrained mod 2. Thus:
H0
holes whose coordinates are all even and whose sum is 2 mod 4.
H1
holes whose coordinates are all odd and whose sum is 1 mod 4.
The four nearest carbons to <0, 0, 0,> are four of the eight corners of the 23 cube described by |x|, |y|, |z| ≤ 1. These cavity centers are at the other four corners. If we add <1, 1, −1> to the coordinates of a carbon we get the coordinates of a cavity, and conversely. This forms a 1-1 correspondence between the carbons and the cavities.
The density of diamond is (atoms / cell)(mass of carbon atom)/(cell edge)3. The cell mentioned above is not the fundamental repeating cell, however. The nature of the specifications of the locations of the atoms indicates a 43 fundamental cell. Here we hear that the C-C distance is 0.154 nm in a diamond crystal!

There are these 8 atoms in the half open cell [0, 4)3:
<0, 0, 0>, <2, 2, 0>, <2, 0, 2>, <0, 2, 2>, <1, 1, 1>, <3, 3, 1>, <3, 1, 3>, <1, 3, 3>.
The length units used above are 0.154/√(3) nm and the cell edge is 4 times that. Thus density =
(8(atoms/cell))(19.925 10−24g/atom)/((4(0.154)/√(3))3 (nm3/cell)) = 3.54 10−21 g/nm3 = 3.54 g/cm3
(It is reported as 3.52 g/cm.)

Lets work out the Voronoi zone of <0, 0, 0>. We seek the facets. Each nearest neighbor contributes a facet. x+y+z = 3/2 is the plane midway between <0, 0, 0> and <1, 1, 1> and x−y−z = 3/2 is half way to <1, −1, −1>, etc. The interior is defined by x+y+z<3/2 and x−y−z<3/2 etc. These four planes belong to a regular tetrahedron. It is tempting to guess that the tetrahedron is the zone, but (Voronoi) zones fill space and regular tetrahedra can’t. The vertices of the tetrahedron are <−3/2, −3/2, −3/2>, <3/2, 3/2, −3/2> etc.

The neighbor’s 12 neighbors are at <±2, ±2, 0>, etc. The zone facets due to those are |x|+|y|<2, |x|+|z|<2, and |y|+|z|<2. These are facets for the Rhombic dodecahedron. The vertices of this are at <±1, ±1, ±1> and <±2, 0, 0> etc. The short diagonals of the rhombi are the edges of the cube |x|, |y|, |z| < 1.

Of the vertices of the dodecahedron, only <−1, −1, −1> and <1, 1, −1> etc. lie inside the tetrahedron. None of the tetrahedron vertices lie in the dodecahedron.

From the center of a cavity there are four closest atoms. The zone of each of these four must reach the center. These four zones come together, each touching the other three−each with three facets coming together at the center point. If we removed population of C1 atoms from the crystal, the rhombic dodecahedron would be the Voronoi zone. As it is, it is twice too big.

The intersection between x+y+z = 3/2 and x−y−z = 3/2 is the line x=3 & y=−z. <3, 0, 0> is on that line but that point is closer to the atom <4, 0, 0>. This suggests that <4, 0, 0> contributes a facet. But first the 12 neighbor’s neighbors, such as <2, 2, 0>, the facet for which is the plane x+y=2. That plane and x+y+z = 3/2 intersect at z=−1/2 & x+y=2. Now I will guess and try to verify. The complementary diamond crystal has atoms in the cavities of the real crystal. The bonds of the complementary crystal form the edges of the Voronoi zones of the real crystal. The zones cover the space and thus cover the center of each cavity. Indeed the zones of four atoms meet at that center.

<−1, −1, −1> is an H1 cavity center adjoining <0, 0, 0>. <1, −1, −1> is C1 atom adjoining the same cavity. The bisecting plane between <0, 0, 0> and <1, −1, −1> is x−y−z=3/2.

The other neighbors of <1, 1, 1> are <2, 2, 0>, <2, 0, 2> and <0, 2, 2> (see above), and the planes midway between these and <0, 0, 0> are x+y=2, x+z=2 and y+z=2.

See Repasky’s page for geometry of many familiar molecules.