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11, 24\}$¶

We will show that a $Q$-polynomial association scheme with Krein array $\{24, 20, 36/11; 1, 30/11, 24\}$ does not exist.

In [1]:

%display latex
import drg

Such a scheme would have $225$ vertices.

In [2]:

p = drg.QPolyParameters([24, 20, 36/11], [1, 30/11, 24])
p.order()

Out[2]:

$\newcommand{\Bold}[1]{\mathbf{#1}}225$

The scheme has two $Q$-polynomial orderings $(0, 1, 2, 3)$ and $(0, 3, 2, 1)$, so we have $q^3_{11} = q^1_{13} = q^1_{31} = q^1_{33} = q^3_{13} = q^3_{31} = 0$.

In [3]:

p.kreinParameters()

Out[3]:

$\newcommand{\Bold}[1]{\mathbf{#1}}\begin{aligned}0: &\ \left(\begin{array}{rrrr} 1 & 0 & 0 & 0 \\ 0 & 24 & 0 & 0 \\ 0 & 0 & 176 & 0 \\ 0 & 0 & 0 & 24 \end{array}\right) \\ 1: &\ \left(\begin{array}{rrrr} 0 & 1 & 0 & 0 \\ 1 & 3 & 20 & 0 \\ 0 & 20 & 132 & 24 \\ 0 & 0 & 24 & 0 \end{array}\right) \\ 2: &\ \left(\begin{array}{rrrr} 0 & 0 & 1 & 0 \\ 0 & \frac{30}{11} & 18 & \frac{36}{11} \\ 1 & 18 & 139 & 18 \\ 0 & \frac{36}{11} & 18 & \frac{30}{11} \end{array}\right) \\ 3: &\ \left(\begin{array}{rrrr} 0 & 0 & 0 & 1 \\ 0 & 0 & 24 & 0 \\ 0 & 24 & 132 & 20 \\ 1 & 0 & 20 & 3 \end{array}\right) \\\end{aligned}$

The intersection numbers are integral and nonnegative.

In [4]:

p.pTable()

Out[4]:

$\newcommand{\Bold}[1]{\mathbf{#1}}\begin{aligned}0: &\ \left(\begin{array}{rrrr} 1 & 0 & 0 & 0 \\ 0 & 88 & 0 & 0 \\ 0 & 0 & 88 & 0 \\ 0 & 0 & 0 & 48 \end{array}\right) \\ 1: &\ \left(\begin{array}{rrrr} 0 & 1 & 0 & 0 \\ 1 & 43 & 32 & 12 \\ 0 & 32 & 32 & 24 \\ 0 & 12 & 24 & 12 \end{array}\right) \\ 2: &\ \left(\begin{array}{rrrr} 0 & 0 & 1 & 0 \\ 0 & 32 & 32 & 24 \\ 1 & 32 & 43 & 12 \\ 0 & 24 & 12 & 12 \end{array}\right) \\ 3: &\ \left(\begin{array}{rrrr} 0 & 0 & 0 & 1 \\ 0 & 22 & 44 & 22 \\ 0 & 44 & 22 & 22 \\ 1 & 22 & 22 & 3 \end{array}\right) \\\end{aligned}$

The parameters do not exceed the absolute bound.

In [5]:

p.check_absoluteBound()

Out[5]:

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

Let $w, x, y, z$ be vertices such that $z$ is in relation $1$ with $x$ and $y$, and $w, x, y$ are mutually in relation $3$. Note that we have $p^3_{11} = 22$ and $p^3_{33} = 3$, so such vertices must exist. We first compute the triple intersection numbers with respect to $x, y, z$. The parameters $\alpha$ and $\beta$ will denote the numbers of vertices in relations $(2, 1, 1)$ and $(2, 2, 1)$, respectively, to $x, y, z$.

In [6]:

p.tripleEquations(3, 1, 1, params={"alpha": (2, 1, 1), "beta": (2, 2, 1)})

Out[6]:

$\newcommand{\Bold}[1]{\mathbf{#1}}\begin{aligned}0: &\ \left(\begin{array}{rrrr} 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \end{array}\right) \\ 1: &\ \left(\begin{array}{rrrr} 0 & 0 & 0 & 0 \\ 1 & -\frac{7}{2} \, \alpha - \beta + 97 & \alpha - 16 & \frac{5}{2} \, \alpha + \beta - 60 \\ 0 & \alpha & \beta + 8 & -\alpha - \beta + 36 \\ 0 & \frac{5}{2} \, \alpha + \beta - 54 & -\alpha - \beta + 40 & -\frac{3}{2} \, \alpha + 36 \end{array}\right) \\ 2: &\ \left(\begin{array}{rrrr} 0 & 0 & 0 & 0 \\ 0 & \alpha & \beta + 8 & -\alpha - \beta + 36 \\ 0 & \beta & -\alpha - \frac{7}{2} \, \beta + 52 & \alpha + \frac{5}{2} \, \beta - 30 \\ 0 & -\alpha - \beta + 32 & \alpha + \frac{5}{2} \, \beta - 28 & -\frac{3}{2} \, \beta + 18 \end{array}\right) \\ 3: &\ \left(\begin{array}{rrrr} 0 & 1 & 0 & 0 \\ 0 & \frac{5}{2} \, \alpha + \beta - 54 & -\alpha - \beta + 40 & -\frac{3}{2} \, \alpha + 36 \\ 0 & -\alpha - \beta + 32 & \alpha + \frac{5}{2} \, \beta - 28 & -\frac{3}{2} \, \beta + 18 \\ 0 & -\frac{3}{2} \, \alpha + 33 & -\frac{3}{2} \, \beta + 12 & \frac{3}{2} \, \alpha + \frac{3}{2} \, \beta - 42 \end{array}\right) \\\end{aligned}$

From $[1\ 1\ 3] \ge 0$ and $[3\ 3\ 2] \ge 0$, it follows that $60 - 5\alpha/2 \le \beta \le 8$. Using this and $[3\ 3\ 1] \ge 0$, we obtain $20.8 \le a \le 22$, but since $\alpha$ and $\beta$ must clearly be even integers, it follows that there are two solutions with $\alpha = 22$ and $\beta \in \{6, 8\}$. In both cases we have $[3\ 3\ 1] = 0$, implying that $w$ cannot be in relation $1$ with $z$ for any choice of $w, x, y, z$ as above.

We now compute the triple intersection numbers with respect to $w, x, y$. The parameters $\gamma$ and $\delta$ will denote the numbers of vertices in relations $(1, 1, 2)$ and $(1, 2, 2)$, respectively, to $w, x, y$.

In [7]:

p.tripleEquations(3, 3, 3, params={"gamma": (1, 1, 2), "delta": (1, 2, 2)})

Out[7]:

$\newcommand{\Bold}[1]{\mathbf{#1}}\begin{aligned}0: &\ \left(\begin{array}{rrrr} 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \end{array}\right) \\ 1: &\ \left(\begin{array}{rrrr} 0 & 0 & 0 & 0 \\ 0 & -\delta - \frac{7}{2} \, \gamma + 66 & \gamma & \delta + \frac{5}{2} \, \gamma - 44 \\ 0 & \gamma & \delta & -\delta - \gamma + 44 \\ 0 & \delta + \frac{5}{2} \, \gamma - 44 & -\delta - \gamma + 44 & -\frac{3}{2} \, \gamma + 22 \end{array}\right) \\ 2: &\ \left(\begin{array}{rrrr} 0 & 0 & 0 & 0 \\ 0 & \gamma & \delta & -\delta - \gamma + 44 \\ 0 & \delta & -\frac{7}{2} \, \delta - \gamma + 66 & \frac{5}{2} \, \delta + \gamma - 44 \\ 0 & -\delta - \gamma + 44 & \frac{5}{2} \, \delta + \gamma - 44 & -\frac{3}{2} \, \delta + 22 \end{array}\right) \\ 3: &\ \left(\begin{array}{rrrr} 0 & 0 & 0 & 1 \\ 0 & \delta + \frac{5}{2} \, \gamma - 44 & -\delta - \gamma + 44 & -\frac{3}{2} \, \gamma + 22 \\ 0 & -\delta - \gamma + 44 & \frac{5}{2} \, \delta + \gamma - 44 & -\frac{3}{2} \, \delta + 22 \\ 1 & -\frac{3}{2} \, \gamma + 22 & -\frac{3}{2} \, \delta + 22 & \frac{3}{2} \, \delta + \frac{3}{2} \, \gamma - 42 \end{array}\right) \\\end{aligned}$

From $[1\ 3\ 3] \ge 0$ and $[2\ 3\ 3] \ge 0$, it follows that $\gamma, \delta \le 44/3$, while $[3\ 3\ 3] \ge 0$ implies $\gamma + \delta \ge 28$. Since $\gamma$ and $\delta$ must clearly be even integers, it follows that there is only one solution with $\gamma = \delta = 14$ and therefore $[1\ 1\ 1] = 3$. This would imply the existence of a vertex $z$ as above that is in relation $1$ with $w$ - a contradiction! We thus conclude that a $Q$-polynomial association scheme with Krein array $\{24, 20, 36/11; 1, 30/11, 24\}$ does not exist.