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!


1 Answer 1


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 $\pmatrix{1&0\\0&1},\pmatrix{1&1\\1&0},\pmatrix{0&1\\1&1}$.

  • $\begingroup$ Thank you! I'm not familiar with automorphic functions on covers of the modular curve. Why does the even/odd order of vanishing determine whether the square root of a function has a branch point and is holomorphic? Is there an explicit way to construct the double cover? And is there a good textbook on automorphic functions on covering spaces of the modular curve, perhaps with relations to polynomials in the field of modular functions on SL(2,Z) (such as the second equation in my question)? $\endgroup$
    – El Rafu
    Nov 5, 2020 at 16:54
  • $\begingroup$ If $C$ is a double cover of $X_0(1)$ then $[\Bbb{C}(C):\Bbb{C}(X_0(1))]=2$, $\Bbb{C}(X_0(1))=\Bbb{C}(j)$ so $\Bbb{C}(C)=\Bbb{C}(j)[t]/(t^2+a(j)t+b(j))=\Bbb{C}(j)[t]/(t^2-f(j))$ where $f(j)\in \Bbb{C}(j)$ is not a square. If $\sqrt{f(j)}$ is modular then $f(j(\tau))$ may have odd order poles and zeros only at (the points of $X_0(1)$) $SL_2(Z)i\infty$ and $SL_2(Z) i$, which implies that $\Bbb{C}(C)=\Bbb{C}(j,\sqrt{j-1728})$. $\endgroup$
    – reuns
    Nov 5, 2020 at 20:31
  • $\begingroup$ $j(\tau)= (\tau-e^{2i\pi /3})^3 h(\tau)$ with $h$ analytic non-zero, $\sqrt{j(\tau)}=(\tau-e^{2i\pi /3})^{3/2} h(\tau)^{1/2}$ with $h(\tau)^{1/2}$ analytic at $e^{2i\pi/3}$ ie. $\sqrt{j(\tau)}$ has a branch point there. $\endgroup$
    – reuns
    Nov 5, 2020 at 20:36
  • $\begingroup$ Thanks. I understand the argument with the branch point now. But why is it enough to extend the field $\mathbb C(j)$ by $\sqrt{j-1728}$ and not also $\sqrt{j}$? Is $\sqrt{j}$ in $\mathbb C(j,\sqrt{j-1728})$? $\endgroup$
    – El Rafu
    Nov 5, 2020 at 21:49
  • $\begingroup$ $\sqrt{j}$ is not modular. The non-modular covers can be interpreted in term of covers of the Rimeann sphere (the image of $j(\tau)$) $\endgroup$
    – reuns
    Nov 5, 2020 at 22:13

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