The Ramanujan-Sato series $$j^*(\tau)=432\frac{\sqrt{ j(\tau)}+\sqrt{j(\tau)-1728}}{\sqrt{ j(\tau)}-\sqrt{j(\tau)-1728}}=432\frac{E_4(\tau)^{\frac32}+E_6(\tau)}{E_4(\tau)^{\frac32}-E_6(\tau)} \\ = \frac{1}{q}-120+10260 q-901120 q^2+91676610 q^3+\mathcal O\left(q^{4}\right)$$ of level 1 generalises Ramanujan's formula for $\frac1\pi$, where $q=e^{2\pi i\tau}$. Here, $j$ is the Klein $j$-invariant and $E_k$ are the Eisenstein series. Due to the square roots, it does not immediately seem to be a modular function for a congruence subgroup of $SL(2,\mathbb Z)$. However, as it relates to the Klein $j$-invariant by $$j=\frac{(j^*+432)^2}{j^*},$$ it seems to satisfy a modular polynomial equation in $j$ of degree $2$, suggesting that $j^*$ is modular for an index 2 subgroup of $SL(2,\mathbb Z)$. Is this true? And what would be the invariance group?
Any recommendation on the literature would be most helpful, many thanks!
It is $$j^*(\tau)= 432\frac{(\sqrt{ j(\tau)}+\sqrt{j(\tau)-1728})^2}{1728}=432\frac{2 j(\tau)-1728+2\sqrt{j(\tau)}\sqrt{j(\tau)-1728}}{1728}$$ So we are looking at the modularity of $$\sqrt{j(\tau)}\sqrt{j(\tau)-1728}$$
$j(\tau)-1728$ is holomorphic non-zero at $e^{2i\pi /3}$ while $j(\tau)$ has a zero of order $3$ (not $6$!!!) at $e^{2i\pi /3}$, thus $\sqrt{j(\tau)}\sqrt{j(\tau)-1728}$ and hence $j^*(\tau)$ have a branch point at $e^{2i\pi/3}$, they are not modular.
They are automorphic on a double cover of the modular curve, though.
Note that $f(\tau)=\sqrt{j(\tau)-1728}$ is holomorphic because the only zeros of $j(\tau)-1728$ are double at $SL_2(\Bbb{Z})i$. It is $2$-periodic and $f(\gamma(\tau))=\chi(\gamma)f(\tau)$ where $\chi$ is the character $SL_2(\Bbb{Z})\to \pm 1$ defined by $\chi(\pmatrix{1&1\\0&1})=\chi(\pmatrix{0&1\\-1&0})=-1$ which factors through $SL_2(\Bbb{Z/2Z})$, in fact $\ker(\chi)$ are the matrices whose reduction $\bmod 2$ is $\pmatrix{1&0\\0&1},\pmatrix{1&1\\1&0},\pmatrix{0&1\\1&1}$.