Entry 178

We go back to the first four McKay-Thompson series of the Monster. We conjecture that the complete elliptic integral of the first kind \(K(k_m)\) can be given by the first McKay-Thomson series \(j_{1A}(\tau) = j(\tau)\) or simply the j-function as,

$$K(k_m) = \frac{\pi}2  \frac{(2^{1/4}G_m)^2}{\big({-j(\tau)}\big)^{1/12}}\sqrt{\sum_{n=0}^\infty \frac{(6n)!}{(3n)!\,n!^3} \frac1{\big(j(\tau)\big)^n}}$$

with Ramanujan G-function \(G_m\) for integer \(m\geq7\) and \(\tau = \frac{1+\sqrt{-m}}2\), 

$$2^{1/4}G_m = \frac{\eta^2(\sqrt{-m})}{\eta(\tfrac12\sqrt{-m})\,\eta(2\sqrt{-m})} = \zeta_{48}\frac{\eta(\tau)}{\eta(2\tau)}$$and \(48\)th root of unity \(\zeta_{48} = e^{2\pi i/48}\). Let odd discriminant \(d=m\). Examples for class number \(h(-d) = 1\),

$$K(k_{7}) = \frac{\pi}2 \frac{2}{\big({15^3}\big)^{1/12}}\sqrt{\sum_{n=0}^\infty \frac{(6n)!}{(3n)!\,n!^3} \frac1{\big({-15^3}\big)^n}}$$

while for higher Heegner numbers, \(x\) is a root of a simple cubic

$$K(k_{11}) = \frac{\pi}2 \left(\frac{1}{T}+1\right)^2\frac1{\big({32^3}\big)^{1/12}}\sqrt{\sum_{n=0}^\infty \frac{(6n)!}{(3n)!\,n!^3} \frac1{\big({-32^3}\big)^n}}$$

with \(T\) the tribonacci constant, the real root of \(T^3-T^2-T-1=0\). For the largest

$$K(k_{163}) = \frac{\pi}2 \frac{x^2}{\big({640320^3}\big)^{1/12}}\sqrt{\sum_{n=0}^\infty \frac{(6n)!}{(3n)!\,n!^3} \frac1{\big({-640320^3}\big)^n}}$$

where \(x\) is the real root of \(x^3-6x^2+4x-2 = 0\). 

P.S. The non-fundamental \(d=27\) also has class number \(1\) hence its j-function is an integer

$$K(k_{27}) = \frac{\pi}2 \frac{(2+2^{4/3}+2^{5/3})^{2/3}}{\big({{-3}\cdot160^3}\big)^{1/12}}\sqrt{\sum_{n=0}^\infty \frac{(6n)!}{(3n)!\,n!^3} \frac1{\big({-3\cdot160^3}\big)^n}}$$