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

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

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

$X_1 = r C(t) – \frac{1}{3}$
$X_2 = -r C(-t) – \frac{1}{3}$
$X_3 = r C(-t) – r C(t) – \frac{1}{3}$

where $-t = r^{-3} (1 + 2r cos (3 \theta))$. Note that for $r = 1$, $t = 0$ when $\theta = \frac{2 \pi}{9}$ and $C(t) = \sqrt{3}$. 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 $cos 3 \theta < 0.5$ and then

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

Observe that the solution condition states that $\theta > \frac{\pi}{9}$. Now observe that $\frac{2}{9} \frac{\pi}{9}$ 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 $D$, $t = D + 2 cos (3 \theta))$ and the solution condition says that $cos (3 \theta) < \frac{2 – D}{2}$ which is less restrictive if $D$ is small.

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out /  Change )

Google photo

You are commenting using your Google account. Log Out /  Change )

Twitter picture

You are commenting using your Twitter account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )

Connecting to %s

%d bloggers like this: