I'm an undergraduate student and I'm currently working on my end-of-degree-project. The main goal of this project is studying the $A_4$ root lattice, the geometry of its Voronoï complex, and using the dualization method and Klotz construction to obtain 2D patterns. In particular, one such pattern would be the Penrose pattern.

The article I'm mainly following is this one by M. Baake et. al.

Given this context, I've already proven that the projection of the 2-facets of the Voronoï domains within the Voronoï complex $\mathcal{V}^{(2)}$ are, in fact, Penrose rhombi. And the vertices of these rhombi are the projection of the vertices of each $P\in\mathcal{V}^{(2)}$.

These vertices can be devided in two classes, called $q_1^*$ and $q_2^*$ in the article. If the projection of these two classes of vertices is studied, all the rest of the vertices can be obtained by trasnlations via lattice vectors and symmetries.

So I need to study the projection of the dual Voronoï cells centered in these two points ($q_1^*$ and $q_2^*$) and study their $2$-facets. The case with $V^*(q_1^*)$ is easy to deal with, for the polytope is a simplex. But I'm having problems when working with $V^*(q_2^*)$.

I have found the 10 vertices of $V^*(q_2^*)$, which are: $$ (0,0,0,0),(1,-1,0,0),(1,0,0,-1),(2,1,1,1),(0,-1,1,0),(0,0,1,-1),(1,1,2,1),(1,-1,1,-1),(2,0,2,1),(2,1,2,0) $$

I understand that the next steps would be finding the convex hull of this set of vertices, which will describe the polytope, finding its $2$-facets (which I know are triangles) and then projecting them, thus obtaining the projection image of the cell (as shown in the article in figure 5.1).

The problem I'm facing is: how can I find the $2$-facets of the 4-polytope? Is there an algorithm or method to do so?


1 Answer 1


Right in the middle of p.2251 in that cited article it is stated that there is a close relationship between the 5D (primitive) hypercubical lattice and the 4D $A_4$ lattice. In fact the latter is being obtained as a cross-section of the former perpendicular to the body-diagonal of the respective hypercubes.

From this relation it follows directly that the asked for cells $V^*(q_i^*)$ are nothing but the according cross-sections of that hypercube.

Now, when representing $\tt x4o3o3o3o$ (hypercube) within $A_1 \times A_4$ subsymmetry you'll get the following sequence of sections: $\tt o3o3o3o$ (point), $\tt x3o3o3o$ (simplex), $\tt o3x3o3o$ (rectified simplex), $\tt o3o3x3o$ (inverted rectified simplex), $\tt o3o3o3x$ (dual simplex), $\tt o3o3o3o$ (opposite point).

And as you could work out, the rectified simplex indeed has 10 vertices, just as you already described. Its facets (cells) will be 5 tetrahedra and 5 octahedra. Thus the (2D) faces indeed all are triangles.

--- rk

  • $\begingroup$ Thank you very much! This is really helpful. In addition to figuring out the shapes of the facets, I was also interested in identifying which vertices define each one. From this point, how could it be done? $\endgroup$
    – ElenaPT
    Sep 3, 2022 at 10:14
  • $\begingroup$ The answer to this further question of yours has already been given on p.2228 in eq.2.45: shallow holes (with simplex Delone cells) correspond to $q_i^*$ with $i=1,4$, while deep holes (with rectified simplex Delone cells) correspond to those with $i=2,3$. $\endgroup$ Sep 3, 2022 at 21:05

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .