CKM Rules IV

Let's consider how the double Fourier transform acts on CKM type operators. Since the three dimensional operator $F_{3}$ cycles the 3d MUBs, and these operators (along with their 2-circulant duals) may be used to form a basis for a $3 \times 3$ matrix $A$, then it follows that $F^{4}$ (the fourth iteration of the transform) is the identity. This is made explicit by the following.

Observe that This 2-circulant transformation acts on a general $3 \times 3$ matrix $A$ to give and we see that the quadruple transform returns the matrix $A$. For the measured CKM values, the triple transform must therefore return the inverse of the $SU(2) \times U(1)$ form of the matrix.

Now the two dimensional operator $F_{2}$, which forms the other component of the neutrino mixing matrix, acts on the Pauli basis for $SU(2)$ to simply cycle the three basis operators. Therefore, the triple application of the MNS transform $(F_{3}F_{2})$ to the CKM matrix yields an inverse also. It follows that

$U T^{3} = T^{3} C$

where $U$ is the CKM group element, $C$ the CKM matrix, and $T$ is the tribimaximal mixing matrix $T = F_{3}F_{2}$. Thus $T$ is a very special operator that acts this way for any (magic) unitary matrix.