Introduction
A complex Hadamard matrix is regular when the entrywise quotient of any 2 rows splits into prime cycles: rotated copies of the full set of p p p p p p
Equivalence consists of row and column permutations and unit scalings; conjugation and transposition are separate operations. At order 7 the quotient types are 7, 5 + 2, and 3 + 2 + 2. There are 6 affine 3-row families and 45 isolated 3-row classes. Completing them gives precisely the Fourier matrix F 7 F_7 F 7 ω = e 2 π i / 3 \omega=e^{2\pi i/3} ω = e 2 π i /3 q ∈ T q\in\mathbb{T} q ∈ T
P 7 q = ( − q q ω 1 ω 1 ω q − q ω 1 1 ω ω ω ω − ω 1 ω ω 1 1 1 1 − 1 ω ω ω ω 1 ω ω − q ‾ ω q ‾ ω 1 1 ω ω ω q ‾ ω − q ‾ ω 1 ω ω 1 ω 1 1 − 1 ) R 7 q = ( 1 1 1 1 1 1 1 1 1 1 ω − 1 − 1 ω 2 1 − ω 2 ω 2 ω 2 q − q ω 1 ω − 1 1 q ω 2 − q ω 2 ω 2 1 − 1 − ω − 1 − q q − ω 2 1 ω 2 ω q ω q − q ω − q 1 ω 2 ω − q ω q ω − q q ) P_7^q=\begin{pmatrix}
-q & q & \omega& 1 & \omega& 1 & \omega\\
q & -q & \omega& 1 & 1 & \omega& \omega\\
\omega& \omega& -\omega& 1 & \omega& \omega& 1\\
1 & 1 & 1 & -1 & \omega& \omega& \omega\\
\omega& 1 & \omega& \omega& -\overline{q}\omega& \overline{q}\omega& 1\\
1 & \omega& \omega& \omega& \overline{q}\omega& -\overline{q}\omega& 1\\
\omega& \omega& 1 & \omega& 1 & 1 & -1
\end{pmatrix}
\qquad
R_7^q=\begin{pmatrix}
1 & 1 & 1 & 1 & 1 & 1 & 1\\
1 & 1 & 1 & \omega& -1 & -1 & \omega^2\\
1 & -\omega^2 & \omega^2 & \omega^2 & q & -q & \omega\\
1 & \omega& -1 & 1 & q\omega^2 & -q\omega^2 & \omega^2\\
1 & -1 & -\omega& -1 & -q & q & -\omega^2\\
1 & \omega^2 & \omega& q\omega& q & -q\omega& -q\\
1 & \omega^2 & \omega& -q\omega& q\omega& -q & q
\end{pmatrix} P 7 q = − q q ω 1 ω 1 ω q − q ω 1 1 ω ω ω ω − ω 1 ω ω 1 1 1 1 − 1 ω ω ω ω 1 ω ω − q ω q ω 1 1 ω ω ω q ω − q ω 1 ω ω 1 ω 1 1 − 1 R 7 q = 1 1 1 1 1 1 1 1 1 − ω 2 ω − 1 ω 2 ω 2 1 1 ω 2 − 1 − ω ω ω 1 ω ω 2 1 − 1 q ω − q ω 1 − 1 q q ω 2 − q q q ω 1 − 1 − q − q ω 2 q − q ω − q 1 ω 2 ω ω 2 − ω 2 − q q The family R 7 q R_7^q R 7 q P 7 q P_7^q P 7 q
At order 8 the possible cycle lengths in a row quotient are A = 2 + 2 + 2 + 2 A=2+2+2+2 A = 2 + 2 + 2 + 2 B = 3 + 3 + 2 B=3+3+2 B = 3 + 3 + 2 C = 5 + 3 C=5+3 C = 5 + 3 A , B A,B A , B C C C
Table 1.
The 9 continuous families are displayed below. All parameters range independently over the unit circle; the transpose families are written out explicitly.
F 8 , 5 a ( a , b , c , d , e ) = ( 1 1 1 1 1 1 1 1 1 − 1 − b d b d − b d b d 1 − 1 1 − 1 c − b d a d b d a − d − 1 1 c 1 1 − 1 − 1 − 1 − 1 1 1 1 − 1 c b d a − d − b d a d − 1 1 c 1 − 1 b d − b d b d − b d 1 − 1 1 1 c e a e b − e − a e b − 1 − 1 c 1 1 c − e − a e b e a e b − 1 − 1 c ) F_{8,5a}(a,b,c,d,e)=
\begin{pmatrix}
1&1&1&1&1&1&1&1\\
1&-1&-bd&bd&-bd&bd&1&-1\\
1&-\frac{1}{c}&-\frac{bd}{a}&d&\frac{bd}{a}&-d&-1&\frac{1}{c}\\
1&1&-1&-1&-1&-1&1&1\\
1&-\frac{1}{c}&\frac{bd}{a}&-d&-\frac{bd}{a}&d&-1&\frac{1}{c}\\
1&-1&bd&-bd&bd&-bd&1&-1\\
1&\frac{1}{c}&e&\frac{ae}{b}&-e&-\frac{ae}{b}&-1&-\frac{1}{c}\\
1&\frac{1}{c}&-e&-\frac{ae}{b}&e&\frac{ae}{b}&-1&-\frac{1}{c}
\end{pmatrix} F 8 , 5 a ( a , b , c , d , e ) = 1 1 1 1 1 1 1 1 1 − 1 − c 1 1 − c 1 − 1 c 1 c 1 1 − b d − a b d − 1 a b d b d e − e 1 b d d − 1 − d − b d b a e − b a e 1 − b d a b d − 1 − a b d b d − e e 1 b d − d − 1 d − b d − b a e b a e 1 1 − 1 1 − 1 1 − 1 − 1 1 − 1 c 1 1 c 1 − 1 − c 1 − c 1 F 8 , 5 b ( a , b , c , d , e ) = ( 1 1 1 1 1 1 1 1 1 − 1 − b d b d − b d b d 1 − 1 1 − 1 c − b d a d b d a − d − 1 1 c 1 1 c − b d a − d b d a d − 1 − 1 c 1 1 − 1 − 1 − 1 − 1 1 1 1 − 1 b d − b d b d − b d 1 − 1 1 e b d a c d e − b d a − c d e − 1 − e 1 − e b d a − c d e − b d a c d e − 1 e ) F_{8,5b}(a,b,c,d,e)=
\begin{pmatrix}
1&1&1&1&1&1&1&1\\
1&-1&-bd&bd&-bd&bd&1&-1\\
1&-\frac{1}{c}&-\frac{bd}{a}&d&\frac{bd}{a}&-d&-1&\frac{1}{c}\\
1&\frac{1}{c}&-\frac{bd}{a}&-d&\frac{bd}{a}&d&-1&-\frac{1}{c}\\
1&1&-1&-1&-1&-1&1&1\\
1&-1&bd&-bd&bd&-bd&1&-1\\
1&e&\frac{bd}{a}&cde&-\frac{bd}{a}&-cde&-1&-e\\
1&-e&\frac{bd}{a}&-cde&-\frac{bd}{a}&cde&-1&e
\end{pmatrix} F 8 , 5 b ( a , b , c , d , e ) = 1 1 1 1 1 1 1 1 1 − 1 − c 1 c 1 1 − 1 e − e 1 − b d − a b d − a b d − 1 b d a b d a b d 1 b d d − d − 1 − b d c d e − c d e 1 − b d a b d a b d − 1 b d − a b d − a b d 1 b d − d d − 1 − b d − c d e c d e 1 1 − 1 − 1 1 1 − 1 − 1 1 − 1 c 1 − c 1 1 − 1 − e e F 8 , 5 c ( a , b , c , d , e ) = ( 1 1 1 1 1 1 1 1 1 − 1 a d − a d − b d b d 1 − 1 1 − 1 c d d − d − d − 1 1 c 1 1 − 1 − 1 − 1 − 1 1 1 1 − 1 c − d − d d d − 1 1 c 1 − 1 − a d a d b d − b d 1 − 1 1 1 c e − e b e a − b e a − 1 − 1 c 1 1 c − e e − b e a b e a − 1 − 1 c ) F_{8,5c}(a,b,c,d,e)=
\begin{pmatrix}
1&1&1&1&1&1&1&1\\
1&-1&ad&-ad&-bd&bd&1&-1\\
1&-\frac{1}{c}&d&d&-d&-d&-1&\frac{1}{c}\\
1&1&-1&-1&-1&-1&1&1\\
1&-\frac{1}{c}&-d&-d&d&d&-1&\frac{1}{c}\\
1&-1&-ad&ad&bd&-bd&1&-1\\
1&\frac{1}{c}&e&-e&\frac{be}{a}&-\frac{be}{a}&-1&-\frac{1}{c}\\
1&\frac{1}{c}&-e&e&-\frac{be}{a}&\frac{be}{a}&-1&-\frac{1}{c}
\end{pmatrix} F 8 , 5 c ( a , b , c , d , e ) = 1 1 1 1 1 1 1 1 1 − 1 − c 1 1 − c 1 − 1 c 1 c 1 1 a d d − 1 − d − a d e − e 1 − a d d − 1 − d a d − e e 1 − b d − d − 1 d b d a b e − a b e 1 b d − d − 1 d − b d − a b e a b e 1 1 − 1 1 − 1 1 − 1 − 1 1 − 1 c 1 1 c 1 − 1 − c 1 − c 1 F 8 , 4 ( a , b , c , d ) = ( 1 1 1 1 1 1 1 1 1 − 1 a d − a d − b d b d 1 − 1 1 − 1 c d d − d − d − 1 1 c 1 1 c d − a d a d − 1 − 1 1 − 1 c d 1 1 c a d 2 − a d 2 b d 2 − b d 2 − 1 − 1 c 1 d − d − d − b d 2 b d 2 − 1 d 1 − 1 c d − 1 − 1 b d − b d 1 1 c d 1 − d − a d 2 a d 2 d d − 1 − d ) F_{8,4}(a,b,c,d)=
\begin{pmatrix}
1&1&1&1&1&1&1&1\\
1&-1&ad&-ad&-bd&bd&1&-1\\
1&-\frac{1}{c}&d&d&-d&-d&-1&\frac{1}{c}\\
1&\frac{1}{cd}&-ad&ad&-1&-1&1&-\frac{1}{cd}\\
1&\frac{1}{c}&ad^{2}&-ad^{2}&bd^{2}&-bd^{2}&-1&-\frac{1}{c}\\
1&d&-d&-d&-bd^{2}&bd^{2}&-1&d\\
1&-\frac{1}{cd}&-1&-1&bd&-bd&1&\frac{1}{cd}\\
1&-d&-ad^{2}&ad^{2}&d&d&-1&-d
\end{pmatrix} F 8 , 4 ( a , b , c , d ) = 1 1 1 1 1 1 1 1 1 − 1 − c 1 c d 1 c 1 d − c d 1 − d 1 a d d − a d a d 2 − d − 1 − a d 2 1 − a d d a d − a d 2 − d − 1 a d 2 1 − b d − d − 1 b d 2 − b d 2 b d d 1 b d − d − 1 − b d 2 b d 2 − b d d 1 1 − 1 1 − 1 − 1 1 − 1 1 − 1 c 1 − c d 1 − c 1 d c d 1 − d F 8 , 4 t ( a , b , c , d ) = ( 1 1 1 1 1 1 1 1 1 − 1 − 1 c 1 c d 1 c d − 1 c d − d 1 a d d − a d a d 2 − d − 1 − a d 2 1 − a d d a d − a d 2 − d − 1 a d 2 1 − b d − d − 1 b d 2 − b d 2 b d d 1 b d − d − 1 − b d 2 b d 2 − b d d 1 1 − 1 1 − 1 − 1 1 − 1 1 − 1 1 c − 1 c d − 1 c d 1 c d − d ) F^{t}_{8,4}(a,b,c,d)=\begin{pmatrix}
1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 \\
1 & -1 & -\frac{1}{c} & \frac{1}{cd} & \frac{1}{c} & d & -\frac{1}{cd} & -d \\
1 & ad & d & -ad & ad^{2} & -d & -1 & -ad^{2} \\
1 & -ad & d & ad & -ad^{2} & -d & -1 & ad^{2} \\
1 & -bd & -d & -1 & bd^{2} & -bd^{2} & bd & d \\
1 & bd & -d & -1 & -bd^{2} & bd^{2} & -bd & d \\
1 & 1 & -1 & 1 & -1 & -1 & 1 & -1 \\
1 & -1 & \frac{1}{c} & -\frac{1}{cd} & -\frac{1}{c} & d & \frac{1}{cd} & -d
\end{pmatrix} F 8 , 4 t ( a , b , c , d ) = 1 1 1 1 1 1 1 1 1 − 1 a d − a d − b d b d 1 − 1 1 − c 1 d d − d − d − 1 c 1 1 c d 1 − a d a d − 1 − 1 1 − c d 1 1 c 1 a d 2 − a d 2 b d 2 − b d 2 − 1 − c 1 1 d − d − d − b d 2 b d 2 − 1 d 1 − c d 1 − 1 − 1 b d − b d 1 c d 1 1 − d − a d 2 a d 2 d d − 1 − d F 8 , 2 ( a , b ) = ( 1 1 1 1 1 1 1 1 1 − 1 b − b − b a b a a − a 1 − a b a b − b a − b − 1 a 1 1 a b a 2 − b a b a − b a 2 − 1 a − 1 1 1 − 1 − 1 − 1 − 1 1 1 1 a − b b a − b a b − a − 1 1 1 a − b a b a 2 b a b a 2 − 1 1 a 1 − 1 − b a 2 b a 2 b a − b a 1 a − 1 a ) F_{8,2}(a,b)=\begin{pmatrix}
1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 \\
1 & -1 & b & -b & -\frac{b}{a} & \frac{b}{a} & a & -a \\
1 & -a & \frac{b}{a} & b & -\frac{b}{a} & -b & -1 & a \\
1 & \frac{1}{a} & \frac{b}{a^{2}} & -\frac{b}{a} & \frac{b}{a} & -\frac{b}{a^{2}} & -\frac{1}{a} & -1 \\
1 & 1 & -1 & -1 & -1 & -1 & 1 & 1 \\
1 & a & -b & \frac{b}{a} & -\frac{b}{a} & b & -a & -1 \\
1 & \frac{1}{a} & -\frac{b}{a} & \frac{b}{a^{2}} & \frac{b}{a} & \frac{b}{a^{2}} & -1 & \frac{1}{a} \\
1 & -1 & -\frac{b}{a^{2}} & \frac{b}{a^{2}} & \frac{b}{a} & -\frac{b}{a} & \frac{1}{a} & -\frac{1}{a}
\end{pmatrix} F 8 , 2 ( a , b ) = 1 1 1 1 1 1 1 1 1 − 1 − a a 1 1 a a 1 − 1 1 b a b a 2 b − 1 − b − a b − a 2 b 1 − b b − a b − 1 a b a 2 b a 2 b 1 − a b − a b a b − 1 − a b a b a b 1 a b − b − a 2 b − 1 b a 2 b − a b 1 a − 1 − a 1 1 − a − 1 a 1 1 − a a − 1 1 − 1 a 1 − a 1 F 8 , 2 t ( a , b ) = ( 1 1 1 1 1 1 1 1 1 − 1 − a 1 a 1 a − 1 a − 1 1 b b a b a 2 − 1 − b − b a − b a 2 1 − b b − b a − 1 b a − b a 2 b a 2 1 − b a − b a b a 2 − 1 − b a b a − b a 1 b a − b − b a 2 − 1 b b a 2 − b a 1 a − 1 − 1 a 1 − a − 1 1 a 1 − a a − 1 1 − 1 1 a − 1 a ) F^{t}_{8,2}(a,b)=\begin{pmatrix}
1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 \\
1 & -1 & -a & \frac{1}{a} & 1 & a & -\frac{1}{a} & -1 \\
1 & b & \frac{b}{a} & \frac{b}{a^{2}} & -1 & -b & -\frac{b}{a} & -\frac{b}{a^{2}} \\
1 & -b & b & -\frac{b}{a} & -1 & \frac{b}{a} & -\frac{b}{a^{2}} & \frac{b}{a^{2}} \\
1 & -\frac{b}{a} & -\frac{b}{a} & \frac{b}{a^{2}} & -1 & -\frac{b}{a} & \frac{b}{a} & -\frac{b}{a} \\
1 & \frac{b}{a} & -b & -\frac{b}{a^{2}} & -1 & b & \frac{b}{a^{2}} & -\frac{b}{a} \\
1 & a & -1 & -\frac{1}{a} & 1 & -a & -1 & \frac{1}{a} \\
1 & -a & a & -1 & 1 & -1 & \frac{1}{a} & -\frac{1}{a}
\end{pmatrix} F 8 , 2 t ( a , b ) = 1 1 1 1 1 1 1 1 1 − 1 b − b − a b a b a − a 1 − a a b b − a b − b − 1 a 1 a 1 a 2 b − a b a 2 b − a 2 b − a 1 − 1 1 1 − 1 − 1 − 1 − 1 1 1 1 a − b a b − a b b − a − 1 1 − a 1 − a b − a 2 b a b a 2 b − 1 a 1 1 − 1 − a 2 b a 2 b − a b − a b a 1 − a 1 K 8 , 2 ( a , b ) = ( 1 1 1 1 1 1 1 1 1 a − b 2 − b 2 a − 1 − a b 2 a b 2 1 a b − 1 b a 2 − 1 − a b − b a − 1 1 − b b b 2 a − b − 1 b a − b 2 a 1 − a − b b − a b a b − 1 a 1 − 1 b − b a − b − b − b 2 a b 2 a 1 a b − b − 1 b a − b b − a 1 b b 2 − b − 1 b − b − b 2 ) K_{8,2}(a,b)=\begin{pmatrix}
1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 \\
1 & a & -b^{2} & -\frac{b^{2}}{a} & -1 & -a & \frac{b^{2}}{a} & b^{2} \\
1 & \frac{a}{b} & -1 & \frac{b}{a^{2}} & -1 & -\frac{a}{b} & -\frac{b}{a} & -1 \\
1 & -b & b & \frac{b^{2}}{a} & -b & -1 & \frac{b}{a} & -\frac{b^{2}}{a} \\
1 & -a & -b & b & -\frac{a}{b} & \frac{a}{b} & -1 & a \\
1 & -1 & b & -\frac{b}{a} & -b & -b & -\frac{b^{2}}{a} & \frac{b^{2}}{a} \\
1 & \frac{a}{b} & -b & -1 & \frac{b}{a} & -b & b & -a \\
1 & b & b^{2} & -b & -1 & b & -b & -b^{2}
\end{pmatrix} K 8 , 2 ( a , b ) = 1 1 1 1 1 1 1 1 1 a b a − b − a − 1 b a b 1 − b 2 − 1 b − b b − b b 2 1 − a b 2 a 2 b a b 2 b − a b − 1 − b 1 − 1 − 1 − b − b a − b a b − 1 1 − a − b a − 1 b a − b − b b 1 a b 2 − a b a b − 1 − a b 2 b − b 1 b 2 − 1 − a b 2 a a b 2 − a − b 2 M 8 , 1 ( a ) = ( 1 1 1 1 1 1 1 1 1 − w a 2 a w a − 1 w a 2 − a − w a 1 a − a a 3 − a 3 − a 4 a 4 − 1 1 w a − w a − w a w a w 2 − 1 − w 2 1 − 1 w a − a 3 w a 2 − w a 2 a 3 − w a 1 w a 2 w − w a 4 − w a 2 − w 2 a 4 − a 4 w 2 1 − a − w − 1 − w a w a w a 1 − w a − 1 w a 4 a 3 − w a 4 − a 3 w a ) M_{8,1}(a)=\begin{pmatrix}
1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 \\
1 & -wa^{2} & a & wa & -1 & wa^{2} & -a & -wa \\
1 & a & -a & a^{3} & -a^{3} & -a^{4} & a^{4} & -1 \\
1 & wa & -\frac{w}{a} & -wa & \frac{w}{a} & w^{2} & -1 & -w^{2} \\
1 & -1 & \frac{w}{a} & -a^{3} & wa^{2} & -wa^{2} & a^{3} & -\frac{w}{a} \\
1 & wa^{2} & w & -wa^{4} & -wa^{2} & -w^{2}a^{4} & -a^{4} & w^{2} \\
1 & -a & -w & -1 & -\frac{w}{a} & w & a & \frac{w}{a} \\
1 & -wa & -1 & wa^{4} & a^{3} & -wa^{4} & -a^{3} & wa
\end{pmatrix} M 8 , 1 ( a ) = 1 1 1 1 1 1 1 1 1 − w a 2 a w a − 1 w a 2 − a − w a 1 a − a − a w a w w − w − 1 1 w a a 3 − w a − a 3 − w a 4 − 1 w a 4 1 − 1 − a 3 a w w a 2 − w a 2 − a w a 3 1 w a 2 − a 4 w 2 − w a 2 − w 2 a 4 w − w a 4 1 − a a 4 − 1 a 3 − a 4 a − a 3 1 − w a − 1 − w 2 − a w w 2 a w w a For the 4 isolated matrices, the prime denotes logarithmic form: H i j = ω H i j ′ H_{ij}=\omega^{H'_{ij}} H ij = ω H ij ′ ω = e π i / 6 \omega=e^{\pi i/6} ω = e π i /6 I 8 , 12 I_{8,12} I 8 , 12 ω = e π i / 15 \omega=e^{\pi i/15} ω = e π i /15
I 8 , 12 ′ = ( 0 0 0 0 0 0 0 0 0 0 0 3 6 6 6 9 0 1 5 9 1 5 9 6 0 4 8 6 10 2 3 9 0 6 2 11 9 8 3 5 0 6 11 5 3 11 9 5 0 7 6 1 7 3 9 1 0 10 6 7 4 9 3 1 ) K 8 , 30 ′ = ( 0 0 0 0 0 0 0 0 0 0 6 10 12 18 20 24 0 6 13 28 27 21 12 15 0 10 28 20 16 4 10 22 0 12 27 16 1 15 24 9 0 18 21 4 15 19 6 3 0 20 12 10 24 6 0 18 0 24 15 22 9 3 18 7 ) I'_{8,12}=\begin{pmatrix}
0&0&0&0&0&0&0&0\\
0&0&0&3&6&6&6&9\\
0&1&5&9&1&5&9&6\\
0&4&8&6&10&2&3&9\\
0&6&2&11&9&8&3&5\\
0&6&11&5&3&11&9&5\\
0&7&6&1&7&3&9&1\\
0&10&6&7&4&9&3&1
\end{pmatrix}
\qquad
K'_{8,30}=\begin{pmatrix}
0&0&0&0&0&0&0&0\\
0&0&6&10&12&18&20&24\\
0&6&13&28&27&21&12&15\\
0&10&28&20&16&4&10&22\\
0&12&27&16&1&15&24&9\\
0&18&21&4&15&19&6&3\\
0&20&12&10&24&6&0&18\\
0&24&15&22&9&3&18&7
\end{pmatrix} I 8 , 12 ′ = 0 0 0 0 0 0 0 0 0 0 1 4 6 6 7 10 0 0 5 8 2 11 6 6 0 3 9 6 11 5 1 7 0 6 1 10 9 3 7 4 0 6 5 2 8 11 3 9 0 6 9 3 3 9 9 3 0 9 6 9 5 5 1 1 K 8 , 30 ′ = 0 0 0 0 0 0 0 0 0 0 6 10 12 18 20 24 0 6 13 28 27 21 12 15 0 10 28 20 16 4 10 22 0 12 27 16 1 15 24 9 0 18 21 4 15 19 6 3 0 20 12 10 24 6 0 18 0 24 15 22 9 3 18 7 L 8 , 30 ′ = ( 0 0 0 0 0 0 0 0 0 0 6 10 12 18 20 24 0 6 21 2 19 17 4 15 0 10 17 25 8 2 15 23 0 12 4 23 27 15 19 8 0 18 2 17 15 3 1 16 0 20 19 15 4 16 5 1 0 24 15 8 23 1 16 9 ) L 8 , 30 t ′ = ( 0 0 0 0 0 0 0 0 0 0 6 10 12 18 20 24 0 6 21 17 4 2 19 15 0 10 2 25 23 17 15 8 0 12 19 8 27 15 4 23 0 18 17 2 15 3 16 1 0 20 4 15 19 1 5 16 0 24 15 23 8 16 1 9 ) L'_{8,30}=\begin{pmatrix}
0&0&0&0&0&0&0&0\\
0&0&6&10&12&18&20&24\\
0&6&21&2&19&17&4&15\\
0&10&17&25&8&2&15&23\\
0&12&4&23&27&15&19&8\\
0&18&2&17&15&3&1&16\\
0&20&19&15&4&16&5&1\\
0&24&15&8&23&1&16&9
\end{pmatrix}
\qquad
L^{t'}_{8,30}=\begin{pmatrix}
0&0&0&0&0&0&0&0\\
0&0&6&10&12&18&20&24\\
0&6&21&17&4&2&19&15\\
0&10&2&25&23&17&15&8\\
0&12&19&8&27&15&4&23\\
0&18&17&2&15&3&16&1\\
0&20&4&15&19&1&5&16\\
0&24&15&23&8&16&1&9
\end{pmatrix} L 8 , 30 ′ = 0 0 0 0 0 0 0 0 0 0 6 10 12 18 20 24 0 6 21 17 4 2 19 15 0 10 2 25 23 17 15 8 0 12 19 8 27 15 4 23 0 18 17 2 15 3 16 1 0 20 4 15 19 1 5 16 0 24 15 23 8 16 1 9 L 8 , 30 t ′ = 0 0 0 0 0 0 0 0 0 0 6 10 12 18 20 24 0 6 21 2 19 17 4 15 0 10 17 25 8 2 15 23 0 12 4 23 27 15 19 8 0 18 2 17 15 3 1 16 0 20 19 15 4 16 5 1 0 24 15 8 23 1 16 9 The maximum regular local moduli dimension is 5. The 4 isolated classes are exactly those listed above, and transposition preserves regularity. Different families can intersect, and parameters need not be unique.
The finite classifications are computer-assisted, using exact integer, rational and cyclotomic arithmetic. All code is available at https://github.com/teobanica.
Regular Hadamard matrices
Cycles and equivalence
We use
T = { z ∈ C : ∣ z ∣ = 1 } , e ( t ) = e 2 π i t , μ l = { e 2 π i j / l : 0 ≤ j < l } . \mathbb{T}=\{z\in\mathbb{C}:|z|=1\},\qquad e(t)=e^{2\pi i t},\qquad\mu_l=\{e^{2\pi i j/l}:0\le j<l\}. T = { z ∈ C : ∣ z ∣ = 1 } , e ( t ) = e 2 π i t , μ l = { e 2 π ij / l : 0 ≤ j < l } . Roots denoted by w w w ω \omega ω w w w H ∈ M K × N ( T ) H\in M_{K\times N}(\mathbb{T}) H ∈ M K × N ( T ) H H ∗ = N I K HH^*=NI_K H H ∗ = N I K
D ( H ) i j = H i j H 11 H i 1 H 1 j . D(H)_{ij}=\frac{H_{ij}H_{11}}{H_{i1}H_{1j}}. D ( H ) ij = H i 1 H 1 j H ij H 11 . Thus the 1st row and column become 1, the all-one vector. We write u / v = u ‾ v u/v=\overline{u}v u / v = u v B ( N , l ) B(N,l) B ( N , l ) [10 ] .
Definition 1.1. A multiset is a collection in which repeated values are counted separately; for example, { 1 , 1 , − 1 , − 1 } \{1,1,-1,-1\} { 1 , 1 , − 1 , − 1 } prime cycle is a rotated copy λ μ p \lambda\mu_p λ μ p p p p p p p λ ∈ T \lambda\in\mathbb{T} λ ∈ T
A cycle of composite length n = m p n=mp n = m p p p p m m m p p p w = e 2 π i / n w=e^{2\pi i/n} w = e 2 π i / n
λ μ n = ⨆ r = 0 m − 1 λ w r μ p . \lambda\mu_n=\bigsqcup_{r=0}^{m-1}\lambda w^r\mu_p. λ μ n = r = 0 ⨆ m − 1 λ w r μ p . For example, a 6-cycle splits into 2 triangles. This agrees with the cycle convention of [4 , 6 ] .
Proposition 1.2. Regularity is preserved by equivalence, conjugation, deletion of rows, and tensor products.
Proof. Scalings rotate cycles, permutations reorder them, and conjugation reverses them. For tensor products, use cycles from a factor where the 2 rows differ, fixing each coordinate of the other factor.
For l = ∏ j = 1 s p j a j l = \prod_{j=1}^{s} p_j^{a_j} l = ∏ j = 1 s p j a j [20 ] identify the lengths of vanishing sums of l l l ∑ j p j N \sum_j p_j\mathbb{N} ∑ j p j N s ≤ 2 s \le2 s ≤ 2 s = 3 s = 3 s = 3 w = e π i / 15 w = e^{\pi i/15} w = e π i /15
S = w 5 + w 6 + w 12 + w 18 + w 24 + w 25 = 0. S = w^5 + w^6 + w^{12} + w^{18} + w^{24} + w^{25} = 0. S = w 5 + w 6 + w 12 + w 18 + w 24 + w 25 = 0. The terms of exponents 6, 12, 18, 24 sum to − 1 -1 − 1
( 1 1 1 1 1 1 w 5 w 6 w 12 w 18 w 24 w 25 ) \begin{pmatrix}
1 & 1 & 1 & 1 & 1 & 1 \\
w^5 & w^6 & w^{12} & w^{18} & w^{24} & w^{25}
\end{pmatrix} ( 1 w 5 1 w 6 1 w 12 1 w 18 1 w 24 1 w 25 ) is a nonregular partial Butson matrix.
Examples and the classification through order 6
The Fourier matrix of Z N \mathbb{Z}_N Z N F N = ( w i j ) F_N = (w^{ij}) F N = ( w ij ) w = e 2 π i / N w = e^{2\pi i/N} w = e 2 π i / N w = e 2 π i / 3 w = e^{2\pi i/3} w = e 2 π i /3
F 2 = ( 1 1 1 − 1 ) , F 3 = ( 1 1 1 1 w w 2 1 w 2 w ) , F 4 = ( 1 1 1 1 1 i − 1 − i 1 − 1 1 − 1 1 − i − 1 i ) . F_2 = \begin{pmatrix} 1 & 1 \\ 1 & -1 \end{pmatrix}, \qquad
F_3 = \begin{pmatrix} 1 & 1 & 1 \\ 1 & w & w^2 \\ 1 & w^2 & w \end{pmatrix}, \qquad
F_4 = \begin{pmatrix} 1 & 1 & 1 & 1 \\ 1 & i & -1 & -i \\ 1 & -1 & 1 & -1 \\ 1 & -i & -1 & i \end{pmatrix}. F 2 = ( 1 1 1 − 1 ) , F 3 = 1 1 1 1 w w 2 1 w 2 w , F 4 = 1 1 1 1 1 i − 1 − i 1 − 1 1 − 1 1 − i − 1 i . The matrix F 4 F_4 F 4
F 4 q = ( 1 1 1 1 1 − 1 1 − 1 1 q − 1 − q 1 − q − 1 q ) . F_4^q =
\begin{pmatrix}
1 & 1 & 1 & 1 \\
1 & -1 & 1 & -1 \\
1 & q & -1 & -q \\
1 & -q & -1 & q
\end{pmatrix}. F 4 q = 1 1 1 1 1 − 1 q − q 1 1 − 1 − 1 1 − 1 − q q . At N = 5 N = 5 N = 5 F 5 F_5 F 5 N = 6 N = 6 N = 6 w = e 2 π i / 3 w = e^{2\pi i/3} w = e 2 π i /3 [4 , 12 , 14 , 33 ] :
F 23 r s = ( 1 1 1 1 1 1 1 w w 2 1 w w 2 1 w 2 w 1 w 2 w 1 r s − 1 − r − s 1 w r w 2 s − 1 − w r − w 2 s 1 w 2 r w s − 1 − w 2 r − w s ) , F 32 r s = ( 1 1 1 1 1 1 1 − 1 1 − 1 1 − 1 1 r w w r w 2 w 2 r 1 − r w − w r w 2 − w 2 r 1 s w 2 w 2 s w w s 1 − s w 2 − w 2 s w − w s ) . F_{23}^{rs} =
\begin{pmatrix}
1 & 1 & 1 & 1 & 1 & 1 \\
1 & w & w^2 & 1 & w & w^2 \\
1 & w^2 & w & 1 & w^2 & w \\
1 & r & s & -1 & -r & -s \\
1 & wr & w^2s & -1 & -wr & -w^2s \\
1 & w^2r & ws & -1 & -w^2r & -ws
\end{pmatrix},
\qquad
F_{32}^{rs} =
\begin{pmatrix}
1 & 1 & 1 & 1 & 1 & 1 \\
1 & -1 & 1 & -1 & 1 & -1 \\
1 & r & w & wr & w^2 & w^2r \\
1 & -r & w & -wr & w^2 & -w^2r \\
1 & s & w^2 & w^2s & w & ws \\
1 & -s & w^2 & -w^2s & w & -ws
\end{pmatrix}. F 23 r s = 1 1 1 1 1 1 1 w w 2 r w r w 2 r 1 w 2 w s w 2 s w s 1 1 1 − 1 − 1 − 1 1 w w 2 − r − w r − w 2 r 1 w 2 w − s − w 2 s − w s , F 32 r s = 1 1 1 1 1 1 1 − 1 r − r s − s 1 1 w w w 2 w 2 1 − 1 w r − w r w 2 s − w 2 s 1 1 w 2 w 2 w w 1 − 1 w 2 r − w 2 r w s − w s . H 6 q = ( 1 1 1 1 1 1 1 − 1 i i − i − i 1 i − 1 − i q − q 1 i − i − 1 − q q 1 − i q ˉ − q ˉ i − 1 1 − i − q ˉ q ˉ − 1 i ) , T 6 = ( 1 1 1 1 1 1 1 1 w w w 2 w 2 1 w 1 w 2 w 2 w 1 w w 2 1 w w 2 1 w 2 w 2 w 1 w 1 w 2 w w 2 w 1 ) . H_6^q =
\begin{pmatrix}
1 & 1 & 1 & 1 & 1 & 1 \\
1 & -1 & i & i & -i & -i \\
1 & i & -1 & -i & q & -q \\
1 & i & -i & -1 & -q & q \\
1 & -i & \bar{q} & -\bar{q} & i & -1 \\
1 & -i & -\bar{q} & \bar{q} & -1 & i
\end{pmatrix},
\qquad
T_6 =
\begin{pmatrix}
1 & 1 & 1 & 1 & 1 & 1 \\
1 & 1 & w & w & w^2 & w^2 \\
1 & w & 1 & w^2 & w^2 & w \\
1 & w & w^2 & 1 & w & w^2 \\
1 & w^2 & w^2 & w & 1 & w \\
1 & w^2 & w & w^2 & w & 1
\end{pmatrix}. H 6 q = 1 1 1 1 1 1 1 − 1 i i − i − i 1 i − 1 − i q ˉ − q ˉ 1 i − i − 1 − q ˉ q ˉ 1 − i q − q i − 1 1 − i − q q − 1 i , T 6 = 1 1 1 1 1 1 1 1 w w w 2 w 2 1 w 1 w 2 w 2 w 1 w w 2 1 w w 2 1 w 2 w 2 w 1 w 1 w 2 w w 2 w 1 . Their quotients split into antipodal pairs or 3-cycles for every parameter. The 2 Fourier families are related by transposition and equivalence.
Theorem 1.3. The regular Hadamard matrices at orders N = 2 , 3 , 4 , 5 , 6 N=2,3,4,5,6 N = 2 , 3 , 4 , 5 , 6
N = 2 N=2 N = 2 F 2 F_2 F 2
N = 3 N=3 N = 3 F 3 F_3 F 3
N = 4 N=4 N = 4 F 4 q F_4^q F 4 q
N = 5 N=5 N = 5 F 5 F_5 F 5
N = 6 N=6 N = 6 F 23 r s F_{23}^{rs} F 23 r s F 32 r s F_{32}^{rs} F 32 r s H 6 q H_6^q H 6 q T 6 T_6 T 6
Proof. At N = 2 , 3 N=2,3 N = 2 , 3 F 2 , F 3 F_2,F_3 F 2 , F 3 N = 4 N=4 N = 4 F 4 q F_4^q F 4 q
For N = 5 N=5 N = 5 5 5 5 3 + 2 3+2 3 + 2 2 × 2 2\times2 2 × 2 H H H
H = ( A B C D ) , A = ( 1 1 1 − 1 ) , B ⋅ j = ( 1 a w j ) , C i ⋅ = ( 1 , b w i ) , 0 ≤ i , j < 3 , H=\begin{pmatrix} A & B \\ C & D \end{pmatrix},\qquad A=\begin{pmatrix} 1 & 1 \\ 1 & -1 \end{pmatrix},\qquad B_{\cdot j}=\begin{pmatrix} 1 \\ aw^j \end{pmatrix},\qquad C_{i\cdot}=(1,bw^i),\qquad0\le i,j<3, H = ( A C B D ) , A = ( 1 1 1 − 1 ) , B ⋅ j = ( 1 a w j ) , C i ⋅ = ( 1 , b w i ) , 0 ≤ i , j < 3 , where a , b ∈ T a,b\in\mathbb{T} a , b ∈ T w = e 2 π i / 3 w=e^{2\pi i/3} w = e 2 π i /3
∑ j = 0 2 D i j = − ( 1 + b w i ) , ∑ j = 0 2 D i j w − j = − a ( 1 − b w i ) . \sum_{j=0}^{2}D_{ij}=-(1+bw^i),\qquad\sum_{j=0}^{2}D_{ij}w^{-j}=-a(1-bw^i). j = 0 ∑ 2 D ij = − ( 1 + b w i ) , j = 0 ∑ 2 D ij w − j = − a ( 1 − b w i ) . These determine 2 Fourier coefficients of each row of D D D z ∈ C z\in\mathbb{C} z ∈ C
D i j = − 1 + a w j + b w i − a b w i + j − z w 2 ( i + j ) 3 . D_{ij}=-\frac{1+aw^j+bw^i-abw^{i+j}-zw^{2(i+j)}}{3}. D ij = − 3 1 + a w j + b w i − ab w i + j − z w 2 ( i + j ) . Since the entries of D D D ∣ 3 D i j ∣ 2 = 9 \lvert3D_{ij}\rvert^2=9 ∣ 3 D ij ∣ 2 = 9 ∣ z ∣ 2 = 5 \lvert z\rvert^2=5 ∣ z ∣ 2 = 5 ( 1 , 1 ) (1,1) ( 1 , 1 ) ( 1 , 2 ) (1,2) ( 1 , 2 )
z − a b z ‾ = ( a b ) 2 , z + a b z ‾ = b 2 a . z-ab\overline{z}=(ab)^2,\qquad z+ab\overline{z}=b^2a. z − ab z = ( ab ) 2 , z + ab z = b 2 a . Adding these equations gives ∣ z ∣ ≤ 1 \lvert z\rvert\le1 ∣ z ∣ ≤ 1 3 + 2 3+2 3 + 2 5 5 5 F 5 F_5 F 5 σ \sigma σ Z 5 \mathbb{Z}_5 Z 5 σ ( 0 ) = 0 \sigma(0)=0 σ ( 0 ) = 0 σ − id \sigma-\operatorname{id} σ − id
( σ ( 1 ) , σ ( 2 ) , σ ( 3 ) , σ ( 4 ) ) = ( 2 , 4 , 1 , 3 ) , ( 3 , 1 , 4 , 2 ) , ( 4 , 3 , 2 , 1 ) . (\sigma(1),\sigma(2),\sigma(3),\sigma(4))=(2,4,1,3),\ (3,1,4,2),\ (4,3,2,1). ( σ ( 1 ) , σ ( 2 ) , σ ( 3 ) , σ ( 4 )) = ( 2 , 4 , 1 , 3 ) , ( 3 , 1 , 4 , 2 ) , ( 4 , 3 , 2 , 1 ) . These are precisely the remaining Fourier rows. Haagerup’s theorem gives the same classification without regularity [14 ] .
At N = 6 N=6 N = 6 [4 ] and Theorem 11.2 classifies triples of rows and then uses their compatibility to complete the matrix.
If all quotients admit ternary decompositions, the 3-row calculation gives, up to equivalence, either a matrix over μ 3 \mu_3 μ 3
( 1 1 1 1 1 1 1 w w 2 r w r w 2 r 1 w 2 w s w 2 s w s ) . \begin{pmatrix}
1 & 1 & 1 & 1 & 1 & 1\\
1 & w & w^2 & r & wr & w^2r\\
1 & w^2 & w & s & w^2s & ws
\end{pmatrix}. 1 1 1 1 w w 2 1 w 2 w 1 r s 1 w r w 2 s 1 w 2 r w s . Here w = e 2 π i / 3 w=e^{2\pi i/3} w = e 2 π i /3 μ 3 \mu_3 μ 3 T 6 T_6 T 6 [4 ] , [4 Lemmas 8.1–8.2 and Theorem 8.3] .
If all quotients admit binary decompositions, the 3-row calculation puts the entries in { ± 1 , ± i , ± q , ± i q } \{\pm1,\pm i,\pm q,\pm iq\} { ± 1 , ± i , ± q , ± i q } 3 × 3 3\times3 3 × 3 2 × 2 2\times2 2 × 2 H 6 q H_6^q H 6 q [4 ] , [4 Lemma 9.1 and Theorem 9.2] .
:::
In the mixed case, regard the rows as vertices of a complete graph and label its edges by the chosen decompositions. The restrictions on 4 and 5 rows leave 2 arrangements: the binary edges form a perfect matching and the others are ternary, or the ternary edges form 2 disjoint triangles and the others are binary [4 Section 10] . Inserting the 3-row normal form determines the remaining rows with 2 free unit parameters, giving F 23 r s F_{23}^{rs} F 23 r s F 32 r s F_{32}^{rs} F 32 r s
Let us discuss as well some basic examples of non-regular matrices, which clearly show the extra complexity which appears, when looking at the non-regular case.
The Björck–Fröberg matrix [8 , 9 ] , which is circulant and nonregular, is
B F 6 = ( 1 i a − a − i − a ˉ i a ˉ i a ˉ 1 i a − a − i − a ˉ − a ˉ i a ˉ 1 i a − a − i − i − a i a ˉ 1 i a − a − a − i − a ˉ i a ˉ 1 i a i a − a − i − a ˉ i a ˉ 1 ) , a 2 + ( 3 − 1 ) a + 1 = 0. BF_6 =
\begin{pmatrix}
1 & ia & -a & -i & -\bar{a} & i\bar{a} \\
i\bar{a} & 1 & ia & -a & -i & -\bar{a} \\
-\bar{a} & i\bar{a} & 1 & ia & -a & -i \\
-i & -a & i\bar{a} & 1 & ia & -a \\
-a & -i & -\bar{a} & i\bar{a} & 1 & ia \\
ia & -a & -i & -\bar{a} & i\bar{a} & 1
\end{pmatrix},
\qquad a^2+(\sqrt{3}-1)a+1=0. B F 6 = 1 i a ˉ − a ˉ − i − a ia ia 1 i a ˉ − a − i − a − a ia 1 i a ˉ − a ˉ − i − i − a ia 1 i a ˉ − a ˉ − a ˉ − i − a ia 1 i a ˉ i a ˉ − a ˉ − i − a ia 1 , a 2 + ( 3 − 1 ) a + 1 = 0. We can see its nonregularity directly. We have a + a − 1 = 1 − 3 a+a^{-1}=1-\sqrt{3} a + a − 1 = 1 − 3 a a a 1 + 3 1+\sqrt{3} 1 + 3 [ − 2 , 2 ] [-2,2] [ − 2 , 2 ]
( − a , a 2 , − a , − a − 1 , a − 2 , − a − 1 ) . (-a,a^2,-a,-a^{-1},a^{-2},-a^{-1}). ( − a , a 2 , − a , − a − 1 , a − 2 , − a − 1 ) . The ratio of any 2 distinct values here is ± a k \pm a^k ± a k k k k
Another class of examples, this time parametric, is provided by the self-adjoint 6 × 6 6 \times6 6 × 6 [7 ] . These matrices, which are generically non-regular, are as follows:
B N 6 q = ( 1 1 1 1 1 1 1 − 1 x ˉ − y − x ˉ y 1 x − 1 t − t − x 1 − y ˉ t ˉ − 1 y ˉ − t ˉ 1 − x − t ˉ y 1 z ˉ 1 y ˉ − x ˉ − t z 1 ) , BN_6^q =
\begin{pmatrix}
1 & 1 & 1 & 1 & 1 & 1 \\
1 & -1 & \bar{x} & -y & -\bar{x} & y \\
1 & x & -1 & t & -t & -x \\
1 & -\bar{y} & \bar{t} & -1 & \bar{y} & -\bar{t} \\
1 & -x & -\bar{t} & y & 1 & \bar{z} \\
1 & \bar{y} & -\bar{x} & -t & z & 1
\end{pmatrix}, B N 6 q = 1 1 1 1 1 1 1 − 1 x − y ˉ − x y ˉ 1 x ˉ − 1 t ˉ − t ˉ − x ˉ 1 − y t − 1 y − t 1 − x ˉ − t y ˉ 1 z 1 y − x − t ˉ z ˉ 1 , where y = q y=q y = q Re q ≤ ( 3 − 1 ) / 2 \operatorname{Re}q\leq(\sqrt{3}-1)/2 Re q ≤ ( 3 − 1 ) /2
x = 1 + 2 q + q 2 − 2 1 + 2 q + 2 q 3 + q 4 1 + 2 q − q 2 , z = 1 + 2 q − q 2 q ( − 1 + 2 q + q 2 ) , t = x y z . x=\frac{1+2q+q^2-\sqrt{2}\sqrt{1+2q+2q^3+q^4}}{1+2q-q^2},
\qquad
z=\frac{1+2q-q^2}{q(-1+2q+q^2)},
\qquad
t=xyz. x = 1 + 2 q − q 2 1 + 2 q + q 2 − 2 1 + 2 q + 2 q 3 + q 4 , z = q ( − 1 + 2 q + q 2 ) 1 + 2 q − q 2 , t = x y z . Affine realization
An affine Hadamard deformation has the form
H ( t ) i j = H i j exp ( 2 π i t R i j ) , R ∈ M K × N ( R ) , H(t)_{ij}=H_{ij}\exp(2\pi i tR_{ij}),\qquad R\in M_{K\times N}(\mathbb{R}), H ( t ) ij = H ij exp ( 2 π i t R ij ) , R ∈ M K × N ( R ) , for t t t t ∈ R t\in\mathbb{R} t ∈ R
T ~ H = { R ∈ M K × N ( R ) : ∑ k = 1 N H i k H ‾ j k ( R i k − R j k ) = 0 for all i ≠ j } . \widetilde{T}_H=\left\{R\in M_{K\times N}(\mathbb{R}):\sum_{k=1}^{N}H_{ik}\overline{H}_{jk}(R_{ik}-R_{jk})=0\text{ for all }i\ne j\right\}. T H = { R ∈ M K × N ( R ) : k = 1 ∑ N H ik H j k ( R ik − R j k ) = 0 for all i = j } . The enveloping defect d e n v = dim R T ~ H d_{\mathrm{env}} = \dim_{\mathbb{R}} \widetilde{T}_{H} d env = dim R T H K + N − 1 K+N-1 K + N − 1 d d e p = d e n v − ( K + N − 1 ) d_{\mathrm{dep}} = d_{\mathrm{env}}-(K+N-1) d dep = d env − ( K + N − 1 ) [12 , 31 , 32 ] .
Theorem 1.4 (Affine realization and finite cover). For fixed K , N K,N K , N
H i j = B i j ∏ s = 1 d q s V i j ( s ) , H_{ij}=B_{ij}\prod_{s=1}^{d}q_s^{V_{ij}^{(s)}}, H ij = B ij s = 1 ∏ d q s V ij ( s ) , with B B B V ( s ) V^{(s)} V ( s ) q s ∈ T q_s\in\mathbb{T} q s ∈ T
Proof. Choose a cycle decomposition of every row quotient. On a block of length p p p k 0 , … , k p − 1 k_0,\ldots,k_{p-1} k 0 , … , k p − 1 1 , w , … , w p − 1 1,w,\ldots,w^{p-1} 1 , w , … , w p − 1 w = e 2 π i / p w=e^{2\pi i/p} w = e 2 π i / p H i j = e ( x i j ) H_{ij}=e(x_{ij}) H ij = e ( x ij ) x i j x_{ij} x ij 1 ≤ a < p 1\leq a<p 1 ≤ a < p x i k a − x j k a − x i k 0 + x j k 0 x_{ik_a}-x_{jk_a}-x_{ik_0}+x_{jk_0} x i k a − x j k a − x i k 0 + x j k 0 a / p a/p a / p
Thus all chosen blocks give linear conditions with integer coefficients on the phases, and with rational constants; phase values are considered modulo integers. We add the dephasing conditions when needed. Integer row and column operations (Smith normal form) change variables so that the conditions separate into equations of the form d j y j = c j ′ d_jy_j=c'_j d j y j = c j ′ d j d_j d j c j ′ c'_j c j ′ y j y_j y j
There are only finitely many choices of cycle blocks and their orderings. For each choice, the solutions form a finite union of closed tori, so the regular locus is a finite union of closed tori. In the Smith normal form coordinates, each constrained coordinate is rational and the free coordinates vary on circles. Joining a point to a rational point by a straight line in the free phase coordinates gives the regular path
H ( t ) i j = e ( x 0 , i j + t ( x i j − x 0 , i j ) ) , t ∈ R , H(t)_{ij}=e\bigl(x_{0,ij}+t(x_{ij}-x_{0,ij})\bigr),\qquad t\in\mathbb{R}, H ( t ) ij = e ( x 0 , ij + t ( x ij − x 0 , ij ) ) , t ∈ R , where x 0 x_0 x 0 □ \square □
Different cycle decompositions can give intersecting, nonmaximal families. The theorem preserves the chosen cycles; classification still requires completion and equivalence analysis.
The regularity conjecture
Conjecture 1.5 (Regularity). Every square Butson matrix is regular.
Lam–Leung proves this at levels with at most 2 prime divisors. The partial 2 × 6 2\times6 2 × 6 B N \mathcal{B}_N B N B ‾ N \overline{\mathcal{B}}_N B N R N \mathcal{R}_N R N A N \mathcal{A}_N A N
Proposition 1.6. We have
R N ⊆ A N ⊆ B ‾ N . \mathcal{R}_N \subseteq\mathcal{A}_N \subseteq\overline{\mathcal{B}}_N. R N ⊆ A N ⊆ B N . The following are equivalent:
B N ⊆ R N , R N = A N , R N = B ‾ N . \mathcal{B}_N \subseteq\mathcal{R}_N,\qquad\mathcal{R}_N = \mathcal{A}_N,\qquad\mathcal{R}_N = \overline{\mathcal{B}}_N. B N ⊆ R N , R N = A N , R N = B N . Under these conditions transposition preserves regularity.
Proof. The 1st inclusion is Theorem 1.4 . Write H = B ∘ e ( R ) H = B \circ e(R) H = B ∘ e ( R ) B ∘ e ( t R ) B \circ e(tR) B ∘ e ( tR ) R i k − R j k R_{ik} - R_{jk} R ik − R j k
∑ k ∈ E B i k B ‾ j k = 0 \sum_{k \in E} B_{ik}\overline{B}_{jk} = 0 k ∈ E ∑ B ik B j k = 0 on every frequency block E E E S S S
S i k − S j k = S i l − S j l ( k , l ∈ E ) S_{ik} - S_{jk} = S_{il} - S_{jl} \qquad(k,l \in E) S ik − S j k = S i l − S j l ( k , l ∈ E ) for every such block and row pair. These integer-coefficient equations define a rational linear space containing R R R S n → R S_n \to R S n → R B ∘ e ( S n ) B \circ e(S_n) B ∘ e ( S n ) H H H
We also have B N ⊆ A N \mathcal{B}_N \subseteq\mathcal{A}_N B N ⊆ A N R N \mathcal{R}_N R N Theorem 1.4 . Together with the inclusions just proved, these facts give the equivalences. Transposition preserves B N \mathcal{B}_N B N
The regular classification at order 7
At order 7, the possible quotient types are A = 7 A = 7 A = 7 B = 5 + 2 B = 5 + 2 B = 5 + 2 C = 3 + 2 + 2 C = 3 + 2 + 2 C = 3 + 2 + 2
Some of this section continues and finishes previous N = 7 N = 7 N = 7 [4 , 6 ] . That work was exploratory at the time. Here we arrange the cases into a systematic proof, first classifying the 3-row matrices and then studying how each can be extended by additional rows.
For the extension step, we impose the cycle conditions for each new row and analyze the resulting phase equations. We treat affine families and isolated classes separately. For the families we follow the phase parameters through the extension equations, including exceptional values; for isolated classes we determine the compatible fourth rows and then continue the surviving candidates through the larger row sizes. Common supports and special configurations organize these checks. Finally, we compare the completed matrices up to equivalence. This leaves precisely the Fourier matrix F 7 F_7 F 7
The proof combines elementary reductions with finite exact calculations. The preliminary type restrictions and the B B B BBB B B B A A A AAA AAA C C C CCC C C C A C C ACC A C C B C C BCC B C C
Cycle types and phase equations
For H = ( 1 ; u ; v ) ∈ M 3 × 7 ( T ) H=(1;u;v)\in M_{3\times7}(\mathbb{T}) H = ( 1 ; u ; v ) ∈ M 3 × 7 ( T ) u u u v v v u v ‾ \overline{uv} uv A = 7 A=7 A = 7 B = 5 + 2 B=5+2 B = 5 + 2 C = 3 + 2 + 2 C=3+2+2 C = 3 + 2 + 2
Proposition 2.1. Every regular length-7 vector has a unique type A A A B B B C C C
( A , A , A ) , ( A , C , C ) , ( B , B , B ) , ( B , C , C ) , ( C , C , C ) . (A,A,A),\quad(A,C,C),\quad(B,B,B),\quad(B,C,C),\quad(C,C,C). ( A , A , A ) , ( A , C , C ) , ( B , B , B ) , ( B , C , C ) , ( C , C , C ) . Proof. The prime partitions of 7 give the list of types. Their maximum numbers of disjoint antipodal pairs are respectively 0, 1, 2: an odd cycle has no antipodal pair; 2 pairs in a 5 + 2 5+2 5 + 2
Exchanging rows permutes the 3 types. If 2 quotients are 7-cycles, normalize them into μ 7 \mu_7 μ 7 A A A A B AB A B B B C BBC B B C
Lemma 2.2. For odd primes p = q + 2 p=q+2 p = q + 2 p p p q + 2 q+2 q + 2 B B C BBC B B C
Proof. Normalize the 2 vectors as
u = ( 1 , ω , … , ω q − 1 , a , − a ) , v = ( w q r 0 , … , w q r q − 1 , w q s , w q t ) , u=(1,\omega,\ldots,\omega^{q-1},a,-a),\qquad v=(w^{qr_0},\ldots,w^{qr_{q-1}},w^{qs},w^{qt}), u = ( 1 , ω , … , ω q − 1 , a , − a ) , v = ( w q r 0 , … , w q r q − 1 , w q s , w q t ) , where w = e 2 π i / ( p q ) w=e^{2\pi i/(pq)} w = e 2 π i / ( pq ) ω = w p \omega=w^p ω = w p r 0 , … , r q − 1 , s , t r_0,\ldots,r_{q-1},s,t r 0 , … , r q − 1 , s , t Z / p \mathbb{Z}/p Z / p
A k = ∑ j = 0 q − 1 w k j w − q r j ( 0 ≤ k < q ) , B = w − q t − w − q s , D = ∣ B ∣ 2 . A_k=\sum_{j=0}^{q-1}w^{kj}w^{-qr_j}\quad(0\le k<q),\qquad B=w^{-qt}-w^{-qs},\qquad D=|B|^2. A k = j = 0 ∑ q − 1 w k j w − q r j ( 0 ≤ k < q ) , B = w − q t − w − q s , D = ∣ B ∣ 2 . Orthogonality gives ∣ A 1 ∣ 2 = D |A_1|^2=D ∣ A 1 ∣ 2 = D w q w^q w q ∣ A k ∣ 2 = D |A_k|^2=D ∣ A k ∣ 2 = D 1 ≤ k < q 1\le k<q 1 ≤ k < q A 0 = − w − q s − w − q t A_0=-w^{-qs}-w^{-qt} A 0 = − w − q s − w − q t
∑ k = 0 q − 1 ∣ A k ∣ 2 = ∑ j , l = 0 q − 1 w q ( r l − r j ) ∑ k = 0 q − 1 w k ( j − l ) = q 2 . \sum_{k=0}^{q-1}|A_k|^2=\sum_{j,l=0}^{q-1}w^{q(r_l-r_j)}\sum_{k=0}^{q-1}w^{k(j-l)}=q^2. k = 0 ∑ q − 1 ∣ A k ∣ 2 = j , l = 0 ∑ q − 1 w q ( r l − r j ) k = 0 ∑ q − 1 w k ( j − l ) = q 2 . Thus q 2 = ( 4 − D ) + ( q − 1 ) D = 4 + ( q − 2 ) D q^2=(4-D)+(q-1)D=4+(q-2)D q 2 = ( 4 − D ) + ( q − 1 ) D = 4 + ( q − 2 ) D D = q + 2 = p D=q+2=p D = q + 2 = p D ≤ 4 < p D\le4<p D ≤ 4 < p
For B B C BBC B B C u , v u,v u , v μ 5 \mu_5 μ 5 U , V U,V U , V 3 + ∣ U ∩ V ∣ 3+|U\cap V| 3 + ∣ U ∩ V ∣ μ 5 \mu_5 μ 5 C C C U ∩ V = ∅ U\cap V=\varnothing U ∩ V = ∅ U ∪ V U\cup V U ∪ V − μ 5 -\mu_5 − μ 5 U , V U,V U , V − μ 5 -\mu_5 − μ 5 μ 5 \mu_5 μ 5 μ 5 \mu_5 μ 5
The same count excludes ( p + 2 , p + 2 , q + 2 + 2 ) (p+2,p+2,q+2+2) ( p + 2 , p + 2 , q + 2 + 2 ) p = q + 2 p=q+2 p = q + 2 μ p \mu_p μ p q + ∣ U ∩ V ∣ q+|U\cap V| q + ∣ U ∩ V ∣ q = 3 q=3 q = 3
Lemma 2.3. Put w = e 2 π i / 3 w=e^{2\pi i/3} w = e 2 π i /3 a , b , c ∈ T a,b,c\in\mathbb{T} a , b , c ∈ T a + b + c = 0 a+b+c=0 a + b + c = 0 ( a , b , c ) = ( a , a ω , a ω 2 ) (a,b,c)=(a,a\omega,a\omega^2) ( a , b , c ) = ( a , aω , a ω 2 ) ω ∈ { w , w 2 } \omega\in\{w,w^2\} ω ∈ { w , w 2 } ω \omega ω
Table 4.
The signs in the last row are correlated.
Proof. If 1 + x + y = 0 1+x+y=0 1 + x + y = 0 ∣ x ∣ = ∣ y ∣ = 1 |x|=|y|=1 ∣ x ∣ = ∣ y ∣ = 1
1 = ∣ 1 + x ∣ 2 = 2 + x + x ˉ , Re ( x ) = − 1 2 . 1=|1+x|^2=2+x+\bar{x},\qquad\operatorname{Re}(x)=-\frac{1}{2}. 1 = ∣1 + x ∣ 2 = 2 + x + x ˉ , Re ( x ) = − 2 1 . so ( x , y ) = ( ω , ω 2 ) (x,y)=(\omega,\omega^2) ( x , y ) = ( ω , ω 2 ) a 3 = 1 a^3=1 a 3 = 1 a 2 = 1 a^2=1 a 2 = 1
The 3-row classification
We first state the full 3-row classification. We prove its C C C CCC C C C A A A AAA AAA B B B BBB B B B A C C ACC A C C B C C BCC B C C
Theorem 2.4. Every regular 3 × 7 3 \times7 3 × 7 C C C CCC C C C F 1 , … , F 6 \mathcal{F}_1,\ldots,\mathcal{F}_6 F 1 , … , F 6
The families have dimensions 3, 2, 2, 1, 1, 1 and none can be omitted. The isolated representatives are pairwise inequivalent and lie outside the families. Family images may intersect, and parameters need not be unique.
Proof of the C C C CCC C C C We begin with the families and isolated representatives for this part. The other types are treated in Theorems 2.5 and 2.6 , which complete the proof of the full statement.
Put w = e 2 π i / 3 w=e^{2\pi i/3} w = e 2 π i /3 T \mathbb{T} T
F 1 ( a , b , c ) = ( 1 1 1 1 1 1 1 1 w w 2 a − a a − a 1 w 2 w b c − b − c ) . \mathcal{F}_1(a,b,c)=
\begin{pmatrix}
1&1&1&1&1&1&1\\
1&w&w^2&a&-a&a&-a\\
1&w^2&w&b&c&-b&-c
\end{pmatrix}. F 1 ( a , b , c ) = 1 1 1 1 w w 2 1 w 2 w 1 a b 1 − a c 1 a − b 1 − a − c . F 2 ( a , b ) = ( 1 1 1 1 1 1 1 − 1 w w 2 1 1 a − a − 1 w 2 w b − b 1 1 ) . \mathcal{F}_2(a,b)=
\begin{pmatrix}
1&1&1&1&1&1&1\\
-1&w&w^2&1&1&a&-a\\
-1&w^2&w&b&-b&1&1
\end{pmatrix}. F 2 ( a , b ) = 1 − 1 − 1 1 w w 2 1 w 2 w 1 1 b 1 1 − b 1 a 1 1 − a 1 . F 3 ( a , b ) = ( 1 1 1 1 1 1 1 − a b w w 2 1 − a a a b − a b w 2 w b a b 1 − b ) . \mathcal{F}_3(a,b)=
\begin{pmatrix}
1&1&1&1&1&1&1\\
-ab&w&w^2&1&-a&a&ab\\
-ab&w^2&w&b&ab&1&-b
\end{pmatrix}. F 3 ( a , b ) = 1 − ab − ab 1 w w 2 1 w 2 w 1 1 b 1 − a ab 1 a 1 1 ab − b . F 4 ( a ) = ( 1 1 1 1 1 1 1 1 w w 2 w a − w a a − a 1 w w 2 a w a − a − w a ) . \mathcal{F}_4(a)=
\begin{pmatrix}
1&1&1&1&1&1&1\\
1&w&w^2&wa&-wa&a&-a\\
1&w&w^2&a&wa&-a&-wa
\end{pmatrix}. F 4 ( a ) = 1 1 1 1 w w 1 w 2 w 2 1 w a a 1 − w a w a 1 a − a 1 − a − w a . F 5 ( a ) = ( 1 1 1 1 1 1 1 1 w w 2 w / a − w / a w 2 a − w 2 a 1 − w a a w w 2 w a − a ) . \mathcal{F}_5(a)=
\begin{pmatrix}
1&1&1&1&1&1&1\\
1&w&w^2&w/a&-w/a&w^2a&-w^2a\\
1&-wa&a&w&w^2&wa&-a
\end{pmatrix}. F 5 ( a ) = 1 1 1 1 w − w a 1 w 2 a 1 w / a w 1 − w / a w 2 1 w 2 a w a 1 − w 2 a − a . F 6 ( a ) = ( 1 1 1 1 1 1 1 1 w w 2 a − a a − 1 − a − 1 1 − w 2 / a w a w − w a w 2 / a w 2 ) . \mathcal{F}_6(a)=
\begin{pmatrix}
1&1&1&1&1&1&1\\
1&w&w^2&a&-a&a^{-1}&-a^{-1}\\
1&-w^2/a&wa&w&-wa&w^2/a&w^2
\end{pmatrix}. F 6 ( a ) = 1 1 1 1 w − w 2 / a 1 w 2 w a 1 a w 1 − a − w a 1 a − 1 w 2 / a 1 − a − 1 w 2 . For the isolated C C C CCC C C C w = e π i / 12 w=e^{\pi i/12} w = e π i /12 [ r ; s ] w = ( 1 ; w r ; w s ) [r;s]_w=(1;w^r;w^s) [ r ; s ] w = ( 1 ; w r ; w s )
r A = ( 0 , 8 , 16 , 1 , 13 , 2 , 14 ) , r_A=(0,8,16,1,13,2,14), r A = ( 0 , 8 , 16 , 1 , 13 , 2 , 14 ) , r B = ( 0 , 8 , 16 , 1 , 13 , 10 , 22 ) , r_B=(0,8,16,1,13,10,22), r B = ( 0 , 8 , 16 , 1 , 13 , 10 , 22 ) , r C = ( 0 , 8 , 16 , 2 , 14 , 6 , 18 ) . r_C=(0,8,16,2,14,6,18). r C = ( 0 , 8 , 16 , 2 , 14 , 6 , 18 ) . the representatives are
Table 6.
The last 3 use 12th roots. The family decompositions are given below; 123|45|67 denotes the indicated triple and pairs.
Table 7.
Put w = e 2 π i / 3 w=e^{2\pi i/3} w = e 2 π i /3
u = ( 1 , w , w 2 , a , − a , b , − b ) . u=(1,w,w^2,a,-a,b,-b). u = ( 1 , w , w 2 , a , − a , b , − b ) . The cyclic permutations of the triple, flips of the pairs, and exchange of the pairs form a group of order 24. Its orbits on the ( 7 3 ) ⋅ 3 = 105 \binom{7}{3}\cdot3=105 ( 3 7 ) ⋅ 3 = 105 v v v
Table 8.
For each support, rotate v v v a , b a,b a , b u v ˉ u\bar{v} u v ˉ 9 ⋅ 2 ⋅ 105 ⋅ 2 = 3780 9\cdot2\cdot105\cdot2=3780 9 ⋅ 2 ⋅ 105 ⋅ 2 = 3780
θ j = γ j + m j x , γ j ∈ 1 6 Z / Z , m j ∈ Z 4 . \theta_j=\gamma_j+m_jx,\qquad\gamma_j\in\frac{1}{6}\mathbb{Z}/\mathbb{Z},\quad m_j\in\mathbb{Z}^4. θ j = γ j + m j x , γ j ∈ 6 1 Z / Z , m j ∈ Z 4 . and a support imposes
θ i 1 − θ i 0 = ε / 3 ( 1 ) , θ i 2 − θ i 0 = 2 ε / 3 ( 1 ) , \theta_{i_1}-\theta_{i_0}=\varepsilon/3\ (1),\qquad\theta_{i_2}-\theta_{i_0}=2\varepsilon/3\ (1), θ i 1 − θ i 0 = ε /3 ( 1 ) , θ i 2 − θ i 0 = 2 ε /3 ( 1 ) , θ p 1 − θ p 0 = 1 / 2 ( 1 ) , θ q 1 − θ q 0 = 1 / 2 ( 1 ) . \theta_{p_1}-\theta_{p_0}=1/2\ (1),\qquad\theta_{q_1}-\theta_{q_0}=1/2\ (1). θ p 1 − θ p 0 = 1/2 ( 1 ) , θ q 1 − θ q 0 = 1/2 ( 1 ) . Thus every solution is obtained from integer congruences A x = b Ax=b A x = b
For completeness, an equation on a branch x = c + B t x=c+Bt x = c + B t r t = h rt=h r t = h r V = ( g , 0 , … , 0 ) rV=(g,0,\ldots,0) r V = ( g , 0 , … , 0 ) g g g
( h + k ) / g , k = 0 , … , g − 1. (h+k)/g,\qquad k=0,\ldots,g-1. ( h + k ) / g , k = 0 , … , g − 1. and leave the remaining coordinates free. A zero row is checked directly. Induction proves exhaustiveness. Independently, choose an invertible rank minor M M M
x P = M − 1 ( b I − A I , F x F ) + g , g ∈ M − 1 Z / Z . x_P=M^{-1}(b_I-A_{I,F}x_F)+g,\qquad g\in M^{-1}\mathbb{Z}/\mathbb{Z}. x P = M − 1 ( b I − A I , F x F ) + g , g ∈ M − 1 Z / Z . Enumerate the ∣ det M ∣ |\det M| ∣ det M ∣
Table 9.
There are 1876 branches and 838 distinct dephased tori.
Each resulting torus admits a checked integral coordinate chart
y = k + C t , C P = I , k P = 0. y=k+Ct,\qquad C_P=I,\qquad k_P=0. y = k + C t , C P = I , k P = 0. For a source y = s + D t y=s+Dt y = s + D t
s − k − C s P ∈ Z 14 , D = C D P . (1) s-k-Cs_P\in\mathbb{Z}^{14},\qquad D=CD_P. \tag*{(1)} s − k − C s P ∈ Z 14 , D = C D P . (1) After row and column permutations and dephasing, these identities assign every source torus to a displayed representative. The assignments recorded in coverage.json have counts
Table 10.
Table 11.
These are coverage counts, not numbers of moduli components. Direct checks of all 3 ! 7 ! 3!7! 3 ! 7 ! independent_verification.json gives the exceptional cycle partitions.
The AAA and BBB classes
Theorem 2.5. With w = e 2 π i / 7 w=e^{2\pi i/7} w = e 2 π i /7
A l = ( 1 1 1 1 1 1 1 1 w w 2 w 3 w 4 w 5 w 6 w π l ( 0 ) w π l ( 1 ) w π l ( 2 ) w π l ( 3 ) w π l ( 4 ) w π l ( 5 ) w π l ( 6 ) ) , A_l=\begin{pmatrix}
1&1&1&1&1&1&1\\
1&w&w^2&w^3&w^4&w^5&w^6\\
w^{\pi_l(0)}&w^{\pi_l(1)}&w^{\pi_l(2)}&w^{\pi_l(3)}&w^{\pi_l(4)}&w^{\pi_l(5)}&w^{\pi_l(6)}
\end{pmatrix}, A l = 1 1 w π l ( 0 ) 1 w w π l ( 1 ) 1 w 2 w π l ( 2 ) 1 w 3 w π l ( 3 ) 1 w 4 w π l ( 4 ) 1 w 5 w π l ( 5 ) 1 w 6 w π l ( 6 ) , For w = e 2 π i / 5 w=e^{2\pi i/5} w = e 2 π i /5
B s = ( 1 1 1 1 1 1 1 1 w w 2 w 3 w 4 s i − s i − s i w − 1 w − 2 w − 3 w − 4 s i 1 ) , s ∈ { 1 , − 1 } . B_s=\begin{pmatrix}
1&1&1&1&1&1&1\\
1&w&w^2&w^3&w^4&si&-si\\
-si&w^{-1}&w^{-2}&w^{-3}&w^{-4}&si&1
\end{pmatrix},\qquad s\in\{1,-1\}. B s = 1 1 − s i 1 w w − 1 1 w 2 w − 2 1 w 3 w − 3 1 w 4 w − 4 1 s i s i 1 − s i 1 , s ∈ { 1 , − 1 } . Conjugation fixes A 1 , A 2 A_1,A_2 A 1 , A 2 A 3 , A 4 A_3,A_4 A 3 , A 4 B + , B − B_+,B_- B + , B −
Proof. For AAA, put w = e 2 π i / 7 w=e^{2\pi i/7} w = e 2 π i /7 u j = w j u_j=w^j u j = w j v j = w π ( j ) v_j=w^{\pi(j)} v j = w π ( j ) π ( 0 ) = 0 \pi(0)=0 π ( 0 ) = 0 π \pi π π − id \pi-\operatorname{id} π − id Z / 7 \mathbb{Z}/7 Z /7 6 ! 6! 6 !
Table 1 (PDF p. 14).
Choose ordered rows ( b , r , s ) (b,r,s) ( b , r , s ) a a a
f ( j ) = e r ( j ) − e b ( j ) , g ( j ) = e s ( j ) − e b ( j ) . f(j)=e_{r(j)}-e_{b(j)},\qquad g(j)=e_{s(j)}-e_{b(j)}. f ( j ) = e r ( j ) − e b ( j ) , g ( j ) = e s ( j ) − e b ( j ) . The complete equivalence action is
π ′ ( f ( j ) − f ( a ) ) = g ( j ) − g ( a ) ( 7 ) . (2) \pi'(f(j)-f(a))=g(j)-g(a)\ (7). \tag*{(2)} π ′ ( f ( j ) − f ( a )) = g ( j ) − g ( a ) ( 7 ) . (2) Indeed, the chosen row and anchor remove all phases, and the 2nd row forces the remaining column order. The 6 ⋅ 7 6\cdot7 6 ⋅ 7 π ( j ) ↦ − π ( − j ) \pi(j)\mapsto-\pi(-j) π ( j ) ↦ − π ( − j )
For BBB, put w = e 2 π i / 5 w=e^{2\pi i/5} w = e 2 π i /5 u , v u,v u , v μ 5 \mu_5 μ 5 U , V , W U,V,W U , V , W u , v , u v ‾ u,v,\overline{uv} u , v , uv O = ( U ∪ V ) c O=(U\cup V)^c O = ( U ∪ V ) c Q = W c Q=W^c Q = W c Q Q Q O O O Q Q Q μ 5 \mu_5 μ 5 U = V U=V U = V U U U Q , W Q,W Q , W μ 5 \mu_5 μ 5 − μ 5 -\mu_5 − μ 5 U ∩ V = ∅ U\cap V=\varnothing U ∩ V = ∅ W W W U U U V V V − μ 5 ∖ { − 1 } -\mu_5\setminus\{-1\} − μ 5 ∖ { − 1 } Q Q Q − μ 5 -\mu_5 − μ 5 W W W U = { i , k } U=\{i,k\} U = { i , k } V = { j , k } V=\{j,k\} V = { j , k }
u i = − a , u k = a , u j = α ∈ μ 5 , v j = − b , v k = b , v i = β ∈ μ 5 . u_i=-a,\quad u_k=a,\quad u_j=\alpha\in\mu_5,\qquad v_j=-b,\quad v_k=b,\quad v_i=\beta\in\mu_5. u i = − a , u k = a , u j = α ∈ μ 5 , v j = − b , v k = b , v i = β ∈ μ 5 . X = z i = − a / β X=z_i=-a/\beta X = z i = − a / β Y = z j = − α / b Y=z_j=-\alpha/b Y = z j = − α / b Z = z k = a / b Z=z_k=a/b Z = z k = a / b X Y / Z = α / β ∈ μ 5 XY/Z=\alpha/\beta\in\mu_5 X Y / Z = α / β ∈ μ 5
If W W W O O O X , Y , Z X,Y,Z X , Y , Z μ 5 \mu_5 μ 5 W W W { i , j , k } \{i,j,k\} { i , j , k } W = { i , k } W=\{i,k\} W = { i , k } { j , k } \{j,k\} { j , k } b = β b=\beta b = β a = α a=\alpha a = α μ 5 \mu_5 μ 5
U = { i , k } , V = { j , k } , W = { i , j } . U=\{i,k\},\qquad V=\{j,k\},\qquad W=\{i,j\}. U = { i , k } , V = { j , k } , W = { i , j } . Products over the 5-cycles give ( a / b ) ( β / α ) = 1 (a/b)(\beta/\alpha)=1 ( a / b ) ( β / α ) = 1 X = − Y X=-Y X = − Y a b = − α β ab=-\alpha\beta ab = − α β
a α = b β = t , t 2 = − 1 , t = s i ( s = ± 1 ) . (3) \frac{a}{\alpha}=\frac{b}{\beta}=t,\qquad t^2=-1,\qquad t=si\ (s=\pm1). \tag*{(3)} α a = β b = t , t 2 = − 1 , t = s i ( s = ± 1 ) . (3) After scaling by α − 1 , β − 1 \alpha^{-1},\beta^{-1} α − 1 , β − 1 j , O , k , i j,O,k,i j , O , k , i
\begin{pmatrix}
1&1&1&1&1&1&1\\
1&w&w^2&w^3&w^4&t&-t\\
-t&w^{\pi(1)}&w^{\pi(2)}&w^{\pi(3)}&w^{\pi(4)}&t&1
\end{pmatrix}. \tag*{(4)}\end{pmatrix}
The permutations of { 1 , 2 , 3 , 4 } \{1,2,3,4\} { 1 , 2 , 3 , 4 }
Row exchanges induce r ↦ r − 1 r\mapsto r^{-1} r ↦ r − 1 r ↦ 1 − r r\mapsto1-r r ↦ 1 − r t t t r = 2 , 3 , 4 r=2,3,4 r = 2 , 3 , 4 B s B_s B s
The antipodal supports are unique, and
t = H 2 k H 1 j H 1 k H 2 j , { k } = U ∩ V , { j } = V ∩ W . t=\frac{H_{2k}H_{1j}}{H_{1k}H_{2j}},\qquad\{k\}=U\cap V,\qquad\{j\}=V\cap W. t = H 1 k H 2 j H 2 k H 1 j , { k } = U ∩ V , { j } = V ∩ W . is unchanged by phases, column permutations, and the generating row swaps, which replace t t t t t t − 1 / t = t -1/t=t − 1/ t = t
The ACC and BCC classes
We next treat the mixed types. We choose the all-ones row at the common vertex of the 2 edges of type C C C A A A B B B
The normal form below makes the last step particularly transparent. Both nonconstant rows already have type C C C A C C ACC A C C B C C BCC B C C
Put w = e 2 π i / 21 w=e^{2\pi i/21} w = e 2 π i /21 ω = w 7 \omega=w^7 ω = w 7
M ( x , y , z , t ) = ( 1 1 1 1 1 1 1 1 ω ω 2 ω z − ω z ω 2 t − ω 2 t 1 ω / x ω 2 / y ω − ω / x ω 2 − ω 2 / y ) . (5) M(x,y,z,t)=
\begin{pmatrix}
1&1&1&1&1&1&1\\
1&\omega&\omega^2&\omega z&-\omega z&\omega^2t&-\omega^2t\\
1&\omega/x&\omega^2/y&\omega&-\omega/x&\omega^2&-\omega^2/y
\end{pmatrix}. \tag*{(5)} M ( x , y , z , t ) = 1 1 1 1 ω ω / x 1 ω 2 ω 2 / y 1 ω z ω 1 − ω z − ω / x 1 ω 2 t ω 2 1 − ω 2 t − ω 2 / y . (5) Its last 2 rows have type C C C
u v ‾ = ( 1 , x , y , z , x z , t , y t ) . (6) u\overline{v}=(1,x,y,z,xz, t,yt). \tag*{(6)} u v = ( 1 , x , y , z , x z , t , y t ) . (6) Theorem 2.6. There are exactly 12 A C C ACC A C C B C C BCC B C C A C C ACC A C C w = e 2 π i / 21 w=e^{2\pi i/21} w = e 2 π i /21 ω = w 7 \omega=w^7 ω = w 7 ( x , y , z , t ) = ( w 3 p , w 3 q , w 3 r , w 3 s ) (x,y,z,t)=(w^{3p},w^{3q},w^{3r},w^{3s}) ( x , y , z , t ) = ( w 3 p , w 3 q , w 3 r , w 3 s )
For B C C BCC B C C w = e 2 π i / 15 w=e^{2\pi i/15} w = e 2 π i /15 ω = w 5 \omega=w^5 ω = w 5
( j , k ) ∈ S = { ( 1 , 2 ) , ( 1 , 3 ) , ( 2 , 1 ) , ( 3 , 1 ) } , ε ∈ { 1 , − 1 } , h = ε i w − 3 ( j + k ) (j,k)\in S=\{(1,2),(1,3),(2,1),(3,1)\},\qquad\varepsilon\in\{1,-1\},\qquad h=\varepsilon i w^{-3(j+k)} ( j , k ) ∈ S = {( 1 , 2 ) , ( 1 , 3 ) , ( 2 , 1 ) , ( 3 , 1 )} , ε ∈ { 1 , − 1 } , h = ε i w − 3 ( j + k ) and take both placements
( x , y , z , t ) = ( h , w 3 j , − h , w 3 k ) or ( w 3 j , h , w 3 k , − h ) . (7) (x,y,z,t)=(h,w^{3j},-h,w^{3k})\quad\text{or}\quad(w^{3j},h,w^{3k},-h). \tag*{(7)} ( x , y , z , t ) = ( h , w 3 j , − h , w 3 k ) or ( w 3 j , h , w 3 k , − h ) . (7) The representatives are pairwise inequivalent. They use respectively 42nd and 60th roots.
Proof. Normalize u = ( 1 , ω , ω 2 , a , − a , b , − b ) u=(1,\omega,\omega^2,a,-a,b,-b) u = ( 1 , ω , ω 2 , a , − a , b , − b ) v v v A A A 6 ! = 720 6!=720 6 ! = 720 B B B ( 4 2 ) 4 ! = 504 \binom{4}{2}4!=504 ( 2 4 ) 4 ! = 504 q j = c j + B j θ q_j=c_j+B_j\theta q j = c j + B j θ
( B j − B i ) θ = δ − c j + c i (8) (B_j-B_i)\theta=\delta-c_j+c_i \tag*{(8)} ( B j − B i ) θ = δ − c j + c i (8) The 2 congruence solvers above process 12960 and 9072 systems. Their exact results are
Table 1 (PDF p. 16).
Table 2.
All consistent systems have rank 4. Only support 8 survives, with the same triple orientation as u u u
u = ( 1 , ω , ω 2 , a , − a , b , − b ) , v = ( 1 , c , d , ω , − c , ω 2 , − d ) . u=(1,\omega,\omega^2,a,-a,b,-b),\qquad v=(1,c,d,\omega,-c,\omega^2,-d). u = ( 1 , ω , ω 2 , a , − a , b , − b ) , v = ( 1 , c , d , ω , − c , ω 2 , − d ) .
The substitutions x = ω / c x=\omega/c x = ω / c y = ω 2 / d y=\omega^{2}/d y = ω 2 / d z = a / ω z=a/\omega z = a / ω t = b / ω 2 t=b/\omega^{2} t = b / ω 2 (5) . With w = e 2 π i / 21 w=e^{2\pi i/21} w = e 2 π i /21 ω = w 7 \omega=w^{7} ω = w 7 A C C ACC A C C
{ 0 , p , q , r , p + r , s , q + s } = Z / 7 , ( x , y , z , t ) = ( w 3 p , w 3 q , w 3 r , w 3 s ) . (9) \{0,p,q,r,p+r,s,q+s\}=\mathbb{Z}/7,\qquad(x,y,z,t)=(w^{3p},w^{3q},w^{3r},w^{3s}). \tag*{(9)} { 0 , p , q , r , p + r , s , q + s } = Z /7 , ( x , y , z , t ) = ( w 3 p , w 3 q , w 3 r , w 3 s ) . (9) With w = e 2 π i / 15 w=e^{2\pi i/15} w = e 2 π i /15 ω = w 5 \omega=w^{5} ω = w 5 B C C BCC B C C (7) with all 8 ordered pairs j , k ∈ { 1 , 2 , 3 , 4 } j,k\in\{1,2,3,4\} j , k ∈ { 1 , 2 , 3 , 4 } j ≠ k j\ne k j = k j + k ≠ 0 j+k\ne0 j + k = 0 ( h , − h ) (h,-h) ( h , − h )
1 , w 3 j , w 3 k , w 3 ( j + k ) , − h 2 = w − 6 ( j + k ) . 1,\quad w^{3j},\quad w^{3k},\quad w^{3(j+k)},\quad-h^{2}=w^{-6(j+k)}. 1 , w 3 j , w 3 k , w 3 ( j + k ) , − h 2 = w − 6 ( j + k ) . This verifies validity without a root-order assumption in the search.
In each type-C C C μ 6 \mu_{6} μ 6
u : 123 ∣ 45 ∣ 67 , v : 146 ∣ 25 ∣ 37. (10) u:123\mid45\mid67,\qquad v:146\mid25\mid37. \tag*{(10)} u : 123 ∣ 45 ∣ 67 , v : 146 ∣ 25 ∣ 37. (10) The all-ones row is the unique vertex incident to both C C C u , v u,v u , v ( 1 , 4 , 6 , 2 , 5 , 3 , 7 ) (1,4,6,2,5,3,7) ( 1 , 4 , 6 , 2 , 5 , 3 , 7 )
Φ ( x , y , z , t ) = ( z − 1 , t − 1 , x − 1 , y − 1 ) . (11) \Phi(x,y,z,t)=(z^{-1},t^{-1},x^{-1},y^{-1}). \tag*{(11)} Φ ( x , y , z , t ) = ( z − 1 , t − 1 , x − 1 , y − 1 ) . (11) A fixed point would give x z = y t = 1 xz=yt=1 x z = y t = 1 (6) , impossible for either type. Thus every orbit has size 2. On the B C C BCC B C C Φ \Phi Φ ( j , k ) (j,k) ( j , k ) ( − k , − j ) (-k,-j) ( − k , − j )
Together these classifications prove Theorem 2.4: different types are inequivalent. Each type locus is compact and hence closed in the monomial quotient; the finite type loci and the 11 remaining C C C CCC C C C
Common supports and special configurations
Lemma 2.7. Put w = e 2 π i / 3 w=e^{2\pi i/3} w = e 2 π i /3 ω ∈ { w , w 2 } \omega\in\{w,w^{2}\} ω ∈ { w , w 2 } u u u v v v u v ‾ \overline{uv} uv C C C u , v u,v u , v
\begin{pmatrix}
1&\omega&\omega^{2}&a&-a&a&-a\\
1&\omega^{2}&\omega&b&c&-b&-c
\end{pmatrix}. \tag*{(12)}\end{pmatrix}
\begin{pmatrix}
1&\omega&\omega^{2}&a&-a&ab&-ab\\
1&\omega^{2}&\omega&c&-c&-bc&bc
\end{pmatrix}. \tag*{(13)}\end{pmatrix}
\begin{pmatrix}
1&\omega&\omega^{2}&\omega a&-\omega a&a&-a\\
1&\omega&\omega^{2}&a&\omega a&-a&-\omega a
\end{pmatrix}. \tag*{(14)}\end{pmatrix}
Here all parameters are unit phases. Conversely every displayed pair has this property.
Proof. Normalize u = ( 1 , ω , ω 2 , a , − a , b , − b ) u=(1,\omega,\omega^{2},a,-a,b,-b) u = ( 1 , ω , ω 2 , a , − a , b , − b )
Table 12.
For opposite triple orientations, their contribution to the inner product is 0. The respective equations are
2 a c ‾ + 2 b d ‾ = 0 , d = − b c a . 2a\overline{c}+2b\overline{d}=0,\qquad d=-\frac{bc}{a}. 2 a c + 2 b d = 0 , d = − a b c . a c ‾ − a d ‾ − b c ‾ + b d ‾ = ( a − b ) ( c ‾ − d ‾ ) = 0. a\overline{c}-a\overline{d}-b\overline{c}+b\overline{d}=(a-b)(\overline{c}-\overline{d})=0. a c − a d − b c + b d = ( a − b ) ( c − d ) = 0. They give (13) and (12) ; in the case c = d c=d c = d
For equal orientations, the quotient starts with 1 , 1 , 1 1,1,1 1 , 1 , 1 { ω , ω 2 , − 1 , − 1 } \{\omega,\omega^2,-1,-1\} { ω , ω 2 , − 1 , − 1 }
( s 1 , s 2 , s 3 , s 4 ) = ( a c ‾ , − a d ‾ , − b c ‾ , b d ‾ ) . (s_1,s_2,s_3,s_4)=(a\overline{c},-a\overline{d},-b\overline{c},b\overline{d}). ( s 1 , s 2 , s 3 , s 4 ) = ( a c , − a d , − b c , b d ) . Then s 1 s 4 = s 2 s 3 s_1s_4=s_2s_3 s 1 s 4 = s 2 s 3 { ω , ω 2 } \{\omega,\omega^2\} { ω , ω 2 } { − 1 , − 1 } \{-1,-1\} { − 1 , − 1 } ( 12 ) ( 34 ) , ( 13 ) ( 24 ) (12)(34),(13)(24) ( 12 ) ( 34 ) , ( 13 ) ( 24 )
( s 1 , s 2 , s 3 , s 4 ) = ( ω , − 1 , − 1 , ω 2 ) . (s_1,s_2,s_3,s_4)=(\omega,-1,-1,\omega^2). ( s 1 , s 2 , s 3 , s 4 ) = ( ω , − 1 , − 1 , ω 2 ) . Hence c = ω 2 a c=\omega^2a c = ω 2 a d = a d=a d = a b = c b=c b = c (14) . Substitution verifies all 3 converses.
For any odd prime p p p
u = ( 1 , w , w 2 , … , w p − 1 , i , − i ) , u=(1,w,w^2,\ldots,w^{p-1},i,-i), u = ( 1 , w , w 2 , … , w p − 1 , i , − i ) , v = ( − i , w − 1 , w − 2 , … , w − ( p − 1 ) , i , 1 ) , v=(-i,w^{-1},w^{-2},\ldots,w^{-(p-1)},i,1), v = ( − i , w − 1 , w − 2 , … , w − ( p − 1 ) , i , 1 ) , u v ‾ = ( i , w 2 , w 4 , … , w 2 ( p − 1 ) , 1 , − i ) . u\overline{v}=(i,w^2,w^4,\ldots,w^{2(p-1)},1,-i). u v = ( i , w 2 , w 4 , … , w 2 ( p − 1 ) , 1 , − i ) . with w = e 2 π i / p w=e^{2\pi i/p} w = e 2 π i / p ( p + 2 , p + 2 , p + 2 ) (p+2,p+2,p+2) ( p + 2 , p + 2 , p + 2 ) p p p
Put w = e π i / 12 w=e^{\pi i/12} w = e π i /12 ω = w W \omega=w^W ω = w W W ∈ { 8 , 16 } W\in\{8,16\} W ∈ { 8 , 16 } E 1 , … , E 8 E_1,\ldots,E_8 E 1 , … , E 8
( 1 ω ω 2 w r − ω 1 + e w 2 r − w r ω 1 + e w 2 r 1 − ω 2 w − r ω 1 + e w 2 r ω ω 2 ω 2 w − r − ω 1 + e w 2 r ) , \begin{pmatrix}
1 & \omega& \omega^2 & w^r & -\omega^{1+e}w^{2r} & -w^r & \omega^{1+e}w^{2r}\\
1 & -\omega^2w^{-r} & \omega^{1+e}w^{2r} & \omega& \omega^2 & \omega^2w^{-r} & -\omega^{1+e}w^{2r}
\end{pmatrix}, ( 1 1 ω − ω 2 w − r ω 2 ω 1 + e w 2 r w r ω − ω 1 + e w 2 r ω 2 − w r ω 2 w − r ω 1 + e w 2 r − ω 1 + e w 2 r ) , w 4 r = − ω , r ∈ Z / 24 , e ∈ { 0 , 1 } . w^{4r}=-\omega,\qquad r\in\mathbb{Z}/24,\quad e\in\{0,1\}. w 4 r = − ω , r ∈ Z /24 , e ∈ { 0 , 1 } . The 16 choices give
Table 13.
For w = e 2 π i / 3 w=e^{2\pi i/3} w = e 2 π i /3 s , t ∈ { 1 , − 1 } s,t\in\{1,-1\} s , t ∈ { 1 , − 1 }
Y s , t = ( 1 1 1 1 1 1 1 1 w w 2 s i t i − s i − t i 1 s i w 2 t i w w w 2 − s i w 2 − t i w ) Y_{s,t}=
\begin{pmatrix}
1 & 1 & 1 & 1 & 1 & 1 & 1\\
1 & w & w^2 & si & ti & -si & -ti\\
1 & siw^2 & tiw & w & w^2 & -siw^2 & -tiw
\end{pmatrix} Y s , t = 1 1 1 1 w s i w 2 1 w 2 t i w 1 s i w 1 t i w 2 1 − s i − s i w 2 1 − t i − t i w instead belong to continuous families:
Y s , − s = F 5 ( − s i w ) , Y s , s ( 5 7 ) = F 6 ( s i ) , Y_{s,-s}=\mathcal{F}_5(-siw),\qquad Y_{s,s}^{(5\,7)}=\mathcal{F}_6(si), Y s , − s = F 5 ( − s i w ) , Y s , s ( 5 7 ) = F 6 ( s i ) , The superscript exchanges columns 5 and 7. For the other cube-root orientation ω = w 2 \omega=w^2 ω = w 2 F 6 ( s i ) \mathcal{F}_6(si) F 6 ( s i ) ( 1 , 3 , 2 , 5 , 6 , 7 , 4 ) (1,3,2,5,6,7,4) ( 1 , 3 , 2 , 5 , 6 , 7 , 4 ) F 5 ( s i w ) \mathcal{F}_5(siw) F 5 ( s i w ) ( 1 , 3 , 2 , 5 , 4 , 7 , 6 ) (1,3,2,5,4,7,6) ( 1 , 3 , 2 , 5 , 4 , 7 , 6 ) code/ccc/special_example_checks.json.
The square families
Put w = e 2 π i / 3 w=e^{2\pi i/3} w = e 2 π i /3 q ∈ T q\in\mathbb{T} q ∈ T [26 , 29 ] and its conjugate in Aalto catalogue normalization [21 ] are
P 7 q = ( − q q w 1 w 1 w q − q w 1 1 w w w w − w 1 w w 1 1 1 1 − 1 w w w w 1 w w − q ˉ w q ˉ w 1 1 w w w q ˉ w − q ˉ w 1 w w 1 w 1 1 − 1 ) . P_7^q=\begin{pmatrix}
-q & q & w & 1 & w & 1 & w\\
q & -q & w & 1 & 1 & w & w\\
w & w & -w & 1 & w & w & 1\\
1 & 1 & 1 & -1 & w & w & w\\
w & 1 & w & w & -\bar{q}w & \bar{q}w & 1\\
1 & w & w & w & \bar{q}w & -\bar{q}w & 1\\
w & w & 1 & w & 1 & 1 & -1
\end{pmatrix}. P 7 q = − q q w 1 w 1 w q − q w 1 1 w w w w − w 1 w w 1 1 1 1 − 1 w w w w 1 w w − q ˉ w q ˉ w 1 1 w w w q ˉ w − q ˉ w 1 w w 1 w 1 1 − 1 . R 7 q = ( 1 1 1 1 1 1 1 1 1 1 w − 1 − 1 w 2 1 − w 2 w 2 w 2 q − q w 1 w − 1 1 q w 2 − q w 2 w 2 1 − 1 − w − 1 − q q − w 2 1 w 2 w q w q − q w − q 1 w 2 w − q w q w − q q ) . R_7^q=\begin{pmatrix}
1 & 1 & 1 & 1 & 1 & 1 & 1\\
1 & 1 & 1 & w & -1 & -1 & w^2\\
1 & -w^2 & w^2 & w^2 & q & -q & w\\
1 & w & -1 & 1 & qw^2 & -qw^2 & w^2\\
1 & -1 & -w & -1 & -q & q & -w^2\\
1 & w^2 & w & qw & q & -qw & -q\\
1 & w^2 & w & -qw & qw & -q & q
\end{pmatrix}. R 7 q = 1 1 1 1 1 1 1 1 1 − w 2 w − 1 w 2 w 2 1 1 w 2 − 1 − w w w 1 w w 2 1 − 1 q w − q w 1 − 1 q q w 2 − q q q w 1 − 1 − q − q w 2 q − q w − q 1 w 2 w w 2 − w 2 − q q . We now state the square classification. Its proof occupies the rest of this section, through Subsection 2.10.
Theorem 2.8. Every regular complex Hadamard matrix of order 7 is equivalent to
Conjugation is not included in equivalence; parameter uniqueness is not asserted.
We first verify the displayed families, then treat the isolated triples and reduce the affine cases to the remaining completions in Subsection 2.10. The argument there closes the proof. Code and exact certificates are in code002_work/N7/, restored as explained in Subsection 6.7.
Proposition 2.9. Both families are regular for every q q q
r = ( 1 , 2 , 7 , 3 , 4 , 6 , 5 ) , c = ( 4 , 7 , 3 , 5 , 2 , 1 , 6 ) . r=(1,2,7,3,4,6,5),\qquad c=(4,7,3,5,2,1,6). r = ( 1 , 2 , 7 , 3 , 4 , 6 , 5 ) , c = ( 4 , 7 , 3 , 5 , 2 , 1 , 6 ) . we have
R 7 q = D ( ( P 7 q ‾ ) r , c ) , q ∈ T . (15) R_7^q=D\left((\overline{P_7^q})_{r,c}\right),\qquad q\in\mathbb{T}. \tag*{(15)} R 7 q = D ( ( P 7 q ) r , c ) , q ∈ T . (15) The seeds P 7 1 P_7^1 P 7 1 R 7 1 R_7^1 R 7 1
Proof. For this proof put w = e π i / 3 w=e^{\pi i/3} w = e π i /3 w E i j q D i j w^{E_{ij}}q^{D_{ij}} w E ij q D ij q q q (15) . Exhaustive row orders and dephasing anchors distinguish the seeds. The linearized equations
∑ j = 1 7 H i j H ‾ k j ( X i j − X k j ) = 0 ( i < k ) , X i j ∈ R . \sum_{j=1}^{7}H_{ij}\overline{H}_{kj}(X_{ij}-X_{kj})=0\quad(i<k),\qquad X_{ij}\in\mathbb{R}. j = 1 ∑ 7 H ij H k j ( X ij − X k j ) = 0 ( i < k ) , X ij ∈ R . give, by rational row reduction,
Table 14.
Subtracting the 13 trivial phase directions gives defect 3; it need not equal the dimension of the displayed affine family.
Extensions of the isolated classes
Theorem 2.10. Among the 45 isolated 3-row classes, only A 1 A_1 A 1 A 2 A_2 A 2 w = e 2 π i / 7 w=e^{2\pi i/7} w = e 2 π i /7 f k = ( 1 , w k , … , w 6 k ) f_k=(1,w^k,\ldots,w^{6k}) f k = ( 1 , w k , … , w 6 k )
Table 15.
Their square completion is F 7 F_7 F 7 K × 7 K \times7 K × 7 4 ≤ K ≤ 7 4 \le K \le7 4 ≤ K ≤ 7
Proof. All seed entries have orders dividing 840. Write r j ( i ) = e 2 π i d i j / 840 r_j^{(i)}=e^{2\pi i d_{ij}/840} r j ( i ) = e 2 π i d ij /840 x j = e 2 π i X j x_j=e^{2\pi i X_j} x j = e 2 π i X j X 1 = 0 X_1=0 X 1 = 0
X j − X k = d i j − d i k 840 + l p ( 1 ) , p ∈ { 2 , 3 , 5 , 7 } . X_j-X_k=\frac{d_{ij}-d_{ik}}{840}+\frac{l}{p}\quad(1),\qquad p\in\{2,3,5,7\}. X j − X k = 840 d ij − d ik + p l ( 1 ) , p ∈ { 2 , 3 , 5 , 7 } . There are respectively 720, 504, 210 support systems of types A A A B B B C C C
The unpruned check first tests every normalized A A A 714 2 714^2 71 4 2 B , C B,C B , C
Table 16.
The only survivors are the 8 Fourier rows above. Weighted graph elimination with the established type exclusions independently gives the same result; see code/isolated45/. If a larger matrix has an isolated triple, every further row is therefore Fourier. Otherwise Theorem 2.4 puts all triples in the affine families.
Completing the family F4
Put w = e 2 π i / 3 w=e^{2\pi i/3} w = e 2 π i /3 x 1 = 1 x_1=1 x 1 = 1
Proposition 2.11. For every a ∈ T a\in\mathbb{T} a ∈ T
F 4 ( a ) = ( 1 1 1 1 1 1 1 1 w w 2 w a − w a a − a 1 w w 2 a w a − a − w a ) \mathcal{F}_4(a)=\begin{pmatrix}
1&1&1&1&1&1&1\\
1&w&w^2&wa&-wa&a&-a\\
1&w&w^2&a&wa&-a&-wa
\end{pmatrix} F 4 ( a ) = 1 1 1 1 w w 1 w 2 w 2 1 w a a 1 − w a w a 1 a − a 1 − a − w a has exactly the 7 regular additional rows
1 2 3 4 5 6 7 x ( 1 ) 1 w 2 − 1 a w 2 w 1 − a w 2 x ( 2 ) 1 w 2 − w 2 a w 2 w − a x ( 3 ) 1 − w w a w w 2 w − a w x ( 4 ) 1 − 1 w a w 2 1 w 2 − a w 2 x ( 5 ) 1 1 1 − 1 w w 2 − 1 x ( 6 ) 1 − 1 − w 2 − a w − w − 1 a w x ( 7 ) 1 − w − 1 − a − 1 − w 2 a \begin{array}{c|rrrrrrr}
&1&2&3&4&5&6&7\\
x^{(1)}&1&w^2&-1&aw^2&w&1&-aw^2\\
x^{(2)}&1&w^2&-w^2&a&w^2&w&-a\\
x^{(3)}&1&-w&w&aw&w^2&w&-aw\\
x^{(4)}&1&-1&w&aw^2&1&w^2&-aw^2\\
x^{(5)}&1&1&1&-1&w&w^2&-1\\
x^{(6)}&1&-1&-w^2&-aw&-w&-1&aw\\
x^{(7)}&1&-w&-1&-a&-1&-w^2&a
\end{array} x ( 1 ) x ( 2 ) x ( 3 ) x ( 4 ) x ( 5 ) x ( 6 ) x ( 7 ) 1 1 1 1 1 1 1 1 2 w 2 w 2 − w − 1 1 − 1 − w 3 − 1 − w 2 w w 1 − w 2 − 1 4 a w 2 a a w a w 2 − 1 − a w − a 5 w w 2 w 2 1 w − w − 1 6 1 w w w 2 w 2 − 1 − w 2 7 − a w 2 − a − a w − a w 2 − 1 a w a Their maximal compatible subsets are
{ 1 , 3 , 5 , 6 } , { 2 , 4 , 5 , 7 } . \{1,3,5,6\},\qquad\{2,4,5,7\}. { 1 , 3 , 5 , 6 } , { 2 , 4 , 5 , 7 } . They give respectively 7 7 7 12 12 12 8 8 8 2 2 2 4 4 4 5 5 5 6 6 6 7 7 7 2 2 2
Proof. By the isolated-class theorem, only C C C CCC C C C
X i − X j = c + n θ ( 1 ) , c ∈ 1 840 Z , n ∈ Z . X_i-X_j=c+n\theta\ (1),\qquad c\in\frac{1}{840}\mathbb{Z},\quad n\in\mathbb{Z}. X i − X j = c + n θ ( 1 ) , c ∈ 840 1 Z , n ∈ Z . A spanning forest expresses the row phases in terms of θ \theta θ n θ + c = 0 n\theta+c=0 n θ + c = 0 ∣ n ∣ |n| ∣ n ∣ n ≠ 0 n\ne0 n = 0 210 2 210^2 21 0 2 a ∈ μ 6 a\in\mu_6 a ∈ μ 6 C C C CCC C C C code/f4/ verifies the list independently.
Put U = ( 1 , 3 , 6 ) U=(1,3,6) U = ( 1 , 3 , 6 ) V = ( 2 , 4 , 7 ) V=(2,4,7) V = ( 2 , 4 , 7 ) U , V U,V U , V
( ⟨ x ( i ) , x ( j ) ⟩ ) i ∈ U , j ∈ V = ( 1 + 3 w 2 1 − 4 w 2 5 + w 2 1 − 4 w 2 1 + 3 w 2 5 + w 2 5 + w 2 5 + w 2 − 3 − 2 w 2 ) . \left(\langle x^{(i)},x^{(j)}\rangle\right)_{i\in U,j\in V}
=
\begin{pmatrix}
1+3w^2&1-4w^2&5+w^2\\
1-4w^2&1+3w^2&5+w^2\\
5+w^2&5+w^2&-3-2w^2
\end{pmatrix}. ( ⟨ x ( i ) , x ( j ) ⟩ ) i ∈ U , j ∈ V = 1 + 3 w 2 1 − 4 w 2 5 + w 2 1 − 4 w 2 1 + 3 w 2 5 + w 2 5 + w 2 5 + w 2 − 3 − 2 w 2 . All entries are nonzero. The 12 orthogonal pairs have fixed C C C K 4 K_4 K 4
Appending rows ( 1 , 3 , 5 , 6 ) (1,3,5,6) ( 1 , 3 , 5 , 6 ) ( 2 , 4 , 5 , 7 ) (2,4,5,7) ( 2 , 4 , 5 , 7 ) H + ( a ) , H − ( a ) H_+(a),H_-(a) H + ( a ) , H − ( a )
H + ( a ) = D ( P 7 − w 2 a ) r + , c + , H − ( a ) = D ( R 7 − a ) r − , c − . H_+(a)=D\left(P_7^{-w^2a}\right)_{r_+,c_+},\qquad
H_-(a)=D\left(R_7^{-a}\right)_{r_-,c_-}. H + ( a ) = D ( P 7 − w 2 a ) r + , c + , H − ( a ) = D ( R 7 − a ) r − , c − . r + = ( 5 , 1 , 2 , 3 , 4 , 6 , 7 ) , c + = ( 7 , 4 , 3 , 5 , 2 , 1 , 6 ) , r_+=(5,1,2,3,4,6,7),\qquad c_+=(7,4,3,5,2,1,6), r + = ( 5 , 1 , 2 , 3 , 4 , 6 , 7 ) , c + = ( 7 , 4 , 3 , 5 , 2 , 1 , 6 ) , r − = ( 1 , 6 , 7 , 3 , 4 , 2 , 5 ) , c − = ( 1 , 3 , 2 , 6 , 4 , 7 , 5 ) . r_-=(1,6,7,3,4,2,5),\qquad c_-=(1,3,2,6,4,7,5). r − = ( 1 , 6 , 7 , 3 , 4 , 2 , 5 ) , c − = ( 1 , 3 , 2 , 6 , 4 , 7 , 5 ) . These are entrywise monomial identities. Thus every square containing an F 4 \mathcal{F}_4 F 4
The affine extension census
Put
w = e π i / 3 , ω = w 2 , Ω = μ 6 , S = Ω ∖ { 1 , − 1 } , U = { w , w − 1 } , T = { − 1 , ω , ω 2 } . w = e^{\pi i/3},\quad\omega= w^2,\quad\Omega= \mu_6,\quad S = \Omega\setminus\{1,-1\},\quad U = \{w,w^{-1}\},\quad T = \{-1,\omega,\omega^2\}. w = e π i /3 , ω = w 2 , Ω = μ 6 , S = Ω ∖ { 1 , − 1 } , U = { w , w − 1 } , T = { − 1 , ω , ω 2 } . Theorem 2.12. The joint spaces of seed parameters and normalized regular 4th rows have the following maximal connected affine components:
Every component is a point or circle. The seed parameters admitting a 4th row are precisely
F 1 a ∈ μ 3 , b ∈ Ω ∖ { a 2 , − a 2 } , c ∈ T , \mathcal{F}_1\quad a \in\mu_3,\ b \in\Omega\setminus\{a^2,-a^2\},\ c \in\mathbb{T}, F 1 a ∈ μ 3 , b ∈ Ω ∖ { a 2 , − a 2 } , c ∈ T ,
or a ∈ − μ 3 , c ∈ Ω ∖ { a 2 , − a 2 } , b ∈ T a \in-\mu_3,\ c \in\Omega\setminus\{a^2,-a^2\},\ b \in\mathbb{T} a ∈ − μ 3 , c ∈ Ω ∖ { a 2 , − a 2 } , b ∈ T
F 2 a ∈ S \mathcal{F}_2\quad a \in S F 2 a ∈ S b ∈ S b \in S b ∈ S
F 3 a ∈ U \mathcal{F}_3\quad a \in U F 3 a ∈ U b ∈ U b \in U b ∈ U a b ∈ { ω , ω 2 } , ab \in\{\omega,\omega^2\}, ab ∈ { ω , ω 2 } ,
or ( a , b ) ∈ T 2 , a ≠ b (a,b) \in T^2,\ a \ne b ( a , b ) ∈ T 2 , a = b
F 4 a ∈ T \mathcal{F}_4\quad a \in T F 4 a ∈ T
F 5 , F 6 a ∈ Ω \mathcal{F}_5,\mathcal{F}_6\quad a \in\Omega F 5 , F 6 a ∈ Ω
The fibres of F 4 , F 6 \mathcal{F}_4,\mathcal{F}_6 F 4 , F 6 F 5 \mathcal{F}_5 F 5 a = ± ω a = \pm\omega a = ± ω
Proof. Only CCC supports are required by Theorem 2.10 . Their equations are
X i − X j = c + n ⋅ θ ( 1 ) , c ∈ 1 6 Z / Z , n ∈ Z d . X_i-X_j=c+n\cdot\theta\ (1),\qquad c\in\tfrac{1}{6}\mathbb{Z}/\mathbb{Z},\quad n\in\mathbb{Z}^d. X i − X j = c + n ⋅ θ ( 1 ) , c ∈ 6 1 Z / Z , n ∈ Z d . A forest eliminates the row phases; the 2 exact congruence solvers then retain every torsion branch in the seed parameters. Enumerating 210 2 210^2 21 0 2
The 264600 initial pairs and 2597070 restrictions produce 1409 raw components; exact torus containment leaves 333. Both solvers agree on the complete loci. The file affine4_catalogue.txt lists every seed entry and row entry as w k t e w^{k}t^{e} w k t e e ∈ { − 1 , 0 , 1 } e \in\{-1,0,1\} e ∈ { − 1 , 0 , 1 } code/affine4/ verify the formulas and cycle certificates.
These counts concern fixed-seed extension spaces; components can intersect and need not be inequivalent as 4-row matrices. For F 3 \mathcal{F}_3 F 3 ( w , w ) , ( w − 1 , w − 1 ) (w,w),(w^{-1},w^{-1}) ( w , w ) , ( w − 1 , w − 1 )
Proposition 2.13. A regular square outside F 7 \mathcal{F}_7 F 7 P 1 q P_1^q P 1 q R 7 q R_7^q R 7 q F 2 \mathcal{F}_2 F 2 F 3 ( w , w ) \mathcal{F}_3(w,w) F 3 ( w , w ) F 3 ( w − 1 , w − 1 ) \mathcal{F}_3(w^{-1},w^{-1}) F 3 ( w − 1 , w − 1 )
Proof. A square requires 4 distinct added rows. For F 1 \mathcal{F}_1 F 1 F 4 \mathcal{F}_4 F 4 F 4 \mathcal{F}_4 F 4 F 5 \mathcal{F}_5 F 5 F 6 \mathcal{F}_6 F 6 Proposition 2.11 treats F 4 \mathcal{F}_4 F 4
Each of the 6 seed circles for F 3 \mathcal{F}_3 F 3 Off ( E ) = { ( a , b ) ∈ E 2 : a ≠ b } \operatorname{Off}(E)=\{(a,b)\in E^2:a\ne b\} Off ( E ) = {( a , b ) ∈ E 2 : a = b }
D 1 = ( { 1 } × S ) ∪ ( S × { 1 } ) ∪ { ( w , ω ) , ( ω , w ) , ( w 2 , w − 1 ) , ( w − 1 , w 2 ) } , D_1= (\{1\}\times S)\cup(S\times\{1\})
\cup\{(w,\omega),(\omega,w),(w^2,w^{-1}),(w^{-1},w^2)\}, D 1 = ({ 1 } × S ) ∪ ( S × { 1 }) ∪ {( w , ω ) , ( ω , w ) , ( w 2 , w − 1 ) , ( w − 1 , w 2 )} , D 2 = Off ( T ) ∪ Off ( { − 1 , w , w − 1 } ) , D_2=\operatorname{Off}(T)\cup\operatorname{Off}(\{-1,w,w^{-1}\}), D 2 = Off ( T ) ∪ Off ({ − 1 , w , w − 1 }) , D 3 = { ( w , w ) , ( w − 1 , w − 1 ) } . D_3=\{(w,w),(w^{-1},w^{-1})\}. D 3 = {( w , w ) , ( w − 1 , w − 1 )} . with the following exact counts:
Table 17.
Only D 3 D_3 D 3 F 4 \mathcal{F}_4 F 4 t ∈ μ 6 t\in\mu_6 t ∈ μ 6
The remaining completions
Proposition 2.14. Every square extending an admissible F 2 \mathcal{F}_2 F 2 F 3 \mathcal{F}_3 F 3 F 4 \mathcal{F}_4 F 4
Proof. For F 2 \mathcal{F}_2 F 2
E = { ( a , b ) ∈ Ω 2 : a ∈ S or b ∈ S } , ∣ E ∣ = 32. E=\{(a,b)\in\Omega^2:a\in S\text{ or }b\in S\},\qquad|E|=32. E = {( a , b ) ∈ Ω 2 : a ∈ S or b ∈ S } , ∣ E ∣ = 32. they exhaust the fibre. Restricting the full census to E E E
Table 18.
All 40 exceptional completions specialize the 8 circle completions. Append the records below from affine4_catalogue.txt, in increasing order with prefix F2-. The indicated ordered submatrix dephases to F 4 ( q ) \mathcal{F}_4(q) F 4 ( q )
Table 19.
Table 20.
For F 3 ( α , α ) \mathcal{F}_3(\alpha,\alpha) F 3 ( α , α ) α = w \alpha=w α = w w − 1 w^{-1} w − 1 Y i ( t ) Y_i(t) Y i ( t )
1 t w w 5 − 1 − 1 − t 1 t − 1 w 5 w − 1 − t 1 w t − 1 − t w 5 − 1 1 − 1 t − 1 − t w 5 w \begin{array}{rrrrrrr}
1 & t & w & w^5 & -1 & -1 & -t\\
1 & t & -1 & w^5 & w & -1 & -t\\
1 & w & t & -1 & -t & w^5 & -1\\
1 & -1 & t & -1 & -t & w^5 & w
\end{array} 1 1 1 1 t t w − 1 w − 1 t t w 5 w 5 − 1 − 1 − 1 w − t − t − 1 − 1 w 5 w 5 − t − t − 1 w for α = w \alpha=w α = w
1 t − 1 − 1 − t w w 5 1 t w 5 − 1 − t w − 1 1 − 1 t w w 5 − 1 − t 1 w 5 t w − 1 − 1 − t \begin{array}{rrrrrrr}
1 & t & -1 & -1 & -t & w & w^5\\
1 & t & w^5 & -1 & -t & w & -1\\
1 & -1 & t & w & w^5 & -1 & -t\\
1 & w^5 & t & w & -1 & -1 & -t
\end{array} 1 1 1 1 t t − 1 w 5 − 1 w 5 t t − 1 − 1 w w − t − t w 5 − 1 w w − 1 − 1 w 5 − 1 − t − t for α = w − 1 \alpha=w^{-1} α = w − 1
⟨ Y i ( t ) , Y i ( s ) ⟩ = 5 + 2 t s ˉ , ∣ 5 + 2 t s ˉ ∣ ≥ 3. \langle Y_i(t),Y_i(s)\rangle=5+2t\bar{s},\qquad\lvert5+2t\bar{s}\rvert\geq3. ⟨ Y i ( t ) , Y i ( s )⟩ = 5 + 2 t s ˉ , ∣ 5 + 2 t s ˉ ∣ ≥ 3. a square takes exactly 1 row from each circle. The first 3 orthogonality conditions are
1 + t 1 t 2 = 0 , 1 + α t 3 + ( 1 + α ˉ ) t 1 = 0 , 1 + α t 4 − ( 1 + α ˉ ) t 1 = 0. 1+\frac{t_1}{t_2}=0,\qquad\frac{1+\alpha}{t_3}+(1+\bar{\alpha})t_1=0,\qquad\frac{1+\alpha}{t_4}-(1+\bar{\alpha})t_1=0. 1 + t 2 t 1 = 0 , t 3 1 + α + ( 1 + α ˉ ) t 1 = 0 , t 4 1 + α − ( 1 + α ˉ ) t 1 = 0. and, as ( 1 + α ) / ( 1 + α ˉ ) = α (1+\alpha)/(1+\bar{\alpha})=\alpha ( 1 + α ) / ( 1 + α ˉ ) = α
( t 1 , t 2 , t 3 , t 4 ) = ( t , − t , − α / t , α / t ) , t ∈ T . (t_1,t_2,t_3,t_4)=(t,-t,-\alpha/t,\alpha/t),\qquad t\in\mathbb{T}. ( t 1 , t 2 , t 3 , t 4 ) = ( t , − t , − α / t , α / t ) , t ∈ T . The other inner products vanish identically, with fixed C C C ( Y 1 ( t ) , Y 3 ( − α / t ) , Y 4 ( α / t ) ) (Y_1(t),Y_3(-\alpha/t),Y_4(\alpha/t)) ( Y 1 ( t ) , Y 3 ( − α / t ) , Y 4 ( α / t )) F 4 ( q ) \mathcal{F}_4(q) F 4 ( q )
Table 21.
All divisions use unit parameters or nonzero constants, so no exceptional parameter is omitted. The checker in code/completion/ verifies these identities and the finite graphs exactly.
The isolated-class theorem, the affine census and this proposition put every non-Fourier square in a Petrescu branch. Conversely, F 7 F_7 F 7 Theorem 2.8 .
Order 8: supports and completion criteria
The order-8 classification proceeds from normalized 3-row matrices to completed squares. A row quotient has type A (4 2-cycles), B (2 3-cycles and 1 2-cycle), or C (a 5-cycle and a 3-cycle). We first record where the cycles sit, then impose the phase conditions that make those supports into actual row quotients. Exact extension equations tell us which further rows can be added and, ultimately, which triples extend to squares.
Sections 3 and 4 carry out this local analysis. Section 3 enumerates the possible supports and their phase solutions, develops the exact test for adding rows, and treats the AAA triples, including the parameter values where their completion behavior changes. Section 4 passes from these triple-level results to the generic square families and then checks every exceptional parameter locus. This is where continuous families, isolated squares, and lower-dimensional intersections are distinguished.
The global step uses a dichotomy. If a square contains a C-pair, then its C-pairs form disjoint cliques, and we classify the possible C-containing squares. If no quotient has type C, every quotient admits A or B; we prove that such a square contains an AAA triple and use the AAA completion analysis. Section 5 combines these branches into the full classification and compares it with the finite catalogue. Section 6 records the finite registries, residue lists, square and cycle certificates, and code needed to check the constructions and exclusions.
Table 1 (PDF p. 25).
Profiles and support equations
The prime partitions of 8 give
A = 2 + 2 + 2 + 2 , B = 3 + 3 + 2 , C = 5 + 3. A = 2 + 2 + 2 + 2,\qquad B = 3 + 3 + 2,\qquad C = 5 + 3. A = 2 + 2 + 2 + 2 , B = 3 + 3 + 2 , C = 5 + 3. Unlike order 7, profiles can overlap: with w = e 2 π i / 3 w = e^{2\pi i/3} w = e 2 π i /3 ( 1 , w , w 2 , − 1 , − w , − w 2 , 1 , − 1 ) (1, w, w^2, -1, -w, -w^2, 1, -1) ( 1 , w , w 2 , − 1 , − w , − w 2 , 1 , − 1 )
A square is A-admitting when every row quotient admits A; additional B decompositions are allowed. A chosen support specifies the cycle positions. Support classes, completions of a fixed triple, and monomial classes of squares are counted separately. Conjugation and transposition remain separate operations.
Even a profile need not determine its support. In a vector with four entries 1 and four entries − 1 -1 − 1 4 ! = 24 4! = 24 4 ! = 24
There are
N A = 8 ! ( 2 ! ) 4 4 ! = 105 , N B = ( 8 2 ) 1 2 ( 6 3 ) = 280 , N C = ( 8 3 ) = 56. N_A = \frac{8!}{(2!)^4 4!} = 105,\qquad N_B = \binom{8}{2}\frac{1}{2}\binom{6}{3} = 280,\qquad N_C = \binom{8}{3} = 56. N A = ( 2 ! ) 4 4 ! 8 ! = 105 , N B = ( 2 8 ) 2 1 ( 3 6 ) = 280 , N C = ( 3 8 ) = 56. undecorated supports. For supports P P P Q Q Q R R R u u u v v v u / v u/v u / v F p \mathbb{F}_p F p p p p
α P ( j ) − β Q ( j ) − γ R ( j ) = ∑ p ∈ { 2 , 3 , 5 } − a p j + b p j + c p j p (16) \alpha_{P(j)}-\beta_{Q(j)}-\gamma_{R(j)}=\sum_{p\in\{2,3,5\}}\frac{-a_{pj}+b_{pj}+c_{pj}}{p} \tag*{(16)} α P ( j ) − β Q ( j ) − γ R ( j ) = p ∈ { 2 , 3 , 5 } ∑ p − a p j + b p j + c p j (16) or, with m m m
L z = r ( 1 ) , L ∈ M 8 × m ( Z ) . (17) Lz=r \quad(1), \qquad L\in M_{8\times m}(\mathbb{Z}). \tag*{(17)} L z = r ( 1 ) , L ∈ M 8 × m ( Z ) . (17) The 2 free row-phase gauges give dephased dimension
d = m − rank L − 2. (18) d=m-\operatorname{rank}L-2. \tag*{(18)} d = m − rank L − 2. (18) Lemma 3.1. Let K K K L L L
K ( − a p + b p + c p ) = 0 ( p ) , p = 2 , 3 , 5. K(-a_p+b_p+c_p)=0 \quad(p), \qquad p=2,3,5. K ( − a p + b p + c p ) = 0 ( p ) , p = 2 , 3 , 5. For a feasible label lift, if U L V = D ULV=D U L V = D d 1 , … , d s d_1,\ldots,d_s d 1 , … , d s ∏ i ∣ d i ∣ \prod_i |d_i| ∏ i ∣ d i ∣ (18) .
Proof. Writing b = U r b=Ur b = U r z = V y z=Vy z = V y
y i = b i + n i d i ( 1 ) , 0 ≤ n i < ∣ d i ∣ , y_i=\frac{b_i+n_i}{d_i} \quad(1), \qquad0\le n_i<|d_i|, y i = d i b i + n i ( 1 ) , 0 ≤ n i < ∣ d i ∣ , The zero rows give K r = 0 Kr=0 K r = 0
The Smith factors also explain why the number of components can exceed the number of label lifts. For example, 2 s = 0 2s=0 2 s = 0 s = 0 s=0 s = 0 s = 1 / 2 s=1/2 s = 1/2 t t t s = 0 s=0 s = 0
Theorem 3.2. The canonical chosen-support census is
Table 3.2.
Thus ACC and BCC are impossible. Components include all label lifts and Smith torsion, before residual automorphisms or alternative decompositions are identified.
Proof. Fix the first support and enumerate the other 2. The raw counts are The intersection tensor n i j k = ∣ P i ∩ Q j ∩ R k ∣ n_{ijk} = |P_i \cap Q_j \cap R_k| n ij k = ∣ P i ∩ Q j ∩ R k ∣ U L V = D ULV = D U L V = D
Table 22.
Setting free coordinates to 0 gives the following seed orders and multiplicities:
Table 23.
Their least common multiple is 120. These are chosen seed orders, not minimal Butson levels or counts of inequivalent families.
Proposition 3.3. In a regular order-8 partial matrix, C-edges form disjoint cliques.
Proof. If r / s r/s r / s s / t s/t s / t r / t r/t r / t r / t r/t r / t
A new row is therefore C-related to all or none of a CCC triple, and to at most 1 row of an AAA triple.
Exact extension and compatibility
For an unknown row w j = e ( x j ) w_j = e(x_j) w j = e ( x j ) h j = e ( y j ) h_j = e(y_j) h j = e ( y j ) p p p
x j − x i = y j − y i + k j − k i p (19) x_j - x_i = y_j - y_i + \frac{k_j-k_i}{p} \tag*{(19)} x j − x i = y j − y i + p k j − k i (19) There are 105 decorated A supports, 1120 B supports and 2688 C supports per quotient. Allowing all 3913 choices at each old row is exhaustive. A weighted forest on the columns expresses each x j x_j x j c c c x 1 = 0 x_1=0 x 1 = 0 c − 1 c-1 c − 1
For a prescribed triple at fixed base parameters, normalized additional rows form the vertices of a graph; an edge means a regular quotient of any profile. A square completion is a 5-clique. On a varying base, both row-appearance and pair-compatibility loci must be included. Refine unsuccessful strata by these loci and their intersections, retaining disconnected row families. Their free phases remain variables until the mutual quotient conditions have also been solved. A symbolic 5-clique settles existence on its whole stratum; unsuccessful surfaces and curves require their lower-dimensional events.
Lemma 3.4. If the base entries lie in μ M \mu_M μ M μ lcm ( M , 30 ) \mu_{\operatorname{lcm}(M,30)} μ lcm ( M , 30 ) C C C
Proof. Edge weights belong to the stated rational phase group. Summing them along paths proves the connected assertion; choosing root phases in that group proves the existence assertion for disconnected graphs.
Thus a finite-root search can disprove row existence without connectedness, but complete row counts require connectedness or explicit free phases. Independent finite generators enumerate cycle partitions and rotations; every accepted row passes an exact multiplicity test. Finite-field root sums serve only as rejection filters.
Affine entries are written as
H i j ( x ) = e ( c i j + a i j 1 x 1 + ⋯ + a i j d x d ) , c i j ∈ Q / Z , a i j k ∈ Z . (20) H_{ij}(x)=e(c_{ij}+a_{ij1}x_1+\cdots+a_{ijd}x_d),\qquad c_{ij}\in\mathbb{Q}/\mathbb{Z},\quad a_{ijk}\in\mathbb{Z}. \tag*{(20)} H ij ( x ) = e ( c ij + a ij 1 x 1 + ⋯ + a ij d x d ) , c ij ∈ Q / Z , a ij k ∈ Z . (20) Their dimension is the rank of the dephased coefficient matrix. For proposed row and column orders p , q p,q p , q
h p ( i ) , q ( j ) − h p ( 1 ) , q ( j ) − h p ( i ) , q ( 1 ) + h p ( 1 ) , q ( 1 ) . (21) h_{p(i),q(j)}-h_{p(1),q(j)}-h_{p(i),q(1)}+h_{p(1),q(1)}. \tag*{(21)} h p ( i ) , q ( j ) − h p ( 1 ) , q ( j ) − h p ( i ) , q ( 1 ) + h p ( 1 ) , q ( 1 ) . (21) Generic computations treat base phases as independent variables, retaining base-only equations only when identically satisfied. Free extension phases are still solved and may have special values.
The canonical registry has 23 feasible AAA supports and 38 affine components; its labels 0 , … , 22 0,\ldots,22 0 , … , 22
Theorem 3.5. Every component of the 19 classes
U = { 0 , 1 , 2 , 3 , 4 , 5 , 6 , 7 , 8 , 9 , 10 , 12 , 13 , 14 , 15 , 16 , 17 , 18 , 20 } \mathcal{U}=\{0,1,2,3,4,5,6,7,8,9,10,12,13,14,15,16,17,18,20\} U = { 0 , 1 , 2 , 3 , 4 , 5 , 6 , 7 , 8 , 9 , 10 , 12 , 13 , 14 , 15 , 16 , 17 , 18 , 20 } has an A A A
Table 1 (PDF p. 28).
All listed surfaces and circles admit A A A A / B A/B A / B
Proof of the uniform part. Smith reduction reconstructs all 38 components. On the 32 components in U \mathcal{U} U Theorem 3.9 for classes 11, 19. Together these give the full statement.
The 2 maximal triple classes have 4 base parameters. The following completions have an additional parameter t t t
H A ( 0 ) = ( 1 1 1 1 1 1 1 1 1 − 1 x − x x − x 1 − 1 1 − y z − w − z w − 1 y 1 t t z / y w − t z / y − w − 1 − t 1 − t − t z / y w t z / y − w − 1 t 1 y − z − w z w − 1 − y 1 1 − 1 − 1 − 1 − 1 1 1 1 − 1 − x x − x x 1 − 1 ) (22) H_A^{(0)} =
\begin{pmatrix}
1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 \\
1 & -1 & x & -x & x & -x & 1 & -1 \\
1 & -y & z & -w & -z & w & -1 & y \\
1 & t & tz/y & w & -tz/y & -w & -1 & -t \\
1 & -t & -tz/y & w & tz/y & -w & -1 & t \\
1 & y & -z & -w & z & w & -1 & -y \\
1 & 1 & -1 & -1 & -1 & -1 & 1 & 1 \\
1 & -1 & -x & x & -x & x & 1 & -1
\end{pmatrix} \tag*{(22)} H A ( 0 ) = 1 1 1 1 1 1 1 1 1 − 1 − y t − t y 1 − 1 1 x z t z / y − t z / y − z − 1 − x 1 − x − w w w − w − 1 x 1 x − z − t z / y t z / y z − 1 − x 1 − x w − w − w w − 1 x 1 1 − 1 − 1 − 1 − 1 1 1 1 − 1 y − t t − y 1 − 1 (22) H A ( 1 ) = ( 1 1 1 1 1 1 1 1 1 − 1 − x x y − y 1 − 1 1 − z − w − w w w − 1 z 1 z t − t t y / x − t y / x − 1 − z 1 z − t t − t y / x t y / x − 1 − z 1 − z w w − w − w − 1 z 1 1 − 1 − 1 − 1 − 1 1 1 1 − 1 x − x − y y 1 − 1 ) (23) H_A^{(1)} =
\begin{pmatrix}
1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 \\
1 & -1 & -x & x & y & -y & 1 & -1 \\
1 & -z & -w & -w & w & w & -1 & z \\
1 & z & t & -t & ty/x & -ty/x & -1 & -z \\
1 & z & -t & t & -ty/x & ty/x & -1 & -z \\
1 & -z & w & w & -w & -w & -1 & z \\
1 & 1 & -1 & -1 & -1 & -1 & 1 & 1 \\
1 & -1 & x & -x & -y & y & 1 & -1
\end{pmatrix} \tag*{(23)} H A ( 1 ) = 1 1 1 1 1 1 1 1 1 − 1 − z z z − z 1 − 1 1 − x − w t − t w − 1 x 1 x − w − t t w − 1 − x 1 y w t y / x − t y / x − w − 1 − y 1 − y w − t y / x t y / x − w − 1 y 1 1 − 1 − 1 − 1 − 1 1 1 1 − 1 z − z − z z 1 − 1 (23) Proposition 3.6. For arbitrary x , y , z , w , t ∈ T x,y,z,w,t \in\mathbb{T} x , y , z , w , t ∈ T
Proof. Each quotient splits into 4 opposite Laurent-monomial pairs. For example, the quotient of rows 2 and 3 in H A ( 0 ) H_A^{(0)} H A ( 0 )
( 1 , y − 1 , x / z , x / w , − x / z , − x / w , − 1 , − y − 1 ) . (1, y^{-1}, x/z, x/w, -x/z, -x/w, -1, -y^{-1}). ( 1 , y − 1 , x / z , x / w , − x / z , − x / w , − 1 , − y − 1 ) . The symbolic checker verifies all 28 row pairs and 28 column pairs of each square.
:::
The real square W r , c = ( − 1 ) r ⋅ c W_{r,c}=(-1)^{r\cdot c} W r , c = ( − 1 ) r ⋅ c r , c ∈ F 2 3 r,c\in\mathbb{F}_2^3 r , c ∈ F 2 3
Proposition 3.7. *The generic completions of the 2 maximal AAA components require 3 5-parameter family types and the additional 4-parameter type F 8 , 4 F_{8,4} F 8 , 4
Proof. The components have respectively 45 and 23 connected rows, plus 3 and 1 free row families. Their connected graphs have 21 and 8 5-cliques. Each free family has a compatible fixed triangle; 2 opposite free phases complete it. The resulting 4 templates give 3 equivalence types.
Solving all pair equations at special free phases adds no row to the first component, but adds 2 rows to the second. Its augmented graph has 25 vertices and 10 5-cliques; 9 lie in its 5-parameter template and the last gives F 8 , 4 F_{8,4} F 8 , 4 F 8 , 4 F_{8,4} F 8 , 4
For the upper rows
( 1 , − 1 , − x , x , y , − y , 1 , − 1 ) , ( 1 , − z , − w , − w , w , w , − 1 , z ) . (1,-1,-x,x,y,-y,1,-1),\qquad(1,-z,-w,-w,w,w,-1,z). ( 1 , − 1 , − x , x , y , − y , 1 , − 1 ) , ( 1 , − z , − w , − w , w , w , − 1 , z ) . the new 4-parameter completion appends
( 1 − z w x − x − 1 − 1 1 z w 1 z x w − x w y w − y w − 1 − z 1 − w w w − y w y w − 1 − w 1 z w − 1 − 1 − y y 1 − z w 1 w − x w x w − w − w − 1 w ) \begin{pmatrix}
1 & -\frac{z}{w} & x & -x & -1 & -1 & 1 & \frac{z}{w}\\
1 & z & xw & -xw & yw & -yw & -1 & -z\\
1 & -w & w & w & -yw & yw & -1 & -w\\
1 & \frac{z}{w} & -1 & -1 & -y & y & 1 & -\frac{z}{w}\\
1 & w & -xw & xw & -w & -w & -1 & w
\end{pmatrix} 1 1 1 1 1 − w z z − w w z w x x w w − 1 − x w − x − x w w − 1 x w − 1 y w − y w − y − w − 1 − y w y w y − w 1 − 1 − 1 1 − 1 w z − z − w − w z w Every row quotient admits A; dephased coefficient rank is 4. Section 4 gives the full square formulas and generic classification.
The 2-parameter exceptional classes
For ϵ = ± 1 \epsilon= \pm1 ϵ = ± 1
T 21 , ϵ ( x , y ) = ( 1 1 1 1 1 1 1 1 1 − 1 − x y x y − x x ϵ y − ϵ y 1 − x / y − ϵ x x x / y − x − 1 ϵ x ) . T_{21,\epsilon}(x,y)=
\begin{pmatrix}
1 & 1 & 1 & 1 & 1 & 1 & 1 & 1\\
1 & -1 & -xy & xy & -x & x & \epsilon y & -\epsilon y\\
1 & -x/y & -\epsilon x & x & x/y & -x & -1 & \epsilon x
\end{pmatrix}. T 21 , ϵ ( x , y ) = 1 1 1 1 − 1 − x / y 1 − x y − ϵ x 1 x y x 1 − x x / y 1 x − x 1 ϵy − 1 1 − ϵy ϵ x . T 22 , ϵ ( x , y ) = ( 1 1 1 1 1 1 1 1 1 − 1 y − y − y / x y / x − ϵ y ϵ y 1 − 1 / x − ϵ y / x 1 / x − y / x − 1 ϵ ) . T_{22,\epsilon}(x,y)=
\begin{pmatrix}
1 & 1 & 1 & 1 & 1 & 1 & 1 & 1\\
1 & -1 & y & -y & -y/x & y/x & -\epsilon y & \epsilon y\\
1 & -1/x & -\epsilon& y/x & 1/x & -y/x & -1 & \epsilon
\end{pmatrix}. T 22 , ϵ ( x , y ) = 1 1 1 1 − 1 − 1/ x 1 y − ϵ 1 − y y / x 1 − y / x 1/ x 1 y / x − y / x 1 − ϵy − 1 1 ϵy ϵ . Theorem 3.8. A regular square completion exists precisely on the following loci:
21 x = ± 1 or y = ± 1 or x = ± y or x y = ± 1 , 22 x = ± 1 or y = ± 1 or x = ± y or x y = 1 or x = y 2 or y = − x 2 . \begin{aligned}
21\quad&x=\pm1\ \text{or}\ y=\pm1\ \text{or}\ x=\pm y\ \text{or}\ xy=\pm1,\\
22\quad&x=\pm1\ \text{or}\ y=\pm1\ \text{or}\ x=\pm y\ \text{or}\ xy=1\\
&\text{or}\ x=y^2\ \text{or}\ y=-x^2.
\end{aligned} 21 22 x = ± 1 or y = ± 1 or x = ± y or x y = ± 1 , x = ± 1 or y = ± 1 or x = ± y or x y = 1 or x = y 2 or y = − x 2 . Both signs have the same criterion. These are respectively 8 and 9 circles, all with A-admitting completions.
Proof. The 16 and 18 symbolic witnesses indexed in Subsection 6.4 prove sufficiency.
For class 21, use x = e t − s x=e^{t-s} x = e t − s y = e t y=e^t y = e t
t = 0 , 1 2 ; s = 0 , 1 2 ; s − t = 0 , 1 2 ; s − 2 t = 0 , 1 2 . t=0,\frac{1}{2};\qquad s=0,\frac{1}{2};\qquad s-t=0,\frac{1}{2};\qquad s-2t=0,\frac{1}{2}. t = 0 , 2 1 ; s = 0 , 2 1 ; s − t = 0 , 2 1 ; s − 2 t = 0 , 2 1 . and
t = 1 4 , 3 4 ; s − t = 1 4 , 3 4 ; s + t = 0 , 1 2 ; 2 s − t = 0 , 1 2 . t=\frac{1}{4},\frac{3}{4};\qquad s-t=\frac{1}{4},\frac{3}{4};\qquad s+t=0,\frac{1}{2};\qquad2s-t=0,\frac{1}{2}. t = 4 1 , 4 3 ; s − t = 4 1 , 4 3 ; s + t = 0 , 2 1 ; 2 s − t = 0 , 2 1 . There are also 34 isolated A-support loci. Off events there are 6 regular rows and 1 edge; on unsuccessful circles there are 7 rows and clique number at most 3. Full decorated-support projection excludes C rows everywhere and confines B rows to points. All lower row loci, pair events and intersections leave 64 residual points per component:
Table 24.
All surviving supports there are connected. The modulus lcm ( 6 , base-entry orders ) \operatorname{lcm}(6,\text{base-entry orders}) lcm ( 6 , base-entry orders )
For class 22, write x = e ( a ) x=e(a) x = e ( a ) y = e ( b ) y=e(b) y = e ( b ) a = s a=s a = s b = t + 1 / 2 b=t+1/2 b = t + 1/2 a + b = 1 / 2 a+b=1/2 a + b = 1/2 a − 2 b = 1 / 2 a-2b=1/2 a − 2 b = 1/2 2 a − b = 0 2a-b=0 2 a − b = 0 ( a , b ) = ( 1 / 6 , 1 / 3 ) , ( 5 / 6 , 2 / 3 ) (a,b)=(1/6,1/3),(5/6,2/3) ( a , b ) = ( 1/6 , 1/3 ) , ( 5/6 , 2/3 ) μ 12 \mu_{12} μ 12
The 3-parameter exceptional classes
The normal forms are
T 19 ( x , y , z ) = ( 1 1 1 1 1 1 1 1 1 − 1 x y / z − x y / z − y / z y / z x − x 1 − 1 / z y / z − x / z 1 / z − y / z − 1 x / z ) . T_{19}(x,y,z)=
\begin{pmatrix}
1 & 1 & 1 & 1 & 1 & 1 & 1 & 1\\
1 & -1 & xy/z & -xy/z & -y/z & y/z & x & -x\\
1 & -1/z & y/z & -x/z & 1/z & -y/z & -1 & x/z
\end{pmatrix}. T 19 ( x , y , z ) = 1 1 1 1 − 1 − 1/ z 1 x y / z y / z 1 − x y / z − x / z 1 − y / z 1/ z 1 y / z − y / z 1 x − 1 1 − x x / z . T 11 ( x , y , z ) = ( 1 1 1 1 1 1 1 1 1 − 1 − x y / z x y / z y / z − y / z y − y 1 − 1 / z − y / z − y / z 1 / z y / z − 1 y / z ) . T_{11}(x,y,z)=
\begin{pmatrix}
1 & 1 & 1 & 1 & 1 & 1 & 1 & 1\\
1 & -1 & -xy/z & xy/z & y/z & -y/z & y & -y\\
1 & -1/z & -y/z & -y/z & 1/z & y/z & -1 & y/z
\end{pmatrix}. T 11 ( x , y , z ) = 1 1 1 1 − 1 − 1/ z 1 − x y / z − y / z 1 x y / z − y / z 1 y / z 1/ z 1 − y / z y / z 1 y − 1 1 − y y / z . with Smith coordinates ( x , y , z ) = ( e ( s ) , e ( t ) , e ( u ) ) (x,y,z)=(e(s),e(t),e(u)) ( x , y , z ) = ( e ( s ) , e ( t ) , e ( u ))
Theorem 3.9. The completion locus of T 19 T_{19} T 19
x = ± 1 , y = ± z , z = ± x y , y = ± x z , x = y z . x=\pm1,\qquad y=\pm z,\qquad z=\pm xy,\qquad y=\pm xz,\qquad x=yz. x = ± 1 , y = ± z , z = ± x y , y = ± x z , x = y z . The completion locus of T 11 T_{11} T 11
y = ± 1 , z = ± 1 , z = ± y , z = − y 2 ; y=\pm1,\qquad z=\pm1,\qquad z=\pm y,\qquad z=-y^{2}; y = ± 1 , z = ± 1 , z = ± y , z = − y 2 ; the 10 circles
x = ± 1 , y = z 2 , x = ± 1 , y z = 1 , y = ± x , z = x 2 , y = − x 2 , z = ± x , y z = − 1 , z = ± x ; \begin{aligned}
x&=\pm1,\qquad y=z^{2},\\
x&=\pm1,\qquad yz=1,\\
y&=\pm x,\qquad z=x^{2},\\
y&=-x^{2},\qquad z=\pm x,\\
yz&=-1,\qquad z=\pm x;
\end{aligned} x x y y y z = ± 1 , y = z 2 , = ± 1 , y z = 1 , = ± x , z = x 2 , = − x 2 , z = ± x , = − 1 , z = ± x ; and the 32 points below. With w = e π i / 6 w=e^{\pi i/6} w = e π i /6 ( x , y , z ) = ( w i , w j , w k ) (x,y,z)=(w^{i},w^{j},w^{k}) ( x , y , z ) = ( w i , w j , w k )
Table 25 (PDF p. 31).
All surfaces and circles admit A-admitting completions. The 32 points lie outside them and admit mixed A/B completions only.
Proof. We begin with an example: at ( x , y , z ) = ( w , w 4 , w 3 ) (x,y,z)=(w,w^4,w^3) ( x , y , z ) = ( w , w 4 , w 3 ) H i j = w E i j H_{ij}=w^{E_{ij}} H ij = w E ij
E = ( 0 0 0 0 0 0 0 0 0 6 8 2 1 7 4 10 0 3 7 7 9 1 6 1 0 9 4 10 3 1 6 7 0 6 4 1 7 9 10 3 0 6 0 6 6 3 0 9 0 0 8 8 4 7 1 4 0 0 1 4 10 6 7 6 ) . E=\begin{pmatrix}
0&0&0&0&0&0&0&0\\
0&6&8&2&1&7&4&10\\
0&3&7&7&9&1&6&1\\
0&9&4&10&3&1&6&7\\
0&6&4&1&7&9&10&3\\
0&6&0&6&6&3&0&9\\
0&0&8&8&4&7&1&4\\
0&0&1&4&10&6&7&6
\end{pmatrix}. E = 0 0 0 0 0 0 0 0 0 6 3 9 6 6 0 0 0 8 7 4 4 0 8 1 0 2 7 10 1 6 8 4 0 1 9 3 7 6 4 10 0 7 1 1 9 3 7 6 0 4 6 6 10 0 1 7 0 10 1 7 3 9 4 6 . Its 28 row quotients comprise 20 A-only, 4 A-and-B, and 4 B-only pairs, as do the other 31 witnesses.
Full support projection gives 20 A-row surfaces for class 11 and 22 for class 19; respectively 7 and 9 have symbolic square witnesses. Off proper row loci there are only 4 regular additional rows. Each of the remaining 13 surfaces has no generic 5-clique after all profiles are admitted.
Outside the certified surfaces every feasible row support is connected. Refine by all lower row loci, pair events and intersections. This gives 126 curves for class 11 and 168 for class 19; exactly the 10 displayed class-11 circles have square witnesses. B rows occur on curves, and class 19 also has AAC rows at 384 points. No C row occurs outside the successful class-11 surfaces. These mixed cases are all retained.
After refining unsuccessful curves and removing successful loci, the finite residues contain 616 and 1504 points:
Table 1 (PDF p. 32).
Explicit monomial transformations reduce the 2120 base triples to 132 finite triples; each transformed seed and row is checked entrywise. Independent finite-root generation reproduces every complete row list. Exact cycle recognition gives precisely the 32 class-11 5-cliques and none in class 19. At each successful point only 4 rows are A-related to all 3 seed rows, so an A-admitting square is impossible. The 26 surface/circle witnesses and the 32 point witnesses are independently verified. Their code and certificate index is in § 7 .
This proves Theorem 3.5 . Completion existence is now settled for every AAA support; the next section classifies all completions over the successful loci.
Order 8: generic and exceptional completions
The affine families
The parameter index in F 8 , 5 a , … , M 8 , 1 F_{8,5a},\ldots,M_{8,1} F 8 , 5 a , … , M 8 , 1 I 8 , 12 I_{8,12} I 8 , 12 F 5 a , F 5 b , F 5 c , F 4 , F 2 , K 2 , M 1 , I 12 F_{5a},F_{5b},F_{5c},F_{4},F_{2},K_{2},M_{1},I_{12} F 5 a , F 5 b , F 5 c , F 4 , F 2 , K 2 , M 1 , I 12
Theorem 4.1. The following 9 affine families are regular, generically locally maximal in the dephased regular locus, and generically distinct under equivalence:
F 8 , 5 a , F 8 , 5 b , F 8 , 5 c , F 8 , 4 , F 8 , 2 t , F 8 , 2 t , K 8 , 2 , M 8 , 1 F_{8,5a},F_{8,5b},F_{8,5c},F_{8,4},F_{8,2}^{\mathrm{t}},F_{8,2}^{\mathrm{t}},K_{8,2},M_{8,1} F 8 , 5 a , F 8 , 5 b , F 8 , 5 c , F 8 , 4 , F 8 , 2 t , F 8 , 2 t , K 8 , 2 , M 8 , 1
Their dimensions are 5 , 5 , 5 , 4 , 4 , 2 , 2 , 2 , 1 5,5,5,4,4,2,2,2,1 5 , 5 , 5 , 4 , 4 , 2 , 2 , 2 , 1 A A A M 8 , 1 M_{8,1} M 8 , 1 A A A B B B
Proof. We begin with the matrices. All parameters in the following dephased formulas range independently over T \mathbb{T} T
F 8 , 5 a ( a , b , c , d , e ) = ( 1 1 1 1 1 1 1 1 1 − 1 − b d b d − b d b d 1 − 1 1 − 1 c − b d a d b d a − d − 1 1 c 1 1 − 1 − 1 − 1 − 1 1 1 1 − 1 c b d a − d − b d a d − 1 1 c 1 − 1 b d − b d b d − b d 1 − 1 1 1 c e a e b − e a e b − 1 − 1 c 1 1 c − e − a e b e a e b − 1 − 1 c ) F_{8,5a}(a,b,c,d,e)=
\begin{pmatrix}
1&1&1&1&1&1&1&1\\
1&-1&-bd&bd&-bd&bd&1&-1\\
1&-\frac{1}{c}&-\frac{bd}{a}&d&\frac{bd}{a}&-d&-1&\frac{1}{c}\\
1&1&-1&-1&-1&-1&1&1\\
1&-\frac{1}{c}&\frac{bd}{a}&-d&-\frac{bd}{a}&d&-1&\frac{1}{c}\\
1&-1&bd&-bd&bd&-bd&1&-1\\
1&\frac{1}{c}&e&\frac{ae}{b}&-e&\frac{ae}{b}&-1&-\frac{1}{c}\\
1&\frac{1}{c}&-e&-\frac{ae}{b}&e&\frac{ae}{b}&-1&-\frac{1}{c}
\end{pmatrix} F 8 , 5 a ( a , b , c , d , e ) = 1 1 1 1 1 1 1 1 1 − 1 − c 1 1 − c 1 − 1 c 1 c 1 1 − b d − a b d − 1 a b d b d e − e 1 b d d − 1 − d − b d b a e − b a e 1 − b d a b d − 1 − a b d b d − e e 1 b d − d − 1 d − b d b a e b a e 1 1 − 1 1 − 1 1 − 1 − 1 1 − 1 c 1 1 c 1 − 1 − c 1 − c 1 F 8 , 5 b ( a , b , c , d , e ) = ( 1 1 1 1 1 1 1 1 1 − 1 − b d b d − b d b d 1 − 1 1 − 1 c − b d a d b d a − d − 1 1 c 1 1 c − b d a − d b d a d − 1 − 1 c 1 1 − 1 − 1 − 1 − 1 1 1 1 − 1 b d − b d b d − b d 1 − 1 1 e b d a c d e − b d a − c d e − 1 − e 1 − e b d a − c d e − b d a c d e − 1 e ) F_{8,5b}(a,b,c,d,e)=
\begin{pmatrix}
1&1&1&1&1&1&1&1\\
1&-1&-bd&bd&-bd&bd&1&-1\\
1&-\frac{1}{c}&-\frac{bd}{a}&d&\frac{bd}{a}&-d&-1&\frac{1}{c}\\
1&\frac{1}{c}&-\frac{bd}{a}&-d&\frac{bd}{a}&d&-1&-\frac{1}{c}\\
1&1&-1&-1&-1&-1&1&1\\
1&-1&bd&-bd&bd&-bd&1&-1\\
1&e&\frac{bd}{a}&cde&-\frac{bd}{a}&-cde&-1&-e\\
1&-e&\frac{bd}{a}&-cde&-\frac{bd}{a}&cde&-1&e
\end{pmatrix} F 8 , 5 b ( a , b , c , d , e ) = 1 1 1 1 1 1 1 1 1 − 1 − c 1 c 1 1 − 1 e − e 1 − b d − a b d − a b d − 1 b d a b d a b d 1 b d d − d − 1 − b d c d e − c d e 1 − b d a b d a b d − 1 b d − a b d − a b d 1 b d − d d − 1 − b d − c d e c d e 1 1 − 1 − 1 1 1 − 1 − 1 1 − 1 c 1 − c 1 1 − 1 − e e F 8 , 5 c ( a , b , c , d , e ) = ( 1 1 1 1 1 1 1 1 1 − 1 a d − a d − b d b d 1 − 1 1 − 1 c d d − d − d − 1 1 c 1 1 − 1 − 1 − 1 − 1 1 1 1 − 1 c − d − d d d − 1 1 c 1 − 1 − a d a d b d − b d 1 − 1 1 1 c e − e b e a − b e a − 1 − 1 c 1 1 c − e e − b e a b e a − 1 − 1 c ) F_{8,5c}(a,b,c,d,e)=
\begin{pmatrix}
1&1&1&1&1&1&1&1\\
1&-1&ad&-ad&-bd&bd&1&-1\\
1&-\frac{1}{c}&d&d&-d&-d&-1&\frac{1}{c}\\
1&1&-1&-1&-1&-1&1&1\\
1&-\frac{1}{c}&-d&-d&d&d&-1&\frac{1}{c}\\
1&-1&-ad&ad&bd&-bd&1&-1\\
1&\frac{1}{c}&e&-e&\frac{be}{a}&-\frac{be}{a}&-1&-\frac{1}{c}\\
1&\frac{1}{c}&-e&e&-\frac{be}{a}&\frac{be}{a}&-1&-\frac{1}{c}
\end{pmatrix} F 8 , 5 c ( a , b , c , d , e ) = 1 1 1 1 1 1 1 1 1 − 1 − c 1 1 − c 1 − 1 c 1 c 1 1 a d d − 1 − d − a d e − e 1 − a d d − 1 − d a d − e e 1 − b d − d − 1 d b d a b e − a b e 1 b d − d − 1 d − b d − a b e a b e 1 1 − 1 1 − 1 1 − 1 − 1 1 − 1 c 1 1 c 1 − 1 − c 1 − c 1 F 8 , 4 ( a , b , c , d ) = ( 1 1 1 1 1 1 1 1 1 − 1 a d − a d − b d b d 1 − 1 1 − 1 c d d − d − d − 1 1 c 1 1 c d − a d a d − 1 − 1 1 − 1 c d 1 1 c d a d 2 − a d 2 b d 2 − b d 2 − 1 − 1 c 1 1 d − d − d − b d 2 b d 2 − 1 d 1 − 1 c d − 1 − 1 b d − b d 1 1 c d 1 − d − a d 2 a d 2 d d − 1 − d ) F_{8,4}(a,b,c,d)=
\begin{pmatrix}
1&1&1&1&1&1&1&1\\
1&-1&ad&-ad&-bd&bd&1&-1\\
1&-\frac{1}{c}&d&d&-d&-d&-1&\frac{1}{c}\\
1&\frac{1}{cd}&-ad&ad&-1&-1&1&-\frac{1}{cd}\\
1&\frac{1}{cd}&ad^{2}&-ad^{2}&bd^{2}&-bd^{2}&-1&-\frac{1}{c}\\
1&\frac{1}{d}&-d&-d&-bd^{2}&bd^{2}&-1&d\\
1&-\frac{1}{cd}&-1&-1&bd&-bd&1&\frac{1}{cd}\\
1&-d&-ad^{2}&ad^{2}&d&d&-1&-d
\end{pmatrix} F 8 , 4 ( a , b , c , d ) = 1 1 1 1 1 1 1 1 1 − 1 − c 1 c d 1 c d 1 d 1 − c d 1 − d 1 a d d − a d a d 2 − d − 1 − a d 2 1 − a d d a d − a d 2 − d − 1 a d 2 1 − b d − d − 1 b d 2 − b d 2 b d d 1 b d − d − 1 − b d 2 b d 2 − b d d 1 1 − 1 1 − 1 − 1 1 − 1 1 − 1 c 1 − c d 1 − c 1 d c d 1 − d F 8 , 4 t ( a , b , c , d ) = ( 1 1 1 1 1 1 1 1 1 − 1 − 1 c 1 c d 1 c d − 1 c d − d 1 a d − a d − a d − 1 − d − 1 − a d 2 1 − a d d a d − a d 2 − d − 1 a d 2 1 − b d − d − 1 b d 2 − b d 2 b d d 1 b d − d − 1 − b d 2 b d 2 − b d d 1 1 − 1 1 − 1 − 1 1 − 1 1 − 1 1 c − 1 c d − 1 c d 1 c d − d ) F^{t}_{8,4}(a,b,c,d)=\begin{pmatrix}
1 & 1 & 1 & 1 & 1 & 1 & 1 & 1\\
1 & -1 & -\frac{1}{c} & \frac{1}{cd} & \frac{1}{c} & d & -\frac{1}{cd} & -d\\
1 & ad & -ad & -ad & -1 & -d & -1 & -ad^{2}\\
1 & -ad & d & ad & -ad^{2} & -d & -1 & ad^{2}\\
1 & -bd & -d & -1 & bd^{2} & -bd^{2} & bd & d\\
1 & bd & -d & -1 & -bd^{2} & bd^{2} & -bd & d\\
1 & 1 & -1 & 1 & -1 & -1 & 1 & -1\\
1 & -1 & \frac{1}{c} & -\frac{1}{cd} & -\frac{1}{c} & d & \frac{1}{cd} & -d
\end{pmatrix} F 8 , 4 t ( a , b , c , d ) = 1 1 1 1 1 1 1 1 1 − 1 a d − a d − b d b d 1 − 1 1 − c 1 − a d d − d − d − 1 c 1 1 c d 1 − a d a d − 1 − 1 1 − c d 1 1 c 1 − 1 − a d 2 b d 2 − b d 2 − 1 − c 1 1 d − d − d − b d 2 b d 2 − 1 d 1 − c d 1 − 1 − 1 b d − b d 1 c d 1 1 − d − a d 2 a d 2 d d − 1 − d F 8 , 2 ( a , b ) = ( 1 1 1 1 1 1 1 1 1 − 1 b − b − b a b a a − a 1 − a b a b − b a − b − 1 a 1 1 a b a 2 − b a b a − b a 2 − 1 a − 1 1 1 − 1 − 1 − 1 − 1 1 1 1 a − b b a − b a b − a − 1 1 − 1 a − b a − b a 2 b a b a 2 − 1 1 a 1 − 1 − b a 2 b a 2 b a − b a 1 a − 1 a ) F_{8,2}(a,b)=\begin{pmatrix}
1 & 1 & 1 & 1 & 1 & 1 & 1 & 1\\
1 & -1 & b & -b & -\frac{b}{a} & \frac{b}{a} & a & -a\\
1 & -a & \frac{b}{a} & b & -\frac{b}{a} & -b & -1 & a\\
1 & \frac{1}{a} & \frac{b}{a^{2}} & -\frac{b}{a} & \frac{b}{a} & -\frac{b}{a^{2}} & -\frac{1}{a} & -1\\
1 & 1 & -1 & -1 & -1 & -1 & 1 & 1\\
1 & a & -b & \frac{b}{a} & -\frac{b}{a} & b & -a & -1\\
1 & -\frac{1}{a} & -\frac{b}{a} & -\frac{b}{a^{2}} & \frac{b}{a} & \frac{b}{a^{2}} & -1 & \frac{1}{a}\\
1 & -1 & -\frac{b}{a^{2}} & \frac{b}{a^{2}} & \frac{b}{a} & -\frac{b}{a} & \frac{1}{a} & -\frac{1}{a}
\end{pmatrix} F 8 , 2 ( a , b ) = 1 1 1 1 1 1 1 1 1 − 1 − a a 1 1 a − a 1 − 1 1 b a b a 2 b − 1 − b − a b − a 2 b 1 − b b − a b − 1 a b − a 2 b a 2 b 1 − a b − a b a b − 1 − a b a b a b 1 a b − b − a 2 b − 1 b a 2 b − a b 1 a − 1 − a 1 1 − a − 1 a 1 1 − a a − 1 1 − 1 a 1 − a 1 F 8 , 2 t ( a , b ) = ( 1 1 1 1 1 1 1 1 1 − 1 − a 1 a 1 a − 1 a − 1 1 b b a b a 2 − 1 − b − b a − b a 2 1 − b b − b a − 1 b a − b a 2 b a 2 1 − b a − b a b a − 1 − b a b a b a 1 b a − b − b a 2 − 1 b b a 2 − b a 1 a − 1 − 1 a 1 − a − 1 1 a 1 − a a − 1 1 − 1 1 a − 1 a ) F^{t}_{8,2}(a,b)=\begin{pmatrix}
1 & 1 & 1 & 1 & 1 & 1 & 1 & 1\\
1 & -1 & -a & \frac{1}{a} & 1 & a & -\frac{1}{a} & -1\\
1 & b & \frac{b}{a} & \frac{b}{a^{2}} & -1 & -b & -\frac{b}{a} & -\frac{b}{a^{2}}\\
1 & -b & b & -\frac{b}{a} & -1 & \frac{b}{a} & -\frac{b}{a^{2}} & \frac{b}{a^{2}}\\
1 & -\frac{b}{a} & -\frac{b}{a} & \frac{b}{a} & -1 & -\frac{b}{a} & \frac{b}{a} & \frac{b}{a}\\
1 & \frac{b}{a} & -b & -\frac{b}{a^{2}} & -1 & b & \frac{b}{a^{2}} & -\frac{b}{a}\\
1 & a & -1 & -\frac{1}{a} & 1 & -a & -1 & \frac{1}{a}\\
1 & -a & a & -1 & 1 & -1 & \frac{1}{a} & -\frac{1}{a}
\end{pmatrix} F 8 , 2 t ( a , b ) = 1 1 1 1 1 1 1 1 1 − 1 b − b − a b a b a − a 1 − a a b b − a b − b − 1 a 1 a 1 a 2 b − a b a b − a 2 b − a 1 − 1 1 1 − 1 − 1 − 1 − 1 1 1 1 a − b a b − a b b − a − 1 1 − a 1 − a b − a 2 b a b a 2 b − 1 a 1 1 − 1 − a 2 b a 2 b a b − a b a 1 − a 1 K 8 , 2 ( a , b ) = ( 1 1 1 1 1 1 1 1 1 a − b 2 − b 2 a − 1 − a b 2 a b 2 1 b a − 1 b a 1 − b a − b a − 1 1 − b b b 2 a − b a − 1 b a − b 2 a 1 − a − b b − b a b a − 1 a 1 − 1 a b − b a b a − b − b 2 a b 2 a 1 − a b − b − 1 a b a b − a 1 b b 2 − b − 1 b − b − b 2 ) K_{8,2}(a,b)=\begin{pmatrix}
1 & 1 & 1 & 1 & 1 & 1 & 1 & 1\\
1 & a & -b^{2} & -\frac{b^{2}}{a} & -1 & -a & \frac{b^{2}}{a} & b^{2}\\
1 & \frac{b}{a} & -1 & \frac{b}{a} & 1 & -\frac{b}{a} & -\frac{b}{a} & -1\\
1 & -b & b & \frac{b^{2}}{a} & -\frac{b}{a} & -1 & \frac{b}{a} & -\frac{b^{2}}{a}\\
1 & -a & -b & b & -\frac{b}{a} & \frac{b}{a} & -1 & a\\
1 & -\frac{1}{a} & b & -\frac{b}{a} & \frac{b}{a} & -b & -\frac{b^{2}}{a} & \frac{b^{2}}{a}\\
1 & -\frac{a}{b} & -b & -1 & \frac{a}{b} & a & b & -a\\
1 & b & b^{2} & -b & -1 & b & -b & -b^{2}
\end{pmatrix} K 8 , 2 ( a , b ) = 1 1 1 1 1 1 1 1 1 a a b − b − a − a 1 − b a b 1 − b 2 − 1 b − b b − b b 2 1 − a b 2 a b a b 2 b − a b − 1 − b 1 − 1 1 − a b − a b a b b a − 1 1 − a − a b − 1 a b − b a b 1 a b 2 − a b a b − 1 − a b 2 b − b 1 b 2 − 1 − a b 2 a a b 2 − a − b 2 With w = e 2 π i / 3 w=e^{2\pi i/3} w = e 2 π i /3
M 8 , 1 ( a ) = ( 1 1 1 1 1 1 1 1 1 − w a 2 a w a − 1 w a 2 − a − w a 1 a − a a 3 − a 3 − a 4 a 4 − 1 1 w a − w a − w a w a w 2 − 1 − w 2 1 − 1 w a − a 3 w a 2 − w a 2 a 3 − w a 1 w a 2 w − w a 4 − w a 2 − w 2 a 4 − a 4 w 2 1 − a − w − 1 − w a w a w a 1 − w a − 1 w a 4 a 3 − w a 4 − a 3 w a ) M_{8,1}(a)=\begin{pmatrix}
1 & 1 & 1 & 1 & 1 & 1 & 1 & 1\\
1 & -wa^{2} & a & wa & -1 & wa^{2} & -a & -wa\\
1 & a & -a & a^{3} & -a^{3} & -a^{4} & a^{4} & -1\\
1 & wa & -\frac{w}{a} & -wa & \frac{w}{a} & w^{2} & -1 & -w^{2}\\
1 & -1 & \frac{w}{a} & -a^{3} & wa^{2} & -wa^{2} & a^{3} & -\frac{w}{a}\\
1 & wa^{2} & w & -wa^{4} & -wa^{2} & -w^{2}a^{4} & -a^{4} & w^{2}\\
1 & -a & -w & -1 & -\frac{w}{a} & w & a & \frac{w}{a}\\
1 & -wa & -1 & wa^{4} & a^{3} & -wa^{4} & -a^{3} & wa
\end{pmatrix} M 8 , 1 ( a ) = 1 1 1 1 1 1 1 1 1 − w a 2 a w a − 1 w a 2 − a − w a 1 a − a − a w a w w − w − 1 1 w a a 3 − w a − a 3 − w a 4 − 1 w a 4 1 − 1 − a 3 a w w a 2 − w a 2 − a w a 3 1 w a 2 − a 4 w 2 − w a 2 − w 2 a 4 w − w a 4 1 − a a 4 − 1 a 3 − a 4 a − a 3 1 − w a − 1 − w 2 − a w w 2 a w w a Formal cycle recognition verifies every row and column quotient of each displayed matrix. Coefficient ranks give the asserted dimensions.
For a quotient q q q m ( q ) m(q) m ( q ) n 1 , n 1 , … , n s , n s n_1,n_1,\ldots,n_s,n_s n 1 , n 1 , … , n s , n s
m ( q ) = ∏ j = 1 s n j ! . m(q)=\prod_{j=1}^{s} n_j!. m ( q ) = j = 1 ∏ s n j ! . Set m ( q ) = 0 m(q)=0 m ( q ) = 0
Table 26.
Here m [ r ] m^{[r]} m [ r ] r r r m m m
Near a fixed dephased square, choose continuous real lifts of its 49 free entry phases. For each decorated support holding at the square, subtract its phase equations at that point; the resulting conditions are homogeneous linear equations in the phase increments. Every failing decorated support remains absent in a sufficiently small neighborhood, since there are only finitely many. The regular locus is therefore locally a finite union of these exact linear spaces.
For a generic A quotient, write P , N P,N P , N
∑ j ∈ P δ q j − ∑ j ∈ N δ q j = 0. \sum_{j\in P}\delta q_j-\sum_{j\in N}\delta q_j=0. j ∈ P ∑ δ q j − j ∈ N ∑ δ q j = 0. Stacking these common equations over row pairs gives ranks 44, 45, 47 for the 5-, 4-, and 2-dimensional families. Their kernels equal the displayed coefficient spaces. No B or C support holds generically, so the local regular locus equals the family. For mixed supports, intersect their constraint row spaces to obtain common equations and then check the remaining alternatives. For M 8 , 1 M_{8,1} M 8 , 1
The transpose identifications are
F 8 , 5 a t ∼ F 8 , 5 a , F 8 , 5 b t ∼ F 8 , 5 c , K 8 , 2 t ∼ K 8 , 2 , M 8 , 1 t ∼ M 8 , 1 , I 8 , 12 t ∼ I 8 , 12 . F^{t}_{8,5a}\sim F_{8,5a},\qquad F^{t}_{8,5b}\sim F_{8,5c},\qquad K^{t}_{8,2}\sim K_{8,2},\qquad M^{t}_{8,1}\sim M_{8,1},\qquad I^{t}_{8,12}\sim I_{8,12}. F 8 , 5 a t ∼ F 8 , 5 a , F 8 , 5 b t ∼ F 8 , 5 c , K 8 , 2 t ∼ K 8 , 2 , M 8 , 1 t ∼ M 8 , 1 , I 8 , 12 t ∼ I 8 , 12 . Their exact monomial witnesses include parameter substitutions. The 2 transpose pairs displayed separately are generically inequivalent.
To verify a family embedding, solve its target parameters from a full-rank coefficient minor, retaining all torsion choices, and check all 64 dephased entries. The archive has 1519 such generic witnesses, including the initial targets. Matching counts prune permutations but never replace the entrywise check.
Generic completion of prescribed triples
Theorem 4.2. Every regular square completion over a generic triple in any of the 38 AAA components belongs to
F 8 , 5 a , F 8 , 5 b , F 8 , 5 c , F 8 , 4 , F 8 , 4 t , F 8 , 2 . F_{8,5a},\quad F_{8,5b},\quad F_{8,5c},\quad F_{8,4},\quad F^{t}_{8,4},\quad F_{8,2}. F 8 , 5 a , F 8 , 5 b , F 8 , 5 c , F 8 , 4 , F 8 , 4 t , F 8 , 2 .
Every such square is A A A
Proof. Apply all 3913 decorated supports per quotient. Weighted component equations retain every free added-row phase; a base-only congruence survives precisely when identically satisfied. Thus every generic regular row occurs. Remove a contained row family only after an explicit affine substitution, determined by its anchor entries and checked in all 8 coordinates.
For every pair of retained families, enumerate all quotient supports and solve their joint congruences. Choose 5 families, allowing repeated choices with independent phases, and intersect all 10 pair conditions. Discard only empty branches, branches restricting the generic base, and loci contained in retained larger loci. Every square follows one such sequence of its realized supports; conversely every surviving branch has all 28 regular quotients.
Smith splitting of
A y + B x + c = 0 (24) Ay + Bx + c = 0 \tag*{(24)} A y + B x + c = 0 (24) retains every torsion residue and free phase. Independent rational reconstruction and cycle recognition give 1512 parameterizations:
All have exact monomial embeddings into the 6 stated families, and all row quotients admit A A A
The isolated square and the mixed circle
Theorem 4.3. Each of the 32 isolated successful class-11 triples has exactly 1 square completion, up to phases and order of the added rows. All give the same monomial class I 8 , 12 I_{8,12} I 8 , 12
20 A A A A A A B B B B B B
Proof. We begin with a canonical representative. With w = e π i / 6 w=e^{\pi i/6} w = e π i /6 I 8 , 12 = ( w E i j ) I_{8,12}=(w^{E_{ij}}) I 8 , 12 = ( w E ij )
( 0 0 0 0 0 0 0 0 0 0 0 3 6 6 6 9 0 1 5 9 1 5 9 6 0 4 8 6 10 2 3 9 0 6 2 11 9 8 3 5 0 6 11 5 3 11 9 5 0 7 6 1 7 3 9 1 0 10 6 7 4 9 3 1 ) . \begin{pmatrix}
0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 3 & 6 & 6 & 6 & 9 \\
0 & 1 & 5 & 9 & 1 & 5 & 9 & 6 \\
0 & 4 & 8 & 6 & 10 & 2 & 3 & 9 \\
0 & 6 & 2 & 11 & 9 & 8 & 3 & 5 \\
0 & 6 & 11 & 5 & 3 & 11 & 9 & 5 \\
0 & 7 & 6 & 1 & 7 & 3 & 9 & 1 \\
0 & 10 & 6 & 7 & 4 & 9 & 3 & 1
\end{pmatrix}. 0 0 0 0 0 0 0 0 0 0 1 4 6 6 7 10 0 0 5 8 2 11 6 6 0 3 9 6 11 5 1 7 0 6 1 10 9 3 7 4 0 6 5 2 8 11 3 9 0 6 9 3 3 9 9 3 0 9 6 9 5 5 1 1 . Each complete list of 7 candidate rows has exactly 1 5-clique. Canonicalization over every pivot row and column and all remaining column orders, followed by row sorting, identifies the 32 squares. All transformations are checked entrywise. The dephased representative has a primitive 12th-root entry, hence a primitive cross ratio; every equivalent Butson level is divisible by 12.
The equations common to every support have rank 48. For a selected B-only quotient there are 4 feasible supports, each killing the remaining direction: adjoining an appropriate constraint gives determinants 1, − 1 -1 − 1 − 1 -1 − 1
The exact profile locations and triple census are
Table 27.
Table 28.
The B-admitting graph is the union of cycles (1, 3, 2, 4) and (5, 8, 6, 7), with B-only and AB edges alternating. Triples within a cycle give 8 mixed-profile cases; the other triples give 16 of each remaining kind. Counting admitted types with overlap gives 32 AAA, 40 AAB, 8 ABB and no BBB. The transpose has the same census. The exponent matrix is entry 739, indexed from 0, in the Aalto catalogue file BH-8-12.txt.
Proposition 4.4. The family M 8 , 1 ( a ) M_{8,1}(a) M 8 , 1 ( a ) is A-admitting exactly when a 12 = 1 a^{12}=1 a 12 = 1 Generically the 4 pairs
( 1 , 6 ) , ( 2 , 5 ) , ( 3 , 4 ) , ( 7 , 8 ) . (1, 6), \qquad(2, 5), \qquad(3, 4), \qquad(7, 8). ( 1 , 6 ) , ( 2 , 5 ) , ( 3 , 4 ) , ( 7 , 8 ) .
are B-only and the other 24 are A-only.
Proof. For each of the 4 quotients, enumerate all 105 antipodal matchings. Their equations have the form
k t = c ( 1 ) , a = e ( t ) , k ∈ Z , c ∈ Q / Z . (25) kt=c\ (1), \qquad a=e(t), \qquad k\in\mathbb{Z},\ c\in\mathbb{Q}/\mathbb{Z}. \tag*{(25)} k t = c ( 1 ) , a = e ( t ) , k ∈ Z , c ∈ Q / Z . (25) Solving every nonzero k k k ∣ k ∣ \lvert k\rvert ∣ k ∣ a ∈ μ 12 a\in\mu_{12} a ∈ μ 12
At a = e ( 1 / 18 ) a=e(1/18) a = e ( 1/18 )
( 0 0 0 0 0 0 0 0 0 17 1 7 9 8 10 16 0 1 10 3 12 13 4 9 0 7 14 16 5 12 9 3 0 9 5 12 8 17 3 14 0 8 6 1 17 7 13 12 0 10 15 9 14 6 1 5 0 16 9 10 3 1 12 7 ) \begin{pmatrix}
0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
0 & 17 & 1 & 7 & 9 & 8 & 10 & 16 \\
0 & 1 & 10 & 3 & 12 & 13 & 4 & 9 \\
0 & 7 & 14 & 16 & 5 & 12 & 9 & 3 \\
0 & 9 & 5 & 12 & 8 & 17 & 3 & 14 \\
0 & 8 & 6 & 1 & 17 & 7 & 13 & 12 \\
0 & 10 & 15 & 9 & 14 & 6 & 1 & 5 \\
0 & 16 & 9 & 10 & 3 & 1 & 12 & 7
\end{pmatrix} 0 0 0 0 0 0 0 0 0 17 1 7 9 8 10 16 0 1 10 14 5 6 15 9 0 7 3 16 12 1 9 10 0 9 12 5 8 17 14 3 0 8 13 12 17 7 6 1 0 10 4 9 3 13 1 12 0 16 9 3 14 12 5 7 It has 24 A-only and 4 B-only row quotients and minimal level 18, since the dephased entry a a a K 8 , 2 K_{8,2} K 8 , 2
Completion on all closed parameter loci
For an affine triple T ( t ) = ( 1 ; u ( t ) ; v ( t ) ) T(t)=(1;u(t);v(t)) T ( t ) = ( 1 ; u ( t ) ; v ( t ))
{ L t + c = 0 ( 1 ) , w j = e ( a j + b j ⋅ t + d j ⋅ s ) , s ∈ ( R / Z ) f . (26) \begin{cases}
L t+c=0\ (1),\\
w_j=e(a_j+b_j\cdot t+d_j\cdot s),\quad s\in(\mathbb{R}/\mathbb{Z})^f. \tag*{(26)}
\end{cases} { L t + c = 0 ( 1 ) , w j = e ( a j + b j ⋅ t + d j ⋅ s ) , s ∈ ( R / Z ) f . (26) Here closing edges impose the base lattice, while disconnected components supply all free row phases s s s
The generic calculation describes rows that exist throughout the open part of a base component; it does not settle what can happen where an additional phase equation becomes true. Such an equation may cut out a circle, surface or finite set on which new rows appear or previously separate row families become compatible. We therefore treat each proper locus as a new base stratum and repeat the extension analysis there. An all-parameter classification therefore requires the exceptional strata and their intersections to be exhausted as well, including loci where existing rows become compatible without any new row appearing.
Theorem 4.5. These records give the complete A/B 4th-row cover. Recursively splitting proper base intersections into connected Smith components, and solving every 5-row compatibility system on each closed stratum, terminates and gives every A/B square completion.
Proof. Every regular added row realizes an enumerated triple of supports. Tree edges, closing equations and free component phases describe exactly its solutions. Merging identical records or removing a contained record by a checked affine substitution preserves the cover.
On a connected base stratum, retain records whose base equations hold identically. If a square uses a row outside this list, its base lies in a proper intersection. Split that intersection by Smith form and recurse; dimension strictly decreases. On the first stratum where all 5 rows occur, choose their families, allowing repetition with independent phases. Enumerate all A/B supports for the 10 mutual quotients and intersect their equations, including every base restriction. Smith splitting retains every connected branch. Conversely, each branch supplies the 3 original, 15 old-new and 10 new-new regular quotients. Independent cycle recognition verifies the output formulas.
All reductions between base strata use certified equivalences of triples. Circle substitutions retain orientation, translation and every finite branch; alternative AAA matchings retain their support tensors. Conjugation and transposition are not used as equivalences.
Completeness over the AAA locus
Theorem 4.6. Every A/B regular square containing AAA belongs to the 9 continuous families of Theorem 4.1 or to I 8 , 12 I_{8,12} I 8 , 12 I 8 , 12 I_{8,12} I 8 , 12
Proof. The existence criteria restrict the 4 exceptional classes to their successful surfaces, circles and 32 isolated points. Apply the complete row cover and closed-stratum procedure to these loci. The exact exhaustion gives
Table 29.
The second calculation uses 16 surfaces, 128 circles and 824 point presentations, reduced by triple equivalences to 177 searches; 50 further cases reduce to classes 21, 22. All 1953 formulas pass cycle checks and embed into the first 8 families:
Table 1 (PDF p. 39).
The 32 isolated points give I 8 , 12 I_{8,12} I 8 , 12 Theorem 4.3 .
For the 32 uniformly completable components, retain all base equations in 548310 checked row records. Recursive intersection gives 15865 nodes:
Table 2 (PDF p. 39).
Certified point and circle equivalences remove 12168 and 3247 aliases. Alternative AAA supports reduce 50 further problems to the exceptional classes, leaving 400 searches. Their 845666 lattice loci give 610840 Smith branches after lattice deduplication and then 10048 distinct square formulas:
Table 3.
Every formula has a checked embedding, with assignment counts
Table 30 (PDF p. 40).
Only 8 formulas are not A-admitting: 4 represent I 8 , 12 I_{8,12} I 8 , 12 M 8 , 1 M_{8,1} M 8 , 1
The 12001 formulas are a complete parametrized cover before equivalence, not pairwise inequivalent families. The C-containing sector is treated next.
Order 8: the complete classification
The maximal BBB branch
The unique 2-dimensional BBB support class has 4 label lifts and 2 Smith branches per lift. Its 8 components reduce by certified monomial transformations and unimodular parameter changes to the following family or its conjugate, 4 components each. Put w = e π i / 6 w=e^{\pi i/6} w = e π i /6 ω = w 4 \omega=w^4 ω = w 4
u ( a ) = ( 1 , ω , ω 2 , a ω 2 , a , a ω , w 11 , w 5 ) , v ( b ) = ( 1 , ω 2 , w 7 , b w 7 , b w 3 , b w 11 , ω , w ) , \begin{aligned}
u(a)&=(1,\omega,\omega^2,a\omega^2,a,a\omega,w^{11},w^5),\\
v(b)&=(1,\omega^2,w^7,bw^7,bw^3,bw^{11},\omega,w),
\end{aligned} u ( a ) v ( b ) = ( 1 , ω , ω 2 , a ω 2 , a , aω , w 11 , w 5 ) , = ( 1 , ω 2 , w 7 , b w 7 , b w 3 , b w 11 , ω , w ) , a , b ∈ T a,b\in\mathbb{T} a , b ∈ T
The triple ( 1 ; u ( a ) ; v ( b ) ) (1;u(a);v(b)) ( 1 ; u ( a ) ; v ( b ))
P = 123 ∣ 456 ∣ 78 , Q = 127 ∣ 456 ∣ 38 , R = 128 ∣ 456 ∣ 37. P=123|456|78,\qquad Q=127|456|38,\qquad R=128|456|37. P = 123∣456∣78 , Q = 127∣456∣38 , R = 128∣456∣37. for u u u v v v u / v u/v u / v
u / v = ( 1 , w 8 , w , ( a / b ) w , ( a / b ) w 9 , ( a / b ) w 5 , w 7 , w 4 ) . u/v=(1,w^8,w,(a/b)w,(a/b)w^9,(a/b)w^5,w^7,w^4). u / v = ( 1 , w 8 , w , ( a / b ) w , ( a / b ) w 9 , ( a / b ) w 5 , w 7 , w 4 ) . verifies the third decomposition.
Theorem 5.1. This triple admits a regular 4th row exactly on the 9 circles
a 3 = − 1 or b 3 = − i or ( a / b ) 3 = i a^3=-1\quad\text{or}\quad b^3=-i\quad\text{or}\quad(a/b)^3=i a 3 = − 1 or b 3 = − i or ( a / b ) 3 = i and 54 additional points in μ 120 2 \mu_{120}^2 μ 120 2 Subsection 6.2. It never admits a regular 5th row. The same holds for the conjugate components.
Proof. Full A/B/C support projection gives 9 row circles and 135 finite special points, with every feasible graph connected. The only surviving profiles are
Table 1 (PDF p. 41).
Each circle has 1 generic row; deleting points on the circles leaves 54 points. The complete special fibres have counts
Table 2 (PDF p. 41).
Independent finite-root generation reproduces every list. Exact cycle recognition rejects all 378 pairs of candidates. Elsewhere there is at most 1 added row, so no 5th row exists. Conjugation preserves cycles.
Subsection 6.5 gives the 9 row formulas in the coordinates of (5.1).
Every A/B square contains AAA
A pure-B triple has B-admitting quotients, none admitting A.
Theorem 5.2. A pure-B triple cannot extend to an A/B regular 5 × 8 5 \times8 5 × 8 K × 8 K \times8 K × 8 K ≥ 6 K \ge6 K ≥ 6
Proof. Enumerate all normalized labels on the 39 feasible BBB supports. The 652 consistent lifts split into 1040 connected Smith components. Of these, 392 have all 3 edges A-admitting identically and 32 have at least 1 such edge; none contains a pure-B triple. The remaining components are
548 fixed triples, 60 circles, 8 two-dimensional components.
The fixed triples reduce to 34 equivalence classes, giving 102 scans. The complete A/B row cover has 3830 records, all with no free added-row phase. For every pair of raw records, including equal indices, impose both base lattices and all 1225 quotient supports. No consistent pair locus occurs, including at special base parameters. Every hypothetical 5-row extension would realize such a pair, a contradiction.
Independently, closed-stratum recursion gives 1230 nodes; after removing identically A-admitting bases and certified aliases, 110 strata remain. Only 44 have rows: 20 circles with 1 generic row, 16 point fibres with 2 rows, 6 with 3 rows and 2 with 8 rows. All 90 finite candidate pairs fail exact A/B/C recognition. This reproduces the obstruction.
Now color a row pair red if it admits A, blue otherwise. Blue means B-only. Every coloring of K 6 K_6 K 6 [28 ] : among the 5 edges from a vertex, 3 share a color; either one edge among their endpoints has that color or all 3 have the opposite color. A blue triangle would extend to 5 rows, contradicting the first assertion. Hence a red, AAA triangle exists.
The maximal CCC branch
The 6 feasible CCC support classes have dimensions 0 , 0 , 0 , 0 , 1 , 2 0, 0, 0, 0, 1, 2 0 , 0 , 0 , 0 , 1 , 2 w = e π i / 15 w = e^{\pi i/15} w = e π i /15 ω = w 10 \omega= w^{10} ω = w 10
H C ( a , b ) = ( 1 1 1 1 1 1 1 1 1 ω ω 2 a a w 6 a w 12 a w 18 a w 24 1 ω 2 ω b b w 12 b w 24 b w 6 b w 18 ) . (27) H_C(a,b)=
\begin{pmatrix}
1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 \\
1 & \omega& \omega^2 & a & aw^6 & aw^{12} & aw^{18} & aw^{24} \\
1 & \omega^2 & \omega& b & bw^{12} & bw^{24} & bw^6 & bw^{18}
\end{pmatrix}. \tag*{(27)} H C ( a , b ) = 1 1 1 1 ω ω 2 1 ω 2 ω 1 a b 1 a w 6 b w 12 1 a w 12 b w 24 1 a w 18 b w 6 1 a w 24 b w 18 . (27) Lemma 5.3. Every component of the maximal CCC support class is equivalent to a member of H C ( a , b ) H_C(a,b) H C ( a , b )
Proof. Normalize the triangle labels to ( 0 , 1 , 2 ) (0,1,2) ( 0 , 1 , 2 ) ( 0 , 2 , 1 ) (0,2,1) ( 0 , 2 , 1 ) ( 0 , 1 , 2 , 3 , 4 ) (0,1,2,3,4) ( 0 , 1 , 2 , 3 , 4 ) k = 2 , 3 , 4 k=2,3,4 k = 2 , 3 , 4
k f ( 0 ) , … , f ( 4 ) f ( j ) − j ( 5 ) 2 0 2 4 1 3 0 1 2 3 4 3 0 3 1 4 2 0 2 4 1 3 4 0 4 3 2 1 0 3 1 4 2 \begin{array}{c|c|c}
k & f(0),\ldots,f(4) & f(j)-j\ (5)\\
2 & 0\quad2\quad4\quad1\quad3 & 0\quad1\quad2\quad3\quad4\\
3 & 0\quad3\quad1\quad4\quad2 & 0\quad2\quad4\quad1\quad3\\
4 & 0\quad4\quad3\quad2\quad1 & 0\quad3\quad1\quad4\quad2
\end{array} k 2 3 4 f ( 0 ) , … , f ( 4 ) 0 2 4 1 3 0 3 1 4 2 0 4 3 2 1 f ( j ) − j ( 5 ) 0 1 2 3 4 0 2 4 1 3 0 3 1 4 2 Swapping the nontrivial rows sends slope 3 to 3 − 1 = 2 3^{-1}=2 3 − 1 = 2 1 − 4 = 2 1-4=2 1 − 4 = 2 a , b a,b a , b
Theorem 5.4. The triple H C ( a , b ) H_C(a,b) H C ( a , b )
a ∈ μ 15 , b = ω a 2 or b = ω 2 a 2 . \boxed{a\in\mu_{15},\qquad b=\omega a^2\ \text{or}\ b=\omega^2a^2.} a ∈ μ 15 , b = ω a 2 or b = ω 2 a 2 . At each of these 30 points there are 10 normalized rows: 8 A A A C C C K 5 ⊔ K 5 K_5\sqcup K_5 K 5 ⊔ K 5 A A A C C C
Proof. Full decorated-support projection retains only AAA and CCC new-row profiles, exactly at the stated points. Every feasible graph is connected; the complete fibres have sizes 8 and 2. Independent finite-root enumeration reproduces them. Exact pair tests give the 2 disjoint cliques.
Table 31.
The fixed-triple extension counts are
For example, at ( a , b ) = ( ω 2 , ω 14 ) (a,b)=(\omega^2,\omega^{14}) ( a , b ) = ( ω 2 , ω 14 )
u = ( 0 , 10 , 20 , 2 , 8 , 14 , 20 , 26 ) , v = ( 0 , 20 , 10 , 14 , 26 , 8 , 20 , 2 ) . u=(0,10,20,2,8,14,20,26),\qquad v=(0,20,10,14,26,8,20,2). u = ( 0 , 10 , 20 , 2 , 8 , 14 , 20 , 26 ) , v = ( 0 , 20 , 10 , 14 , 26 , 8 , 20 , 2 ) . and the complete added-row list is
Table 32.
Its 2 components are { 1 , 3 , 6 , 8 , 10 } \{1,3,6,8,10\} { 1 , 3 , 6 , 8 , 10 } { 2 , 4 , 5 , 7 , 9 } \{2,4,5,7,9\} { 2 , 4 , 5 , 7 , 9 }
All 60 squares have the representative ( K 8 , 30 ) i j = w M i j (K_{8,30})_{ij}=w^{M_{ij}} ( K 8 , 30 ) ij = w M ij
M = ( 0 0 0 0 0 0 0 0 0 0 6 10 12 18 20 24 0 6 13 28 27 21 12 15 0 10 28 20 16 4 10 22 0 12 27 16 1 15 24 9 0 18 21 4 15 19 6 3 0 20 12 10 24 6 0 18 0 24 15 22 9 3 18 7 ) . M=\begin{pmatrix}
0&0&0&0&0&0&0&0\\
0&0&6&10&12&18&20&24\\
0&6&13&28&27&21&12&15\\
0&10&28&20&16&4&10&22\\
0&12&27&16&1&15&24&9\\
0&18&21&4&15&19&6&3\\
0&20&12&10&24&6&0&18\\
0&24&15&22&9&3&18&7
\end{pmatrix}. M = 0 0 0 0 0 0 0 0 0 0 6 10 12 18 20 24 0 6 13 28 27 21 12 15 0 10 28 20 16 4 10 22 0 12 27 16 1 15 24 9 0 18 21 4 15 19 6 3 0 20 12 10 24 6 0 18 0 24 15 22 9 3 18 7 . Theorem 5.5. The maximal CCC branch yields exactly the class K 8 , 30 K_{8,30} K 8 , 30 K 4 ⊔ K 4 K_4\sqcup K_4 K 4 ⊔ K 4
Proof. For each of the 60 squares, canonicalization over all pivots and column orders, followed by row sorting, gives M M M
D i j = E i j − E r j − E i c + E r c ( 30 ) . D_{ij}=E_{ij}-E_{rj}-E_{ic}+E_{rc}\quad(30). D ij = E ij − E r j − E i c + E r c ( 30 ) . Arbitrary equivalence phases cancel in these dephased cross ratios. For the 2 example cliques, respective pivots ( 1 , 2 ) , ( 1 , 7 ) (1,2),(1,7) ( 1 , 2 ) , ( 1 , 7 )
Table 33.
The cross ratio w 13 w^{13} w 13
Each A quotient has 1 support and each of the 12 C quotients has 2; no B support occurs. For all 2 12 2^{12} 2 12
θ i j − θ k j − θ i l + θ k l = 0 ( j , l in the same quotient block ) . \theta_{ij}-\theta_{kj}-\theta_{il}+\theta_{kl}=0\qquad(j,l\text{ in the same quotient block}). θ ij − θ k j − θ i l + θ k l = 0 ( j , l in the same quotient block ) . have rank 49 modulo 2, hence over Q \mathbb{Q} Q
Direct cycle checks give C-cliques { 1 , 2 , 4 , 7 } \{1,2,4,7\} { 1 , 2 , 4 , 7 } { 3 , 5 , 6 , 8 } \{3,5,6,8\} { 3 , 5 , 6 , 8 }
Table 34.
Every triple contains a C-edge, excluding AAA.
All squares containing a C-pair
This branch is reduced in stages. A chosen C-pair is normalized to P ( z ) P(z) P ( z )
We begin with the 3 matrices that will occur in the classification, then establish the 2 finite-root reductions. With ω = e π i / 15 \omega= e^{\pi i/15} ω = e π i /15 ω = w 10 \omega= w^{10} ω = w 10 H i j = w E i j H_{ij}=w^{E_{ij}} H ij = w E ij
E ( K 8 , 30 ) = ( 0 0 0 0 0 0 0 0 0 0 6 10 12 18 20 24 0 6 13 28 27 21 12 15 0 10 28 20 16 4 10 22 0 12 27 16 1 15 24 9 0 18 21 4 15 19 6 3 0 20 12 10 24 6 0 18 0 24 15 22 9 3 18 7 ) . E(K_{8,30})=
\begin{pmatrix}
0&0&0&0&0&0&0&0\\
0&0&6&10&12&18&20&24\\
0&6&13&28&27&21&12&15\\
0&10&28&20&16&4&10&22\\
0&12&27&16&1&15&24&9\\
0&18&21&4&15&19&6&3\\
0&20&12&10&24&6&0&18\\
0&24&15&22&9&3&18&7
\end{pmatrix}. E ( K 8 , 30 ) = 0 0 0 0 0 0 0 0 0 0 6 10 12 18 20 24 0 6 13 28 27 21 12 15 0 10 28 20 16 4 10 22 0 12 27 16 1 15 24 9 0 18 21 4 15 19 6 3 0 20 12 10 24 6 0 18 0 24 15 22 9 3 18 7 . E ( L 8 , 30 ) = ( 0 0 0 0 0 0 0 0 0 0 6 10 12 18 20 24 0 6 21 2 19 17 4 15 0 10 17 25 8 2 15 23 0 12 4 23 27 15 19 8 0 18 2 17 15 3 1 16 0 20 19 15 4 16 5 1 0 24 15 8 23 1 16 9 ) . E(L_{8,30})=
\begin{pmatrix}
0&0&0&0&0&0&0&0\\
0&0&6&10&12&18&20&24\\
0&6&21&2&19&17&4&15\\
0&10&17&25&8&2&15&23\\
0&12&4&23&27&15&19&8\\
0&18&2&17&15&3&1&16\\
0&20&19&15&4&16&5&1\\
0&24&15&8&23&1&16&9
\end{pmatrix}. E ( L 8 , 30 ) = 0 0 0 0 0 0 0 0 0 0 6 10 12 18 20 24 0 6 21 17 4 2 19 15 0 10 2 25 23 17 15 8 0 12 19 8 27 15 4 23 0 18 17 2 15 3 16 1 0 20 4 15 19 1 5 16 0 24 15 23 8 16 1 9 . E ( L 8 , 30 t ) = ( 0 0 0 0 0 0 0 0 0 0 6 10 12 18 20 24 0 6 21 17 4 2 19 15 0 10 2 25 23 17 15 8 0 12 19 8 27 15 4 23 0 18 17 2 15 3 16 1 0 20 4 15 19 1 5 16 0 24 15 23 8 16 1 9 ) . E(L_{8,30}^{t})=
\begin{pmatrix}
0&0&0&0&0&0&0&0\\
0&0&6&10&12&18&20&24\\
0&6&21&17&4&2&19&15\\
0&10&2&25&23&17&15&8\\
0&12&19&8&27&15&4&23\\
0&18&17&2&15&3&16&1\\
0&20&4&15&19&1&5&16\\
0&24&15&23&8&16&1&9
\end{pmatrix}. E ( L 8 , 30 t ) = 0 0 0 0 0 0 0 0 0 0 6 10 12 18 20 24 0 6 21 2 19 17 4 15 0 10 17 25 8 2 15 23 0 12 4 23 27 15 19 8 0 18 2 17 15 3 1 16 0 20 19 15 4 16 5 1 0 24 15 8 23 1 16 9 . Normalize a chosen C-pair to
P ( z ) = ( 1 1 1 1 1 1 1 1 1 ω ω 2 z z ω 6 z ω 12 z ω 18 z ω 24 ) , z ∈ T . (28) P(z)=
\begin{pmatrix}
1&1&1&1&1&1&1&1\\
1&\omega&\omega^2&z&z\omega^6&z\omega^{12}&z\omega^{18}&z\omega^{24}
\end{pmatrix},
\qquad z\in\mathbb{T}. \tag*{(28)} P ( z ) = ( 1 1 1 ω 1 ω 2 1 z 1 z ω 6 1 z ω 12 1 z ω 18 1 z ω 24 ) , z ∈ T . (28) This uses only row and column phases and permutations, with z z z
Lemma 5.6. If a regular partial matrix containing P ( z ) P(z) P ( z ) z ∈ μ 30 z\in\mu_{30} z ∈ μ 30 P ( 1 ) P(1) P ( 1 ) P ( − 1 ) P(-1) P ( − 1 )
Proof. Put z = e ( t ) z=e(t) z = e ( t ) 3913 2 = 15311569 3913^2=15311569 391 3 2 = 15311569
k t + a 30 = 0 ( m o d 1 ) . (29) kt+\frac{a}{30}=0\pmod{1}. \tag*{(29)} k t + 30 a = 0 ( mod 1 ) . (29) Independent graph traversal solves each equation directly, retaining every torsion value. Both solvers give
Table 1 (PDF p. 44).
and 14178 row/locus records before containment. Every nontrivial base locus lies in μ 30 \mu_{30} μ 30
r s ( v ) = ( 1 , ω 2 , ω , v , v w 6 s , v w 12 s , v w 18 s , v w 24 s ) , s = 2 , 3 , 4 , v ∈ T . (30) r_s(v) = (1,\omega^2,\omega,v,vw^{6s},vw^{12s},vw^{18s},vw^{24s}),\qquad s=2,3,4,\ v\in\mathbb{T}. \tag*{(30)} r s ( v ) = ( 1 , ω 2 , ω , v , v w 6 s , v w 12 s , v w 18 s , v w 24 s ) , s = 2 , 3 , 4 , v ∈ T . (30) Different slopes have inner product 3; for equal slopes, ⟨ r s ( v ) , r s ( t ) ⟩ = 3 + 5 v t ‾ ≠ 0 \langle r_s(v),r_s(t)\rangle=3+5v\overline{t}\ne0 ⟨ r s ( v ) , r s ( t )⟩ = 3 + 5 v t = 0
Cyclic triangle and pentagon reordering changes the exponent of z z z 6 b − 10 a 6b-10a 6 b − 10 a 1 , − 1 1,-1 1 , − 1
The complete normalized 3rd-row cover over these pairs is
Table 1 (PDF p. 45).
For P ( 1 ) P(1) P ( 1 ) r j ( u ) k = e ( c k / 30 + m k u ) r_j(u)_k=e(c_k/30+m_ku) r j ( u ) k = e ( c k /30 + m k u )
Table 2 (PDF p. 45).
For P ( − 1 ) P(-1) P ( − 1 ) (30) . No retained fixed row lies on a circle; the P ( 1 ) P(1) P ( 1 ) code002_work/N8/n8_C_sector/.
Lemma 5.7. Every regular partial matrix with at least 5 rows and a C C C P ( 1 ) P(1) P ( 1 ) P ( − 1 ) P(-1) P ( − 1 ) μ 60 \mu_{60} μ 60
Proof. Enumerate every decorated quotient support between 2 added families, allowing a circle twice with independent parameters. At P ( 1 ) P(1) P ( 1 )
Table 35.
At P ( − 1 ) P(-1) P ( − 1 ) ± 1 \pm1 ± 1 μ 30 \mu_{30} μ 30 μ 60 \mu_{60} μ 60
u − v + c 30 = 0 ( 1 ) . (31) u-v+\frac{c}{30}=0\ (1). \tag*{(31)} u − v + 30 c = 0 ( 1 ) . (31) with the 18 entries
Table 36.
A self-line pairs opposite phases, so at most 2 rows from the same circle can be compatible through self-lines. The 6 cross-pairs of circles are disjoint, ruling out 3 distinct circles. If 2 rows belong to one circle and a third to its partner, the cross-line fixes the same phase difference for both pairs, forcing the first 2 phases to coincide. This contradicts their being opposite. Hence 3 added rows cannot be mutually compatible using only these lines.
If one added row is fixed, compatibility forces every occurring circle phase into μ 30 \mu_{30} μ 30 μ 60 \mu_{60} μ 60 μ 30 \mu_{30} μ 30 μ 60 \mu_{60} μ 60
The finite-root reductions now give the classification and the uniqueness of completion.
Theorem 5.8. Every regular order-8 square containing a C-pair belongs to exactly 1 of
K 8 , 30 , L 8 , 30 , L 8 , 30 t . K_{8,30}, \qquad L_{8,30}, \qquad L^{t}_{8,30}. K 8 , 30 , L 8 , 30 , L 8 , 30 t . These classes are distinct, regularly isolated, and have minimal Butson level 30. Every regular K × 8 K \times8 K × 8 5 ≤ K ≤ 7 5 \le K \le7 5 ≤ K ≤ 7
Proof. Evaluate the circles at all 60th roots, add fixed rows and deduplicate. Independent recursive cycle recognition gives the graphs For at least 3 added rows, Lemma 5.7 makes this grid exhaustive. The 16 6-cliques have disjoint 2-element subcliques, and
Table 37 (PDF p. 47).
16 ( 6 3 ) = 320 , 16 ( 6 4 ) = 240 , 16 ( 6 5 ) = 96. 16\binom{6}{3}=320,\qquad16\binom{6}{4}=240,\qquad16\binom{6}{5}=96. 16 ( 3 6 ) = 320 , 16 ( 4 6 ) = 240 , 16 ( 5 6 ) = 96. exhaust the smaller clique counts. Thus every 5-, 6-, or 7-row partial matrix belongs to exactly 1 square. The negative pair stops at 4 rows.
Independently, the affine completion engine intersects all pair conditions for 6 added families. Its 14 loci split into 16 points at P ( 1 ) P(1) P ( 1 ) P ( − 1 ) P(-1) P ( − 1 ) μ 30 \mu_{30} μ 30
Complete pivot-and-permutation canonicalization gives multiplicities 12, 2, 2 for the 3 displayed, distinct matrices. Exact transformations certify all equalities. Each representative contains a primitive 30th-root cross ratio, proving minimal level. A chosen C quotient remains outside the closed A and B loci in a neighborhood, so nearby regular matrices remain in this finite C-sector. The 3 compact monomial orbits are disjoint, proving regular isolation.
The lower-row boundary is also explicit: at P ( 1 ) P(1) P ( 1 ) P ( − 1 ) P(-1) P ( − 1 ) 1492 − 720 = 772 1492-720=772 1492 − 720 = 772
The row and column cycle censuses are
Table 38 (PDF p. 47).
The C-cliques of K 8 , 30 K_{8,30} K 8 , 30 L 8 , 30 L_{8,30} L 8 , 30 { 1 , 2 } \{1,2\} { 1 , 2 } { 3 , 8 } \{3,8\} { 3 , 8 } { 4 , 7 } \{4,7\} { 4 , 7 } { 5 , 6 } \{5,6\} { 5 , 6 }
The complete family cover
Theorem 5.9. Every regular complex Hadamard matrix of order 8 is ordinarily monomially equivalent to a member of
F 8 , 5 a , F 8 , 5 b , F 8 , 5 c , F 8 , 4 , F 8 , 4 t , F 8 , 2 , F 8 , 2 t , K 8 , 2 , M 8 , 1 , (32) F_{8,5a},\ F_{8,5b},\ F_{8,5c},\ F_{8,4},\ F^{t}_{8,4},\ F_{8,2},\ F^{t}_{8,2},\ K_{8,2},\ M_{8,1}, \tag*{(32)} F 8 , 5 a , F 8 , 5 b , F 8 , 5 c , F 8 , 4 , F 8 , 4 t , F 8 , 2 , F 8 , 2 t , K 8 , 2 , M 8 , 1 , (32) or to 1 of the 4 isolated matrices
I 8 , 12 , K 8 , 30 , L 8 , 30 , L 8 , 30 t . (33) I_{8,12},\qquad K_{8,30},\qquad L_{8,30},\qquad L^{t}_{8,30}. \tag*{(33)} I 8 , 12 , K 8 , 30 , L 8 , 30 , L 8 , 30 t . (33) Conversely all these matrices are regular. The continuous dimensions are 5, 5, 5, 4, 4, 2, 2, 2, 1; their generic types are distinct and locally maximal in the regular locus. The isolated classes are pairwise inequivalent. Conjugation and transposition are not included in equivalence.
Proof. A C-pair puts the square in Theorem 5.8 . Otherwise every quotient admits A or B; Theorem 5.2 supplies AAA, and Theorem 4.6 gives the remaining cover. The explicit cycle certificates prove the converse. The matching invariants and local support ranks distinguish the generic continuous types. The C-classes are mutually distinct; I 8 , 12 I_{8,12} I 8 , 12
Corollary 5.10. At order 8:
(i) Every regular matrix is an affine deformation of a Butson matrix, up to equivalence.
(ii) The maximum regular local moduli dimension is 5; the isolated classes are exactly (33) .
(iii) K 8 , 30 K_{8,30} K 8 , 30
(iv) Transposition preserves regularity. It fixes the classes of I 8 , 12 I_{8,12} I 8 , 12 K 8 , 30 K_{8,30} K 8 , 30 L 8 , 30 L_{8,30} L 8 , 30 L 8 , 30 t L^{t}_{8,30} L 8 , 30 t (32) and the transpose identities of Subsection 4.1.
Proof. Set all family parameters to 1 to obtain Butson seeds; the isolated matrices are constant deformations. The finite affine cover has maximum rank 5, unchanged by the finite equivalence action after dephasing. Every point on a positive-dimensional family has inequivalent nearby points; the other 4 classes are isolated. The AAA assertion follows from the C-census and the A/B theorem. Finally, every displayed family and isolated square has regular column quotients; the explicit transpose witnesses give the identifications.
The finite catalogue comparison
Theorem 5.11. All 5469 entries of the Aalto B ( 8 , q ) B(8,q) B ( 8 , q ) q = 2 , 4 , 6 , 8 , 10 , 12 , 14 , 16 q = 2, 4, 6, 8, 10, 12, 14, 16 q = 2 , 4 , 6 , 8 , 10 , 12 , 14 , 16 I 8 , 12 I_{8,12} I 8 , 12
Proof. Using the Aalto catalogue [21 ] and its format conventions, exact cycle recognition tests all quotients and all 56 row triples. Independent substitution checks every embedding:
Table 5.
At level 12 only I 8 , 12 I_{8,12} I 8 , 12 K 8 , 2 K_{8,2} K 8 , 2 M 8 , 1 M_{8,1} M 8 , 1 I 8 , 12 I_{8,12} I 8 , 12
The catalogue’s completeness is the published input; entries at different levels may repeat classes. A C quotient in B ( 8 , q ) B(8,q) B ( 8 , q ) 3 ∣ q 3 \mid q 3 ∣ q 5 ∣ q 5 \mid q 5 ∣ q
Order 8: certificates and verification
The AAA registry
The first support is P = 12 ∣ 34 ∣ 56 ∣ 78 P = 12|34|56|78 P = 12∣34∣56∣78 Q Q Q R R R
Table 1 (PDF p. 49).
The natural coordinates equal the Smith exponentials for classes 11 and 19. For class 21, x = e ( t − s ) x=e(t-s) x = e ( t − s ) y = e ( t ) y=e(t) y = e ( t ) x = e ( s ) x=e(s) x = e ( s ) y = e ( t + 1 / 2 ) y=e(t+1/2) y = e ( t + 1/2 )
Finite residues
The class-21 residual set has ( s , t ) = ( i / 24 , j / 24 ) (s,t)=(i/24,j/24) ( s , t ) = ( i /24 , j /24 ) j j j i i i
Table 2 (PDF p. 49).
It has 64 points in each component, outside the successful circles. For class 22, write ( x , y ) = ( e ( i / 12 ) , e ( j / 12 ) ) (x,y)=(e(i/12),e(j/12)) ( x , y ) = ( e ( i /12 ) , e ( j /12 ))
The same parameter sets apply to both signs; their row lists are checked separately. For the BBB family (5.1), put w = e π i / 60 w=e^{\pi i/60} w = e π i /60 ( a , b ) = ( w i , w j ) (a,b)=(w^i,w^j) ( a , b ) = ( w i , w j )
Conjugate both parameters and entries for the conjugate components.
Uniform square certificates
The matrix H ( c , b ) H^{(c,b)} H ( c , b ) c c c b b b
H ( 0 , 0 ) ( a , b , c , d ) = ( 1 1 1 1 1 1 1 1 1 − 1 − b d b d − b d b d 1 − 1 1 − c − 1 − a − 1 b d d a − 1 b d − d − 1 c − 1 1 − a c − 1 b d − d − b d d − 1 a c − 1 1 − 1 b d − b d b d − b d 1 − 1 1 a c − 1 − b d − d b d d − 1 − a c − 1 1 1 − 1 − 1 − 1 − 1 1 1 1 c − 1 a − 1 b d d − a − 1 b d − d − 1 − c − 1 ) H^{(0,0)}(a,b,c,d)=
\begin{pmatrix}
1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 \\
1 & -1 & -bd & bd & -bd & bd & 1 & -1 \\
1 & -c^{-1} & -a^{-1}bd & d & a^{-1}bd & -d & -1 & c^{-1} \\
1 & -ac^{-1} & bd & -d & -bd & d & -1 & ac^{-1} \\
1 & -1 & bd & -bd & bd & -bd & 1 & -1 \\
1 & ac^{-1} & -bd & -d & bd & d & -1 & -ac^{-1} \\
1 & 1 & -1 & -1 & -1 & -1 & 1 & 1 \\
1 & c^{-1} & a^{-1}bd & d & -a^{-1}bd & -d & -1 & -c^{-1}
\end{pmatrix} H ( 0 , 0 ) ( a , b , c , d ) = 1 1 1 1 1 1 1 1 1 − 1 − c − 1 − a c − 1 − 1 a c − 1 1 c − 1 1 − b d − a − 1 b d b d b d − b d − 1 a − 1 b d 1 b d d − d − b d − d − 1 d 1 − b d a − 1 b d − b d b d b d − 1 − a − 1 b d 1 b d − d d − b d d − 1 − d 1 1 − 1 − 1 1 − 1 1 − 1 1 − 1 c − 1 a c − 1 − 1 − a c − 1 1 − c − 1 H ( 1 , 0 ) ( a , b , c , d ) = ( 1 1 1 1 1 1 1 1 1 − 1 a d − a d − b d b d 1 − 1 1 − c − 1 d d − d − d − 1 c − 1 1 − 1 − a d a d b d − b d 1 − 1 1 − c − 1 − d − d d d − 1 c − 1 1 1 − 1 − 1 − 1 − 1 1 1 1 c − 1 − a d a d − b d b d − 1 − c − 1 1 c − 1 a d − a d b d − b d − 1 − c − 1 ) H^{(1,0)}(a,b,c,d)=
\begin{pmatrix}
1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 \\
1 & -1 & ad & -ad & -bd & bd & 1 & -1 \\
1 & -c^{-1} & d & d & -d & -d & -1 & c^{-1} \\
1 & -1 & -ad & ad & bd & -bd & 1 & -1 \\
1 & -c^{-1} & -d & -d & d & d & -1 & c^{-1} \\
1 & 1 & -1 & -1 & -1 & -1 & 1 & 1 \\
1 & c^{-1} & -ad & ad & -bd & bd & -1 & -c^{-1} \\
1 & c^{-1} & ad & -ad & bd & -bd & -1 & -c^{-1}
\end{pmatrix} H ( 1 , 0 ) ( a , b , c , d ) = 1 1 1 1 1 1 1 1 1 − 1 − c − 1 − 1 − c − 1 1 c − 1 c − 1 1 a d d − a d − d − 1 − a d a d 1 − a d d a d − d − 1 a d − a d 1 − b d − d b d d − 1 − b d b d 1 b d − d − b d d − 1 b d − b d 1 1 − 1 1 − 1 1 − 1 − 1 1 − 1 c − 1 − 1 c − 1 1 − c − 1 − c − 1 H ( 2 , 0 ) ( a , b , c ) = ( 1 1 1 1 1 1 1 1 1 − 1 − a b a b − a b a b 1 − 1 1 − c − 1 − a a a − a − 1 c − 1 1 − a a 2 c − a − a 2 c a − 1 a 1 − 1 a b − a b a b − a b 1 − 1 1 a − a 2 c − a a 2 c a − 1 − a 1 1 − 1 − 1 − 1 − 1 1 1 1 c − 1 a a − a − a − 1 − c − 1 ) H^{(2,0)}(a,b,c)=
\begin{pmatrix}
1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 \\
1 & -1 & -ab & ab & -ab & ab & 1 & -1 \\
1 & -c^{-1} & -a & a & a & -a & -1 & c^{-1} \\
1 & -a & a^{2}c & -a & -a^{2}c & a & -1 & a \\
1 & -1 & ab & -ab & ab & -ab & 1 & -1 \\
1 & a & -a^{2}c & -a & a^{2}c & a & -1 & -a \\
1 & 1 & -1 & -1 & -1 & -1 & 1 & 1 \\
1 & c^{-1} & a & a & -a & -a & -1 & -c^{-1}
\end{pmatrix} H ( 2 , 0 ) ( a , b , c ) = 1 1 1 1 1 1 1 1 1 − 1 − c − 1 − a − 1 a 1 c − 1 1 − ab − a a 2 c ab − a 2 c − 1 a 1 ab a − a − ab − a − 1 a 1 − ab a − a 2 c ab a 2 c − 1 − a 1 ab − a a − ab a − 1 − a 1 1 − 1 − 1 1 − 1 1 − 1 1 − 1 c − 1 a − 1 − a 1 − c − 1 H ( 2 , 1 ) ( a , b , c ) = ( 1 1 1 1 1 1 1 1 1 − 1 − a b a b a b − a b 1 − 1 1 − c − 1 a a − a − a − 1 c − 1 1 − a − a 2 c − a a a 2 c − 1 a 1 − 1 a b − a b − a b a b 1 − 1 1 a a 2 c − a a − a 2 c − 1 − a 1 1 − 1 − 1 − 1 − 1 1 1 1 c − 1 − a a − a a − 1 − c − 1 ) H^{(2,1)}(a,b,c)=
\begin{pmatrix}
1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 \\
1 & -1 & -ab & ab & ab & -ab & 1 & -1 \\
1 & -c^{-1} & a & a & -a & -a & -1 & c^{-1} \\
1 & -a & -a^{2}c & -a & a & a^{2}c & -1 & a \\
1 & -1 & ab & -ab & -ab & ab & 1 & -1 \\
1 & a & a^{2}c & -a & a & -a^{2}c & -1 & -a \\
1 & 1 & -1 & -1 & -1 & -1 & 1 & 1 \\
1 & c^{-1} & -a & a & -a & a & -1 & -c^{-1}
\end{pmatrix} H ( 2 , 1 ) ( a , b , c ) = 1 1 1 1 1 1 1 1 1 − 1 − c − 1 − a − 1 a 1 c − 1 1 − ab a − a 2 c ab a 2 c − 1 − a 1 ab a − a − ab − a − 1 a 1 ab − a a − ab a − 1 − a 1 − ab − a a 2 c ab − a 2 c − 1 a 1 1 − 1 − 1 1 − 1 1 − 1 1 − 1 c − 1 a − 1 − a 1 − c − 1 H ( 3 , 0 ) ( a , b , c ) = ( 1 1 1 1 1 1 1 1 1 − 1 − 1 1 − 1 1 1 − 1 1 − c − 1 − a − 1 b − 1 c − 1 − b − 1 − 1 a − 1 1 − a c − 1 1 − 1 a c − 1 − 1 1 − 1 1 − 1 1 − b − 1 − 1 b − 1 − 1 1 1 a c − 1 − 1 − 1 − a c − 1 − 1 1 1 1 1 − 1 − b − 1 1 b − 1 − 1 − 1 1 c − 1 a − 1 b − 1 − c − 1 − b − 1 − 1 − a − 1 ) H^{(3,0)}(a,b,c)=
\begin{pmatrix}
1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 \\
1 & -1 & -1 & 1 & -1 & 1 & 1 & -1 \\
1 & -c^{-1} & -a^{-1} & b^{-1} & c^{-1} & -b^{-1} & -1 & a^{-1} \\
1 & -ac^{-1} & 1 & -1 & ac^{-1} & -1 & 1 & -1 \\
1 & -1 & 1 & -b^{-1} & -1 & b^{-1} & -1 & 1 \\
1 & ac^{-1} & -1 & -1 & -ac^{-1} & -1 & 1 & 1 \\
1 & 1 & -1 & -b^{-1} & 1 & b^{-1} & -1 & -1 \\
1 & c^{-1} & a^{-1} & b^{-1} & -c^{-1} & -b^{-1} & -1 & -a^{-1}
\end{pmatrix} H ( 3 , 0 ) ( a , b , c ) = 1 1 1 1 1 1 1 1 1 − 1 − c − 1 − a c − 1 − 1 a c − 1 1 c − 1 1 − 1 − a − 1 1 1 − 1 − 1 a − 1 1 1 b − 1 − 1 − b − 1 − 1 − b − 1 b − 1 1 − 1 c − 1 a c − 1 − 1 − a c − 1 1 − c − 1 1 1 − b − 1 − 1 b − 1 − 1 b − 1 − b − 1 1 1 − 1 1 − 1 1 − 1 − 1 1 − 1 a − 1 − 1 1 1 − 1 − a − 1 H ( 4 , 0 ) ( a , b , c ) = ( 1 1 1 1 1 1 1 1 1 − 1 − a c − 1 a c − 1 b c − 1 − b c − 1 1 − 1 1 − c − 1 − c − 1 − c − 1 c − 1 c − 1 − 1 c − 1 1 − 1 a c − 1 − a c − 1 − b c − 1 b c − 1 1 − 1 1 − c − 1 c − 1 c − 1 − c − 1 − c − 1 − 1 c − 1 1 1 − 1 − 1 − 1 − 1 1 1 1 c − 1 − a c − 1 a c − 1 − b c − 1 b c − 1 − 1 − c − 1 1 c − 1 a c − 1 − a c − 1 b c − 1 − b c − 1 − 1 − c − 1 ) H^{(4,0)}(a,b,c)=
\begin{pmatrix}
1&1&1&1&1&1&1&1\\
1&-1&-ac^{-1}&ac^{-1}&bc^{-1}&-bc^{-1}&1&-1\\
1&-c^{-1}&-c^{-1}&-c^{-1}&c^{-1}&c^{-1}&-1&c^{-1}\\
1&-1&ac^{-1}&-ac^{-1}&-bc^{-1}&bc^{-1}&1&-1\\
1&-c^{-1}&c^{-1}&c^{-1}&-c^{-1}&-c^{-1}&-1&c^{-1}\\
1&1&-1&-1&-1&-1&1&1\\
1&c^{-1}&-ac^{-1}&ac^{-1}&-bc^{-1}&bc^{-1}&-1&-c^{-1}\\
1&c^{-1}&ac^{-1}&-ac^{-1}&bc^{-1}&-bc^{-1}&-1&-c^{-1}
\end{pmatrix} H ( 4 , 0 ) ( a , b , c ) = 1 1 1 1 1 1 1 1 1 − 1 − c − 1 − 1 − c − 1 1 c − 1 c − 1 1 − a c − 1 − c − 1 a c − 1 c − 1 − 1 − a c − 1 a c − 1 1 a c − 1 − c − 1 − a c − 1 c − 1 − 1 a c − 1 − a c − 1 1 b c − 1 c − 1 − b c − 1 − c − 1 − 1 − b c − 1 b c − 1 1 − b c − 1 c − 1 b c − 1 − c − 1 − 1 b c − 1 − b c − 1 1 1 − 1 1 − 1 1 − 1 − 1 1 − 1 c − 1 − 1 c − 1 1 − c − 1 − c − 1 H ( 5 , 0 ) ( a , b ) = ( 1 1 1 1 1 1 1 1 1 − 1 − 1 1 − 1 1 1 − 1 1 − b − 1 − a − 1 b − 1 b − 1 − b − 1 − 1 a − 1 1 − a b − 1 1 − 1 a b − 1 − 1 1 − 1 1 − 1 1 − b − 1 − 1 b − 1 − 1 1 1 a b − 1 − 1 − 1 − a b − 1 − 1 1 1 1 1 − 1 − b − 1 1 b − 1 − 1 − 1 1 b − 1 a − 1 b − 1 − b − 1 − b − 1 − 1 − a − 1 ) H^{(5,0)}(a,b)=
\begin{pmatrix}
1&1&1&1&1&1&1&1\\
1&-1&-1&1&-1&1&1&-1\\
1&-b^{-1}&-a^{-1}&b^{-1}&b^{-1}&-b^{-1}&-1&a^{-1}\\
1&-ab^{-1}&1&-1&ab^{-1}&-1&1&-1\\
1&-1&1&-b^{-1}&-1&b^{-1}&-1&1\\
1&ab^{-1}&-1&-1&-ab^{-1}&-1&1&1\\
1&1&-1&-b^{-1}&1&b^{-1}&-1&-1\\
1&b^{-1}&a^{-1}&b^{-1}&-b^{-1}&-b^{-1}&-1&-a^{-1}
\end{pmatrix} H ( 5 , 0 ) ( a , b ) = 1 1 1 1 1 1 1 1 1 − 1 − b − 1 − a b − 1 − 1 a b − 1 1 b − 1 1 − 1 − a − 1 1 1 − 1 − 1 a − 1 1 1 b − 1 − 1 − b − 1 − 1 − b − 1 b − 1 1 − 1 b − 1 a b − 1 − 1 − a b − 1 1 − b − 1 1 1 − b − 1 − 1 b − 1 − 1 b − 1 − b − 1 1 1 − 1 1 − 1 1 − 1 − 1 1 − 1 a − 1 − 1 1 1 − 1 − a − 1 H ( 5 , 1 ) ( a , b ) = ( 1 1 1 1 1 1 1 1 1 − 1 − 1 1 1 − 1 1 − 1 1 b − 1 − a − 1 b − 1 − b − 1 − b − 1 − 1 a − 1 1 − a b − 1 − 1 − 1 − 1 a b − 1 1 1 1 − 1 1 − b − 1 b − 1 − 1 − 1 1 1 − b − 1 a − 1 b − 1 − b − 1 b − 1 − 1 − a − 1 1 a b − 1 1 − 1 − 1 − a b − 1 1 − 1 1 1 − 1 − b − 1 b − 1 1 − 1 − 1 ) H^{(5,1)}(a,b)=
\begin{pmatrix}
1&1&1&1&1&1&1&1\\
1&-1&-1&1&1&-1&1&-1\\
1&b^{-1}&-a^{-1}&b^{-1}&-b^{-1}&-b^{-1}&-1&a^{-1}\\
1&-ab^{-1}&-1&-1&-1&ab^{-1}&1&1\\
1&-1&1&-b^{-1}&b^{-1}&-1&-1&1\\
1&-b^{-1}&a^{-1}&b^{-1}&-b^{-1}&b^{-1}&-1&-a^{-1}\\
1&ab^{-1}&1&-1&-1&-ab^{-1}&1&-1\\
1&1&-1&-b^{-1}&b^{-1}&1&-1&-1
\end{pmatrix} H ( 5 , 1 ) ( a , b ) = 1 1 1 1 1 1 1 1 1 − 1 b − 1 − a b − 1 − 1 − b − 1 a b − 1 1 1 − 1 − a − 1 − 1 1 a − 1 1 − 1 1 1 b − 1 − 1 − b − 1 b − 1 − 1 − b − 1 1 1 − b − 1 − 1 b − 1 − b − 1 − 1 b − 1 1 − 1 − b − 1 a b − 1 − 1 b − 1 − a b − 1 1 1 1 − 1 1 − 1 − 1 1 − 1 1 − 1 a − 1 1 1 − a − 1 − 1 − 1 H ( 6 , 0 ) ( a , b , c ) = ( 1 1 1 1 1 1 1 1 1 − 1 − b c b c − b c b c 1 − 1 1 − b − 1 − a − 1 b c c a − 1 b c − c − 1 b − 1 1 − a b − 1 b c − c − b c c − 1 a b − 1 1 − 1 b c − b c b c − b c 1 − 1 1 a b − 1 − b c − c b c c − 1 − a b − 1 1 1 − 1 − 1 − 1 − 1 1 1 1 b − 1 a − 1 b c c − a − 1 b c − c − 1 − b − 1 ) H^{(6,0)}(a,b,c)=
\begin{pmatrix}
1&1&1&1&1&1&1&1\\
1&-1&-bc&bc&-bc&bc&1&-1\\
1&-b^{-1}&-a^{-1}bc&c&a^{-1}bc&-c&-1&b^{-1}\\
1&-ab^{-1}&bc&-c&-bc&c&-1&ab^{-1}\\
1&-1&bc&-bc&bc&-bc&1&-1\\
1&ab^{-1}&-bc&-c&bc&c&-1&-ab^{-1}\\
1&1&-1&-1&-1&-1&1&1\\
1&b^{-1}&a^{-1}bc&c&-a^{-1}bc&-c&-1&-b^{-1}
\end{pmatrix} H ( 6 , 0 ) ( a , b , c ) = 1 1 1 1 1 1 1 1 1 − 1 − b − 1 − a b − 1 − 1 a b − 1 1 b − 1 1 − b c − a − 1 b c b c b c − b c − 1 a − 1 b c 1 b c c − c − b c − c − 1 c 1 − b c a − 1 b c − b c b c b c − 1 − a − 1 b c 1 b c − c c − b c c − 1 − c 1 1 − 1 − 1 1 − 1 1 − 1 1 − 1 b − 1 a b − 1 − 1 − a b − 1 1 − b − 1 H ( 7 , 0 ) ( a , b ) = ( 1 1 1 1 1 1 1 1 1 − 1 − 1 1 1 − 1 1 − 1 1 − b − 1 − a − 1 − b − 1 b − 1 a − 1 − 1 b − 1 1 − a b − 1 1 − 1 − 1 − 1 1 a b − 1 1 − 1 1 b − 1 − b − 1 1 − 1 − 1 1 a b − 1 − 1 − 1 − 1 1 1 − a b − 1 1 1 − 1 b − 1 − b − 1 − 1 − 1 1 1 b − 1 a − 1 − b − 1 b − 1 − a − 1 − 1 − b − 1 ) H^{(7,0)}(a,b)=
\begin{pmatrix}
1&1&1&1&1&1&1&1\\
1&-1&-1&1&1&-1&1&-1\\
1&-b^{-1}&-a^{-1}&-b^{-1}&b^{-1}&a^{-1}&-1&b^{-1}\\
1&-ab^{-1}&1&-1&-1&-1&1&ab^{-1}\\
1&-1&1&b^{-1}&-b^{-1}&1&-1&-1\\
1&ab^{-1}&-1&-1&-1&1&1&-ab^{-1}\\
1&1&-1&b^{-1}&-b^{-1}&-1&-1&1\\
1&b^{-1}&a^{-1}&-b^{-1}&b^{-1}&-a^{-1}&-1&-b^{-1}
\end{pmatrix} H ( 7 , 0 ) ( a , b ) = 1 1 1 1 1 1 1 1 1 − 1 − b − 1 − a b − 1 − 1 a b − 1 1 b − 1 1 − 1 − a − 1 1 1 − 1 − 1 a − 1 1 1 − b − 1 − 1 b − 1 − 1 b − 1 − b − 1 1 1 b − 1 − 1 − b − 1 − 1 − b − 1 b − 1 1 − 1 a − 1 − 1 1 1 − 1 − a − 1 1 1 − 1 1 − 1 1 − 1 − 1 1 − 1 b − 1 a b − 1 − 1 − a b − 1 1 − b − 1 H ( 7 , 1 ) ( a , b ) = ( 1 1 1 1 1 1 1 1 1 − 1 1 − 1 − 1 1 1 − 1 1 b − 1 a − 1 − b − 1 − b − 1 − a − 1 − 1 b − 1 1 − a b − 1 − 1 − a b − 1 a b − 1 − 1 1 a b − 1 1 − a b − 1 1 a b − 1 b − 1 − 1 − 1 − b − 1 1 − b − 1 − a − 1 b − 1 − b − 1 a − 1 − 1 b − 1 1 a b − 1 − 1 − a b − 1 b − 1 1 − 1 − b − 1 1 a b − 1 − 1 a b − 1 − a b − 1 − 1 1 − a b − 1 ) H^{(7,1)}(a,b)=
\begin{pmatrix}
1&1&1&1&1&1&1&1\\
1&-1&1&-1&-1&1&1&-1\\
1&b^{-1}&a^{-1}&-b^{-1}&-b^{-1}&-a^{-1}&-1&b^{-1}\\
1&-ab^{-1}&-1&-ab^{-1}&ab^{-1}&-1&1&ab^{-1}\\
1&-ab^{-1}&1&ab^{-1}&b^{-1}&-1&-1&-b^{-1}\\
1&-b^{-1}&-a^{-1}&b^{-1}&-b^{-1}&a^{-1}&-1&b^{-1}\\
1&ab^{-1}&-1&-ab^{-1}&b^{-1}&1&-1&-b^{-1}\\
1&ab^{-1}&-1&ab^{-1}&-ab^{-1}&-1&1&-ab^{-1}
\end{pmatrix} H ( 7 , 1 ) ( a , b ) = 1 1 1 1 1 1 1 1 1 − 1 b − 1 − a b − 1 − a b − 1 − b − 1 a b − 1 a b − 1 1 1 a − 1 − 1 1 − a − 1 − 1 − 1 1 − 1 − b − 1 − a b − 1 a b − 1 b − 1 − a b − 1 a b − 1 1 − 1 − b − 1 a b − 1 b − 1 − b − 1 b − 1 − a b − 1 1 1 − a − 1 − 1 − 1 a − 1 1 − 1 1 1 − 1 1 − 1 − 1 − 1 1 1 − 1 b − 1 a b − 1 − b − 1 b − 1 − b − 1 − a b − 1 H ( 8 , 0 ) ( a , b , c ) = ( 1 1 1 1 1 1 1 1 1 − 1 a c − a c − b c b c 1 − 1 1 − b − 1 c c − c − c − 1 b − 1 1 − 1 − a c a c b c − b c 1 − 1 1 − b − 1 − c − c c c − 1 b − 1 1 1 − 1 − 1 − 1 − 1 1 1 1 b − 1 − a c a c − b c b c − 1 − b − 1 1 b − 1 a c − a c b c − b c − 1 − b − 1 ) H^{(8,0)}(a,b,c)=\begin{pmatrix}
1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 \\
1 & -1 & ac & -ac & -bc & bc & 1 & -1 \\
1 & -b^{-1} & c & c & -c & -c & -1 & b^{-1} \\
1 & -1 & -ac & ac & bc & -bc & 1 & -1 \\
1 & -b^{-1} & -c & -c & c & c & -1 & b^{-1} \\
1 & 1 & -1 & -1 & -1 & -1 & 1 & 1 \\
1 & b^{-1} & -ac & ac & -bc & bc & -1 & -b^{-1} \\
1 & b^{-1} & ac & -ac & bc & -bc & -1 & -b^{-1}
\end{pmatrix} H ( 8 , 0 ) ( a , b , c ) = 1 1 1 1 1 1 1 1 1 − 1 − b − 1 − 1 − b − 1 1 b − 1 b − 1 1 a c c − a c − c − 1 − a c a c 1 − a c c a c − c − 1 a c − a c 1 − b c − c b c c − 1 − b c b c 1 b c − c − b c c − 1 b c − b c 1 1 − 1 1 − 1 1 − 1 − 1 1 − 1 b − 1 − 1 b − 1 1 − b − 1 − b − 1 H ( 9 , 0 ) ( a , b ) = ( 1 1 1 1 1 1 1 1 1 − 1 − a b a b − a b a b 1 − 1 1 − b − 1 − a a a − a − 1 b − 1 1 − a a 2 b − a − a 2 b a − 1 a 1 − 1 a b − a b a b − a b 1 − 1 1 a − a 2 b − a a 2 b a − 1 − a 1 1 − 1 − 1 − 1 − 1 1 1 1 b − 1 a a − a − a − 1 − b − 1 ) H^{(9,0)}(a,b)=\begin{pmatrix}
1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 \\
1 & -1 & -ab & ab & -ab & ab & 1 & -1 \\
1 & -b^{-1} & -a & a & a & -a & -1 & b^{-1} \\
1 & -a & a^{2}b & -a & -a^{2}b & a & -1 & a \\
1 & -1 & ab & -ab & ab & -ab & 1 & -1 \\
1 & a & -a^{2}b & -a & a^{2}b & a & -1 & -a \\
1 & 1 & -1 & -1 & -1 & -1 & 1 & 1 \\
1 & b^{-1} & a & a & -a & -a & -1 & -b^{-1}
\end{pmatrix} H ( 9 , 0 ) ( a , b ) = 1 1 1 1 1 1 1 1 1 − 1 − b − 1 − a − 1 a 1 b − 1 1 − ab − a a 2 b ab − a 2 b − 1 a 1 ab a − a − ab − a − 1 a 1 − ab a − a 2 b ab a 2 b − 1 − a 1 ab − a a − ab a − 1 − a 1 1 − 1 − 1 1 − 1 1 − 1 1 − 1 b − 1 a − 1 − a 1 − b − 1 H ( 9 , 1 ) ( a , b ) = ( 1 1 1 1 1 1 1 1 1 − 1 − a b a b a b − a b 1 − 1 1 b − 1 a a − a − a − 1 − b − 1 1 − a − a a 2 b − a 2 b a − 1 a 1 − 1 a b − a b − a b a b 1 − 1 1 − b − 1 a − a a − a − 1 b − 1 1 a − a − a 2 b a 2 b a − 1 − a 1 1 − 1 − 1 − 1 − 1 1 1 ) H^{(9,1)}(a,b)=\begin{pmatrix}
1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 \\
1 & -1 & -ab & ab & ab & -ab & 1 & -1 \\
1 & b^{-1} & a & a & -a & -a & -1 & -b^{-1} \\
1 & -a & -a & a^{2}b & -a^{2}b & a & -1 & a \\
1 & -1 & ab & -ab & -ab & ab & 1 & -1 \\
1 & -b^{-1} & a & -a & a & -a & -1 & b^{-1} \\
1 & a & -a & -a^{2}b & a^{2}b & a & -1 & -a \\
1 & 1 & -1 & -1 & -1 & -1 & 1 & 1
\end{pmatrix} H ( 9 , 1 ) ( a , b ) = 1 1 1 1 1 1 1 1 1 − 1 b − 1 − a − 1 − b − 1 a 1 1 − ab a − a ab a − a − 1 1 ab a a 2 b − ab − a − a 2 b − 1 1 ab − a − a 2 b − ab a a 2 b − 1 1 − ab − a a ab − a a − 1 1 1 − 1 − 1 1 − 1 − 1 1 1 − 1 − b − 1 a − 1 b − 1 − a 1 H ( 10 , 0 ) ( a , b , c ) = ( 1 1 1 1 1 1 1 1 1 − 1 a c − a c − a − 1 b a − 1 b 1 − 1 1 a − 1 c − c − a − 1 − a − 1 − 1 a − 1 1 − 1 − a c a c − 1 − 1 1 1 1 − a − 1 − a − 1 − a − 1 a − 1 a − 1 − 1 a − 1 1 − a − 1 a − 1 a − 1 a − 2 b − a − 2 b − 1 − a − 1 1 1 − 1 − 1 a − 1 b − a − 1 b 1 − 1 1 a − 1 − c c − a − 2 b a − 2 b − 1 − a − 1 ) H^{(10,0)}(a,b,c)=\begin{pmatrix}
1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 \\
1 & -1 & ac & -ac & -a^{-1}b & a^{-1}b & 1 & -1 \\
1 & a^{-1} & c & -c & -a^{-1} & -a^{-1} & -1 & a^{-1} \\
1 & -1 & -ac & ac & -1 & -1 & 1 & 1 \\
1 & -a^{-1} & -a^{-1} & -a^{-1} & a^{-1} & a^{-1} & -1 & a^{-1} \\
1 & -a^{-1} & a^{-1} & a^{-1} & a^{-2}b & -a^{-2}b & -1 & -a^{-1} \\
1 & 1 & -1 & -1 & a^{-1}b & -a^{-1}b & 1 & -1 \\
1 & a^{-1} & -c & c & -a^{-2}b & a^{-2}b & -1 & -a^{-1}
\end{pmatrix} H ( 10 , 0 ) ( a , b , c ) = 1 1 1 1 1 1 1 1 1 − 1 a − 1 − 1 − a − 1 − a − 1 1 a − 1 1 a c c − a c − a − 1 a − 1 − 1 − c 1 − a c − c a c − a − 1 a − 1 − 1 c 1 − a − 1 b − a − 1 − 1 a − 1 a − 2 b a − 1 b − a − 2 b 1 a − 1 b − a − 1 − 1 a − 1 − a − 2 b − a − 1 b a − 2 b 1 1 − 1 1 − 1 − 1 1 − 1 1 − 1 a − 1 1 a − 1 − a − 1 − 1 − a − 1 H ( 12 , 0 ) ( a , b ) = ( 1 1 1 1 1 1 1 1 1 − 1 − b b − 1 1 b − b 1 − b − 1 − a − 1 b b − 1 b − 1 − 1 − 1 a − 1 b 1 − a b − 1 b − 1 a b − 1 − 1 1 − b 1 − 1 b − b − 1 1 − b b 1 a b − 1 − b − 1 − a b − 1 − 1 1 b 1 1 − 1 − 1 1 1 − 1 − 1 1 b − 1 a − 1 b 1 − b − 1 − 1 − 1 − a − 1 b ) H^{(12,0)}(a,b)=\begin{pmatrix}
1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 \\
1 & -1 & -b & b & -1 & 1 & b & -b \\
1 & -b^{-1} & -a^{-1}b & b^{-1} & b^{-1} & -1 & -1 & a^{-1}b \\
1 & -ab^{-1} & b & -1 & ab^{-1} & -1 & 1 & -b \\
1 & -1 & b & -b & -1 & 1 & -b & b \\
1 & ab^{-1} & -b & -1 & -ab^{-1} & -1 & 1 & b \\
1 & 1 & -1 & -1 & 1 & 1 & -1 & -1 \\
1 & b^{-1} & a^{-1}b & 1 & -b^{-1} & -1 & -1 & -a^{-1}b
\end{pmatrix} H ( 12 , 0 ) ( a , b ) = 1 1 1 1 1 1 1 1 1 − 1 − b − 1 − a b − 1 − 1 a b − 1 1 b − 1 1 − b − a − 1 b b b − b − 1 a − 1 b 1 b b − 1 − 1 − b − 1 − 1 1 1 − 1 b − 1 a b − 1 − 1 − a b − 1 1 − b − 1 1 1 − 1 − 1 1 − 1 1 − 1 1 b − 1 1 − b 1 − 1 − 1 1 − b a − 1 b − b b b − 1 − a − 1 b H ( 12 , 1 ) ( a , b ) = ( 1 1 1 1 1 1 1 1 1 − 1 b − b 1 − 1 − b b 1 b − 1 a − 1 b − 1 − b − 1 1 − 1 − a − 1 b 1 − b a − 1 b 2 b − b − 1 b − a − 1 b 2 1 − 1 − a − 1 b − 1 − 1 1 1 a − 1 b 1 − b − 1 − 1 1 b − 1 1 − 1 − 1 1 b − b − b b − 1 b − b 1 1 − a − 1 b 2 b − 1 − 1 − b a − 1 b 2 ) H^{(12,1)}(a,b)=\begin{pmatrix}
1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 \\
1 & -1 & b & -b & 1 & -1 & -b & b \\
1 & b^{-1} & a^{-1}b & -1 & -b^{-1} & 1 & -1 & -a^{-1}b \\
1 & -b & a^{-1}b^{2} & b & -b & -1 & b & -a^{-1}b^{2} \\
1 & -1 & -a^{-1}b & -1 & -1 & 1 & 1 & a^{-1}b \\
1 & -b^{-1} & -1 & 1 & b^{-1} & 1 & -1 & -1 \\
1 & b & -b & -b & b & -1 & b & -b \\
1 & 1 & -a^{-1}b^{2} & b & -1 & -1 & -b & a^{-1}b^{2}
\end{pmatrix} H ( 12 , 1 ) ( a , b ) = 1 1 1 1 1 1 1 1 1 − 1 b − 1 − b − 1 − b − 1 b 1 1 b a − 1 b a − 1 b 2 − a − 1 b − 1 − b − a − 1 b 2 1 − b − 1 b − 1 1 − b b 1 1 − b − 1 − b − 1 b − 1 b − 1 1 − 1 1 − 1 1 1 − 1 − 1 1 − b − 1 b 1 − 1 b − b 1 b − a − 1 b − a − 1 b 2 a − 1 b − 1 − b a − 1 b 2 H ( 13 , 0 ) ( a , b ) = ( 1 1 1 1 1 1 1 1 1 − 1 − a b − 1 a b − 1 1 − 1 1 − 1 1 − b − 1 − b − 1 − b − 1 b − 1 b − 1 − 1 b − 1 1 − 1 a b − 1 − a b − 1 − 1 − 1 1 1 1 − b − 1 b − 1 b − 1 − b − 1 b − 1 − 1 − b − 1 1 1 − 1 − 1 − 1 1 1 − 1 1 b − 1 − a b − 1 a b − 1 − 1 − b − 1 − 1 1 1 b − 1 a b − 1 − a b − 1 1 − b − 1 − 1 − 1 ) H^{(13,0)}(a,b)=\begin{pmatrix}
1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 \\
1 & -1 & -ab^{-1} & ab^{-1} & 1 & -1 & 1 & -1 \\
1 & -b^{-1} & -b^{-1} & -b^{-1} & b^{-1} & b^{-1} & -1 & b^{-1} \\
1 & -1 & ab^{-1} & -ab^{-1} & -1 & -1 & 1 & 1 \\
1 & -b^{-1} & b^{-1} & b^{-1} & -b^{-1} & b^{-1} & -1 & -b^{-1} \\
1 & 1 & -1 & -1 & -1 & 1 & 1 & -1 \\
1 & b^{-1} & -ab^{-1} & ab^{-1} & -1 & -b^{-1} & -1 & 1 \\
1 & b^{-1} & ab^{-1} & -ab^{-1} & 1 & -b^{-1} & -1 & -1
\end{pmatrix} H ( 13 , 0 ) ( a , b ) = 1 1 1 1 1 1 1 1 1 − 1 − b − 1 − 1 − b − 1 1 b − 1 b − 1 1 − a b − 1 − b − 1 a b − 1 b − 1 − 1 − a b − 1 a b − 1 1 a b − 1 − b − 1 − a b − 1 b − 1 − 1 a b − 1 − a b − 1 1 1 b − 1 − 1 − b − 1 − 1 − 1 1 1 − 1 b − 1 − 1 b − 1 1 − b − 1 − b − 1 1 1 − 1 1 − 1 1 − 1 − 1 1 − 1 b − 1 1 − b − 1 − 1 1 − 1 H ( 13 , 1 ) ( a , b ) = ( 1 1 1 1 1 1 1 1 1 − 1 − a b − 1 a b − 1 − 1 1 1 − 1 1 b − 1 − b − 1 − b − 1 − b − 1 b − 1 − 1 b − 1 1 − 1 − 1 − 1 1 − 1 1 1 1 − 1 a b − 1 − a b − 1 b − 1 1 − 1 − b − 1 1 − b − 1 b − 1 b − 1 − b − 1 − b − 1 − 1 b − 1 1 1 − a b − 1 a b − 1 b − 1 − 1 − 1 − b − 1 1 1 a b − 1 − a b − 1 − 1 − 1 1 − 1 ) H^{(13,1)}(a,b)=\begin{pmatrix}
1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 \\
1 & -1 & -ab^{-1} & ab^{-1} & -1 & 1 & 1 & -1 \\
1 & b^{-1} & -b^{-1} & -b^{-1} & -b^{-1} & b^{-1} & -1 & b^{-1} \\
1 & -1 & -1 & -1 & 1 & -1 & 1 & 1 \\
1 & -1 & ab^{-1} & -ab^{-1} & b^{-1} & 1 & -1 & -b^{-1} \\
1 & -b^{-1} & b^{-1} & b^{-1} & -b^{-1} & -b^{-1} & -1 & b^{-1} \\
1 & 1 & -ab^{-1} & ab^{-1} & b^{-1} & -1 & -1 & -b^{-1} \\
1 & 1 & ab^{-1} & -ab^{-1} & -1 & -1 & 1 & -1
\end{pmatrix} H ( 13 , 1 ) ( a , b ) = 1 1 1 1 1 1 1 1 1 − 1 b − 1 − 1 − 1 − b − 1 1 1 1 − a b − 1 − b − 1 − 1 a b − 1 b − 1 − a b − 1 a b − 1 1 a b − 1 − b − 1 − 1 − a b − 1 b − 1 a b − 1 − a b − 1 1 − 1 − b − 1 1 b − 1 − b − 1 b − 1 − 1 1 1 b − 1 − 1 1 − b − 1 − 1 − 1 1 1 − 1 1 − 1 − 1 − 1 1 1 − 1 b − 1 1 − b − 1 b − 1 − b − 1 − 1 H ( 14 , 0 ) ( a , b ) = ( 1 1 1 1 1 1 1 1 1 − 1 − a a 1 − 1 b − b 1 − b − 1 − 1 − 1 b − 1 1 − 1 1 1 − a a − a − a − 1 a a 1 − 1 1 1 − 1 1 − 1 − 1 1 a a − a a − 1 − a − a 1 1 − a a − 1 − 1 − b b 1 b − 1 − 1 − 1 − b − 1 1 1 − 1 ) H^{(14,0)}(a,b)=\begin{pmatrix}
1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 \\
1 & -1 & -a & a & 1 & -1 & b & -b \\
1 & -b^{-1} & -1 & -1 & b^{-1} & 1 & -1 & 1 \\
1 & -a & a & -a & -a & -1 & a & a \\
1 & -1 & 1 & 1 & -1 & 1 & -1 & -1 \\
1 & a & a & -a & a & -1 & -a & -a \\
1 & 1 & -a & a & -1 & -1 & -b & b \\
1 & b^{-1} & -1 & -1 & -b^{-1} & 1 & 1 & -1
\end{pmatrix} H ( 14 , 0 ) ( a , b ) = 1 1 1 1 1 1 1 1 1 − 1 − b − 1 − a − 1 a 1 b − 1 1 − a − 1 a 1 a − a − 1 1 a − 1 − a 1 − a a − 1 1 1 b − 1 − a − 1 a − 1 − b − 1 1 − 1 1 − 1 1 − 1 − 1 1 1 b − 1 a − 1 − a − b 1 1 − b 1 a − 1 − a b − 1 H ( 14 , 1 ) ( a , b ) = ( 1 1 1 1 1 1 1 1 1 − 1 a − a − 1 1 − b b 1 b − 1 1 1 − b − 1 − 1 − 1 − 1 1 − 1 − a a − 1 − 1 1 1 1 − b − 1 − a a b − 1 1 − 1 − 1 1 − b − 1 a − a b − 1 − 1 b − b 1 1 − 1 − 1 1 − 1 − b b 1 b − 1 − 1 − 1 − b − 1 1 b − b ) H^{(14,1)}(a,b)=\begin{pmatrix}
1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 \\
1 & -1 & a & -a & -1 & 1 & -b & b \\
1 & b^{-1} & 1 & 1 & -b^{-1} & -1 & -1 & -1 \\
1 & -1 & -a & a & -1 & -1 & 1 & 1 \\
1 & -b^{-1} & -a & a & b^{-1} & 1 & -1 & -1 \\
1 & -b^{-1} & a & -a & b^{-1} & -1 & b & -b \\
1 & 1 & -1 & -1 & 1 & -1 & -b & b \\
1 & b^{-1} & -1 & -1 & -b^{-1} & 1 & b & -b
\end{pmatrix} H ( 14 , 1 ) ( a , b ) = 1 1 1 1 1 1 1 1 1 − 1 b − 1 − 1 − b − 1 − b − 1 1 b − 1 1 a 1 − a − a a − 1 − 1 1 − a 1 a a − a − 1 − 1 1 − 1 − b − 1 − 1 b − 1 b − 1 1 − b − 1 1 1 − 1 − 1 1 − 1 − 1 1 1 − b − 1 1 − 1 b − b b 1 b − 1 1 − 1 − b b − b H ( 15 , 0 ) ( a , b ) = ( 1 1 1 1 1 1 1 1 1 − 1 − a b a b − a b a b 1 − 1 1 − 1 a 2 a a − a 2 − 1 1 1 − a b a 2 b 2 − a b a b − a 2 b 2 a b − 1 1 − 1 a − a − a a − 1 1 1 a b − a 2 b 2 − a b a b a 2 b 2 − a b − 1 1 1 − 1 − 1 − 1 − 1 1 1 1 1 a b a b − a b − a b − 1 − 1 ) H^{(15,0)}(a,b)=\begin{pmatrix}
1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 \\
1 & -1 & -ab & ab & -ab & ab & 1 & -1 \\
1 & -1 & a^{2} & a & a & -a^{2} & -1 & 1 \\
1 & -ab & a^{2}b^{2} & -ab & ab & -a^{2}b^{2} & ab & -1 \\
1 & -1 & a & -a & -a & a & -1 & 1 \\
1 & ab & -a^{2}b^{2} & -ab & ab & a^{2}b^{2} & -ab & -1 \\
1 & 1 & -1 & -1 & -1 & -1 & 1 & 1 \\
1 & 1 & ab & ab & -ab & -ab & -1 & -1
\end{pmatrix} H ( 15 , 0 ) ( a , b ) = 1 1 1 1 1 1 1 1 1 − 1 − 1 − ab − 1 ab 1 1 1 − ab a 2 a 2 b 2 a − a 2 b 2 − 1 ab 1 ab a − ab − a − ab − 1 ab 1 − ab a ab − a ab − 1 − ab 1 ab − a 2 − a 2 b 2 a a 2 b 2 − 1 − ab 1 1 − 1 ab − 1 − ab 1 − 1 1 − 1 1 − 1 1 − 1 1 − 1 H ( 15 , 1 ) ( a , b ) = ( 1 1 1 1 1 1 1 1 1 − 1 − a b a b − a b a b − 1 1 1 1 − a a a − a − 1 − 1 1 − a b − 1 − 1 a b a b 1 − a b 1 − 1 a b − a b a b a b − 1 1 1 − 1 1 1 − 1 − 1 1 − 1 1 a b − 1 − 1 − a b − a b 1 a b 1 1 a − a − a a − 1 − 1 ) H^{(15,1)}(a,b)=\begin{pmatrix}
1&1&1&1&1&1&1&1\\
1&-1&-ab&ab&-ab&ab&-1&1\\
1&1&-a&a&a&-a&-1&-1\\
1&-ab&-1&-1&ab&ab&1&-ab\\
1&-1&ab&-ab&ab&ab&-1&1\\
1&-1&1&1&-1&-1&1&-1\\
1&ab&-1&-1&-ab&-ab&1&ab\\
1&1&a&-a&-a&a&-1&-1
\end{pmatrix} H ( 15 , 1 ) ( a , b ) = 1 1 1 1 1 1 1 1 1 − 1 1 − ab − 1 − 1 ab 1 1 − ab − a − 1 ab 1 − 1 a 1 ab a − 1 − ab 1 − 1 − a 1 − ab a ab ab − 1 − ab − a 1 ab − a ab ab − 1 − ab a 1 − 1 − 1 1 − 1 1 1 − 1 1 1 − 1 − ab 1 − 1 ab − 1 H ( 15 , 2 ) ( a , b ) = ( 1 1 1 1 1 1 1 1 1 − 1 − a b a b a b − a b 1 − 1 1 − 1 a a − a − a − 1 1 1 − a b a 2 b 2 − a 2 b 2 − a b a b a b − 1 1 − 1 − a − a a a − 1 1 1 a b − a 2 b 2 a 2 b 2 − a b a b − a b − 1 1 1 − 1 − 1 − 1 − 1 1 1 1 1 a b − a b a b − a b − 1 − 1 ) H^{(15,2)}(a,b)=\begin{pmatrix}
1&1&1&1&1&1&1&1\\
1&-1&-ab&ab&ab&-ab&1&-1\\
1&-1&a&a&-a&-a&-1&1\\
1&-ab&a^{2}b^{2}&-a^{2}b^{2}&-ab&ab&ab&-1\\
1&-1&-a&-a&a&a&-1&1\\
1&ab&-a^{2}b^{2}&a^{2}b^{2}&-ab&ab&-ab&-1\\
1&1&-1&-1&-1&-1&1&1\\
1&1&ab&-ab&ab&-ab&-1&-1
\end{pmatrix} H ( 15 , 2 ) ( a , b ) = 1 1 1 1 1 1 1 1 1 − 1 − 1 − ab − 1 ab 1 1 1 − ab a a 2 b 2 − a − a 2 b 2 − 1 ab 1 ab a − a 2 b 2 − a a 2 b 2 − 1 − ab 1 ab − a − ab a − ab − 1 ab 1 − ab − a ab a ab − 1 − ab 1 1 − 1 ab − 1 − ab 1 − 1 1 − 1 1 − 1 1 − 1 1 − 1 H ( 15 , 3 ) ( a , b ) = ( 1 1 1 1 1 1 1 1 1 − 1 − a b a b a b − a b − 1 1 1 1 − a a − a − a − 1 − 1 1 − a b − 1 a b − 1 a b 1 − a b 1 − 1 a b − a b − a b a b − 1 1 1 − 1 1 − 1 1 − 1 1 − 1 1 a b − 1 − a b − 1 − a b 1 a b 1 1 − a − a a a − 1 − 1 ) H^{(15,3)}(a,b)=\begin{pmatrix}
1&1&1&1&1&1&1&1\\
1&-1&-ab&ab&ab&-ab&-1&1\\
1&1&-a&a&-a&-a&-1&-1\\
1&-ab&-1&ab&-1&ab&1&-ab\\
1&-1&ab&-ab&-ab&ab&-1&1\\
1&-1&1&-1&1&-1&1&-1\\
1&ab&-1&-ab&-1&-ab&1&ab\\
1&1&-a&-a&a&a&-1&-1
\end{pmatrix} H ( 15 , 3 ) ( a , b ) = 1 1 1 1 1 1 1 1 1 − 1 1 − ab − 1 − 1 ab 1 1 − ab − a − 1 ab 1 − 1 − a 1 ab a ab − ab − 1 − ab − a 1 ab − a − 1 − ab 1 − 1 a 1 − ab − a ab ab − 1 − ab a 1 − 1 − 1 1 − 1 1 1 − 1 1 1 − 1 − ab 1 − 1 ab − 1 H ( 16 , 0 ) ( a , b ) = ( 1 1 1 1 1 1 1 1 1 − 1 − b b − a − 1 b a − 1 b a − a 1 − a − a − 1 b a − 1 b − a − 1 b − b − 1 a 1 − a a − 1 b − b b − a − 1 b a − 1 1 − 1 b − b − a − 1 b a − 1 b − a a 1 a − a − 1 b − b a − 1 b b − 1 − a 1 a a − 1 b b − b − a − 1 b − a − 1 1 1 − 1 − 1 − 1 − 1 1 1 ) H^{(16,0)}(a,b)=\begin{pmatrix}
1&1&1&1&1&1&1&1\\
1&-1&-b&b&-a^{-1}b&a^{-1}b&a&-a\\
1&-a&-a^{-1}b&a^{-1}b&-a^{-1}b&-b&-1&a\\
1&-a&a^{-1}b&-b&b&-a^{-1}b&a&-1\\
1&-1&b&-b&-a^{-1}b&a^{-1}b&-a&a\\
1&a&-a^{-1}b&-b&a^{-1}b&b&-1&-a\\
1&a&a^{-1}b&b&-b&-a^{-1}b&-a&-1\\
1&1&-1&-1&-1&-1&1&1
\end{pmatrix} H ( 16 , 0 ) ( a , b ) = 1 1 1 1 1 1 1 1 1 − 1 − a − a − 1 a a 1 1 − b − a − 1 b a − 1 b b − a − 1 b a − 1 b − 1 1 b a − 1 b − b − b − b b − 1 1 − a − 1 b − a − 1 b b − a − 1 b a − 1 b − b − 1 1 a − 1 b − b − a − 1 b a − 1 b b − a − 1 b − 1 1 a − 1 a − a − 1 − a 1 1 − a a − 1 a − a − 1 1 H ( 16 , 1 ) ( a , b ) = ( 1 1 1 1 1 1 1 1 1 − 1 b − b − a − 1 b a − 1 b a − a 1 − a a − 1 b b − a − 1 b − b − 1 a 1 − 1 − a − 2 b a − 2 b a − 1 b − a − 1 b a − 1 − a − 1 1 − a − 1 − a − 1 b − a − 2 b a − 1 b a − 2 b − 1 a − 1 1 a − b a − 1 b − a − 1 b b − a − 1 1 1 − 1 − 1 − 1 − 1 1 1 1 a − 1 a − 2 b − a − 1 b a − 1 b − a − 2 b − a − 1 − 1 ) H^{(16,1)}(a,b)=\begin{pmatrix}
1&1&1&1&1&1&1&1\\
1&-1&b&-b&-a^{-1}b&a^{-1}b&a&-a\\
1&-a&a^{-1}b&b&-a^{-1}b&-b&-1&a\\
1&-1&-a^{-2}b&a^{-2}b&a^{-1}b&-a^{-1}b&a^{-1}&-a^{-1}\\
1&-a^{-1}&-a^{-1}b&-a^{-2}b&a^{-1}b&a^{-2}b&-1&a^{-1}\\
1&a&-b&a^{-1}b&-a^{-1}b&b&-a&-1\\
1&1&-1&-1&-1&-1&1&1\\
1&a^{-1}&a^{-2}b&-a^{-1}b&a^{-1}b&-a^{-2}b&-a^{-1}&-1
\end{pmatrix} H ( 16 , 1 ) ( a , b ) = 1 1 1 1 1 1 1 1 1 − 1 − a − 1 − a − 1 a 1 a − 1 1 b a − 1 b − a − 2 b − a − 1 b − b − 1 a − 2 b 1 − b b a − 2 b − a − 2 b a − 1 b − 1 − a − 1 b 1 − a − 1 b − a − 1 b a − 1 b a − 1 b − a − 1 b − 1 a − 1 b 1 a − 1 b − b − a − 1 b a − 2 b b − 1 − a − 2 b 1 a − 1 a − 1 − 1 − a 1 − a − 1 1 − a a − a − 1 a − 1 − 1 1 − 1 H ( 17 , 0 ) ( a , b , c ) = ( 1 1 1 1 1 1 1 1 1 − 1 − b c b c − c c b − b 1 − a − 1 b − a − 1 b c c a − 1 b c − c − 1 a − 1 b 1 − a b − 1 a b − 1 c − c c − a b − 1 c a b − 1 − 1 1 − 1 b c − b c − c c − b b 1 a b − 1 − a b − 1 c − c c a b − 1 c − a b − 1 − 1 1 1 − 1 − 1 − 1 − 1 1 1 1 a − 1 b a − 1 b c c − a − 1 b c − c − 1 − a − 1 b ) H^{(17,0)}(a,b,c)=\begin{pmatrix}
1&1&1&1&1&1&1&1\\
1&-1&-bc&bc&-c&c&b&-b\\
1&-a^{-1}b&-a^{-1}bc&c&a^{-1}bc&-c&-1&a^{-1}b\\
1&-ab^{-1}&ab^{-1}c&-c&c&-ab^{-1}c&ab^{-1}&-1\\
1&-1&bc&-bc&-c&c&-b&b\\
1&ab^{-1}&-ab^{-1}c&-c&c&ab^{-1}c&-ab^{-1}&-1\\
1&1&-1&-1&-1&-1&1&1\\
1&a^{-1}b&a^{-1}bc&c&-a^{-1}bc&-c&-1&-a^{-1}b
\end{pmatrix} H ( 17 , 0 ) ( a , b , c ) = 1 1 1 1 1 1 1 1 1 − 1 − a − 1 b − a b − 1 − 1 a b − 1 1 a − 1 b 1 − b c − a − 1 b c a b − 1 c b c − a b − 1 c − 1 a − 1 b c 1 b c c − c − b c − c − 1 c 1 − c a − 1 b c c − c c − 1 − a − 1 b c 1 c − c − a b − 1 c c a b − 1 c − 1 − c 1 b − 1 a b − 1 − b − a b − 1 1 − 1 1 − b a − 1 b − 1 b − 1 1 − a − 1 b H ( 18 , 0 ) ( a , b ) = ( 1 1 1 1 1 1 1 1 1 − 1 − a b a b − a a b − b 1 − 1 − a a a − a − 1 1 1 − a a 2 − a a − a 2 a − 1 1 − 1 a b − a b − a a − b b 1 a − a 2 − a a a 2 − a − 1 1 1 − 1 − 1 − 1 − 1 1 1 1 1 a a − a − a − 1 − 1 ) H^{(18,0)}(a,b)=\begin{pmatrix}
1&1&1&1&1&1&1&1\\
1&-1&-ab&ab&-a&a&b&-b\\
1&-1&-a&a&a&-a&-1&1\\
1&-a&a^2&-a&a&-a^2&a&-1\\
1&-1&ab&-ab&-a&a&-b&b\\
1&a&-a^2&-a&a&a^2&-a&-1\\
1&1&-1&-1&-1&-1&1&1\\
1&1&a&a&-a&-a&-1&-1
\end{pmatrix} H ( 18 , 0 ) ( a , b ) = 1 1 1 1 1 1 1 1 1 − 1 − 1 − a − 1 a 1 1 1 − ab − a a 2 ab − a 2 − 1 a 1 ab a − a − ab − a − 1 a 1 − a a a − a a − 1 − a 1 a − a − a 2 a a 2 − 1 − a 1 b − 1 a − b − a 1 − 1 1 − b 1 − 1 b − 1 1 − 1 H ( 18 , 1 ) ( a , b ) = ( 1 1 1 1 1 1 1 1 1 − 1 − a b a b − a a − b b 1 1 a a − a − a − 1 − 1 1 − a − a a 2 a − a 2 a − 1 1 − 1 a b − a b − a a b − b 1 − 1 a − a a − a − 1 1 1 a − a − a 2 a a 2 − a − 1 1 1 − 1 − 1 − 1 − 1 1 1 ) H^{(18,1)}(a,b)=\begin{pmatrix}
1&1&1&1&1&1&1&1\\
1&-1&-ab&ab&-a&a&-b&b\\
1&1&a&a&-a&-a&-1&-1\\
1&-a&-a&a^2&a&-a^2&a&-1\\
1&-1&ab&-ab&-a&a&b&-b\\
1&-1&a&-a&a&-a&-1&1\\
1&a&-a&-a^2&a&a^2&-a&-1\\
1&1&-1&-1&-1&-1&1&1
\end{pmatrix} H ( 18 , 1 ) ( a , b ) = 1 1 1 1 1 1 1 1 1 − 1 1 − a − 1 − 1 a 1 1 − ab a − a ab a − a − 1 1 ab a a 2 − ab − a − a 2 − 1 1 − a − a a − a a a − 1 1 a − a − a 2 a − a a 2 − 1 1 − b − 1 a b − 1 − a 1 1 b − 1 − 1 − b 1 − 1 1 H ( 20 , 0 ) ( a , b ) = ( 1 1 1 1 1 1 1 1 1 − 1 a − a − a a b − b 1 − 1 a b − 1 a − a b − 1 − a − 1 1 1 − 1 − a − a a a − b b 1 − 1 − a b − 1 a a b − 1 − a 1 − 1 1 1 − a − 1 − a − 1 a a 1 1 − 1 1 − 1 1 − 1 − 1 1 1 a − 1 a − 1 − a − a ) H^{(20,0)}(a,b)=\begin{pmatrix}
1&1&1&1&1&1&1&1\\
1&-1&a&-a&-a&a&b&-b\\
1&-1&ab^{-1}&a&-ab^{-1}&-a&-1&1\\
1&-1&-a&-a&a&a&-b&b\\
1&-1&-ab^{-1}&a&ab^{-1}&-a&1&-1\\
1&1&-a&-1&-a&-1&a&a\\
1&1&-1&1&-1&1&-1&-1\\
1&1&a&-1&a&-1&-a&-a
\end{pmatrix} H ( 20 , 0 ) ( a , b ) = 1 1 1 1 1 1 1 1 1 − 1 − 1 − 1 − 1 1 1 1 1 a a b − 1 − a − a b − 1 − a − 1 a 1 − a a − a a − 1 1 − 1 1 − a − a b − 1 a a b − 1 − a − 1 a 1 a − a a − a − 1 1 − 1 1 b − 1 − b 1 a − 1 − a 1 − b 1 b − 1 a − 1 − a H ( 20 , 1 ) ( a , b ) = ( 1 1 1 1 1 1 1 1 1 − 1 − a a − a a − b b 1 1 a b − 1 a − a b − 1 − a − 1 − 1 1 − 1 − a a b a − a b b − b 1 − 1 a − a b − a a b b − b 1 − 1 a − a a − a − b b 1 1 − a b − 1 − a a b − 1 a − 1 − 1 1 1 − 1 − 1 − 1 − 1 1 1 ) H^{(20,1)}(a,b)=\begin{pmatrix}
1&1&1&1&1&1&1&1\\
1&-1&-a&a&-a&a&-b&b\\
1&1&ab^{-1}&a&-ab^{-1}&-a&-1&-1\\
1&-1&-a&ab&a&-ab&b&-b\\
1&-1&a&-ab&-a&ab&b&-b\\
1&-1&a&-a&a&-a&-b&b\\
1&1&-ab^{-1}&-a&ab^{-1}&a&-1&-1\\
1&1&-1&-1&-1&-1&1&1
\end{pmatrix} H ( 20 , 1 ) ( a , b ) = 1 1 1 1 1 1 1 1 1 − 1 1 − 1 − 1 − 1 1 1 1 − a a b − 1 − a a a − a b − 1 − 1 1 a a ab − ab − a − a − 1 1 − a − a b − 1 a − a a a b − 1 − 1 1 a − a − ab ab − a a − 1 1 − b − 1 b b − b − 1 1 1 b − 1 − b − b b − 1 1 Exceptional completion certificates
The following files give the positive witnesses for the exceptional loci. Paths in the table are relative to code002_work/N8/proofs/; each record specifies the base substitution and 5 added rows. Their parameter maps cover whole connected tori. Substitution into the prescribed triples and exact opposite-monomial pairing verify every surface/circle square, including intersections. For example, x = 1 x=1 x = 1 ϵ = 1 \epsilon=1 ϵ = 1
Table 1 (PDF p. 57).
( 1 − y − 1 1 − 1 − y − 1 − 1 y − 1 y − 1 1 − 1 y − y 1 1 − 1 − 1 1 y − 1 − y − 1 − y − 1 y − 1 − 1 1 − 1 1 y − 1 y − 1 y − 1 − y − 1 − 1 − y − 1 − y − 1 1 1 − 1 − 1 − 1 1 − y y ) . \begin{pmatrix}
1 & -y^{-1} & 1 & -1 & -y^{-1} & -1 & y^{-1} & y^{-1}\\
1 & -1 & y & -y & 1 & 1 & -1 & -1\\
1 & y^{-1} & -y^{-1} & -y^{-1} & y^{-1} & -1 & 1 & -1\\
1 & y^{-1} & y^{-1} & y^{-1} & -y^{-1} & -1 & -y^{-1} & -y^{-1}\\
1 & 1 & -1 & -1 & -1 & 1 & -y & y
\end{pmatrix}. 1 1 1 1 1 − y − 1 − 1 y − 1 y − 1 1 1 y − y − 1 y − 1 − 1 − 1 − y − y − 1 y − 1 − 1 − y − 1 1 y − 1 − y − 1 − 1 − 1 1 − 1 − 1 1 y − 1 − 1 1 − y − 1 − y y − 1 − 1 − 1 − y − 1 y . For class 22 at x = 1 x=1 x = 1 ϵ = 1 \epsilon=1 ϵ = 1
( 1 − 1 1 − 1 y − 1 1 − y 1 − 1 − y 1 − 1 1 y − 1 1 1 − 1 − y y y − 1 − y 1 1 y y − 1 − y − y − 1 1 1 − y − 1 − y − 1 y y ) . \begin{pmatrix}
1 & -1 & 1 & -1 & y & -1 & 1 & -y\\
1 & -1 & -y & 1 & -1 & 1 & y & -1\\
1 & 1 & -1 & -y & y & y & -1 & -y\\
1 & 1 & y & y & -1 & -y & -y & -1\\
1 & 1 & -y & -1 & -y & -1 & y & y
\end{pmatrix}. 1 1 1 1 1 − 1 − 1 1 1 1 1 − y − 1 y − y − 1 1 − y y − 1 y − 1 y − 1 − y − 1 1 y − y − 1 1 y − 1 − y y − y − 1 − y − 1 y . These rectangular witnesses illustrate the verification; the complete lists are in the files above.
BBB circle rows
For (5.1), each displayed pair ( a , b ) (a,b) ( a , b ) w = e π i / 6 w=e^{\pi i/6} w = e π i /6 q ∈ T q\in\mathbb{T} q ∈ T
a = q , b = w 11 : ( 1 w 4 w 10 q w 2 q − q w 7 q w 8 w q ) a=q,\quad b=w^{11}:\quad\left(1\quad w^4\quad w^{10}q\quad w^2q\quad-q\quad w^7q\quad w^8\quad wq\right) a = q , b = w 11 : ( 1 w 4 w 10 q w 2 q − q w 7 q w 8 w q ) a = q , b = w 7 : ( 1 w 4 − q w 2 q w 3 q w 10 q w 8 w 9 q ) a=q,\quad b=w^7:\quad\left(1\quad w^4\quad-q\quad w^2q\quad w^3q\quad w^{10}q\quad w^8\quad w^9q\right) a = q , b = w 7 : ( 1 w 4 − q w 2 q w 3 q w 10 q w 8 w 9 q ) a = q , b = w 3 : ( 1 w 4 w 2 q w 11 q − q w 10 q w 8 w 5 q ) a=q,\quad b=w^3:\quad\left(1\quad w^4\quad w^2q\quad w^{11}q\quad-q\quad w^{10}q\quad w^8\quad w^5q\right) a = q , b = w 3 : ( 1 w 4 w 2 q w 11 q − q w 10 q w 8 w 5 q ) a = w 10 , b = q : ( 1 w 8 w 4 w q w 9 q w 2 q w 5 q w 8 q ) a=w^{10},\quad b=q:\quad\left(1\quad w^8\quad w^4\quad wq\quad w^9q\quad w^2q\quad w^5q\quad w^8q\right) a = w 10 , b = q : ( 1 w 8 w 4 w q w 9 q w 2 q w 5 q w 8 q ) a = − 1 , b = q : ( 1 w 8 w 4 w 10 q w 9 q w 5 q w q w 4 q ) a=-1,\quad b=q:\quad\left(1\quad w^8\quad w^4\quad w^{10}q\quad w^9q\quad w^5q\quad wq\quad w^4q\right) a = − 1 , b = q : ( 1 w 8 w 4 w 10 q w 9 q w 5 q w q w 4 q ) a = w 2 , b = q : ( 1 w 8 w 4 w q − q w 5 q w 9 q q ) a=w^2,\quad b=q:\quad\left(1\quad w^8\quad w^4\quad wq\quad-q\quad w^5q\quad w^9q\quad q\right) a = w 2 , b = q : ( 1 w 8 w 4 w q − q w 5 q w 9 q q ) a = w 9 q , b = q : ( 1 1 w 9 q − 1 w 3 − 1 − 1 w 3 q − 1 w 9 ) a=w^9q,\quad b=q:\quad\left(1\quad1\quad w^9q^{-1}\quad w^3\quad-1\quad-1\quad w^3q^{-1}\quad w^9\right) a = w 9 q , b = q : ( 1 1 w 9 q − 1 w 3 − 1 − 1 w 3 q − 1 w 9 ) a = w 5 q , b = q : ( 1 1 w 5 q − 1 − 1 − 1 w 3 w 11 q − 1 w 9 ) a=w^5q,\quad b=q:\quad\left(1\quad1\quad w^5q^{-1}\quad-1\quad-1\quad w^3\quad w^{11}q^{-1}\quad w^9\right) a = w 5 q , b = q : ( 1 1 w 5 q − 1 − 1 − 1 w 3 w 11 q − 1 w 9 ) a = w q , b = q : ( 1 1 w q − 1 − 1 w 3 − 1 w 7 q − 1 w 9 ) a=wq,\quad b=q:\quad\left(1\quad1\quad wq^{-1}\quad-1\quad w^3\quad-1\quad w^7q^{-1}\quad w^9\right) a = w q , b = q : ( 1 1 w q − 1 − 1 w 3 − 1 w 7 q − 1 w 9 ) Cycle certificates for the square families
A string such as 12 34 56 78 gives antipodal pairs; 123 456 78 gives 2 triangles and a pair. These supports refer to the exact formulas in Subsection 4.1. On each block, the quotient is a common monomial times the indicated prime roots, so every certificate is an identity in all parameters.
Table 39. F 8 , 5 a F_{8,5a} F 8 , 5 a
Table 40. F 8 , 5 b F_{8,5b} F 8 , 5 b
Table 41. F 8 , 5 c F_{8,5c} F 8 , 5 c
Table 42. F 8 , 4 F_{8,4} F 8 , 4
Table 43. F 8 , 4 t F^{t}_{8,4} F 8 , 4 t
Table 44. F 8 , 2 F_{8,2} F 8 , 2
Table 45. F 8 , 2 t F^{t}_{8,2} F 8 , 2 t
Table 46. K 8 , 2 K_{8,2} K 8 , 2
Table 47. M 8 , 1 M_{8,1} M 8 , 1
Code and certificate index
The accompanying code002.zip contains the Python and C++ sources, exact inputs, saved outputs and reproduction instructions. Its source folders are
Table 48.
The files N7_data.zip and N8_data.zip retain the inputs and saved results. After extracting code002.zip, run python3 run.py prepare from its folder to create the sibling directory code002_work/. This restores the paths used by the original programs and by the certificate references in this paper. Within § 3 , paths beginning with code/ are relative to code002_work/N7/. The file SOURCE_INDEX.md maps restored source paths to the folders above.
Use Python 3.10 or later and install requirements.txt, including SymPy 1.14.0. The command python3 run.py check runs the 16 supplied certificate checks. These verify saved enumeration records and identities; some final checkers also use saved verification records. This command does not rerun the exhaustive searches. The file README.md lists their separate reproduction commands and the GNU C++17 requirements. Assertions must remain enabled.
The JSON certificates retain row supports, congruences, square formulas and equivalence witnesses. In the A/B completion stages, completion_squares.json records the affine squares, completion_embeddings.json their entrywise identifications, and final_verification.json the completed checks.
We end by explaining how to read the classification and what remains to be understood. Theorem 5.9 gives a complete family cover: every regular square of order 8 belongs, up to equivalence, to one of the 9 continuous families or to one of the 4 isolated classes. The formulas are valid at every unit-modulus parameter value. In particular, the exceptional values are part of the classification, together with the generic points. The support calculations in § 4 –§ 6 account for the extra rows and square completions that can appear when a parameter satisfies an additional relation.
A family cover does not give unique coordinates on the space of equivalence classes. Parameters within one family can describe equivalent matrices, and different families can meet at special values. The matching distributions distinguish the generic types, while the saved monomial witnesses identify particular coincidences exactly. A fuller description would determine all parameter symmetries within each family and all intersections between families. For example, an identification at one parameter value does not by itself identify a neighborhood of that point; the local regular branches must also be compared. The local phase calculations and equivalence witnesses retained here provide starting data for this question. Such coincidences concern the identifications within the proved cover, and cannot add a family outside it.
We should make the same distinction for the labels of partial matrices. A chosen cycle support records how a row quotient is decomposed, and a matrix may admit several such decompositions. Thus the support registry organizes the equations to be solved, but its labels need not distinguish the underlying matrices. Likewise, finding one square above a prescribed triple settles its completion existence, whereas describing every square above that triple requires the full compatibility calculation. This is why the successful surfaces and circles were treated with all their special parameter values retained. A formula that supplies a completion everywhere on a circle need not supply every completion at its exceptional points.
The separation between the A/B and C cases also explains the reach of the triple method. The pure-B obstruction and the Ramsey argument force an AAA triple in every A/B square, so the prescribed-AAA calculation covers that whole sector. Among the C-containing squares, L 8 , 30 L_{8,30} L 8 , 30 L 8 , 30 t L^{t}_{8,30} L 8 , 30 t K 8 , 30 K_{8,30} K 8 , 30
The finite catalogue comparison has a different role. Each of its 5469 entries has an exact identification with the displayed cover, but the first family found by a search need not be its only presentation. Consequently, counting those identifications by target family would not give a disjoint count of equivalence classes in the families. Moreover, none of the tested root orders is divisible by 15, so these catalogues cannot test the C-containing part of the classification. The example M 8 , 1 M_{8,1} M 8 , 1
Finally, the complete regular classification leaves the corresponding questions without the regularity assumption open. We have classified all regular matrices at orders 7 and 8, and the affine descriptions explain how their continuous families arise from Butson matrices. Deciding whether every Butson matrix is regular, as in Conjecture 1.5, and classifying arbitrary complex Hadamard matrices are separate questions. Within the regular locus, an explicit description of the family intersections would complement the present list by explaining how its continuous pieces fit together. The formulas and certificates above make both kinds of comparison possible without repeating the completion classification.
References 