The Jacobi product formula for the discriminant

Put $P_0(q)=1$ for the empty product, $P_N(q)={\prod }_{1\le n\le N}(1-q^n{)}^{24}$ for $N\ge 1$, $P(q):={\lim }_N{P}_N(q)$ and $F(\tau ):=qP(q)$. For $\mathrm{\mid }q\mathrm{\mid }\le r<1$ we have $\mathrm{\mid }(1-q^n{)}^{24}-1\mathrm{\mid }\le (2^{24}-1)r^n$, a summable majorant independent of $q$, so the product is normally convergent on the disc and [F1] makes $P$ holomorphic and zero-free there with $P(0)=1$; hence $F$ is holomorphic and zero-free on $\mathfrak{H}$ and $F(\tau )=q+O(q^2)$ (as a function of $q$). By [F2] the logarithmic derivative of $P$ may be computed from the finite products and the geometric series [F4]: $\frac{1}{2\pi i}\frac{F^{\mathrm{\prime }}}{F}=1-24{\sum }_{n\ge 1}\frac{n{q}^n}{1-q^n}=1-24{\sum }_{n\ge 1}{\sum }_{r\ge 1}n{q}^{nr}=1-24{\sum }_{m\ge 1}{\sigma }_1(m)q^m=E_2(\tau )$, the regrouping being justified by the bound ${\sum }_{n,r}n\mathrm{\mid }q{\mathrm{\mid }}^{nr}\le (1-r{)}^{-1}{\sum }_{n}n{r}^n<\mathrm{\infty }$ on $\mathrm{\mid }q\mathrm{\mid }\le r<1$ [F4]; the last series converges by the ratio test.

For $\gamma =\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)\in S{L}_2(\mathbb{Z})$ put $R_{\gamma }(\tau ):=F(\gamma \tau )\mathrm{/}[(c\tau +d{)}^{12}F(\tau )]$; this is holomorphic and zero-free on $\mathfrak{H}$. Its logarithmic derivative, divided by $2\pi i$, equals $\frac{E_2(\gamma \tau )}{(c\tau +d{)}^2}-\frac{12c}{2\pi i(c\tau +d)}-E_2(\tau )$ by 1.1, and by the $E_2$ transformation law this is $[E_2(\tau )-\frac{6ic}{\pi (c\tau +d)}]-[-\frac{6ic}{\pi (c\tau +d)}]-E_2(\tau )=0$ (using $\frac{12c}{2\pi i(c\tau +d)}=-\frac{6ic}{\pi (c\tau +d)}$). Hence $R_{\gamma }$ is constant, $C_{\gamma }$, by [F3], and $C_{\gamma \delta }=C_{\gamma }{C}_{\delta }$ because $(c_{\gamma \delta }\tau +d_{\gamma \delta })=(c_{\gamma }\delta \tau +d_{\gamma })(c_{\delta }\tau +d_{\delta })$. For $T$ we have $F(\tau +1)=F(\tau )$, so $C_T=1$; for $S$ we have $R_S(\tau )=F(-1\mathrm{/}\tau )\mathrm{/}[{\tau }^{12}F(\tau )]$ and at $\tau =i$, where $-1\mathrm{/}i=i$ and $i^{12}=1$, this gives $C_S=F(i)\mathrm{/}F(i)=1$. Since $S,T$ generate $S{L}_2(\mathbb{Z})$ [F5] and $(-I)$ acts with factor $(-1{)}^{12}=1$, multiplicativity gives $C_{\gamma }=1$ for every $\gamma$; thus $F(\gamma \tau )=(c\tau +d{)}^{12}F(\tau )$ for all $\gamma$, and since $F=q+O(q^2)$ it satisfies the cusp condition. Hence $F\in S_{12}$ by [F6].

By [F5], $S_{12}$ is one-dimensional and contains the nonzero $\mathrm{\Delta }=q-24q^2+O(q^3)$; so $F=c\mathrm{\Delta }$ for some $c\in \mathbb{C}$. Comparing $q$-coefficients, $1=c\cdot 1$, so $F=\mathrm{\Delta }$, which is the product formula.

Integrality of the coefficients: the coefficient of $q^N$ in $P$ equals the coefficient of $q^N$ in the finite product ${\prod }_{n\le N}(1-q^n{)}^{24}$, because all factors with $n>N$ are congruent to $1$ modulo $q^{N+1}$; the finite product has integer coefficients by [F4], and its coefficients agree with those of the analytic expansion of $P$ because the finite products converge locally uniformly, with all derivatives, to $P$ near $0$ by [F1] and [F2]. Hence every Taylor coefficient of $P$ is an integer, and the coefficients of $\mathrm{\Delta }=qP$ are integers as well; the simple zero at the cusp is ${\mathrm{ord}}_{\mathrm{\infty }}(\mathrm{\Delta })=1$ from the expansion $q-24q^2+\cdots$, consistent with [F5].