
\bigbreak
{\nobigbf 7. $P_{96}$}

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In this section, we compute the spectrum of $P_{96}$
(Coxeter's $s\{3,4,3\}$),
one of the three semi-regular polytopes of dimension 4.
Each of its 96 vertices is surrounded by
three icosahedra and five tetrahedra,
arranged according to the vertex figure shown in Figure 8.9A of [C];
each vertex has nine neighbours.
Our main task is to obtain a group $\Gamma$ of isometries of $P_{96}$
acting simply transitively on vertices.

We remind the reader of a construction for $P_{96}$,
detailed in [C], section 8.4.
Edges of $\{3,4,3\}$ can be oriented so that,
given any vertex $p$, there are four edges pointing outwards from $p$,
no two of which belong to the same (triangular) 2-cell.
Now, divide each oriented edge in two segments $a$ and $b$
(in this order) satisfying $b/a = \tau$;
the points thus obtained are the vertices of a $P_{96}$.
Alternatively, the 120 vertices of $\{3,3,5\}$
are the disjoint union of the 24 vertices of a $\{3,4,3\}$
and the 96 vertices of a $P_{96}$;
from the coordinates for $\{3,3,5\}$ in the previous section,
we thus obtain, as in [C] (section 8.7) coordinates for $P_{96}$.

The finite group $(Q_{24} \times Q_{24})/(-1,-1)$
acts on $\{3,4,3\}$ by isometries preserving edge orientation,
therefore acting also on $P_{96}$.
This group is too large to act simply transitively on the vertices
of $P_{96}$:
the subgroup $\Gamma = (Q_{24} \times Q_8)/(-1,-1)$ has the right order.
Clearly, $\Gamma$ acts transitively on the vertices of $\{3,4,3\}$
by elements of the form $(q,1)$.
Also, $\Gamma$ acts transitively on oriented edges:
by the transitivity on vertices, it is enough to show that it acts
transitively on the four edges starting from $1$,
which is done by elements of the form $(r,r)$.
Adding up, $\Gamma$ acts simply transitively
on the vertices of $P_{96}$
and this polytope therefore admits a Cayley structure.
The set $H$ of generators has 9 elements;
we explain how to obtain it.
We begin by identifying vertices of $P_{96}$
with oriented edges of $\{3,4,3\}$ as in the construction above.
Let $p_0$ be the vertex $(1,(1+i+j+k)/2)$:
by simple geometric considerations, its nine neighbours are
$(1,(1+i-j-k)/2)$, $(1,(1-i+j-k)/2)$, $(1,(1-i-j+k)/2)$,
$((1+i+j-k)/2,1)$, $((1+i-j+k)/2,1)$, $((1-i+j+k)/2,1)$,
$((1+i+j+k)/2, (1+i+j-k)/2)$, $((1+i+j+k)/2, (1+i-j+k)/2)$ and
$((1+i+j+k)/2, (1-i+j+k)/2)$.
The (unique) elements of $\Gamma$
taking $p_0$ to its nine neighbours are
$(i,i)$, $(j,j)$, $(k,k)$,
$((1+i-j+k)/2,k)$, $((1-i+j+k)/2,j)$, $((1+i+j-k)/2,i)$,
$((-1+i+j-k)/2,i)$, $((-1+i-j+k)/2,k)$ and $((-1-i+j+k)/2,j)$.
Thus, if we choose $p_0$ to be the identity
for the Cayley structure, the nine elements of $\Gamma$ above
are the nine elements of $H$.
From the previous section, we have explicit matrices
for the 19 irreducible representations of $\Gamma$;
their dimensions are at most 4.
The spectrum of the adjacency matrix $X$ can now
be computed as in section 4.
