
\bigbreak
{\nobigbf 8. $P_{720}$}

\nobreak\bigskip\nobreak

Following Coxeter ([C], sections 8.1 and 8.9),
we take for vertices of $P_{720} = \left\{{3 \atop 3,5}\right\}$
the midpoints of the edges of the regular polytope $\{3,3,5\}$.
Each vertex of $P_{720}$ is surrounded by two icosahedra
and five octahedra; its vertex figure is a pentagonal prism.

We now describe the group $G_{14400} \subset O(4)$
of all isometries of $P_{720}$.
The group $G_{7200}$ of orientation preserving isometries
of $\{3,3,5\}$ has order $120 \times 60$,
since it is transitive on the 120 vertices
and the subgroup of such isometries keeping a given vertex fixed
equals the group of orientation preserving isometries
of the vertex figure, an icosahedron.
Thus, $G_{7200} = (Q_{120} \times Q_{120})/(-1,-1)$.
The group $G_{14400}$ is generated by $G_{7200}$
together with a reflection on a hyperplane preserving
the vertices of $\{3,3,5\}$.

Unfortunately, the technique of the previous sections
does not apply directly.

{\nobf Proposition:}
{\sl The polytope $P_{720}$ admits no Cayley structure.}

{\nobf Proof:}
Let $G_{720}$ be an arbitrary subgroup of order 720 of $G_{14400}$:
we prove that $G_{720}$ does not
act simply transitively on edges of $\{3,3,5\}$.
Let $G_{25}$ be a 5-Sylow subgroup of $G_{720}$.
Clearly, $G_{25}$ is contained in
$G_{7200} = (Q_{120} \times Q_{120})/(-1,-1)$
and, by lifting,
we obtain a 5-Sylow subgroup of $Q_{120} \times Q_{120}$
which is conjugate to
$\{1,q,q^2,q^3,q^4\} \times \{1,q,q^2,q^3,q^4\}$,
where $q$ is some quaternion of order 5.
Thus, $G_{720}$ contains some element $g$ conjugate to $(q,q)$,
whose action keeps some vertex $v \in \{3,3,5\}$ fixed
(since $(q,q)$ does).
The element $g$ permutes the 12 neighbours of $v$,
splitting them into orbits of size 1 or 5;
hence, there is a neighbour $w$ of $v$, and hence an edge $vw$,
which are kept fixed under $g$, as we wanted to show.
\qed

Lov\'asz describes a method ([L]) to reduce the problem
of computing the spectrum of a graph with a transitive group
$\Gamma$ of isomorphisms to each irreducible decomposition of $\Gamma$,
which could be applied to this example.
As before, instead of counting paths,
we prefer to work with the matrices for representations
in a modified version of Lov\'asz's technique.
For simplicity, we start by applying the procedure
to $\langle 3,5,3,5\rangle$, which also admits no Cayley structure,
since its isometry group, $A_5 \oplus \ZZ/(2)$,
has no subgroup of order 30 (the number of vertices).

As illustrated in Figure 8.1, 
the vertices of $\langle 3,5,3,5\rangle$ are the midpoints
of edges of a dodecahedron.
If instead of taking midpoints of edges we take two
suitably spaced points per edge, we obtain $\langle 3,10,10\rangle$,
which admits a Cayley structure described in section 5.
Recall that $a = (12345)$, $b = (253)$ and $H = \{b,b^{-1},ab\}$.
Edges between decagons correspond to the generator $ab = (ab)^{-1}$.
For each vertex of $\langle 3,5,3,5\rangle$ there are
two elements of $A_5$ (counting the identity) which keep it fixed.

\midinsert\smallskip

\centerline{\psfig{file=fig81.ps,width=3.0in}}

\smallskip

\centerline{\eightbf Figure 8.1}

\smallskip\endinsert

Define a linear injection
$A_1: V_{\langle 3,5,3,5\rangle} \to V_{\langle 3,10,10\rangle}$
such that the value of $A_1(v)$ at a vertex $p$
of $\langle 3,10,10\rangle$
is the value of $v$ at the midpoint of the only edge
between decagons containing $p$
(recall that $V_P$ is the set of complex valued
functions on the vertices of the polytope $P$).
Conversely, define
$A_2: V_{\langle 3,10,10\rangle} \to V_{\langle 3,5,3,5\rangle}$
so that the value of $A_2(w)$ at a vertex $q$ of
$\langle 3,5,3,5\rangle$ is the sum of the values of $w$
at the two ends of the edge of $\langle 3,10,10\rangle$ containing $q$.
Thus, $A_2 A_1 = 2I$ and $A_1 A_2 = I + R_{ab}$,
where $R$ is the right multiplication action.
We claim that
$$X_{3535} = A_2(R_b + R_{b^{-1}})A_1;$$
Figure 8.2 illustrates the equality
at a basis vector of $V_{\langle 3,5,3,5\rangle}$.
The matrix $Y = (R_b + R_{b^{-1}})A_1A_2 =
(R_b + R_{b^{-1}})(I + R_{ab})$ has the same spectrum
as $X_{3535}$, up to 30 extra zero eigenvalues.
It is now clear that $Y$ splits into the irreducible representations
and its spectrum is computed in the usual manner.


\midinsert\smallskip

\centerline{\psfig{file=fig82.ps,width=3.0in}}

\smallskip

\centerline{\eightbf Figure 8.2}

\smallskip\endinsert

We are ready to consider $P_{720}$.
As with $\langle 3,5,3,5\rangle$,
take two points in each edge of $\{3,3,5\}$
to obtain a (non-semi-regular) polytope $P_{1440}$ with 1440 vertices.
Call edges of $P_{1440}$
contained in edges of $\{3,3,5\}$ {\sl special}.
The group $G_{7200}$ of orientation preserving
isometries of $\{3,3,5\}$ does not act simply transitively
on the vertices of $P_{1440}$.
Happily, its subgroup $G_{1440} = (Q_{120} \times Q_{24})/(-1,-1)$ does;
in other words, $P_{1440}$ admits a Cayley structure.
Indeed, the first factor $Q_{120}$ guarantees that
$G_{1440}$ acts transitively on the vertices of $\{3,3,5\}$.
The subgroup of $G_{1440}$ keeping the vertex 1 fixed
consists of the 12 elements of the form $(q,q)$, where $q \in Q_{24}$.
These act simply transitively on the 12 neighbours in $\{3,3,5\}$
of the vertex 1,
as can be checked using the coordinate system in Section 6.
The identity for the Cayley structure of $P_{1440}$
is chosen to be the vertex
between 1 and $(\tau + i - \taub j)/2$ which is closer to 1.
Using coordinates and quaternion multiplication,
the reader may check that
$H = \{g_0,g_1,g_2,g_3,g_4,g_s\}$, where
$g_0 = (i,i)$, $g_1 = ((1+i+j+k)/2,(1+i+j+k)/2)$,
$g_2 = ((1+i+j-k)/2,(1+i+j-k)/2)$,
$g_3 = ((1-i-j+k)/2,(1-i-j+k)/2)$,
$g_4 = ((1-i-j-k)/2,(1-i-j-k)/2)$ and
$g_s = ((-\taub i - j + \tau k)/2, k)$;
special edges correspond to $g_s$.
Notice that $g_0^{-1} = g_0$,
$g_1^{-1} = g_4$, $g_2^{-1} = g_3$ and $g_s^{-1} = g_s$.

Again, define linear transformations
$A_1: V_{P_{720}} \to V_{P_{1440}}$ and
$A_2: V_{P_{1440}} \to V_{P_{720}}$:
$A_1(v)$ at a vertex $p$ of $P_{1440}$
is the value of $v$ at the midpoint of the special edge containing $p$
and $A_2(w)$ at a vertex $q$ of
$P_{720}$ is the sum of the values of $w$
at the two ends of the special edge containing $q$.
Thus, $A_2 A_1 = 2I$ and $A_1 A_2 = I + R_{g_s}$.
Also, $$X_{P_{720}} =
A_2(R_{g_0} + R_{g_1} + R_{g_2} + R_{g_3} + R_{g_4})A_1.$$
In order to prove this equality,
we relate the adjacencies of $P_{720}$ and $P_{1440}$.
A vertex $p$ of $P_{720}$ has 10 neighbours
and is the midpoint of a special edge of $P_{1440}$
with vertices $q$ and $q'$.
The vertex $q$ has six neighbours: $qg_0,\ldots,qg_4$ and $q' = qg_s$.
Similarly, the six neighbours of $q'$ are $q'g_0,\ldots,q'g_4$ and
$q = q'g_s$.
Omitting the repetitions of $q$ and $q'$,
the special edges containing the remaining 10 points
have as midpoints the 10 neighbours $p_0,\ldots,p_9$ of $p$.
The process $p \mapsto \{q,q'\} \mapsto
\{qg_0,\ldots,qg_4,q'g_0,\ldots,q'g_4\} \mapsto
\{p_0,\ldots,p_9\}$ corresponds the successive application on 
a basis vector of $V_{720}$ of the transformations
$A_1$, $R_{g_0} + \cdots + R_{g_4}$ and $A_2$,
finishing the proof of the equality.
The rest is routine by now: 
$Y = (R_{g_0} + \cdots + R_{g_4})(I + R_{g_s})$
has the same spectrum as $X_{P_{720}}$,
up to 720 extra zero eigenvalues.
Finally, split $Y$ into the 32 irreducible representations
of $G_{1440}$ to compute its spectrum.


