M Theory Lesson 107

First let $Y=r{e}^{i\theta }$. Continuing our conversation on determinant cubics, for a complex circulant with $Z=\overline{Y}$ we can solve the cubic

$X^3-3r^{2}X+(1+Y^3+{\overline{Y}}^3)=0$

for $X=X(Y,\overline{Y})$ with Chebyshev radicals, under certain restrictions on $Y$. In terms of the Chebyshev root function $C$ (omitting the subscript) the solutions are

$X_1=rC\left(t\right)-\frac{1}{3}$
$X_2=-rC(-t)-\frac{1}{3}$
$X_3=rC(-t)-rC\left(t\right)-\frac{1}{3}$

where $-t=r^{-3}(1+2rcos(3\theta \left)\right)$. Note that for $r=1$, $t=0$ when $\theta =\frac{2\pi }{9}$ and $C\left(t\right)=\sqrt{3}$. But for the case $r=1$ (which might not be interesting since we have used the determinant to renormalise the matrix) this solution set makes sense only provided and then

$C\left(t\right)=2cos\left(\frac{co{s}^{-1}(0.5t)}{3}\right)$

Observe that the solution condition states that $\theta >\frac{\pi }{9}$. Now observe that so this method is not directly helpful in analysing the lepton type matrix with $r=1$. On the other hand, $\frac{1}{2}>\frac{\pi }{9}$ so the neutrino type cubic has three real solutions for $X$, but only $X_1=1.5644$ is positive. For a general positive real determinant $D$, $t=D+2cos\left(3\theta \right))$ and the solution condition says that which is less restrictive if $D$ is small.