Greg Egan on Nostr: Here are the coordinates for all the vertices on one horosphere. As conventional ...

Here are the coordinates for all the vertices on one horosphere.

As conventional 4-vectors in the hyperboloid model, (t,x,y,z), define:

\[\begin{array}{l}V(j,k)=\\(\frac{j^2+j k+k^2+5}{2 \sqrt{6}},\frac{1}{2} \sqrt{\frac{3}{2}} j,\frac{j+2 k}{2 \sqrt{2}},\frac{j^2+j k+k^2-1}{2 \sqrt{6}})\end{array}\]

For any real values of j and k this 4-vector satisfies:

\[\begin{array}{rcl} g(V(j,k),V(j,k))&=&-1\\
g(V(j,k),(1,0,0,1))&=&-\sqrt{\frac{3}{2}}\\
g(V(j,k),V(j+1,k))&=&-\frac{5}{4}\\ g(V(j,k),V(j,k+1))&=&-\frac{5}{4}\\ g(V(j,k+1),V(j+1,k))&=&-\frac{5}{4}\end{array}\]

So this lets us embed a Euclidean \(A_2\) lattice in the horosphere, and getting the vertices of a hexagonal honeycomb is just a matter of omitting some points.

The equivalent 2×2 Hermitian matrices are:

\[\frac{1}{\sqrt{6}}\left(
\begin{array}{cc}
j^2+j k+k^2+2 & \omega (j+2 k)+2 j+k \\
-\omega (j+2 k)+j-k & 3 \\
\end{array}
\right)\]

where \(\omega\) is the cube root of 1 with positive imaginary part.

We omit points where \(j-k=3n.\) We get the hexagon centres by taking those points, summing their 6 nearest neighbours and normalising.

\[\begin{array}{l}C(j,n)=\\
(\frac{1}{2} \left(j^2+3 j n+3 n^2+2\right),\frac{j}{2},\frac{1}{2} \sqrt{3}
(j+2 n),\\\frac{1}{2} \left(j^2+3 j n+3 n^2\right))
\end{array}\]

or in matrix form:

\[\left(
\begin{array}{cc}
j^2+3 j n+3 n^2+1 & \omega (j+2 n)+j+n \\
-\omega (j+2 n)-n & 1 \\
\end{array}\right)\]

These hexagon centres lie on a parallel horosphere, with:

\[\begin{array}{rcl}
g(C(j,n),(1,0,0,1))&=&-1\\
g(C(j,n),C(j+1,n))&=&-\frac{3}{2}\\ g(C(j,n),C(j-1,n+1))&=&-\frac{3}{2}\\ g(C(j-1,n+1),C(j+1,n))&=&-\frac{3}{2}\end{array}\]