The robustness of Fourier forms for the tribimaximal mixing matrix is evident when we consider the up quark analogue of the Fourier operator $F_3$. Using a sum over the colour index, and for the $12$th root $\phi$, this is given by One readily checks that $U_3 F_2$ is once again the tribimaximal matrix. In fact, we now see that two phases $\phi_1$ and $\phi_2$ will always give the same mixing matrix, so long as they satisfy the rules
1. $\phi_1 + \phi_2$ is a phase
2. $\phi_1 - \phi_2$ has norm $\sqrt{3}$
Meanwhile, the down quark gives the Fourier operator which obviously also returns the tribimaximal matrix.
14 years ago
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