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2\}$ and $r \equiv 3 \pmod{4}$¶

We show that a $Q$-polynomial association scheme with Krein array \begin{multline*} \{(r^2+1)/2, (r^2-1)/2, (r^2+1)^2/(2r(r+1)), (r-1)(r^2+1)/(4r), (r^2+1)/(2r); \\ 1, (r-1)(r^2+1)/(2r(r+1)), (r+1)(r^2 + 1)/(4r), (r-1)(r^2+1)/(2r), (r^2+1)/2\} \end{multline*} and $r \equiv 3 \pmod{4}$ does not exist. For $r \equiv 1 \pmod{4}$ a prime power, this Krein array is realized by a scheme derived by Moorhouse and Williford from a double cover of the $C_2(r)$ dual polar graph.

In [1]:

%display latex
import drg

Such a scheme would have $2(r+1)(r^2+1)$ vertices.

In [2]:

r = var("r")
p = drg.QPolyParameters([(r^2+1)/2, (r^2-1)/2, (r^2+1)^2/(2*r*(r+1)), (r-1)*(r^2+1)/(4*r), (r^2+1)/(2*r)],
                        [1, (r-1)*(r^2+1)/(2*r*(r+1)), (r+1)*(r^2 + 1)/(4*r), (r-1)*(r^2+1)/(2*r), (r^2+1)/2])
p.order(factor=True, simplify=True)

Out[2]:

$\newcommand{\Bold}[1]{\mathbf{#1}}2 \, {\left(r^{2} + 1\right)} {\left(r + 1\right)}$

This scheme is $Q$-bipartite, so it has $q^h_{ij} = 0$ whenever $h+i+j$ is odd, or $h, i, j$ do not satisfy the triangle inequality. Additionally, we have $q^4_{55} = q^5_{45} = q^5_{54} = 0$.

In [3]:

p.kreinParameters(factor=True, simplify=2)

Out[3]:

$\newcommand{\Bold}[1]{\mathbf{#1}}\begin{aligned}0: &\ \left(\begin{array}{rrrrrr} 1 & 0 & 0 & 0 & 0 & 0 \\ 0 & \frac{1}{2} \, r^{2} + \frac{1}{2} & 0 & 0 & 0 & 0 \\ 0 & 0 & \frac{1}{2} \, r^{3} + r^{2} + \frac{1}{2} \, r & 0 & 0 & 0 \\ 0 & 0 & 0 & r^{3} + r & 0 & 0 \\ 0 & 0 & 0 & 0 & \frac{1}{2} \, r^{3} + \frac{1}{2} \, r & 0 \\ 0 & 0 & 0 & 0 & 0 & \frac{1}{2} \, r^{2} + \frac{1}{2} \end{array}\right) \\ 1: &\ \left(\begin{array}{rrrrrr} 0 & 1 & 0 & 0 & 0 & 0 \\ 1 & 0 & \frac{1}{2} \, r^{2} - \frac{1}{2} & 0 & 0 & 0 \\ 0 & \frac{1}{2} \, r^{2} - \frac{1}{2} & 0 & \frac{1}{2} \, r^{3} + \frac{1}{2} \, r^{2} + \frac{1}{2} \, r + \frac{1}{2} & 0 & 0 \\ 0 & 0 & \frac{1}{2} \, r^{3} + \frac{1}{2} \, r^{2} + \frac{1}{2} \, r + \frac{1}{2} & 0 & \frac{1}{2} \, r^{3} - \frac{1}{2} \, r^{2} + \frac{1}{2} \, r - \frac{1}{2} & 0 \\ 0 & 0 & 0 & \frac{1}{2} \, r^{3} - \frac{1}{2} \, r^{2} + \frac{1}{2} \, r - \frac{1}{2} & 0 & \frac{1}{2} \, r^{2} + \frac{1}{2} \\ 0 & 0 & 0 & 0 & \frac{1}{2} \, r^{2} + \frac{1}{2} & 0 \end{array}\right) \\ 2: &\ \left(\begin{array}{rrrrrr} 0 & 0 & 1 & 0 & 0 & 0 \\ 0 & \frac{r^{3} - r^{2} + r - 1}{2 \, {\left(r^{2} + r\right)}} & 0 & \frac{r^{4} + 2 \, r^{2} + 1}{2 \, {\left(r^{2} + r\right)}} & 0 & 0 \\ 1 & 0 & \frac{r^{4} + 4 \, r^{3} - 4 \, r - 1}{4 \, r} & 0 & \frac{r^{4} + 2 \, r^{2} + 1}{4 \, r} & 0 \\ 0 & \frac{r^{4} + 2 \, r^{2} + 1}{2 \, {\left(r^{2} + r\right)}} & 0 & \frac{r^{5} + r^{3} - r^{2} - 1}{r^{2} + r} & 0 & \frac{r^{4} + 2 \, r^{2} + 1}{2 \, {\left(r^{2} + r\right)}} \\ 0 & 0 & \frac{r^{4} + 2 \, r^{2} + 1}{4 \, r} & 0 & \frac{r^{4} - 1}{4 \, r} & 0 \\ 0 & 0 & 0 & \frac{r^{4} + 2 \, r^{2} + 1}{2 \, {\left(r^{2} + r\right)}} & 0 & \frac{r^{3} - r^{2} + r - 1}{2 \, {\left(r^{2} + r\right)}} \end{array}\right) \\ 3: &\ \left(\begin{array}{rrrrrr} 0 & 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & \frac{r^{3} + r^{2} + r + 1}{4 \, r} & 0 & \frac{r^{3} - r^{2} + r - 1}{4 \, r} & 0 \\ 0 & \frac{r^{3} + r^{2} + r + 1}{4 \, r} & 0 & \frac{r^{4} + r^{3} - r - 1}{2 \, r} & 0 & \frac{r^{3} + r^{2} + r + 1}{4 \, r} \\ 1 & 0 & \frac{r^{4} + r^{3} - r - 1}{2 \, r} & 0 & \frac{r^{4} - r^{3} + 2 \, r^{2} - r + 1}{2 \, r} & 0 \\ 0 & \frac{r^{3} - r^{2} + r - 1}{4 \, r} & 0 & \frac{r^{4} - r^{3} + 2 \, r^{2} - r + 1}{2 \, r} & 0 & \frac{r^{3} - r^{2} + r - 1}{4 \, r} \\ 0 & 0 & \frac{r^{3} + r^{2} + r + 1}{4 \, r} & 0 & \frac{r^{3} - r^{2} + r - 1}{4 \, r} & 0 \end{array}\right) \\ 4: &\ \left(\begin{array}{rrrrrr} 0 & 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & \frac{r^{3} - r^{2} + r - 1}{2 \, r} & 0 & \frac{r^{2} + 1}{2 \, r} \\ 0 & 0 & \frac{r^{4} + 2 \, r^{3} + 2 \, r^{2} + 2 \, r + 1}{4 \, r} & 0 & \frac{r^{4} + 2 \, r^{3} - 2 \, r - 1}{4 \, r} & 0 \\ 0 & \frac{r^{3} - r^{2} + r - 1}{2 \, r} & 0 & \frac{r^{4} - r^{3} + 2 \, r^{2} - r + 1}{r} & 0 & \frac{r^{3} - r^{2} + r - 1}{2 \, r} \\ 1 & 0 & \frac{r^{4} + 2 \, r^{3} - 2 \, r - 1}{4 \, r} & 0 & \frac{r^{4} - 2 \, r^{3} + 2 \, r^{2} - 2 \, r + 1}{4 \, r} & 0 \\ 0 & \frac{r^{2} + 1}{2 \, r} & 0 & \frac{r^{3} - r^{2} + r - 1}{2 \, r} & 0 & 0 \end{array}\right) \\ 5: &\ \left(\begin{array}{rrrrrr} 0 & 0 & 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 & \frac{1}{2} \, r^{2} + \frac{1}{2} & 0 \\ 0 & 0 & 0 & \frac{1}{2} \, r^{3} + \frac{1}{2} \, r^{2} + \frac{1}{2} \, r + \frac{1}{2} & 0 & \frac{1}{2} \, r^{2} - \frac{1}{2} \\ 0 & 0 & \frac{1}{2} \, r^{3} + \frac{1}{2} \, r^{2} + \frac{1}{2} \, r + \frac{1}{2} & 0 & \frac{1}{2} \, r^{3} - \frac{1}{2} \, r^{2} + \frac{1}{2} \, r - \frac{1}{2} & 0 \\ 0 & \frac{1}{2} \, r^{2} + \frac{1}{2} & 0 & \frac{1}{2} \, r^{3} - \frac{1}{2} \, r^{2} + \frac{1}{2} \, r - \frac{1}{2} & 0 & 0 \\ 1 & 0 & \frac{1}{2} \, r^{2} - \frac{1}{2} & 0 & 0 & 0 \end{array}\right) \\\end{aligned}$

The intersection numbers can be checked to be nonnegative and integral for all odd values of $r \ge 3$.

In [4]:

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

Out[4]:

$\newcommand{\Bold}[1]{\mathbf{#1}}\begin{aligned}0: &\ \left(\begin{array}{rrrrrr} 1 & 0 & 0 & 0 & 0 & 0 \\ 0 & {\left(r + 1\right)} r & 0 & 0 & 0 & 0 \\ 0 & 0 & r^{3} & 0 & 0 & 0 \\ 0 & 0 & 0 & r^{3} & 0 & 0 \\ 0 & 0 & 0 & 0 & {\left(r + 1\right)} r & 0 \\ 0 & 0 & 0 & 0 & 0 & 1 \end{array}\right) \\ 1: &\ \left(\begin{array}{rrrrrr} 0 & 1 & 0 & 0 & 0 & 0 \\ 1 & \frac{1}{2} \, r - \frac{1}{2} & r^{2} & 0 & \frac{1}{2} \, r - \frac{1}{2} & 0 \\ 0 & r^{2} & \frac{1}{2} \, {\left(r - 1\right)} r^{2} & \frac{1}{2} \, {\left(r - 1\right)} r^{2} & 0 & 0 \\ 0 & 0 & \frac{1}{2} \, {\left(r - 1\right)} r^{2} & \frac{1}{2} \, {\left(r - 1\right)} r^{2} & r^{2} & 0 \\ 0 & \frac{1}{2} \, r - \frac{1}{2} & 0 & r^{2} & \frac{1}{2} \, r - \frac{1}{2} & 1 \\ 0 & 0 & 0 & 0 & 1 & 0 \end{array}\right) \\ 2: &\ \left(\begin{array}{rrrrrr} 0 & 0 & 1 & 0 & 0 & 0 \\ 0 & r + 1 & \frac{1}{2} \, {\left(r + 1\right)} {\left(r - 1\right)} & \frac{1}{2} \, {\left(r + 1\right)} {\left(r - 1\right)} & 0 & 0 \\ 1 & \frac{1}{2} \, {\left(r + 1\right)} {\left(r - 1\right)} & \frac{1}{2} \, {\left(r + 1\right)} {\left(r - 1\right)} r & \frac{1}{2} \, {\left(r - 1\right)}^{2} r & \frac{1}{2} \, {\left(r + 1\right)} {\left(r - 1\right)} & 0 \\ 0 & \frac{1}{2} \, {\left(r + 1\right)} {\left(r - 1\right)} & \frac{1}{2} \, {\left(r - 1\right)}^{2} r & \frac{1}{2} \, {\left(r + 1\right)} {\left(r - 1\right)} r & \frac{1}{2} \, {\left(r + 1\right)} {\left(r - 1\right)} & 1 \\ 0 & 0 & \frac{1}{2} \, {\left(r + 1\right)} {\left(r - 1\right)} & \frac{1}{2} \, {\left(r + 1\right)} {\left(r - 1\right)} & r + 1 & 0 \\ 0 & 0 & 0 & 1 & 0 & 0 \end{array}\right) \\ 3: &\ \left(\begin{array}{rrrrrr} 0 & 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & \frac{1}{2} \, {\left(r + 1\right)} {\left(r - 1\right)} & \frac{1}{2} \, {\left(r + 1\right)} {\left(r - 1\right)} & r + 1 & 0 \\ 0 & \frac{1}{2} \, {\left(r + 1\right)} {\left(r - 1\right)} & \frac{1}{2} \, {\left(r - 1\right)}^{2} r & \frac{1}{2} \, {\left(r + 1\right)} {\left(r - 1\right)} r & \frac{1}{2} \, {\left(r + 1\right)} {\left(r - 1\right)} & 1 \\ 1 & \frac{1}{2} \, {\left(r + 1\right)} {\left(r - 1\right)} & \frac{1}{2} \, {\left(r + 1\right)} {\left(r - 1\right)} r & \frac{1}{2} \, {\left(r - 1\right)}^{2} r & \frac{1}{2} \, {\left(r + 1\right)} {\left(r - 1\right)} & 0 \\ 0 & r + 1 & \frac{1}{2} \, {\left(r + 1\right)} {\left(r - 1\right)} & \frac{1}{2} \, {\left(r + 1\right)} {\left(r - 1\right)} & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 & 0 \end{array}\right) \\ 4: &\ \left(\begin{array}{rrrrrr} 0 & 0 & 0 & 0 & 1 & 0 \\ 0 & \frac{1}{2} \, r - \frac{1}{2} & 0 & r^{2} & \frac{1}{2} \, r - \frac{1}{2} & 1 \\ 0 & 0 & \frac{1}{2} \, {\left(r - 1\right)} r^{2} & \frac{1}{2} \, {\left(r - 1\right)} r^{2} & r^{2} & 0 \\ 0 & r^{2} & \frac{1}{2} \, {\left(r - 1\right)} r^{2} & \frac{1}{2} \, {\left(r - 1\right)} r^{2} & 0 & 0 \\ 1 & \frac{1}{2} \, r - \frac{1}{2} & r^{2} & 0 & \frac{1}{2} \, r - \frac{1}{2} & 0 \\ 0 & 1 & 0 & 0 & 0 & 0 \end{array}\right) \\ 5: &\ \left(\begin{array}{rrrrrr} 0 & 0 & 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 & {\left(r + 1\right)} r & 0 \\ 0 & 0 & 0 & r^{3} & 0 & 0 \\ 0 & 0 & r^{3} & 0 & 0 & 0 \\ 0 & {\left(r + 1\right)} r & 0 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 & 0 & 0 \end{array}\right) \\\end{aligned}$

By the absolute bound for $(1, 1)$, such a scheme is only feasible when $r \ge 5$.

In [5]:

p.check_absoluteBound(expand=True, factor=True)

Out[5]:

$\newcommand{\Bold}[1]{\mathbf{#1}}\left\{{\left(1, 1\right) : \frac{1}{8} \, {\left(r^{2} + 1\right)} {\left(r + 1\right)} {\left(r - 5\right)}}, {\left(1, 2\right) : \frac{1}{4} \, {\left(r^{3} + 2 \, r^{2} - 3 \, r - 2\right)} {\left(r^{2} + 1\right)}}, {\left(1, 3\right) : \frac{1}{2} \, {\left(r^{4} - 2 \, r - 1\right)} r}, {\left(1, 4\right) : \frac{1}{4} \, {\left(r^{2} + 1\right)} {\left(r + 1\right)}^{2} {\left(r - 2\right)}}, {\left(1, 5\right) : \frac{1}{4} \, {\left(r^{2} + 1\right)} {\left(r - 1\right)}^{2}}, {\left(2, 2\right) : \frac{1}{8} \, {\left(r^{3} + 3 \, r^{2} + 2 \, r - 8\right)} {\left(r^{2} + 1\right)} {\left(r + 1\right)}}, {\left(2, 3\right) : \frac{1}{2} \, {\left(r^{2} + 2 \, r + 2\right)} {\left(r^{2} + 1\right)} {\left(r + 1\right)} {\left(r - 1\right)}}, {\left(2, 4\right) : \frac{1}{4} \, {\left(r^{5} + 2 \, r^{4} + 2 \, r^{3} - 2 \, r^{2} - 3 \, r - 4\right)} r}, {\left(2, 5\right) : \frac{1}{4} \, {\left(r^{3} + 2 \, r^{2} - 3 \, r - 2\right)} {\left(r^{2} + 1\right)}}, {\left(3, 3\right) : \frac{1}{2} \, {\left(r^{4} + r^{2} - r - 2\right)} {\left(r^{2} + 1\right)}}, {\left(3, 4\right) : \frac{1}{2} \, {\left(r^{4} + r^{2} - 2 \, r - 2\right)} {\left(r^{2} + 1\right)}}, {\left(3, 5\right) : \frac{1}{2} \, {\left(r^{4} - 2 \, r - 1\right)} r}, {\left(4, 4\right) : \frac{1}{8} \, {\left(r^{2} + r + 4\right)} {\left(r^{2} + 1\right)} {\left(r + 1\right)} {\left(r - 2\right)}}, {\left(4, 5\right) : \frac{1}{4} \, {\left(r^{2} + 1\right)} {\left(r + 1\right)}^{2} {\left(r - 2\right)}}, {\left(5, 5\right) : \frac{1}{8} \, {\left(r^{2} + 1\right)} {\left(r + 1\right)} {\left(r - 5\right)}}\right\}$

Let us check that the smallest admissible case with $r \equiv 3 \pmod{4}$, namely $r = 7$, satisfies the known feasibility conditions. We skip the family nonexistence check since the Krein array of the members of family with $r \equiv 3 \pmod{4}$ is already included.

In [6]:

p.subs(r == 7).check_feasible(skip=["family"])

We now compute the triple intersection numbers with respect to three vertices $x, y, z$ mutually in relation $1$. Note that we have $p^1_{11} = (r-1)/2 > 0$ for all $r \ge 5$, so such triples must exist.

In [7]:

p.dualEigenmatrix(expand=True, factor=True, simplify=True)
p.tripleEquations(1, 1, 1)

Out[7]:

$\newcommand{\Bold}[1]{\mathbf{#1}}\begin{aligned}0: &\ \left(\begin{array}{rrrrrr} 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \end{array}\right) \\ 1: &\ \left(\begin{array}{rrrrrr} 0 & 1 & 0 & 0 & 0 & 0 \\ 1 & \frac{1}{4} \, r - \frac{5}{4} & 0 & 0 & \frac{1}{4} \, r - \frac{1}{4} & 0 \\ 0 & 0 & r^{2} & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & \frac{1}{4} \, r - \frac{1}{4} & 0 & 0 & \frac{1}{4} \, r - \frac{1}{4} & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \end{array}\right) \\ 2: &\ \left(\begin{array}{rrrrrr} 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & r^{2} & 0 & 0 & 0 \\ 0 & r^{2} & \frac{1}{4} \, r^{3} - \frac{5}{4} \, r^{2} & \frac{1}{4} \, r^{3} - \frac{1}{4} \, r^{2} & 0 & 0 \\ 0 & 0 & \frac{1}{4} \, r^{3} - \frac{1}{4} \, r^{2} & \frac{1}{4} \, r^{3} - \frac{1}{4} \, r^{2} & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \end{array}\right) \\ 3: &\ \left(\begin{array}{rrrrrr} 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & \frac{1}{4} \, r^{3} - \frac{1}{4} \, r^{2} & \frac{1}{4} \, r^{3} - \frac{1}{4} \, r^{2} & 0 & 0 \\ 0 & 0 & \frac{1}{4} \, r^{3} - \frac{1}{4} \, r^{2} & \frac{1}{4} \, r^{3} - \frac{5}{4} \, r^{2} & r^{2} & 0 \\ 0 & 0 & 0 & r^{2} & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \end{array}\right) \\ 4: &\ \left(\begin{array}{rrrrrr} 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & \frac{1}{4} \, r - \frac{1}{4} & 0 & 0 & \frac{1}{4} \, r - \frac{1}{4} & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & r^{2} & 0 & 0 \\ 0 & \frac{1}{4} \, r - \frac{1}{4} & 0 & 0 & \frac{1}{4} \, r - \frac{5}{4} & 1 \\ 0 & 0 & 0 & 0 & 1 & 0 \end{array}\right) \\ 5: &\ \left(\begin{array}{rrrrrr} 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \end{array}\right) \\\end{aligned}$

We see that there is a single solution, which can only be integral if $r \equiv 1 \pmod{4}$. We thus conclude that a $Q$-polynomial association scheme with Krein array \begin{multline*} \{(r^2+1)/2, (r^2-1)/2, (r^2+1)^2/(2r(r+1)), (r-1)(r^2+1)/(4r), (r^2+1)/(2r); \\ 1, (r-1)(r^2+1)/(2r(r+1)), (r+1)(r^2 + 1)/(4r), (r-1)(r^2+1)/(2r), (r^2+1)/2\} \end{multline*} and $r \equiv 3 \pmod{4}$ does not exist.