
\bigbreak
{\nobigbf 5. Archimedean solids II}

\nobreak\bigskip\nobreak

In this section we compute the spectra of the six remaining
Archimedean solids using the methods of the previous section.

As a first example, let
$P = \langle 4,6,6\rangle$.
The cube and $P$ have the same group of symmetries (of order 48);
the subgroup $\Gamma$ of orientation preserving isometries
acts simply transitively on the vertices of $P$,
thus giving $P$ a Cayley structure.
As is well known, $\Gamma$ is isomorphic to $S_4$,
the permutations of $\{1,2,3,4\}$
(number pairs of antipodal hexagons $1, 2, 3, 4$ as in Figure 5.1).
With this notation for $\Gamma$, $H = \{ (12), (1234), (1432) \}$:
the first generator corresponds in Figure 5.1
to the edge joining $e$ to $a$,
the second to the edge going southwest from $e$ to $d$
and the third to the remaining edge starting from $e$.
Generators for the five irreducible representations
of $S_4$ are given in Table 5.2.
The corresponding $X_i = M_{(12),i} + M_{(1234),i} + M_{(1432),i}$,
with their spectra, are listed in Table 5.3.

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\centerline{\psfig{file=fig51.ps,width=3.0in}}

\smallskip

\centerline{\eightbf Figure 5.1}

\smallskip\endinsert

\midinsert\smallskip

$$\matrix{
M_{(12),1} = 1& M_{(1234),1} = 1\cr
&\cr
M_{(12),2} = -1& M_{(1234),2} = -1\cr
&\cr
M_{(12),3} = \pmatrix{0 & 1 \cr 1 & 0 \cr} &
M_{(1234),3} = \pmatrix{-1 & 0 \cr -1 & 1 \cr} \cr
&\cr
M_{(12),4} =
\pmatrix{0 & 1 & 0 \cr 1 & 0 & 0 \cr 0 & 0 & 1 \cr}&
M_{(1234),4} =
\pmatrix{0 & 0 & -1 \cr 1 & 0 & -1 \cr 0 & 1 & -1 \cr}\cr
&\cr
M_{(12),5} = - M_{(12),4}&
M_{(1234),5} = - M_{(1234),4}\cr
}$$

\smallskip

\centerline{\eightbf Table 5.2}

\smallskip\endinsert

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$$\matrix{
X_1 = 3& \sigma(X_1) = \{3\}\cr
&\cr
X_2 = -3& \sigma(X_2) = \{-3\}\cr
&\cr
X_3 = \pmatrix{-2 & 1 \cr -1 & 2 \cr}&
\sigma(X_3) = \{\pm \sqrt{3}\}\cr
&\cr
X_4 = \pmatrix{-1 & 2 & -1 \cr 1 & 0 & 0 \cr -1 & 1 & 0 \cr}&
\sigma(X_4) = \{1, -1 \pm \sqrt{2}\}\cr
&\cr
X_5 = - X_4&
\sigma(X_5) = \{-1, 1 \pm \sqrt{2}\}\cr
}$$

\smallskip

\centerline{\eightbf Table 5.3}

\smallskip\endinsert

Consider now $\langle 3,3,3,3,4\rangle$,
which admits a Cayley structure
for its full symmetry group $\Gamma = S_4$.
Comparing Figures 5.1 and 5.4, we see two additional generators
$(234)$ and $(243)$, so
$H = \{ (12), (1234), (1432), (234), (243) \}$.
From Table 5.2 we have
$M_{(12),i}$ and $M_{(1234),i}$ and thus we have also
$M_{(234),i} = M_{(12),i}M_{(1234),i}$ and
$M_{(243),i} = M_{(234),i}^{-1}$.
The restrictions $X_i$ and their spectra follow immediately.

\midinsert\smallskip

\centerline{\psfig{file=fig56.ps,width=3.0in}}

\smallskip

\centerline{\eightbf Figure 5.4}

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To handle $\langle 3,4,5,4\rangle$,
consider the planes containing the triangular faces:
they form the faces of a circumscribed icosahedron.
The group $\Gamma$ of orientation preserving isometries
of these two polyhedra is therefore the same.
It is well known that this group can be identified
with $A_5$, the even permutations of $\{1,2,3,4,5\}$.
The polytope $\langle 3,4,5,4\rangle$ admits a Cayley structure
since $A_5$ acts simply transitively on its 60 vertices.
With an appropriate identification of $A_5$ with
the group of isometries of the icosahedron,
$H = \{a, a^{-1}, b, b^{-1}\}$, for $a = (12345)$ and $b = (253)$.
Generators for the five irreducible representations of $A_5$
are given in Table 5.5 where,
again, $\tau = (1 + \sqrt{5})/2$ and $\taub = (1 - \sqrt{5})/2$.
The matrices $X_i$ and their spectra are easily obtained.

\midinsert\smallskip

$$\matrix{
M_{a,1} = 1& M_{b,1} = 1\cr
&\cr
M_{a,2} = {1 \over 2}
\pmatrix{1 & \tau - 1 & -\tau \cr
\tau - 1 & \tau & 1 \cr \tau & -1 & \tau - 1 \cr}&
M_{b,2} = {1 \over 2}
\pmatrix{-1 & \tau - 1 & -\tau \cr
- \tau + 1 & \tau & 1 \cr \tau & 1 & - \tau + 1 \cr}\cr
&\cr
M_{a,3} = {1 \over 2}
\pmatrix{1 & \taub - 1 & -\taub \cr
\taub - 1 & \taub & 1 \cr \taub & -1 & \taub - 1 \cr}&
M_{b,3} = {1 \over 2}
\pmatrix{-1 & \taub - 1 & -\taub \cr
- \taub + 1 & \taub & 1 \cr \taub & 1 & - \taub + 1 \cr}\cr
&\cr
M_{a,4} =
\pmatrix{0 & 0 & 0 & -1 \cr 1 & 0 & 0 & -1 \cr
0 & 1 & 0 & -1 \cr 0 & 0 & 1 & -1 \cr}&
M_{b,4} =
\pmatrix{1 & -1 & 0 & 0 \cr 0 & -1 & 1 & 0 \cr
0 & -1 & 0 & 0 \cr 0 & -1 & 0 & 1 \cr}\cr
&\cr
M_{a,5} =
\pmatrix{0 & 0 & 0 & 0 & 1 \cr 1 & 0 & 0 & 0 & 0 \cr
0 & 1 & 0 & 0 & 0 \cr 0 & 0 & 1 & 0 & 0 \cr 0 & 0 & 0 & 1 & 0}&
M_{b,5} =
\pmatrix{-1 & 0 & 0 & 0 & 1 \cr -1 & 0 & 1 & 0 & 0 \cr
-1 & 0 & 0 & 1 & 0 \cr -1 & 1 & 0 & 0 & 0 \cr -1 & 0 & 0 & 0 & 0 \cr}\cr
}$$

\smallskip

\centerline{\eightbf Table 5.5}

\smallskip\endinsert

The three solids $\langle 5,6,6\rangle$,
$\langle 3,10,10\rangle$ and $\langle 3,3,3,3,5\rangle$
admit Cayley structures for the same $\Gamma = A_5$
with $H$ in each case being
$\{a, a^{-1}, ab\}$, $\{b, b^{-1}, ab\}$ and
$\{a, a^{-1}, b, b^{-1}, ab\}$.
The rest of the procedure is analogous.

The polyhedron $\langle 4,6,10\rangle$,
unlike the previous four examples,
has 120 vertices and admits a Cayley structure for its {\sl full}
isometry group, which is isomorphic to $A_5 \oplus \ZZ/(2)$,
where the generator of $\ZZ/(2)$ is the (orientation reversing)
antipodal map. The characters and representations for this
larger group are trivially obtained from those of $A_5$ ([Se])
and the ten $X_i$ matrices are thus easily obtained.
We prefer, however, to consider $X^2$,
the square of the adjacency matrix.
Since $\langle 4,6,10\rangle$ is bipartite,
$X^2$ has two obvious invariant
subspaces corresponding to the white and black vertices.
The white vertices are naturally identified
with those of $\langle 3,3,3,3,5\rangle$
and admit a simply transitive action of $A_5$.
Let $Y$ be the restriction of $X^2$ to the white vertices:
in the notation of section 4,
$Y = 3 I + R_a + R_{a^{-1}} + R_b + R_{b^{-1}} + 2 R_{ab}$.
Thus, the computation of $\sigma(Y)$ reduces as usual
to the computations of spectra of matrices in each irreducible
representation of $A_5$.
Finally, $\sigma(X)$ contains the two square roots
of each element of $\sigma(Y)$.

