M Theory Lesson 212
The $3 \times 3$ democratic matrix is useful in many ways. Here we see it acts as a cyclic shift operator for three 1-circulants that additively act like modular 3 arithmetic. 
But observe that $(2 \cdot \textrm{id})^{2} = \textrm{id}$ and, similarly, the square of $(312) + (231)$ is twice $(312) + (231)$. That is, we have the relations
$A^{2} = A$
$B^{2} = 2A$
$C^{2} = \frac{1}{2} C = \textrm{id}$
$A + B + C = 0$
Exponentiating the third relation yields the multiplicative honeycomb rule $A \cdot B \cdot C = 1$. However, using ordinary matrix multiplication one obtains $ABC = 2 \cdot \textrm{id}$, so the correct normalisation factor for all four matrices is the reciprocal of the cubed root of 2, namely 1.25992. Alternatively, since 2 is really $-1$, $C^{2} = C$ and $B^{2} = - A$ and the correct normalisation is a cubed root of $-1$, that is a 6th root of unity.
But observe that $(2 \cdot \textrm{id})^{2} = \textrm{id}$ and, similarly, the square of $(312) + (231)$ is twice $(312) + (231)$. That is, we have the relations$A^{2} = A$
$B^{2} = 2A$
$C^{2} = \frac{1}{2} C = \textrm{id}$
$A + B + C = 0$
Exponentiating the third relation yields the multiplicative honeycomb rule $A \cdot B \cdot C = 1$. However, using ordinary matrix multiplication one obtains $ABC = 2 \cdot \textrm{id}$, so the correct normalisation factor for all four matrices is the reciprocal of the cubed root of 2, namely 1.25992. Alternatively, since 2 is really $-1$, $C^{2} = C$ and $B^{2} = - A$ and the correct normalisation is a cubed root of $-1$, that is a 6th root of unity.
posted by Kea | 8:04 PM
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