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Nonexistence of a $Q$-polynomial association scheme with Krein array $\{2r^2-1, 2r^2-2, r^2+1; 1, 2, r^2-1\}$ and $r$ odd¶

We will show that a $Q$-polynomial association scheme with Krein array $\{2r^2-1, 2r^2-2, r^2+1; 1, 2, r^2-1\}$ and $r$ odd does not exist. This Krein array is feasible for all $r \ge 2$. For $r = 2^j$, this Krein array is realized by the duals of Kasami codes with minimum distance $5$.

In [1]:

%display latex
import drg

Such a scheme would have $4r^4$ vertices.

In [2]:

r = var("r")
p = drg.QPolyParameters([2*r^2-1, 2*r^2-2, r^2+1], [1, 2, r^2-1])
p.order(expand=True, factor=True)

Out[2]:

$\newcommand{\Bold}[1]{\mathbf{#1}}4 \, r^{4}$

Since the scheme is $Q$-polynomial, we have $q^1_{13} = q^1_{31} = q^3_{11} = 0$. Additionally, we have $q^1_{11} = 0$.

In [3]:

[p.q[1, 1, 1], p.q[1, 1, 3], p.q[1, 3, 1], p.q[3, 1, 1]]

Out[3]:

$\newcommand{\Bold}[1]{\mathbf{#1}}\left[0, 0, 0, 0\right]$

The intersection numbers can be checked to be nonnegative and integral for both even and odd values of $r \ge 2$.

In [4]:

p.pTable(factor=True, simplify=True)

Out[4]:

$\newcommand{\Bold}[1]{\mathbf{#1}}\begin{aligned}0: &\ \left(\begin{array}{rrrr} 1 & 0 & 0 & 0 \\ 0 & \frac{1}{2} \, {\left(2 \, r^{2} - 1\right)} {\left(r + 1\right)} r & 0 & 0 \\ 0 & 0 & {\left(2 \, r^{2} - 1\right)} {\left(r^{2} + 1\right)} & 0 \\ 0 & 0 & 0 & \frac{1}{2} \, {\left(2 \, r^{2} - 1\right)} {\left(r - 1\right)} r \end{array}\right) \\ 1: &\ \left(\begin{array}{rrrr} 0 & 1 & 0 & 0 \\ 1 & \frac{1}{4} \, {\left(r + 3\right)} {\left(r + 2\right)} {\left(r - 1\right)} r & \frac{1}{2} \, {\left(r^{2} + 1\right)} {\left(r + 2\right)} {\left(r - 1\right)} & \frac{1}{4} \, {\left(r + 1\right)} {\left(r - 1\right)} {\left(r - 2\right)} r \\ 0 & \frac{1}{2} \, {\left(r^{2} + 1\right)} {\left(r + 2\right)} {\left(r - 1\right)} & {\left(r^{2} + 1\right)} r^{2} & \frac{1}{2} \, {\left(r^{2} + 1\right)} {\left(r - 1\right)} r \\ 0 & \frac{1}{4} \, {\left(r + 1\right)} {\left(r - 1\right)} {\left(r - 2\right)} r & \frac{1}{2} \, {\left(r^{2} + 1\right)} {\left(r - 1\right)} r & \frac{1}{4} \, {\left(r + 2\right)} {\left(r - 1\right)}^{2} r \end{array}\right) \\ 2: &\ \left(\begin{array}{rrrr} 0 & 0 & 1 & 0 \\ 0 & \frac{1}{4} \, {\left(r + 2\right)} {\left(r + 1\right)} {\left(r - 1\right)} r & \frac{1}{2} \, {\left(r + 1\right)} r^{3} & \frac{1}{4} \, {\left(r + 1\right)} {\left(r - 1\right)} r^{2} \\ 1 & \frac{1}{2} \, {\left(r + 1\right)} r^{3} & {\left(r^{2} + 2\right)} {\left(r + 1\right)} {\left(r - 1\right)} & \frac{1}{2} \, {\left(r - 1\right)} r^{3} \\ 0 & \frac{1}{4} \, {\left(r + 1\right)} {\left(r - 1\right)} r^{2} & \frac{1}{2} \, {\left(r - 1\right)} r^{3} & \frac{1}{4} \, {\left(r + 1\right)} {\left(r - 1\right)} {\left(r - 2\right)} r \end{array}\right) \\ 3: &\ \left(\begin{array}{rrrr} 0 & 0 & 0 & 1 \\ 0 & \frac{1}{4} \, {\left(r + 1\right)}^{2} {\left(r - 2\right)} r & \frac{1}{2} \, {\left(r^{2} + 1\right)} {\left(r + 1\right)} r & \frac{1}{4} \, {\left(r + 2\right)} {\left(r + 1\right)} {\left(r - 1\right)} r \\ 0 & \frac{1}{2} \, {\left(r^{2} + 1\right)} {\left(r + 1\right)} r & {\left(r^{2} + 1\right)} r^{2} & \frac{1}{2} \, {\left(r^{2} + 1\right)} {\left(r + 1\right)} {\left(r - 2\right)} \\ 1 & \frac{1}{4} \, {\left(r + 2\right)} {\left(r + 1\right)} {\left(r - 1\right)} r & \frac{1}{2} \, {\left(r^{2} + 1\right)} {\left(r + 1\right)} {\left(r - 2\right)} & \frac{1}{4} \, {\left(r + 1\right)} {\left(r - 2\right)} {\left(r - 3\right)} r \end{array}\right) \\\end{aligned}$

Let us check that the case $r = 3$ satisfies the known feasibility conditions. We skip the family nonexistence check since the Krein array of the members of family with $r$ odd is already included.

In [5]:

p.subs(r == 3).check_feasible(skip=["family"])

We now compute the triple intersection numbers with respect to three vertices $x, y, z$ such that $x$ is in relation $1$ with $y$ and $z$, and $y$ is in relation $2$ with $z$. Note that we have $p^2_{11} = r(r+2)(r^2-1)/4 > 0$ for all $r \ge 2$, so such triples must exist. The parameters $\alpha, \beta, \gamma, \delta$ will denote the number of vertices in relations $(1, 3, 1), (2, 3, 3), (3, 1, 1), (3, 3, 3)$, respectively, to $x, y, z$.

In [6]:

S112 = p.tripleEquations(1, 1, 2, params={'alpha': (1, 3, 1), 'beta': (2, 3, 3), 'gamma': (3, 1, 1), 'delta': (3, 3, 3)})
S112[1, 2, 3]

Out[6]:

$\newcommand{\Bold}[1]{\mathbf{#1}}-\frac{1}{2} \, r^{4} + 2 \, r^{2} + \alpha + 3 \, \beta + 4 \, \delta - \gamma$

We see that the above triple intersection number can only be integral when $r$ is even. We thus conclude that a $Q$-polynomial association scheme with Krein array $\{2r^2-1, 2r^2-2, r^2+1; 1, 2, r^2-1\}$ and $r$ odd does not exist.