Jacobi theta triple product and nonvanishing of the theta constant

Let $K$ be a compact subset of the product, so $\mathrm{Im}\tau \ge y_0>0$ and $\mathrm{\mid }\mathrm{Im}z\mathrm{\mid }\le M$ on $K$. Put $r:=e^{-\pi y_0}<1$. Then $\mathrm{\mid }Q\mathrm{\mid }\le r$ and $\mathrm{\mid }{Q}^{m^2}{e}^{2\pi imz}\mathrm{\mid }\le r^{m^2}{e}^{2\pi M\mathrm{\mid }m\mathrm{\mid }}$ for every $(m,z,\tau )$ under consideration; the bounding series converges by [F8], since the ratio of consecutive terms is $r^{2m+1}{e}^{2\pi M}\to 0$. The $M$-test [F1] therefore gives uniform convergence on $K$; each term is entire in $z$ and holomorphic in $\tau$ by [F6] and [F7], so [F2] shows that $\mathrm{\Theta }$ is entire in $z$ for fixed $\tau$, holomorphic in $\tau$, and that the series converges locally uniformly in each variable, uniformly on compact products. Termwise index shifts, licensed by absolute convergence [F8], give $\mathrm{\Theta }(z+1\mid \tau )=\mathrm{\Theta }(z\mid \tau )$, since $e^{2\pi im}=1$, and, using $Q^{m^2}{e}^{2\pi im\tau }=Q^{(m+1{)}^2-1}$, also $\mathrm{\Theta }(z+\tau \mid \tau )=Q^{-1}{e}^{-2\pi iz}\mathrm{\Theta }(z\mid \tau )$.

Put $\mathrm{\Pi }(z\mid \tau ):={\prod }_{n\ge 1}(1-Q^{2n})(1+Q^{2n-1}{e}^{2\pi iz})(1+Q^{2n-1}{e}^{-2\pi iz})$ and $g_n(z,\tau ):=(1-Q^{2n})(1+Q^{2n-1}{e}^{2\pi iz})(1+Q^{2n-1}{e}^{-2\pi iz})$. For finite grouped products put ${\mathrm{\Pi }}_0=1$ (the empty product) and ${\mathrm{\Pi }}_N={\prod }_{1\le n\le N}{g}_n$ for $N\ge 1$; normal convergence concerns this explicit sequence including its empty prefix. On $K$ as above, with $B:=e^{2\pi M}$, the estimate $\mathrm{\mid }{g}_n-1\mathrm{\mid }\le (1+r^{2n})(1+B{r}^{2n-1}{)}^2-1\le C_K{r}^{2n-1}$ shows ${\sum }_n{\sup }_K\mathrm{\mid }{g}_n-1\mathrm{\mid }<\mathrm{\infty }$, so $\mathrm{\Pi }$ is normally convergent and [F3] makes it holomorphic in each variable with the displayed product. The factor $1-Q^{2n}$ never vanishes because $\mathrm{\mid }{Q}^{2n}\mathrm{\mid }<1$. By [F6], $1+Q^{2n-1}{e}^{2\pi iz}=0$ is equivalent to $e^{2\pi iz}=-Q^{-(2n-1)}=e^{\pi i(1-(2n-1)\tau )}$, that is to $z\equiv \frac{1}{2}-\frac{2n-1}{2}\tau (\mathrm{m}\mathrm{o}\mathrm{d}\mathbb{Z})$, and likewise $1+Q^{2n-1}{e}^{-2\pi iz}=0$ is equivalent to $z\equiv -\frac{1}{2}+\frac{2n-1}{2}\tau (\mathrm{m}\mathrm{o}\mathrm{d}\mathbb{Z})$. Writing a solution of the first type as $\frac{1+\tau }{2}+a+b\tau$ with $a=k$ and $b=-n\le -1$, and one of the second type with $a=k-1$ and $b=n-1\ge 0$, shows that the union of all solutions over $n\ge 1$ is exactly the coset $\frac{1+\tau }{2}+\mathbb{Z}+\tau \mathbb{Z}$, each of whose points arises from exactly one $n$ and one factor because the representation $a+b\tau$ with $a,b\in \mathbb{Z}$ is unique ($\mathrm{Im}\tau >0$). At such a solution the derivative of the vanishing factor with respect to $z$ is $\pm 2\pi i{Q}^{2n-1}{e}^{\pm 2\pi iz}\mathrm{\ne }0$ by [F6] and [F7], so every zero of $\mathrm{\Pi }$ is simple and is a zero of exactly one factor. Finally, cancelling the shifted factors in the normally convergent product, with the reindexings $n\mapsto n+1$ in the second product and $n\mapsto n\pm 1$ in the third, gives $\mathrm{\Pi }(z+1\mid \tau )=\mathrm{\Pi }(z\mid \tau )$ and $\mathrm{\Pi }(z+\tau \mid \tau )=\frac{1+Q^{-1}{e}^{-2\pi iz}}{1+Q{e}^{2\pi iz}}\mathrm{\Pi }(z\mid \tau )=Q^{-1}{e}^{-2\pi iz}\mathrm{\Pi }(z\mid \tau )$.

At $z_0:=\frac{1+\tau }{2}$ we have $\mathrm{\Theta }(z_0\mid \tau )={\sum }_{m\in \mathbb{Z}}(-1{)}^m{Q}^{m^2+m}$; the terms with indices $m$ and $-m-1$ are negatives of each other, and the series is absolutely convergent by [F8], so it sums to $0$. The shift laws of 1.1 then propagate the vanishing to all points of $z_0+\mathbb{Z}+\tau \mathbb{Z}$, the multiplier $Q^{-1}{e}^{-2\pi iz}$ being nonzero by [F6]. By 1.2 that coset is exactly the zero set of $\mathrm{\Pi }$, all its zeros are simple, and $\mathrm{\Theta }$ is entire in $z$ by 1.1; hence $F_{\tau }:=\mathrm{\Theta }(\cdot \mid \tau )\mathrm{/}\mathrm{\Pi }(\cdot \mid \tau )$ is holomorphic off the coset and, by the Taylor series of its vanishing numerator and the simple-zero factorisation [F5] of its denominator at each zero, extends holomorphically across it to an entire function. The shift laws of 1.1 and 1.2 give $F_{\tau }(z+1)=F_{\tau }(z)$ and $F_{\tau }(z+\tau )=F_{\tau }(z)$ off the coset, hence everywhere. Every $z\in \mathbb{C}$ is $w+m+n\tau$ with $m,n\in \mathbb{Z}$ and $w$ in the compact parallelogram $\{s+t\tau :0\le s,t\le 1\}$ (take $m,n$ the integer parts of the coefficients of $z$ in the basis $1,\tau$), so $z\mapsto F_{\tau }(z)$ is bounded on $\mathbb{C}$ by its supremum on that compact set; [F4] gives $F_{\tau }\equiv c(\tau )$ for a number $c(\tau )$ depending only on $\tau$.

Evaluating the constant of 2.1 at $z=\frac{1}{4}$, where no factor of $\mathrm{\Pi }$ vanishes, gives $c(\tau )=\mathrm{\Theta }(\frac{1}{4}\mid \tau )\mathrm{/}\mathrm{\Pi }(\frac{1}{4}\mid \tau )$. The series identity is $\mathrm{\Theta }(\frac{1}{4}\mid \tau )={\sum }_m{Q}^{m^2}{i}^m={\sum }_k(-1{)}^k{Q}^{4k^2}=\mathrm{\Theta }(\frac{1}{2}\mid 4\tau )$: the odd-$m$ terms cancel in the pairs $m\leftrightarrow -m$, and the even ones $m=2k$ contribute $i^{2k}{Q}^{4k^2}=(-1{)}^k{Q}^{4k^2}$, all licensed by [F8]. The product identity is $\mathrm{\Pi }(\frac{1}{4}\mid \tau )={\prod }_n(1-Q^{2n})(1+Q^{4n-2})={\prod }_k(1-Q^{4k})(1-Q^{8k-4})={\prod }_j(1-Q^{8j})(1-Q^{8j-4}{)}^2=\mathrm{\Pi }(\frac{1}{2}\mid 4\tau )$, obtained from ${\prod }_n(1+Q^{4n-2})={\prod }_n(1-Q^{8n-4})\mathrm{/}{\prod }_n(1-Q^{4n-2})$ and from splitting ${\prod }_k(1-Q^{4k})={\prod }_j(1-Q^{8j}){\prod }_j(1-Q^{8j-4})$; all rearrangements are legitimate by normal convergence [F3]. Hence $c(\tau )=c(4\tau )$, and iteration gives $c(\tau )=c(4^k\tau )$ for every $k\ge 0$. Writing $Q_k:=Q^{4^k}\to 0$, the evaluation of 2.1 at $z=\frac{1}{2}$ gives $c(4^k\tau )=\mathrm{\Theta }(\frac{1}{2}\mid 4^k\tau )\mathrm{/}\mathrm{\Pi }(\frac{1}{2}\mid 4^k\tau )$ with $\mathrm{\Theta }(\frac{1}{2}\mid 4^k\tau )=1+2{\sum }_{m\ge 1}(-1{)}^m{Q}_k^{m^2}$ and $\mathrm{\Pi }(\frac{1}{2}\mid 4^k\tau )={\prod }_n(1-Q_k^{2n})(1-Q_k^{2n-1}{)}^2$. Since $\mathrm{\mid }{Q}_k\mathrm{\mid }\to 0$, $\mathrm{\mid }\mathrm{\Theta }(\frac{1}{2}\mid 4^k\tau )-1\mathrm{\mid }\le 2\mathrm{\mid }{Q}_k\mathrm{\mid }\mathrm{/}(1-\mathrm{\mid }{Q}_k\mathrm{\mid })\to 0$, and with $\rho =\mathrm{\mid }{Q}_k\mathrm{\mid }$, $\mathrm{\mid }\mathrm{\Pi }(\frac{1}{2}\mid 4^k\tau )-1\mathrm{\mid }\le \exp (\rho (2+\rho )\mathrm{/}(1-{\rho }^2))-1\to 0$ by the elementary product bound $\prod (1+a_n)\le \exp \sum a_n$. Therefore $c(\tau )={\lim }_{k}c(4^k\tau )=1$.

By 3.1, $\mathrm{\Theta }(z\mid \tau )=\mathrm{\Pi }(z\mid \tau )$ for all $z$, so by 2.1 the zeros of $\mathrm{\Theta }$ in $z$ are exactly the simple zeros $z=\frac{1+\tau }{2}+a+b\tau$, $a,b\in \mathbb{Z}$. Moreover $\mathrm{\Theta }(0\mid \tau )=\mathrm{\Pi }(0\mid \tau )={\prod }_n(1-Q^{2n})(1+Q^{2n-1}{)}^2\mathrm{\ne }0$, since $\mathrm{\mid }{Q}^{2n}\mathrm{\mid }<1$ and $\mathrm{\mid }{Q}^{2n-1}\mathrm{\mid }<1$ make every factor nonzero. This is the zero-free theta constant $\theta (\tau )$, and $Q^2=q$ by the addition law [F6].