Let
S be a Delsarte coclique for
\(\Gamma \) and let
B be the outer distribution of
S. By a theorem of Delsarte [
5, Theorem 3.1], for all vertices
x of
\(\Gamma \), and for all
\(j\in \{1, 3, 4, 6, 7, 8\}\),
$$\begin{aligned} \sum _{i=0}^d \frac{P_{ji}}{P_{0i}} B_{x,i}=0. \end{aligned}$$
(1)
Fix an element
\(x\notin S\). Recall that there are 22 elements of
S adjacent to
x (see Sect.
3.2 above), and so we can write
\(B_{x,i}=(0, y_1, 22, y_3, y_4, y_5, y_6, y_7, y_8, 330 - y_1 - y_3 - y_4 - y_5 - y_6 - y_7 - y_8)\) for some
\(y_i\). Then, we have the following equations (arising from (
1)):
1 | \(220 + y_1 - (y_3 + y_4 + 2 y_5 + y_6 + y_7) = 0\) |
3 | \(y_3 + y_5 + y_8 = 110\) |
4 | \(y_1 + y_3 + 3 y_4 - y_6 - y_7=0\) |
6 | \(y_1 + y_3 + y_6 - y_4 - y_7=0\) |
7 | \(y_1 + y_3 + y_7 - y_4 - y_6=0\) |
8 | \(2 (y_1 + y_3) - y_8=0\) |
These equations reduce:
\(2y_4=y_6=y_7 = y_8\),
\(y_3 =110 - y_5 - y_8\),
\(y_1 = y_5 + \tfrac{3}{2} y_8-110\). So
$$\begin{aligned} B_{x,i}=(0, y_5 + \tfrac{3}{2} y_8-110, 22, 110 - y_5 - y_8, \tfrac{1}{2}y_8, y_5, y_8, y_8, y_8, 330 - y_5 - 4 y_8). \end{aligned}$$
Let
a be the inner distribution of
S. Now
\(A_2\) is the adjacency matrix of
\(\Gamma \), and so
\(\mathbb {1}_{S}A_2\mathbb {1}_{S}^\top =0\). Hence we can write the inner distribution of
S as
\(a=(1,x_1,0,x_3,x_4,x_5,x_6,x_7,x_8,351-x_1-x_3-x_4-x_5-x_6-x_8)\), where the
\(x_i\) are indeterminate. Now multiply by
Q to yield the MacWilliams transform of
S:
$$\begin{aligned} aQ=32(&11, \quad \tfrac{1}{2}(- x_1 + x_3 + x_4 + 2 x_5 + x_6 + x_7-234), \quad x_1 + x_3 + x_4 + x_6 + x_7 + x_8-234, \\&117 - x_3 - x_5 - x_8,\quad \tfrac{1}{2}(x_1 + x_3 + 3 x_4 - x_6 - x_7), \quad 2 x_1 - x_3 + x_5,\\&x_1 + x_3 - x_4 + x_6 - x_7, \quad x_1 + x_3 - x_4 - x_6 + x_7,\quad -2 (x_1 + x_3) + x_8, \\&351 - 3 x_1 - x_4 - x_5 - x_6 - x_7 - x_8). \end{aligned}$$
Recall that if
\(E:=\sum _{j\in \{1, 3, 4, 6, 7, 8\}}E_j\), then
\(\mathbb {1}_SE=0\). Consider
\(E_j\) where
\(j\in \{1, 3, 4, 6, 7, 8\}\). Then
\(E_j=EE_j\) and so
\(\mathbb {1}_SE_j=\mathbb {1}_SEE_j=0\). So we have
\((aQ)_j=0\) for
\(j\in \{1, 3, 4, 6, 7, 8\}\), and hence
$$\begin{aligned} 2x_4=x_6=x_7 = x_8, \quad x_3 = 117 - x_5 - x_8,\quad x_5 = x_1-\tfrac{3}{2} x_8+117. \end{aligned}$$
Therefore,
$$\begin{aligned} a&=\left( 1,x_1,0,\frac{x_8}{2}-x_1,\frac{x_8}{2},x_1-\frac{3 x_8}{2}+117,x_8,x_8,x_8,-x_1-\frac{5 x_8}{2}+234\right) ,\\ aQ&=\left( 352, 0, 64 ( 2 x_8-117), 0, 0, 32 (117 + 4 x_1 - 2 x_8), 0, 0, 0, 64 (117-2 x_1 - x_8)\right) . \end{aligned}$$
We now take a different fusion scheme yielding a strongly regular graph
\({\mathcal {F}}\). Let
\(A= \sum _{i \in \{2,7,8\}}A_i\) where the
\(A_i\) are adjacency matrices of
\({\mathcal {A}}\), ordered according to the matrix
P above. Then
A is the adjacency matrix of a strongly regular graph
\({\mathcal {F}}\) with parameters (4096, 1638, 662, 650). The matrix of eigenvalues for
\({\mathcal {F}}\) is
$$\begin{aligned} P_{\mathcal {F}}=\begin{bmatrix} 1&{}1638&{}2457\\ 1&{}38&{}-39\\ 1&{}-26&{}25 \end{bmatrix} \end{aligned}$$
and the matrix of dual eigenvalues
\(Q_{\mathcal {F}}\) is exactly the same as
\(P_{\mathcal {F}}\). From the inner distribution
a for
S, it follows that
\(|{\mathcal {F}}(v)\cap S|=2x_8\) for all
\(v\in S\), where
\({\mathcal {F}}(v)\) denotes the neighbourhood of
v in
\({\mathcal {F}}\). From the outer distribution of
S, it follows that
\(|{\mathcal {F}}(v)\cap S|=2y_8\) for all
\(v\notin S\). Therefore,
$$\begin{aligned} \mathbb {1}_SA=2x_8\mathbb {1}_S +2y_8 (\mathbb {1}-\mathbb {1}_S) \end{aligned}$$
where
\(\mathbb {1}\) is the ‘all ones’ vector, and so
\((2x_8-2y_8-1638)\mathbb {1}_S+2y_8\mathbb {1}\) is an eigenvector for
A. In particular,
\((2x_8-2y_8-1638)\mathbb {1}_S+2y_8\mathbb {1}\) is annihilated by one of the non-principal idempotents
D of
\({\mathcal {F}}\), and so
\(\mathbb {1}_SD=0\) as
\(\mathbb {1}D=0\). The inner distribution for
S, with respect to
\({\mathcal {F}}\), is
\(a_{\mathcal {F}}:=(1,2x_8,351-2x_8)\) and therefore, its MacWilliams transform is
$$\begin{aligned} a_{\mathcal {F}}Q_{\mathcal {F}}=\left( 352, 64 (2x_8-117), 32(351 - 4 x_8)\right) . \end{aligned}$$
Since
\(\mathbb {1}_S\) is annihilated by one of the non-principal minimal idempotents,
\((a_{\mathcal {F}}Q_{\mathcal {F}})_{j}=0\) for either
\(j=1\) or
\(j=2\). So there are two cases to consider.
Case 1: \((a_{\mathcal {F}}Q_{\mathcal {F}})_{1}=0\)
Here we have
\(x_8=117/2\) and so
$$\begin{aligned} a&=\left( 1, x_1, 0, \tfrac{117}{4} - x_1, \tfrac{117}{4}, \tfrac{117}{4} + x_1, \tfrac{117}{2}, \tfrac{117}{2},\tfrac{117}{2}, \tfrac{351}{4} -x_1\right) ,\\ aQ&=(352, 0, 0, 0, 0, 128 x_1,0,0,0, 32 (117 - 4 x_1)). \end{aligned}$$
This implies that
S is design-orthogonal to the subfield design given in Sect.
3.1. So by Roos’ Theorem
2.1,
$$\begin{aligned} |S\cap \mathbb {F}_{4^3}|=\frac{|S||\mathbb {F}_{4^3}|}{|\Gamma |}=\frac{352\cdot 64}{4096}=\frac{11}{2}. \end{aligned}$$
This is a contradiction as
\(|S\cap \mathbb {F}_{4^3}|\) is an integer.
Case 2: \((a_{\mathcal {F}}Q_{\mathcal {F}})_{2}=0\)
Here we have
\(x_8=351/4\) and
$$\begin{aligned} a&=\left( 1, x_1, 0, \tfrac{351}{8} - x_1, \tfrac{351}{8}, x_1-\tfrac{117}{8}, \tfrac{351}{4}, \tfrac{351}{4}, \tfrac{351}{4}, \tfrac{117}{8} - x_1\right) ,\\ aQ&=\left( 352, 0, 3744, 0, 0, 32 (4x_1-\tfrac{117}{2}),0,0,0,32(\tfrac{117}{2}-4x_1)\right) . \end{aligned}$$
In particular,
\(aQ\geqslant 0\) implies that
\(x_1=\tfrac{117}{8}\) and hence
$$\begin{aligned} aQ=(352, 0, 3744, 0, 0, 0, 0, 0, 0, 0) \end{aligned}$$
and we have
\(\mathbb {1}_S\in V_0\perp V_2\). So
S is design-orthogonal to
\(S^\tau \), and so by Roos’ Theorem
2.1,
$$\begin{aligned} |S\cap S^\tau |=\frac{|S||S^\tau |}{|\Gamma |}=\frac{352\cdot 352}{4096}=\frac{121}{4}. \end{aligned}$$
This is a contradiction as
\(|S\cap S^\tau |\) is an integer.
Both cases lead to a contradiction, and so there is no Delsarte coclique.
\(\square \)