Modular discriminant ($\Delta$)

The modular discriminant is the normalized weight-twelve level-one cusp form

The modular discriminant has a simple zero at infinity and no zero in the upper half-plane, by the valence formula for the modular group. Multiplication by it identifies weight-$k$ modular forms with weight-$(k+12)$ cusp forms. Conversely dividing a cusp form by it preserves interior holomorphy and removes one order of vanishing at infinity.

The effective group ${\Gamma }_0(11)$ is torsion-free, has index twelve and has two regular cusps. If its weight-two cusp form space is one-dimensional, any nonzero member has order one at each cusp and no interior zeros by the regular-cusp valence formula on a torsion-free modular curve. Its twelfth power and $\Delta (\tau )\Delta (11\tau )$ have the same weight and divisor. Their ratio is a holomorphic function without poles on the compactified modular curve, hence a nonzero constant. Normalizing the leading Fourier coefficient to one proves the displayed formula, with the holomorphic root chosen to start with $q$. It equals $q{\prod }_{n\ge 1}(1-q^n{)}^2(1-q^{11n}{)}^2$.

Put $S_j={\sum }_{n\ge 1}{\sigma }_j(n)q^n$. Expanding $E_4=1+240S_3$ and $E_6=1-504S_5$ gives $\Delta =(5S_3+7S_5)/12+100S_3^2+8000S_3^3-147S_5^2$. For every integer $d$, $5d^3+7d^5$ is divisible by twelve, by separate congruences modulo three and four. Consequently every coefficient of the modular discriminant is integral, without using a product expansion.

The Ramanujan tau function gives the Fourier coefficients at positive indices of the modular discriminant: $\Delta (z)={\sum }_{n\ge 1}\tau (n)q^n$, with $\tau (1)=1$. The Serre derivative and vanishing of weight-fourteen level-one cusp forms give $qd\Delta /dq=E_2\Delta$ and hence the convolution recurrence

The modular discriminant satisfies $1728\Delta =E_4^3-E_6^2$, while the Serre derivative gives $2\Theta E_6=E_2{E}_6-E_4^2$. Their integral Fourier expansions have $E_4\equiv 1$ and $E_2\equiv E_6(\mathrm{mod}5)$, because $d^5\equiv d(\mathrm{mod}5)$. Reducing the two identities gives $\Delta \equiv \Theta E_6(\mathrm{mod}5)$. The coefficient of $q^n$ in $\Theta E_6$ is $-504n{\sigma }_5(n)\equiv n{\sigma }_5(n)$.

The Fourier expansion of a normalized Eisenstein series and the one-dimensional weight-twelve cusp form space give $691E_{12}=691E_4^3-432000\Delta$. The coefficient of $q^n$, for $n\ge 1$, reduces modulo $691$ to $566{\sigma }_{11}(n)=566\tau (n)$. Since $566$ is invertible modulo $691$, this proves the congruence. Integrality follows from $E_4=1+240\sum {\sigma }_3(n)q^n$ and the product $\Delta =q{\prod }_{n\ge 1}(1-q^n{)}^{24}$.

For every real $s$, the rapidly convergent integralis the analytic continuation of $(2\pi {)}^{-s}\Gamma (s)L(\Delta ,s)$. The product for $\Delta$ makes the integrand positive, so $L(\Delta ,s)>0$ for real $s>0$.