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I can't find a reference for a set of generating vectors for the Tetrahedral-octahedral honeycomb lattice. I would like to know the "canonical" set and if possible a more general set described by angles (I think it would take three angles to completely describe it but I'm not sure on this either.)

Will Jagy
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tacos_tacos_tacos
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1 Answers1

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Here you go. A3 or D3, called face centered cubic

http://www.math.rwth-aachen.de/~Gabriele.Nebe/LATTICES/A3.html

Note that an extra dimension is required to avoid square roots in the basis coordinates. So, the three basis vectors are in $\mathbb R^4.$

$$ \langle 1,-1,0,0 \rangle $$ $$ \langle 0,1,-1,0 \rangle $$ $$ \langle 0,0, 1,-1 \rangle $$

EEEEEEEDDDDDDIIIIIIIITTTTTTTTTT: Here is a somewhat cleaner answer, after I checked it in Schiemann's reduction. The lattice you want is the one I asked about, all integer points in $\mathbb R^3$ such that $$ x+y+z \; \; \; \; \mbox{is even.} $$

A basis can be given in the same dimension by $$ \langle 0,1,1 \rangle $$ $$ \langle 1,0,1 \rangle $$ $$ \langle 1,1, 0 \rangle $$ Notice that the squared lengths are 2, and the inner products are 1. So the matrix of inner products, the Gram matrix is

$$ \left( \begin{array}{rrr} 2 & 1 & 1 \\ 1 & 2 & 1 \\ 1 & 1 & 2 \end{array} \right) $$ The twelve lattice points nearest the origin, in this lattice, are $$ (0,1,1), (1,0,1), (1,1,0), $$ $$ (0,-1,1), (-1,0,1), (-1,1,0), $$ $$ (0,1,-1), (1,0,-1), (1,-1,0), $$ $$ (0,-1,-1), (-1,0,-1), (-1,-1,0) $$ for kissing number 12, as they are all the same distance from the origin, $\sqrt 2.$

Will Jagy
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  • Cool, thank you... Is it $<2,-1,0>,<-1,2,-1><0,-1,2>$? – tacos_tacos_tacos Apr 22 '13 at 03:31
  • @tacos_tacos_tacos, no, that is the matrix of inner products, called the Gram matrix. I put a basis in the answer, three vectors in four dimensions, each squared length 2, inner products between are either 0 or $-1.$ – Will Jagy Apr 22 '13 at 03:35
  • Yeah, I was wondering about that. Ok that makes sense - thank you – tacos_tacos_tacos Apr 22 '13 at 03:40
  • @tacos_tacos_tacos, I edited in a second version that will probably be more useful to you. No square roots required, but all happening in $\mathbb R^3.$ – Will Jagy Apr 22 '13 at 04:31
  • Very nice. Is it typical to multiply them by $\frac{1}{\sqrt{2}}$ to get them to have norm $1$? This doesn't change any of the properties of the lattice right? – tacos_tacos_tacos Apr 22 '13 at 04:37
  • @tacos_tacos_tacos, if you want you can multiply by a constant, no change. If you want the radii of the packed spheres to be $1,$ you want the centers apart by 2, so multiply the basis vectors by $\sqrt 2$ to accomplish that. Do as seems useful. – Will Jagy Apr 22 '13 at 04:45
  • Note that the determinant of the matrix $\begin{pmatrix} 0&1&1\ 1&0&1\ 1&1&0 \end{pmatrix}$ is $2$ (this is a special case of a frequently asked question ;-) which means that the parallelepiped the vectors span has volum $2$, as the fundamental region for this lattice, containing half the points of $\Bbb Z^3$, obviously should. – Marc van Leeuwen Apr 22 '13 at 04:57
  • @MarcvanLeeuwen, well, yes. – Will Jagy Apr 22 '13 at 05:09
  • @MarcvanLeeuwen (and WillJagy) is the only requirement for a lattice to determine a valid sphere packing that they have to be the same norm away from eachother as they are from the origin? – tacos_tacos_tacos Apr 22 '13 at 05:44
  • Every lattice defines a regular stacking of spheres whose diameter equals the minimal distance between the lattice points. Calling it a sphere packing suggests that one cannot (easily) increase the density by deforming the lattice while retaining the diameter; intuitively having touching spheres in many different directions gives an obstruction to such deformations. In the case of FCC, there are $12$ spheres touching any given one, and this is the largest number one can obtain in dimension $3$. – Marc van Leeuwen Apr 22 '13 at 06:39