Idempotent Nilpotent II
A nicer way to represent the neutrino tetrahedron group with $3 \times 3$ operators is to choose 
since this also obeys $T^{3} = 1$ and $(TS)^{3} = 1$ but $T$ looks a lot more like the circulant $S$ than a diagonal operator. Moreover, the quantum Fourier diagonal still appears in the relation 
from which it also follows that $T = SD$ and $T^{2} = S \overline{D} = S D^{2}$ where $\overline{D}$ is just the rotation in the opposite direction in the plane. Observe how the squaring of $T$ shifts the horizontal phase factors to vertical ones.
since this also obeys $T^{3} = 1$ and $(TS)^{3} = 1$ but $T$ looks a lot more like the circulant $S$ than a diagonal operator. Moreover, the quantum Fourier diagonal still appears in the relation 
from which it also follows that $T = SD$ and $T^{2} = S \overline{D} = S D^{2}$ where $\overline{D}$ is just the rotation in the opposite direction in the plane. Observe how the squaring of $T$ shifts the horizontal phase factors to vertical ones.
posted by Kea | 6:16 PM
2 Comments:
Wow, maybe all physicists should experience a little starvation every now and then. I guess if you have an academic job, this is the time of year where you take a break. Things are very quiet.
You know, it really is kind of odd that nobody seems interested in your post about meson masses.
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