User blog comment:Ikosarakt1/Ideas of going past psi(K)/@comment-5029411-20140617213739/@comment-24509095-20140618014858

Nope. Nope. Nope. Nope.

That's just his definition for his other blog post.

$$\chi(0,\alpha) = I_\alpha$$

$$\chi(1,0)$$ is, according to Deedlit11's blog post, the diagonalizer of the function $$\psi_{\chi(1,0)}$$

$$\chi(\alpha,0)$$ is the diagonalizer of the function $$\psi_{\chi(\alpha,0)}$$