User blog comment:King2218/The Theta Function/@comment-5529393-20140326131700

Some corrections:

$$\theta_{I+1}$$ diagonalizes over the ordinal $$\Omega_{I+1}$$, so

$$\theta_{I+1} (\Omega_{I+1}) = \theta_{I+1} (\theta_{I+1} (\ldots))$$.

$$\theta_{I_2}(0) = \Omega_{\Omega_{\cdots_{I+1}}}$$.

$$\theta_{I_2}(0, \alpha) = $$ the $$I + 1 + \alpha$$th fixed point of $$ \beta \mapsto \Omega_\beta$$.

$$\theta_{I_2}(I_2, \alpha) = $$ the $$ \alpha$$th fixed point of $$\beta \mapsto \theta_{I_2}(\beta)$$.

$$\theta_{I(1,0)}(0) = I_{I_{\cdots}} $$.

$$\theta_{I(1,0,0)}(0) = I(I(\cdots,0),0) $$.

M diagonalizes over the variables of I, e.g.

$$\chi(\alpha) = I_\alpha$$.

$$\chi(M*\alpha + \beta) = I(\alpha, \beta)$$.

$$\chi(M^2 * \alpha + M * \beta + \gamma) = I(\alpha,\beta,\gamma)$$,

and so on.