Difference between revisions of "Questions"

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(Created page with "==On $(A_2,A_7)$== ===sol 1=== $$ \begin{bmatrix} z_0^{(1)} & z_0^{(2)} \\ z_1^{(1)} & z_1^{(2)} \\ z_2^{(1)} & z_2^{(2)} \\ z_3^{(1)} & z_3^{(2)} \\ z_4^{(1)} & z_4^{(2)...")
 
 
(2 intermediate revisions by the same user not shown)
Line 1: Line 1:
==On $(A_2,A_7)$==
+
==On $(A_2,A_6)$==
 
===sol 1===
 
===sol 1===
 
$$
 
$$
Line 23: Line 23:
 
  1. & 1.
 
  1. & 1.
 
\end{bmatrix}
 
\end{bmatrix}
 +
$$
 +
Here $2.24698\cdots$ is a solution of $x^3-2 x^2-x+1=0$.
 +
 +
This solution can be obtained from $g\in SU(3)$ given by
 +
$$
 +
g=\left(
 +
\begin{array}{ccc}
 +
e^{\frac{2 i \pi }{7}} & 0 & 0 \\
 +
0 & 1 & 0 \\
 +
0 & 0 &  e^{-\frac{2 i \pi }{7}} \\
 +
\end{array}
 +
\right)
 
$$
 
$$
  
Line 48: Line 60:
 
  1. & 1.
 
  1. & 1.
 
\end{bmatrix}
 
\end{bmatrix}
 +
$$
 +
where $1.61803\cdots$ is the Golden ratio.
 +
 +
This solution can be obtained from $g\in SU(3)$ given by
 +
$$
 +
g=\left(
 +
\begin{array}{ccc}
 +
e^{\frac{2 i \pi }{5}} & 0 & 0 \\
 +
0 & 1 & 0 \\
 +
0 & 0 & e^{-\frac{2 i \pi }{5}} \\
 +
\end{array}
 +
\right)
 
$$
 
$$

Latest revision as of 13:11, 30 September 2013

On $(A_2,A_6)$

sol 1

$$ \begin{bmatrix}

z_0^{(1)} & z_0^{(2)} \\
z_1^{(1)} & z_1^{(2)} \\
z_2^{(1)} & z_2^{(2)} \\
z_3^{(1)} & z_3^{(2)} \\
z_4^{(1)} & z_4^{(2)} \\
z_5^{(1)} & z_5^{(2)} \\
z_6^{(1)} & z_6^{(2)} \\
z_7^{(1)} & z_7^{(2)}

\end{bmatrix} = \begin{bmatrix}

1. & 1. \\
2.24698 & 2.24698 \\
2.80194 & 2.80194 \\
2.24698 & 2.24698 \\
1. & 1. \\
0 & 0 \\
0 & 0 \\
1. & 1.

\end{bmatrix} $$ Here $2.24698\cdots$ is a solution of $x^3-2 x^2-x+1=0$.

This solution can be obtained from $g\in SU(3)$ given by $$ g=\left( \begin{array}{ccc}

e^{\frac{2 i \pi }{7}} & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 &  e^{-\frac{2 i \pi }{7}} \\

\end{array} \right) $$

sol 2

$$ \begin{bmatrix}

z_0^{(1)} & z_0^{(2)} \\
z_1^{(1)} & z_1^{(2)} \\
z_2^{(1)} & z_2^{(2)} \\
z_3^{(1)} & z_3^{(2)} \\
z_4^{(1)} & z_4^{(2)} \\
z_5^{(1)} & z_5^{(2)} \\
z_6^{(1)} & z_6^{(2)} \\
z_7^{(1)} & z_7^{(2)}

\end{bmatrix} = \begin{bmatrix}

1. & 1. \\
1.61803 & 1.61803 \\
1. & 1. \\
0 & 0 \\
0 & 0 \\
1. & 1. \\
1.61803 & 1.61803 \\
1. & 1.

\end{bmatrix} $$ where $1.61803\cdots$ is the Golden ratio.

This solution can be obtained from $g\in SU(3)$ given by $$ g=\left( \begin{array}{ccc}

e^{\frac{2 i \pi }{5}} & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & e^{-\frac{2 i \pi }{5}} \\

\end{array} \right) $$