The vertex figure of the tiling is a tesseract (with 8 cubical facets); so the dual tiling fills 4-space with their duals, which are cross-polytopes (with 16 tetrahedral facets). WFL On 3/10/15, Dan Asimov <dasimov@earthlink.net> wrote:
And it's easy to see that a nice fundamental domain for the Hurwitz integers arises from the Voronoi region of any one of them, like 0, and is a 24-cell. So this gives the tiling of R^4 by 24-cells.
The Standard Mistake is to assume that, since the 24-cell is self-dual, that its tiling of R^4 is also self-dual (guilty). Hmm, let's see, what exactly is the dual tiling?
--Dan
On Mar 10, 2015, at 9:15 AM, Dan Asimov <dasimov@earthlink.net> wrote:
Think of R^4 as chopped into cubes of side = 1/2, each corresponding to the quaternion at its (+,+,+,+) vertex. Every unit cube in 4-space is 16 of these side-(1/2) cubes, so has 16 of these (+,+,+,+) vertices. The Hurwitz quaternions correspond to 2 of these vertices, and the Lipschitz quaternions to just 1 of them. So it would appear that the index is 2.
--Dan
On Mar 10, 2015, at 8:38 AM, Fred Lunnon <fred.lunnon@gmail.com> wrote:
https://en.wikipedia.org/wiki/Hurwitz_quaternion
<< The Lipschitz quaternions L form an index 2 sublattice of H. >>
"index 3" , I think ?! See eg. Conway & Smith sect. 5.5 .
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