M Theory Lesson 195
Recall from Mulase's lectures on the modular group $PSL (2, \mathbb{Z})$ that the generators are given by 
where $T$ represents a translation by 1 in the complex plane. Note that $S$ really does square to the identity because $\pm 1$ are identified, but $T$ is quite distinct from the tetrahedral generator $(1, \omega , \omega^{2})$ used in neutrino mixing, which is more naturally associated with the quantum Fourier transform.
Consider how the diagonal $(\omega , \omega^{2})$ acts on $z$. As a modular transformation it would act via
$z \mapsto \frac{az + b}{cz + d} = \omega^{-1}z$
that is, a rotation by $\frac{2 \pi}{3}$ in the plane. This is like the action of $TS$, which also rotates a third of a circle but fixing instead the point $z = e^{\frac{\pi i}{3}}$ which is a vertex of the Grothendieck ribbon graph for the notorious j invariant.
where $T$ represents a translation by 1 in the complex plane. Note that $S$ really does square to the identity because $\pm 1$ are identified, but $T$ is quite distinct from the tetrahedral generator $(1, \omega , \omega^{2})$ used in neutrino mixing, which is more naturally associated with the quantum Fourier transform. Consider how the diagonal $(\omega , \omega^{2})$ acts on $z$. As a modular transformation it would act via
$z \mapsto \frac{az + b}{cz + d} = \omega^{-1}z$
that is, a rotation by $\frac{2 \pi}{3}$ in the plane. This is like the action of $TS$, which also rotates a third of a circle but fixing instead the point $z = e^{\frac{\pi i}{3}}$ which is a vertex of the Grothendieck ribbon graph for the notorious j invariant.
posted by Kea | 10:10 AM
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