doubt about the equation of truncated icosahedron

Question

I would like to obtain the inner region of a truncated icosahedron [https://en.wikipedia.org/wiki/Truncated_icosahedron][1], in this web site the equations are defined like even permutations.

v =

$(0,\pm 1 , \pm 3\phi)$

$(\pm 1,\pm (2+\phi) , \pm 2\phi)$

$(\pm \phi,\pm 2 , \pm (2\phi+1))$

The equation of inner surface is defined by \vec{v}\cdot (x,y,z)<1. Where v are the the vector defined below. How I can calculate the even permuation of v?

Answer 1

The truncated dodecahedron (soccer ball see representation on Fig. 1) has 12 pentagons on which one can base our reasoning.

Its mode of generation can be understood by associating three approaches :

• graphical (see Fig. 2 below),

• algebraic (linear algebra) and as well

• computational, meaning by that, the important complementary understanding brought by following the steps of the program (see below) which has generated this figure.

The $20$ initial vertices you give (depicted with red stars $\color{red}{\star}$), can be considered as the vertices of a first family of 4 pentagons. The two other families of 4 pentagons (with $\color{green}{\star}$ and $\color{blue}{\star}$ stars resp.) can be obtained from this first family by rotations (= isometries with determinant $1$). There are three of them :

$$I=\begin{pmatrix}1&0&0\\0&1&0\\0&0&1\end{pmatrix}, \ \ P_1=\begin{pmatrix}0&0&1\\1&0&0\\0&1&0\end{pmatrix} \ \ \text{and} \ \ P_2=\begin{pmatrix}0&1&0\\0&0&1\\1&0&0\end{pmatrix}.$$

Of course, matrix $I$ hasn't to be activated (it has already been "used" for the first family !).

More precisely,$\{I,P_1,P_2\}$ constitute the matrix version of the subgroup of the 3 even permutations (those with determinant $1$) within the group of permutations on 3 objects (with $3!=6$ elements).

Remark : the correspondence between the classical representation of permutations and matrices is easy. Let us take the example of $P_1$ ; if $e_1, e_2, e_3$ is the canonical basis,

$$P_1(e_1)=e_2, \ \ P_1(e_2)=e_3, \ \ P_1(e_3)=e_1 \ \ \ \leftrightarrow \ \ \ \binom{1 \ \ 2 \ \ 3}{2 \ \ 3 \ \ 1}.$$

Fig. 1 : A figure generated by the software given in Remark 3 below.

Fig 2 : The axis of rotation (in black) is shared by $P_1$ (angle of rotation $2 \pi/3$) and $P_2$ (angle of rotation $-2 \pi/3$, due to the fact that $P_2=P_1^{-1}$). The effect of these rotations is understandable on the hexagon at the top (its center belonging to the axis of rotation), adjacent to three pentagons, a blue, a green and a red one, images one of the other by a $k 2 \pi/3$ rotation.

Remarks :

1) The three other permutation matrices (these ones with determinant $-1$) are transposition matrices :

$$\begin{pmatrix}1&0&0\\0&0&1\\0&1&0\end{pmatrix}, \ \ \begin{pmatrix}0&1&0\\1&0&0\\0&0&1\end{pmatrix} \ \ \text{and} \ \ \begin{pmatrix}0&0&1\\0&1&0\\1&0&0\end{pmatrix}.$$

2) The truncated dodecahedron belongs to a particular family, the Goldberg polyhedra, sharing with it the fact that they all possess exactly 12 pentagons.

3) See the wonderful site https://levskaya.github.io/polyhedronisme/ where you can in particular generate all truncated polyhedra you can imagine ; for example, in our case, type tI to get ... a truncated Icosahedron, or ttI if you want the twice truncated icosahedron...

Matlab program for the generation of fig. 2:

    clear all;close all;hold on;axis equal
    p=(1+sqrt(5))/2; % golden ratio
    V1=diag([0,1,  3*p]); % initial vertices...
    V2=diag([1,2+p,2*p]); % ...presented as diagonal...
    V3=diag([p,2,  2*p+1]); % ...matrices for "technical" reasons
    G=[1 , 1,  1,  1
       1 , 1, -1, -1
       1 ,-1,  1, -1]; % signs combinations
    W=[V1*G,V2*G,-V2*G,V3*G,-V3*G];% the 20 basic vertices
    I=eye(3);% identity matrix
    P1=I([2,3,1],:); % The 2 other even...
    P2=I([3,1,2],:); % ...permutation matrices
    W=[W,P1*W,P2*W];% All 3 x 20 = 60 vertices 
    view([48,28]);
    z=1:20; plot3(W(1,z),W(2,z),W(3,z),'pr'); % red stars (1st family)
    z=21:40;plot3(W(1,z),W(2,z),W(3,z),'pg'); % green stars
    z=41:60;plot3(W(1,z),W(2,z),W(3,z),'pb'); % blue stars
    for k1=1:60
        for k2=k1+1:60
            if norm(W(:,k1)-W(:,k2))<2.01
                I=[k1,k2];
                plot3(W(1,I),W(2,I),W(3,I),'k') % edge drawing
            end;
        end;
    end;

Answer 2

The even permutations of $(a,b,c)$ are simply $(a,b,c)$, $(b,c,a)$, and $(c,a,b)$. The odd permutations are $(b,a,c)$, $(a,c,b)$, and $(c,b,a)$.

So, for example, you need $(0,\pm1,\pm3\phi)$, $(\pm1,\pm3\phi,0)$, and $(\pm3\phi,0,\pm1)$.

In general, a permutation is even if it results from an even number of transpositions, and odd if it results from an odd number of transpositions.