### M Theory Lesson 229

Using the three dimensional Fourier operator $F$, and a generator $M_{1}$ on which it acts, one obtains the cycle of four MUB operators satisfying

$F^{\dagger} M_{1} F = M_{2}$

$F^{\dagger} M_{2} F = M_{3}$

$F^{\dagger} M_{3} F = M_{4}$

$F^{\dagger} M_{4} F = M_{1}$

where factors of 3 and $\sqrt{3}$ are ignored as usual (this is no worse than the habit of insisting that $c = 1$ all the time). Observe how the set of four matrices naturally factors into two sets of two, just like the number 4. For example, $M_{3}$ and $M_{4}$ are related by a simple two dimensional map. If we divide all entries by $\omega$, this component of $M_{i}$ is just $\sigma_{Z}$.

$F^{\dagger} M_{1} F = M_{2}$

$F^{\dagger} M_{2} F = M_{3}$

$F^{\dagger} M_{3} F = M_{4}$

$F^{\dagger} M_{4} F = M_{1}$

where factors of 3 and $\sqrt{3}$ are ignored as usual (this is no worse than the habit of insisting that $c = 1$ all the time). Observe how the set of four matrices naturally factors into two sets of two, just like the number 4. For example, $M_{3}$ and $M_{4}$ are related by a simple two dimensional map. If we divide all entries by $\omega$, this component of $M_{i}$ is just $\sigma_{Z}$.

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